A few math problems, I hope the big brother and big sister will help me solve 20

1.Set the water down by x centimeters according to the title.

15*15*x=20*15*10

Solution. x=40/3

2.Let the number in the tenth place be x, and the number in the single place be y, according to the title.

2 (600 + 10x + y) - 13 = 100 x + 10 y + 3 simplification. 80x+8y=1184 (Who can solve it!.)~3.Set the white x block according to the title.

12*5=3*x

Solution. x=20

I did my best! It's no wonder there's something wrong!

2.Let the original three-digit number = 300+x, where x is a two-digit number. The three-digit number after the transposition becomes 10x+3

Solving the equation 10x+3=2*(300+x)-13 yields x=83 The original three-digit number is 383

3。Each black block is co-edged with 5 white blocks, and there are a total of 12 x 5 adjacent edges of black and white.

Each white block is co-edged with 3 black blocks, so that the adjacent edges of black and white have a total of x*3 (with x white blocks).

12*5=x*3

x = 20 white has 20 pieces.

1) Solution: Set down x centimeters, according to the topic.

15*15*x=20*15*10

x=40/3

2) Solution: Let the number on the ten digit be x, and the number on the single digit be y, according to the topic.

2（300+10x+y）-13=100x+10y+3

1) 20 15 10 (15 15) = 3000 225 = cm When the tin box is poured, the height of the water in the cuboid container drops by centimeters.

2) Set ten digits n and single digit m

300+10n+m)*2-13=3+100n+10m3) A pentagon has five sides and is not adjacent.

So only the edges of the hexagonal row are adjacent to it.

So 12*5=60

60 6 = 10 (pcs), i.e. there are 10 hexagons.

1. Set down xcm

15*15*x=20*15*10

x=40/3 （cm）

2. Let the original three-digit number = 300 + x, where x is a two-digit number. The three-digit number after the transposition becomes 10x+3

10x+3=2*(300+x)-13

x=73 The original three-digit number was 373

3. Set the x block in white.

12*5=x*3

x=20 (block).

Because the structure of a soccer ball is a circle of hexagons around a pentagon, and each hexagon has three pentagons around three hexagons, the answer can be found by dividing the total number of sides bordered by the pentagons and hexagons by the number of sides that each hexagon borders the pentagon.

1.Let x be dropped, and the height of the box filled with water will be (10+x)15 15 (10+x)=20 15 10 will give x=10 3

2.If the ten digits and the single digit are x, the original number is 300+x

2（300+x）-13=10x+3

x = 73 The original three-digit number is 373

3.There are x white blocks.

Each black block is co-edged with 5 white blocks, and there are a total of 12 x 5 adjacent edges of black and white.

Each white block is co-edged with 3 black blocks, so that the adjacent edges of black and white have a total of x*3 12*5=x*3

x = 20 white has 20 pieces.

I only have a clue about the second problem, I'm ashamed, the equation: 2 (100x+10y+z)-13=100y+10z+x

x 3 These two conditions are not able to find the three unknowns, is there a lack of conditions, or I know not enough, hehe,

The area of pasture that is converted into a forest farm is x hectares.

54-x=（108+x）×20%

270-5x=108+x

6x=162

x=27 hectares,3,(108+54) (1+20%)-108=27 hectares,2,The area of pasture that is changed to a forest farm is x hectares(108+x)*20%=54-x

So x=27,2, let the pasture area of the forest farm be x, then the equation: (54-x) (108+x)=20%, and the solution is: x=27 hectares.

1. Let the pasture area of the forest farm be x hectares 54-x 108 x 20 100x=27,0, please help me solve a math problem, a team has a forest farm of 108 hectares and a pasture of 54 hectares, and now we want to cultivate a new fruit tree, and change part of the pasture into a forest farm, so that the pasture area only accounts for 20% of the forest farm area, how many hectares is the pasture area of the forest farm?

1）x=882） n/12=2

n=24 You're a bit stumped on me, and I don't even know how to get to you.

1.Let the length of the swimming pool be ab=50, the deepest ad=, and the shallowest ac=The original question can be understood as:

The right-angle trapezoidal ABCD is known, the angle A=90°, AB=50, AD=, AC=, and the 2-angle ADC is found.

Drawing the diagram shows that the TAN angle ADC=AB (AD-AC)=50, and the angle ADC=ACRTAN50

The slope of the slope is 2-acrtan50

2.The upper bottom of the trapezoid is 6, the lower bottom = 6 + 10 + 10 3 = 46, and the height is 6 The area = (6 + 46) 6 2 = 156

Let the number of cow heads be x, pig y, sheep z

Equation x+y+z=100

1000x+300y+50z=10000

z=100-y-x

1000x+300y+50(100-x-y)=1000019x+5y=100

There is only one set of integer solutions: x=5

y=1z=94

1 b 5 divided by (5 plus 35) equals.

2 20% 20 divided by 100 equals.

-(35 divided by 40)=, so choose (b).

2. (100-20) divided by 100=80%.

A: 20% price reduction.

If the price is reduced by a few percent, it is good to divide the reduced money by the original price.

20% price reduction