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3 years ago

Find the volume of this cylinder. As soon as possible please

Mathematics

2 answers:

lozanna [386]3 years ago

7 0

Answer:

36 inches cube

Step-by-step explanation:

radius=diameter/2=2/2=1

Volume of cylinder=πr^2h

3×1^2×12

Gnom [1K]3 years ago

4 0

Answer:

Step-by-step explanation:

pi (3) x 1^2 x 12 = 3 x 12 = 36

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A recipe for potato casserole calls for 7/4 cups of cheese. Write the fractions as a mixed number.

alexira [117]

7/4

4 goes into 7 one time.
3 left over.

1 3/4

3 0

3 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

Read more about the center of mass.

brainly.com/question/27549055

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8 0

2 years ago

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Aliun [14]

As

$\tan x=\dfrac{\sin x}{\cos x}$
$\cot x=\dfrac{\cos x}{\sin x}$

we have

$\sin x(\tan x\cos x-\cot x\cos x)=\sin x\left(\sin x-\dfrac{\cos^2x}{\sin x}\right)=\sin^2x-\cos^2x$

Recalling that $\cos2x=\cos^2x-\sin^2x$ , we end up with

$-\cos2x=1-2\cos2x$
$1=\cos2x$
$\implies2x=2n\pi$

(where $n$ is any integer)

$\implies x=n\pi$