Answer:
There are 10⁻⁷ hydrogen ions in one-liter water
Step-by-step explanation:
Hi there!
a. If the pH = 7.0, then:
-log(H) = 7.0
multiply both sides of the equation by -1
log(H) = -7.0
Apply 10ˣ to both sides
10^(log (H)) = 10⁻⁷
(H) = 10⁻⁷
There are 10⁻⁷ hydrogen ions in one-liter water.
Why 10^(log (H)) = (H)?
Let´s consider this function:
y = 10^(log x)
(Apply log to both sides)
log y = log 10^(log x)
(Apply logarithmic property: log xᵃ = a · log x)
log y = log x · log 10
(log 10 = 1)
log y = log x
y = x
(since y = 10^(log x) and y = x):
10^(log x) = x
That´s why 10^(log (H)) = (H)
Once you combine it, you get 10w + 18
Current amount in account
P=36948.61
Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n
Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11
We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000
We can solve this by trial and error.
The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000. It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.
So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.
So it takes 15.00 years to reach the goal of 280000 years.
Answer:
Length of the line segment with endpoints (11,−4) and (−12,−4) is 23 units
Step-by-step explanation:
Given:
Endpoints are (11,−4) and (−12,−4)
To Find:
The length of the line = ?
Solution:
The length of the line can be found by using the distance formula
Here
= 11
= -12
= -4
= -4
Substituting the values
Length of the line
=>
=>
=>
=>
=>23
