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.
Step-by-step explanation:
hello,
the shape is a right trapezium.
you use the following formula:
area of a trapezium = ½ × sum of parallel sides× height.
or simply,
a = ½ ( a + b ) h
where a and b are the parallel sides
and h is the height.
thus,
a = ½ × ( 4 + 6 )× 4
a = ½ ( 10 ) × 4
a = 5 × 4
a = 20
have a great day! au revoir !
Answer:
b
Step-by-step explanation:
I hope this helps and have a nice day
Answer:C
Step-by-step explanation:
the line above AB signifies that its a line segment
there are points for A and B
He used an incorrect time ratio converting hours to minutes.