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VARVARA [1.3K]
3 years ago
9

Which is equivalent to (9 y squared minus 4 x)(9 y squared + 4 x), and what type of special product is it? 81 y Superscript 4 Ba

seline minus 16 x squared, a perfect square trinomial 81 y Superscript 4 Baseline minus 16 x squared, the difference of squares 81 y Superscript 4 Baseline minus 72 x y squared minus 16 x squared, a perfect square trinomial 81 y Superscript 4 Baseline minus 72 x y squared minus 16 x squared, the difference of squares
Mathematics
2 answers:
Nikitich [7]3 years ago
7 0

Answer:

Option B 81 y Superscript 4 Baseline minus 16 x squared, the difference of squares is correct.

Step-by-step explanation:

We need to solve:

(9y^2 -4x)(9y^2+4x)

Finding the product:

9y^2(9y^2+4x) -4x(9y^2+4x)

81y^4+36y^2x-36y^2x-16x^2

81y^4 - 16x^2

<u>Type of Special product</u>

The product is differences of square because

81y^4 = (9y^2)^2

and 16x^2 = (4x)^2

So, 81y^4 - 16x^2 =  (9y^2)^2 - (4x)^2

Hence Option B 81 y Superscript 4 Baseline minus 16 x squared, the difference of squares is correct.

Elodia [21]3 years ago
6 0

Answer:

Option B 81 y Superscript 4 Baseline minus 16 x squared, the difference of squares is correct.

Step-by-step explanation:

We need to solve:

(9y^2 -4x)(9y^2+4x)

Finding the product:

9y^2(9y^2+4x) -4x(9y^2+4x)

81y^4+36y^2x-36y^2x-16x^2

81y^4 - 16x^2

Type of Special product

The product is differences of square because

81y^4 = (9y^2)^2

and 16x^2 = (4x)^2

So, 81y^4 - 16x^2 =  (9y^2)^2 - (4x)^2

Hence Option B 81 y Superscript 4 Baseline minus 16 x squared, the difference of squares is correct.

Read more on Brainly.com - brainly.com/question/13155327#readmore

Step-by-step explanation:

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