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kenny6666 [7]
3 years ago
14

Which is the equation of a line that passes through the point (-1,8) and has a y-intercept of 15?

Mathematics
1 answer:
Stolb23 [73]3 years ago
5 0
D. You can substitute the ordered pair in the equation
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The formulat for finding the slop is

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so if we put the values of our two given points

m = (-13 -12) / (5 - 5)  = -25/0 = undefined

so our slop is undefined which means it is a vertical line


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Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1
MA_775_DIABLO [31]

Answer:

\int_R 15x+15y dA = \frac{8}{16875}

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where \frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}.

We will calculate the jacobian of the change variables. That is

\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right| (we are calculating the determinant of this matrix). The matrix is

\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75} (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v

We will use the change of variables theorem. So,

\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}

5 0
3 years ago
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