T then V
There are five letters between B and H, four between that and M, then three, then two...
Lxi h,xx50 mov a,m inx h mov b,m mvi ,ffh inr c sub b jnc xx08 add b inx h mov m,c inx h mov m,a hlt
what comes next I S W M L E S
M and N
The next letter in the sequence AFAHMAS is "A." The pattern alternates between the letter "A" and the first letters of the words in the phrase "For Every letter" (F, H, M, S). Therefore, after "S," the sequence returns to "A."
you use mx+b m is slope. m always has to be next to x b is y intercept
the next letter would be 'b'
It's CDL
m
H. H. B. M. Thomas has written: 'The calculation of the rotary lateral stability derivatives of a jet-flapped wing'
Lxi h,xx50 mov a,m inx h mov b,m mvi ,ffh inr c sub b jnc xx08 add b inx h mov m,c inx h mov m,a hlt
In order to add 32 bit numbers in the 8085, you need to add them 8 bits at a time, tracking the carrys between each add. LXI B,first_number LXI H,second_number LXI D,result LDAX B ;first byte - no carry in ADD M STAX D INX B; point to next byte INX D INX H LDAX B ;second byte - carry in ADC M ;note the ADC instead of ADD STAX D INX B; point to next byte INX D INX H LDAX B ;third byte - carry in ADC M STAX D INX B; point to next byte INX D INX H LDAX B ;fourth - carry in ADC M STAX D
e öX` ñneu ào_ H$Wm e Hw$R>o hmoVmg Vy, Vwbm A¸$b Amho H$m,Jobo XmoZ Vmg _r gmaIr Vwbm H$m°b H$aVo`, ~K VwÂ`m _mo~mB©b da -{_g-H$m°b AgVrb.Vwbm H$er ao OamgwÕm _mPr H$miOr Zmhr. H$m` g_OVmog Vy H$moU ñdVmbm? AJ hmo hmo hmo, {H$Vr AmoaS>erb _r Var H$m
3 blind mice
what comes next I S W M L E S
"F". The sequence follows the alphabet in reverse order, skipping three letters each time: a (skip z) z (skip y) e (skip o) b (skip n) I (skip h) y (skip v) o (skip m), then the next letter is "f".
A=m*h m=(a+b)/2
M. B. Subrahmanyam has written: 'Finite horizon H [infinity] and related control problems' -- subject(s): H