To determine if 12m, 4m, and 2m can form a right triangle, we apply the Pythagorean theorem, which states that for a right triangle, the square of the length of the longest side (hypotenuse) must equal the sum of the squares of the other two sides. Here, the longest side is 12m. Calculating, (12^2 = 144) and (4^2 + 2^2 = 16 + 4 = 20). Since 144 does not equal 20, these lengths cannot form a right triangle.
4m
first, multiply the base and height then divide it into 2 example: base:4m height:2m multiply 4 by 2= 8 then divide 8 by 2= 4 4m squared
6m + 5n - 4m + 7n = 2m + 12n
2m + 13n
To determine if 12m, 4m, and 2m can form a right triangle, we apply the Pythagorean theorem, which states that for a right triangle, the square of the length of the longest side (hypotenuse) must equal the sum of the squares of the other two sides. Here, the longest side is 12m. Calculating, (12^2 = 144) and (4^2 + 2^2 = 16 + 4 = 20). Since 144 does not equal 20, these lengths cannot form a right triangle.
It is: 2m times 2m is equivalent to 4m^2
The triangle with side lengths of 2m, 4m, and 7m does not form a valid triangle. In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side according to the Triangle Inequality Theorem. In this case, 2m + 4m is less than 7m, violating the theorem. Therefore, a triangle with these side lengths cannot exist in Euclidean geometry.
4m
Well, darling, if the length of your rectangle is 4m and the width is 2m, then the perimeter is simply calculated by adding up all the sides, which gives you 4m + 4m + 2m + 2m, totaling 12 meters. Voilà, there's your answer!
6m + 7 - 2m + 1 = 4m + 8 If you meant: (6m + 7) - (2m + 1) = 4m + 6
40m^3
first, multiply the base and height then divide it into 2 example: base:4m height:2m multiply 4 by 2= 8 then divide 8 by 2= 4 4m squared
40 cubic meters
6m + 5n - 4m + 7n = 2m + 12n
None. By definition a hole is an empty space.
2m + 13n