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Karo-lina-s [1.5K]
3 years ago
9

What does the line 6x+2y=8 look like

Mathematics
2 answers:
sdas [7]3 years ago
7 0
To determine what the line looks like, simply change it to y = mx + b form. Then simply either make a table of values. And substitute any points into the equation and obtain points to plot on the graph. Or you can simply plot the graph by using the slope of the equation.

Y = 8 - 6x/2
Y = 4 - 3x
Y = -3x + 4.

This would be the rearranged equation.
Ugo [173]3 years ago
5 0
You can rewrite the Sum as follows:
(6x+2y=8)/2 = (3x+y=4) This is because 2 is a common factor of both sides

y = 4-3x (moving the 3x to the other side)

Then you can tell that the line will have a negitive gradient because you are taking away the 3x from the right hand side.
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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Csqrt%7Bx%7D%20" id="TexFormula1" title="f(x) = \sqrt{x} " alt="f(x) =
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<u>Step-by-step explanation:</u>

\lim_{h \to 0} f(x)=\dfrac{f(x+h)-f(x)}{h}

f(x) = \sqrt x\\

f(x+h) = \sqrt{x+h}

\lim_{h \to 0} f(x)=\dfrac{\sqrt{x+h}-\sqrt x}{h}

                   =\dfrac{\sqrt{x+h}-\sqrt x}{h}\bigg(\dfrac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\bigg)

                   =\dfrac{(x + h)-(x)}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{h}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{1}{\sqrt{x+h}+\sqrt x}

                  =\dfrac{1}{\sqrt{x+0}+\sqrt x}

                  =\dfrac{1}{2\sqrt x}

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