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The fourth shoe will guarantee that there is a matching pair.

What is the number?

If  (a + b) / b  =  5     --->     a/b + b/b  =  5     --->     a/b + 1  =  5     --->     a/b  =  4

If  (b + c) / c  =  4     --->     b/c + c/c  =  4     --->      b/c + 1  =  4      --->     b/c  =  3     --->     c/b  =  1/3

a/b + c/b  =  4 + 1/3  =  13/3     and     a/b + c/b  =  (a + c)/b     --->     (a + c)/b  =  13/3

Since  b/c  =  3     --->     b  =  3c

(a + c)/b  =  13/3     and    b  =  3c     --->     (a + c)/(3c)  =  13/3     --->     (a + c)/c  =  13  

ah ok, I get it now. Thanks.

x³ - x² - 21x + 45  can be factored into  (x + 5)(x -3)²

so multiply each term by   (x + 5)(x -3)² 

You'll end with:      1  =  A(x - 3)² + B(x - 3)(x + 5) + C(x + 5)

                              1  =  A(x² - 6x + 9) + B(x² + 2x - 15) + C(x - 5)

                              1  =  Ax² - 6Ax + 9A + Bx² + 2Bx - 15B + Cx - 5C

             0x² + 0x + 1  =  Ax² + Bx²  - 6Ax + 2Bx + Cx + 9A - 15B - 5C

So:            Ax² + Bx²  =  0x²   --->   (A + B)x²  =  0x²   --->    A + B  =  0

       -6Ax + 2Bx + Cx  =  0x     --->   (-6A + 2B + C)x  =  0x   --->   -6A + 2B + C  =  0

           9A - 15B - 5C  =  1   

Now, solve these three equations simultaneously:         A +     B          =  0

                                                                                     -6A +   2B +  C  =  0

                                                                                       9A - 15B - 5C   =  1   

There are 300 spiders and millipedes, of which 10% are spiders:  0.10 · 300  =  30.

He has 30 spiders and 270 millipedes.

He wants a group of millipedes and spiders to contain 30 spiders and these 30 spiders 

are 20% of the group.

Let  x  be the total number of millipedes and spiders:  0.20 · x  =  30

Dividing both sides by 0.20:                                                    x  =  30 / 0.20  =  150.

He'll end with 30 spiders and 120 millipedes.

He'll need to remove  270 - 120  =  150  millipedes.

Where's the diagram?

Did you enter the problem correctly?

As geno3141 says, here is another way:

4! / [3!*1!*2^4] * 2 =48 / 96 = 1 / 2

Every line that is parallel to the line  x + 3y  =  -5  has the form  x + 3y  =  k

where diffierent values of k will give you different (but parallel) lines.

Placing the point  (-7, 2)  into  x + 3y  =  k     --->   (-7) + 3(2)  =  -7 + 6  =  -1   --->   k  =  -1.

So, the equation of the line that passes through  (-7, 2)  that is parallel to  x + 3y  =  -5

      is  x + 3y  =  -1.

To get this equation to have the form  Ax + By  =  3

                                             instead of  x + 3y  =  -1

we'll need to multiply the bottom equation by  -3:

     -3 [ x + 3y  =  -1 ]     --->     -3x + -9y  =  3

This makes  A = -3  and  B = -9.

The smallest a = 363

Your instinct is correct!

Call that whose expression on the left side of the equal sign "expression" so you have

   [expression]  =  9

Square both sides  --  you'll get:

  x + [expression]  =  81

Since  [expression]  =  9  --  by substituting  '9'  for  [expression]  --  you'll now have:

  x + 9  =  81

Therefore,  x  =  72

(AB)² = 3.872983346² = 15 

I have drawn these two circles and realized that the distance from the center of a larger circle to AB is 3.5, and the distance from the smaller circle's center to AB is 2.5. After that, it was easy to calculate the length of AB.  

(AB)² = 3.872983346² = 15 

I have drawn these two circles and realized that the distance from the center of a larger circle to AB is 3.5, and the distance from the smaller circle's center to AB is 2.5. After that, it was easy to calculate the length of AB.

we first find that q=2r and substitute p-2q into p-4r where we add p+r to get 2p-3r=5  and we also get that p=2q so p=4r and we get 8r-3r=5 so r=1 and then q=2 and p=4 so the ordered pair is (4,2,1).

well you need 100-1=99 cuts to do to make 100 of 1m pieces from a 100m cloth so we then do 99*3=297 so she needs 297 seconds

450m/s is 45000cm/s or is that what you're asking for or not?

I do Python,JavaScript,HTML, and CSS but I don't do C sorry :(

\binom makes a binom like this \(\binom xy\)

also try \begin{document} and \end{document}

I even saw some of his answers on his profile a few hours ago so he is back

He takes breaks sometimes and eventeully comes back so he is going to come back

In fact he just came back a few hours ago

He probably just went on vacation he's going to come back