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Alexxx [7]
3 years ago
6

Given that f(x) = 3x + 5, find f(x+2)

Mathematics
2 answers:
julsineya [31]3 years ago
8 0

Answer:

f(x)=3(x)+5

f(x+2)=3(x+2)+5

=3x+6+5

=3x+11

Andre45 [30]3 years ago
5 0
F(x+2) = 3(x+2) + 5
= 3x + 6 + 5
= 3x + 11
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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
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
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)
a_0=-\dfrac{2L^2}3

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
4 0
3 years ago
Denise is selling roses. She sold 3 for $5 and 5 for $7. What will Denise charge for 12 roses if she keeps selling roses at the
Vlada [557]

Answer:

14$

Step-by-step explanation:

3 0
3 years ago
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Please answer and explain
sashaice [31]
8=2*2*2=2^3
32=2*2*2*2*2=2^5
<span>[8^(2/3)] / [32^(2/5)] = [</span>(2^3)^(2/3)] / [(2^5)^(2/5)] =
= [2^(3*2/3)] / [2^(5*2/5)]= [2^2] / [2^2] = 4/4 = 1
Answer: 1
4 0
3 years ago
Define fn : [0,1] --&gt; R by the
sasho [114]

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

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f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

8 0
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Ronch [10]

Answer: 3.0 moles of water.

3 0
2 years ago
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