2 years ago

Is it possible for association that is not linear

1 answer:

4 0

Answer: yes

Step-by-step explanation:

You might be interested in

exis [7]

They both bought 1 bottle each.

4 0

2 years ago

DIA [1.3K]

For this case we have that by definition, the point-slope equation of a line is given by:

$y-y_ {0} = m (x-x_ {0})$

We have the following points:

$(x_ {1}, y_ {1}) :( 2,6)\\(x_ {2}, y_ {2}) :( 4,0)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-6} {4-2} = \frac {-6} {2} = -3$

We chose a point:

$(x_ {0}, y_ {0}) :( 4,0)$

Substituting in the equation we have:

$y-0 = -3 (x-4)\\y = -3 (x-4)$

Finally, the equation is: $y = -3 (x-4)$

Answer:

OPTION C

6 0

3 years ago

Nutka1998 [239]

865:2678 is the answer

8 0

2 years ago

Paraphin [41]

That's easy
5x+4x–x=8x

8 0

3 years ago

Xelga [282]

The answer to your question is w =40(2)n [n is in exponent form I just couldn’t type it that way ]

4 0

2 years ago