2 years ago

What is the linear equation that has slope of 5 and pass through the point (-1, -3)?

1 answer:

4 0

What you would do is that you would use the y=mx+b linear equation
m=5
x=-1
y=-3
b=?
hence, you must solve b to have the equation

-3=(5)(-1)+b
-3=-5+b
-3+5=b
2=b
therefore the equation is
y=5x+2

for standard form you just move everything to equal 0 because there is no fraction next to the x
thus

-5x+y-2=0

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Suppose that weight of adolescents is being studied by a health organization and that the accompanying tables describes the prob

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Answer:

<em>hello your question lacks some information attached below is the complete question</em>

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Step-by-step explanation:

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3 years ago

Thee vertex of a parabola is (-5,2), and its focus is (-1,2). What is the standard form of the parabola?

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Check the picture below. So the parabola looks more or less like so.

$\bf \textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \begin{cases} h=-5\\ k=2\\ p=4 \end{cases}\implies 4(4)[x-(-5)]=[y-2]^2\implies 16(x+5)=(y-2)^2 \\\\\\ x+5=\cfrac{1}{16}(y-2)^2\implies x = \cfrac{1}{16}(y-2)^2-5$