In order to find the value of \(\displaystyle d\), we first need to find the equation for the line with a slope \(\displaystyle m=\frac{1}{2}\) that passes through \(\displaystyle (3,6)\). 

\(\displaystyle y-6=\frac{1}{2}(x-3)\)

\(\displaystyle y-6=\frac{1}{2}x-\frac{3}{2}\)

\(\displaystyle y=\frac{1}{2}x+\frac{9}{2}\)

Plugging in \(\displaystyle (2,d)\) and solving for \(\displaystyle d\):

\(\displaystyle d=\frac{1}{2}(2)+\frac{9}{2}\)

\(\displaystyle d=\frac{11}{2}\)

In order to find the value of \(\displaystyle a\), we first need to find the equation for the line with a slope \(\displaystyle m=4\) that passes through \(\displaystyle (1,3)\). 

\(\displaystyle y-3=4(x-1)\)

\(\displaystyle y-3=4x-4\)

\(\displaystyle y=4x-1\)

Plugging in \(\displaystyle (7,a)\) and solving for \(\displaystyle a\):

\(\displaystyle a=4(7)-1\)

\(\displaystyle a=28-1\)

\(\displaystyle a=27\)

The domain of the function specifies the values that \(\displaystyle x\) can take.  Here, \(\displaystyle 4-x^{2}\) is defined for every value of \(\displaystyle x\), so the domain is all real numbers. 

To find the domain, we need to decide which values \(\displaystyle x\) can take.  The \(\displaystyle x\) is under a square root sign, so \(\displaystyle x\) cannot be negative.  \(\displaystyle x\) can, however, be 0, because we can take the square root of zero.  Therefore the domain is \(\displaystyle x\geq 0\).

To find the domain, we must find the interval on which \(\displaystyle \sqrt{4-x^{2}}\) is defined.  We know that the expression under the radical must be positive or 0, so \(\displaystyle \sqrt{4-x^{2}}\) is defined when \(\displaystyle x^{2}\leq 4\).  This occurs when \(\displaystyle x \geq -2\) and \(\displaystyle x \leq 2\).  In interval notation, the domain is \(\displaystyle \left [ -2,2 \right ]\).

The domain of \(\displaystyle (f+g)\) is the intersection of the domains of \(\displaystyle f\) and \(\displaystyle g\). \(\displaystyle f\) and \(\displaystyle g\) are each restricted to all values of \(\displaystyle x\) that allow the radicand \(\displaystyle x-1\) to be nonnegative - that is, 

\(\displaystyle \small \small x - 1\geq 0\), or 

\(\displaystyle \small \small \small x \geq 1\)

Since the domains of \(\displaystyle f\) and \(\displaystyle g\) are the same, the domain of \(\displaystyle (f+g)\) is also the same. In interval form the domain of \(\displaystyle (f+g)\) is \(\displaystyle \small \small [1, \infty )\)

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression \(\displaystyle \sqrt[3]{x -27}\) is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which 

\(\displaystyle \sqrt[3]{x -27} = 0\)

\(\displaystyle \left (\sqrt[3]{x -27} \right )^{3}= 0^{3}\)

\(\displaystyle x -27 = 0\)

\(\displaystyle x = 27\)

27 is the only number excluded from the domain.

Since the expression \(\displaystyle x^{2}+3x-4\) is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which \(\displaystyle x^{2}+3x-4 =0\). We solve for \(\displaystyle x\) by factoring the polynomial, which we can do as follows:

\(\displaystyle x^{2}+3x-4 =0\)

\(\displaystyle (x+?)(x+?)=0\)

Replacing the question marks with integers whose product is \(\displaystyle -4\) and whose sum is 3:

\(\displaystyle (x+4)(x-1)=0\)

\(\displaystyle x + 4 = 0 \Rightarrow x = -4\)

\(\displaystyle x -1 = 0 \Rightarrow x = 1\)

Therefore, the domain excludes these two values of \(\displaystyle x\).

The only restriction on the domain of \(\displaystyle f\) is that the denominator cannot be 0. We set the denominator to 0 and solve for \(\displaystyle x\) to find the excluded values:

\(\displaystyle x ^{2}-25 =0\)

\(\displaystyle x ^{2}= 25\)

\(\displaystyle x = -5 \textrm{ or }x = 5\)

The domain is the set of all real numbers except those two - that is, 

\(\displaystyle (-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)\).

Set \(\displaystyle y = 0\) and solve:

\(\displaystyle y = \log _{2}(x+ 4)\)

\(\displaystyle \log _{2}(x+ 4) = 0\)

\(\displaystyle 2 ^ {\log _{2}(x+ 4) }= 2 ^ { 0}\)

\(\displaystyle x + 4 = 1\)

\(\displaystyle x + 4 - 4 = 1 - 4\)

\(\displaystyle x = -3\)

The  \(\displaystyle x\)-intercept is \(\displaystyle (-3,0)\).