uuid stringlengths 36 36 | subject stringclasses 6
values | has_image bool 2
classes | image stringclasses 160
values | problem_statement stringlengths 32 784 | golden_answer stringlengths 7 1.13k |
|---|---|---|---|---|---|
9d959d14-7b9a-4159-a162-048d3b44d728 | integral_calc | false | null | Solve the integral:
$$
\int \frac{ -8 \cdot \cos(-4 \cdot x)^3 }{ 5 \cdot \sin(-4 \cdot x)^9 } \, dx
$$ | $\int \frac{ -8 \cdot \cos(-4 \cdot x)^3 }{ 5 \cdot \sin(-4 \cdot x)^9 } \, dx$ = $C-\frac{2}{5}\cdot\left(\frac{1}{3}\cdot\left(\cot(4\cdot x)\right)^6+\frac{1}{4}\cdot\left(\cot(4\cdot x)\right)^4+\frac{1}{8}\cdot\left(\cot(4\cdot x)\right)^8\right)$ |
9d9a08e8-cb1b-4e73-bdb1-5c8f16ce0448 | algebra | false | null | Find the zeros of the polynomial function with complex roots: $f(x) = 2 \cdot x^3 - 2 \cdot x^2 + 5 \cdot x - 5$. | $x$ = $1$, $i\cdot\sqrt{\frac{5}{2}}$, $-i\cdot\sqrt{\frac{5}{2}}$ |
9d9a2009-a680-4c90-b633-7abf18e2f9e0 | differential_calc | false | null | Calculate the integral:
$$
\int \frac{ 1 }{ \sqrt[4]{(x-1)^3 \cdot (x+2)^5} } \, dx
$$ | $\int \frac{ 1 }{ \sqrt[4]{(x-1)^3 \cdot (x+2)^5} } \, dx$ = $\frac{4}{3}\cdot\sqrt[4]{\frac{x-1}{x+2}}+C$ |
9e0bf1a0-877c-4582-bfb0-0f1dd3945485 | sequences_series | false | null | For a sequence with the general term $x_{n} = \frac{ 3 \cdot n - 5 }{ 9 \cdot n + 4 }$, it is known that $\lim_{n \to \infty} \left(x_{n}\right) = \frac{ 1 }{ 3 }$.
Find the number of points $x_{n}$ lying outside the open interval $\left(\frac{ 1 }{ 3 } - \frac{ 1 }{ 1000 }, \frac{ 1 }{ 3 } + \frac{ 1 }{ 1000 }\right)... | The final answer: $703$ |
9e33d4d5-111e-40d8-947f-1e6745799fb6 | algebra | false | null | Find two consecutive even numbers whose product is $624$.
Use the method of factoring, the square root principle, or the Quadratic Formula. | The final answer: $24$, $26$, $-24$, $-26$ |
9e64c3be-6d67-467d-ba24-30c5fb0b90dc | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAACC9klEQVR4nO3dd5iU5b3G8XvYwrosSwcFQQQREKT3tgIC0gSkiViiEgu2qFESrLHGgkaNqBhQo2ABUekdlg5SpChNkbpIhy2wfc4fHonD84KwO/M+U76f68p1HX8zzN45WWFv3qd4vF6vVwDOyYgRIzRixAhJ0sMPP6yHH37YciIAAIDQUsR2AAAAAACRgwICAAAAwDUUEAAAAACuoYAAAAAAcA0FBAAAAIBrKCAAAAAAXEMBAQAAAOAaCg... | Let $S$ be the region bounded by the graphs of $y = -\sin(\pi \cdot x)$ and $y = -x^3 + 4 \cdot x$, as shown in the figure above. Find the area of $S$. | The area of $S$ is $4$ units². |
9e8c6804-bff6-4a53-9a3d-aafd24edb916 | multivariable_calculus | false | null | Evaluate the integral $\int_{0}^1 \int_{x^3}^{x^2} (x+y) \, dy \, dx$ by reversing the order of integration. | The integral corresponding to the reversed order of integration: $\int_0^1\int_{\sqrt{y}}^{\sqrt[3]{y}}(x+y)dxdy$
The value of the given integral: $\frac{11}{140}$ |
9eaefe5e-2a89-4c91-8807-8a51a4682bd8 | sequences_series | false | null | Find the Fourier series of the function $f(x) = \frac{ -1 }{ 2 } \cdot x$ in the interval $[-2,2]$. | The Fourier series is: $\sum_{n=1}^\infty\left(\frac{2\cdot(-1)^n}{\pi\cdot n}\cdot\sin\left(\frac{\pi\cdot n\cdot x}{2}\right)\right)$ |
9ed2cf0b-2cab-4481-8a1f-d82d0592618b | multivariable_calculus | false | null | Find the moment of inertia of one arch of the cycloid $x = 3 \cdot a \cdot \left(\frac{ t }{ 2 } - \sin\left(\frac{ t }{ 2 }\right)\right)$, $y = 3 \cdot a \cdot \left(1 - \cos\left(\frac{ t }{ 2 }\right)\right)$ relative to the x-axis. | Moment of Inertia: $\frac{1152}{5}\cdot a^3$ |
9eedef68-3b7f-472b-80eb-594924624379 | precalculus_review | false | null | Find zeros of $f(x) = 2 \cdot x^4 - 13 \cdot x^3 + 24 \cdot x^2 - 13 \cdot x + 2$. | The final answer: $x_1=2+\sqrt{3}$, $x_2=2-\sqrt{3}$, $x_3=2$, $x_4=\frac{1}{2}$ |
9eee7969-14e8-4e46-af9e-f5abe22824c9 | sequences_series | false | null | Let $P_{n}(x)$ be the n-th order Taylor polynomial of $f(x) = \cos(x)$ about $a = 0$. Find the minimum $n$ such that the approximation $f(x) \approx P_{n}(x)$ is accurate to within $0.00001$ for every value of $x$ satisfying $|x| \le 0.1$. | The final answer: $n=3$ |
9f291789-5366-4314-8c23-34f9d11ce96d | integral_calc | false | null | Compute the volume of the solid formed by rotating about the x-axis the area bounded by the axes and the parabola $x^{\frac{ 1 }{ 2 }}+y^{\frac{ 1 }{ 2 }}=3^{\frac{ 1 }{ 2 }}$. | Volume = $\pi\cdot\frac{9}{5}$ |
9f754564-1291-4160-9cb2-349689537534 | differential_calc | false | null | Solve to four decimal places using Newton's method: $x^3 + (x + 1)^3 = 10^3$. Choose any initial guess $x_0$ that is not the exact root. | $x$ ≈ $x=7.4055$ |
a05cd0c6-c878-437a-8b28-5137e7036859 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ 3 }{ 4 \cdot x^2 \cdot \sqrt{5 \cdot x^2-2 \cdot x+1} } \, dx
$$ | Answer is: $\frac{3}{4}\cdot\left(C-\sqrt{5+\frac{1}{x}^2-\frac{1\cdot2}{x}}-\ln\left(\left|\frac{1}{x}+\sqrt{5+\frac{1}{x}^2-\frac{1\cdot2}{x}}-1\right|\right)\right)$ |
a0840748-bc14-4a3b-b024-7db40ab0ae4a | differential_calc | false | null | Use implicit differentiation to find an equation of the tangent line to the curve
$x^4 - x^3 \cdot y^2 - y^3 \cdot x + 5 \cdot y^4 = 17$ at the point
$P(-2,-1)$. | The line tangent to the curve $x^4 - x^3 \cdot y^2 - y^3 \cdot x + 5 \cdot y^4 = 17$ at the point $(-2,-1)$ is: $y+1=-\frac{43}{30}\cdot(x+2)$ |
a0aa7803-36b6-4822-a264-43c4b180b36f | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAA0YAAAE0CAIAAACHM7SsAABBpElEQVR4nO3deWBTZaI28DfJyZ50S7qmdC8tLbSUpeyUvawygMgyyqAfAg64IYq7jAN4r9cFvaCAIMo4M+wIwrBZULDsUFpooXvT0n1J2uzr+f6I9HbQgS6hp6c8v7+Sk5PjU0OSJ+9ZXg5N0wQAAAAA2IzLdAAAAAAA6ChUOgAAAADWQ6UDAAAAYD1UOgAAAADWQ6UDAAAAYD1UOgAAAADWQ6UDAAAAYD1UOgAAAADWQ6UDAAAAYD1UOg... | Given $m(x) = g(x)^2$, find $m'(-2)$ using the table below: | $m'(-2)$ = $-418$ |
a112f370-d3c2-4094-b750-8d9886059273 | multivariable_calculus | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAWQAAACyCAIAAAD6aSLxAAAYDUlEQVR4nO2dz28TV/fG73zVLVROukIUVZl0UXWRqnIkhPJGohJxSrsAAXIAqULqgo4XbEChMlRdNNDGFV10kTiLStmAXSmLqqojAlIWJGKBsrDFohtsRSjqyial/AHzXZyXeQdnfOfM+M6d6/T5rBzHiZ87c+bMnV/PY7muKwAAIIz/S1sAAGAwQLMAALB4y3tlWVaKOgAAZuKdqXgr8N39imVZ+36MACjEP4fAYQgAgAWaBQCABZoFAI... | Use the midpoint rule with $m=2$, $n=2$ to estimate $\int \int f(x,y) \, dx \, dy$, where the values of the function $f$ on $R = [8,10] \times [9,11]$ are given in the following table: | $\int \int f(x,y) \, dx \, dy$ is approximately: $21.3$ |
a14e11ca-bae3-4aa2-9e08-d7b73a0e3f5a | integral_calc | false | null | Calculate the integral:
$$
\int_{-\sqrt{2}}^{\sqrt{2}} \frac{ 23 \cdot x^7+7 \cdot x^6-130 \cdot x^5-72 \cdot x^3-112 \cdot x^2+4 \cdot x+7 }{ x^2+4 } \, dx
$$ | $\int_{-\sqrt{2}}^{\sqrt{2}} \frac{ 23 \cdot x^7+7 \cdot x^6-130 \cdot x^5-72 \cdot x^3-112 \cdot x^2+4 \cdot x+7 }{ x^2+4 } \, dx$ = $7\cdot\arctan\left(\frac{1}{\sqrt{2}}\right)-\frac{392\cdot\sqrt{2}}{15}$ |
a1cfa2ef-8b1f-409b-bffe-7e18847ddd71 | multivariable_calculus | false | null | Find the first derivative $y_{x}'$ of the function:
$$
x = \arcsin\left(\frac{ t }{ \sqrt{1+t^2} }\right), \quad y = \arccos\left(\frac{ 1 }{ \sqrt{1+t^2} }\right), \quad t \ge 0
$$ | $y_{x}'$ = $1$ |
a20500e2-a8d0-47d3-a626-99e954939c82 | multivariable_calculus | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAVQAAAFFCAIAAAA8eoLcAACvnUlEQVR4nOxdd3wUxRd/b66l995DgBB6700QUEEsSFEMUkWKArafXUQEVBBUwN4L9oIVBZROgJBKeu+997ub9/tjdi9HU0IuyQXz/WhIruzO7s6befX7kIigC13own8PrKMH0IUudKFj0CX81w7+WYnrUvG6cAG6hP+/AkQ0/rNrLehCl/BfO0DEKxfpC9aCLvwH0SX81wji4+N37tx5+PBhIf9ZWVk7d+48deoU59zwmdzc3Pz8/I4bYx... | The graph of the polar rectangular region $D$ is given. Express the region $D$ in polar coordinates. | 1. The interval of $r$ is $[3,5]$
2. The interval of $\theta$ is $\left[\frac{\pi}{4},\frac{5}{4}\cdot\pi\right]$ |
a2102909-944e-4b02-bcfb-74f3d7ebf95b | integral_calc | false | null | Compute the integral:
$$
3 \cdot \int x^{-8} \cdot \left(9+x^2\right)^{\frac{ 1 }{ 2 }} \, dx
$$ | $3 \cdot \int x^{-8} \cdot \left(9+x^2\right)^{\frac{ 1 }{ 2 }} \, dx$ = $C+\frac{2}{1215}\cdot\left(1+\frac{9}{x^2}\right)^2\cdot\sqrt{1+\frac{9}{x^2}}-\frac{1}{729}\cdot\left(1+\frac{9}{x^2}\right)\cdot\sqrt{1+\frac{9}{x^2}}-\frac{1}{1701}\cdot\left(1+\frac{9}{x^2}\right)^3\cdot\sqrt{1+\frac{9}{x^2}}$ |
a2746873-77c8-49fc-97b9-98a814c7f0f1 | algebra | false | null | Perform the indicated operation and express the result as a simplified complex number:
$$
\frac{ 4+5 \cdot i }{ 4-5 \cdot i }
$$ | The final answer: $-\frac{9}{41}+\frac{40}{41}\cdoti$ |
a29dedad-3c66-4fb6-a142-1d724cd3f3f0 | sequences_series | false | null | For which $p$ does the series $\sum_{n=1}^\infty \left(\frac{ p^{n^2} }{ 2^n }\right)$ converge? | Converges for : $p\le1$ (leave empty, if the series diverges for any $p$). |
a328709a-3973-45ce-ac0c-956e736146fb | precalculus_review | false | null | A rental car company rents cars for a flat fee of $\$20$ and an hourly charge of $\$10.25$. Therefore, the total cost $C$ to rent a car is a function of the hours $t$ the car is rented plus the flat fee.
1. Write the formula for the function that models this situation.
2. Find the total cost to rent a car for 2 days a... | 1. The formula for the function is $C(t)$ = $10.25\cdot t+20$
2. The total cost to rent a car is $ $583.75$
3. The time of car rental in hours is $40.27$ |
a3940e7d-d8a4-47f5-a5b8-fd8850fea3e8 | differential_calc | false | null | For the function $r = \frac{ 1 }{ 2 } \cdot \arctan\left(\frac{ 3 }{ \beta }\right) + \arccot\left(3 \cdot \cot\left(\beta\right)\right)$, find the derivative $r'(0)$ and $r'(2 \cdot \pi)$. Submit as your final answer:
1. $r'(0)$
2. $r'(2 \cdot \pi)$ | 1. $r'(0)$ = $\frac{1}{6}$
2. $r'(2 \cdot \pi)$ = $\frac{9+8\cdot\pi^2}{54+24\cdot\pi^2}$ |
a3b43f93-43b4-4b70-90f5-548457398733 | multivariable_calculus | false | null | Find the tangential and normal components of acceleration if $\vec{r}(t) = \left\langle 6 \cdot t, 3 \cdot t^2, 2 \cdot t^3 \right\rangle$ | $a_{T}$ = $\frac{12\cdot t^3+6\cdot t}{\sqrt{t^4+t^2+1}}$ ; $a_{N}$ = $\frac{6\cdot\sqrt{t^4+4\cdot t^2+1}}{\sqrt{t^4+t^2+1}}$ |
a3bd66e6-c94c-4f2f-8571-02904921c0b1 | differential_calc | false | null | Find $\frac{ d y }{d x}$, given $y=\tan(2 \cdot v)$ and $v=\arctan(2 \cdot x-1)$. | The final answer: $\frac{dy}{dx}=\frac{2\cdot x^2-2\cdot x+1}{2\cdot\left(x-x^2\right)^2}$ |
a3d65618-5fb7-4674-96dd-b23aa1ecfe58 | algebra | false | null | A sky diver jumps from a reasonable height above the ground. The air resistance she experiences is proportional to her velocity, and the constant of proportionality is $0.24$. It can be shown that the downward velocity of the sky diver at time $t$ is given by
$$
v(t) = 180 \cdot \left(1 - e^{-0.24 \cdot t}\right)
$$
wh... | 1. The initial velocity of the sky diver is $0$.
2. The velocity after $4$ seconds is $111.1$.
3. The terminal velocity of the sky diver is $180$. |
a3d66c89-c806-451b-8a24-006d0d179cf7 | algebra | false | null | Determine the point(s) where the line $y = m \cdot x$ intersects the circle $x^2 + y^2 = 4$. | The final answer: $P\left(\sqrt{\frac{4}{1+m^2}},\sqrt{\frac{4\cdot m^2}{1+m^2}}\right)$, $P\left(-\sqrt{\frac{4}{1+m^2}},-\sqrt{\frac{4\cdot m^2}{1+m^2}}\right)$ |
a4a45e5a-b4ee-43fd-813e-09deaedcad2b | multivariable_calculus | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAPwAAAEjCAIAAACZ1vt+AABQeklEQVR4nO2deXwURdrHn6d6ZnJf5CIhhCMBQoJAuE9RFLkEZL0QxBVddN11V1f3VVfd1WU91mtFxftEdr1FEBCUQzAcAuEICeEKCZD7ziRzz3Q97x/VMxlyEWByTNLfjx+Z9HRXV/f8uvqpp556CokI2gsiQsTW7NOaPVVULg3WnidrTsfuD56qdZW2pj1E3+BlUlVVtXz58ueee85qtYotzz333HvvvVdVVeXax2Kx7Nq1qx3qptINaQ... | Find the area of the region $D$: | The area is $1$ |
a4e37a73-6d8c-4d88-b69a-73d95d6eb24c | multivariable_calculus | false | null | Let $z = e^{1 - x \cdot y}$, $x = t^{\frac{ 1 }{ 3 }}$, $y = t^3$. Find $\frac{ d z }{ d t }$. | $\frac{ d z }{ d t }$ = $\frac{-(10\cdot e)}{3}\cdot e^{-t^3\cdot\sqrt[3]{t}}\cdot t^2\cdot\sqrt[3]{t}$ |
a5188065-d6c7-41b1-875b-d6511217d604 | algebra | false | null | Consider the equation $25 + 8x + x^2 + y^2 - 10y = 0$.
1. Complete the square to write the equation in the standard form of a circle.
2. Identify the center and the radius of the circle.
*Note: The standard form of a circle is $(x-a)^2 + (y-b)^2 = r^2$.* | 1. Standard form: $(x+4)^2+(y-5)^2=16$
2. Center $(a,b) = $ $P\left(-4,\ 5\right)$
3. Radius $r = $ $4$ |
a5621a7e-3e22-421b-9363-cae4747d5c39 | algebra | false | null | To convert from $x$ degrees Celsius to $y$ degrees Fahrenheit, we use the formula $f(x) = \frac{ 9 }{ 5 } \cdot x + 32$. Find the inverse function, if it exists. If it doesn't exist, write $\text{None}$. | Inverse function: $g(x)=\frac{5}{9}\cdot(x-32)$ |
a674c2d7-a34b-478c-a2db-568882e13a76 | integral_calc | false | null | The chain of length $200$ rises up, winding on the winch. Compute the work of the weight force when lifting the chain, neglecting the size of the winch, if the running meter of the chain weighs $50$ kg. | $W$ = $-1000000$ |
a6a2e299-2748-478f-b471-75c40da9448b | algebra | false | null | Rewrite the quadratic expression $2 + z - 6 \cdot z^2$ by completing the square. | $2 + z - 6 \cdot z^2$ = $-6\cdot\left(z-\frac{1}{12}\right)^2+\frac{49}{24}$ |
a6d4ebc2-4ce7-401f-bafd-bb7e50fe8ee7 | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAATMAAAEwCAIAAABUrgGnAAA5lElEQVR4nO2deXxTZfb/z5M2adOkS7oXutFCW8raQin7IrKDCCqbfhUdvjOOM+j3Jzpu428c5+vMy2XUGeUno6iIKIygwJSyF4qUxdLSBVq6722a7s2+3vP746axtqUtTdrkps/79RLT3OTk3Nx87rOd5xyCiEChUBwMnr0doFAofUCVSaE4IlSZHIaORJwYqkwOQwgBqk8nhSqTM+Tl5aWmpsrlcgCoq6s7ceJEVVUVABBCEFGpVMpkMl... | Use the graph of the function $y = f(x)$ shown here to find $\lim_{x \to 0} \left(f(x)\right)$, if possible. Estimate when necessary. | $\lim_{x \to 0} \left(f(x)\right)$ = None |
a747b4a2-1af3-4522-83ed-b5d654c1b27d | sequences_series | false | null | Compute $\lim_{x \to 0}\left(\frac{ \cos(x)+2 }{ x^3 \cdot \sin(x) }-\frac{ 3 }{ x^4 }\right)$. Use the expansion of the function in the Taylor series. | The final answer: $\frac{1}{60}$ |
a77b8f2c-9d90-4790-8e66-137b7346b88c | integral_calc | false | null | Compute the integral:
$$
\int \frac{ -\sin(2 \cdot x)^4 }{ \cos(2 \cdot x) } \, dx
$$ | $\int \frac{ -\sin(2 \cdot x)^4 }{ \cos(2 \cdot x) } \, dx$ = $\frac{1}{2}\cdot\left(C+\frac{1}{3}\cdot\left(\sin(2\cdot x)\right)^3+\sin(2\cdot x)-\frac{1}{2}\cdot\ln\left(\frac{\left|1+\sin(2\cdot x)\right|}{\left|\sin(2\cdot x)-1\right|}\right)\right)$ |
a80b7fea-f3d2-4cee-a773-0468cbb62864 | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAXUAAAE+CAIAAAD9ABcwAACFXUlEQVR4nO39d3xc2XXni/7WPpWQc86BBBhBMKdm6m42Oyq3ZcnKsjWW7aeR9fm8++7csT1znz8e3zf+2B7HUcuW1Yod1LmbqZkTCJIgQSKDIEDknHPVOXu9P/apYhEAu0FkkOcrNlQoVO2z69Q+66y9IjEz5h91FCJiZt9P9fyUD3yzUk9aWFgsR8TCHIaIfOJDPeP/YIJMgSVWLCweCRZIvtTV1b377rt1dXVKjtTX1x8+fLigoMBfjr... | Evaluate the integral of the functions graphed using the formula for areas of triangles, and subtracting the areas below the $x$-axis: | The final answer: $6$ |
a84a369a-f932-4e1b-81d4-3ba80b0674d9 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ \sin(x)^2 \cdot \cos(x) }{ \sin(x) + \cos(x) } \, dx
$$ | $\int \frac{ \sin(x)^2 \cdot \cos(x) }{ \sin(x) + \cos(x) } \, dx$ = $C+\frac{1}{4}\cdot\left(\ln\left(1+\tan(x)\right)+\ln\left(\left|\cos(x)\right|\right)\right)-\frac{1}{4}\cdot\left(1+\tan(x)\right)\cdot\left(\cos(x)\right)^2$ |
a8f0eaa3-b9da-4124-8a5f-fb474f6d858d | differential_calc | false | null | Compute the limit:
$$
\lim_{x \to 3} \left( \frac{ x }{ x-3 } - \frac{ 1 }{ \ln\left( \frac{ x }{ 3 } \right) } \right)
$$ | $\lim_{x \to 3} \left( \frac{ x }{ x-3 } - \frac{ 1 }{ \ln\left( \frac{ x }{ 3 } \right) } \right)$ = $\frac{1}{2}$ |
a8f44f48-ca13-4087-8c8f-a3a62866f729 | differential_calc | false | null | Compute the derivative of the function $y = \sqrt[4]{\frac{ x^5 \cdot \left(5 \cdot x^2+1\right) }{ \sqrt[3]{2+3 \cdot x} }}$ by taking the natural log of both sides of the equation. | Derivative: $y'=\frac{50\cdot x^3+35\cdot x^2+14\cdot x+10}{30\cdot x^4+20\cdot x^3+12\cdot x^2+8\cdot x}\cdot\sqrt[4]{\frac{x^5\cdot\left(5\cdot x^2+1\right)}{\sqrt[3]{2+3\cdot x}}}$ |
a9032c60-861d-40cc-9d18-5703fc391b80 | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAWUAAADuCAIAAABu54FDAABhAklEQVR4nO2deSBU3/v4z4ytLFmikEgoJWVNuyRKpZI2lfaN9n0P7dJOSkX7Sohs2bNmKyLZ96XsO2Nm7u+P833f3/3MSENmzIz7+mvOcsdz7nM9c89zznkeAoIg4D+ioqJu3Ljh5OQEBirTp0+PiYnpbymYgoWFhbOzs7i4eH8L0j9cuXJFT0/PwMCgvwXhMB49eiQmJrZv3z5Y5MW2DRs2bPDgwSNGjOgPwdgCXl5ebh2+gICAjIzM0K... | The data in the following table are used to estimate the average power output produced by Peter Sagan for each of the last $18$ sec of Stage $1$ of the $2012$ Tour de France.
Average Power Output:
Estimate the net energy used in kilojoules (kJ), noting that $1 \cdot W = 1 \cdot \frac{ J }{ s }$ and the average pow... | $17$ kJ. |
a91364ec-3bee-4ace-a8e8-c9d004878f3f | sequences_series | false | null | Compute $\lim_{x \to 0}\left(\frac{ 3 \cdot \cos(x)+6 }{ 4 \cdot x^3 \cdot \sin(x) }-\frac{ 9 }{ 4 \cdot x^4 }\right)$. Use the expansion of the function in the Taylor series. | The final answer: $\frac{1}{80}$ |
a91e763c-994d-43f5-bee0-b4348dcf9e16 | precalculus_review | false | null | Solve $2 \cdot z = |z| + 2 \cdot i$. | The final answer: $z=\frac{1}{\sqrt{3}}+i$ |
a92b0179-4b43-4f1b-844d-303642f4a04f | multivariable_calculus | false | null | Find $f(x,y) = e^{x^2} + \sqrt{y}$ at $P(0.1,9.1)$. Find the linear approximation for this function. | The final answer:
1. $f\left(x_{0},y_{0}\right)$: $4.02667079$
2. $L(x,y)$: $4+\frac{1}{6}\cdot(y-9)$ |
a9475aa9-2408-495d-8644-3b969183c644 | differential_calc | false | null | Compute the derivative of the implicit function:
$$
2 \cdot y^4 - \frac{ 3 \cdot x + 3 \cdot y }{ 2 \cdot x - 2 \cdot y } = 0
$$ | $y'$: $y'=-\frac{y^2}{2\cdot\left(x^2-y^2\right)-x\cdot y}$ |
a984873a-826c-47e9-827c-626afd748f10 | sequences_series | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAATYAAADgCAIAAAAGzt14AAAnyElEQVR4nO3deVwU9f8H8M/MXiy7wALLjYAcKgKZQkmapOVtfPvaYVHZofV9lGdfssv4dlumX7vs0A6V/KYhoqZ5X6CJKeKNKQJCHHIty8LCHnP9/vj8mu9+gQgQZmbx/fyDBzu7O/ve4zWfz8x8ZobgOA4BAKSKFLsAAEBnIKIASBpEFABJg4gCIGkQUQAkDSIKgKRBRAGQNIgoAJIGEQVA0iCiAEgaRBQASYOIAiBpEFEAJA0iCoCkQU... | Find the Fourier series expansion of the function $f(x) = \begin{cases} x, & -\pi \le x < 0 \\ \pi, & 0 \le x \le \pi \end{cases}$ with the period $2 \cdot \pi$ on the interval $[-\pi, \pi]$. | The Fourier series is: $f(x)=\frac{\pi}{4}+\sum_{n=1}^\infty\left(\frac{1+(-1)^{n+1}}{\pi\cdot n^2}\cdot\cos(n\cdot x)+\frac{(-1)^{n+1}\cdot2+1}{n}\cdot\sin(n\cdot x)\right)$ |
a9c9f912-3c8e-4269-b099-4025c1d4eabb | integral_calc | false | null | Compute the integral:
$$
\int \frac{ 2 \cdot x+\sqrt{x-2} }{ \sqrt[4]{x-2}+\sqrt[4]{(x-2)^3} } \, dx
$$ | $\int \frac{ 2 \cdot x+\sqrt{x-2} }{ \sqrt[4]{x-2}+\sqrt[4]{(x-2)^3} } \, dx$ = $C+20\cdot\sqrt[4]{x-2}+\frac{8\cdot1}{5}\cdot\sqrt[4]{x-2}^5-20\cdot\arctan\left(\sqrt[4]{x-2}\right)-\frac{1\cdot4}{3}\cdot\sqrt[4]{x-2}^3$ |
a9dffc8e-7ced-4002-b02a-9baebbe9603d | precalculus_review | false | null | Solve $\sqrt{3 \cdot x^2+5 \cdot x+8}-\sqrt{3 \cdot x^2+5 \cdot x+1}=1$. | The final answer: $x=-\frac{8}{3} \lor x=1$ |
aa1d7802-dd5b-476c-b14f-4a42e7c3e978 | algebra | false | null | Find the zeros of the polynomial function with complex roots: $f(x) = 2 \cdot x^3 + 4 \cdot x^2 + 3 \cdot x + 6$. | $x$ = $-i\cdot\sqrt{\frac{3}{2}}$, $i\cdot\sqrt{\frac{3}{2}}$, $-2$ |
aaa1aca4-7a50-4e00-8119-c633ec60663a | differential_calc | false | null | For the function $y = \frac{ x }{ x^2 - 1 }$ find the derivative $y^{(n)}$. | The General Form of the Derivative of $y = \frac{ x }{ x^2 - 1 }$: $y^{(n)}=\frac{1}{2}\cdot(-1)^n\cdot\left(n!\right)\cdot\left((x+1)^{-(n+1)}+(x-1)^{-(n+1)}\right)$ |
aaf96d92-2e6d-4d51-9302-ebdab0643835 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ 4 \cdot x-\sqrt[3]{36 \cdot x^2}-\sqrt[6]{6 \cdot x} }{ x \cdot \left(1+\sqrt[3]{6 \cdot x}\right) } \, dx
$$ | $\int \frac{ 4 \cdot x-\sqrt[3]{36 \cdot x^2}-\sqrt[6]{6 \cdot x} }{ x \cdot \left(1+\sqrt[3]{6 \cdot x}\right) } \, dx$ = $C+5\cdot\ln\left(\left|1+\sqrt[6]{6\cdot x}^2\right|\right)+\sqrt[6]{6\cdot x}^4-5\cdot\sqrt[6]{6\cdot x}^2-6\cdot\arctan\left(\sqrt[6]{6\cdot x}\right)$ |
ab489b92-5b36-40ad-9a82-491fad3195f7 | differential_calc | false | null | Given $y = 3 \cdot x^5 + 10 \cdot x^4 - 20$, find where the function is
1. concave up,
2. concave down, and
3. point(s) of inflection. | 1. Concave up: $(-2,0)$, $(0,\infty)$
2. Concave down: $(-\infty,-2)$
3. Point(s) of Inflection: $P(-2,44)$ |
ab60fd32-4edd-4d98-9274-20131190ef32 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ x^3 }{ \sqrt{x^2+x+1} } \, dx
$$ | $\int \frac{ x^3 }{ \sqrt{x^2+x+1} } \, dx$ = $\left(\frac{1}{3}\cdot x^2-\frac{5}{12}\cdot x-\frac{1}{24}\right)\cdot\sqrt{x^2+x+1}+\frac{7}{16}\cdot\ln\left(\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|\right)+C$ |
ab849658-cfda-452a-9ade-f2e14d96048a | integral_calc | false | null | Compute the integral:
$$
\int \frac{ -3 }{ e^{4 \cdot x} + \sqrt{1 + e^{8 \cdot x}} } \, dx
$$ | $\int \frac{ -3 }{ e^{4 \cdot x} + \sqrt{1 + e^{8 \cdot x}} } \, dx$ = $C-\frac{1}{4}\cdot\left(\frac{3}{e^{4\cdot x}+\sqrt{1+e^{8\cdot x}}}+3\cdot\ln\left(\frac{e^{4\cdot x}+\sqrt{1+e^{8\cdot x}}-1}{1+e^{4\cdot x}+\sqrt{1+e^{8\cdot x}}}\right)\right)$ |
aba7e397-b652-4286-99f3-01d44606e733 | sequences_series | false | null | Since $1 + x^3 + x^6 + \cdots + x^{3 \cdot n} + \cdots = \frac{ 1 }{ 1 - x^3 }$, when $|x| < 1$, then find the sum of the series:
$$
\sum_{n=1}^\infty \left(3 \cdot n \cdot x^{3 \cdot n-1}\right)
$$ | The final answer: $\frac{3\cdot x^2}{\left(1-x^3\right)^2}$ |
ac4b6e12-42b6-4dbe-a5fd-4258b6db282b | differential_calc | false | null | Find the derivative of $y = x \cdot \sin(x) + 2 \cdot x \cdot \cos(x) - 2 \cdot \sin(x) + \ln(\sin(x)) + c^2$. | The final answer: $y'=\frac{\left(\sin(x)\right)^2+x\cdot\sin(x)\cdot\cos(x)-2\cdot x\cdot\left(\sin(x)\right)^2-\cos(x)}{\sin(x)}$ |
ac5571b3-1a5f-4cb5-b874-0ec90e7dc14b | algebra | false | null | Evaluate the expression $\frac{ (3+i)^2 }{ (1+2 \cdot i)^2 }$ and write the result as a simplified complex number. | $\frac{ (3+i)^2 }{ (1+2 \cdot i)^2 }$ = $-2\cdoti$ |
ace5f20b-07ce-4e6f-ac92-1de254bede66 | precalculus_review | false | null | Use properties of logarithms to write the expression as a sum, difference, and/or product of logarithms:
$$
\log_{4}\left(\frac{ \sqrt[3]{x \cdot y} }{ 64 }\right)
$$ | $\log_{4}\left(\frac{ \sqrt[3]{x \cdot y} }{ 64 }\right)$ = $-3+\frac{1}{3}\cdot\log_4(x)+\frac{1}{3}\cdot\log_4(y)$ |
ad240384-461f-42ca-bfb9-6227380f6674 | precalculus_review | false | null | Find the domain of the function $f(x) = \arccos\left(\frac{ 3 }{ 4+2 \cdot \sin(x) }\right)$ | The final answer: $-\frac{\pi}{6}+2\cdot k\cdot\pi\le x\le\frac{7\cdot\pi}{6}+2\cdot k\cdot\pi$ |
ad9dc540-5423-4bf8-86c4-a43e8a8dcdc7 | sequences_series | false | null | Given that $\frac{ 1 }{ 1-x } = \sum_{n=0}^\infty x^n$, use term-by-term differentiation or integration to find a power series for the function $f(x) = \ln\left(1+x^2\right)$ centered at $x=0$. | $\ln\left(1+x^2\right)$ = $\sum_{n=0}^\infty\left((-1)^n\cdot\frac{x^{2\cdot n}}{n+1}\cdot x^2\right)$ |
ae0786f8-f333-4c38-acea-8be761157570 | differential_calc | false | null | Find the derivative of the function: $y = -3 \cdot x^{\sqrt[3]{2 \cdot x}}$ | $\frac{ d y }{d x}$ = $-\left(\frac{3\cdot\sqrt[3]{2}}{x^{\frac{2}{3}}}+\frac{\sqrt[3]{2}\cdot\ln(x)}{x^{\frac{2}{3}}}\right)\cdot x^{\sqrt[3]{2}\cdot\sqrt[3]{x}}$ |
ae2e969d-3eb4-4b7d-aeb1-9816a02b2670 | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAABnXklEQVR4nO3deXhU9dn/8c9AyEIWICwBVKAoxhQQBTVRCIgoglQRW0RNAClLAlos8Ij+qFgFoajgUiRCFBUNqI9a60JBQfaYoCAEFQh7RAhrAoQkZD2/Pyh5ms5kCJCcc2bm/bquXMXvOZnc3B0m85n7LA7DMAzZSHJysiRp1KhRFldiT/THPfpTPfTJPfrj3qxZszRr1ixJ0oQJEzRhwgSLK7Innkfu0Z/qoU/u0R/37NqfOlYXAAAAAM... | Graph of $f$:
The function $g$ is defined as $g(x) = 2 \cdot x^2 + \int_{-1}^{x^2} f(t) \, dt$ on $[-2,1]$. The graph of $f$ consists of four line segments and a semi-circle, and is shown above. What is the absolute maximum of $g$ on the interval $[-2,1]$? | The absolute maximum of $g$ on the interval $[-2,1]$ is $1+\frac{ \pi }{ 4 }$. |
ae7253eb-638d-404d-8764-16d3eca93f52 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ 1 }{ \left(\cos(2 \cdot x)\right)^3 } \, dx
$$ | $\int \frac{ 1 }{ \left(\cos(2 \cdot x)\right)^3 } \, dx$ = $C+\frac{\sin(2\cdot x)}{4\cdot\left(\cos(2\cdot x)\right)^2}+\frac{1}{4}\cdot\ln\left(\left|\tan\left(\frac{1}{2}\cdot\left(2\cdot x+\frac{\pi}{2}\right)\right)\right|\right)$ |
aee39c3c-776c-45db-a4b1-78f279287461 | integral_calc | false | null | Compute the volume of the solid formed by rotating about the x-axis the area bounded by the axes and the parabola $x^{\frac{ 1 }{ 2 }}+y^{\frac{ 1 }{ 2 }}=2^{\frac{ 1 }{ 2 }}$. | Volume = $\frac{8\cdot\pi}{15}$ |
af144f31-5706-4ad7-84ab-94a14799bbb6 | precalculus_review | false | null | Find all values of $x$ that satisfy the following equation:
$$
\left|\left(x^4-4\right)-\left(x^2+2\right)\right|=\left|x^4-4\right|-\left|x^2+2\right|
$$ | The final answer: $|x|\ge\sqrt{3}$ |
af573af8-742d-47e3-9086-1c954a95c86b | multivariable_calculus | false | null | Use the method of Lagrange multipliers to find the maximum and minimum values of the function $f(x,y,z) = x + y + z$ subject to the constraint $\frac{ 1 }{ x }+\frac{ 1 }{ y }+\frac{ 1 }{ z }=1$. | Minimum: $1$
Maximum: $9$ |
af740483-5226-4f0a-b027-4762d47d868b | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAASgAAAEhCAIAAABDaqqGAAAwcklEQVR4nO3deXhb1Z038N+RLNuSLcuLvMn7EtuJE6+J4+wLTiANZIEQptAwtClLocPDZIb3YTp9n/I+fWdgeBkGAgydshQIbUIKJQkhi53EcZzYceLY8e543+RFljfJsnad948rC+EstrXkXoXfp32CfSUdHcn66px77rnnEkopIITuLh7bFUDoxwiDhxALMHgIsQCDhxALMHieBEfC7hkYPE9CCLl5I6bRE2HwuMtgMDQ3N2s0Gubn1t... | Use the graph of the function $y = g(x)$ shown here to find $\lim_{x \to 0^{+}}\left(g(x)\right)$, if possible. Estimate when necessary. | $\lim_{x \to 0^{+}}\left(g(x)\right)$ = $0$ |
af865556-4ec7-4cd3-8fff-592c30074325 | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAABWfElEQVR4nO3deZydZX3w/+9MZs1M9n0jBEJYZI0ssqhUgdpHsSxSERGCSlurrRWwBdciVtyoWqvtI31EK60LUBIXFAEVfkiFIIisCoSE7HsmmZnMZLbfH/GczuRMkpnJzH2f+5z3+/XKC3LNmTPfoUjnk+u677uip6enJzLgtNNOi+XLl/dZe+ihh+Lggw9OZR7K20033RQ33XRTRERcffXVcfXVV6c8EQBANlSmPcBAVVdXF6x1dHSkMA... | Let $R$ be the shaded region bounded by the graphs of $y=\frac{ 1 }{ x+2 }$ and $y=-\frac{ 1 }{ 2 } \cdot x+3$, as shown above.
Find the volume of the solid generated when $R$ is rotated about the vertical line $x=-3$. | The volume of the solid is $292.097$ units³. |
afb51ba2-1ec2-4e8b-b6ca-eee46a61357f | differential_calc | false | null | Make full curve sketching of $y = \frac{ 1 }{ 2 } \cdot \ln\left(\left|\frac{ x-4 }{ x+4 }\right|\right)$. Submit as your final answer:
1. The domain (in interval notation)
2. Vertical asymptotes
3. Horizontal asymptotes
4. Slant asymptotes
5. Intervals where the function is increasing
6. Intervals where the function ... | 1. The domain (in interval notation) $(-\infty,-4)\cup(-4,4)\cup(4,\infty)$
2. Vertical asymptotes $x=-4$, $x=4$
3. Horizontal asymptotes $y=0$
4. Slant asymptotes None
5. Intervals where the function is increasing $(-\infty,-4)$, $(-4,0)$
6. Intervals where the function is decreasing $(-4,4)$
7. Intervals where the fu... |
afc947d1-b4ea-4c8e-9540-0fc72f85b060 | algebra | false | null | Solve the following equation for $x$:
$3 \cdot x - (x - 4) = 10 + 5 \cdot (2 \cdot x + 2)$ | $x$ = $-2$ |
afd775c9-9303-4cfc-a781-3399caf2e0b1 | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAQUAAAEGCAIAAADjYBdoAABKO0lEQVR4nO2dZ1wU19fH753ZvsDSBSkKClYQMBpL/rHEEqypJiaxoUFUihQBiYpSNIlKUUqCJXajxscSxRITUy3Yu0YsCEhnF5btO3OfF6NIaILO7CLO94Uf3J2558zs/Ob2cyBCCLCwsAAAAMCM7QALSxuC1QMLyzNYPbCwPIPVAwvLM1g9sLA8g2NsB55AkuSjR48ePXrUtWvXjh07Uh/m5t4tL6/o0KGDi4tL3YNlMhmEUCKRGMNTlv... | Use the following figure to find the indicated derivatives, if they exist (enter 'undefined' in case, if the derivative doesn't exist):
Let $h(x) = f(x) \cdot g(x)$. Find:
1. $h'(1)$
2. $h'(3)$
3. $h'(4)$ | 1. $h'(1)$ = $2$
2. $h'(3)$ = None
3. $h'(4)$ = $2.5$ |
afeb0054-83bd-4632-a3f9-8cf9685452d7 | multivariable_calculus | false | null | The integral has been converted to polar coordinates. Verify that the identity is true and choose the easiest way to evaluate the integral, in rectangular or polar coordinates:
$$
\int_{0}^1 \int_{x^2}^x \frac{ y }{ \sqrt{x^2+y^2} } \, dy \, dx = \int_{0}^{\frac{ \pi }{ 4 }} \int_{0}^{\sec(\varphi) \cdot \tan(\varphi)}... | $I$ = $\frac{1}{6}\cdot\left(2-\sqrt{2}\right)$ |
b031545d-ffe2-41b9-ae2f-5c5e1b989168 | differential_calc | false | null | The position function of a ball dropped from the top of a $200$-meter tall building is given by $s(t) = 200 - 4.9 \cdot t^2$, where position $s$ is measured in meters and time $t$ is measured in seconds. Compute the average velocity of the ball over the given time intervals. Round your answer to eight significant digit... | 1. $-48.951000$ m/sec.
2. $-49.049000$ m/sec.
3. $-48.995100$ m/sec.
4. $-49.004900$ m/sec. |
b04b5d6e-8250-4931-b2cd-f93c3b884ea5 | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAPEAAAEkCAIAAABxLUB2AABEJElEQVR4nO29WVQcV5bvfU5EZETO8wgJCQkksyQk0ICsAUl2yWrZZbvKdavLX1X3vV91Vfd97LV69XO/9WOv1V/3uqt69e3qdtllt6pctmVLsmQkYSHJQkgIEDPJTDIk5DwPcb6Hk6RSDJKAnID4LRslSWTEiYh/7tjnnH32hgghwMGxiyBy3QAOjjTDaZpjt8FpOs1wvlzO4TSdZiCE+MW64uYUnwU4TWcKLO5VIk4qniNzULluwA6DZV... | Using the graph, find each limit:
1. $\lim_{x \to -1}\left(f(x)\right)$
2. $\lim_{x \to 1}\left(f(x)\right)$
3. $\lim_{x \to 0^{+}}\left(f(x)\right)$
4. $\lim_{x \to 2}\left(f(x)\right)$ | 1. $\lim_{x \to -1}\left(f(x)\right)$ = None
2. $\lim_{x \to 1}\left(f(x)\right)$ = $1$
3. $\lim_{x \to 0^{+}}\left(f(x)\right)$ = $0$
4. $\lim_{x \to 2}\left(f(x)\right)$ = None |
b0d28b67-d169-4f97-9f81-a001177c77df | algebra | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAZgAAAE3CAIAAAAYE8A5AAAKMWlDQ1BJQ0MgUHJvZmlsZQAAeJydlndUU9kWh8+9N71QkhCKlNBraFICSA29SJEuKjEJEErAkAAiNkRUcERRkaYIMijggKNDkbEiioUBUbHrBBlE1HFwFBuWSWStGd+8ee/Nm98f935rn73P3Wfvfda6AJD8gwXCTFgJgAyhWBTh58WIjYtnYAcBDPAAA2wA4HCzs0IW+EYCmQJ82IxsmRP4F726DiD5+yrTP4zBAP+flLlZIjEAUJiM5/... | Use the graph to write an equation for the function: | $f(x)$ = $-8\cdot\frac{(x-3)}{(x+3)\cdot(x-4)}$ |
b0f6581a-bda8-4027-96db-3dfb44273449 | multivariable_calculus | false | null | Calculate the second-order partial derivatives. (Treat $A$,$B$,$C$,$D$ as constants.)
$$
f(x,y) = \frac{ A \cdot x + B \cdot y }{ C \cdot x + D \cdot y }
$$ | $f_{xx}(x,y)$ = $\frac{2\cdot C\cdot(B\cdot C-A\cdot D)\cdot y}{(C\cdot x+D\cdot y)^3}$
$f_{xy}(x,y)$ = $f_{yx}(x,y)$ = $\frac{-(B\cdot C-A\cdot D)\cdot(C\cdot x-D\cdot y)}{(C\cdot x+D\cdot y)^3}$
$f_{yy}(x,y)$ = $-2\cdot D\cdot\frac{B\cdot C\cdot x-A\cdot D\cdot x}{(C\cdot x+D\cdot y)^3}$ |
b1cbb170-d4d2-4f07-81f3-ee4094350ade | integral_calc | false | null | Compute the integral:
$$
\int \frac{ 3 \cdot x+\sqrt[3]{9 \cdot x^2}+\sqrt[6]{3 \cdot x} }{ x \cdot \left(4+\sqrt[3]{3 \cdot x}\right) } \, dx
$$ | $\int \frac{ 3 \cdot x+\sqrt[3]{9 \cdot x^2}+\sqrt[6]{3 \cdot x} }{ x \cdot \left(4+\sqrt[3]{3 \cdot x}\right) } \, dx$ = $C+3\cdot\arctan\left(\frac{\sqrt[6]{3}}{2}\cdot\sqrt[6]{x}\right)+36\cdot\ln\left(\left|4+\sqrt[3]{3}\cdot\sqrt[3]{x}\right|\right)+\frac{\left(3\cdot3^{\frac{2}{3}}\right)}{2}\cdot x^{\frac{2}{3}}... |
b1d83dc5-da1a-4559-b8f4-c8a0ac9c7e7d | multivariable_calculus | false | null | The integral has been converted to polar coordinates. Verify that the identity is true and choose the easiest way to evaluate the integral, in rectangular or polar coordinates.
$$
\int_{0}^1 \int_{x^2}^x \frac{ 1 }{ \sqrt{x^2+y^2} } \, dy \, dx = \int_{0}^{\frac{ \pi }{ 4 }} \int_{0}^{\tan(\varphi) \cdot \sec(\varphi)}... | $I$ = $\sqrt{2}-1$ |
b26a668e-b0d8-497f-be3e-9e980702227d | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAQYAAAHRCAIAAAAHbID8AABxAUlEQVR4nO29e5RtXVUf+Jtr73NOVd3vxUsUfEIAsRUQSfsgIQZ8AwJRUYhtDJ3WJG0bsB3RpPM0xnRGp9NJdERjHnZifIBRE4dD0diokYBCfAQJj483AkHlg+9x7606e++1Zv8xz/rV3PtUnVO37qmq81i/8Y37nTpn77XX3nvONd9zSUoJOwNRIAEBXUgRCKhqVUkJQFdVCggQEIMCADQAggAAbduORqOUUgjB/lXVq7yTggtDuOoJXD... | The region bounded by the parabola $y^2 = 2 \cdot p \cdot x$ and the line AB is revolved about the Y-axis. The line AB passes through the focus of the parabola and is perpendicular to the X-axis. Find the volume of this solid of revolution using integration with respect to $y$. Use $p = 6$. | Volume: $\frac{432}{5}\cdot\pi$ |
b27a2ed1-91ab-4864-b50e-4aebf48d87b1 | differential_calc | false | null | Consider a wire 4 ft long cut into two pieces. One piece forms a circle with radius $r$ and the other forms a square of side $x$. Choose $x$ to maximize the sum of their areas. | $x$ = $0$ |
b27c8456-96d3-4d5b-891a-f756b95ca624 | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAACZ1klEQVR4nOzdd3hUddrG8TvNQIBQpAhKkSZBURFRXFlUFGV1BRGsgEiR3nvvvUrvPQKCiBTBAoqooLACglQhNAWEUAwQIPX9g3eOczgJJGEyZ8r3c125nPyYmfPMLmRyz/MrAcnJyckCACCNxowZozFjxkiSOnXqpE6dOtlcEQDAmwTaXQAAAAAA/0EAAQAAAOA2BBAAAAAAbkMAAQAAAOA2BBAAAAAAbkMAAQAAAOA2BBAAAAAAbkMAAQ... | Let $R$ be the region enclosed by the graphs of $f(x) = 1 - \cos(3 \cdot x)$ and $g(x) = e^{0.3 \cdot x}$, as shown in the figure above. Find the volume of the solid generated when $R$ is revolved about the $x$-axis. | The volume of the solid is $2.655$ units³. |
b27efb6c-ef58-438a-9b61-007ea74acd47 | multivariable_calculus | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAgAAAADQCAIAAAD+u06rAAB9jElEQVR4nO39d3wd53Xnj5/zTL39olz0QgJgAUmwUyKpQlLdKlaxZMtxj205suOvk9jZjRPvL4mdZBOleOPNOllb9lqWLduyZKtYXaQKZVFi7wUkAILo/fYy5Tm/P2buxSUAQiAJkCA579eLIDB3Zu7U5zzPec75HCQicHBwcHC48mAX+wAcHBwcHC4OjgFwcHBwuEJxDICDg4PDFYpjABwcHByuUMSLfQAXCQIAAiQARgBIAAijPyaGAz... | Two children are playing with a ball. The girl throws the ball to the boy. The ball travels in the air, moves $3$ ft to the right, and falls $5$ ft away from the girl (see the following figure). If the plane that contains the trajectory of the ball is perpendicular to the ground, find its equation. | The equation of the plane: $3\cdot y-5\cdot x=0$
(If it is not perpendicular, just click submit.) |
b28ab25c-39a1-415e-8c6c-5a8cef4b18f1 | algebra | false | null | Solve the following equations:
1. $\frac{ 2 }{ 3 } (6 x+18)=20$
2. $\frac{ 6 }{ 5 } (10 z+15)=-6$
3. $\frac{ 3 z }{ 8 }-4=5$
4. $\frac{ 1 }{ 2 } x+\frac{ 3 }{ 4 }=-\frac{ 7 }{ 8 }$
5. $3\frac{1}{4} p+4 \left(\frac{ 3 }{ 4 } p-12\right)=-79.25$
6. $3\frac{1}{2} w-2 \left(\frac{ 5 }{ 6 } w+2\right)=-\frac{ 1 }{ 3 }$ | The solutions to the given equations are:
1. $x=2$
2. $z=-2$
3. $z=24$
4. $x=-\frac{ 13 }{ 4 }$
5. $p=-5$
6. $w=2$ |
b2e51a16-da23-4146-b609-b4083ae88786 | precalculus_review | false | null | Use the Rational Zero Theorem to find all real zeros of the following polynomial:
$p(x) = 2 \cdot x^3 - 5 \cdot x^2 + 9 \cdot x - 9$ | The real zeros are $\frac{3}{2}$ |
b2f618b5-8496-4cb1-9cfa-bb7ec9d0cd99 | precalculus_review | false | null | 1. Find the inverse function of $f(x) = \sqrt[3]{x-4}$
2. Find the domain of the inverse function
3. Find the range of the inverse function | 1. The inverse function is: $x^3+4$
2. The domain of the inverse function is: $(-\infty,\infty)$
3. The range of the inverse function is: $(-\infty,\infty)$ |
b3072b54-bb84-41dd-b303-7dd7249ba64d | integral_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAACLzUlEQVR4nOzdd3hUdfr+8XsmPYReAqGG3nsVlGIBpImCvcAqothQXPtaUexr737FtuqCIl2aKDX0kpAAIfQWegkhpM3vD37MJpwDpEzOmfJ+XZeXyTNnZh4iZuaeT3O4XC6XAAAAAMACTrsbAAAAABA4CCAAAAAALEMAAQAAAGAZAggAAAAAyxBAAAAAAFiGAAIAAADAMgQQAAAAAJYhgAAAAACwDAEEAAAAgGUIIAAAAAAsQwABAAAAYB... | The region $R$ is the region in the first and second quadrants bounded by the graphs of $y = 5 \cdot \cos\left(\frac{ x^2 }{ 4 }\right)$ and $y = 2 + \sqrt[3]{x}$, as shown in the diagram above.
Find the volume of the solid generated from rotating region $R$ about the horizontal line $y = -2$. | The volume of the solid is $336.509$ units³. |
b318ad71-d185-4390-acca-0699ca019773 | precalculus_review | false | null | Find zeros of $f(x) = \sin(x) + \sin(2 \cdot x) + 2 \cdot \sin(x) \cdot \sin(2 \cdot x) - 2 \cdot \cos(x) - \cos(2 \cdot x)$. | The final answer: $x_1=-\frac{\pi}{2}+2\cdot\pi\cdot n$, $x_2=-\frac{2\cdot\pi}{3}+2\cdot\pi\cdot n$, $x_3=\frac{2\cdot\pi}{3}+2\cdot\pi\cdot n$, $x_4=(-1)^n\cdot\frac{\pi}{6}+\pi\cdot n$ |
b3573d9a-dceb-44bb-bd9a-009f13c4cc13 | integral_calc | false | null | Compute the integral:
$$
\int \frac{ \tan(2 \cdot x) }{ \sqrt{\sin(2 \cdot x)^4+\cos(2 \cdot x)^4} } \, dx
$$ | $\int \frac{ \tan(2 \cdot x) }{ \sqrt{\sin(2 \cdot x)^4+\cos(2 \cdot x)^4} } \, dx$ = $C+\frac{1}{4}\cdot\ln\left(\sqrt{1+\tan(2\cdot x)^4}+\tan(2\cdot x)^2\right)$ |
b39f8c2e-5c3c-474e-a7e1-adce52e0feaf | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAyAAAAMgCAYAAADbcAZoAAB1rElEQVR4nO3deXiU5aH+8TtkQoCEYRMimoiCiBASl1AxI2610aDFwoEmp6dYTpVYa3sUjKWtrSDYYy0YRetSAbW2sW1SWbQ9NRqr4JIgLVWDMSqCxkQxLiyThABJyO8PfjMlzDtZJjPzLvP9XFcuyTNL7oyZTO55nud94zo6OjoUglWrVkmSrr322lBubpqtW7dKkrKyskxOEhq75ie3OXieRpddc/tYNf+aNWu0ePHiTmPXXHONbr/9dk... | The graph of the function $f$ is shown above. What are all values of $x$ at which $f$ has a jump discontinuity?
[Image_0] | The function $f$ has a jump discontinuity at $x$ = $-2$ |
b3ccc40f-c39b-46f5-8e7a-93f91b06f532 | multivariable_calculus | false | null | Let $u(x,y,z) = x \cdot y + y \cdot z + x \cdot z$, where $x = r \cdot \cos(\theta)$, $y = r \cdot \sin(\theta)$, and $z = r \cdot \theta$.
Evaluate $\frac{\partial w}{\partial r}(r,\theta)$ and $\frac{\partial w}{\partial \theta}(r,\theta)$ for $r = 1$, $\theta = \frac{ 3 \cdot \pi }{ 2 }$, where $w(r,\theta) = u(x(r... | The final answer:
$\frac{\partial w}{\partial r}(1,\frac{ 3 \cdot \pi }{ 2 })$ is equal to $-3\cdot\pi$ .
$\frac{\partial w}{\partial \theta}(1,\frac{ 3 \cdot \pi }{ 2 })$ is equal to $\frac{3\cdot\pi}{2}-2$ . |
b3f1345e-4df9-4e1e-b9f5-f0b644753869 | sequences_series | false | null | Find the Fourier series of the function $u = \left| \frac{ \sin\left( \frac{ x }{ 2 } \right) }{ 2 } \right|$ in the interval $[-2 \cdot \pi,2 \cdot \pi]$. | The Fourier series is: $\frac{1}{\pi}+\sum_{k=1}^\infty\left(-\frac{2}{\pi\cdot\left(4\cdot k^2-1\right)}\cdot\cos(k\cdot x)\right)$ |
b414e258-7f9a-4c2a-aad3-5529aead66c2 | sequences_series | false | null | Find the Fourier series of the function $\psi(x) = e^{-x}$ in the interval $(-2 \cdot \pi, 2 \cdot \pi)$. | The Fourier series is: $e^{-x}=\frac{\left(e^{2\cdot\pi}-e^{-2\cdot\pi}\right)}{\pi}\cdot\left(\frac{1}{4}+\sum_{n=1}^\infty\left(\frac{(-1)^n}{4+n^2}\cdot\left(2\cdot\cos\left(\frac{n}{2}\cdot x\right)+n\cdot\sin\left(\frac{n}{2}\cdot x\right)\right)\right)\right)$ |
b46190d9-ebed-4e4b-b32e-6167579d4c2b | differential_calc | false | null | If it takes $6$ years for an exponentially growing quantity to triple in size, then by what percentage does it increase every year? Round your answer to two decimal places, and omit the % sign. | The final answer: $20.09$ |
b4947d95-c61d-4939-9adb-119988a7c1c1 | precalculus_review | false | null | 1. Find the inverse function of $f(x) = x^2 - 4$, for $x \ge 0$.
2. Find the domain of the inverse function.
3. Find the range of the inverse function. | 1. The inverse function is: $\sqrt{x+4}$
2. The domain of the inverse function is: $x\ge-4$
3. The range of the inverse function is: $y\ge0$ |
b55f6e2f-3e64-466f-9524-b9bcc1986c46 | differential_calc | true | data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAVwAAAFpCAYAAADQoZAyAAAYKGlDQ1BJQ0MgUHJvZmlsZQAAeJyVeQk4Vd/X/z733MnlmudZZjLPJPM8z0Mq1zzTNUWRkAyVZEghhUSKRlNChpRkylCKFEKpVIZMeQ+q7+//ff/v8z7vfp597ueuvdban7332nufdS8AHKykkJAAFC0AgUFhZGsDbV5HJ2de3DiAABEQgBrgI7mHhmhZWpqC/7EsDSHaSHkhseXrf9b7/xY6D89QdwAgSwS7eYS6ByL4HgBodvcQchgAmF... | The graph of the function $f(x)$ is shown below. Determine the interval(s) where $\frac{ d f }{d x}>0$. | The final answer: $(1,4)$ |
b5c10639-0db9-4bd6-8b11-e245dac06e15 | integral_calc | false | null | Solve the integral:
$$
\int \frac{ 4 \cdot \cos(6 \cdot x)^3 }{ 9 \cdot \sin(6 \cdot x)^9 } \, dx
$$ | $\int \frac{ 4 \cdot \cos(6 \cdot x)^3 }{ 9 \cdot \sin(6 \cdot x)^9 } \, dx$ = $C-\frac{2}{27}\cdot\left(\frac{1}{3}\cdot\left(\cot(6\cdot x)\right)^6+\frac{1}{4}\cdot\left(\cot(6\cdot x)\right)^4+\frac{1}{8}\cdot\left(\cot(6\cdot x)\right)^8\right)$ |
b6943d35-373f-44b1-8886-587ed5656553 | integral_calc | false | null | $\int \frac{ 7+2 \cdot x-4 \cdot x^2 }{ 2 \cdot x^2+x-3 } \, dx$ | $\int \frac{ 7+2 \cdot x-4 \cdot x^2 }{ 2 \cdot x^2+x-3 } \, dx$ = $-2\cdot x+\ln\left(\left|2\cdot x^2+x-3\right|\right)+C$ |
b726f1ca-7edd-47ed-918f-b9fc63c3a1d9 | integral_calc | false | null | Solve the integral:
$$
\int \left(\frac{ x+1 }{ x-1 }\right)^{\frac{ 3 }{ 2 }} \, dx
$$ | $\int \left(\frac{ x+1 }{ x-1 }\right)^{\frac{ 3 }{ 2 }} \, dx$ = $C+\sqrt{\frac{x+1}{x-1}}\cdot(x-5)-3\cdot\ln\left(\left|\frac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}+\sqrt{x+1}}\right|\right)$ |
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Differential Calc Problems
Retrieves specific math problems related to differential calculus, providing basic filtering but limited analytical value beyond finding relevant entries.