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2 years ago

The (period,amplitude,frequence) of y=1/4cos1/2x is 4pie. one of these is the answer need help

Mathematics

1 answer:

Sidana [21]2 years ago

7 0

Answer:

period

Step-by-step explanation:

The attached graph shows a period of the function. Its amplitude is 1/4 (vertical distance from centerline to peak). Its period is 4π (horizontal distance from peak to peak). Its frequency is the reciprocal of the period, so would be 1/(4π).

The value 4π is the period of the function.

You might be interested in

Find the distance between the given points (2,8) and (7,7)

Ray Of Light [21]

Answer:

$d=\sqrt{26}\approx5.0990$

Step-by-step explanation:

To find the distance between any two points, we can use the distance formula.

The distance formula is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let (2,8) be x₁ and y₁ and let (7,7) be x₂ and y₂. Thus:

$d=\sqrt{(7-2)^2+(7-8)^2}$

Simplify:

$d=\sqrt{5^2+(-1)^2}$

Square:

$d=\sqrt{25+1}$

Add:

$d=\sqrt{26}\approx5.0990$

And that's our answer :)

6 0

3 years ago

Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y

katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with

center (h , k) and transverse axis parallel to the x-axis is

(x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are (h ± a , k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is 2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

4 0

3 years ago

Read 2 more answers

Does anyone know the answer

murzikaleks [220]

Answer:

20,160

Step-by-step explanation:

6 0

2 years ago

~**Will mark brainliest **~<br> For the correct answers to all three questions

aleksandr82 [10.1K]

$\dfrac{42}{65}\cdot\dfrac{25}{36}\cdot\dfrac{26}{49}=\dfrac{2\cdot3\cdot7}{5\cdot13}\cdot\dfrac{5\cdot5}{2\cdot2\cdot3\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\dfrac{7}{13}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\\\\\\=\dfrac{1}{1}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{1}{7}=\dfrac{5}{21}$

$\dfrac{21}{32}\cdot\dfrac{39}{120}\cdot\dfrac{40}{65}=\dfrac{21}{32}\cdot\dfrac{3\cdot13}{2\cdot2\cdot2\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot2\cdot5}{5\cdot13}=\\\\\\=\dfrac{3\cdot7}{32}\cdot\dfrac{13}{2\cdot2\cdot2\cdot5}\cdot\dfrac{2\cdot2\cdot2}{13}=\dfrac{21}{32}\cdot\dfrac{1}{5}\cdot\dfrac{1}{1}=\dfrac{21}{160}$

$\dfrac{15}{90}\cdot\dfrac{36}{75}\cdot\dfrac{27}{42}=\dfrac{3\cdot5}{2\cdot3\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot3\cdot3}{3\cdot5\cdot5}\cdot\dfrac{3\cdot3\cdot3}{2\cdot3\cdot7}=\\\\\\=\dfrac{1}{2\cdot3}\cdot\dfrac{2\cdot2\cdot3}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{1}\cdot\dfrac{2}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{25}\cdot\dfrac{9}{7}=\dfrac{9}{175}$

6 0

3 years ago

FREEE POINTSSSSSSSSSSS

ira [324]

Answer:

hehc dhcbshhceucuiifvbrifrjvrjv

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers