What are you suppose to do - solve it or reduce it?
In order to understand if the inequalities are always, never or sometimes true, you need to perform the calculations:
A) <span>9(x+2) > 9(x-3)
9x + 18 > 9x - 27
the two 9x cancel out and you get:
+18 > -27
which is always true.
B) <span>6x-13 < 6(x-2)
6x - 13 < 6x - 12</span>
</span><span>the two 6x cancel out and you get:
- 13 < -12
which is always true
C) </span><span>-6(2x-10) + 12x ≤ 180
-12x +60 +12x </span>≤ 180
-12x and +<span>12x cancel out and you get:
60 </span><span>≤ 180
which is always true.
All three cases are always true.</span>
Answer:
6 to the second is 36 and 2 to the second is 4
Step-by-step explanation:
Both of those combined would equal 40. Are you looking for the algebraic expression, or is this good enough? Let me know! Bya!
Let the equal sides of the isosceles Δ ABC be x.
Given that the perimeter of Δ ABC = 50m.
Therefore, 2x + AC = 50 --- (1)
It is also given that the perimeter of Δ ABD = 40m.
Therefore, x + BD + AD = 40
BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.
So AD = CD.
Or, AD = 
AC
Therefore, 
Multiply both sides by 2.
2x + 2BD + AC = 80
From (1), 2x + AC = 50.
Therefore, 2BD + 50 = 80
2BD = 80 - 50
2BD = 30
BD = 15m.
