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igor_vitrenko [27]

3 years ago

15

Find an equation of the line with the slope m= 3/2 that passes through the point (8,-3). Write the equation in the form Ax+By =

C.

1 answer:

mr Goodwill [35]3 years ago

4 0

Answer:

work is shown and pictured

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Find any values of x that must be excluded from the domain. This part really confuses me.

gregori [183]

The only restriction on the domain is that division by zero is undefined, or "not allowed", so x cannot equal zero.

So the domain is x=(-oo,0),(0,+oo)

3 0

3 years ago

Read 2 more answers

Find the altitude of triangle whose vertex is a(1,-2) and the equation of the base is X+y=3.

Sophie [7]

Answer:

$2\sqrt {2}$

Step-by-step explanation:

<u>Distance From a Point to a Line</u>

Given a line with an equation :

ax + by + c = 0

And the point $(x_0,y_0)$

The distance from the line to the point is given by

$\displaystyle d=\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}$

The triangle has a vertex at (1,-2) and the base lies on the equation

x + y = 3

Rearranging:

x + y - 3 = 0

The altitude of the triangle is the distance from the point to the line.

The values to use in the formula of the distance are: a=1, b=1, c=-3, xo=1, yo=-2:

$\displaystyle d=\frac {|1*1+1*(-2)-3|}{\sqrt {1^{2}+1^{2}}}$

$\displaystyle d=\frac {|1-2-3|}{\sqrt {2}}$

$\displaystyle d=\frac {4}{\sqrt {2}}$

Rationalizing:

$\displaystyle d=\frac {4}{\sqrt {2}}\cdot\frac{\sqrt {2}}{\sqrt {2}}$

$\displaystyle d=2\sqrt {2}$

The altitude of the triangle is $\mathbf{4\sqrt {2}}$

6 0

2 years ago

Solve using substitution<br><br> x-y= 6<br> 4x+2y=36

Elza [17]

X=6+y
4(6+y) +2y=36
24+6y=36
6y=12
Y=2
X=8

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2 years ago

What is the slope of a line parallel to the line whose equation is 12x+15y=225

faltersainse [42]

Y= -4/5 + 25
.....................

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3 years ago

How do I simplify

Vsevolod [243]

X^2•y^12/xy^5 = x^a•y^b

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3 years ago