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3 years ago

Which graph represents the reflection in the x-axis of the curve y = cos x

Mathematics

1 answer:

Ber [7]3 years ago

3 0

Answer:

The curve represented by Green color Dotted Lines in the attached figure represents the Reflection in x-axis of the curve $y = cos x$ .

Step-by-step explanation:

As we know, the value of $cosx$ lies between (-1,1) and its graph is a curve. Its value keeps on varying between -1 and 1 only depending on the value of $x$ .

The value of $y$ is $\text{0}$ at $x = \frac{\pi}{2}$ . Few values are labeled in the figure attached in the answer. The value $y$ is 1 at $x = 0$ .

The curve shown in blue color solid line represents the curve of $y = cos x$ .

And the curve shown in the dotted green color in the attached figure represents the reflection in the x-axis.

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Numbers along the outside of frequency tables, calculated from row and column totals, are __________.

Agata [3.3K]

Answer:

The answer is B.

Step-by-step explanation:

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3 years ago

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5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9)

Irina-Kira [14]

Label the points A,B,C

A = (1,2)
B = (4,5)
C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

$A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\$

Segment AB is exactly $3\sqrt{2}$ units long.

Now let's find the distance from B to C

$B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\$

Segment BC is exactly $4\sqrt{2}$ units long.

Adding these segments gives

$AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}$

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

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

$A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\$

AC is exactly $7\sqrt{2}$ units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

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2 years ago

Variance and standard deviation

kirza4 [7]

Answer: 8.3

Step-by-step explanation:

To find the variance, you want to find the mean of the data.

$\frac{3+3+8+1+9+6}{6} =5$

Now that we have the mean, we find the difference between each data point and the mean.

3-5=-2

8-5=3

1-5=-4

9-5=4

6-5=1

With the difference, you square each and find the average.

(-2)²=4

3²=9

(-4)²=16

4²=16

1²=1

$\frac{4+4+9+16+16+1}{6} =\frac{25}{3}$

The variance is 8.3.

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3 years ago

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Darya [45]

The answer would be 6 (This is just to get 20 caracters NOT PART OF ANSWER)

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3 years ago

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