Brett lives 3/10 mile from school. If he rides his bike to and from school each day, Monday to Friday, how many miles does he ri
2 answers:
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Answer/Step-by-step explanation:
Equation to represent the daily rental cost for each type of truck can be written as follows:
Daily rental cost for Trucks-A-Lot = 42 + 0.72m
Daily rental cost for Move-in-Truckers = 70 + 0.12m
Where, m = Emily's mileage
To determine the number of miles for which the truck cost the same amount, set both equations equal to each other and solve for m.
Collect like terms
Divide both sides by 0.6
At approximately 47 miles, both trucks would cost the same amount.
Check:
Daily rental cost for Trucks-A-Lot = 42 + 0.72m
Plug in the value of x = 47
= 42 + 0.72(47) = $75.84 ≈ $76
Daily rental cost for Move-in-Truckers = 70 + 0.12m
Plug in the value of x = 47
= 70 + 0.12(47) = $75.64 ≈ $76
Answer:
B is the answer.
Step-by-step explanation:
=
Answer:
1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.
2) See attached picture.
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Step-by-step explanation:
1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:
and we set the first derivative equal to zero so we get:
95-18t=0
and solve for t
-18t=-95
t=5.28s
so now we calculate the position of the car after 5.28 seconds, so we get:
so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.
2) For answer 2 I take the second derivative of the function so I get:
j"(t)=-18
and then we graph them. (See attached picture)
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
