9th grade math problem reward, rush a 9th grade math problem for a correct answer to a bounty

Solution: First of all, let the two-digit number be x+10y, that is, the single digit is x, and the ten-digit number is y, which is known by the title: y=x+2

10x+y)(10x+y)=(x+10y)+138 Substituting the first formula into the second formula yields:

11x+2)(11x+2)=(11x+20)+138 Solving this system of equations yields: x=-11 14 (rounded) or x=1 so y=3 so the tens are 31

If you want to learn how to solve this kind of problem, you will be able to do it the next time you meet it.

If the single digit is x, then the ten digit is x+2, according to the title.

10x+x+2) squared - (10(x+2)+x)=138 solution x=1, the single digit is 1, the ten digit is 3 is a number, and the other is 31.

Let the original tens of digits be a, then this number can be expressed as: 10a+a-2=11a-2, the number after the two sides are changed, can be expressed as: (a-2)*10+a=11a-20 according to the conditional equation.

11a-20) squared - (11a-2) = 138 gives a = 3, so the original number is 31

Let the ten digits be a and the single digit be b.

So there is ab

There is also b=(a-2).

There can be ba*ba-ab=138

Set a=3 again

That's 13*13=169-31=138

Establish. So this two-digit number is 31

Set the single digit as a and the ten digit as A+2

10*a+a+2)^2=138+10(a+2)+a

The solution is a=1, and the original 2 digits are 31

Such a simple question is also to be asked.

Now these children ...

I can't tell you, or I'll hurt you.

Hint: Let the single digit be x and the ten digit be y, and list a system of equations.

By dividing this iron plate into many 20cm squares, you can cut more into it.

It can be divided into 82*55 (20*20) = 11 squares with sides of 20 cm, which are then cut into circles with a radius of 10 cm.

There are a total of 5*8=40 kinds.

Cut 5 round parts at the width of the rectangular iron plate and 8 round parts at the long part.

You list the quadratic functions analytically, and then find the vertex coordinates and pull it!

9 problems, x two solutions, one 3 and one 4, the diagonal of the rectangle is equal to 5, the Pythagorean theorem.

A square minus B is equal to A plus B times A minus B

30 degrees, the opposite right-angled edge is half of the hypotenuse. When that is a right angle, divide the hypotenuse into thirds. i.e. ec' is one-third and two-thirds of the hypotenuse is long.

You see, bc is equal to the root number (3)+1, and b is again 30°, so you can wittily guess that be is equal to the root number (3) and ec is equal to 1, and then ec' = ec = 1, which is exactly right, because c'e : be : bc' =1:

Root number(3): 2. Then you know bc'= 2.

Easy to get be = root number 3 ec', again ec'=ec, so ec + root number 3 ec=bc, and bc = root number 3 1, so ec=1, so bc'=2

Swap the letters and notice that the equality of the angles gives the equality of the edges, and gives gf=ge

So 80+60=2(40+x) and therefore x=30°