3-SAT ≤p TSP
If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
both the statements are true.
<u>Explanation:</u>
- 3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
- If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
- If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
Answer:
Ig =7.2 +j9.599
Explanation: Check the attachment
Answer:
I'm completely sure that the answer is: The most important rating for batteries is the ampere-hour rating. Ampere-hour is the battery discharge rating. It's used as a measure of charge in your device. It indicates how long your device will work without charging.
Explanation:
Hope this helped!
Answer:3.47 m
Explanation:
Given
Temperature(T)=300 K
velocity(v)=1.5 m/s
At 300 K


And reynold's number is given by



x=3.47 m
Answer:
d. 2.3 ohms (5.3 amperes)
Explanation:
The calculator's 1/x key makes it convenient to calculate parallel resistance.
Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms
This corresponds to answer choice D.
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<em>Additional comment</em>
This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.