An oblique rectangular prism with a square base has a volume of 539 cubic units. the edges of the prism measure 7 by 7 by 14 units. how many units longer is the slanted edge length of the prism, 14, compared to its perpendicular height? 1 unit 2 units 3 units 4 units

In the oblique rectangular prism with a base square, the slanted edge length is 3 units longer compared to its perpendicular height.

Answer. Third option: 3 units

Solution:

Volume of the oblique rectangular prism: V=539 cubic units

Square base, with edge a=7 units

Slanted edge length: s=14 units

How many units longer is the slanted edge length of the prism, 14, compared to its perpendicular height?

s-h=?

Perpendicular height of the prism: h=14 units

V=Ab h

Area of the base: Ab

Ab=a^2

Replacing a by 7 units in the formula above:

Ab=(7 units)^2

Ab=49 square units

V= Ab h

Replacing V by 539 cubic units and Ab by 49 square units in the formula above:

539 cubic units = (49 square units) h

Solving for h: Dividing both sides of the equation by 49 square units:

(539 cubic units) / (49 square units) = (49 square units) h / (49 square units)

11 units = h

h= 11 units

s-h=14 units-11 units

s-h=3 units

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! : )

cretdit to source

i’ll just s take the points thxs