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2 years ago

Which equation shows a line with a slope of 3 that passes through the point ((1,-2)

Mathematics

1 answer:

qaws [65]2 years ago

7 0

Answer:

y = 3x - 5

Step-by-step explanation:

Slope m = 3

Using (1,-2)

Slope-intercept: y = mx + b

-2 = 3(1) + b

-2 = 3 + b

b = -5

then y = 3x - 5

You might be interested in

Csc x= -√2 for π≤x≤3π/2

love history [14]

Answer:

Step-by-step explanation:

We can write this as Sinx by "flipping" the $-\sqrt{2}$ .

So we will have: $Sin(x)=-\frac{1}{\sqrt{2} }$

From basic trigonometry, we know the value of $\frac{1}{\sqrt{2}}$ of sine is of the angle $\frac{\pi}{4}$

But when is sine negative? Either in 3rd or 4th quadrant. But the answer has to be between 0 and $\frac{3\pi}{2}$ , so we disregard 4th quadrant.

To get the angle in 3rd quadrant, we add π to the acute angle of the first quadrant (which is π/4 in our case). Thus we have:

$\frac{\pi}{4}+\pi\\=\frac{\pi +4\pi}{4}\\=\frac{5\pi}{4}$

A is the right answer.

7 0

3 years ago

Can someone help me identify c(6)=

Vanyuwa [196]

Answer and Step-by-step explanation:

To solve, simply plug 6 in for m in the equation.

C(6) = 50(6) + 79.50

6 times 50 is 300.

300 + 79.50 = 379.50 = C(6)

The answer is 379.5

(Looking at the question again, I don't think what I said is what it is asking for, soo)

The slope of the function given represents the amount that needs to be paid every month. The y-intercept (79.50) represents the initial cost (starting amount) for paying the membership.

#teamtrees #PAW (Plant And Water)

3 0

3 years ago

Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units? the value of b t

Artist 52 [7]

The value of b to 2.
By doing that you are applying a vertical translation to the function

3 0

3 years ago

Read 2 more answers

Which conic section is represented by the following equation? 9x^2 – 5x – 3y^2 – 2y + 4 = 0

stealth61 [152]

Answer:

the awnser is B. Elipse

Step-by-step explanation:

its right on edge 2021

3 0

3 years ago

Read 2 more answers

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