What is the only solution of 2x^2+8x=x^2-16

Move all terms to one side

2x^2 + 8x - x^2 + 16 = 0

Simplify 2x^2 + 8x - x^2 + 16 to x^2 + 8x + 16

x^2 + 8x + 16 = 0

Rewrite x^2 + 8x + 16 in the form a^2 + 2ab + b^2, where a = x and b = 4

x^2 + 2(x)(4) + 4^2 = 0

Use the Square of Sum: (a + b) = a^2 + 2ab + b^2

(x + 4)^2 = 0

Take the square root of both sides

x + 4 = 0

Subtract 4 from both sides

x = -4

  q

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.

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.