Radical probelm can someone please check my answers?

[zelda1850]

anyone mind checking my homework and help me with the problems i got wrong

1)-5$$\sqrt{3}$$ - 3$$\sqrt{3}$$ = -8$$\sqrt{3}$$

2)2$$\sqrt{8}$$ - $$\sqrt{8}$$ = 4$$\sqrt{2}$$ - 2$$\sqrt{2}$$ =
2$$\sqrt{2}$$

3)-4$$\sqrt{6}$$ - $$\sqrt{6}$$ = -4$$\sqrt{6}$$

4)-3$$\sqrt{5}$$ + 2$$\sqrt{5}$$ = $$\sqrt{5}$$

5)-3$$\sqrt{27}$$ - 3$$\sqrt{27}$$ - 3$$\sqrt{27}$$ = -9$$\sqrt{27}$$

6)-3$$\sqrt{12}$$ + 3$$\sqrt{3}$$ + 3$$\sqrt{20}$$ = -6$$\sqrt{3}$$ + 3$$\sqrt{3}$$ + 6$$\sqrt{5}$$

7)-2$$\sqrt{45}$$ - 3$$\sqrt{20}$$ - 2$$\sqrt{6}$$ = -6$$\sqrt{5}$$ - 6$$\sqrt{5}$$ - 2$$\sqrt{6}$$ = $$\sqrt{5}$$ - 2$$\sqrt{6}$$

8)$$\sqrt{6}$$ * $$\sqrt{2}$$ = $$\sqrt{12}$$

9)$$\sqrt{5}$$ * $$\sqrt{3}$$ = $$\sqrt{15}$$

10)$$\sqrt[3]{3}$$ * $$\sqrt[3]{9}$$ = $$\sqrt[3]{27}$$ = 3

11)$$\sqrt[3]{-20}$$ * $$\sqrt[3]{3}$$ = $$\sqrt[3]{60}$$

 

[micromass]

3) is wrong, probably a typo...
4) is wrong: the sign is incorrect
6) is correct, but you can simplify it further: I see two $$\sqrt{3}$$ there...
7) the second step is wrong, I really don't see how you got there
8) can be simplified further...
11) your sign is wrong

 

[zelda1850]

can u tell me how i can fix number 11 and number 7?

 

[micromass]

For number 11, you just need that the cube root of a negative is a negative. Thus $$\sqrt[3]{-20}=-\sqrt[3]{20}$$.

For number 7:

$$-6\sqrt[3]{5}-5\sqrt[3]{5}=(-6-6)\sqrt[3]{5}=-12\sqrt[3]{5}$$

 

[collinsmark]

Also, 5) is correct, but you can simplify it a little more too.