Answer:
 x= 1 , y=4
Step-by-step explanation:
 x-y= -3 => Equation 1
 x+5y= 21 => Equation 2
<u>Substitut</u><u>ion</u><u> </u><u>Method</u><u>:</u>
<u>Substitu</u><u>te</u><u> </u><u>Equation</u><u> </u><u>1</u>=> 
x=y-3 <= Equation 3
Put x=y-3 in Equation 2:
 x+5y=21
( y-3)+5y=21
 y-3+5y=21
 6y-3=21
 6y=21+3
 6y=24
 y=24÷6
 y=4
Put y=4 in Equation 1:
 x-y= -3
 x-4=-3
 x=4-3
 x=1
Hope this helps :)
 
        
                    
             
        
        
        
Hello from MrBillDoesMath!
Answer:
2(x+8) = 40
Discussion:
Rewrite the original Question as
"
Twice the (sum of a number and 8) is 40"   =>
2                ( x                              +    8 ) = 40 =>
2 (x+8) = 40
which is the third bullet point from the top
Thank you,
MrB
 
        
             
        
        
        
To find your constant of variation, you just need to figure out what you multiply the x by in order to get f(x). In other words, what do you multiply the 3 by to get 6? What do you multiply the 7 by to get 14? Another way to think of it is to divide f(x) by the x to get your constant. 
        
             
        
        
        
Answer:
<em>15c + 25s ≤ 1500</em>
Step-by-step explanation:
Let the number of stools be s
Let the number of chairs bought be c
If each tools cost $15, s stools will cost $15c
If a chairs cost $25, c chairs will cost $25s
Total costs = 15c + 25s
If the center cannot spend more than $1500, then the required equation will be 15c + 25s ≤ 1500
<em>The inequality sign will be less than and equal to since they cannot spend an amount greater than $1500.</em>
<em>Hence the correct equation is  15c + 25s ≤ 1500</em>
 
        
             
        
        
        
Answer:
f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)
Step-by-step explanation:
f(x) = eˣ
Expand using first Taylor Polynomial based around b = 0
The Taylor's expansion based around any point b, is given by the infinite series
f(x) = f(b) + xf'(b) + (x²/2!)f"(b) + (x³/3!)f'''(b) + ....= Σ (xⁿfⁿ(b)/n!) (Summation from n = 0 to n = ∞)
Note: f'(x) = (df/dx)
So, expanding f(x) = eˣ based at b=0
f'(x) = eˣ
f"(x) = eˣ
fⁿ(x) = eˣ
And e⁰ = 1
f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)