RiemannIntegralzzz

Cool short manipulation way:

You can simplify $\frac{1}{a}+\frac{1}{b}$ as $\frac{a+b}{ab}$.(If you don't believe me, expand out $\frac{a+b}{ab}$ yourself and simplify. So, now we just plug in $a*b=\frac{9}{18}=\boxed{\frac{1}{2}}$.

Long grind way:

Solving $a+b=9$, we get $a=9-b$. Substituting in, we get $(9-b)(b)=18$. Expanding out:

$(9-b)(b)=18$

$9b-b^2=18$

$-b^2+9b-18=0$

$b^2-9b+18=0$.

$(b-6)(b-3)=0$.

Now we have $b=6$ or $b=3$, and because $a+b=9$, we have $a=3$ if $b=6$, or $a=6$ if $b=3$.

Note that this does not matter because $\frac{1}{a}+\frac{1}{b}$ is symmetric. I will go with $a=3$ and $b=6$, although it does not matter.

Substituting into $\frac{1}{a}+\frac{1}{b}$ gives $\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$

He can mail $\frac{150}{9}=16$ $\frac{6}{9}$, so he can mail $\boxed{16}$ packages.

He has $\boxed{6}$ stamps left over, because the remainder of the division was $6$.

To mail out $17$ packages, he would need $\boxed{3}$ additional stamps to get the total of $9$ for the last package.

1)

Let the original amount of sacks of rice be $s$.

After the first day, we have that $s-(1/2s+4)=1/2s-4$ sacks remaining.

The second day, they had $(1/2s-4)-(1/2(1/2s-4)-3)=1/4s+1$ sacks remaining.

The third day, they had $s/4+1+30=s/4+31$ sacks remaining, because they ordered 30 sacks.

So, we have $s/4+31=50$.

$s/4=19$.

$\boxed{s=76}$

2)

Let the original amount of cell phones be $s$.

After the first month, we have that $s-(1/2s+20)=1/2s-20$ cell phones remaining.

After the second month, we have that $(1/2s-20)-(1/2(1/2s-20)+15)=1/4s-25$ cell phones remaining.

We have now that $1/4s-25=75$.

$1/4s=100$.

$\boxed{s=400}$

Thank you!

First, let's solve for $g(x)$. We have $g(x)=f(f(f(f(x))))=f(f(f(3x-2)))=f(f(3(3x-2)-2=9x-8))=f(3(9x-8)-2=27x-26)=3(27x-26)-2=81x-80$.

So, the lower range is $81(0)-80=-80$.

Also, the upper range is $81(2)-80=82$.

So, the range is $\boxed{[-80, 82]}$, or $\boxed{-80 \leq y \leq 82}$

Wait you were right, I just realized that I actually factored the quadratic the wrong way :)) the irony. Thank you for helping out!

Ohhh, thank you so much @Cphill, I see where I went wrong, I forgot to divide by 4 when I simplified the square roots, so I got 6 instead of $\frac{6}{4}$. Thanks for helping out!

How did you get 11? See my solution for how I got 12. I think you might have factored the quadratic the wrong way as $(n+12)(n-11)=0$ or something.

You have the sequence $1+2+3+4+5+...+(n-1)+n$, so the sum of this will be $\frac{n(n+1)}{2}$.

So, you are looking for $\frac{n(n+1)}{2}=66$.

$n(n+1)=132$

$n^2+n-132=0$

$(n+12)(n-11)=0$.

This means that $n=11$ or $n=-12$. $n$ obviously cannot be negative, so the only solution is $n=11$, so at the $\boxed{11^{th}}$ stop.

$2x^2-6x+9=0$.

Using the quadratic formula, you get $x=\frac{6+\sqrt{36-72}}{4}$ and $x=\frac{6+\sqrt{36-72}}{4}$. These simplify to $x=\frac{3}{2}+6i$ and $x=\frac{3}{2}-6i$.

So mn=$\frac{3}{2}*6=3*3=\boxed{9}$