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2 years ago

Correct answers only please!

Mathematics

2 answers:

Mariulka [41]2 years ago

8 0

I got 71%, by finding out the individual percents and adding them up. Don’t know if that’s how you solve it, but it’s an answer choice, so makes enough sense to me.

Travka [436]2 years ago

5 0

Answer:

1st Question

A. 85

2nd Question

B. 71

Step-by-step explanation:

1st Question

{90 + 94 + 96 + 90} = 370/4 = 92.5

{95 + 99} = 194/2 = 97

{87 + 75 + 82 + 85} = 329/4 = 82.25

{71} = 71/1 = 71

.20(92.5) +.25(97) +.30(82.25) +.25(71) = 85.175 Rounded-up = 85

____________________________________________________

2nd Question

.10(95) +.20(80) +.50(67) +.20(60) = 71

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Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence

Citrus2011 [14]

Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn,so he needs no fence on that side.

Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).

one side of the length is not counted for perimeter because one side of length will be against the barn.

Perimeter = 400 ft

Perimeter of rectangle = L + W + W

400 = L + 2W

L = 400 - 2W

Area = L * W

Replace L by 400 - 2W

A(W) = (400 - 2W) * W

$A(W) = -2W^2 + 400W$

Now we find out x coordinate of vertex to find the width that maximize the area

$W = \frac{-b}{2a}$

a= -2 and b = 400

$W = \frac{-(400)}{2(-2)}=100$

The width w would maximize the area is w = 100ft

To find maximum area we plug in 100 for W in A(W)

$A(W) = -2W^2 + 400W$

$A(W) = -2(100)^2 + 400(100)= 20000$

the maximum area is 20,000 square feet

6 0

3 years ago

The area of the triangle is eqaula to the sqare show that x²-3x-2=0

agasfer [191]

Answer:

Step-by-step explanation:

Area of square = area of triangle

$side*side = \dfrac{1}{2}b*h\\\\\\x*x=\dfrac{1}{2}*(x+1)(x+2)\\x^{2}=\dfrac{1}{2}[x*x +x*2+1*x+1*2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+2x+x+2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+3x+2]\\\\\\Multiply \ both \ sides \ by \ 2\\\\2x^{2}=x^{2}+3x+2\\\\2x^{2}-x^{2}-3x-2 = 0$