Laziness is human nature, so people will not sketch out the figure for this unless you do that job 

? is \(\dfrac{386}{862}\).

It's easy!

The sum of the deominator and the numerator is constantly 1807.

So, \(?\) is \(\dfrac{15}{1792}\).

\(3x^n.\left(4x^{n-1}-1\right)-2x^{n+1}\left(6x^{n-2}-1\right)\\ =12x^{n+n-1}-3x^n-12x^{n+1+n-2}+2x^{n+1}\\ =2x^{n+1}-3x^n\)

When written in base 7, the number is a.7² + b.7 + c

When written in base 6, the number is (a+1).6² + (b+1).6 + c +1

We get: a.7² + b.7 + c = (a+1).6² + (b+1).6 + c +1, it implies 13a + b = 43.

This yields a = 3, b = 4.

When written in base 10, the largest number possible: 3.7² + 4.7 + 9 = 184

This can be more simply illustrated using Vevtors and magnitude of vectors since they all lie on a line. 

"Thương" in English is "Quotient".

I think he meant to simplify the expression bro 

We have: 5 = 1 + 0 + 4 = 1 + 1 + 3 = 1 + 2 + 2 = 2 + 0 + 3 = 5 + 0 + 0

With 1, 0, 4, we can have 4 three-digit numbers including these 3 numbers

(104, 140, 401, 410)

With 1, 1, 3, we can have 3 three-digit numbers including these 3 numbers

(113, 131, 311)

With 1, 2, 2, we can have 3 three-digit numbers including these 3 numbers

(122, 212, 221)

With 2, 0, 3, we can have 4 three-digit numbers including these 3 numbers

(203, 230, 302, 320)

With 5, 0, 0, we can have 1 three-digit number including these 3 numbers

(500)

In total, we have 4 + 3 + 3 + 4 + 1 = 15 (numbers)

Let S = {0, -1, k}.

1. Necessary condition: S is closed under multiplication if  \((-1)\times (-1)\in S\) . It implies \(1\in S\). This means \(k=1\).

2. Sufficient condition: when \(k=1\) , we observe that \(0\times0;0\times-1;-1\times-1;-1\times1;1\times0;1\times1 \)are all in \(S\).
Therefore, \(k=1\).

The transformation is incorrect from the step: \(\left(\dfrac{1}{10}\right)^2\)$=\(10^2\) ¢. It is because the unit of currency do not remain $ and ¢ anymore when squaring both side of \(\left(\dfrac{1}{10}\right)\)$=\(10\)¢. 

Hi Mr Puppy, you should add tag to your question.

\(AB = (45+50) \times \dfrac{3}{4} =71.25\)(km)

Your question is not clear, please check again!

There is no triangle with such sides. Check your problem again.

We normally use "side" rather than "edge" when referring AB, BC and CA.

Hi Phương, you should use "latex" or "formula" to express your question.

Your question should be presented as: \(x! = 1+2+x\).

About your question, I think \(x \) should be a natural number.

After some transformation, we get:

\(x.[(x-1)!-1]=3\)

So \(x\) is a divisor of 3. We can easily find \(x=3.\)

You should add tags to your question.

Let the capacity of bottle A and the capacity of bottle B be \(x\left(ml\right)\), \(y\left(ml\right)\) respectively.

We can deduce simutaneous equations:

\(\left\{{}\begin{matrix}x+y=600\\0.15\times x+0.4\times y=0.3\times600\end{matrix}\right.\)

From the first equation, we have: (1)  \(y=600-x.\)

Substituting (1) to the secon equation we get: \(0.15\times x+0.4\times\left(600-x\right)=180.\).

We can easily find \(x=240.\)

You should add tag to your question guy.

For instance, your question should be tagged: geometry, measure, ...

Since \(\widehat{ABC}=\widehat{BDC} \) and \(\hat{C}\) is the common angle, then \(\Delta ABC \sim \Delta BDC \quad (a.a)\).

It implies that: \(\dfrac{BD}{AB}=\dfrac{BC}{AC}=\dfrac{DC}{BC}\). From that we have: \(BC^2=AC.DC=144\Rightarrow BC=12\left(cm\right)\).

Then \(BD=\dfrac{AB.BC}{AC}=\dfrac{24.12}{16}=18\left(cm\right).\)

Sory, I mistyped. It should've been \(720^o\) not \(540^o.\)