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ziro4ka [17]
3 years ago
10

Find the value of the missing coefficient in the factored form of 8f^3-216g^3

Mathematics
1 answer:
adoni [48]3 years ago
7 0

Answer:

The value of the missing coefficient is 12

Step-by-step explanation:

* Lets explain how to factorize the difference of two cubes

- The factorization of the difference of two cubes like a³ - b³, is a

  product of a binomial and trinomial

- The binomial is the cube root of the first term and the second term

∵ The ∛a³ = a and ∛b³ = b

∴ The binomial is (a - b)

- We will find the trinomial from the binomial by square the 1st term

 of the binomial and multiply the 1st term and the 2nd term of the

 binomial with opposite sign of the binomial and square the 2nd

 term of the binomial

∴ The trinomial is (a² + ab + b²

∴ The factorization of (a³ - b³) is (a - b)(a² + ab + b²)

* Lets solve the problem

∵ 8f³ - 216g³ is the difference of two cubes

∵ ∛(8f³) = 2f

∵ ∛(216g³) = 6g

∴ The binomial is (2f - 6g)

- Lets make the trinomial

∵ (2f)² = 4f²

∵ (2f)(6g) = 12fg

∵ (6g)² = 36g²

∴ The trinomial = (4f² + 12fg + 36g²)

∴ The factorization of 8f³ - 216g³ = (2f - 6g)(4f² + 12fg + 36g²)

∴ The value of the missing coefficient is 12

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Diya spent 2/5 of her money on a dress and 1/2 of the reminder on a doll. She spent $8 more o the dress than the doll. How much
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Answer:

$24

Step-by-step explanation:

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The magnitude and direction of two vectors are shown in the diagram. What is the magnitude of their sum? ​
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Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that

<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>

<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>

where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.

Then the sum is

<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>

and its magnitude is

||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))

… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

… = √(16 + 16 cos(135° - 45°) + 4)

… = √(20 + 16 cos(90°))

… = √20 = 2√5

5 0
3 years ago
Read 2 more answers
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