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anonymous
 3 years ago
PLEASE CAN YOU HELP ME? I'm stuck
Simplify the radical expressions
sqrt 7 (sqrt 14 + sqrt 3)
sqrt 45n^5
anonymous
 3 years ago
PLEASE CAN YOU HELP ME? I'm stuck Simplify the radical expressions sqrt 7 (sqrt 14 + sqrt 3) sqrt 45n^5

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on the first one I got until here sqrt 7 * sqrt 14 + sqrt 7 * sqrt 3 sqrt 7 * sqrt 7x2 + ......

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and on the second one until here: sqrt 5*9*n^2*n^2*n = sqrt 5 * sqrt 9 * sqrt n^2 * sqrt n^2 * sqrt n

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.0the 1st question is simply \[\sqrt{7} \times \sqrt{14} + \sqrt{7} \times \sqrt{3} = \sqrt{ 7 \times 14} + \sqrt{7 \times 3}\] you need to simplify it the 2nd question \[\sqrt{45n^5} = \sqrt{9n^4 \times 5n} = \sqrt{9n^4} \times \sqrt{5n}\] just simplify

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so on the fist one was I right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just missed the sqrt 7x3 right?

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.0going back to the 1st question \[\sqrt{7 \times 14} = \sqrt{98} = \sqrt{49 \times 2}\] which can be simplified.

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.0well \[\sqrt{49} \times \sqrt{2} = 7\sqrt{2}\] so the 1st question is \[7\sqrt{2} + \sqrt{21}\] the 2nd question \[\sqrt{9n^4} \times \sqrt{5n} = 3n^2\sqrt{5n}\] hope this makes sense.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0second question final form. is it. \[3n ^{2} \sqrt{5n}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh! ok I busy making the equation haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes on the second yes it does

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok! got the firt one
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