RiemannIntegralzzz

Sorry @Melody and @Guest, they where both wrong. The solution it gave was:

Dennis gains $77-55=20$ miles on Theo every hour. Theo had a $55*2=110$ hour headstart. So it will take $\frac{110}{20}=\boxed{11/2}$ or $5$ $\frac{1}{2}$ hours for Dennis to catch up to Theo.

Writing out your arithmetic progression in summation notation, we have \(\sum_{n=1}^{c}\frac{3n-15}{2}\) where c is the number of terms, what we are trying to solve for.

Pulling the 1/2 out from the summation and breaking it up gives $1/2\sum_{n=1}^{c}3n+1/2\sum_{n=1}^{c}-15$.

Now, we first solve the summation on the left. Pulling out the three, and using the appropriate technique for a sum of $n$, we get $\frac{3}{2}\frac{c(c+1)}{2}$, and combining gives $\frac{3c^2+3c}{4}$.

Now, we solve the summation on the right. This is easy, for any constant c summed from 1 to n, the result is nc. So we have $1/2\sum_{n=1}^{c}-15=\frac{-15c}{2}$

Now, we have an equation for c!

$\frac{3c^2+3c}{4}-\frac{15c}{2}=84$

You should be able to find c(the number of terms) from here.

HINT: Expand all the terms and bring everything to the left side. Look familiar?

You can write this equation, given $x=888...883$, as $x(x+2)-(x-1)(x+3)=x^2+2x-(x^2-x+3x-3)=x^2+2x-(x^2+2x-3)=x^2-x^2+2x-2x+3$. Hey, it all cancels! We are left with $\boxed{3}$ (Note how this is true for all x, that $x(x+2)-(x-1)(x+3)=3$)

If it is perpendicular to the line $y=x-5$, it means the line's slope must be $-1$, hence $y=-x+c$ for some c. plugging in $(5, 4)$ gives $4=-(5)+c, 9=c$, so the equation for the line is $\boxed{y=-x+9}$

Let us say that, because $f(x)$ is an exponential function, you write it as $f(x)=a*b^x$. To get from $a*b^3$ to $a*b^{7.5}$, you multiply by $\frac{10}{3}$. So that means to go up another 4.5 x, to 12, you multiply by another $\frac{10}{3}$. So $f(12)=60*\frac{10}{3}=20*10=\boxed{200}$.

I am not one hundred percent sure about this solution because it says round to the nearest hundredth and my number is a whole integer, but try it anyways. I hope it works :)

Sorry, it says that was incorrect, but I have one more try. Can you please show your work so I can look for errors?

Thank you, it was correct!

There is an algebraic way through manipulation:

Let $x=0.\overline{09}$

Multiply by 100 to get $100x=9.\overline{09}$

Subtract x from 100x to get $100x-x=9.\overline{09}-0.{09}$ which cancels out the infinite decimals leaving $99x=9$

Dividing both sides by 99 gives $x=\frac{9}{99}$, which simplifies to $x=\frac{1}{11}$. The product of the numerator and denominator is $1*11=\boxed{11}$

If you are familiar with geometric series there is also another way:

x=0*10^-1+9*10^-2+0*10^-3+9*10^-4+0*10^-5+...

simplifying gives x=9*10^-2+9*10^-4+9*10^-6+... (you got lucky the 0's canceled out, there is a trick to dealing with them when there is no 0's to cancel out)

We can see this is an infinite geometric series with first term 9*10^-2 and common ration 10^-2. Using the formula for the sum of an infinite geometric series gives:

$S=\frac{9*10^-2}{1-10^-2}$

$S=\frac{\frac{9}{100}}{\frac{99}{100}}$

$S=\frac{9}{99}$

$S=\frac{1}{11}$

The product is 11 :) These are the two algebraic solutions

Thank you!