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Pachacha [2.7K]
4 years ago
15

Devon purchased one bag of dog food and three bags of cat food. Each bag of dog food weighs x pounds and each bag of cat food we

ighs half as much as the bag of dog food. The total weight of Devon's purchase is 32.5 pounds. The equation x+3(x/2)=32.5 can be used to determine the weight of the dog food and cat food. What is the weight in pounds of one bag of cat food?
Mathematics
1 answer:
likoan [24]4 years ago
6 0

Answer:

A pound of cat food will weigh 6.5 pounds

Step-by-step explanation:

Firstly, we need to get the value of x. Thus, let’s solve the equation at hand;

x + 3x/2 = 32.5

x + 1.5x = 32.5

2.5x = 32.5

x = 32.5/2.5

x = 13

Now from the question, we are made to know that a pound of cat food measures half of what a pound of dog food measures

Since we calculate a pound of dog food to weigh 13 pounds, a pound of cat food will weigh = 13/2 = 6.5 pounds

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I need help thank you
IRISSAK [1]

Answer:

1) k = 3

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Step-by-step explanation:

Jared used the equation;

k + (k + 1) + (k + 2) = 4k

Let's find k.

3k + 3 = 4k

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Thus,the integers will be;

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He is looking for consecutive even integers but yet got 3 and 5 which are odd numbers. However, he should have used the format;

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Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

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Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

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3 years ago
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