Answer:
the number of of litres of paint to the nearest litre for 1000 bars is 1 litre
Step-by-step explanation:
Given that:
1000 cylindrical metal bars have :
a length of 120 m
and a diameter of 0.2 m
The volume of the cylindrical bar can be calculated by the formula:

where ;
radius 

r = 0.1 m
the length of the cylindrical metal bars is equal to the height of the metal bar = 120 m
Thus the volume can now be calculated as follows:

V = 3.77 m³
So; from the given information.
One litre of paint will cover 4m³
and we want to find how many litres of paint will be used for the 1000 cylindrical bars.
If 1 litre = 4m³
Then x litre = 3.77 m³
4x = 1 × 3.77

x = 0.9425 litres
x = 1 litres to the nearest litres.
Thus ; the number of of litres of paint to the nearest litre for 1000 bars is 1 litre
Answer:
C) sin 55
Step-by-step explanation:
Answer: 7208 remainder 5
Step-by-step explanation:
Answer:
15+6+2=23
Step-by-step explanation:
![\sqrt[3]{\frac{9\sqrt{5}}{2\sqrt{3}}\cdot\frac{5\sqrt{2}}{8\sqrt{2}}}\\=\sqrt[3]{\frac{9\sqrt{5}}{2\sqrt{3}}\cdot\frac{5}{8}}\\=\sqrt[3]{\frac{45\sqrt{5}}{16\sqrt{3}}}\\=\sqrt[3]{\frac{45\sqrt{15}}{16}}\\=\sqrt[3]{\frac{3\cdot\sqrt{15^3}}{16}}\\=\frac{\sqrt[3]{3}\cdot\sqrt{15}}{2\cdot\sqrt[3]{2}}\\=\frac{\sqrt[3]{6}\cdot\sqrt{15}}{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B9%5Csqrt%7B5%7D%7D%7B2%5Csqrt%7B3%7D%7D%5Ccdot%5Cfrac%7B5%5Csqrt%7B2%7D%7D%7B8%5Csqrt%7B2%7D%7D%7D%5C%5C%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B9%5Csqrt%7B5%7D%7D%7B2%5Csqrt%7B3%7D%7D%5Ccdot%5Cfrac%7B5%7D%7B8%7D%7D%5C%5C%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B45%5Csqrt%7B5%7D%7D%7B16%5Csqrt%7B3%7D%7D%7D%5C%5C%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B45%5Csqrt%7B15%7D%7D%7B16%7D%7D%5C%5C%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%5Ccdot%5Csqrt%7B15%5E3%7D%7D%7B16%7D%7D%5C%5C%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B3%7D%5Ccdot%5Csqrt%7B15%7D%7D%7B2%5Ccdot%5Csqrt%5B3%5D%7B2%7D%7D%5C%5C%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B6%7D%5Ccdot%5Csqrt%7B15%7D%7D%7B2%7D)
we want it in the form of
, so a = 15, b=6, c=2
also this is the minimum possible value, as we cannot simplify it further
Use photomath because it is very helpful bro