For question 11, you essentially need to find when h(t) = 0, since that is when the height of the ball reaches 0 (ie touches the ground).
For question 12, it is asking for a maximum height, so you need to find when dh/dt = 0 and taking the second derivative to prove that there is maximum at t. That will find you the time at which the ball will hit a maximum height.
Rinse and repeat question 12 for question 13
Answer:
y = x(4+z)+6
y-6 = x(4+z)
x=(y-6)/(4+z)
Step-by-step explanation:
Answer:
sin ∅ = 16 /34
cos ∅ = 30 /34
tan ∅ = 16 /30
Step-by-step explanation:
SOH CAH TOA
sin = opp/hyp
cos = adj/hyp
tan = opp/adj
the hypotenuse is always across from the right angle and has the longest side
Answer:
1.918
Step-by-step explanation:
1 then do 459/500 x 2 to make 918/1000 then add and convert
Answer:
∠A = 107
∠B = 73
Step-by-step explanation:
∠A , ∠B are supplementary. So, ∠A + ∠B = 180
5x + 27 + 5x - 7 = 180
5x + 5x + 27 - 7 = 180 {Combine like terms}
10x + 20=180
10x = 180 -20
10x = 160
x = 160/10
x = 16
∠A = 5x + 27 = 5*16 + 27 = 80 + 27 = 107
∠B = 5x - 7 = 5*16 - 7 = 80 - 7 = 73