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Kisachek [45]
2 years ago
13

Easy maths question 20 points and brainliest

Mathematics
1 answer:
Ymorist [56]2 years ago
6 0

Answer:

a = 2; b = 1 and c =3

Step-by-step explanation:

2(q +p) = 1 + 5q

2q + 2p = 1 + 5q

2p = 1 + 3q

2p - 1 = 3q

\frac{2p -1}{3} = q

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Help me with this please
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4 years ago
If f(x) = 2x^3 - 14x^2 + 38x -26 and x-1 is a factor of f(x) find all of the zeros of f(x) algebraically.
Anestetic [448]

Answer:

Step-by-step explanation:

First confirm that x = 1 is one of the zeros.

f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26

f(1) = 2 - 14 + 38 - 26

f(1) = -12 + 38 = + 26

f(1) = 26 - 26

f(1) = 0

=========================

next perform a long division

x -1  || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26

          2x^3 - 2x^2

          ===========

                    -12x^2 + 28x

                     -12x^2 +12x

                     ==========

                                  26x -26

                                  26x - 26

                                 ========

                                      0

Now you can factor 2x^2 - 12x + 26

                                 2(x^2 - 6x + 13)

The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16

So you are going to get a complex result.

x = -(-6) +/- sqrt(-16)

     =============

                 2

x  = 3 +/- 2i

f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)

The zeros are

1

3 +/- 2i

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Answer:

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Step-by-step explanation:

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3 years ago
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Anna11 [10]

Answer:

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Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
What is the value of 12C4? <br> A. 3 <br> B. 48 <br> C. 495 <br> D. 11,880
mars1129 [50]
The answer to this is C
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3 years ago
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