# Inverse proportion help? Year 9 work...

#1
2a) Two variables, a and b, are known to be proportional to each other.
When a= 2, b= 4.

Find the constant of proportion, k if:

i) a∝b^2

My working:
a=kb^2
2=kx16

Is k 1/8?

ii) a∝1 over square root of b (sorry, don't know how to type it out)

My working:
2= k over square root of 4
k= 2x2

Is k 4?

This is the part I don't know how to do...

b) y is inversely proportional to the cube root of x.

When y= 10, x= 8.
Find the value of y when x= 125.

0
8 years ago
#2
(Original post by no_name123)

This is the part I don't know how to do...

b) y is inversely proportional to the cube root of x.

When y= 10, x= 8.
Find the value of y when x= 125.

If p and q are inversely proportional then pq=k

Yes to the first 2 parts
0
#3
(Original post by TenOfThem)
If p and q are inversely proportional then pq=k

Yes to the first 2 parts
So y x 5 = k ?

I don't understand the cube root part of the question.
0
8 years ago
#4
(Original post by no_name123)
2a) Two variables, a and b, are known to be proportional to each other.
When a= 2, b= 4.

Find the constant of proportion, k if:

i) a∝b^2

My working:
a=kb^2
2=kx16

Is k 1/8?

ii) a∝1 over square root of b (sorry, don't know how to type it out)

My working:
2= k over square root of 4
k= 2x2

Is k 4?

This is the part I don't know how to do...

b) y is inversely proportional to the cube root of x.

When y= 10, x= 8.
Find the value of y when x= 125.

Can you rearrange for k?

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0
#5
(Original post by majmuh24)

Can you rearrange for k?

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k = y*symbol for cube root of x*?
0
8 years ago
#6
(Original post by no_name123)
k = y*symbol for cube root of x*?
Yep, so now you have all your numbers, so just sub back in This is supposed to be the easy part

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0
#7
(Original post by majmuh24)
Yep, so now you have all your numbers, so just sub back in This is supposed to be the easy part

Posted from TSR Mobile
Ahh, I get it now, thank you so much!!!

(Original post by TenOfThem)
If p and q are inversely proportional then pq=k

Yes to the first 2 parts
Thanks to you too
0
8 years ago
#8
(Original post by no_name123)
Ahh, I get it now, thank you so much!!!

Thanks to you too
No problem

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