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3 years ago

3x^2 - 4x + 7 (Find the domain of the expression)

Mathematics

1 answer:

Jobisdone [24]3 years ago

3 0

x1 = −(−4) − √ (−4)2 − 4×3×7 2×3 = 4 − 2 × √ 17 i 6 ≈ 0.666667 − 1.374369ix2 = −(−4) + √ (−4)2 − 4×3×7 2×3 = 4 + 2 × √ 17 i 6 ≈ 0.666667 + 1.374369i

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Translate and solve: eight more than the product of three and a number is -13

Murrr4er [49]

Ok, so let’s take the given phrase.
Eight more than the product of 3 and a number is -13. A translation to numbers looks like this:
3x + 8 = -13
Now, let’s solve for x. We can start the process by subtracting 8 from both sides.
3x = -21
now, we will divide each side by 3 to isolate the variable, thus solving for x.
x = -7

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3 years ago

What would x be equal to?

Vilka [71]

Answer:

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Step-by-step explanation:

Given

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3 years ago

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sergij07 [2.7K]

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2 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
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$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
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$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago

Is V 64 rational or irrational

Marizza181 [45]

Answer:

rational

Step-by-step explanation:

enjoy the great answers

4 0

2 years ago

Read 2 more answers