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liberstina [14]
3 years ago
10

Need help with question 11 and 12

Mathematics
1 answer:
qaws [65]3 years ago
4 0

Answer:

  11.  your choices are correct

  12.  132°

Step-by-step explanation:

11. The given equations can be (re)written to slope-intercept form. Only those lines with a slope of 2 will be parallel to the given line.

  A: y = 2x -8 . . . . . slope = 2, an answer

  B: y = -2x +1 . . . . . slope = -2, not an answer

  C: y = -2x +7 . . . . .slope = -2, not an answer

  D: y = 2x +2 . . . . . slope = 2, an answer

  E: y = -2x -9 . . . . . slope = -2, not an answer

__

12. The marked angles are supplementary, so ...

  (3x -6)° +(x +2)° = 180°

  4x -4 = 180 . . . . . . . . . . . divide by °, collect terms

  x -1 = 45 . . . . . . . divide by 4

  x = 46 . . . . . . . . . add 1

Now, we can find the angle of interest.

∠PQR = (3x -6)° = (3·46 -6)°

∠PQR = 132°

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First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

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We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

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

Let's compute dy/dx. We'll use f(x) as defined earlier.

y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\

If you need a refresher on the quotient rule, then

\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\

where P and Q are functions of x.

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

This then means

\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

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

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form  ( f(x) )^2 in the numerator to cancel out with the denominator.

\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\

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Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

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