Question 7 of 10 Consider the inequality 3 (-1-2) <-26 – 3.3.
The solution set of the inequality is:
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The graph of the solution set is
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Pleaseen help meee
Ayudenmee

   203

 ———  < 0  

    10  

Step-by-step explanation:

Changes made to your input should not affect the solution:  

(1): "3.3" was replaced by "(33/10)".

Rearrange the equation by subtracting what is to the right of the greater than a sign from both sides of the inequality :

               3*(-1-2)-(-26-(33/10))<0

                  33

Simplify   ——

                  10

                   33

 -9 -  (-26 -  ——)  < 0  

                    10

Subtracting a fraction from a whole

Rewrite the whole as a fraction using  10  as the denominator :

           -26     -26 • 10

    -26 =  ———  =  ————————

            1         10    

Equivalent fraction: The fraction thus generated looks different but has the same value as the whole

the Common denominator: The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

         -26 • 10 - (33)                        -293

———————————————  =  ————

               10                                        10

           -293

 -9 -  ————  < 0  

              10

Subtracting a fraction from a whole

Rewrite the whole as a fraction using  10  as the denominator :

             -9              -9 • 10

    -9 =  ——  =  ———————

              1                     10  

Adding up the two equivalent fractions

        -9 • 10 - (-293)                           203

————————————————  =  ———

              10                                           10

   203

  ———  < 0  

    10  

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.