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2 years ago

8

HELP PLEASE ASAP I will mark brainliest if you explain

1 answer:

scoray [572]2 years ago

3 0

Answer:

y - 5, (x, y) - (0,5)

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victus00 [196]

Answer:

F the point is 3.16

Explantion:

the point is 3.16, which is close to 3.2

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3 years ago

Let f(x) = x^3 + 2x^2 + 3x + 1 and g(x)= 4x-5 Find f(x) x g(x)

Brums [2.3K]

$▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪$

Given :

$f(x) = {x}^{3} + 2 {x}^{2} + 3x + 1$

$g(x) = 4x - 5$

let's solve for f(x) × g(x) :

$( {x}^{3} + 2x {}^{2} + 3x + 1)(4x - 5)$

$4x {}^{4} - 5 {x}^{3} + 8 {x}^{3 } - 10 {x}^{2} + 12 {x}^{2} - 15x + 4x - 5$

$4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 11x - 5$

therefore, correct choice is : D

8 0

2 years ago

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kolezko [41]

It’s the first one.

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3 years ago

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andrew11 [14]

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2 years ago

Read 2 more answers

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