Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
Answer- D-apex hope i’m right i’m not sure
Maybe u should do inverse operation on both sides idk but just try it
<h3>
Answer: addition property of inequality</h3>
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Explanation:
These are the steps to focus on
step 3: -6x - 8 < -2
step 4: -6x < 6
The move from the third step to the fourth step has us adding 8 to both sides. Therefore, we use the addition property of inequality.
That property has four forms
- If
then 
- If
then 
- If
then 
- If
then 
It's similar to the idea of starting with a = b, then adding c to both sides to get a+c = b+c
We add the same thing to both sides to keep things balanced.
24+15+11=50, the total number of times the markers were chosen.
Since 11 of these 50 tries resulted in a red and blue marker, the probability is 11/50. This is equal to 0.22 or 22%.
