problem stringlengths 22 5.08k | solution stringlengths 41 7.62k | answer stringlengths 0 550 | domain stringclasses 6
values | prompt stringlengths 72 5.13k | completion stringlengths 98 7.72k |
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Mr. John works at a shopping mall and receives a monthly income of $2000. He spends a certain percentage of this amount on public transport to and from the mall in a month. After deducting his monthly transport fare, he has $1900 left. What percentage of his income does he spend on public transport? | To find the percentage of his income that Mr. John spends on public transport, we first need to determine the amount he spends on transport.
He has $1900 left after spending on transport, and his original income is $2000. So, the amount spent on transport is:
$2000 (original income) - $1900 (remaining after transpor... | 5\% | prealgebra | <|im_start|>user
Mr. John works at a shopping mall and receives a monthly income of $2000. He spends a certain percentage of this amount on public transport to and from the mall in a month. After deducting his monthly transport fare, he has $1900 left. What percentage of his income does he spend on public transport?<|i... | To find the percentage of his income that Mr. John spends on public transport, we first need to determine the amount he spends on transport.
He has $1900 left after spending on transport, and his original income is $2000. So, the amount spent on transport is:
$2000 (original income) - $1900 (remaining after transpor... |
In a space with 1989 points, where no three points are collinear, these points are divided into 30 groups with different numbers of points in each group. To maximize the total number of triangles formed by choosing one point from each of any three different groups as the vertices, how should the points be distributed a... | We need to divide 1989 points in space, such that no three points are collinear, into 30 groups where each group has a different number of points and maximize the total number of triangles formed by choosing one vertex from any three different groups.
1. **Setting up the problem:**
Let the number of points in the 3... | 51, 52, \ldots, 55, 56, 58, 59, \ldots, 81 | geometry | <|im_start|>user
In a space with 1989 points, where no three points are collinear, these points are divided into 30 groups with different numbers of points in each group. To maximize the total number of triangles formed by choosing one point from each of any three different groups as the vertices, how should the points... | We need to divide 1989 points in space, such that no three points are collinear, into 30 groups where each group has a different number of points and maximize the total number of triangles formed by choosing one vertex from any three different groups.
1. **Setting up the problem:**
Let the number of points in the 3... |
Given $n > 1$ monic quadratic polynomials $x^2 - a_{1} x + b_{1}, \ldots, x^2 - a_{n} x + b_{n}$, and all $2n$ numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ are distinct.
Can it happen that each of the numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ is a root of one of these polynomials? | We are given \( n \) quadratic trinomials \( x^2 - a_i x + b_i \) for \( 1 \leq i \leq n \) where all \( 2n \) numbers \( a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \) are distinct. Let us check if each of these \( 2n \) numbers can be the root of one of the quadratic trinomials.
1. **Assumption:**
\[
\text{... | \text{It cannot happen.} | algebra | <|im_start|>user
Given $n > 1$ monic quadratic polynomials $x^2 - a_{1} x + b_{1}, \ldots, x^2 - a_{n} x + b_{n}$, and all $2n$ numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ are distinct.
Can it happen that each of the numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ is a root of one of these polynomials?<|... | We are given \( n \) quadratic trinomials \( x^2 - a_i x + b_i \) for \( 1 \leq i \leq n \) where all \( 2n \) numbers \( a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \) are distinct. Let us check if each of these \( 2n \) numbers can be the root of one of the quadratic trinomials.
1. **Assumption:**
\[
\text{... |
a jogger running at 9 km / hr along side a railway track is 250 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | First, let's convert the speeds from km/hr to m/s because the distance given is in meters.
Speed of jogger = 9 km/hr = 9 * (1000 m / 1 km) * (1 hr / 3600 s) = 2.5 m/s
Speed of train = 45 km/hr = 45 * (1000 m / 1 km) * (1 hr / 3600 s) = 12.5 m/s
Relative speed of the train with respect to the jogger = Speed of train -... | 37 | precalculus_calculus | <|im_start|>user
a jogger running at 9 km / hr along side a railway track is 250 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?<|im_end|>
<|im_start|>assistant
| First, let's convert the speeds from km/hr to m/s because the distance given is in meters.
Speed of jogger = 9 km/hr = 9 * (1000 m / 1 km) * (1 hr / 3600 s) = 2.5 m/s
Speed of train = 45 km/hr = 45 * (1000 m / 1 km) * (1 hr / 3600 s) = 12.5 m/s
Relative speed of the train with respect to the jogger = Speed of train -... |
In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 81 more candidates got selected than State A. What was the number of candidates appeared from each state? | Let's denote the number of candidates appeared from each state as X.
In State A, 6% of the candidates got selected. So, the number of candidates selected from State A is 0.06X.
In State B, 7% of the candidates got selected. So, the number of candidates selected from State B is 0.07X.
According to the problem, State ... | 8100 | algebra | <|im_start|>user
In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 81 more candidates got selected than State A. What was the number of candidates appeared from each state?<|im_en... | Let's denote the number of candidates appeared from each state as X.
In State A, 6% of the candidates got selected. So, the number of candidates selected from State A is 0.06X.
In State B, 7% of the candidates got selected. So, the number of candidates selected from State B is 0.07X.
According to the problem, State ... |
Find the number of moles of KHSO4 formed on combining 2 moles of KOH and some moles of H2SO4, if 2 moles of KHSO4 are formed. | To find the number of moles of H2SO4 needed to form 2 moles of KHSO4 when combined with 2 moles of KOH, we need to look at the balanced chemical equation for the reaction between KOH and H2SO4 to form KHSO4.
The balanced chemical equation is:
KOH + H2SO4 → KHSO4 + H2O
From the equation, we can see that 1 mole of KOH... | 2 | algebra | <|im_start|>user
Find the number of moles of KHSO4 formed on combining 2 moles of KOH and some moles of H2SO4, if 2 moles of KHSO4 are formed.<|im_end|>
<|im_start|>assistant
| To find the number of moles of H2SO4 needed to form 2 moles of KHSO4 when combined with 2 moles of KOH, we need to look at the balanced chemical equation for the reaction between KOH and H2SO4 to form KHSO4.
The balanced chemical equation is:
KOH + H2SO4 → KHSO4 + H2O
From the equation, we can see that 1 mole of KOH... |
In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. How many cows were there in the group? | Let's use algebra to solve this problem. Let the number of cows be \( c \) and the number of chickens be \( h \).
We know that cows have 4 legs each and chickens have 2 legs each. The total number of legs in the group is then \( 4c + 2h \).
We also know that the total number of heads is the sum of the number of cows ... | 7 | algebra | <|im_start|>user
In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. How many cows were there in the group?<|im_end|>
<|im_start|>assistant
| Let's use algebra to solve this problem. Let the number of cows be \( c \) and the number of chickens be \( h \).
We know that cows have 4 legs each and chickens have 2 legs each. The total number of legs in the group is then \( 4c + 2h \).
We also know that the total number of heads is the sum of the number of cows ... |
The parabola $y=2(x-3)^{2}$ can be seen as a transformation of the parabola $y=2x^{2}$. Determine the type of transformation:
A: Shift left by $3$ units
B: Shift right by $3$ units
C: Shift up by $3$ units
D: Shift down by $3$ units | Given that the vertex of the parabola $y=2(x-3)^{2}$ is at $(3,0)$, and the vertex of the parabola $y=2x^{2}$ is at $(0,0)$,
We can conclude that the parabola $y=2(x-3)^{2}$ is obtained by shifting the parabola $y=2x^{2}$ right by $3$ units.
Thus, the answer is $\boxed{\text{B}}$.
The translation of a parabola can b... | \text{B} | precalculus_calculus | <|im_start|>user
The parabola $y=2(x-3)^{2}$ can be seen as a transformation of the parabola $y=2x^{2}$. Determine the type of transformation:
A: Shift left by $3$ units
B: Shift right by $3$ units
C: Shift up by $3$ units
D: Shift down by $3$ units<|im_end|>
<|im_start|>assistant
| Given that the vertex of the parabola $y=2(x-3)^{2}$ is at $(3,0)$, and the vertex of the parabola $y=2x^{2}$ is at $(0,0)$,
We can conclude that the parabola $y=2(x-3)^{2}$ is obtained by shifting the parabola $y=2x^{2}$ right by $3$ units.
Thus, the answer is $\boxed{\text{B}}$.
The translation of a parabola can b... |
Given that $\log_{10} \sin x + \log_{10} \cos x = -0.7$ and that $\log_a (\sin x + \cos x) = \frac{1}{2} (\log_a n - 1)$, find $n$ with $a = 5.$ | From the first equation, using logarithm properties, we have:
\[
\log_{10} (\sin x \cos x) = -0.7 \implies \sin x \cos x = 10^{-0.7} \approx \frac{1}{5}.
\]
Now, solving the second equation:
\[
\log_5 (\sin x + \cos x) = \frac{1}{2} (\log_5 n - 1) \implies \log_5 (\sin x + \cos x) = \log_5 \sqrt{\frac{n}{5}}.
\]
This s... | 7 | precalculus_calculus | <|im_start|>user
Given that $\log_{10} \sin x + \log_{10} \cos x = -0.7$ and that $\log_a (\sin x + \cos x) = \frac{1}{2} (\log_a n - 1)$, find $n$ with $a = 5.$<|im_end|>
<|im_start|>assistant
| From the first equation, using logarithm properties, we have:
\[
\log_{10} (\sin x \cos x) = -0.7 \implies \sin x \cos x = 10^{-0.7} \approx \frac{1}{5}.
\]
Now, solving the second equation:
\[
\log_5 (\sin x + \cos x) = \frac{1}{2} (\log_5 n - 1) \implies \log_5 (\sin x + \cos x) = \log_5 \sqrt{\frac{n}{5}}.
\]
This s... |
Two natural numbers have a difference of 11, and the sum of the digits of each number is divisible by 11. What is the smaller of the smallest two natural numbers that meet these requirements? | :
Given two natural numbers with a difference of 11, where the digit sum of each number is divisible by 11. We need to find the smallest of the two natural numbers.
Let's denote the smaller number by \(a\). The digit sum of \(a\) is a multiple of 11, say \(11n\).
1. Since the difference between the two numbers is 11... | 189999999999 | number_theory | <|im_start|>user
Two natural numbers have a difference of 11, and the sum of the digits of each number is divisible by 11. What is the smaller of the smallest two natural numbers that meet these requirements?<|im_end|>
<|im_start|>assistant
| :
Given two natural numbers with a difference of 11, where the digit sum of each number is divisible by 11. We need to find the smallest of the two natural numbers.
Let's denote the smaller number by \(a\). The digit sum of \(a\) is a multiple of 11, say \(11n\).
1. Since the difference between the two numbers is 11... |
If $9:y^3 = y:81$, what is the value of $y$? | Given the equation $\frac{9}{y^3} = \frac{y}{81}$, we need to solve for $y$. Cross-multiplying gives $9 \cdot 81 = y^4$, which simplifies to $729 = y^4$.
Taking the fourth root on both sides, we find $y = \sqrt[4]{729}$.
Since $729 = 3^6$, we have $y = \sqrt[4]{3^6} = 3^{6/4} = 3^{1.5} = 3 \cdot \sqrt{3}$.
Thus, $y = ... | 3 \cdot \sqrt{3} | algebra | <|im_start|>user
If $9:y^3 = y:81$, what is the value of $y$?<|im_end|>
<|im_start|>assistant
| Given the equation $\frac{9}{y^3} = \frac{y}{81}$, we need to solve for $y$. Cross-multiplying gives $9 \cdot 81 = y^4$, which simplifies to $729 = y^4$.
Taking the fourth root on both sides, we find $y = \sqrt[4]{729}$.
Since $729 = 3^6$, we have $y = \sqrt[4]{3^6} = 3^{6/4} = 3^{1.5} = 3 \cdot \sqrt{3}$.
Thus, $y = ... |
What is the modulo $11$ residue of $325 + 3 \cdot 66 + 8 \cdot 187 + 6 \cdot 23?$ | First, simplify each term under modulo 11:
- $325 \equiv 6 \pmod{11}$ (since $325 = 29 \cdot 11 + 6$),
- $66 \equiv 0 \pmod{11}$ (since $66 = 6 \cdot 11$),
- $187 \equiv 0 \pmod{11}$ (since $187 = 17 \cdot 11$),
- $23 \equiv 1 \pmod{11}$ (since $23 = 2 \cdot 11 + 1$).
Now, compute the sum:
- $325 + 3 \cdot 66 + 8 \cdo... | 1 | number_theory | <|im_start|>user
What is the modulo $11$ residue of $325 + 3 \cdot 66 + 8 \cdot 187 + 6 \cdot 23?$<|im_end|>
<|im_start|>assistant
| First, simplify each term under modulo 11:
- $325 \equiv 6 \pmod{11}$ (since $325 = 29 \cdot 11 + 6$),
- $66 \equiv 0 \pmod{11}$ (since $66 = 6 \cdot 11$),
- $187 \equiv 0 \pmod{11}$ (since $187 = 17 \cdot 11$),
- $23 \equiv 1 \pmod{11}$ (since $23 = 2 \cdot 11 + 1$).
Now, compute the sum:
- $325 + 3 \cdot 66 + 8 \cdo... |
The line graph represents the price of a collectible coin during the first six months of 1994. Determine the month with the greatest monthly drop in price.
$\text{(A)}\ \text{January} \qquad \text{(B)}\ \text{March} \qquad \text{(C)}\ \text{April} \qquad \text{(D)}\ \text{May} \qquad \text{(E)}\ \text{June}$ | To find the month with the greatest monthly drop in price for the collectible coin, evaluate the price changes each month:
1. **January**: The price change is $-1.00$.
2. **February**: The price change is $+2.50$.
3. **March**: The price change is $0.00$ (no change).
4. **April**: The price change is $-3.00$.
5. **May... | \text{(C)}\ \text{April} | combinatorics_probability | <|im_start|>user
The line graph represents the price of a collectible coin during the first six months of 1994. Determine the month with the greatest monthly drop in price.
$\text{(A)}\ \text{January} \qquad \text{(B)}\ \text{March} \qquad \text{(C)}\ \text{April} \qquad \text{(D)}\ \text{May} \qquad \text{(E)}\ \text{... | To find the month with the greatest monthly drop in price for the collectible coin, evaluate the price changes each month:
1. **January**: The price change is $-1.00$.
2. **February**: The price change is $+2.50$.
3. **March**: The price change is $0.00$ (no change).
4. **April**: The price change is $-3.00$.
5. **May... |
A train travels 125 km in 2.5 hours and 270 km in 3 hours. What is the average speed of the train? | To find the average speed of the train, we need to calculate the total distance traveled and the total time taken, and then divide the total distance by the total time.
First, let's calculate the total distance:
125 km (first part) + 270 km (second part) = 395 km
Now, let's calculate the total time:
2.5 hours (first ... | 71.82 | prealgebra | <|im_start|>user
A train travels 125 km in 2.5 hours and 270 km in 3 hours. What is the average speed of the train?<|im_end|>
<|im_start|>assistant
| To find the average speed of the train, we need to calculate the total distance traveled and the total time taken, and then divide the total distance by the total time.
First, let's calculate the total distance:
125 km (first part) + 270 km (second part) = 395 km
Now, let's calculate the total time:
2.5 hours (first ... |
In triangle \( \triangle ABC \), given that
$$
\angle A = 30^{\circ}, \quad 2 \overrightarrow{AB} \cdot \overrightarrow{AC} = 3 \overrightarrow{BC}^2,
$$
find the cosine of the largest angle of \( \triangle ABC \). | 1. Begin by noting the given information. In $\triangle ABC$, we know:
\[
\angle A = 30^\circ \quad \text{and} \quad 2 \overrightarrow{A B} \cdot \overrightarrow{A C} = 3 \overrightarrow{B C}^2
\]
Let $AB = c$, $AC = b$, and $BC = a$.
2. Use the given vector dot product condition:
\[
2 \overrightarro... | -\frac{1}{2} | geometry | <|im_start|>user
In triangle \( \triangle ABC \), given that
$$
\angle A = 30^{\circ}, \quad 2 \overrightarrow{AB} \cdot \overrightarrow{AC} = 3 \overrightarrow{BC}^2,
$$
find the cosine of the largest angle of \( \triangle ABC \).<|im_end|>
<|im_start|>assistant
| 1. Begin by noting the given information. In $\triangle ABC$, we know:
\[
\angle A = 30^\circ \quad \text{and} \quad 2 \overrightarrow{A B} \cdot \overrightarrow{A C} = 3 \overrightarrow{B C}^2
\]
Let $AB = c$, $AC = b$, and $BC = a$.
2. Use the given vector dot product condition:
\[
2 \overrightarro... |
Working alone, printers X, Y, and Z can do a certain printing job, consisting of a large number of pages, in 15, 10, and 20 hours, respectively. What is the ratio of the time it takes printer X to do the job, working alone at its rate, to the time it takes printers Y and Z to do the job, working together at their indiv... | First, let's find out how much of the job each printer can do in one hour.
Printer X can do the job in 15 hours, so in one hour, it can do 1/15 of the job.
Printer Y can do the job in 10 hours, so in one hour, it can do 1/10 of the job.
Printer Z can do the job in 20 hours, so in one hour, it can do 1/20 of the job.
... | 9:4 | prealgebra | <|im_start|>user
Working alone, printers X, Y, and Z can do a certain printing job, consisting of a large number of pages, in 15, 10, and 20 hours, respectively. What is the ratio of the time it takes printer X to do the job, working alone at its rate, to the time it takes printers Y and Z to do the job, working togeth... | First, let's find out how much of the job each printer can do in one hour.
Printer X can do the job in 15 hours, so in one hour, it can do 1/15 of the job.
Printer Y can do the job in 10 hours, so in one hour, it can do 1/10 of the job.
Printer Z can do the job in 20 hours, so in one hour, it can do 1/20 of the job.
... |
A plumber charges $30 to fix a sink, $40 to fix a shower, and a certain amount to fix a toilet. He is trying to decide which of three different jobs he should take to make the most money. The first job involves fixing three toilets and three sinks at an apartment complex. The second involves two toilets and five sinks ... | Let's denote the amount the plumber charges to fix a toilet as T.
For the first job, the plumber would earn 3T (for three toilets) + 3 * $30 (for three sinks), which equals 3T + $90.
For the second job, the plumber would earn 2T (for two toilets) + 5 * $30 (for five sinks), which equals 2T + $150.
For the third job,... | \$50 | combinatorics_probability | <|im_start|>user
A plumber charges $30 to fix a sink, $40 to fix a shower, and a certain amount to fix a toilet. He is trying to decide which of three different jobs he should take to make the most money. The first job involves fixing three toilets and three sinks at an apartment complex. The second involves two toilet... | Let's denote the amount the plumber charges to fix a toilet as T.
For the first job, the plumber would earn 3T (for three toilets) + 3 * $30 (for three sinks), which equals 3T + $90.
For the second job, the plumber would earn 2T (for two toilets) + 5 * $30 (for five sinks), which equals 2T + $150.
For the third job,... |
Two different numbers are selected simultaneously and at random from the set $\{1, 2, 3, 4, ..., 10\}$. What is the probability that the positive difference between the two numbers is $3$ or greater? Express your answer as a common fraction. | First, we determine the total number of ways to choose two different numbers from the set $\{1, 2, 3, ..., 10\}$. The size of the set is $10$, and the number of ways to choose two numbers is given by the combination formula $\binom{n}{2}$ where $n$ is the size of the set. Thus, $\binom{10}{2} = \frac{10 \times 9}{2} = ... | \frac{28}{45} | combinatorics_probability | <|im_start|>user
Two different numbers are selected simultaneously and at random from the set $\{1, 2, 3, 4, ..., 10\}$. What is the probability that the positive difference between the two numbers is $3$ or greater? Express your answer as a common fraction.<|im_end|>
<|im_start|>assistant
| First, we determine the total number of ways to choose two different numbers from the set $\{1, 2, 3, ..., 10\}$. The size of the set is $10$, and the number of ways to choose two numbers is given by the combination formula $\binom{n}{2}$ where $n$ is the size of the set. Thus, $\binom{10}{2} = \frac{10 \times 9}{2} = ... |
In triangle $XYZ$, where it is known that $XY = 5$, $\angle YXZ = 30^\circ$ and $\angle XZY$ is less than $60^\circ$. A point $Q$ on $\overline{XY}$ is chosen such that $\angle ZQY = 2\angle ZYQ$ and $ZQ = 2$. Find the ratio $\frac{XQ}{QY}$ in the form $s + t\sqrt{u}$, where $s$, $t$, and $u$ are positive integers. | Let $W$ be the circumcenter of triangle $XYZ$, and let the second intersection of $ZQ$ with the circumcircle of $XYZ$ be $V$. Then $\angle{ZWV} = 2\angle ZYQ = \angle ZQY = \angle ZVW$, meaning triangle $ZWV$ is isosceles with $ZW = WV = r$ (the circumradius).
Also consider points where $\overline{ZW}$ intersects $YQ$... | 30 | geometry | <|im_start|>user
In triangle $XYZ$, where it is known that $XY = 5$, $\angle YXZ = 30^\circ$ and $\angle XZY$ is less than $60^\circ$. A point $Q$ on $\overline{XY}$ is chosen such that $\angle ZQY = 2\angle ZYQ$ and $ZQ = 2$. Find the ratio $\frac{XQ}{QY}$ in the form $s + t\sqrt{u}$, where $s$, $t$, and $u$ are posit... | Let $W$ be the circumcenter of triangle $XYZ$, and let the second intersection of $ZQ$ with the circumcircle of $XYZ$ be $V$. Then $\angle{ZWV} = 2\angle ZYQ = \angle ZQY = \angle ZVW$, meaning triangle $ZWV$ is isosceles with $ZW = WV = r$ (the circumradius).
Also consider points where $\overline{ZW}$ intersects $YQ$... |
Determine the range of the function $$G(x) = |x+2| - 2|x-2|.$$ Express your answer in interval notation. | Consider the function $$ G(x) = |x+2| - 2|x-2|.$$
We analyze this function by considering the behavior of \( x \) relative to the points \( -2 \) and \( 2 \) where the absolute values change their nature:
- For \( x < -2 \), both \( x+2 \) and \( x-2 \) are negative. Thus, \( G(x) = -(x+2) - 2(x-2) = -x-2 - 2x + 4 = -3... | [-8, 8] | precalculus_calculus | <|im_start|>user
Determine the range of the function $$G(x) = |x+2| - 2|x-2|.$$ Express your answer in interval notation.<|im_end|>
<|im_start|>assistant
| Consider the function $$ G(x) = |x+2| - 2|x-2|.$$
We analyze this function by considering the behavior of \( x \) relative to the points \( -2 \) and \( 2 \) where the absolute values change their nature:
- For \( x < -2 \), both \( x+2 \) and \( x-2 \) are negative. Thus, \( G(x) = -(x+2) - 2(x-2) = -x-2 - 2x + 4 = -3... |
Find all values of \( a \) such that for each \( x \), the largest of the two numbers \( x^{3}+3x+a-9 \) and \( a+2^{5-x}-3^{x-1} \) is positive. | 1. **Analyzing the functions:**
- We have two functions of $x$:
\[
f(x) = x^3 + 3x + a - 9
\]
\[
g(x) = a + 2^{5-x} - 3^{x-1}
\]
- It's noticeable that $f(x)$ is strictly increasing because its derivative
\[
f'(x) = 3x^2 + 3 > 0 \quad \text{for all } x
\]
- On the ... | a > -5 | precalculus_calculus | <|im_start|>user
Find all values of \( a \) such that for each \( x \), the largest of the two numbers \( x^{3}+3x+a-9 \) and \( a+2^{5-x}-3^{x-1} \) is positive.<|im_end|>
<|im_start|>assistant
| 1. **Analyzing the functions:**
- We have two functions of $x$:
\[
f(x) = x^3 + 3x + a - 9
\]
\[
g(x) = a + 2^{5-x} - 3^{x-1}
\]
- It's noticeable that $f(x)$ is strictly increasing because its derivative
\[
f'(x) = 3x^2 + 3 > 0 \quad \text{for all } x
\]
- On the ... |
When the binary number $101011100101_2$ is divided by 8, what is the remainder (give your answer in base 10)? | The binary number is formed as $2^k b_k + 2^{k-1} b_{k-1} + \dots + 2^2 b_2 + 2^1 b_1 + 2^0 b_0$. To find the remainder when dividing by 8, we only consider the last three digits due to $8 = 2^3$.
Here, the last three digits are $101_2$. We express this in decimal for the remainder:
\[1 \cdot 2^2 + 0 \cdot 2^1 + 1 \c... | 5 | number_theory | <|im_start|>user
When the binary number $101011100101_2$ is divided by 8, what is the remainder (give your answer in base 10)?<|im_end|>
<|im_start|>assistant
| The binary number is formed as $2^k b_k + 2^{k-1} b_{k-1} + \dots + 2^2 b_2 + 2^1 b_1 + 2^0 b_0$. To find the remainder when dividing by 8, we only consider the last three digits due to $8 = 2^3$.
Here, the last three digits are $101_2$. We express this in decimal for the remainder:
\[1 \cdot 2^2 + 0 \cdot 2^1 + 1 \c... |
Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$. The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?... | 1. **Determine the total number of products in the table**:
The multiplication table ranges from $0 \times 0$ to $12 \times 12$. Since each factor (both rows and columns) ranges from $0$ to $12$, there are $13$ options for each factor (including $0$). Therefore, the total number of entries in the table is:
\[
... | \textbf{(A)} \, 0.21 | number_theory | <|im_start|>user
Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$. The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of t... | 1. **Determine the total number of products in the table**:
The multiplication table ranges from $0 \times 0$ to $12 \times 12$. Since each factor (both rows and columns) ranges from $0$ to $12$, there are $13$ options for each factor (including $0$). Therefore, the total number of entries in the table is:
\[
... |
Let \( Q(z) \) and \( R(z) \) be the unique polynomials such that \[ z^{2023} + 1 = (z^2 + 1)Q(z) + R(z) \] and the degree of \( R \) is less than 2. What is \( R(z)? \)
A) \( -z \)
B) \( -1 \)
C) \( z + 1 \)
D) \( 2z + 1 \)
E) \( z - 1 \) | 1. **Understanding the Problem**: We need to find \( R(z) \), where \( z^{2023} + 1 = (z^2 + 1)Q(z) + R(z) \), and \( R(z) \) is a polynomial of degree less than 2.
2. **Polynomial Division**: As \( R(z) \) must have a degree less than 2, it can be expressed as \( az + b \).
3. **Using Roots of Unity**: The roots of ... | C | precalculus_calculus | <|im_start|>user
Let \( Q(z) \) and \( R(z) \) be the unique polynomials such that \[ z^{2023} + 1 = (z^2 + 1)Q(z) + R(z) \] and the degree of \( R \) is less than 2. What is \( R(z)? \)
A) \( -z \)
B) \( -1 \)
C) \( z + 1 \)
D) \( 2z + 1 \)
E) \( z - 1 \)<|im_end|>
<|im_start|>assistant
| 1. **Understanding the Problem**: We need to find \( R(z) \), where \( z^{2023} + 1 = (z^2 + 1)Q(z) + R(z) \), and \( R(z) \) is a polynomial of degree less than 2.
2. **Polynomial Division**: As \( R(z) \) must have a degree less than 2, it can be expressed as \( az + b \).
3. **Using Roots of Unity**: The roots of ... |
On an island, there are only knights, who always tell the truth, and liars, who always lie. There are at least two knights and at least two liars. One day, each islander pointed to each of the others in turn and said either "You are a knight!" or "You are a liar!". The phrase "You are a liar!" was said exactly 230 time... | 1. Let the number of knights be denoted by $r$ and the number of liars be denoted by $\ell$.
2. Recall the behavior of knights and liars:
- Knights always tell the truth, so a knight will say "You are a knight!" to another knight.
- Knights will say "You are a liar!" to a liar.
- Liars always lie, so a lia... | 526 | algebra | <|im_start|>user
On an island, there are only knights, who always tell the truth, and liars, who always lie. There are at least two knights and at least two liars. One day, each islander pointed to each of the others in turn and said either "You are a knight!" or "You are a liar!". The phrase "You are a liar!" was said... | 1. Let the number of knights be denoted by $r$ and the number of liars be denoted by $\ell$.
2. Recall the behavior of knights and liars:
- Knights always tell the truth, so a knight will say "You are a knight!" to another knight.
- Knights will say "You are a liar!" to a liar.
- Liars always lie, so a lia... |
Doughnuts, cookies and muffins in a pastry shop are in the ratio of 5 : 3 : 1. If there are 50 doughnuts, how many muffins are there in the shop? | If the ratio of doughnuts to cookies to muffins is 5:3:1, this means that for every 5 doughnuts, there are 3 cookies and 1 muffin.
Since there are 50 doughnuts, we can use the ratio to find out how many muffins there are. First, we need to determine how many "sets" of the ratio there are in 50 doughnuts.
To do this, ... | 10 | prealgebra | <|im_start|>user
Doughnuts, cookies and muffins in a pastry shop are in the ratio of 5 : 3 : 1. If there are 50 doughnuts, how many muffins are there in the shop?<|im_end|>
<|im_start|>assistant
| If the ratio of doughnuts to cookies to muffins is 5:3:1, this means that for every 5 doughnuts, there are 3 cookies and 1 muffin.
Since there are 50 doughnuts, we can use the ratio to find out how many muffins there are. First, we need to determine how many "sets" of the ratio there are in 50 doughnuts.
To do this, ... |
In 1945, Anna was tasked to memorize her multiplication facts from $0 \times 0$ to $15 \times 15$. The multiplication table she used had rows and columns labeled with the factors, and the products made up the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?
$\t... | 1. **Determine the total number of products in the table**:
The multiplication table ranges from $0 \times 0$ to $15 \times 15$. Since each factor (both rows and columns) ranges from $0$ to $15$, there are $16$ options for each factor. Therefore, the total number of entries in the table is:
\[
(15+1) \times (... | \textbf{B} | number_theory | <|im_start|>user
In 1945, Anna was tasked to memorize her multiplication facts from $0 \times 0$ to $15 \times 15$. The multiplication table she used had rows and columns labeled with the factors, and the products made up the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the t... | 1. **Determine the total number of products in the table**:
The multiplication table ranges from $0 \times 0$ to $15 \times 15$. Since each factor (both rows and columns) ranges from $0$ to $15$, there are $16$ options for each factor. Therefore, the total number of entries in the table is:
\[
(15+1) \times (... |
In triangle $DEF$, we have $\angle D = 90^\circ$ and $\sin E = \frac{3}{5}$. Find $\cos F$. | In a right triangle $DEF$ with $\angle D = 90^\circ$, the angles $E$ and $F$ are complementary, meaning that $\angle E + \angle F = 90^\circ$. Hence, the cosine of one angle in a right triangle is equal to the sine of its complementary angle. Therefore, $\cos F = \sin E$.
Given $\sin E = \frac{3}{5}$, we find:
\[
\cos... | \frac{3}{5} | geometry | <|im_start|>user
In triangle $DEF$, we have $\angle D = 90^\circ$ and $\sin E = \frac{3}{5}$. Find $\cos F$.<|im_end|>
<|im_start|>assistant
| In a right triangle $DEF$ with $\angle D = 90^\circ$, the angles $E$ and $F$ are complementary, meaning that $\angle E + \angle F = 90^\circ$. Hence, the cosine of one angle in a right triangle is equal to the sine of its complementary angle. Therefore, $\cos F = \sin E$.
Given $\sin E = \frac{3}{5}$, we find:
\[
\cos... |
Find the smallest positive period of the function $f(x)=\sqrt{2}\sin(\frac{x}{2}+\frac{π}{3}) (x\in\mathbb{R})$.
A. $\frac{π}{2}$
B. $π$
C. $2π$
D. $4π$ | The smallest positive period of the function $f(x)=\sqrt{2}\sin(\frac{x}{2}+\frac{π}{3})$ is calculated as follows:
$$T=\frac{2π}{ω}=\frac{2π}{\frac{1}{2}}=4π$$
So, the correct answer is option D, $\boxed{4π}$.
This answer is derived from the formula for the smallest positive period of a sine function of the form $y... | 4π | precalculus_calculus | <|im_start|>user
Find the smallest positive period of the function $f(x)=\sqrt{2}\sin(\frac{x}{2}+\frac{π}{3}) (x\in\mathbb{R})$.
A. $\frac{π}{2}$
B. $π$
C. $2π$
D. $4π$<|im_end|>
<|im_start|>assistant
| The smallest positive period of the function $f(x)=\sqrt{2}\sin(\frac{x}{2}+\frac{π}{3})$ is calculated as follows:
$$T=\frac{2π}{ω}=\frac{2π}{\frac{1}{2}}=4π$$
So, the correct answer is option D, $\boxed{4π}$.
This answer is derived from the formula for the smallest positive period of a sine function of the form $y... |
Twelve balls numbered from 1 to 12 are placed in a jar. Jason removes one ball at random, followed by Mary who also removes another ball, ensuring it is different from the one Jason removed. What is the probability that the sum of the numbers on the two balls they removed is even?
A) $\frac{1}{2}$
B) $\frac{5}{11}$
C) ... | 1. **Total Number of Outcomes**:
- Jason has 12 options when he picks the first ball.
- Mary then has 11 options as one ball is already taken.
- Total outcomes = $12 \times 11 = 132$.
2. **Favorable Outcomes for an Even Sum**:
- **Both Balls Even**: There are 6 even-numbered balls (2, 4, 6, 8, 10, 12). Fi... | B | combinatorics_probability | <|im_start|>user
Twelve balls numbered from 1 to 12 are placed in a jar. Jason removes one ball at random, followed by Mary who also removes another ball, ensuring it is different from the one Jason removed. What is the probability that the sum of the numbers on the two balls they removed is even?
A) $\frac{1}{2}$
B) $... | 1. **Total Number of Outcomes**:
- Jason has 12 options when he picks the first ball.
- Mary then has 11 options as one ball is already taken.
- Total outcomes = $12 \times 11 = 132$.
2. **Favorable Outcomes for an Even Sum**:
- **Both Balls Even**: There are 6 even-numbered balls (2, 4, 6, 8, 10, 12). Fi... |
Given sets A={0, 1} and B={-1, 1, 3}, find A ∩ B (A intersect B) equals to ( )
A: {0}
B: {1}
C: {0, 1}
D: {0, 1, 3} | Solution:
Since set A={0, 1} and B={-1, 1, 3},
A ∩ B = {1}.
Hence, the answer is $\boxed{\text{B}}$.
By using the definition of intersection, we can directly solve the problem.
This question tests the ability to find the intersection of two sets and understanding of basic concepts such as set definitions and inequal... | \text{B} | precalculus_calculus | <|im_start|>user
Given sets A={0, 1} and B={-1, 1, 3}, find A ∩ B (A intersect B) equals to ( )
A: {0}
B: {1}
C: {0, 1}
D: {0, 1, 3}<|im_end|>
<|im_start|>assistant
| Solution:
Since set A={0, 1} and B={-1, 1, 3},
A ∩ B = {1}.
Hence, the answer is $\boxed{\text{B}}$.
By using the definition of intersection, we can directly solve the problem.
This question tests the ability to find the intersection of two sets and understanding of basic concepts such as set definitions and inequal... |
In triangle $ABC$, the sides opposite angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively, and it is given that $2a\sin B - \sqrt{5}b\cos A = 0$.
(1) Find $\cos A$;
(2) If $a = \sqrt{5}$ and $b = 2$, find the area of triangle $ABC$. | (1) In triangle $ABC$, with sides $a$, $b$, and $c$ opposite to angles $A$, $B$, and $C$, respectively, simplify the equation $2a\sin B - \sqrt{5}b\cos A = 0$ using the sine rule. We get:
$$2a\sin B - \sqrt{5}b\cos A = 0$$
By the law of sines, $\frac{\sin A}{a} = \frac{\sin B}{b}$, so we can rewrite this as:
$$2\sin A ... | \sqrt{5} | geometry | <|im_start|>user
In triangle $ABC$, the sides opposite angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively, and it is given that $2a\sin B - \sqrt{5}b\cos A = 0$.
(1) Find $\cos A$;
(2) If $a = \sqrt{5}$ and $b = 2$, find the area of triangle $ABC$.<|im_end|>
<|im_start|>assistant
| (1) In triangle $ABC$, with sides $a$, $b$, and $c$ opposite to angles $A$, $B$, and $C$, respectively, simplify the equation $2a\sin B - \sqrt{5}b\cos A = 0$ using the sine rule. We get:
$$2a\sin B - \sqrt{5}b\cos A = 0$$
By the law of sines, $\frac{\sin A}{a} = \frac{\sin B}{b}$, so we can rewrite this as:
$$2\sin A ... |
An author writes a book containing 55,000 words. The author spent a total of 120 hours working, but 20 hours were spent on breaks and not writing. Calculate the average words per hour that the author wrote during the actual writing time. | To find out how many hours were spent actually writing, subtract the break hours from the total hours:
$$ 120\text{ hours} - 20\text{ hours} = 100\text{ hours} $$
Next, we calculate the average number of words written per hour during the actual writing time:
$$ \frac{55,\!000\text{ words}}{100\text{ hours}} = 550\text... | 550 | prealgebra | <|im_start|>user
An author writes a book containing 55,000 words. The author spent a total of 120 hours working, but 20 hours were spent on breaks and not writing. Calculate the average words per hour that the author wrote during the actual writing time.<|im_end|>
<|im_start|>assistant
| To find out how many hours were spent actually writing, subtract the break hours from the total hours:
$$ 120\text{ hours} - 20\text{ hours} = 100\text{ hours} $$
Next, we calculate the average number of words written per hour during the actual writing time:
$$ \frac{55,\!000\text{ words}}{100\text{ hours}} = 550\text... |
A cubic block with dimensions $n$ by $n$ by $n$ is made up of a collection of $1$ by $1$ by $1$ unit cubes. What is the smallest value of $n$ so that if the outer layer of unit cubes are removed from the block, more than half the original unit cubes will still remain? | 1. **Calculate the total number of unit cubes in the cubic block:**
The total number of unit cubes in an \( n \times n \times n \) cubic block is given by:
\[
n^3
\]
2. **Determine the number of unit cubes in the outer layer:**
The outer layer consists of all the unit cubes that are on the surface of th... | 10 | number_theory | <|im_start|>user
A cubic block with dimensions $n$ by $n$ by $n$ is made up of a collection of $1$ by $1$ by $1$ unit cubes. What is the smallest value of $n$ so that if the outer layer of unit cubes are removed from the block, more than half the original unit cubes will still remain?<|im_end|>
<|im_start... | 1. **Calculate the total number of unit cubes in the cubic block:**
The total number of unit cubes in an \( n \times n \times n \) cubic block is given by:
\[
n^3
\]
2. **Determine the number of unit cubes in the outer layer:**
The outer layer consists of all the unit cubes that are on the surface of th... |
The third and sixth terms of an arithmetic sequence are 23 and 29, respectively. What is the ninth term? | Let's denote the first term of the arithmetic sequence as \( a \) and the common difference as \( d \). The third term is \( a + 2d = 23 \) and the sixth term is \( a + 5d = 29 \).
1. Find the common difference:
\[
(a + 5d) - (a + 2d) = 3d \implies 29 - 23 = 3d \implies 3d = 6 \implies d = 2
\]
2. Use the co... | 35 | combinatorics_probability | <|im_start|>user
The third and sixth terms of an arithmetic sequence are 23 and 29, respectively. What is the ninth term?<|im_end|>
<|im_start|>assistant
| Let's denote the first term of the arithmetic sequence as \( a \) and the common difference as \( d \). The third term is \( a + 2d = 23 \) and the sixth term is \( a + 5d = 29 \).
1. Find the common difference:
\[
(a + 5d) - (a + 2d) = 3d \implies 29 - 23 = 3d \implies 3d = 6 \implies d = 2
\]
2. Use the co... |
In how many ways can 5 books be selected from a shelf of 10 books if the order in which the books are selected does not matter? | - The problem states that the order of the books does not matter, so combinations are used, not permutations.
- We need to choose 5 books out of 10 which can be computed using the binomial coefficient $\binom{10}{5}$.
Calculate $\binom{10}{5}$:
\[
\binom{10}{5} = \frac{10!}{5! \times (10 - 5)!} = \frac{10 \times 9 \ti... | 252 | combinatorics_probability | <|im_start|>user
In how many ways can 5 books be selected from a shelf of 10 books if the order in which the books are selected does not matter?<|im_end|>
<|im_start|>assistant
| - The problem states that the order of the books does not matter, so combinations are used, not permutations.
- We need to choose 5 books out of 10 which can be computed using the binomial coefficient $\binom{10}{5}$.
Calculate $\binom{10}{5}$:
\[
\binom{10}{5} = \frac{10!}{5! \times (10 - 5)!} = \frac{10 \times 9 \ti... |
Let the sequence $\{a_n\}$ be a geometric sequence with each term being positive, and let $S_n$ be the sum of its first $n$ terms. Given that $a_2a_4=16$ and $$\frac{a_4+a_5+a_8}{a_1+a_2+a_5}=8,$$ find $S_5$.
A: 40
B: 20
C: 31
D: 43 | Let the common ratio of the geometric sequence $\{a_n\}$ be $q>0$. Since $a_2a_4=16$ and $$\frac{a_4+a_5+a_8}{a_1+a_2+a_5}=8,$$ we have $$a_1^2q^4=16$$ and $q^3=8$. Solving these, we get $q=2$ and $a_1=1$.
Therefore, $$S_5=\frac{2^5-1}{2-1}=31.$$
Hence, the correct answer is $\boxed{C}$.
This problem can be solve... | C | combinatorics_probability | <|im_start|>user
Let the sequence $\{a_n\}$ be a geometric sequence with each term being positive, and let $S_n$ be the sum of its first $n$ terms. Given that $a_2a_4=16$ and $$\frac{a_4+a_5+a_8}{a_1+a_2+a_5}=8,$$ find $S_5$.
A: 40
B: 20
C: 31
D: 43<|im_end|>
<|im_start|>assistant
| Let the common ratio of the geometric sequence $\{a_n\}$ be $q>0$. Since $a_2a_4=16$ and $$\frac{a_4+a_5+a_8}{a_1+a_2+a_5}=8,$$ we have $$a_1^2q^4=16$$ and $q^3=8$. Solving these, we get $q=2$ and $a_1=1$.
Therefore, $$S_5=\frac{2^5-1}{2-1}=31.$$
Hence, the correct answer is $\boxed{C}$.
This problem can be solve... |
Arlene has three ropes whose lengths are 28 inches, 45 inches, and 63 inches. She wants to cut the ropes into pieces of equal length for her crafts, and she wants to ensure that no rope material is wasted. What is the greatest possible length for each piece? | To find the greatest possible length for each piece that Arlene can cut from all ropes without wasting any material, we need to determine the greatest common divisor (GCD) of the lengths of the ropes.
- The prime factors of 28 are $2^2 \cdot 7$.
- The prime factors of 45 are $3^2 \cdot 5$.
- The prime factors of 63 ar... | 7 \text{ inches} | number_theory | <|im_start|>user
Arlene has three ropes whose lengths are 28 inches, 45 inches, and 63 inches. She wants to cut the ropes into pieces of equal length for her crafts, and she wants to ensure that no rope material is wasted. What is the greatest possible length for each piece?<|im_end|>
<|im_start|>assistant
| To find the greatest possible length for each piece that Arlene can cut from all ropes without wasting any material, we need to determine the greatest common divisor (GCD) of the lengths of the ropes.
- The prime factors of 28 are $2^2 \cdot 7$.
- The prime factors of 45 are $3^2 \cdot 5$.
- The prime factors of 63 ar... |
There is a cube $A B C D-E F G H$, with three square faces passing through point $B$ painted red. The cube is then cut along several of its edges to form a net. If none of the three red squares share a common edge in the net, how many different ways are there to make the cuts? (Two ways are considered the same if they ... | 1. To find how many different ways we can cut the edges of a given cube and obtain a net without any red squares sharing an edge, we need to consider the cutting process systematically.
2. When unfolding a cube to form a net, we need to cut 7 out of its 12 edges.
3. Each red face borders one specific vertex of the ... | 54 | combinatorics_probability | <|im_start|>user
There is a cube $A B C D-E F G H$, with three square faces passing through point $B$ painted red. The cube is then cut along several of its edges to form a net. If none of the three red squares share a common edge in the net, how many different ways are there to make the cuts? (Two ways are considered ... | 1. To find how many different ways we can cut the edges of a given cube and obtain a net without any red squares sharing an edge, we need to consider the cutting process systematically.
2. When unfolding a cube to form a net, we need to cut 7 out of its 12 edges.
3. Each red face borders one specific vertex of the ... |
Calculate the value of $$\sqrt {2}\cdot \sqrt[3]{4}\cdot \sqrt[6]{32}+\lg \frac {1}{100}-3^{\log_{3}2}$$. | Solution: The original expression can be rewritten as $$2^{ \frac {1}{2}}$$×$$2^{ \frac {2}{3}}$$×$$2^{ \frac {5}{6}}$$+(-2)-2=$$2^{ \frac {1}{2}+ \frac {2}{3}+ \frac {5}{6}}$$-4=4-4=0.
Therefore, the answer is $\boxed{0}$.
By converting the radicals into fractional powers of 2, and then using the properties of exp... | 0 | precalculus_calculus | <|im_start|>user
Calculate the value of $$\sqrt {2}\cdot \sqrt[3]{4}\cdot \sqrt[6]{32}+\lg \frac {1}{100}-3^{\log_{3}2}$$.<|im_end|>
<|im_start|>assistant
| Solution: The original expression can be rewritten as $$2^{ \frac {1}{2}}$$×$$2^{ \frac {2}{3}}$$×$$2^{ \frac {5}{6}}$$+(-2)-2=$$2^{ \frac {1}{2}+ \frac {2}{3}+ \frac {5}{6}}$$-4=4-4=0.
Therefore, the answer is $\boxed{0}$.
By converting the radicals into fractional powers of 2, and then using the properties of exp... |
$A$ and $B$ together can complete a job in $3$ days; $B$ and $C$ can do it in $3$ days; and $A$ and $C$ can complete it in $1.5$ days. Determine the number of days required for $C$ to do the job alone.
$\textbf{(A)}\ 1.5 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 4.5$ | 1. **Define the rates of work**:
Let $r_A$, $r_B$, and $r_C$ be the rates at which $A$, $B$, and $C$ respectively can complete the job per day.
2. **Set up the equations**:
- $A$ and $B$ together in 3 days:
\[
r_A + r_B = \frac{1}{3}
\]
- $B$ and $C$ together in 3 days:
\[
r_B + r_C =... | \textbf{(C)}\ 3 | algebra | <|im_start|>user
$A$ and $B$ together can complete a job in $3$ days; $B$ and $C$ can do it in $3$ days; and $A$ and $C$ can complete it in $1.5$ days. Determine the number of days required for $C$ to do the job alone.
$\textbf{(A)}\ 1.5 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textb... | 1. **Define the rates of work**:
Let $r_A$, $r_B$, and $r_C$ be the rates at which $A$, $B$, and $C$ respectively can complete the job per day.
2. **Set up the equations**:
- $A$ and $B$ together in 3 days:
\[
r_A + r_B = \frac{1}{3}
\]
- $B$ and $C$ together in 3 days:
\[
r_B + r_C =... |
A train running at the speed of 60 km/hr crosses a pole in 12 seconds. What is the length of the train in meters? | To find the length of the train, we can use the formula:
Distance = Speed × Time
First, we need to convert the speed from km/hr to m/s because the time is given in seconds. We know that 1 km = 1000 meters and 1 hour = 3600 seconds.
So, to convert 60 km/hr to m/s:
Speed in m/s = (Speed in km/hr × 1000 meters/km) / 3... | 200.04 | precalculus_calculus | <|im_start|>user
A train running at the speed of 60 km/hr crosses a pole in 12 seconds. What is the length of the train in meters?<|im_end|>
<|im_start|>assistant
| To find the length of the train, we can use the formula:
Distance = Speed × Time
First, we need to convert the speed from km/hr to m/s because the time is given in seconds. We know that 1 km = 1000 meters and 1 hour = 3600 seconds.
So, to convert 60 km/hr to m/s:
Speed in m/s = (Speed in km/hr × 1000 meters/km) / 3... |
A box contains tiles, numbered 1, 2, 3, ..., 49, 50. Only tiles which are marked with a number congruent to $1 \pmod{4}$ are blue. One tile is chosen randomly from the box. What is the probability that the tile is blue? | Positive integers congruent to $1 \pmod{4}$ are of the form:
$$1 + 4k$$
where $k$ is a non-negative integer. To determine the largest integer $k$ such that $1 + 4k \leq 50$, we solve the inequality:
$$1 + 4k \leq 50 \Rightarrow 4k \leq 49 \Rightarrow k \leq \frac{49}{4} = 12.25$$
Thus, the largest integer $k$ is $k = 1... | \frac{13}{50} | number_theory | <|im_start|>user
A box contains tiles, numbered 1, 2, 3, ..., 49, 50. Only tiles which are marked with a number congruent to $1 \pmod{4}$ are blue. One tile is chosen randomly from the box. What is the probability that the tile is blue?<|im_end|>
<|im_start|>assistant
| Positive integers congruent to $1 \pmod{4}$ are of the form:
$$1 + 4k$$
where $k$ is a non-negative integer. To determine the largest integer $k$ such that $1 + 4k \leq 50$, we solve the inequality:
$$1 + 4k \leq 50 \Rightarrow 4k \leq 49 \Rightarrow k \leq \frac{49}{4} = 12.25$$
Thus, the largest integer $k$ is $k = 1... |
Two people, A and B, simultaneously work on the same mathematics competition problem. If they solve it correctly, the teacher awards a prize of 190 yuan. The reward goes to the one who solves it correctly, and if both solve it correctly, the prize is split equally. Given that the probability of A solving it correctly i... | To solve for the most reasonable allocation of the prize money, we need to determine the expected prize allocation for both 甲 and 乙 based on the probabilities of their correct answers. The following steps provide a detailed solution:
1. **Analyze the Probabilities:**
Define the events:
- Let \( P(\text{甲 correct... | 100 | combinatorics_probability | <|im_start|>user
Two people, A and B, simultaneously work on the same mathematics competition problem. If they solve it correctly, the teacher awards a prize of 190 yuan. The reward goes to the one who solves it correctly, and if both solve it correctly, the prize is split equally. Given that the probability of A solvi... | To solve for the most reasonable allocation of the prize money, we need to determine the expected prize allocation for both 甲 and 乙 based on the probabilities of their correct answers. The following steps provide a detailed solution:
1. **Analyze the Probabilities:**
Define the events:
- Let \( P(\text{甲 correct... |
Martha starts with 3 cards. She gets 76 more from Emily. How many cards does Martha end with? | Martha starts with 3 cards and gets 76 more from Emily. To find out how many cards she ends with, you add the number of cards she starts with to the number of cards she gets:
3 (starting cards) + 76 (cards from Emily) = 79 cards
Martha ends with $\boxed{79}$ cards. | 79 | combinatorics_probability | <|im_start|>user
Martha starts with 3 cards. She gets 76 more from Emily. How many cards does Martha end with?<|im_end|>
<|im_start|>assistant
| Martha starts with 3 cards and gets 76 more from Emily. To find out how many cards she ends with, you add the number of cards she starts with to the number of cards she gets:
3 (starting cards) + 76 (cards from Emily) = 79 cards
Martha ends with $\boxed{79}$ cards.
Therefore, the final answer is: $\boxed{79}$<|im_e... |
If the height of a parallelogram with an area of 108 square centimeters is 9 centimeters (cm), what is the length of the base in centimeters (cm)? | The area of a parallelogram can be calculated using the formula:
Area = Base × Height
We are given the area (108 square centimeters) and the height (9 centimeters), and we need to find the base. We can rearrange the formula to solve for the base:
Base = Area / Height
Plugging in the given values:
Base = 108 cm² / ... | 12 | geometry | <|im_start|>user
If the height of a parallelogram with an area of 108 square centimeters is 9 centimeters (cm), what is the length of the base in centimeters (cm)?<|im_end|>
<|im_start|>assistant
| The area of a parallelogram can be calculated using the formula:
Area = Base × Height
We are given the area (108 square centimeters) and the height (9 centimeters), and we need to find the base. We can rearrange the formula to solve for the base:
Base = Area / Height
Plugging in the given values:
Base = 108 cm² / ... |
The breadth of a rectangular hall is two-thirds of its length. If the area of the hall is 2400 sq meters, what is the length of the hall in meters? | Let the length of the hall be \( L \) meters and the breadth be \( B \) meters. According to the problem, the breadth is two-thirds of the length, so we can write:
\[ B = \frac{2}{3}L \]
The area \( A \) of the rectangle is given by the product of its length and breadth:
\[ A = L \times B \]
We are given that the a... | 60 | algebra | <|im_start|>user
The breadth of a rectangular hall is two-thirds of its length. If the area of the hall is 2400 sq meters, what is the length of the hall in meters?<|im_end|>
<|im_start|>assistant
| Let the length of the hall be \( L \) meters and the breadth be \( B \) meters. According to the problem, the breadth is two-thirds of the length, so we can write:
\[ B = \frac{2}{3}L \]
The area \( A \) of the rectangle is given by the product of its length and breadth:
\[ A = L \times B \]
We are given that the a... |
For four distinct points A, B, C, and D in space, consider the following five propositions:
1. If lines AB and CD are coplanar, then lines AC and BD are coplanar as well.
2. If lines AB and CD are skew lines (not coplanar), then lines AC and BD are skew lines as well.
3. If $AB=AC$ and $DB=DC$, then $AD \perp BC... | 1. If $AB$ and $CD$ are coplanar, then by observing basic geometric principles, we note that any four points within the same plane can be connected via lines that are also within the plane. Therefore, lines $AC$ and $BD$ must be coplanar as well, making proposition 1 correct.
2. Similarly, if $AB$ and $CD$ are skew li... | 1, 2, 3, 4 | geometry | <|im_start|>user
For four distinct points A, B, C, and D in space, consider the following five propositions:
1. If lines AB and CD are coplanar, then lines AC and BD are coplanar as well.
2. If lines AB and CD are skew lines (not coplanar), then lines AC and BD are skew lines as well.
3. If $AB=AC$ and $DB=DC$, ... | 1. If $AB$ and $CD$ are coplanar, then by observing basic geometric principles, we note that any four points within the same plane can be connected via lines that are also within the plane. Therefore, lines $AC$ and $BD$ must be coplanar as well, making proposition 1 correct.
2. Similarly, if $AB$ and $CD$ are skew li... |
Shirley went to the store planning to buy $120$ balloons for $10$ dollars. When she arrived, she was surprised to nd that the balloons were on sale for $20$ percent less than expected. How many balloons could Shirley buy for her $10$ dollars? | 1. Initially, Shirley planned to buy 120 balloons for $10. This implies that the expected price per balloon is:
\[
\text{Expected price per balloon} = \frac{10 \text{ dollars}}{120 \text{ balloons}} = \frac{1}{12} \text{ dollars per balloon}
\]
2. The balloons were on sale for 20% less than expected. Therefor... | 150 | combinatorics_probability | <|im_start|>user
Shirley went to the store planning to buy $120$ balloons for $10$ dollars. When she arrived, she was surprised to nd that the balloons were on sale for $20$ percent less than expected. How many balloons could Shirley buy for her $10$ dollars?<|im_end|>
<|im_start|>assistant
| 1. Initially, Shirley planned to buy 120 balloons for $10. This implies that the expected price per balloon is:
\[
\text{Expected price per balloon} = \frac{10 \text{ dollars}}{120 \text{ balloons}} = \frac{1}{12} \text{ dollars per balloon}
\]
2. The balloons were on sale for 20% less than expected. Therefor... |
For a certain natural number $n$, $n^3$ gives a remainder of 3 when divided by 7, and $n^4$ gives a remainder of 2 when divided by 7. What remainder does $n$ give when divided by 7? | We'll use similar reasoning as earlier by looking at modular relationships:
1. We know $n^3 \equiv 3 \pmod{7}$ and $n^4 \equiv 2 \pmod{7}$. Multiplying $n^3$ by $n$, we get $n^4$. So we have the equation $n \cdot 3 \equiv 2 \pmod{7}$.
2. We need to find $n$ such that this condition is met:
- Multiplying both sides ... | 6 | number_theory | <|im_start|>user
For a certain natural number $n$, $n^3$ gives a remainder of 3 when divided by 7, and $n^4$ gives a remainder of 2 when divided by 7. What remainder does $n$ give when divided by 7?<|im_end|>
<|im_start|>assistant
| We'll use similar reasoning as earlier by looking at modular relationships:
1. We know $n^3 \equiv 3 \pmod{7}$ and $n^4 \equiv 2 \pmod{7}$. Multiplying $n^3$ by $n$, we get $n^4$. So we have the equation $n \cdot 3 \equiv 2 \pmod{7}$.
2. We need to find $n$ such that this condition is met:
- Multiplying both sides ... |
Given an odd function $f(x)$ defined on $R$ such that $f(x) = f(2-x)$. When $1 \leq x \leq 2$, $f(x) = \log_{2}(x+7)$. Find $f(2021)$.
A: $3$
B: $-3$
C: $-5$
D: $5$ | Given the function $f(x)$ is odd and defined on $\mathbb{R}$ with the property $f(x) = f(2-x)$, we can explore the implications of these conditions step by step.
1. **Understanding the Odd Function Property**: For an odd function, we have the property $f(-x) = -f(x)$. This is a fundamental characteristic of odd functi... | A | prealgebra | <|im_start|>user
Given an odd function $f(x)$ defined on $R$ such that $f(x) = f(2-x)$. When $1 \leq x \leq 2$, $f(x) = \log_{2}(x+7)$. Find $f(2021)$.
A: $3$
B: $-3$
C: $-5$
D: $5$<|im_end|>
<|im_start|>assistant
| Given the function $f(x)$ is odd and defined on $\mathbb{R}$ with the property $f(x) = f(2-x)$, we can explore the implications of these conditions step by step.
1. **Understanding the Odd Function Property**: For an odd function, we have the property $f(-x) = -f(x)$. This is a fundamental characteristic of odd functi... |
Wes wants to place a large planter pot at each corner of his rectangle-shaped pool. Each planter will have a large palm fern that is $15.00 per plant, 4 creeping jennies that costs $4.00 per plant and 4 geraniums that cost $3.50 per plant. How much will it cost to fill all the pots? | To calculate the cost to fill all the pots for Wes's rectangle-shaped pool, we break down the cost per plant type and then sum these costs for one pot. Finally, we multiply by the number of pots needed, which corresponds to the number of corners in a rectangle.
1. **Cost of Creeping Jennies per Pot:**
- Each pot co... | \$180.00 | geometry | <|im_start|>user
Wes wants to place a large planter pot at each corner of his rectangle-shaped pool. Each planter will have a large palm fern that is $15.00 per plant, 4 creeping jennies that costs $4.00 per plant and 4 geraniums that cost $3.50 per plant. How much will it cost to fill all the pots?<|im_end|>
<|im_st... | To calculate the cost to fill all the pots for Wes's rectangle-shaped pool, we break down the cost per plant type and then sum these costs for one pot. Finally, we multiply by the number of pots needed, which corresponds to the number of corners in a rectangle.
1. **Cost of Creeping Jennies per Pot:**
- Each pot co... |
For any three real numbers $a$, $b$, and $c$, with $b\neq c$, the operation $\otimes$ is defined by:
\[\otimes(a,b,c)=\frac{a}{b-c}\]
What is $\otimes(\otimes(1,2,3),\otimes(2,3,1),\otimes(3,1,2))$?
$\mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}... | 1. **Calculate each $\otimes$ operation inside the main operation**:
- $\otimes(1,2,3) = \frac{1}{2-3} = \frac{1}{-1} = -1$
- $\otimes(2,3,1) = \frac{2}{3-1} = \frac{2}{2} = 1$
- $\otimes(3,1,2) = \frac{3}{1-2} = \frac{3}{-1} = -3$
2. **Substitute these values into the main $\otimes$ operation**:
- $\otime... | \mathrm{(B)}\ -\frac{1}{4} | prealgebra | <|im_start|>user
For any three real numbers $a$, $b$, and $c$, with $b\neq c$, the operation $\otimes$ is defined by:
\[\otimes(a,b,c)=\frac{a}{b-c}\]
What is $\otimes(\otimes(1,2,3),\otimes(2,3,1),\otimes(3,1,2))$?
$\mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm... | 1. **Calculate each $\otimes$ operation inside the main operation**:
- $\otimes(1,2,3) = \frac{1}{2-3} = \frac{1}{-1} = -1$
- $\otimes(2,3,1) = \frac{2}{3-1} = \frac{2}{2} = 1$
- $\otimes(3,1,2) = \frac{3}{1-2} = \frac{3}{-1} = -3$
2. **Substitute these values into the main $\otimes$ operation**:
- $\otime... |
At the beginning of every session of the Math Club, Mr. Euler selects two students at random to solve a challenging puzzle together. There are 12 students in the club and sessions occur three times a week. How many different pairs of student sequences can be selected in one week? | 1. **Calculation for one session**: Since Mr. Euler picks two students out of 12, and the order in which they are picked matters, this is a permutation of 12 students taken 2 at a time.
2. **Number of ways to pick 2 students in one session**: $P(12,2) = 12 \times 11 = 132$ (since the first student can be any of the 12 ... | 2,299,968 | combinatorics_probability | <|im_start|>user
At the beginning of every session of the Math Club, Mr. Euler selects two students at random to solve a challenging puzzle together. There are 12 students in the club and sessions occur three times a week. How many different pairs of student sequences can be selected in one week?<|im_end|>
<|im_start|>... | 1. **Calculation for one session**: Since Mr. Euler picks two students out of 12, and the order in which they are picked matters, this is a permutation of 12 students taken 2 at a time.
2. **Number of ways to pick 2 students in one session**: $P(12,2) = 12 \times 11 = 132$ (since the first student can be any of the 12 ... |
Let A be able to complete a work in 10 days, B in 15 days, C in 20 days, and D in 30 days. If they all work together for 5 days, then what is the fraction of the work that is left? | First, let's find out how much work each person can complete in one day.
A can complete the work in 10 days, so in one day, A can complete 1/10 of the work.
B can complete the work in 15 days, so in one day, B can complete 1/15 of the work.
C can complete the work in 20 days, so in one day, C can complete 1/20 of the ... | 0, | number_theory | <|im_start|>user
Let A be able to complete a work in 10 days, B in 15 days, C in 20 days, and D in 30 days. If they all work together for 5 days, then what is the fraction of the work that is left?<|im_end|>
<|im_start|>assistant
| First, let's find out how much work each person can complete in one day.
A can complete the work in 10 days, so in one day, A can complete 1/10 of the work.
B can complete the work in 15 days, so in one day, B can complete 1/15 of the work.
C can complete the work in 20 days, so in one day, C can complete 1/20 of the ... |
In a bag, there are 7 blue chips, 5 red chips, and 4 yellow chips. One chip is drawn and then replaced before drawing a second chip. What is the probability that the two chips drawn are of different colors? Express your answer as a common fraction. | 1. Calculate the total number of chips:
$$ 7 \text{ (blue)} + 5 \text{ (red)} + 4 \text{ (yellow)} = 16 \text{ chips} $$
2. Calculate the probability of drawing each color first and then a different color:
- If the first chip is blue (7/16), the probability it is not blue (red or yellow) next is:
$$ \frac{5... | \frac{83}{128} | combinatorics_probability | <|im_start|>user
In a bag, there are 7 blue chips, 5 red chips, and 4 yellow chips. One chip is drawn and then replaced before drawing a second chip. What is the probability that the two chips drawn are of different colors? Express your answer as a common fraction.<|im_end|>
<|im_start|>assistant
| 1. Calculate the total number of chips:
$$ 7 \text{ (blue)} + 5 \text{ (red)} + 4 \text{ (yellow)} = 16 \text{ chips} $$
2. Calculate the probability of drawing each color first and then a different color:
- If the first chip is blue (7/16), the probability it is not blue (red or yellow) next is:
$$ \frac{5... |
A can finish a work in 9 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work? | First, let's find out how much work A and B can do in one day.
A can finish the work in 9 days, so A's work rate is \( \frac{1}{9} \) of the work per day.
B can finish the work in 15 days, so B's work rate is \( \frac{1}{15} \) of the work per day.
Now, let's find out how much work B has done in 10 days.
B's work in... | 3 | prealgebra | <|im_start|>user
A can finish a work in 9 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?<|im_end|>
<|im_start|>assistant
| First, let's find out how much work A and B can do in one day.
A can finish the work in 9 days, so A's work rate is \( \frac{1}{9} \) of the work per day.
B can finish the work in 15 days, so B's work rate is \( \frac{1}{15} \) of the work per day.
Now, let's find out how much work B has done in 10 days.
B's work in... |
(The full score of this question is 12 points) Given the function $f(x) = x^3 - x^2 + bx + c$.
(1) If $f(x)$ is an increasing function on $(-\infty, +\infty)$, find the range of values for $b$;
(2) If $f(x)$ takes an extreme value at $x = 1$, and $f(x) < c^2$ always holds when $x \in [-1,2]$, find the range of values f... | Solution:
(1) $f'(x) = 3x^2 - x + b$. Since $f(x)$ is an increasing function on $(-\infty, +\infty)$, then $f'(x) \geq 0$, i.e., $3x^2 - x + b \geq 0$,
Therefore, $b \geq x - 3x^2$ always holds on $(-\infty, +\infty)$.
Let $g(x) = x - 3x^2$.
When $x = \frac{1}{6}$, $g(x)_{\text{max}} = \frac{1}{12}$, therefore $b \geq ... | (-\infty, -1) \cup (2, +\infty) | combinatorics_probability | <|im_start|>user
(The full score of this question is 12 points) Given the function $f(x) = x^3 - x^2 + bx + c$.
(1) If $f(x)$ is an increasing function on $(-\infty, +\infty)$, find the range of values for $b$;
(2) If $f(x)$ takes an extreme value at $x = 1$, and $f(x) < c^2$ always holds when $x \in [-1,2]$, find the ... | Solution:
(1) $f'(x) = 3x^2 - x + b$. Since $f(x)$ is an increasing function on $(-\infty, +\infty)$, then $f'(x) \geq 0$, i.e., $3x^2 - x + b \geq 0$,
Therefore, $b \geq x - 3x^2$ always holds on $(-\infty, +\infty)$.
Let $g(x) = x - 3x^2$.
When $x = \frac{1}{6}$, $g(x)_{\text{max}} = \frac{1}{12}$, therefore $b \geq ... |
Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is an Ace and the second card is a $\diamondsuit$? | We consider two cases because the first card being an Ace could be a $\diamondsuit$ or not a $\diamondsuit$.
1. **Case 1: The first card is the Ace of Diamonds**:
- Probability that the Ace of Diamonds is drawn first: $\dfrac{1}{52}$.
- Probability that a Diamond (excluding the Ace of Diamonds) is drawn second: ... | \dfrac{1}{52} | combinatorics_probability | <|im_start|>user
Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is an Ace and the second card is a $\diamondsuit$?<|im_end|>
<|im_start|>assistant
| We consider two cases because the first card being an Ace could be a $\diamondsuit$ or not a $\diamondsuit$.
1. **Case 1: The first card is the Ace of Diamonds**:
- Probability that the Ace of Diamonds is drawn first: $\dfrac{1}{52}$.
- Probability that a Diamond (excluding the Ace of Diamonds) is drawn second: ... |
Tommy has 3 different types of crates. Crate A can hold 20 kilograms of tomatoes and 10 kilograms of oranges, and Tommy bought 2 of Crate A for $220. Crate B can hold 25 kilograms of tomatoes, 15 kilograms of oranges, and 5 kilograms of apples, and Tommy bought 3 of Crate B for $375. Crate C can hold 30 kilograms of to... | First, let's calculate the total cost of the crates, transportation, and packing materials:
Cost of Crate A: $220
Cost of Crate B: $375
Cost of Crate C: $180
Transportation: $50
Packing materials: $30
Total cost = $220 + $375 + $180 + $50 + $30 = $855
Now, let's calculate the total weight of the fruits that Tommy ca... | \$179 | prealgebra | <|im_start|>user
Tommy has 3 different types of crates. Crate A can hold 20 kilograms of tomatoes and 10 kilograms of oranges, and Tommy bought 2 of Crate A for $220. Crate B can hold 25 kilograms of tomatoes, 15 kilograms of oranges, and 5 kilograms of apples, and Tommy bought 3 of Crate B for $375. Crate C can hold 3... | First, let's calculate the total cost of the crates, transportation, and packing materials:
Cost of Crate A: $220
Cost of Crate B: $375
Cost of Crate C: $180
Transportation: $50
Packing materials: $30
Total cost = $220 + $375 + $180 + $50 + $30 = $855
Now, let's calculate the total weight of the fruits that Tommy ca... |
Evaluate the sum and product of the sequence: $1342 + 2431 + 3124 + 4213$, then multiply the sum by $3$.
$\textbf{(A)}\: 33{,}000\qquad\textbf{(B)} \: 33{,}110\qquad\textbf{(C)} \: 33{,}210\qquad\textbf{(D)} \: 33{,}300\qquad\textbf{(E)} \: 33{,}330$ | To find the answer, first sum the numbers $1342$, $2431$, $3124$, and $4213$. Like the original method, align the numbers vertically and add them column by column:
\[
\begin{array}{c}
\phantom{+}1342 \\
+2431 \\
+3124 \\
+4213 \\
\hline
\end{array}
\]
Calculate column by column:
1. **Units column**: $2 + 1 + 4 + 3 =... | \textbf{(E)} \: 33{,}330 | number_theory | <|im_start|>user
Evaluate the sum and product of the sequence: $1342 + 2431 + 3124 + 4213$, then multiply the sum by $3$.
$\textbf{(A)}\: 33{,}000\qquad\textbf{(B)} \: 33{,}110\qquad\textbf{(C)} \: 33{,}210\qquad\textbf{(D)} \: 33{,}300\qquad\textbf{(E)} \: 33{,}330$<|im_end|>
<|im_start|>assistant
| To find the answer, first sum the numbers $1342$, $2431$, $3124$, and $4213$. Like the original method, align the numbers vertically and add them column by column:
\[
\begin{array}{c}
\phantom{+}1342 \\
+2431 \\
+3124 \\
+4213 \\
\hline
\end{array}
\]
Calculate column by column:
1. **Units column**: $2 + 1 + 4 + 3 =... |
In the sequence \( \left\{a_{n}\right\}_{n=1}^{+\infty} \), where \( x_{1}=1 \), for any positive integer \( n \), \( x_{n+1}=x_{n}+\left[\sqrt{x_{n}}\right] \), where the symbol \( [x] \) denotes the greatest integer not exceeding \( x \), find the smallest positive integer \( n \) such that \( x_{n} \geqslant 2019 \)... | 1. Start with the initial conditions of the sequence \( \{ x_n \}_{n=1}^{\infty} \) where \( x_1 = 1 \).
2. Apply the recurrence relation \( x_{n+1} = x_n + \lfloor \sqrt{x_n} \rfloor \):
\[
\begin{aligned}
x_1 &= 1, \\
x_2 &= x_1 + \lfloor \sqrt{x_1} \rfloor = 1 + 1 = 2, \\
x_3 &= x_2 + \lfloor \sqrt{x_... | 95 | number_theory | <|im_start|>user
In the sequence \( \left\{a_{n}\right\}_{n=1}^{+\infty} \), where \( x_{1}=1 \), for any positive integer \( n \), \( x_{n+1}=x_{n}+\left[\sqrt{x_{n}}\right] \), where the symbol \( [x] \) denotes the greatest integer not exceeding \( x \), find the smallest positive integer \( n \) such that \( x_{n} ... | 1. Start with the initial conditions of the sequence \( \{ x_n \}_{n=1}^{\infty} \) where \( x_1 = 1 \).
2. Apply the recurrence relation \( x_{n+1} = x_n + \lfloor \sqrt{x_n} \rfloor \):
\[
\begin{aligned}
x_1 &= 1, \\
x_2 &= x_1 + \lfloor \sqrt{x_1} \rfloor = 1 + 1 = 2, \\
x_3 &= x_2 + \lfloor \sqrt{x_... |
Let $f(x)$ be a function from positive real numbers to real numbers satisfying the equation
\[ xf(y) - 2yf(x) = f \left( \frac{x}{y} \right) \]
for all positive real numbers $x$ and $y.$ Determine all possible values of $f(50).$ | - **Step 1: Set $x = y$:**
\[ xf(x) - 2xf(x) = f(1) \implies -xf(x) = f(1) \]
\[ \text{which gives us } f(1) = 0. \]
- **Step 2: Set $y = 1$:**
\[ xf(1) - 2f(x) = f(x) \]
\[ 0 - 2f(x) = f(x) \]
\[ -2f(x) = f(x) \]
\[ -3f(x) = 0 \]
\[ f(x) = 0 \text{ for all } x. \]
- **Step 3: Verify the solution by plugging $f(x) = ... | 0 | precalculus_calculus | <|im_start|>user
Let $f(x)$ be a function from positive real numbers to real numbers satisfying the equation
\[ xf(y) - 2yf(x) = f \left( \frac{x}{y} \right) \]
for all positive real numbers $x$ and $y.$ Determine all possible values of $f(50).$<|im_end|>
<|im_start|>assistant
| - **Step 1: Set $x = y$:**
\[ xf(x) - 2xf(x) = f(1) \implies -xf(x) = f(1) \]
\[ \text{which gives us } f(1) = 0. \]
- **Step 2: Set $y = 1$:**
\[ xf(1) - 2f(x) = f(x) \]
\[ 0 - 2f(x) = f(x) \]
\[ -2f(x) = f(x) \]
\[ -3f(x) = 0 \]
\[ f(x) = 0 \text{ for all } x. \]
- **Step 3: Verify the solution by plugging $f(x) = ... |
Find the value of $k$ so that the line $4x + 7y + k = 0$ is tangent to the parabola $y^2 = 16x.$ | First, express $x$ from the line equation:
\[ x = -\frac{7y + k}{4}.\]
Substitute this into the equation of the parabola:
\[ y^2 = 16 \left(-\frac{7y + k}{4}\right).\]
\[ y^2 = -28y - 4k. \]
\[ y^2 + 28y + 4k = 0. \]
This is a quadratic equation in terms of $y$. For the line to be tangent to the parabola, there must ... | 49 | algebra | <|im_start|>user
Find the value of $k$ so that the line $4x + 7y + k = 0$ is tangent to the parabola $y^2 = 16x.$<|im_end|>
<|im_start|>assistant
| First, express $x$ from the line equation:
\[ x = -\frac{7y + k}{4}.\]
Substitute this into the equation of the parabola:
\[ y^2 = 16 \left(-\frac{7y + k}{4}\right).\]
\[ y^2 = -28y - 4k. \]
\[ y^2 + 28y + 4k = 0. \]
This is a quadratic equation in terms of $y$. For the line to be tangent to the parabola, there must ... |
Our volleyball team features 12 members. I want to choose a starting team consisting of 6 players, one of whom will be the libero (a specialized defensive position in volleyball). How many ways can I select my starting lineup, considering the libero needs to be one of the players and the other 5 positions don't have a ... | 1. First, choose the libero from the 12 players. This can be done in 12 ways.
2. Next, choose 5 other players from the remaining 11 players. Since order does not matter for these 5 players, we need to determine the number of combinations which are given by $\binom{11}{5}$ (the number of ways to choose 5 players out of ... | 5544 | combinatorics_probability | <|im_start|>user
Our volleyball team features 12 members. I want to choose a starting team consisting of 6 players, one of whom will be the libero (a specialized defensive position in volleyball). How many ways can I select my starting lineup, considering the libero needs to be one of the players and the other 5 positi... | 1. First, choose the libero from the 12 players. This can be done in 12 ways.
2. Next, choose 5 other players from the remaining 11 players. Since order does not matter for these 5 players, we need to determine the number of combinations which are given by $\binom{11}{5}$ (the number of ways to choose 5 players out of ... |
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