Dimitristhym

No its :

\(x - (-9) = -74 <=>x +9 = -74 \) we remove from both parts the 9 so \(x +9 -9 = -74 - 9 <=> x=-83\)

Now you understand it? 

This means x - y= The difference so 

\(x - (-9) = -74 <=> x= -74 - 9 = -83 \)

so the number is \(-83\)

Hope this helps! 

Why? Where is my false? 

1) \(sinθ=\frac{1}{6}\)

and  \(tanθ=\frac{sinθ}{cosθ}\) <=> \(cosθ=\frac{sinθ}{tanθ}=\frac{\frac{1}{6}}{\frac{-\sqrt{35}}{35}}=\frac{-\sqrt{35}}{6 }\) 

so \(cosθ= \frac{-\sqrt{35}}{6}\)

2) \(cosθ=cos(-θ)=\frac{\sqrt{3}}{4}\)

and \(cos^2θ + sin^2θ =1 \) 

so  \(sinθ =\sqrt{1-cos^2θ} = \sqrt{\frac{16}{16}-\frac{3}{16}}=\sqrt{\frac{13}{16}}=\frac{\sqrt{13}}{-4}\)

so \(sinθ=-\frac{\sqrt{13}}{4}\)

Hope this helps! 

\(sin^2θ +cos^2θ =1\)

so \(sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}\)

so \(tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}\)

Hope this helps! 

Generally its: 

\(sin^2θ + cos^2θ=1 \)

so \(cosθ=\sqrt{1 - sin^2θ }=\sqrt{1 - \frac{4}{25} }=\sqrt{\frac{25}{25}-\frac{4}{25}}=\sqrt{\frac{21}{25}}=\frac{\sqrt{21}}{5}\)

Sorry if you don't understand my solution sent me message to explain it step by step 

2 fast :p

\(\sqrt{128r^3}=\sqrt{64 \times 2 \times r^2 \times r }=\sqrt{64} \sqrt{r^2} \sqrt{2r}=8r\sqrt{2r}\)

Hope this helps!

\(\sqrt[3]{64\times4\times x^2 \times x^2 \times x }\) = \(\sqrt[3]{64}\sqrt[3]{\ x^6}\sqrt[3]{ 4\times x }\) = \(4x^2\sqrt[3]{4x}\)

Hope this helps!