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Orlov [11]
3 years ago
9

WHAT ARE THE EXCLUDED VALUES?

Mathematics
1 answer:
hichkok12 [17]3 years ago
5 0
Since the denominator cannot be zero, you can set 4z+1=0 to find the excluded values. 
4z+1=0
4z=-1
z=-1/4
And thus, z cannot equal -1/4
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What is the answer to this problem?
finlep [7]

Hello from MrBillDoesMath!

Answer:

The correct answer is p >= -5/3

but I don't see that answer as a provided choice  (?)

Discussion:

Solve:

2(6p-5) >= 3(p-8) - 1   =>

2*6p - 2*5 >= 3*p - 3*8 -1  =>

12p -10 >= 3p - 24 - 1         =>   (subtract 3p from both sides)

12p-3p - 10 >= -25              =>   (add 10 to both sides)

9p               >=  -25 + 10 = -15    =>

p >= -15/9 = -5/3


Regards,  

MrB


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