Math problems, math problems, afternoon homework

Solution: Let A weigh 3x then B weighs 10x

According to the proposition: (3x+20) (10x-20)=7 6 solution: x=5

So A originally weighed 15 kilograms, and B originally weighed 50 kilograms.

The total amount of 13 parts remains the same, so A has 4 more parts and B has 4 fewer parts, 20 4 = 5 kg per portion.

Original A: 3*5=15 kg.

Original B: 10*5=50 kg.

A: 15kg B: 50kg

Let A and B have 3x and 10xkg respectively, then (3x+20):(10x-20)=7:6, and cross-multiply to get 52x=260, so x=5Then the original A and B each have 15kg and 50kg

Let A be x kilograms and B be y kilograms.

x/y=3/10 （x+20）/（y-20）=7/6x=15 y=50

That is, the original A is 15 kilograms, and the original B is 50 kilograms.

Let the weight of bag A be x the weight of bag B and the weight of rice be y

3：10=x:y 7:6=(x+20):(y-20）

This gives x=15 y=50

The total weight of A and B remains unchanged, A was originally 3 13 of the total weight, and after A increased 20 kg, it is now the total weight.

The total weight of A and B is 20 (4 13) = 65 (kg) A (3) = 15 (kg) and B (13) = 15 (kg) and B (10 13) = 50 (kg)

Solution: Let this steel bar be x meters long.

6+6-x=1/4x

5/4x=12

x = A: This rebar is in meters.

Because the steel bar is less than 10 meters long, there can only be one case, and B is on the left side of A (a mark is made from one end to 6 meters with a meter ruler, and a mark is made from the left side when measured to 6 meters from the other end, and B is remembered from the right).

So if the length is x meters, there is:

6+6-x/4=x

Therefore x=48 5m

6 + 6 = 12 (m).

12 (5 4) = m).

Therefore this rebar is meters long.

1.Solution: If B has x kilograms, then A has kilograms.

x=24÷x=48

Kilogram A: It turns out that A has 72 kilograms and B has 48 kilograms.

Solution: When two bags weigh the same weight, there are x kilograms each.

Then it turns out that A has x+12 kg, and B has x-12 kg, according to the title x+12=

x+12=x=60

60 + 12 = 72 kg.

60-12 = 48 kg.

A: It turns out that A has 72 kilograms and B has 48 kilograms.

2.Solution: Set up x liters.

52:2=x:10

2x=52×10

x=260A: It can hold 260 liters.

Solution: Set up x liters.

52:x=2:10

x=52×10÷2

x=260A: It can hold 260 liters.

Suppose B is xkg and A is.

x=48(kg)

Method 2: If the same weight is assumed, the average weight of bag A and B is xkgx+12=(x-12)*x+12=

x=60(kg)

So B original weight = 60-12 = 48kg

2.Bottom area of the container = oil volume Oil height = 52 2 = 26dm Volume of the container = bottom area * depth of the container = 26 * 10 = 260 liters Method 2:

52 liters can increase the height of the container by 2dm, and to increase the height by 10dm, 5 52 liters are needed, that is, (10 2) * 52 = 260 liters.

Hope it helps.

If you have any questions, you can follow up.

Thank you for adopting!

1: There are two bags of rice, bag A is twice the size of belt B, take 12kg from bag A to bag B, the two bags are the same weight, how much does bag B weigh?

Solution: Let it be that B has x

x=48Solution 2: Let B now have x

x=60 turns out that B has: 60-12=48

2: Put 52 liters of oil into a rectangular container, the oil is 2dm high, and only 10dm deep in this container, how many liters of oil can this container hold.

Solution: Set the container to be able to hold oil x

2÷10)x=52

x=260Solution 2: Let the bottom area of the container be x

2x=52x=26

This container can hold oil: 10x26=260

Set the weight of the second bag x kg.

x=48(kg)

The bag weighs x kg.

x-12=x/

x = 72 B bag weighs 72

10 (2 52) = 260 (liters).

Let this container be able to hold x liters of oil.

52/2=x/10

x = 260 (liters).

1.Let B be x, then.

12 = x + 12 or.

x =12+12

Solution x = 48

2.52 x 10 = 270 (liters) 52 x (10 2) = 270 (liters).

(1) If the original weight of bag B is xkg, then bag A is based on the title: solution x=48 (that is, the original weight of bag B).

Another is to let the last two bags weigh akg, then a+12= solution a=60, so the original weight of bag B is 60-12=48

2) Method 1 52 2*10=260 liters Method 2 52 (10 2)=260 liters.

1.(1) Set the original bag A x kg, the original bag B y kg, x y= x-12=y+12 x=72, y=48, so bag B weighs 48 kg.

2) Let the two bags weigh the same x kg, x+12 x-12 = solve x=60, so bag B weighs 60-12=48 kg.

cubic decimeter Because 52 liters are loaded with 2dm, the cuboid has 10 2 = 5 52 liters, so 52 * 5 = 260

The bottom area of the cuboid is 52 2 = 26, so the volume of the cuboid is 26x10 = 260 cubic decimeters = 260l

1. Solution: Set the original weight of B bag of rice x kg.

Method 1: x=48kg

Method 2: x=48kg

2.The second question feels too simple, directly 52*(10 2)=260

Column equation solution. Question 1 takes A as x, or B as x. The second question is that you didn't see the conditions clearly.

Let a=(x,y) then.

x²+y²=52①

a⊥ba·b=-2x+3y=0②

Solution: x=4, y=6; x=-4, y=-6, so it is (4,6) or (-4,-6).

Solution: (30-3-20)*(100+5*3)=7*115

805 (yuan).

Answer: According to this, the profit of one week can be 805 yuan.

Hope you are satisfied!

The effect of the price reduction of 3 yuan is:

1. This week's profit is 30-3-20 = 7 yuan per piece 2, this week sells 5 * 3 = 15 pieces, that is, this week the merchant sells 115 construction goods, so this week's profit is 7 * 115 = 805 yuan.

(-7 and 3/8) (

(-7 and 3/8) (

- 7 and 3/8) 1

(-7 and 3/8) (

5/11 (-2/9) (2/5) (1/4) (1/4) (approximate simple arithmetic).

5/11 (-2/9) (11/5) (9/2) = [5/11 (-11/5)] 2/9) (9/2)].

(-7 and 3/8) (

- 7 and 3/8) 1

(-7 and 3/8) (

5/11 (-2/9) (1/5 of 2) (1/4 of 2) = 5/11 (-2/1 of 5) [2/2 of 9] (1/4 of 2)].

(-7 and 3/8) (

(-7 and 3/8) (

(-7 and 3/8) (

5/11 (-2/9) (2/5) (1/4) = -5 11 11 5 2 9 9 2 = -1

(-7 and 3/8) (=-7 and 3/8 (the last two terms are multiplied to equal 1).

5/11 (-2/9) (2/5) (1/4) (1/4 1/2) = 5 11*2 9*11 5*9 2=1 (1 3, 2 4 terms multiplied equals 1).

(-7 and 3/8) (= -59 8 (1 2 2) = - 59 8

5/11 (-2/9) (2 and 1/5) (4 and 1/2) = [5/11 (-2 and 1/5)] 2/9) (1/4 and 1/2) = (-1) (1) = 1

-7 (3 8) minus seven and three-eighths.

1. The solution of both problems is to turn the fraction (or decimals) into a false fraction, for example: 5/11 and -2 and 1/5, and the false fraction is: 5/11/11 and -5/11

Khan, now elementary school students are used to know! Poor us post-80s.

8+ (3/3).

8+4 (-3/4).

5 (-3/2) (1 and 1/3) (1-4/1) = (-3 2) (4 3) (5 4) = 5 2

3 and 3/8 - (-5/6) (7/24) = 3 and 3/8 - 5 6* 24 7

81) 2 and 1/4 4/9 (-1/16) = (81) (4 9) (4 9) (1 16) = -1

8+(3/3) = 8+4 (-3 4) = 8-3=5

3/2) (1/1 3) (1-4/1) = (-3 2) (4 3) (5 4) = 5 2

3 and 3/8 - (-5/6) (7/24) = 27 8-(-5 6) (24 7) = 27 8-20 7

81) 2 and 1/4 4/9 (-1/16) = 81 9 4 4 9 (-1 16) = 81 4 9 4 9 (-1 16) = -1

This is very simple, calculate slowly, as long as you are careful.

In junior high school, you must study by yourself, and you can't rely on others, otherwise you won't even be able to do the most basic calculations in the future.