hairyberry

\(f(x) = \frac{4x - 1}{5x} + 2\), so \(f(a) = \frac{4a - 1}{5a} + 2\).

Setting this to a, 

\(a = \frac{4a - 1}{5a} + 2\).

\(a-2=\frac{4a-1}{5a}\).

\(5{a}^{2}-10a=4a-1\)

\(5{a}^2-14a+1=0\).

Applying the quadratic formula,

We get two values of a,  \(a_{1, 2}=\frac{7\pm2\sqrt{11}}{5}\).

The general form of an exponential function is usually \(y=a\cdot{b}^{x}\). (Notice I didn't set it as \(y=a\cdot{b}^{x}+k\).

We plug our values in,

 \(\begin{cases} 3=a\cdot{b}^{1} \\ \frac{1}{3}=a\cdot{b}^{2} \end{cases} \)

\(\begin{cases} ab=3 \\ a{b}^{2}=\frac{1}{3} \end{cases} \)

Subbing 1 into 2, 

\(3b=\frac{1}{3}\)

\(b=\frac{1}{9}\)

Subbing b back into equation 1, we get

\(a=27\)

So our exponential fuction is \(y=27\cdot{(\frac{1}{9})}^{x}\).

But note, there are infinitely many solutions to this equation, this is just one of the solutions, in the standard form without translations.

\( \lfloor t\rfloor = 3t + 4 - \lfloor 2t \rfloor\)

Looking at this equation, what can we conclude about t? \(\lfloor t \rfloor, 4, \lfloor 2t \rfloor\), are all integers, because that is the property of floors.

Here is a little trick, now we know 3t is an integer, so t = some integer / 3.

\(\because\)  \(t\equiv 0 \pmod 3\), or \(t \equiv 1 \pmod3\) or \(t \equiv 2 \pmod3\)

\(\therefore\) t either has a decimal part of \(0, \frac{1}{3}, \frac{2}{3}\).

Lets split this in three cases:

Case 1:

t has a decimal part of 0 or  \(t=\lfloor t \rfloor\).

This case is realtively simple, replacing all floors with t gives us,

\(t = 3t+4-2t\), or \(t=t+4\). We see that this is impossible because \(0\neq4\).

Case 2:

t has a decimart part of 1/3.

We see that \(t=\lfloor t\rfloor+\frac{1}{3}\). or \(\lfloor t\rfloor=t-\frac{1}{3}\)

Since the decimal part of t is 1/3, then the decimal part of 2t doesn't exceed 1, so\(\lfloor 2t\rfloor = 2t-\frac{2}{3}\)

Substituting, \(\lfloor t\rfloor=3t+4-2t+\frac{2}{3}\). Subbing our conversion in, and simplifying, \(t-\frac{1}{3}=t+\frac{14}{3}\). \(0\neq5\), we see that this case has no solution too.

Case 3:

t has a decimal part of 2/3.w

\(t=\lfloor t\rfloor+\frac{2}{3}\), or \(\lfloor t\rfloor=t-\frac{2}{3}\).

We need to be careful here, because \(2t=\lfloor 2t\rfloor+\frac{4}{3}\), and 4/3 exceeds the next number so \(\lfloor 2t\rfloor=2t-\frac{1}{3}\).

Substituting, 

\(t-\frac{2}{3}=3t+4-2t+\frac{1}{3}\), \(t-\frac{2}{3}=t+\frac{13}{3}\), \(0\neq5\). We once again see there is no solution.

This means there is no solution, to \( \lfloor t\rfloor = 3t + 4 - \lfloor 2t \rfloor\).

P.S, This can also be seen through a graph, where \(y = \lfloor x\rfloor + \lfloor 2x \rfloor\), and \(y=3x+4\)

Set this as r.

\(r = \sqrt{x - \sqrt{x}}\)

\({r}^{2} = x - \sqrt{x}\).

We see that because x is an integer, and r is an interger, then x must be a perfect square.

Suppose \(x={a}^{2}, a\ge0\)

\({r}^{2}={a}^{2}-a\).

Correct me if I'm wrong, but I don't think this is possible. 

We see a is larger than r, because you subtract from a larger number to get a smaller number.

Lets suppose a general case where a = r + n, where n > 0, because we know a > r.

Lets compare these two. 

When comparing, and dealing with positives like we are now we generally use if \(\frac{a}{b}>1\), then \(a>b\), if the opposite the opposite is true.

So we want to see if \(\frac{{r}^{2}}{{(r+n)}^{2}-(r+n)}\) is greater than 1 or not.

Simplifying, \(\frac{{r}^{2}}{{r}^{2}+rn}\). We see that this is always less than one, because r and n and both positive, so therefore \({r}^{2}<{(r+n)}^{2}-(r+n)\), so therefore substituting a back in, we see \({r}^{2}<{a}^2-a\), so it can't be equal.

So unless you are counting 0, as a 3 digit number, there is no solution.

Simplify, getting \({x}^{2}-23x+17=0\). We what to find \(\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{{b}^{2}+{a}^{2}}{{(ab)}^{2}}\).

We see that what we want simplfies into \(\frac{(a+b)^2-2ab}{(ab)^2}\). By vieta's this is \(\frac{{23}^{2}-2*17}{{17}^{2}}=\frac{495}{289}\).

The sum of the infinite geometric series listed is \(\frac{a}{1-r}\), and this equals four, so \(\frac{a}{1-r}=4\).

The series made up of all the cubes of these terms is \({a}^3+{a}^{3}{r}^{3}+{a}^3{r}^{6}+{a}^{3}{r}^{9}...\). Because \(r<1\), then \({r}^{3}<1\). So applying our formula again, \(\frac{{a}^3}{1-{r}^3}=10\). Now we have \(a=4(1-r)\), and \({a}^{3}=10(1-{r}^{3})\). We can substitute the first equation in the second one, \(64{(1-r)}^{3}=10(1-{r}^{3})\). 

\(64(1-3r+3{r}^{2}-{r}^{3})=10-10{r}^{3}\)

\(54-192r+192{r}^{2}-54{r}^{3}=0\).

\(9-32r+32{r}^{2}-9{r}^{3}=0\)

This is factored into

\(-(r-1)(9{r}^{2}-23r+9)=0\).

We know r is not 1, so

\(9{r}^2-23r+9=0\).

We solve and discard one solution, getting​\(\frac{23-\sqrt{205}}{18}\)as the common ratio.

We move everything to one side, \({n}^{2}+40n-144<0\).

We don't see a good way to factor, so we apply the quadratic formula, \(n_{1, 2} = -20 \pm 4\sqrt{34}\). This is also \(n_{1, 2} = -20 \pm \sqrt{544}\). The solutions shouls be in between these two roots because a > 0 in this quadratic. Square root of 544 is a little more than 23, so we know in the range, there is -20, 23 integers greater than -20, 23 integers less than -20. So there is 23*2 + 1 = 47 total integers in this range.

This at first feels like it will be a really nasty equation, but lets just start writing it out.

Suppose the original fraction is \(\frac{a+1}{a}\). The next one is \(\frac{a+2}{a+1}\). The third one is \(\frac{a+3}{a+2}\). Multiplying these, we get \(\frac{a+1}{a}*\frac{a+2}{a+1}*\frac{a+3}{a+2}=\frac{a+3}{a}\). We see these cancel out!

This equals five, so \(\frac{a+3}{a}=5\), 5a = a+3, 4a=3. a = 3/4. The original fraction written is \(\frac{\frac{3}{4}+1}{\frac{3}{4}}=\frac{7}{3}\). BUT NOTE!! A better answer might be better LEAVING IT IN THE FORM \(\frac{\frac{3}{4}+1}{\frac{3}{4}} \) or \(\frac{\frac{7}{4}}{\frac{3}{4}}\), BECAUSE \(\frac{7+1}{3+1} \neq \frac{\frac{7}{4}+1}{\frac{3}{4}+1}\).

This is pretty simple partial fraction decomposition. 

Our first indication that partial fraction decomposition works is \((x-2)(x+1)={x}^{2}-x-2\)

\(\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\), and we make the denominators equal, \(\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}\), and \(x+17=A(x+1)+B(x-2)\). To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.\(\)

The first ball doesn't matter, because no matter what ball they draw, the only condition is the next one needs to be the opposite color

After Rainas draws the first one, there is 11 balls, 6 of the opposite color, so the probability the next ball is the opposite color is \(\frac{6}{11}\). If this happens, there is now 10 balls, 5 of each color, so now there is a \(\frac{1}{2}\) chance of drawing a different colored ball.

So the probability is \(\frac{6}{11}*\frac{1}{2}=\frac{3}{11}\).