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Ivahew [28]
3 years ago
12

There are three numbers for the combination to the store’s safe. The first number is 14.The other two numbers can be multiplied

together to give a product of 24.What are all the possibilities of the other two numbers?Write your answers as multiplication number sentence, and then list all the possible combinations to the safe.
Mathematics
1 answer:
Oduvanchick [21]3 years ago
3 0
1, 24; 2, 12; 3, 8; 4, 6. mix and match for the codes but that's like 100+ combinations and I don't have that kind of time sorry
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Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based
Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for \mathrm{sinc}\,x about x=0:

\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}

Then the Taylor series for

f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt

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f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

We have f(0)=0, so C=0 and so

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

which converges by the ratio test if the following limit is less than 1:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}

Like in the linked problem, the limit is 0 so the series for f(x) converges everywhere.

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3 years ago
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4 0
2 years ago
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valina [46]
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