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Alisiya [41]
3 years ago
7

The flagpole at school casts a shadow that is 18 feet long. A student nearby is 5.5 feet tall and casts a shadow that is 3 feet

long. How tall is the flagpole?
Mathematics
1 answer:
azamat3 years ago
5 0

Answer:

33 feet.

Step-by-step explanation:

If the ratio of the student to their shadow and flagpole to its shadow are equal, then you can use the proportion

5.5/3 = x/18 and cross multiply to find x.

5.5 × 18 = 99

3 × x = 3x

3x/3 = x, 99/3 = 33.

x = 33.

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By computing x=3600 * 1/ 57.1= 63.047 miles per hour
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Which expression is equivalent to 3^4 x 3^-9?
serious [3.7K]
1/3^5

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5. Jackson has 5 more CDs than Amal. They have a total of 95 CDs. How many CDs does Amal have?
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Answer:


Step-by-step explanation:

Jackson= 5 + Amal  

Jackson + Amal= 95  

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5+2a=95  

subtract both sides by 5  

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8 0
3 years ago
A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =
GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

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Differentiating with respect to x,

y'=-\frac{1}{20}(x-50)

∵ arc length of a curve is,

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Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx

Put -\frac{1}{20}(x-50)=tan\theta

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\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta

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=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta

By integration by parts,

=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}

If x = 80, tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}

sec \theta = \frac{\sqrt{13}}{2}

\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}

sec \theta = \frac{\sqrt{29}}{2}

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Step-by-step explanation:

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