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2 years ago

Write a sine function that has a midline of 5, an amplitude of 4 and a period of 7/5

Mathematics

1 answer:

rusak2 [61]2 years ago

3 0

Answer:

$y=4sin(\frac{10\pi}{7}x)+5$

Step-by-step explanation:

Recall formula

$y=asin(bx+c)+d\\\\y=4sin(bx)+5$

Determine b from period

$\frac{2\pi}{b}=\frac{7}{5}\\ \\ 10\pi=7b\\\\b=\frac{10\pi}{7}$

Final Equation

$y=4sin(\frac{10\pi}{7}x)+5$

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590. Since 2 is under 5 you round it down

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3 years ago

The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil

satela [25.4K]

The length of pencil A is 5 cm

Solution:

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

The total length of pencils A, B and C is 29 cm

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

Pencil A is 11 cm shorter then pencil B

x = y - 11 ------- eqn 2

Pencil B is twice as long a pencil C

y = 2z

$z = \frac{y}{2}$ ------ eqn 3

Substitute eqn 2 and eqn 3 in eqn 1

$y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16$

Substitute y = 16 in eqn 2

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0

3 years ago

Convert 120 Ibs to kilograms

posledela

Answer:

54.431 kilograms

Step-by-step explanation:

don't have step by step but 1kg is 2.205lbs

7 0

3 years ago

PLEASE HELP ASAP!!!!!!!!!!!

Serga [27]

$8x^2y^{-2}=\dfrac{8x^2}{y^2}\\\\\\\dfrac{8\cdot(-1)^2}{2^2}=\dfrac{8\cdot1}{4}=\dfrac{8}{4}=\boxed{2}$

Answer A)

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4 years ago

A set of 200 test scores are normally distributed with a mean of 70 and a standard deviation of 5. Approximately how many studen

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Answer: About 191 students scored between a 60 and an 80.

Step-by-step explanation:

Given : A set of 200 test scores are normally distributed with a mean of 70 and a standard deviation of 5.

i.e. $\mu=70$ and $\sigma=5$

let x be the random variable that denotes the test scores.

Then, the probability that the students scored between a 60 and an 80 :

$P(60$

The number of students scored between a 60 and an 80 = 0.9544 x 200

= 190.88 ≈ 191

Hence , about 191 students scored between a 60 and an 80.

8 0

4 years ago