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alexdok [17]
2 years ago
12

1. Subtract the linear expressions.

Mathematics
1 answer:
jeyben [28]2 years ago
7 0

(1)\\\\(45x -5) -(13x +4) = 45x-5-13x -4 = 32x -9\\\\\\(2)\\\\-2(4n) -3 + (-7)(8n) +2 = -8n-56n -1 = -64n -1 = -(64n+1)\\\\(3)\\\\(-3.6x+9.4) - (1.9x +3.6) = -3.6x +9.4 -1.9x -3.6 =-5.5x+5.8\\\\\\(4)\\\\-3x+4-(5x-2) = -3x+4-5x +2=-8x +6

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Step-by-step explanation:

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Assume that the annual household expenditure on sugar is approximately normally distributed with a mean of $8.22 and a standard
irina1246 [14]

 To solve this problem, we will have to make use of the z statistic. The formula for the z score is given as:

z = (x – u) / s

where,

x is the sample value = more than $10.00

u is the sample mean = $8.22

s is the standard deviation = $1.10

 

Substituting the values into the equation to solve for z:

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3 years ago
consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros
Bezzdna [24]

Answer:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

Step-by-step explanation:

The given quadratic form is of the form

q(x,y,z)=ax^2+by^2+dxy+exz+fyz.

Where a=11,b=-1,c=-4,d=-16,e=8,f=-4.Every quadratic form of this kind can be written as

q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)

Observe that A is a symmetric matrix. So A is orthogonally diagonalizable, that is to say,  D=Q^{T}AQ where Q is an orthogonal matrix and D is a diagonal matrix.

In our case we have:

A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)

The eigenvalues of A are \lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5.

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)

D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)

Now, we have to do the change of variables {\bf x}=Q{\bf y} to obtain

q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2

Which can be written as:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

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