Good job, Jver!!! Your award will be sent to you via PayPal later today!
Your lucidity and level of detail is astonishing! The resolution of the other problems posted by help-needing members is also […]

Let √x = y. Then y^2 = x and 2y(dy) = (dx)
∫3(√x)(x-1)(dx)
= 3∫(√x)(x-1)(dx)
= 3∫(y)(y^2-1)(2y)(dy)
= 6∫(y^2)(y^2-1)(dy)
= 6∫(y^4 – y^2)(dy)
= 6[(y^5)/5 - (y^3)/3] + C
= 6(y^5)/5 – 2(y^3) + C

Suppose we have a circle with radius r. The circumference C of the circle is of the formula 2r(pi) while the area A of the circle is of the formula (pi)r^2. If we want the ratio of the square of the circle’s […]

A cone with slant height h = 4 inches can be unfolded into a semi-circular region with radius r = h = 4 inches. Since the unfolded figure is a semi-circle, the length of its curved edge is half the circumference […]

You have 9 dials with 5 settings (0,1,2,3,4) each.
To produce an arrangement where adjacent dials do not have the same settings:
1) choose a setting for the left most dial (5 choices)
2) choose a setting for the […]

In the diagram that I attached, we have the circle with radius R and center O that is inscribed in the square ABCD.
Let points E and F be the points on the circle that are tangent to sides AB and AD respectively.<br […]

x^2+x-12≥ 0
For starters, this is the same as
(x+4)(x-3)≥ 0
The graph intersects the y axis at x=-4 and x=3.
This is a parabola in a “vertical position opening in the upward direction”. Sorry I can’t graph it […]

Which is (1±3√2)/5
Thanks TheMathProf! Thanks a lot! If you don’t mind, please copy and paste this excellent answer to the same problem on our Facebook page, posted by the same member! And thanks for being a […]

I am not sure that the problem is stated properly. Square root property is used when solving quadratic equations. There is no 2nd degree term here.

I am sorry. I had to correct my answer. My apologies. The Facebook answer was the correct answer.