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3 years ago

Which expression is equivalent to the expression below? Select all that apply. –2/5(15 – 20d + 5c)

1 answer:

7 0

$Use\ distributive\ property:\\\\-\dfrac{2}{5}(15-20d+5c)=\left(-\dfrac{2}{5}\right)(15)+\left(-\dfrac{2}{5}\right)(-20d)+\left(-\dfrac{2}{5}\right)(5c)\\\\=(-2)(3)+(-2)(-4d)+(-2)(c)\\\\=\boxed{-6+8d-2c}$

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An insurance policy with comprehensive coverage is dubbed as insurance coverage for God’s acts which includes theft, vandalism, weather, etc.

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When the car is stolen and is never recovered
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4 years ago

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3 years ago

Read 2 more answers

A page should have perimeter of 42 inches. The printing area within the page

Tamiku [17]

Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .

<u>Step-by-step explanation:</u>

We have , A page should have perimeter of 42 inches. The printing area within the page would be determined by top and bottom margins of 1 inch from each side, and the left and right margins of 1.5 inches from each side. let's assume width of the page be x inches and its length be y inches So,

Perimeter = 42 inches

⇒ $2(x+y) = 42\\x+y = 21\\y = 21-x$

width of printed area = x-3 & length of printed area = y-2:

area = $length(width)$

$area = (x-3)(y-2)\\area = (x-3)(21-x-2)\\area = (x-3)(19-x)\\area = -x^{2} + 22x -57$

Let's find $\frac{d(area)}{dx}$ :

$\frac{d(area)}{dx}$ = $\frac{d(-x^{2}+22x-57)}{dx} = -2x +22$ , for area to be maximum $\frac{d(area)}{dx}$ = 0

⇒ $-2x+22 = 0\\2x =22\\x=11 inches$

And ,

$y = 21-x\\y = 21-11\\y = 10 inches$

∴ Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .

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3 years ago

What is 2 multiplied by 4

pishuonlain [190]

2 × 4 = 8 easy peasy

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3 years ago

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Please help me just wait for me to put the pictures up but this time WAIT

Ganezh [65]

1.A.
The base lengths of the trapezoid are 35cm and 19cm. The height is 15cm.
1.B.
The formula for the area of a trapezoid is
A=(b1+b2)/2*h
plug in the data we know
A=(35+19)/2*15
A=54/2*15
A=27*15
A=405cm²
since there are two trapezoids, we double that number 405*2=810cm²
Answer=810cm²

1.C.
To solve for this, we need to find the perimeter of the trapezoid
P=35+19+17+17=88cm
Answer=No. The edges of the trapezoid are longer than 80cm.

2.A.
3units. Since their x values are the same, you just subtract the y values to calculate distance 5-2=3
Answer=3

2.B
A right triangle. Since one coordinate is directly above, and the other is directly to the right of the bench

3.A
The two triangles would be...
A=1/2b*h=1/2*30*72=15*72=1080ft² each
and the center rectangle would be
A=l*w=20*72=1440ft²
Add them all together and you get
Total area=3600ft²

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3 years ago