100% = 88.57 60% : 55.14 = 100% : n 60%n/60% = (53.14 * 100%)/60% n = 88.57
Let the number be N, then (N/100) x 7 = 60 or (7N/100) = 60 or 7N = 6000 or N = 6000/7 = 857.1428571
n = 6050 = 100 Using the criss cross multiplication trick we get 50X60=3000 and 100 X n=100n 3000 = 100n 100 = 100 So 3000÷100=30 and 100n÷100=n. n=30 or 60% of 50 is 30
10% of 60.0 is 6 (just move the decimal point one place to the left)60% = 6x10% = 6x6 = 36%Better still:6/10 of 60 [or 3/5 of 60] can be expressed as 0.6 x 60.00 = 36.00.The double word typo led original anwer to misread the last figure, 60, as 60%.
Suppose the length and width are L and W Perimeter = 2(L+W) = 200 so L+W = 100 so L= 100-W Area = L*W = 2400 so (100-W)*W = 2400 W2 - 100W + 2400 = 0 W = (100 +/- sqrt(10000 - 4*2400))/2 where the minus sign will give W and the plus sign will give L So W = (100 - sqrt(400))/2 = (100 - 20)/2 = 80/2 = 40 feet So L = 100-40 = 60 feet
60 N 100 W is on the Canadian border between Manitoba and Nunavut.
60 N 100 W is on the Canadian border between Manitoba and Nunavut.
60 N 100 W is on the Canadian border between Manitoba and Nunavut.
South America
100% = 88.57 60% : 55.14 = 100% : n 60%n/60% = (53.14 * 100%)/60% n = 88.57
266.67 160 : 60% = n : 100% 60%n/60% = (160 * 100%)/60% n = 1600/6 n = 266.67
60 N 100 W is Nueltin Lake, Canada.So the continent is North America.
I believe your question is : Can you use a 100 W rating LED in a 60 W power source (meant for lighting) If so, the answer is NO. Since 60 W power source can provide only 60 W. However if power source is 100 w capacity, you can use LED or any type lamp that has 100 W or less.
60 N, 55 W is in the water between Greenland and Newfoundland. 55 N, 60 W is in Canada.
Cape farewell is the close (59.7792° N, 43.9117° W)
100 n 25 w 400 w 100 w
Let the number be N, then (N/100) x 7 = 60 or (7N/100) = 60 or 7N = 6000 or N = 6000/7 = 857.1428571