CPhill

This is an equation relating to the Gateway Arch in St.Louis.......I don't know if the poster wanted an answer or was just curious as to what it was (I wouldn't know how to give one, anyway).....but I guess if he/she were interested, they could look at this page:

The volume of clay is given by volume of the outer hemisphere less volume of the inner hemisphere =  (2/3)*pi (8.2^3 - 7.7^3) ≈ 198.6cm³ = pi^3 * k

Dividing this by pi^3, k ≈ 6.41cm³

I'm assuming that this topic relates to the volume of spheres ???

5/9 is larger than 1/3...so it has to be more than 100% of it

1/3 = 3/9     so dividing, we have

5/9 / 3/9   = 5/3  = 1.66 ≈ 166%

We might be able to use some kind of sum or difference identity, but it would be tough trying to find two such angles with known cosine and sine values.....we might know one, but the second would be as unknown as cos(80). For example, we know cos(90) and sin(90), but what's cos(10) and sin(10)??

My advice.....just put it into the calculator and evaluate....make sure you're in degree mode.

i^3 = i^2 * i   = (-1) * i  = -i

So......

-i + x = 3        add i  to both sides

x = 3 + i

Note that Charo's line of sight is 5 ft up the pole, so we have that the angle of elevation = opposite side/ adjacent side = 168/50

What function relates the opposite and adjacent??

Take the inverse of this function to find your answer.......

See if you can find the solution.....if you have trouble.....let us know

The limit of any constant is just that constant......

(³√64)^3/2 +2X-3x = 62

Taking it in steps, we have

(64^(1/3))^3/2 +2X-3x = 62

(64)^(1/2)  - x = 62

√64 - x = 62

8 - x = 62

x = 8 - 62  =  -54

Very, very nice explanation, NinjaDevo !!!   One of the best I've seen on this site for the surface area of a cylinder. I'd give you 6pts (If I could), but alas, I'm limited to just 3 and a BIG "Thumbs Up."

P.S.  .....I've recommended that this one be highlighted in Melody's "End of the Day" wrap!!

arc BC = 2(48) = 96...remember....an inscribed angle measures 1/2 of its intercepted arc

Now...I'm going to ask you some questions so you can finish this...

What is the measure of the arc intercepted by the inscribed angle B..isn't it 110 degrees??? (What is the name of this arc ???) And....what did I just tell you about the measure of an inscribed angle??

And arc AB must be 360 -arc BC - arc AC = ????

So, angle C would be 1/2 of arc AB....so angle C = ????

Post again if you're having trouble with this one.......but try to solve some of it, if you can....