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3 years ago

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Is this correct????????

1 answer:

Pavel [41]3 years ago

5 0

Answer:

i dont see photo

Step-by-step explanation:

i don't no..

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Anybody wanna help please?

Lisa [10]

Basically all you got to do is at each point go left and down 6 times!

Answer: (-9,0) (-3,0) (-2,-1) (-4,-2)

5 0

3 years ago

The equations that must be solved for maximum or minimum values of a differentiable function w=f(x,y,z) subject to two constrai

sammy [17]

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$ , then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$ , then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

but there are no real solutions for $y$ , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)

5 0

4 years ago

Find the remaining trig values of θ if <br> cotθ = − 3/5 and csc θ < 0 .

EastWind [94]

Answer:

1 Answer. Aviv S. The value of tan(π3) is √3 .

Step-by-step explanation:

3 0

3 years ago

The account balance of each of three children at the end of a month is shown below: John has −$2.75 Cheryl has −$3.00 Andrew has

Shalnov [3]

Answer:

Step-by-step explanation:

I really dont know

3 0

3 years ago

What is the approximate value of x in this figure?<br> 19 POINTS

Sladkaya [172]

Answer:

4.69 inches (2 decimal places)

Step-by-step explanation:

Refer to image attached to this answer:

Using Pythagoras Theorem:

$AB^2+AC^2=BC^2$

$2^2+3^2=BC^2$

$4+9=BC^2$

$BC^2=13$

-------------------------

Using Pythagoras Theorem,

$BC^2+CD^2=BD^2\ (x)$

$13+3^2=BD^2$

$13+9=BD^2$

$BD=\sqrt{22} = 4.69\ (2d.p.)$

7 0

3 years ago

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