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:
This statement is sometimes true
If you break it up:
Three-eighths = 3/8
Sixteen times a number, we'll call the unknown number X, so 16X.
Twenty Four = 24
The sum of 16X and 24 = 16X + 24
We want 3/8 of that sum so we put the sum in parenthesis and we get:
Y = 3/8(16X + 24) where Y is the answer youd get if you knew what X was.
Hello There!
2y x 2y = 4y²
2y x 8 = 16y.
Put it together and you get the answer of:
4y² + 16y.
Hope This Helps You!
Good Luck :)
- Hannah ❤