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

4 0

2 years ago

Use natural logarithms to solve the equation. Round to the nearest thousandth. 3e^2x +5=27

puteri [66]

Answer:

x=0.996

Step-by-step explanation:

$3e^{2x} +5=27$

To take natural log ln , we need to get e^2x alone

Subtract 5 on both sides

$3e^{2x}=22$

Now we divide both sides by 3

$e^{2x}=\frac{22}{3}$

Now we take 'ln' on both sides

$ln(e^{2x})=ln(\frac{22}{3})$

As per log property we can move exponent 2x before ln

$(2x)ln(e)=ln(\frac{22}{3})$

The value of ln(e) = 1

$2x=ln(\frac{22}{3})$

Divide both sides by 2

$x=\frac{ln(\frac{22}{3})}{2}$

x= 0.996215082

Round to nearest thousandth

x=0.996

6 0

3 years ago