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svetoff [14.1K]
3 years ago
10

Does anyone get this? The answers are A - 22.8 B - 25.5 C - 14.1 D - 31.2

Mathematics
2 answers:
Sedbober [7]3 years ago
6 0
Yeah i think you are suppose to add
Rom4ik [11]3 years ago
5 0

Answer:

Step-by-step explanation:

I think your supposed to add all those numbers together

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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
3 years ago
The quotient of the opposite of a number squared and three
Mars2501 [29]

Answer:

"The quotient of the opposite of a number squared and 3"  

Take "the opposite of a number squared" and call it y.  

So you get "The quotient of y and 3"  

This is y/3.  

Now what is y? "The opposite of a number squared"  

Take "The opposite of a number" and call it z.  

So y is "z squared"  

Replacing y, we get z^2 / 3  

But what is z? "The opposite of a number"  

Call "a number" x.  

The opposite of x is -x.  

So z is "-x"  

Replacing z, we get (-x)^2 / 3



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