1. 14

2.2(x-2)^2-5

3.3/2

4. 1 1: 54

5. It doesn't exist

Answer:

1.

Because BC = a2-12a+52 = [(a-6) 2]+42 = [(A-6)2]+[ (b+c)/2] 2, that is, [(a-6) 2]+[

2. y = 2x2-4x-5 = 2 [(x-1) 2]-7, shift it by 3 units to the left, and then shift it by 2 units to get y = 2 [(x- 1+3) 2]-7+2 = 2]. Its parabola symmetric about the Y axis is 2 (x-2) 2-5.

3. Make two auxiliary lines, extending ED to F, making DF=ED, making FG parallel to BC and AC parallel to G..

Since D is the midpoint between AB and EF, it is concluded that the quadrilateral AEBF is a parallelogram, so BF is parallel to AC, BF=AE, FG is parallel to BC, so BCGF is also a parallelogram, so AE=BF=CG, and the angle FGC= 180 degrees-angle C; AED=90 degrees+half of angle c, so angle FEG=90 degrees-half of angle c, so angle EFG=90 degrees-half of angle c = angle FEG, so triangle EFG is isosceles, EG=FG=BC=4, so AE=(7-4)/2=3/2.

4. When X cars leave the library, the garage is empty; Then the integer part of x =15+1+[(6x-3)/8] is obtained, so

x- 16 & lt； =(6X-3)/8 and x-15 > (6X-3)/8, the minimum satisfying condition of X=59 can be obtained by solving these two inequalities, so the garage will be empty after 59.6 minutes, which is 5 hours and 54 minutes, so it is 1 1: 54.

Step 5 write next time