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Sloan [31]
3 years ago
10

The speed of the current in a creek is 2mph. A person can kayak 10 mi upstream in the same time that it takes him to kayak 14 mi

downstream. How long will it take the person to kayak 28 mi downstream?
Mathematics
1 answer:
ad-work [718]3 years ago
5 0

Answer:

  2 hours

Step-by-step explanation:

The relation between time, speed, and distance is ...

  time = distance/speed

For a kayak speed of k miles per hour in still water, the speed upstream is (k-2) and the speed downstream is (k+2). Over the distances given, the times are the same, so we can write ...

  10/(k-2) = 14/(k+2)

Multiplying by (k+2)(k-2) gives ...

  10(k+2) = 14(k -2)

  10k +20 = 14k -28 . . . . . eliminate parentheses

  48 = 4k . . . . . . . . . . . . . . add 28-10k

  12 = k

Then the time for 28 miles downstream is ...

  time = 28 / (12 +2) = 2 . . . . hours

It will take the person 2 hours to kayak 28 miles downstream.

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3 years ago
parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm. point F is on RS exactly 5 cm from S. let T be the intersection of PF
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<u>Solution-</u>

Given that,

In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.

Then considering ΔPQT and ΔSTF,

1-    ∠FTS ≅ ∠PTQ            ( ∵ These two are vertical angles)

2-   ∠TFS ≅ ∠TPQ            ( ∵ These two are alternate interior angles)

3-   ∠TSF ≅ ∠TQP            ( ∵ These two are also alternate interior angles)

<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>

∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.

\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS} =\frac{TP}{TF}

As QS = TQ + TS = 10 (given)

If TS is x, then TQ will be 10-x. Then putting these values in the equation

\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS}

\Rightarrow \frac{8}{5} =\frac{10-x}{x}

\Rightarrow x=3.85

∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm




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