Mathematics in the first year of junior high school, everyone help

Scheme 1: Set up x tons of processed milk flakes and 90-x tons of fresh milk, then:

x/1=4.

So x=4, fresh milk 90-4 tons=86 tons.

The profit is 2000 * 4 + 500 * 86 = 51000 yuan.

Option 2 can be distributed as follows: tons of fresh milk are needed to make yogurt days, and tons of fresh milk are needed to make milk tablets days, which is exactly 90 tons.

The total profit is.

75 1200 15 2000 120000 (yuan) So choose option 2

If you make preserved fruits all day, the profit is 30 * 1000 = 30,000 yuan.

Make juice all day, and the profit is 10 * 1500 = 10,000 yuan.

Plan 1, make juice for 4 days, and sell the rest, and the profit is:

4*10,000 yuan. The second plan is to make juice for x days and preserved fruit for 4-x days.

After the juice is used to remove the fresh fruit, the remaining fresh fruit just meets the production capacity of the preserved fruit.

Problems become demands.

90-10x)/30=4-x

3-x/3=4-x

2/3)x=1

x = i.e., the day of production of fruit juice, the day of production of preserved fruit is the most profitable. For. 10,000 yuan

Solution: If there are x first prize students, then there are 22-x second prize students, so 200x+50(22-x)=1400

The solution is x=2, so there are 2 students who won the first prize.

There are x number of students who won the first prize.

200x+50（22-x）=1400

x=2 There are 2 students who won the first prize.

There are x students in the first class and y people in the second class.

x + y = 22

200x + 50y = 1400

Get x = 2 and y = 20

Solution: If there are x first-prize students, then there will be 22-x second-prize students: 200x+50(22-x)=1400

200x+1100-50x=1400

150x=300

x=2A: There are 2 students who won the first prize.

Set: Project B will be completed in x days after it is added.

3/12+ (1/15+1/12)x=1x=5

If the project is 1, then A will do 1 12 per day, and B will have to do 1 15 per day and complete it in x days.

3 12 + (1 12 + 1 15) * x=1 Solution: x=5

That is, it will be completed in five days.

From the question, it is obtained: A does it alone, completes 1 12 every day, B does it alone, completes 1 15 every day, A does 3 12 if it does 3 days, A and B do it together, and completes (1 12 + 1 15) every day, from which it can be obtained: (1-3 12) (1 12 + 1 15) = 5 days.

Boss,This seems to be an application problem for elementary school students.。。。 Thoroughly sweated.

Solution: From the meaning of the question: a+3=0, a=-3 b-4=0 b=4

So the square of a + the square of b = 9 + 16 = 25

a=-3b=4

The value of the square of a + the square of b = 9 + 16 = 25

Because the square of a+3 is not less than 0 and the absolute value of b-4 is not less than 0, then a=-3 and b=4 result is 25

1.Solution: 3x+2-(4x-9)=10

3x+2-4x+9=10

x=1 means that when x=1, the algebraic formula 3x+2 is 10 greater than the value of 4x-9.

2.A middle school plans to buy two types of small blackboards A and B from Roewe Company, and after negotiation, it takes 20 yuan more to buy a small blackboard than a small blackboard of type B, and it costs 820 yuan to buy 5 small blackboards of type A and 4 small blackboards of type B, so it costs yuan to buy a small blackboard of type A.

Solution: If it takes $x to buy a small A-shaped blackboard, then:

5x+4(x-20)=820

5x+4x-80=820

x = 100 means that it costs 100 yuan to buy a small A-shaped blackboard.

4.A bookstore can still make a 20% profit by discounting a new book at 9% off the list price**. If the purchase price of the book is 21 yuan, then the price is yuan?

Solution: Set the price to be $x, then:

90%*x=21*(1+20%)

x = that is, the list price is 28 yuan.

1) It is +6 km away from the departure point of Xisun Town.

2) When returning to Kaibo in the afternoon: (15 2+2 2+5 2+1 2+10 2+3 2+12 2+5 2+6 2).

15-2+5-1+10-3-2+12+4-5+6=39 39 km east of the place of departure.

Jing Kuan = l

Solving Wang Zhi: (1) From the diagram, it can be seen that Xiao Li's last trip was +6 kilometers, and the distance between the mausoleum and the departure place in the afternoon was 6 kilometers.

2) Fuel consumption is liter kilometer, so all trips in the afternoon = (15 + 2 + 5 + 1 + 10 + 3 + 2 + 12 + 4 + 5) * 2 + 6 = 120 (km) fuel consumption sail sensitivity = 120 * liters).

Solution: From x=1/2, y=-1/5, then [(y-2x)(-2x-y)-(2x-3y)] (2y)=[-(y-2x)(y+2x)-(4x to the 2nd power-12xy+9y's 2nd power] (2y).

- (2 power of y - 2 power of 4x) -2 power of 4x + 2 power of 12xy - 9y power of 2] (2y).

- (2 power of y - 2 power of 4x) -2 power of 4x + 2 power of 12xy - 9y power of 2] (2y).

12xy-10y to the 2nd power) (2y).

6x-5y）×（2y）÷（2y）

6x-5y6 (1/2)-5 (-1/5).