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MrRa [10]
2 years ago
14

just please help 6 through 8 thank you please help me its a homework but i really dont get it that much please help me​

Mathematics
1 answer:
masha68 [24]2 years ago
5 0

Problem 6

It's not clear what the instructions are, but I'm assuming your teacher wants you to rationalize the denominator.

If so, then we could have these steps

\sqrt{\frac{3\text{x}}{7}}\\\\\frac{\sqrt{3\text{x}}}{\sqrt{7}}\\\\\frac{\sqrt{3\text{x}}*\sqrt{7}}{\sqrt{7}*\sqrt{7}}\\\\\frac{\sqrt{3\text{x}*7}}{(\sqrt{7})^2}\\\\\frac{\sqrt{21\text{x}}}{7}\\\\

In the third step, I multiplied top and bottom by \sqrt{7} so that the denominator can get rid of the square root.

<h3>Answer: \frac{\sqrt{21\text{x}}}{7}\\\\</h3>

======================================================

Problem 7

We'll follow the same idea for this problem.

\sqrt{\frac{5m^2}{2}}\\\\\frac{\sqrt{5m^2}}{\sqrt{2}}\\\\\frac{\sqrt{5m^2}*\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\\frac{\sqrt{5m^2*2}}{(\sqrt{2})^2}\\\\\frac{\sqrt{10m^2}}{2}\\\\\frac{\sqrt{10}*\sqrt{m^2}}{2}\\\\\frac{m\sqrt{10}}{2}\\\\

Where m is nonnegative.

In the second to last step, I split up the root so I could use the rule \sqrt{x^2} = x \text{ when } x \ge 0

<h3>Answer:  \frac{m\sqrt{10}}{2}\\\\</h3>

======================================================

Problem 8

\sqrt{\frac{12\text{x}^3}{5}}\\\\\frac{\sqrt{12\text{x}^3}}{\sqrt{5}}\\\\\frac{\sqrt{12\text{x}^3}*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\\frac{\sqrt{12\text{x}^3*5}}{(\sqrt{5})^2}\\\\\frac{\sqrt{60\text{x}^3}}{5}\\\\\frac{\sqrt{4\text{x}^2*15\text{x}}}{5}\\\\\frac{\sqrt{4\text{x}^2}*\sqrt{15\text{x}}}{5}\\\\\frac{\sqrt{(2\text{x})^2}*\sqrt{15\text{x}}}{5}\\\\\frac{2x\sqrt{15\text{x}}}{5}\\\\

where x is nonnegative.

<h3>Answer: \frac{2x\sqrt{15\text{x}}}{5}\\\\</h3>
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[ -1 2 3 ]

[ 1 1 4 ]

The determinant of this is 1.

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| 2 3 | |-1 3 | |-1 2 |

| 1. 4. | |1 4 | |1. 1 |

|-1. -1 | |1 -1 | |1 -1

| 1. 4 | |1. 4| |1 1|

|-1. -1 | |1 -1 | |1. -1

|2. 3| |-1. 3| |-1 2|

After Evaluating The Determinant of each 2x 2 Matrix

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Reflect this along the diagonal( Keep 5,5 -2)

Then switching positions of other value

No need of Multiplying by the determinant because its value is 1 from calculation.

After this

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THIS IS OUR INVERSE.

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