Answer:
can you include the graph so we can know what you are talking about?
Step-by-step explanation:
Arc SCD is two times of angle SED, so arc SCD is 8x+22. Now set 8x+22, 5x-8, and 11x+10 to 360 and solve for x
Answer:
C
Step-by-step explanation:
![7^{\tfrac 14} = \sqrt[4]{7}](https://tex.z-dn.net/?f=7%5E%7B%5Ctfrac%2014%7D%20%3D%20%5Csqrt%5B4%5D%7B7%7D)
1) We calculate the volume of a metal bar (without the hole).
volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²
Volume=(0.935 m²)(2 m)=1.871 m³
2) we calculate the volume of the parallelepiped
Volume of a parallelepiped= area of the section x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³
3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar - volume of a parallelepiped
Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³
4) we calculate the mass of the metal bar
density=mass/ volume ⇒ mass=density *volume
Data:
density=8.10³ kg/m³
volume=1.551 m³
mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg
answer: The mas of the metal bar is 12.408x10³ kg or 12408 kg
