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3 years ago

What are the solutions to the following system?

Mathematics

2 answers:

RoseWind [281]3 years ago

7 0

$\bf \begin{cases}
-2x^2+y=-5\\
\boxed{y}=-3x^2+5\\
----------\\
-2x^2+\left( \boxed{-3x^2+5} \right)=-5
\end{cases}
\\\\\\
-5x^2+5=-5\implies -5(x^2-1)=-5\implies x^2-1=\cfrac{-5}{-5}
\\\\\\
x^2-1=1\implies x^2=2\implies x=\pm\sqrt{2}$

and since we know what "x" is, then let's substitute it on say the 1st equation

$\bf -2x^2+y=-5\implies -2\left( \pm \sqrt{2} \right)^2+y=-5\implies -4+y=-5
\\\\\\
y=-1\\\\
-------------------------------\\\\
\left( \sqrt{2}~,~-1 \right)\qquad \left( -\sqrt{2}~,~-1 \right)$

NeX [460]3 years ago

6 0

Answer:

Step-by-step explanation:

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

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Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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