Given: in the previous year, the table shows the prices of various amount of maize bushels in the same store.
A diagram displays the number of x-axis bussels and the price of maize in y axis dollars for 2012.
Search: The rate of change in this year's maize bushel, Part B: How many dollars is the present year's price of a maize bushel higher than in the previous year's maize bushel
Finding:
Earlier Year Panel
Bushel Number Corn prices
Shift rate of the previous year's maize bushel = (20-10)/(4 – 2) = = 10/2 = 5 Graph 2, 14, 4, 28, 6, 42 and 8, 56 and 10, 70, and 12, 84 Current year This year's rate of adjustment of the maize bushel, = (28 - 14)/(4 -2) = 14/2 = 7 7 - 5= 2 USD is more than the price of the maize bushel in the current year than the pre-year price of the maize bushel.
Answer:
(x, y) = (-4, 4)
Step-by-step explanation:
You find the values of x and y by working the problem using any of the methods of solving simultaneous equations that you have been taught.
Here, you observe that the y-coefficients are the same in each equation. That means you can cancel the y-terms by subtracting one equation from the other.
(7x +6y) -(4x +6y) = (-4) -(8)
3x = -12 . . . . simplify
x = -4 . . . . . . divide by 3
Now, you can substitute this into either equation to find y.
4(-4) +6y = 8
6y = 24 . . . . . . . . add 16
y = 4 . . . . . . . . divide by 6
The solution to this system of equation is ...
The cost of one drink is $2.50 and cost of one popcorn is $3.75
<u><em>Explanation</em></u>
Suppose, the cost of one drink is
dollar and cost of one popcorn is
dollar.
Given that, total cost for 2 drinks and 2 popcorns is $12.50 and total cost for 6 drinks and 5 popcorns is $33.75
So, the system of equations will be......

Multiplying equation (1) by -3 , we will get.....

Now, adding equation (2) and equation (3) , we will get ............

Plugging this
into equation (1) ......

So, the cost of one drink is $2.50 and cost of one popcorn is $3.75

so, let's first change the bounds from 1 to 9, so we'll be dropping the indices by 1, in order to do the variable "n".
however, let's use a variable hmmm say "k", where "k" is the value of "n" beginning at 1, thus
k = n - 1.
however, if k = n - 1, then
k + 1 = n. So let's use those fellows then,