MaxWong

Driti1, that's the elementary method, thank you for providing it. I am going to share a method different to yours.

If you want to find the remainder of 7 ÷ 5, type 7 mod 5 in your calculator or the calculator on this website because normal calculators doesn't have mod function.

Example: 6 mod 6 = 0 because 6 ÷ 6 = 1 r 0

237 mod 4 = 1 because 237 ÷ 4 = 59 r 1.

It doesn't matter what the quotient is, it just output the remainder.

\(\sqrt{2x}\times \sqrt{64x}\\ = \sqrt{2}\times \sqrt{x}\times \sqrt{64}\times \sqrt{x}\\ = 8\times \sqrt{2}\times (\sqrt{x})^2\\ =8\sqrt2\;x\)

Same as asinus's answer.

Perfect cube is a number whose any cube root is an positive integer.

\(\sqrt[3]{27}=3\) Therefore 27 is a perfect cube.

Same as the above Guest's answer.

These examples may help you.

f(-2)=√(-2)+6

f(x)=√x + 6

:D

sin(45-x)=cos(45-x)

(sin45)(cos x) - (cos45)(sin x) = (cos 45)(cos x)+(sin 45)(sin x)

cos x - sin x = cos x + sin x

Notes: sin 45 = cos 45

- sin x = sin x

2 sin x = 0

sin x = 0

x = sin^-1 0 = 0

Graph : Draw the curved part of a semi-circle (a half circle) with center at (0,0) and 1 as radius.

OR

We first work out the derivative of sinx cosx

Using product rule: 

\(\dfrac{d}{dx}\sin x \cos x \\=(\cos x)(\cos x)+(-\sin x)(\sin x)\\=\cos 2x\)

Using chain rule:

\(y = \sqrt u, u=\sin x\cos x\)

\(\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}=\dfrac{1}{2\sqrt{u}}\times \cos 2x=\dfrac{\cos 2x}{2\sqrt{\sin x\cos x}}\)

\(256^{\frac{-3}{4}}\\ =(\sqrt[4]{256})^{-3}\\ =\sqrt[4]{4^4}^{-3}\\ = 4^{-3}\\ =\dfrac{1}{4^3}\\ =\dfrac{1}{64}\)

Wow asinus. That was not exponential equation. Oops. Great job though. I didn't know that ln(-7) = 1.94591015 + πi