Answer:
a(n) = (3/5)a(n-1), where a(1) = 10 and n is 2 or greater.
Step-by-step explanation:
determine the common factor...The first term (10) by 3/5 results in the given second term (6); the third term is 3/5 of the second term (6), resulting in 18/5 (equivalent to 36/10 or 3.6/1, or just 3.6. Resulting in a(n) = (3/5)a(n-1).
the third term is 3.6, which, if mult. by (3/5), produces 2.16, as expected.
Thus, the recursive formula for this geometric sequence is a(n) = (3/5)a(n-1).
This is good only for a(1) = 10 and n = 2, 3, ....
Hope this helps you
Is that all the info ? If so say 128 voted
1800
Step-by-step explanation:
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Answer:
We conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Step-by-step explanation:
Given the expression
![\sqrt[3]{200k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
so the expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
first solving
![\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
Apply radical rule: ![\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![\sqrt[3]{k^{15}}=k^{\frac{15}{3}}=k^5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D%3Dk%5E%7B%5Cfrac%7B15%7D%7B3%7D%7D%3Dk%5E5)
then solving
![\sqrt[3]{200}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200%7D)
prime factorization: 200: 2³ · 5²
![=\sqrt[3]{2^3\cdot \:5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%5C%3A5%5E2%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
![=\sqrt[3]{2^3}\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%7D%5Csqrt%5B3%5D%7B5%5E2%7D)
Apply radical rule:
![\sqrt[n]{a^n}=a,\:\quad \:a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%2C%5C%3A%5Cquad%20%5C%3Aa%5Cge%200)
so
![=2\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D2%5Csqrt%5B3%5D%7B5%5E2%7D)
Thus, the main expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
![=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Therefore, we conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Answer:
adjacent angles and complementry
Step-by-step explanation:
I think