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3 years ago

Find the area of the region that is not shaded

Mathematics

2 answers:

AfilCa [17]3 years ago

8 0

A = h(b/2)

So the height is 8 and the base is 5

A = 8(5/2)

A = 20

I hope this helps! :)

~Ayden

IrinaK [193]3 years ago

8 0

The area of what is not shaded should be 20 Square Units. If you want to know how to do that you just multiply the base of the triangle by the height of the triangle then multiply that number by 0.5 and you should get your answer.

The problem:

solve for bh=?

5×8=40

then multiply by .5

40×.5=20

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2 years ago

A school is building a rectangular soccer field that has an area of 6000 square yards. The soccer field must be 40 yards longer

kogti [31]

Area is 6000 yd2

LW = 6000

field must be 40 yards longer than its width

L = W + 40

Replace L with W+40 in 1st equation to solve for Width

(W+40)(W) = 6000

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This is a quadratic but is factorable

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3 years ago

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3 years ago

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3 years ago

What is the perimeter of angle DEF to the nearest tenth of a unit?<br> helppppp<br> tysm

hammer [34]

<h3>Answer: 36.4 units (choice A)</h3>

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

Explanation:

Let's use the distance formula to find the distance from D to E

$D = (x_1,y_1) = (-7,2) \text{ and } E = (x_2, y_2) = (3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-7-3)^2 + (2-4)^2}\\\\d = \sqrt{(-10)^2 + (-2)^2}\\\\d = \sqrt{100 + 4}\\\\d = \sqrt{104}\\\\d = \sqrt{4*26}\\\\d = \sqrt{4}*\sqrt{26}\\\\d = 2\sqrt{26}\\\\d \approx 10.198\\\\$

Note: uppercase D refers to the point, while lowercase d is the distance from D to E.

The length of segment DE is roughly 10.198 units long.

----------------

Repeat for the distance from E to F.

$E = (x_1,y_1) = (3,4) \text{ and } F = (x_2, y_2) = (5,-7)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-5)^2 + (4-(-7))^2}\\\\d = \sqrt{(3-5)^2 + (4+7)^2}\\\\d = \sqrt{(-2)^2 + (11)^2}\\\\d = \sqrt{4 + 121}\\\\d = \sqrt{125}\\\\d = \sqrt{25*5}\\\\d = \sqrt{25}*\sqrt{5}\\\\d = 5\sqrt{5}\\\\d \approx 11.1803\\\\$

Segment EF is roughly 11.1803 units long.

----------------

Repeat for the distance from F to D.

$F = (x_1,y_1) = (5,-7) \text{ and } D = (x_2, y_2) = (-7,2)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(5-(-7))^2 + (-7-2)^2}\\\\d = \sqrt{(5+7)^2 + (-7-2)^2}\\\\d = \sqrt{(12)^2 + (-9)^2}\\\\d = \sqrt{144 + 81}\\\\d = \sqrt{225}\\\\d = 15\\\\$

Unlike the others, this result is exact.

----------------

Add up the three segment lengths to get the perimeter

DE + EF + FD

10.198 + 11.1803 + 15

36.3783

The perimeter is approximately 36.3783 units which rounds to <u>36.4</u>

The answer has been confirmed with GeoGebra.

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2 years ago