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r-ruslan [8.4K]
2 years ago
15

After running the following pseudocode, what will the value of VARIABLE be? 1. Set value of VARIABLE to 5 2. Increase the value

of VARIABLE by 3 3. If the value of VARIABLE is odd, increase its value by 1 4. If the value of VARIABLE is even, increase its value by 1 5. If the value of VARIABLE is odd, increase its value by 1​
Computers and Technology
1 answer:
Nutka1998 [239]2 years ago
8 0

Answer:

10

Explanation:

you start with 5       5

you add 3                8

the variable is not odd so skip         8

the variable is even so add 1            9

the variable is odd so add 1              10

10

hope this helps <3

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Enter a formula in cell B10 to return a value of 35000 if the Net Profit After Tax (cell B9) is greater than or equal to 470000
FrozenT [24]

Statement to be written in cell B10 :

IF ( B9 >= 470000, 35000, 1000)

Formula:

IF ( logical_test ,  [value_if_true] ,  [value_if_false] )

Explanation:

  1. logical_test = Net Profit After Tax (cell B9)
  2. value_if_true = 35000

<em>(if the Net Profit After Tax (cell B9) is greater than or equal to 47000 )</em>

<em>     </em>3<em>.</em> value_if_false = 1000

<em>(if the Net Profit After Tax (cell B9) is lesser than 47000 )</em>

Each value should be separated by comma.

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2 years ago
The major result of treating 1-butyne with 6M aqueous NaOH would be:_______.A. the production of an alkene.B. the production of
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D. nothing, as the alkyne would not react to an appreciable extent.

Explanation:

Nothing, as the alkyne would not react to an appreciable extent.

6 0
2 years ago
9. Which of the following are true for all regular languages and all homomorphisms? (a) h (L1 ∪ L2)= h (L1) ∩ h (L2). (b) h (L1
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Answer:

h (L1 ∪ L2)= h (L1) ∩ h (L2).

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There will be w ∈  (L1 ∪ L2) for any s ∈ h (L1 ∪ L2) in such a way that s=h(w)

we can assume that w ∈ L1

So In this case h(w) ∈ L (S1). Hence s ∈ L(S1)

for any s ∈ h (L1) U h(L2)

We can assume that s ∈ L(S1)

There exists w ∈ L1 such that s= h(w)

In this case it is w ∈ L1 U L2 as well.

Hence , s ∈ h (L1 U L2)

Explanation

consider  = 0,1 and  = a,b and h(0) = a , h(1) = ab

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   Now L1 ∪ L2 = 00,01,10,11

   h (L1 ∪ L2) = h(00) , h(01) , h(10) , h(11) = h(0)h(0) , h(0)h(1) , h(1)h(0) , h(1)h(1)

  = aa, aab , aba , abab

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Here h (L1) = h(10) , h(01) = h(1)h(0) , h(0)h(1) = aba , aab

Hence h (L1) = aba , aab .

Here h (L2) = h(00) , h(11) = h(0)h(0), h(1)h(1) = aa, abab

Hence h(L2) = aa, abab.

Finally Hence , h (L1 ∪ L2)= h (L1) ∩ h (L2).

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2 years ago
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