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enot [183]
4 years ago
11

3(5y - 4) + 31 - 3y) = 3(2y - 3) How to solve it

Mathematics
1 answer:
kumpel [21]4 years ago
3 0

After solving 3(5y - 4) + 31 - 3y = 3(2y - 3) the value of y is y=-\frac{14}{3}\,\,or\,\,y= -4.66

Step-by-step explanation:

We need to solve the expression:

3(5y - 4) + 31 - 3y = 3(2y - 3)

and find the value of y

Solving:

3(5y - 4) + 31 - 3y = 3(2y - 3)\\15y-12+31-3y=6y-9\\Combining\,\,like\,\,terms:\\15y-3y-6y=-9+12-31\\15y-9y=12-40\\6y=-28\\y=-\frac{28}{6}\\y=-\frac{14}{3}\\or\,\,y= -4.66

So, after solving 3(5y - 4) + 31 - 3y = 3(2y - 3) the value of y is y=-\frac{14}{3}\,\,or\,\,y= -4.66

Keywords: Solving the expressions

Learn more about Solving the expressions at:

  • brainly.com/question/5147732
  • brainly.com/question/4046256
  • brainly.com/question/2586096

#learnwithBrainly

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Answer : 236
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How many times greater is the value of the fidget 2 in 234567 than the value of the digit 2 in 765432
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Answer:

The value of 2 in the first digit is 100,000 times greater than the value of 2 in the second digit

Step-by-step explanation:

Firstly, we need to know the value of the 2 in 234,567 and that is 200,000

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So the number of times the value of the first 2 is greater than the value of the second 2 is;

200,000/2 = 100,000

So the value of 2 in the first digit is 100,000 times greater than the value of 2 in the second digit

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16 units is the correct answer
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saveliy_v [14]

Answer: The negations are:

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Step-by-step explanation:

Hi!

De Morgan's laws for two propositions P and Q are:

1. ¬(P ∨ Q) = (¬P) ∧ (¬Q)

2. ¬(P ∧ Q) = (¬P) ∨ (¬Q)

where the symbols are,

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∨ = or

∧ = and

1. In this case the proposition is P ∨  Q, with

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Then by law 1:  ¬(P ∨ Q) =  "The train is not late and my watch is not fast"

2. In this case the proposition is P  ∧ Q, with

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