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:
Okay so I am not positive but I will try to help you out. Okay so the line of best fit would positive. Because it is all going up. The approximate slope is 3/4 Because slope is change in y over change in x. The y intercept of the line of best fit would be (0,0). This is because when drawing the line you can see that it would intercept the y-axis at zero. My steps taken were basically just looking at it and peicing everything together. If you want to get more involved with slope you could pick to coordinates which I will include and do change in y over change in x.
My points I will pick are (1,1) and (5,2)
2-1/5-1=3/4
therefore approximate slope is 3/4
Like I said I am not positive but I am just trying to help out in some way.
If something is incorrect please inform me. If this helps also please inform me.
Perpendicular lines, C is the answer.
To solve this equation, Simplify the first and then the second.
First will be solved like: -y=3x+3, y=-3x-3. The second will be simplified as: 9y=3x+2, y=1/3x+2/9. Then you divide by 3.t.
Answer:
-s = -4
Step-by-step explanation:
-9s - 4 = -8s = -9s + 8s = -4 = -s= -4
Answer:
15
Step-by-step explanation:
Let x represent the number of hours it takes the larger hose to fill the pool. Then each hour, it contributes 1/x of the pool volume. Similarly, the other hose contributes a volume of 1/(1.5x) each hour.
The two hoses together contribute a volume of 1/9 of the pool volume each hour:
1/x +1/(1.5x) = 1/9
2.5/(1.5x) = 1/9 . . . . combine the fractions
22.5/1.5 = x . . . . . . multiply by 9x
15 = x
It takes the larger hose 15 hours to fill the pool alone.
