Answer:
-2\7 I think
Step-by-step explanation:
mafffffffffgfgggggfffs
Answer:
A, C, D and K
Step-by-step explanation:
First, plot the solution set to the system of two inequalities.
The inequality
determines solid line and bottom shaded region under this line.
The inequality
determines dotted line and top shaded region under this line.
The intersection of these two regions is the solution set to the system of two inequalities.
Points A, C, D and K belong to this region.
Answer:
![A=81(\pi-1)\ in^2](https://tex.z-dn.net/?f=A%3D81%28%5Cpi-1%29%5C%20in%5E2)
Step-by-step explanation:
step 1
Find the area of the plate
The area of a circle is given by the formula
![A=\pi r^{2}](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E%7B2%7D)
we have
---> the radius is half the diameter
substitute
![A=\pi (9)^{2}\\A=81\pi\ in^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20%289%29%5E%7B2%7D%5C%5CA%3D81%5Cpi%5C%20in%5E2)
step 2
Find the area of the square napkin folded (is a half of the area of the square napkin)
we know that
The diagonal of the square is the same that the diameter of the plate
Applying Pythagorean theorem
![D^2=2b^2](https://tex.z-dn.net/?f=D%5E2%3D2b%5E2)
where
b is the length side of the square
we have
![D=18\ in](https://tex.z-dn.net/?f=D%3D18%5C%20in)
substitute
![18^2=2b^2](https://tex.z-dn.net/?f=18%5E2%3D2b%5E2)
solve for b^2
-----> is the area of the square
Divide by 2
![162/2=81\ in^2](https://tex.z-dn.net/?f=162%2F2%3D81%5C%20in%5E2)
step 3
Find the area of the space on the plate that is NOT covered by the napkin
we know that
The area of the space on the plate that is NOT covered by the napkin, is equal to subtract the area of the square napkin folded (is a half of the area of the square napkin) from the area of the plate
so
![A=(81\pi-81)\ in^2](https://tex.z-dn.net/?f=A%3D%2881%5Cpi-81%29%5C%20in%5E2)
simplify
![A=81(\pi-1)\ in^2](https://tex.z-dn.net/?f=A%3D81%28%5Cpi-1%29%5C%20in%5E2)
Answer:
62>(9*116)
Step-by-step explanation: