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irakobra [83]
3 years ago
9

Andrew is flying a kite at a height of 120ft. While the wind is carrying the kite horizontally away from her at the rate of 24ft

/sec. how fast is the kite string being let out when there is 200ft of line out?

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

The string of the kite is being let out at the rate of 19.2 ft/sec.

Step-by-step explanation:

See the diagram attached.

The vertical height (BK) of the kite is 120 ft.

The kite string is of length (AK) of 200 ft.

Therefore, the horizontal distance (AB) from Andrew to kite is = \sqrt{200^{2} - 120^{2}}  = 160 ft.  

Now, applying the Pythagoras Theorem,

AB² + BK² = AK²

⇒ l² + h² = s² {Where, l is the horizontal length, h is the height and s is the string length}

Differentiating with respect to time t both sides

2l\frac{dl}{dt} + 0 = 2s\frac{ds}{dt} {Since, height of kite is constant}

⇒ 2 \times 160 \times 24 = 2 \times 200 \times \frac{ds}{dt}

⇒\frac{ds}{dt} = 19.2 ft per sec.

Therefore, when the wind is carrying the kite horizontally at the rate of 24 ft/sec, then the string of the kite is being let out at the rate of 19.2 ft/sec. (Answer)

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Answers:

  • 7:    20 miles
  • 8:    28 miles

  • 9:   1 error
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  • 11:   365 words in 5 mins

Step-by-step explanation:

  • #7 and #8

Our equation is: f=2.25+0.20(m-1)   With f meaning "fare" (price) and m meaning "miles".

Question 7 - Juan's fare for his ride costs $6.05. We must solve for m.

Step 1. Substitute - <em>Substitute the fare for f in the equation.</em>

6.05=2.25+0.20(m-1)

Step 2. Simplify/Solve - Solve for m.

6.05=2.25+0.20(m-1)

<em>- Distribute</em>

6.05=2.25+0.2m-0.2

<em>- Subtract 0.2 from 2.25</em>

6.05=2.05+0.2m

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<em>- Divide both sides by 0.2</em>

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<u>And you have your answer of 20 miles.</u>

Question 8 - Same equation, different fare.

Step 1. Substitute

7.65=2.25+0.20(m-1)

Step 2. Solve

7.65=2.25+0.20(m-1)\\7.65=2.25+0.20m-0.20\\7.65=2.05+0.20m\\5.6=0.20m\\28=m

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<em>The equation given is S=\frac{1}{5} (w-10e). S =  typing speed, w = words per 5 mins, and e= errors.</em>

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<em>We are given this information: S = 55, W  = 285. We are solving for e</em>.

Substitute -

55=\frac{1}{5} (285 -10e)

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55=\frac{1}{5} (285 -10e)\\55=\frac{1}{5} (285 -10e)\\55= 57-2e\\-2=-2e\\1=e

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<em>Information given: 300 = w, 5=e. We are solving for S</em>

Substitute -

S= \frac{1}{5} (300-10(5))

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S= \frac{1}{5} (300-10(5))\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (250)\\S= 50

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<em>Information given: S = 65, e = 4. We are solving for w.</em>

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So w = 365.

  • Hoped this helped!~

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