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1 year ago

the variables x and y have a proportional relatinship and y=2/5 when x=5/8 which equation represents this relationship

Mathematics

2 answers:

mafiozo [28]1 year ago

6 0

The equation is y = 16/25 x

lets find the proportional relationship,

y = kx

2/5 = k * 5/8

k = (2/5) / (5/8)

k = 16/25

so if k, constant is 16/25

equation is:

y = 16/25 x

<h3>What are proportional relationships?</h3>

Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other. That constant is know as the "constant of proportionality".

<h3>How do you find the proportional relationship in an equation?</h3>

The equation that represents a proportional relationship, or a line, is y = k x , where is the constant of proportionality. Use k = y x from either a table or a graph to find k and create the equation.

To learn more about proportional relationship from the given link

brainly.com/question/2143065

#SPJ4

Evgesh-ka [11]1 year ago

4 0

Answer:

y = 16/25 x

Step-by-step explanation:

Formula: y = kx

2/5 = k(5/8)

(multiplying each side by 40)

16 = k(25)

k = 16/25

y = 16/25 x

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Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

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The distance of a point (x, y) and the origin is calculated by:

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We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

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The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

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