What is the range of the function y = 2sin x?

1 answer:

5 0

The standard form of equation for the sin function is y = Asin(Bx-C) where A is amplitude, Period is 2π/B, and the phase shift is equal to Bx-C set to zero. C is then obtained in this equation.

For given problem: A is equal to 2, period=2π/B=2π/1=2π, where B is equal to 1. Period then is equal to A. 2pi

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Simplify the expression<br> (3x4 + 9x3 – 7x + 15) + (-6X4 – 8x2 + 5x – 3)

elena-s [515]

Answer:

If the numbers next to the "x" are "carroted" ( ^ ) then the answer is

-3x^4 + 9x^3 -8x^2 -2x +12

Step-by-step explanation:

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5 0

3 years ago

A particle is moving along the curve y= 4sqrt(5x+11) . As the particle passes through the point (5,24) , its -coordinate increas

kupik [55]

The rate of change of the distance from the particle to the origin at this instant is 3 units per second.

<h3>What is the rate of change?</h3>

The instantaneous rate of change is the rate of change at a particular instant.

A particle is moving along the curve

$y= 4\sqrt{5x+11} .$

The rate of change of y is given as:

dy / dx = 2

by differentiating both sides,

$\dfrac{dy}{dx} = 4\dfrac{1}{2\sqrt{5x+11} } 5\dfrac{dx}{dt} \\\\\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\$

From the question, we have:

(x, y) = (5,24)

Substitute 5 for x and dy / dx = 2

$\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\\\5= \dfrac{10}{\sqrt{5(5)+11} }\dfrac{dx}{dt}\\\\\\\sqrt{5(5)+11} = 2\dfrac{dx}{dt}\\\\\\\dfrac{dx}{dt} = 6/ 2 = 3$