Answer:
<h2>14mph</h2>
Step-by-step explanation:
Given the gas mileage for a certain vehicle modeled by the equation m=−0.05x²+3.5x−49 where x is the speed of the vehicle in mph. In order to determine the speed(s) at which the car gets 9 mpg, we will substitute the value of m = 9 into the modeled equation and calculate x as shown;
m = −0.05x²+3.5x−49
when m= 9
9 = −0.05x²+3.5x−49
−0.05x²+3.5x−49 = 9
0.05x²-3.5x+49 = -9
Multiplying through by 100
5x²+350x−4900 = 900
Dividing through by 5;
x²+70x−980 = 180
x²+70x−980 - 180 = 0
x²+70x−1160 = 0
Using the general formula to get x;
a = 1, b = 70, c = -1160
x = -70±√70²-4(1)(-1160)/2
x = -70±√4900+4640)/2
x = -70±(√4900+4640)/2
x = -70±√9540/2
x = -70±97.7/2
x = -70+97.7/2
x = 27.7/2
x = 13.85mph
x ≈ 14 mph
Hence, the speed(s) at which the car gets 9 mpg to the nearest mph is 14mph
Answer: 25.7495 full: 25.75 rounded
Step-by-step explanation: Multiply 16*0.62, simple. From that you get 25.7495 Then, since there is a 9 after the four, you would round the 4 up to a 5.
Step-by-step explanation:
as the velocity is not constant over time, we actually have to integrate v(t) over the interval 0<=t<=5 to get the distance.
v(t) = 60×ln(t + 1)
V(t) = 60 × integral(ln(t + 1)) between 0 and 5.
integral(ln(t + 1)) = (t + 1)ln(t + 1) - t + C
V(t) = 60 × ((t + 1)ln(t + 1) - t + C)
the distance traveled between t = 0 and t = 5 is then
60 × ((5 + 1)ln(5 + 1) - 5 + C) - 60 × ((0 + 1)ln(0 + 1) - 0 + C) =
= 60×(6ln(6) - 5 + C) - 60×(1ln(1) + C) =
= 60×(5.750556815... + C) - 60×C =
= 60×5.750556815... = 345.0334089... ≈ 345 km
Difference is 2.71 and the estimate is 3
