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Sophie [7]
4 years ago
9

Solve the inequality. Show your work. |4r + 8| ≥ 32

Mathematics
3 answers:
saw5 [17]4 years ago
7 0

|4r + 8| ≥ 32

4r + 8 ≥ 32         or         4r + 8 ≤ -32      <em> inside of absolute value could be + or -</em>

4r       ≥ 24         or          4r       ≤ -40       <em>subtracted 8 from both sides</em>  

r       ≥  6           or            r       ≤ -10         <em>divided both sides by 4</em>

Graph: ←---------------- -10        6 ------------------→

Interval Notation: (-∞, -10] U [6, ∞)

finlep [7]4 years ago
3 0

Okay here are the steps to solving this inequality:

First- <u><em>Break down the problem into these 2 equations</em></u>

               4r + 8 ≥ 32    →    - (4r + 8) ≥ 32


Second- <u>Solve the 1st equation: </u><u><em>4r + 8 ≥ 32</em></u>

                    r ≥ 6


Third- <u>Solve the 2nd equation: - (4r + 8) ≥ 32</u>

                    r ≤ -10


Lastly- <u>Collect all solutions </u>

<h2>           r ≥ 6   and   r ≤ -10</h2>

       

                         

                         

lord [1]4 years ago
3 0

Okay here are the steps to solving this inequality:

First- <u><em>Break down the problem into these 2 equations</em></u>

               4r + 8 ≥ 32    →    - (4r + 8) ≥ 32


Second- <u>Solve the 1st equation: </u><u><em>4r + 8 ≥ 32</em></u>

                    r ≥ 6


Third- <u>Solve the 2nd equation: - (4r + 8) ≥ 32</u>

                    r ≤ -10


Lastly- <u>Collect all solutions </u>

<h2>           r ≥ 6   and   r ≤ -10</h2>

       

                         

                         

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<u>Finding</u><u> the</u><u> </u><u>inverse</u><u> of</u><u> </u><u>f(</u><u>x)</u><u>:</u>

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divide both sides by a:

\displaystyle  y =   \frac{{x}^{2}  - 1 }{a}

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<u>finding</u><u> the</u><u> </u><u>inverse</u><u> of</u><u> </u><u>g(</u><u>x)</u><u>:</u>

\displaystyle g(x) =   \frac{x + 1}{x}

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\displaystyle y=   \frac{x + 1}{x}

interchange:

\displaystyle  \frac{y + 1}{y}  =x

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\displaystyle  y + 1= xy

cancel 1 from both sides

\displaystyle  y  - xy= - 1

factor out y:

\displaystyle  y(1  - x)= - 1

divide both sides by 1-x:

\displaystyle  y=    -  \frac{1}{ 1 - x}

substitute g^-1 for y:

\displaystyle  g ^{ - 1} (x)=    -  \frac{1}{ 1 - x}

remember that

\displaystyle (f   \circ g)x = f(g(x))

therefore we obtain:

\rm \displaystyle   (f ^{ - 1} \circ g ^{ - 1} ) (3) =   \frac{{  \bigg(- \dfrac{1}{1 - 3} }  \bigg)^{2}  - 1 }{a}

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\rm \displaystyle    \frac{{  \bigg(- \dfrac{1}{1 - 3} }  \bigg)^{2}  - 1 }{a} =   - \frac{3}{8}

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simplify complex fraction:

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