Help me solve this math problem!! Please write down the reason, thank you

6 hours. The 8 a.m. and 9 a.m. hour and minute hands should coincide between 8:40 a.m. and 8:45 a.m., without counting the exact minute or second of the coincidence; The 2 p.m. and 3 p.m. hour and minute hands coincide between 2:10 p.m. and 15 p.m., and they don't count the exact minute and second of the coincidence.

The hour hand travels 1 2 degrees per minute, and the minute hand travels 6 degrees per minute.

The minute hand of the clock is set at 8 o'clock and x minutes when the hour hand coincides between 8 a.m. and 9 a.m.

So 240+1 2x=6x

Get x = 480 for 11 minutes.

The clock is clocked between 2 p.m. and 3 p.m., and 2 o'clock is exactly 180 degrees.

So 60+1 2y=6y-180

Get y=480 for 11 min.

So a total of 14 + 480 11-8-480 11 = 6 hours have passed.

It's as simple as taking the wall clock on the wall and turning it around.

6 hours.

It was half past eight in the morning.

The end is half past two, and you can try it by looking at the clock.

It's very simple, after a small block + 30 degrees (a small block is 5, for an hour) 180 30 = 6

6 hours.

This student, this question mainly understands the positive real number a>0, this condition is easy to solve, use the separation variable method to find the range, I hope it can help you!

Let g(x)=a(x+1), and let f(x)=e x, then: f'The tangent equation for (x)=e x on f(x)=e x (b, e b) is y-e b=e b·(x-b).

When the tangent is fixed (-1,0), 0-e b=e b·(-1-b) e b·b=0

The tangent equation for b=0 over f(x)=e x (0,1) is y-1=x, i.e.: y=x+1

That is, when a=1, e x a(x+1) (when x=0, equal) combine the image, to make e x a(x+1), then: 0 a 1

Let's say the price is reduced by $x.

40-x）*（20+2x)=1200

I'll solve the quadratic equation myself.

The answer is x=10 negative rounding.

The second question does not need to be explained, basic nature.

1c 2c

1.Calculate the square of the ABC and compare the sizes.

Solution: 1-m x 1+m from x2-2x+1-m2 0(m 0) so q:a=

by |1-x-13|2, get -2 x 10

Hence p:b=

p is a sufficient but not necessary condition for q.

1-m≥-21+m≤10​

The value range of 0 m 3 real number m is 0 m 3

(1) Let's assume that box x is damaged during transportation. Then from the question can be deduced:

350 - x) *10 - 50 * x = 3200 to get x = 5

Because there are 6 in a box, there are 5 * 6 = 30 damaged water bottle galls.

2) If B has $x, then A has $4x.

Then after the first time A is spent, the remaining 4x(1-1 3)=8x 3 yuan, and after the second spending, the remaining 8x 3 (1-1 3)=16x 9 yuan, so 16x 9-7=x+7

7x/9=14

x=18, so it turns out that A has 18 4=72 yuan.

I wish you success in your studies! ~

1 solution Because of 350 boxes of water bottle galls! In the case of no damage, the logistics company should get 3,500 yuan, and in fact, the logistics company gets 3,200 yuan, so the logistics company's loss cost (including no freight and compensation) is 3,500-3,200 = 300 yuan.

Set up damage to x box of water bottle bile

x（10＋50）＝300

x=5 and because there are 6 pcs per carton.

Therefore, 30 water bottle gall bladders were damaged (because the water bottle gall was damaged, so the logistics company did not pay for the freight, and there was compensation, which is equivalent to the loss of freight + compensation costs, so use 50 + 10.)

2 Solution A has x yuan.

x-1/3x）*（1-1/3）-7=1/4x+7x=72

(1) Logistics for the company to transport 350 boxes of water bottles, 6 in each box. The contract Guidong freight is 10 yuan per box, and if a box is damaged, not only will the freight not be paid, but also 50 yuan will be compensated. At the time of settlement, the logistics company has to pay 3,200 yuan for freight. How many water bottles have been damaged?

30 pcs. 2) A and B each have a certain amount of money, and A has 4 times as much money as B. When A spends one-third and then spends the remaining one-third, if this is 7 yuan from A to B, the amount of money A and B are exactly equal. How many dollars did A have?

72 yuan.

1. Count the twentieth to the boss.

2. The sixth count goes to the boss.

3. First, the diagonal is stacked to get an isosceles triangle.

Then the two base angles of the equal triangle are stacked opposite each other to get a small isosceles triangle, and finally the two base angles of the small equal triangle are stacked opposite each other, and cut along the polyline position to get four small squares.

I didn't understand the first question.

Second: The square is folded in half along the diagonal to get an isosceles triangle. Cut a single knife down the vertical midline along the vertices of the triangle to get four squares.

Question 1: Since a is a root of the equation, a can be substituted into the equation: a 2-3a = 1

1/2 a^2- 3/2 a +4=1/2(a^2 -3a)+4= 9/2

Question 2: 1).Since the equation is a univariate one-dimensional equation, the coefficient before x 2 is 0, and the coefficient before x cannot be zero, i.e. k 2-1=0, and k+1 is not equal to 0, so k = 1, and the root of the equation, x = 1

2) Since the equation is a one-dimensional quadratic equation, the coefficients in front of x 2 cannot be 0, that is, k 2-1 is not equal to 0, so when k is not equal to plus or minus 1, the equation is a one-dimensional quadratic equation.

Question 3: 1) Since the equation is a one-dimensional quadratic equation, m+ 3 is not equal to 0, and m2-1=2, so it can be obtained that when m= 3, the equation is a one-dimensional quadratic equation.

2) Since the equation is a univariate linear equation, m+ 3 = 0, and m-1 is not equal to 0, so when m=- 3, the equation is a univariate linear equation.