All categories

3 years ago

LESS THAN 5 MIN!!! WILL GIVE BRAINLIESTRaise to the power:(–xn)^4

Mathematics

2 answers:

jonny [76]3 years ago

8 0

Answer:

$n^{4} x^{4}$

I really hope this helps!

Serhud [2]3 years ago

6 0

Answer:

x^4n^4

Step-by-step explanation:

You might be interested in

Does this have infinite solutions or no solutions?

Papessa [141]

Answer:

when they cross u should get a perfect solution of (2,6)

8 0

3 years ago

Read 2 more answers

Am i Correct???????????????????????????????????????

aksik [14]

Answer:

Pretty sure its cylinder.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Given that ab=16 bc=5 and cd = 3 which is the length of secant de

jolli1 [7]

Given that AB = 16, BC = 5, and CD = 3, which is the length of secant DE?It might be useful to see ...

3 0

4 years ago

Can a triangle have three acute angles? Justify your answer

dybincka [34]

Yes because each side can be measured at atleast 90 degrees.

8 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

4 years ago