MaxWong

$\text{product of 10 and 20 is }10\times 20\\ \text{one half of something is }\dfrac{1}{2}\times \text{something}\\ \text{Therefore one half of the product on 10 and 20 is }\dfrac{1}{2}\times 10\times 20$

$\cos\left(\dfrac{3\pi}{2}-\Theta\right)\\=-\sin \Theta$

$7e \geq 21\\ e\geq 3$

Graphing the solution......

I bet you mean x for e.

Graph x = 3 and it should be a vertical line. Shade all the parts right hand side of the vertical line.

No. We can't do it in negative 4 words. By the way what's negative 4 words????

Guest #1, a little mistake.

$b + 4 = 2.4\\ b + 4 - 4 = 2.4 - 4\\ b = -1.6$

Large fonts and different colours......

Best way to get kindergarten students learning!!

P(more grandsons than granddaughters or more granddaughters than grandsons)

= 1 - P(equal amount of granddaughters and grandsons)

= 1 - $\left(\dfrac{1}{2}\right)^{10}$

= $\dfrac{1023}{1024}$

By looking at the question, there are clues that you come from China too(Mr WONG!?!?IS IT ME???? LOL). Where you from? Me from Hong Kong.

Each orbit takes:

$644 \div 7\\ = (630)\div 7 + 14 \div 7\\ =90 + 2\\ = 92\text{ minutes}$

The first step is a clever method to do division that I found by myself in Primary School.(or Elementary School you call it.)

$\sqrt[4]{t^8}=\pm t^2$

A system:

$ax + by = c\\ dx + ey = f$

Changing any of the equations of subject x. I would choose the first one

$ax + by = c\\ ax = c - by\\ x = \dfrac{c-by}{a}$

Then substitute into the second one if you chose first one, and substitute into the first one if you chose second. In this case I will sub it into second one.

$d(\dfrac{c-by}{a})+ey = f\\ \dfrac{cd}{a}-\dfrac{bdy}{a}+ey = f\\ ey - \dfrac{bdy}{a} = f - \dfrac{cd}{a}\\ y (e-\dfrac{bd}{a}) = f - \dfrac{cd}{a}\\ y = \dfrac{e-\frac{bd}{a}}{f-\frac{cd}{a}}=\dfrac{ae-bd}{af-cd}$

There we solved y. Then we will solve x.

$x = \dfrac{c-by}{a} = \dfrac{c-b}{a}\cdot \dfrac{ae-bd}{af-cd}=\dfrac{ace-bcd-abe+b^2d}{a^2f-acd}$

And this is the general formula for any equation systems in the form:

$ax+by=c\\ dx+ey=f$