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WITCHER [35]
2 years ago
11

How many distinct diagonals does an undecagon have?

Mathematics
1 answer:
Evgesh-ka [11]2 years ago
7 0

Answer:

-45

Step-by-step explanation:

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Answer:

we have:

8x³ + mx² - 6x + n

= 8x³ - 8x² + (m + 8)x²- (m + 8)x + (m + 2)x - (m + 2) + m + 2+ n

= 8x²(x - 1) + (m + 8)x(x - 1) + (m + 2)(x - 1) + (m + n + 2)

= (x - 1)[8x² + (m + 8)x + m + 2] + (m + n + 2)

because the remainder if divided by (x-1) is 2

=> m + n + 2 = 2

⇔ m + n = 0 (1)

we also have:

8x³ + mx² - 6x + n

= 8x³ - 12x² + (m + 12)x² - 3/2.x.(m + 12) + ( 12 + 3/2.m)x - (9/4.m +  18) + n +9/4m + 18              

= 4x²(2x - 3) + 1/2.(m + 12)x(2x - 3) + (3/2m + 12).1/2.(2x - 3) + 9/4m + n + 18

= (2x - 3)(4x² + (m + 12)/2.x + 3/4m + 6) + 9/4m + n + 18

 because the remainder if divided by  (2x - 3) is 8

=> 9/4m + n + 18 = 8

⇔ 9/4m + n = -10 (2)

from (1) and (2), we have:

m + n = 0

9/4m + n = -10

=> m = -8

      n = 8

Step-by-step explanation:

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Step-by-step explanation:

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Answer:

See Explanation

Step-by-step explanation:

In\: \triangle ABD \: \&\: \triangle ACB\\\angle BAD \cong \angle CAB... (common \: \angle 's) \\\angle ABD \cong \angle ACB... (each\: 45\degree) \\\therefore \triangle ABD \: \sim\: \triangle ACB... (By\: AA\: postulate) \\

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