No, since 0 doesn't belong to the domain.
Answer:
The value of the expression given is:
Step-by-step explanation:
First, you must divide the expression in three, and to the final, you can multiply it:
- [(3^8)*(2^(-5))*(9^0)]^(-2)
- [(2^(-2))/(3^3)]^4
- 3^28
Now, we can solve each part one by one:
<em>First part.
</em>
- 3^8 = 6561
- 2^(-5) = 0.03125
- 9^0 = 1 (Whatever number elevated to 0, its value is 1)
- (6561 * 0,03125 * 1) = 205.03125
And we elevate this to -2:
- 205.03125^-2 = 2<u>.378810688*10^(-5)</u> or <u>0.00002378810688
</u>
<em>Second part.
</em>
- 2^(-2) = 0.25
- 3^3 = 27
- 0.25 / 27 = 9.259259259 * 10^(-3) or 0.00925925925925
And we elevate this to 4:
- 0.00925925925925^4 = <u>7.350298528 * 10^(-9)</u> or <u>0.000000007350298528
</u>
<em>Third Part.
</em>
- 3^28 = <u>2.287679245 * 10^13</u> or <u>22876792450000</u>
At last, we multiply all the results obtained:
- 0.00002378810688 * 0.000000007350298528 * 22876792450000 = <u>3.999999999999999999</u> approximately <u>4</u>
<u><em>We approximate the value because the difference to 4 is minimal, which could be obtained if we use all the decimals in each result</em></u>.
9.252.063 rounded to the nearest hundredth is 9.252.060
To find:
An irrational number that is greater than 10.
Solution:
Irritation number: It cannot be expression in the form of 
, where, 
are integers.
For example: 
.
We know that square of 10 is 100. So, square root of any prime number is an example of an irrational number that is greater than 10.
First prime number after 100 is 101.
Required irrational number 
Therefore, 
is an irrational number that is greater than 10.
