Including all real and imaginary zeros and taking multiplicities into account, every equation of degree n has exactly n roots
Answer:
∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and lies between AB and KM and BK is the transversal line)
m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)
Step-by-step explanation:
Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK
TO prove: KM ║AB
Now, As given in figure 1,
In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)
Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)
Now ∵ ∠MBK = ∠BKM
and ∠ABK = ∠KBM
∴ ∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)
Hence proved.
Answer:
B is the right answer and if you want learn this, than learn algebric expression
Answer:

Step-by-step explanation:

Combine all like terms.


Multiply both sides by -1.

Let no red ribbons=x
no. Of blue ribbons=x+3
No. Of yellow ribb9ns=x+7
Totsl ribbons=28
28=x+x+3+x+7
28=3x+10
3x=28-10
3x=18
X=18/3
X=6
No .of red ribbons=6
No. Of blue ribbons=x+3=9
No. Of yellow ribbons=x+7=13