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3 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]3 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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Step-by-step explanation:

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3 years ago

Segment FG begins at point F(-2, 4) and ends at point G(-2, -3). The segment is translated 3 units left and 2 units up, then ref

Alexus [3.1K]

Answer:

$F'G' = 7$

Step-by-step explanation:

Given

$F = (-2,4)$

$G = (-2,-3)$

Required

Distance of F'G'

The transformation that give rise to F'G' from FG are:

Translation
Reflection

The above transformations are referred to as rigid transformation, and as such the side lengths remain unchanged.

i.e.

$F'G' = FG$

Calculating FG, we have:

$FG = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Where:

$F = (-2,4)$ --- $(x_1,y_1)$

$G = (-2,-3)$ --- $(x_2,y_2)$

$FG = \sqrt{(-2 - -2)^2 + (4 - -3)^2}$

$FG = \sqrt{(0)^2 + (7)^2}$

$FG = \sqrt{0 + 49}$

$FG = \sqrt{49}$

Take positive square root

$FG = 7$

Recall that:

$F'G' = FG$

$F'G' = 7$

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3 years ago