Answer:
The area of the Iris is 4 times greater.
Step-by-step explanation:
Firstly, we have to realize that we cannot solve this problem without knowing the radius of the iris. because from the information given, the radius of the iris could well be the size of the galaxy and contract from width of 4mm to 2mm and we wouldn't know!
The average radius of the iris is 6mm, so we take this value.
Now, initially the width of the iris is 4mm, that means the radius of the pupil is:
![6mm-4mm=2mm](https://tex.z-dn.net/?f=6mm-4mm%3D2mm)
Therefore it's area
is:
![A=\pi r^2=\pi (2)^2=4\pi](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D%5Cpi%20%282%29%5E2%3D4%5Cpi)
When the iris contracts to 2mm, the radius of the pupil becomes:
![6mm-2mm=4mm](https://tex.z-dn.net/?f=6mm-2mm%3D4mm)
Then it's area is
:
![A=\pi r^2=\pi (4)^2=16\pi](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D%5Cpi%20%284%29%5E2%3D16%5Cpi)
To find how many times greater this final area is than the initial area, we just divide it by the initial area:
![\frac{16\pi }{4\pi } =4](https://tex.z-dn.net/?f=%5Cfrac%7B16%5Cpi%20%7D%7B4%5Cpi%20%7D%20%3D4)
This is 4 times greater than the initial area.
<span>The contradiction term for "at most one obtuse angle" is "no obtuse angles
so therefore i conclude that the </span> statement she will most likely use as an assumption is
Let each angle of a triangle be acute
so option A is correct
hope it helps
List the coordinates,
Coordinates are shown as (x,y)
If 2+2+7+7 is 16
You make those your side lengths
Coordinates are (9,7) (2,7) (2,5) (9,5)
they maybe slightly off because of the broken peice of paper
Answer:5 by 8 by 5 by 8
Step-by-step explanation: the perimeter is 26 and one side if 5 so you know that another side is also 5. So add 5 and 5 together you get 10. Then subtract 10 from the perimeter(witch we know is 26) and you end up with 16. we know that the last two side lengths are the same so divide 16 by two and the sidle length is 8.