Answer:
The maximum height of the ball is 256m
Step-by-step explanation:
Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t
At maximum height, the velocity of the ball is zero.
velocity = dh/dt
velocity = -32t + 128
Since v = 0 at maximum height
0 = -32t+128
32t = 128
t = 128/32
t = 4seconds
The maximum height can be gotten by substituting t = 4 into the modelled equation.
h = -16t² + 128t
h = -16(4)²+128(4)
h = -16(16)+512
h = -256+512
h = 256m
Answer:
0.24
Step-by-step explanation:
also i see what you did there
Answer:
The plane PRS passes through the points P, R and S. So it contains the line RS. Also the plane QRS passes through the points Q, R and S. So it contains the line RS as well. Since both the planes contain the line RS, the line RS must be the intersection of plane PRS and QRS
Step-by-step explanation:
Answer:
A)30705
Step-by-step explanation:
3* 10^4 + 7 x 10^2 + 5 x 10^0
Evaluate the exponent
3⋅10^4+7⋅10^2+5⋅10^0
3⋅10000+7⋅10^2+5⋅10^0
Multiply the numbers
3⋅10000+7⋅10^2+5⋅10^0
30000+7⋅10^2+5⋅10^0
Evaluate the exponent
30000+7⋅10^2+5⋅10^0
30000+7⋅100+5⋅10^0
Multiply the numbers
30000+7⋅100+5⋅10^0
30000+700+5⋅10^0
Evaluate the exponent
30000+700+5⋅10^0
30000+700+5⋅1
Multiply the numbers
30000+700+5⋅1
30000+700+5
Add the numbers
30000+700+5
30705
Answer:
x = -12/5
Step-by-step explanation:
hope this helps you
