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Sagot :
using the rule that √a times √b = √(a times b)
√(x^10) times √(x^4) = √(x^10 times x^4)
using the rule that x^a times x^b = x^(a+b)
√(x^10 times x^4) = √x^(10 + 4) = √x^14
to cancel out √ with ^14, divide the number 14 by 2 (if it is ∛ then you would divide 14 by 3, and so on, but in this case it is the square root)
√x^14 = x^(14/2) = x^7
so the answer is D. x^7
√(x^10) times √(x^4) = √(x^10 times x^4)
using the rule that x^a times x^b = x^(a+b)
√(x^10 times x^4) = √x^(10 + 4) = √x^14
to cancel out √ with ^14, divide the number 14 by 2 (if it is ∛ then you would divide 14 by 3, and so on, but in this case it is the square root)
√x^14 = x^(14/2) = x^7
so the answer is D. x^7
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