the number 144 is multiplied with
2^(-3 ), 12^(-1) and 3^(-2) consecutively
find the final product
144 * 1/8 * 1/12 * 1/9 = ?
Where it is
To find the final product, we need to multiply the number 144 consecutively with each of the given exponents.
Step 1: Multiply 144 by 2^(-3)
2^(-3) can be written as 1/2^3 = 1/8
144 * (1/8) = 18
Step 2: Multiply 18 by 12^(-1)
12^(-1) can be written as 1/12
18 * (1/12) = 3/2 = 1.5
Step 3: Multiply 1.5 by 3^(-2)
3^(-2) can be written as 1/3^2 = 1/9
1.5 * (1/9) = 1/6 = 0.1667 (rounded to four decimal places)
Therefore, the final product is approximately 0.1667.
To find the final product of these consecutive multiplications, we need to multiply 144 by each of the given powers separately and then multiply the results together.
Let's start by calculating each power individually:
1. 2^(-3):
To calculate this, we need to find the reciprocal of 2^3 (2 raised to the power of 3). The reciprocal of a number is obtained by dividing 1 by that number. So, the reciprocal of 2^3 is 1/(2^3), which is 1/8.
2. 12^(-1):
Similar to the previous step, we need to find the reciprocal of 12^1. The reciprocal of a number is obtained by dividing 1 by that number. So, the reciprocal of 12^1 is 1/12.
3. 3^(-2):
Again, we need to find the reciprocal of 3^2. The reciprocal of 3^2 is 1/(3^2), which is 1/9.
Now, let's multiply these individual powers with 144:
144 * (2^(-3)) = 144 * (1/8) = 18
18 * (12^(-1)) = 18 * (1/12) = 3/2
(3/2) * (3^(-2)) = (3/2) * (1/9) = 1/6
Therefore, the final product is 1/6.