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mafiozo [28]
3 years ago
12

Caleb and Eva are running around a track. They both start at the same time. It takes Caleb 18 minutes to complete 1 lab. It take

s Eva 14 minutes to complete 1 lab. How many labs will each of them have run before they meet at the start again?
Mathematics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

For x=7 and y=9 Caleb and Eva will meet again. That is, Caleb will have made 7 labs and Eva 9 labs.

Step-by-step explanation:

In order to find the number of labs that Caleb and Eva must complete, to meet again at the start, you can use the following equation:

18x=14y      (1)

where x is the number of labs made by Caleb and y the number of labs made by Eva. The equation (1) means that for the specific values of x and y, Caleb and Eva will spend the same amount of minutes on the labs.

Then, you need to find the least integers values of x and y.

First, you solve the equation (1) for y:

y=\frac{18}{14}x      (2)

By inspection you can find that the least value of x that makes possible that y is an integer, is 7

In fact, you have:

y=\frac{18}{14}(7)=9

Hence, for x=7 and y=9 Caleb and Eva will meet again. That is, Caleb will have made 7 labs and Eva 9 labs.

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Step-by-step explanation:

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Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
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Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

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(x-\frac{1}{2})^2

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The first inequality gives us the solution

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The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

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