Demand
D(q)=(12-q)^(1/2)
Revenue,
R(q)=q*D(q)=q*sqrt(12-q)
At maximum revenue,
R'(q)=0
where
R'(q)=sqrt(12-q)-q/[2(sqrt(12-q))]=0
Solve for q=8.
Hence, quantity=8, price=sqrt(12-q)=sqrt(12-8)=sqrt(4)=2
Let the number be x.
2*(x-4) < (x+5).
2x -8 < x +5
2x-x < 5 +8
x<13.
The number is less than 13.
That's it. Cheers.
RemarkThe very first thing you should do with a question like this is get the graph. Then you will know what you are looking for. I have provided you with such a graph. Your table should center around -2 ≤ x ≤ -1
So let's set up a table and see what we get. Start with y = x + 5
x x + 5
-1 4
-1.25 3.75
-1.5 3.5
-1.75 3.25
-2 3
Do the same thing for - (3)^x + 4 See below to see how this is entered your calculator
x -(3)^x + 4
-1 3.66
-1.25 3.75
-1.5 3.8
-1.75 3.85
-2 3.89
ConclusionWhen x = - 1.25 y = 3.75 for both graphs. <<<< Answer
FootnoteYou may not be familiar with how to put this in your calculator. This is the way I would do it. I'm only doing it for y = -(3^x) + 4
Let x = - 1.25
3
^ Note your calculator might have x^y or y^x. You'll have 1 of the three.
1.25
+/-
=
X
1
+/-
= At this point you should have -0.25
+
4
=
That gives you 3.75
Answer:
B) 4y - 2 - 3y = 6
Step-by-step explanation:
x = 4y - 2 Substitute this into the second solution
x - 3y = 6
4y - 2 - 3y = 6
1-5/8=3/8
If he had 5/8 leftover, then he used 3/8.