hectictar

Let the center of the circle be  (h, 0)

distance between  (h, 0)  and  (3, -2)   =   distance between (h, 0)  and  (-2, 3)

\(\begin{array}{r} \sqrt{(-2-0)^2+(3-h)^2}&\ =&\ \sqrt{(-2-h)^2+(3-0)^2}\\~\\ (-2-0)^2+(3-h)^2&=&(-2-h)^2+(3-0)^2\\~\\ (-2-0)^2-(3-0)^2&=&(-2-h)^2-(3-h)^2 \end{array}\)

Notice here that if  h = 0  both sides of the equation are identical.

If you want to, you can expand both sides to also find that  h = 0.

So the center of the circle is  (0, 0)

And the radius  =  \(\sqrt{(-2-0)^2+(3-0)^2}\ =\ \sqrt{4+9}\ =\ \sqrt{13}\)

Let  n  be the value of the smallest addend. Let's find the sums of the first few consecutive integers starting at  n .

When the number of addends is  2 ,   the sum  =  2n + 1

When the number of addends is  3 ,   the sum  =  3n + 2 + 1

When the number of addends is  4 ,   the sum  =  4n + 3 + 2 + 1

When the number of addends is  5 ,   the sum  =  5n + 4 + 3 + 2 + 1

When the number of addends is  6 ,   the sum  =  6n + 5 + 4 + 3 + 2 + 1

When the number of addends is  7 ,   the sum  =  7n + 6 + 5 + 4 + 3 + 2 + 1  =  7n + 7 + 7 + 7  =  7n + 7(7 - 1)/2

When the number of addends is  a ,   the sum  =  an + a(a - 1) / 2

So we want to find the integer solutions to this equation:

an + a(a - 1) / 2   =   60

I don't know how to do that........so I had to use WolframAlpha to solve it. See:

The equation of a vertical parabola with a vertex of  (3, 2)  in vertex form is

y  =  A(x - 3)² + 2

We know that  (1, 0)  is a solution to the equation. We can use that to find  A

0  =  A(1 - 3)² + 2

0  =  A( -2 )² + 2

0  =  4A + 2

-4A  =  2

A  =  -\(\frac12\)

So the equation of the parabola is

y  =  -\(\frac12\)(x - 3)² + 2         Now we just have to get this equation into the form  ax² + bx + c

y  =  -\(\frac12\)(x - 3)(x - 3) + 2

y  =  -\(\frac12\)(x² - 6x + 9) + 2

y  =  -\(\frac12\)x² + 3x - \(\frac92\) + 2

y  =  -\(\frac12\)x² + 3x - \(\frac{5}{2}\)

Now we can see that     (a,  b,  c)  =  ( -\(\frac12\),  3,  -\(\frac{5}{2}\) )

Here's another way...

Let's extend

WZ to point  A,

XY  to point  B,

ZY  to point  C,

WX to point  D,  and

YX  to point  E

Like this:

Add the two equations together to get

2a  =  8b

a  =  4b

Substitute this value in for  a  in the first given equation.

4b + (4b)b²  =  40b

4b + 4b³  =  40b

                            Subtract  40b  from both sides of the equation.

4b³ - 36b  =  0

                            Factor  b  out of the terms on the left side.

b(4b² - 36)  =  0

                                     Factor  4b² - 36  as a difference of squares.

b(2b + 6)(2b - 6)  =  0

                                     Set each factor equal to zero and solve for  b

If     b  =  0     then     a  =  4b  =  4(0)  =  0          so     (0, 0)  is a solution.

If     b  =  -3     then     a  =  4b  =  4(-3)  =  -12          so     (-12, -3)  is a solution.

If     b  =  3     then     a  =  4b  =  4(3)  =  12          so     (12, 3)  is a solution.

\(a^5b^5\ =\ 7776\\~\\ (ab)^5\ =\ 7776\\~\\ ab\ =\ \sqrt[5]{7776}\\~\\ ab\ =\ 6\\~\\ b\ =\ \frac6a\)

We can substitute this value in for  b  in the other given equation.

\(ab^4\ =\ 12\\~\\ a(\frac6a)^4\ =\ 12\\~\\ \frac{6^4}{a^3}\ =\ \frac{12}{1}\\~\\ 12a^3\ =\ 6^4\\~\\ a^3\ =\ \frac{6^4}{12}\\~\\ a^3\ =\ 108\\~\\ a\ =\ \sqrt[3]{108}\)_

Notice that     m∠PBQ  =  90° - 60°  =  30°     so   △PBQ  is a  30-60-90 triangle. So

BQ  =  x√3

And notice     m∠PCQ  =  45°     so   △PCQ is a  45-45-90  triangle. So

QC  =  x

And

BQ + QC  =  4

                                  Substitute   x√3   in for  BQ   and   x   in for  QC

x√3 + x  =  4

                                  Factor  x  out of the terms on the left side.

x(√3 + 1)  =  4

                                  Divide both sides of the equation by   (√3 + 1)

x  =  \(\frac{4}{\sqrt3+1}\)

                                  Multiply the numerator and denominator by  √3 - 1  and simplify

x  =  \(\frac{4}{\sqrt3+1}\cdot\frac{\sqrt3-1}{\sqrt3-1}\)

x  =  \(\frac{4\sqrt3-4}{3-1}\)

x  =  \(\frac{4\sqrt3-4}{2}\)

x  =  \(2\sqrt3-2\)

First we will need to find the length of ZX using the Pythagorean Theorem.

Do you know how to use the Pythagorean Theorem to find the length of ZX?

There are 26 different letters and 10 different digits.

number of possibilities with 3 letters and 3 digits  =  26³ * 10³

number of possibilities with 4 letters and 2 digits  =  26⁴ * 10²

the positive difference between them  =  26⁴ * 10² - 26³ * 10³

the positive difference between them  =  26³ * 10² * (26 - 10)

the positive difference between them  =  26³ * 10² * 16

the positive difference between them  =  28121600