The range is all positive values...
R{y | y>0}
Range > "Y" such that y is greater than 0
Answer: 6x-9-9x-4=2x+12
6x-9x-9-4-2x-12=0
-5x-25=0
-5x=25
X=-5
Step-by-step explanation:
I'm going to be making the following assumptions
Assumption 1) The expression M^6+3/Y should be M^6 = 3/Y
Assumption 2) For the equation M^5 = Y^2/6, only the 2 is in the exponent. So it should be written as M^5 = (Y^2)/6
If any of those assumptions are incorrect, then please let me know
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Based on those assumptions, we can divide M^6 over M^5 to get...
(M^6)/(M^5) = M^(6-5) = M^1 = M
So in short, (M^6)/(M^5) = M
We can also say
(M^6)/(M^5) = (3/Y) divided by (Y^2)/6
(M^6)/(M^5) = (3/Y)*(6/(Y^2))
(M^6)/(M^5) = (3*6)/(Y*Y^2)
(M^6)/(M^5) = 18/(Y^3)
Therefore,
M = 18/(Y^3)
So the answer is choice A assuming choice A is saying 18 over Y^3
(again this all hinges on if the assumptions above are correct)