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

0 < k < 2

Step-by-step explanation:

In the first quadrant both of x and y get positive values, so

And replacing x with 2k in the second equation:

Since y > 0:

Combining both k > 0 and k < 2, we get:

  0 < k < 2

Step-by-step explanation:

The intersection will lie in the first quadrant when the solution has the characteristics: x > 0, y > 0.

For x > 0, we require ...

  x = 2k

  x > 0

  2k > 0 . . . . substitute for x

  k > 0 . . . . . divide by 2

__

For y > 0, we require ...

  y = (12 -3x)/2

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

  y > 0

  6 -3k > 0 . . . . . . substitute for y

  2 -k > 0 . . . . . . . divide by 3

  k < 2 . . . . . . . . . add k

So, the values of k that result in the intersection being in the first quadrant are ...

  0 < k < 2

c

step-by-step explanation:

i think because what else can be a sample of every member

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.