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3 years ago

Any helppp pLeAsE? T^T

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

6 0

answer:

c = 7.6 kilometers

step-by-step explanation:

you can use the pythagorean theorem for this since this is a right triangle
know the formula: a^2 + b^2 = c^2

now plug in what we know, also remember:

a = leg
b = the other leg
c = hypotenuse

a^2 + b^2 = c^2

3^2 + 7^2 = c^2

9 + 49 = c^2

58 = c^2

c = 7.6157

rounded: c = 7.6 kilometers

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

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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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3 years ago