12/18
6/9
2/3
Simply in common factor you know and keep doing it
12/18
BCD:6
2/3
Answer :
no. of days used to repair 1250m of road = 5
no. of days used to repair 1m of road = 5/1250
no. of days used to repair 8000m of road (8*1000m ) = 5/1250*8000= 32 days
If m ACD = 30 => m DCB = 60.
In triangle ACD:
![sin 30^{o}=sin \frac{AD}{AC} =\ \textgreater \ \frac{1}{2} = \frac{8}{AC} =\ \textgreater \ AC=16 ](https://tex.z-dn.net/?f=sin%2030%5E%7Bo%7D%3Dsin%20%20%5Cfrac%7BAD%7D%7BAC%7D%20%3D%5C%20%5Ctextgreater%20%5C%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%3D%20%20%5Cfrac%7B8%7D%7BAC%7D%20%3D%5C%20%5Ctextgreater%20%5C%20%20AC%3D16%0A)
![CD= 8 \sqrt{3}](https://tex.z-dn.net/?f=CD%3D%208%20%5Csqrt%7B3%7D%20)
![BC= 16\sqrt{3}](https://tex.z-dn.net/?f=BC%3D%2016%5Csqrt%7B3%7D%20)
AC^{2}+CB^{2}=AB^{2} => AB=
![\sqrt{1024}](https://tex.z-dn.net/?f=%20%5Csqrt%7B1024%7D%20)
=32.
=> P=16+32+
![16 \sqrt{3}](https://tex.z-dn.net/?f=16%20%5Csqrt%7B3%7D%20)
=48 +
![16 \sqrt{3}](https://tex.z-dn.net/?f=16%20%5Csqrt%7B3%7D%20)
.
1 meter is 100 centimeters so do 200÷10 what does that leavr you with ? 20
Answer: 6.40 = c
Step-by-step explanation:
a^2 + b^2 = c^2
5^2 + 4^2 = c^2
25 + 16 = c^2
41 = c^2
____
_/ 41 = c
6.40 = c