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Bess [88]
1 year 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]1 year 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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harina [27]
32 is the answer  I know this really well 
5 0
3 years ago
Hello I'm stuck on this homework question and need help please thank you
Alexxx [7]

We need to find the surface area of the solid.

Since it's formed by cubes with edges of 1 meter, each square on the surface of the solid has an area equal to:

1\text{ meter }\times1\text{ meter }=1\text{ meter}^2

Thus, to find its surface area, we need to count the number of squares on its surface, and then multiply this number by 1 meter².

We can see that this solid has two equal latera surfaces (right and left). Each one of them has 17 squares.

Also, the number of squares on the horizontal surfaces is the same on the top and bottom of the solid. Each one of them has 10 squares.

And the vertical surfaces on the front and back of the solid have the same number of squares: 8 squares each.

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6 0
1 year ago
Please answer this question now
kotegsom [21]

Answer:

The midpoint is (8,9.5)

Step-by-step explanation:

The midpoint is found by averaging the x coordinates and the y coordinates

x = (x1+x2)/2     y = (y1+y2)/2

x = (13+3)/2 = 16/2 = 8

y = (5+14)/2 = 19/2 = 9.5

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3 0
3 years ago
Find an equation of the circle tangent to the lines x=1, x=9, y=0
Karolina [17]

Answer:

(x − 5)² + (y − 4)² = 4²

(x − 5)² + (y + 4)² = 4²

Step-by-step explanation:

x=1 and x=9 are vertical lines.  If both are tangent to the circle, then the circle has a diameter of 8, or a radius of 4, and the center of the circle is on the line x=5.

y=0 is the x-axis.  Since the circle is tangent to that, the center of the circle is either 4 units above the x-axis or 4 units below.

So two possible equations of the circle are:

(x − 5)² + (y − 4)² = 4²

(x − 5)² + (y + 4)² = 4²

Here's a graph: desmos.com/calculator/9e4lxx731u

5 0
3 years ago
Read 2 more answers
Show me how to <br>x+2/y-1=2
Romashka-Z-Leto [24]
X+2 / y-1 =2
to leave y form

x+2 / y-1 =2
x y-1          x y-1
x+2 = 2y-2
 +2          +2
x+4 = 2y
/2          /2
x/2+2=y
7 0
3 years ago
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