Answer:
First of all you need to find the total ratio in order to find the number of seeds in each garden.
Total ratio = 3 + 5+7
= 15
The number of seed in garden a = ratio of garden a/ total ratio x 1500
= 3/15 x 1500
= 300 seeds
since each of the seeds in garden a cost 10 cents , then the cost of 300 seeds = 300 x 10 = 3000 cents
= $ 30 ( 100 cents = $1 )
The number of seeds in garden c = 7/15 x 1500
= 700 seeds
each seed in garden b cost 5 cent, cost of 500 = 700 x 5 = 3500 cents
= $35
The number of seeds in garden b = 5 / 15 x 1500 = 500 seeds
It was said that $ 110 was pent in all, so the amount spent on c = $110 - ( $30 + $35)
= $110 - $65
= $45
Step-by-step explanation:
Answer:
walmart
Step-by-step explanation:
Step-by-step explanation:
First put the values of t
= 6×2-2(2)+6÷2
= 12-4+3
= 11
7. Correct
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9. By the ratio test, the series will converge if
![\displaystyle\lim_{n\to\infty}\left|\frac{\frac{b^{n+1}(x-a)^{n+1}}{\ln(n+1)}}{\frac{b^n(x-a)^n}{\ln n}}\right|](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%7C%5Cfrac%7B%5Cfrac%7Bb%5E%7Bn%2B1%7D%28x-a%29%5E%7Bn%2B1%7D%7D%7B%5Cln%28n%2B1%29%7D%7D%7B%5Cfrac%7Bb%5En%28x-a%29%5En%7D%7B%5Cln%20n%7D%7D%5Cright%7C%3C1)
The limit reduces to
![\displaystyle|b(x-a)|\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=b|x-a|](https://tex.z-dn.net/?f=%5Cdisplaystyle%7Cb%28x-a%29%7C%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%5Cln%20n%7D%7B%5Cln%28n%2B1%29%7D%3Db%7Cx-a%7C)
where
![|b|=b](https://tex.z-dn.net/?f=%7Cb%7C%3Db)
because
![b>0](https://tex.z-dn.net/?f=b%3E0)
is given. So the series converges when
![b|x-a|](https://tex.z-dn.net/?f=b%7Cx-a%7C%3C1%5Cimplies%7Cx-a%7C%3C%5Cdfrac1b)
This means the radius of convergence is
![\dfrac1b](https://tex.z-dn.net/?f=%5Cdfrac1b)
.
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10. By the ratio test, the series converges if
![\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2}}{\frac{x^{2n}}{n(\ln n)^2}}\right|](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%7C%5Cfrac%7B%5Cfrac%7Bx%5E%7B2%28n%2B1%29%7D%7D%7B%28n%2B1%29%28%5Cln%28n%2B1%29%29%5E2%7D%7D%7B%5Cfrac%7Bx%5E%7B2n%7D%7D%7Bn%28%5Cln%20n%29%5E2%7D%7D%5Cright%7C%3C1)
The limit is
![\displaystyle|x^2|](https://tex.z-dn.net/?f=%5Cdisplaystyle%7Cx%5E2%7C%3C1%5Cimplies%7Cx%7C%3C1)
and so the radius of convergence is 1.
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11. Incorrect. By the root test, the series converges for
![\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\frac{(x-2)^n}{n^n}\right|}=\lim_{n\to\infty}\frac{|x-2|}n=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cleft%7C%5Cfrac%7B%28x-2%29%5En%7D%7Bn%5En%7D%5Cright%7C%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B%7Cx-2%7C%7Dn%3D0%3C1)
which means the series converges for all
![x](https://tex.z-dn.net/?f=x)
, and so the interval of convergence is
![(-\infty,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%5Cinfty%29)
.
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For 14 and 16, it'll probably be too late to edit this post by the time you see this. You can try posting the remaining problems in a new question.