Answer:
A whole number with 4 digits is always greater than a whole number with 3 digits.
This requires that both numbers be positive.
Explanation.
In the decimal system, we count 0,1,2, ..., 9 (1st digit).
When we reach 10, we shift left and count in tens from 10,20,30, ..., 90 (2nd digit).
When we reach 100, we shift left and count in hundreds from 100,200,300, ..., 900 (3rd digit).
Then we count in thousands as 1000,2000, ... ( 4th digit).
Because the 4th digit counts in thousands and the 3rd digit counts in hundreds, a whole number in the 4th digit will be greater than a whole number in the 3rd digit.
The only thing you need to do is divide 56 by -7. You know the d must be a negative number because negative x negative= positive. So 56 / -7 is -8.
First expression
=> 3 x 3 x 3 x 3 x 3 – In this expression, we multiplied 3 5 times and the product is 243
Second expression
=> 3 ^ 5 , in where ^ read as raised to the power. , the product is also 243
Third expression
=> 3^2 x 3 ^3
=> (3 x 3) x (3 x 3 x 3)
=> 9 x 27, the product is also equals to 243.
Answer:
I assume you mean 16^(1/3) i.e. the cube root of 16
I am also assuming you mean the real cube root because, as you may know, every non-zero real number has three cube root - one real and two complex conjugates.
Since 16 = 8 x 2 and 8 = 2³ then (2³ x 2)^1/3 = (2³)^1/3 x 2^1/3 = 2 x 2^1/3
You might check that the cube root of 16 is about 2.52 which is twice the cube root of 2
Step-by-step explanation:
