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
The first and only stop of the flight can be Boise, Omaha or Chicago. This means there are 3 possibilities where to stop. From thee the flight connects to New York City, either to La Guardia or to JFK Airport.
1 possible route: Seattle-Boise-La Guardia
<span>2 possible route: Seattle-Omaha-La Guardia
</span><span>3 possible route: Seattle-Chicago-La Guardia
</span><span>4 possible route: Seattle-Boise-NYC
</span><span>5 possible route: Seattle-Omaha-NYC
</span><span>6 possible route: Seattle-Chicago-NYC
</span>
Or in other words: There are two airports where the flight can arrive and for each of them there are three possible routes. So, in total there are 2*3=6 possible routes.
Answer: All real numbers :)
Answer:
d
Step-by-step explanation:
