Help with cone/frustum volume maths question? GCSE?
I have a picture link. I have tried to do this so many times on mymaths but I just don't understand what I'm supposed to do?! Here's the pic: http://i48.tinypic.com/10n5cw6.png Thank you
Grade 10 Math Question Help!?
A rectangular picture frame measures 20cm by 30cm. a new frame is to be made by increasing each side length by the same amount. The resulting enclosed area is to be 1064cm. Find the dimensions of the new garden. Make an equation to solve the problem.
What is the Integral of sec^3(3x) with all steps included?
⌠ sec^3(3x) dx = ⌠ sec(3x) sec^2(3x) dx by parts: u = sec(3x) du = 3 sec(3x)tan(3x) dx dv = sec^2(3x) dx v = (1/3)tan(3x) = sec(3x)[(1/3)tan(3x)] - ⌠ [(1/3)tan(3x)][3 sec(3x)tan(3x)] dx = (1/3)sec(3x)tan(3x) - ⌠ tan(3x)sec(3x)tan(3x)] dx = (1/3)sec(3x)tan(3x) - ⌠ tan²(3x)sec(3x) dx = (1/3)sec(3x)tan(3x) - ⌠ [sec²(3x) - 1 ]sec(3x) dx = (1/3)sec(3x)tan(3x) - ⌠ [sec3(3x)dx +⌠sec(3x) dx so .. ⌠ sec^3(3x) dx = (1/3)sec(3x)tan(3x) - ⌠ [sec3(3x)dx +⌠sec(3x) dx transposing ⌠ [sec3(3x)dx from the RHS to LHS 2 ⌠ sec^3(3x) dx = (1/3)sec(3x)tan(3x) + ⌠sec(3x) dx 2 ⌠ sec^3(3x) dx = (1/3)sec(3x)tan(3x) + ⌠sec(3x)[(sec(3x)+tan(3x))/(sec(3x)+tan(... dx 2 ⌠ sec^3(3x) dx = (1/3)sec(3x)tan(3x) + ⌠{sec²(3x) + sec(3x)tan(3x))/(tan(3x)+sec(3x))] dx u = tan(3x)+sec(3x) du = [3sec²(3x) + 3tan(3x)sec(3x)] dx 2 ⌠ sec^3(3x) dx = (1/3)sec(3x)tan(3x) + (1/3)ln[tan(3x)+sec(3x)] + c therefore ⌠ sec^3(3x) dx = (1/6)sec(3x)tan(3x) + (1/6)ln[tan(3x)+sec(3x)] + C
Help with math question involving the frustum of a cone?
Find the total area of a frustum of a cone if the smaller circle on top has a radius of 4, and the larger bottom circle has a radius of 8 and if the height is 5. picture just in case of a plain one: http://www.analyzemath.com/Geometry/coni... thanks in advance. 10 points for person who explains step by step =)
The graph below shows the solution to which system of inequalities? PIC INCLUDED!?
The answer is B. The line on the top is y = (x/6) + 2 The line on the top is y = (x/4) + 1 For inequalities you need to plug in (0,0) for (x,y).....If the statement is true then the solution (shaded region) is towards the origin of the cartesian coordinate plane, If the statement is false then the solution goes away from the origin. For y ≤ (x/6) + 2..........0 ≤ 2---> True For y > (x/4) + 1.........0 > 1----> False Therefore the solution must be in between both lines.