Sixth Grade Mathematics. After that, for sure. Hurry!

Question 1: 360x3% = 10,000.

Question 2: 840x17%=10,000.

Question 3: 5000x(1-20%)=4000 yuan.

Question 4: Mother: (1800-1600) x 5% = 10 yuan.

Dad: (2500-1600) x10%=90 RMB.

Question 5: 1x (1 + is about equal to 10,000 yuan.)

Question 6: Interest due: 2000 x 30% x yuan.

Take-home interest: x (1-5%)=RMB.

The person who wrote the question is a pig! The questions don't conform to common sense, isn't this a pit for children!

Question 1; 3,600,000 yuan 3,600,000 yuan 3,600,000 3% 10,800 yuan.

Question 2; 840 17% 142,800 yuan.

Question 3; 5000 20% 1000 yuan 5000-1000 4000 yuan.

4 questions; Mom 1800 10% 180 yuan Dad 2500 15% 375 yuan.

5 questions; 10,000 yuan 10,000 yuan 10,000 yuan 522 3 1,566 yuan 10,000 + 1,566 = 11,566 yuan.

6 questions; 2000 30% is equal to 600 yuan interest 600 3,87% = real interest; 232,2 5%=116,1 yuan 232,2 116,1=116,1 yuan.

360 times 3% = 10,000 yuan.

840 times 17% = 10,000 yuan.

5000--5000 multiplied by 20% = 4000 yuan.

1800-1600 = 200, 200 times 5% = 10 yuan.

2500-1600 = 900, 900-500 = 400, 500 times 5% = 25 yuan, 400 times 10% = 40 yuan, 25 + 40 = 65 yuan.

The principal amount is 10,000 yuan. Interest 10,000 times the dollar.

2000 multiplied by 30% = 600 yuan. 600 times yuan, multiplied by 5% = yuan, yuan.

4.Mother: (1800*1600)*5%=10 Dad: 500*5%+400*10%=65

Keep two decimal places).

After payment: The above answers pay attention to the plus unit, oh.

3 all in one package:

Start by putting a bunch according to conventional thinking.

The height is 30 divided by this.

And then there's an L-shaped space left.

Put the side first. It can hold 30-19cm books, that is, 18 books.

In this way, the front side is full, and there is still 30-25=5cm space in the back for 8 books.

Such a package can put 18 + 8 + 50 = 76 copies.

300 76= 4 parcels in the first method.

First of all, the person above wrote it wrong, we did it.

Length: 25 cm. Width: 19 cm Thick: 50 copies. 50x1x1=50 (Ben) (this is the first question).

30 6 = 5 (pcs).

The teacher criticized me right.

The volume of the cone is divided by to find the base area of the cylinder

The volume of the cone is v1 = 1 3 ( r 2) = 1 3 ( *6 2 * 3) = 36

After putting the cone into the water, the volume of the discharged water is 36 converted into the volume of the cylinder v2= 48 2*h (h is the height of the discharged water, that is, the height of the water surface descending).

v1=v2 gives 36 =48 2h h=1 64cm

The volume of the cone after removal is divided by the base area of the cylinder.

The volume of the cone is v1 = 1 3 ( r 2) = 1 3 ( *6 2 * 3) = 36

v1=v2 gives 36 =48 2h h=1 64cm

Calculate the volume of the cone, and use it to divide the base area, which I wrote about similarly, and this is how it is calculated.

Because the shadow area = , eo=3, and oc=3 is obtained, the shadow triangle is an isosceles direct triangle, and because ab cd, the angle abo= angle oec = 45°

So bo=ao, and because the quadrilateral abcd is a parallelogram, s=base*height=(5+3)*5=40 square centimeters.

The shadow triangle is a right-angled triangle with area = oe*oc 2 so oc=

The two angles of the triangle ABO and the triangle OEC correspond to equal, so the two triangles are similar, and the corresponding sides of the similar triangles are proportional, so OA=ob=5, so the area of the parallelogram ABCD=(ob+OC)*OA=(5+3)*5=40

High is the middle divider.

So the quadrilateral area is:

A batch of goods, 18 trucks are used to load, 54 are used to load agricultural vehicles, and it is known that each truck is 4 tons more than each agricultural vehicle, how many tons of this batch of goods are there? Arithmetic:

108 tons.

This is an engineering problem that treats the total amount of goods as "1", so the column formula is:

4 (1 18 - 1 54) = 108 tons.

Set up agricultural vehicles to load x tons each.

18*(x+4)=54*x

54x=18x+72

54x-18x=72

36x=72

x = 2 how many tons of this shipment = 2x54 = 108 (tons).

Set up x tons for agricultural vehicles and x + 4 tons for trucks.

18（x+4）=54x

x = 2 cargo tonnage is: 54x = 108 tons.

Set up agricultural vehicles to load x tons each.

54x=18（x+4）

x = 2 This shipment has 54 2 = 108 tons.

Solution: Set up agricultural vehicles to load x tons each.

18(x+4)=54x

18x+72=54x

36x=72

x=22x54=108 (tons) A: This shipment has 108 tons.

Set up agricultural vehicles to install x, trucks x+4There is 18(x+4)=54x, and the total tonnage of the solution is obtained by solving x offspring.

Idea: This is a typical equation, let's find their relationship, find the tonnage, so we want to how to let both sides express the tonnage at the same time, we know that the agricultural vehicle is 4 tons less than the truck, and the agricultural vehicle x tons, the truck x 4 tons, (of course, you can also set the truck to x tons, and the agricultural vehicle is x 4 tons), and then use x and x 4 to find the relation, and then solve it step by step.

Solution: Set up x tons for agricultural vehicles and x 4 tons for trucks.

18x＋72＝54x

36x＝72

x 22 times 54 108

The side length of the small square is 1, the side length of the large square is 2, and the small square is the large square that is reduced in a ratio of 2:1, (large: small).

Large squares are small squares that are enlarged in a 1:2 ratio. (Small: Large).

A: 20%.

B: (1-20%) 25%=20%.

C: (1-20%-20%) 40% less 100=6 25 less 100 so total:

30,000 pcs.

Let the total number of parts be x

20%x+25%x+40%x-100+3700=x3600=

The solution is x=24000