Math, asked by vidhikagarg2007, 7 months ago

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how much does 8xcube-6xsquare+9x-10 exceed 4xcube+2xsquare+7x-3

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Answers

Answered by ishika16629
1

Answer:

4x3 – 8x2 + 2x – 7

Step-by-step explanation:

8x3 – 6x2 + 9x – 10 exceeds 4x3 + 2x2 + 7x – 3

= (8x3 – 6x2 + 9x – 10) – (4x3 + 2x2 + 7x – 3)

= 8x3 – 6x2 + 9x – 10 – 4x3 – 2x2 – 7x + 13

= 8x3 – 4x3 – 6x2 – 2x2 + 9x – 7x – 10 + 3

= 4x3 – 8x2 + 2x – 7

Answered by padigarbhavani
0

Answer:

4x³ - 8x² + 2x - 7

Step-by-step explanation:

(8x³ - 6x² + 9x - 10) - (4x³ + 2x² + 7x - 3)

8x³ - 6x² + 9x - 10 - 4x³ - 2x² - 7x + 3

4x³ - 8x² + 2x - 7

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