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gulaghasi [49]
2 years ago
9

Eliza’s parents are installing an above ground swimming pool in their backyard. The pool is a right circular cylinder with a dia

meter of 18 feet and a height of 7 feet.
Once her parents fill the pool, it will have a depth of 6 feet. What is the volume of water, to the nearest cubic foot, that will be in the pool once it is filled.
(The volume, V, of a right circular cylinder with radius r and height h is given by the formula)
F. 170
G. 1,527
H. 1,781
j. 2,375
K. 6,107

Mathematics
1 answer:
steposvetlana [31]2 years ago
6 0

Answer:

H = 1781

Step-by-step explanation:

First, find the radius which is 9; 18/2. Then use cyclinder formula pi*r^2*h. Substitute values, but replace height with 6 because you are finding volume of water. Equation: pi*9^2*6

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F(x) = 4x - 1
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    (f + g)(x) = 2x² + 4x + (-1 + 3)
    (f + g)(x) = 2x² + 4x + 2
    Domain: {x| -∞ < x < ∞}, (-∞, ∞)

2. (f - g)(x) = (4x + 1) - (2x² + 3)
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    Domain: {x| -∞ < x < ∞}, (-∞, ∞)

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    (f · g)(x) = 8x³ + 12x + 2x² + 3
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6. (\frac{g}{f})(x) = \frac{2x^{2} + 3}{4x - 1}
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6 0
2 years ago
Which of the following rational functions is graphed below?
AlekseyPX

Answer:

The rational function that is graphed is B

6 0
2 years ago
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6-column table with 5 rows. The 1st column is labeled + x squared with entries +x, +x, +x, +x, +x. The 2nd column is labeled +x
denis-greek [22]

Answer:

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Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche
aliya0001 [1]

Answer:

The water level rising when the water is 4 inches deep is \frac{3}{8\times \pi} inch/s.

Step-by-step explanation:

Rate of water pouring out in the cone = R=\frac{3}{2} inch^3/s

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}

r = h/2

Volume of the cone = V=\frac{1}{3}\pi r^2h

V=\frac{1}{3}\pi r^2h

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}

\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}

\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

h = 4 inches

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2

\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s

The water level rising when the water is 4 inches deep is \frac{3}{8\times \pi} inch/s.

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Two figures are similar if one is the scaled version of the other.

This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.

To be more precise, you only need two steps to prove that every two circles are similar:

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Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.

Recapping, we have:

  1. Start with circle X and radius r
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So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.

8 0
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