Aball is shot out of a cannon at ground level, we know that its height h in feet after t sec given by the function h(t) 144t-16t com is a. find h(3), h(6), h(4), and h(5). why are some of the outputs equal? h(3) feet

H(3) = 288 feet

H(4) = 320 feet

H(5) = 320 feet

H(6) = 288 feet

Step-by-step explanation:

A ball is shot out of a cannon at ground level so the ball will follow a parabolic path.

Since height H and time t of the ball have been described by a function H(t) = 144t - 16t²

Then we have to find the values of H(3), H(4), H(5) and H(6).

H(3) = 144×3 - 16(3)²

       = 432 - 144

       = 288 feet

H(4) = 144×4 - 16(4)²

       = 576 - 256

       = 320 feet

H(5) = 144×5 - 16(5)²

       = 720 - 400

       = 320 feet

H(6) = 144×6 - 16(6)²

       = 864 - 576

       = 288 feet

Here we are getting the value like H(3), H(6) and H(4), H(5) are same because in a parabolic path ball first increase in the height above the ground then after the maximum height it decreases.

Therefore, after t = 3 and t = 6 heights of the canon ball are same. Similarly after t = 4 and t = 5 heights of the canon above the ground are same.

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm