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Oduvanchick [21]
3 years ago
13

How do you find the remainder of (x^3-6x^2-9x+3) / (x-3)

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
4 0
 x^3-6x^2-9x+3     | x-3
-x^3+3x^2               x^2-3x-18
 \\\\\ -3x^2-9x+3
      +3x^2-9x
         \\\\\\ -18x+3 
                 +18x-54
                  \\\\\ -51

Q(x) = x^2-3x-18
R(x) = -51
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Help please. this is really hard for me to understand.
bogdanovich [222]

Answer:

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3 years ago
the legs of a right triangle have lengths of 28 meters and 21 meters. the hypotenuse has a length of 5x meters. what is the valu
Hitman42 [59]
We have a right triangle with
a = one leg = 28
b = other leg = 21
c = hypotenuse = 5x

use the pythagorean theorem (a^2+b^2 = c^2) and solve for x

a^2+b^2 = c^2
28^2+21^2 = (5x)^2
784+441 = 25x^2
1225 = 25x^2
25x^2 = 1225
(25x^2)/25 = 1225/25
x^2 = 49
sqrt(x^2) = sqrt(49)
|x| = 7
x = 7 or x = -7

Since the hypotenuse has length 5x meters, this means
5x = 5*7 = 35
or
5x = 5(-7) = -35

Toss out the negative length as it makes no sense. 

The only value of x is x = 7

Therefore the final answer is x = 7
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3 years ago
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