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2 years ago

Ten years ago Victors age WAs half of the age he will be in 10 years how old is he now?

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

3 0

By solving a linear equation, we will see that Victor is 30 years old.

<h3>How old is Victor now?</h3>

Let's say that Victor is x years old.

We know that 10 years ago, (x - 10), his age was half of what it will be in 10 years (x + 10).

Then we can write the linear equation:

(x - 10) = (x + 10)/2

Now we can solve that equation for x:

2*(x - 10) = x + 10

2x - 20 = x + 10

2x - x = 10 + 20

x = 30

So we conclude that Victor is 30 years old.

If you want to learn more about linear equations:

brainly.com/question/1884491

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dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

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If $x=0$ , then $y=0$ - critical point at (0, 0).
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The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

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and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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