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ElenaW [278]
3 years ago
13

A car is traveling on a small highway and is either going 55 miles per hour or 35 miles per hour, depending on the speed limits,

until it reaches its destination 200 miles away. Letting x represent the amount of time in hours that the car is going 55 miles per hour, and y being the time in hours that the car is going 35 miles per hour, an equation describing the relationship is:
55x + 35y = 200

If the car spends 2.5 hours going 35 miles per hour on the trip, how long does it spend going 55 miles per hour?

Round your answer to the nearest hundredths decimal place.
Mathematics
1 answer:
elena-s [515]3 years ago
6 0

9514 1404 393

Answer:

  2.05 hours

Step-by-step explanation:

Using the given equation with the given time, we have ...

  55x +35(2.5) = 200

  55x = 112.5 . . . . . . . . subtract 87.5

  x = 2.05 . . . . . . . . . . divide by 55

The car spends 2.05 hours at 55 mph.

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Use f’( x ) = lim With h ---&gt; 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²
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<h2>Answer:</h2>

The derivative of the function f(x) is:

                 f'(x)=-2x

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

f(x)=5-x^2

We have:

f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)

( Since,

(a+b)^2=a^2+b^2+2ab )

Hence, we get:

f(x+h)=5-x^2-h^2-2xh

Also, by using the definition of f'(x) i.e.

f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}

Hence, on putting the value in the formula:

f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x

      Hence, the derivative of the function f(x) is:

          f'(x)=-2x

3 0
3 years ago
Read 2 more answers
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