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natta225 [31]
3 years ago
6

Drag the tiles to the correct boxes to complete the pairs.

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0

Answer:

x² - 16 ⇒ (x + 4)(x - 4)

(2x + 1)³ ⇒ 8x³ + 12x² + 6x + 1

(2x + 3y)² ⇒ 4x² + 12xy + 9y²

x³ + 8y³ ⇒ (x + 2y)(x² - 2xy + 4y²)

Step-by-step explanation:

* Lets explain how to solve the problem

# x² - 16

∵ x² - 16 is a difference of two squares

- Its factorization is two brackets with same terms and different

 middle signs

- To factorize it find the square root of each term

∵ √x² = x and √16 = 4

∴ The terms of each brackets are x and 4 and the bracket have

   different middle signs

∴ x² - 16 = (x + 4)(x - 4)

* x² - 16 ⇒ (x + 4)(x - 4)

# (2x + 1)³

- To solve the bracket we will separate (2x + 1)³ to (2x + 1)(2x + 1)²

∵ (2x + 1)² = (2x)(2x) + 2(2x)(1) + (1)(1) = 4x² + 4x + 1

∴ (2x + 1)³ = (2x + 1)(4x² + 4x + 1)

∵ (2x + 1)(4x² + 4x + 1) = (2x)(4x²) + (2x)(4x) + (2x)(1) + (1)(4x²) + (1)(4x) + (1)(1)

∴ (2x + 1)(4x² + 4x + 1) = 8x³ + 8x² + 2x + 4x² + 4x + 1 ⇒ add like terms

∴ (2x + 1)(4x² + 4x + 1) = 8x³ + (8x² + 4x²) + (2x + 4x) + 1

∴ (2x + 1)(4x² + 4x + 1) = 8x³ + 12x² + 6x + 1

∴ (2x + 1)³ = 8x³ + 12x² + 6x + 1

* (2x + 1)³ ⇒ 8x³ + 12x² + 6x + 1

# (2x + 3y)²

∵ (2x + 3y)² = (2x)(2x) + 2(2x)(3y) + (3y)(3y)

∴ (2x + 3y)² = 4x² + 12xy + 9y²

* (2x + 3y)² ⇒ 4x² + 12xy + 9y²

# x³ + 8y³

∵ x³ + 8y³ is the sum of two cubes

- Its factorization is binomial and trinomial

- The binomial is cub root the two terms

∵ ∛x³ = x and ∛8y³ = 2y

∴ The binomial is (x + 2y)

- We will make the trinomial from the binomial

- The first term is (x)² = x²

- The second term is (x)(2y) = 2xy with opposite sign of the middle

  sign in the binomial

- The third term is (2y)² = 4y²

∴ x³ + 8y³ = (x + 2y)(x² - 2xy + 4y²)

* x³ + 8y³ ⇒ (x + 2y)(x² - 2xy + 4y²)

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2) i don't know answer for 2, but i know that it isn't right triangle because from Pythagoras' Theory we know that for right triangle: a^2 + b^2 = c^2 but in this triangle 30^2 + 38^2 ≠ 54^2

3) from Pythagoras' theory we know that the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides: a^2 + b^2 = c^2

we have to check this in this triangle:

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