To solve the expression (3.5 - 0.03 \times 0.71), first calculate (0.03 \times 0.71), which equals (0.0213). Then, subtract this value from (3.5):
[3.5 - 0.0213 = 3.4787].
Thus, the solution is approximately (3.4787).
35% of 60= 35% * 60= 0.35 * 60= 21
The answer is 35%. Solution: 70/200x100 =0.35x100 =35%
x = 10% solution y = 35% solution .1x+.35y=.20*12 x+y=12 y=12-x .1x+.35(12-x)=.20*12 .1x+4.2-.35x=2.4 -.25x=-1.8 .25x=1.8 x=7.2 gallons y=4.8 gallons
71/1000
8 + x = .35(26.66 +x) 8+x =9.33 +.35x x-.35x=9.33-8 .65x =1.333 x=2.05 kg of salt This should be checked out. By adding 2.05 kg of salt the total salt is now 8+2.05=10.05 kg salt and the total solution is now 26.666 kg + 2.05 kg or a total of 28.7166, but is it a 35% solution.........lets see (.35)(28.7166)=10.05kg
The 10 sugar solution has a lower concentration of sugar compared to the 35 sugar solution. This means that in the 10 solution, there are fewer sugar molecules per unit volume than in the 35 solution. As a result, the 10 sugar solution is less sweet and has a lower osmotic pressure than the 35 sugar solution.
35% of 60= 35% * 60= 0.35 * 60= 21
35÷59.85 Solution:- 35/59.85 0.585
The answer is 35%. Solution: 70/200x100 =0.35x100 =35%
If 6x-7=35 then 6x=35+7=42 and x=7
a = 16
35% off Merchant Solution Packages
n=30, a value under 35.
105/6 = 35/2
- freezing point for a solution of 35 g/L NaCl: -2 deg. Celsius- density for a solution of 35 g/L NaCl: 1,025 g/cm3- thermal conductivity for a solution of 35 g/L NaCl: 0,6 W/m.K
This is an unsaturated solution.
x=5