50 points
the function h(t) = −16t2 + 48t + 12 shows the height h(t), in feet, of a cannon ball after t seconds. a second cannon ball moves in the air along a path represented by g(t) = 10 + 15.2t, where g(t) is the height, in feet, of the object from the ground at time t seconds.
part a: create a table using integers 0 through 3 for the 2 functions. between what 2 seconds is the solution to h(t) = g(t) located? how do you know? (6 points)
part b: explain what the solution from part a means in the context of the problem. (4 points)

Check the picture below.

how do we know?  well, notice h(t), starts off at 12, up up up reaches 47.84 then down down down, which is pretty much the trajectory of a flying object, by the time it gets to 44, is still going down.

now, let's look at g(t), starts off at 10, and goes up up up, never down, by the time it gets to 41, is still going up,

so at second 2, h(t) is 44 and going down, g(t) is 41 and going up, at 2.2 h(t) is 40.16, and g(t) is 44.1, between that lapse, h(t) became 44, 43, 42, 41, in the same lapse g(t) became 41, 42, 43, 44, so somewhere in those values h(t) = g(t).

what does the solution mean?  It's the seconds or the instant lapse when the first cannon ball was at the same height as the second cannonball.

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2