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2 years ago

Find the least common multiple of three and 21

Mathematics

2 answers:

Sati [7]2 years ago

7 0

Answer:

Because:

7 × 3 = 21

1 × 21 = 21

NeTakaya2 years ago

3 0

Answer:

Step-by-step explanation:

3: 3,6,9,12,15,18,21...

21: 21...

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Please answer, thank you

JulijaS [17]

Answer:

C) 5

Step-by-step explanation:

5 0

2 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

Simplify 3(2f+6)+7 thanks

Naya [18.7K]

{Expand} 3(2f+6): 6f+18

6f+18+7

18+7=25

= 6f+25

4 0

3 years ago

Every year, the value of a property in singapore appreciates by 15% of its value in the previous year. If the value of the prope

tensa zangetsu [6.8K]

Answer:

629 510$

Step-by-step explanation:

$899300 - (30% of $899300)

cuz its two years away from 2010 so it will have gone up 30 % so to rewind time, just go back 30%

4 0

2 years ago

Read 2 more answers

Solve. for. c m=c-s

podryga [215]

Answer:

s-c= m c

Step-by-step explanation:

8 0

2 years ago