Summer looked at the formula for the pythagorean theorem, a2 + b2 = c2, she said that it didn’t matter which sides of a right triangle were a, b, and c. is she correct? a. yes, she is correct.
b. no, a must be the shortest side, then b, and c is the longest side.
c. no, c must be the shortest side, then b, and a is the longest side.
d. no, a and b are interchangable, and c is the longest side.

D. No, a and b are interchangable, and c is the longest side.

Step-by-step explanation:

Given that Summer looked at the formula for the Pythagorean Theorem,

, she said that it didn’t matter which sides of a right triangle were a, b, and c

She is wrong because c should be the hypotenuse in this case. Hypotenuse being the largest side of a right triangle has the square as the sum of squares of other legs.

But a and b the legs can be interchanged here, but a, c or a, b cannot be interchanged.

Hence we have out of 4 options only one option is right

Option D . No, a and b are interchangable, and c is the longest side.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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