The area of a rectangle is given by A(x) = x2 + 5x with x being the height and x + 5 being the base. If the area is 14, what is the only viable solution for the height? Why are there not two solutions?

See below.

Step-by-step explanation:

A(x) = x^2 + 5x

A(x) = x^2 + 5x = 14

x^2 + 5x - 14 = 0

(x + 7)(x - 2) = 0

x + 7 = 0  or  x - 2 = 7

x = -7 or x = 2

x is the height of a rectangle.

The only viable solution is x = 2 because the height of a rectangle cannot be a negative number such as -7.

hi friend

here is your answer

there are two angles that are alternate interior angle with angle 9

the angle 15 and angle 7 are alternate interior angle with angle 9.

hope it is for you so mark my answer as a brainliest

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.