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1 Question:

(1) The height of the cylinder (also the circumference of the bottom): 2.5× 3.14× 2 =15.7 (cm).

(2) Surface area:

2π×2.5× 15.7+2π2.5×2.5

= 78.5π+ 12.5π ..........................................................................................................................................................

=9 1π

=285.74 square centimeters

Two questions:

(1) Let the height of the original cylinder be h and the radius be r, then the radius will be (4r).

(2) Volume expansion: (later volume ÷ original volume = expansion volume)

(π× 4r× 4r× h) ÷ (π rrh) ............

= 16

If the height of the cylinder remains the same, the radius of the bottom surface will be expanded by 4 times and the volume will be expanded by (16) times.

Three questions:

(1) * * oil: 24 ÷ (1-70%) = 80 (L) =80(dm)

(2) Height of oil drum: (volume/bottom area)

80 (3.14× 2× 2) =1000/157 (Deutsche Mark)

Four questions:

(1) Volume of the rising part of water in the cylinder (also the volume of the cone):

12×2=24 (cubic centimeter)

(2) The height of the cone (3× the area of the bottom of the volume)

3×24÷6= 12 (cm)

Five questions:

(1) Height of the cylinder (equal to the circumference of the bottom): 2×3. 14×5=3 1.4 (cm).

(2) lateral area: 2× 3.14× 31.4 =197.192 (square centimeter).

(3) Surface area:197.192+2× 3.14× 5 = 354.192 (square centimeter).

(4) Volume: 3.14× 5× 5× 31.4 = 246.49 (cubic centimeter)