Answer:
line mo
Step-by-step explanation:
Because the triangle are congrument
RST and MNO are the triangle
RT is a line in the RST triqangle
it is equal to the MO line in MNO
Answer:
A = 1/2 B H area = 1/2 base X height
B = 2 A / H = 2 * 99 / 12 = 16.5 cm
dA / d t = 1/2 * (B dH / dt + H dB / dt)
dB / dt = (2 dA / dt - B dH / dt) / H
dB / dt = (2 * 1.5 cm^2 / min - 16.5 cm * 2 cm / min) / 12 cm
dB / dt = (3 - 33) / 12 cm/min = -2.5 cm/min
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
Answer:
all work is shown and pictured
Answer:
3x-8y+40z
Step-by-step explanation:
