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frez [133]
3 years ago
6

Simplify completely sqrt of 20Y^8

Mathematics
1 answer:
brilliants [131]3 years ago
3 0

Answer:

Final answer is 2y^4 \sqrt{5}.

Step-by-step explanation:

Given expression is \sqrt{20y^8}.

Now we need to smplify the given expression \sqrt{20y^8}.

So begin with factoring 20 into two parts so that one part is perfect square number.

\sqrt{20y^8}

=\sqrt{4 \times 5y^8}

=\sqrt{4 y^8} \times \sqrt{5}

=2y^4 \times \sqrt{5}

Hence final answer is 2y^4 \sqrt{5}.

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4 years ago
The first term in a geometric sequence is 54, and the 5th term is 2 / 3. Find an explicit form for the geometric sequence.
quester [9]

Answer:  ar^{n-1}=54(\dfrac{1}{3})^{n-1}

Step-by-step explanation:

The nth term for a geometric sequence is given by :-

a_n=ar^{n-1}    (1)

We are given that The first term in a geometric sequence is 54. i.e. a=54

5th term=\dfrac{2}{3}   (2)

Put n=5 and a= 54 in (1), we get

a_5=(54)r^{4}=     (3)

From (2) and (3), we have

\Rightarrow(54)r^{4}=\dfrac{2}{3}\\\\\Rightarrow r^4=\dfrac{2}{3}\times\dfrac{1}{54}=\dfrac{1}{3\times27}=\dfrac{1}{81}\\\\\Rightarrow\ r=(\dfrac{1}{81})^{\frac{1}{4}}=\dfrac{1}{3}

Explicit form for the geometric sequence: ar^{n-1}=54(\dfrac{1}{3})^{n-1}

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