PLEASE HELP I DONT GET THIS!! If you answer both, you get 20 points! If you answer one, you get 10 points!

Mathematics

1 answer:

Tatiana [17]3 years ago

3 0

17. y = .5x + 2

Coordinates:
(0, 2)
(1,2.5)
(2,3)
(3,3.5)
(4,4)
(5,4.5)

18. y = x + 1

Coordinates:
(0,1)
(1,2)
(2,3)
(3,4)
(4,5)
(5,6)

This one should just be a straight diagonal line, so it'll be easy to graph.

You have to do the plotting, I don't have a tool for that.

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<h2> The first and the third</h2>

Step-by-step explanation:

$x^2+\frac bax+\frac ca=0\\\\ ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\qquad\quad x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\\\x_1+x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}+\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{-2b}{2a}=\dfrac{-b}a\\\\\\x_1\cdot x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\cdot\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\\\\{}\ \ =\dfrac{b^2-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac}-(\sqrt{b^2-4ac})^2}{2a}=\dfrac{b^2-(b^2-4ac)}{4a^2}=\\\\{}\ \ =\dfrac{b^2-b^2+4ac}{4a^2}=\dfrac{4ac}{4a^2}=\dfrac{c}{a}$

$x_1-x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-2\sqrt{b^2-4ac}}{2a}=\frac{-\sqrt{b^2-4ac}}{a}\\\\\\x_1\div x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\div\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-b-\sqrt{b^2-4ac}}{2a}\,\cdot\,\frac{2a}{-b+\sqrt{b^2-4ac}}=\\\\=\frac{-b-\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}=\frac{b+\sqrt{b^2-4ac}}{b-\sqrt{b^2-4ac}}=\frac{b^2+2\sqrt{b^2-4ac}+b^2-4ac}{b^2-b^2+4ac}=\frac{2b^2+2\sqrt{b^2-4ac}-4ac}{4ac}=$

$=\frac{b^2+\sqrt{b^2-4ac}-2ac}{2ac}$

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