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Natali5045456 [20]
3 years ago
13

A surveyor holds a laser range finder 3 feet above the ground and determines that the range finder is 92 feet from a building an

d 106 feet from the top of the building. What is the height h of the building? Round your answer to the nearest tenth. The building is about feet tall.
Mathematics
1 answer:
Ivahew [28]3 years ago
8 0

Answer:

55.64

Step-by-step explanation:

The range finder is 92 feet from the building and 106 feet from the top of the building i.e. hypotheses

We will use the Pythagorean theorem:

x^2 = 106^2 - 92^2

= 11,236 -  8,464

= 2,772

= √2,772

x = 52.64978632

The height of the building:

h = x + 3

= 52.64978632 + 3

= 55.64978632

or

≈ 55.64 feet

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SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

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Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

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The parametrization of the line segment from (1,1,1) to (2,2,2) is

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\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

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\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

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