why the number pin12 of j41 on nano devkit is 79?
the number of gpio is calculated with blow formula

value = offset+ (port number* 8 + pin value)

where offset is 320 if the group is main, and offset is 256 if the group is aon.

the pin12 of j41 on nano devkit is i2s0\_sclk(gpio3\_pj.07). so its gpio number shoule be (320 + 9*8 + 7) = 320 + 79 = 39…
79 is correct for sysfs port.

the formula is (port * 8) + offset

the pin is j7

tegra\_gpio\_port\_j = 9 so ( 9 * 8 ) + 7 = 79

from gpioinfo…

line 79: “i2s0\_sclk” unused input active-high
the octal numbering system is very similar in principle to the previous hexadecimal numbering system except that in octal, a binary number is divided up into groups of only 3 bits, with each group or set of bits having a distinct value of between 000 (0) and 111 ( 4+2+1 = 7 ). the octal numeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7. octal number system is one the type of number representation techniques, in which there value of base is 8. the number system whose base is 8 is known as the octal number system. octal number system is a base-8 number system where the range of numbers is 0,1,2,3,4,5,6,7. octal number system watch more videos at https://www. dear student
red is coded as 2076
which means
r=20
e=7
d=6
now lets us number the alphabets as follows
a=1
b=2
c= 3
d= 4
e= 5




r=18


y=25
z=26

now according to the question
r=20=18+2
e=7=5+2
d=6=4+2

therefore we can say that
g = 7+2 =9
a=1+2=3


gag will be coded as 939


regards
â€‹â€‹â€‹â€‹â€‹â€‹â€‹ > 
> I want to know if I can use any other port number other than 4370
> 
> 
> 


That depends on the configuration of your biometric device. > You might have missed a bracket somewhere... Only such a mistake causes so many insignificant errors.
> 
> 


And to find it, [format your code](http://www.gidnetwork.com/b-38.html) properly. It's your first debugging tool. (a) for digit



expression : 5n + 1

where n = no of digits

for 5 digits

no of segments required = 5(5) + 1 = 25 + 1 = 26

for 10 digits

no of segments required = 5(10) + 1 = 50 + 1 = 51

for 100 digits

no of segments required = 5(100) + 1 = 500 + 1 = 501

(b) for digit



expression : 3n + 1

where n = no of digits

for 5 digits

no of segments required = 3(5) + 1 = 15 + 1 = 16

for 10 digits

no of segments required = 3(10) + 1 = 30 + 1 = 31

for 100 digits

no of segments required = 3(100) + 1 = 300 + 1 = 301

(c) for digit



expression : 5n + 2

where n = no of digits

for 5 digits

no of segments required = 5(5) + 2 = 25 + 2 = 27

for 10 digits

no of segments required = 5(10) + 2 = 50 + 2 = 52

for 100 digits

no of segments required = 5(100) + 2 = 500 + 2 = 502 we know that dividend = divisor × quotient + remainderaccording to the problem :dividend 1 = 245dividend 2 = 1029dividend - remainder = divisor × quotientso dividend 1- remainder = 240 = divisor × quotient 1prime factor of 240 = 24 × 3 × 5dividend 2 – remainder = 1024 = divisor × quotient 2prime factor of 1024 = 24 × 26since, the divisor is common for both the numbers we need to find the highest common factor between both the numbers. from the prime factors, we find thehighest common factor between the two numbers is 24 = 16 since, atomic number of an atom = the number of protons.hence, the number of protons is 17.now, the number of protons = the number of electrons.∴ the number of electrons is also 17.given: mass number = 35we know that mass number = number of protons + number of neutrons.∴ number of protons + number of neutrons = 35⇒ 17 + number of neutrons = 35⇒ number of neutrons = 35 – 17⇒ number of neutrons = 18 Hi Ajay, thank you for leaving a question. You can find the answer about where to find your IU number through here: <https://www.motorist.sg/ask-motorists/2090/how-to-find-my-car-iu-number> a pattern with twelve data and twelve greedydata is going to be ridiculously expensive if it fails to match. it may well exceed the 30 second timeout. i would suggest you match multiple patterns to pick out parts of the message

grok {

break\_on\_match => false

match => {

… [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v