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TEA [102]
4 years ago
12

WHICH EXPRESSION IS NOT EQUIVALENT?

Mathematics
2 answers:
s344n2d4d5 [400]4 years ago
3 0

Answer:

C

Step-by-step explanation:

First, factor the original expression:

2x^2+10x+12\\2(x^2+5x+6)\\2(x+3)(x+2)

As we can see, D is the same as above. Eliminate D.

Go through each of the answer choices.

A:

(2x+4)(x+3)\\=2(x+2)(x+3)

This is equivalent to what we factored. Eliminate A.

B:

(2x+6)(x+2)\\=2(x+3)(x+2)

This is again equivalent to what we factored. Eliminate B.

C:

(2x+3)(x+4)\\

This cannot be simplified and it is not equivalent to what we have previously. C is not the equivalent expression.

Stella [2.4K]4 years ago
3 0

Answer:

The expression that is not equivalent to 2x²+ 10x + 12 is c)

Step-by-step explanation:

Hello!

To resolve these equations you have to do the following steps:

For example, you have  (a+b)(c+d) you have to multiply the first term included in the first parenthesis with the terms of the second parenthesis and add them:

a*c + a*d

Then do the same with the second term:

b*c + b*d

Finally, you add them

ac+ad+bc+bd

If there are common terms, you have to add them.

a)

(2x+4)(x+3)

First you multiply 2x by the terms contained in the second parenthesis.

2x*x + 2x*3= 2x²+6x

Then you do the same with 4

4*x + 4*3= 4x + 12

Now you put it all together:

2x²+6x + 4x + 12

and add common terms 6x + 4x= 10x

2x²+ 10x + 12

b)

(2x+6)(x+2)= 2x²+4x +6x +12= 2x²+ 10x + 12

c)

(2x+3)(x+4)= 2x²+8x + 3x + 12= 2x² + 11x + 12

d)

2(x+3)(x+2)

In this case you have three terms in the equation "2" "(x+3)" and "(x+2)"

First you have to resolve the multiplication between the parenthesis and then you can multiply it by two

First:

(x+3)(x+2)= x*x+x*2+3*x+3*2= x²+2x+3x+6= x²+ 5x + 6

Now you can multiply it by two:

2(x²+ 5x + 6)= 2*x²+ 2*5x + 2*6= 2x² + 10x + 12

The expression that is not equivalent to 2x²+ 10x + 12 is c)

I hope this helps!

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=============================================

Explanation:

The domain is the set of all allowed x inputs. Visually, the left-most point is when x = -3, so -3 is the smallest allowed x input. We have a closed circle at this endpoint, so this endpoint x value is included. In contrast, if you had an open circle, then you would exclude the endpoint.

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