Answer:
There are infinitely many solutions to the system because the equations represent the same line
Answer:
cat food a
Step-by-step explanation:
6.5/10 = 0.65 unit price
14/20 = 0.70 unit price
Answer:
x=5, y=2
Step-by-step explanation:
You simply write down what you see:
3x - 4y + 4x - 5y = 17
2x - 3y + 3x - 5y = 9
Then you simplify by grouping the x and the y terms:
7x - 9y = 17
5x - 8y = 9
Now you want to get equal amounts of x or y. Let's go for x. If we multiply the top eq with 5 and the bottom one with -7, we get:
35x - 45y = 85
-35x + 56y = -63
-------------------------+ now we can add them:
11y = 22 divide by -11
y = 2
So 7x - 18 = 17, x =5
Finally, it makes sense to double check by filling in these answers in the original equations, just to see if no mistakes were made:
3*5 - 4*2 = 7
4*5 - 5*2 = 10
7+10 = 17: ok!
2*5 - 3*2 = 4
3*5 - 5*2 = 5
4+5 = 9: ok!
<span>The number of dollars collected can be modelled by both a linear model and an exponential model.
To calculate the number of dollars to be calculated on the 6th day based on a linear model, we recall that the formula for the equation of a line is given by (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), where (x1, y1) = (1, 2) and (x2, y2) = (3, 8)
The equation of the line representing the model = (y - 2) / (x - 1) = (8 - 2) / (3 - 1) = 6 / 2 = 3
y - 2 = 3(x - 1) = 3x - 3
y = 3x - 3 + 2 = 3x - 1
Therefore, the amount of dollars to be collected on the 6th day based on the linear model is given by y = 3(6) - 1 = 18 - 1 = $17
To calculate the number of dollars to be calculated on the 6th day based on an exponential model, we recall that the formula for exponential growth is given by y = ar^(x-1), where y is the number of dollars collected and x represent each collection day and a is the amount collected on the first day = $2.
8 = 2r^(3 - 1) = 2r^2
r^2 = 8/2 = 4
r = sqrt(4) = 2
Therefore, the amount of dollars to be collected on the 6th day based on the exponential model is given by y = 2(2)^(5 - 1) = 2(2)^4 = 2(16) = $32</span>
The answer is...B. ways of maintaining relationships