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mrs_skeptik [129]
3 years ago
15

2x-y=2s 2x-2y=4s The line whose equations are shown intersect at which point?

Mathematics
1 answer:
sergejj [24]3 years ago
7 0
There are three variables, so it really doesn't make sense. But, if you don't solve for s, and make it stay how it is, then the intersect would be (0, -2s)
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9:6, 12:8, and 15:10 are all ratios equivalent to 3:2

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Please help me w the answer
evablogger [386]

Answer:

\frac{2(x-6)(x-10)}{(x-4)(x-5)}

Step-by-step explanation:

In essence, one needs to work their way backwards to solve this problem. Use the information to construct the function.

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\frac{()}{(x-4)(x-5)}

The function has x-intercepts at (6, 0), and (0, 10). This means that the numerator must equal (0) when (x) is (6) or (10). Using similar logic that was applied to find the denominator, one can conclude that the numerator must be ((x - 6)(x-10)). Now one has this much of the function assembled

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\frac{2(x-6)(x-10)}{(x-4)(x-5)}

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\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})


so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}


just a quick note on notation:

\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}

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