No clue I’m not great at math but you will figure it ou
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
If you’re trying to find y the answer should be anything that equals more than negative 7 for example y could be 10
H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
46/100
31/100
2 20/100
4 32/100
