MaxWong

Algebra means a part of maths which involves variables.

Algebra was first invented by Babylonians. You should blame them. Not us. LOL.

\(\cos^{-1}(0.23702273156)\approx 76.28911407\)

Try each of them

If it's A,

L.H.S.=4

R.H.S.=12-3(0)=12

LHS not equal to RHS

If it's B,

LHS = 4

RHS = 12 - 3(-4)= 24

LHS not equal to RHS

If it's C,

LHS = 3

RHS = 12 - 3(4)=0

LHS not equal to RHS

If it's D,

LHS = 0

RHS = 12 - 3(4)=0

LHS = RHS

\(\therefore \text{D is the answer.}\)

9012x099123-12323

That 0 is unnecessary LOL.

9012x 99123 - 12323

= 893284153

Calculators are your best friends, but only in Maths exam, Maths text, or doing Maths homework xD

I recommend this calculator in the forum, it can do many things that all Hong Kong calculators can't do.(Because I live in Hong Kong :P)

\(\text{Area of square cloth}=8^2 \space \text{sq. in.}=64\space\text{sq. in.}\)

\(\text{Area of the circle hole :P = }\pi \times(\frac{7}{4})^2=\frac{49\pi}{16}\space\text{sq. in.}\)

\(\text{Area that remains}= 64 - \frac{49\pi}{16}\approx 54.3788725\)

I personally think this is a trick question :P

Always, if you buy more, the average cost will be lower.

So if you buy 8 bottle of 1.5L water will it be cheaper?

\(\frac{x+1}{2}+\frac{x+4}{3}=\frac{x-1}{6}\\ \text{That's what Melody have answered.}\\ \frac{1}{2(x+1)}+\frac{1}{3(x+4)}=\frac{1}{6(x-1)}\\ \text{And I am going to answer this.}\)

\(\frac{1}{2(x+1)}+\frac{1}{3(x+4)}=\frac{1}{6(x-1)}\\ \frac{3x+12+2x+2}{2(3)(x+1)(x+4)}=\frac{1}{6}\times \frac{1}{x-1}\\ \frac{5x+14}{(x+1)(x+4)}=\frac{1}{x-1}\\ (5x+14)(x-1)=(x+1)(x+4) \\5x^2+9x-14=x^2+5x+4\\ 4x^2+4x-16=0\\ x^2+x-4=0\\ x=\frac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}\\ \;\;=\frac{-1\pm\sqrt{17}}{2}\)

\(\text{The answer is}\mathbf{\frac{-1+\sqrt{17}}{2}}\text{and}\mathbf{\frac{-1-\sqrt{17}}{2}}\)

\((\frac{a}{b})^2\)

= \(\frac{a^2}{b^2}\) According to Power Rule of Indices.

\((\frac{a}{b})^2\)

= \(\frac{a^2}{b^2}\) According to Power Rule of Indices.

\(\frac{1}{2}\div \frac{1}{56}= \frac{1}{2}\times 56=28\)

Just the same as Guest #1's answer