Application will be a few to do a few! After the points! A cone-shaped beach, the perimeter of the bottom is 18.84 meters, and the height is 1.5 meters. How many cubic meters of sand is there? 2. There is a rectangular stone, 30 cm long and 15 cm high. When processing, the length, width and height are chiseled away by 2 cm. How much is the volume reduced? What is the surface area? 3 as shown in the figure, in a square ABCD, the area of the shadow part is twice the area of the triangle bef, and the area of the shadow part is 6? There is a kind of drink which is cylindrical (not including bottle neck). The volume of the bottle is 40cm & sup3;. Now there are some drinks in the bottle. When the bottle is placed upright, the liquid is 20cm high and the rest is 5cm high. Please calculate the volume of the drink in the bottle 5. There are two kinds of salt water solution, the ratio of salt content is 2:3, the ratio of water content is 1:2, and the ratio of solution weight is 40:77 Two is eighteen I can't figure out

Application will be a few to do a few! After the points! A cone-shaped beach, the perimeter of the bottom is 18.84 meters, and the height is 1.5 meters. How many cubic meters of sand is there? 2. There is a rectangular stone, 30 cm long and 15 cm high. When processing, the length, width and height are chiseled away by 2 cm. How much is the volume reduced? What is the surface area? 3 as shown in the figure, in a square ABCD, the area of the shadow part is twice the area of the triangle bef, and the area of the shadow part is 6? There is a kind of drink which is cylindrical (not including bottle neck). The volume of the bottle is 40cm & sup3;. Now there are some drinks in the bottle. When the bottle is placed upright, the liquid is 20cm high and the rest is 5cm high. Please calculate the volume of the drink in the bottle 5. There are two kinds of salt water solution, the ratio of salt content is 2:3, the ratio of water content is 1:2, and the ratio of solution weight is 40:77 Two is eighteen I can't figure out


1.1/3×PI*(18.84/2/PI)^2*1.5
2.30×18×15-28×16×13=2276.2×(30×18+30×15+18×15-28×16-28×13-16×13)=480
There is no picture
4. If the bottom area is a, then 20A = 40-5a, a = 1.6 and the volume is 32cm ^ 2
2m + M = 40 3M + 2n = 77 M = 3 N = 34 concentration = 2 × 3 / 40 = 15%



Help me to do some physical density calculation problems
1. After a bottle is filled with water, the total mass of the bottle and the water is 500 g. if several metal particles are put into the bottle, the mass of the spilled water is 100 g. at this time, the total mass of the remaining water bottle and metal particles is 670 G, Find out the length of this coil of copper wire (assuming that the copper wire is uniform in thickness, the density of copper is 8.9x10 cubic kilogram per cubic meter, and the oil barrel just holds 200 kg of kerosene. If this kind of oil barrel is used to refit water, how many oil barrels need to be used to hold 6 tons of water? (the density of kerosene is 0.8x10 cubic kilogram per cubic meter)


1, 1 because the bottle is full of water, so the volume of spilled water is the volume of input metal, which can be calculated according to the mass = density x volume. 2, if the water spills 100g and there is 400g left, the metal mass is 670-400g. 3, the metal density can be obtained by dividing the above volume by the mass. 2, the copper wire is a cylinder, and its volume = cross-sectional area X length, So length = mass divided by the product of density and cross-sectional area