anonymous
  • anonymous
will give medal for answer and explanation pls help What is the constant of variation for the quadratic variation? 9y = 4x2 A.-2 B.4/9 C.2/3 D.9/4
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
and your thoughts?
anonymous
  • anonymous
im thinking b?
anonymous
  • anonymous
mathway?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Okay, so since y is being multiplied by 9 and you want y by itself, divide both sides by 9.
anonymous
  • anonymous
1 and 2
anonymous
  • anonymous
9/9.1 4/9.2
anonymous
  • anonymous
Remember that y=kx^2. And you want to know k.
anonymous
  • anonymous
|dw:1436053521059:dw|
anonymous
  • anonymous
You're correct, though.
anonymous
  • anonymous
Do you know what k is? :)
anonymous
  • anonymous
thanks for your help can you help me with one more
anonymous
  • anonymous
Sure. :)
anonymous
  • anonymous
k
anonymous
  • anonymous
If f(x) varies directly with x2, and f(x) = 75 when x = 5, find the value of f(4). A. 100 B. 50 C. 48 D. 25
anonymous
  • anonymous
Would be better if you'll open a new question. :)
anonymous
  • anonymous
ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.