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2 years ago

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Question 1 (1 point)

1 answer:

Natalija [7]2 years ago

5 0

Answer: OPTION B.

Step-by-step explanation:

Given the equation $48=-8(\frac{3b}{4}+2)$ you can find an equivalent equation by simplifying it.

You need to remember the multiplication of signs:

$(-)(-)=+\\(+)(+)=+\\(+)(-)=-$

Now you can apply the Distributive property:

$48=\frac{(-8)(3b)}{4}+(-8)(2)\\\\48=\frac{-24b}{4}-16$

Simplifying the equation, you get:

$48=-6b-16$

You can observe that this equivalent equation matches with the equation provided in option B.

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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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yuradex [85]

Given:

The function is

$f(x)=\dfrac{4x}{x^2-16}$

To find:

The asymptotes and zero of the function.

Solution:

We have,

$f(x)=\dfrac{4x}{x^2-16}$

For zeroes, f(x)=0.

$\dfrac{4x}{x^2-16}=0$

$4x=0$

$x=0$

Therefore, zero of the function is 0.

For vertical asymptote equate the denominator of the function equal to 0.

$x^2-16=0$

$x^2=16$

Taking square root on both sides, we get

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So, vertical asymptotes are x=-4 and x=4.

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