All categories

2 years ago

Solve the equation for [0,2pi] cos2x=1/sqrt2

1 answer:

5 0

$cos(2x) = \frac{1}{\sqrt{2}}$
$cos(2x) = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$cos(2x) = \frac{\sqrt{2}}{2}$
$cos(2x) = cos(45)$
$cos(x) = cos(22.5)$
$x = 22.5$

You might be interested in

Which expression is equivalent to (- 2c / d^2)^3

nikdorinn [45]

-8c^3/d^6 is equivalent to the expression

4 0

3 years ago

Read 2 more answers

A caterer is preparing an estimate for a customer. There are 285 people attending a banquet at $7.89 per person.Round the per pe

VikaD [51]

Sent a pic of the solution (s).

6 0

3 years ago

Hi why do we need to keep watching ads tbh im very tired of it

Nady [450]

Answer:

cs sometimes we needa pass a test n dis is the only shi dat b righh fr

Step-by-step explanation:

8 0

2 years ago

Mr. Sonny's science class is calculating the average number of blinks per minute. Jan blinks 72 times in 4 minutes. What is her

serg [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Quadrilateral TRAM is rotated -90 degrees about the origin. Draw the image of this rotation.

SpyIntel [72]

Answer:

Please find attached the image of the quadrilateral TRAM after a rotation of -90 degrees, created with MS Excel

Step-by-step explanation:

The given coordinates of the vertices of the quadrilateral TRAM are;

T(-5, 1), R(-7, 7), A(-1, 7), M(-5, 4)

By a rotation of -90 degrees = Rotation of 90 degrees clockwise, we get;

The coordinates of the preimage before rotation = (x, y)

The coordinates of the image after rotation = (y, -x)

Therefore, we get for the the quadrilateral T'R'A'M', by rotating TRAM -90 degrees as follows;

T(-5, 1) → T'(1, 5)

R(-7, 7) → R'(7, 7)

A(-1, 7) → A'(7, 1)

M(-5, 4) → M'(4, 5)

The image of TRAM after -90 degrees rotation is created by plotting the derived points of the quadrilateral T'R'A'M' on MS Excel and joining the corresponding points as presented in the attached diagram.

7 0

3 years ago