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3 years ago

Determine the equation of the line that passes through the points (-3. -7) and (5, 2)

Mathematics

1 answer:

a_sh-v [17]3 years ago

7 0

Step-by-step explanation:

Given points

( -3 , -7) and ( 5, 2)

Now slope (m) 

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

= ( 2 + 7) / ( 5+ 3)

= 9 / 8

Now the equation of line is 

y - y1 = m ( x - x1) 

y + 7 = 9/8 ( x + 3)

8 ( y + 7) = 9 ( x + 3)

8y + 56 = 9x + 27

9x - 8y +27 - 56= 0

9x - 8y -29 = 0

Which is the required equation 

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2 years ago

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Solve for A in the equation AB + CD + EFG = Z.

Semenov [28]

Answer:

A = (Z-CD - EFG)/B

Step-by-step explanation:

AB + CD + EFG = Z.

Subtract CD and EFG from each side

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AB = Z-CD - EFG

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3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$