Maths homework?

Question

Ok need desperate help as its in for tommorow and dont have a clue where to start.
q1)
y is proprtional to the cube of x find the when x=10 y=1
find y when x=20
find x when y=5
and also q2)
y is inversely proprtional to the sqaure of x
what happens to y when x is doubled?
y is multiplied by 25 what happens to x?

Answer 1

Just don't pay any attention to moshetoshe. He's even more clueless than you are ☺

y is proportional to x^3 means y = ax^3 where a is some constant. If y = 1 when x = 10 then 1 = 2*10^3 = a*1000 so that a = .001 (or 1/1000) What happens when x = 20?
y = (1/1000)*20^3 = (1/1000)*8000 = 8. When y = 5 then 5 = (1/1000)*x^3 so x = cube root of 5000 = 17.099

If y varies inversly as the square of x, then y = a/x² where a is some constant. If x doubles then y = a/(2x)² or
y = a/4x² = (1/4)*(a/x²) so that y becomes 1/4 of it's original value. If y is multiplied by 25 then x must be multiplied by √25 = 5.

Now get yer butt in gear and *learn* this stuff, 'cause I am **not** gonna show up to do your tests ☺


Doug

Answer 2

Your question is a little unclear, but I think you're saying that when x = 10, y = 1. Since y is proportional to the cube of x, doubling x to 10 would increase y by a factor of 2^3 = 8. If you increase y from 1 to 5, that's a factor of five, so x would be increased by 5^(1/3), because it has the inverse relationship with y compared to what y has with x.

For the second question, y is inversely proportional to the square x, so doubling x would multiply y by 1/(2^2) = 1/4. Squared because it's related to the square, but in the denominator because it's an inverse relationship. If you multiply y by 25, x must have been multiplied by 1/(25^(1/2)) = 1/5. Hopefully you see what the pattern is.

Answer 3

q1)i can't follow the first line after q1)

if y is proportional to the cube of x,we can say

y=ax^3 ..............(1) where a is a

constant

given initial conditions, x=10,y=1

from (1) we have 1=a*10^3 >>>>>>>>> giving a=1/1000

now we have y=x^3/1000.............(2)

when x=20, y= 20^3/1000 =8 (from (2))

from (2), x=(1000y)^(1/3) ...........(3)

when y=5, x= (1000*5)^(1/3) = 17.1 approx from (3)

q2) if y is inversely proportional to the square of x,we can say

y=b/x^2 where b is a constant............(4)

b= y x^2 is a constant

so we can say, ( y1) (x1)^2 =(y2) (x2)^2 ...........(5)

when x is doubled, it becomes 2x >>>> (x2)=2(x1)

sub into (5), we have y1(x1)^2 = y2(2(x1))^2
= 4(y2) (x1)^2
>>>>>>>> (y1) = 4(y2) (divide by (x1)^2 )

>>>>>>>> (y2) = (1/4)(y1)

so, when x is doubled, y is only a quarter of its original value

when y is mult by 25, (y2)=25 (y1)

sub back into (5) we have (y1)(x1)^2 = 25 (y1)(x2)^2

>>>>>>>> (x1)^2 = 25 (x2)^2

>>>>>>>> (x1)/(x2) =(25)^(1/2) =5

so,when y is mult by 25 x is reduced to 1/5 of it's original value

i hope this helps

Answer 4

q1)
y = kx^3
Sub x=10 and y = 1, k = 1/1000
So, y = (1/1000)x^3

When x=20, y = (1/1000)(20^3) = 8
When y=5, 5 = (1/1000)x^3 => 5000 = x^3 => x = 17.1

q2)
y = k/x^2

When x is doubles, i.e becomes 2x,
y = k/(2x)^2
y = k/4x^2
y is reduced by 4 times.

When y is multiplied by 25,
25y = k/x^2
y = k/(25x^2)
y = k/(5x)^2
x is increased by 5 times.

Answer 5

q1 y=cx^3, c=0.001 x=20 y=8; y=5 x^3=5000
q2 y=c/(x^2), x*2, y/4; y*25, x/5

Answer 6

Maths was never my strong point and it will never be yours either if you always get someone else to do it for you. Especially if its not explained to you how to do it

Answer 7

q1)
a)2
b)50
q2)
a)it is also doubled
b)it is also multiplyed by 25

Answer 8

do your own homework it will make you a better person later in life :)

Answer 9

guess you should've listened in class...