The prime number is:23,37,43,47,73 so we have 5 prime numbers

The number of composite numbers is:

  4 x 3 x 2 x 1 = 24 (composite numbers)

So the answer is:\(\dfrac{5}{24}\)

The number of marblesTricia gave to Barbara is:

   \(24.\dfrac{1}{3}=8\) (marbles)

We look the picture,we see the diagonal of the square is:AB

So the edge of the square is:BC=4

So the area of this square is:BC²=4²=16

The volume of smaller object is:

 216:240 x 60=54(cubic centimeters)

We have:1 USD=100 cent so 3,75 USD=3,75 . 100=375 cent

So the answer is:\(\dfrac{375}{40}\approx9,4\)

Because x<8 so \(\sqrt{2+4+6+x}< \sqrt{2+4+6+8}=\sqrt{20}\le4\) and \(\sqrt{2+4+6+x}\ge0\)

\(\Rightarrow\sqrt{2+4+6+x}\in\left\{0,1,2,3,4\right\}\)\(\Rightarrow2+4+6+x\in\left\{0,1,4,9,16\right\}\)

\(\Rightarrow x\in\left\{-12,-11,-8,-3,4\right\}\)

So the answer is:\(\left(-12\right)+\left(-11\right)+\left(-8\right)+\left(-3\right)+4=-30\)

1 minutes Kevin can mow the number of the part of lawn is:

  \(1:40=\dfrac{1}{40}\)(part)

1 minutes they can mow the number of the part of lawn is:

 \(1:15=\dfrac{1}{15}\)(part)

1 minutes Josh can mow the number of the part of lawn is:

 \(\dfrac{1}{15}-\dfrac{1}{40}=\dfrac{1}{24}\)(part)

So the answer is:\(1:\dfrac{1}{24}=24\)(minutes)

Because x<8 so \(\sqrt{2+4+6+x}< \sqrt{2+4+6+8}=\sqrt{20}\le4\) and \(\sqrt{2+4+6+x}\ge0\)

\(\Rightarrow\sqrt{2+4+6+x}\in\left\{0,1,2,3,4\right\}\)\(\Rightarrow2+4+6+x\in\left\{0,1,4,9,16\right\}\)

\(\Rightarrow x\in\left\{-12,-11,-8,-3,4\right\}\)

So the answer is:\(\left(-12\right)+\left(-11\right)+\left(-8\right)+\left(-3\right)+4=-30\)

The number of times to arrang is:

  \(7.6=42\)(times)

So the answer is:\(\dfrac{4}{42}=\dfrac{2}{21}\)

We have:\(64x+16x^2-4x^3-x^4=0\)

\(\Leftrightarrow16x\left(x+4\right)-x^3\left(x+4\right)=0\)

\(\Leftrightarrow\left(16x-x^3\right)\left(x+4\right)=0\)

\(\Leftrightarrow x\left(x+4\right)\left(16-x^2\right)=0\)

\(\Leftrightarrow x\left(4-x\right)\left(x+4\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\4-x=0\\x+4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Help you solve math,I Math English:Không làm thì thôi đừng spam bậy bạ

P/s:cho em nói TV ạ.Dao Trong Luan:Làm như You vậy hình như không ổn thì phải

Dao Trong Luan:a²-ab+b²\(⋮6\) so what

Short 1 case

Because the role of three numbers a,b,c is the same so suppose \(c\le b\le a\)

With \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) then \(a=b=c=0\)(satisfy)

With a,b\(\ne\)0 we have:\(a^2b^2\le3a^2\Rightarrow b^2\le3\)\(\Rightarrow b\in\left\{-1,1\right\}\)

With b=-1 then \(a^2+1+c^2=a^2\Rightarrow c^2+1=0\)(unsatisfactory)

With b=1 then \(a^2+1+c^2=a^2\Rightarrow c^2+1=0\)(unsatisfactory)

Conclude:Only have a=b=c=0 satisfy the problem

Dao Trong Luan:Not right

\(-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)\)

\(=-3\left(a-b\right)\left(b-c\right)\left(a-c\right)=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

I'm sorry .Let a,b,c............

I'm sorry.Let a,b>0....

Help you solve math:CTV nào cũng nói là mày gian lận cả

Continue Dao Trong Luan'answer:

\(\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\)

\(=\left(x-1\right)\left(x^2-4x+4\right)\)

\(=\left(x-1\right)\left(x-2\right)^2\)

We have:\(\left(x+1\right)\left(x-3\right)-\left(x+5\right)\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow x^2-2x-3-\left(x^2-25\right)\left(x-2\right)=0\)

\(\Leftrightarrow x^2-2x-3-x^3+2x^2+25x-50=0\)

\(\Leftrightarrow3x^2-x^3+23x-53=0\)

\(\Leftrightarrow x^2\left(3-x\right)-23\left(3-x\right)+16=0\)

\(\Leftrightarrow\left(x^2-23\right)\left(3-x\right)+16=0\)

\(\Rightarrow x^2-23\in\left\{-16,-8,-4,-2,-1,1,2,4,8,16\right\}\)

\(\Rightarrow x^2\in\left\{7,15,19,21,22,24,25,27,31,39\right\}\)

Because 3-x is a integer number so x is a integer number so \(x^2=25\) and 3-x=-8

\(\Rightarrow\) x=\(\pm\)5 and x=11 (unsatisfactory)

So not have x satisfy