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Alex Ar [27]
3 years ago
5

Simplify this expression.

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0

<em>Answer:</em>

<em></em>\frac{6w^2 y^4}{x}<em></em>

<em>Step-by-step explanation:</em>

<em></em>\frac{12x^3 y^6 w^5}{2x^4 y^2 w^3}<em></em>

<em></em>= \frac{12w^5 x^3 y^6}{2w^3 x^4 y^2}<em></em>

<em></em>= \frac{12w^2 y^4}{2x}<em></em>

<em></em>= \frac{6w^2 y^4}{x}<em></em>

<em>Hope this helped you!</em>

Margarita [4]3 years ago
7 0

Answer

\frac{ 6   {y}^{4}  {w}^{2} }{{x}}

Step-by-step Explanation

\huge \frac{12 {x}^{3} {y}^{6}   {w}^{5} }{ 2{x}^{4} {y}^{2}   {w}^{3} }  \\  \\  \huge= \frac{6 {x}^{3} {y}^{6}   {w}^{5} }{{x}^{4} {y}^{2}   {w}^{3} }  \\  \\ \huge = 6 {x}^{3 - 4}  {y}^{6 - 2}  {w}^{5 - 3}  \\  \\ \huge= 6 {x}^{ - 1}  {y}^{4}  {w}^{2}  \\  \\ \huge =  \frac{ 6   {y}^{4}  {w}^{2} }{{x}^{1}} \\  \\ \huge =  \frac{ 6   {y}^{4}  {w}^{2} }{{x}}

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2. Use the Distributive Property to determine which of the following statements is equivalent to 5(6-2).
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Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

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

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0
3 years ago
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