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3 years ago

Please help me I need to submit this asap

Mathematics

1 answer:

frez [133]3 years ago

8 0

Answer:

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

Here { a, b ] = [ - 1, 2 ], thus

f(b) = 4 ← from (2, 4 )

f(a) = $\frac{1}{2}$ ← from (- 1, $\frac{1}{2}$ ) , thus

average rate of change = $\frac{4-\frac{1}{2} }{2-(-1)}$ = $\frac{\frac{7}{2} }{3}$ = $\frac{7}{6}$ → D

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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 $\mathsf{AG}$ :

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