# need help on maths questions

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#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
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1 month 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 ...
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1 month ago
#3
waiting...
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#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?
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1 month 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
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1 month 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 month ago
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#8
ohh thank you
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1 month ago
#9
(Original post by Mia 15)
ohh thank you
Does that make sense okay?
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#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!
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