What is the slope of a line that is parallel to y=2

Mathematics

1 answer:

avanturin [10]3 years ago

6 0

Answer:

Step-by-step explanation:

All lines that are parallel to y=−2 have the same slope of 0 .

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WILL GIVE 32 POINTS FOR WHOEVER GIVES CORRECT ANSWER

Brilliant_brown [7]

Graph the line that goes through the point (3,-1) and has a y-intercept of -4:

You can simply graph two points to create the line for this problem. The points are (3, -1) and (0, -4). The slope for this line is 1/1 too, which means that if you need more points, you would start at either of the two given points and go up one unit and over one unit to the right.

Select three points that the line would go through (y = x - 3):

In the equation, the y-intercept is given, which is one point - (0,-3). Then, we can go up one unit and right one unit to find a second point - (1,-2). And also down one unit and left one unit to find a third point - (-1,4).

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First, let's find the y-intercept, which is (0,6). Then, we can find other points where the line crosses and count how many units up and to the right that would be, which is 1/1. Therefore, the correct equation is B - y = x + 6.

Which of the following graphs represents y = x - 2:

The y-intercept is (0,-2) and the slope is positive and 1/1. We are looking for a graph that crosses through the point (0,-2), points up and to the right, and increases by 1 unit up and 1 unit right each time. Therefore, the correct graph is B.

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2 years ago

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Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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2 years ago

Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(

BlackZzzverrR [31]

We can parameterize this part of a cone by

$\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle$

with $0\le u\le a$ and $0\le v\le2\pi$ . Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv$

The area of this surface (call it $\mathcal S$ ) is then

$\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi$