C is also the best answer
Answer:
15
Step-by-step explanation:
Rate of change or slope or gradient of a line passes two points (x1, y1) and (x2, y2) could be calculated by:
(y2 - y1)/(x2 - x1)
=> Rate of change of the line passing (1,15) and (3,45):
(45 - 15)/(3 - 1) = 30/2 = 15
<span>Last
year, Shantell bought a car for 24 000 dollars. It decreases to 21 000 dollars
this current year.
Let’s find for the percentage that it decreased.
=> in 24 000 the 50% is 12 000 as we all know since it’s divided by 2. So
meaning, the possible answer is less than 50%.
=> now 25% of 24 000 is 6 000, and the amount that we’re looking is 3000,
thus, the answer is 12.5%
=> 24 000 * 0.125 = 3000
=> 24 000 – 3000 = 21 000</span>
Solve for one of the variables in the first equation (in this case, I will solve for X):
X = Y + 3
Then use that value of X in the second equation to solve for Y:
X + 3Y = 9
(Y + 3) + 3Y = 9
4Y + 3 = 9
4Y = 6
Y = 1.5
Use the value of Y we just found in the X equation we created:
X = Y + 3
X = 1.5 + 3
X = 4.5
Therefore X = 4.5 or 9/2 and Y = 1.5 or 3/2.
We have

So

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

and this happens if <em>a</em> = 1 and <em>b</em> = -1.
To confirm, we have

as required.