) What is the solution to the system of equations?

Mathematics

1 answer:

yaroslaw [1]2 years ago

4 0

$\begin{cases} y = -\cfrac{1}{3}x+2\\[1em] y = x + 2 \end{cases}\qquad \implies \stackrel{y}{-\cfrac{1}{3}x+2}~~=~~\stackrel{y}{\cfrac{}{}x + 2} \\\\\\ -\cfrac{1}{3}x=x\implies -x=3x\implies 0=4x\implies \cfrac{0}{4}=x\implies \boxed{0=x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{y = x + 2\implies y = 0+2\implies }\boxed{y = 2} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (0~~,~~2)~\hfill$

You might be interested in

I need help finding the mean, median, and mode(s) I need help with number 7

natta225 [31]

Answer:

the mean is 57.5 the median is 44.5 and the mode is 47

Step-by-step explanation:

the mode is 47 cuz it appeared twice. the median is 57.5 bc 47+147+24+47+38+42=345 and 345/6 is 57.5

7 0

2 years ago

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

valina [46]

Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

$P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}$