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liq [111]
3 years ago
6

Which table contains data with a nonproportional relationship?

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

Answer:

Step-by-step explanation:

Natali5045456 [20]3 years ago
4 0

Answer:

The second table, counting from the top.

X      Y

2      0

4       2

6       4

Step-by-step explanation:

A proportional relationship is written as:

y = k*x

where k is called the constant of proportionality, is a real number different than zero.

Here we have different tables, if for two pairs (x, y) of the table we have the same value of k, then the relation is proportional. Also you can see that if x = 0, we have:

y = k*0 = 0

then a proportional relationship always has the point (0, 0)

A)

X       Y

1        6

2       12

3       18

Let's use the first two pairs:

(1, 6)

6 = k*1

6/1 = k = 6

(2, 12)

12 = k*2

12/2 = k = 6

in both cases we have k = 6, then this is a proportional relationship.

B)

X      Y

2      0

4       2

6       4

Here we can see the pair (2, 0)

This means that when x = 2, we have y = 0

Then this can not be a proportional relationship, because:

0 = k*2

k = 0

This is not a proportional relationship.

C)

X    Y

1     2

2    4

3    6

Let's find k for the first two points.

(1, 2)

2 = k*1

2/1 = k = 2

(2, 4)

4 = k*2

4/2 = k = 2

In both cases we have k = 2, then this is a proportional relationship.

D)

X   Y

2   -4

4   -8

6   -12

Let's find k for the first two points:

(2, -4)

-4 = k*2

-4/2 = k = -2

(4, -8)

-8 = k*4

-8/4 = k = -2

In both cases we have k = -2, then this is a proportional relationship.

 

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Find sin4a and cos4α, if tanα=3
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<h2>See the explanation.</h2>

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Similarly Cos4α can be written as Cos2(2α).

Sin 4\alpha = 2Sin 2\alpha \times Cos 2\alpha = 2Sin 2\alpha \times(Cos^{2} \alpha - Sin^{2} \alpha ) = 4Sin \alpha \times Cos \alpha \times (Cos^{2} \alpha - Sin^{2} \alpha )

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Sin4α = 4\times \sqrt{(\frac{9}{10} )} \times  \sqrt{(\frac{1}{10} )} \times (\frac{1 - 9}{10} ) = -\frac{4\times8\times3}{100} = -\frac{96}{100}.

Cos4α = Cos^{2} 2\alpha  - Sin^{2} 2\alpha = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - Sin^2 {2\alpha } = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - (2Sin \alpha Cos \alpha )^{2} = (-\frac{8}{10} )^{2} - \frac{4\times9}{100} = \frac{64 - 36}{100} = \frac{28}{100} = \frac{7}{25}

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