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2 years ago

Rewrite 64^18z^12 as a power of a product.

Mathematics

2 answers:

just olya [345]2 years ago

5 0

Yeah get Photomath to help you I don’t know if I spelled that right

tamaranim1 [39]2 years ago

4 0

Get photo math, easy questions like that will be easy

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Find the difference- (ab+3a+7)- (-5ab-2)

coldgirl [10]

3 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

3 years ago

Solve for n: n/80 = 3/24

Sedaia [141]

Answer:

n=10

Step-by-step explanation:

simplify both side

6 0

2 years ago

The graph shows the path that Karis bicycles on Thursday. Each unit represents 1 mile. She bicycles north, then east, then south

Ulleksa [173]

Yes Karis is correct I hope this helps and if its wrong let me know but I'm sure I'm correct

6 0

3 years ago

Kelsey drew a rectangle with the same area as the square. The length of Kelsey's rectangle is 8 inches. What is the perimeter in

Nutka1998 [239]

Answer: The answer is 20 inches.

Step-by-step explanation: As given in the question and shown in the attached figure, Keyley drew the rectangle 'R' and square 'S' with equal areas.

Also, the length of Keyley's rectangle 'R' = 8 inches. Let the breadth be 'b' inches. Again, let 'l' be the length of a side of square 'S'.

Then, we must have

$8\times b=\ell^2.$