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LuckyWell [14K]
3 years ago
14

Im gonna need help plz

Mathematics
2 answers:
guapka [62]3 years ago
5 0

Answer:

its 4/64

if u want to simplfy its 1/16 :)

Step-by-step explanation:

i promise i did the math

docker41 [41]3 years ago
4 0
4/64 I think owa owa
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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

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

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
2 years ago
Solve for x in the following equation:
Feliz [49]

Answer:

x = 17

Step-by-step explanation:

1) Make the bases the same then rewrite.

4 = 2²

2^2(2x - 5) = 2^3x + 7

2) Since the bases are the same, set the exponents equal to each other.

2(2x -5) = 3x + 7

4x - 10 = 3x + 7

4x - 3x = 7 + 10

x = 17

4 0
1 year ago
Four less that the quotient of six and two
san4es73 [151]
(6/2) - 4 would be your answer
3 0
2 years ago
Read 2 more answers
Pleaseeee help on question 3 and 5 pleaseeeee wuick thank youuuuuuuuu
elena55 [62]

Answer:

3a, 13cm 5, 1:200

Step-by-step explanation:

In the question 3 you just have to divide all of those numbers with that 5000, maybe also make them cm so it makes more sense. So 650m = 65000cm:5000= 13cm

then for the question 5, make the km to cm, so 4.2km would be 420000cm. Then you have to divide that 420000cm with the 21 cm, so 420000:21= 200. Then the scale should be 1:200

I had difficulties with these as well. If you need extra help with the task 3 please pm me :)

7 0
3 years ago
Expreessions represent The sum of 3 and n
weeeeeb [17]

Answer:

3+n

Step-by-step explanation:

Simple questions but it can be technical atimes, according to the question we are to solve in terms of n, simple...

The sum of 3 and n can be written as 3+n

This cannot be solved further because there are two different variables and it is impossible to add them together

Therefore the final answer is 3+n

But in case a value is given for n we can then substitute and solve further

Hope this will help you

6 0
3 years ago
Read 2 more answers
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