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kherson [118]
3 years ago
12

Each set of (x, y) values represent a set of equivalent ratios. Find the ratio of x to y for each set to determine which set des

cribes the proportional relationship: 40.5 ounces of nuts to 48.6 ounces of raisins.
A) (5, 7)(60, 84)(47, 65.8)
B) (36, 48.6)(2, 2.7)(12, 16.2)
C) (3, 3.9)(65, 84.5)(24, 31.2)
D) (11, 13.2)(99, 118.8)(20, 24)
Mathematics
2 answers:
Ivan3 years ago
7 0

Solution:

\frac{x}{y}=\frac{40.5}{48.6}=\frac{45}{54}=\frac{5}{6}

If you will choose among the options

Option A contains (5,7) as the ratio is (5,6) , so incorrect option.

Option B contain (2,2.7) not equivalent to (5,6), so Incorrect option.

Option C contain (3,3.9) which is not equivalent to (5,6), so incorrect option.

Now coming to option (D) , we see that

\frac{11}{13.2}=\frac{5}{6},\frac{99}{118.8}=\frac{90}{108}=\frac{5}{6},\frac{20}{24}=\frac{5}{6}

All of them are equivalent to (5,6) i.e 40.5 ounces of nuts to 48.6 ounces of raisins.

So , Option (D) (11, 13.2)(99, 118.8)(20, 24) is correct option.

Assoli18 [71]3 years ago
5 0
Answer: The correct set is choice D.

We are looking for the same ratio with the given set of measurements. So lets start by determine the ratio of nuts to raisins. It is 40.5 to 48.6 or reduced it is 5 to 6. If you divide the fraction, you get the decimal 0.83333.

Only one of the relationships given has the same ratio. It is choice D.  11 to 13.32 can be divided to get 0.83333.
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Convert the given system of equations to matrix form
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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]

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I also need help with this math problem.
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Answer:

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