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2 years ago

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Mathematics

1 answer:

viktelen [127]2 years ago

3 0

4) The first and second terms for both ratios need to be in the same order.

Step-by-step explanation:

Given ratio is:

16:36

In order to find any ratio equivalent to given ratio, the ratio can be divided by a number or multiplied to a number.

The equivalent ratio that is given: 72:32

If we multiply the given ratio by 2: We get 32:72

So,

Looking at the options we can conclude that the right answer is

4) The first and second terms for both ratios need to be in the same order.

Keywords: Ratio, Fractions

Learn more about ratios at:

#LearnwithBrainly

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a) $A=\left[\begin{array}{ccc}1&2&3\\1&-1&1\end{array}\right]$

$b=\left[\begin{array}{ccc}0\\1\end{array}\right]$

b) $||Ax-b||^{2} =(-bx_{2}+4)^{2} (-4x_{1} +3x_{2} -1)^{2} +(x_{1} +8x_{2} -3)^{2}$

c) $A=\left[\begin{array}{ccc}0&6\sqrt{2} &0\\\sqrt{3} &3\sqrt{3} &0\\2&-16&0\end{array}\right]$

$x=\left[\begin{array}{ccc}x_{1} \\x_{2} \\x_{3} \end{array}\right]$

$b=\left[\begin{array}{ccc}-\sqrt{2} \\\sqrt{3} \\6\end{array}\right]$

Step-by-step explanation:

a) considering the equation:

Minimize $x_{1}^{2} +2x_{2}x^{2} +3x_{3}^{2}+(x_{1} -x_{2} +x_{3} -1)^{2} +(-x_{1} -4x_{2} +2)^{2}$

$A=\left[\begin{array}{ccc}1&2&3\\1&-1&1\end{array}\right]$ (matrix A)

vector b

$b=\left[\begin{array}{ccc}0\\1\end{array}\right]$

b) If Pxn is matrix B and p-vector d, we have:

minimize $(-6x_{2}+4)^{2} +(-4x_{1} +3x_{2} -1)+(x_{1} +8x_{2} -3)^{2}$

$Ax=\left[\begin{array}{ccc}0&-6&0\\-4&3&0\\1&8&0\end{array}\right]$

$\left[\begin{array}{ccc}x_{1} \\x_{2} \\x_{3} \end{array}\right]$

$b=\left[\begin{array}{ccc}-4\\1\\3\end{array}\right]$

$Ax-b=\left[\begin{array}{ccc}-bx_{2}+4 \\-4x_{1}+3x_{2}-1 \\x_{1}+8x_{2}-3 \end{array}\right] =1$

$||Ax-b||^{2} =(-bx_{2}+4)^{2} (-4x_{1} +3x_{2} -1)^{2} +(x_{1} +8x_{2} -3)^{2}$

c) minimize $2(-bx_{2}+4)^{2} +3(-4x_{1} +3x_{2} -1)^{2} +4(x_{1} -x_{2} -3)^{2} -(6\sqrt{2}x_{2} +4\sqrt{2} )^{2} +(-4\sqrt{3} x_{1} +3\sqrt{3}x_{2} -\sqrt{3})^{2} +(2x_{1} -16x_{2} -6)^{2}$

in matrix:

$A=\left[\begin{array}{ccc}0&6\sqrt{2} &0\\\sqrt{3} &3\sqrt{3} &0\\2&-16&0\end{array}\right]$

$x=\left[\begin{array}{ccc}x_{1} \\x_{2} \\x_{3} \end{array}\right]$

$b=\left[\begin{array}{ccc}-\sqrt{2} \\\sqrt{3} \\6\end{array}\right]$

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2 years ago