C*h=area/2
type i am going to say acute triangle
Answer:
I am not sure but for
CAE it is 65˚
CBD it is 65˚
Step-by-step explanation:
x+40=3x-10
Move the 3x to the other side by subtracting by 3x on both sides
x-3x+40=3x-3x-10
The equation now looks like this:
-2x+40=-10
Move 40 to the other side by subtracting 40 both sides:
-2x+40-40=-10-40
-2x=-50
Divide by -2 on both sides:
-2x/-2=-50/-2
x=25
Since we found out what x is we can replace x in both CAE and CBD:
For CBD: 25+40 is 65
For CAE: 3(25)-10 is 65
The volume of a sphere is calculated like so:
v = 4*pi*r^3/3
so half sphere will have half that volume:
v = (<span>4*pi*r^3/3)(1/2)
</span>v = <span>4*pi*r^3/6
</span>plug in the data:
v = <span>4*pi*(10)^3/6
</span>v = 2094.4
that is the volume of the half sphere
Answer: u=7
Step-by-step explanation:
-6= -2u+4(u-5)
-6= -2u+4u-20
-6= 2u-20
14= 2u
7=u
Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
