Correct question is;
Consider the following operations on the number 8.85 x 10^(-2). Without using a calculator, decide which would give a significantly smaller value than 8.85 x 10^(-2), which would give a significantly larger value, or which would give essentially the same value.
A) 8.85 × 10^(-2) + 6.69 × 10^(5)
B) 8.85 x 10^(-2) - 6.69 x 10^(5)
C) 8.85 x 10^(-2) x 6.69 x 10^(5)
D) 8.85 × 10^(-2) ÷ 6.69 × 10^(5)
Answer:
Option A & B yield essentially the same value.
Option C yields a significantly smaller value
Option D yields a significantly larger value
Step-by-step explanation:
Since given value for the operations is in 10^(-2) indices, then let's convert the operations numbers from 10^(5) indices to 10^(-2).
Thus;
6.69 × 10^(5) = (6.69 × 10^(-7)) × 10^(-2)
Thus;
A) 8.85 × 10^(-2) + 6.69 × 10^(5) is now;
8.85 × 10^(-2) + ((6.69 × 10^(-7)) × 10^(-2))
This gives;
[8.85 + (6.69 × 10^(-7))] × 10^(-2)
The 10^(-7) indices indicates that the initial value of 8.85 × 10^(-2) would not change much and would essentially be the same
B) 8.85 x 10^(-2) - 6.69 x 10^(5) is now;
8.85 × 10^(-2) - ((6.69 × 10^(-7)) × 10^(-2))
This gives;
[8.85 - (6.69 × 10^(-7))] × 10^(-2)
Again the 10^(-7) indices indicates that the initial value of 8.85 × 10^(-2) would not change much and would essentially be the same.
C) 8.85 x 10^(-2) x 6.69 x 10^(5) is now;
(8.85 × 10^(-2)) × ((6.69 × 10^(-7)) × 10^(-2))
This gives;
[8.85 × (6.69 × 10^(-7))] × 10^(-2)
When we multiply 8.85 by 6.69 × 10^(-7), the value gotten will be way lower than the initial value of 8.85.
Thus,this operation will give a significantly smaller value.
D) (8.85 × 10-2) ÷ (6.69 × 10^5) is now;
(8.85 × 10^(-2)) ÷ ((6.69 × 10^(-7)) × 10^(-2))
This gives;
[8.85 ÷ (6.69 × 10^(-7))] × 10^(-2)
When we carry out the operation:
8.85 ÷ (6.69 × 10^(-7)), it is essentially equal to (8.85 × 10^(7))/6.69.
This means that the new value would be significantly larger than the initial value of 8.85.
Thus,this operation yields a significantly larger value.
(Straightforward)
(log to base of itself is always 1)
(Using log power property)