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Veseljchak
1 month ago
7

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

Mathematics
2 answers:
tester [12.3K]1 month ago
5 0

Answer:

  0 < k < 2

Step-by-step explanation:

The intersection will occur in the first quadrant if the solution satisfies these conditions: x > 0, y > 0.

To ensure x > 0, we need...

  x = 2k

  x > 0

  2k > 0.... substituting for x

  k > 0..... dividing by 2

__

Next, for y > 0, we need...

  y = (12 -3x)/2

  y = (12 -3(2k))/2 = 6 -3k..... substitute for x and simplify

  y > 0

  6 -3k > 0...... substituting for y

  2 -k > 0....... dividing by 3

  k < 2......... adding k

Thus, the values for k that ensure the intersection is in the first quadrant are...

  0 < k < 2

AnnZ [12.3K]1 month ago
3 0

Answer:

0 < k < 2

Step-by-step explanation:

In the first quadrant, both x and y are positive, thus

  • x > 0 and y > 0
  • x = 2k, where k > 0

Substituting x with 2k into the second equation gives:

  • 3x + 2y = 12
  • 3*2k + 2y = 12
  • 2y = 12 - 6k
  • y = 6 - 3k

Given that y > 0:

  • 6 - 3k > 0
  • 2 - k > 0
  • k < 2

By combining k > 0 and k < 2, we obtain:

  • 0 < k < 2
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