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Arte-miy333 [17]
3 years ago
10

What linear function equation can be represented by the set of ordered pairs? {(−4, −6), (0, −3), (4, 0), (12, 6)} Enter your an

swer in the box.
Mathematics
2 answers:
Anika [276]3 years ago
7 0

Answer:

The equation of the line is f(x) = 3/4x - 3. Checked this answer and it is 100% correct.

dalvyx [7]3 years ago
5 0

Answer:

The equation of the line is y = 3/4x - 3

Step-by-step explanation:

To find the equation of the line, pick any two points and find the slope.

m(slope) = (y2 - y1)/(x2 - x1)

m = (6 - 0)/(12 - 4)

m = 6/8

m = 3/4

Now that we have the slope, we can use the slope and point-slope form to find the equation in slope intercept form.

y - y1 = m(x -x1)

y - 6 = 3/4(x - 12)

y - 6 = 3/4x - 9

y = 3/4x - 3

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

6 0
3 years ago
Which expression is equivalent to √4j^4/9k^8?
solmaris [256]
\frac{ 2j^{2} }{ 3^{4} }
SQRT each individual bit: 4 andj^{4} and 9 and k^{8}
6 0
4 years ago
Read 2 more answers
How can you decompose the composite figure to determine its area?
Anna007 [38]

For this case the first thing you should do is observe that the diameter of the four semicircles is the same.

Therefore, we can decompose the figure as follows:


1) We draw the diameters of the four semicircles to form a square.


2) We divide the figure into a square and four semicircles


3) The total area is the sum of the area of the square, plus the area of the 4 semicircles.


Answer:

c)as a square and four semicircles


5 0
3 years ago
Read 2 more answers
ILL BRAINLIEST YOU PLEASE HELP ME
gtnhenbr [62]

Answer:

Option C

The length of RT is 14

Step-by-step explanation:

For a given Right Triangle

m∠R = 90°, RG = 17, TG = 22

Now, <u>By Pythagoras Theorem</u>

(RT)² + (RG)² = (TG)²

(RT)² = (TG)² - (RG)²

(RT)² = (22)² - (17)²

(RT)² = 484 - 289

(RT)² = 195

RT = √195

RT = √195 = 13.96 = 14

Thus, The length of RT is 14

<u>-TheUnknownScientist</u>

4 0
3 years ago
The hypotenuse of a right triangle is twice the length of one of its legs. Th length of the other leg is three feet. Find the le
shutvik [7]

Step-by-step explanation:

go for it this is your questions answer

3 0
3 years ago
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