anonymous
  • anonymous
http://img862.imageshack.us/i/page596.jpg/ I need help # 37,38
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Which one do you want to tackle first?
anonymous
  • anonymous
37 first
anonymous
  • anonymous
Ok, what do you know about the length of AD?

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anonymous
  • anonymous
the answer is d, How I can find the length?
anonymous
  • anonymous
What can you tell me about the length of AD by looking at the picture?
anonymous
  • anonymous
the question is what the measure of AD?
anonymous
  • anonymous
Yes I know. I'm asking what you think about the problem
anonymous
  • anonymous
can you see the picture?
anonymous
  • anonymous
This is math. Some thinking is required ;)
anonymous
  • anonymous
Yes I can see it. I'm asking what does it tell you? I know what it's telling me.
anonymous
  • anonymous
yes it math
anonymous
  • anonymous
given the lenghth AB, BC, CD, I don't get what you asking
anonymous
  • anonymous
I wonder do you know how to get answer find length AD
anonymous
  • 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.
anonymous
  • anonymous
Nobody cares if I can do it.
anonymous
  • anonymous
I don't know
anonymous
  • anonymous
if given radius is may easy for my to do
anonymous
  • anonymous
I been thinking many hours but I can't solve
anonymous
  • anonymous
if you can help me ,
anonymous
  • anonymous
Good! You know how to solve it with the radius?
anonymous
  • anonymous
yes, I can do if I know radius or given tangents or secants
anonymous
  • 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.
anonymous
  • anonymous
what side?
anonymous
  • anonymous
AB + CD = AD + BC
anonymous
  • anonymous
great, thank
anonymous
  • anonymous
What do you think about 38?
anonymous
  • anonymous
let me think
anonymous
  • anonymous
I think H
anonymous
  • anonymous
Why?
anonymous
  • anonymous
w is the weight the total weight have to be less or equal
anonymous
  • anonymous
r no more than 12
anonymous
  • anonymous
r no more than 12 comes from what statement?
anonymous
  • anonymous
If I tell you, there are 4 reams of paper in the box, what is the weight w?
anonymous
  • anonymous
the given r is the box can carry no more than 12 reams of paper
anonymous
  • anonymous
right.
anonymous
  • anonymous
so my r is less than 12 it's 4, what is w?
anonymous
  • anonymous
4.4
anonymous
  • anonymous
if the box itself weighs 1.3 and each ream weighs 4.4 and there are 4 reams, what is the total weight?
anonymous
  • anonymous
total; 18.9 w
anonymous
  • anonymous
You sure the weight isn't 1 ?
anonymous
  • anonymous
22.8w
anonymous
  • anonymous
I think 18.9 was correct actually.
anonymous
  • anonymous
But wait, why isn't the weight just 3.1 or 1.3 or 10
anonymous
  • anonymous
Aren't all those < 1.3 + 4.4r ?
anonymous
  • anonymous
w<1.3 +4.4r, r<12
anonymous
  • anonymous
Right. So if I tell you that r=1, what is the weight? It's 1.1111111111111 right?
anonymous
  • anonymous
because 1.111111 < 1.3 + 4.4
anonymous
  • anonymous
r=1? w<5.7
anonymous
  • 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?
anonymous
  • 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)
anonymous
  • 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.
anonymous
  • anonymous
and I said there was 1 ream, so the box can weigh anything less than 5.7
anonymous
  • anonymous
And you say "No dummy the box weighs 5.7"
anonymous
  • anonymous
;)
anonymous
  • anonymous
ok, thank you so much
anonymous
  • anonymous
Wait. What? Did you find the solution?
anonymous
  • anonymous
yes I think w<1.3 + 4.4 r , r<12
anonymous
  • anonymous
no.. that's what I'm trying to explain
anonymous
  • anonymous
Once we pick an r, we know exactly how much the box weighs. The restriction is on r not on the weight.
anonymous
  • 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
anonymous
  • anonymous
the key not reqite r
anonymous
  • anonymous
hrm? I didn't follow that. The answer is G.
anonymous
  • 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.
anonymous
  • anonymous
err w will be exactly 1.3 + 4.4r
anonymous
  • anonymous
Ok , you said w = 1.3+4.4r ,r<12
anonymous
  • anonymous
yes. Because once you pick how many reams in the box, the box has an exact wieght.
anonymous
  • 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.
anonymous
  • anonymous
OK, it help
anonymous
  • anonymous
I understand now
anonymous
  • anonymous
thank you
anonymous
  • anonymous
no problem
anonymous
  • anonymous
It late I go to the bed now bye

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