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luda_lava [24]
3 years ago
13

Another math problem help please?

Mathematics
2 answers:
kozerog [31]3 years ago
5 0
The answer is the second option, or Ray BE is a bisector of angle ABE.
alukav5142 [94]3 years ago
3 0
The correct answer to this question is D.
If you look at the angles, you can see that angle CBE is 60 degrees, which is half of the total for angle CBA. This means that angles EBD and DBA must total 60 degrees. Since DBA is 30 degrees, EBD must also be 30 degrees. This means that BD is a bisector of EBA, meaning that the ray splits the angle perfectly in half.
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ch4aika [34]

Answer:

P= 6 \frac{1}{7}\\\\Q = 4 \frac{2}{7}

Step-by-step Explanation

\frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} }  = P+Q\sqrt 2 \\  \\  \frac{(5 + 3 \sqrt{2}) }{(5 - 3 \sqrt{2}) }  \times  \frac{(5 + 3 \sqrt{2}) }{(5   +  3 \sqrt{2} )} = P+Q\sqrt 2 \\  \\  \frac{ {(5 + 3 \sqrt{2} )}^{2} }{ {(5)}^{2}  - (3 \sqrt{2})^{2}  }  = P+Q\sqrt 2 \\  \\ \frac{ {5}^{2} +  {(3 \sqrt{2} )^{2}  + 2.5.3 \sqrt{2} } }{25 - 9 \times 2} = P+Q\sqrt 2 \\  \\\frac{ 25 + 18  + 30 \sqrt{2}  }{25 - 18} = P+Q\sqrt 2 \\  \\\frac{ 43  + 30 \sqrt{2}  }{7} = P+Q\sqrt 2 \\  \\ \frac{43}{7}  +  \frac{30 }{7} \sqrt{2}  = P+Q\sqrt 2 \\  \\ equating \: like \: terms \: on \: both \: sides \\  \\ P =  \frac{43}{7}   \\  \\ \huge \red { \boxed{ P= 6 \frac{1}{7}}}  \\  \\ Q =  \frac{30}{7}  \\  \\ \huge  \orange { \boxed{ Q = 4 \frac{2}{7} }}

4 0
3 years ago
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