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2 years ago

Solving the math function

Mathematics

2 answers:

Sati [7]2 years ago

8 0

Answer:

C) $f(x)=(x-1)^\frac{1}{2}$

Step-by-step explanation:

The graph shows a radical function that starts at (0,1). This must mean that the original function had been translated to the right one unit. Therefore, $f(x)=(x-1)^\frac{1}{2}$ is correct.

Sunny_sXe [5.5K]2 years ago

3 0

Answer: the answer is C

Step-by-step explanation: trust me took the final yesterday ‍♀️

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A car is 160 inches long.

sladkih [1.3K]

Answer:

Length of truck = 160 +7% of 160 = 160 +11.2 = 171.2inches

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3 years ago

Determine the sum of the first 9 terms in the geometric series where r = -0.25 and a1 = 256

Yuri [45]

The answer would ve B. -204.8 the sum

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3 years ago

Read 2 more answers

Correct answers only!

malfutka [58]

Answer:

$60.5

Step-by-step explanation:

Simply plug in all known values to the equation. You know that p is 50, r is 0.10, and t is 2, for two years. After you plug in all those values, you can solve for this equation, and get 60.5 as your answer.

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2 years ago

The brackets for a shelf are in the shape of a triangle. Find the angles of the triangle if the

zimovet [89]

Answer:

Step-by-step explanation:

2x+3x+5x = 180°

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6 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago