1, 5, 9, 4 down the left side.
1, 3, 8, 7 down the right side.
4, 2, 6, 7 across the bottom.
Other combinations are possible.
Put the two bases of the triangle together. To form a square. Place the squares side by side to form a rectangle
Then you have a tie, for either first place or last place.
6
Answered this yesterday... Top: 3-17-7, middle 13-9-5, bottom 11-1-15
maybe if some place were in a triangle shape and if you had a sort out the length of the place they use it
Put the two bases of the triangle together. To form a square. Place the squares side by side to form a rectangle
Then you have a tie, for either first place or last place.
The sides of a triangle are its lengths are cannot be negative. However, you could place a triangle on coordinate system and some points where the vertices are could be negative numbers.
I would place this triangle in the category of isosceles triangles, because the 10m side and the 10m side have equal lengths.
Sure, place a triangle's hypotenuse (longest side) on the other triangle's hypotenuse, that will give either a square or a rectangle. Then place the square on one end of the rectangle. For this to work though, the length of the square's side HAS to equal the length of the triangles hypotenuses, and likewise each triangle's hypotenuse much equal the length of a side of the square. Hope this is clear.
6
Answered this yesterday... Top: 3-17-7, middle 13-9-5, bottom 11-1-15
Top row: 3-17-7 Middle row: 13-9-5 Bottom row: 11-1-15
Yes. Bermuda triangle is a real place
the Bermuda triangle
This is only possible if one of the digits is equal to zero. There are 90 3-digit numbers with a zero in the 10's place, and 90 3-digit numbers with a zero in the 1's place - and 9 numbers that have both a zero in the 10's place and a zero in the 1's place; these would be counted double if you just add the first two. So, you get: 90 + 90 - 9 such numbers.
I think its even because in the ones place it has a 2 in the ones place