Help me please!!!!!!!!!!!!!!!!!!
1 answer:
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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.
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:
76 ft
Step-by-step explanation:
Use the circumference formula, C = 2r
Plug in the radius:
C = 2r
C = 2(12.1)
C = 76.0
So, the circumference is 76 ft
Answer:
see below
Step-by-step explanation:
I enter the equation into a graphing calculator and let it do the graphing.
___
If you're graphing this by hand, you start by looking for the parent function. Here, it is |x|. That has a vertex of (0, 0) and a slope of +1 to the right of the vertex and a slope of -1 to the left of the vertex.
Here, the function is multiplied by -3/2, so will open downward and have slopes of magnitude 3/2 (not 1). The graph has been translated 5 units upward, so the vertex is (0, 5).
I'd start by plotting the vertex point at (0, 5), then identifying points with slope ±3/2 either side of it. To the left, it is left 2 and down 3 to (-2, 2). The points on the right of the vertex are symmetrically located about the y-axis, so one of them will be (2, 2).
Of course, you don't plot any function values for x > 4.
