Question

plz help me solve this!

Answer 1

\[\sqrt{27} - \sqrt{12} + \sqrt{8} X \sqrt{2}\]

Answer 2

Decompose each number into its prime-power factors:\[\sqrt{3^3}-\sqrt{2^2\cdot3}+\sqrt{2^3}\cdot\sqrt{2}=3\sqrt{3}-2\sqrt{3}+2\sqrt{2}\cdot\sqrt{2}=\sqrt{3}+4\]

Answer 3

ohh! Thanks!

Answer 4

Can you help me with the rest?

Answer 5

Just post them here and we will help you.

Answer 6

Thanks! \[\sqrt[3]{x}^{5}\]

Answer 7

That expression can't be reduced further:\[\sqrt[3]{x^5}=x^{5/3}\text{, gcd}(5,3)=1.\]

Answer 8

Well, here's the possible answers:

Answer 9

You can always rewrite it a thousand other ways, however, like\[\sqrt[3]{x^5}=x^2\sqrt[3]{x}\text{ and }x\sqrt[3]{x^3}.\]

Answer 10

so wait, which one is it? I'm confused..

Answer 11

The answer you're looking for is\[x^2\sqrt[3]{x}.\]

Answer 12

ohh ok thank you!

Answer 13

How do I figure that one out?

Answer 14

You have to use the Pythagorean theorem:\[a^2+b^2=c^2.\]However, since you have a square, we have that\[2a^2=c^2.\]Moreover, you are told that \(c=8\sqrt{2}\). So, \(c^2=128\). It follows that\[2a^2=128,\]\[a^2=64,\]\[a=8.\]Hence, the length of the side of the square is \(8\).

Answer 15

ohhh ok :D Thank you!

Answer 16

check this...

\[\sqrt[3]{x^5}=x^2\sqrt[3]{x}\text{ and }x\sqrt[3]{x^3}\]

is not right.

Answer 17

@dpalnc is right: I overlooked the cube.

Answer 18

oh so..it should be...

Answer 19

The last question is: find a number between 3/4 and 3/5 and I chose 13/20, is that right?