Find rates of change until you find a constant.
dy/dx=1,2,3,4,5,6
d2y/dx2=1,1,1,1,1
So the acceleration, d2y/d2x, is constant. This means that this is a quadratic sequence of the form a(n)=an^2+bn+c. So we can set up a system of equations to solve for the values of a,b, and c. Using the first three points, (1,1), (2,2), and (3,4) we have:
9a+3b+c=4, 4a+2b+c=2, and a+b+c=1 getting the differences...
5a+b=2 and 3a+b=1 and getting this difference...
2a=1, so a=1/2 making 5a+b=2 become:
2.5+b=2, so b=-1/2, making a+b+c=1 become:
1/2-1/2+c=1, so c=1 so the rule is:
a(n)=0.5x^2-0.5x+1 or if you prefer to not have decimals
a(n)=(x^2-x+2)/2
Answer:
(-2, 5)
Step-by-step explanation:
I will assume that your system is
y = -4x - 3
y = -2x + 1
Multiplying the second equation by -2 results in
y = -4x - 3
-2y = 4x - 2
combining these two equations eliminates x temporarily:
-y = -5, so that y must be 5.
Substituting 5 for y in the second equation, we get:
5 = -2x + 1, or
4 = -2x, which results in x = -2
and so the solution is (-2, 5)
Answer:
The fish population will be in 2050 = 27,123
Step-by-step explanation:
Given,
In 2005, the total number of fish were introduced (fish population), PV = 750
The growth rate, i = 8.3% = 0.083
The difference between 2050 and 2005 is (2050 - 2005), n = 45 years
We need to solve this problem with the help of future value of compounding. We know, FV = PV (1+i)^n, where, PV = present value, i = interest rate, n = period.
Therefore, the fish population will be in 2050 = 750 x (1 + 0.083)^45
The fish population will be in 2050 = 27,123.26
2050 fish population = 27,123 (Nearest to whole number).
<span>.0833333333 with repeating 3 is (2/24) or (1/12) <- There's you're answer.
<em>(Just put this up here since I answered it in a comment)</em></span>
Answer:
59
Solution
Equation 7x^2 + 2y=0
Where x=1,y=5
Put the value of x and y in equation
⟹(7×1)^2 + 2×5
⟹49+10
=59
