@WhySoSerious is probably right.

\(67,05+4,2-56,27=71,25-56,27=14,98\)

\(\dfrac{a^{12}b^{15}c^6}{\left(a^2\right)^5b^3.b^8.c^5}=\dfrac{a^{12}b^{15}c^6}{a^{10}b^{11}c^5}=a^2b^4c^1\) => x = 2; y = 4 and z = 1.

So the value is: \(2+4+1=7\)

The distance that he had drive is: \(\left(48-25\right):0,08=287,5\left(miles\right)\)

So the answer is: 287,5 miles

She used: \(29-19=10\left(eggs\right)\)

So the answer is: 10 eggs

\(x^2+2xy+y^2=81\)

=> \(\left(x+y\right)^2=81\)

=> \(x+y=\sqrt{81}=9\)

=> The mean of x,y is: \(9:2=4,5\)

Yuki Judai, why do you answer when you order the question?

\(=36-9,05=26,95\)

\(=72+648+12,97=732,97\)

\(\dfrac{15}{32}+\dfrac{14}{69}+\dfrac{7}{2}=\dfrac{1483}{2208}+\dfrac{7}{2}=\dfrac{9211}{2208}\)

\(x-\dfrac{2}{3}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}+\dfrac{2}{3}=\dfrac{9}{6}=\dfrac{3}{2}\)

Sorry, but there isn't enough information to answer.

P/s: This question is asked before.

The number of pairs is: \(\dfrac{\left(1000-1\right)+1}{2}=500\left(pairs\right)\)

The sequence of the first term and the second term goes in the 6th term: \(64and60\). So the least value of n is: 6.

The number of combinations are: \(20.\left(4.3.2.1\right)=480\left(combinations\right)\)

Call the edge of the normal cube: a, the edge of the increased cube: b.

The normal edge of a cube is: \(100\%=1\). The increased edge is: \(100\%+50\%=150\%=1,5\)

The volume of the normal cube is: \(a.a.a\left(1.1.1\right)=a^3\left(1\right)\)

The volume of the increased cube is: \(b.b.b\left(1,5.1,5.1,5\right)=b^3\left(3,375\right)\)

So the percent is: \(3,375=337,5\%\)

Multiplies of 3,4,5 => Multiplies of 60 (3.4.5)

So the numbers satisfy is: \(60,120,180,240,300,360,420,480.\)So there are 8 positive integers satisfy the question.

Call the students before the addition is: A. The students after the addition: B. We have:

\(\dfrac{3}{2}=\dfrac{B}{A}\) (Arrcoding to the ratio). So the students before class is: \(5.2=10\left(students\right)\)

Don't keep asking questions in Vietnamese 

P/s: I knew you're copiers and spammers.