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3 years ago

2. Kim is x years old. Jordan is 7 years older than Kim. Four times Jordan’s age is equal to 200.

Mathematics

2 answers:

Rom4ik [11]3 years ago

7 0

a:200/4= 50

50-7=43

b:43

c:50

lesantik [10]3 years ago

6 0

Answer:

Kim is 43 years old, and Jordan (7 years older) is 50 years old.

Step-by-step explanation:

Representation:

Kim's age is x

Jordan's is x + 7

Then 4(x + 7) = 200, and 4x + 28 = 200.

Simplifying, 4x = 172. Dividing both sides by 4 to isolate x, we get:

x = 172/4 = 43

Thus, Kim is 43 years old, and Jordan (7 years older) is 50 years old.

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Step-by-step explanation:

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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:

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

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

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3 years ago

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VARVARA [1.3K]

Answer:

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Step-by-step explanation:

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Subtracting a function by a constant x is the same as moving the function down x units.

In this question:

$f(x) = 2^x$

$g(x) = 2^x - 4$

Thus, the graph of g(x) is the graph of f(x) moved down 4 units.

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3 years ago