Answer:
It's X^1 = 3, x^2 =6
Step-by-step explanation:
The answer would be idk, maybe 1(5(^^22
A solution with a pH of 11 has a [H+] of 
Option B is correct option.
Step-by-step explanation:
A solution with a pH of 11 has a [H+] of= ?
pH=?
Concentration of H+ [H+]= ?
Formula used: ![pH=-log[H+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D)
Putting values and finding [H+]
![pH=-log[H+]\\\,[H+]\,=10^{-pH}\\\,[H+]\,=10^{-11}\\\,[H+]\,=1.0\times 10^{-11}](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D%5C%5C%5C%2C%5BH%2B%5D%5C%2C%3D10%5E%7B-pH%7D%5C%5C%5C%2C%5BH%2B%5D%5C%2C%3D10%5E%7B-11%7D%5C%5C%5C%2C%5BH%2B%5D%5C%2C%3D1.0%5Ctimes%2010%5E%7B-11%7D)
So, A solution with a pH of 11 has a [H+] of
Option B is correct option.
I think is c that what i think
Answer: 11 year
P(1) = 37,100
P(4) = 58,400
The linear equation (for x ≥ 1)
P(x) = 37,100 + a(x-1)
For x = 4
58,400 = 37,100 + a(4-1)
58,400 - 37,100 = 3a
21300 = 3a
a = 7100
So, the linear equation:
P(x) = 37100 + 7100*(x-1)
P(x) = 37100 + 7100x - 7,100
P(x) = 7100x + 30000
To find when the profit should reach 108100, we can substitute P(x) by 108100.
108100 = 7100x + 30000
108100 - 30000 = 7100x
78100 = 7100x
x = 78100/7100
x = 11
Answer: 11 year
