Can you please also make videos on calculas (the early trascidentals by james Stewart) your video are very informative and interesting as well as east to understand
@@BelanPro8 Sorry if i am late to answer you For 3rd subnet we generated , See the starting range it is 150.15.4.0 so the starting IP is always called as Network address and the end IP is called as Broadcast address So Question 1 : What is the Network address of 3rd Subnet it is 150.15.4.0 (Starting IP of 3rd subnet ). Question 2 : What is the broadcast address of 2nd Subnet it is 150.15.3.255 (Ending IP of 2nd subnet ).
if you convert 500(hosts) in binary you will get 111110100 which are in total 9 bits another example is that if you have 52(hosts)in binary is 110100 which are 6 bits in total.
The host requirements. (500 host we need) Now convert 500 into binary and it is = (111110100). Now count all the bits of above binary digits. NOTE : Ignore leading zeros in binary and then write only.
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Answer to H.W. : Network address of 3rd Subnet = 150.15.4.0 and Broadcast Address of 2nd Subnet is = 150.15.3.255
@@neelprajapati6690 That would be for all the subnets combined.
I'm from SriLanka I'm studying Advance Level ICT. Your videos are very helpful ❤😩 thank you so much 😩🔥
Wonderful tutorial. Thanks for always making learning easy
Amazing explanation
Can you please also make videos on calculas (the early trascidentals by james Stewart) your video are very informative and interesting as well as east to understand
Homework answer:
Network address of third Subnet => 150.15.4.0
Broadcast address of second subnet => 150.15.3.255
Was my address correct sir ?
Yes !
thank you!
Why we have to change 4th octet sir .
network address of third subnet: 150.15.4.0, broadcast address of second subnet: 150.15.3.255
Answer is: Network address of 3rd subnet is 150.15.4.0 and Broadcast address of 2nd subnet is 150.15.3.255
Why like that I'm not understanding
@@BelanPro8 Sorry if i am late to answer you
For 3rd subnet we generated , See the starting range it is 150.15.4.0 so the starting IP is always called as Network address and the end IP is called as Broadcast address
So
Question 1 : What is the Network address of 3rd Subnet it is 150.15.4.0 (Starting IP of 3rd subnet ).
Question 2 : What is the broadcast address of 2nd Subnet it is 150.15.3.255 (Ending IP of 2nd subnet ).
since the octet position is 3 why is the 4th octet also being changed from 0 to 255?
Same question. Let me know if u know the ans now?
If someone knows please tell me why 4th octet is changed to 255
First subnet ranges from 150.15.0.0 to 150.15.1.255 as asked in the question we need 500 ip address
Please check again.
4th octet is not changing.
Check only left side sticky Network IP address.
Don't look at right side IP.
3rd subnet network of network address->150.15.4.0
2nd subnet network of broadcast address->150.15.3.255
what is the last range is it 150.15.254.0 to 150.15.255.255
yes
How to write u subnet mask it is a class b so 255.0.0.0 but ur write to 255.255.0.0 how explain
255.0.0. is class A, 255.255.0.0 is class B
class a 255.0.0.0
class b 255.255.0.0
class c 255.255.255.0.0 and so on you have to follw this
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Or else
Some of the paid lectures for free?
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1st comment
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How are you finding 9 bits
if you convert 500(hosts) in binary you will get 111110100 which are in total 9 bits
another example is that if you have 52(hosts)in binary is 110100 which are 6 bits in total.
The host requirements. (500 host we need)
Now convert 500 into binary and it is = (111110100).
Now count all the bits of above binary digits.
NOTE : Ignore leading zeros in binary and then write only.