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2 years ago

10

Please help! x^3 +y^3=9

1 answer:

Crank2 years ago

7 0

Step-by-step explanation:

Taking the derivative of both sides with respect to x, we get

$\dfrac{d}{dx}(x^3 + y^3) = \dfrac{d}{dx}(3)$

$\Rightarrow 3x^2 + 3y^2\dfrac{dy}{dx} = 0$

Solving for $\frac{dy}{dx}$ , we get

$\dfrac{dy}{dx} = -\dfrac{3x^2}{3y^2} = -\dfrac{x^2}{y^2}$

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Need help with this 1 math problem.

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A juice shop uses 1/2 of a bag of oranges to make 5 glasses of orange juice. What fraction of a bag of oranges did the shop use

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insens350 [35]

$\\ \sf\longmapsto 2x+y=2$

$\\ \sf\longmapsto 2x=2-y$

$\\ \sf\longmapsto x=\dfrac{2-y}{2}\dots(1)$

And

$\\ \sf\longmapsto x+1=y+2$

$\\ \sf\longmapsto x=y+2-1$

$\\ \sf\longmapsto x=y+1$

Put the value

$\\ \sf\longmapsto \dfrac{2-y}{2}=y+1$

$\\ \sf\longmapsto 2-y=2(y+1)$

$\\ \sf\longmapsto 2-y=2y+2$

$\\ \sf\longmapsto 2-2=2y+y$

$\\ \sf\longmapsto 3y=0$

$\\ \sf\longmapsto y=\dfrac{0}{3}$

$\\ \sf\longmapsto y=\infty$

Put in eq(1)

$\\ \sf\longmapsto x=\dfrac{2-y}{2}$

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3 years ago

Read 2 more answers

A N S W E R Q U I C K P L E A S E

chubhunter [2.5K]

Answer:

1. A

2. D

3. D

Step-by-step explanation:

The standard form of a parabola is

$y=\frac{1}{4p}(x-h)^2+k$ ..... (1)

Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.

1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .

$k-p=-8$ ...(a)

$(h,k+p)=(-2,-6)$

$k+p=-6$ .... (b)

$h=-2$

On solving (a) and (b), we get k=-7 and p=1.

Put h=-2, k=-7 and p=1 in equation (1).

$y=\frac{1}{4(1)}(x-(-2))^2+(-7)$

$y=\frac{1}{4}(x+2)^2-7$

Therefore option A is correct.

2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .

$k-p=5$ ...(c)

$(h,k+p)=(2,1)$

$k+p=1$ .... (d)

$h=2$

On solving (c) and (d), we get k=3 and p=-2.

Put h=2, k=3 and p=-2 in equation (1).

$y=\frac{1}{4(-2)}(x-(2))^2+(3)$

$y=-\frac{1}{8}(x-2)^2+3$

Therefore option D is correct.

3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .

$k-p=-3$ ...(e)

$(h,k+p)=(0,-2)$

$k+p=-2$ .... (f)

$h=0$

On solving (e) and (f), we get k=-2.5 and p=0.5.

Put h=0, k=-2.5 and p=0.5 in equation (1).

$y=\frac{1}{4(0.5)}(x-(0))^2+(-2.5)$

$y=\frac{1}{2}(x)^2-2.5$

$y=\frac{1}{2}(x)^2-\frac{5}{2}$

Therefore option D is correct.

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3 years ago