if you are adding a/b + c/d the answer is (a * d + c * b)/(b*d)
4d3 multiplications add in powers.
A add 4= 8 a=? A = 4 algibra is pretty much the missing number but instead of ? add 4 = 8 they put a letter in heres some more b add 7=10 b=? c add 9 = 27 c=? d add 62= 122 d=? Answeres A = 4 b = 3 c = 18 d = 60 Hope this helped :) XD
With no knowledge about A, B, C and D or the relation between them, it is impossible to give an answer.
The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
if you are adding a/b + c/d the answer is (a * d + c * b)/(b*d)
#include<stdio.h> #include,conio.h> void main() int a,b,d; pintf("enter the number"); scanf("%d%d",&a,&b); d=add(a,b); pintf("sum=%d",d); } add(a,b) { int sum; sum=a+b; return (sum); getch(0; }
d, e, d, g, g, a, b, a, d', b, a, b, e, e,d. you do that 2 times. then it is: b, g, b, b, a, e, a, a, g, e, g, g, g, g, e. you do that 2 times there is a high part to which is d',c',b,b,b,b c' d' c' a g you finished!!!
The product of the means equals the product of the extremes. In other words, if A is to B as C is to D, then B times C equals A times D, so... A = B x C ÷ D B = A x D ÷ C C = A x D ÷ B D = B x C ÷ A
#include<stdio.h> void main() { int a,b,t; printf(enter the value of a and b"); scanf("%d%d",a,b); t=a; a=b; b=t; printf("a=%d/nb=%d",a,b); getch(); }
4d3 multiplications add in powers.
bcdbabag bcdbabag eeffgeaeag aeagaeaeg reapet as many times as you like
A add 4= 8 a=? A = 4 algibra is pretty much the missing number but instead of ? add 4 = 8 they put a letter in heres some more b add 7=10 b=? c add 9 = 27 c=? d add 62= 122 d=? Answeres A = 4 b = 3 c = 18 d = 60 Hope this helped :) XD
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
What the butterfly method does is to make the fractions equivalent fractions with a new denominator (bottom number) the product of the original denominators (bottom numbers). Then to compare/add/subtract the fractions, the numerators (top numerators) can be compared/added/subtracted.Let the two fractions be a/b and c/d, then the new denominator is bd (b times d). Then the equivalent fractions are:a/b ⇒ (a x d)/(b x d) ⇒ ad/bdc/d ⇒ (b x c)/(b x d) ⇒ bc/bd
With no knowledge about A, B, C and D or the relation between them, it is impossible to give an answer.
I figured out the flute solo (if that's what you mean) and the notes are: B,B,B,B,B,C,D,G,B,C,D,G x2 F x11 many times then G. :D