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bogdanovich [222]
3 years ago
12

V=1/3Πr^2(2r+h)

Mathematics
2 answers:
Lemur [1.5K]3 years ago
7 0

Answer:

h =  \frac{3v  - 2\pi \: r{ }^{3} }{\pi \: r {}^{2} }

Step-by-step explanation:

Step 1 add -2/3Πr^3 to both sides

v + -2/3Πr^3 = 1/3hpr^2 + 2/3Πr^3 + -2/3Πr^3

Step 2 divide both sides by (Πr^2)/3

(v + -2/3Πr^3)/(Πr^2)/3 = (1/3hΠr^2)/(Πr^2)/3

(3v - 2Πr^3)/Πr^2 = h

hope this helps....

stiv31 [10]3 years ago
3 0

Answer:

h= \frac{−2.094395r3+v}{1.047198r2}

Step-by-step explanation:

v=(  \frac{1}{3} (3.141593))(r2)(2r+h)

Step 1: Flip the equation.

1.047198hr2+2.094395r3=v

Step 2: Add -2.094395r^3 to both sides.

1.047198hr2=−2.094395r3+v

Step 3: Divide both sides by 1.047198r^2.

h= \frac{−2.094395r3+v}{1.047198r2}

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I need help with algebra 2 pleasee
Aleonysh [2.5K]

Answer:

x=\pm \sqrt{21}/7

Step-by-step explanation:

We can add 9 to both sides of the equation to get 21x^2=9.

We can divide both sides of the equation by 21 to get x^2=9/21.

We take the square root of both sides of the equation to get x=\pm \sqrt{9/21}.

However, that isn't in the form we need.

We know that \sqrt{9/21}=\sqrt{9}/\sqrt{21}=3/\sqrt{21}=3\sqrt{21}/(\sqrt{21}*\sqrt{21})=3\sqrt{21}/21=\sqrt{21}/7

so the answer is really x=\pm \sqrt{21}/7 and we're done!

5 0
2 years ago
At an ocean depth of 10 meters, a buoy bobs up and then down 6 meters from the ocean's depth. Ten seconds pass from the time the
Ivanshal [37]

The sine function for the given scenario is y  = 6sin ( (π / 10) x )  -  10.

<u>Step-by-step explanation:</u>

The standard formula for sine function is Asin(Bx)+C.

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∴Amplitude A= 6.

Period is the distance in the x-axis that makes one full oscillation.

We know that the time taken buoy to move from its highest point to lowest point (half oscillation) is 10 seconds.

∴The period for full oscillation is 20 seconds.

Also, Period =2π divided by B.

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B=2π /20.

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Applying all the values in the formula,

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The first point is x=0,

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ira [324]

F(h+9)

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olga_2 [115]

Answer:

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