
If you only had 4:16 One Hot Decoder and an OR gate with the number of inputs of your choosing, fill in the blanks to explain how you would implement the function !x3x2 + x2x0 + !x3!x1x0 with the hardware you were provided. There are 4 inputs for this function, x3-x0. 1. How many inputs an OR gate would connect to? 2. For input x3,x2,x1,x0, I would connect it input to which options: (0, 1, 2,3) of the decoder. 3. Which outputs of the decoder will not use to connect to the OR gate.
![1. For output 0000 of the decoder,
2. For output 0001 of the decoder,
[Select]
[Select]
I would not use it
I would connect it to the OR gate](https://content.bartleby.com/qna-images/question/a8a6433a-daca-4307-8eb1-311fecdc6fbb/0c13b7b4-505e-42be-9d60-b78e0e450d2a/nzfjjb3_thumbnail.png)

Here's how you would implement the given function !x3x2 + x2x0 + !x3!x1x0 using the provided hardware:
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The number of inputs the OR gate would connect to depends on the number of terms in the Boolean expression. In this case, there are three terms, so we would need three inputs for the OR gate.
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For the input x3x2x1x0, we would connect it to the input option 11 (corresponding to binary 1100) of the decoder. This is because this input corresponds to the combination of x3, x2, and x1 being true (i.e., equal to 1), and x0 being false (i.e., equal to 0). This input is used to implement the first term of the Boolean expression, !x3x2.
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