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Trava [24]
3 years ago
5

Algebra trig 42 need help asap

Mathematics
1 answer:
Anon25 [30]3 years ago
7 0
To do this you first have to find the measure of the angle that is inside that pie-shaped part of the circle.  The measure around the circle is 360 which is the same as 2pi.  Subtract 5pi/3 from 2pi and you get pi/3, which happens to be a 60 degree angle.  What you are looking for is arc length (not arc measure which is measured in degrees, not inches, feet, etc.) The formula for that is this: \frac{ \beta }{360}* \pi d, where beta is the arc measure in question and d is the diameter of the circle.  If the central angle we found measured 60 degrees, then the arc it intercepts also measures 60 degrees, therefore s measures 300 (360-60).  So now our formula looks like this: \frac{300}{360} * \pi 18.  When you do all that math, multiplying pi in, you get an arc length of 47.1, the first choice.
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  • Question 1a. i) x=1.8m

  • Question 1a. ii)   Volume=27.6m^3

  • Question 1b) Volume=65.9m^3

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<u><em>Question 1 a. i) Find the value of x.</em></u>

         tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}

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         Volume=(1/3)\pi \times radius^2\times height

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         Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3

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        Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3

  • Subtract the volume of the small cone from the volume of the big cone

        Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

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           2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m

        Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)

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<u>iii) Add the volume of the two frustrums</u>

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