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nevsk [136]
4 years ago
11

Find the missing value. Show your work. please

Mathematics
1 answer:
Sophie [7]4 years ago
5 0
#1
sin(51^o)= \frac{y}{12} \ \ \to \ \ y=12*sin(51^o)=12* 0.7771 \approx 9.33

#2
\theta=arccos( \frac{5^2+13^2-12^2}{2*5*13}) = arccos( \frac{25+169-144}{130})=arccos( \frac{50 }{130})= \\ \\ arccos(0.3846) \approx67.4^o

#3
tan(13^o)= \frac{x}{24} \ \ \to \ \ x=24*tan(13^o)=24* 0.2309 \approx 5.54

#4
sin(20^o)= \frac{10}{x} \ \ \to \ \ x= \frac{10}{sin(20^o)} = \frac{10}{0.342}  \approx 29.24
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Step-by-step explanation:

Let P be the polynomial that has to be found

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(5x^2-3x-9) + P = x^2-5x+6\\P = x^2-5x+6 - (5x^2-3x-9)\\P = x^2-5x+6-5x^2+3x+9\\P = x^2-5x^2-5x+3x+6+9\\P = -4x^2-2x+15

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Keywords: Polynomials, expressions

Learn more about polynomials at:

  • brainly.com/question/1704778
  • brainly.com/question/1695461

#LearnwithBrainly

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