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Mathematics 15 Online
OpenStudy (anonymous):

http://img862.imageshack.us/i/page596.jpg/ I need help # 37,38

OpenStudy (anonymous):

Which one do you want to tackle first?

OpenStudy (anonymous):

37 first

OpenStudy (anonymous):

Ok, what do you know about the length of AD?

OpenStudy (anonymous):

the answer is d, How I can find the length?

OpenStudy (anonymous):

What can you tell me about the length of AD by looking at the picture?

OpenStudy (anonymous):

the question is what the measure of AD?

OpenStudy (anonymous):

Yes I know. I'm asking what you think about the problem

OpenStudy (anonymous):

can you see the picture?

OpenStudy (anonymous):

This is math. Some thinking is required ;)

OpenStudy (anonymous):

Yes I can see it. I'm asking what does it tell you? I know what it's telling me.

OpenStudy (anonymous):

yes it math

OpenStudy (anonymous):

given the lenghth AB, BC, CD, I don't get what you asking

OpenStudy (anonymous):

I wonder do you know how to get answer find length AD

OpenStudy (anonymous):

I know how to get the answer yes. I want to know what you know? That way I can see how to help you.

OpenStudy (anonymous):

Nobody cares if I can do it.

OpenStudy (anonymous):

I don't know

OpenStudy (anonymous):

if given radius is may easy for my to do

OpenStudy (anonymous):

I been thinking many hours but I can't solve

OpenStudy (anonymous):

if you can help me ,

OpenStudy (anonymous):

Good! You know how to solve it with the radius?

OpenStudy (anonymous):

yes, I can do if I know radius or given tangents or secants

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

what side?

OpenStudy (anonymous):

AB + CD = AD + BC

OpenStudy (anonymous):

great, thank

OpenStudy (anonymous):

What do you think about 38?

OpenStudy (anonymous):

let me think

OpenStudy (anonymous):

I think H

OpenStudy (anonymous):

Why?

OpenStudy (anonymous):

w is the weight the total weight have to be less or equal

OpenStudy (anonymous):

r no more than 12

OpenStudy (anonymous):

r no more than 12 comes from what statement?

OpenStudy (anonymous):

If I tell you, there are 4 reams of paper in the box, what is the weight w?

OpenStudy (anonymous):

the given r is the box can carry no more than 12 reams of paper

OpenStudy (anonymous):

right.

OpenStudy (anonymous):

so my r is less than 12 it's 4, what is w?

OpenStudy (anonymous):

4.4

OpenStudy (anonymous):

if the box itself weighs 1.3 and each ream weighs 4.4 and there are 4 reams, what is the total weight?

OpenStudy (anonymous):

total; 18.9 w

OpenStudy (anonymous):

You sure the weight isn't 1 ?

OpenStudy (anonymous):

22.8w

OpenStudy (anonymous):

I think 18.9 was correct actually.

OpenStudy (anonymous):

But wait, why isn't the weight just 3.1 or 1.3 or 10

OpenStudy (anonymous):

Aren't all those < 1.3 + 4.4r ?

OpenStudy (anonymous):

w<1.3 +4.4r, r<12

OpenStudy (anonymous):

Right. So if I tell you that r=1, what is the weight? It's 1.1111111111111 right?

OpenStudy (anonymous):

because 1.111111 < 1.3 + 4.4

OpenStudy (anonymous):

r=1? w<5.7

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

\(\pi\) 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.

OpenStudy (anonymous):

and I said there was 1 ream, so the box can weigh anything less than 5.7

OpenStudy (anonymous):

And you say "No dummy the box weighs 5.7"

OpenStudy (anonymous):

;)

OpenStudy (anonymous):

ok, thank you so much

OpenStudy (anonymous):

Wait. What? Did you find the solution?

OpenStudy (anonymous):

yes I think w<1.3 + 4.4 r , r<12

OpenStudy (anonymous):

no.. that's what I'm trying to explain

OpenStudy (anonymous):

Once we pick an r, we know exactly how much the box weighs. The restriction is on r not on the weight.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the key not reqite r

OpenStudy (anonymous):

hrm? I didn't follow that. The answer is G.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

err w will be exactly 1.3 + 4.4r

OpenStudy (anonymous):

Ok , you said w = 1.3+4.4r ,r<12

OpenStudy (anonymous):

yes. Because once you pick how many reams in the box, the box has an exact wieght.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

OK, it help

OpenStudy (anonymous):

I understand now

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

no problem

OpenStudy (anonymous):

It late I go to the bed now bye

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