An object is thrown upward at a speed of 152 feet per second by a machine from a height of 9 feet off the ground. the height h of the object after t seconds can be found using the equation h=−16t^2+152t+9
when will the height be 231 feet?
when will the object reach the ground?

First part:

Set h(t) = 231and solve for t.

-16t²+ 152t + 9= 231

-16t² + 152t - 222= 0

Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 231 on the way up, and the higher value is the time to reach 231 again, on the way down.

Second part:

Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12