S ={x|x are numbers completely divisible by 5 and x<100}

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

100 POINTS HEY YALL IS THIS LINE OF BEST FIT A LINE OF BEST FIT LOL

telo118 [61]

Answer:

try and angle it more upwards so that it goes more through the second dot from the top and those 2 dots towards the bottom of the graph above the line.

Step-by-step explanation:

8 0

3 years ago

Simplifying exponents (picture provided)

Hitman42 [59]

Answer:

<em>3^2 ; Option B</em>

Step-by-step explanation:

We are given the equation 3^ -6 * ( 3^4 / 3^0 )^2, which can be solved through the application of exponential rules;

$3^{ - 6 } * ( 3^{ 4 } / 3^{ 0 } )^{2} - take 3^{ 0 } as 1,\\\\3^{ - 6 } * ( 3^{ 4 } / 1 )^{2} - take 3^{4} to power of 2, ( * of exponents ),\\\\3^{ - 6 } * ( 3^{ 8 } ) - combine powers, adding exponents,\\\\3^{ - 6 + 8 } - add like terms,\\\\3^{2} \\\\Solution; 3^{2}$

<em>3^2 ; Option B</em>

7 0

4 years ago

Read 2 more answers

How to solve an expression with variables in the exponents?

gtnhenbr [62]

Answer:

use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

__

You will note that this approach works well enough for ...

a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

__

Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0

3 years ago

Graph the function f(x)=(1/3)^x+2-6

Hitman42 [59]

I got this picture from photomath, hope it helps

3 0

3 years ago