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This is an area problem. The key words are 120 square feet and 12 feet longer.
And of course width is a key word when you are reading this.
Formula
Area = L * W
Givens
W = W
L = W + 12
Substitute and Solve
Area = L* W
120 = W*(W + 12)
W^2 + 12W = 120 square feet
w^2 + 12w - 120 = 0
This does not factor easily. I would have thought that a graph might help but not if the dimension has to be to the nearest 1/100 of a foot. The only thing we can do is use the quadratic formula.
a = 1
b = 12
c = - 120
w = [ -b +/- sqrt(b^2 - 4ac) ]/(2a)
w = [-12 +/- sqrt(12^2 - 4*(1)(-120)] / 2*1
w = [-12 +/- sqrt(144 - (-480)]/2
w = [-12 +/- sqrt(624)] / 2
w = [- 12 +/- 24.979992] / 2 The minus root has no meaning whatever.
w = (12.979992) / 2
w = 6.489995 I'll round all this when I get done
L = w + 12
L = 6.489995 + 12
L = 18.489995
check
Area = L * W
Area = 6.489995*18.489995
Area = 119.999935 The difference is a rounding error
Answer
L = 18.489995 = 18.49 feet
W = 6.489995 = 6.49 feet
Note: in the check if you round first to the answer, LW = 120.0001 when you find the area for the check. Kind of strange how that nearest 1/100th makes a difference.
6(x + y) + (x - y)
<em>use distributive property</em>
= 6x + 6y + x - y
<em>combine like terms</em>
= (6x + x) + (6y - y)
<h3>= 7x + 6y</h3>
Hello,
The formula for finding the area of a circular region is:
then:
With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.
Now we can find its area:
![A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)](https://tex.z-dn.net/?f=A_%7B2%7D%3D2%2A%20%5Cfrac%7Bb%2Ah%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20A_%7B2%7D%3D%20%5Br%2Asen%2840%29%5D%5Br%2Acos%2840%29%5D%5C%5C%20%20%5C%5CA_%7B2%7D%3D%20r%5E%7B2%7D%2Asen%2840%29%2Acos%2840%29)
The subtraction of the two areas is 100cm^2, then:
Answer: r= 1.59cm
Answer is: Radius = 8.6 cm
In a pack of 52 cards, there are 3 square numbers - 1, 4, and 9.
1 × 1 = 1
2 × 2 = 4
3 × 3 = 9
