A box is 1 meter wider than its height and 2 meters longer than his width.
1 answer:
Let the height be
Hence, the width is $x+1$ and length is $(x+1)+2=x+3$
The volume is given to be 2 m³
Hence, $2=x(x+1)(x+3)$
Solving,
$x^3+4x^2+3x-2=0$
$(x+2)(x^2+2x-1)=0$
$\implies x+2=0$ or $x^2+2x-1=0$
But $x=-2$ cannot be possible
Hence, $x^2+2x-1=0$
$\implies x=-1-\sqrt 2$ or $x=-1+\sqrt 2$
Reject the negative solution.
Hence, the height is $\sqrt 2-1$,
width is $\sqrt 2$ and length is $\sqrt 2 +2$
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We must find the value of the constant k.
To do this, we substitute the values of x1 = 12, y1 = 15.
We have then:
Clearing k we have:
Substituting the value of k we have that the inverse variation is:
For y2 = 3 we have:
Clearing x2 we have:
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