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1 year ago

What is the range of the function graphed below?

Mathematics

1 answer:

Nadya [2.5K]1 year ago

3 0

Answer:

Step-by-step explanation:

Range is the 'y' values a graph can have

you can see that the 'y' values can be anythig from - inf to + 2

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What is the equation of the line in point-slope form?

Alisiya [41]

Answer: y=-2x+2/4

Step-by-step explanation:

6 0

3 years ago

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BRAINLIEST AWARDED TO FIRST CORRECT ANSWER NEED ASAP! AND 20 POINTS!! The table shows the cost for taxi rides of various distanc

Elza [17]

Answer:

b because it cost u more for a certain distance

Step-by-step explanation:

7 0

3 years ago

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A quadratic function is defined by

ella [17]

Answer:

1. The vertex of the graph of the function is (-4, 7)

2. The vertex represents a minimum value

3. The equation represented by the new graph is g(x) = (x + 4)² + 1

Step-by-step explanation:

The given quadratic function is given in vertex form, y = a·(x - h)² + k, as follows;

g(x) = (x + 4)² + 7

1. By comparing the given quadratic function and the vertex form of a quadratic equation, we have;

a = 1, h = -4, and k = 7

The vertex of the graph of the function, (h, k) = (-4, 7)

2. Given that a = 1 > 0, the graph of the quadratic function opens upwards and the vertex represents a minimum value

3. Shifting the graph 6 units down from where it is now will give;

The vertex = (h, k - 6) = (-4, 7 - 6) = (-4, 1)

h = -b/(2·a), k = (-b²/(4·a) + c = -a·h² + c

c = k + a·h²

Therefore, initial value of the constant term, c = -1×(-4)² + 7 = 23

After the shifting the graph 6 units down, c = 23 - 6 = 17

∴ k = 1 = -a·(-4)² + 17

∴ -16/16 = -1 = -a

a = 1

Therefore, the equation represented by the new graph is g(x) = (x + 4)² + 1.

5 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

Solve the equation15=2y-5

Lesechka [4]

15 = 2Y - 5
20 = 2Y
10 = Y

5 0

3 years ago