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oksano4ka [1.4K]
3 years ago
15

Thomasina is asked to factorise y^2 - 2y + 35.

Mathematics
2 answers:
stepan [7]3 years ago
8 0

Answer:

Step-by-step explanation:

hello :

(y + 7)(y - 5) = y² -5y+7y-35=y²+2y-35 no   y²-2y-35

Irina18 [472]3 years ago
4 0

Answer:

  there is no factorization of y^2 -2y +35 in real numbers

Step-by-step explanation:

Thomasina correctly determined two factors of 35 that differ by 2, but that is not what is needed here. For this problem, she needs two factors of 35 that have a <em>sum</em> of 2.* (There aren't any real-number factors that meet that condition.)

__

The factorization in complex numbers would be ...

  = (y -1 -i√34)(y -1 +i√34)

_____

* Technically, she needs two factors of 35 with a sum of -2. Both factors would have to be negative.

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Given equation x = 6 + i, w = -1 + 5i and z = 4 - 8i, determine each of the following in the form of a + bi
Temka [501]

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i) 28 - 30i

ii) 36 + 28i

Step-by-step explanation:

i) x = 6 + i ⇒2x = 2(6 + i) = 12 + 2i

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ii) w = -1 + 5i and z = 4 - 8i

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3 0
3 years ago
Which expressions are equivalent to 4(x+1) + 7(x+3)? Select two answers.
Natalija [7]

Answer:

9(x+3) + 2(x+3) or 10(x+2) +(x+5)

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