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3 years ago

8

With her motorboat at full speed Dawn gets to her fishing hole, which is 21 miles upstream, in 2 hours. The return trip takes 1.

5 hours. How fast could her motorboat go in still water? What is the rate of the current?

1 answer:

Vikentia [17]3 years ago

3 0

Answer:

Speed_still = 12.25 mi/h

Speed_current = 1.75 mi/h

Step-by-step explanation:

In the first trip, the motorboat goes upstream

Speed_boat = Speed_still - Speed_stream

The speed is defined as

Speed = Distance / time

This means

Speed_boat = 21 miles/ 2 h = 10.5 mi/h

Then

Speed_still - Speed_stream = 10.5 mi/h

In the second trip:

Speed_boat = Speed_still + Speed_stream

Speed_boat = 21 miles/ 1.5 h = 14 mi/h

Speed_still + Speed_stream = 14 mi/h

Then, the ystem of equations result

Speed_still + Speed_stream = 14 mi/h

Speed_still - Speed_stream = 10.5 mi/h

If we add them together

2*Speed_still = 24.5 mi/h

Speed_still = 12.25 mi/h

Speed_stream = 14 mi/h - 12.25 mi/ h = 1.75 mi/h

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Answer:

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Step-by-step explanation:

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Expand:

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After 5 games Adrian Peterson has rushed for 498 yards. If he gains 4.5 yards per carry, how many carries will it take for him t

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Answer:

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Step-by-step explanation:

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3 years ago

Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1]

Alekssandra [29.7K]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

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can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

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3 years ago

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sertanlavr [38]

<span>ABCD is an isosceles trapezoid with legs AB and CD
∴ AB = CD
</span>
<span>AB = 6y +5
</span>
<span>CD= 2y +1

∴ 6y + 5 = 2y + 1
6y - 2y = 5 - 1
4y = 4

∴ y = 4/4 = 1
</span>

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