To simplify, you need to distribute the -6 out into both the 2 and the t to get:
![2^{-6} * t^{-6}](https://tex.z-dn.net/?f=%202%5E%7B-6%7D%20%2A%20%20t%5E%7B-6%7D%20)
Now, if you only want positive exponents, you need to write the expression in this form:
![\frac{1}{ 2^{6}* t^{6} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%202%5E%7B6%7D%2A%20t%5E%7B6%7D%20%20%7D%20)
Which would simplify to:
Answer: first option
Step-by-step explanation:
To form an arithmetic sequence, you have that for the sequence
:
![6k-7=22-6k](https://tex.z-dn.net/?f=6k-7%3D22-6k)
Therefore, to calculate the value of k to form an arithmetic sequence, you must solve for k, as following:
- Add like terms:
![6k+6k=22+7\\12k=29](https://tex.z-dn.net/?f=6k%2B6k%3D22%2B7%5C%5C12k%3D29)
- Divide both sides by 12. Then you obtain;
![k=\frac{29}{12}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B29%7D%7B12%7D)
Answer:
2.3,2.03,2.003
Step-by-step explanation:
i need points
Let <em>V</em> be the volume of the tank. The inlet pipe fills the tank at a rate of
<em>V</em> / (5 hours) = 0.2<em>V</em> / hour
and the outlet pipe drains it at a rate of
<em>V</em> / (8 hours) = 0.125<em>V</em> / hour
With both valves open, the net rate of water entering the tank is
(0.2<em>V</em> - 0.125<em>V </em>) / hour = 0.075<em>V</em> / hour
If <em>t</em> is the time it takes for the tank to be full, then
(0.075<em>V</em> ) / hour • <em>t</em> = <em>V</em>
Solve for <em>t</em> :
<em>t</em> = <em>V</em> / ((0.075<em>V</em> ) / hour)
<em>t</em> = 1/0.075 hour
<em>t</em> ≈ 13.333 hours
Answer:
Number 1:3.75 gallons
Step-by-step explanation: