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melamori03 [73]
2 years ago
12

What is the value of 8-|2(-3.5)-5|

Mathematics
2 answers:
ololo11 [35]2 years ago
6 0

\huge\text{Hey there!}

\huge\textbf{Equation:}

\mathbf{8-|2(-3.5)-5|}

\huge\textbf{Solving:}

\mathbf{8-|2(-3.5)-5|}

\mathbf{= 8 - |-7 - 5|}

\mathbf{= -8 -  |-12|}

\mathbf{= 8 - 12}

\mathbf{= -4}

\huge\textbf{Answer:}

\huge\boxed{\mathsf{-4}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>~\frak{Amphitrite1040:)}</h3>
levacccp [35]2 years ago
5 0

Answer:

-4

Step-by-step explanation:

resolve the absolute value

I 2(-3.5) -5 I = I -7 -5 I = I -12 I = 12 The absolute value of any number is always positive

Then:

8 - 12 = -4

Hope this helps

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What is the common denominator of (5/x^2-4) - (2/x+2) in the complex fraction (2/x-2) - (3/x^2-4)/(5/x^2-4) - (2/x+2)
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9514 1404 393

Answer:

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  • simplified complex fraction: (2x +1)/(9 -2x)

Step-by-step explanation:

It is helpful to remember the factoring of the difference of squares:

  a² -b² = (a -b)(a +b)

__

Your denominator of (x² -4) factors as (x -2)(x +2). You will note that one of these factors is the same as the denominator in the other fraction.

It looks like you want to simplify ...

  \dfrac{\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}\right)}{\left(\dfrac{5}{x^2-4}-\dfrac{2}{x+2}\right)}=\dfrac{\left(\dfrac{2(x+2)}{(x-2)(x+2)}-\dfrac{3}{(x-2)(x+2)}\right)}{\left(\dfrac{5}{(x-2)(x+2)}-\dfrac{2(x-2)}{(x-2)(x+2)}\right)}\\\\=\dfrac{2(x+2)-3}{5-2(x-2)}=\boxed{\dfrac{2x+1}{9-2x}}

3 0
3 years ago
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