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gulaghasi [49]
3 years ago
5

Please answer number 14 I already tried but don’t understand

Mathematics
1 answer:
Rudiy273 years ago
8 0
  • Answer:
  • < abc =  < ebc \\  < acb =  < ecd \\ if \: two \: angles \: in  \:  the \:two \: different \:  triangles \: are \: equal \: then \: the \: third \: angle  \: in \: the \: triangle\: must \: be  \:  equal. \\ using \: pythogres \: theorum   \\  ab = \sqrt( ac ^{2}  - bc^{2} )\\  =  \sqrt (15 ^{2}  - 12 ^{2})  \\  =  \sqrt( 225 - 144 )\\  =   \sqrt(81) \\  ab = 9 \\ sides \: of \: similar \: triangle \: are \: propotional \\   ac  \div ec = dc \div bc \\ 15 \div x = 12 \div 6\\ x = 15 \times 6\div 12\\ x = 15  \div  2 \\ x = 7.5 \\ so \: ec \:  = 7.5 \\ you \: can \: check \: this \: by \: applying \: pythogores \: theorum. \\ i \: hope \: this  \:  helps

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3 years ago
(60 points)
natka813 [3]

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Step-by-step explanation:

2.\\(a-b)^2=a^2-2ab+b^2\\\\x^2-16x=-8\\\\x^2-2(x)(8)=-8\\\\\text{We have}\ 2ab=2(x)(8).\ \text{Therefore}\ b=8.\\\\x^2-2(x)(8)=-8\qquad\text{add}\ 8^2=64\ \text{to both sides}\\\\x^2-2(x)(8)+8^2=-8+64\\\\(x-8)^2=56

5.\\2x^2+x-1=2\qquad\text{subtract 2 from both sides}\\\\2x^2+x-3=0\\\\2x^2+3x-2x-3=0\\\\x(2x+3)-1(2x+3)=0\\\\(2x+3)(x-1)=0\iff 2x+3=0\ \vee\ x-1=0\\\\2x+3=0\qquad\text{subtract 3 from both sides}\\2x=-3\qquad\text{divide both sides by 2}\\\boxed{x=-\dfrac{3}{2}}\\\\x-1=0\qquad\text{add 1 to both sides}\\\boxed{x=1}

6.\\2x^2-4x=0\qquad\text{divide both sides by 2}\\\\x^2-2x=0\\\\x(x-2)=0\iff x=0\ \vee\ x-2=0\\\\\boxed{x=0}\\\\x-2=0\qquad\text{add 2 to both sides}\\\boxed{x=2}

8 0
3 years ago
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