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vladimir1956 [14]
2 years ago
15

Estimate the perimeter and the area of the shaded figure to the nearest whole number

Mathematics
1 answer:
ollegr [7]2 years ago
7 0

Answer:

See below ~

Step-by-step explanation:

<u>Perimeter</u>

  • 2(6 + 8)
  • 2(14)
  • <u>28</u> units

<u>Area</u>

  • 2(4 x 2) + 2(8)
  • 4(8)
  • <u>32</u> square units
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b

Step-by-step explanation:

5(0)-3(-4)=12 that simply's to 0+12=12

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A contractor is building a new subdivision on the outside of a city. He has started work on the first street and is planning for
sergeinik [125]

Answer:

x + y = 6

Step-by-step explanation:

Standard Form:

y - y1 = m(x - x1)

In this question, the coordinate is (1, 5) with a slope of -1 (the slope is the same as street 1 since street 2 is going to be parallel.

Insert the coordinate and slope into standard form and you'll get this:

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5 0
4 years ago
Cos^2 x+4sin^2 x/2=1
lana [24]

Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2

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\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}

6 0
3 years ago
Will give brainliest to whoever answers first
BigorU [14]

Answer:

The area of the triangle on left is 9

in², the area of the triangle on the right is 9

in², and the area of the rectangle is 108

in².

The area of the trapezoid is the sum of these areas, which is 126

in².

6 0
3 years ago
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