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madreJ [45]
3 years ago
8

(sum by gcf ) what is the gcf of 44+40? 44+40=4 (__+__)

Mathematics
2 answers:
kati45 [8]3 years ago
6 0
44|2
22|2
11|11
1

40|2
20|2
10|2
5|5
1

GCF(40,44)=2^2=4

44+40=4\cdot11+4\cdot10=4(11+10)
Illusion [34]3 years ago
4 0


The answer is 4

|________ you put 44+40 in the box

  |_______ then you see what goes into both and it is 4 so then divide 44 and 40 by 4

then put 11+10 in the box and nothing goes into 11 and 10 so 4 is the only thing that goes into 44 +40

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7 0
3 years ago
The coordinates of the endpoints of AB and CD are A(2,
Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

m= \frac{y_2-y_1}{x_2-x_1}

Formula to Find Distance between two points

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

mAB ( represents , Slope of AB )

1. mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}

2. mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}

3. mCD=\frac{5-2}{6-5}=\frac{3}{1}=3

4. AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

5. BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

mAC = mBC  , and C is common point , hence these three are collinear points  making a straight line whole slope is -\frac{1}{3}

mAB=-\frac{1}{3}

mCD=3

mAB \times mCD = -\frac{1}{3} \times 3 = -1

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0
3 years ago
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