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iris [78.8K]
3 years ago
5

What is the product of 6x – y and 2x – y + 2? A=8x2 – 4xy + 12x + y2 – 2y B=12x2 – 8xy + 12x + y2 – 2y C=8x2 + 4xy + 4x + y2 – 2

y D=12x2 + 8xy + 4x + y2 + 2y
Mathematics
2 answers:
aksik [14]3 years ago
6 0

Answer: B: 12x^2-8xy+12x+y^2-2y

Step-by-step explanation:

(6x-y)*(2x-y+2)

12x^2-6xy+12x-2xy+y^2-2y

12x^2+y^2+12x-2y-8xy

Schach [20]3 years ago
5 0

Answer:

Should be Product of two expression is (6x-y)(2x-y)=12x^2-8xy+y^2

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Answer:

  • D. 4x - 6

Step-by-step explanation:

  • f(x) = 4x - 2
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Correct option is D

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3 years ago
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Marianna [84]

Answer:

answer 1 b

Step-by-step explanation:

answer 2 d

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2 years ago
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Strike441 [17]
The answer is 5 and 0

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2 years ago
Algebra I<br> Evaluate each function. h ( n ) = 2 n − 5 ; Find h ( 2 )
slavikrds [6]

Answer:

h ( 2 ) = - 1

Step-by-step explanation:

<u>Two-Variable Functions</u>

A function expresses the relation between two variables in such a way that for each input for the independent variable n, there one and only one value for the function h(n). If it's explicitly given as an equation, then we can use values for n as we wish, and compute the different values of h(n).

The question provides the following function

h ( n ) = 2 n - 5

We are required to find h(2), which can be computed by replacing n by the value of 2

h ( 2 ) = 2 (2) - 5

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h ( 2 ) = - 1

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2 years ago
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trapecia [35]

Answer:

See below

Step-by-step explanation:

Let the  2 roots be  A and A^2.

Then A^3 = c/a and A + A^2 = -b/a.

Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-

A^3 + (A^2)^3 =  ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2

= (A + A^2)^3 - 3A*3( A + A^2)

Substituting:-

c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)

Multiply through by a^3:-

a^2c + ac^2 = -b^3 + 3abc

Factoring:-

ac(a + c) = 3abc - b^3

This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.

a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3

c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.

These are equal so we have

ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3

The 3acb^2 cancel out so we have:-

a^3c - ac^3 =  3a^2bc - 3abc^2 + b^3c - ab^3

ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)

ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)

Divide through by (a - c):-

ac(a + c) = 3abc - b^3 , which is the result we got earlier.

This completes the proof.

 

3 0
3 years ago
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