Let's call this line y=mx+C, whereby 'm' will be its gradient and 'C' will be its constant.
If this line is parallel to the line you've just mentioned, it will have a gradient 2/3. We know this, because when we re-arrange the equation you've given us, we get...
![y-4=\frac { 2 }{ 3 } \left( x-3 \right) \\ \\ y-4=\frac { 2 }{ 3 } x-2\\ \\ y=\frac { 2 }{ 3 } x-2+4\\ \\ y=\frac { 2 }{ 3 } x+2](https://tex.z-dn.net/?f=y-4%3D%5Cfrac%20%7B%202%20%7D%7B%203%20%7D%20%5Cleft%28%20x-3%20%5Cright%29%20%5C%5C%20%5C%5C%20y-4%3D%5Cfrac%20%7B%202%20%7D%7B%203%20%7D%20x-2%5C%5C%20%5C%5C%20y%3D%5Cfrac%20%7B%202%20%7D%7B%203%20%7D%20x-2%2B4%5C%5C%20%5C%5C%20y%3D%5Cfrac%20%7B%202%20%7D%7B%203%20%7D%20x%2B2)
So, at the moment, our parallel line looks like this...
y=(2/3)*x + C
However, you mentioned that this line passes through the point Q(1, -2). If this is the case, for the line (almost complete) above, when x=1, y=-2. With this information, we can figure out the constant of the line we want to find.
-2=(2/3)*(1) + C
Therefore:
C = - 2 - (2/3)
C = - 6/3 - 2/3
C = - 8/3
This means that the line you are looking for is:
y=(2/3)*x - (8/3)
Let's find out if this is truly the case with a handy graphing app... Well, it turns out that I'm correct.
2•14 =28, 2• 270=540, 2•141=282, 2 is the GCF
Answer:
5 scoops
Step-by-step explanation:
If the feeder already has 1/2 cups in it, Porter will need to put 5 scoops in because 5/10= 1/2. 5 is half of ten
Hope this helps!
Answer:
(-2, 0.5) first
Step-by-step explanation:
X (-4, 1) → X' (-4*0.5, 1*0.5)
X'(-2, .5)