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aksik [14]
3 years ago
7

Fran swims at a speed of 2.8 mph in still water. The Lazy River flows at a speed of 0.6 mph. How long will it take Fran to swim

2.2 mi? upstream? 2.2 mi? downstream?
Mathematics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

1. Upstream Time:  1 hour (or 60 minutes)

2. Downstream Time: 0.65 hour (or 39 minutes)

Step-by-step explanation:

The time it will take goes by the distance formula:

D=RT, where D is the distance, R is the rate (speed), and T is the time.

For Upstream, you are going <em>against</em> the current, so your still water speed slows down. So upstream rate is still water rate - stream rate. So 2.8 - 0.6 = 2.2mph

Time it will take him to swim 2.2 miles, is:

D=RT\\2.2=2.2(T)\\T=\frac{2.2}{2.2}=1

So, 1 hour

For Downstream, you are going <em>with</em> the current, so your still water speed increases. So downstream rate is still water rate + stream rate. So 2.8 + 0.6 = 3.4 mph

Time it will take him to swim 2.2 miles, is:

D=RT\\2.2=3.4(T)\\T=\frac{2.2}{3.4}=0.65

So, 0.65 hours (or 0.65*60=39 minutes)

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Step-by-step explanation:

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We can see that -1 and 7 are solutions, but make sure they are not extraneous by substituting them in the original equation:

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The square root of 49 equals 7, but the square root of -1 is an imaginary number.

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umka2103 [35]

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

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Substitute:

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For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

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Answer:

Step-by-step explanation:

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7nadin3 [17]
The answer to this question is:
You walk an average of 3.2 miles per hour on the way to school and at an average of 4 miles per hour on the way home the total walking time is 18 minutes how far away is the school Round your answer to the nearest hundredth of a mile
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Hope This Helped, Markadoodle
Your Welcome :)
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