Mathematics College

Answer pllllllleeeaaaaasssss

Answers

Answer 1

(3.1) … … …

[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2x-y}{x-2y}[/tex]

Multiply the right side by x/x :

[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{2-\dfrac yx}{1-\dfrac{2y}x}[/tex]

Substitute y(x) = x v(x), so that dy/dx = x dv/dx + v :

[tex]x\dfrac{\mathrm dv}{\mathrm dx} + v = \dfrac{2-v}{1-2v}[/tex]

This DE is now separable. With some simplification, you get

[tex]x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{2-2v+2v^2}{1-2v}[/tex]

[tex]\dfrac{1-2v}{2-2v+2v^2}\,\mathrm dv = \dfrac{\mathrm dx}x[/tex]

Now you're ready to integrate both sides (on the left, the denominator makes for a smooth substitution), which gives

[tex]-\dfrac12\ln\left|2v^2-2v+2\right| = \ln|x| + C[/tex]

Solve for v, then for y (or leave the solution in implicit form):

[tex]\ln\left|2v^2-2v+2\right| = -2\ln|x| + C[/tex]

[tex]\ln(2) + \ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]

[tex]\ln\left|v^2-v+1\right| = \ln\left(\dfrac1{x^2}\right) + C[/tex]

[tex]v^2-v+1 = e^{\ln\left(1/x^2\right)+C}[/tex]

[tex]v^2-v+1 = \dfrac C{x^2}[/tex]

[tex]\boxed{\left(\dfrac yx\right)^2 - \dfrac yx+1 = \dfrac C{x^2}}[/tex]

(3.2) … … …

[tex]y' + \dfrac yx = \dfrac{y^{-3/4}}{x^4}[/tex]

It may help to recognize this as a Bernoulli equation. Multiply both sides by [tex]y^{\frac34}[/tex] :

[tex]y^{3/4}y' + \dfrac{y^{7/4}}x = \dfrac1{x^4}[/tex]

Substitute [tex]z(x)=y(x)^{\frac74}[/tex], so that [tex]z' = \frac74 y^{3/4}y'[/tex]. Then you get a linear equation in z, which I write here in standard form:

[tex]\dfrac47 z' + \dfrac zx = \dfrac1{x^4} \implies z' + \dfrac7{4x}z=\dfrac7{4x^4}[/tex]

Multiply both sides by an integrating factor, [tex]x^{\frac74}[/tex], which gives

[tex]x^{7/4}z'+\dfrac74 x^{3/4}z = \dfrac74 x^{-9/4}[/tex]

and lets us condense the left side into the derivative of a product,

[tex]\left(x^{7/4}z\right)' = \dfrac74 x^{-9/4}[/tex]

Integrate both sides:

[tex]x^{7/4}z=\dfrac74\left(-\dfrac45\right) x^{-5/4}+C[/tex]

[tex]z=-\dfrac75 x^{-3} + Cx^{-7/4}[/tex]

Solve in terms of y :

[tex]y^{4/7}=-\dfrac7{5x^3} + \dfrac C{x^{7/4}}[/tex]

[tex]\boxed{y=\left(\dfrac C{x^{7/4}} - \dfrac7{5x^3}\right)^{7/4}}[/tex]

(3.3) … … …

[tex](\cos(x) - 2xy)\,\mathrm dx + \left(e^y-x^2\right)\,\mathrm dy = 0[/tex]

This DE is exact, since

[tex]\dfrac{\partial(-2xy)}{\partial y} = -2x[/tex]

[tex]\dfrac{\partial\left(e^y-x^2\right)}{\partial x} = -2x[/tex]

are the same. Then the general solution is a function f(x, y) = C, such that

[tex]\dfrac{\partial f}{\partial x}=\cos(x)-2xy[/tex]

[tex]\dfrac{\partial f}{\partial y} = e^y-x^2[/tex]

Integrating both sides of the first equation with respect to x gives

[tex]f(x,y) = \sin(x) - x^2y + g(y)[/tex]

Differentiating this result with respect to y then gives

[tex]-x^2 + \dfrac{\mathrm dg}{\mathrm dy} = e^y - x^2[/tex]

[tex]\implies\dfrac{\mathrm dg}{\mathrm dy} = e^y \implies g(y) = e^y + C[/tex]

Then the general solution is

[tex]\sin(x) - x^2y + e^y = C[/tex]

Given that y (1) = 4, we find

[tex]C = \sin(1) - 4 + e^4[/tex]

so that the particular solution is

[tex]\boxed{\sin(x) - x^2y + e^y = \sin(1) - 4 + e^4}[/tex]

Related Questions

Need the answer please, soon as possible

Answers

9514 1404 393

Answer:

(d) 27.4%

Step-by-step explanation:

The desired percentage is ...

(juniors for Kato)/(total juniors) × 100%

= 129/(129 +194 +147) × 100%

= (129/470) × 100% ≈ 27.4%

About 27.4% of juniors voted for Kato.

find the exact value of tan(165°) using a difference of two angles

Answers

Answer: [tex]-2+\sqrt{3}[/tex]

=========================================================

Work Shown:

Apply the following trig identity

[tex]\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\[/tex]

Now let's rationalize the denominator

[tex]\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\[/tex]

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

[tex]\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949[/tex]

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as [tex]\sqrt{3}-2[/tex] but it's the same as above.

What is the equation of the line through (4, -3) and (0, -2)?
O A. y = -x + 2
B. Y= -1x – 2
O C. y= - 4x –
D. y = 4x + 2

Answers

Answer:

Step-by-step explanation:

Pull in -4 and 0 into the variable x. and the 4 and -2 in the variable y and solve to get an equal answer.

Raul is a manager at a local restaurant. He earns $18.50 per hour. How many hours per week does Raul
work if he earns $740 per week?

Answers

Answer: 40 hours
Just divide the total amount (740) by the rate per hour (18.50) and you get 40

When A = 200, solve the equation x2 - 40x + A=0 using the quadratic formula. Show all your working and give your answers correct to 2 decimal places.

Answers

Answer:

Solution given:

equation is:

x²-40x+A=0

when A=200

equation becomes

x²-40x+200=0

Comparing above equation with ax²+bx+c=0 we get

a=1

b=-40

c=200

By using quadratic equation formula

x=[tex]\displaystyle \frac{-b±\sqrt{b²-4ac}}{2a}[/tex]

substituting value

x=[tex]\displaystyle \frac{-*-40±\sqrt{(-40)²-4*1*200}}{2*1}[/tex]

x=[tex]\displaystyle \frac{40±\sqrt{800}}{2}[/tex]

x=[tex]\displaystyle \frac{40±20\sqrt{2}}{2}[/tex]

taking positive

x=[tex]\displaystyle \frac{40+20\sqrt{2}}{2}[/tex]

x=34.14

taking negative

x=[tex]\displaystyle \frac{40-20\sqrt{2}}{2}[/tex]

x=5.86

x=34.14 or 5.86

The firm had 15 billion VND of earnings before interest and tax (EBIT), corporate tax is 20%. The market price of stock is 60.000đ. Knowing that net income will be held 40% before using it for dividend. How much of the net income can be divided for shareholders?