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andreyandreev [35.5K]
3 years ago
9

Enter an inequality that represents the graph in the box.

Mathematics
2 answers:
Slav-nsk [51]3 years ago
7 0
Since y=-1 when x=0 and we can write the equation (we'll turn it into an inequality later) as y=mx-1 from y=mx+b by plugging (0,1) in. Next, y equals 2 when x=1, so we plug those in to get 2=m*1-1. Adding 1 to both sides, we get 3=m, making our equation y=3x-1 (since y and x stay variables). Lastly, we turn it into an inequality, It seems to be inclusive to the line, so it's either 
3x-1≤y or 3x-1≥y. Finding a random point in the inequality (4, 1), we plug it in to get 12-1=11, which is clearly larger than 1, so we get 3x-1≥y.
ziro4ka [17]3 years ago
6 0

The correct answer is

y <= 3x -1

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Step-by-step explanation:

The system of linear equations are , (i) y = 2x and (ii) y = x + 1 . We will find some coordinates and then we will plot its graph. The point where both Graphs will meet will be the solution of the graph .

<u>•</u><u> </u><u>Findin</u><u>g</u><u> </u><u>coordinates</u><u> </u><u>of </u><u>Equ</u><u>ⁿ</u><u> </u><u>(</u><u>i</u><u>)</u><u> </u><u>:</u><u>-</u>

Step 1 : <u>Put</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>0</u><u> </u><u>:</u><u>-</u>

\tt y = 2x = 2(0) = \red{0}

Step 2: <u>Put</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

\tt y = 2x = 2(1) = \red{2}

\boxed{\begin{array}{c|c|c} \bf x & \tt 0 &\tt 1 \\ \bf y &\tt 0 & \tt 2 \end{array}}

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<u>•</u><u> </u><u>Finding</u><u> </u><u>coord</u><u>inates</u><u> </u><u>of</u><u> </u><u>equⁿ</u><u> </u><u>(</u><u>ii</u><u>)</u><u> </u><u>:</u><u>-</u>

Step 1 : Put x = 0 :-

\tt y = x + 1  = 1 + 0  = \red{1}

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\tt y = x + 1 = 1 + 1 = \red{2}

\boxed{\begin{array}{c|c|c} \bf x & \tt 0 &\tt 1 \\ \bf y &\tt 1 & \tt 2\end{array}}

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