posted by .

4b^7/5t^2 x 25t^4/16b
Could someone help me with this?

  • math -

    4b^7/5t^2 x 25t^4/16b , I will assume you mean
    = (4b^7/(5t^2) (25t^4/(16b)
    = (5b^6 t^2/4

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. mat117

    Factor each expression a^2(b-c)-16b^2(b-c) Help show me how (b-c) appears in both terms and can be factored out, giving you (b-c)(a^2-16b^2) Now note that the second term can also be factored since it is the difference of two perfect …
  2. Math

    How would I do: sqrt32a^8b + sqrt50a^16b Are the exponents 8b and 16b or 8 and 16?
  3. math

    I've worked these problems but I was just wondering if someone could check over them to see if they're correct...if not please help! (u-2)(u+7) my answer: 7u^2-49u (5t-4)^2 my answer: 25t^2-40+16 Thank you :)
  4. Math

    Simplify: [(a^2-16b^2)/(2a-8b)] divided by [4a+16b)/(8a+24b)] Please Help!!!! Thanks.
  5. MATH!!!!!!

    16b^2 - 25z^2 = A- (4b-5z)^2 B- (4b+5z)^2 C- (4b-5z)(4b + 5z) D- (16b-25z)^2 E- (5z-4b)(5z+4b)
  6. 7th grade math

  7. math binomial expansion

    binomial expansion (3a^2-2b^3)^4 is this right?
  8. alg 2

    The path that a football takes can be described by the euation h=25t-5t^2 where h is the height in meters, of the football, at t, in seconds. How high is the ball after 3 seconds?
  9. Calculus

    The particular solution of the differential equation dy/dt=y/4 for which y(0) = 20 is y = 20e^-0.25t y = 19 + e^0.25t y = 20 e^0.25t y = 20^e4t
  10. math

    The particular solution of the differential equation dy/dt equals y over 4 for which y(0) = 20 is y = 20e−0.25t y = 19 + e0.25t y = 20 e0.25t - my answer y = 20e4t

More Similar Questions