Find the slope of the line through the points \((-2, 4)\) and \((5, 7)\text{.}\)
Apply the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\text{.}\)
\begin{align*} m \amp= \frac{7 - 4}{5 - (-2)}\\ \amp= \frac{3}{7} \end{align*}The slope is \(\frac{3}{7}\text{.}\)
Let \(f(x) = x^2 - 5\text{.}\) Find and simplify \(f(t + 3)\text{.}\)
Substitute \(t + 3\) everywhere \(x\) appears, then expand.
\begin{align*} f(t+3) \amp= (t+3)^2 - 5\\ \amp= t^2 + 6t + 9 - 5\\ \amp= t^2 + 6t + 4 \end{align*}
A plumber charges a flat trip fee plus an hourly rate. A one-hour visit costs $75 and a four-hour visit costs $195. What is the hourly rate?
The hourly rate is the slope of the cost line through \((1, 75)\) and \((4, 195)\text{.}\)
\begin{align*} m \amp= \frac{195 - 75}{4 - 1}\\ \amp= \frac{120}{3} = 40 \end{align*}The plumber charges $40 per hour.
Rewrite \(\log_2\left(\frac{1}{8}\right) = -3\) in exponential form.
The statement \(\log_b a = c\) means \(b^c = a\text{.}\) Here the base is \(2\text{,}\) the exponent is \(-3\text{,}\) and the result is \(\frac{1}{8}\text{.}\)
\begin{gather*} 2^{-3} = \frac{1}{8} \end{gather*}
A bacteria population starts at \(500\) and doubles every \(3\) hours. How many bacteria are there after \(12\) hours?
In \(12\) hours the population doubles \(12 \div 3 = 4\) times.
\begin{align*} P \amp= 500 \cdot 2^{4}\\ \amp= 500 \cdot 16\\ \amp= 8000 \end{align*}There are \(8000\) bacteria.
Find the vertex of \(f(x) = -2x^2 + 12x - 7\text{.}\)
The vertex occurs at \(x = -\frac{b}{2a}\text{,}\) with \(a = -2\) and \(b = 12\text{.}\)
\begin{gather*} x = -\frac{12}{2(-2)} = 3 \end{gather*}Evaluate \(f\) at that input.
\begin{align*} f(3) \amp= -2(9) + 12(3) - 7\\ \amp= -18 + 36 - 7 = 11 \end{align*}The vertex is \((3, 11)\text{.}\)
In a right triangle, one acute angle measures \(30^\circ\) and the hypotenuse has length \(12\text{.}\) Find the length of the side opposite the \(30^\circ\) angle.
Sine relates the opposite side to the hypotenuse.
\begin{align*} \sin(30^\circ) \amp= \frac{\text{opposite}}{12}\\ \text{opposite} \amp= 12\sin(30^\circ)\\ \amp= 12 \cdot \frac{1}{2} = 6 \end{align*}
Solve \(5e^x = 40\) for \(x\text{.}\) Give the answer in terms of a logarithm.
Isolate the exponential, then take the natural logarithm of both sides.
\begin{align*} e^x \amp= 8\\ \ln(e^x) \amp= \ln(8)\\ x \amp= \ln(8) \end{align*}
Solve \(\cos(t) = -\frac{\sqrt{3}}{2}\) on the interval \([0, 2\pi]\text{.}\)
The reference angle satisfying \(\cos = \frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\text{.}\) Cosine is negative in quadrants II and III.
\begin{align*} t \amp= \pi - \frac{\pi}{6} = \frac{5\pi}{6}\\ t \amp= \pi + \frac{\pi}{6} = \frac{7\pi}{6} \end{align*}
Find the zeros of \(f(x) = x^2 - 5x + 6\text{.}\)
Factor, then set each factor equal to zero.
\begin{align*} x^2 - 5x + 6 \amp= (x - 2)(x - 3) = 0 \end{align*}The zeros are \(x = 2\) and \(x = 3\text{.}\)
Find the \(x\)-intercept and the \(y\)-intercept of \(3x + 5y = 15\text{.}\)
For the \(x\)-intercept, set \(y = 0\text{.}\)
\begin{gather*} 3x = 15 \quad \Rightarrow \quad x = 5 \end{gather*}For the \(y\)-intercept, set \(x = 0\text{.}\)
\begin{gather*} 5y = 15 \quad \Rightarrow \quad y = 3 \end{gather*}The intercepts are \((5, 0)\) and \((0, 3)\text{.}\)