Question

1 point
Find mzJKL.
(3x + 15) (2x + 10)°
J
K
M

Answer 1

∠JKL = 108°

Explanation:

1.) Add 3x + 15 and 2x + 10 for 180

→ (3x + 15) + (2x + 10) = 180

2.) Add common factors

→ 5x + 25 = 180

3.) Solve for x

→ x = 31

4.) Replace x with 31

→ 3(31) + 15 = 108

∠JKL = 108°

TO CHECK

1.) Replace x with 31

→ 2(31) + 10

2.) Add 108 and 72 to get 180

→ 108 + 72 = 180

Answer 2

m<jkl =0

Explanation:

step 1: keep it ir mind that angle jkl and angle lkm are equal

step 2: since they are equal thus:

m<jkl= m<lkm

3x + 15= 2x + 10

3x-2x= 10-15

x = -5

Step 3: substitute x= -5 in m<jkl

3(-5)+15

-15+15 = 0

therefore your answer for both anhle should be 0