Answer:
8) x = 27
Angle P = x - 12 (15)
Angle Q = 5x - 27 (108)
Angle R = 2x + 3 (57)
Equation: (x - 12) + (5x - 27) + (2x + 3) = 180
x = 27
9) x = 14
Angle A = 7x - 6 (92)
Angle B = 3x - 5 (37)
Angle C = 4x - 5 (51)
Equation: (7x - 6) + (3x - 5) + (4x - 5) = 180
x = 14
10) x = 20
Angle K = x (20)
Angle L = x + 13 (33)
Angle J = 6x + 7 (127)
Equation: x + (x + 13) + (6x + 7) = 180
x = 20
11) x = 41
Angle Y = x (41)
Angle W = 2x - 5 (77)
Angle X = x + 21 (62)
Equation: x + (2x - 5) + (x + 21) = 180
x = 41
Answer:
7.6 seconds
Step-by-step explanation:
To solve this problem we can use the following equation:
S = So + 113*t + at^2/2
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.
In this problem, we have that S = 0, So = 65, Vo = 113 and a = -32 ft/s2 (The acceleration of gravity)
So we have that:
0 = 65 + 113t - 16t2
16t2 - 113t - 65 = 0
Using Bhaskara's formula, we have:
Delta = 113^2 + 4*65*16 = 16929
sqrt(Delta) = 130.11
t = (113 + 130.11) / (2*16) = 7.597 seconds
Rounding to nearest tenth, we have t = 7.6 seconds
just divide 20 by coefficient of x
x < 20/2
x < 10
