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schepotkina [342]
4 years ago
6

The stopping distance of a car is modeled by the function d = 0.05r(r + 2) where d is the stopping distance of the car measured

in feet and r is the speed of the car in miles per hour. If skid marks left on the road are 75 feet long, how fast was the car traveling?
Mathematics
1 answer:
adell [148]4 years ago
3 0

Answer:

The car was moving approximately at a speed of 37.74 miles per hour.

Step-by-step explanation:

We are given the following in the question:

The stopping distance of a car is given by

d = 0.05r(r + 2)

where d is the stopping distance in feet and r is speed of the car in miles per hour.

The stopping distance is 75 feet, we have to find the speed of the car,

Putting d = 75 in the equation, we get,

75 = 0.05r(r + 2)\\r^2 + 2r = 1500\\r^2 + 2r-1500 = 0\\\\r =\dfrac{-2\pm \sqrt{4-(4)(1)(-1500)}}{2}\\\\r\approx -39.74,37.74

Thus, the car was moving approximately at a speed of 37.74 miles per hour.

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If cos(xy) = 3x+1 , find dy/dx
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867070

_______________


          dy
Find  ——  for an implicit function:
          dx

cos(xy) = 3x + 1.


First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

\mathsf{\dfrac{d}{dx}\big[cos(xy)\big]=\dfrac{d}{dx}(3x+1)}\\\\\\&#10;\mathsf{-\,sin(xy)\cdot \dfrac{d}{dx}(xy)=\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(1)}


Apply the product rule to differentiate that term at the left-hand side:

\mathsf{-\,sin(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=3+0}\\\\\\&#10;\mathsf{-\,sin(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=3}\\\\\\&#10;\mathsf{-\,sin(xy)\cdot \left[y+x\cdot \dfrac{dy}{dx}\right]=3}

   

Now, multiply out the terms to get rid of the brackets at the left-hand
                                       dy
side, and then isolate  —— :
                                       dx

\mathsf{-\,sin(xy)\cdot y-sin(xy)\cdot x\cdot \dfrac{dy}{dx}=3}\\\\\\&#10;\mathsf{-\,y\,sin(xy)-x\,sin(xy)\cdot \dfrac{dy}{dx}=3}\\\\\\&#10;\mathsf{-\;x\,sin(xy)\cdot \dfrac{dy}{dx}=3+y\,sin(xy)}\\\\\\\\&#10;\therefore~~\mathsf{\dfrac{dy}{dx}=\dfrac{3+y\,sin(xy)}{-\;x\,sin(xy)}\qquad\quad for~~x\,sin(xy)\ne 0\qquad\quad\checkmark}


and there it is.


I hope this helps. =)


Tags:  <span><em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
</span>
4 0
4 years ago
PLEASE HELP! WILL MARK BRAINLIEST!
AfilCa [17]
Number 3 is the correct answer 
5 0
4 years ago
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