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Nostrana [21]
3 years ago
7

Converting 4.9 kilograms would give you 11 of which US customary system unit? A. Pounds B. Stones C. Teaspoons D. Ounces Mark fo

r review (Will be highlighted on the review page)
Mathematics
2 answers:
Studentka2010 [4]3 years ago
4 0

Hello!

One kilogram is about 2.2 pounds. Let's convert below.

4.9(2.2)=10.78

One kilogram is about 1.5 stones, but as that is less than 2.2, it won't get us any closer to 11.

A kilogram would be way more than 11 teaspoons, along with 11 ounces.

Therefore, our closest answer is A) Pounds.

I hope this helps!

Usimov [2.4K]3 years ago
3 0

Kilograms is a type of measurement for weight, and so the converted measure should also be weight (which leaves only oz. and lbs.)

The converting ratio of kilograms to pounds is: 1 = 2.2

The converting ratio of kilograms to ounces is: 1 = 35.274

This means that pounds, or (B) is your closest answer


hope this helps

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Solve the lenear system by using the inverse of the coefficient matrix:
EleoNora [17]

Answer:

The solution of this system is x=9/4, y=5/2, and z=-13/8

Step-by-step explanation:

<em>1. Writing the equations in matrix form</em>

The system of linear equations given can be written in matrix form as

\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}-1&2&1\end{array}\right]

Writing

A = \left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]

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

B = \left[\begin{array}{c}-1&2&1\end{array}\right]

we have

AX=B

This is the matrix form of the simultaneous equations.

<em>2. Solving the simultaneous equations</em>

<em>Given</em>

AX=B

we can multiply both sides by the inverse of A

A^{-1}AX=A^{-1}B

We know that A^{-1}A=I, the identity matrix, so

X=A^{-1}B

All we need to do is calculate the inverse of the matrix of coefficients, and finally perform matrix multiplication.

<em>3. Calculate the inverse of the matrix of coefficients</em>

A = \left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\2&-1&0&0&1&0\\0&3&4&0&0&1\end{array}\right]

  • <em>Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 2 from row 2</em>

\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&3&4&0&0&1\end{array}\right]

  • <em>Make zeros in column 2 except the entry at row 2, column 2. Add row 2 multiplied by 3 to row 3</em>

\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&0&-8&-6&3&1\end{array}\right]

  • <em>Multiply row 2 by −1</em>

\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&-8&-6&3&1\end{array}\right]

  • <em>Make zeros in column 3 except the entry at row 3, column 3. Divide row 3 by −8</em>

\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right]

  • <em>Subtract row 3 multiplied by 2 from row 1</em>

\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right]

  • <em>Subtract row 3 multiplied by 4 from row 2</em>

\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&0&-1&1/2&1/2\\0&0&1&3/4&-3/8&-1/8\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

A^{-1} = \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right]

<em>4. Find the solution X=A^{-1}B</em>

X= \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right]\cdot \left[\begin{array}{c}-1&2&1\end{array}\right] = \left[\begin{array}{c}9/4&5/2&-13/8\end{array}\right]

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4 years ago
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