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pogonyaev
3 years ago
7

-5(1 - r) - 2 Find r

Mathematics
1 answer:
Grace [21]3 years ago
7 0

Answer:

-5+5r-2 -->

--> 5r-7 <--

Explanation --

Distribute from the parentheses, of course excluding the -2 because it isn't involved, and you'll get --> -5 + 5r -2

Simplify by putting the like terms together (subtract) -->

5r -7 (you get -7 by doing -5-2, and you leave 5r by itself since it doesn't have anything more to simplify with or has any other like term)

So, the final, simplified answer -->

5r-7

Hope this helps!! Have a nice day! :)

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How to solve part ii and iii
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\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}

when x=1 this reduces to

\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

2 \tan^{-1}(y) = \dfrac\pi3

\tan^{-1}(y) = \dfrac\pi6

\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)

\implies \boxed{y = \dfrac1{\sqrt3}}

(ii) Differentiate \tan^{-1}(xy) implicitly with respect to x. By the chain and product rules,

\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}

(iii) Differentiating both sides of the given equation leads to

\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0

where we use the result from (ii) for the derivative of \tan^{-1}(xy).

Solve for \frac{dy}{dx} :

\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0

\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)

\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}

\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}

\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}

From part (i), we have x=1 and y=\frac1{\sqrt3}, and substituting these leads to

\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}

\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}

\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}

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