Answer:
Step-by-step explanation:
h = -6
+20t + 4
I will use calculus, maybe that's not how you're supposed to do this
-12t +20 =0
12t = 20
t = 20 /12
t = 1 
t = 1 
there will be a max at 1.6666666666 seconds
-6*
+ 20 * 1.6666666666 +4
= 16.666666666666 + 33.333333333333 + 4
= - 16 + 33 
+ 4
= 20 feet max height ( not too high, for a rocket)
time of flight:
0 = -6 +20t + 4
use quadratic formula to find t
-20 +- sqrt [ 
- 4*(-6)*4 ] / 2*(-6)
-20 +- sqrt [400 + 96 ] / -12
-20 +- sqrt [496 ] / -12
-20 +- 22.27105 / -12
try the negative option 1st
-42.27105 / -12
3.522 seconds. time of flight
when will the rocket be at 12' ? :
12 = -6 +20t + 4
0 = -6 +20t -8
use quadratic formula again to find t
-20 +- sqrt [ - 4*(-6)*(-8) ] / 2*(-6)
-20 +- sqrt [ 400 - 192 ] / -12
-20 +- sqrt [208 ] / -12
-20 - 14.4222 / -12
-34.4222 / -12
2.8685 seconds ( on the way down)
and
-20 + 14.4222 / -12
-5.578 / - 12
0.4648 seconds ( on the way up )
Answer:
Choice C: (x-2)^2 + (y-1)^2 = 25
9514 1404 393
Answer:
maximum difference is 38 at x = -3
Step-by-step explanation:
This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.
__
Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.
h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2
Then the derivative is ...
h'(x) = 3x² +4x -15 = (x +3)(3x -5)
This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.
The maximum difference between f(x) and g(x) is 38 at x = -3.
