$$ABC=acute\ triangle\triangle$$
Side AB = 7
Side AC = 9
Side X = BC

Question => How many possible integers of x do we have?

$$In\ any\ acute\ angle\ triangle\ c^2<a^2+b^2$$
Where A, B, and C are the 3 sides of the triangle and C is the longest/greatest side
The maximum value of x is when x = C i.e when x is the longest/greatest side, and when x is not = C, the only value of C is 9

$$Max\ \left(x\right)=x^2<9^2+7^2$$
$$=x^2<81+49$$
$$=x<\sqrt{130}$$
$$=x<11.4$$

$$Min\ \left(x\right)=9^2<7^2+x^2$$
$$x^2>9^2-7^2$$ $$x^2>9^2-7^2$$
$$x^2>81-49$$
$$x>\sqrt{32}$$
$$x>5.7$$

Therefore, x > 5.7 and x < 11.4
= 5.7 < x < 11.4
All possible integers of x = 6, 7, 8, 9, 10, and 11
= 6 possible integers

Well, from the expression above, I got 6 as the number of possible integers of x.

=>

Since 7, 9 and x are the sides of a triangle, we have 9 – 7 = 2 < x < 16 < 9 + 7.
Case 1: x is the largest length ( x ≥ 9 )
Since the triangle is acute, we have x^2 < 7^2 + 9^2 = 49 + 81 = 130 or x < √130 < 12.
Case 2: 9 is the largest length ( x < 9 )
Since the triangle is acute, we have 7^2 + x^2 > 9^2 or x^2 > 81 - 49 = 32 or x > √32 > 5.

Then we have 6, 7, …, 11 as the possible values of x.
Thus, the number of possible values of x is 11 – 6 + 1 = 6.

Therefore, A is the answer.
Answer: A