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Nookie1986 [14]
3 years ago
7

It cost $660 to put on the sschool play.How many tickets must be sold at $6 apiece in order to make a profit?

Mathematics
1 answer:
SVEN [57.7K]3 years ago
3 0
111 tickets, because 660 divided by 6 equals 110 which is the number of tickets you need to make no profit and if you sell one more ticket you make a profit.

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icang [17]
20 cm^3 : 1280 cm^3 =
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Y_Kistochka [10]

Answer:

The volume is V=\frac{64}{15}

Step-by-step explanation:

The General Slicing Method is given by

<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

V=\int\limits^b_a {A(x)} \, dx

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve y = 2 - x^2 and the lower bounding curve y = x^2

The width of each square is given by

w=(2-x^2)-x^2=2-2x^2

This means that the area of the square cross section at the point x is

A(x)=(2-2x^2)^2

The intersection points of the two bounding curves satisfy 2 - x^2=x^2, which has solutions x = ±1.

2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}

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3 years ago
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3 years ago
20 increased by twice Greg's savings<br> Use the variable g to represent Greg's savings.
Sophie [7]
40g
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4 0
2 years ago
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