CPhill

I don't believe this might be correct, Melody.......your drawing has "A" at the bottom left, and, in order clockwise from this point, we have "B," "C" and "D."  And the problem states that "B," "C" and "D" lie on the horizontal plane. So this triangle must form the "base." And, by Heron's formula, its area is

A = √(s(s-BD)(s-BC)(s-CD))  where s = [10 + 6 + 6]/ 2   = 11 

So we have

A = √((11)(1)(5)(5)) = √(275) = 16.583 units  (???)

Actually....I'm not sure we have anything, here......If BCD lies on the horizontal plane, DAB is just some triangular "flap" that shares a common side with BCD.... ("A" doesn't touch the plane, according to the problem)

Maybe I'm seeing this incorrectly, or interpreting it incorrectly???

We can find one angle using the tangent inverse....we have

tan^-1(30/.9375) = about 88.21°

So...the other angle is just (90 - 88.21)°  =  about 1.79°

The volume of the water and the marble added together will provide a clue.

V_water = pi(r^2) h = pi(4cm)^2(6cm)  = 96pi cm^3

V_marble = (4/3)pi(1.2cm)^3 = 2.304pi cm^3

So.....the volume of the two things added together is 98.304pi cm^3

So we have the volume of the two things occupying some space in a cylinder =

98.304pi cm^3 = pi(4cm)^2(h)    and solving for h, we have

98.304pi cm^3 / (4cm)^2 = h = 6.144 cm

So the water will rise .144 cm

Sorry....I didn''t see your train pull into the station....!!!!

I think it was the sunglasses...!!!

Sorry...I should have been a little more explicit...we have

2 ≥  1/x^2     multiply both sides by x^2

2x^2 ≥ 1       divide both sides by 2

x^2 ≥ 1/2      subtract 1/2 from both sides

x^2 - 1/2 ≥ 0

Does that help ??

Smart aleck.......

Very clever, Melody........that initial "trick" of mutiplying the top and bottom by  2√x  was the key, wasn't it....???....I could have looked for hours and never discovered that......

(Maybe I should have bought some Coronas, too!!)

Think of it like this......10% of 33 is 3.3 ....so.... 20% must be twice as much.....

Yeah.....it seemed like too much.....I'm a little suspicious about my answer.....there's probably a more direct way to do it.......but my brain is too fried to want to think of another way!!!

We can write L as

y = mx + 1

So, let us find the intersection of these two functions

mx + 1 = x^2 - 2x + 2

X^2 - 2x + 2 - mx - 1 = 0

x^2 -(2+m)x + 1 = 0

So, using the quadratic formula, we have

x = [( 2 + m) ± √[(2+m)^2 - 4]/2 = [(2 +m) ±√(m^2 + 2m)] /2

So, the y coordinates of M, N lie on line L and are given by

y₁ = m[(2+m) + √(m^2 + 2m)] /2 + 1    and

y₂ = m[(2+m) - √(m^2 + 2m)] /2 + 1 

Adding these two we have

[m[(2+m) + √(m^2 + 2m)] /2 + 1 ] + [ m[(2+m) - √(m^2 + 2m)] /2 + 1 ] =

[m(2 + m) + 2]

And dividing this by 2, we get the midpoint =

(m/2)(2 + m) + 1 =

(1/2)m^2 + m + 1