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Answer:
Okay so I am not positive but I will try to help you out. Okay so the line of best fit would positive. Because it is all going up. The approximate slope is 3/4 Because slope is change in y over change in x. The y intercept of the line of best fit would be (0,0). This is because when drawing the line you can see that it would intercept the y-axis at zero. My steps taken were basically just looking at it and peicing everything together. If you want to get more involved with slope you could pick to coordinates which I will include and do change in y over change in x.
My points I will pick are (1,1) and (5,2)
2-1/5-1=3/4
therefore approximate slope is 3/4
Like I said I am not positive but I am just trying to help out in some way.
If something is incorrect please inform me. If this helps also please inform me.
F(x) is vertically stretched by a factor of 5/4 matches w/ f(x)-5/4
f(x) reflects the x axis matches w/ -f(x)
f(x) is translated 5/4 units up matches with f(x)+5/4
f(x) translated 5/4 to the left matches w/ f(x)-5/4
I didn't really know about this one but it seemed kind of easy, what was your first guesses and did you forget to put one more or was the question like that.
Hope this is right :) or if someone else has a different answer but in the meantime, good luck :)
Given :
- Area of the trapezium is 140 cm².
- Its height is 10 cm.
- One of the parallel side is 16 cm.
To Find :
Solution :
- Let the other parallel side be x
We know that,

Now, Substituting the given values in the formula :







Therefore,
- The other parallel side of the trapezium is 12 cm.
Step-by-step explanation:

By simplifying the right side of the equation, we come up with
, or D