Answer:
D
Explanation:
We are going to define π = 3 (π is actually ≈ 3.14159...)
Why would I define π = 3? This is because we know that π > 3 and after we evaluate the value, we will check using 3.14159..
From A choice; 12/3 is 4. We know that if denominator keeps increasing, the evaluated value will be decreasing.
That means if 12/3 = 4 then 12/3.14159 would be less than 4.
Thus 12/π > 4 is FALSE.
Choice B;
π+3 < 6 — let π = 3
3+3 < 6
6 < 6
Now 3.14159 +3 would be > 6 because π ≈ 3.14 > 3
Thus, π+3<6 is FALSE.
Choice C;
6-π>3
6-3>3
3>3
Now we know that if we subtract 6 with 3.14, the value would be less than 3.
6-3.14 would be around 2.7 to 2.8
Therefore 6-π>3 is also FALSE.
Choice D;
8π>24
8×3>24
24>24
Now if we let π ≈ 3.14, we know that π > 3 and 8 times 3.14 would be greater than 24.
Therefore, 8π>23 is TRUE.