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2 years ago

Sin tita + sin ^2 tita = 0

Mathematics

1 answer:

AnnyKZ [126]2 years ago

3 0

Answer:

tita = 0 , 180, 270, 360. ( tita = 0 ≤ x ≤ 360).

Step-by-step explanation:

sin tita + sin ^2 tita = 0

sin tita( sin tita + 1) = 0

sin tita = 0, - 1.

For angles between 0 and 360 inclusive sin = 0 gives 0, 180 and 360 degrees. The angle whose sine is -1 is 270 degrees.

tita = 0 , 180, 270, 360.

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Solve the equation. Then check your solution

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Answer:

Option B is correct i.e. 21

Step-by-step explanation:

In the question we're provided with an equation that is :

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$\longrightarrow \: \frac{v}{ \cancel{7} }\times \cancel{7} = 3 \times 7$

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Putting value of v :

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By dividing 21 with 7 , we get :

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

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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.

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