The equation ℎ = −162 + 32 + 9 gives the height of a ball, h, in feet above the ground, at t seconds after the ball is thrown upward. How many seconds after the ball is thrown will it reach its maximum height? What is the max height?

time = 1secs

Max height = 25metres

Step-by-step explanation:

Let the equation that models the height of the ball be expressed as;

h(t) = −16t^2 + 32t + 9

t is the time in seconds

The velocity of the ball at its maximum height is zero i.e dh/dt = 0

dh/dt = -32t + 32

0 = -32t + 32

32t = 32

t = 32/32

t = 1 seconds

The ball will reach its maximum height after 1 seconds.

Get the maximum height reached. Substitute t = 1 into the modeled function;

h(t) = −16t^2 + 32t + 9

h(1) = −16(1)^2 + 32(1) + 9

h(1) = −16 + 32 + 9  

h(1)= 16+9

h(1) = 25

Hence the maximum height reached is 25metres

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632