Thank you for helping me solve 2 math problems

1. If the circumference difference is 8 cm, then the side length difference is 2 cm, and the side length of the small square is x, then the equation is obtained: (x+2)square-xsquare=20, that is, 4x+4=20, x=4

then large: area (4+2) square = 36, side length 6

Small: area 4 square = 16, side length 4

2. Four to six times.

First of all, there are five piles on both sides of the scale, and the light side is singled out;

Then there are two piles on each side of the scale, and if they are the same weight, the remaining pile contains; If it is not the same weight, then the light one contains, and then put the two piles on both sides of the scale, and you can weigh them. I used it two or three times, and in the same way I picked out the light bag two or three times. It is used four to six times.

1.Let the side lengths of the large square and the small square be b and a respectively, then there is.

4（b-a）=8

b^2-a^2=20

Solve the equation to get b=6, a=4

Large square area = 36, small square 16

2.At least twice, the first time is two piles of laundry, one of which happens to be the pile of missing catties.

The second time was to find the bag from the pile that was missing two pounds. Just two bags, one of them is exactly the other one. That's it.

1.Let the side length of the large square be x, and the small square be y, 4x-4y=8x2 (x's square)-y2 (y's square)=20

x=6,y=4

2.At least twice, put five generations at each end of the scale, there must be a light side, take out the light one and put four generations on the side, if balanced, the generation taken out is unqualified.

Otherwise, do it again.

1:15 = ( ) hours.

3/8 days = (9) hours.

7/5 = (1) hour (24).

Hour = (2) hour (18) minute.

Year 2010 = (20) century (10) years.

1.Time.

9 o'clock at 2 o'clock. 1:24 a.m.

10 years of the 21st century.

2.September 28, 2008, 17:37.

1.Let one be x and the other (x+1).

x(x+1)=272

The solution yields x = -17 or 16

So x+1 = -16 or 17

Therefore, these two numbers -17, -16 or 17, let the original two-digit single digit be x, then, according to the meaning of the problem, the equation can be listed.

10(x+2)+x+138=(10x+x+2) Solve x1=-14 11 (not in place, discard).

x2=1 So, this digit is 31

1。Solution: Let the smaller number be x

x（x+1）=272

x2+x-272=0

x—16）（x+17）=0

x1=16，x2=-172.If the original number of single digits x is dissolved, then the ten digits x+2 will be swapped.

10x+(x+

2)=11x+

2 According to the title: 11x

10(x+2)

x]=138

That is, the equation: 11x 2+3x

Solve x11, x2

14 11 (discarded).

So the original two-digit number is 31

Solution: Let the smaller of these two consecutive integers be a, then the other is (a+1), a(a+1)=272

a²+a-272=0

a+17)(a-16)=0

a = -17 or a 16

When a=-17, a+1=-16

When a=16, a+1=17

So these two numbers are -17, -16, or (2).

I get a headache when I see something like this.

<> please be good at Bu Yu and your friends to roll the shortcomings.

1.Because sin cab = four out of five.

The smiling stool is resistant to an angle of 90°

And because ac + bc = ab rough (Pythagorean theorem) 225 + bc = 625

bc²=400

bc=20, so bc=20

Because the points a and e are on the circle.

So ac=ce=15

And because ce=20

So be=5

s aeb=ac be half = 15 5 This little brother trouble next time you turn the ** upside down!! My neck hurts!!