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What else can I help you with?
When cross multiplying do you add then or subtract?
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 ).
How do you add fractions that don't have the same denominator?
if you are adding a/b + c/d the answer is (a * d + c * b)/(b*d)
What is 4d times d times d expressed in index form?
4d3 multiplications add in powers.
Show me algibra?
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 why is their product also less than 1?
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 do you add then or subtract?
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 ).
How do you add fractions that don't have the same denominator?
if you are adding a/b + c/d the answer is (a * d + c * b)/(b*d)
Write a c program as an example for call by value and call by reference?
#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; }
How do you play skye boat on recorder?
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!!!
How do you work a proportional problem?
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
How you add two no without any airthmaic opertor?
#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(); }
What is 4d times d times d expressed in index form?
4d3 multiplications add in powers.
What are the notes for yellow submarine with the recorder?
bcdbabag bcdbabag eeffgeaeag aeagaeaeg reapet as many times as you like
Show me algibra?
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 why is their product also less than 1?
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.
How do you do addition of 32 bit numbers in 8085?
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 property is D equals to A times B and C?
With no knowledge about A, B, C and D or the relation between them, it is impossible to give an answer.
