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9966 [12]
2 years ago
8

Pls help with 7 and 8 pls thanks

Mathematics
1 answer:
Nastasia [14]2 years ago
4 0
-3  1/12 should be your main answer :3
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This table shows data from a key to a map of ohio
Serga [27]
Where is the map??????
8 0
3 years ago
The graph shows the relationship between the number of hours that Michelle has been driving and the distance that she has left t
nikitadnepr [17]

Answer:

(b) For each hour that Michelle drove, she traveled an additional 50 miles.

Step-by-step explanation:

Given

(x_1,y_1) = (0,350) ---- start

(x_2,y_2) = (7,0) --- end

Required

Which is true

(a): Journey = 6 hours

This is false, because:

x = x_2 - x_1

x = 7-0

x = 7 ---- 7 hours

(b): The average rate is 50 miles per hour

To do this, we calculate the slope (m) using:

m = \frac{y_2 -y_1}{x_2- x_1}

m = \frac{0 - 350}{7-0}

m = -\frac{350}{7}

m = -50

This means that the rate is 50 miles driven in 1 hour.

<em>(b) is correct</em>

<em></em>

<em>Others are incorrect</em>

7 0
3 years ago
during Saturday's 12-mile hike Jim walk 5 miles in the first hour she then walk three more miles during the next hour what fract
Korvikt [17]
5 miles + 3 miles = 8 miles

8 miles out of 12 equals 8/12 = 1/3

so, your answer would be “he completed 1/3 of the total hike so far”

8 0
3 years ago
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:

  • <em>Convert the given system of equations to matrix form</em>

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).

so

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]

  • <em>Use row operations to put the augmented matrix in echelon form.</em>

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]

  • <em>Find the solutions set and put in vector form.</em>

<u>Interpret the reduced row echelon form:</u>

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 <em>z:</em>

<em>z=\frac{2}{3}(u+w+3)</em>

and replace this value into the other two equations

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

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:

<em>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</em>

where <em>u</em> and <em>w</em> 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 <em>u</em> out of the first vector and <em>w</em> 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]

7 0
3 years ago
Whats the slope for (-17,12) (-8,3)
Nitella [24]

Answer:

-9/11

Step-by-step explanation:

slope is change in y/change in x

change in y is 9 (12-3)

change in x is -11 (-17--8)

slope is -9/11

5 0
2 years ago
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