Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
<em>4.3 Print count</em>
<em>4.4 count = 0</em>
End the outer loop
4.5 End i loop
End the algorithm
5. End
Answer:
The condition is true when their voltage and current specifications with their impedance are matched or complementary to each other.
Explanation:
Solution
Yes it is possible or true to interface an IC with a different technology like the TTL to HCS12 ports. but the condition is that their current and voltage specifications should be matched and their impedance and power also should be matched.
What this implies is that both their voltage and current requirements should be complementary to each other so as their impedance.
Answer:
Connect the test light in series with the negative post, and start pulling feed wires. The first to check is the heavy charging wire from the alternator. A bad or leaky diode in an alternator is a very common source of overnight battery drain. Connect wires one at a time to see what lead is drawing current.
Answer: A first class lever in static equilibrium has a 50lb resistance force and 15lb effort force. The lever's effort force is located 4 ft from the fulcrum.
Explanation:
