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3 years ago

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Anyone understand the answer

1 answer:

oksian1 [2.3K]3 years ago

7 0

$\bf 2\frac{1}{2}\left( 8x-\cfrac{4}{5} \right)=-62\implies \cfrac{5}{2}\left( 8x-\cfrac{4}{5} \right)=-62\implies \stackrel{\textit{distributing}}{\cfrac{5\cdot 8}{2}x-\cfrac{5\cdot 4}{2\cdot 5}}=-62 \\\\\\ 20x-2=-62\implies 20x=-60\implies x=\cfrac{-60}{20}\implies x=-3$

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Use separation of variables to solve dy dx − tan x = y2 tan x with y(0) = √3. Find the value of c in radians, not degrees

a_sh-v [17]

Answer:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

Step-by-step explanation:

Rewrite the equation as:

$\frac{dy(x)}{dx}-tan(x)=y(x)^{2} *tan(x)$

Isolating $\frac{dy}{dx}$

$\frac{dy}{dx} =tan(x)+tan(x)*y^{2}$

Factor:

$\frac{dy}{dx} =tan(x)*(1+y^{2} )$

Dividing both sides by $(1+y^{2} )$ and multiplying them by $dx$

$\frac{dy}{1+y^{2} } =tan(x)dx$

Integrate both sides:

$\int\ \frac{dy}{1+y^{2} } = \int\ tan(x) dx$

Evaluate the integrals:

$arctan(y)=-log(cos(x))+C_1$

Solving for y:

$y(x)=tan(-log(cos(x))+C_1)$

Evaluating the initial condition:

$y(0)=\sqrt{3} =tan(-log(cos(0))+C_1)=tan(-log(1)+C_1)=tan(0+C_1)$

$\sqrt{3} =tan(C_1)\\arctan(\sqrt{3} )=C_1\\60=C_1$

Converting 60 degrees to radians:

$60degrees*\frac{\pi }{180degrees} =\frac{\pi }{3}$

Replacing $C_1$ in the diferential equation solution:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

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BaLLatris [955]

Since 980÷2=490÷2=245÷5=49÷7=7÷7=1
980=2×2×5×7×7
the greatest perfect square will be
7*2*7*2
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