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If there are x students directly, then there are (x-2) students who actually go.

Originally, each person was burdened with 600 x yuan, but now each person is burdened with 600 x + yuan, and equations can be listed.

x-2)×[600/x)+

Multiply x on both sides at the same time

x-2)(600+

x²-2x-2400=0

Solution: x=50 or x=-48 (rounded).

50-2 48 people.

Later, 48 students participated in the spring outing.

Take advantage of the nature of rectangles: the diagonals are equal and bisected from each other.

q1q2=p1p2=10

q1o=10÷2=5

Let the coordinates of q1 be (m,-m).

then m + (m) = 2m = 5

The solution yields m = 5 (root number 2) 2

So the coordinates of Q1 and Q2 are respectively.

5 (root number 2) 2, -5 (root number 2) 2) or (-5 (root number 2) 2, 5 (root number 2) 2).

m-0)²+m-0)²=5²

This step uses the formula for the distance between two points.

m²+m²=25

2m²=25

m²=25/2

Both sides are opened.

m|=5 (root number 2).

m= 5 (root number 2).

The denominator makes it rational.

m = 5 (root number 2) 2

It's really tiring to change it around

If x number of students participated in this spring outing, then the students who actually went were (x-2) who originally paid 600 x yuan per person, but now they have 600 x+ yuan per person, and the equation x-2) [600 x)+" can be listed

Multiply x on both sides at the same time

x-2)(600+

x²-2x-2400=0

Solution: x1 = 50 x 2 = -48 rounds.

It turned out that a total of 50 students participated in the spring outing.

A: Later, a total of 50 students participated in the spring outing.

P1P2 has a distance of 10

On y=-x, we find out that q1q2 is symmetrical about point o and the coordinates of 10qiq2 are (root number 2 5 times 2, root number 2 5 times negative 2) and (root number 2 5 times negative 2, root number 2 5 times 2).

The proof principle is that the diagonals of the rectangle are bisected and equal to each other.

1.Later, a total of x number of students participated in this spring outing.

Then the question can be listed: (600 (x+2)+

Simplification: x 2 + 2x - 2400 = 0

Solution: x1 48, x2 50 (rounded).

So a total of 48 students participated in this spring outing.

2.Let q1 be (x0,y0), then y0=-x0 Since p1q1 is perpendicular to p2q1, the product of their slopes is -1, i.e.

y0-4)/(x0-3)*(y0+4)/(x0+3)=-1y0=-x0

So. x0^2=25/2

x0 = + (-5 root number 2 2

y0=-x0=-(+5 root number 2 2.)

Thereupon. Q1 (5 root No. 2 2, -5 root No. 2 2), Q2 (-5 root No. 2 2, 5 root No. 2 2).

Or. Q1 (-5 root number 2 2, 5 root number 2 2).

Q2 (5 root number 2 2, -5 root number 2 2).

Coordinates are points on a plane, and it is normal for coordinates to contain open squared numbers (irrational numbers) because the entire axis of real numbers is full of irrational numbers.

Don't say I'm not kind.

Solution: Later, a total of x students participated in this spring outing.

From the inscription: (600 (x+2)+

Solution: x1 48, x2 50 (rounded).

So a total of 48 students participated in this spring outing.

P1P2 has a distance of 10

On y=-x, we find out that q1q2 is symmetrical about point o and the coordinates of 10qiq2 are (root number 2 5 times 2, root number 2 5 times negative 2) and (root number 2 5 times negative 2, root number 2 5 times 2).

principle, the diagonal lines of the rectangle are bisected and equal to each other.

(x+2)=600, the amount of money spent by each person in the past = the amount of money spent by everyone now.

Solve x=48, x=-50 rounded.

2. P1Q1P2Q2 is a rectangle, the diagonal lines of the rectangle are equal, and Y=-X passes through the origin, the points P1 and P2 are symmetrical about the origin, then the origin O is the diagonal intersection point, the points Q1 and Q2 are also symmetrical about the origin, the length of the line segment P1P2 = the length of the line segment Q1Q2, then Q1(M,-M),Q2(-M,M).

According to the formula for the distance between two points: (m 2 + (-m)) after the two sides are squared, 2m 2=25 is obtained, and the m value can be solved.

Later, a total of x students participated in the spring outing, and if they paid 600 x per person, x+2 students began to participate in the activity, and each student paid 600 (x+2) to get: 600 (x+2) + x=48

q1q2=p1p2=10

q1o=10÷2=5

It's just that Q1O=Q2O=10 2=5 is used

Because it is necessary to satisfy that the quadrilateral p1q1p2q2 is rectangular, and there is o as the origin to get q1o=q2o=10 2=5, q1o,q2o can be regarded as m,|-m|A right-angled triangle with a right-angled side is: m + (m) = 5

Get: 2m = 5

Later, a total of x students participated, and the equation was: 600 (x+2)+

The solution is: x=48

1.If you learn fractional equations, let's start with x number of participants.

600/x+

x-1)^2=2401

x=502.You can also use binary, set up x people to participate at the beginning, and each person pays y yuan.

xy=600

x-2)(y+ xy-2y+

y=x/x(x/ x^2-2x=2400

x=50

Let's say there are x people.

600 x This is the unit price per person, 600 x +

The unit price after the withdrawal of two people"

Solve x on it.

The original number of participants is obtained, and subtracting 2 is what is requested.

X students were set up to participate, and each person paid Y yuan.

x+2）y=600

x(y+ solution. y=1/4x

Then put in the formula.

x-48）(x+50)=0

The solution is x=48 and cannot be negative.

So went 48 people.

The number of people is x and the unit price is yThat is, xy=600, (x-2) (y+solve.

There are x people after the set.

It should have been 600 (x 2) for one person

After the shortage of people. One person pays 600 x

Rule. 600/（x＋2）+

x=50 with 50 participants.

There are x students in this class.

then the cost per person is 600 x yuan.

600/x）+

Later, a total of x number of students participated in this spring outing.

600/x＝600/（x+2）+

x 48 (person).

Later, a total of x number of students participated in this spring outing.

600/(x+2)-600/x=

x=48

The velocity in the vein is 50*2 5 = 20 cm seconds.

The velocity in the capillaries is 20*1 40 = cm seconds.

Combined formula: 50*2 5*1 40=cms.

1) 1 difference 3 + 1 8 + 1 6 = 15 24 = 5 8, accounting for about five-eighths of the time of the socks in a day.

2) 1-5 8=3 8 Sleep time accounts for about three-eighths of the day.

The money brought by A is half of the positive amount of money brought by the other three people, indicating that A's money is 1 of the total amount of money 3 The money brought by B is 1/3 of the amount of money brought by the other three people, and the money brought by C's money is 1 4 of the total amount of money, and the amount of money brought by C is 1/4 of the total amount of money, indicating that C's money is 1 5 of the total amount of money

Therefore, Ding's money is 1-1 3-1 4-1 5 = 13 60, so the total amount of money carried by the four people is 910 (13 60) = 4200 yuan.

The total amount of money carried by the four people is: 910 (1-1 3-1 4-1 5) = 4200 (yuan).

Planned = 150 5 6 = 1.8 million yuan.

The actual investment is less than planned: 180-150 = 300,000 yuan.

1.You can think of it as 50 people (21 males, 29 females) 20 people A (14 males, 6 females), so (29+6) (50+20)=1/2

2.The number A is 5 3 times the number of A 5, so it is 1 3 5 3 = 5 9 of the sum of A and B

Question 1: The proportion of A to the total number of students in the two schools is: 2 5 2 5 +1) = 2 7 The proportion of female students in school A to the total number of students:

3 10 2 7) =6 70 The proportion of B in the total number of students in the two schools is: 1 1 + 2 5) =5 7 The proportion of girls in the two schools is (1 - 21 50) 5 7) =29 70 The proportion of girls in the two schools is the total number of students:

6 70 + 29 70 = 35 70 = 1 2 Question 2:

A and B. Originally 1 0

First time 1 2 1 roll this 2

The second time 2 Daxun 3 1 3

The third time 1 2 1 2

Fourth 3 5 2 5

Fifth 1 2 1 2

From the above table, it can be seen that when the number of pourings is an odd number, there are 1 2 bottles of water in A and B, so when pouring the 31st time, the water in the two bottles is 1 2 of one bottle of water.