2015

Mathematics

POST UTME

find the value of $3\sqrt\frac{n^9}{27}$ if $n$ $= 3$

A.

1

B.

3

C.

9

D.

27 Correct Answer: 9

Explanation

$\frac{n^{9/3}}{3}$ $= \frac{n^3}{3}$ when n = 3

we have $\frac{3^3}{3} = \frac{27}{3} = 9$

$\frac{n^{9/3}}{3}$ $= \frac{n^3}{3}$ when n = 3

we have $\frac{3^3}{3} = \frac{27}{3} = 9$

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2015

Mathematics

POST UTME

find the value of ﻿3n9273\sqrt\frac{n^9}{27} 327n9​​﻿ if ﻿nnn﻿ ﻿=3 = 3=3﻿

A.

1

B.

3

C.

9

D.

27

Correct Answer: 9

Explanation

﻿n9/33\frac{n^{9/3}}{3}3n9/3​﻿ ﻿=n33= \frac{n^3}{3}=3n3​﻿ when n = 3 we have ﻿333=273=9\frac{3^3}{3} = \frac{27}{3} = 9333​=327​=9﻿

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find the value of $3\sqrt\frac{n^9}{27}$ if $n$ $= 3$

$\frac{n^{9/3}}{3}$ $= \frac{n^3}{3}$ when n = 3

we have $\frac{3^3}{3} = \frac{27}{3} = 9$