Lane ORCCA (2020–2021): Open Resources for Community College Algebra

1. We know the formula to calculate a square's area is \(A=l^2\text{,}\) where \(A\) stands for a square's area, and \(l\) is the length of the square's side. To build a function to calculate \(l\text{,}\) we solve for \(l\) in the formula:

   \begin{align*} A\amp=l^2 \end{align*}
   \begin{align*} l\amp=\sqrt{A}\amp\amp\text{or} \amp l\amp=-\sqrt{A} \end{align*}

   The formula is \(l=\sqrt{A}\text{.}\) We don't consider the negative solution in this context since negative length doesn't make sense. Since \(l\) depends on \(A\text{,}\) we can use function notation and write \(f(A)=\sqrt{A}\text{.}\) We will make a table, plot points, and look at the graph of \(l=\sqrt{A}\text{.}\)

   \(A\)	\(f(A)\)	Points on the Curve
   \(0\)	\(\sqrt{0}=0\)	\((0,0)\)
   \(1\)	\(\sqrt{1}=1\)	\((1,1)\)
   \(4\)	\(\sqrt{4}=2\)	\((4,2)\)
   \(6.25\)	\(\sqrt{6.25}=2.5\)	\((6.25,2.5)\)
   \(9\)	\(\sqrt{9}=3\)	\((9,3)\)

   

2. The point \((6.25,2.5)\) implies that a square plate with an area of \(6.25\) square feet would have a length of \(2.5\) feet on each side.

3. According to the graph, the function's domain is \([0,\infty)\) because the graph goes forever to the right from \(A=0\text{.}\) This should make some sense because you cannot take the square root of a negative number. The function's range is \([0,\infty)\) because the graph seems to go up forever starting at \(l=0\text{.}\) This should also make sense because a square root never gives you a negative number as an answer.

Using Fact 13.1.7 to find the function's domain, we set the radicand greater than or equal to zero and solve:

The function's domain is \([2,\infty)\) in interval notation.

To find the function's range, we use point plotting to graph the function and then look at its graph. The graph shows that the function's range is \([1,\infty)\text{.}\) The graph also verifies the function's domain is indeed \([2,\infty)\text{.}\)

To find the function's domain, we set the radicand to be greater than or equal to zero:

From the graph, we can see that the range is all numbers below (or equal to) the \(y\)-value \(2\text{.}\) In interval notation, this would be written \((-\infty,2]\text{.}\)

1. To find how long it will take for the object to hit the ground, we need to plug in \(100\) for the distance that it will fall,\(d\) and then solve for \(t\text{.}\)

   \begin{align*} t=\sqrt{\frac{d}{16}}\\ t\amp=\sqrt{\frac{100}{16}}\\ t\amp=\frac{\sqrt{100}}{sqrt{16}}\\ t\amp=\frac{10}{4}\\ t\amp=2.5 \end{align*}

   This means it will take the beanbag \(2.5\) seconds to hit the ground if it's released from the top of a \(100\)-foot-tall building.

2. This time we substitute \(3\) for \(t\) and solve for \(d\text{:}\)

   \begin{align*} t=\sqrt{\frac{d}{16}}\\ 3\amp=\sqrt{\frac{d}{16}}\\ 3^2\amp=\left(\sqrt{\frac{d}{16}}\right)^2\\ 9\amp=\frac{d}{16}\\ 16\cdot 9\amp=\frac{d}{16}\cdot 16\\ 144\amp=d \end{align*}

   This means the beanbag will fall approximately \(144\) feet in \(3\) seconds, so the second building is approximately \(144\) feet tall.

First, we will sketch a graph of those two points.

To calculate the distance between \((2,3)\) and \((5,7)\text{,}\) we sketch a right triangle as in the figure and then use the Pythagorean Theorem:

\begin{align*} d^2\amp=(5-2)^2+(7-3)^2\\ d^2\amp=3^2+4^2\\ d^2\amp=9+16\\ d^2\amp=25\\ d\amp=\sqrt{25}\\ d\amp=5 \end{align*}

In conclusion, the distance between \((2,3)\) and \((5,7)\) is \(5\text{.}\) Note that in our calculations, we didn't need to show \(d=\pm\sqrt{25}\) because distance must have a positive value.

First, we will sketch a graph of those two points. We will put the image from last example side by side with the new image, so it's clear that we are using the same method.

To calculate the distance between \(\highlight{(x_1,y_1)}\) and \(\highlight{(x_2,y_2)}\text{,}\) we sketch a right triangle as in the figure and then use Pythagorean Theorem:

To calculate the distance between \((-2,4)\) and \((5,-20)\text{,}\) we use the distance formula. It's good practice to mark each value with the corresponding variables in the formula. Again, \((x_1,y_1)\) stands for the first point's coordinates, and \((x_2,y_2)\) stands for the second point's coordinates:

We have:

The distance between \((-2,4)\) and \((5,-20)\) is \(25\) units.

First, to find the radius we will substitute \(\substitute{2}\) in for \(V\text{,}\) and we have:

The spheres will have a radius of approximately 0.78 in.

When we calculate \(\sqrt[3]{\frac{3(2)}{4\pi}}\) with a calculator, we enter (3*2/(4π))^(1/3). To find the diameter, we multiply the radius by \(2\) to get 1.56 in. Nasim can advertise the spheres to be 1.56 inches in diameter.

1. First, note that the index of this function is \(4\text{.}\) By Fact 13.1.27, to find the domain of this function we must set the radicand greater or equal to zero and solve.

   \begin{align*} 10-5x\amp\ge0\\ -5x\amp\ge\subtractright{10}\\ x\amp\mathbin{\highlight{\le}}\divideunder{-10}{-5}\\ x\amp\le2 \end{align*}

   So, the domain of \(g\) must be \((-\infty,2]\text{.}\)

2. First, note that the index of the function \(h\) for \(h(x)=4\sqrt[5]{2x-5}+1\) is \(5\text{.}\) By Fact 13.1.27, the domain is \((-\infty,\infty)\text{.}\)