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Sophie [7]
3 years ago
7

An internet service provider changes an initial fee of $65 and $18 per hour of labor for office internet installation. Which equ

ation represents the total cost, C, of the installation after h hours of labor?
a C = 18 + 65h
b C = 65 - 18h
c C = 18h - 65
d C = 65 + 18h
Mathematics
2 answers:
mart [117]3 years ago
3 0

Answer:

D

Step-by-step explanation:

65 is your "b" value in this linear equation

18 is your slope

wolverine [178]3 years ago
3 0
The answers is D c=65+18h
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ivanzaharov [21]
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Given the exponential equation 4x = 64, what is the logarithmic form of the equation in base 10?
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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}

4 0
3 years ago
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