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3 years ago

Find equation of a line that is perpendicular to y= -1/2x+9 and passes through (4,12).

Mathematics

1 answer:

Georgia [21]3 years ago

7 0

Answer:

y=2x+4

Step-by-step explanation:

m=2 because it is perpendicular to (-1/2)

y-12=2(x-4)

y=2x+4

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Factorise this equation

pashok25 [27]

$x^{2} - 2x +35$

To factor we must first find factors of 35

The only factors of 35 are : 5 & 7

-2x is what 5 and 7 make

but only if 7 is negative and 5 is positive

so our final factors are

(x - 7) (x + 5)

______________________________________________

$2x^{2} +7x +6$

For the second equation, were going to multiple the coefficient of $x^{2}$ ( which is 2 ) by 6

We get 12

What factors of 12 add up to 7?

3 and 4

So our final factors are

(2x+3)(x+2)

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3 years ago

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Alex Ar [27]

The answer is 2. The numerator adds to -8 while the denominator adds to -2. Divide them and get 4.

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3 years ago

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:

Distance From a Point to a Line

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 $\mathbf{4\sqrt {2}}$

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ivanzaharov [21]

Answer:

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