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ASHA 777 [7]
3 years ago
9

The problem on the left has been solved INCORRECTLY.

Mathematics
2 answers:
lbvjy [14]3 years ago
7 0

The student used the wrong trig ratio. They should use cosine instead.

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(x) = \frac{9}{29}\\\\x = \cos^{-1}\left(\frac{9}{29}\right)\\\\x \approx 71.9^{\circ}\\\\

If the student wanted to use sine, then they would need to use the pythagorean theorem to find the opposite side first. I recommend using cosine since it's faster. Make sure your calculator is in degree mode.

const2013 [10]3 years ago
7 0

Answer:

x = 71.9 degrees

Step-by-step explanation:

cos x= 9÷29

cos x= 0.31

x =cos-1(0.31)

x = 71.9 degrees

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What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

                                           = 5.2 feet

r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

                            = 3.14 x (2.6)^{2}

                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

                                           = 21.2264 - 19.625

                                           = 1.6014

The cross sectional area of the cement pipe is 1.60 ft^{2}.

6 0
3 years ago
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3 years ago
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Answer:

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Step-by-step explanation:

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<em>Hope that helps! :)</em>

<em></em>

<em>-Aphrodite</em>

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