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2 years ago

Please help whatbis the answer

Mathematics

1 answer:

Irina-Kira [14]2 years ago

6 0

It’s the last one, cause Susan charges 20 per photo

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A sine function has an amplitude of 3, a period of pi, and a phase shift of pi/2. What is the y-intercept of the function?

jarptica [38.1K]

Based on the calculations, we can logically deduce that the y-intercept of this sine function is equal to: A. 3.

<u>Given the following data:</u>

Amplitude = 3

Period = π

Phase shift = π/2

<h3>How to determine the y-intercept of this function?</h3>

Mathematically, a sine function is modeled by this equation:

y = Asin(ωt + ø)

<u>Where:</u>

A represents the amplitude.

ω represents angular velocity.

t represents the period.

ø represents the phase shift.

Also, the period of a sine wave is given by:

t = 2π/ω

2 = 2π/ω

ω = 2

Substituting the given parameters into the equation, we have;

y = 3sin(2t + π/2)

At t = 0, we have:

y = 3sin(2(0) + π/2)

y = 3sin(π/2)

y = 3sin(90)

y = 3 × 1

y = 3.

In conclusion, we can logically deduce that the y-intercept of this sine function is equal to 3.

Read more on phase shift here: brainly.com/question/27692212

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5 0

1 year ago

What is the difference between d and 7?

Goryan [66]

Answer:

d - 7

Step-by-step explanation:

The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.

3 0

5 months ago

Solve 4x+2y=-6 for y

mel-nik [20]

Answer:

y=-2x-3

Step-by-step explanation:

4x+2y=-6

2y=-6-4x

y=-6/2-4/2x

y=-3-2x

y=-2x-3

8 0

2 years ago

Read 2 more answers

Find the average value of f over the region D. f(x, y) = 3xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 9)

MakcuM [25]

The average value of f over the region D is 243/4

To answer the question, we need to know what the average value of a function is

<h3>What is the average value of a function?</h3>

The average value of a function f(x) over an interval [a,b] is given by

$\frac{1}{b - a} \int\limits^b_a {f(x)} \, dx$

Now, given that we require the average value of f(x,y) = 3xy over the region D where D is the triangle with vertices (0, 0), (1, 0), and (1, 9).

x is intergrated from x = 0 to 1 and the interval is [0,1] and y is integrated from y = 0 to y = 9

So, $\frac{1}{b - a} \int\limits^b_a {f(x,y)} \, dA = \frac{1}{1 - 0} \int\limits^1_0 \int\limits^9_0 {3xy} \, dxdy \\= \frac{3}{1} \int\limits^1_0 {x} \,dx\int\limits^9_0 {y} \,dy\\ = \frac{3}{1} [\frac{x^{2} }{2} ]^{1}_{0}[\frac{y^{2} }{2} ]^{9}_{0} \\= 3[\frac{1^{2} }{2} - \frac{0^{2}}{2} ] [\frac{9^{2} }{2} - \frac{0^{2}}{2} ] \\= 3[\frac{1}{2} - 0 ][\frac{81}{2} - 0 ]\\= \frac{81}{2} X3 X \frac{1}{2} \\= \frac{243}{4}$

So, the average value of f over the region D is 243/4

Learn more about average value of a function here:

brainly.com/question/15870615

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8 0

1 year ago

(8.4x10^3)-(6.4x10^2)<br> Write anserr in scientific notation

Alex_Xolod [135]

I think the answer is 2x10^1

7 0

2 years ago

Read 2 more answers