Given the sets M = : -2 x 4\ and N = : 0 < x < 6 we need to find their intersection M N.

The set $M$ includes all real numbers $x$ such that $x$ is greater than or equal to -2 and less than or equal to 4. In interval notation, this is $[-2, 4]$.
The set $N$ includes all real numbers $x$ such that $x$ is greater than 0 and less than 6. In interval notation, this is $(0, 6)$.

The intersection $M \cap N$ consists of all values of $x$ that are present in both sets.
For $x$ to be in both sets, it must satisfy both conditions:
$-2 \le x \le 4$
AND
$0 < x < 6$

To find the combined range:

• The lower bound for $x$ must be greater than the largest of the two lower bounds. Comparing $-2$ and 0, the larger is 0. Since $x > 0$ in $N$, the intersection will have $x > 0$.
• The upper bound for $x$ must be less than the smallest of the two upper bounds. Comparing 4 and 6, the smaller is 4. Since $x \le 4$ in $M$, the intersection will have $x \le 4$.

Combining these, we get $0 < x \le 4$.

The correct option is A.

To subtract the terms on the left side, we need a common denominator, which is $\sqrt{3}$.

To isolate $x$, divide both sides by $\sqrt{3}$:

The correct option is C.

For a geometric progression, the common ratio $r$ is found by dividing any term by its preceding term.

Given:
First term ($a$) = $\sqrt{3} - 2$
Second term ($ar$) = $7 - 4\sqrt{3}$

To simplify this expression, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{3} - 2$ is $\sqrt{3} + 2$.

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

The correct option is D.

We know that $k = \log_b (b^k)$. So, $5 = \log_2 (2^5)$.