Solving three mathematical problems 1. The relationship between the area s (CM & # 178;) and radius R (CM) of a circle is s = π R & # 178;. First, we need to make a circular part with an area of 49cm & # 178;, what is the radius of the part? 2. It is known that x, y and Z satisfy 2Y + Z + (Z-1 / 2) &# 178; = 0 under | 4x-4y + 1 | + 1 / 5 radical The part under the root sign is only 2Y + Z 3. The area of the small meeting room of the school is 27m. Xiao Ming counted the tiles on the ground, which are exactly 300 squares of the same size. Do you know the side length of the tiles in the small meeting room? There must be a process!

Solving three mathematical problems 1. The relationship between the area s (CM & # 178;) and radius R (CM) of a circle is s = π R & # 178;. First, we need to make a circular part with an area of 49cm & # 178;, what is the radius of the part? 2. It is known that x, y and Z satisfy 2Y + Z + (Z-1 / 2) &# 178; = 0 under | 4x-4y + 1 | + 1 / 5 radical The part under the root sign is only 2Y + Z 3. The area of the small meeting room of the school is 27m. Xiao Ming counted the tiles on the ground, which are exactly 300 squares of the same size. Do you know the side length of the tiles in the small meeting room? There must be a process!

one
49=πR²
R²=49/π
R²=49/π
R²=15.6
R = 3.95 (CM)
two
From the meaning of the title, there are
4x-4y+1=0
2y+z=0
z-1/2=0
therefore
z=1/2,y=-z/2=-1/4,x=y-1/4=-1/2
three
The area of each small square = 27 / 300 = 0.09m & # 178;
So the side length of small square = root 0.09 = 0.3m
Hope to help you
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3. A passenger car and a freight car leave from Party A and Party B at the same time, and meet at the place 20 kilometers away from the midpoint. When they meet, the distance ratio of the passenger car and the freight car is 5:3. How many kilometers is the distance between Party A and Party B?
Add two questions
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