Length: 2w + 59
width: w
diagonal: (2w + 59) + 2 = 2w + 61
Length² + width² = diagonal²
(2w + 59)² + (w)² = (2w + 61)²
(4w² + 118w + 3481) + w² = 4w² + 122w + 3721
5w² + 118w + 3481 = 4w² + 122w + 3721
w² + 118w + 3481 = 122w + 3721
w² - 4w + 3481 = 3721
w² - 4w - 240 = 0
a = 1, b = -4, c = -240
w = ![[-(b) +/- \sqrt{(b)^{2} - 4(a)(c) }]/2(a)](https://tex.z-dn.net/?f=%5B-%28b%29%20%2B%2F-%20%5Csqrt%7B%28b%29%5E%7B2%7D%20%20-%204%28a%29%28c%29%20%7D%5D%2F2%28a%29)
= ![[-(-4) +/- \sqrt{(-4)^{2} - 4(1)(-240) }]/2(1)](https://tex.z-dn.net/?f=%5B-%28-4%29%20%2B%2F-%20%5Csqrt%7B%28-4%29%5E%7B2%7D%20%20-%204%281%29%28-240%29%20%7D%5D%2F2%281%29)
= ![[4 +/- \sqrt{(16 + 960) }]/2](https://tex.z-dn.net/?f=%5B4%20%2B%2F-%20%5Csqrt%7B%2816%20%20%2B%20960%29%20%7D%5D%2F2)
= ![[4 +/- \sqrt{(976) }]/2](https://tex.z-dn.net/?f=%5B4%20%2B%2F-%20%5Csqrt%7B%28976%29%20%7D%5D%2F2)
= ![[2 +/- 4\sqrt{(61) }]/2](https://tex.z-dn.net/?f=%5B2%20%2B%2F-%204%5Csqrt%7B%2861%29%20%7D%5D%2F2)
= 
since width cannot be negative, disregard 1 - 2√61
w = 1 + 2√61 ≈ 16.62
Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24
Answer: width = 16.62 in, length = 92.24 in
I think it’s C. Sorry if I’m incorrect
Answer:
The correct option is;
a. F is the midpoint of 
because line 
bisects
Step-by-step explanation:
Here, since we have that the triangle is reflected across EG therefore the location of the point F which is along EG bisects the line as the dimensions of the line from A to F must be equal to the dimension of the line that extends from A' to F
Therefore the point F is the midpoint of because line bisects .
Answer:
P = 2x^3 - 4x^2 - 16x
Step-by-step explanation:
Only 1 step: divide both sides by x
D)4 is the answer jcidifodofif
