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.
Factor the equation so...
(r^2-pr) and (p^2q-pqr)
Factor out (r^2-pr) = r(r-p)
Factor out (p^2q-pqr) = pq(p-r)
Add a negative to r(r-p) to make it -r(p-r)
(pq-r)(p-r) is the answer... I'm sorry I can't explain things well, but I tried.
If A and B are supplementary angles, then:
A+B=180º
In this case:
A=44º
44º+B=180º
B=180º-44º=136º
Anser: the measure of the other angle is 136º
Misha would combine 3x-2x and-8-7
We need to convert 300 feet to yards.
We know,
3 feet = 1 yard
Hence,
300 feet * 1 yard/3 feet = 300/3 = 100 yards
He should buy 100 yards
