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Mathematics 21 Online
OpenStudy (anonymous):

simplify the radical expression. 5. √250h^4k^5 6. √14q x 2√4q 7. √63x^15y^9/7xy^11 8. √3+4√3 5√6 5√3 3√6 3√3 9. 2√6+3√96 14√6 14√96 5√96 50√6 10. (8 +√11 )(8 –√11 ) 53 75 + 16√11 –57 64 + √11

OpenStudy (callisto):

For (5), is this the question? \[\sqrt{250h^4k^5}\]

OpenStudy (anonymous):

yea

OpenStudy (chaise):

Simple foil out question 11, remembering that: \[\sqrt{11}\times \sqrt{11}=11\]

OpenStudy (callisto):

\[\sqrt{250h^4k^5} = \sqrt{25 \times 10 (h^2)^2(k^2)^2k} = \sqrt{(5h^2k^2)^2\times 10k}\]\[ =\sqrt{(5h^2k^2)^2}\times \sqrt{10k} =? \]

sam (.sam.):

You should remember that\[\sqrt{a} \times \sqrt{b}=\sqrt{ab}\] and \[\sqrt{a} \times \sqrt{a}=a\]

OpenStudy (anonymous):

25hk√10k

OpenStudy (callisto):

Nope ._. \[\sqrt{a^2} =a\] Can you try again?

OpenStudy (anonymous):

5√10h^4k^5

OpenStudy (callisto):

Nope.... try a =5h^2k^2 what do you get?

OpenStudy (anonymous):

hk√125

OpenStudy (callisto):

Incorrect again.... \[\sqrt{(5h^2k^2)^2} \times \sqrt{10k} = 5h^2k^2 \sqrt{10k}\] Since 5h^2k^2 are ALL squared... Got it?

OpenStudy (anonymous):

so did i get it right

OpenStudy (callisto):

Nope ._.

OpenStudy (anonymous):

5h^2k^2√10k

OpenStudy (callisto):

Yes. this is right

OpenStudy (anonymous):

thanks

OpenStudy (callisto):

For (6) is this the question? \[\sqrt{14q} \times 2\sqrt{4q}\]

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

I have 5 and 6.

OpenStudy (callisto):

\[\sqrt{14q} \times 2\sqrt{4q}=\sqrt{14q} \times 2\sqrt{2^2q}=\sqrt{14q} \times 4\sqrt{q}\]\[=4\sqrt{(14q)\times q} = 4\sqrt{14q^2} = ?\]

OpenStudy (anonymous):

can you the answers

OpenStudy (anonymous):

give the

OpenStudy (callisto):

Can you solve it yourself first?

OpenStudy (callisto):

I didn't mean I didn't want to help you.. but I think you should at least try to do it once first...

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