How would I write T varies jointly as r and n?
This is what I've got T=krn Not sure if that's right
Well that makes sense.
Okay so the key concept of combined variations is : z varies jointly with x and y Equation Form: z=kxy
So I was right
Yes you are are correct :)
Ant other q's?
Yes. One second. Thanks
What's the format of a power function? I'm given four options and I'm not sure which one it is because I'm unsure of the format
A power function is a function of the form f(x) = x^a, where a is any real number.
Is f(x)=9x^(2/3) one?
I've got it narrowed down to that or f(x)=9x^(2/3)+27
Yes, I believe it would be the first one according to the general form of the function. Now later you learn of negative and positive derivates, if this is the case then the equation could be altered, but If your just learning this I would go w/ the first choice :)
Thanks. Do you have any problem helping me with some more? I've got my answers, just unsure if they're right. I missed this day at school
Go ahead, and if we can't finish just post another open q, that others can help u w/ :)
Thanks a bunch! If I have y=8x^2, when x is halved, y is...
I have y gets multiplied by 4
y=(8x^2) / 2 ? that would be y= 4x^2 , but if you talking about the value of x, what happens to one half of the equation has to be balanced in accordance to the other.
Then, yes you would be correct
No, if the value of x is halved.
Oh, okay, good.
So in g(x)=6x^4 If x doubles in value... y is multiplied by 16?
Oh wait, divided by 16, no?
I'm confused
Remember as I stated before what happens to one half of the equation has to be balanced in accordance to the other. You can't subtract 4 from one side without subtracting from the other. So actually I'm not sure but i think you would multiply by 16 look you might want another opinion, but to solve x+4= 6 You would subtract 4 from both sides, not do the opposite of. -4 -4
Ok, this is why I'm confused, would doubling the value of x have the same result as halving it?
@radar @robtobey @jhonyy9 @Loser66 can you help??
Sorry g2g, but your on the right track :)
One of these people will come to help :) They should help u understand ur q's
Alright thanks
Anyone?
@radar
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