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lozanna [386]
3 years ago
7

if the value of x varies directly as the value of y , and x = 16 when y = 3/4 , what is the value of x when y = 3/8 ?

Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

x=8

Step-by-step explanation:

If the value of x varies directly as the value of y we can say they are directly proportional. This implies that when y increases x will increase too, and when y decreases x will also decrease.

If x=16 when y=3/4 we can write their relation as:

x/y = 16/(3/4) = 16*(4/3) = 64/3

So, for every value of y, the ratio x/y must maintain equal to 64/3 because both values are proportional.

Now we can see what happens to x when y varies. If y=3/8:

x/(3/8) = 64/3

Multiply both sides by 3/8 to eliminate the 3/8 in the denominator.

x/(3/8) * (3/8) = (64/3) * (3/8)

See that we have a 3 multiplying and another dividing, we can eliminate them.

x = 64/8

x = 8

So, if y decreases to 3/8 x will decrease to 8. Notice that y is decreasing in a half as 3/8 is the half of 3/4, and x is also decreasing in a half from 16 to 8, what makes sense.

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