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zaharov [31]
4 years ago
6

What is the value of n in the equation 0=18+128n^2?

Mathematics
1 answer:
lbvjy [14]4 years ago
7 0

Answer:

\large\boxed{\text{no soltion in the set of real numbers}}\\\boxed{n=-\dfrac{3}{8}i\ \vee\ n=\dfrac{3}{8}i\ \text{in the set of complex numbers}}

Step-by-step explanation:

18+128n^2=0\qquad\text{subtract 18 from both sides}\\\\128n^2=-18\qquad\text{divide both sides by 128}\\\\n^2=-\dfrac{18}{128}\\\\n^2=-\dfrac{18:2}{128:2}\\\\n^2=-\dfrac{9}{64}

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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
3 years ago
1.34 expressed as a percent
alexandr1967 [171]
1.34 = 134/100 = 134% (/100 =%)
6 0
3 years ago
Read 2 more answers
Simplify the answer -12 ÷ 3 × (-8 + (-4)² - 6) + 2 I need the steps for the equation on how to solve it.
aleksley [76]

Answer:

-6

Step-by-step explanation:

We use PEMDAS to solve this,

so P stands for parentheses, so that's where we start.

We first, square the innermost parentheses with the exponent which is the E in PEMDAS, then then the outer parentheses

-12/3*(-8+16-6)+2

-12/3*(2)+2

Now we divide as in Division in PEMDAS.

-4*2+2

Now we multiply as in Multiplication in PEMDAS.

-8+2

Now we add as in A for Addition

-6

In PEMDAS, Multiplication doesn't always come before division, and same for addition and subtraction.

7 0
3 years ago
Which table represents the function?
horrorfan [7]
D because they all increase by the same number (+1)
7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216%5C%5C%7D" id="TexFormula1" title="\sqrt[3]{216\\}" alt="\sqrt[3]{216\\}"
aivan3 [116]

Answer:

6

Step-by-step explanation:

\sqrt[3]{216} = \sqrt[3]{8} · \sqrt[3]{27}  =  \sqrt[3]{2}³ · \sqrt[3]{3}³ = 2 x 3  = 6

7 0
3 years ago
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