(x, y) --> (x + 5, y - 1)
A(3, -1) --> A'(3 + 5, -1 - 1) --> A'(8, -2)
B(6, 1) --> B'(6 + 5, 1 -1) --> B'(11, 0)
C(2, 4) --> C'(2 + 5, 4 - 1) --> C'(7, 3)
D(-1, 3) --> D'(-1 + 5, 3 - 1) --> D'(4, 2)
8x²+3y²=24
8x²+3y²-24=0
Let x be 1 ,
Therefore
8+3y²-24=0
3y²-16=0
3y²=16
y²=16/3
y=4/1.7
y=40/17
<em><u>Take</u></em><em><u> </u></em><em><u>different</u></em><em><u> </u></em><em><u>values</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>&</u></em><em><u>y</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>plot</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>graph</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em>
X-int:let y=0
3x-2(0)=18
3x=18
divide both sides by 3
therefore x=6
y-int:let x=0
3(0)-2y=18
-2y=18
we divide both side by -2
therefore y=-9
It is considered categorical data. Categorical data is the edible plants that are encountered.
