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Huairou Ermo Mathematics Answer 20 17
(1) Let the function expression y 1=kx+b be the first charge and y2=k 1x be the second charge.

Substituting (0,6), (100, 16) into y 1=kx+b, we get

b=6 100k+b= 16,

The solution is k = 0. 1b = 6,

∴y 1=0. 1x+6(x≥0 integer),

Substitute (100, 12) into y2=k 1x,

Solution: k 1=0. 12,

∴y2=0. 12x(x≥0 integer);

∴y 1=0. 1x+6(x≥0 integer), y2=0. 12x(x≥0 integer).

(2) From the meaning of the question, you can get

When y 1 > y2, 0. 1x+6 > 0. 12x, x < 300

When y1= y2,0.1x+6 = 0.12x, x = 300.

When y 1 < y2, 0. 1x+6 < 0. 12x, x > 300;

When x is in the range of 320 ~ 350, it is more cost-effective to choose the first method.