MaxWong

\(\text{Base length}=2\times\sqrt{13^3-12^2}\\ \;\;\qquad\qquad\;\;=2\times 5 \\\;\;\qquad\qquad\;\;= 10\)

\(\text{Surface area}=\text{Base length}^2+2\times \text{Base length}\times \text{Slant height}\\\qquad\qquad\quad\;=10^2+2\times 10 \times 12\\\qquad\qquad\quad\;=340 \text{cm}^2\)

233.44

Calculators are your best friends remember :D

\(2\sqrt2\times2\sqrt2\\=2\times2\times{\sqrt2}^2\\=2\times2\times2\\=8\)

No. The slope is -3. Just look for the coefficient of x.

I suppose you mean 5+2x = 17

\(5+2x=17\\2x=12\\x=6\)

\(\text{Do you actually mean }(30 + A) + (10A + B)= 10B\;\text{?}\)

If so we will have to try each positive 1-digit integers.

First substitute A = 1

31 + 1B = B0

B have to be 9 so A = 1 didn't work.

Substitute A = 2

32 + 2B = B0

B have to be 8, so A = 2 didn't work.

Substitute A = 3

33 + 3B = B0

B have to be 7. We did it.

33 + 37 = 70.

B = 7

LOL xD

\(\text{Your answer is:}\sqrt{1-(\frac{1}a)^2}\\ \text{But the answer is:}\frac{\sqrt{a^2-1}}{a}\)

\(\sin\theta = \frac{1}{a}\\\text{Opposite side of }\theta = 1\\\text{Hypotenuse}=a \\\therefore \text{Adjacent side of }\theta = \sqrt{a^2-1}\text{<---- Pythagoras' Theorem}\\ \therefore\cos\theta = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}}\\\qquad\quad=\dfrac{\sqrt{a^2-1}}{a}\)

\(\cos{56}=\frac{10}{x}\\ x=\frac{10}{\cos56}\\ \;\;\approx17.8829\)

OK the real purpose of this thread because I can't edit my post the 2nd time. Anyone seen some funny math jokes? I've got one from a Guest of the forum but I forgot which page did I get it. If you have some really funny math jokes please comment here :D

\(\sqrt{-1}\space 8\space \sum\space \pi\) and it's yummy.