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3 years ago

12

Mr. Howard and Ms. Egan enjoy running on the weekends. They both usually run at a steady pace and try to meet in the middle. The

y are 14 miles apart. Mr. Howard can run 3 miles in 30 minutes. Ms. Egan can run 4 miles in 30 minutes.

1 answer:

antoniya [11.8K]3 years ago

3 0

Given :

Distance between them, D = 14 miles .

Mr. Howard can run 3 miles in 30 minutes.

Ms. Egan can run 4 miles in 30 minutes.

To Find :

Time taken by them to meet.

Solution :

Mr Howard speed, $v_H=\dfrac{3}{0.5}=6\ mph$ .

Mr Egan speed, $v_e=\dfrac{4}{0.5}=- 8\ mph$ . [ Since they run in opposite direction ]

Their relative speed :

$v=v_E-v_H\\\\v = 8-(-6) \ mph \\\\v = 14 \ mph$

Time taken by them to meet :

$T = \dfrac{14}{14}\ hours\\\\T = 1 \ hours$

Hence, this is the required solution.

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ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
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Now let's factor $\sqrt{18x}$
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So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

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We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
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We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
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All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

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