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...

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.
Answer:
Easy lol mzvcn=48
Step-by-step explanation:
Answer:
Step-by-step explanation:
![(\frac{5}{2})^{x}+(\frac{5}{2})^{(x+3)}=(\frac{5}{2})^{x}+(\frac{5}{2})^{x}*(\frac{5}{2})^{3}\\\\=(\frac{5}{2})^{x}*[1+(\frac{5}{2})^{3}]](https://tex.z-dn.net/?f=%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7Bx%7D%2B%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7B%28x%2B3%29%7D%3D%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7Bx%7D%2B%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7Bx%7D%2A%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7B3%7D%5C%5C%5C%5C%3D%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7Bx%7D%2A%5B1%2B%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7B3%7D%5D)
Therefore,

Answer:-47 3/4
Step-by-step explanation:
The correct answer would be -47 3/4
You would use the equation
-32 1/4 - 15 1/2 = -47 3/4
to find your answer
Answer: A. There were 500 liters in the tank when the leak started, and it is decreasing by 9 liters per minute.
Step-by-step explanation:
This equation is in the form; y =mx + c.
<em>c</em> in this equation is the value of y when x is at 0.
<em>c</em> in the above formula is 500 which means that the number of litres (y) at zero minutes(x) is 500 litres.
<em>m</em> is the rate of change as a result of a change in x.
<em>m</em> in this equation is -9 which means that the water is reducing by 9 litres every minute (x).