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MaRussiya [10]
3 years ago
14

A chili recipe calls for 2 pounds of ground pork and 3.5 pounds of ground beef. The cost of ground beef per pound is $0.20 less

than twice the cost of ground pork. If the total cost of meat for the recipe is $15.05, what is the cost per pound of ground pork?
Mathematics
1 answer:
Savatey [412]3 years ago
3 0

Answer:

$1.87

Step-by-step explanation:

Given the amount of ground pork =2 pounds and

the amount of ground beef =3.5 pounds.

Let the cost of ground pork= $x per pound

So, the cost of the 2 pounds pork = $2x

As the cost of ground beef per pound is $0.20 less than twice the cost of ground pork, so

the cost of ground beef =$ 2x-0.20 per pound

So, the cost of 3.5 pounds ground beef =$ 3(2x-0.20)

Total cost of meat = $15.05

\Righatarrow 2x +3(2x-0.20)=15.05

\Righatarrow 8x-0.60=15.05

\Rightarrow 8x=15.05-0.20=14.45

\Rightarrow x=14.95/8=1.87

Hence, the cost of ground pork= $1.87 per pound

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Step-by-step explanation:

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We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c
Kisachek [45]
We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

\frac{dT}{dt}=k(T-a)

We now separate the time and the temperature variables as follows,

\frac{dT}{T-a}=kdt

Integrating both sides of the differential equation, we obtain;

ln(T-a)=kt +c

This natural logarithmic equation can be rewritten as;

T-a=e^{kt +c}

Applying the laws of exponents, we obtain,

T-a=e^{kt}\times e^{c}

T-a=e^{c}e^{kt}

We were given the initial conditions,

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Let us apply this condition to obtain;

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Now our equation, becomes

T-a=-16e^{kt}

or

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T=20-16e^{kt}

b) We were also given that,

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Let us apply this condition again to find k.

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This implied

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After 15 minutes, the temperature will be,


T=20-16e^{-0.2877\times 15}

T=20-0.21376


T=19.786

After 15 minutes, the temperature is approximately 20°C


6 0
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