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3 years ago

Write a simplified polynomial expression in standard form to represent the area of the rectangle below.

1 answer:

4 0

Formula for area of rectangle:
A= Length x width
Put values
A=(x-4)(5x+2)
Expand
A=5x²+2x-20x-8
Add like terms
A=5x²-18x-8

Answer: Option C is correct.

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If there were 300 accidents in 1967 and 380 in 1968, what is the percent increase?

VikaD [51]

First find the amount of increase: 380-300=80. Then divide the amount of increase by the original amount: 80/300=.266666667. Normally you round to the tenths place: .27. Then make it a percent: .27=27%. The percent of increase is 27% :) hope this helps

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What does x/4=3/2 equal?

MrRissso [65]

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3 years ago

Use the picture to find each of the following

goldfiish [28.3K]

$\large{ \tt{☄ANSWERS \: + \: REASONS}} :$

$\large{ \tt{ \angle \: 7 = \boxed{ \tt{38 \degree}}}} \: ( \text{Being \: alternate \: angles)}$

$\large{ \tt{38 \degree + \angle 4 = 180 \degree}}( \text{Sum \: of \: co - interior \: angles}$

$\: \: \: \: \: \large{ \tt{ ↦\angle \: 4 = 180 \degree - 38 \degree = \boxed{ \tt{142 \degree}}}}$

$\large{ \tt{ \angle \: 1 = \angle \: 4 = \boxed{ \tt{142 \degree}( \text{Being \: vertical \: angles)}}}}$

$\large{ \tt{ \angle \: 8 = \angle \: 4 = \boxed{ \tt{142 \degree} \text{(Being \: corresponding \: angles)}}}}$

$\large{ \tt{ \angle \: 6 = \angle \: 1 = \boxed{ \tt{142 \degree} \text{(Being \:corresponding \: angles) }}}}$

$\large {\tt{ \angle \: 2 = \boxed{\tt{38 \degree}}\text{(Being \: corresponding \: angles)}}}$

$\large{ \tt{ \angle \: 3 = \boxed{\tt{38 \degree} }\text{(Being \: alternate \: angles)}}}$

$\large{ \tt{✺ \: TIPS} }:$

If you have planned for study at 8:00 and now it's 8:05 then don't wait for 9 : 00 , Start studying that subject right now ! ♪

$\large{ \tt{❃ \: LEARN \: ABOUT \: ALTERNATE \: , CORRESPONDING\:AND \: CO - INTERIOR \: ANGLES}} :$

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8 0

3 years ago

Evaluating integral by net area

N76 [4]

The graph of $f(x)=2$ is a horizontal line. So the area under $f(x)$ on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.

The graph of $g(x)=\sqrt{64-x^2}$ is the upper half of a circle with radius $\sqrt{64}=8$ . It's symmetric about $x=0$ , so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is $\frac{\pi 8^2}4=16\pi$ .

Then use the fact that the definite integral is distributive over sums, meaning

$\displaystyle\int_0^8f(x)+g(x)\,\mathrm dx=\int_0^8f(x)\,\mathrm dx+\int_0^8g(x)\,\mathrm dx=16+16\pi$

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3 years ago