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3 years ago

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