HELP NEEDED !!! find the size of unknown angles

Mathematics

1 answer:

Karolina [17]3 years ago

8 0

Explanation :

The line DF is parallel to OB, therefore Angle x = Angle y.

( because, they for corresponding angle pair which are equal. )

And the angle made by an arc at center is always double the angle made by the same arc at its boundary, hence

x = 22° × 2

x = 44°

y = 44° ( cuz, x = y )

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Simplify the expression \[\frac{6}{\sqrt{36} + \sqrt{27}} + \frac{6}{\sqrt{27} + \sqrt{18}} + \frac{6}{\sqrt{18} + \sqrt{9}}.\]

Viktor [21]

Answer:

= 18 ( 3 + 2(\sqrt{3} + \sqrt{2} ))

Step-by-step explanation:

$6(\sqrt{36} +\sqrt{27}) + 6(\sqrt{27} +\sqrt{18}) + 6(\sqrt{18} + \sqrt{9} )\\= 6( {6} +\sqrt{9*3}) + 6( \sqrt{9*3} +\sqrt{9*2}) + 6(\sqrt{9*2} + 3 )\\= 6( {6} +3\sqrt{3}) + 6(3\sqrt{3} +3\sqrt{2}) + 6(3\sqrt{2} + 3 )\\=36 + 18\sqrt{3} + 18\sqrt{3} + 18\sqrt{2}+ 18\sqrt{2} + 18\\= 36+18 +18( \sqrt{3}+ \sqrt{3} +\sqrt{2} + \sqrt{2})\\= 54 + 18 (2\sqrt{3} +2\sqrt{2})\\=54 + 36\sqrt{3} + 36\sqrt{2}\\= 18 ( 3 + 2(\sqrt{3} + \sqrt{2} ))$