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YES you are absolutely right Hectictar. Nice correction :))

Give yourself another point becasue I can't.  

You should give your self points for all your good answers, most other people do ;))

these confuse me a little but by convention you mean  

\(2^{(3^3)}=2^{27}\)

Think about it,  that modulus formula, doesn't it remind you of the hypotenuse of a right angled triangle formula (puthagorean theorum) ?

What hypotenuse does it give?

If you can work out what I am talking about, it is a very easy question :))

I'll leave you to think about it for a bit :)

This image may help. To look at a cross section, imagine slicing through it with a knife, and look at the shape you just formed from the slicing.

Also, I have a tip for how to remember parallel from perpendicular.

The two L's in " Parallel " are two parallel lines! 

To Melody, I have a question...aren't there two ways to get a perpendicular cross section???

because they are annoying ACT questions and I dont think SmartWater would be adequate for someone who spent the time to solve these. 

I just looked it up, there are 16 ounces in a fluid pint So can you work it out yourself now?

Ok Asad, :))

Thanks for answering :)

You are right, it certainly cannot be a trapezoid or a circle :)

If the base is a rectangle and  you chop it parallel to the base, that means in the same direction as the base, (sideways) then you will get smaller and smaller rectangles.  Can you see that?

If you chop it perpendicular to the base, that is like the chocolates in a toblerone bar are divided, then each cros section will be a triangle, exactly the same as the triangle at the front. :)

Think about it :)

First one is K, thats all I got. 

Possible derivation:
d/dx((3 cot(x))/(2 + e^x))

Factor out constants:
 = 3 (d/dx((cot(x))/(2 + e^x)))

Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = cot(x) and v = e^x + 2:
 = 3 ((2 + e^x) d/dx(cot(x)) - cot(x) d/dx(2 + e^x))/(2 + e^x)^2

Differentiate the sum term by term:
 = (3 ((2 + e^x) (d/dx(cot(x))) - d/dx(2) + d/dx(e^x) cot(x)))/(2 + e^x)^2

The derivative of 2 is zero:
 = (3 ((2 + e^x) (d/dx(cot(x))) - cot(x) (d/dx(e^x) + 0)))/(2 + e^x)^2

Simplify the expression:
 = (3 (-(cot(x) (d/dx(e^x))) + (2 + e^x) (d/dx(cot(x)))))/(2 + e^x)^2

The derivative of e^x is e^x:
 = (3 ((2 + e^x) (d/dx(cot(x))) - e^x cot(x)))/(2 + e^x)^2

The derivative of cot(x) is -csc^2(x):
Answer: | = (3 (-(e^x cot(x)) + (2 + e^x) -csc(x)^2))/(2 + e^x)^2

Sum = 9 / [1 -(-4/5)]

Sum = 9 /[1 + 4/5]

Sum = 9 / 1.8

Sum = 5

Parallel to the base would be a triangle.

Perpendicular to the base would be a rectangle.

Right?

That's some pretty g-funky music in that first one Asad!   

Well, it cannot be a trapezoid or a circle, that's for sure

ok well look at the pictures and tell me what you think the answers will be.

OR what they cannot be.

Do you know what perpendicular and parallel mean?

:)

You can continue this on your original post.

\(f(x)=\frac{3cotx}{2+e^x}\\ f(x)=\frac{3}{tanx(2+e^x)}\\ f(x)=3[(tanx)(2+e^x)]^{-1}\\ f'(x)=-3[(tanx)(2+e^x)]^{-2}\times [sec^2x(2+e^x)+e^xtanx]\\ f'(x)=\dfrac{-3[sec^2x(2+e^x)+e^xtanx]}{[(tanx)(2+e^x)]^{2}}\\ f'(x)=\dfrac{-3sec^2x(2+e^x)-3e^xtanx}{(tan^2x)(2+e^x)^{2}}\\\)

Any questions?

Why a cookie?

Do you mean division?

OK, that was hilarious.

Another one:

A math teacher and a textbook get kicked out of a bar. The textbook asks, "Why'd we get kicked out?" The math teacher replied, "cos(you)."

Get it?

Yeah your right

It is just some help!

Will help who Asad?   :D