The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.
We know there will be a constant downward acceleration of 32.15 ft/s^2.
Use the following equation:
v_final^2 = v_initial^2 + 2ah
v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.
Answer:
x = 59,53445508
Tan(x) = 34/20
Step-by-step explanation:
Step-by-step explanation:
First, replace f(x) with y . ...
Replace every x with a y and replace every y with an x .
Solve the equation from Step 2 for y . ...
Replace y with f−1(x) f − 1 ( x ) . ...
Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
Answer:
6.384 rounded to one decimal place is 6.4
Step-by-step explanation:
