Find the values for k so that the intersection of x = 2k and 3x + 2y = 12 lies in the first quadrant. PLEA

0< k < 2

Step-by-step explanation:

x = 2k

3x + 2y = 12

Substitute the first equation into the second equation

3(2k) +2y = 12

6k+2y = 12

Subtract 6k from each side

2y = 12 -6k

Divide by 2

y = 6 - 3k

For this to be in the first quadrant y > 0

6 - 3k > 0

6 > 3k

Divide by 3

2 > k

The lower limit is when it reaches the axis or when k=0

0 <k <2

Checking for x

x = 2k

If 0<k <2 then it will be in the first quadrant

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.