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svlad2 [7]
3 years ago
7

Anyone understand how to do both of these?

Mathematics
1 answer:
Evgen [1.6K]3 years ago
6 0

for the first one set 3y+y=180 and y+x=180


the second one set w+y=180, 42+x=180, y+20=180, and 87+v

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3 years ago
Help me class8 (surds)​
Mashcka [7]

Answer:

  • \cfrac{991\sqrt{2} }{80}

Step-by-step explanation:

For a start simplify each of the roots:

  • \sqrt{72} =\sqrt{36*2} =6\sqrt{2}
  • \sqrt{50} =\sqrt{25*2} =5\sqrt{2}
  • \sqrt{128} =\sqrt{64*2} =8\sqrt{2}
  • \sqrt{98} =\sqrt{49*2} =7\sqrt{2}

Now simplify the expression in steps:

\sqrt{72}-\cfrac{48}{\sqrt{50} }  -\cfrac{45}{\sqrt{128} }  +2\sqrt{98} =

6\sqrt{2}-\cfrac{48}{5\sqrt{2} }  -\cfrac{45}{8\sqrt{2} }  +2*7\sqrt{2} =

6\sqrt{2}-\cfrac{48*8+45*5}{5*8\sqrt{2} }  +14\sqrt{2} =

20\sqrt{2}-\cfrac{609}{40\sqrt{2} } =

20\sqrt{2}-\cfrac{609*\sqrt{2} }{40\sqrt{2} *\sqrt{2} } =

20\sqrt{2}-\cfrac{609\sqrt{2} }{40*2 } =

20\sqrt{2}-\cfrac{609\sqrt{2} }{80 } =

\sqrt{2} (20-7\cfrac{49}{80} )=

\sqrt{2} *12\cfrac{31}{80} =

\cfrac{991\sqrt{2} }{80}

3 0
2 years ago
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Delicious77 [7]

Answer:

-16 (t+3) (t-3)

Step-by-step explanation:

6 0
3 years ago
Factorise fully<br>14x – 8​
andriy [413]

Answer:

2(7x-4)

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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