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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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Step-by-step explanation:

A triangle has 3 sides. Let the length of the three sides be a, b, c.

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a+b+c = 18... 1

If the sum of the squares of the three side lengths is 128, this means;

a²+b²+c² = 128... 2

Since the triangle is also right angled, according to Pythagoras theorem;

c² = a²+b²... 3 (taking c as the hypotenuse)

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Substituting c = 8 into 1

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Substituting c = 8 into 2

a²+b²+c² = 128

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a²+b²+64 = 128

a²+b²= 64..... 4

Solve equation 3 and 4 simultaneously;

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a²+b²= 64

From 4; a = 10-b

Substitute a = 10-b into 4;

(10-b)²+b² = 64

100-20b+b²+b² = 64

2b²-20b+100-64 = 0

2b²-20b+36 = 0

Divide through by 2:

b²-10b+18 = 0

b = 10±√10²-4(18)/2

b = 10±√100-72/2

b =( 10±√28)/2

b =(10±√4×7)/2

b = (10±2√7)/2

b = 5±√7

Hence b = 5+√7

If a+ b = 10

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a = 10-(5+√7)

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Area of the triangle A = 1/2 ab

A = 1/2 × (5+√7) × (5-√7)

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A = 1/2 × (25-7)

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Hence the are of the triangle is 9units²

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