Consider the equation below and select two equations that are equivalent to the given equation. 5(3a - 1) - 2(3a + 2) = 3(a + 2) + v
a) -a - 10 B) 3(5a + 2) C) 6a - 7 D) 3( 2a - 5) E) 6a - 15 F) 3(a + 2)
G) 6(a - 15) H) -6a - 4

PICK TW PLEAS

D and E

Step-by-step explanation:

Our equation is: 5(3a - 1) - 2(3a + 2) = 3(a + 2) + v. Let's distribute all the parentheses:

5 * 3a + 5 * (-1) + (-2) * 3a + (-2) * 2 = 3 * a + 3 * 2 + v

15a - 5 - 6a - 4 = 3a + 6 + v

Combine like terms (all the constants and all the terms with a in them):

15a - 6a - 3a = 6 + v + 5 + 4

6a = 15 + v

Let's solve for v:

v = 6a - 15

Ah, notice that this already matches choice E, so click that one. Notice that we can also factor out a 3 from both terms:

v = 3 * (6/3a - 15/3) = 3(2a - 5), and this matches choice D.

Our answers are thus D and E.

~ an aesthetics lover

an altitude of 15,000 feet

step-by-step explanation:

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.