thelizzybeth

Let the inner circle's radius be x. We are solving for x in terms of r.

So connect the centers of two adjacent circles with radii r and the center of the small circle with radius x.

This is a isoceles right triangle with sides r+x, r+x, and 2r. 

We know that through the Pythagorean Theorem, the ratio of a side to the hypotenuse is \(1:\sqrt 2\).

So \(\sqrt 2(r+x)=2r\). Divide both sides by \(\sqrt 2\) and rationalize to get \(r+x=\sqrt 2 r\). Minus both sides by r and factor out r to get our answer: \(x = (\sqrt 2-1)r\)

No problem! Glad it helped :)

The definition of a horizontal asymptote is a horizontal line to which the graph becomes closer and closer as x becomes a very large or very small number, but never actually reaches it. So it's a y value that's undefined. A vertical asymptote is the same thing, but an undefined value for x. 

a) Since the vertical asymptote is x=3, we can see that c = -3, since placing x=3 into \(\frac{ax+b}{x-3}\) is undefined. (The denominator of a fraction is undefined for 0). 

Let f(x) equal y. Now we can find a by solving this equation for x.

\(x=\frac{3y-b}{a-y}\). Using the same thought process as above, we find that a = -4

Now we can use our values c = -3 and a = -4 and plug them in:

\(y=\frac{-4x+b}{x-3}\)

All we have to do is find b. We still have on piece of information we haven't used: the point (1, 0). Plug these x and y values in. 

\(0=\frac{-4+b}{1-3}\)

Now we can solve for b and find that b = 4. 

So putting everything together, we have our final answer: \(f(x)=\frac{-4x+4}{x-3}\)

You can do the same thing for part b.

There actually are square roots for negative numbers- just not real numbers. Read the problem. It specifically says "complex numbers". Complex numbers have imaginary parts to it. 

Let's first combine the top two fractions in the numerator.

\(\frac{x(x+1)-x(x-1)}{x^2-1}=\frac{x^2+x-x^2+x}{x^2-1}=\frac{2x}{x^2-1}\)

This is what our fraction looks like now:
\(\frac{\frac{2x}{x^2-1}}{\frac{x}{x^2-1}}\)

We can multiply top and bottom by \(x^2-1\), which cancels to \(\frac{2x}{x}=2\)

So let the width, height, and length be x, y, and z. We are trying to find the volume, which is xyz. 

xy = 6

xz = 8

yz = 12

Multiply all these equations together. 

\((xyz)^2 =12*8*6\)

Take the square root of both sides, which is our final answer.

\(xyz=24\)

I believe the answer is 6. After testing out a few cases, it becomes obvious that there is always a multiple of 2 and a multiple of 3 in three consecutive integers. 

I'm don't know a rigorous way to prove it though.

Solve this as a polynomial, complete the square, and factorize. 

First write it as \(z^4+4=0\)

Complete the square. \(z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z+2)\)

We can factorize these two quadratics using \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) to find our answer (four complex numbers):
\(-1+i, -1-i, 1+i, 1-i\) or \(-1\pm i, 1\pm i\)

Let x and y be our two numbers, with y > x. Write two equations.

y - x = 13

3y + 2x = 129

Multiply the first equation by 2. 

2y - 2x = 26

Add this to our second equation to get

5y = 155

y = 155/5 = 31

I edited my answer.