Fifth grade on the mathematics book (people's Education Press) exercise 18 how to write?

Yellow flower is a quarter of the area, so is safflower, and half is green grass

People's education press English compulsory 1 page 4 topic how to do?

1、 1. Trust 2. Calm dowm5. Be crazy about 6. Set down7. Go through 8

People's education press, the answer to question 1 of exercise 16.3 for Grade 8 Mathematics

1.（1）X=3/4
(2)X=7/6
(3) X = 1 / 2, but 2x-1 = 0, so there is no solution
（4）X=4
（5）X=-3
(6) X=1
(7) X=-6/7
(8) X=10/9

Given that a, B, C satisfy AB + A + B = AC + C + a = 3, find the value of (a + 1) (B + 1) (c + 1)

If AB + A + B = BC + B + C = AC + A + C = 3, then AB + A + B + 1 = BC + B + C + 1 = AC + A + C + 1 = 4, then a (B + 1) + B + 1 = (a + 1) (B + 1) = 4

1. Students asked the teacher: "teacher, how old are you?" the teacher said wittily: "when I was your age, you were born; when you reached my age, I was 37 years old." how old were the teachers and students? 2. A and B practice running on a 400 meter circular track. If they start from opposite directions, they meet every 2.5 minutes. If they start in the same direction, they meet every 10 minutes. Assuming that the speed of two people is constant and a is slower than B, what is the speed of a and B? 3. A tunnel is 3460 meters long. A car passes through the tunnel. It takes 200.5 seconds for the car to enter the tunnel and drive out of the tunnel safely, while the whole vehicle takes 200 seconds in the tunnel? 4. There are 150 students taking part in the mathematics competition in a school. The average score is 55. The average score of the students who have passed the test is 77, and the average score of the students who fail is 47? 1 problem directly ball out the answer, 4 questions as long as the list of equations, and give the final answer to it

If the teacher is x years old and the student is y years old, then
Y - (X-Y) = 1 x = 25
X = (X-Y) = 37, y = 13
A: the teacher is now 25 years old
Therefore, fill in 25
Let a speed x m / min, B speed Y M / min,
Then, 2.5x + 2.5Y = 400
10x-10y=400
X = 60, y = 100
A: speed a is 60 m / min, B is 100 m / min
There are x passing students and Y failing students
x+y=150
77X + 47y = 150 * 55
x=40 y=110
A: there are 40 passing students and 110 failing students

For example, the "characteristic point" of equation 3x = - 4 is (3, - 4), and the "characteristic point" of equation (A-2) x = 5 is (a-2,5) (1) The solution of the equation whose "characteristic point" is (3, A-2) is a + 4 (2) If C (m, n) is a point in the coordinate plane, and the "characteristic point" is a (2, M + 2) equation, the solution of equation is M-1, "characteristic point" is B (3,2n + 4), and the solution of equation is n + 1

1. Put the characteristic point into the equation 3x = A-2 and get a = - 7
2. Substitute x = M-1 into 2x = m + 2 to get m = 4
Similarly, n = 1
A (2,6) B (3,6) C (4,1) was obtained
Two fifths is the area

The third question after class of the little match girl, lesson 14, people's education press

The first is the fireplace, the second is the roast duck, the third is the Christmas tree. It can be seen that she was very hungry, cold and lonely at that time. 2. (1) that is the meaning of death. But here to say death is to "where there is no cold, no hunger, and no pain" is to expose the people's suffering, to describe death as a kind of relief, to put

A balloon rises from the ground at a constant speed of 10m / s. when it rises to 18m above the ground, a stone is thrown vertically from the ground below the balloon at the initial speed of 30m / s. regardless of the air resistance, try to analyze whether the stone can hit the balloon?

If it can be encountered after T seconds, then
10t+18=x1
30+1/2*10*tˇ=x2
x1=x2
10t+18=30+1/2*10*tˇ
Solve the equation of two variables of T. If there is a solution, you can hit it. One person just happens to encounter it. Two solutions are met once on a stone. Another time when you fall down·
It's just that there's no solution·
This is the standard solution to this problem. I hope you can remember it·

1. It is known that the maximum acceleration and take-off speed can be 5.0 M / S 2 and 50 m / s when the fighter plane accelerates on the runway. If the aircraft is to taxi for 100 m and then take off, what initial speed must the aircraft have? Assuming that an aircraft carrier is not equipped with an ejection system, but the aircraft is required to take off normally on it, what is the minimum length of the ship? (root number can be retained) 2. A and B practice 4 × 100m relay running on the straight track, and they have the same maximum speed when running. B needs to run 25m to reach the maximum speed from static to full running. This process can be seen as uniform variable speed movement. Now a runs to B with the stick at the maximum speed, and B is waiting for the opportunity to run out in the relay area (1) How much distance does B have to run in the relay area? (2) How far should b start from a? 3. The speed limit of a tunnel is 36 km / h. a train is 100 m long and runs at a speed of 72 km / h. when it reaches 50 m away from the tunnel, it starts to make uniform deceleration movement and passes through the tunnel at a speed not higher than the speed limit (1) The minimum acceleration of train in uniform deceleration motion (2) The shortest time for a train to pass through a tunnel 4. The U.S. plane flies at 200 altitude and at ultra-low altitude. After leaving the aircraft, the American soldiers first fall freely. After moving for a period of time, the parachute is opened immediately. After the deployment of the parachute, the average acceleration of 14 m / S2 is used to decelerate. For safety requirements, the landing speed of American soldiers should not exceed 4 m / s, (g = 10 m / S2). The ground searchlight of Iraq is scanned once every 10 seconds, Can American soldiers use the safe landing interval 5. In a 100m long tunnel, a road mender suddenly found a train appeared at 200m away from the right tunnel entrance. The position of the road mender is just in the position that can safely escape from danger no matter left or right. What is the distance from this position to the right exit of the tunnel? How many times the minimum speed of his running should be at least the speed of the train? 6. If an astronaut releases a heavy object freely from a 32m high place on a planet, and the measured distance in the last 1s is 14m, then what is the falling time of the heavy object and what is the gravitational acceleration of the planet? 7. The maximum speed that a motorcycle can reach is 30 m / S. in order to catch up with the car running at a constant speed of 20 m / s at the front 100 m along a straight road from standstill within 3 min, what acceleration must the motorcycle start with? 8. The "acceleration distance" and "vertical height" of human jumping on the spot are 0.50m and 1.0m respectively, and the "acceleration distance" and "vertical height" of fleas are 0.00080m and 0.10m respectively?

Let V T = 50 m / s, a = 5.0 M / S 2, s = 100 mV T ^ 2 represents the square of the final velocity, so we can solve it by introducing the data from the acceleration formula 2As = (VT ^ 2-v0 ^ 2)

1. For the sake of traffic safety in the city, the driving speed of the vehicle is limited to no more than 30km / h. If the reaction time of the driver is 0.3s, the dynamic friction coefficient between the vehicle and the road surface is 0.75. The safe distance of the vehicle running at 30km / h is calculated. (G is 10m / S ^ 2) 2. When the car runs at the speed of 15m / s on the horizontal highway, due to the sharp turning of the road, it is found that there is a pedestrian moving forward at the speed of 5m / s at 30m ahead after turning. The reaction time of the driver is 0.4s. In order to avoid collision, what is the acceleration of the car at least? The problem didn't tell the quality of the car

One
v=30*1000/3600 = 25/3 m/s
Deceleration a = μ g = 0.75 * 10 = 7.5m/s ^ 2
Safety distance S = S1 + S2 = VT '+ V ^ 2 / (2a) = 25 / 3 * 0.3 + (25 / 3) ^ 2 / (2 * 7.5) = 385 / 54 ≈ 7.13 M
Two
If the braking deceleration is a, it must be ensured that the vehicle will not collide when decelerating to the pedestrian speed
Time required for deceleration to be equal to pedestrian speed t = (v2-v1) / a = (15-5) / a = 10 / A
Reaction time t '= 0.4s
S person = V1 (T + T ')
S car = v2t '+ (V2 ^ 2-v1 ^ 2) / 2a)
S person + 30 ≥ s vehicle
v1(t+t')+30 ≥ v2t'+(v2^2-v1^2)/(2a)
5*(10/a+04) + 30 ≥ 15*0.4 + (15^2-5^2)/(2a)
50/a +2 + 30 ≥ 6 + 100/a
50/a ≤ 26
a ≥ 25/13 ≈ 1.923m/s^2

Fifth grade on the mathematics book (people's Education Press) exercise 18 how to write?