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Mrac [35]
3 years ago
7

The length of a rectangle is 3 inches longer then twice its width, where w is the width of the rectangle the area of the rectang

le is 90 square inches. Write an equation that represents the area and find the width and length of the rectangle.
Mathematics
1 answer:
vaieri [72.5K]3 years ago
6 0
A = w(2w + 3)
90 = 2w^2 + 3w
2w^2 +3w - 90 = 0
(w-6)(2w+15) = 0        (TRINOMIAL FACTORING)
w = 6 inch                   ( it can't be -15/2 because lengths can't be negative)
l = 2w + 3
  = 15 inch
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Trapezoid ABCD is graphed in a coordinate plane.
Ann [662]

Answer:

12

Step-by-step explanation:

he coordinates of the vertices of the trapezoid ABCD are:

A=(-5,-2)=(xa,ya)→xa=-5, ya=-2

B=(-1,2)=(xb,yb)→xb=-1, yb=2

C=(0,-1)=(xc,yc)→xc=0, yc=-1

D=(-2,-3)=(xd,yd)→xd=-2, yd=-3

yi x(i+1)        xi     yi     xi y(i+1)

                 -5    -2

(-2)(-1)=2     -1      2     (-5)(2)=-10

(2)(0)=0       0     -1     (-1)(-1)=1

(-1)(-2)=2    -2    -3     (0)(-3)=0

(-3)(-5)=15   -5    -2     (-2)(-2)=4

S1=-10+1+0+4→S1=-5

S2=2+0+2+15→S2=19

Area: A=(1/2) Absolute value (S1-S2)

A=(1/2) Absolute value (-5-19)

A=(1/2) Absolute value (-24)

A=(1/2) (24)

A=24/2

A=12

The area of the trapezoid is 12 square units

8 0
3 years ago
If f(x)= <br><img src="https://tex.z-dn.net/?f=x%20-%20%7Bx%7D%5E%7B2%7D%20" id="TexFormula1" title="x - {x}^{2} " alt="x - {x}^
astra-53 [7]

Answer:

f(2+h)=-(h+2/3)^2+1/4

f(x+h)=-(x+h-1/2)^2+1/4

Step-by-step explanation:

1. f(2+h)=(2+h)-(2+h)^2=2+h-4-4h-h^2=-h^2-3h-2=-(h^2+3h+2)

=-(h+2/3)^2+1/4

2. Let (x+h)=a, then rewrite the equation into f(a)=a-a^2.

a-a^2=-(a^2-a)=-[(a-1/2)^2-1/4]=-(a-1/2)^2+1/4.

Insert a=x+h, f(x+h)=-(x+h-1/2)^2+1/4

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