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3 years ago

What are the six trigonometric functions and how are their values determined using the three sides of a right triangle?

Mathematics

1 answer:

algol [13]3 years ago

4 0

Explanation:

Six trigonometric functions are ratio of sides with their angle in right angled triangle.

Six trigonometric functions are:

Sinθ = Perpendicular / Hypotenuse

Cosθ = Base / Hypotenuse

Tanθ = Perpendicular / Base

Cotθ = Base / Perpendicular

Secθ = Hypotenuse / Base

Cosecθ = Hypotenuse / Perpendicular

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Convert the given system of equations to matrix form

yuradex [85]

Answer:

The matrix form of the system of equations is $\left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right] \left[\begin{array}{c}x&y&w&z&u\end{array}\right] =\left[\begin{array}{c}5&4&3\end{array}\right]$

The reduced row echelon form is $\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

The vector form of the general solution for this system is $\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Step-by-step explanation:

Convert the given system of equations to matrix form

We have the following system of linear equations:

$x+y+w+z-3u=5\\x-y-2w+z+2u=4\\2x+w-z+u=3$

To arrange this system in matrix form (Ax = b), we need the coefficient matrix (A), the variable matrix (x), and the constant matrix (b).

$A= \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right]$

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]$

$b=\left[\begin{array}{c}5&4&3\end{array}\right]$

Use row operations to put the augmented matrix in echelon form.

An augmented matrix for a system of equations is the matrix obtained by appending the columns of b to the right of those of A.

So for our system the augmented matrix is:

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\1&-1&-2&1&2&4\\2&0&1&-1&1&3\end{array}\right]$

To transform the augmented matrix to reduced row echelon form we need to follow this row operations:

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\2&0&1&-1&1&3\end{array}\right]$

add -2 times the 1st row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\0&-2&-1&-3&7&-7\end{array}\right]$

multiply the 2nd row by -1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&-2&-1&-3&7&-7\end{array}\right]$

add 2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&2&-3&2&-6\end{array}\right]$

multiply the 3rd row by 1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&1&-3/2&1&-3\end{array}\right]$

add -3/2 times the 3rd row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 3rd row to the 1st row

$\left[\begin{array}{ccccc|c}1&1&0&5/2&-4&8\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 2nd row to the 1st row

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

Find the solutions set and put in vector form.

Interpret the reduced row echelon form:

The reduced row echelon form of the augmented matrix is

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

which corresponds to the system:

$x+1/4\cdot z=3\\y+9/4\cdot z-4u=5\\w-3/2\cdot z+u=-3$

We can solve for z:

$z=\frac{2}{3}(u+w+3)$

and replace this value into the other two equations

$x+1/4 \cdot (\frac{2}{3}(u+w+3))=3\\x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}$

$y+9/4 \cdot (\frac{2}{3}(u+w+3))-4u=5\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}$

No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The system has infinitely many solutions:

$x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}\\z=\frac{2u}{3}+\frac{2w}{3}+2$

where u and w are free variables.

We put all 5 variables into a column vector, in order, x,y,w,z,u

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=\left[\begin{array}{c}-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}&\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}&w&\frac{2u}{3}+\frac{2w}{3}+2&u\end{array}\right]$

Next we break it up into 3 vectors, the one with all u's, the one with all w's and the one with all constants:

$\left[\begin{array}{c}-\frac{u}{6}&\frac{5u}{2}&0&\frac{2u}{3}&u\end{array}\right]+\left[\begin{array}{c}-\frac{w}{6}&-\frac{3w}{2}&w&\frac{2w}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Next we factor u out of the first vector and w out of the second:

$u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

The vector form of the general solution is

$\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

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4 years ago

Select the correct answer from each drop-down menu.

kati45 [8]

Answer:

3, 3, -60

Step-by-step explanation:

The area of a rectangle is equal to the product of its length and width. So, to create an equation representing the area of this rectangle, multiply the length, , and the width, , and set the product equal to the area, 54.

Then rewrite the equation using the distributive property, and move all terms to the left side of the equation.

So, the equation to find the value of n is .

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Point A lies on the circle and has an x-coordinate of 1.

stellarik [79]

Answer:

$\sqrt{(0-y)^{2}+(0-1)^{2}}=2$