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Rudik [331]
3 years ago
10

The product of the polynomials (2ab + b) and (a2 – b2) is . If this product is multiplied by (2a + b), the result is a polynomia

l with terms.
Mathematics
1 answer:
alex41 [277]3 years ago
6 0
" the product " means multiply

(2ab + b)(a^2 - b^2) = 2a^3b - 2ab^3 + a^2b - b^3

when multiplied : (2a + b)(2a^3b - 2ab^3 + a^2b - b^3) =
 4a^4b - 4a^2b^3 + 2a^3b - 2ab^3 + 2a^3b^2 - 2ab^4 + a^2b^2 - b^4
...result is a polynomial with 8 terms
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Find the limit of the function by using direct substitution.
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lim_{x \to \: 0}( {x}^{2} - 1)  =  - 1

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lim_{x \to \: 0}( {x}^{2} - 1)  =  {0}^{2}  - 1

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4 years ago
Meldie made a cube whose side is (3x - 2) inches. Find its volume in cubic inches.
irinina [24]

Answer:

x^2(3x-2) cubic inches OR in^3

OR

3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 cubic inches OR in^3

I AM UNAWARE IF YOU ASKED THAT ONE SIDE IS (3X-2) OR ALL. I WILL ANSWER BOTH PARTS

<em>-</em>

<em>NOTE</em><em>:</em><em> </em><em>'</em><em>^</em><em>'</em><em> </em><em>MEANS</em><em> </em><em>TO</em><em> </em><em>THE</em><em> </em><em>POWER</em><em> </em><em>OF</em><em>.</em><em>.</em>

<em>-</em>

Volume = v, abc = 3 sides of cube (height, width, length)

Using the formula for volume in a cube,

v = abc

We can solve this.

If one side is (3x-2)in,

  • (3x-2)(x)(x) = v.... x are the other two sides
  • x^2(3x-2) = v

x^2(3x-2) cubic inches OR in^3

If all sides are (3x-2)in,

Use the formula,

(a - b) ^{3}  =  {a}^{3}  + 3ab(b - a) -  {b}^{3}

We can solve this.

  • (3x-2)(3x-2)(3x-2) = v
  • (3x-2)^3 = v.... 3x = a and -2 = b
  • (3x)^3 + [(3)(3x)(2)][2-3x] - (2)^3 = v
  • 27x^3 + 18x(2-3x) -8 = v
  • (27x^3 + 36x - 54x^2) - 8 = v.. Terms inside brackets - take 3x as common and leave out 8
  • 3x(9x^2 -18x +12) = v... Take 3 as common again in the brackets
  • 3x [ 3 ([3x^2 -6x] + 4) -8 = v....Take 3x common in the terms in square brackets
  • 3x [ 3 [ 3x (x-2) + 4 ]] - 8 = v
  • 3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 = v

3x { 3 [ 3x ( x - 2 ) + 4 ] } - 8 cubic inches OR in^3

___

If you have any questions regarding formulas or anything, comment and I will get back to you asap.

___

5 0
1 year ago
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