Find the difference (3-2x+2x^2)-(4x-5+3x^2)

An apartment complex on Ferenginar with 250 units currently has 223 occupants. The current rent for a unit is 892 slips of Gold-

Triss [41]

Answer:

Step-by-step explanation:

Given data

Total units = 250

Current occupants = 223

Rent per unit = 892 slips of Gold-Pressed latinum

Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum

After increase in the rent, then the rent function becomes

Let us conside 'y' is increased in amount of rent

Then occupants left will be [223 - y]

Rent = [892 + 2y][223 - y] = R[y]

To maximize rent =

$\frac{dR}{dy}=0\\=2(223-y)-(892+2y)=0\\=446-2y-892-2y=0\\=-446-4y=0\\y=\frac{-446}{4}=-111.5$

Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.

Since there are only 250 units available;

$y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units$

Optimal rent - 838 slips of Gold-Pressed latinum

8 0

3 years ago

Four pets in “Among Us” cost $16. Mr. Maxine says, at that rate, he will be able to purchase all 10 pets for $35!!! Is he right?

Rus_ich [418]

Answer:

No he is not lol 16÷4=4, To buy 10, 10×4=40. 40-35=5 he needs 5 dollars more

7 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago

John works a fortnightly wage of $750.00. If his basic week is 40 hours. what is his: Weekly wage basic rate and basic fortnight

Vladimir [108]

Answer:

Basic wage rate = $9.375 per hour

Step-by-step explanation:

Given :

Fortnight wage = $750

Basic week, hours worked per week = 40 hours

Weekly basic wage rate :

Fortnight = 2 weeks

Hence, weekly wage = fortnight wage / 2 = 750 /2 = $375

Weekly basic wage rate = $375 / Number of hours worked per week = $375 / 40 = $9.375 per hour

4 0

2 years ago

Which of the following is not a valid probability? A. 1.3 B. 0 C. 0.4 D. 1

kozerog [31]

Answer:

A 1.3

Step-by-step explanation:

probabilities lie between 0 and 1 and cannot be more or less

7 0

2 years ago