Melody

$$\begin{pmatrix}
1 \\
2
\end{pmatrix} %
+%
\begin{pmatrix}
3 \\
4
\end{pmatrix} %
=%
\begin{pmatrix}
1+3 \\
2+4
\end{pmatrix} %
=%
\begin{pmatrix}
4 \\
6
\end{pmatrix}$$

$$\\(-2x + \frac{2}{3} y^2 x-x) * (x^2 2y 2x^3)\\\\
=(-2x-x + \frac{2}{3} y^2 x) * (2*2*x^2*x^3*y)\\\\
=(-3x + \frac{2}{3} y^2 x) * (4*x^{2+3}*y)\\\\
=(-3x + \frac{2}{3} y^2 x) * (4x^5y)\\\\
=(4x^5y)(-3x + \frac{2}{3} y^2 x) \\\\
=(4x^5y)(-3x) +(4x^5y)(\frac{2}{3} y^2 x ) \\\\
=-12x^6y +\frac{8}{3} y^3 x ^6 \\\\$$

Just multiply both sides by -6

You can do that on the site calc but I don't think that it works on the forum calc version.  

Is this really your homework Dragon.

Most of what you have dones really good but you made a mistake with the numbers 4divided by 4=1

Sp there is no 4 on the bottom.

After this it gets far to difficult for you!!!

$$\\\frac{3abb(3-b)}{1(bb+b-12)}\\\\
=\frac{3ab^2(3-b)}{b^2+b-12}\\\\
$ and that is as far as your teacher could expect you to get - the rest is FAR too difficult. I'll finish if for you just this time$\\\\
=\frac{-3ab^2(b-3)}{b^2+b-12}\\\\
=\frac{-3ab^2(b-3)}{(b+4)(b-3)}\\\\
=\frac{-3ab^2}{b+4}\\\\
=-\frac{3ab^2}{b+4}\\\\$$

This is a sticky topic in the bottom right of your page.  Look there. 

Yes.

You may be able to do more than one cancellation at the same time but it might be better to do it slowly like I have until you get the hang of it.  

It will be positive when the numerator and the denominator are both positive 

so you solve for that.

then

it will also be positive when the numerator and the denominator are both negative so you solve that.

Then you include both those answers.  

$$\dfrac{6a^5b^4}{9a^3b^7}\\\\
=\dfrac{2a^5b^4}{3a^3b^7}\qquad $ I have cancelled top and bottom by 3$\\\\
=\dfrac{2aaaaabbbb}{3aaabbbbbbb}\qquad $ I have expanded, there are invisable multiply signs between the letters$\\\\
=\dfrac{2aabbbb}{3bbbbbbb} \qquad $ I have cancelled out aaa top and bottom $\\\\
=\dfrac{2aa}{3bbb} \qquad $ I have cancelled out bbbb top and bottom $\\\\
=\dfrac{2a^2}{3b^3}$$

The density of iron is 7.87 g/mL. What is the mass of a 3.00cm × 4.00 cm  1.00 cm block of iron?

is this what you mean?

The density of iron is 7.87 g/cm^3. What is the mass of a 3.00cm × 4.00 cm x 1.00 cm block of iron?

Volume=12cm^3

$$\\\frac{7.87g}{cm^3}\times \frac{12cm^3}{1}\\\\
=\frac{7.87*12g}{cm^3}\\\\
=94.99g$$