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college_math.PRECALCULUS
|
exercise.9.2.32
|
Express the repeating decimal as a fraction of integers: $-5.8 \overline{67}$
|
$-\frac{5809}{990}$
|
Creative Commons License
|
college_math.precalculus
|
-\frac{5809}{990}
|
college_math.PRECALCULUS
|
exercise.10.7.91
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-2 \pi \leq x \leq 2 \pi$: $\sin (2 x) \geq \sin (x)$
|
$\left[-2 \pi,-\frac{5 \pi}{3}\right] \cup\left[-\pi,-\frac{\pi}{3}\right] \cup\left[0, \frac{\pi}{3}\right] \cup\left[\pi, \frac{5 \pi}{3}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[-2 \pi,-\frac{5 \pi}{3}\right] \cup\left[-\pi,-\frac{\pi}{3}\right] \cup\left[0, \frac{\pi}{3}\right] \cup\left[\pi, \frac{5 \pi}{3}\right]
|
college_math.PRECALCULUS
|
exercise.10.7.84
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-\pi \leq x \leq \pi$: $\sin ^{2}(x)<\frac{3}{4}$
|
$\left[-\pi,-\frac{\pi}{4}\right] \cup\left(0, \frac{3 \pi}{4}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[-\pi,-\frac{\pi}{4}\right] \cup\left(0, \frac{3 \pi}{4}\right]
|
college_math.PRECALCULUS
|
exercise.6.1.35
|
Evaluate the expression: $\ln \left(e^{5}\right)$
|
$\ln \left(e^{5}\right)=5$
|
Creative Commons License
|
college_math.precalculus
|
\ln \left(e^{5}\right)=5
|
college_math.PRECALCULUS
|
exercise.6.3.10
|
Solve the equation analytically: $5^{-x}=2$
|
$x=-\frac{\ln (2)}{\ln (5)}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{\ln (2)}{\ln (5)}
|
college_math.PRECALCULUS
|
exercise.3.4.23
|
Simplify the given power of $i$: $i^{15}$
|
$i^{15}=\left(i^{4}\right)^{3} \cdot i^{3}=1 \cdot(-i)=-i$
|
Creative Commons License
|
college_math.precalculus
|
i^{15}=\left(i^{4}\right)^{3} \cdot i^{3}=1 \cdot(-i)=-i
|
college_math.PRECALCULUS
|
exercise.10.7.57
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (6 x)+\sin (x)=0$
|
$x=0, \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}, \frac{8 \pi}{7}, \frac{10 \pi}{7}, \frac{12 \pi}{7}, \frac{\pi}{5}, \frac{3 \pi}{5}, \pi, \frac{7 \pi}{5}, \frac{9 \pi}{5}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}, \frac{8 \pi}{7}, \frac{10 \pi}{7}, \frac{12 \pi}{7}, \frac{\pi}{5}, \frac{3 \pi}{5}, \pi, \frac{7 \pi}{5}, \frac{9 \pi}{5}
|
college_math.PRECALCULUS
|
exercise.6.4.17
|
Solve the equation analytically: $\log _{169}(3 x+7)-\log _{169}(5 x-9)=\frac{1}{2}$
|
$x=2$
|
Creative Commons License
|
college_math.precalculus
|
x=2
|
college_math.PRECALCULUS
|
exercise.10.7.40
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\csc ^{3}(x)+\csc ^{2}(x)=4 \csc (x)+4$
|
$x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}
|
college_math.PRECALCULUS
|
exercise.3.4.13
|
Simplify the quantity $\sqrt{-25}\sqrt{-4}$
|
-10
|
Creative Commons License
|
college_math.precalculus
|
-10
|
college_math.PRECALCULUS
|
exercise.4.3.4
|
Solve the rational equation: $\frac{2 x+17}{x+1}=x+5$
|
$x=-6, x=2$
|
Creative Commons License
|
college_math.precalculus
|
x=-6, x=2
|
college_math.PRECALCULUS
|
exercise.2.3.23
|
The height $h$ in feet of a model rocket above the ground $t$ seconds after lift-off is given by $h(t)=-5 t^{2}+100 t$, for $0 \leq t \leq 20$. When does the rocket reach its maximum height above the ground? What is its maximum height?
|
The rocket reaches its maximum height of 500 feet 10 seconds after lift-off.
|
Creative Commons License
|
college_math.precalculus
|
The rocket reaches its maximum height of 500 feet 10 seconds after lift-off.
|
college_math.PRECALCULUS
|
exercise.2.2.13
|
Solve the equation: $|x|=x^{2}$
|
$x=-1, x=0$ or $x=1$
|
Creative Commons License
|
college_math.precalculus
|
x=-1, x=0$ or $x=1
|
college_math.PRECALCULUS
|
exercise.7.3.19
|
The mirror in Carl's flashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror?
|
The bulb should be placed 0.625 centimeters above the vertex of the mirror. (As verified by Carl himself!)
|
Creative Commons License
|
college_math.precalculus
|
The bulb should be placed 0.625 centimeters above the vertex of the mirror. (As verified by Carl himself!)
|
college_math.PRECALCULUS
|
exercise.6.4.13
|
Solve the equation analytically: $6-3 \log _{5}(2 x)=0$
|
$x=\frac{25}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{25}{2}
|
college_math.PRECALCULUS
|
exercise.10.7.55
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (5 x)=\sin (3 x)$
|
$x=0, \frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \pi, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}, \frac{15 \pi}{8}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \pi, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}, \frac{15 \pi}{8}
|
college_math.PRECALCULUS
|
exercise.9.1.5
|
Write out the first four terms of the given sequence: $\left\{\frac{x^{n}}{n^{2}}\right\}_{n=1}^{\infty}$
|
$x, \frac{x^{2}}{4}, \frac{x^{3}}{9}, \frac{x^{4}}{16}$
|
Creative Commons License
|
college_math.precalculus
|
x, \frac{x^{2}}{4}, \frac{x^{3}}{9}, \frac{x^{4}}{16}
|
college_math.PRECALCULUS
|
exercise.6.3.36
|
Solve the inequality analytically: $2^{\left(x^{3}-x\right)}<1$
|
$(-\infty,-1) \cup(0,1)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-1) \cup(0,1)
|
college_math.PRECALCULUS
|
exercise.1.1.28
|
Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them: $(2 \sqrt{45}, \sqrt{12}),(\sqrt{20}, \sqrt{27})$.
|
$d=\sqrt{83}, M=\left(4 \sqrt{5}, \frac{5 \sqrt{3}}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
d=\sqrt{83}, M=\left(4 \sqrt{5}, \frac{5 \sqrt{3}}{2}\right)
|
college_math.PRECALCULUS
|
exercise.6.1.19
|
Evaluate the expression: $\log _{6}\left(\frac{1}{36}\right)$
|
$\log _{6}\left(\frac{1}{36}\right)=-2$
|
Creative Commons License
|
college_math.precalculus
|
\log _{6}\left(\frac{1}{36}\right)=-2
|
college_math.PRECALCULUS
|
exercise.10.7.33
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (2 x)=\tan (x)$
|
$x=0, \pi, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \pi, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}
|
college_math.PRECALCULUS
|
exercise.10.1.41
|
Convert the angle from radian measure into degree measure: $\frac{\pi}{3}$
|
$60^{\circ}$
|
Creative Commons License
|
college_math.precalculus
|
60^{\circ}
|
college_math.PRECALCULUS
|
exercise.3.4.52
|
Create a polynomial $f$ that is degree 5 and has the following characteristics:
- $x=6$, $x=i$, and $x=1-3i$ are zeros of $f$
- As $x \rightarrow -\infty$, $f(x) \rightarrow \infty$
|
$f(x)=a(x-6)(x-i)(x+i)(x-(1-3 i))(x-(1+3 i))$ where $a$ is any real number, $a<0$
|
Creative Commons License
|
college_math.precalculus
|
f(x)=a(x-6)(x-i)(x+i)(x-(1-3 i))(x-(1+3 i))$ where $a$ is any real number, $a<0
|
college_math.PRECALCULUS
|
exercise.11.4.23
|
Convert the point from polar coordinates into rectangular coordinates: $\left(9, \frac{7 \pi}{2}\right)$
|
$(0,-9)$
|
Creative Commons License
|
college_math.precalculus
|
(0,-9)
|
college_math.PRECALCULUS
|
exercise.10.2.39
|
Find all of the angles which satisfy the given equation: $\cos (\theta)=-1.001$
|
$\cos (\theta)=-1.001$ never happens
|
Creative Commons License
|
college_math.precalculus
|
\cos (\theta)=-1.001$ never happens
|
college_math.PRECALCULUS
|
exercise.6.3.43
|
Use your calculator to help you solve the inequality: $e^{-x}-x e^{-x} \geq 0$
|
$(-\infty, 1]$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty, 1]
|
college_math.PRECALCULUS
|
exercise.5.3.20
|
Solve the equation or inequality: $3 x+\sqrt{6-9 x}=2$
|
$x=-\frac{1}{3}, \frac{2}{3}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{1}{3}, \frac{2}{3}
|
college_math.PRECALCULUS
|
exercise.10.7.79
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 \leq x \leq 2 \pi$: $\sec (x) \leq \sqrt{2}$
|
$\left[0, \frac{\pi}{4}\right] \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \cup\left[\frac{7 \pi}{4}, 2 \pi\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[0, \frac{\pi}{4}\right] \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \cup\left[\frac{7 \pi}{4}, 2 \pi\right]
|
college_math.PRECALCULUS
|
exercise.3.3.41
|
Find the real solutions of the polynomial equation $x^{3}+x^{2}=\frac{11 x+10}{3}$.
|
$x= \pm \sqrt{3}$
|
Creative Commons License
|
college_math.precalculus
|
x= \pm \sqrt{3}
|
college_math.PRECALCULUS
|
exercise.6.2.7
|
Expand the given logarithm and simplify: $\log _{\sqrt{2}}\left(4 x^{3}\right)$
|
$3 \log _{\sqrt{2}}(x)+4$
|
Creative Commons License
|
college_math.precalculus
|
3 \log _{\sqrt{2}}(x)+4
|
college_math.PRECALCULUS
|
exercise.2.3.32
|
Solve the quadratic equation $y^{2}-4 y=x^{2}-4$ for $x$.
|
$x= \pm(y-2)$
|
Creative Commons License
|
college_math.precalculus
|
x= \pm(y-2)
|
college_math.PRECALCULUS
|
exercise.10.2.50
|
Approximate the given value to three decimal places: $\cos (-2.01)$
|
$\cos (-2.01) \approx-0.425$
|
Creative Commons License
|
college_math.precalculus
|
\cos (-2.01) \approx-0.425
|
college_math.PRECALCULUS
|
exercise.10.7.32
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\cos (x) \csc (x) \cot (x)=6-\cot ^{2}(x)$
|
$x=\frac{\pi}{6}, \frac{7 \pi}{6}, \frac{5 \pi}{6}, \frac{11 \pi}{6}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{6}, \frac{7 \pi}{6}, \frac{5 \pi}{6}, \frac{11 \pi}{6}
|
college_math.PRECALCULUS
|
exercise.6.4.19
|
Solve the equation analytically: $2 \log _{7}(x)=\log _{7}(2)+\log _{7}(x+12)$
|
$x=6$
|
Creative Commons License
|
college_math.precalculus
|
x=6
|
college_math.PRECALCULUS
|
exercise.10.2.63
|
If $\theta=5^{\circ}$ and the hypotenuse has length 10 , how long is the side adjacent to $\theta$ ?
|
The side adjacent to $\theta$ has length $10 \cos \left(5^{\circ}\right) \approx 9.962$.
|
Creative Commons License
|
college_math.precalculus
|
The side adjacent to $\theta$ has length $10 \cos \left(5^{\circ}\right) \approx 9.962$.
|
college_math.PRECALCULUS
|
exercise.6.3.17
|
Solve the equation analytically: $70+90 e^{-0.1 t}=75$
|
$t=\frac{\ln \left(\frac{1}{18}\right)}{-0.1}=10 \ln (18)$
|
Creative Commons License
|
college_math.precalculus
|
t=\frac{\ln \left(\frac{1}{18}\right)}{-0.1}=10 \ln (18)
|
college_math.PRECALCULUS
|
exercise.8.7.13
|
Solve the system of nonlinear equations: $\left\{\begin{array}{rr}y & =x^{3}+8 \\ y & =10 x-x^{2}\end{array}\right.$
|
$(-4,-56),(1,9),(2,16)$
|
Creative Commons License
|
college_math.precalculus
|
(-4,-56),(1,9),(2,16)
|
college_math.PRECALCULUS
|
exercise.6.1.17
|
Evaluate the expression: $\log _{6}(216)$
|
$\log _{6}(216)=3$
|
Creative Commons License
|
college_math.precalculus
|
\log _{6}(216)=3
|
college_math.PRECALCULUS
|
exercise.3.3.35
|
Find the real solutions of the polynomial equation $9 x^{3}=5 x^{2}+x$.
|
$x=0, \frac{5 \pm \sqrt{61}}{18}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{5 \pm \sqrt{61}}{18}
|
college_math.PRECALCULUS
|
exercise.6.2.14
|
Expand the given logarithm and simplify: $\log _{\frac{1}{2}}\left(\frac{4 \sqrt[3]{x^{2}}}{y \sqrt{z}}\right)$
|
$-2+\frac{2}{3} \log _{\frac{1}{2}}(x)-\log _{\frac{1}{2}}(y)-\frac{1}{2} \log _{\frac{1}{2}}(z)$
|
Creative Commons License
|
college_math.precalculus
|
-2+\frac{2}{3} \log _{\frac{1}{2}}(x)-\log _{\frac{1}{2}}(y)-\frac{1}{2} \log _{\frac{1}{2}}(z)
|
college_math.PRECALCULUS
|
exercise.3.3.44
|
Find the real solutions of the polynomial equation $2 x^{5}+3 x^{4}=18 x+27$.
|
$\left\{-\frac{1}{2}\right\} \cup[1, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
\left\{-\frac{1}{2}\right\} \cup[1, \infty)
|
college_math.PRECALCULUS
|
exercise.1.1.22
|
Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them: $(1,2),(-3,5)$
|
$d=5, M=\left(-1, \frac{7}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
d=5, M=\left(-1, \frac{7}{2}\right)
|
college_math.PRECALCULUS
|
exercise.9.1.1
|
Write out the first four terms of the given sequence: $a_{n}=2^{n}-1, n \geq 0$
|
$0,1,3,7$
|
Creative Commons License
|
college_math.precalculus
|
0,1,3,7
|
college_math.PRECALCULUS
|
exercise.9.1.10
|
Write out the first four terms of the given sequence: $c_{0}=-2, c_{j}=\frac{c_{j-1}}{(j+1)(j+2)}, j \geq 1$
|
$-2,-\frac{1}{3},-\frac{1}{36},-\frac{1}{720}$
|
Creative Commons License
|
college_math.precalculus
|
-2,-\frac{1}{3},-\frac{1}{36},-\frac{1}{720}
|
college_math.PRECALCULUS
|
exercise.11.4.52
|
Convert the equation from rectangular coordinates into polar coordinates: $x^{2}+y^{2}=x$
|
$\left(\frac{1}{3}, \pi+\arctan (2)\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(\frac{1}{3}, \pi+\arctan (2)\right)
|
college_math.PRECALCULUS
|
exercise.9.2.29
|
Express the repeating decimal as a fraction of integers: $0 . \overline{7}$
|
$\frac{7}{9}$
|
Creative Commons License
|
college_math.precalculus
|
\frac{7}{9}
|
college_math.PRECALCULUS
|
exercise.10.7.66
|
Solve the equation: $9 \arccos ^{2}(x)-\pi^{2}=0$
|
$x=-1,0$
|
Creative Commons License
|
college_math.precalculus
|
x=-1,0
|
college_math.PRECALCULUS
|
exercise.6.3.21
|
Solve the equation analytically: $\frac{150}{1+29 e^{-0.8 t}}=75$
|
$t=\frac{\ln \left(\frac{1}{29}\right)}{-0.8}=\frac{5}{4} \ln (29)$
|
Creative Commons License
|
college_math.precalculus
|
t=\frac{\ln \left(\frac{1}{29}\right)}{-0.8}=\frac{5}{4} \ln (29)
|
college_math.PRECALCULUS
|
exercise.10.7.86
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-\pi \leq x \leq \pi$: $\cos (x) \geq \sin (x)$
|
$\left[-\frac{3 \pi}{4}, \frac{\pi}{4}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[-\frac{3 \pi}{4}, \frac{\pi}{4}\right]
|
college_math.PRECALCULUS
|
exercise.6.4.27
|
Solve the inequality analytically: $10 \log \left(\frac{x}{10^{-12}}\right) \geq 90$
|
$\left[10^{-3}, \infty\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left[10^{-3}, \infty\right)
|
college_math.PRECALCULUS
|
exercise.6.3.4
|
Solve the equation analytically: $4^{2 x}=\frac{1}{2}$
|
$x=-\frac{1}{4}$
|
Creative Commons License
|
college_math.precalculus
|
x=-\frac{1}{4}
|
college_math.PRECALCULUS
|
exercise.6.3.34
|
Solve the inequality analytically: $e^{x}>53$
|
$(\ln (53), \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(\ln (53), \infty)
|
college_math.PRECALCULUS
|
exercise.10.7.103
|
Express the domain of the function using the extended interval notation: $f(x)=\csc (2 x)$
|
$\bigcup_{k=-\infty}^{\infty}\left(\frac{k \pi}{2}, \frac{(k+1) \pi}{2}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\bigcup_{k=-\infty}^{\infty}\left(\frac{k \pi}{2}, \frac{(k+1) \pi}{2}\right)
|
college_math.PRECALCULUS
|
exercise.10.4.80
|
Write the given sum as a product: $\cos (3 \theta)+\cos (5 \theta)$
|
$2 \cos (4 \theta) \cos (\theta)$
|
Creative Commons License
|
college_math.precalculus
|
2 \cos (4 \theta) \cos (\theta)
|
college_math.PRECALCULUS
|
exercise.3.3.46
|
Solve the polynomial inequality $x^{4}-9 x^{2} \leq 4 x-12$ and state your answer using interval notation.
|
$\{2\} \cup[4, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
\{2\} \cup[4, \infty)
|
college_math.PRECALCULUS
|
exercise.11.4.47
|
Convert the equation from rectangular coordinates into polar coordinates: $x=3 y+1$
|
$\left(10, \arctan \left(\frac{4}{3}\right)\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(10, \arctan \left(\frac{4}{3}\right)\right)
|
college_math.PRECALCULUS
|
exercise.8.7.15
|
Solve the system of nonlinear equations: $\left\{\begin{aligned} x^{2}+y^{2} & =25 \\ 4 x^{2}-9 y & =0 \\ 3 y^{2}-16 x & =0\end{aligned}\right.$
|
$(3,4)$
|
Creative Commons License
|
college_math.precalculus
|
(3,4)
|
college_math.PRECALCULUS
|
exercise.10.7.78
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 \leq x \leq 2 \pi$: $\cos (3 x) \leq 1$
|
$[0,2 \pi]$
|
Creative Commons License
|
college_math.precalculus
|
[0,2 \pi]
|
college_math.PRECALCULUS
|
exercise.6.3.9
|
Solve the equation analytically: $3^{2 x}=5$
|
$x=\frac{\ln (5)}{2 \ln (3)}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\ln (5)}{2 \ln (3)}
|
college_math.PRECALCULUS
|
exercise.1.1.32
|
Find all of the points on the $x$-axis which are 2 units from the point $(-1,1)$.
|
$(-1+\sqrt{3}, 0),(-1-\sqrt{3}, 0)$
|
Creative Commons License
|
college_math.precalculus
|
(-1+\sqrt{3}, 0),(-1-\sqrt{3}, 0)
|
college_math.PRECALCULUS
|
exercise.10.7.74
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 \leq x \leq 2 \pi$: $\sin \left(x+\frac{\pi}{3}\right)>\frac{1}{2}$
|
$\left(0, \frac{\pi}{3}\right] \cup\left[\frac{2 \pi}{3}, \pi\right) \cup\left(\pi, \frac{4 \pi}{3}\right] \cup\left[\frac{5 \pi}{3}, 2 \pi\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(0, \frac{\pi}{3}\right] \cup\left[\frac{2 \pi}{3}, \pi\right) \cup\left(\pi, \frac{4 \pi}{3}\right] \cup\left[\frac{5 \pi}{3}, 2 \pi\right)
|
college_math.PRECALCULUS
|
exercise.3.4.18
|
Simplify the quantity $-\sqrt{(-9)}$
|
$-3 i$
|
Creative Commons License
|
college_math.precalculus
|
-3 i
|
college_math.PRECALCULUS
|
exercise.11.4.62
|
Convert the equation from polar coordinates into rectangular coordinates: $\theta=\pi$
|
$\theta=\frac{\pi}{3}$
|
Creative Commons License
|
college_math.precalculus
|
\theta=\frac{\pi}{3}
|
college_math.PRECALCULUS
|
exercise.6.3.2
|
Solve the equation analytically: $3^{(x-1)}=27$
|
$x=4$
|
Creative Commons License
|
college_math.precalculus
|
x=4
|
college_math.PRECALCULUS
|
exercise.10.7.88
|
Solve the inequality. Express the exact answer in interval notation, restricting your attention to $-2 \pi \leq x \leq 2 \pi$: $\cos (x) \leq \frac{5}{3}$
|
$[-2 \pi, 2 \pi]$
|
Creative Commons License
|
college_math.precalculus
|
[-2 \pi, 2 \pi]
|
college_math.PRECALCULUS
|
exercise.6.4.4
|
Solve the equation analytically: $\log _{5}\left(18-x^{2}\right)=\log _{5}(6-x)$
|
$x=-3,4$
|
Creative Commons License
|
college_math.precalculus
|
x=-3,4
|
college_math.PRECALCULUS
|
exercise.6.1.34
|
Evaluate the expression: $\log _{36}\left(36^{216}\right)$
|
$\log _{36}\left(36^{216}\right)=216$
|
Creative Commons License
|
college_math.precalculus
|
\log _{36}\left(36^{216}\right)=216
|
college_math.PRECALCULUS
|
exercise.8.5.17
|
Find the inverse of the given matrix: $B=\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]$
|
$B^{-1}=\left[\begin{array}{rr}3 & 7 \\ 5 & 12\end{array}\right]$
|
Creative Commons License
|
college_math.precalculus
|
B^{-1}=\left[\begin{array}{rr}3 & 7 \\ 5 & 12\end{array}\right]
|
college_math.PRECALCULUS
|
exercise.6.1.56
|
Find the domain of the function: $f(x)=\frac{\sqrt{-1-x}}{\log _{\frac{1}{2}}(x)}$
|
No domain
|
Creative Commons License
|
college_math.precalculus
|
No domain
|
college_math.PRECALCULUS
|
exercise.2.2.10
|
Solve the equation: $|2 x-1|=x+1$
|
$x=0$ or $x=2$
|
Creative Commons License
|
college_math.precalculus
|
x=0$ or $x=2
|
college_math.PRECALCULUS
|
exercise.6.3.29
|
Solve the equation analytically: $e^{2 x}=e^{x}+6$
|
$x=\ln (3)$
|
Creative Commons License
|
college_math.precalculus
|
x=\ln (3)
|
college_math.PRECALCULUS
|
exercise.10.7.49
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sqrt{2} \cos (x)-\sqrt{2} \sin (x)=1$
|
$x=\frac{\pi}{12}, \frac{17 \pi}{12}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{12}, \frac{17 \pi}{12}
|
college_math.PRECALCULUS
|
exercise.9.4.17
|
Simplify the power of a complex number: $\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} i\right)^{4}$
|
-1
|
Creative Commons License
|
college_math.precalculus
|
-1
|
college_math.PRECALCULUS
|
exercise.6.4.23
|
Solve the equation analytically: $(\log (x))^{2}=2 \log (x)+15$
|
$x=10^{-3}, 10^{5}$
|
Creative Commons License
|
college_math.precalculus
|
x=10^{-3}, 10^{5}
|
college_math.PRECALCULUS
|
exercise.4.3.11
|
Solve the rational inequality and express your answer using interval notation: $\frac{x^{2}-x-12}{x^{2}+x-6}>0$
|
$(-\infty,-3) \cup(-3,2) \cup(4, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
(-\infty,-3) \cup(-3,2) \cup(4, \infty)
|
college_math.PRECALCULUS
|
exercise.8.2.22
|
Solve the following system of linear equations: $\left\{\begin{aligned} x-3 y-4 z & =3 \\ 3 x+4 y-z & =13 \\ 2 x-19 y-19 z & =2\end{aligned}\right.$
|
$\left(\frac{19}{13} t+\frac{51}{13},-\frac{11}{13} t+\frac{4}{13}, t\right)$ for all real numbers $t$
|
Creative Commons License
|
college_math.precalculus
|
\left(\frac{19}{13} t+\frac{51}{13},-\frac{11}{13} t+\frac{4}{13}, t\right)$ for all real numbers $t
|
college_math.PRECALCULUS
|
exercise.6.3.1
|
Solve the equation analytically: $2^{4 x}=8$
|
$x=\frac{3}{4}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{3}{4}
|
college_math.PRECALCULUS
|
exercise.6.4.28
|
Solve the inequality analytically: $5.6 \leq \log \left(\frac{x}{10^{-3}}\right) \leq 7.1$
|
$\left[10^{2.6}, 10^{4.1}\right]$
|
Creative Commons License
|
college_math.precalculus
|
\left[10^{2.6}, 10^{4.1}\right]
|
college_math.PRECALCULUS
|
exercise.6.3.42
|
Use your calculator to help you solve the equation: $e^{\sqrt{x}}=x+1$
|
$x=0$
|
Creative Commons License
|
college_math.precalculus
|
x=0
|
college_math.PRECALCULUS
|
exercise.6.4.29
|
Solve the inequality analytically: $2.3<-\log (x)<5.4$
|
$\left(10^{-5.4}, 10^{-2.3}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left(10^{-5.4}, 10^{-2.3}\right)
|
college_math.PRECALCULUS
|
exercise.8.4.2
|
Find the inverse of the matrix or state that the matrix is not invertible: $B=\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]$
|
$B^{-1}=\left[\begin{array}{rr}3 & 7 \\ 5 & 12\end{array}\right]$
|
Creative Commons License
|
college_math.precalculus
|
B^{-1}=\left[\begin{array}{rr}3 & 7 \\ 5 & 12\end{array}\right]
|
college_math.PRECALCULUS
|
exercise.6.2.29
|
Use the properties of logarithms to write the expression as a single logarithm: $\log _{2}(x)+\log _{\frac{1}{2}}(x-1)$
|
$\log _{2}\left(\frac{x}{x-1}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\log _{2}\left(\frac{x}{x-1}\right)
|
college_math.PRECALCULUS
|
exercise.6.3.35
|
Solve the inequality analytically: $1000(1.005)^{12 t} \geq 3000$
|
$\left[\frac{\ln (3)}{12 \ln (1.005)}, \infty\right)$
|
Creative Commons License
|
college_math.precalculus
|
\left[\frac{\ln (3)}{12 \ln (1.005)}, \infty\right)
|
college_math.PRECALCULUS
|
exercise.6.2.25
|
Use the properties of logarithms to write the expression as a single logarithm: $\log _{7}(x)+\log _{7}(x-3)-2$
|
$\log _{7}\left(\frac{x(x-3)}{49}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\log _{7}\left(\frac{x(x-3)}{49}\right)
|
college_math.PRECALCULUS
|
exercise.10.7.21
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (2 x)=\cos (x)$
|
$x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$
|
Creative Commons License
|
college_math.precalculus
|
x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{3 \pi}{2}
|
college_math.PRECALCULUS
|
exercise.10.2.59
|
If $\theta=12^{\circ}$ and the side adjacent to $\theta$ has length 4 , how long is the hypotenuse?
|
The hypotenuse has length $\frac{4}{\cos \left(12^{\circ}\right)} \approx 4.089$.
|
Creative Commons License
|
college_math.precalculus
|
The hypotenuse has length $\frac{4}{\cos \left(12^{\circ}\right)} \approx 4.089$.
|
college_math.PRECALCULUS
|
exercise.10.2.19
|
Find the exact value of the cosine and sine of the given angle: $\theta=\frac{10 \pi}{3}$
|
$\cos \left(\frac{10 \pi}{3}\right)=-\frac{1}{2}, \sin \left(\frac{10 \pi}{3}\right)=-\frac{\sqrt{3}}{2}$
|
Creative Commons License
|
college_math.precalculus
|
\cos \left(\frac{10 \pi}{3}\right)=-\frac{1}{2}, \sin \left(\frac{10 \pi}{3}\right)=-\frac{\sqrt{3}}{2}
|
college_math.PRECALCULUS
|
exercise.1.1.34
|
Let's assume for a moment that we are standing at the origin and the positive $y$-axis points due North while the positive $x$-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be?
|
(-3, -4), 5 miles, $(4,-4)$
|
Creative Commons License
|
college_math.precalculus
|
(-3, -4), 5 miles, $(4,-4)
|
college_math.PRECALCULUS
|
exercise.8.2.8
|
Solve the following system of linear equations: $\left\{\begin{aligned} x+y+z & =3 \\ 2 x-y+z & =0 \\ -3 x+5 y+7 z & =7\end{aligned}\right.$
|
$(-3,20,19)$
|
Creative Commons License
|
college_math.precalculus
|
(-3,20,19)
|
college_math.PRECALCULUS
|
exercise.1.1.26
|
Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them: $\left(\frac{24}{5}, \frac{6}{5}\right),\left(-\frac{11}{5},-\frac{19}{5}\right)$.
|
$d=\sqrt{74}, M=\left(\frac{13}{10},-\frac{13}{10}\right)$
|
Creative Commons License
|
college_math.precalculus
|
d=\sqrt{74}, M=\left(\frac{13}{10},-\frac{13}{10}\right)
|
college_math.PRECALCULUS
|
exercise.10.7.8
|
Find all of the exact solutions of the equation and then list those solutions which are in the interval $[0,2 \pi)$: $\cos (9 x)=9$
|
No solution
|
Creative Commons License
|
college_math.precalculus
|
No solution
|
college_math.PRECALCULUS
|
exercise.6.4.33
|
Solve the equation or inequality using your calculator: $\ln \left(x^{2}+1\right) \geq 5$
|
$\approx(-\infty,-12.1414) \cup(12.1414, \infty)$
|
Creative Commons License
|
college_math.precalculus
|
\approx(-\infty,-12.1414) \cup(12.1414, \infty)
|
college_math.PRECALCULUS
|
exercise.10.2.7
|
Find the exact value of the cosine and sine of the given angle: $\theta=\pi$
|
$\cos (\pi)=-1, \sin (\pi)=0$
|
Creative Commons License
|
college_math.precalculus
|
\cos (\pi)=-1, \sin (\pi)=0
|
college_math.PRECALCULUS
|
exercise.10.2.53
|
Approximate the given value to three decimal places: $\sin \left(\pi^{\circ}\right)$
|
$\sin \left(\pi^{\circ}\right) \approx 0.055$
|
Creative Commons License
|
college_math.precalculus
|
\sin \left(\pi^{\circ}\right) \approx 0.055
|
college_math.PRECALCULUS
|
exercise.8.4.11
|
Use one matrix inverse to solve the following system of linear equations:
$\left\{\begin{aligned} 3 x+7 y & =-7 \\ 5 x+12 y & =5\end{aligned}\right.$
|
$\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{r}-7 \\ 5\end{array}\right]=\left[\begin{array}{r}-119 \\ 50\end{array}\right]$ So $x=-119$ and $y=50$.
|
Creative Commons License
|
college_math.precalculus
|
\left[\begin{array}{rr}12 & -7 \\ -5 & 3\end{array}\right]\left[\begin{array}{r}-7 \\ 5\end{array}\right]=\left[\begin{array}{r}-119 \\ 50\end{array}\right]$ So $x=-119$ and $y=50$.
|
college_math.PRECALCULUS
|
exercise.8.4.5
|
Find the inverse of the matrix or state that the matrix is not invertible: $E=\left[\begin{array}{rrr}3 & 0 & 4 \\ 2 & -1 & 3 \\ -3 & 2 & -5\end{array}\right]$
|
$E^{-1}=\left[\begin{array}{rrr}-1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3\end{array}\right]$
|
Creative Commons License
|
college_math.precalculus
|
E^{-1}=\left[\begin{array}{rrr}-1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3\end{array}\right]
|
college_math.PRECALCULUS
|
exercise.6.2.24
|
Use the properties of logarithms to write the expression as a single logarithm: $3-\log (x)$
|
$\log \left(\frac{1000}{x}\right)$
|
Creative Commons License
|
college_math.precalculus
|
\log \left(\frac{1000}{x}\right)
|
college_math.PRECALCULUS
|
exercise.8.2.23
|
Solve the following system of linear equations: $\left\{\begin{aligned} x+y+z & =4 \\ 2 x-4 y-z & =-1 \\ x-y & =2\end{aligned}\right.$
|
Inconsistent
|
Creative Commons License
|
college_math.precalculus
|
Inconsistent
|
college_math.PRECALCULUS
|
exercise.10.7.20
|
Solve the equation, giving the exact solutions which lie in $[0,2 \pi)$: $\sin (2 x)=\sin (x)$
|
$x=0, \frac{\pi}{3}, \pi, \frac{5 \pi}{3}$
|
Creative Commons License
|
college_math.precalculus
|
x=0, \frac{\pi}{3}, \pi, \frac{5 \pi}{3}
|
college_math.PRECALCULUS
|
exercise.3.4.25
|
Simplify the given power of $i$: $i^{117}$
|
$i^{117}=\left(i^{4}\right)^{29} \cdot i=1 \cdot i=i$
|
Creative Commons License
|
college_math.precalculus
|
i^{117}=\left(i^{4}\right)^{29} \cdot i=1 \cdot i=i
|
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