The value of g after solving the expression, 5/2-2g = 5g/2-3+g, is 1.
According to the question,
We have the following expression:
5/2-2g = 5g/2-3+g
Now, moving the terms with the same variables on one side and the integers on another side.
Note that when we move any number of term from one side to another which is in addition or subtraction will result in change of the sign from minus to plus and plus to minus.
5/2+3 = 2g+5g/2+g
Taking 2 as the least common factor on the left hand side and right hand side:
(5+6)/2 = (4g+2g+5g)/2
11/2 = 11g/2
Now, 11/2 on the left hand side can be divided by 11/2 on the right hand side:
g = 1
Hence, the value of g is 1.
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Answer:
Expected project duration is determined as a direct result of computing the earliest starting and finishing times for the activities of a project network.
Step-by-step explanation:
In project management, early start is the earliest moment at which a specific activity may start and early finish the earliest moment at which it can end. Early finish (es) is computed as:
EF = ES + d
where d is the activity duration and ES early start
Early start is computed as
ES: Max (EF-1)
or the maximum early start of predecessory activities.
Once we have calculated these values in the network for each activity , the early finish of last activity corresponds to the expected project duration, the earliest time in which we may finish the project if there are no issues.
Max project duration is open question as we could have infinite delays or never finish the project. Time variance in project duration may only be estimated once we have actual execution times of our project.
Slack time and critical path are obtained after obtaining the ES and EF but this info alone is not sufficient. We require either the late start or the late finish of the activities to calculate slack, Zaero slack activities, those that cannot be delayed form critical path and can only be obtained after having ES. EF. LS and LF
Answer:
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Which pair shows equivalent expressions?
A.2(2/5x + 2)=2 2/5x + 1
B.2(2/5x + 2)=4/5x + 4
C.2(2/5x + 4)=4/5x + 2
D.2(2/5x + 4)=2 2/5x + 8
Solution:
Let us distribute 2 inside the parenthesis.
That is, we use distributive property:
a(b+c)=ab+ac
So,
Answer:Option (b)
Applying distributive property, a(b+c)=ab+ac
So, Option (B) is correct.