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mr Goodwill [35]

3 years ago

10

WHY CAN'T ANYONE HELP ME PLEASE?? The Pool Fun Company has learned that, by pricing a newly released Fun Noodle at $3, sales wi

ll reach 8000 Fun Noodles per day during the summer. Raising the price to $6 will cause the sales to fall to 5000 Fun Noodles per day. a. Assume that the relationship between sales price, x, and number of Fun Noodles sold, y, is linear. Write an equation in slope-intercept form describing this relationship. Use ordered pairs of the form (sales price, number sold).

1 answer:

Zinaida [17]3 years ago

6 0

Answer:

y = -1000x +11000

Step-by-step explanation:

<u>Given:</u>

(x, y) = (sales price, number sold) = (3, 8000), (6, 5000)

<u>Find</u>:

slope-intercept equation for a line through these points

<u>Solution</u>:

When given two points, it often works well to start with the 2-point form of the equation for a line.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

Filling in the given points, you have ...

y = (5000 -8000)/(6 -3)/(x -3) +8000

y = (-3000/3)(x -3) +8000

y = -1000x +3000 +8000 . . . . eliminate parentheses

y = -1000x +11000 . . . . the desired equation

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Answer:

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Step-by-step explanation:

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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.

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2 years ago

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Answer:

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Step-by-step explanation:

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So according to the given options, it is the sample where each member have the known chance to be selected

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