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Compare the formulae for the volume of a cylinder and of a cone.  

Volume of a cylinder is        π • r² • h  

Volume of a cone is     1/3 • π • r² • h   

A cone has 1/3 the volume of a cylinder that has the same base and same height.  

So the cone in the question has a volume of 1/3 • 54π cm³  or  18π cm³  

Please don't answer over me when I have given a teaching answer.

If a guest needs more help, please let him/her ask for it first.

I do not think that it is usually best for people to be given full answers. 

This should not be just a cheat site.

The sum can be $4,6,8,12$. Some casework:

4 can be written as $2+2, 1+3, 3+1$

6 can be written as $1+5, 5+1, 4+2, 2+4, 3+3$

8 can be written as $2+6, 6+2, 3+5, 5+3, 4+4$

12 can be written as $6+6$

The total number of ways is $14$. There are $36$ possibilities. The answer is $\frac{14}{36}=\frac{7}{18}$.

2 7     7 2    2 9    9 2     4 7     7 4            six

no radius was provided, no

*** edit ***

Sorry, I got the vertex coordinates backwards  (as this is a sideways parabola)

   should be   vertex =  (  36/5 , 7 )

We have ADB and C can go anywhere. C can be before the A, in the middle of A&D, in the middle of D&B, after B. That's $4$ ways.

This is $\frac{2010\cdot 2011}{2}$, which is an odd multiple of $1005$, hence it's residue is $1005$.

Volume of cylinder = pi 2r h = 54pi

                                  rh = 54/2 = 27

                                    h = 27/r

volume of cone = 1/3 pi r^2 h

                              1/3 pi r^2 27/r = 9 pi r          do you have the radius ?

Taking mod 10, it's just $201(1^{2009}+2^{2009}+...+9^{2009})+1^{2009}+2^{2009}\equiv 1^{2009}+...+9^{2009}+1+2^{2009}\pmod{10}$. Now note that $9^{2009}=-1^{2009}$, etc. so it is just $5^{2009}+2^{2009}+1\pmod{10}$. But $5$ is always $5$ mod 10, and $2$ cycles as $2,4,8,6$. Hence the answer is $8$.

We want the value that $f(x)$ cannot be. I claim this is $-1$, can you see why?

Use the multiplication rule. There are $26$ possibilities for each letter, and $5$ possibilities for an even digit. The answer is $26^5\cdot 5^2$.

26 x 26 x 26 x 5 x 5 x 26 x 26 = 26⁵ * 5² = _____________ combos

Using the discriminant, the possible values of n are 5, 6, 7, 8, 9, 10, so the probability is 6/10 = 3/5.

Find the vertex of the graph of the equation x - y^2 + 8y = 13 + 6x - 6y.

Hello Guest!

$x - y^2 + 8y = 13 + 6x - 6y\\ x-6x+6y+8y-y^2-13=0\\ -y^2+14y-13=5x\\ x=f(y)=-\frac{1}{5}y^2+\frac{14}{5}y-\frac{13}{5}$

$\frac{df(y)}{dy}=-0.4y+2.8=0\\ y_v=\frac{-2.8}{-0.4}\\ \color{blue}y_v=7\\ x_v=-\frac{1}{5}y^2+\frac{14}{5}y-\frac{13}{5}$

$x_v=-\frac{1}{5}\cdot 7^2+\frac{14}{5}\cdot 7-\frac{13}{5}\\ \color{blue}x_v=7.2$

$The\ vertex\ of\ the\ parabola\ is\ \color{blue}P\ [7.2,\ 7]$

  !

Use Newton's sums.

It's not that hard, I'll do the first one for you. $P_1$ has x value 1, y value 2, so $2t+1=1, 3t+2=2\to t_1=0$.

Need to 'massage' this around to vertex form of a parabola

x - y^2 + 8y = 13 + 6x - 6y

x - 6x = y^2 -14y+13

-5x = y^2 -14y+13     complete the square for y

-5x = ( y-7)^2 - 49 + 13

-5x = (y-7)^2 - 36

x = - 1/5 (y-7)^2 + 36/5            vertex is   7  , 36/5

2^(x²-3x-2)  = (2^2)^x-4  * 2^3x     equate the exponents

x^2 -3x -2 = 2x-8+3x

x^2 -8x +6     Use quadratic formula to find   x =   4 +- sqrt(10)      summed = 8

2^(x^2 - 3x - 2) = 2^(2*(x-4)) * 2^(3x)

2^(x^2 - 3x - 2) = 2^(5x - 8)

x^2 - 3x - 2 = 5x - 8

x^2 - 8x + 6 = 0

Using vieta's, -b/a = --8/1 = 8.

=^._.^=

f(x) = 3 cos x + 9     Here is a desmos graph    it is cos amplitude 3   shifted up 9

https://www.desmos.com/calculator/wvyynsm13g

Assuming your guage works better than mine:

3/4 x   -   3/8x = 6

3/8 x = 6

x = 6 * 8/3 =.........gal

5 ' x 6 '   x   .75 '  = volume of water in ft³    (.75 ft is 9 inches)

the water will rise by 1 ft³     which will be    1 ft³  =  5' * 6' * x'       x = 1/30 ft = .4 inch

a second cube will raise the level by the same amount   .4 inch

adding all of those cubes raises the wate level to the top of the cubes    =  1 ft deep

The questions asking for greater than 1 are different from this question