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Sedaia [141]
3 years ago
10

If a test consists of ten multiple choice questions with each permitting a possible answers. How many ways are there in which a

student give his answer?
Mathematics
1 answer:
dangina [55]3 years ago
5 0

The question isn't correctly given, a possible format is in the comment below, however, the explanation will cover then concept which can be applied to different but similar

Answer:

10048576 ways

Step-by-step explanation:

We are given a question, from which we can choose aby of 4 options, this gives ua 4 possible choices for 1.

This will also apply if we have more than 1 question with the same number of options.

This can be called the product rule, as each possibility is the same of each question given :

Therefore, given 10 questions:

. We have

4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 = 4^10 = 10048576 ways

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(side 1)² + (side 2)² = (diagonal)²

(9.5² + (14)² = (diagonal)²

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286.25 = (diagonal)²

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Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al
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\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5

\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}

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