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Which one do you want to tackle first?
Ok, what do you know about the length of AD?
the answer is d, How I can find the length?
What can you tell me about the length of AD by looking at the picture?
the question is what the measure of AD?
Yes I know. I'm asking what you think about the problem
can you see the picture?
This is math. Some thinking is required ;)
Yes I can see it. I'm asking what does it tell you? I know what it's telling me.
yes it math
given the lenghth AB, BC, CD, I don't get what you asking
I wonder do you know how to get answer find length AD
I know how to get the answer yes. I want to know what you know? That way I can see how to help you.
Nobody cares if I can do it.
I don't know
if given radius is may easy for my to do
I been thinking many hours but I can't solve
if you can help me ,
Good! You know how to solve it with the radius?
yes, I can do if I know radius or given tangents or secants
Ok, well for this we don't actually need the radius. This is called a tangential quadrilateral. It has the property that the two opposite sides will add to the same length.
AB + CD = AD + BC
What do you think about 38?
let me think
I think H
w is the weight the total weight have to be less or equal
r no more than 12
r no more than 12 comes from what statement?
If I tell you, there are 4 reams of paper in the box, what is the weight w?
the given r is the box can carry no more than 12 reams of paper
so my r is less than 12 it's 4, what is w?
if the box itself weighs 1.3 and each ream weighs 4.4 and there are 4 reams, what is the total weight?
total; 18.9 w
You sure the weight isn't 1 ?
I think 18.9 was correct actually.
But wait, why isn't the weight just 3.1 or 1.3 or 10
Aren't all those < 1.3 + 4.4r ?
w<1.3 +4.4r, r<12
Right. So if I tell you that r=1, what is the weight? It's 1.1111111111111 right?
because 1.111111 < 1.3 + 4.4
Sure, and 1.1111 is less than 5.7, so 1.1111 is the weight? But 0.1274 is also less than 5.7, so which one is the weight?
Or is the weight both 1.1111 and 0.1274 or maybe the weight is \(\pi\). I do like \(\pi\). (Here is where you tell me I'm being dumb and the weight is 5.7)
\(\pi\) is less than 5.7 so I think \(\pi\) is the weight. Cause you said anything less than 1.3 + 4.4r was a solution for the weight of a box.
and I said there was 1 ream, so the box can weigh anything less than 5.7
And you say "No dummy the box weighs 5.7"
ok, thank you so much
Wait. What? Did you find the solution?
yes I think w<1.3 + 4.4 r , r<12
no.. that's what I'm trying to explain
Once we pick an r, we know exactly how much the box weighs. The restriction is on r not on the weight.
w is not any number less than 1.3 + 4.4r. If I tell you r =1, w is not any number less than 5.7. It is 5.7
the key not reqite r
hrm? I didn't follow that. The answer is G.
If we have r, we know exactly what w is. Not that it's less than something, we know exactly. The only restriction we have is that r must be less than or equal to 12. But will be exactly 1.3 + 4.4r for any r you pick that is allowed.
err w will be exactly 1.3 + 4.4r
Ok , you said w = 1.3+4.4r ,r<12
yes. Because once you pick how many reams in the box, the box has an exact wieght.
if you pick 1 ream goes in the box, the box will wiegh 5.7. It will not weigh 1.1 or 0.213. If w < 5.7 were the definition then w could be any of those.
OK, it help
I understand now
It late I go to the bed now bye