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Mathematical driving problem
(1) Let y=kt+b 1.

Bring (0,25) (2,9) into the equation and get:

25=b 1

9=2k+25

Solution: b 1=25, k=-8.

So: y= -8t+25.

(2) Since the two straight lines are parallel, the equation of the straight line after refueling is: y= -8t+b2.

Bring point (2, 30) into the equation:

2= -8x30+b2

Solution: b2=46, so the equation is y= -8t=46.

According to the meaning of the question: when y=0, bring it into the equation and get t=5.75 (hours).

(3) When t = 500 ÷ 100 = 5 (hours) is substituted into the equation, we get:

Y= -8x5+46=6 (length)