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3 years ago

12

Add the following numbers and use the checking method (add down and then add up) to make sure your answer is correct. (Copy care

fully on scratch paper to work the problem.)
35
+47

1 answer:

Serggg [28]3 years ago

7 0

35+47=82, to check it, 82-35=47 :)

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How to work the problem 2u²=3u+2

vlabodo [156]

U= -1/2, 2

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3 years ago

Read 2 more answers

3(2+q)+153(2+q)+15 simplified is

blondinia [14]

6 + 3q + 306 + 153q + 15
156q + 327

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3 years ago

The admission prices for a small fair are $1.50 for children and $4.00 for adults. In one day there was $5050 collected. If we k

aliina [53]

Answer: 475 adults paid admission

Step-by-step explanation:

1.50c+4a=5050 (1.50 per child and 4 per adult)

1.50(2100)+4c=5050 (2100 children times 1.5 each)

3130+4a=5050

4a=1900

a=475

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3 years ago

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

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

is obtained by integrating the series above:

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

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.

7 0

3 years ago

Find the value of the expression 9.16 + k for k = 9.

maria [59]

Answer:

18.16

Step-by-step explanation:

6 0

2 years ago