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ANEK [815]
2 years ago
13

What is the constant of proportionality in the equation y = 2.5x?

Mathematics
2 answers:
photoshop1234 [79]2 years ago
8 0

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form \frac{y}{x}=k or y=kx

where

k is the constant of proportionality

In this problem we have

y=2.5x

so

the constant of proportionality k is equal to

k=2.5

therefore

<u>the answer is</u>

k=2.5

Nostrana [21]2 years ago
4 0

The constant of proportionality in the equation y = 2.5x is \boxed{2.5}.

Further explanation:

The relation is defined as the relationship between the input values and output values.

The x coordinates are the domain of the function and the y coordinates are the range of the function.

Given:

The equation is y = 2.5x.

Explanation:

The proportionality equation can be expressed as follows,

y \propto x

The value of y changes as x changes. If the value of x increases then the value of y is also increasing.

The equation can be expressed as follows,

y = kx

The inversely proportional relationship can be expressed as,

y \propto \dfrac{1}{x}

Here, k is the proportionality constant.

The given equation is y = 2.5x.

The y is the independent variable and x is the dependent variable.

The proportionality constant is 2.5.

The constant of proportionality in the equation y = 2.5x is \boxed{2.5}.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Ratio and proportion

Keywords: proportional, directly proportional, constant, proportionality equation, y=0.41x, constant of proportionality.

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f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

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