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3 years ago

3 4orths and 2 5ths

Mathematics

1 answer:

ikadub [295]3 years ago

4 0

Answer:

so you have 3/4 × 2/5= 6/20

Step-by-step explanation:

so you multiply three fourths and two fitness and you get the answer of 6/20 you could simply but I forgot the number but it can be a lower number.

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Evaluate 5÷9(F - 32) when f=50

solmaris [256]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Help?? I seriously don’t understand?

Anvisha [2.4K]

Answer:

Step-by-step explanation:

3(x-2)

Replace x as 12

3(12-2)

3(10)

3 times 10

7 0

3 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

−5/8+3/4 equals what, In fraction form.

kipiarov [429]

How to get answer by Mimiwhatsup (In Decimal Form):

$-5/8+3/4\\\mathrm{Divide\:the\:numbers:}\:5\div \:8=0.625\\=-0.625+3\div \:4\\\mathrm{Divide\:the\:numbers:}\:3\div \:4=0.75\\=-0.625+0.75\\\mathrm{Add/Subtract\:the\:numbers:}\:-0.625+0.75=0.125\\=0.125$