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2 years ago

John drove from house to work at an average speed of 30 mph, trip took 40 minutes. How far in meters is his house to work?

Mathematics

1 answer:

AysviL [449]2 years ago

5 0

Answer:

20 meters

Step-by-step explanation:

Firstly we need to know that distance = speed * time

In this particular question, we have the time and the speed, we are only told to find the distance.

Now as we can see, the units are not consistent. While the average speed is 30mph, the time is not in hours but minutes. So practically, we have to convert the time to hours.

Since there are 60 minutes in an hour, the number of hours in 40 minutes would be 40/60 = 2/3 hours.

Now we can proceed to find the distance. We simply multiply the average speed with the time.

Hence, that is 30 * 2/3 = 20 meters

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Question 21

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Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals of one another. slope 1/2 perpendicular will be -2/1=-2

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3 years ago

If f and g are differentiable functions for all real values of x such that f(1) = 4, g(1) = 3, f '(3) = −5, f '(1) = −4, g '(1)

belka [17]

Answer:

h'(1)=0

Step-by-step explanation:

We use the definition of the derivative of a quotient:

If $h(x)=\frac{f(x)}{g(x)}$ , then:

$h'(x)=\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}$

Since in our case we want the derivative of $h(x)$ at the point x = 1, which is indicated by: h'(1), we need to evaluate the previous expression at x = 1, that is:

$h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}$

which, by replacing with the given numerical values:

$f(1) =4\\g(1)=3\\f'(1)=-4\\g'(1)=-3$

becomes:

$h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}=\\=\frac{-4*3-4*(-3)}{(3)^2}=\frac{-12+12}{9} =\frac{0}{9} =0$

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3 years ago