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What LCM of 2 3 and 5 and why?
The LCM of 2, 3 and 5 is 30. The LCM of any three relatively prime numbers a, b and c is a*b*c, and here, 2 is a, 3 is b and 5 is c. Since 2, 3 and 5 are relatively prime, the LCM is 2*3*5=30.
How can you find the percent of one number is of another For example what perecnt of 35 is 7?
question: find the percentage A is of B or, 7 (A) is what percente of 35 (B) (B-A)/B x 100 =C. then subtract C from 100 B-A 35-7=28 /B 28/35=0.8 x 100 =C 0.8 x 100=80 - C 100-80=20. so, 7 is 20 percent of 35
What are the LCM of 4 8 and 88?
88 b/c 4x22=88 8x11=88 and88x1=88
What is the definition associative property multiplication?
The property states that for all real numbers a, b, and c, their product is always the same, regardless of their grouping: (a . b) . c = a . (b . c) Example: (6 . 7) . 8 = 6 . (7 . 8) The associative property also applies to complex numbers. Also, as a consequence of the associative property, (a . b) . c and a . (b . c) can both be written as a . b . c without ambiguity.
What is the LCM of 7 and 49?
The LCM of 7 and 49 is 49.
How can you find the Lowest common denominator of four integers?
lcm(a,b,c,d) = lcm(lcm(a,b,c),d) = lcm(lcm(a,b),lcm(c,d))
What is the LCM of A B and C?
That depends on the values of A, B and C.
What is the LCM of 6 7 and C?
If C is co-prime with 6 and with 7, then LCM(6, 7, C) = 42*C If not, the answer depends on the value of C.
What is the LCM Of a to the second B and B to the second c?
The LCM of a^2b and b^2c is a^2b^2c
What is the least common multiple of 7 5 and 13?
If a, b & c are prime numbers then their LCM is equal to their product i.e. a x b x c.Here, 5, 7 and 13 are prime numbers.Therefore, LCM(7,5,13) = 7 x 5 x 13 = 455.
What is the least common multiple of decimals and fractions in C?
To calculate the least common multiple (lcm) of decimals (integers) and fractions you first need to calculate the greatest common divisor (gcd) of two integers: int gcd (int a, int b) { int c; while (a != 0) { c = a; a = b % a; b = c; } return b; } With this function in place, we can calculate the lcm of two integers: int lcm (int a, int b) { return a / gcd (a, b) * b; } And with this function in place we can calculate the lcm of two fractions (a/b and c/d): int lcm_fraction (int a, int b, int c, int d) { return lcm (a, c) / gcd (b, d); }
What LCM of 2 3 and 5 and why?
The LCM of 2, 3 and 5 is 30. The LCM of any three relatively prime numbers a, b and c is a*b*c, and here, 2 is a, 3 is b and 5 is c. Since 2, 3 and 5 are relatively prime, the LCM is 2*3*5=30.
How do you calculate LCM of three numbers by pesudo code?
For this you will need a couple of helper algorithms. The first is the GCD (greatest common divisor) which is expressed as follows:procedure GCD (a, b) isinput: natural numbers a and bwhile ab doif a>blet a be a-belselet b be b-aend ifend whilereturn aThe second algorithm is the LCM (least common multiple) of two numbers:procedure LCM (a, b) isinput: natural numbers a and b return (a*b) / GCD (a, b)Now that you can calculate the GCD and LCM of any two natural numbers, you can calculate the LCM of any three natural numbers as follows:procedure LCM3 (a, b, c) isinput: natural numbers a, b and c return LCM (LCM (a, b), c)Note that the LCM of three numbers first calculates the LCM of two of those numbers (a and b) and then calculates the LCM of that result along with the third number (c). That is, if the three numbers were 8, 9 and 21, the LCM of 8 and 9 is 72 and the LCM of 72 and 21 is 504. Thus the LCM of 8, 9 and 21 is 504.
How do you find the LCM and GCF of two no in c?
#include<stdio.h> main() { int a,b,i,lcm,gcf; printf("\n Enter two numbers"); scanf("%d%d",&a,&b); for(i=0;i<=a;i++) { if((b%i==0)&&(a%i==0)) { gcf=i; } } lcm=a*b/gcf; printf("\n GCF is %d and LCM is %d",gcf,lcm); }
What is the LCM of 4 and 7 and 9 and 8 in c?
1008
What is the LCM of 4b2 and 6b3 A 2b2 B 12b2 C 12b3 D 24b3?
C. 12b3
If A is less than B and B plus C equals 10 and none of them equal zero then which of the following must be true?
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
