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rewona [7]
2 years ago
11

Erome has an app that tracks his running times and distances. He records his distances and times in a table.

Mathematics
1 answer:
Ber [7]2 years ago
8 0

Answer:

IM SMART

Step-by-step explanation:

You might be interested in
what is equivalent to −3.5(2 − 3n) − 2.5n? A. −7 − 8n B. −7 + 8n C. −7 − 13n D. −7 − n(10.5 − 2.5) E. −7 + n(10.5 − 2.5)
inna [77]

−3.5(2 − 3n) − 2.5n

distribute

-7 -10.5n -2.5n

-7 -13n

Choice C

5 0
3 years ago
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

=========================================================

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.

-------------------

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.

-------------------

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.

-------------------

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.

-------------------

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.

-------------------

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
3 years ago
ASAP what is the quotient of 1 divided by the fraction 4/3
Virty [35]

Answer:

0.75

Step-by-step explanation:

6 0
3 years ago
14
mr_godi [17]
<span>radius=1/2 diameter
 are=radius^2*3.14
Hope I helped, and good luck!</span>
7 0
3 years ago
Read 2 more answers
Whats the solution for y+5x=17
Travka [436]
This is my answer/how I did it :
y+5x=17 can also be 5x+y=17
5x+y=17
   -y      -y
5x=17-y
+17       +17
5x+17=y (answer)

5 0
3 years ago
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