A.) Find a general linear equation Ax+By+C=0 of the straight line that passes through the point (2,1) and is vertical.

1 answer:

5 0

the straight line passes through the point (2,1) and is vertical

x=2 (vertical), so we can have the value of B and C, x-2=0 so B=0 and C=-2
the main equation is x - 2 = 0

inear equation Ax+By+C=0 of the straight line that passes through the point (-2,3) and is horizontal.
y=3, (horizontal) y-3=0, so A=0, B=1 and C=-3 the main equation is y-3=0

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Simplify the expression (6x^3/3x^4)^2 A. 12x^5/6x^6 B. 4/x^2 C. 2x^6/x^8 D. 2/x

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The Answer is B.
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8 0

3 years ago

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The vertex of this parabola is at (2,-1) when the y value is 0 and then x value is 5 what is the coefficient of the squared term

Darina [25.2K]

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}$

now, let's expand the squared term to get the standard form of the quadratic.

$\bf y=\cfrac{1}{9}(x-2)^2-1\implies y=\cfrac{1}{9}(x^2-4x+4)-1 \\\\\\ y=\cfrac{1}{9}x^2-\cfrac{4}{9}x+\cfrac{4}{9}-1\implies \stackrel{its~coefficient}{y=\stackrel{\downarrow }{\cfrac{1}{9}}x^2-\cfrac{4}{9}x-\cfrac{5}{9}}$