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2 years ago

11

Jordan volunteered for the same amount of time each day for 7 days. If he volunteered for 329 minutes altogether, how many minut

es did he volunteer for each day?

1 answer:

AVprozaik [17]2 years ago

3 0

Answer:

47

Step-by-step explanation:

Just divide 329 by 7

That gives you 47

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Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr

Pani-rosa [81]

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$ .

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
$L_\lambda=x^2+y^2-1$

Set each partial derivative to 0 to find the possible critical points within the disk $D$ . Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
$xL_y=x^2+2xy+2\lambda xy=0$
$L_\lambda=0\implies x^2+y^2=1$

$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$ , we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$ , or $x^2=y^2$ , we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
$f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32$

It should be clear enough which of these correspond to the absolute extrema of $f$ over $D$ .

8 0

3 years ago

Someone please help<br> Me

Ne4ueva [31]

Answer:

dude r u in a cc rn for math?

Step-by-step explanation:

3 0

3 years ago

Someone help ASAP for 5 points

prisoha [69]

Answer:

it may be 27

Step-by-step explanation:

12^2+16^2=400

square root 400= 20

36/16=2.25

12*2.25=27

6 0

3 years ago

Jason went shopping.

melomori [17]

53.55 / 85 = 0.63

0.63 * 100 = 63

63 - 38 = 25

watch was 25 before discount

8 0

2 years ago

Two numbers 60 and n have LCM of 420. If n is a multiple of 6 less than 90,find the possible value of n

Ber [7]

Lcm of 60 and n = 420
Stated: n <90 and a multiple of 6

420/60 = 7 thus n must be a multiple of 7
Therefore n =42 or 84

3 0

3 years ago