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)
you cant do that because they are not like terms
A. The population would be 120
You would find 20 percent of 100
To do that, you do 100× 0.2 and add that to 100
B. The population doubles
10×2= 20
20×2= 40, not 30
hope this helps!
The answer is 440.
What I do is just take the -11 and -15 out of the parentheses so it is -11*-15 and the answer is 165. Then, you take -11 and -25 out of the parentheses and find the answer.
Add them together.
Hope that helped :)
