<h2><em>hope</em><em> it</em><em> helps</em></h2>
H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t
A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).
B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.
15% is equal to 15/100
15 divided by 100 = 0.15
0.15 is your answer
hope this helps
Answer:
1.
centre(h,k)=(-13,9)
radius (r)=6
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x+13)²+(y-9)²=6²
x²+26x+169+y²-18y+81=36
x²+y²+26x-18y+169+81-36=0
x²+y²+26x-18y +214=0
is a required equation of the circle.
2.
centre(h,k)=(1,-1)
radius (r)=11
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x-1)²+(y+1)²=11²
x²-2x+1+y²+2y+1=121
x²+y²-2x+2y=121-2
x²+y²-2x+2y=119
is a required equation of the circle.
