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2 years ago

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One vegetable warehouse had 210 tons of potatoes, another one had 180 tons. 90 tons of potatoes were delivered to the first and

120 tons to the second warehouse every day. In how many days will the first warehouse have 1.2 times less potatoes than the second warehouse?

1 answer:

vivado [14]2 years ago

4 0

Answer:

6 days

Step-by-step explanation:

The equation for first warehouse would be:

210 + 90x

The equation for 2nd warehouse would be:

180 + 120x

Where x is he number of days.

<em><u>In how many days will the first warehouse have 1.2 times less potatoes than the second warehouse?</u></em>

This means the first warehouse expression TIMES 1.2 would give us 2nd warehouse expression. Equating and solve:

$180+120x=1.2(210+90x)\\180+120x=252+108x\\12x=72\\x=\frac{72}{12}\\x=6$

So it would require 6 days

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Required result : (a) Divergent. (b) Convergent. (c) Divergent, (d) Not exists. (e) Not exists.

Step-by-step explanation:

By the definition of integral test the series for a integer n and a continuous function f(x) which is monotonic decreasing in $[n,\infty)$ then the infinite series $\sum_{n}^{\infty}$ converges or diverges if and only if the improper integral $\int_{n}^{\infty}$ converges or diverges.

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$I=\int 9^nn^9$

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$I=\dfrac{\left(2\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(\ln\left(3\right)n\left(2\ln\left(3\right)n-5\right)+10\right)-15\right)+15\right)-15\right)\mathrm{e}^{2\ln\left(3\right)n}}{8\ln^6\left(3\right)}$

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(b) $\sum_{n=1}^{\infty}ne^{-8n}\hfill (3)$

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$\int_{1}^{\infty}ne^{-8n}$

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$=4.72\times 10^{-5}$

Thus given series (3) is convergent.

(c) $\sum_{n=1}^{\infty}ne^{8n}\hfill (4)$

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$\int_{1}^{\infty}ne^{-8n}$

Now let,

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Applying integral calculator we get,

$I=\dfrac{\left(8n-1\right)\mathrm{e}^{8n}}{64}$

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(d) $\sun_{n=1}^{\infty}ln(3n)^n$

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$\int_{1}^{\infty}\ln (3n)^n$

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(e) $\sum_{n=1}^{\infty}n+7(-4)^n$

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$\int_{1}^{\infty}n+7^{(-4)^n}$

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$I=\int _{1}^{b}n+7^{(-4)^n}$

Using integral calculator we get,

$I=\frac{1}{2\ln(-4)}\Big[\ln(-4)b^2-2\Gamma(0,-\ln(7)(-4)^b)+2\Gamma(0,4\ln(7))-\ln(-4)\Big]$

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$\lim_{b\to \infty}I$ not exists.

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