Question down below please I need help!!

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Question down below please I need help!!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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what trouble are you having?
identifying the coordinates for C and D?

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Yes I don't know how to solve the problem overall
|dw:1437368521661:dw| y=b is a horizontal line any point falling on it will have it's y coordinate as b y=0 is a horizontal line any point falling on it will have it's y coordinate as 0
|dw:1437368652986:dw| what I'm saying is I know 2a comes after a sometime you could have picked something else I'm sure calling it x=c would have been fine we would just assume c>a
anyways x=a is a vertical line any point falling on it will have it's x coordinate as a x=2a is a vertical line any point falling on it will have it's x coordinate as 2a
so what coordinates are you going to assign to C?
notice the lines goes through point C above on the picture are x=2a and y=b and I did say any point on x=2a will have x-coordinate 2a and I also said any point on y=b will have y-coordinate b so we could give the coordinates of C as?
2a?
c has cordinates of 2a, b?
yes C has coordinate (2a,b) or that is what we are assigning C anyways it is appropriate since 2a does come after a but as I said before we could have pick another number like c and make the assumption c is greater than a but anyways |dw:1437369322715:dw| now we have to assign coordinates to D
notice D lays on the horizontal line y=0 any point falling on y=0 has y-coordinate 0
|dw:1437369388989:dw| what number are you going to put in place of that ?
umm 2a squared
I guess you could say 2a^2 the spacing between B's x-coordinate and C's x-coordinate seem pretty close to the spacing between C's x-coordinate and D's x-coordinate I think a better number might be 3a but I think 2a^2 would be great if we don't care about scales
by the way that is according to your picture I have drawn my D a little closer to my C than the pic actually shows
|dw:1437369720333:dw|
ok
now we want to verify that is a trapezoid
So we need to show the slope of line segment BC is the same as the slope of the line segment AD
This is because a trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides that are parallel. To show two lines are parallel you just show that their slopes are equal.
\[\text{ slope of a line } y=mx+b \text{ with coordinates } (x_1,y_1) \text{ and } (x_2,y_2) \\ \text{ is } \frac{y_2-y_1}{x_2-x_1}\]
so line BC has points (a,b) and (2a,b) so it's slope is?
I don't know actually.
you can also just observe from the picture the slope of line segment BC is zero since it is a horizontal line segment but you can also do it algebraically using the formula I gave you have: \[(x_1,y_1)=(a,b) \text{ and } (x_2,y_2)=(2a,b) \\ \text{ so } x_1=a \\ y_1=b \\ x_2 =2a \\ y_2=b \\ \text{ so entering into the the formula } \\ \frac{y_2-y_1}{x_2-x_1}=\frac{b-b}{2a-a}\] we know that b-b=0 and 2a-a=a and 0/a is 0 (since we also know a is not 0)
oh ok
anyways we also need the slope of the side that is opposite to BC
if we can show that the side opposite to BC has the same slope I would say we are done since obviously the shape is 4-sided flat shape
What did I say the slope of any horizontal line is?
0?
so slope of BC is 0 and slope of AD is 0 |dw:1437370464744:dw| the opposite sides called BC and AD are parallel since they both have the same slope I think they asked you to find the coordinates to use the slope formula though (but honestly that is kind of weird since the line segments BC and AD obviously appear horizontal ) \[\text{ slope of } BC=\frac{b-b}{2a-a}=\frac{0}{a}=0 \\ \\ \text{ slope of } AC=\frac{0-0}{3a-0}=...\] I will let you finish that if you want to.
anyways that is done we have shown this 4-sided flat shape does consist of opposite sides that are parallel
0/3a
and 0/(3a) is 0 since 0/anything =0 (exlcuding the anything from being 0) (we know a isn't 0 since according to the picture a is greater than 0)
in short 0/(3a)=0
thank you
np do you have anymore question about this question?
I'm about to vanish into thin air
no thank you though
abracadabra

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