The system of equations can be used to determine how much of product x and product y the store owner bought is x + y = 4,000
0.10x + 0.04y = 352
<h3>Simultaneous equation</h3>
- product x
- Product y
- Total units of x and y = 4,000 units
- Cost of shipping each product x = $0.10
- Cost of shipping each product y = $0.04
- Total cost of shipping = $352
The equation:
x + y = 4,000
0.10x + 0.04y = 352
Therefore, the system of equations can be used to determine how much of product x and product y the store owner bought is x + y = 4,000
0.10x + 0.04y = 352
Learn more about simultaneous equation:
brainly.com/question/16863577
#SPJ4
It would show the average time spent on the homework.
And the average distance between the mean and the lowest and highest numbers in the data.
What we are going to do is use the DISTRIBUTIVE PROPERTY. This distributive the numbers, to make it easier to solve.
So we will break 64 up.
64 = 60 + 4
43 = 40 + 3
60 + 40 = 100
4 + 3 = 7
100 + 7 = 107
So, 64 + 43 = 107.
Hope I helped ya!!!!!!!
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
