anonymous
  • anonymous
find the volume of the given solid: under the plane x+2y-z=0, and above the region bounded by y=x and y x^4
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

TuringTest
  • TuringTest
where are you stuck?
anonymous
  • anonymous
find the value for x
anonymous
  • anonymous
finding*

Looking for something else?

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

More answers

TuringTest
  • TuringTest
where do the graphs of y=x and y=x^4 intersect?
anonymous
  • anonymous
at 1?@TuringTest
TuringTest
  • TuringTest
that's one of the intersection points, yes and the other?
anonymous
  • anonymous
16
anonymous
  • anonymous
|dw:1360707284131:dw|
TuringTest
  • TuringTest
x=16 is not an intersection, set the two equations equal and solve:\[x=x^4\]\[x^4-x=0\]\[x^3(x-1)=0\]so the intersections are x=1 and x=?
TuringTest
  • TuringTest
typo, should be\[x^4=x\]\[x^4-x=0\]\[x(x^3-1)=0\]
anonymous
  • anonymous
-1 to 1
TuringTest
  • TuringTest
\[x(x^3-1)=0\implies x=0\text{ or }x^3-1=0\implies x=1\]so\[x=\{0,1\}\]try drawing the graphs out if you doubt that

Looking for something else?

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