P = A.e^(r.t) is the population growth formula, where A is the initial population in a given year, exponent "r" =growth rate and exponent t = time
A₍₁₉₉₅₎ = 184 Mil. after t = 7 years, P₍₂₀₀₂₎ = 188 Mil
Let's calculate te growth rate:
P₍₂₀₀₂₎ = A₍₁₉₉₅₎.e^rt
188 = 184.e(^7r)
188/184 = e^(7r)
ln(188/184) = 7.r.lne
0.0215 = 7.r and r = 0.0030 = 0.3%
Population in 2006
P₍₂₀₀₆₎ = P₍₂₀₀₂₎.e⁽⁰.⁰⁰³ˣ⁴
P₍₂₀₀₆₎ = 188.e⁰.⁰¹² = 190.2 Mil
{2×[20×5]+68} could be your answer because 20×5 is 100 and 2×100 is 200. 200+68 is 268
Answer:
x = -1
Step-by-step explanation:
The equation is -2x + 3 = 2x +7
Group the terms together. Don't forget to change the sign
-2x -2x = 7-3
-4x = 4
Divide both sides by 4
-x = 1
Flip it over
x = -1
Answer:
The solution is similar to the 2-point form of the equation for a line:
y = (y2 -y1)/(x2 -x1)·x + (y1) -(x1)(y2 -y1)/(x2 -x1)
Step-by-step explanation:
Using the two points, write two equations in the unknowns of the equation of the line.
For example, you can use the equation ...
y = mx + b
Then for the points (x1, y1) and (x2, y2) you have two equations in m and b:
b + (x1)m = (y1)
b + (x2)m = (y2)
The corresponding augmented matrix for this system is ...
![\left[\begin{array}{cc|c}1&x1&y1\\1&x2&y2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26x1%26y1%5C%5C1%26x2%26y2%5Cend%7Barray%7D%5Cright%5D)
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The "b" variable can be eliminated by subtracting the first equation from the second. This puts a 0 in row 2 column 1 of the matrix, per <em>Gaussian Elimination</em>.
0 + (x2 -x1)m = (y2 -y1)
Dividing by the value in row 2 column 2 gives you the value of m:
m = (y2 -y1)/(x2 -x1)
This value can be substituted into either equation to find the value of b.
b = (y1) -(x1)(y2 -y1)/(x2 -x1) . . . . . substituting for m in the first equation