\(x^2+y^2=10^2 <=> x^2+y^2=100 <=> x^2=100-y^2 <=> x=\sqrt{100-y^2}\)
So :\(x=\sqrt{100-y^2}\)
Congrats!!!
because can be 0 or infinitive or a number its different the result every time!
21!!!
\(\lim_{x\rightarrow 2} [(χ^2-5χ+6)/(χ-2)] \)Can you find it?(it's easy i trust you)
If f(x) = Log(x),
then, f ^-1(x) = AntiLog (x).
Also, AntiLog (x) = 10x
\({8}^{2}=64 \)
So 8 is the square root of 64!!!
5.7 meters long
but more simply 15 -> 25 χ=15*100/25 <=>x=1500/25 <=>x=60 so 60%
χ ->100
(25-15)/25*100%=10/25*100%=40% so 100%-40%=60%
