Solution: from the meaning of the question:

( 1)k & lt； 0, straight line crossing point (-3, 13), (4, 8).

Then: k=(-8- 13)/(4+3)=-3.

Namely: -3x-y+b=0.

Substitute the point (-3, 13) into 9- 13+b=0 and b=4.

The linear equation is: -3x-y+4=0, that is: 3x+y-4=0.

(2)k & gt； 0, straight line crossing point (-3, -8), (4, 13).

Then: k=( 13+8)/(4+3)=3.

That is: 3x-y+b=0.

Substitute the point (-3, -8) into: -9+8+b=0, and get: b= 1.

The linear equation is 3x-y+ 1=0.

To sum up, the linear equation is 3x+y-4=0 or 3x-y+ 1=0.

22、

Certificate:

(1) In the triangle ABC, AC=3, BC=4 and AB=5.

AB？ =AC？ +BC？

Therefore, triangle ABC is a right triangle,

So, AC⊥BC

(2) From (1): AC⊥BC

Side CC'⊥ bottom ABC

AC is in the ABC plane.

So, CC'⊥AC

AC⊥BC, AC⊥CC', BC∩CC'=C, BC, CC' are all in the face BCC'B'

Therefore, AC⊥ surface BCC'B'

Therefore, the angle between the straight line AC' and the plane bcc' b' is ∠ AC' C.

In Rt△ACC', AC=3, CC'=AA'=4.

So tan∠AC'C=AC/CC'=3/4.

That is to say, the tangent of the angle formed by the straight line AC' and the plane BCC'b' is 3/4.

(3) Let the intersection of BC' and CB' be point O and connect OD.

The side BCC'B' is rectangular, so o is the midpoint of BC'

And because D is the midpoint of AB, in the triangle ABC', OD is the center line.

OD∨AC', OD is on the plane of CDB', AC' is not on the plane of CDB'

So, the AC '∑ plane CDB'

23、

Solution:

( 1)y 1 =( 10-m)x-20，& ltx≤200

y2= 10x-0.05x？ -40，& ltx≤ 120

(2)y 1 is a linear function about x, y10-m >; 0, so it's increasing,

Therefore, when x=200, y 1 has the maximum value.

In other words, product A produces 200 pieces.

y2=-0.05x？ + 10x-40，& ltx≤ 120

It is a quadratic function about x, with the opening downward and the axis of symmetry x= 100.

Therefore, when x= 100, y2 has a maximum value.

That is to say, product B produces 100 pieces.

Have fun! I hope I can help you. If you don't understand, please ask. I wish you progress in your study! O(∩_∩)O