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ss7ja [257]
3 years ago
10

Jonathan forgot his math homework on the kitchen table and left for school without it. He biked at the speed of 15 mph. His moth

er saw the homework when he was already 1 mile from home, and started chasing him. She drove at a speed of 25 mph and caught him at the school entrance. How far is the school from Jonathan's home?
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
6 0

Answer:

2.5 miles

Step-by-step explanation:

We know that: time = distance/speed

Now, if the distance of the school from Jonathan's home is y.

At speed of 15 mph, Jonathans time is; time = y/15 hours

At 25 mph, his mother's time was;

time = y/25 hours

We are told that Jonathan was 1 mile from home when he saw his mother.

Thus means that difference in time is;

Time difference = 1/15 hours

Thus;

y/15 hours - y/25 hours = 1/15 hours

Multiply through by 75 to get;

5y - 3y = 5

2y = 5

y = 5/2

y = 2.5 miles

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The angle from a stake in the ground to the top of a tree is 60 degrees. If the height of the tree is 40 feet, what is the dista
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23.09 ft

Step-by-step explanation:

Here we are given that the angle from stake in ground to the top of the tree is 60° and the height of the tree is 40ft . We are interested in finding out the distance from stake to the bottom of the tree . For figure refer to the attachment .

From the figure , BC = 40ft which is the perpendicular of the right angled triangle ABC . Also assume that AB = x ft , which is the base of the triangle .

Now since we have base and perpendicular , we should use the ratio of tangent as ,

\rm\longrightarrow tan\theta =\dfrac{perpendicular}{base}=\dfrac{p}{b}

And here \theta = 60° . On substituting the respective values , we have ;

\rm\longrightarrow tan 60^o =\dfrac{40\ ft}{x \  }\\

And the value of tan60° = √3 , so that ;

\rm\longrightarrow \sqrt3 =\dfrac{40\ ft}{x}

Cross multiply ,

\rm\longrightarrow x =\dfrac{40}{\sqrt3}ft.

The value of √3 is 1.732 approximately .

\rm\longrightarrow x =\dfrac{40}{1.732}ft.

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\rm\longrightarrow \underline{\underline{\red{{x =23.09\ ft }}}}

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