inverse proportions

[homeschool mom]

Could someone help me solve this problem ?

Suppose that a^2 varies inversely with b^3. if a=7 when b=3 find the value of a when b=6?

 

[MarkFL]

Hello, and welcome to FMH!

I would expect that you have likely already written something like:

[MATH]a^2=\frac{k}{b^3}[/MATH]
Have you in fact done this?

 

[homeschool mom]

yes

 

[homeschool mom]

ultimately I got a = 220.5 but I was unsure if I was doing it properly as its a topic I'm learning for the first time

 

[MarkFL]

yes

Okay, so given:

[MATH](a,b)=(7,3)[/MATH]
We find:

[MATH]k=a^2b^3=(7^2)(3^3)=1323[/MATH]
And so we have:

[MATH]a^2=\frac{1323}{b^3}[/MATH]
When \(b=6\), what do you get for \(a\)?

 

[homeschool mom]

a= +or - 7sqrt2 /4 ?

 

[MarkFL]

a= +or - 7sqrt2 /4 ?

Yes, I get the same.

 

[Steven G]

Could someone help me solve this problem ?

Suppose that a^2 varies inversely with b^3. if a=7 when b=3 find the value of a when b=6?

I like to think that varies inversely with means =k/

So a^2 varies inversely with b^3 becomes a^2 =k/b^3

 

[homeschool mom]

Yes, I get the same.

Thank you so very much for your help i truely appriciate it