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Veronika [31]
3 years ago
10

The score on a trivia game is obtained by subtracting the number of incorrect answers from twice the number of correct answers.

If a player answered 40 questions and obtained a score of 50 how many questions did the player answer correctly?
Mathematics
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer: the player answered 30 questions correctly.

Step-by-step explanation:

Let x represent the number of questions that the player answered correctly.

Let y represent the number of questions that the player answered incorrectly.

The number of questions that the player answered is 40. It means that

x + y = 40- - - - - - - - - - -1

The score on the trivia game is obtained by subtracting the number of incorrect answers from twice the number of correct answers. If the player obtained a score of 50, It means that

2x - y = 50

y = 2x - 50

Substituting y = 2x - 50 into equation 1, it becomes

x + 2x - 50 = 40

3x - 50 = 40

3x = 40 + 50 = 90

x = 90/3

x = 30

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1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

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