Answer:  x>15
Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:
let cp be x
sp= 115x/100 ---(when 15%gain)
sp = 88x/100------(when 12% loss
according to question
115x/100 - 88x/100 = 81
or, (115x-88x)= 8100
or, 27x = 8100
or, x= 8100/27
x = 300
hence cp is RS 300
 
        
             
        
        
        
The total surface area of the remaining solid is 48(4+✓3) centimeters square.
<h3>How to calculate the surface area?</h3>
Through a regular hexagonal prism whose base edge is 8 cm and the height is 12 cm, a hole in the shape of a right prism.
The formula for the total surface area will be:
= Total surface area=2(area of the base)+ parameter of base × height
where,
Height= 8cm
Parameter of base=12(2) = 24
Area of the base= 6×✓3/4×4² = 64✓3/4
The surface area of the remaining solid will be:
= 2(64✓3/4) + 24 × 8
= 2(64✓3/4 + 192
With the hole is a rhombus prism with the following parameters:
diagonal 1 = 6, diagonal 2 = 8, height = 12 
The volume is: 
V1 =0.5 × d1 × d2 × h
V1 = 0.5 × 6 × 8 × 12
V1 = 96
The dimensions of the hexagonal prism are:
Base edge (a) = 8
Height (h) = 12
The volume is
V2 = (3✓(3)/2)a²h
V2 = (3✓3)/2) × 8² × 12
V2 = 1152✓3
The remaining volume is
V = V2 -V1
V = 1152✓3 - 96
Learn more about the hexagonal prism on:
brainly.com/question/27127032
#SPJ4
 
        
             
        
        
        
Answer:
We just need to evaluate and get f(2i)=0, f(-2i)=0.
Step-by-step explanation:
Since  , then
, then  , and we can apply this when we evaluate
, and we can apply this when we evaluate  for 2i and -2i.
 for 2i and -2i.
First we have:

Which shows that 2i is a zero of f(x).
Then we have:

Which shows that -2i is a zero of f(x).
 
        
             
        
        
        
<h3>
♫ - - - - - - - - - - - - - - - ~<u>
Hello There</u>
!~ - - - - - - - - - - - - - - - ♫</h3>
➷ final = original x multiplier^n
n is the number of years
Substitute the values in:
final = 3800 x 1.054^3
final = 4449.440763
The closest answer is A
<h3><u>
✽</u></h3>
➶ Hope This Helps You!
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