# need help on maths questions

Announcements
#1
I used a random method and got 2/3 of the questions right but the third question doesn't work so the method is probably wrong. Can someone please tell me how to work it out. I'll upload the questions and my workings below 0
1 year ago
#2
(Original post by Mia 15)
I used a random method and got 2/3 of the questions right but the third question doesn't work so the method is probably wrong. Can someone please tell me how to work it out. I'll upload the questions and my workings below Here to help ...
0
1 year ago
#3
waiting...
0
#4
Sorry, I tried to upload the pictures earlier but it wouldn't load, I'll just type it out.

Q6) The quantity for y is directly proportional to the square root of x for x > 0.
When x = 1/2 , y = 1
(i) find the equation connecting y and x

Q8) y is inversely proportional to the square of x, find the equation of the curve when x=1 and y =3
if it's inversely proportional then y = 1/x^2
and since y= 3, I just tried to join the two y values together like in functions so I multiplied 3 by 1/x^2 to get y = 3/x^2

Q9) The quantity of y is inversely proportional to a power of x for x >0
when x = 1, y = square root 2
when x=8, y=1/2
like above I just tried to join them together and got y= square root of (2x1 / x^2)

I got the right answers but the method doesn't really make sense because I was working back from the answers and trying to find out how to get to the answer. Can someone tell me a better method or explain how to to it?
0
1 year ago
#5
6) you can say

y = k√x

8) you can say

y = k/x2

9) you can say

y = k/xn
1
#6
(Original post by the bear)
6) you can say

y = k√x

8) you can say

y = k/x2

9) you can say

y = k/xn
I know the answers but I don't know what method to use to get there thank you anyway 0
1 year ago
#7
(Original post by Mia 15)
I know the answers but I don't know what method to use to get there thank you anyway What you have missed is that when it says it is inversely proportional you can’t just skip to

Y=1/x^2.
-there needs to be a coefficient in there also called k.
Y=k/x^2
You then put the points in that you’re given to work out the value of k.

And once you know the value of k, you can put it into the initial equation y=k/x^2
Last edited by GabiAbi84; 1 year ago
0
#8
ohh thank you 0
1 year ago
#9
(Original post by Mia 15)
ohh thank you Does that make sense okay?
0
#10
(Original post by GabiAbi84)
Does that make sense okay?
yes I forgot you had to work out k first, I can't rate you because it says 'please rate other members before you rate this member again' but thanks so much! 0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (130)
65.99%
No (45)
22.84%
I didn't use it to prepare (22)
11.17%