An easier way of doing this would be to subtract the initial attitude from the final attitude and divide by the rate of increase
= 30000 - 6000÷4000
= 24000÷4000
=6mins
Area=(Theta/2) * r^2 This should be it, obviously Theta is your angle and you should use radiants in theory.
Answer:
7 hours
Step-by-step explanation:
12hrs=1116kms
1hr=1116÷12
1hr=93kms
Xhrs=651kms
Xhrs=(651÷93) ×1
=7×1
=7hrs
Answer:
I'm getting stuck by
Sorry I can't help you out, if you find the answer through FP lmk
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
