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Vesnalui [34]
3 years ago
15

A holiday park is digging a swimming pool.

Mathematics
1 answer:
OLga [1]3 years ago
4 0
I think it’s A
Hope this helps:)
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How to solve part ii and iii
iragen [17]

(i) Given that

\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}

as required.

3 0
2 years ago
3. The points (0, 1) and (1,2) lie on all three of a linear, an exponential, and a quadratic function.
svlad2 [7]

Answer:

  exponential

Step-by-step explanation:

The exponential function has the largest value for large values of x. Here, "large" means "greater than 4.3". A graph shows this to be true.

For x=10, the exponential function will have a value of 2^10 = 1024.

The quadratic function will have a value of 10^2+1 = 101.

The linear function will have a value of 10+1 = 11.

The exponential function has the largest value.

7 0
3 years ago
Let v=-9i+j and w=-i-6j find 8v-6w
hram777 [196]

Answer:

78i+52j

Step-by-step explanation:

8(9i+2j)-6(-i-6j)

72i+16j+6i+36j

=78i+52j

7 0
3 years ago
What is the value of y when x = 0? When x = 0, y =
ra1l [238]

Answer:

The point in which the graph crosses the x-axis is called the x-intercept and the point in which the graph crosses the y-axis is called the y-intercept.

Step-by-step explanation:

The x-intercept is found by finding the value of x when y = 0, (x, 0), and the y-intercept is found by finding the value of y when x = 0, (0, y).

4 0
3 years ago
Read 2 more answers
copy central prints black and white or colored copies. Yesterday, they printed 9 black and white copies for every 2 colored copi
nalin [4]
\bf \cfrac{bw}{color}\qquad 9:2\qquad \cfrac{9}{2}\qquad \qquad \cfrac{bw}{200}=\cfrac{9}{2}\implies bw=\cfrac{200\cdot 9}{2}
4 0
3 years ago
Read 2 more answers
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