ksaimouli
  • ksaimouli
find intersection in terms of y of y^2-4x=4and 4x-y=16
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!
ksaimouli
  • ksaimouli
\[y^2-4x=4 and 4x-y=16\]
ksaimouli
  • ksaimouli
@ZeHanz
Hero
  • Hero
Hint: y = 4x - 16

Looking for something else?

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

More answers

ksaimouli
  • ksaimouli
i did but i dont know whre i was wrong
ksaimouli
  • ksaimouli
i know dude to solve for x and set it eual and then solve for x
ksaimouli
  • ksaimouli
|dw:1357491125167:dw|
ksaimouli
  • ksaimouli
|dw:1357491191134:dw|
Hero
  • Hero
Don't make it complicated bro
ksaimouli
  • ksaimouli
|dw:1357491240591:dw|
ksaimouli
  • ksaimouli
solve for y am i right this is to find area between curve sooo i must to with respect y
Hero
  • Hero
Actually, that's a good approach after all.
ksaimouli
  • ksaimouli
but something is wrong in solving hmm i am not getting the answer
ksaimouli
  • ksaimouli
multiply both sides by 4
Hero
  • Hero
What did you get for the answer?
ksaimouli
  • ksaimouli
.08 and -12.08
Hero
  • Hero
That's what you got for y values?
ksaimouli
  • ksaimouli
nope
ksaimouli
  • ksaimouli
i gote it 5 and -4
Hero
  • Hero
Okay, so what's the problem?
ksaimouli
  • ksaimouli
arithmetic mistake lol
ksaimouli
  • ksaimouli
now i have to integrate thats it
ksaimouli
  • ksaimouli
|dw:1357491558149:dw|
ksaimouli
  • ksaimouli
but not sure which one goes top and which goes bottom with respect to y
Hero
  • Hero
Here's another way to do it to avoid confusion: \[\frac{y^2 - 4}{4} - \frac{16 + y}{4} = 0\]
ksaimouli
  • ksaimouli
ya i tried that worked well
Hero
  • Hero
Bro, a is always the lowest value. b is always the highest value.
ksaimouli
  • ksaimouli
i mean the function not the points (curves)
ksaimouli
  • ksaimouli
\[x=\frac{ y^2-4 }{ 4 }\]
Hero
  • Hero
Are you sure you're supposed to be using y values for a and b?
ksaimouli
  • ksaimouli
or\[x=\frac{ 16+y }{ 4 }\]
ksaimouli
  • ksaimouli
ya i am damn sure
Hero
  • Hero
I remember these. They can be tricky. What you do is just turn your graph sideways. And it will be the same as like when you're working with x-axis.
ksaimouli
  • ksaimouli
if \[\sqrt{y}\]
ksaimouli
  • ksaimouli
|dw:1357491948129:dw|
ksaimouli
  • ksaimouli
?
ksaimouli
  • ksaimouli
alright thx
Hero
  • Hero
Bro, I graphed it as \[\frac{16 + x}{4}\] and \[\frac{x^2 - 4}{4}\] You get the same area when integrating from a = -4 to b = -5 The only difference is you're using variable y instead of variable x
Hero
  • Hero
@ksaimouli
ksaimouli
  • ksaimouli
ya i noticed it thx

Looking for something else?

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