Answer:
option C
Step-by-step explanation:
Answer is A
Step-by-step explanation:
hope that helps
Force 1, F1 = 90, angle 30°
Force 2, F2 = 50, angle 160°
F1 = 90 cos(30) i + 90 sin (30) j
F2 = 50 cos (160) i + 50 sin (160) j
F1 = 90*0.866 i + 90*0.5 j
F2 = 50*(- 0.940) i + 50*0.342 j
F1 = 77.94 i + 45 j
F2 = -47 i +17.10 j
Resultant force, Fr = F1 + F2
Fr = [77.94 - 47.00] i +[45 +17.10]j = 30.94 i + 62.10 j
Magnitude = √[30.94 ^2 + 62.10^2] = 69.38 pounds
Direction = arctan[62.10/30.94] = 63.52 °
Answer:
28
Step-by-step explanation:
AB + BC = AC
14 + 3x-4 = 4x+4
Combine like terms
10 +3x = 4x+4
Subtract 3x from each side
10+3x-3x = 4x+4-3x
10 = x+4
Subtract 4 from each side
10-4 =x-4+4
6 =x
We want AC
AC = 4x+4 = 4*6+4 = 24+4 = 28
Answer:
A) g(t) = - 16*t² + 80*t
B) domain is ( 0 ≤ t ≤ 5 )
Step-by-step explanation:
A) for the model rocket
h = 0 and v = 80 feet/sec
Then the equation g(t) = -16*t² + v*t + h
became
g(t) = - 16*t² + 80*t + 0 g(t) = - 16*t² + 80*t
B) The equation g(t) = - 16*t² + 80*t is defined for all real numbers then the domain interval for t is ( -∞ , ∞ ). Now in our case, a model rocket was launched is not possible to get negative values for g(t) then the biggest value for t = 5 from
-16*t² + 80*t ≥ 0
-16*t ≥ -80 16*t ≤ 80
t ≤ 5
And the domain is ( 0 ≤ t ≤ 5 )
