Answer:
33/36
Step-by-step explanation:
This on the web one-way conversion tool converts volume or capacity units from milliliters ( ml ) into cubic feet ( ft 3 , cu ft ) instantly online. 1 milliliter ( ml ) = 0.000035 cubic feet ( ft 3 , cu ft ). How many cubic feet ( ft 3 , cu ft ) are in 1 milliliter ( 1 ml )? How much of volume or capacity from milliliters to cubic feet, ml to ft<sup>3</sup> , cu ft?
hope this helps!
Answer:
The solution is x = e⁶
Step-by-step explanation:
Hi there!
First, let´s write the equation
ln(x⁶) = 36
Apply logarithm property: ln(xᵃ) = a ln(x)
6 ln(x) = 36
Divide both sides of the equation by 6
ln(x) = 6
Apply e to both sides
e^(ln(x)) = e⁶
x = e⁶
The solution is x = e⁶
Let´s prove why e^(ln(x)) = x
Let´s consider this function:
y = e^(ln(x))
Apply ln to both sides of the equation
ln(y) = ln(e^(ln(x)))
Apply logarithm property: ln(xᵃ) = a ln(x)
ln(y) = ln(x) · ln(e) (ln(e) = 1)
ln(y) = ln(x)
Apply logarithm equality rule: if ln(a) = ln(b) then, a = b
y = x
Since y = e^(ln(x)), then x =e^(ln(x))
Have a nice day!
-18/35
is the answer hope I am helpful
The 15th term will be 71. Why? Well, see below for an explanation!
By subtracting all of these numbers by the term that comes prior to them, we will find that all of them result in 5. Because of this, we know that each time the term increases, 5 is being added to the numbers. Additionally, I noticed that all of the numbers in this arithmetic sequence only end in a 1 or a 6. Because of this, we can apply the same principle when adding 5 each time:
First term: 1
Second term: 6
Third term: 11
Fourth term: 16
Fifth term: 21
Sixth term: 26
Seventh term: 31
Eighth term: 36
Ninth term: 41
Tenth term: 46
Eleventh term: 51
Twelfth term: 56
Thirteenth term: 61
Fourteenth term: 66
Fifteenth term: 71
By adding 5 each time and keeping in mind that the digits all end in only 1 or 6, we will find that the fifteenth term results in 71. Therefore, the 15th term is 71.
Your final answer: The 15th term of this arithmetic sequence comes down to be 71. If you need extra help, let me know and I will gladly assist you.
