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steposvetlana [31]
3 years ago
12

En un programa de televisión se hacen 30 preguntas. Por cada respuesta correcta se suman 8 puntos, por cada respuesta errónea se

restan 5 puntos y por aquellas preguntas que no se contesten no se suman ni se restan puntos. Si un participantes obtuvo 13 puntos, ¿Cuantas respuestas erróneas pudo tener?
Mathematics
1 answer:
nikdorinn [45]3 years ago
8 0

Answer:

7 wrong answers.

Step-by-step explanation:

We know that when you answer a question well, you earn 8 points and a bad one loses 5 points, this means that when you answer 1 good and 1 bad, there is a total of 3 points (8-5).

This means that when answering 5 good and 5 bad, in total it would be 3 * 5 points, that is to say 15 points, that is, it goes through 2 points. To subtract 2 points, you would have to answer 1 good and 2 bad, (8 - 2 * 5), which turns out to be -2 points.

In total there would be 6 questions right and 7 questions wrong, like this:

8 * 6 + 5 * 7 = 13

13 points, in 13 questions (6 correct + 7 incorrect)

The rest of the questions were not answered so that the score does not go up or down.

Which means you got 7 wrong answers.

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3 years ago
Help. <br>Please its urgent show workings.<br>​
laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

\frac{dy}{dx} = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × \frac{d}{dx} (x + 4) ← chain rule

       = 2(x + 4) × 1

        = 2(x + 4)

Then

\frac{dy}{dx} = (2x - 1). 2(x + 4) + (x + 4)². 2

    = 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

    = 2(x + 4) (2x - 1 + x + 4)

    = 2(x + 4)(3x + 3) ← factor out 3

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--------------------------------------------------------------------------

(b)

y =  x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × \frac{d}{dx} (x² - 1) ← chain rule

        = 3(x² - 1)² × 2x

        = 6x(x² - 1)²

Then

\frac{dy}{dx} = x. 6x(x² - 1)² + (x² - 1)³. 1

    = 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

    = (x² - 1)² (6x² + x² - 1)

     = (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

\frac{dy}{dx} = (x² - 1). 3x² + (x³ + 1), 2x

   = 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

   = x[3x(x² - 1) + 2(x³ + 1) ]

   = x(3x³ - 3x + 2x³ + 2)

   = x(5x³ - 3x + 2) ← distribute

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--------------------------------------------------------------------

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y = 3x³(x² + 4)²

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g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × \frac{d}{dx}(x² + 4) ← chain rule

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Then

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    = 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

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3 years ago
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