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4 years ago

I need help someone help asap please I'm late I have been needing help thanks in advance and I appreciate it

Mathematics

1 answer:

Nadusha1986 [10]4 years ago

6 0

Answer: I believe it would be tan(-5pi/6)

Step-by-step explanation:

Since tan is he length of the side opposite the angle divided by the length of the adjacent side.

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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

The point (-3,-1) lies in the quadrant: A.

xenn [34]

Answer:

$\displaystyle QUADRANT\:III$

Step-by-step explanation:

$\displaystyle [POSITIVE, NEGATIVE] → QUADRANT\:IIII \\ [NEGATIVE, NEGATIVE] → QUADRANT\:III \\ [NEGATIVE, POSITIVE] → QUADRANT\:II \\ [POSITIVE, POSITIVE] → QUADRANT\:I$

I am joyous to assist you at any time.

3 0

3 years ago

Sandy works at a clothing store. She makes $9 per hour plus earns 10% commission on her sales. She worked 74 hours over the last

Rina8888 [55]

C. $206
2,057*.10=205.7

3 0

3 years ago

Can someone be so freaking awesome and help me out with the correct answer please :( !?!?!?!?!???!!! 30 points!!!

Sindrei [870]

$\bf 7~~,~~\stackrel{7+6}{13}~~,~~\stackrel{13+6}{19}~~,~~\stackrel{19+6}{25}\qquad \impliedby \qquad \textit{common difference "d" is 6}$

we know all it's doing is adding 6 over again to each term to get the next one, so then

$\bf \stackrel{\textit{Recursive Formula}}{\stackrel{\textit{nth term}}{f(n)}~~=~~\stackrel{\textit{the term before it}}{f(n-1)}~~~~\stackrel{\textit{plus 6}}{+~~~~6}}$

now for the explicit one

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61$

3 0

3 years ago

6b x 3 =<br><br> PLEAS HELP ME

MArishka [77]

18 bytes looked it up online lol

4 0

3 years ago