Answer:
he area that is un-shaded is (x-3)(x-6).
The entire area is (2x+2)(3x-4).
Area of the shaded region is:
(2x+2)(3x-4) - (x-3)(x-6) =
(6x2-2x-8) - (x2-9x+18) =
6x2-2x-8-x2+9x-18 =
5x2+7x-26 hopes this help btw (;
Answer: -10
Step-by-step explanation:
Simplifying
-3(y + 5) = 15
Reorder the terms:
-3(5 + y) = 15
(5 * -3 + y * -3) = 15
(-15 + -3y) = 15
Solving
-15 + -3y = 15
Solving for variable 'y'.
Move all terms containing y to the left, all other terms to the right.
Add '15' to each side of the equation.
-15 + 15 + -3y = 15 + 15
Combine like terms: -15 + 15 = 0
0 + -3y = 15 + 15
-3y = 15 + 15
Combine like terms: 15 + 15 = 30
-3y = 30
Divide each side by '-3'.
y = -10
Simplifying
Answer:
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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
