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Aliun [14]
3 years ago
7

Given that (in picture)...

Mathematics
1 answer:
GuDViN [60]3 years ago
7 0
<h3>Answer:  a = -3 and b = 5</h3>

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

Work Shown:

Multiply top and bottom by \sqrt{2} to rationalize the denominator

\frac{10-\sqrt{18}}{\sqrt{2}}\\\\\frac{\sqrt{2}(10-\sqrt{18})}{\sqrt{2}*\sqrt{2}}\\\\\frac{\sqrt{2}*10-\sqrt{2}*\sqrt{18}}{\sqrt{2*2}}\\\\\frac{\sqrt{2}*10-\sqrt{2*18}}{\sqrt{4}}\\\\\frac{10\sqrt{2}-\sqrt{36}}{2}\\\\\frac{10\sqrt{2}-6}{2}\\\\\frac{-6+10\sqrt{2}}{2}\\\\\frac{2(-3+5\sqrt{2})}{2}\\\\-3+5\sqrt{2}\\\\

We end up with something in the form a+b\sqrt{2} where a = -3 and b = 5

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Solution:      [  - ∞    <    x    <   ∞ ]

Interval Notation:   [ (- ∞,  ∞) ]


Range of:

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Axis interception points of:

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