Square both sides to eliminate the radical.

\(\displaystyle (\sqrt{-3x-5}) ^2= 7^2\)

\(\displaystyle -3x-5 = 49\)

Add five on both sides.

\(\displaystyle -3x-5 +5= 49+5\)

\(\displaystyle -3x = 54\)

Divide by negative three on both sides.

\(\displaystyle \frac{-3x}{-3} = \frac{54}{-3}\)

The answer is:  \(\displaystyle -18\)

Substitute the value of negative three as \(\displaystyle x\).

\(\displaystyle f(-3) = 2\sqrt{-3}-\sqrt{(-3)-3}\)

\(\displaystyle 2\sqrt{-3}-\sqrt{-6} = 2\sqrt{-1}\cdot \sqrt{3} - \sqrt{-1}\cdot \sqrt{6}\)

The terms will be imaginary.  We can factor out an \(\displaystyle i = \sqrt{-1}\) out of the right side.  Replace them with \(\displaystyle i\).

The answer is:  \(\displaystyle 2i\sqrt3 -i\sqrt6\)

When the baseball hits the ground, the height is 0, so we set \(\displaystyle h\left ( t\right ) = 0\). and solve for \(\displaystyle t\).

\(\displaystyle -16t^{2}+60t + 100 = 0\)

This can be done using the quadratic formula:

\(\displaystyle t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

Set \(\displaystyle a = -16, b = 60, c = 100\):

\(\displaystyle t = \frac{-60 \pm \sqrt{60^{2}-4(-16)100}}{2(-16)}\)

\(\displaystyle t = \frac{-60 \pm \sqrt{3,600- (- 6,400)}}{-32}\)

\(\displaystyle t = \frac{-60 \pm \sqrt{3,600+6,400}}{-32}\)

\(\displaystyle t = \frac{-60 \pm \sqrt{10,000}}{-32}\)

\(\displaystyle t = \frac{-60 \pm 100}{-32}\)

One possible solution:

\(\displaystyle t = \frac{-60 + 100}{-32} = \frac{40}{-32} = -1.25\) 

We throw this out, since time must be positive.

The other:

\(\displaystyle t = \frac{-60 - 100}{-32} = \frac{-160}{-32} = 5\)

This solution, we keep. The baseball hits the ground in 5 seconds.

The period of the graph of a sine function \(\displaystyle f(x)= A \sin Bx\) is \(\displaystyle \frac{2 \pi}{B}\), or \(\displaystyle 2 \pi \div B\).

Since \(\displaystyle B= 8\),

\(\displaystyle 2 \pi \div 8 = \frac{\pi }{4}\).

This answer is not among the given choices.

Evaluate each trig function at the specified angle.

\(\displaystyle sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)

\(\displaystyle cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)

Replace the terms into the function.

\(\displaystyle f(\frac{\pi}{4})=2sin(\frac{\pi}{4})+cos(\frac{\pi}{4}) = 2(\frac{\sqrt{2}}{2})+\frac{\sqrt{2}}{2}\)

Combine like-terms.

The answer is:  \(\displaystyle \frac{3\sqrt2}{2}\)

\(\displaystyle (f \circ g) \left ( x\right ) = f (g(x))\) by definition.

\(\displaystyle (f \circ g) \left ( 3\right ) = f (g (3))\)

\(\displaystyle g(x) = 3x\) on the set \(\displaystyle [3, \infty)\), so 

\(\displaystyle g(3) = 3 \cdot 3 = 9\).

\(\displaystyle f(x) = 4-x\) on the set \(\displaystyle [0 , \infty)\), so

\(\displaystyle f (g(3)) = f(9) = 4-9 = -5\).

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If \(\displaystyle x < 0\), then \(\displaystyle f(x) = 2x - 5\). To find the range of \(\displaystyle f\) on the interval \(\displaystyle (-\infty, 0)\), we note:

\(\displaystyle x < 0\)

\(\displaystyle 2x < 2 \cdot 0\)

\(\displaystyle 2x < 0\)

\(\displaystyle 2x- 5 < 0 - 5\)

\(\displaystyle 2x- 5 < - 5\)

\(\displaystyle f(x)< - 5\)

The range of this portion of \(\displaystyle f\) is \(\displaystyle (-\infty, -5)\).

If \(\displaystyle x \ge 0\), then \(\displaystyle f(x) = -3x\). To find the range of \(\displaystyle f\) on the interval \(\displaystyle [0, \infty)\), we note:

\(\displaystyle x \ge 0\)

\(\displaystyle -3x \le -3 \cdot 0\)

\(\displaystyle -3x \le 0\)

\(\displaystyle f(x)\le 0\)

The range of this portion of \(\displaystyle f\) is \(\displaystyle (-\infty, 0 ]\)

The union of these two sets is \(\displaystyle (-\infty, 0 ]\), so this is the range of \(\displaystyle f\) over its entire domain.

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If \(\displaystyle x < -1\), then \(\displaystyle f(x)=x-7\). 

To find the range of \(\displaystyle f\) on the interval \(\displaystyle (-\infty, -1)\), we note:

\(\displaystyle x < -1\)

\(\displaystyle x-7 < -1-7\)

\(\displaystyle f(x)< -8\)

The range of \(\displaystyle f\) on \(\displaystyle (-\infty, -1)\) is \(\displaystyle (-\infty, -8)\).

If \(\displaystyle x > 1\), then \(\displaystyle f(x)=6-x\). 

To find the range of \(\displaystyle f\) on the interval \(\displaystyle (1, \infty )\), we note:

\(\displaystyle x > 1\)

\(\displaystyle -x < - 1\)

\(\displaystyle 6-x < 6- 1\)

\(\displaystyle f(x)< 5\)

The range of \(\displaystyle f\) on \(\displaystyle (1, \infty )\) is \(\displaystyle (-\infty, 5)\).

The range of \(\displaystyle f\) on its entire domain is the union of these sets, or \(\displaystyle (-\infty, 5)\).

\(\displaystyle (f \circ g) \left ( 2\frac{1}{3}\right )=f\left ( g \left ( 2\frac{1}{3}\right ) \right )\)

First, we evaluate  \(\displaystyle g \left ( 2\frac{1}{3}\right )\). Since \(\displaystyle 2\frac{1}{3} \ge 0\), the definition of \(\displaystyle g\) for \(\displaystyle x \ge 0\) is used, and 

\(\displaystyle g(x)= x-3\)

\(\displaystyle g \left ( 2\frac{1}{3}\right ) = 2\frac{1}{3} - 3 = -\frac{2}{3}\)

Since 

\(\displaystyle f(x)= 4x+7\), then

\(\displaystyle (f \circ g) \left ( 2\frac{1}{3}\right )\)

\(\displaystyle =f\left ( g \left ( 2\frac{1}{3}\right ) \right )\)

\(\displaystyle =f\left ( -\frac{2}{3} \right )\)

\(\displaystyle = 4 \left ( -\frac{2}{3} \right )+7\)

\(\displaystyle = -2\frac{2}{3} +7\)

\(\displaystyle = 4\frac{1}{3}\)

\(\displaystyle (f \circ g)(3) = f(g(3))\)

First, evaluate \(\displaystyle g(3)\) using the definition of \(\displaystyle g\) for \(\displaystyle x > 1\):

\(\displaystyle g(x)= x-2\)

\(\displaystyle g(3)= 3-2 = 1\)

Therefore, 

\(\displaystyle (f \circ g)(3) = f(g(3)) = f(1)\)

However, \(\displaystyle x = 1\) is not in the domain of \(\displaystyle f\). 

Therefore, \(\displaystyle (f \circ g)(3)\) is an undefined quantity.