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lutik1710 [3]
3 years ago
8

You oversee the community garden. You have 3 teams who want plots in your garden, and all gardens are in the shape of a square.

Team A wants a plot which is 36 square meters, Team B wants a plot which is 64 square meters and Team C wants a plot which is 49 square meters. You need to determine the length and width of your garden. You also need to determine which team gets which plot. a. Outline the dimensions of the garden and insert the plots for the 3 teams. You may use the graph below or use technology. If you use technology, either insert a picture of your graph below or attach it after page 2 of the test. (2 points) b.) Calculate the dimensions of each side and show your work (6 points) Team A. Team B: Team C: c.) What are the minimum dimensions of your garden plot? (2 points) Length: Width:​
Mathematics
1 answer:
Alika [10]3 years ago
7 0

Step-by-step explanation:

just draw a square that's big enough to fit all wanted plots (10 by 10 is recommended)

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If f(x)= 3/x+2 - square root of x-3, the domain for f(x) is all real numbers ___ than or equal to 3
aksik [14]

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the diameter of a cylindrical construction pipe is 4 ft. if the pipe is 19 ft long, what s it’s volume?
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Answer:

76\pi or 238.76 units^{3}

Step-by-step explanation:

The volume of a cylinder is V = \pi r^{2} h

Here the radius is  r = \frac{1}{2} d = \frac{1}{2}(4)=2

The height is given as 19 ft

Then plug in all the numbers into the equation

V = \pi (2^{2})(19) = 76\pi = 238.76 units^{3}

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3 years ago
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Step-by-step explanation:

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A) Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given cur
Leno4ka [110]

(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of

2<em>π</em> (radius)² (height) = 4<em>πy</em>

Then the volume of the solid is obtained by integrating over [2, 4]:

\displaystyle 4\pi \int_2^4 y\,\mathrm dy = 2\pi y^2\bigg|_{y=2}^{y=4} = 2\pi (4^2-2^2) = \boxed{24\pi}

(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of

2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)

so that the overall volume of the solid would be

\displaystyle 2\pi \int_0^2 (648-144x+8x^2-81x^3+18x^4-x^5)\,\mathrm dx = \boxed{\frac{24296\pi}{15}}

I leave the details of integrating to you.

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Answer:

Whats the question

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