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y is directly proportional to the cube of x.
When x =2, y = 56
Find the value of y when x=5
HELP ME
(edited 4 years ago)
If y is directly proportional to x^3 it means that y and x^3 increase at the same rate. So if you double the value for y, the corresponding value for x^3 is doubled. This means that y = k* x^3, where k is any constant.

So now you have a general equation of y = k * x^3, and you have a pair of values for x and y. Can you find the value for k? Try find the value for k to form a specific equation relating x to y.
(edited 4 years ago)
Reply 2
Original post by Maximus 190
If y is directly proportional to x^3 it means that y and x^3 increase at the same rate. So if you double the value for y, the corresponding value for x^3 is doubled. This means that y = k* x^3, where k is any constant.

So now you have a general equation of y = k * x^3, and you have a pair of values for x and y. Can you find the value for k? Try find the value for k to form a specific equation relating x to y.

That’s just confused me more?? Where did k come from.
Reply 3
Original post by KlpDonnelly
y is directly proportional to the cube of x.
When x =2, y = 56
Find the value of y when x=5


y=kx^3

56=k(2)^3
56=8k

divide both sides by 8 to get k on its own
k=7

plug into formula

y=7(5)^3
y=7(125)
y=875
Original post by hudz1e
y=kx^3

[redacted]


On TSR you aren’t meant to straight up post solutions, it doesn’t help anyone.
(edited 4 years ago)
Original post by KlpDonnelly
That’s just confused me more?? Where did k come from.


It’s just the general formula for directly proportional. If two things are directly proportional, for example if a and b are directly proportional, then a=kb. The k comes as it’s the gradient of b.

So if y= k*x^3, k is the gradient of x^3. If you drew a line, then k would be how steep the line is on the graph
Original post by hudz1e
y=kx^3

DO NOT post solutions - it's against the rules :smile:
Reply 7
Original post by Maximus 190
On TSR you aren’t meant to straight up post solutions, it doesn’t help anyone.


relax it's just one question. op can do more questions now after learning all the steps
Reply 8
omg whys everyone popping a vein.. if a teacher doesn't show u an example in school how are u ever gonna learn how to do it?
Original post by hudz1e
relax it's just one question. op can do more questions now after learning all the steps


Oh I was freaking out that you gave this guy the solution so thanks for telling me to relax, I’m now going to take 30 minutes out to meditate in my head space.
Original post by Maximus 190
On TSR you aren’t meant to straight up post solutions, it doesn’t help anyone.

None of it is helping to be honest I don’t understand any of it
Reply 11
Original post by KlpDonnelly
None of it is helping to be honest I don’t understand any of it


i'd recommend watching a video on mathsgenie until you get it
Original post by KlpDonnelly
None of it is helping to be honest I don’t understand any of it


Try find some YouTube videos on ‘directly proportional relationships’. You will find a better explanation there than on here, and the problems for it are all very similar so I’m sure they will go through some examples of this similar problem.
Original post by KlpDonnelly
y is directly proportional to the cube of x.
When x =2, y = 56
Find the value of y when x=5
HELP ME

If two variables are directly proportional, this means their ratio is equal to a constant (if you don't know what a constant is, just think of it as a number). Remember that the ratio of a a and b b is ab \dfrac{a}{b}.

so if a a is directly proportional to b b, then,
ab=k \dfrac{a}{b} = k, where k is a constant.
This can be thought of in another way if we multiply both sides of the above equation by b b, we get a=k×b a = k \times b

So we can think of proportionality as one variable is always the same constant multiple of another variable. That constant is called the constant of proportionality.


So when you have variable 1\mathrm{variable\ 1} is directly proportional to variable 2\mathrm{variable\ 2} this can be written as variable 1=constant×variable 2\mathrm{variable\ 1} = \mathrm{constant} \times \mathrm{variable \ 2} (commit this to memory).

Try drawing a graph of y = mx (notice how this defines y and x to be directly proportional when m is fixed) and play around with different values of m to get an intuition for it.

Now if y is proportional to the cube of x (x^3) what do you think the equation that governs the relationship between y and x^3 will be?

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