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Let a, b and c be positive numbers satisfying ab = 11.25; ac = 20; and bc = 36. What is the value of c?

Nov 3, 2019

#1
0

a = 5/2,  b = 9/2,  c = 8,

So, c = 8

Nov 3, 2019
#2
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Without the workings, this is fucking useless!

Guest Nov 3, 2019
#6
+22114
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Note that most 'Answerers' are less inclined to provide methods of finding an answer to 'guest' posters and may just post  an answer.....

Ever thought of signing up as a registered user?    ~EP

ElectricPavlov  Nov 3, 2019
#5
+22114
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b=11.25/a     (from first equation)

c=36/b = 36a/11.25

ac=20

a  * [36a / 11.25] = 20

solve for a =15/6

c= 36a/11.25 = 36(15/6) /11.25 = 8

Nov 3, 2019
#7
+1

Cool!! Thankx EP!

Guest Nov 3, 2019
#10
+109740
+1

ab  =  11.25       ⇒  b  =  11.25 / a       (1)

ac  = 20   ⇒  c  = 20  / a        (2)

bc  = 36         (3)

Sub (1)  and (2)   into (3)

(11.25 /a ) (20 / a)  = 36

225/ a^2  =  36       rearrange as

a^2  =  225/36       take the positive root

a  =  15 / 6    =   5 / 2

So....using (2).....

c =  20 / ( 5/2)   =   8

Nov 3, 2019
#11
+1

Really cool! Thanks CPhill!  Man Im glad I found this math site

What do you mean by positive root.

Guest Nov 3, 2019
#12
+109740
+1

225 / 36     has two   roots  : (-15/6)     and  (15/6)

If we square both of these.....we get   225 /36

But.....since    "a"  is positive....we need   15 / 6    =   5   / 2

Does that make sense  ???

CPhill  Nov 3, 2019
#13
+1

Oh yea! I remember the 2 roots thing from quadratic equations but I didnt know that one was a minus number. I remember some them had two plus numbers. How do you know which one to use if they are both plus numbers?

Guest Nov 3, 2019
#14
+109740
+1

The only way that we can get two  positive roots is if we have something like this :

(x  - a)^2   =   0             take both roots

x  - a   =  0    [  both roots of   0   are  -0   and  + 0   =  0   ]

x  = a           (assuming that  a  >  0  )

CPhill  Nov 3, 2019
edited by CPhill  Nov 3, 2019
#15
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I put numbers in there to see what you mean but I only understand this a little. I will work on it some more later when I have time.

Thanks CPhill with your help I’ll pass the quiz tomorrow and maybe I’ll pass the class.

Guest Nov 3, 2019