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spin [16.1K]
3 years ago
11

Kristen is 64 inches tall, and she stands 12 feet away from a streetlight. If she casts an 82-inch-long shadow, how tall is the

streetlight rounded to the nearest hundredth of a foot?
Mathematics
1 answer:
o-na [289]3 years ago
6 0

Answer:

176.39 inches or

14.70 feet

Step-by-step explanation:

Consider the right triangle made by Kristen, ground and shadow.

This triangle has one leg as 64 inches.

Next consider the right triangle formed by street light, ground upto shadow tip.

The two triangles have common angle of elevation and also another angle as 90 degrees.

Hence the two triangles would be similar

Also if A is the angle made by hypotenuse of both triangles with the ground we have

tanx=\frac{64}{82}

This value also equals by bigger triangle as

tanx=\frac{h}{82+12(12)}=\frac{h}{226}

From these two we get

h = height of street light =\frac{226(64)}{82} =176.39

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A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet
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Answer:

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Step-by-step explanation:

We are given that

Distance between poles=35 feet

h(x)=10+0.1(x^{1.5})

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}

s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx

s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx

s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx

Let 1+0.0225x=t

dx=\frac{1}{0.0225}dt

s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt

s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}

s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}

s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)

s=28.21

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4 years ago
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Answer:

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Step-by-step explanation:

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