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AlladinOne [14]
1 year ago
8

The local hamburger restaurant is famous for how quickly they can assemble a burger. The restaurant advertises that its constant

rate of assembling burgers is t=11.3h, where t is the total amount of time in seconds and h is the number of hamburgers assembled

Mathematics
1 answer:
melisa1 [442]1 year ago
3 0

The gievn equation is ,

t=11.3h\ldots(1)

where t is the time in seconds and h is the no.of hamburgers assembled.

put h = 2 in equation (1).

\begin{gathered} t=11.3\text{ (2)} \\ t=22.6 \end{gathered}

blank A assembles 2 hamburges in 22.6 seconds.

put h = 3 in equation (1)

\begin{gathered} t=11.3(3) \\ t=33.9 \end{gathered}

blank B assembles 3 hamburgers in 33.9 seconds.

put h = 5 in equation (1)

\begin{gathered} t=11.3(5) \\ t=56.5 \end{gathered}

blank C assembles 5 hamburgers in 56.5 seconds.

put h = 8 in equation (1)

\begin{gathered} t=11.3(8) \\ t=90.4 \end{gathered}

blank D assembles 8 hamburgers in 90.4 seconds.

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Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.​
Vedmedyk [2.9K]

Compute the gradient of f.

\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle

Set this equal to the zero vector and solve for the critical points.

3x^2-6y = 0 \implies x^2 = 2y

-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}

\implies x^2 = \pm2\sqrt{2x}

\implies x^4 = 8x

\implies x^4 - 8x = 0

\implies x (x-2) (x^2 + 2x + 4) = 0

\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0

\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0

The last case has no real solution, so we can ignore it.

Now,

x=0 \implies 0^2 = 2y \implies y=0

x=2 \implies 2^2 = 2y \implies y=2

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}

Check the sign of the determinant of the Hessian at each of the critical points.

\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0

which indicates a saddle point at (0, 0);

\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0

We also have f_{xx}(2,2) = 12 > 0, which together indicate a local minimum at (2, 2).

3 0
2 years ago
What are the slope and y-intercept of the graph of the given equation?
tia_tia [17]

Answer:

The slope is 9/8, and the y-intercept is -10/3

Step-by-step explanation:

y = mx + b

in this case:

m = 9/8

b = -10/3

and the m in the equation is the slope while the b is the y-intercept

5 0
3 years ago
URGENT NEED HELP
Gelneren [198K]

Emma:

Number of hours = 4

Total cost = 18

Where x is the cost per hour n and y is the flat fee

18 = 4x + y

Louise:

number of hours = 7

total cost = 25.50

25.50 = 7x + y

We have the system of equation:

18 = 4x + y

25.50 = 7x + y

Solve the first equation for y

18-4x = y

y = 18-4x

Replace y on the second equation; and solve for x

25.50 = 7x + (18-4x)

25.50 = 7x + 18 -4x

25.50 = 3x + 18

25.50 - 18 = 3x

7.5 = 3x

7.5/3 = x

2.5 = x

a. Cost per hour = $2.5

Now, replace x on any equation and solve for y

18 = 4x + y

18 = 4 (2.5) + y

18 = 10 +y

18 - 10 = y

8 = y

b. Flat fee = $8

c. Use the equation:

Cost = flat fee + cost per hourx (number of hour)

Cost = 8 + 2.5 (2)

Cost = 8 + 5 = $13

5 0
3 years ago
Solve the following system by the comparison method.
Brut [27]
<span>2p + q = 1
9p + 3q + 3 = 0

q = 1 - 2p
replace </span>q = 1 - 2p into 9p + 3q + 3 = 0

9p + 3(1 - 2p) + 3 = 0
9p + 3 - 6p + 3 = 0
3p + 6 =0
3p = -6
p = -2

q = 1 - 2p
q = 1 -2(-2)
q = 1 + 4
q = 5

(-2 , 5)

answer 
<span>{(-2, 5)} first one</span>
6 0
4 years ago
Read 2 more answers
Leanne runs a lap in 12 minutes. Kory runs a lap in 9 minutes. If they
klemol [59]

Kory and Leanne will meet each other again after 36 minutes from starting i.e. at 10:36 am .

<u>Step-by-step explanation:</u>

Here we have , Leanne runs a lap in 12 minutes. Kory runs a lap in 9 minutes. We need to find out that If they  both start running at 10 am, at what time will they both cross the start  together again . Let's find out:

Kory runs a lap in 9 minutes and , Leanne in 12 minutes , in order to find at what time will they both cross the start  together again we will find LCM of 9 & 12 :

⇒ \frac{9(4)}{12(4)}

⇒ \frac{36}{36}

So ,LCM of 9 & 12 is 36 .That means Kory and Leanne will meet each other again after 36 minutes from starting i.e. at 10:36 am .

In clear way , Kory will run race as :

1st lap = 9 minutes

2nd lap = 18 minutes

3rd lap = 27 minutes

4th lap = 36 minutes , i.e. at 10:36 am he'll complete his 4th lap

For , Leanne

1st lap = 12 minutes

2nd lap = 24 minutes

3rd lap = 36 minutes i.e. at 10:36 am he'll complete his 3rd lap .

This means Kory & Leanne will meet at 10:36 am completing there 4th and 3rd lap respectively .

6 0
4 years ago
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