Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units
R= 5
You divide -7 by both sides to 1. Cancel out the -7 2. To simplify the -35. So simplified -35/-7=5
Alrighty, so 280.05 rounded to the nearest integer is 280. 280.05 rounded to tenths is 280.1. add those two together and you get 560.1.
The answer would be the first choice 4+b
