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2 years ago

Find the equation of the linear function represented by the table below in slope-intercept form.

1 answer:

5 0

well, to get the equation of a line all we need is two points, so let's simply pick two points off the table, hmmmm say (1 , -3) and hmmm (11 , -43)

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{11}~,~\stackrel{y_2}{-43}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-43}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{11}-\underset{x_1}{1}}}\implies \cfrac{-43+3}{10}\implies \cfrac{-40}{10}\implies -4$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-4}(x-\stackrel{x_1}{1}) \\\\\\ y+3=-4x+4\implies y=-4x+1$

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3 years ago

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BabaBlast [244]

Answer:

$\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1$

Step-by-step explanation:

<u>Algebra Simplifying</u>

We are given the expression:

$\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }$

We need to repeatedly use the following identity:

$(a^2-b^2)=(a-b)(a+b)$

For example, the first numerator has a difference of squares thus we factor as:

$(a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)$

The first denominator can be factored also:

${a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)$

Applying the same procedure to all the expressions:

$\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}$

Simplifying all the fractions:

$\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}$

Since all the denominators are equal:

$\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}$

Simplifying:

$\displaystyle T=\frac{a +c+b}{a+b+c}$

Simplifying again:

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Thus:

$\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}$

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3 years ago

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