The expression (y^20) (y^-5)^2 is equivalent to y^n. What is the value of n? n = 27 n = 10 n = -200 n = 17

Carolina is mowing lawns for a summer job. For every mowing job, she charges an initial fee plus \$6$6 for each hour of work. He

GenaCL600 [577]

Answer:

y=8+6t

Step-by-step explanation:

4 0

3 years ago

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For f(x)=0.01(2)power of x, find the average rate of change from x=3 to x=8

sineoko [7]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= 0.01(2)^x \qquad 
\begin{cases}
x_1=3\\
x_2=8
\end{cases}\implies \cfrac{f(8)-f(3)}{8-3}
\\\\\\
\cfrac{0.01(2)^8-0.01(2)^3}{8-3}$

4 0

3 years ago

What is the measurement of the longest line segment in a right rectangular prism that is 26 inches long, 2 inches wide, and 2 in

EastWind [94]

Answer:

$6\sqrt{19} \approx 26.153$ inches.

Step-by-step explanation:

The longest line segment in a right rectangular prism is the diagonal that connects two opposite vertices. On the first diagram attached, the green line segment connecting A and G is one such diagonals. The goal is to find the length of segment $\mathsf{AG}$ .

In this diagram (not to scale,) $\mathsf{AB} = 26$ (length of prism,) $\mathsf{AC} = 2$ (width of prism,) $\mathsf{AE} = 2$ (height of prism.)

Pythagorean Theorem can help find the length of $\mathsf{AG}$ , one of the longest line segments in this prism. However, note that this theorem is intended for right triangles in 2D, not the diagonal in a 3D prism. The workaround is to simply apply this theorem on two different right triangles.

Start by finding the length of line segment $\mathsf{AD}$ . That's the black dotted line in the diagram. In right triangle $\triangle\mathsf{ABD}$ (second diagram,)

Segment $\mathsf{AD}$ is the hypotenuse.
One of the legs of $\triangle\mathsf{ABD}$ is $\mathsf{AB}$ . The length of $\mathsf{AB}$ is $26$ , same as the length of this prism.
Segment $\mathsf{BD}$ is the other leg of this triangle. The length of $\mathsf{BD}$ is $2$ , same as the width of this prism.

Apply the Pythagorean Theorem to right triangle $\triangle\mathsf{ABD}$ to find the length of $\mathsf{AB}$ , the hypotenuse of this triangle:

$\mathsf{AD} = \sqrt{\mathsf{AB}^2 + \mathsf{BD}^2} = \sqrt{26^2 + 2^2}$ .

Consider right triangle $\triangle \mathsf{ADG}$ (third diagram.) In this triangle,

Segment $\mathsf{AG}$ is the hypotenuse, while
$\mathsf{AD}$ and $\mathsf{DG}$ are the two legs.

$\mathsf{AD} = \sqrt{26^2 + 2^2}$ . The length of segment $\mathsf{DG}$ is the same as the height of the rectangular prism, $2$ (inches.) Apply the Pythagorean Theorem to right triangle $\triangle \mathsf{ADG}$ to find the length of the hypotenuse $\mathsf{AG}$ :

$\begin{aligned}\mathsf{AG} &= \sqrt{\mathsf{AD}^2 + \mathsf{GD}^2} \\ &= \sqrt{\left(\sqrt{26^2 + 2^2}\right)^2 + 2^2}\\ &= \sqrt{\left(26^2 + 2^2\right) + 2^2} \\&= 6\sqrt{19} \\&\approx 26.153\end{aligned}$ .