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Nady [450]
2 years ago
14

Part 1 (Unit 2): Subtract Polynomials:

Mathematics
1 answer:
Alinara [238K]2 years ago
8 0
Part 1
 We have the following polynomials:
 (3-6n5-8n4)
 (-6n4-3n-8n5)
 Subtracting the polynomials we have:
 (3-6n5-8n4) - (- 6n4-3n-8n5)
 n5 (-6 + 8) + n4 (-8 + 6) + 3n + 3
 Rewriting:
 2n5 - 2n4 + 3n + 3

 
Part 2
 
For this case we have the following polynomial:
 4x2-2x-5 = 0
 Using resolver we have:
 x = (- b +/- root (b ^ 2 - 4 * a * c)) / (2 * a)
 x = (- (- 2) +/- root ((- 2) ^ 2 - 4 * 4 * (- 5))) / (2 * 4)
 x = (2 +/- root (4 + 80)) / (8)
 x = (2 +/- root (84)) / (8)
 x = (2 +/- root (4 * 21)) / (8)
 x = (2 +/- 2raiz (21)) / (8)
 x = (1 +/- root (21)) / (4)
 The roots are:
 
x1 = (1 + root (21)) / (4)
 
x2 = (1 - root (21)) / (4)
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10. 9

In the first question we can solve using the Pythagorean theorem. It states that the hypotenuse in a right triangle or the longest side of the triangle, squared is equal to the other 2 sides, squared. Its expressed as so: C^2 = A^2 + b^2 where c is the hypotenuse and a and b are the other sides oft eh triangle.

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__________________________________________________________

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__________________________________________________________

12.

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Step-by-step explanation:

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