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Alenkasestr [34]
3 years ago
8

A group of 8 friends (5 girls and 3boys ) plan to watch a movie, but they have only 5 tickets. How many different combinations o

f 5 friends could possibly receive the tickets ?
Mathematics
1 answer:
Molodets [167]3 years ago
8 0
The answer will be 56 different combinations
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\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}}

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