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3 years ago

What is the area, in square feet, of the shaded part of the rectangle below?

Mathematics

1 answer:

likoan [24]3 years ago

8 0

<h3>Answer: 208</h3>

===============================================

Work Shown:

A = area of the rectangle

A = length*width

A = 16*17

A = 272 sq ft

B = area of triangle

B = base*height/2

B = 16*8/2 ..... note that 17-9 = 8

B = 16*4

B = 64 sq ft

C = area of shaded trapezoid

C = A - B

C = 272 - 64

C = 208 sq ft

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Alternative Method:

Apply the trapezoid area formula

b1 = 17 and b2 = 9 are the parallel bases

h = 16 is the height perpendicular to either base

A = h*(b1+b2)/2

A = 16*(17+9)/2

A = 16*(26)/2

A = 16*13

A = 208 sq ft

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Which of the following is the reciprocal of x2 – 2x – 8?

Ann [662]

Hi again

x² - 2x - 8
Factor
= (x + 2)(x -4)
Answer : A is the correct option.

If you wanna check if A is the right answer, I will solve it for you and it will you give you the exact answer which will be x² - 2x - 8
(x+2)(x-4)
Distribute
(x)(x) + (x)(-4)+(2)(x)+(2)(-4)
= x² - 4x + 2x - 8
= x² - 2x - 8
Here it is, we find the exam answer.

Thus, (x+2)(x-4) is the reciprocal of x² -2x - 8

I hope that's help !

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3 years ago

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adell [148]

1) 43

2) 21

step by step:

Both triangles are right triangle which means they're equal to 90.

1) 23+24+X= 90

47+x= 90

x= 90-47

X= 43

2) 30+39+X= 90

69+X= 90

X= 90-69

X= 21

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3 years ago

How do you complete problem number 4?

brilliants [131]

(I'm going to translate y' to dy/dx as it makes it easier to read for me, you could change it back if you wanted.)
$x \frac{dy}{dx} -y=0$
$\frac{dy}{dx} = \frac{y}{x}$
$\int \frac{1}{y} dy= \int \frac{1}{x} dx$ (separate the variables)
$\ln y=\ln x +c$
$\ln y = \ln kx$ (by letting c = ln k and using log laws)
$y=kx$ (raise everything to power e)
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3 years ago

Need help <br> Correct answers only.

disa [49]

The diameter of the circle is the same as the side length of the bounding square. The radius of the circle is 8 in, so the side length of the square is 16 in.

Then the area between the square and circle is obtained by subtracting the area of the circle from the area of the square:

$A_{\rm circle}=\pi(8\,\mathrm{in})^2=64\pi\,\mathrm{in}^2$