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Cerrena [4.2K]
4 years ago
12

I NEED HELP ASAP PLEASE!!

Mathematics
1 answer:
Makovka662 [10]4 years ago
8 0

Answer:

(2x-3) (2x+3)

zeros, x intercepts:  -3/2, 3/2

Step-by-step explanation:

4x^2 -9

We know the difference of squares is a^2 -b^2

This factors into (a-b) (a+b)

Let 4x^2 =a^2  

Taking the square root

2x =a

Let b^2 =9

Taking the square root

b= 3

(4x^2-9 ) = (2x-3) (2x+3)

To find the zeros, we set the equation equal to zero

(4x^2-9 ) = (2x-3) (2x+3) =0

Using the zero product property

2x-3 =0   and 2x+3 =0

2x-3+3 = 0+3                    2x+3-3 = 0-3

2x=3                                  2x=-3

Divide by 2

2x/2 = 3/2                               2x/2 = -3/2

x = 3/2                                       x = -3/2

These are the zeros of the equation (which are also the x intercepts)

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-84(-3 - 5p)=-176 help please
vagabundo [1.1K]
-84 = 3(5n + 2)
 
Multiply everything in the parenthesis.

-84 = 3(5n + 2)
          3*5n + 3*2

-84 = 15n + 6 
Subtract 6 from both sides

-84 = 15n + 6 
-6                -6

-90 = 15n
Divide 15 from both sides

-90 = 15n
--------------
15       15

-6 = n

-6 is your answer.
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