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Citrus2011 [14]
3 years ago
15

What is the value of a+b?

Mathematics
1 answer:
MissTica3 years ago
6 0

Answer:

125

Step-by-step explanation:

a =  \frac{1}{2}  \times 90 \degree \\  \\ a = 45 \degree \\  \\ b =\frac{1}{2} (360\degree - 200\degree) \\  \\ b = 80\degree \\  \\ a + b = 45 + 80 \\  \\ a + b = 125 \degree

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Solve the proportion = 2x+?<br> 2x+2 for x.
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No proportion here. Repost your question.

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3 years ago
Simplify the following surds :<br><br> 1) 2√21 × √27 ÷ √343<br> 2) 7√5 × √125 ÷ 2√27
ludmilkaskok [199]

Answer:

1) \frac{18}{7}

2) \frac{175\sqrt{3}}{18}

Step-by-step explanation:

* Lets explain how to simplify a square root

1)

∵ 2\sqrt{21} × \sqrt{27} ÷ \sqrt{343}

∵ \sqrt{21}=\sqrt{3} × \sqrt{7}

∴ 2\sqrt{21} = 2\sqrt{3} × \sqrt{7}

∵ \sqrt{27} = \sqrt{3} × \sqrt{3} × \sqrt{3}

∵ \sqrt{3} × \sqrt{3} = 3

∴ \sqrt{27} = 3\sqrt{3}

∴ 2\sqrt{21} × \sqrt{27} =

  2\sqrt{3} × \sqrt{7} × 3\sqrt{3}

∵ \sqrt{3} × \sqrt{3} = 3

∵ 2 × 3 × 3 = 18

∴ 2\sqrt{21} × \sqrt{27} = 18\sqrt{7}

∵ \sqrt{343} = \sqrt{7} × \sqrt{7} × \sqrt{7}

∵ \sqrt{7} × \sqrt{7} = 7

∴ \sqrt{343} = 7\sqrt{7}

∵ 2\sqrt{21} × \sqrt{27} ÷ \sqrt{343} =

  18\sqrt{7} ÷ 7\sqrt{7}

∵ \sqrt{7} ÷ \sqrt{7} = 1

∴ 2\sqrt{21} × \sqrt{27} ÷ \sqrt{343} =

  \frac{18}{7}

2)

∵ 7\sqrt{5} × \sqrt{125} ÷ 2\sqrt{27}  

∵ \sqrt{125} = \sqrt{5} × \sqrt{5} × \sqrt{5}

∵ \sqrt{5} × \sqrt{5} = 5

∴ \sqrt{125} = 5\sqrt{5}

∴ 7\sqrt{5} × \sqrt{125} =

  7\sqrt{5} × 5\sqrt{5}

∵ \sqrt{5} × \sqrt{5} = 5

∴ 7\sqrt{5} × \sqrt{125} = 7 × 5 × 5 = 175

∵ 2\sqrt{27} = 2\sqrt{3} × \sqrt{3} × \sqrt{3}

∵ \sqrt{3} × \sqrt{3} = 3

∴ 2\sqrt{27} = 6\sqrt{3}

∴ 7\sqrt{5} × \sqrt{125} ÷ 2\sqrt{27} =

  175 ÷ 6\sqrt{3} = \frac{175}{6\sqrt{3}}

∵ \frac{175}{6\sqrt{3}} not in the simplest form because

  the denominator has square root

∴ Multiply up and down by \sqrt{3}

∴  \frac{175}{6\sqrt{3}} = \frac{175\sqrt{3}}{6\sqrt{3}*\sqrt{3}}

∴  \frac{175}{6\sqrt{3}} = \frac{175\sqrt{3}}{18}

∴ 7\sqrt{5} × \sqrt{125} ÷ 2\sqrt{27} =

  \frac{175\sqrt{3}}{18}

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