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3 years ago

10

What’s the equation for the order pairs (1,-5) and (-2,5)

1 answer:

liq [111]3 years ago

6 0

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})\qquad
(\stackrel{x_2}{-2}~,~\stackrel{y_2}{5})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-(-5)}{-2-1}\implies \cfrac{5+5}{-2-1}\implies \cfrac{10}{-3}$

$\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-(-5)=-\cfrac{10}{3}(x-1)\implies y+5=-\cfrac{10}{3}x+\cfrac{10}{3}
\\\\\\
y=-\cfrac{10}{3}x+\cfrac{10}{3}-5\implies y=-\cfrac{10}{3}x-\cfrac{5}{3}$

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We have

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Read 2 more answers

Find all the zeroes of the equation(with simple steps).

uysha [10]

<u>Answer-</u>

<em>The zeros are, $5,\ -5,\ 4i,\ -4i$ </em>

<u>Solution-</u>

$\Rightarrow -3x^4+27x^2+1200=0$

$\Rightarrow -3(x^2)^2+27(x^2)+1200=0$

Here,

a = -3, b = 27, c = 1200

So,

$x^2=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}$

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So,

$\Rightarrow x^2=25,\ -16$

$\Rightarrow x=\sqrt{25},\ \sqrt{-16}$

$\Rightarrow x=5,\ -5,\ 4i,\ -4i$

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