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mel-nik [20]
3 years ago
7

Find the 15th term of the geometric sequence 10, 20, 40, ...

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

70,110160,220,290,370,460 i thinks

step by step explanation

first no they added 10 next they added 20 so they are increasing 10

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Answer:

\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}

Step-by-step explanation:

16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}

=2\left(\dfrac{1}{a^3}\right)=\dfrac{2}{a^3}

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What is the sum of 6 and g
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Sum = addition
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Read 2 more answers
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