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3 years ago

If xy + y² = 6, then the value of dy/dx at x= -1 is

Mathematics

1 answer:

mylen [45]3 years ago

5 0

Hi there!

$\large\boxed{\text{At (-1, -2), }\frac{dy}{dx} = -\frac{2}{5}}}$

$\large\boxed{\text{At (-1, 3), }\frac{dy}{dx} = -\frac{3}{5}}}$

We can calculate dy/dx using implicit differentiation:

xy + y² = 6

Differentiate both sides. Remember to use the Product Rule for the "xy" term:

(1)y + x(dy/dx) + 2y(dy/dx) = 0

Move y to the opposite side:

x(dy/dx) + 2y(dy/dx) = -y

Factor out dy/dx:

dy/dx(x + 2y) = -y

Divide both sides by x + 2y:

dy/dx = -y/x + 2y

We need both x and y to find dy/dx, so plug in the given value of x into the original equation:

-1(y) + y² = 6

-y + y² = 6

y² - y - 6 = 0

(y - 3)(y + 2) = 0

Thus, y = -2 and 3.

We can calculate dy/dx at each point:

At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.

At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.

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Step-by-step explanation:

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<em>Comment on this solution</em>

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2 years ago

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3 years ago

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

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3 years ago

9th grade

il63 [147K]

Answer:

May be the second one

Step-by-step explanation:

hope it'll help ☺️

4 0

3 years ago

Thanks for the help!

NikAS [45]

Answer:

Step-by-step explanation:

So we have the piecewise function:

$f(x)= \begin{cases}\sqrt{2x} & \text{if }x$

And we want to find f(8).

Since our input value is 8, choose the equation that fits our input.

The first equation demand x to be less than 3. 8 is not less than 3, so we won't use that.

The second equation demand x to e greater than or equal to 3 and less than 8. 8 is <em>not</em> less than 8. So, we won't use that.

The third equation demands x to be greater than or equal to 8. 8 is greater than or equal to 8, so we'll use the third equation.

So:

$f(x)=42\text{ if }x\geq 8$

$f(8)=42$

There're nothing more to do, we're done :)

The answer is A.

Edit: Improved Format

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3 years ago

If xy + y² = 6, then the value of dy/dx at x= -1 is​

If xy + y² = 6, then the value of dy/dx at x= -1 is