25. if CD 1 EF, m_HCE = 40 and mcHCD = 5x, find each missing value.
K
a) =
b) mZECH =
c) mZHCD =
d) nZGCF =
e) ZECG =
1 MZGCD =
D
H

Easy question

x=10

mECH=40

mHCD=50

mGCF=40

mECG=140

mGCD=13

Hence we found the solution.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!