X1. Y1. X2. Y2
A= (-5,-4), B=(-3,3)
Midpoint formula = y2-y1/ x2-x1
3-(-4)/ 3-(-5) = 7/8
The value of 6 is 6 hundreths
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
We want to get b alone so I would start by multiplying 2 to get it on the other side of the = sign. So then I have 2L=a-b. I would subtract a on both sides and then I would have 2L-a= -b. We don't want a -b so I would multiply everything by -1 so then the answer would then be a positive b= -2L+a
<span>So we want to know what is the area of the shaded reagion that consists of a circle inside a square If the side of the square is a=10cm. For that we simply calculate both areas and subtract the area of a circle from the area of a square: area of a square A=10^2=100 cm^2, area of a circle B=pi*r^2 where r=a/2. So: B=3.14*5^2=78.5cm^2: A-B=100-78.5=21.5cm^2 So the correct answer is D</span>