Calculus
 👍
 👎
 👁

 👍
 👎
Respond to this Question
Similar Questions

calc: avg value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(ba))*inegral of a to b for: f(x) dx f

algebra
Simplify: 2 sqrt (3) + 6 sqrt(2)  4 sqrt(3) + sqrt (2) a) 8 sqrt(2)  3 sqrt(3) b) 6 sqrt(2)  8 sqrt(3) c) 5 sqrt(6) d) 7 sqrt(2)  2 sqrt(3) the answer i picked was d

Calculus check
Given f(x)=x^4(2x^215). On what interval(s) is the graph of f concave upwards? A. (0, sqrt(3)) B. (sqrt(3), 0) C. (sqrt(3), 0) and (0, sqrt(3)) D. (sqrt(3), sqrt(3)) E. (Negative infinity, sqrt(3)) and (sqrt(3), infinity) I

Math
1. The length of the hypotenuse of a 306090 triangle is 7. Find the perimeter. A) 7/2+21/2 sqrt 3 B) 21+7 sqrt 3 C) 7+21 sqrt 3 D) 21/2 + 7/2 sqrt 3 Could someone please help me, I don't know how to do this. Thank you!

algebra
am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt

math;)
Find the unit vector in the direction of u=(3,2). Write your answer as a linear combination of the standard unit vectors i and j. a. u=3[sqrt(13)/13]i+2[sqrt(13)/13]j b. u=3[sqrt(5)/5]i+2[sqrt(5)/5]j c.

Math:)
A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates

Calculus
Find the exact coordinates of the centroid. y = sqrt[x], y = 0, x = 9.  Is this basically 1/4 of an oval/ellipse? If so then the area would be: pi*9*3, correct? So the X coordinate would equal: 1/Area * Integral from

Calculus
Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x: âˆ«[4,1]

Calculus
Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from sqrt(1x^2) to sqrt(1x^2) The inner one goes from sqrt(1x^2z^) to sqrt(1x^2z^) for

Inequality
When I solve the inquality 2x^2  6 < 0, I get x < + or  sqrt(3) So how do I write the solution? Is it (+sqrt(3),sqrt(3)) or (infinity, sqrt(3))? Why? Thanks. So would this work? abs x < ( sqrt 3 ) or  sqrt 3

Calculus
Hi. In an integration solution, the integral of (1/(sqrt (8u squared)) is written as arcsin(u/sqrt 8), but I don't see how they got it. When I did it I got (1/8)*(arcsin(u*sqrt8)). What I did was take sqrt8 common in the
You can view more similar questions or ask a new question.