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2 years ago

Which line is parallel to 8x+2y=12

Mathematics

2 answers:

Nat2105 [25]2 years ago

4 0

x=3/2-1/4

8x+2y=12

move the variable to the right side and change the sign.

8x=12-2y

divide both sides of the equation by 8

x=3/2-1/4y

the final solution is

x=3/2-1/4y

salantis [7]2 years ago

4 0

Answer: x= 14y+32

Step-by-step explanation:

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Step-by-step explanation:

Draw a triangle and label the sides (picture below):

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2 years ago

Write the equation of the line that contains the given point and has the given slope.

ArbitrLikvidat [17]

Hey there!!

What is slope-intercept form :

... y = mx + b

( a ) Given :

... ( 2 , -2 ) and slope=1.4

... y = mx + b

... -2 = 2×1.4 + b

... -2 = 2.8 + b

... -4.8 = b

The slope-intercept form :

... y = 1.4x - 4.8

( b ) Given :

... ( -1 , 4 ) and slope = -3.

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... 4 = -3×-1 + b

... 4 = 3 + b

... 1 = b

The slope-intercept form :

... y = -3x + 1

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2 years ago

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X-1/x-2+x+3/x-4=2/(x-2).(4-x)

Varvara68 [4.7K]

The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

According to the statement

we have given that the equation and we have to prove that the given answer is a correct answer for those equivalent equation.

So, The given expression are:

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

And we have to prove the answer.

So, For this

$\frac{x-1}{x-2} +\frac{x+3}{x-4}$

$\frac{({x-1}) ({x-4}) +({x+3})({x-2})} {(x-2) (x-4)}$

Then the equation become

$\frac{x^{2} -4x -x +4 + x^{2} -2x + 3x -6 }{(x-2) (x-4)}$

Now solve it then

$2x^{2} - 4x -2 / (x-2) (x-4)$

Now take 2 common from answer then equation become

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

Hence proved.

So, The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

Learn more about equations here

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3 0

10 months ago

Create the equation of a quadratic function with a vertex of (5,6) and a y-intercept of -69

Elena L [17]

If the y-intercept is at -69, meaning the point is (0, -69), thus x = 0, y = -69

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
vertex~(5,6)\quad 
\begin{cases}
x=5\\
y=6
\end{cases}\implies y=a(x-5)^2+6
\\\\\\
\textit{we also know that }
\begin{cases}
x=0\\
y=-69
\end{cases}\implies -69=a(0-5)^2+6
\\\\\\
-75=25a\implies \cfrac{-75}{25}=a\implies a=-3
\\\\\\
therefore\qquad \boxed{y=-3(x-5)^2+6}$

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2 years ago