Factor the trinomial completely.<br> 3n2

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in $R^{2}$ . To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

3) Finally, the last vector leads to the treasure from the Old Oak Tree, i.e. the resultant.

So, let's calculate the norm which is the length of the each vector.

1) Graphing them we can find the points, then the components and then calculate the norm, the length of each vector.

Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

$\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle$

Now, let's calculate each vector length by calculating the norm.

$\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86$

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.

8 0

3 years ago

A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per

HACTEHA [7]

Answer:

A) Revenue function = R(x) = (580x - 10x²)

Marginal Revenue function = (580 - 20x)

B) Fixed Cost = 900

Marginal Cost function = (300 + 50x)

C) Profit function = P(x) = (-35x² + 280x - 900)

D) The quantity that maximizes profit = 4

Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

8 0

2 years ago

Factor the trinomial completely. 3n2 - 48n + 192