Answer:
99.38
Step-by-step explanation:
the area of all the circles by using the formula A=ℼr2 then filling in r with 3.2 A=ℼ3.22=32.2
Knowing that one circle is 32.2 4 circles would have an area of 128.8
Two circles have a combined length and height of 12.8 because the circumference of one circle is 6.4 so two circles reach the sides of the square. So all the sides of the square are 12.8. So finding the area of the square knowing what the sides are, so the area is a=163.84. To find the shaded area you subtract the volume of the square to volume of all 4 circles which is 163.84-128.8 which equals 35.04. Then you find the volume of the semi circle which is 64.34. Then the combination of both shaded regions is 99.38
Answer:
hmmm I get 216
Step-by-step explanation:
(10×7)+(6×8) the front n back triangle +(6×7)+(8×7)
Answer:
1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle
Step-by-step explanation:
Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is
V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³
the mass in that volume would be m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)
The density of an alpha particle is ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be
ρ= m/V
since both should be equal ρ=ρa , then
ρa= m/V =N*L/V → N =ρa*V/L
replacing values
N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg = 1.544*10⁹ Linebackers
N=1.544*10⁹ Linebackers
Answer:
5 = 5
5 < 6
7 > 6
Step-by-step explanation:
The two slashes is the equal sign, comparing two numbers that are of the same value. The sign below that means "less than," and 5 is less than 6. The sign below the "less than" sign is the "greater than" sign, and in the given example, 7 is GREATER THAN 6.
Answer:
Note that both relations and functions have domains and ranges. The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements "used" by the relation or function constitute the range.
