Mathematically, t can take any value, so the domain would be R.
In reality (physics), I would expect t≥0.
Mathematically, the range is h≤98/5, and will become infinitely negative.
In reality, h is probably ≥0, making the range 0 ≤ h ≤ 98/5.
So it depends on whether you want to take the real-world limitations into account. Seen as how they are mentioned in the question prominently, I would do so.
Answer:
There is no solution.
Hope this helps!
PLEASE let me know if this is correct.
The problem is that ∃xp(x) is true, and <span>∃xq(x) is true means that for some x=x1, p(x) is true, and for some x=x2, q(x) is true. There is no guarantee that x1=x2, although it is not impossible.
However, </span><span>∃x(p(x) ∧ q(x)) means that for the same x=x0, p(x0) is true, and q(x0) is true. This cannot be implied from the first proposition, therefore
</span>
<span>∃xp(x) ∧ ∃xq(x) => ∃x(p(x) ∧ q(x)) is a false statement.</span>
Divisibility by 2: the last digit must be even. The last digit of 15114 is 4, which is even, so 15114 is divisible by 2.
Divisibility by 3: the sum of the digits must be divisible by 3. The sum of the digits of 15114 is 12, which is divisible by 3, so 15114 is divisible by 3.
Divisibility by 5: the last digit must be 0 or 5. The last digit of 15114 is 4, so 15114 is not divisible by 5
Divisibility by 9: the sum of the digits must be divisible by 9. The sum of the digits of 15114 is 12, which is not divisible by 9, so 15114 is not divisible by 9.
3
This is because the area of a rhombus is A=sh
since the height is 4 mm, and the area is 12mm^2 the length is 3