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)
[ (12 - n)/7 ] = -1
12- n = -7
-n = -19
n = 19
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Explanation:</h2><h2>
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The complete question is in the attached file. So we have to choose between two graphs. On of them is a linear model while the other is an exponential model. From the statements, we have a relationship between time and the number of teams registered. So we can establishes variables in the following form:

We also know that each week 6 teams register to participate, so:

As you can see, as x increases one week, y increases at a constant ratio of 6. Therefore, this can be modeled by a linear function given by the form:

In conclusion, <em>the linear model (first graph below) is the one that bests represents the relationship between time and the number of teams registered.</em>
Assuming independence,
prob=P(late,early)+P(early,late)=(1/10)(2/5)+(2/5)(1/10)
Answer:
2/52
Step-by-step explanation:
Because a normal deck of cards only has one eight of spades and only one of seven of heart. So that means the probability would be 2/52. Hope you get it!