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/Step-by-step explanation:
Given:
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To solve, collect like terms.
Subtract 5w from both sides
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(incorrect)
The inequality given has no solution or an information is missing.
I'm not sure if that is supposed to be 7 or -7
If it's -7:
1<2n-5<-7
+5 +5 +5 Add 5 to all parts of the inequality get 2n by itself
6<2n<-2
/2 /2 /2 Divide all parts by 2 to isolate the variable
3
Answer:bc you asked her do she got dat wap
Step-by-step explanation:but dam
Answer: B, C, and D
Step-by-step explanation:
Just did it and got the answers wrong so these are the correct ones