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3 years ago

Determine the value of each variable. Enter your answers as decimals. X: Y: Z:

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

Part 1) $x=28.75\°$

Part 2) $z=44.5\°$

Part 3) $y=135.5\°$

Step-by-step explanation:

Part 1) Determine the value of x

we know that

$(\frac{6}{5}x+10)\°=(2x-13)\°$ ------> by corresponding angles

Solve for x

Multiply by 5 both sides to remove the fraction

$5(\frac{6}{5}x+10)\°=5(2x-13)\°$

$6x+50=10x-65$

$10x-6x=50+65$

$4x=115$

Divide by 4 both sides

$x=28.75\°$

Part 2) Determine the value of z

we know that

$z=(\frac{6}{5}x+10)\°$ ------> by corresponding angles

we have

$x=28.75\°$

substitute the value of x and solve for z

$z=(\frac{6}{5}(28.75)+10)=44.5\°$

Part 3) Determine the value of y

we know that

$y+z=180\°$ ------> by supplementary angles

we have

$z=44.5\°$

substitute and solve for y

$y+44.5\°=180\°$

$y=180\°-44.5\°=135.5\°$

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Read 2 more answers

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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3 years ago

A square has a diagonal of 10 cm. What are the lengths of the sides?

Katen [24]

Answer:

5√2 cm

Step-by-step explanation:

Let x = the length of a side of the square

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x2 + x2 = 102

2x2 = 100

x2 = 50

x = √50

x = 5√2 cm

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