Simplify.
5(x+2) + 2(x-5)

Α.7x-3
B. 7x
C 7x+ 20
D.0
Ε. 2x +4

B=7x

Step-by-step explanation:

5(x+2)+2(x-5)

5x+10+2x-10

since 10-10=0, it just won't show, there's no need to say "0"

5x+2x+10-10

7x+0

so just

7x

7x

Step-by-step explanation:

5 times x 5 times 2 = 5x + 10

2 times x 2 times -5 = 2x -10

5x +2x = 7x

10- 10= 0

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.