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blsea [12.9K]
3 years ago
8

The time, in seconds, required for an object accelerating at a constant rate of a meters/second to travel a distance of d meters

is given by this
equation
t = va
Which equation expresses din terms of tanda?
OA d = 2at?
OB.
OC. d =
= VA
OD. d = 1
Mathematics
1 answer:
Serhud [2]3 years ago
8 0

Answer:

d=\dfrac{t^2}{a}

Step-by-step explanation:

The time  in seconds, required for an object accelerating at a constant rate of a meters/second to travel a distance of d meters is given by this  equation.

t = va

We need to find the expression for d in terms of t and a.

As velocity = distance/time

t=\dfrac{d}{t}a\\\\t^2=da\\\\d=\dfrac{t^2}{a}

So, the distance d is \dfrac{t^2}{a}.

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The gas mileage for a certain vehicle can be approximated by m=−0.05x2+3.5x−49, where x is the speed of the vehicle in mph. Dete
Whitepunk [10]

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

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