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The first one has 10 options, the second has 9 options, and the final one has 8 options. 

10*9*8

=^._.^=

x^2 - y^2 = (x+y)(x-y)

Plug in the numbers. :))

=^._.^=

(x + y)^2 = x^2 + y^2 + 2xy

Can you take it from here?

=^._.^=

Let PQRSTUVW be a rectangular prism, as shown. The area of face PQRS is 6. The area of face PQUT is 15. The area of face QUVR is 10. Find the volume of the rectangular prism.

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[PQRS] = 6           [PQUT] = 15            [QUVR] = 10

(xz) = 6         x = 6/z          z = 6/x

(yz) = 15       y = 15/z        z = 15/y

(xy) = 10       x = 10/y        y = 10/x

(yz) = (10/x)(6/x) = 15        x = 2          y = 10/2         z = 6/2

V = xyz

Here is a diagram explaining this problem. Essentially, the circles represent students. A and B represent Alice and Bob. We can see that there are 4 students between Alice and Bob. Thus, if there are 12 students shorter than Alice, $12 - 5$ (the students marked on the diagram), means that there should be $7$ students shorter than Bob. Similarly, $14-5$ means there should be $9$ students taller than Alice. 

Adding everything up gives us an answer of $7+6+9 = \boxed{22}$

14 + 12 + 4 + alice + bob = 32?

Let the function $f$ be $f(x)=ax^2+bx+c$.   The zeroes of $f$ is the roots of the quadratic equation $x^2+\frac{b}{a}x+\frac{c}{a}=0$.  We are given that the zeroes of $f$ are at $x=2$ and $x=4$.  So the zeroes are the roots of the equation $(x-2)(x-4) = x^2-6x+8 = 0$.  We are also given that $f(3)=6$.  From these facts we get a system of three equations in three unknowns $a,b,c$:
\begin{eqnarray*}
    \frac{b}{a} &=& -6 \\
    \frac{c}{a} &=& 8 \\
    9a+3b+c &=& 6.
\end{eqnarray*}
This solves to $a=-6$, $b=36$, and $c=-48$.  Thus, $f(x)=-6x^2+36x-48$.

In order to find the percentage of bows that are white, we subtract the percentage of everything else, as it makes sense that the remaining after subtracting would all be white. 

$1 - \frac{1}{5} - \frac{1}{2} - \frac{1}{10} = \frac{1}{5}$. Thus, $\frac{1}{5}$ of the total bows are white. Since there are $\frac{1}{2}$ as many green bows as white bows, we know our answer is $\frac{30}{2} = \boxed{15}$

Let the triangle = $x$ and the circle = $y$.

Thus, the two equations are

$3x + 2y = 21$ and $3y + 2x = 19$. 

Since we want to find $y$, we multiply the second equation by $\frac{3}{2}$ to get $\frac{9y}{2} + 3x = \frac{57}{2}$. 

Subtracting the second and first equations gives $\frac{5y}{2} = \frac{15}{2}$

Thus, $y = 3$, which is the value of the circle. Can you solve it from here? 

Let's take the inside first and rationalize the denominator. (note that the outer square root still exists, it's just more convenient this way).

$$\frac{ \sqrt{5} (\frac{5}{\sqrt{80}} + \frac{ \sqrt{845}}{9} + \sqrt{45})}{\sqrt{5} \cdot \sqrt{5}}$$.

Simplifying the fraction gives

$$\frac{\frac{5 \sqrt{5}}{\sqrt{80}} + \frac{65}{9} + 15}{5}$$

Simplifying the fraction further gives 

$$\frac{\frac{5}{4} + \frac{65}{9} + 15}{5}$$

Doing some basic operations (adding and dividing) gives us 

$$ \frac{169}{36} $$

Finally, we finish by applying the outer square root. This gives us 

$$ \sqrt{\frac{169}{36}} = \boxed{\frac{13}{6}} $$,

which is our final answer.  The $\LaTeX$ on that one was tough!

The only terms in the poynomials that will provide the constant in the product polynomial are the costant terms (which the question states are the same )

so    constant * constant = 81       the constant is 9

This problem can be solved through guessing. 

We know that one side length of each side corresponds to one side length of another side. 

Thus, we can split $6 = 3 \cdot 2$. From here, we know that the $2$ must correspond to the side with an area of $10$, as $2$ can't divide $15$.

As a result, the measurements of the side with an area of $10$ are $5 \cdot 2$. 

Finally, the side lengths of the side with an area of $15$ are $5 \cdot 3$, which also corresponds to what we found previously. 

Our 3 measurements are $2, 3, 5$, meaning the volume is $2 \cdot 3 \cdot 5 = 30$. 

6 C 2 = 15 ways to paint two faces

    only ONE way to paint faces resulting in a product of 4  (...paint the 1 and the 4)

15 -1 = 14 ways to paint two faces that do not result in painted faces = 4

Thank you.

x=2a+1

  = (2)(-1)+1

x = -1

y = [2(-1) - 3(-1)^2 ] / 3

  =  (-2 -3)/3

  = -5/3

The area of the parallelogram is 20.

-(4-5+2)-3-[1-[6+(-3-1)-(-2+4)]+3-4     though a ' ] '   is missing

- (1)    - 3   - [1-[6 + ( -4) - (2) ]  -1

         -4      - [1- [6 -6]  -1

          -4     - 0

           -4                    possibly     depends on the missing  ] 

The quantity inside square root must be non-negative for $g(x)$ to be a real-valued function.  This constraint $(x - 3)^2 - (x - 18)^2 \ge 0$ can be solved for $x$ to $x\ge 10.5$, so 10.5 is the desired minimum.
 

x = beginning value

.9x     after monday      (10 % loss leaves 90 % value)

.9x  * .8    after tuesday

.9x * .8 * .7   after wed    = .504x     is the value on Wed    a loss of   49.6 %

Come to think of it, flipping should not be allowed.   Sorry.

The distance between a point $P=(x_0,y_0)$ and the line $L: ax+by+c = 0$ is $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.

thanks!

This equation can be solved and it gives two possible real values for $a$.  The two possible values of $a$ give two different values of $2a+1$.  Choose the smaller value of $2a+1$ as answer.

Notice that $\frac{x^2 + 1}{x} = x+\frac{1}{x}$ so if you can simplify the given equation to see if it contains $x+\frac{1}{x}$ as a "variable" that should solve it.