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Mekhanik [1.2K]
3 years ago
10

True or false. indicate whether each comparison is true

Mathematics
1 answer:
kicyunya [14]3 years ago
3 0

TRUE

FALSE

TRUE

hope this helps

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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
Si el radio de un circulo es de 3 pulgadas. cual es el area del circulo?<br>​
nata0808 [166]

Answer:

I don’t know

Step-by-step explanation:

4 0
3 years ago
What is the perimeter of a square with a length of 19 meters?
lord [1]
The answer is b 38cm
4 0
3 years ago
Read 2 more answers
Complete 6,7 for 5 points.
seraphim [82]

Answer:

idk the answer

Step-by-step explanation:

4 0
3 years ago
cho tứ giác ABCD. Gọi M,N,P,Q là trung điểm của các cạnh AB, CD, AD, BC. Chứng minh rằng vecto MP = vecto QN, vecto MQ = vecto P
Dmitriy789 [7]

Answer:

Step-by-step explanation:

Xét tam giác DAB có: P là trung điểm AD, M là trung điểm AB

=> MP là đường trung bình của tam giác DAB => MP//BD và MP=\frac{1}{2}BD  (1)

Xét tam giác DBC có: N là trung điểm DC, Q là trung điểm BC

=> QN là đường trung bình của tam giác DBC => QN//BD và QN=\frac{1}{2}BD  (2)

Từ (1) và (2) => vecto MP song song cùng chiều với vecto QN

và độ dài MP = độ dài QN = \frac{1}{2}BD

=> vecto MP = vecto QN

Tương tự xét các tam giác DAC và tam giác ABC => vecto MQ = vecto PN

4 0
3 years ago
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