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Answer: y^4 - 2y^3 + 7y^2 + y - 5</h3>
Work Shown:
h(y) = f(y) + g(y)
h(y) = (y^4 - 3y^3 + y - 3) + (y^3 + 7y^2 - 2)
h(y) = y^4 - 3y^3 + y - 3 + y^3 + 7y^2 - 2
h(y) = y^4 + (-3y^3+y^3) + 7y^2 + y + (-3 - 2)
h(y) = y^4 - 2y^3 + 7y^2 + y - 5
This is a fourth degree polynomial (aka quartic).
<span>If this is an isosceles triangle, then it has two 45 degree angles corresponding to two legs of equal length. Orient the base of this triangle so that it's horizontal, and represent its length by b. Let h represent the height of the triangle. Then the area of this right triangle is 50 square inches = (1/2)(b)(h), or A = (b/2)h = 50 in^2.
Due to the 45 degree angles, the height of this triangle is equal to half the base, or h = b/2. Thus, (b/2)h = 50 becomes (b/2)(b/2) = 50, or b^2=200. Thus, b = 10sqrt(2), and h=(1/2)(10 sqrt(2)), or h = 5sqrt(2).
The length of one of the legs is the sqrt of [5sqrt(2)]^2+[5sqrt(2)]^2, or
sqrt(25(2)+25(2)) = sqrt(100) = 10.
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