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san4es73 [151]
3 years ago
8

Please help me with both of these!!

Mathematics
1 answer:
boyakko [2]3 years ago
4 0
I only know the second one, you would use pemdas to break down the equation doing exponents first which would leave you with 2x x 2x x 2x x 2x which would leave you with 16x and the second one would be 4 x 4 which is also 16x so the new equation would be y = 16x - 16x - 3x + 1 which would leave you with an answer of -3x + 1. Hope this helped
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so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})

\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\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\implies \blacktriangleright  y = 4x+(-1)\blacktriangleleft

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