disstillation project
Collection
66 items • Updated
prompt_id stringlengths 64 64 | question stringlengths 13 3.18k | response stringlengths 2 8.28k | source stringclasses 9
values | source_row int64 60 896k | answer stringlengths 0 261 ⌀ |
|---|---|---|---|---|---|
2fc9277319ca7d9a6a44c3d858cbe77cdf0bd35b8de02a8ce99f9661d537347c | (3 points) (2011•Chongqing) In the expansion of $(1+3x)^n$ (where $n \in \mathbb{N}$ and $n \geq 6$), the coefficients of $x^5$ and $x^6$ are equal. Then, $n=$ ( )
A: 6
B: 7
C: 8
D: 9 | **Analysis of the Problem**: Use the general term formula of the binomial expansion to find the general term of the binomial expansion. Calculate the coefficients of $x^5$ and $x^6$ in the expansion, and set up an equation to solve for $n$.
**Solution**:
The general term of the binomial expansion is $T_{r+1} = \bino... | cn_k12 | 534,603 | B |
fc2fa3be74062ec0945bda62ba0e44b5b18f9d63f0db3ba4d76a3b92c5c4d77f | Given an ellipse $C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$ with the right vertex $A$, the line $x=\frac{a}{2}$ intersects the ellipse $C$ at points $M$ and $N$. If $\overrightarrow{AM} \cdot \overrightarrow{AN}=0$, then the eccentricity of the ellipse $C$ is ____. | Given the ellipse $C: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$, and the right vertex $A$ of the ellipse, we are to find the eccentricity of the ellipse under the condition that the line $x = \frac{a}{2}$ intersects the ellipse at points $M$ and $N$ such that $\overrightarrow{AM} \cdot \overrightarrow{AN... | cn_k12 | 392,683 | \frac{\sqrt{6}}{3} |
af77a6e99c87c5a5ecf7f4b02c530d57725ecc5605150db1b416722d9195474f | Given an ellipse $C$: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$ with eccentricity $\frac{\sqrt{3}}{2}$, where the left and right foci are $F_{1}$ and $F_{2}$ respectively, the top vertex is $P$, and $\overrightarrow{PF_{1}} \cdot \overrightarrow{PF_{2}} = -2$.
$(1)$ Find the equation of the ellipse $C$;
$(2)$ The ... | ### Solution:
#### Part (1): Finding the Equation of Ellipse $C$
Given the eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, and using the standard form of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a>b>0$, we can derive the equation as follows:
- The foci of the ellipse are at $(-c,0)$ and $(c,0)$, an... | cn_k12 | 459,991 | 1 |
226fb9177bac7a848456e89bbf89b75e8e3863f6e0eb31181ccf75123d8fbf45 | James collects all the fruits from his 2 trees. Each tree has 20 plants. Each plant has 1 seed. He plants a certain percentage of those seeds and ends up planting 24 trees. What percentage of seeds did he plant? | James collects seeds from 2 trees, and each tree has 20 plants. So the total number of plants (and therefore seeds, since each plant has 1 seed) is:
2 trees * 20 plants/tree = 40 plants (seeds)
He ends up planting 24 trees. To find the percentage of seeds he planted, we use the formula:
(Seeds planted / Total seeds)... | orca_math | 668,272 | 60\% |
45ca41f1747c2899f2db2f85bbe9de569e92f792afe7cc037b188cea99ea6737 | When $a<0$, the solution set of the inequality $x^2 - 2ax - 3a^2 < 0$ is ______. | Since we want to find the solution set of $x^2 - 2ax - 3a^2 < 0$, let's first find the roots of the equation $x^2 - 2ax - 3a^2 = 0$. This can be done by factoring or using the quadratic formula.
The factored form of the equation is:
$$(x - 3a)(x + a) = 0$$
Setting each factor to zero gives us the roots:
$$x = 3a \qua... | cn_k12 | 320,695 | \{ x | 3a < x < -a \} |
881b0dba553e46f8c932518e698f8267b66548b5528c3bbcd9dae6328c8d7b9c | Given vectors $\overrightarrow{a}=(-3,3)$ and $\overrightarrow{b}=(3,x)$, if $\overrightarrow{a}$ and $\overrightarrow{b}$ are collinear, then $x$ is ______. | Given vectors $\overrightarrow{a}=(-3,3)$ and $\overrightarrow{b}=(3,x)$ are collinear, this implies that there exists a scalar $\lambda$ such that $\overrightarrow{b} = \lambda\overrightarrow{a}$. Writing this in the component form gives us two equations:
1. $3 = \lambda(-3)$
2. $x = \lambda(3)$
From the first equat... | cn_k12 | 436,740 | -3 |
85ac6b526ecb0b6ade28865a6602a62b652b3460b7ddc5c92ea6263de4dcf7a2 | 4. For the acute triangle $\qquad$ $ABC$, find
$$
F=\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}
$$
the range of values for $F$. | Given $0<\angle A \leqslant \angle B \leqslant \angle C<\frac{\pi}{2}$. Then $0<\angle A \leqslant \frac{\pi}{3} \leqslant \angle C<\frac{\pi}{2}$.
Fixing $\angle C$ we get
$F=\frac{2 \cos \frac{C}{2} \cdot \cos \frac{A-B}{2}+\sin C}{2 \sin \frac{C}{2} \cdot \cos \frac{A-B}{2}+\cos C}$.
Let $f(x)=\frac{2 x \cos \frac{C... | olympiads | 132,486 | (1+\frac{\sqrt{2}}{2},2) |
b6a9f2b4bcad893e9b05ff7db79ce5555ab8262f0bb0bdbf3b915fae410a54ed | Let \( y \) be a positive real number. Find the minimum value of \( 9y^3 + 4y^{-6} \). | By AM-GM (Arithmetic Mean - Geometric Mean Inequality),
\begin{align*}
9y^3 + 4y^{-6} &= y^3 + y^3 + y^3 + y^3 + y^3 + y^3 + y^3 + y^3 + y^3 + y^{-6} + y^{-6} + y^{-6} + y^{-6} \\
&\ge 13 \sqrt[13]{(y^3)^9 \cdot (y^{-6})^4} \\
&= 13 \sqrt[13]{y^{27} \cdot y^{-24}} \\
&= 13 \sqrt[13]{y^3} \\
&= 13.
\end{align*}
Equality... | synthetic_math | 763,159 | 13 |
acc373e1a744a621285614557071f6a6ccf7f0a2149e6033672a07028d07f74f | Simplify $(2)(2a)(4a^2)(3a^3)(6a^4)$. | Start by reordering and regrouping the terms for clarity:
\[
(2)(2a)(4a^2)(3a^3)(6a^4) = (2 \cdot 2 \cdot 4 \cdot 3 \cdot 6)(a \cdot a^2 \cdot a^3 \cdot a^4).
\]
Calculate the product of the coefficients:
\[
2 \cdot 2 \cdot 4 \cdot 3 \cdot 6 = 288.
\]
Merge all the powers of \(a\):
\[
a^{1+2+3+4} = a^{10}.
\]
Combine b... | synthetic_math | 871,434 | 288a^{10} |
b355504c046786bfb9d25c8bc15a25b6a986ec243059ee6b9cd3602521b84128 | 12.3.1 * The axial section of a circular cone with vertex $P$ is an isosceles right triangle. $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B$, $O H \perp P B$, with the foot of... | Since $A B \perp O B, A B \perp O P$, therefore $A B \perp P B$, plane $P A B \perp$ plane $P O B$. Also, $O H \perp P B$, so $O H \perp H C, O H \perp P A$. Since $C$ is the midpoint of $P A$, $O C \perp P A$, thus $P C$ is the height of the tetrahedron $P-H O C$, and $P C=2$.
In the right triangle $\triangle O H C$,... | olympiads | 102,964 | \frac{2\sqrt{6}}{3} |
9ffd5795460ff6401bb0f5408539eb091aa4e75ff624e30960a0587a7c40efa2 | $\{a_n\}$ is a positive integer sequence such that $a_{i+2} = a_{i+1} +a_i$ (for all $i \ge 1$).
For positive integer $n$, de fine as $$b_n=\frac{1}{a_{2n+1}}\Sigma_{i=1}^{4n-2}a_i$$
Prove that $b_n$ is positive integer. | Given the sequence $\{a_n\}$ such that $a_{i+2} = a_{i+1} + a_i$ for all $i \ge 1$, we need to prove that $b_n = \frac{1}{a_{2n+1}} \sum_{i=1}^{4n-2} a_i$ is a positive integer.
1. **Expressing $a_i$ in terms of Fibonacci numbers:**
We start by expressing $a_i$ in terms of the Fibonacci sequence. Let $F_i$ be the $... | aops_forum | 242,558 | null |
ae5f8cc353c600c4bdc791f6313bef3eea528c753cf83fdd27c9afd7723606e7 | Topsoil now costs $8$ per cubic foot. Calculate the total cost, in dollars, for 7 cubic yards of topsoil if there is a 10% discount on the total amount for purchases exceeding 100 cubic feet. | First, convert cubic yards to cubic feet. Since $1 \text{ yard}^3 = 27 \text{ feet}^3$, then:
\[ 7 \text{ yd}^3 = 7 \times 27 \text{ ft}^3 = 189 \text{ ft}^3. \]
Next, calculate the cost without any discount. At $8$ per cubic foot:
\[ \text{Cost without discount} = 189 \text{ ft}^3 \times 8 \text{ dollars/ft}^3 = 1512... | synthetic_math | 879,047 | 1360.8 \text{ dollars} |
2cc0e3345e10b88c5166f5eee9779ff9ace33fd299fa2df74c29ce180aab8529 | Some boys went to water trees. There were 29 trees. Each of them watered the equal amount of trees, and 2 trees were left. How many boys went to water the trees? | Let's denote the number of boys as B. Each boy watered an equal number of trees, and since 2 trees were left over, the total number of trees that were watered by the boys is 29 - 2 = 27 trees.
Since each boy watered an equal number of trees, the number of trees each boy watered must be a factor of 27. The factors of 2... | orca_math | 630,528 | 3 |
a3efb0b9eb65f6770ad897b5cdd4450fec947f80ff30c05aeaa2371919e3386c | 5.1.2. A particle moves in the plane $O x y$ of a rectangular coordinate system according to the law (here $x, y$ are the coordinates of the point in meters, $t$ is time in seconds):
$$
x(t)=3 ; y(t)=3+\sin t \cdot \cos t-\sin t-\cos t
$$
In the same plane, a light ray constantly emanates from the origin according to... | Answer: $c \in\left(0 ; \frac{3}{2}\right),\left(\frac{7+2 \sqrt{2}}{6} ; \infty\right)$ | olympiads | 11,392 | \in(0;\frac{3}{2}),(\frac{7+2\sqrt{2}}{6};\infty) |
b70d921f3dbf91a2fda5782d78c34b37ea04055462765cdde3bc92a0b29aedae | Let's find three-digit numbers where the sum of the digits becomes three times smaller when these numbers are tripled. | Let the digits of the number to be found be $a, b, c$; then the numbers are of the form $100a + 10b + c$. In these numbers, the last digit $c$ can only be 7, 8, or 9, because otherwise, the sum of the digits would increase by 3 when 3 is added to the number.
First, let $b < 9$. If $c = 7$, then according to the proble... | olympiads | 173,619 | 108,117,207 |
1fd6dbbf58e22c551c2f5f69ab7615907fec8bafb6b5004ff0d0640a422a36d6 | In a certain condominium, 29 families live, each of which has either one, three, or five cats. The number of families that have only one cat is the same as the number of families that have five cats. How many cats are there in this condominium? | Let $x$ be the number of families that have only one or exactly five cats, and $y$ be the number of families that have exactly three cats. It follows that $x+y+x=29$ and, therefore, $2 x+y=29$. Since the number of cats is $x+3 y+5 x=6 x+3 y$, we obtain
$$
\text { number of cats }=6 x+3 y=3(2 x+y)=3 \times 29=87 \text ... | olympiads | 74,476 | 87 |
351f918bd07a9a3afafe20884ce3993ffeb0e8814a338cd9c938fd50f7715019 | A triangular array of squares has one square in the first row, two in the second row, and in general, $k$ squares in the $k$th row for $1 \leq k \leq 10$. Except for the bottom row, each square's entry is the sum of the entries in the two squares directly below it. In each square of the tenth row, either a $0$ or a $1$... | Label the squares in the bottom row as $x_0, x_1, \ldots, x_8, x_9$.
By a similar inductive and combinatorial reasoning as in the original problem, the top square's value will be given by the sum:
$$ {9\choose0}x_0 + {9\choose1}x_1 + \ldots + {9\choose9}x_9. $$
Analyzing this expression $\mod 3$:
- Binomial coefficien... | synthetic_math | 814,749 | 320 |
6038f0718bd4300ee0d128509fb99fc96131d105b0763ed9c160521817d524ac | A batch of canned food has a standard weight of $454g$ per can. Now, $10$ cans are sampled for testing, and the results are shown in the table. What is the total weight of these $10$ cans?
| Can Number | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ |
|------------|-----|-----|-----|-----|-----|-----|----... | To find the total weight of the $10$ cans, we first calculate the difference between each can's weight and the standard weight of $454g$. This difference can be positive (if the can is heavier than the standard) or negative (if the can is lighter).
For each can, the differences from the standard weight are as follows:... | cn_k12 | 353,166 | 4550g |
79ad219352d42d2232bc380efd36be49b23c0a5789ed41c5fe6091f1c0175780 | $\begin{array}{l}\text { 6. Given } P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+ \\ a_{3} x^{2}+a_{4} x+a_{5} \text {, and when } k=1,2,3,4 \text {, } P(k) \\ =k \cdot 1997 \text {. Then } P(10)-P(-5)=\end{array}$ | 6.75315 .
Let $Q(x)=P(x)-1997 x$. Then for $k=1,2,3,4$, $Q(k)=P(k)-1997 k=0$. Hence, $1,2,3,4$ are roots of $Q(x)=0$. Since $Q(x)$ is a fifth-degree polynomial, we can assume
$$
Q(x)=(x-1)(x-2)(x-3)(x-4)(x-r) \text {. }
$$
Therefore,
$$
\begin{array}{l}
P(10)= Q(10)+1997 \cdot 10 \\
= 9 \cdot 8 \cdot 7 \cdot 6(10-r)+... | cn_contest | 709,920 | 75315 |
91520d21a0e613007fd113609b3f3a79f31912c3f546a37c605a2ee5452ca6fe | 21. In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$, respectively. Given that $\sqrt{3}a\sin C + a\cos C = c + b$:
(1) Find the measure of angle $A$;
(2) If $a = \sqrt{3}$, determine the range of possible values for $b + c$. | (1) Since $a\cos C + \sqrt{3}a\sin C = b + c$,
by the Law of Sines, we have $\sin A\cos C + \sqrt{3}\sin A\sin C = \sin B + \sin C$,
thus, $\sin A\cos C + \sqrt{3}\sin A\sin C = \sin (A + C) + \sin C$,
which leads to $\sqrt{3}\sin A - \cos A = 1$,
therefore, $\sin (A - 30^{\circ}) = \frac{1}{2}$,
hence, $A - ... | cn_k12 | 486,809 | \sqrt{3} < b + c \leqslant 2\sqrt{3} |
12f1790a8c2c74962edf8b65ec2fb548958c254c6b0c1f5471d40691d1388727 | The base-10 numbers 356 and 78 are multiplied. The product is then added to 49, and this sum is written in base-8. What is the units digit of the base-8 representation? | Step 1: Calculate the product of 356 and 78.
\[
356 \times 78 = 27768
\]
Step 2: Add 49 to the product.
\[
27768 + 49 = 27817
\]
Step 3: Find the units digit of 27817 in base-8, which is equivalent to finding the remainder of 27817 when divided by 8.
\[
27817 \div 8 = 3477 \text{ remainder } 1
\]
Therefore, the units d... | synthetic_math | 889,403 | 1 |
7b86bb768e50aaf5fe31e49d1fbfe541c853f6f2e19dd5fd7db9051caabb2dab | Among the following pairs of numbers, the ones that are opposite to each other are:
A: $-\left(-2\right)^{3}$ and $|-2|^{3}$
B: $(-2)^{3}$ and $-2^{3}$
C: $-2^{2}$ and $+\left(-2\right)^{2}$
D: $-\left(-2\right)$ and $|-2|$ | To solve this problem, we evaluate each pair of numbers to determine if they are opposites of each other.
**Option A**: $-\left(-2\right)^{3}$ and $|-2|^{3}$
- For $-\left(-2\right)^{3}$, we first evaluate the exponent, which gives us $-\left(-8\right)$. Since the negative sign outside affects the result of the expon... | cn_k12 | 433,425 | C |
24a717f875717b8de1e8c55d42f80cc3d796ae8a9a7a586d0dbc94597f965d11 | Given $\overrightarrow{a}=(2,-1,3)$, $\overrightarrow{b}=(-4,2,3-x)$. And $\overrightarrow{a}\perp \overrightarrow{b}$, then the value of $x$ is ( ).
A: $3$
B: $-3$
C: $-\dfrac{1}{3}$
D: $\dfrac{1}{3}$ | **Analysis**
This question examines the necessary and sufficient condition for vectors to be perpendicular: their dot product is $0$. Then, it uses the formula for the dot product of vectors, which is a basic problem.
Using the given condition $\overrightarrow{a}\perp \overrightarrow{b}$, we set up the equation $\ove... | cn_k12 | 305,436 | C |
b926d53b07dd41be031fe72b5d0111a908a0959b6fc18ca239319675b7c9026b | Let $P(x)$ be a polynomial with real coefficients. Prove that not all roots of $x^3P(x)+1$ are real. | 1. Let \( P(x) \) be a polynomial with real coefficients. We need to prove that not all roots of \( x^3P(x) + 1 = 0 \) are real.
2. Suppose, for the sake of contradiction, that all roots of \( x^3P(x) + 1 = 0 \) are real. Let these roots be \( r_1, r_2, \ldots, r_n \).
3. Consider the polynomial \( Q(x) = x^3P(x) + 1... | aops_forum | 204,876 | null |
d317666791b4d759c734e95a7f85a2f7b5e4d39686a5973cdf89967b4232ca5d | 5. As shown in Figure 1, in $\triangle A B C$, it is known that $O$ is the midpoint of $B C$, and points $M$ and $N$ are on sides $A B$ and $A C$ respectively, and
$$
\begin{array}{l}
A M=6, M B=4, A N=4, \\
N C=3, \angle M O N=90^{\circ} .
\end{array}
$$
Then $\cos A=$ | 5. $\frac{3}{8}$.
Let $\overrightarrow{A B}=a, \overrightarrow{A C}=b$. Then $|a|=10,|b|=7$.
Since $O$ is the midpoint of $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2} a+\frac{1}{2} b \text {. }
$$
From the problem, we know $\overrightarrow{A M}=\frac{3}{5} a, \overrightarrow{A N}=\frac{4}{7} b$.
$$
\begin{array... | cn_contest | 727,418 | \frac{3}{8} |
a255552e1dbdee60712b381e01f7ea48124eac61b53fb461d745cd99bbc1bcb6 | A train of a certain length is traveling at 45 km/hr and can cross a bridge in 30 seconds. The bridge is 295 meters long. What is the length of the train? | To find the length of the train, we first need to determine the distance the train travels while crossing the bridge. This distance is the sum of the length of the train and the length of the bridge.
The train crosses the bridge in 30 seconds, and its speed is 45 km/hr. We need to convert the speed to meters per secon... | orca_math | 650,809 | 80 |
d7a82d8e23ccdb36fc9c87d0ef96691f7419ba6de96c93330abfb1f199bdf2f5 | 12. Let $S$ be the set composed of the sides and diagonals of a regular pentagon. If any two elements are chosen from it, the probability that these two elements have the same length is ( ).
(A) $\frac{2}{5}$
(B) $\frac{4}{9}$
(C) $\frac{1}{2}$
(D) $\frac{5}{9}$
(E) $\frac{4}{5}$ | 12. B.
$$
\frac{C_{5}^{2}+C_{5}^{2}}{C_{10}^{2}}=\frac{20}{45}=\frac{4}{9} .
$$ | cn_contest | 726,706 | B |
ddbf1de245c0e3462ce7eb0221012d2098324bf2e74d2bfc52078f7c91e3986f | If two distinct members of the set $\{ 2, 4, 8, 10, 12, 15, 20, 50 \}$ are randomly selected and multiplied, what is the probability that the product is a multiple of 200? Express your answer as a common fraction. | To be a multiple of 200, a product needs at least three factors of 2 and two factors of 5. Let's evaluate possible pairs:
1. **Starting with 50 (which has 2 and $5^2$)**:
- Pair with 4, 8, 12, 20 (each having at least three factors of 2 together with 50)
- This results in 4 pairs.
2. **Now consider 20 (which ha... | synthetic_math | 826,689 | \frac{1}{4} |
e69f25f21222bbe168bc8a8eeb81bf6312d7c80c7f420c937fb8e05f75da9dfa | (Geometry Proof Exercise) From a point A outside a circle ⊙O with radius 2, draw a line intersecting ⊙O at points C and D. A line segment AB is tangent to ⊙O at B. Given that AC=4 and AB=$4 \sqrt {2}$, find $tan∠DAB$. | Since the circle ⊙O has a radius of 2, and given AC=4, AB=$4 \sqrt {2}$,
by the Tangent-Secant Theorem, we have $AB^2=AC \cdot AD$
Therefore, $32=4 \cdot AD$
Thus, $AD=8$
Hence, $CD=4$,
which means CD is the diameter.
Draw a perpendicular from O to AB, with the foot at B,
Therefore, $$tan∠DAB=tan∠OAB= \fr... | cn_k12 | 373,439 | \frac { \sqrt {2}}{4} |
0385c084a0ef131d84c03ddeb32f5157cb11a93b9096d5b5bb5d24f9f6071203 | soja finished 2 / 3 of a book . she calculated that she finished 100 more pages than she has yet to read . how long is her book ? | Let's call the total number of pages in the book P. Soja has finished 2/3 of the book, which means she has read 2/3 * P pages. She has 1/3 of the book left to read, which is 1/3 * P pages.
According to the information given, the number of pages Soja has read is 100 more than the number of pages she has left to read. W... | orca_math | 600,507 | 300 |
c8516bf05cda27958f3bf76800a9843aeca497cec53a286fa2dde39153021a28 | Show that for any real number $x$:
\[ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0 . \] | To show that for any real number \( x \):
\[ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0, \]
we will analyze the expression step by step.
1. **Rewrite the expression:**
\[ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} = x^2 (\sin{x} + 1) + x \cos{x} + \frac{1}{2}. \]
2. **Analyze the coefficient of \( x^2 \):**
... | aops_forum | 239,693 | x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0 |
5c0255b4075cffa3c725386965546ca3c7a11427a993ea7c5f40452732e37ba9 | If x + y = 9 and x - y = some number, then x ^ 2 - y ^ 2 = 27. What is the difference between x and y? | Given that \( x + y = 9 \) and \( x^2 - y^2 = 27 \), we can use the difference of squares formula to find the difference between \( x \) and \( y \).
The difference of squares formula is \( a^2 - b^2 = (a + b)(a - b) \).
Let's apply this to \( x^2 - y^2 \):
\( x^2 - y^2 = (x + y)(x - y) \)
We know that \( x + y = 9... | orca_math | 632,849 | 3 |
5a886bbc4b282c09d6839cf07f6cef19637db960120aaa1f55ee91bdbf7053df | Given 6 line segments, each with a length of 2cm, the maximum number of equilateral triangles with a side length of 2cm that can be formed is ____. | To solve this problem, consider that 6 line segments of 2cm each can form a regular tetrahedron. Each face of this tetrahedron is an equilateral triangle with sides of 2cm. Therefore, the maximum number of equilateral triangles with a side length of 2cm that can be formed is 4.
Hence, the answer is $\boxed{4}$. | cn_k12 | 341,334 | 4 |
bdcf97f970772d530294670d7e5ec4f812c5637075c9547563417cc74dab7e09 | How many integers $M$ less than $500$ can be written as the sum of $k$ consecutive positive even integers from exactly 3 values of $k\ge 1$? | Let the first even integer be $2m$, $m\geq 1$. Then the final even integer is $2m + 2(k-1) = 2(m+k-1)$. The even integers form an arithmetic sequence with sum $M = k\left(\frac{2m + 2(m+k-1)}{2}\right) = k(2m+k-1)$. Thus, $k$ is a factor of $M$.
Since $m\geq 1$, it follows that $2m+k-1 \geq k$ and $k\leq \sqrt{M}$.
Sin... | synthetic_math | 877,420 | 9 |
805e0e7679f8becccf82d7901627759151057f515f201f1ca0d13c5fde967bf4 | Given the lines $l_{1}$: $x+y-4=0$, $l_{2}$: $x-y+2=0$, and $l_{3}$: $ax-y+1-4a=0$. $(1)$ If there exists a triangle with its sides represented by the lines $l_{1}$, $l_{2}$, and $l_{3}$, find the range of real numbers for $a$; $(2)$ If a line $l$ passes through the intersection point of $l_{1}$ and $l_{2}$, and the di... | ### Solution:
#### Part (1)
To find the range of real numbers for $a$ such that the lines $l_{1}$, $l_{2}$, and $l_{3}$ form a triangle, we first find the intersection point of $l_{1}$ and $l_{2}$.
- Solving the system $\left\{\begin{array}{l}{x+y-4=0}\\{x-y+2=0}\end{array}\right.$, we find the intersection point $A... | cn_k12 | 535,571 | x=3 \text{ or } 3x+4y-15=0 |
26c1d16ff4d7270f804d3764722f794ecbe05578b377e35a4de73f8c6cc27298 | There are three propositions: ① A pentagon has at least two obtuse angles, ② A dodecagon has a total of 54 diagonals, ③ A polygon whose sum of interior angles equals the sum of exterior angles has 4 sides. The number of correct propositions is ( )
A: 0
B: 1
C: 2
D: 3 | **Solution**: For ①, since the sum of the exterior angles of a pentagon is 360 degrees,
there cannot be 4 or 5 obtuse angles among the 5 exterior angles, meaning at most there can be 3 obtuse exterior angles, which implies there can be at most 3 acute interior angles, and at least 2 obtuse interior angles.
Therefor... | cn_k12 | 464,961 | \text{D} |
8dfb900d95af5b400160864d9537b55da0daaf0c92b9c81607745c21cebba796 | Given that the geometric sequence $\{a\_n\}$ is an increasing sequence, where the sum of the first $n$ terms is $S\_n$. If $a\_3=8$, $S\_3= \int _{ 0 }^{ 2 }(4x+3)dx$, find the common ratio $q=$ \_\_\_\_\_\_. | First, compute the integral to find $S\_3$:
$$S\_3= \int _{ 0 }^{ 2 }(4x+3)dx=2x^{2}+3x| \;\_{ 0 }^{ 2 }=8+6=14$$
Now, we know that the sum of the first three terms of a geometric sequence can be expressed as:
$$S\_3=a\_3(1+ \dfrac {1}{q}+ \dfrac {1}{q^{2}})$$
Given that $a\_3=8$ and $S\_3=14$, substitute these value... | cn_k12 | 312,154 | q=2 |
fc2bb149c4757443f25e266255acdb8b9520ebb72631e6ebe9bab30cb0e59ab9 | Given that ${a_n}$ is a geometric sequence with all positive terms, and $3a_1$, $\frac{1}{2}a_3$, $2a_2$ form an arithmetic sequence, find the value of $\frac{a_{2014} + a_{2015}}{a_{2012} + a_{2013}}=$ ( ).
A: $3$ or $-1$
B: $9$ or $1$
C: $1$
D: $9$ | Let $q > 0$ be the common ratio of the positive geometric sequence ${a_n}$. Since $3a_1$, $\frac{1}{2}a_3$, $2a_2$ form an arithmetic sequence, we can write the following equation:
$$a_3 = 3a_1 + 2a_2.$$
Substituting $a_3 = a_1q^2$ and $a_2 = a_1q$ into the equation above, we get:
$$a_1q^2 = 3a_1 + 2a_1q,$$
which can ... | cn_k12 | 477,351 | D |
fceb4982b6bedb736cb959d5d272b872c720984b68dd9752c0acdc021f547b3e | Two cards are drawn successively from a standard deck of 52 cards without replacement. What is the probability that the first card is a King and the second card is an Ace? | 1. **Probability of drawing a King as the first card**: There are 4 Kings in the deck of 52 cards, so the probability is:
\[
P(\text{King first}) = \frac{4}{52} = \frac{1}{13}
\]
2. **Probability of drawing an Ace as the second card**: After drawing one King, 51 cards remain. There are still 4 Aces in the dec... | synthetic_math | 887,145 | \frac{4}{663} |
b9ff40f57f73f1d836104c8ea5d50c99f151819316a2b9623c07d046f23d1ab0 | An infinite geometric series has a common ratio of $\frac{1}{4}$ and a sum of $48$. What is the second term of the sequence? | Let the first term of the series be $a$. Since the sum of the infinite geometric series is given by the formula $S = \frac{a}{1-r}$, where $r$ is the common ratio, substituting the known values we get:
\[ 48 = \frac{a}{1 - \left(\frac{1}{4}\right)} = \frac{a}{\frac{3}{4}} = \frac{4a}{3} \]
Solving for $a$, we find:
\[... | synthetic_math | 826,673 | 9 |
51b23c7fef26cee8de2ace3b5b53797220fb9f86c68b0d89c1d74ef81bee3c83 | Daniel collects Russian dolls that normally cost $4 each. He saves enough money to buy 15 Russian dolls. However, the price suddenly drops to $3 each. How many Russian dolls can he buy now at the discounted price, given his savings? | Daniel initially saved enough money to buy 15 Russian dolls at $4 each. So, the total amount of money he saved is:
15 dolls * $4/doll = $60
Now that the price has dropped to $3 each, we can calculate how many dolls he can buy with his $60 savings:
$60 / $3/doll = 20 dolls
So, Daniel can now buy $\boxed{20}$ Russia... | orca_math | 566,177 | 20 |
30bf7531b75a4e950bc606cbe9e5bfd4520c20dc3cf711855125e4a1205a5ab7 | In a team of three employees, M, N, and P, they are paid a total of $3,200 per month by their employer. The salaries are distributed as follows:
1. Employee M is paid 20% more than Employee N.
2. Employee N earns exactly the median salary of the entire three-person team.
3. Employee P earns 35% less than Employee M.
... | Let's denote the monthly salary of Employee N as \( N \).
According to the information given:
1. Employee M is paid 20% more than Employee N, so the salary of Employee M, which we'll denote as \( M \), is \( M = N + 0.20N = 1.20N \).
2. Employee N earns the median salary, which means that one employee earns more tha... | orca_math | 675,926 | \$837.38 |
f92866e9bd5df316b43f74e074241a199311abab54ba34f781b41c89e5bd9007 | Given that $α \in (0,π)$, and $\sin α= \frac {3}{5}$, find the value of $\tan (α- \frac {π}{4})$. | We have two cases to consider based on the given interval for $α$:
Case 1: When $α \in (0, \frac {π}{2})$, we know that $\cos α$ is positive. Using the identity $\cos^2 α + \sin^2 α = 1$, we can find $\cos α$:
$$
\cos α = \sqrt{1 - \sin^2 α} = \frac{4}{5}
$$
Now, we can find $\tan α$:
$$
\tan α = \frac{\sin α}{\cos... | cn_k12 | 442,810 | -7 |
e5b67594332facb009d9b1289029f1d6a37e40291de2b4dd064f2f2860f04ed5 | Which of the following functions represents a quadratic function for the two variables involved?
A: The area $S$ of an equilateral triangle and the side length $x$ of the equilateral triangle.
B: The distance $s$ from school when Xiaoxi rides a bike at a constant speed, and the time $t$ Xiaoxi rides the bike.
C: The... | To determine which of the given options represents a quadratic function for the two variables involved, let's analyze each option step by step:
**Option A: The area $S$ of an equilateral triangle and the side length $x$ of the equilateral triangle.**
The formula for the area of an equilateral triangle in terms of its... | cn_k12 | 345,211 | A |
99aea32ed2b0cb71bba5a3e8e5d41ee693d40466bafae6cb76e15627a9c84809 | How many triangles are in the figure provided? The figure is composed of a triangular grid with three rows. The top row has three small equilateral triangles, the second row has two, and the third row has one. Additionally, there is one large triangle inverted relative to the others, sharing a base with the bottom row ... | 1. **Counting small triangles**: There are 1 + 2 + 3 = 6 small equilateral triangles within the grid.
2. **Identifying larger triangles**:
- There is 1 large triangle formed by combining all 6 small triangles.
- There are 2 additional triangles, each formed by combining 3 small triangles (one using the three tria... | synthetic_math | 833,177 | 9 |
a97487a85c35197642f2f1819805bd579536ace484fa0a98fa4090bfcae6f666 | Simplify first, then evaluate: $(1-\frac{2}{{m+1}})\div \frac{{{m^2}-2m+1}}{{{m^2}-m}}$, where $m=\tan 60^{\circ}-1$. | To simplify and then evaluate the given expression with $m=\tan 60^{\circ}-1$, we proceed as follows:
1. **Simplify the Expression:**
First, we rewrite the original expression by breaking it down into simpler parts:
\[
\begin{align*}
Original\ expression &= \left(1-\frac{2}{{m+1}}\right)\div \frac{{{m^2}-2m+1}}{{{m^... | cn_k12 | 289,394 | \frac{3-\sqrt{3}}{3} |
0b8cb147ef77fa47c99efa389d7394bdf85ba5c23309ab783c7c6f664755576d | Given that the equation of a hyperbola is $\dfrac{x^{2}}{a}-y^{2}=1$ where $a>0$, and one of its asymptote lines is $x+2y=0$, what is the eccentricity of the hyperbola?
A: $\dfrac{\sqrt {5}}{2}$
B: $\dfrac{\sqrt {17}}{4}$
C: $\dfrac{\sqrt {3}}{2}$
D: $\dfrac{\sqrt {15}}{4}$ | To find the eccentricity of the hyperbola given by $\dfrac{x^{2}}{a}-y^{2}=1$ with an asymptote line $x+2y=0$, we follow these steps:
1. **Identify the Asymptote Equation**: The general form of the asymptote lines for a hyperbola $\dfrac{x^{2}}{a}-y^{2}=1$ is $y=\pm \dfrac{1}{\sqrt {a}}x$.
2. **Match the Given Asymp... | cn_k12 | 481,973 | \dfrac{\sqrt{5}}{2} |
cb43d3ea98f91439bd125c3e5a758e7d4efb9cb3c794043a84cc91341c290ec5 | Elective 4-4: Coordinate System and Parametric Equations
Take the origin \\(O\\) of the Cartesian coordinate system as the pole, the positive half-axis of the \\(x\\)-axis as the polar axis, and both coordinate systems have the same unit length. It is known that the parametric equation of line \\(l\\) is \\(\begin{ca... | Solution:
\\((1)\\) Since the parametric equation of the line is \\(\begin{cases} & x=\dfrac{\sqrt{2}}{2}t \\\\ & y=3+\dfrac{\sqrt{2}}{2}t \\\\ \end{cases}\\) (where \\(t\\) is the parameter),
Eliminating the parameter \\(t\\) yields the standard equation of line \\(l\\) as \\(x-y+3=0\\).
From the polar equation of... | cn_k12 | 444,301 | 3 |
bf243a79fe6a876b44ddf5c7f268cbe5f9897a79fb452069fe16d4f0089cd455 | We select a real number $\alpha$ uniformly and at random from the interval $(0,500)$. Define \[ S = \frac{1}{\alpha} \sum_{m=1}^{1000} \sum_{n=m}^{1000} \left\lfloor \frac{m+\alpha}{n} \right\rfloor. \] Let $p$ denote the probability that $S \ge 1200$. Compute $1000p$.
[i]Proposed by Evan Chen[/i] | 1. **Switch the order of summation:**
\[
S = \frac{1}{\alpha} \sum_{n=1}^{1000} \sum_{m=1}^{n} \left\lfloor \frac{m+\alpha}{n} \right\rfloor
\]
This can be rewritten as:
\[
S = \frac{1}{\alpha} \sum_{n=1}^{1000} \left[ \sum_{m=1}^{n} \frac{m+\alpha}{n} - \sum_{m=1}^{n} \left\{ \frac{m+\alpha}{n} \righ... | aops_forum | 213,638 | 5 |
b74164f1b05c8b12cce809edf17bd63fc28648a8a0c15ad23caf604ebac5bc5e | Some amount of water is added with 9 ltr of a solution containing 57 % of alcohol in the water. The % of alcohol in the new mixture is 42.75. How much water was added? | Let's denote the amount of water added as \( W \) liters.
Initially, we have 9 liters of a solution with 57% alcohol. This means there is \( 0.57 \times 9 \) liters of alcohol in the original solution.
After adding \( W \) liters of water, the total volume of the new mixture becomes \( 9 + W \) liters.
The percentag... | orca_math | 669,905 | 3 |
17271f8c3b94a45875581fe924ead7bfe2744111249d4db8929280fd44e76e62 | There are several solid and hollow circles arranged in a certain pattern as follows:
●○●●○●●●○●○●●○●●●…
Among the first 2001 circles, the number of hollow circles is ( )
A: 660
B: 665
C: 667
D: 670 | **Solution**: Divide 2001 by 9, we get 222 remainder 3, which means there is 1 more hollow circle among the remaining 3 circles.
Therefore, the number of hollow circles is $222 \times 3 + 1 = 667$.
Hence, the correct choice is $\boxed{C}$. | cn_k12 | 388,568 | C |
337d67105001fcd730269924b750aeefac4c7a2b6c827b751a1ea0a25557d270 | Given $y=f(x)$, where $x \in [0,1]$, and $f'(x) > 0$, which of the following relationships must hold?
A: $f(0) 0$
C: $f(1) > f(0)$
D: $f(1) < f(0)$ | Since $f'(x) > 0$, it indicates that $f(x)$ is monotonically increasing on the interval $[0,1]$. Therefore, $f(1) > f(0)$. The correct choice is $\boxed{\text{C}}$. | cn_k12 | 413,032 | \text{C} |
dfe49eda6d83771b48c2a34cbceb4356c02e3f4f2ffee28c23822a2f20abe450 | In $\triangle ABC$, $AB=17$, $AC=25$, and $BC=28$. Points $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively, and $P$ is a point on $\overline{BC}$. Let $Q$ be the second intersection point of the circumcircles of $\triangle BMP$ and $\triangle CNP$. It is known that as $P$ moves along... | 1. **Identify the given elements and their properties:**
- In $\triangle ABC$, we have $AB = 17$, $AC = 25$, and $BC = 28$.
- Points $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively.
- Point $P$ is a point on $\overline{BC}$.
- $Q$ is the second intersection point of the ci... | aops_forum | 215,653 | 710 |
fa92b38c1451d5e38f01571016dead5e69ff0a96a8ae11dafbbd8e8e037595e2 | Given that the line $x-my+1=0$ intersects the circle $C: \left(x-1\right)^{2}+y^{2}=4$ at points $A$ and $B$, find one value of $m$ that satisfies the condition "$\triangle ABC$ has an area of $\frac{8}{5}$." | To solve for the value of $m$ that satisfies the condition "$\triangle ABC$ has an area of $\frac{8}{5}$," we proceed as follows:
**First Approach:**
1. Given the circle $C: \left(x-1\right)^{2}+y^{2}=4$, we identify its center as $C(1,0)$ and its radius as $r=2$.
2. The area of $\triangle ABC$, $S_{\triangle ABC}$, ... | cn_k12 | 381,105 | 2 |
1342bee4da1ac35481b2ea14b17aef3d4516d78d0af78944328d9d91430af5a9 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 20$, $BC = 26$, and $TX^2 + TY^2 + XY^2 = 1750$. Find $XY^2$. | - Let $O$ be the center of the circumcircle of $\triangle ABC$, with $OT$ intersecting $BC$ at $M$. $P$ and $Q$ are midpoints of $BT$ and $CT$, respectively, implying $MT = 4\sqrt{15}$ using similar triangle properties and Pythagoras theorem.
- Since $\triangle BTX$ and $\triangle CTY$ are right triangles, and $\angle ... | synthetic_math | 835,836 | 950 |
a6ed58d80f6cfe1ef8365e368994bf0b9661bbaeb04d321aa79cdb7651aae0a0 | Moving the factor outside the square root of $\left(a-b\right)\sqrt{-\frac{1}{a-b}}$ into the square root results in ______. | To move the factor outside the square root of $\left(a-b\right)\sqrt{-\frac{1}{a-b}}$ into the square root, we follow these steps:
1. First, we observe that for the square root to be defined, the expression inside must be negative, which implies $-\frac{1}{a-b} > 0$. This leads to the conclusion that $a-b 0$ since $a... | cn_k12 | 340,427 | -\sqrt{b-a} |
3812162b727e921edd162d075b3c5ed8b28842d3990d869afdc58953f948c386 | A shopkeeper sold an article offering a certain discount and earned a profit of 21.6%. If no discount was offered, the percentage of profit earned would have been 28%. What was the discount offered by the shopkeeper? | Let's assume the cost price (CP) of the article is $100.
When the shopkeeper offers a discount and earns a profit of 21.6%, the selling price (SP) with discount is:
SP with discount = CP + Profit
SP with discount = $100 + 21.6% of $100
SP with discount = $100 + $21.60
SP with discount = $121.60
If no discount was of... | orca_math | 636,332 | 5\% |
a0710259e83e3f3a8584f83e1992965539af945ab34095b30be661337580aae2 | Given the function $f(x)=2\sin^2\left(\frac{\pi}{4}+x\right)-\sqrt{3}\cos 2x-1$, where $x\in\mathbb{R}$,
$(1)$ The graph of the function $h(x)=f(x+t)$ is symmetric about the point $\left(-\frac{\pi}{6},0\right)$, and $t\in(0,\pi)$. Find the value of $t$;
$(2)$ For $x\in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, it a... | Solution:
$(1)$ Since $f(x)=2\sin^2\left(\frac{\pi}{4}+x\right)-\sqrt{3}\cos 2x-1=1-\cos\left(\frac{\pi}{2}+2x\right)-\sqrt{3}\cos 2x-1$,
Then $h(x)=f(x+t)=2\sin\left(2x+2t-\frac{\pi}{3}\right)$,
Given that the point $\left(-\frac{\pi}{6},0\right)$ is a center of symmetry for the graph of $h(x)$,
Then $t=\frac{k\pi... | cn_k12 | 355,110 | (-1,4) |
f4870754126c34befb68ce5445677316c5731b369d4524274af2b602e6312c7a | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 320 . how much does c have ? | Let's denote the amount of money that A, B, and C have as A, B, and C respectively.
From the information given, we have the following equations:
1) A + B + C = 500 (since A, B, and C have Rs. 500 between them)
2) A + C = 200 (since A and C together have Rs. 200)
3) B + C = 320 (since B and C have Rs. 320 together)
W... | orca_math | 553,835 | 20 |
769d94c37c5e50f0e8ecaed1c1e2a84a9c7477638d1da80aa7b2be2d60a6ba21 | \section*{Problem 4 - 011244}
Given is a convex plane quadrilateral.
It is to be proven that for the quotient \(q\) of the largest and the smallest of all distances between any two vertices, it always holds that: \(q \geq \sqrt{2}\). | }
For a square, it is obvious that the equality sign holds. All other quadrilaterals have at least one interior angle that is greater than \(90^{\circ}\); let this be \(\alpha\) without loss of generality.
In a convex quadrilateral, one can use the cosine rule for the diagonals:
\[
f^{2}=a^{2}+d^{2}-2 a d \cos \alph... | olympiads | 185,617 | proof |
bf58bcaea5e029e44ee73dee928d65f96b4712b5ae9d7f2963ade03cc59fb90f | If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 52, what is the compound interest on the same sum at the same rate and for the same time? | To find the compound interest, we first need to calculate the principal amount using the simple interest formula:
Simple Interest (SI) = (Principal (P) * Rate (R) * Time (T)) / 100
We are given:
Simple Interest (SI) = Rs. 52
Rate (R) = 5% per annum
Time (T) = 2 years
Plugging these values into the formula, we get:
... | orca_math | 693,966 | 53.30 |
26d5d88dc89327eb7cb417611950ab162826d74d2bd05f34329e3da3bed12117 | In a convex pentagon $ABCDE$, the points $M$, $N$, $P$, $Q$, $R$ are the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EA$, respectively. If the segments $AP$, $BQ$, $CR$ and $DM$ pass through a single point, prove that $EN$ contains that point as well.
[i]Yugoslavia[/i] | 1. Let the four lines \(AP\), \(BQ\), \(CR\), and \(DM\) concur at point \(X\). We need to show that \(EN\) also passes through \(X\).
2. Let \(EX\) meet \(BC\) at \(N'\). We will show that \(N' = N\), where \(N\) is the midpoint of \(BC\).
3. Consider the angles formed by the lines at point \(X\):
\[
\angle BXM ... | aops_forum | 222,745 | null |
b6468982b0e9dcb02d376d99f64ffec7870ab1d8035c0a259862a31e913a30b5 | A group of 4 fruit baskets contains 9 apples, 15 oranges, and 14 bananas in the first three baskets and 2 less of each fruit in the fourth basket. How many fruits are there? | Let's calculate the number of fruits in the first three baskets:
- Apples: 9
- Oranges: 15
- Bananas: 14
Total fruits in the first three baskets = 9 apples + 15 oranges + 14 bananas
Total fruits in the first three baskets = 38
Now, let's calculate the number of fruits in the fourth basket:
- Apples: 9 - 2 = 7
- Ora... | orca_math | 646,427 | 70 |
2857fa60c827820b83be32a04dcf11010c947c2b5797df1787cbca1e081c9a96 | 14. Given two non-decreasing sequences of non-negative integers, with different first terms, and each subsequent term in the sequences is the sum of the two preceding terms. If the seventh term of both sequences is $N$, then the smallest possible value of $N$ is $(\quad)$.
(A) 55
(B) 89
(C) 104
(D) 144
(E) 273 | 14. C.
Let the first two terms of the first sequence be $x_{1}, x_{2}\left(x_{1}, x_{2} \in \mathrm{N}\right)$, and the first two terms of the second sequence be $y_{1}, y_{2}\left(y_{1}, y_{2} \in \mathrm{N}\right)$. Then the seventh terms of the two sequences are $5 x_{1}+8 x_{2}$ and $5 y_{1}+8 y_{2}$, respectively... | cn_contest | 727,357 | C |
622143172ece60efecc8d0234e82613ea2ac034f05ca02ec45fc06801e838f40 | II. (Total 50 points) Let $x_{1}, x_{2}, \cdots, x_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ be two sets of real numbers satisfying the following conditions:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\sum_{i=1}^{n}\left|x_{i}\right|=1$;
(3) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}\left(a_{1}>a_{n}, n \geqslant 2\r... | Let $n=2$, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=0, \\
\left|x_{1}\right|+\left|x_{2}\right|=1 .
\end{array}\right.
\]
Then $x_{1}=\frac{1}{2}, x_{2}=-\frac{1}{2}$ or $x_{1}=-\frac{1}{2}, x_{2}=\frac{1}{2}$.
From $\left|\sum_{i=1}^{n} a_{i} x_{1}\right| \leqslant m\left(a_{1}-a_{n}\right)$,
we get $\frac{1}{2}... | cn_contest | 713,421 | \frac{1}{2} |
b8c6a7a262c0b08061e6bd8c1f444ade780bfc71b1e11485bfabfb08b95166f8 | (a) Let \(a_{0}, a_{1}, \ldots, a_{2024}\) be real numbers such that \(\left|a_{i+1}-a_{i}\right| \leqslant 1\) for \(i=0,1, \ldots, 2023\). Find the minimum possible value of
\[ a_{0} a_{1}+a_{1} a_{2}+\cdots+a_{2023} a_{2024} \]
(b) Does there exist a real number \(C\) such that
\[ a_{0} a_{1}-a_{1} a_{2}+a_{2} a_... | (a) The minimum value is -506. Note that from $\left|a_{i}-a_{i-1}\right| \leq 1$ it follows that
$$
a_{i} a_{i-1}=\frac{\left(a_{i}+a_{i-1}\right)^{2}-\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{1}{4}
$$
Adding this for $i=1,2, \ldots, 2024$, we obtain that
$$
a_{0} a... | olympiads_ref | 195,087 | -506 |
cac3785d0d8ffedc49cb53aee3898154d123ff7b149c1db78d4d1027078a028f | (EGMO 2021 P3) Let $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the points of intersection of the external bisector of angle $\widehat{BAC}$ with the altitudes from vertices $B$ and $C$, respectively. Let $M$ be a point on the segment $[EC]$ such that $\widehat{EMA}=\widehat{ACB}$. Let $N$ be a p... | To plot point $M$, we introduce an intermediate point.
Think of an intermediate point! An interesting intermediate point is the second intersection point of the circumcircles of triangles, one of whose angles is involved in the angle equality.
Here, we consider the point $X$, the second point of intersection of the c... | olympiads | 80,696 | proof |
91efc64429241638a8ce3aa4a6576dd3682aa936fa5af746a136348ad462765b | 5.
The Bank of Russia will ensure the conversion of the "digital ruble" into another form of money (rubles) at a one-to-one ratio.
Correct answers: Stablecoins, backed by cash or gold, are an example of a CBDC., The Bank of Russia will ensure the conversion of the "digital ruble" into another form of money (rubles) ... | Provide the answer in rubles, without spaces and units of measurement. Round the answer to the nearest whole number.
Answer: $\qquad$
Correct answer: 1321412
Question 20
Score: 7.00
Financial Literacy 11th Grade Day 1
Ivan and Petr, twin brothers, went on a vacation by the sea together and purchased two different ... | olympiads | 2,778 | 1321412 |
f6e48e1d3ccf8e85a24e035f11058a45cf6b09fab00ced6e6b0246061d523a5e | Given that $S_{n}$ is the sum of the first $n$ terms of the arithmetic sequence $\{a_{n}\}$, if $a_{1}=-2014$, and $\dfrac{S_{2014}}{2014}- \dfrac{S_{2008}}{2008}=6$, then $S_{2017}=$ ()
A: $1$
B: $2017$
C: $2008$
D: $4034$ | **Analysis**
This question examines the sum formula of an arithmetic sequence. The key to solving the problem is to find the common difference $d$, which is a basic question.
From the given conditions, we can derive $\dfrac{S_{n}}{n}=a_{1}+ \dfrac{n-1}{2}d$. Combining this with $\dfrac{S_{2014}}{2014}- \dfrac{S_{2008... | cn_k12 | 494,966 | D |
d64cf73c7c06806af0be90864b4c2fe068d43b44d4516fc64d7da85eb03d7ea3 | For each real value of $p$, find all real solutions of the equation $\sqrt{x^2 - p}+2\sqrt{x^2-1} = x$. | 1. **Initial Conditions and Substitution:**
Given the equation:
\[
\sqrt{x^2 - p} + 2\sqrt{x^2 - 1} = x
\]
Note that for the square roots to be defined, we must have \(x^2 \geq p\) and \(x^2 \geq 1\). Since \(x^2 \geq 1\), it follows that \(x \geq 1\).
Let \(t = \sqrt{x^2 - 1}\). Then \(t \geq 0\) an... | aops_forum | 252,082 | x = \frac{4 - p}{\sqrt{16 - 8p}} |
81cde1610b71a33c12c0fac900381290c4787c0b7fe7e1d63a4751430100268d | Solution x is 30% chemical a and 70% chemical b by volume. Solution y is 40% chemical a and 60% chemical b by volume. A mixture of x and y has a certain percentage of chemical a. If solution x makes up 80% of the mixture, what percent of the mixture is chemical a? | To find the percentage of chemical a in the mixture, we need to calculate the weighted average of chemical a from both solutions x and y based on their respective proportions in the mixture.
Let's denote the percentage of chemical a in the mixture as P_a.
Since solution x is 30% chemical a and makes up 80% of the mix... | orca_math | 551,909 | 32\% |
ef0997ad00441a4a47763d7866df2a1227331306890fd5bab2b21cdde0c11841 | Let $a$, $n$, and $l$ be real numbers, and suppose that the roots of the equation \[x^4 - 10x^3 + ax^2 - nx + l = 0\] are four distinct positive integers. Compute $a + n + l.$ | By Vieta's formulas, the sum of the roots of this quartic equation is $10$. Considering distinct positive integers that sum to $10$, we can have combinations such as $\{1, 2, 3, 4\}$, since they are the smallest integers whose sum is $10$ and they are distinct.
The roots of the equation are therefore $1$, $2$, $3$, an... | synthetic_math | 814,798 | 109 |
ab82d6978d3383da69da77bd78de8d7f79aad0ec05b51c64c3542e25070a5efc | In $\triangle A B C$, $M$ is the midpoint of $A B$, $P$ is a point inside $\triangle A B C$, $Q$ is the point symmetric to $P$ with respect to $M$, $A P$ intersects $B C$ at point $D$, and $B P$ intersects $A C$ at point $E$. Prove that $A, B, D, E$ are concyclic if and only if
$$
\angle A C P=\angle Q C B .{ }^{[1]}
$... | Prove that if $\angle A C P=\angle Q C B$, as shown in Figure 1, let the extension of $C B$ intersect the extension of $A Q$ at point $N$.
$$
\text { From } \frac{S_{\triangle A C P}}{S_{\triangle D C P}}=\frac{A P}{P D} \Rightarrow \frac{A P}{P D} \cdot \frac{C D}{A C}=\frac{\sin \angle A C P}{\sin \angle D C P} \text... | olympiads | 112,303 | proof |
215aad667722622b7602855d97c03e1124d8d78443e8bcc85a42f88e5c907180 | Given non-zero vectors $\overrightarrow{m}$ and $\overrightarrow{n}$ that satisfy $3|\overrightarrow{m}| = 2|\overrightarrow{n}|$ and $\langle \overrightarrow{m}, \overrightarrow{n} \rangle = 60^{\circ}$. If $\overrightarrow{n} \perp (t\overrightarrow{m} + \overrightarrow{n})$, find the value of the real number $t(\ )$... | Given non-zero vectors $\overrightarrow{m}$ and $\overrightarrow{n}$ that satisfy $3|\overrightarrow{m}| = 2|\overrightarrow{n}|$, thus $|\overrightarrow{m}| = \frac{2}{3}|\overrightarrow{n}|$.
Also, it is given that $\langle \overrightarrow{m}, \overrightarrow{n} \rangle = 60^{\circ}$ and $\overrightarrow{n} \perp (t\... | cn_k12 | 352,161 | -3 |
bbf0b57db5eba8d9b8498bdd15ccaad5b2e831a186dbe2c89c5d171a309793d1 | For any real number sequence $A=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},\cdots \right\}$, define $\Delta A$ as the sequence $\left\{ {{a}_{2}}-{{a}_{1}},{{a}_{3}}-{{a}_{2}},{{a}_{4}}-a{}_{3}^{{}},\cdots \right\}$, whose $n$th term is $\left({{a}_{n+1}}-{{a}_{n}}\right)$. Suppose that all terms in the sequence $\Delta \le... | [Analysis]
This problem primarily tests the properties of a newly defined sequence and examines the formula for the sum of the first $n$ terms and the general term formula of an arithmetic sequence. The sequence defined in the problem is a second-order arithmetic sequence. The definition of a higher-order arithmetic se... | cn_k12 | 295,202 | 1000 |
7e40c469fd0a5e37300edfe07c658780d5830264fac84513476f833eff03343a | 3.42 The lottery has 50 spaces arranged in sequence, and each participant should fill in the 50 natural numbers from 1 to 50 on their lottery ticket without repetition, with the order determined by themselves. The host also fills in these 50 natural numbers on a lottery ticket as the base. In a lottery ticket, if any o... | [Solution] Fill 26 lottery tickets as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50, \\
2,3,4, \cdots, 26,1,27, \cdots, 50, \\
3,4,5, \cdots, 1,2,27, \cdots, 50, \\
\cdots \quad \cdots \quad \cdots \\
25,26,1, \cdots, 23,24,27, \cdots, 50, \\
26,1,2, \cdots, 24,25,27, \cdots, 50,
\end{array}
$$
That... | olympiads | 109,554 | 26 |
2fe33698ecce6abbedbe6b8fda8fc925bac95606e7416453d14b4efa005cbeda | if anangletis defined as 1 percent of 1 degree , then how many anglets are there in a sixth of a circle ? | A full circle is 360 degrees. A sixth of a circle would therefore be:
360 degrees / 6 = 60 degrees
If 1 anglet is defined as 1 percent of 1 degree, then there are 100 anglets in 1 degree (since 1 percent is 1/100th of a whole).
Therefore, in 60 degrees, there would be:
60 degrees * 100 anglets/degree = 6000 anglets... | orca_math | 569,677 | 6000 |
081b53a0431eb94e15e998ad5c45099f03f0f24687f66cae1387979af0d5cc37 | Example 9. Solve the equation
$$
5^{2 x}-2 \cdot 5^{x}-15=0
$$ | Solution. Let $t=5^{x}$. Then $t^{2}-2 t-15=0$. From this, we find $t_{1}=5, t_{2}=-3$. Thus, the given equation is equivalent to the system of equations
$$
5^{x}=5, \quad 5^{x}=-3
$$
## § 3. EXPONENTIAL EQUATIONS
The second equation of this system has no roots, since $-3<0$ for any $x$, while from the first equatio... | olympiads | 40,529 | 1 |
a9da635d0d92de9d2b7f0b72f584ba4477a513c2be36e9648b81961c9e978e0f | Given that the function $f(x)$ is differentiable at $x=x_0$, and $\lim\limits_{\Delta x \to 0} \frac{f(x_0+3\Delta x)-f(x_0)}{\Delta x}=1$, find the value of $f'(x_0)=$ ( ).
A: $1$
B: $0$
C: $3$
D: $\frac{1}{3}$ | We are given that the function $f(x)$ is differentiable at $x=x_0$ and the limit $\lim\limits_{\Delta x \to 0} \frac{f(x_0+3\Delta x)-f(x_0)}{\Delta x}=1$.
First, let's rewrite the given limit using the definition of a derivative:
$$f'(x_0) = \lim\limits_{\Delta x \to 0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$
Now,... | cn_k12 | 334,005 | \frac{1}{3} |
9a8fc7c5716526622736138ded1f38bd0382a57ced7dbb7727f4406e3f67a340 | The sum of the first $n$ terms of a sequence is $S_n = 2n^2 + n + 1$. Find the general formula for the sequence. | Based on $S_n = 2n^2 + n + 1$, we calculate the value of $a_1$ when $n=1$. Then, using the formula $a_n = S_n - S_{n-1}$, we find the relationship for $a_n$ when $n \geq 2$. Finally, we check whether $a_1$ satisfies this relationship.
This problem tests the method of finding the general formula of a sequence given $S_... | cn_k12 | 349,714 | a_n = \begin{cases}
4 & \text{if } n=1 \\
4n - 2 & \text{if } n \geq 2
\end{cases} |
728610a48c54d61918d05c33aac96dd0e05e9097f20184ad0e8fb80fbedf9b1f | B. As shown in Figure 2, in quadrilateral $A B C D$, $A C$ and $B D$ are diagonals, $\triangle A B C$ is an equilateral triangle, $\angle A D C=$ $30^{\circ}, A D=3, B D=5$. Then the length of $C D$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5 | B. B.
As shown in Figure 10, construct an equilateral $\triangle CDE$ with $CD$ as a side, and connect $AE$.
$$
\begin{aligned}
& \text{Given } AC=BC, \\
& CD=CE, \\
& \angle BCD \\
= & \angle BCA + \angle ACD \\
= & \angle DCE + \angle ACD = \angle ACE,
\end{aligned}
$$
we know that $\triangle BCD \cong \triangle AC... | cn_contest | 725,140 | B |
3b75107cce25e3fcbac613af841075a6344d6ffd880f2ec6e5c873a99d70c1ef | Determine all sequences $a_1, a_2, a_3, \dots$ of nonnegative integers such that $a_1 < a_2 < a_3 < \dots$ and $a_n$ divides $a_{n - 1} + n$ for all $n \geq 2$. | To determine all sequences \(a_1, a_2, a_3, \dots\) of nonnegative integers such that \(a_1 < a_2 < a_3 < \dots\) and \(a_n\) divides \(a_{n-1} + n\) for all \(n \geq 2\), we proceed as follows:
1. **Initial Observation**:
Since the sequence \((a_i)\) is strictly increasing and \(a_1 \geq 0\), we have:
\[
a_n... | aops_forum | 218,680 | null |
10936ba0cec5ce5b6ece62c2d060185707b2e16ff49fd579195e419a13c4bae8 | A particle begins at the point $(5,5)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at point $(a,b)$, it moves at random to one of the points $(a-1,b)$, $(a,b-1)$, $(a-1,b-1)$, or $(a-2,b-2)$, with each point having an equal probability of being chosen. Find the probabilit... | To solve this new problem, we can use a similar recursive approach as in the original problem, with a modified probability distribution.
1. **Recursive Probability Function**: Define \( P(x, y) \) to represent the probability of hitting $(0,0)$ starting from $(x,y)$,
\[
P(x,y) = \frac{1}{4} P(x-1,y) + \frac{1}{4... | synthetic_math | 877,669 | \frac{3805}{16384} |
01a57f93ac8fcd0fdcf99c8740c470e513f5aa63815334397438aba9fc6d20ba | The cost of four pencils and three pens is $\$3.70$, and the cost of three pencils and four pens is $\$4.20$. What is the cost of one pencil and one pen? | Let the cost of one pencil be $x$ and the cost of one pen be $y$. From the problem, we can set up the following system of equations:
\[
\begin{align*}
4x + 3y &= 3.70 \\
3x + 4y &= 4.20 \\
\end{align*}
\]
We are trying to find the value of $x + y$. We can solve this system by multiplying the first equation by 4 and the... | synthetic_math | 759,972 | 1.1286 |
7f75b5424dcda43b4a783f3e75d2f805481e184dc3a7d15a87b7a975eaf35988 | 11.2. Find all differentiable functions $f:(0,+\infty) \rightarrow(0,+\infty)$ that satisfy the relations $f\left(\frac{1}{2}\right)=1$ and $f^{\prime}(x)=-f^{3}(x)$. | ## Solution.
We have $\left(f^{-2}(x)\right)^{\prime}=-2 f^{-3}(x) f^{\prime}(x)=2 f^{-3}(x) f^{3}(x)=2$, which means the derivative of the function $f^{-2}(x)$ is a constant. It follows that the function $f^{-2}(x)$ is a linear function, which can be written in the form $f^{-2}(x)=a x+b$. By differentiating, we get $... | olympiads | 85,195 | f(x)=\frac{1}{\sqrt{2x}} |
84c63a02c1f1eb3e11101cd6afd151b369e4ebe6132dfde04a3346cd8648a664 | The Pinedale bus line travels at an average speed of 60 km/h, and has stops at a certain interval along its route. Yahya wants to go from his house to the Pinedale mall, which is 8 stops away and 40 kilometers away from his house. How many minutes apart are the stops along the bus route? | First, let's find out the total time it takes for the bus to travel from Yahya's house to the Pinedale mall.
Since the distance is 40 kilometers and the bus travels at an average speed of 60 km/h, we can use the formula:
Time = Distance / Speed
Time = 40 km / 60 km/h
To convert hours into minutes (since there are 6... | orca_math | 588,259 | 5.71 |
2cd4964dc08c575d9b731befe4514a2eacf38a4ea4aaaa2920f5ff119eab9090 | 6. In the Cartesian coordinate system $x O y$, it is known that points $A(3,5)$, $B(1,3)$, and $C(4,3)$. Now, $\triangle A B C$ is translated to the position of $\triangle A_{1} B_{1} C_{1}$, and the translation direction forms an angle of $30^{\circ}$ with the $x$-axis. If the area of the common part of $\triangle A B... | 6. A.
As shown in Figure 2, let $A_{1} B_{1}$ intersect $B C$ at point $K$, then
$$
\begin{array}{l}
\frac{C K}{C B}=\sqrt{\frac{S_{\text {common }}}{S_{\triangle U C}}}=\sqrt{m}, \\
\frac{B K}{C B}=1-\frac{C K}{C B}=1-\sqrt{m} .
\end{array}
$$
Let the line $l$ passing through points $C$ and $C_{1}$ intersect $A B$ a... | cn_contest | 716,288 | A |
aee1170118afe34268793f5e154f0c4dbbe760d306d3a78e0a02b5e7a9391faa | Determine the molecular weight of 5 moles of Aluminum phosphate and 3 moles of Sodium sulfate, and then find the total molecular weight of the mixture. | First, we need to determine the molecular formula for each compound and then calculate their respective molecular weights.
Aluminum phosphate has the chemical formula AlPO4. To find its molecular weight, we add the atomic weights of all the atoms in the formula:
Aluminum (Al) has an atomic weight of approximately 26.... | orca_math | 664,259 | 1035.90 |
c9ced8091709edf8379c1780b828bac1dc51a42caacc95645575b57536ef1ee3 | Example 1. Two students took four tests, their average scores were different, but both were below 90 and were integers. They then took a fifth test, after which their average scores both increased to 90. What were the scores of the two students on the fifth test? | Let the average score of the first four tests for one of the students be $x$ points, and the score of the fifth test be $y$ points. According to the problem, we have
$\frac{4 x+y}{5}=90$, which means $y=450-4 x$.
However, it is clear that the score of the fifth test is higher than 90 points and no more than 100 points,... | cn_contest | 705,886 | 98 \text{ and } 94 |
15c56b8ecdab8c99af603f1d0a110e19d84c57997ac8b9c70ab342614d234e52 | What is the value of $\left(107_{10} + 93_{10}\right) - 47_{10}$ in base 7? | First, calculate the sum of $107_{10}$ and $93_{10}$, which gives $107 + 93 = 200_{10}$.
Next, subtract $47_{10}$ from $200_{10}$, i.e., $200 - 47 = 153_{10}$.
Now, convert $153_{10}$ to base 7. The largest power of $7$ less than or equal to $153$ is $7^2=49$. The coefficient for $49$ in $153$ is found by dividing $15... | synthetic_math | 851,692 | 306_7 |
2a34dcbb3da8528d25c44a8a35d8e82502302f7aa389e590568b48672a9b6dd7 | There are 9 table tennis players: 5 men and 4 women. The task is to select 2 male and 2 female players to pair up for a mixed doubles match. The total number of different pairing methods is ( ).
A: 60
B: 84
C: 120
D: 240 | To determine the number of different pairing methods for the mixed doubles match, we follow these steps:
1. Select 2 male players from the 5 available male players.
- The number of ways to do this is given by the combination formula $C(n, k)$, where $n$ is the total number of items to choose from and $k$ is the num... | cn_k12 | 320,135 | 120 |
634734c6951760178e5f638f1e6be837c4ebf12340f1bb5c5ddb11e58d00d23a | In the tetrahedron $ABCD$, let $K$ and $L$ be the midpoints of edges $AB$ and $CD$ respectively. Prove: every plane containing the line $FL$ divides the tetrahedron into two equal volume parts. | 6. (Czechoslovakia 3)
Solution: Let $\omega$ denote the plane containing line $A B$ and parallel to $C D$. In space, define the point transformation $f: X \rightarrow X^{\prime}$: If $X \in K L$ then $X^{\prime}=\lambda^{\prime}$, if $X \in K L$, let $p$ be the intersection line of the plane $X K L$ and the plane thro... | cn_contest | 703,818 | proof |
6fc5873f61186004036e4e7af0ad0344d4f1de58ba783e98b6408a25a8f96c4a | The event "tossing a fair coin with the head facing up" is ( )
A: impossible event
B: certain event
C: random event
D: definite event | When we toss a fair coin, there are two equally likely outcomes: the coin landing with the head facing up or the coin landing with the tail facing up. Since both outcomes have an equal chance of occurring and neither outcome is guaranteed nor impossible, we classify this event as a random event.
Therefore, the event ... | cn_k12 | 514,590 | C |
b8df72c74cb0d210f0946394aa4239f43223a8968159da3a91c40c9b471649c5 | Doropov D.V.
In the country, there are several cities, some pairs of which are connected by non-stop flights of one of $N$ airlines, and from each city there is exactly one flight of each of the airlines. It is known that from each city you can fly to any other (possibly with layovers). Due to a financial crisis, $N-1... | Consider some path connecting two cities, possibly including some flights that were closed after the crisis. We will show that in this path, any closed flight can be replaced by a sequence of open flights.
Let's number the airlines from 1 to N. In one of the airlines (let's say the first one), all flights are still op... | olympiads | 57,124 | proof |
cf5083add338ff8990907fe12791d13ea7aa4a8d3b49ec7ad966830aa0936e08 | If the proposition "$\exists x \in \mathbb{R}$, $x^{2}+(a-2)x+1 < 0$" is false, then the range of the real number $a$ is ______. | **Analysis**
This question examines the issue of constant validity in functions through logical expressions. First, negate the proposition, then solve it using the concept of constant validity.
**Solution**
Solution: The negation of the proposition "$\exists x \in \mathbb{R}$, such that $x^{2}+(a-2)x+1 < 0$" is:
"f... | cn_k12 | 285,482 | 0 \leqslant a \leqslant 4 |
15f6c5b032cfb0ee2a80f802e7695a2110b43c5137890d5aae433131bb62fdf2 | Calculate the value of $4 \cdot 7 + 5 \cdot 12 + 6 \cdot 4 + 7 \cdot 5$. | Following the order of operations, we first perform the multiplications:
- $4 \cdot 7 = 28$
- $5 \cdot 12 = 60$
- $6 \cdot 4 = 24$
- $7 \cdot 5 = 35$
Now, we add the results:
\[
4 \cdot 7 + 5 \cdot 12 + 6 \cdot 4 + 7 \cdot 5 = 28 + 60 + 24 + 35
\]
Adding these values together in a stepwise manner:
\[
28 + 60 = 88, \qu... | synthetic_math | 820,340 | 147 |
d40fef6475725155a78a76b0e50ded15409aa014923462afd3dfe3184278ae62 | Example 9 Let $x_{i}, y_{i}, \cdots, z_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, prove:
$$\sum_{i=1}^{n} \sqrt{x_{i}^{2}+y_{i}^{2}+\cdots+z_{i}^{2}} \geqslant \sqrt{\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\left(\sum_{i=1}^{n} y_{i}\right)^{2}+\cdots+\left(\sum_{i=1}^{n} z_{i}\right)^{2}}$$ | Let $a=\sum_{i=1}^{n} x_{i}, b=\sum_{i=1}^{n} y_{i}, \cdots, c=\sum_{i=1}^{n} z_{i}$. Without loss of generality, assume $a^{2}+b^{2}+\cdots+c^{2} \neq 0$. Then by the Cauchy-Schwarz inequality, we have
$$\left(a^{2}+b^{2}+\cdots+c^{2}\right)\left(x_{i}^{2}+y_{i}^{2}+\cdots+z_{i}^{2}\right) \geqslant\left(a x_{i}+b y_{... | inequalities | 738,018 | proof |
55bcb6f02bce2d4779c743b9b8adf1561dc07dee4f4c70498eb538ee4cfcf766 | The number of degrees in the measures of the interior angles of a convex hexagon are six consecutive odd integers. What is the number of degrees in the largest angle? | Suppose we list the angle measures in increasing order, and let $x$ be the measure of the third smallest angle in degrees. Then, the six angles, being consecutive odd integers, have measures $x - 4^\circ$, $x - 2^\circ$, $x$, $x + 2^\circ$, $x + 4^\circ$, and $x + 6^\circ$. The sum of the angle measures in a hexagon is... | synthetic_math | 843,158 | 125^\circ |
56e6b428d2b9f0cda8868b3d00993690fc02595b61eb8be50586507fe2c692bf | Given the relationship $C = \frac{4}{7}(F - 40)$ between the Celsius temperature $C$ and the Fahrenheit temperature $F$, find the value of $F$ when $C = 35$. | Starting with the given equation:
\[ C = \frac{4}{7}(F - 40) \]
Setting \( C=35 \), we substitute and get:
\[ 35 = \frac{4}{7}(F - 40) \]
To isolate \( F \), first multiply both sides by 7:
\[ \ 35 \times 7 = 4(F - 40) \]
\[ 245 = 4(F - 40) \]
Next, divide both sides by 4 to solve for \( F-40 \):
\[ \frac{245}{4} =... | synthetic_math | 876,889 | 101.25 |
1d3d0ab7ce287b6a0b64b060a803a3f8f4aa6a1ca0ee78b9fbba8bf188f4c3f6 | In an examination, 35% of the students passed and 481 students failed. How many students appeared for the examination? | Let's denote the total number of students who appeared for the examination as \( T \).
According to the information given, 35% of the students passed the examination. This means that 65% of the students failed the examination because \( 100\% - 35\% = 65\% \).
We are told that 481 students failed the examination, whi... | orca_math | 587,050 | 740 |
End of preview. Expand in Data Studio
DistillationProject: original NuminaMath 50K pool
The exact 50,000-example prepared pool for the original Qwen2.5-1.5B base SFT experiment. The prepared_50k configuration preserves the original prepared parquet byte for byte. Training stopped with a retained 45K checkpoint. The retained_45k_checkpoint configuration exposes its 45,000 contributing examples in the actual seed-42 loader order, plus the 5,000 examples not represented in that checkpoint. Intermediate Numina checkpoints correspond to 5K prefixes of the ordered 45K view. The remaining 5K view is a provenance split, not a validation benchmark.
Exact original question and response text is preserved. No resampling, solution rewriting or truncation is performed for publication. Source revisions, row counts, order and file SHA256 values are recorded in publication.json and provenance/. Prepared tables retain their original columns. Source validity labels do not establish independently verified mathematical correctness.
Collection: https://huggingface.co/collections/zbeeb/disstillation-project-6ace05e5f40dbe715407516c
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