Answer:
jvjjvscsjcsjsvjscjsvhchhcuhcjwcwh ejcisvsufjdtsydghcwhcjwvhsvjscekfdjsvjsfjsvkdvs
Step-by-step explanation:
30y
 
        
             
        
        
        
Let x be the cost of 1 pen
then cost of 1 notebook = x + 8.20
Let y be the number of pens Tan buys
then number of notebooks Tan buys = y/4
She spent $26 more on books than on pens which means
Cost of notebooks - Cost of pens = 26
(x + 8.20) * y/4 - xy = 26
Sinplifying it
(xy + 8.20y)/4 - xy = 26
(xy + 8.20y - 4xy)/4 = 26
8.20y - 3xy = 104
She spent $394 which means
Cost of notebooks + Cost of pens = 394
(x + 8.20) * y/4 + xy = 394
Simplifying it
(xy + 8.20y)/4 + xy = 394
(xy + 8.20y + 4xy)/4 = 394
8.20y + 5xy = 1576
Now, we have two equations,
(1) 8.20y - 3xy = 104
(2) 8.20y + 5xy = 1576
Now we need to find a third equation with either x or y as the subject of any of both the previous equations.
Let's make y the subject of (2) equation
8.20y + 5xy = 1576
y(8.20 + 5X) = 1576
(3) y = 1576/(8.20 + 5x)
Let's substitute the new value of y from (3) into (1) because we rearranged (2) to from (3)
8.20y - 3xy = 104
y(8.20 - 3x) = 104
y = 104/(8.20 - 3x)
1576/(8.20 + 5x) = 104/(8.20 - 3x)
1576 * (8.20 - 3x) = 104 * (8.20 + 5x)
12923.2 - 4728x = 852.8 + 520x
12923.2 - 852.8 = 4728x + 520x
12070.4 = 5248x
12070.4/5248 = x
x = 2.3
Now find the value of y by substituting the value of x in either equation, preferably (3)
y = 1576/(8.20 + 5x)
y = 1576/(8.20 + 5 * (2.3))
y = 80
Therefore cost of 1 notebook = x + 8.20 = 2.3 + 8.20 = $10.50
        
             
        
        
        
Answer: 6 for $7.50
Step-by-step explanation:
6 * 1.30 = $7.80
6 for $7.50
 
        
             
        
        
        
Area of the square base = 5^2 = 25 cm^2
Area of one of the triangular side = 1/2 * 5 * 8 = 20 cm^2 and there are 4 sides
So the surface area  of while pyramid = 4 * 20 + 25 = 105 cm^2
Volume of the cone = (1/3) pi r^2 h
  = (1/3) pi * 5^2 * 8
  =  209.44  ft^2
        
             
        
        
        
The inclusion/exclusion principle states that

That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).
Therefore, 

 will have the most elements when the sets 

 and 

 are disjoint, i.e. 

, which would mean the most we can can in this case would be

(Note that 

 denotes the cardinality of the set 

.)