The area of a regular hexagon is
... A = (3√3)/2×s² . . . . . where s is the side length
Of course, the 6-sided figure will have a side length that is 1/6 of the perimeter.
... s = (48 in)/6 = 8 in
... A = (3√3)/2×(8 in)² = 96√3 in² . . . area of your regular hexagon
<span>So we want to know how much ball bearings can be made with 5.24 cm^3 if one ball bearing has a diameter of 1 cm. We know that radius r=d/2=0.5cm So the volume V of one ball bearing is: V=(4/3)*pi*r^3 so V=(4/3)*3.14*(0.5)^3cm^3=0.524cm^3. Now we simply divide the volume of steel by the volume of the ball bearing: 5.24/0.524=10. So we can make 10 ball bearings from 5.24 cm^3 of steel</span>
Answer:
16
Step-by-step explanation:
Simply plug in 2 for a, then 6 for b and substitute
(2+6)2
Then do order of operations to simplify PEMDAS
First what's inside the parenthesis
(8)2
Then multiply
16.
Answer:
Step-by-step explanation:
This is a quartic equation with a positive coefficient for x^4 so it is shaped like an M xo ir rises to both the left and the right.
