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hammer [34]
3 years ago
15

Help me simplify this, please!!

Mathematics
1 answer:
OlgaM077 [116]3 years ago
6 0
- \sqrt[3]{2 x^{4} y^{2}  } *3 \sqrt[3]{20 x^{5}y } =-3  \sqrt[3]{2 x^{4} y^{2}20 x^{5}y} =  \\ \\ =-3 \sqrt[3]{40 x^{9}  y^{3} } =-3 \sqrt[3]{2 ^{3}*5* x^{9}  y^{3}  } = \\  \\ =-3*2 ^{ \frac{3}{3} } * x^{ \frac{9}{3} } * y^{ \frac{3}{3} }  \sqrt[3]{5} =-6 x^{3} y \sqrt[3]{5}
Answer: B.
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2 years ago
Can someone help idk if this is right
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Answer:

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You gave the explicit form.

Step-by-step explanation:

You gave the explicit form.

The recursive form is giving you a term in terms of previous terms of the sequence.

So the recursive form of a geometric sequence is a_n=r \cdot a_{n-1} and they also give a term of the sequence; like first term is such and such number. All this says is to get a term in the sequence you just multiply previous term by the common ratio.

r is the common ratio and can found by choosing a term and dividing by the term that is right before it.

So here r=-3 since all of these say that it does:

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18/-6

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If these quotients didn't match, then it wouldn't be geometric.

Anyways the recursive form for this geometric sequence is

a_n=-3 \cdot a_{n-1}

a_1=2

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