The expression is: b^2 +d
When cross multiplying, you do not add or subtract; instead, you multiply. In the equation ( \frac{a}{b} = \frac{c}{d} ), you cross multiply by calculating ( a \times d ) and ( b \times c ). This results in the equation ( a \times d = b \times c ).
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
If two positive fractions are less than 1, it means that both fractions can be expressed as ( a/b ) and ( c/d ), where ( a < b ) and ( c < d ). When you multiply these fractions, the product is ( (a/b) \times (c/d) = (a \times c) / (b \times d) ). Since both ( a ) and ( c ) are less than their respective denominators ( b ) and ( d ), the numerator ( a \times c ) will also be less than the denominator ( b \times d ). Thus, the product remains a positive fraction less than 1.
When cross multiplying, you do not add or subtract; instead, you multiply. In the equation ( \frac{a}{b} = \frac{c}{d} ), you cross multiply by calculating ( a \times d ) and ( b \times c ). This results in the equation ( a \times d = b \times c ).
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
If two positive fractions are less than 1, it means that both fractions can be expressed as ( a/b ) and ( c/d ), where ( a < b ) and ( c < d ). When you multiply these fractions, the product is ( (a/b) \times (c/d) = (a \times c) / (b \times d) ). Since both ( a ) and ( c ) are less than their respective denominators ( b ) and ( d ), the numerator ( a \times c ) will also be less than the denominator ( b \times d ). Thus, the product remains a positive fraction less than 1.
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
With no knowledge about A, B, C and D or the relation between them, it is impossible to give an answer.