What is the area of the two-dimensional cross section that is parallel to face abc ?

 The required area is 84 square feet.

Step-by-step explanation:  We are given to find the area of the two-dimensional cross section that is parallel to face ABC.

Since ABC is a right-angled triangle at ∠B = 90°, so its parallel face will also be a right-angled triangle.

And the area of the parallel face must also be equal to the area of ΔABC.

Applying Pythagoras theorem, we have from ΔABC that

So, area of triangle ABC is

Thus, the area of the parallel face is 84 sq. ft.

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

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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