Supplementary angles , when added, = 180
complimentary angles, when added, = 90
< AQC + < GQC = 180.....supplementary
< BQD + < DQE = 90.......complimentary
< CQE + < EQF = 90.......complimentary
< GQF , < FQE.....neither
< BQC + < DQC = 90....complimentary
< W and < X are supplementary...
if < W = 37, then < X = (180 - 37) = 143
< S and < T are complimentary
if < S = 64, then < T = (90 - 64) = 26
< C and < D are supplementary
if < C = 83, the < D = (180 - 83) = 97
cant read all of the last one.....but if they are complimentary, and
< U = 41, then the other angle is : (90 - 41) = 49
Answer:
f(x) = 16x2 + 32x - 9
To find the "zeros", set f(x) = 0:
0 = 16x2 + 32x - 9
The equation is solvable by factoring or by using the quadratic formula. The factors (use the "ac" method) are:
0 = (4x-1)(4x+9)
x = 1/4 = 0.25 and x = -9/4 = -2.25
Step-by-step explanation:
Answer:
Step-by-step explanation:
The sum of angle measures is ...
(7x +13)° +(83 -2x)° = 141°
5x +96 = 141 . . . . . collect terms, divide by °
5x = 45 . . . . . . . . . subtract 96
x = 9 . . . . . . . . . . . divide by 5
7x +13 = 7·9 +13 = 76 . . . . . find m∠J
m∠J = 76°
m∠K = 141° -76°
m∠K = 65°
