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
Answer:
Step-by-step explanation:
Im notr sure but if we take the chances that the 4 sum has to be either (1.3) (3.1) (2.2) (2.2) that means its chance is 4/36 for each roll which is also 1/9 ,, we can multiply 1/9 times the roll times ( 252 ) we get =28
Answer:
210
Step-by-step explanation: so 5 times 12 120 plus 100 is 210
The perimeter of the a and BC is 14 unitsBecause if you add all of the sides together you get 14 and I need points