The variables x=5/2 and y=5/4 varies directly. How do you write an equation that relates the variables and find x when y=-5?

Answer 1

#x=-10#

#"the initial statement is "ypropx#
#"to convert to an equation multiply by k the constant"# #"of variation"#
#rArry=kx#
#"to find k use the given condition"#
#x=5/2" and "y=5/4#
#y=kxrArrk=y/x=5/4xx2/5=1/2#
#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=x/2)color(white)(2/2)|)))#
#"when "y=-5#
#x=2xx-5=-10#
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Answer 2

The equation that relates the variables x and y when they vary directly is y = kx, where k is the constant of variation. To find k, substitute the given values of x and y: (5/4 = k \times 5/2). Solving for k, we get (k = 5/8). Now, to find x when y = -5, substitute y = -5 and k = 5/8 into the equation: (-5 = 5/8 \times x). Solving for x, we get (x = -40).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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