9p^2-6p5q^2 in standard form it would be: 9p^2+5q^2-6p
The height at time t is
h(t) = 144 - 16t²
When t = 0, then h = 144.
Therefore the height from the ground is 144 when the object is dropped.
When the object reaches the ground, h = 0.
Therefore
144 - 16t² = 0
t² = 144/16 = 9
t = 3 s
Answer:
The object reaches the ground in 3 seconds.
To find the midpoint, Add the two X coordinates together then divde by two and then to the same with the Y coordinates:
X = -4 + 2 = -2 / 2 = -1
Y = 6+7 = 13/2
Midpoint = (-1,13/2)
B or 12 would be the correct choice!
Hope this helps and mark as brainliest please!
y = -1 + 3/8x
2x - 5y = 6
Substitute the first equation into the second equation, since y is already by itself.
2x - 5(-1 + 3/8x) = 6
2x + 5 - 15/8x = 6
2x - 15/8x = 1
16/8x - 15/8x = 1
1/8x = 1 Multiply 8 on both sides to get x by itself
x = 8
Plug x into either of the equations.
y = -1 + 3/8(8)
y = -1 + 3
y = 2
2(8) - 5y = 6
16 - 5y = 6
-5y = -10
y = 2
(8,2)
