The least common multiple or LCM
<span>Perimeter of a rectangle: 2(l + w)
</span>
2(l + w) ≥ 30 in
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Simplify- divide both sides by 2.
l + w ≥ 15 in
Substitute (w + 3) into l so only 1 variable is used.
(w + 3) + w ≥ 15 in
Simplify further- add variables and subtract 3 from both sides.
2w ≥ 12 in
Divide both sides by 2.
w ≥ 6 in
Answer: D</span>
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:
56
Step-by-step explanation:
Lowest Common Multiple= 56
