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.
The 82nd term for the sequence would be -4
Answer:
D
Step-by-step explanation:
I am taking the practice.
Answer:
A shaded region below a dashed line (y <3x-4)
Step-by-step explanation:
For linear inequalities, we have shaded regions below or above a dashed or solid or continuous line represented by an inequality.
Their graphs are called region graphs. Those regions are only enclosed ones if before the sign of inequality it is included both variables. Like x²+y²<1, |x+y|<1, etc.
In this case what we have here is an open region, as the graph below shows.
3x-y>4 =-y>4-3x∴y<3x-4
Setting a value for x
3x-4=0
3x=4
3x/3=4/3
x=4/3 (slope)
(4/3,0)
-2 * 2n -2 equals -4n + 4 > 18n -3n
