(3x-4)(x+5)= I need help with this fast

3x^2+11x-20

Step-by-step explanation:

step 1 3x(x+5)-4(x+5)

step 2 3x(x+5)-4(x+5)= 3x^2+15-4(x+5)

step 3 3x^2+15x-4x+20

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.

i have not got your picture clearly so see it once again