What’s the answer to.... Add j to 1

nydimaria [60]

22,647 inches to m iles will give you 0.35743371 of a mile (this is a repeated decimal since it goes on forever with different numbers.) Using a calculator, 22646.9998656 inches would give 0.35743371 miles (round the inches to the whole number and you will still get the same answer.) Into hundredths, you get 0.36, but changing to tenths would give 0.4.

Hope this helped!

Nate

7 0

3 years ago

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Can someone plz help me with this one problem plzzzzz I’m being timed!!!!

MaRussiya [10]

Answer:

2:28

3:42

4:56

5:70

8 0

3 years ago

Look at the following relation. What is the domain and range?

mylen [45]

Answer & Step-by-step explanation:

The domain is the input, and the range is the output.

The domain is also known as the x values and the output as the y values (x,y).

So:

<em>domain: {4, -1, 0, 2}</em>

<em>range: {-4, 3, 5, 8}</em>

:Done

8 0

3 years ago

Select the correct answer. Solve the equation using the method of completing the square. 2r2 + 16r - 8 = 0

adoni [48]

Answer:

r=-4+2 $\sqrt{5}$ , r=-4-2 $\sqrt{5}$

8 0

3 years ago

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Calculus hw, need help asap with steps.

nikdorinn [45]

Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

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Explanation:

Let $f(n) = \frac{(-1)^{n+1}}{n!}$

The summation given to us represents the following

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\$

There are infinitely many terms to be added.

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The partial sums only care about adding a finite amount of terms.

The partial sum $S_1$ is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say $S_1 = f(1) = 1$

I'm skipping the steps to compute f(1) since you already have done so.

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The second partial sum is when things get a bit more interesting.

We add the first two terms.

$S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\$

The scratch work for computing f(2) is shown in the diagram below.

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We do the same type of steps for the third partial sum.

$S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\$

The scratch work for computing f(3) is shown in the diagram below.

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Now add the first four terms to get the fourth partial sum.

$S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\$

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like $S_3 = S_2 + f(3)$ and $S_4 = S_3 + f(4)$

In general, $S_{n+1} = S_{n} + f(n+1)$ so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

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Let's use that recursive trick to find $S_5$

$S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333$

The scratch work for f(5) is shown below.

7 0

2 years ago