Let's take a look at the <em>relationship</em> between x and y in this function. When we look at the first non-zero pair, (-1, 1), we can see that, in the transition from x to y, our x value was had to have been multiplied by -1. For now, we could tentatively say that y = -x. Unfortunately, the next pair, (-4, 2), debunks this hypothesis pretty quickly, as 2 ≠ -(-4). We'll need to reexamine the relationship between x and y here.
One patter that we can be fairly confident in is that multiplication by -1; the sign is flipped from negative to positive when we go from x to y in each case, so we can hold onto that component of the function. Something else is happening before that sign flip, though, and we'll need to look into that by seeing how else the numbers are connected.
Ignoring the negative signs for a moment, let's take a look at the pairs (1,1) and (4, 2). We can transform 1 to 1 pretty easily; we simply multiply or divide it by 1. From 4 to 2, we can either divide it by 2 or multiply it by 1/2. Let's thing in terms of division for now. We know that
1/1 = 1 (obviously)
4/2 = 2
We might have something here! It looks like the number we divide by and the number we obtain are the same, so let's explore this further. Multiplying both sides by those denominators, we find:
1 = 1²
4 = 2²
Almost there! To finally find the relationship, we just square root both sides to get:
√1 = 1
√4 = 2
So that was the function we'd been looking for - a square root! Putting that together with what we said about changing the sign at the beginning, we have:
y =√(-x)
To verify this, we just need to plug in a few pairs, and we find that indeed:
1 = √-(-1) = √1 = 1
2 = √-(-4) = √4 = 2
For this case we have that by definition, a direct variation is given in the form:
Where:
k: It is the constant of proportionality
According to the statement we have to:
Substituting:
On the other hand we have:
Answer:
Number 6
Answer:
y = (c - ax)/b
Step-by-step explanation:
ax + by = c
by = c - ax
y = (c - ax)/b
<em>I answered number 5 in your last question.</em>
