334 divided by 12 what is the correct answer?

Chef Andy needed to adapt his recipes for his European kitchen staff in Paris. He was 1-pound portions of entrecote, 2 ounces

Ronch [10]

Answer:

1 pound = 0.454 kilograms

1 ounce = 0.0296 liters

1 ounce = 0.0283 kilograms

Step-by-step explanation:

You need to convert the quantities to their corresponding unit.

Convert pounds to kilograms (For the entrecote)

1 pound = 0.454 kilograms

Thus, 1 pound of entrecote = 0.454 kilograms

Ounces to liters (for the Béarnaise, since it is a liquid)

1 ounce = 0.0296 liters

2 ounces of Béarnaise = 0.0592 liters

Ounces to kilograms (for the asparagus)

1 ounce = 0.0283 kilograms

4 ounces of asparagus =0.1132 kilograms

8 0

3 years ago

PLZ HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!

notsponge [240]

Answer:

$72 would be the retail price

Step-by-step explanation:

6 0

3 years ago

For Tran thank you appreciate it

Anestetic [448]

Step-by-step explanation:

Let alpha be the unknown angle. We can set up our sine law as follows:

$\frac{ \sin( \alpha ) }{5 \: m} = \frac{ \sin(42) }{3.6 \: m }$

or

$\sin( \alpha ) = \frac{5 \: m}{3.6 \: m} \times \sin(42) = 0.929$

Solving for alpha,

$\alpha = arcsin(0.929) = 68 \: degrees$

3 0

2 years ago

What is an equation of the line that passes through the point (-7,-6)(−7,−6) and is parallel to the line x-y=7x−y=7?

kotykmax [81]

Answer:

y = x + 1

Step-by-step explanation:

The gradient of a line can be defined by the equation:

m (gradient) = (y1 – y2 ) ÷ (x1 – x2) ----> "1" and "2" should be in subscript

For (-7,-6) we use x2 and y2 (because this point can be anywhere along a line):

x2 = -7, y2 = -6

Plug these values into the formula above:

m = (y-(-6)) ÷ (x-(-7))

m = (y+6) ÷ (x+7)

At this stage, the equation can't be solved as there are two unknowns. Therefore, the gradient must be found another way. Two lines are parallel if they have the same gradient - in their y=mx+c equations, m will be equal.

x - y=7 is the line alluded to in the question. Rearranging this equation into the line equation format gives:

y = x-7 ---> The gradient (coefficient of x) is 1.

Therefore, the gradient of the other parallel line must also be 1.

This can be substituted into the previous equation to give:

1 = (y+6)÷(x+7)

x+7 = y+6

x+1 = y

Therefore, the answer is y=x+1

5 0

2 years ago

The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s

oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is $A(6) = 28$ .

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable $t$ .

Notice that the interval of interest $0 \le t \le 6$ is closed on both ends. In other words, this interval includes both endpoints: $t = 0$ and $t= 6$ . Over this interval, the value of $A(t)$ might be maximized when $t$ is at the following:

One of the two endpoints of this interval, where $t = 0$ or $t = 6$ .
A local maximum of $A(t)$ , where $A^\prime(t) = 0$ (first derivative of $A(t)\!$ is zero) and $A^{\prime\prime}(t)$ (second derivative of $\! A(t)$ is smaller than zero.)

Start by calculating the value of $A(t)$ at the two endpoints:

$A(0) = 10$ .
$A(6) = 28$ .

Apply the power rule to find the first and second derivatives of $A(t)$ :

$\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}$ .

$\displaystyle A^{\prime\prime}(t) = 6\, t - 15$ .

Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$ .

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$ .) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if $A^{\prime\prime}(t) = 0$ ) or a local minimum (if $A^{\prime\prime}(t) > 0$ .)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$ .

Calculate the value of $A(t)$ at this local maximum:

$A(1) = 15.5$ .

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$ . Apparently, $t = 6$ would maximize the value of $A(t)\!$ . That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$ .

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$ .)

4 0

3 years ago