A ball is thrown from an initial height of 1 meter with an initial upward velocity of
1 m/s. The ball's height h
(in meters) after t
seconds is given by the following. h=1+30t-5t^2
Find all values of t
for which the ball's height is 11
meters.

Round your answer(s) to the nearest hundredth.

Step-by-step explanation:

If we are looking for the times that the ball was 11 meters off the ground, we sub in 11 for the height on the left and solve for t:

and

and factor this however it is you are factoring in class to solve for t to get

t = .35 seconds and t = 5.6 seconds

Because the ball reaches this point in its way up and then passes it again on its way down, the ball will have 2 times at this height.

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.

180°

step-by-step explanation:

sum of equilateral triangle=180°

< n=60°

< m=120°°[angle supplementary of equilateral angle]

< m+< r+< x=360°(sum of exterior angle of triangle)