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Bess [88]
2 years ago
8

ppositional angle closure in eyes with narrow angles: comparison between the fellow eyes of acute angle-closure glaucoma and nor

motensive case
Mathematics
1 answer:
skad [1K]2 years ago
8 0

Here is the comparision

Purpose:To compare the topologic features of acute primary angle-closure glaucoma eyes before an attack to those of normotensive eyes, assuming that untreated fellow acute primary angle-closure glaucoma eyes are candidates for an acute attack.

Methods:Under dark-room conditions, ultrasound biomicroscopy was used to examine 50 eyes (12 fellow eyes of acute primary angle-closure glaucoma and 38 normotensive cases with a closure-possible narrow angle). Before any surgical or laser intervention, all eyes were examined and found to have normal pupillary response without the use of any topical drugs. Each eye was examined at four predetermined angle locations. The chamber angle configuration parameters were measured and compared between the two groups.

Result:Appositional angle closures were detected in 27 fellow eyes and 48 normotensive eyes with a closure-possible narrow angle. The incidence differed statistically between the two groups (69.2% in fellow eyes and 48% in normotensive eyes). In the fellow eye group, appositional angle closures beginning at the angle's entrance were more frequently detected. The distance between the iris root and the bottom of the angle varied significantly between groups.

Conclusion:Acute primary angle-closure glaucoma fellow eyes have different topologic features than normotensive narrow-angled eyes, as well as a higher incidence of appositional closure, which may predispose these eyes to an impending acute attack.

Learn more about glaucoma here:

brainly.com/question/1318395

#SPJ4

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Where to put the ( )’s 5+4+3 2=29
Salsk061 [2.6K]

Answer:

It all depends.

Step-by-step explanation:

It depends on what you are trying to do, and what you want to solve for. If you are trying to distribute the 3 into 2, you would place the parentheses around the 2. I hope I was able to help, but there is not much that I can do without more clear instructions!

8 0
3 years ago
I have 31 stamps in total. I have 4 more 1 cent stamps than 8 cent stamps and twice as many one cent stamps as 12 cent stamps. I
Neko [114]

Answer:

There area 14 one cent stamps

Step-by-step explanation:

Let

x -----> the number of 1 cent stamps

y -----> the number of 8 cent stamps

z -----> the number of 12 cent stamps

we know that

0.01x+0.08y+0.12z=1.78 -------> equation A

x=y+4 ------> y=x-4 -----> equation B

x=2z -----> z=0.5x ------> equation C

substitute equation B and equation C in equation A and solve for x

0.01x+0.08(x-4)+0.12(0.5x)=1.78

0.01x+0.08x-0.32+0.06x=1.78

0.15x=1.78 +0.32

0.15x=2.10

x=14

therefore

There area 14 one cent stamps

6 0
3 years ago
Miguel deposited $1,200 into an account that pays simple interest at an annual rate of 4%. What is the value of his initial depo
Angelina_Jolie [31]

Answer:

end of third year = 1,200*(1.02)^3  = 1,273.45 plus

end of second year = 1,200*(1.02)^2 = 1,248.48  plus

end of first year = 1,200*(1.02) = 1,224.00

Total = 3,745.9

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I don’t understand please help?
Lerok [7]

Answer:

D

Step-by-step explanation:

Cuz sqr root of 20 = 2 sqrt 5, -2 sqrt 5

7 0
3 years ago
Elena thinks length BC is 16.5 units. Lin thinks the length of BC is 17.1 units. Do you agree with either of them? Explain or sh
kirill [66]

Answer:

Option D, I agree with neither Elena nor Lin

Step-by-step explanation:

The remaining part of the question is attached herewith

Solution

As we can see in the image  

ED is perpendicular to AB and BC is perpendicular to AB

Thus, as per the Pythagoras theorem,  

AC^2 = AB^2 + BC^2  

14^2 = 6^2 + BC^2

BC^2 = 196-36 = 160

BC = 12.649 OR 12.65

Hence, option D

I agree with neither Elena nor Lin

5 0
3 years ago
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