What is the volume of the cone below? 3.33 3.75 15.97 5.32

Mathematics

1 answer:

EastWind [94]2 years ago

4 0

Answer:

D. 5.32 cubic units

Step-by-step explanation:

Volume of a cone = ⅓πr²h

Radius (r) = 1.52 units

Height (h) = 2.2 units

π = 3.14

Substitute

Volume = ⅓*3.14*1.52²*2.2

Volume = 5.32 cubic units

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You are assigned 32 math exercises for homework. You complete 87.5 of these before dinner. How many do you have left to do after

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Answer:

Step by step:

This is if the question is saying you completed 87.5% of the exercises. Otherwise, I don't know what it is asking.

87.5% of 32=28

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2 years ago

Read 2 more answers

ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ДАЮ 23 ОЧКА!!!

posledela

(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)

Use the shell method. Each shell has a height of 3 - 3/4 y ², radius y, and thickness ∆y, thus contributing an area of 2π y (3 - 3/4 y ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ y ≤ 2, thus given by the integral

$\displaystyle 2\pi \int_0^2 y \left(3-\frac34 y^2\right) \,\mathrm dy = 2\pi \left(\frac32 y^2 - \frac3{16} y^4\right)\bigg|_0^2 = 6\pi$

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 x), thus contributing an area of π (√(4/3 x))² = 4π/3 x. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ x ≤ 3, or by the integral

$\displaystyle \pi \int_0^3 \left(\sqrt{\frac43x}\right)^2 \,\mathrm dx = \frac{2\pi}3 x^2\bigg|_0^3 = 6\pi$

Using either method, the volume is 6π ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...