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a quiz 

Not maybe the fastest way....but.....pick a point in the feasible region  ....let's say....(2,0)

Look at the first system.....

3(2) + 5(0) ≤ 10   

6 + 0 ≤ 10            true

-2 + 0  >  -5     true

0 - 6  >  5(-2)

-6 > -10     true

So....the point makes all  three true....so....this system is the correct one

BTW, Cupcake......another way  is to graph these inequalities and see which look like the given graph....do you know how to do this???

I find Desmos helpful....!!!!

The new generation …

I suspected from the beginning that Chimp Loki used recombinant DNA from the four (4) Mr. BBs to create a mathematician.  The experiment was a partial success –Loki created a PartialMathematician

For those unfamiliar with the act III scenes of Web2.0calc’s Opera Camelot (Camel-Lot,)

here’s a partial synopsis of the BB’s:

LOL!!!!....THX>>>>!!!!!

Let  x be the equal sides of the  triangle...let y be the remaining side

One way of solving this is with a graph...

We have these constraints  

2x + y = 23

And by the Triangle Inequality, we have

2x > y

x + y > x   ⇒  y > 0

Look at the graph of the feasible region here :  https://www.desmos.com/calculator/kyzw7wa8v5

It shows that there are six such triangles

Let n be the number of quarts of Blend X  and (10- n) the number of quarts of Blend Y

Orange concentration in X  * n  +  orange concentration in Y * (10 - n)  =  .10 quarts at 60% orange

Lemon concentration in X * n  +  lemon concentraion in Y * (10-n) = 10 quarts at 40% lemon

We have this system

.80n +  .20 (10 -n)  =  10(.60)

.20n  + .80 (10-n)  = 10(.40)               simplify   [ multiply through by 10 ]

8n + 2(10-n) = 60

2n+ 8 (10 - n) = 40         simplify more

8n + 20 - 2n  = 60

2n+ 80 - 8n   = 40

6n + 20  = 60

-6n + 80 = 40

6n = 40

-6n = -40

Note that  n    =  40/6 qts   makes both equations true = quarts of Blend X = 6 + 2/3 qts

And Blend Y = 3 + 1/3  qts

Let's pick a point n the shaded region of the top left graph  ⇒ (0, - 5) = (x, y)

Put this into the inequalities

(-5)+ 3(0) ≥ 2   not true....so....we can stop here

Pick a point in the shaded region in the top right graph ⇒ (1, 0)  ⇒ (x, y)

(0) + 3(1) ≥ 2   

0 +3 ≥ 2    true

0 + 4 ≥  (1/2) (1)   

0  + 4 ≥ 1/2      true

Since this point makes both inequalities true....this must be the correct graph  !!!!

Thanks!

I apologize. Typo. The Y blend is 20% orange and 80% lemon

Thanks, Guest.....just goes to show that there is genius in simplicity....!!!!

Try solving it as Geometric sequence and see how far you get. You can try solving it an infinite series or Arithmetico - Geometric series and show the result. Remember evry term, except the first, is an arithmetic mean of its left and right terms.

\((f\cdot g)(x) = (x^2+2x-6)(x^4+3)\\ \text{I leave the multiplication to you}\)

One way to do this is to pick a point in the each shaded region  and see if that point satisifies all three inequalities

So....in the top left graph, let's just choose a point, say (1,0)  = (x, y)

So

0 - 1 ≤ 4(1)     is true

1 + 0  > - 4      is true

0 ≤ (1/3) (1)    is true

So....as Penquino found.....the top left graph is correct  !!!!

It didn't say that it was an arithmetic sequence though. I am confused as to what you mean. Couldn't it have been any other kind of sequence?

By the Triangle inequality, we have these possible inequalities

x/2 + 15 >  47

x/2 > 32

x > 64

47 + 15 > x/2

62 > x/2

124 > x

x < 124

x/2 + 47 > 15

x/2 > -32

x > -64

We need to take the most restictive interval that will solve all three inequalities....this is

64 < x < 124

And the number of integers in this interval is   123 - 65 + 1  =  59

\(32=(47-15)<\dfrac x 2 < (47+15) = 62 \\ 64 < x < 124 \\ x = 65, 66, \dots , 123\\ \text{which is 59 values}\)

I believe it's #1. Not completely sure though.

Cannot be done ...... with the two concentrations for X and Y given.

   X  and Y BOTH have lemon concentration as 20%........combining them in ANY ratio will only result in Lemon 20%....signature is 40 % lemon and you cannot make that.

It cannot be negative because the right side of the original equation is positive......thus...x is positive