Help with mid-level algebra? (variation equations)

[Snowbear]

Closed, thanks everyone!

 

[Lean]

http://archive.org/stream/ost-math-ck_12_algebra___basic/CK_12_Algebra___Basic_djvu.txt

 

[Snowbear]

http://archive.org/stream/ost-math-ck_12_algebra___basic/CK_12_Algebra___Basic_djvu.txt

I'm sorry, I found that text hard to follow.. could you perhaps just solve one or two of the above problems with an explanation? I learn really well by seeing a template of the process.

 

[Clashfan]

Come in sythechat and I'll show you how

 

[iAGZzzz]

For inverse relations, you're looking for a constant, let's say k, in the relationship y = (k / x), where in this case y = F and x = d^2.

So, you're solving for k in the equation: F = k / (d^2)

Now you evaluate the function given the parameters already provided, that is "F=30 when d=5"

We have: 30 = k / (5^2), or 30 = k / 25

You can solve for k now; your answer will be in the form of a general equation: F = k / (d^2).

 

[Snowbear]

Closed!

 

[Alex12345]

http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/proportionhirev2.shtml
*should* make things pretty simple

 

[Clashfan]

but for this problem

M varies directly with the square of d and inversely with the square root of x; M=12 when d=4 and x=9

it says the formula is

y = k/x

So that gives me: 12 = k/?

What do I do with the other variable? Do I somehow use two equations?

When it varies directly with one variable and inversely with another you do y = k * (direct variable / inverse variable)
So 12 = k * (4^2 / sqrt(9) )

 

[Bangclaw]

When it varies directly with one variable and inversely with another you do y = k * (direct variable / inverse variable)
So 12 = k * (4^2 / sqrt(9) )

That slipped my mind anyways so thanks for posting!