Before you can construct a ratio, both numbers must have
either the same units, or else NO units.
1 yard = 3 feet
12 yards = 36 feet
So the ratio is (99 feet) / (36 feet) = 11 / 4 .
Answer: OPTION D. ![130\°](https://tex.z-dn.net/?f=130%5C%C2%B0)
Step-by-step explanation:
The missing figure is attached.
As you can observe in the figure AC and BD intersect each other at the point O. Then, you can identify that:
![\angle AOB =\angle COD\\\\\angle AOD =\angle BOC](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%5Cangle%20COD%5C%5C%5C%5C%5Cangle%20AOD%20%3D%5Cangle%20BOC)
Subsitute values:
![\angle AOB =\angle COD\\\\3x - 70=x + 10](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%5Cangle%20COD%5C%5C%5C%5C3x%20-%2070%3Dx%20%2B%2010)
Solve for "x":
![3x-x=70 + 10\\\\2x=80\\\\x=40](https://tex.z-dn.net/?f=3x-x%3D70%20%2B%2010%5C%5C%5C%5C2x%3D80%5C%5C%5C%5Cx%3D40)
Therefore:
![\angle AOB =arc\ AB=3(40) - 70=50\°\\\\\angle COD=arc\ CD=40 + 10 =50\°](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3Darc%5C%20AB%3D3%2840%29%20-%2070%3D50%5C%C2%B0%5C%5C%5C%5C%5Cangle%20COD%3Darc%5C%20CD%3D40%20%2B%2010%20%3D50%5C%C2%B0)
They are 360 degrees in a circle. Then:
![arc\ AB+arc\ CD+arc\ BC+arc\ AD=360\°](https://tex.z-dn.net/?f=arc%5C%20AB%2Barc%5C%20CD%2Barc%5C%20BC%2Barc%5C%20AD%3D360%5C%C2%B0)
Substituting values and solving for arc BC, you get:
![50\°+50\°+arc\ BC+arc\ BC=360\°\\\\arc\ BC=\frac{260\°}{2}\\\\arc\ BC=130\°](https://tex.z-dn.net/?f=50%5C%C2%B0%2B50%5C%C2%B0%2Barc%5C%20BC%2Barc%5C%20BC%3D360%5C%C2%B0%5C%5C%5C%5Carc%5C%20BC%3D%5Cfrac%7B260%5C%C2%B0%7D%7B2%7D%5C%5C%5C%5Carc%5C%20BC%3D130%5C%C2%B0)
Answer:
88
Step-by-step explanation:
6 + 5 = 11
8 ÷ 2 = 4
11 · 4 = 44
44 · 2 = 88
Answer:
3
Step-by-step explanation:
Answer:
1165.73 sq.cm. of newspaper are needed to cover the lateral areas of all five sculptures
Step-by-step explanation:
Diameter of 1 sculpture = 9 cm
Radius of 1 sculpture =![\frac{9}{2}=4.5 cm](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B2%7D%3D4.5%20cm)
Slant height of sculpture = l = 16.5 cm
Lateral surface area of each sculpture = ![\pi r l = \frac{22}{7} \times 4.5 \times 16.5 =233.3571 cm^2](https://tex.z-dn.net/?f=%5Cpi%20r%20l%20%3D%20%5Cfrac%7B22%7D%7B7%7D%20%5Ctimes%204.5%20%5Ctimes%2016.5%20%3D233.3571%20cm%5E2)
Lateral surface area of 5 sculptures =![233.3571 \times 5 =1165.73 cm^2](https://tex.z-dn.net/?f=233.3571%20%5Ctimes%205%20%3D1165.73%20cm%5E2)
Hence 1165.73 sq.cm. of newspaper are needed to cover the lateral areas of all five sculptures