DewdropDancer

Hmmm, I got something similar to what Log got...

\(\frac{y-2}{y^2\:-\:9}\:-\:\frac{y-7}{y^2\:+\:4y\:-\:21}\)

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{y^2+4y-21}\)

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{\left(y-3\right)\left(y+7\right)}\)

\(=\frac{\left(y-2\right)\left(y+7\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}-\frac{\left(y-7\right)\left(y+3\right)}{\left(y-3\right)\left(y+7\right)\left(y+3\right)}\)

\(=\frac{\left(y-2\right)\left(y+7\right)-\left(y-7\right)\left(y+3\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

\(=\frac{9y+7}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

Great job CubeyThePenguin...!

\(​​​​​\left(5+8\div \:4\times \:3\right)\times \:2-7\)

\(x=11\times \:2-7\)

\(=22-7\)

\(=15\)

120 + 130x = 180 * (x+1-2)

120 + 130x = 180x - 180 

300 = 50x

x = 6

You get that x = 6, but there is still one angle that equals 120, so you have a 7 sided polygon.

\(\int _{-2}^3x^2-2x+3dx\)

\(=\int _{-2}^3x^2dx-\int _{-2}^32xdx+\int _{-2}^33dx\)

\(=\frac{35}{3}-5+15\)

\(=\frac{65}{3}\)

Well done CPhill! Here is another way to do it:

 x is the number of members total

8% * 75% * x = 6

0.06x = 6

6x = 600

x = 100

8% * 75 = 0.08 * 75 = 6

100 members 

Congrats on graduating!

(Everyone here is very helpful)

Helping others is not only good for them and a good thing to do, but it also makes us happier and healthier too.

Giving also connects us to others, creating stronger communities and helping to build a happier society for everyone. And it's not all about money - we can also give our time, ideas and energy.

Good job ChowMein...! 

Here is another way to do it: 

\(101^2 - 97^2 + 93^2 - 89^2\)

\(=10201-97^2+93^2-89^2\)

\(=10201-9409+93^2-89^2\)

\(=10201-9409+8649-89^2\)

\(=10201-9409+8649-7921\)

\(=1520\)

(ChowMien's way was much easier though)

\(6t-\frac{2}{3}+\frac{t}{5}=\frac{8+\left(2-t\right)}{3}\)

\(6t\cdot \:15-\frac{2}{3}\cdot \:15+\frac{t}{5}\cdot \:15=\frac{8+2-t}{3}\cdot \:15\)

\(93t-10=5\left(-t+10\right)\)

\(93t-10=-5t+50\)

\(93t-10+10=-5t+50+10\)

\(93t=-5t+60\)

\(93t+5t=-5t+60+5t\)

\(98t=60\)

\(\frac{98t}{98}=\frac{60}{98}\)

\(t=\frac{30}{49}\)