We know that
[area of a regular hexagon]=6*[area of one <span>equilateral triangle]
</span>210.44=6*[area of one equilateral triangle]
[area of one equilateral triangle]=210.44/6-----> 35.07 cm²
[area of one equilateral triangle]=b*h/2
h=7.794 cm
b=2*area/h------> b=2*35.07/7.794------>b= 9 cm
the length side of a regular hexagon is 9 cm
<span>applying the Pythagorean theorem
</span>r²=h²+(b/2)²------>r²=7.794²+(4.5)²------> r²=81--------> r=9 cm
<span>this last step was not necessary because the radius is equal to the hexagon side------> (remember the equilateral triangles)
</span>
the answer is
the radius is 9 cm
1x24, 2x12, 3x8 are 3 ways that get 24
The domain is t=0 to t=3.2 and range is y = 0 to 57.125 feet.
<u>What are Domain and range of a Function?</u>
- The range of values that we are permitted to enter into our function is known as the domain of a function.
- The x values for a function like f make up this set (x). A function's range is the collection of values it can take as input.
- After we enter an x value, the function outputs this sequence of values. The y values are those.
- A function can be compared to a piece of equipment on an assembly line.
- A few screws and nuts are located on one end of the assembly line, while a car is located on the other. The function is the device in the center. The domain is the screws and bolts that we insert into the machine. The car at the other can be thought of as the range.
Given function is : y = -16t^2 + 50t + 4
It hits the ground when y=0
0 = -16t^2 + 50t + 4
8t^2 -25t -2 = 0
t = 25/16 + (1/16)sqrt(625+64) = (25+26.25)/16 = 51.25/16 = 3.2 seconds
domain is t=0 to t=3.2 [3.2]
take the derivative and set it equal to zero to find maximum height
-32t +50 = 0
t = 50/32 = 25/16 = 1.5625 seconds
-16(25/16) + 50(25/16) + 4 = 34(25/16) + 64/16 = (17(25)+32)/8 = 57.125 = maximum height
range is y = 0 to 57.125 feet. the discus goes from 4 feet to 57 1/8 feet and back down to 0 feet.
Know more about Domain and Range with the help of the given link:
brainly.com/question/20349151
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