Problem 18)
The distance from J to L is 30 units. We can count out the number of spaces or do subtraction to get 30-0 = 30. I subtracted the number line coordinate of J and L. The result is positive as distance is never negative.
Take 2/3 of this value to get (2/3)*30 = 20 which means that we start at J and add on 20 units to get 0+20 = 20
So the final answer is 20, which is where this point is located on the number line.
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Problem 19)
The distance from A to B is 3 units because |-5-(-2)| = |-5+2| = |-3| = 3
Double this value to get 2*3 = 6
Now add 6 units to the coordinate of point A to get
-5+6 = 1
The location of this point is at 1 on the number line.
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Problem 20)
From J to I, or vice versa, we travel 5 units.
Multiply this by 1.5 to get 1.5*5 = 7.5
So we subtract 7.5 units from 0 (the coordinate of point J) getting us 0-7.5 = -7.5
Final Answer: -7.5
Note: this is the midpoint of -10 and -5 on the number line
The radius is 4 centimeters
Answer:
37.7 cm
Step-by-step explanation:
C=2πr=2·π·6≈37.69911
Answer:24=3+x i believe this correct
Step-by-step explanation:
