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horsena [70]
3 years ago
9

Multiply (3x^2-2x)(2x^2+3x-1)​

Mathematics
2 answers:
Vladimir [108]3 years ago
6 0

Answer:

\large\boxed{(3x^2-2x)(2x^2+3x-1)==6x^4+5x^3-8x^2+2x}

Step-by-step explanation:

Use the distributive property and (a^n)(a^m=)=a^{n+m}

(3x^2-2x)(2x^2+3x-1)\\\\=(3x^2)(2x^2)+(3x^2)(3x)+(3x^2)(-1)+(-2x)(2x^2)+(-2x)(3x)+(-2x)(-1)\\\\=6x^4+9x^3-3x^2-4x^3-6x^2+2x\qquad\text{combine like terms}\\\\=6x^4+(9x^3-4x^3)+(-3x^2-5x^2)+2x\\\\=6x^4+5x^3-8x^2+2x

____ [38]3 years ago
3 0

Answer:

6x^4+5x^3-9x^2+2x

Step-by-step explanation:

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Step-by-step explanation:

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How many full periods and classes does a 202-digit number have?
DerKrebs [107]

The number of full periods and classes does a 202-digit number have are 1 and 3 respectively.

<h3>What are periods?</h3>

Periods are simply groups of three digits separated by commas when writing numbers in standard form.

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Classes is simply the number of digits in a set.

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Learn more about period here:

https://brainly.in/question/39059156

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