: 88, 87, 89 90, 87, 85, 88, 91, 86, 86, 88, 89<br>what's is the inerquartile range
andreyandreev [35.5K]
I think 88 i don’t really know
3and5 are both divisible by 15
Answer:
Point N(3,0) is equidistant from A and B.
Step-by-step explanation:
In order to check whether the point is equidistant from A and B, it is required to measure the distance of A and B from each point. The formula for distance is:
d= √((x_2-x_(1))^2+(y_2-y_(1))^2 )
For J
AJ= √((-4-0)^2+(-5+4)^2 )
=√((-4)^2+(-1)^2 )
= √(16+1)
= √17 units
BJ=√((-4+2)^2+(-5-0)^2 )
=√((-2)^2+(-5)^2 )
= √(4+25)
= √29 units
Point J is not equidistant from A and B.
For K
AK= √((-3-0)^2+(0+4)^2 )
=√((-3)^2+(4)^2 )
= √(9+16)
= √25 units
=5
BK=√((-3+2)^2+(0-0)^2 )
=√((-1)^2+(0)^2 )
= √(1+0)
= √1
=1 unit
Point K is not equidistant from A and B.
For M
AM= √((0-0)^2+(0+4)^2 )
=√((0)^2+(4)^2 )
= √(0+16)
= √16
=4 units
BM=√((0+2)^2+(0-0)^2 )
=√((2)^2+(0)^2 )
= √(4+0)
= √4
=2 units
Point M is not equidistant from A and B.
For N
AN= √((3-0)^2+(0+4)^2 )
=√((3)^2+(4)^2 )
= √(9+16)
= √25
=5 units
BN=√((3+2)^2+(0-0)^2 )
=√((5)^2+(0)^2 )
= √(25+0)
= √25
=5 units
As point N's distance is equal from A and B, Point N is equidistant from A and B.
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Do what? Using 5th grade strategies??
