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3 years ago

Please help me I don't know how to do these at all.

Mathematics

1 answer:

Arlecino [84]3 years ago

3 0

Answer:

The quotient is (-x³ + 4x² + 4x - 8) and the remainder is 0

Step-by-step explanation:

Look to the attached file

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If 4 Maths books are selected from 6 different Maths different English books, how many ways can the seven om Maths books and 3 E

alexgriva [62]

Answer:

Please see the answer below

Step-by-step explanation:

a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000

b. The number of ways such that the 4 math books remain together

The pattern is as follows: MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM

Where M = Math’s Book and E= English Book.

Number of ways = 4!*8!*4*150= 86400 ways.

c. The number of ways such that math book is at the beginning of the shelf

The number of ways = 6!*4*150 = 432000

d. The number of ways such that math and English books alternate

The number of ways = 150*4!*3! =2160 ways

e. The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.

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3 years ago

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I think that the answer is c

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Step-by-step explanation:

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Step-by-step explanation:

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