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DaniilM [7]
4 years ago
11

There are two spinners. The first spinner has three equal sectors labeled 1, 2, and 3. The second spinner has four equal sectors

labeled 3, 4, 5, and 6. Spinners are spun once.
How many outcomes do not show an odd number on the first spinner and show a 5 on the second spinner?




1

2

3

5
Mathematics
1 answer:
zhuklara [117]4 years ago
7 0
<h3>Answer: A) 1</h3>

=========================

Explanation:

Let's label each pair of outcomes as an ordered pair (x,y)

x = first spinner outcome

y = second spinner outcome

We have these 12 different possible combos

(1,3),   (1,4),   (1,5),   (1,6)

(2,3),  (2,4),  (2,5),  (2,6)

(3,3),  (3,4),   (3,5),  (3,6)

Focus on the third column where y = 5 for each (x,y) pair. Of this column only (2,5) has x be an even number. The other results in that column have x be odd.

Basically (1,5) is the only possible outcome if we want the first spinner to lan on an even number, and the second spinner to land on 5.

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telo118 [61]

In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

Let's write both sides in terms of \sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x) only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)

and

\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)

So, their multiplication is

\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)

So, the left hand side simplifies to

\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

And as you can see, the two sides are equal.

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Answer: x^2

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Answer:

Step-by-step explanation:

multiply first,

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