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dybincka [34]
2 years ago
9

The product of the ages of all teenagers at Joyce’s birthday party last Saturday was 2971987200. The number of 18 year olds who

attended the party was?
Mathematics
2 answers:
Tom [10]2 years ago
8 0

Answer:

7 18-year olds

Step-by-step explanation:

The product is 2,971,987,200. 18^7 is 612,220,032. However, 18^8 is 11,019,960,576, which is too big. So, 7 is the answer

Elenna [48]2 years ago
6 0

Click here to see ALL problems on Age Word Problems

 Question 1128242: The product of the three ages, in years of three teenagers is 4590. None of the teens are the same ages. What are the ages of the three teen ages? (Use guess and check)

A hat and a jacket cost $100. The jacket cost $90 more than the hat. What are the cost of the hat and the cost of the jacket?

Melody picks a number. She doubles the number, squares are result, divide the square by 3 subtracts 30 from the quotient and gets 18. What are the possible numbers that Melody could have picked? What operation does Melody perform that prevents us from knowing 100% certainly which she picked?

(Scroll Down for Answer!)

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Answer by greenestamps(9721) About Me  (Show Source):

You can put this solution on YOUR website!

These are three very different problems. The rules of the forum are one problem per post. Please follow those rules in the future.

(1) Certainly guess and check is the easiest way to solve this problem, since you know the three ages are whole numbers. But you should be able to get to the solution with very few if any wrong "guesses" by making your guesses logically.

The product of the three ages is a multiple of 10, and therefore a multiple of 5; since all three ages are in the teens, one of them has to be 15. Then 4590/15 = 306.

You could do guess and check again from there. However, if you note that both the other ages have to be greater than 15, and you know the last digit of the product has to be "6", then the only possible pair of ages is 17 and 18. And indeed 17*18 is 306.

ANSWER: The ages are 15, 17, and 18.

(2) Here is an informal solution method using logical reasoning.

Take away the "extra" $90 that the jacket costs; you are left with a jacket and a hat that each cost the same, and the total is now $100-$90 = $10. That means each of them costs $5. Now add those extra $90 back on to the price of the jacket.

ANSWER: The jacket costs $95; the hat costs $5.

(3) Melody picks a number: Call it x.

 

She doubles the number: It is now 2x.

She squares are result: It is now (2x)^2 = 4x^2.

She divides the square by 3: It is now 4x^2/3.

She subtracts 30 from the quotient: It is now 4x^2/3-30.

The number she ends up with is 18.

Solve the equation that says 4x^2/3-30 = 18:

4x%5E2%2F3-30+=+18

4x%5E2%2F3+=+48

4x%5E2+=+144

x%5E2+=+36

x+=+6 or x+=+-6

ANSWER: The two possible starting numbers are 6 and -6. We get two possible answers because of the step where she squares the number.

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The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
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Answer:

Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

\frac{K-P(t)}{P(t)} = Ae^{-kt}.

The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

\frac{500-100}{100} = A} and A=4.

So,

\frac{500-P(t)}{P(t)} = 4e^{-0.04t}.

Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

\frac{500-200}{200} = 4e^{-0.04t}.

This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325.

So, the population of rats will be 200 after 25 months.

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