Answer:
The correct answer is the linear model would be y = 500x - 390 where x is the number of swords sold in a month and y is the net monthly profit; B. 4.96 ≈ 5 swords monthly.
Step-by-step explanation:
Let x number of swords are sold per month.
Cost price of the swords per month is $ 195x.
Fixed cost to maintain the website per month is $390.
Total cost incurred per month is $ (195x + 390).
Selling price per katana is $695.
Total selling price of x swords per month is $695x.
Therefore, Net monthly profit y =695x - (195x + 390)
⇒ y = 695x - 195x - 390
⇒ y = 500x - 390
Thus the linear model would look like y = 500x - 390 where x is the number of swords sold in a month and y is the net monthly profit.
B. Now, given monthly profit y = $2090.
Thus the number of swords needed to be sold is
2090 = 500x - 390
⇒ 2480 = 500x
⇒ x = 4.96
A minimum of 5 swords need to be sold to get a monthly profit of more than $2090.
19.36
Reminder I don't know if I'm right, if I'm wrong please tell me. <3
Answer
are we adding or subtracting ? 17 for adding , 13 for subtracting , 30 for times , 7.5 for divideding
Step-by-step explanation:
Yay, implicit differnentiation
when you take the derivitive of y, you multiply it by dy/dx
example
dy/dx y^2=2y dy/dx
for x, the dy/dx dissapears
ok
so differnetiate and solve for dy/dx
3y² dy/dx-(y+x dy/dx)=0
expand
3y² dy/dx-y-x dy/dx=0
3y² dy/dx-x dy/dx=y
dy/dx (3y²-x)=y
dy/dx=y/(3y²-x)
so at (7,2)
x=7 and y=2
dy/dx=2/(3(2)²-7)
dy/dx=2/(3(4)-7)
dy/dx=2/(12-7)
dy/dx=2/5
answer is 2/5
