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Q: What s the LCM of 2 3 and 4?

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4=2*2 12=2*2*3 14=2*7 4 goes into 12, so the lcm of 4,12, and 14 is equal to the lcm of 12 and 14 12=2*2*3 14=2*7 We can cancel the 2's and just have 2*2*3*7=84 84.

The LCM of 4, 8, and 56 is 6. Since 4 and 8 are factors of 56, 56 is the LCM of the set. The answer can be found by dividing the 3 numbers into their prime factors, in this case 4=2x2, 8=2x2x2, and 56=7x2x2x2. The lowest common multiple is the smallest number which has all of these prime factors in. That means it has to have two 2's in (from 4) three 2's in (from 8) and three 2's and a 7 from 56. In this case, all of the prime factors of 4 can be found in 8, and all of the prime factors of 8 are in 56, so the LCM (lowest common multiple) of these 3 is 56.

240 = 2 x 2 x 2 x 2 x 3 x 5 900 = 2 x 2 x 3 x 3 x 5 x 5 Max is four 2's, two 3's and two 5's So LCM is 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600

15 = 3 x 5 6 = 2 x 3 25 = 5 x 5 Therefore, the LCM is the product of a 2, a 3, and two 5's. 2 x 3 x 5 x 5 = 150

Start by factoring both numbers into primes: 210 = 2*3*5*7 450= 2*3*3*5*5 What's the shortest list that includes all the factors of both numbers? One 2, two 3's, two 5's, one 7 That's the recipe for lcm in this problem. lcm = 2*3*3*5*5*7 = 3150

36

4725.189 = 3*63 = 3*3*21 = 3*3*3*7.75 = 3*25 = 3*5*5.The LCM needs 3 3's, 2 5's and a 7. ... (Leave out the 3 from 75 since there are already 3's in 189.)LCM = 3*3*3*5*5*7 = 27*25*7 = 675*7 = 4725.

well do the math: 1's 1*4(# of suits)=4 2's 1*4(# of suits)=4 3's 1*4(# of suits)=4 add them all up and you get 12 so you have 12 1's 2's and 3's in a deck of 52 playing cards.

2 3/4= 2.75

The least common multiple (LCM) has enough factors from each number that you can divide the LCM by the number and get a whole number as a result (no remainder) 12 = 2 x 2 x 3 16 = 2 x 2 x 2 x 2 factor down to prime numbers (2, 3, 5, 7, 11, etc) to see relationships better. in other words, divide the factors until you can't any more without getting a remainder 12 = two of 2 and one 3 16 = four of 2 16 has more 2's than 12, so the LCM needs to have the number of 2's that 16 has. 12 has more 3's than 16, so the LCM needs to have the number of 3's that 12 has. LCM = 2 x 2 x 2 x 2 x 3 = 48 This might be more complicated than other methods, but this is the one that makes the most sense to me.

The LCM of 4 and 15 is 60.The least common multiple is the smallest number that is multiple of two or more numbers.Multiples of 4: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60Multiples of 15: 15, 30, 45, 60The LCM of 4 and 15 is 60.The LCM must include the highest amount of each prime factor in both numbers. The highest power of 2's is 2, of 3's is 1 and of 5's is 1.So the LCM = 2 x 2 x 3 x 5 = 6060 15x1=15 15x2=30 15x3=45 15x4=60

2/3 = 14/21 4/7 = 12/21 So 2/3 IS bigger.

9 = 1, 3, 9 4 = 1, 2, 4

It is: 140

1/4 - 3/4 = -2/4 or -1/2

The LCM of 3, 6 and 7 is 42.

Factor into primes: 8 = 2*2*2 12 = 2*2*3 14 = 2*7 To cover all three, the LCM will need three 2's, one 3, and one 7. So the LCM is 2*2*2*3*7 = 168

Reduce all numbers to prime factors and combine (most of each prime) For example, LCM of 30 and 24 30 is 2 x 3 x 5 24 is 2 x 2 x 2 x 3 The most 2's we have is 3 (in 24) The most 3's we have is 1 The most 5's we have is 1 So LCM is 2 x 2 x 2 x 3 x 5 = 120

4/6 = (2*2)/(3*2) = 2/3 [the common 2's in numerator and denominator cancel]

Suppose the plane faces of the cylinder lie in the planes that are s units above and below the centre of the sphere. Therefore, the height of the cylinder is 2*s units. Also, by Pythagoras's theorem, the radius of the cylinder, x units, is such that x^2 + s^2 = R^2 = 16 Therefore x^2 = 16 - s^2. Volume of cylinder = pi*x^2*2*s = 2*pi*(16-s^2)*s = 2*pi*(16*s - s^3) Then dV/ds = 0 implies that 2*pi*(16 - 3*s^2) = 0 so s^2 = 16/3 and so s = 4/sqrt(3) The second derivative is -2*pi*6*s which is negative and so the volume is a maximum. When s = 4/sqrt(3), Volume of cylinder = 2*pi*[16 - 16/3]*4/sqrt(3) = 2*pi*32/3*4/sqrt(3) = 256*pi/[3*sqrt(3)] = 256*sqrt(3)*pi/9 = 49.2672*pi approx. The volume of the cylinder is approx 0.5774 times that of the sphere.

There are 6 numbers total with 3 4's. Then, the odds is ½.

3! - [(3 + 3)/3] = 6 - 2 = 4 Or (√3 x √3) + (3/3) = 3 + 1 = 4

2 of each, in a standard deck.

6 = 2 x 3 8 = 23 The LCM contains just enough factors to be able to make each of the original numbers, so we need three 2's and one 3: LCM = 23 x 3 = 24

o=-2*3=-6 and k=+1*2=2 s=4/2=+2