Answer:
a) 72.25sec
b) 6.25secs
c) after 10.5secs and 2 secs
Step-by-step explanation:
Given the height reached by the rocket expressed as;
s(t)= -4t^2 + 50t - 84
At maximum height, the velocity of the rocket is zero i.e ds/dt = 0
ds/dt = -8t + 50
0 = -8t + 50
8t = 50
t = 50/8
t = 6.25secs
Hence it will reach the maximum height after 6.25secs
To get the maximum height, you will substitute t - 6.25s into the given expression
s(t)= -4t^2 + 50t - 84
s(6.25) = -4(6.25)^2 + 50(6.25) - 84
s(6.25) = -156.25 + 312.5 - 84
s(6.25) = 72.25feet
Hence the maximum height reached by the rocket is 72.25feet
The rocket will reach the ground when s(t) = 0
Substitute into the expression
s(t)= -4t^2 + 50t - 84
0 = -4t^2 + 50t - 84
4t^2 - 50t + 84 = 0
2t^2 - 25t + 42 = 0
2t^2 - 4t - 21t + 42 = 0
2t(t-2)-21(t-2) = 0
(2t - 21) (t - 2) = 0
2t - 21 = 0 and t - 2 = 0
2t = 21 and t = 2
t = 10.5 and 2
Hence the time the rocket will reach the ground are after 10.5secs and 2 secs
3 + 3/4x > = 15
3/4x > = 15 - 3
3/4x > = 12
x > = 12 / (3/4)
x > = 12 * 4/3
x > = 48/3 = 16 sessions <==
================
-p - 4p > -10
-5p > -10
p < -10/-5
p < 2 <===
===============
-3 - 6(4x + 6) > = 9
-3 - 24x - 36 > = 9
-24x - 39 > = 9
-24x > = 9 + 39
-24x > = 48
x < = -48/24
x < = -2 <===
=================
2x - 2 > = 10
2x > = 10 + 2
2x > = 12
x > = 12/2
x > = 6........so ur solution set is {6,7}
Answer:
I believe the answer would be 20 miles max for one passenger.
Step-by-step explanation:
This is because if you create an equation out of this, knowing that only one passenger will be riding, you will get:
$3.00+$1.25x=$28 ($3 added too $1.25 times the number of miles (x) which equals the total amount you have ($28)).
x being the number of miles (so we need to calculate for x)
$3 + $1.25x = $28
-subtract 3 on both sides-
$1.25x=$28-$3
$1.25x=$25
-divide by 1.25 on both sides-
x=$25/$1.25
x=20 miles
Not sure if all calculations are correct, but I hope this helps :)!
Answer: have you tried to use photomath?
