Let's add triangles to make a square
This is a 5*5 square, so the area is 25sq un
The combined area of all the triangles around the grey triangle is 17 sq un
25-17
The grey triangle: 8 sq un
\(\sin \left(x\right)\left(\csc \left(x\right)-\cot \left(x\right)\cos \left(x\right)\right)\)
\(\sin \left(x\right)\sin \left(x\right)\)
\(\sin ^2\left(x\right)\)
-Vinculum
Please post one question at a time!!!
I'm not sure if this even has a constant term. Also please write in LaTeX so it is easy for anyone to understand.
\(\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)
\(\sum _{i=0}^9\binom{9}{i}\left(2z\right)^{\left(9-i\right)}\left(-\frac{1}{\sqrt{z}}\right)^i\)
\(512z^9-2304z^{\frac{15}{2}}+4608z^6-\frac{5376z^6}{\left(\sqrt{z}\right)^3}+4032z^3-\frac{2016z^4}{\left(\sqrt{z}\right)^5}+672-\frac{144z^2}{\left(\sqrt{z}\right)^7}+\frac{18}{z^3}-\frac{1}{\left(\sqrt{z}\right)^9}\)
\(= 672\)
\(\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}\left(\sqrt{5}\right)^i\)
\(81+108\sqrt{5}+270+60\sqrt{5}+25\)
\(376+168\sqrt{5}\)
\(t^2+8t\left(\sqrt{t}\right)^3+24t^3+32t^3\sqrt{t}+16t^4+\left(\sqrt{t}-2t\right)^4\)
\(t^2+8t\left(\sqrt{t}\right)^3+24t^3+32t^3\sqrt{t}+16t^4+t^2-8t\left(\sqrt{t}\right)^3+24t^3-32t^3\sqrt{t}+16t^4 \)
\(32t^4+48t^3+2t^2\)
Hello
The Answer:
(-2, -6) and (4, 2)
Graph:
Let
f(z) = 2z if z is not real
f(z) = z if z is real
so, 2z + z + z + 2z.
This equals 6z
