3 years ago

16.

1 answer:

o-na [289]3 years ago

3 0

4.5 miles I would say

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VashaNatasha [74]

The answer is $3x^4-13x^3-x^2-11x+6$ .

Solution:

Use algebraic identity: $a^m\times a^n=a^{m+n}$

For example: $x^2\times x=x^{2+1}=x^3$

Given expression $(x^2-5x+2)$ and $(3x^2+2x+3)$ .

To multiply these equations.

$(x^2-5x+2)\times(3x^2+2x+3)$

$=x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)$

$=(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)$

$=3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6$

Combine like terms together.

$=3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6$

$=3x^4-13x^3-x^2-11x+6$

$(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6$

Hence the answer is $3x^4-13x^3-x^2-11x+6$ .

3 0

3 years ago

denis-greek [22]

Answer:

do 24 points i know it very well

8 0

2 years ago

e-lub [12.9K]

It should be:

69(2)=138
138-48=90

5 0

3 years ago

frutty [35]

Pt 1 - how much the value goes down each year
Pt 2 - how much the car was originally worth

3 0

3 years ago

777dan777 [17]

3x-15+2x-2
=5x-17
use the distribution property

5 0

3 years ago