CPhill

So....from my reading, the answer to "a" would be in this set-up??

pi *∫ (h/2)² - y²) dy  =  pi * ∫ (12/2)² - y²) dy =   pi * ∫ (6)² - y²) dy ..... where the limits on the integral are -6 and 6 ......???

Your set-up to "a" is slightly incorrect...it should be

A = 5000(1+.05/12)^12*6  =  $6745.09

For "b', the amount invested doesn't really matter....let's see why......

3A = A (1 + .05/12)^12t      divide through by A....

3 = (1 + .05/12)^12t       (see why "A" doesn't matter??) .......take the log of both sides

log3 = log (1 + .05/12)^12t         and by a property of logs, we can write

log3 = (12t)*log(1 + .05/12)       divide both sides by 12*log(1 + .05/12)

[log3] / (12*log(1 + .05/12)) = t = about 22.018 years

Sorry...I meant to type "function" rather than "equation".....I'll stand by the rest...

The obtuse angle in any triangle will be opposite the longest side. We can use the Law of Cosines to find this.   So we have

11^2  = 9^2 + 5^2 - 2(9)(5)cosx     where x is the angle we're looking for

121 = 81 + 25 - 90cosx          rearranging, we have

90cosx = 81 + 25 - 121          simplify on the right

90cosx = -15        divide both sides by 90

cosx = -15/90   =  -1/6       use the acos function to find x

acos(-1/6) = x ≈ 99.59°

f(x)=9x² -2

This is a parabola that opens upward with a vertex at (0, -2). Thus the range is just

[-2, ∞) and since we can substitute anything in for x, the domain is (∞,∞)

f(x)=√(9x² -2)

Since the result of taking a positive square root is always positive or 0, the range is just [0,  ∞). And since we can't have a negative value under the square root, we can just see what makes 9x² - 2 = 0.

9x² -2 = 0

9x² = 2     divide both sides by 9    

x² = 2/9    take the square root of both sides

x = ±√2/3

So the domain is given by (-∞, -√2/3] U [√2/3, ∞)

2√3 - √27

Note that we can write √27  as √9 * √3   = 3 * √3 = 3√3

So we have

2√3 - 3√3   =   -1√3  or just  -√3

And that's it.....

Translation:  I like achievement of another way to write my question is equal to the root of 1/6 but not the result but the other way to say

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I'm assuming you want to know how to express another way of saying that somethig is raised to the "1/6" power.....

It's the same as saying the "6th root" of that thing.......

I hope that is what you were asking....

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Estoy asumiendo que usted quiere saber cómo expresar otra forma de decir que somethig se eleva a la potencia "1/6" .....

Es lo mismo que decir el "sexto raíz" de esa cosa .......

Espero que sea lo que estabas preguntando ....

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The trig function that relates the opposite and the hypoteneuse is the sine...thus.....

sin(x) = opp/hyp = 4/15

To find this angle, use the arcsin function

arcsin(4/15) = x ≈ 15.47°

10*4 = 40

40/6 = 20/3  = 6+2/3 =  6.6666.......

.....Who ever said I was "trustworthy??"........

On this forum....you pays your money......you takes your chances....!!!!