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3 years ago

What is the value of this expression when X=-1 and Y = 2? 44²7²

Mathematics

1 answer:

jeyben [28]3 years ago

6 0

For the example in the explanation the result is:

$\boxed{E=-1}$

<h2>Explanation:</h2>

Hello! remember to write complete questions in order to find good and exact answers. Here you haven't provided any expression so I'll help you writing an expression. Suppose your expression is:

$E=XY^2+2Y+X$

So in order to know the value of our expression E, we need to substitute both x and y in it, so:

$E=XY^2+2Y+X \\ \\ E=(-1)(2)^2+2(2)-1 \\ \\ E=-4+4-1 \\ \\ \boxed{E=-1}$

<h2>Learn more:</h2>

Distributive property: brainly.com/question/10794694

#LearnWithBrainly

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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

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

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Answer:

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2 x 36

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2 x 2 x 2 x 3 x 3

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