data_source
stringclasses 9
values | question_number
stringlengths 14
17
| problem
stringlengths 14
1.32k
| answer
stringlengths 1
993
| license
stringclasses 8
values | data_topic
stringclasses 7
values | solution
stringlengths 1
991
|
|---|---|---|---|---|---|---|
college_math.PRECALCULUS
|
exercise.8.2.26
|
Solve the following system of linear equations: $\left\{\begin{aligned} x_{1}-x_{3} & =-2 \\ 2 x_{2}-x_{4} & =0 \\ x_{1}-2 x_{2}+x_{3} & =0 \\ -x_{3}+x_{4} & =1\end{aligned}\right.$
|
$(1,2,3,4)$
|
Creative Commons License
|
college_math.precalculus
|
(1,2,3,4)
|
college_math.PRECALCULUS
|
exercise.10.7.58
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\tan (x)=\cos (x)$
|
$x=\arcsin \left(\frac{-1+\sqrt{5}}{2}\right) \approx 0.6662, \pi-\arcsin \left(\frac{-1+\sqrt{5}}{2}\right) \approx 2.4754$
|
Creative Commons License
|
college_math.precalculus
|
x=\arcsin \left(\frac{-1+\sqrt{5}}{2}\right) \approx 0.6662, \pi-\arcsin \left(\frac{-1+\sqrt{5}}{2}\right) \approx 2.4754
|
college_math.PRECALCULUS
|
exercise.10.2.41
|
Solve the equation for $t$: $\sin (t)=-\frac{\sqrt{2}}{2}$
|
$\sin (t)=-\frac{\sqrt{2}}{2}$ when $t=\frac{5 \pi}{4}+2 \pi k$ or $t=\frac{7 \pi}{4}+2 \pi k$ for any integer $k$.
|
Creative Commons License
|
college_math.precalculus
|
\sin (t)=-\frac{\sqrt{2}}{2}$ when $t=\frac{5 \pi}{4}+2 \pi k$ or $t=\frac{7 \pi}{4}+2 \pi k$ for any integer $k$.
|
college_math.PRECALCULUS
|
exercise.10.7.82
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-\pi \leq x \leq \pi$: $\sin (x)>\frac{1}{3}$
|
$\left(\arcsin \left(\frac{1}{3}\right), \pi-\arcsin \left(\frac{1}{3}\right)\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(\arcsin \left(\frac{1}{3}\right), \pi-\arcsin \left(\frac{1}{3}\right)\right)
|
college_math.PRECALCULUS
|
exercise.10.7.41
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $2 \tan (x)=1-\tan ^{2}(x)$
|
$x=\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8}
|
college_math.PRECALCULUS
|
exercise.6.1.26
|
Evaluate the expression: $\log (0.01)$
|
$\log (0.01)=-2$
|
Creative Commons License
|
college_math.precalculus
|
\log (0.01)=-2
|
college_math.PRECALCULUS
|
exercise.6.4.3
|
Solve the equation analytically: $\ln \left(8-x^{2}\right)=\ln (2-x)$
|
$x=-2$
|
Creative Commons License
|
college_math.precalculus
|
x=-2
|
college_math.PRECALCULUS
|
exercise.10.2.15
|
Find the exact value of the cosine and sine of the given angle: $\theta=-\frac{13 \pi}{2}$
|
$\cos \left(-\frac{13 \pi}{2}\right)=0, \sin \left(-\frac{13 \pi}{2}\right)=-1$
|
Creative Commons License
|
college_math.precalculus
|
\cos \left(-\frac{13 \pi}{2}\right)=0, \sin \left(-\frac{13 \pi}{2}\right)=-1
|
college_math.PRECALCULUS
|
exercise.9.1.12
|
Write out the first four terms of the given sequence: $s_{0}=1, s_{n+1}=x^{n+1}+s_{n}, n \geq 0$
|
$1, x+1, x^{2}+x+1, x^{3}+x^{2}+x+1$
|
Creative Commons License
|
college_math.precalculus
|
1, x+1, x^{2}+x+1, x^{3}+x^{2}+x+1
|
college_math.PRECALCULUS
|
exercise.10.1.44
|
Convert the angle from radian measure into degree measure: $\frac{\pi}{2}$
|
$90^{\circ}$
|
Creative Commons License
|
college_math.precalculus
|
90^{\circ}
|
college_math.PRECALCULUS
|
exercise.1.1.25
|
Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them: $\left(-\frac{2}{3}, \frac{3}{2}\right),\left(\frac{7}{3}, 2\right)$
|
$d=\frac{\sqrt{37}}{2}, M=\left(\frac{5}{6}, \frac{7}{4}\right)$
|
Creative Commons License
|
college_math.precalculus
|
d=\frac{\sqrt{37}}{2}, M=\left(\frac{5}{6}, \frac{7}{4}\right)
|
college_math.PRECALCULUS
|
exercise.9.4.9
|
Evaluate: $\left(\begin{array}{c}n \\ n-2\end{array}\right), n \geq 2$
|
$\frac{n(n-1)}{2}$
|
Creative Commons License
|
college_math.precalculus
|
\frac{n(n-1)}{2}
|
college_math.PRECALCULUS
|
exercise.11.8.53
|
A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of $S 68^{\circ} \mathrm{W}$. The river is flowing due east at 8 miles per hour. What is the boat's true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.
|
The current is moving at about 10 miles per hour bearing $\mathrm{N} 54.6^{\circ} \mathrm{W}$.
|
Creative Commons License
|
college_math.precalculus
|
The current is moving at about 10 miles per hour bearing $\mathrm{N} 54.6^{\circ} \mathrm{W}$.
|
college_math.PRECALCULUS
|
exercise.2.4.29
|
Let $L$ be the line $y=2 x+1$. Find a function $D(x)$ which measures the distance squared from a point on $L$ to $(0,0)$. Use this to find the point on $L$ closest to $(0,0)$.
|
$(-\infty, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty, \infty)
|
college_math.PRECALCULUS
|
exercise.10.7.31
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sec (x)=2 \csc (x)$
|
$x=\frac{\pi}{6}, \frac{\pi}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{6}, \frac{\pi}{2}
|
college_math.PRECALCULUS
|
exercise.7.2.14
|
Find the standard equation of the circle which satisfies the given criteria: center $(3,6)$, passes through $(-1,4)$
|
$(x-3)^{2}+(y-6)^{2}=20$
|
Creative Commons License
|
college_math.precalculus
|
(x-3)^{2}+(y-6)^{2}=20
|
college_math.PRECALCULUS
|
exercise.6.5.29
|
Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, RubidiumStrontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation 6.5 to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time $t$ in billions of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates.
|
$A(t)=2.3 e^{-0.0138629 t}$
|
Creative Commons License
|
college_math.precalculus
|
A(t)=2.3 e^{-0.0138629 t}
|
college_math.PRECALCULUS
|
exercise.9.4.4
|
Simplify the expression: $\frac{9 !}{4 ! 3 ! 2 !}$
|
1260
|
Creative Commons License
|
college_math.precalculus
|
1260
|
college_math.PRECALCULUS
|
exercise.2.3.28
|
Find all of the points on the line $y=1-x$ which are 2 units from $(1,-1)$.
|
$\left(\frac{3-\sqrt{7}}{2}, \frac{-1+\sqrt{7}}{2}\right),\left(\frac{3+\sqrt{7}}{2}, \frac{-1-\sqrt{7}}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(\frac{3-\sqrt{7}}{2}, \frac{-1+\sqrt{7}}{2}\right),\left(\frac{3+\sqrt{7}}{2}, \frac{-1-\sqrt{7}}{2}\right)
|
college_math.PRECALCULUS
|
exercise.2.2.12
|
Solve the equation: $|x-4|=x-5$
|
no solution
|
Creative Commons License
|
college_math.precalculus
|
no solution
|
college_math.PRECALCULUS
|
exercise.6.1.45
|
Find the domain of the function: $f(x)=\ln (4 x-20)$
|
$(5, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(5, \infty)
|
college_math.PRECALCULUS
|
exercise.6.4.11
|
Solve the equation analytically: $-\log (x)=5.4$
|
$x=10^{-5.4}$
|
Creative Commons License
|
college_math.precalculus
|
x=10^{-5.4}
|
college_math.PRECALCULUS
|
exercise.1.1.10
|
Write the set using interval notation: $\{x \mid x \neq-3,4\}$
|
$(-\infty,-3) \cup(-3,4) \cup(4, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-3) \cup(-3,4) \cup(4, \infty)
|
college_math.PRECALCULUS
|
exercise.1.1.27
|
Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them: $(\sqrt{2}, \sqrt{3}),(-\sqrt{8},-\sqrt{12})$
|
$d=3 \sqrt{5}, M=\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{3}}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
d=3 \sqrt{5}, M=\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{3}}{2}\right)
|
college_math.PRECALCULUS
|
exercise.10.7.94
|
Solve the given inequality: $3 \arccos (x) \leq \pi$
|
$\left[\frac{1}{2}, 1\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[\frac{1}{2}, 1\right]
|
college_math.PRECALCULUS
|
exercise.6.4.24
|
Solve the equation analytically: $\ln \left(x^{2}\right)=(\ln (x))^{2}$
|
$x=1, x=e^{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=1, x=e^{2}
|
college_math.PRECALCULUS
|
exercise.6.1.21
|
Evaluate the expression: $\log _{36}(216)$
|
$\log _{36}(216)=\frac{3}{2}$
|
Creative Commons License
|
college_math.precalculus
|
\log _{36}(216)=\frac{3}{2}
|
college_math.PRECALCULUS
|
exercise.10.2.40
|
Solve the equation for $t$: $\cos (t)=0$
|
$\cos (t)=0$ when $t=\frac{\pi}{2}+\pi k$ for any integer $k$.
|
Creative Commons License
|
college_math.precalculus
|
\cos (t)=0$ when $t=\frac{\pi}{2}+\pi k$ for any integer $k$.
|
college_math.PRECALCULUS
|
exercise.6.2.18
|
Use the properties of logarithms to write the expression as a single logarithm: $\log _{3}(x)-2 \log _{3}(y)$
|
$\log _{3}\left(\frac{x}{y^{2}}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\log _{3}\left(\frac{x}{y^{2}}\right)
|
college_math.PRECALCULUS
|
exercise.6.3.28
|
Solve the equation analytically: $e^{2 x}-3 e^{x}-10=0$
|
$x=\ln (5)$
|
Creative Commons License
|
college_math.precalculus
|
x=\ln (5)
|
college_math.PRECALCULUS
|
exercise.6.5.14
|
For each isotope:
- Find the decay constant $k$. Round your answer to four decimal places.
- Find a function which gives the amount of isotope $A$ which remains after time $t$. (Keep the units of $A$ and $t$ the same as the given data.)
- Determine how long it takes for $90 \%$ of the material to decay. Round your answer to two decimal places. (HINT: If $90 \%$ of the material decays, how much is left?)
14. Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of 5.27 years.
|
$\bullet k=\frac{\ln (1 / 2)}{5.27} \approx-0.1315$
|
Creative Commons License
|
college_math.precalculus
|
\bullet k=\frac{\ln (1 / 2)}{5.27} \approx-0.1315
|
college_math.PRECALCULUS
|
exercise.6.2.28
|
Use the properties of logarithms to write the expression as a single logarithm: $\log _{2}(x)+\log _{4}(x-1)$
|
$\log _{2}(x \sqrt{x-1})$
|
Creative Commons License
|
college_math.precalculus
|
\log _{2}(x \sqrt{x-1})
|
college_math.PRECALCULUS
|
exercise.5.3.23
|
Solve the equation or inequality: $x^{\frac{2}{3}}=4$
|
$x= \pm 8$
|
Creative Commons License
|
college_math.precalculus
|
x= \pm 8
|
college_math.PRECALCULUS
|
exercise.6.1.36
|
Evaluate the expression: $\log \left(\sqrt[9]{10^{11}}\right)$
|
$\log \left(\sqrt[9]{10^{11}}\right)=\frac{11}{9}$
|
Creative Commons License
|
college_math.precalculus
|
\log \left(\sqrt[9]{10^{11}}\right)=\frac{11}{9}
|
college_math.PRECALCULUS
|
exercise.4.3.2
|
Solve the rational equation: $\frac{3 x-1}{x^{2}+1}=1$
|
$x=1, x=2$
|
Creative Commons License
|
college_math.precalculus
|
x=1, x=2
|
college_math.PRECALCULUS
|
exercise.10.7.85
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-\pi \leq x \leq \pi$: $\cot (x) \geq-1$
|
$\left(-\pi,-\frac{\pi}{4}\right] \cup\left(0, \frac{3 \pi}{4}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left(-\pi,-\frac{\pi}{4}\right] \cup\left(0, \frac{3 \pi}{4}\right]
|
college_math.PRECALCULUS
|
exercise.10.7.73
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 \leq x \leq 2 \pi$: $\cos (2 x) \leq 0$
|
$\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] \cup\left[\frac{5 \pi}{4}, \frac{7 \pi}{4}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] \cup\left[\frac{5 \pi}{4}, \frac{7 \pi}{4}\right]
|
college_math.PRECALCULUS
|
exercise.10.1.34
|
Convert the angle from degree measure into radian measure, giving the exact value in terms of $\pi$: $150^{\circ}$
|
$\frac{5 \pi}{6}$
|
Creative Commons License
|
college_math.precalculus
|
\frac{5 \pi}{6}
|
college_math.PRECALCULUS
|
exercise.10.2.3
|
Find the exact value of the cosine and sine of the given angle: $\theta=\frac{\pi}{3}$
|
$\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}, \sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$
|
Creative Commons License
|
college_math.precalculus
|
\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}, \sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}
|
college_math.PRECALCULUS
|
exercise.6.1.57
|
Find the domain of the function: $f(x)=\ln \left(-2 x^{3}-x^{2}+13 x-6\right)$
|
$(-\infty,-3) \cup\left(\frac{1}{2}, 2\right)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-3) \cup\left(\frac{1}{2}, 2\right)
|
college_math.PRECALCULUS
|
exercise.10.7.35
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (2 x)+\csc ^{2}(x)=0$
|
$x=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}
|
college_math.PRECALCULUS
|
exercise.10.7.46
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (5 x) \cos (3 x)-\sin (5 x) \sin (3 x)=\frac{\sqrt{3}}{2}$
|
$x=0, \frac{\pi}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{2}
|
college_math.PRECALCULUS
|
exercise.10.7.51
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (2 x)-\sqrt{3} \sin (2 x)=\sqrt{2}$
|
$x=0, \pi, \frac{\pi}{3}, \frac{4 \pi}{3}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \pi, \frac{\pi}{3}, \frac{4 \pi}{3}
|
college_math.PRECALCULUS
|
exercise.10.7.70
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 \leq x \leq 2 \pi$: $\tan (x) \geq \sqrt{3}$
|
$x=\left[\frac{\pi}{3}, \frac{\pi}{2}\right) \cup\left[\frac{4 \pi}{3}, \frac{3 \pi}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
x=\left[\frac{\pi}{3}, \frac{\pi}{2}\right) \cup\left[\frac{4 \pi}{3}, \frac{3 \pi}{2}\right)
|
college_math.PRECALCULUS
|
exercise.10.7.9
|
Find all of the exact solutions of the equation and then list those solutions which are in the interval $[0,2 \pi)$: $\sin \left(\frac{x}{3}\right)=\frac{\sqrt{2}}{2}$
|
$x=\frac{3 \pi}{4}+6 \pi k$ or $x=\frac{9 \pi}{4}+6 \pi k ; x=\frac{3 \pi}{4}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{3 \pi}{4}+6 \pi k$ or $x=\frac{9 \pi}{4}+6 \pi k ; x=\frac{3 \pi}{4}
|
college_math.PRECALCULUS
|
exercise.1.1.31
|
Find all of the points on the $y$-axis which are 5 units from the point $(-5,3)$.
|
$(0,3)$
|
Creative Commons License
|
college_math.precalculus
|
(0,3)
|
college_math.PRECALCULUS
|
exercise.10.7.60
|
Solve the equation: $\pi-2 \arcsin (x)=2 \pi$
|
$x=-1$
|
Creative Commons License
|
college_math.precalculus
|
x=-1
|
college_math.PRECALCULUS
|
exercise.8.2.19
|
Solve the following system of linear equations: $\left\{\begin{aligned} x-y+z & =-4 \\ -3 x+2 y+4 z & =-5 \\ x-5 y+2 z & =-18\end{aligned}\right.$
|
$(1,3,-2)$
|
Creative Commons License
|
college_math.precalculus
|
(1,3,-2)
|
college_math.PRECALCULUS
|
exercise.6.3.3
|
Solve the equation analytically: $5^{2 x-1}=125$
|
$x=2$
|
Creative Commons License
|
college_math.precalculus
|
x=2
|
college_math.PRECALCULUS
|
exercise.10.7.50
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sqrt{3} \sin (2 x)+\cos (2 x)=1$
|
$x=\frac{17 \pi}{24}, \frac{41 \pi}{24}, \frac{23 \pi}{24}, \frac{47 \pi}{24}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{17 \pi}{24}, \frac{41 \pi}{24}, \frac{23 \pi}{24}, \frac{47 \pi}{24}
|
college_math.PRECALCULUS
|
exercise.10.1.50
|
A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places.
|
About 30.12 miles per hour
|
Creative Commons License
|
college_math.precalculus
|
About 30.12 miles per hour
|
college_math.PRECALCULUS
|
exercise.10.7.47
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (x)+\cos (x)=1$
|
$x=0, \frac{\pi}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{2}
|
college_math.PRECALCULUS
|
exercise.11.4.76
|
Convert the equation from polar coordinates into rectangular coordinates: $r=1+\sin (\theta)$
|
$r=\sin (\theta)$
|
Creative Commons License
|
college_math.precalculus
|
r=\sin (\theta)
|
college_math.PRECALCULUS
|
exercise.2.1.41
|
A mobile plan charges a base monthly rate of $\$ 10$ for the first 500 minutes of air time plus a charge of $15 \notin$ for each additional minute. Write a piecewise-defined linear function which calculates the monthly cost $C$ (in dollars) for using $m$ minutes of air time.
|
$C(m)=\left\{\begin{array}{rll}10 & \text { if } & 0 \leq m \leq 500 \\ 10+0.15(m-500) & \text { if } & m>500\end{array}\right.$
|
Creative Commons License
|
college_math.precalculus
|
C(m)=\left\{\begin{array}{rll}10 & \text { if } & 0 \leq m \leq 500 \\ 10+0.15(m-500) & \text { if } & m>500\end{array}\right.
|
college_math.PRECALCULUS
|
exercise.8.4.9
|
Use one matrix inverse to solve the following system of linear equations:
$\left\{\begin{array}{r}3 x+7 y=26 \\ 5 x+12 y=39\end{array}\right.$
|
$\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{l}26 \\ 39\end{array}\right]=\left[\begin{array}{r}39 \\ -13\end{array}\right]$ So $x=39$ and $y=-13$.
|
Creative Commons License
|
college_math.precalculus
|
\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{l}26 \\ 39\end{array}\right]=\left[\begin{array}{r}39 \\ -13\end{array}\right]$ So $x=39$ and $y=-13$.
|
college_math.PRECALCULUS
|
exercise.3.4.53
|
Create a polynomial $f$ with the following characteristics:
- The leading term of $f(x)$ is $-2x^3$
- $c=2i$ is a zero
- $f(0)=-16$
|
$f(x)=-2(x-2 i)(x+2 i)(x+2)$
|
Creative Commons License
|
college_math.precalculus
|
f(x)=-2(x-2 i)(x+2 i)(x+2)
|
college_math.PRECALCULUS
|
exercise.10.7.22
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (2 x)=\sin (x)$
|
$x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}
|
college_math.PRECALCULUS
|
exercise.6.1.20
|
Evaluate the expression: $\log _{8}(4)$
|
$\log _{8}(4)=\frac{2}{3}$
|
Creative Commons License
|
college_math.precalculus
|
\log _{8}(4)=\frac{2}{3}
|
college_math.PRECALCULUS
|
exercise.6.5.9
|
How much money needs to be invested now to obtain $\$ 5000$ in 10 years if the interest rate in a CD is $2.25 \%$, compounded monthly? Round your answer to the nearest cent.
|
$P=\frac{5000}{\left(1+\frac{0.025}{12}\right)^{12 \cdot 10}} \approx \$ 3993.42$
|
Creative Commons License
|
college_math.precalculus
|
P=\frac{5000}{\left(1+\frac{0.025}{12}\right)^{12 \cdot 10}} \approx \$ 3993.42
|
college_math.PRECALCULUS
|
exercise.9.4.5
|
Simplify the expression: $\frac{(n+1) !}{n !}, n \geq 0$.
|
$n+1$
|
Creative Commons License
|
college_math.precalculus
|
n+1
|
college_math.PRECALCULUS
|
exercise.10.1.1
|
Convert the angle into the DMS system and round the answer to the nearest second: $63.75^{\circ}$
|
$63^{\circ} 45^{\prime}$
|
Creative Commons License
|
college_math.precalculus
|
63^{\circ} 45^{\prime}
|
college_math.PRECALCULUS
|
exercise.6.4.6
|
Solve the equation analytically: $\log _{\frac{1}{2}}(2 x-1)=-3$
|
$x=\frac{9}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{9}{2}
|
college_math.PRECALCULUS
|
exercise.2.3.29
|
Let $L$ be the line $y=2 x+1$. Find a function $D(x)$ which measures the distance squared from a point on $L$ to $(0,0)$. Use this to find the point on $L$ closest to $(0,0)$.
|
$D(x)=x^{2}+(2 x+1)^{2}=5 x^{2}+4 x+1, D$ is minimized when $x=-\frac{2}{5}$, so the point on $y=2 x+1$ closest to $(0,0)$ is $\left(-\frac{2}{5}, \frac{1}{5}\right)$
|
Creative Commons License
|
college_math.precalculus
|
D(x)=x^{2}+(2 x+1)^{2}=5 x^{2}+4 x+1, D$ is minimized when $x=-\frac{2}{5}$, so the point on $y=2 x+1$ closest to $(0,0)$ is $\left(-\frac{2}{5}, \frac{1}{5}\right)
|
college_math.PRECALCULUS
|
exercise.6.3.15
|
Solve the equation analytically: $2000 e^{0.1 t}=4000$
|
$t=\frac{\ln (2)}{0.1}=10 \ln (2)$
|
Creative Commons License
|
college_math.precalculus
|
t=\frac{\ln (2)}{0.1}=10 \ln (2)
|
college_math.PRECALCULUS
|
exercise.2.2.19
|
Solve the equation: $|4-x|-|x+2|=0$
|
$x=1$
|
Creative Commons License
|
college_math.precalculus
|
x=1
|
college_math.PRECALCULUS
|
exercise.2.2.3
|
Solve the equation: $|4-x|=7$
|
$x=-3$ or $x=11$
|
Creative Commons License
|
college_math.precalculus
|
x=-3$ or $x=11
|
college_math.PRECALCULUS
|
exercise.6.4.25
|
Solve the inequality analytically: $\frac{1-\ln (x)}{x^{2}}<0$
|
$(e, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(e, \infty)
|
college_math.PRECALCULUS
|
exercise.8.4.7
|
Find the inverse of the matrix or state that the matrix is not invertible: $G=\left[\begin{array}{rrr}1 & 2 & 3 \\ 2 & 3 & 11 \\ 3 & 4 & 19\end{array}\right]$
|
$G$ is not invertible
|
Creative Commons License
|
college_math.precalculus
|
G$ is not invertible
|
college_math.PRECALCULUS
|
exercise.10.7.62
|
Solve the equation: $6 \operatorname{arccot}(2 x)-5 \pi=0$
|
$x=-\frac{\sqrt{3}}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{\sqrt{3}}{2}
|
college_math.PRECALCULUS
|
exercise.4.3.1
|
Solve the rational equation: $\frac{x}{5 x+4}=3$
|
$x=-\frac{6}{7}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{6}{7}
|
college_math.PRECALCULUS
|
exercise.8.4.10
|
Use one matrix inverse to solve the following system of linear equations:
$\left\{\begin{aligned} 3 x+7 y & =0 \\ 5 x+12 y & =-1\end{aligned}\right.$
|
$\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{r}0 \\ -1\end{array}\right]=\left[\begin{array}{r}7 \\ -3\end{array}\right] \quad$ So $x=7$ and $y=-3$.
|
Creative Commons License
|
college_math.precalculus
|
\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{r}0 \\ -1\end{array}\right]=\left[\begin{array}{r}7 \\ -3\end{array}\right] \quad$ So $x=7$ and $y=-3$.
|
college_math.PRECALCULUS
|
exercise.10.7.13
|
Find all of the exact solutions of the equation and then list those solutions which are in the interval $[0,2 \pi)$: $\csc (x)=0$
|
No solution
|
Creative Commons License
|
college_math.precalculus
|
No solution
|
college_math.PRECALCULUS
|
exercise.10.7.53
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (3 x)=\cos (5 x)$
|
$x=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}
|
college_math.PRECALCULUS
|
exercise.3.4.24
|
Simplify the given power of $i$: $i^{26}$
|
$i^{26}=\left(i^{4}\right)^{6} \cdot i^{2}=1 \cdot(-1)=-1$
|
Creative Commons License
|
college_math.precalculus
|
i^{26}=\left(i^{4}\right)^{6} \cdot i^{2}=1 \cdot(-1)=-1
|
college_math.PRECALCULUS
|
exercise.10.2.44
|
Solve the equation for $t$: $\cos (t)=\frac{1}{2}$
|
$\cos (t)=\frac{1}{2}$ when $t=\frac{\pi}{3}+2 \pi k$ or $t=\frac{5 \pi}{3}+2 \pi k$ for any integer $k$.
|
Creative Commons License
|
college_math.precalculus
|
\cos (t)=\frac{1}{2}$ when $t=\frac{\pi}{3}+2 \pi k$ or $t=\frac{5 \pi}{3}+2 \pi k$ for any integer $k$.
|
college_math.PRECALCULUS
|
exercise.2.2.8
|
Solve the equation: $\frac{2}{3}|5-2 x|-\frac{1}{2}=5$
|
$x=-\frac{13}{8}$ or $x=\frac{53}{8}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{13}{8}$ or $x=\frac{53}{8}
|
college_math.PRECALCULUS
|
exercise.3.4.20
|
Simplify the given power of $i$: $i^6$
|
$i^{6}=i^{4} \cdot i^{2}=1 \cdot(-1)=-1$
|
Creative Commons License
|
college_math.precalculus
|
i^{6}=i^{4} \cdot i^{2}=1 \cdot(-1)=-1
|
college_math.PRECALCULUS
|
exercise.10.7.48
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (x)+\sqrt{3} \cos (x)=1$
|
$x=\frac{\pi}{2}, \frac{11 \pi}{6}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{2}, \frac{11 \pi}{6}
|
college_math.PRECALCULUS
|
exercise.6.3.14
|
Solve the equation analytically: $e^{-5730 k}=\frac{1}{2}$
|
$k=\frac{\ln \left(\frac{1}{2}\right)}{-5730}=\frac{\ln (2)}{5730}$
|
Creative Commons License
|
college_math.precalculus
|
k=\frac{\ln \left(\frac{1}{2}\right)}{-5730}=\frac{\ln (2)}{5730}
|
college_math.PRECALCULUS
|
exercise.6.3.23
|
Solve the equation analytically: $e^{2 x}=2 e^{x}$
|
$x=\ln (2)$
|
Creative Commons License
|
college_math.precalculus
|
x=\ln (2)
|
college_math.PRECALCULUS
|
exercise.10.7.101
|
Express the domain of the function using the extended interval notation: $f(x)=\sqrt{\tan ^{2}(x)-1}$
|
$\bigcup_{k=-\infty}^{\infty}\left\{\left[\frac{(4 k+1) \pi}{4}, \frac{(2 k+1) \pi}{2}\right) \cup\left(\frac{(2 k+1) \pi}{2}, \frac{(4 k+3) \pi}{4}\right]\right\}$
|
Creative Commons License
|
college_math.precalculus
|
\bigcup_{k=-\infty}^{\infty}\left\{\left[\frac{(4 k+1) \pi}{4}, \frac{(2 k+1) \pi}{2}\right) \cup\left(\frac{(2 k+1) \pi}{2}, \frac{(4 k+3) \pi}{4}\right]\right\}
|
college_math.PRECALCULUS
|
exercise.7.2.16
|
Find the standard equation of the circle which satisfies the given criteria: endpoints of a diameter: $\left(\frac{1}{2}, 4\right),\left(\frac{3}{2},-1\right)$
|
$(x-1)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{2}$
|
Creative Commons License
|
college_math.precalculus
|
(x-1)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{13}{2}
|
college_math.PRECALCULUS
|
exercise.10.7.96
|
Solve the given inequality: $\pi>2 \arctan (x)$
|
$(-\infty, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty, \infty)
|
college_math.PRECALCULUS
|
exercise.6.1.52
|
Find the domain of the function: $f(x)=\sqrt[4]{\log _{4}(x)}$
|
$[1, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
[1, \infty)
|
college_math.PRECALCULUS
|
exercise.8.4.4
|
Find the inverse of the matrix or state that the matrix is not invertible: $D=\left[\begin{array}{rr}2 & -1 \\ 16 & -9\end{array}\right]$
|
$D^{-1}=\left[\begin{array}{cc}\frac{9}{2} & -\frac{1}{2} \\ 8 & -1\end{array}\right]$
|
Creative Commons License
|
college_math.precalculus
|
D^{-1}=\left[\begin{array}{cc}\frac{9}{2} & -\frac{1}{2} \\ 8 & -1\end{array}\right]
|
college_math.PRECALCULUS
|
exercise.6.5.33
|
The current $i$ measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by $i(t)=2-2 e^{-10 t}$ where $t \geq 0$ is the number of seconds after the circuit is switched on. Determine the value of $i$ as $t \rightarrow \infty$. (This is called the steady state current.)
|
The steady state current is 2 amps.
|
Creative Commons License
|
college_math.precalculus
|
The steady state current is 2 amps.
|
college_math.PRECALCULUS
|
exercise.5.3.29
|
Solve the equation or inequality: $2(x-2)^{-\frac{1}{3}}-\frac{2}{3} x(x-2)^{-\frac{4}{3}} \leq 0$
|
$(-\infty, 2) \cup(2,3]$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty, 2) \cup(2,3]
|
college_math.PRECALCULUS
|
exercise.2.2.5
|
Solve the equation: $2|5 x+1|-3=0$
|
$x=-\frac{1}{2}$ or $x=\frac{1}{10}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{1}{2}$ or $x=\frac{1}{10}
|
college_math.PRECALCULUS
|
exercise.5.3.35
|
Solve the equation or inequality: $\frac{2}{3}(x+4)^{\frac{3}{5}}(x-2)^{-\frac{1}{3}}+\frac{3}{5}(x+4)^{-\frac{2}{5}}(x-2)^{\frac{2}{3}} \geq 0$
|
$(-\infty,-4) \cup\left(-4,-\frac{22}{19}\right] \cup(2, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-4) \cup\left(-4,-\frac{22}{19}\right] \cup(2, \infty)
|
college_math.PRECALCULUS
|
exercise.4.3.17
|
Solve the rational inequality and express your answer using interval notation: $\frac{-x^{3}+4 x}{x^{2}-9} \geq 4 x$
|
$(-\infty,-3) \cup[-2 \sqrt{2}, 0] \cup[2 \sqrt{2}, 3)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-3) \cup[-2 \sqrt{2}, 0] \cup[2 \sqrt{2}, 3)
|
college_math.PRECALCULUS
|
exercise.10.2.52
|
Approximate the given value to three decimal places: $\cos \left(207^{\circ}\right)$
|
$\cos \left(207^{\circ}\right) \approx-0.891$
|
Creative Commons License
|
college_math.precalculus
|
\cos \left(207^{\circ}\right) \approx-0.891
|
college_math.PRECALCULUS
|
exercise.5.3.43
|
Find the inverse of $k(x)=\frac{2 x}{\sqrt{x^{2}-1}}$.
|
$k^{-1}(x)=\frac{x}{\sqrt{x^{2}-4}}$
|
Creative Commons License
|
college_math.precalculus
|
k^{-1}(x)=\frac{x}{\sqrt{x^{2}-4}}
|
college_math.PRECALCULUS
|
exercise.10.1.29
|
Convert the angle from degree measure into radian measure, giving the exact value in terms of $\pi$: $0^{\circ}$
|
0
|
Creative Commons License
|
college_math.precalculus
|
0
|
college_math.PRECALCULUS
|
exercise.1.1.15
|
Write the set using interval notation: $\{x \mid x<3$ or $x \geq 2\}$
|
$(-\infty, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty, \infty)
|
college_math.PRECALCULUS
|
exercise.6.1.41
|
Evaluate the expression: $\log _{2}\left(3^{-\log _{3}(2)}\right)$
|
$\log _{2}\left(3^{-\log _{3}(2)}\right)=-1$
|
Creative Commons License
|
college_math.precalculus
|
\log _{2}\left(3^{-\log _{3}(2)}\right)=-1
|
college_math.PRECALCULUS
|
exercise.9.4.16
|
Simplify the power of a complex number: $\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)^{3}$
|
$i$
|
Creative Commons License
|
college_math.precalculus
|
i
|
college_math.PRECALCULUS
|
exercise.4.3.3
|
Solve the rational equation: $\frac{1}{x+3}+\frac{1}{x-3}=\frac{x^{2}-3}{x^{2}-9}$
|
$x=-1$
|
Creative Commons License
|
college_math.precalculus
|
x=-1
|
college_math.PRECALCULUS
|
exercise.6.1.54
|
Find the domain of the function: $f(x)=\ln (\sqrt{x-4}-3)$
|
$(13, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(13, \infty)
|
college_math.PRECALCULUS
|
exercise.11.4.75
|
Convert the equation from polar coordinates into rectangular coordinates: $r=1-2 \cos (\theta)$
|
$r=6 \sin (\theta)$
|
Creative Commons License
|
college_math.precalculus
|
r=6 \sin (\theta)
|
college_math.PRECALCULUS
|
exercise.1.3.45
|
Determine whether or not the equation represents $y$ as a function of $x$: $2 x+3 y=4$
|
Function
|
Creative Commons License
|
college_math.precalculus
|
Function
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.