What is the value of the discriminant for the quadratic equation 0 = 2x2 + x

The telephone company is planning to introduce two new types of executive communications systems that it hopes to sell to its la

cluponka [151]

Answer:

x = 31 hundred dolars and

y = 91/2 = 45.5 hundred dolars

Step-by-step explanation:

Given

R(x) = (40−8x+5y)*x + (50+9x−7y)*y

C(x) = (40−8x+5y)*10 + (50+9x−7y)*29

We can use the equation

P(x) = R(x) - C(x)

where

P(x) is the profit

R(x) is the revenue

and C(x) is the costs

In order to maximize the telephone company's profit, we apply

P'(x) = R(x)' - C(x)' = 0

⇒ R(x)' = ((40−8x+5y)*x + (50+9x−7y)*y)' = (40x-8x²+14xy+50y-7y²)'

⇒ C(x)' = ((40−8x+5y)*10 + (50+9x−7y)*29)' = (1850+181x-153y)'

⇒ P'(x) = -8x²-7y²-141x+203y+14xy-1850

The first-order partial derivatives of these functions are

Px(x,y) = -16x-141+14y

Py(x,y) = -14y+203+14x

Setting these equal to zero and solving we obtain:

-16x+14y-141 = 0

14x-14y+203=0

we get the solution

x = 31 and y = 91/2 = 45.5

Finally, the company should produce 3100 units of the first system, and 4550 units of the second system.

8 0

2 years ago

Estimate the sum of 196 and 482

k0ka [10]

The answer to this problem would be 700. if you need the sum that isnt estimated, its 678.

4 0

2 years ago

An open-top rectangular box is being constructed to hold a volume of 150 in3. The base of the box is made from a material costin

Oksanka [162]

Answer:

minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

Step-by-step explanation:

given data

volume = 150 in³

base material costing = 5 cents/in²

front cost = 10 cents/in²

remainder sides cost = 2 cents/in²

to find out

the dimensions that will minimize the cost

solution

we consider here length = x and breadth = y and height = z

and

area of base = xy

area of front = xz

and area of remaining side = xz + 2yz .....................1

so

cost of base will be = 5xy

cost of front = 10xz

cost of remaining side = 2 ( xz+ 2yz)

and

total cost will be

total cost TC = 5xy + 10xz + 2 ( xz+ 2yz)

total cost TC = 5xy + 10xz + 2xz+ 4yz

total cost TC = 5xy + 12xz + 4yz ....................2

and total volume will be = xyz

150 = xyz

z = $\frac{150}{xy}$ .......................3

now put z value in equation 2

total cost TC = 5xy + 12xz + 4yz

total cost TC = 5xy + 12x $\frac{150}{xy}$ + 4y $\frac{150}{xy}$

total cost TC = 5xy + $\frac{1800}{y}$ + $\frac{600}{x}$ ...........4

now differentiate TC w.r.t x and y

TC (x) = 5y - $\frac{600}{x^2}$

TC (x) = 5x - $\frac{1800}{y^2}$

now equating with 0 these

5y - $\frac{600}{x^2}$ = 0

x² = $\frac{120}{y}$

and

5x - $\frac{1800}{y^2}$ = 0

x = $\frac{360}{y^2}$

solve we get

y = 10.2598

x = 3.4199

now put x and y value in equation 3

z = $\frac{150}{xy}$

z = $\frac{150}{3.4199*10.2598}$

z = 4.2750

so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

4 0

2 years ago

WILL GIVE BRAINLIEST AND 39 POINTS

Mashcka [7]

Part 1)
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hours, the velocity of the car is 59 km/h.

Part 1 a): Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used

Let

A(3,51) B(5,59)

x------ > represent different times

y------ > represent the velocity of the car

Step 1

Find the slope AB

m=(y2-y1)/(x2-x1)------ > m=(59-51)/(5-3)------ > m=8/2---- > m=4

Step 2

With m=4 and point A(3,51) find the equation of the line

y-y1=m*-(x-x1)------ > y-51=4*(x-3)----- > y=4x-12+51----- > y=4x+39

we know that

The standard form of line equation is Ax + By = C

So

y=4x+39----- > y-4x=39------ > this is the standard form

the answer part 1 a) is

y-4x=39

Part 1 b) How can you graph the equation obtained in Part a) for the first six hours?

To graph the equation obtained in Part a) plot the point A and the point B
and join the points to draw the line

To obtain the velocity for the first six hours, substitute the value of x=6 hour in the equation

for x=6 hour

y-4x=39------ > y-4*6=39------ > y=39+24------ > y=63 km/h

using a graph tool

see the attached figure N 1

Part 2)

g(x)=1+1.5^x

step 1

find the equation of the line of f(x)

let

A(-5,3) B(-3,-1)

m=(-1-3)/(-3+5)----- > m=-4/2---- > m=-2

with m=-2 and point A

y-y1=m*(x-x1)------ > y-3=-2*(x+5)---- > y=-2x-10+3----- > y=-2x-7

so

f(x)=-2x-7

step 2

find the equation of the line of p(x)

let

C(0,2) D(-2,-3)

m=(-3-2)/(-2-0)----- > m=-5/-2---- > m=2.5

with m=2.5 and point C

y-y1=m*(x-x1)------ > y-2=2.5*(x-0)---- > y=2.5x+2

so

p(x)=2.5x+2

Part 2 a) What is the solution to the pair of equations represented by p(x) and f(x)?

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 2

The solution is the point (-2,-3)

Part 2 b) Write any two solutions for f(x).

f(x)=-2x-7

for x=0

f(0)=2*0-7---- > f(0)=-7

solution 1 is the point (0,-7)

for x=1

f(1)=2*1-7---- > f(1)=-5

solution 2 is the point (1,-5)

Part 2 c) What is the solution to the equation p(x) = g(x)?

We have

p(x)=2.5x+2

g(x)=1+1.5^x

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 3

The solution are the points (0,2) and (7.3,20.2)

Part 3
)

Part A:There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.

An example of such system of equation is

x < 0

y > 0

The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.

see the attached figure N 4

Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.

Substituting D(-4, 2) into the system

we have:

-4 < 0

2 > 0

as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.

Also,

substituting E(-1, 5) into the system we have:

-1 < 0

5 > 0

as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.

To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.

For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true

For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false

For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false

For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true

For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true

For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true

Therefore,

the farms in which chicken can be raised are the farms at point A, D, E and F.

5 0

2 years ago

Read 2 more answers

Find the volume of a right circular cone that has a height of 12.5 cm and a base with a circumference of 5.8 cm. Round your answ

BaLLatris [955]

Answer:

$V = 11.2 cm^3$

Step-by-step explanation:

Given

Shape: Right Circular cone

Height, h = 12.5 cm

Base circumference, C = 5.8cm

Required:

Calculate Volume

The volume of a cone is calculated as thus;

Volume, $V = \frac{1}{3}\pi r^2h$

Where V, r and h represent the volume, the radius and the height of the cone respectively

But first we need to calculate the radius of the cone.

Using formula of circumference.

$C = 2\pi r$

We have C to be 5.8cm

By substituting this value;

$5.8 = 2\pi r$

Divide through by $2\pi$

$\frac{5.8}{2\pi} = \frac{2\pi r}{2\pi}$

$r = \frac{5.8}{2\pi}$

$r = \frac{2.9}{\pi}$

Now, we can calculate the volume of the cone

By substituting $r = \frac{2.9}{\pi}$ and h = 12.5 cm;

$V = \frac{1}{3}\pi r^2h$ becomes

$V = \frac{1}{3}\pi * (\frac{2.9}{\pi})^2 * 12.5$

$V = \frac{1}{3}\pi * \frac{8.41}{\pi^2} * 12.5$

$V = \frac{1}{3} * \frac{8.41}{\pi} * 12.5$

Take $\pi$ as 3.14

$V = \frac{1}{3} * \frac{8.41}{3.14} * 12.5$

$V = 11.15976$

$V = 11.2 cm^3$ ---- Approximated

Hence, the volume of the cone is; $V = 11.2 cm^3$

4 0

2 years ago

What is the value of the discriminant for the quadratic equation 0 = 2x2 + x – 3?