Open Resources for Community College Algebra

To make a good table for this line, we should have \(x\)-values that are multiples of \(3\) to make sure that the fraction cancels nicely for the outputs.

\(x\)	\(y=\frac{5}{3}x-3\)	Point
\(-3\)	\(\frac{5}{3}(-3)-3=-8\)	\((1,-8)\)
\(0\)	\(\frac{5}{3}(0)-3=-3\)	\((2,-3)\)
\(3\)	\(\frac{5}{3}(3)-3=2\)	\((3,2)\)
\(6\)	\(\frac{5}{3}(6)-3=7\)	\((4,7)\)

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The slope of \(j(x)=\firsthighlight{-\frac{7}{2}}x+\secondhighlight{5}\) is \(\firsthighlight{-\frac{7}{2}}\text{,}\) and the vertical intercept is \((0,\secondhighlight{5})\text{.}\) Starting at \((0,5)\text{,}\) we go down \(7\) units and right \(2\) units to reach more points.

From the graph, we can read that two more points that \(j(x)=-\frac{7}{2}x+5\) passes through are \((2,-2)\) and \((4,9)\text{.}\)

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The slope of \(k(x)=\firsthighlight{\frac{3}{4}}(x-\secondhighlight{2})\mathbin{\secondhighlight{-}}\secondhighlight{5}\) is \(\firsthighlight{\frac{3}{4}}\) and the point on the graph given in the equation is \((\secondhighlight{2},\secondhighlight{-5})\text{.}\) So to graph \(k\text{,}\) start at \((2,-5)\text{,}\) and the go up \(3\) units and right \(4\) units (or down \(3\) left \(4\)) to reach more points.

From the graph, we can read that two more points that \(k(x)=\frac{3}{4}(x-2)-5\) passes through are \((6,-2)\) and \((10,1)\text{.}\)

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Typical wind speeds vary between \(5\) and 20 mph, with gusty conditions up to 60 mph, depending on location. A good way to enter the sixth root into a calculator is to recall that \(\sqrt[6]{x}=x^{\frac{1}{6}}\text{.}\)

With the values in Table 13.8.11, we can sketch the graph.

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The specifics of how to use any one particular technology tool vary. Whether you use an app, a physical calculator, or something else, a table and graph should look like:

\(x\)	\(g(x)\)
\(-1\)	\(-12\)
\(0\)	\(-6\)
\(1\)	\(-2\)
\(2\)	\(0\)
\(2\)	\(0\)
\(3\)	\(0\)
\(4\)	\(-2\)

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Additional features of your technology tool can enhance the graph to help answer these questions. You may be able to make the graph appear like:

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1. The vertex is \((2.5,0.25)\text{.}\)

2. The vertical intercept is \((0,-6)\text{.}\)

3. The horizontal intercepts are \((2,0)\) and \((3,0)\text{.}\)

4. \(g(-1)=-2\text{.}\)

5. The solutions to \(g(x)=-6\) are the \(x\)-values where \(y=6\text{.}\) We graph the horizontal line \(y=-6\) and find the \(x\)-values where the graphs intersect. The solution set is \(\{0,5\}\text{.}\)

6. The solutions are all \(x\)-values where the function below (or touching) the line \(y=-6\text{.}\) The solution set is \((-\infty,0]\cup[5,\infty)\text{.}\)

   

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7. The domain is \((-\infty,\infty)\) and the range is \((-\infty,0.25]\text{.}\)

1. The vertex of the quadratic function \(r(x)=-8(x+\highlight{1})^2+\secondhighlight{7}\) is \((\highlight{-1},\secondhighlight{7})\text{.}\)

2. The vertex of the quadratic function \(u(x)=5(x-\highlight{7})^2\secondhighlight{-3}\) is \((\highlight{7},\secondhighlight{-3})\text{.}\)

3. The formula for a parabola with vertex \((\highlight{-5},\secondhighlight{3})\) and \(a=2\) is \(y=2(x+\highlight{5})^2+\secondhighlight{3}\text{.}\)

4. The formula for a parabola with vertex \((\highlight{1},\secondhighlight{-17})\) and \(a=-4\) is \(y=4(x-\highlight{1})^2\secondhighlight{-17}\text{.}\)

1. The graph of the quadratic function \(s(x)=-8(x+1)^2+7\) is the same as the graph of \(f(x)=x^2\) shifted to the left \(1\) unit and up \(7\) units.

2. The graph of the quadratic function \(v(x)=5(x-7)^2-3\) is the same as the graph of \(f(x)=x^2\) shifted to the right \(7\) units and down \(3\) units.

1. The horizontal intercepts of \(n\) are \((1,0)\) and \((-6,0)\text{.}\)

2. The horizontal intercepts of \(p\) are \((\frac{2}{3},0)\) and \((-\frac{1}{2},0)\text{.}\)

1. To complete the square in the equation \(k^2-18k+1=0\text{,}\) we first we will first move the constant term to the right side of the equation. Then we will use Fact 13.3.3 to find \(\left(\frac{b}{2}\right)^2\) to add to both sides.

   \begin{align*} k^2-18k+1\amp=0\\ k^2-18k\amp=-1 \end{align*}

   In our case, \(b=-18\text{,}\) so \(\left(\frac{b}{2}\right)^2=\left(\frac{-18}{2}\right)^2=81\)

   \begin{align*} k^2-18k\addright{81}\amp=-1\addright{81}\\ (k-9)^2\amp=80 \end{align*}
   \begin{align*} k-9\amp=-\sqrt{80}\amp\text{ or }\amp\amp k-9\amp=\sqrt{80}\\ k-9\amp=-4\sqrt{5}\amp\text{ or }\amp\amp k-9\amp=4\sqrt{5}\\ k\amp=9-4\sqrt{5}\amp\text{ or }\amp\amp k\amp=9+4\sqrt{5} \end{align*}

   The solution set is \(\{9+4\sqrt{5},9-4\sqrt{5}\}\text{.}\)

2. To complete the square in the equation \(4p^2-3p=2\text{,}\) we first divide both sides by \(4\) since the leading coefficient is 4.

   \begin{align*} \divideunder{4p^2}{4}-\divideunder{3p}{4}\amp=\divideunder{2}{4}\\ p^2-\frac{3}{4}p\amp=\frac{1}{2}\\ p^2-\frac{3}{4}p\amp=\frac{1}{2} \end{align*}

   Next, we will complete the square. Since \(b=-\frac{3}{4}\text{,}\) first,

   \begin{equation} \frac{b}{2}=\frac{-\frac{3}{4}}{2}=-\frac{3}{8}\label{equation-completing-square-3-4ths-over-2}\tag{13.8.1} \end{equation}

   and next, squaring that, we have

   \begin{equation} \left(-\frac{3}{8}\right)^2=\frac{9}{64}\text{.}\label{equation-completing-square-3-over-8}\tag{13.8.2} \end{equation}

   So we will add \(\frac{9}{64}\) from Equation (13.8.2) to both sides of the equation:

   \begin{align*} p^2-\frac{3}{4}p\addright{\frac{9}{64}}\amp=\frac{1}{2}\addright{\frac{9}{64}}\\ p^2-\frac{3}{4}p+\frac{9}{64}\amp=\frac{32}{64}+\frac{9}{64}\\ p^2-\frac{3}{4}p+\frac{9}{64}\amp=\frac{41}{64} \end{align*}

   Here, remember that we always factor with the number found in the first step of completing the square, Equation (13.8.1).

   \begin{align*} \left(p-\frac{3}{8}\right)^2\amp=\frac{41}{64} \end{align*}
   \begin{align*} p-\frac{3}{8}\amp=-\frac{\sqrt{41}}{8}\amp\text{ or }\amp\amp p-\frac{3}{8}\amp=\frac{\sqrt{41}}{8}\\ p\amp=\frac{3}{8}-\frac{\sqrt{41}}{8}\amp\text{ or }\amp\amp p\amp=\frac{3}{8}+\frac{\sqrt{41}}{8}\\ p\amp=\frac{3-\sqrt{41}}{8}\amp\text{ or }\amp\amp p\amp=\frac{3+\sqrt{41}}{8} \end{align*}

   The solution set is \(\left\{\frac{3-\sqrt{41}}{8}, \frac{3+\sqrt{41}}{8}\right\}\text{.}\)

Before we can complete the square, we will factor the \(\highlight{4}\) out of the first two terms. Don't be tempted to factor the \(4\) out of the constant term.

Now we will complete the square inside the parentheses by adding and subtracting \(\left(\frac{5}{2}\right)^2=\frac{25}{4}\text{.}\)

Notice that the constant that we subtracted is inside the parentheses, but it will not be part of our perfect square trinomial. In order to bring it outside, we need to multiply it by \(4\text{.}\) We are distributing the \(4\) to that term so we can combine it with the outside term.

Note that The vertex is \(\left(-\frac{5}{2},-1\right)\text{.}\)

To start, we'll note that this function opens downward because the leading coefficient, \(-1\text{,}\) is negative.

Now we will complete the square to find the vertex. We will factor the \(-1\) out of the first two terms, and then add and subtract \(\left(\frac{8}{2}\right)^2=4^2=\highlight{16}\) on the right side.

The vertex is \((-4,1)\) so the axis of symmetry is the line \(x=-4\text{.}\)

To find the \(y\)-intercept, we'll replace \(x\) with \(0\) or read the value of \(c\) from the function in standard form:

The \(y\)-intercept is \((0,-15)\) and we will find its symmetric point on the graph, which is \((-8,-15)\text{.}\)

Next, we'll find the horizontal intercepts. We see this function factors so we will write the factored form to get the horizontal intercepts.

The \(x\)-intercepts are \((-3,0)\) and \((-5,0)\text{.}\)

Now we will plot all of the key points and draw the parabola.

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The equation \(\abs{9-4x}=17\) breaks into two pieces, each of which needs to be solved independently.

The solution set is \(\left\{-2,\frac{13}{2}\right\}\text{.}\)

The equation \(\abs{7-3x}=\abs{6x-5}\) breaks into two pieces, each of which needs to be solved independently.

The solution set is \(\left\{\frac{4}{3},-\frac{2}{3}\right\}\text{.}\)

1. The equation \(5x^2-2x=9\) is a quadratic equation since the variable is being squared (but doesn't have any higher power).

2. The equation \(\pi x-3=4(x+1)\) is a linear equation since the variable is only to the first power.

3. The equation \(8x^2=x+9\) is a quadratic equation since there is a degree-two term.

4. The equation \(\frac{2}{x+2}+\frac{3}{2x+4}=\frac{7}{x+8}\) is a rational equation since the variable exists in the denominator.

5. The equation \(\abs{2x-7}+2=3\) is an absolute value equation since the variable is inside an absolute value.

6. The equation \(\sqrt{x+2}=x-4\) is a radical equation since the variable appears inside the radical.

7. The equation \((3-2x)^2-9=9\) is a quadratic equation since if we were to distribute everything out, we would have a term with \(x^2\text{.}\)

8. The equation \(3=\sqrt[3]{5-2x}\) is a radical equation since the variable is inside the radical.

9. The system

   \begin{align*} \left\{ \begin{alignedat}{4} 5x\amp-{}\amp y \amp={}\amp -6 \\ 2x\amp-{}\amp 3y \amp={}\amp 8 \\ \end{alignedat} \right. \end{align*}

   is a system of linear equations.

10. The equation \(x^x=x-\abs{x-x^2-3}\) is an equation type that we have not covered and is not listed above.

1. Since the equation \(5x^2-2x=9\) is a quadratic we should consider the square root method, the quadratic formula, factoring, and completing the square. In this case, we will start with the quadratic formula. First, note that we should rearrange the terms in equation into standard form.

   \begin{align*} 5x^2-2x\amp=9\\ 5x^2-2x-9\amp=0 \end{align*}

   Note that \(a=\substitute{5}\text{,}\) \(b=\substitute{-2}\text{,}\) and \(c=\substitute{-9}\text{.}\)

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{-2})\pm\sqrt{(\substitute{-2})^2-4(\substitute{5})(\substitute{-9})}}{2(\substitute{5})}\\ x\amp=\frac{2\pm\sqrt{4+180}}{10}\\ x\amp=\frac{2\pm\sqrt{184}}{10}\\ x\amp=\frac{2\pm\sqrt{4\cdot46}}{10}\\ x\amp=\frac{2\pm2\sqrt{46}}{10}\\ x\amp=\frac{1\pm\sqrt{46}}{5} \end{align*}

   The solution set is \(\left\{\frac{1+\sqrt{46}}{5},\frac{1-\sqrt{46}}{5}\right\}\text{.}\)

2. Since the equation \(\pi x-3=4(x+1)\) is a linear equation, we isolate the variable step-by-step.

   \begin{align*} \pi x-3\amp=4(x+1)\\ \pi x-3\amp=4x+4\\ \pi x-4x\amp=7\\ x(\pi-4)\amp=7\\ x\amp=\frac{7}{\pi-4} \end{align*}

   The solution set is \(\left\{\frac{7}{\pi-4}\right\}\text{.}\)

3. Since the equation \(8x^2=x+9\) is a quadratic equation, we again have several options to consider. We will try factoring on this one first after converting it to standard form.

   \begin{align*} 8x^2\amp=x+9\\ 8x^2-x-9\amp=0\\ \end{align*}

   Here, \(ac=-72\) and two numbers that multiply to be \(-72\) but add to be \(-1\) are \(8\) and \(-9\text{.}\)

   \begin{align*} 8x^2\firsthighlight{{}+8x-9x}-9\amp=0\\ \left(8x^2\mathbin{\firsthighlight{+}}\firsthighlight{8x}\right)+\left(\mathbin{\firsthighlight{-}}\firsthighlight{9x}-9\right)\amp=0\\ 8x(x+1)-9(x+1)\amp=0\\ (8x-9)(x+1)\amp=0 \end{align*}
   \begin{align*} 8x-9\amp=0\amp\amp\text{or}\amp x+1\amp=0\\ x\amp=\frac{9}{8}\amp\amp\text{or}\amp x\amp=-1 \end{align*}

   The solution set is \(\left\{\frac{9}{8},-1\right\}\)

4. Since the equation \(\frac{2}{x+2}+\frac{3}{2x+4}=\frac{7}{x+8}\) is a rational we first need to cancel the denominators after factoring and finding the least common denominator.

   \begin{align*} \frac{2}{x+2}+\frac{3}{2x+4}\amp=\frac{7}{x+8}\\ \frac{2}{x+2}+\frac{3}{2(x+2)}\amp=\frac{7}{x+8}\\ \end{align*}

   At this point, we note that the least common denominator is \(2(x+2)(x+8)\text{.}\) We need to multiply every term by this least common denominator.

   \begin{align*} \frac{2}{x+2}\multiplyright{2(x+2)(x+8)}+\frac{3}{2(x+2)}\multiplyright{2(x+2)(x+8)}\amp=\frac{7}{x+8}\multiplyright{2(x+2)(x+8)}\\ \frac{2}{\cancelhighlight{x+2}}\cdot2(\cancelhighlight{x+2})(x+8)+\frac{3}{\cancel{2}(\cancelhighlight{x+2})}\cdot\cancel{2}(\cancelhighlight{x+2})(x+8)\amp=\frac{7}{\secondcancelhighlight{x+8}}\cdot2(x+2)(\secondcancelhighlight{x+8})\\ 2\cdot2(x+8)+3(x+8)\amp=7\cdot2(x+2)\\ 4(x+8)+3(x+8)\amp=14(x+2)\\ 4x+32+3x+24\amp=14x+28\\ 7x+56\amp=14x+28\\ 28\amp=7x\\ 4\amp=x \end{align*}

   We always check solutions to rational equations to ensure we don't have any “extraneous solutions”.

   \begin{align*} \frac{2}{(\substitute{4})+2}+\frac{3}{2(\substitute{4})+4}\amp\stackrel{?}{=}\frac{7}{(\substitute{4})+8}\\ \frac{2}{6}+\frac{3}{12}\amp\stackrel{?}{=}\frac{7}{12}\\ \frac{4}{12}+\frac{3}{12}\amp\stackrel{?}{=}\frac{7}{12}\\ \frac{7}{12}\amp\stackrel{\checkmark}{=}\frac{7}{12} \end{align*}

   So, the solution set is \(\{4\}\text{.}\)

5. Since the equation \(\abs{2x-7}+2=3\) is an absolute value equation, we will first isolate the absolute value and then use Equations with an Absolute Value Expression to solve the remaining equation.

   \begin{align*} \abs{2x-7}+2\amp=3\\ \abs{2x-7}\amp=1 \end{align*}
   \begin{align*} 2x-7\amp=5\amp\amp\text{or}\amp 2x-7=-5\\ 2x\amp=12\amp\amp\text{or}\amp 2x=2\\ x\amp=6\amp\amp\text{or}\amp x=1 \end{align*}

   The solution set is \(\left\{6,1\right\}\text{.}\)

6. Since the equation \(\sqrt{x+2}=x-4\) is a radical equation, we will have to isolate the radical (which is already done), then square both sides to cancel the square root. After that, we will solve whatever remains.

   \begin{align*} \sqrt{x+2}\amp=x-4\\ \sqrt{x+2}\amp=x-4\\ \firsthighlight{(\unhighlight{\sqrt{x+2}})^2}\amp=\firsthighlight{(\unhighlight{x-4})^2}\\ x+2\amp=(x-4)(x-4)\\ x+2\amp=x^2-8x+16\\ 0\amp=x^2-9x+14\\ \end{align*}

   We now have a quadratic equation. We will solve by factoring.

   \begin{align*} 0\amp=(x-2)(x-7) \end{align*}
   \begin{align*} x-2\amp=0\amp\amp\text{or}\amp x-7=0\\ x\amp=2\amp\amp\text{or}\amp x=7 \end{align*}

   Every potential solution to a radical equation should be verified to check for any “extraneous solutions”.

   \begin{align*} \sqrt{\substitute{2}+2}\amp\stackrel{?}{=}\substitute{2}-4\amp\amp\text{or}\amp\sqrt{\substitute{7}+2}\amp\stackrel{?}{=}\substitute{7}-4\\ \sqrt{4}\amp\stackrel{?}{=}-2\amp\amp\text{or}\amp\sqrt{9}\amp\stackrel{?}{=}3\\ 2\amp\stackrel{no}{=}-2\amp\amp\text{or}\amp3\amp\stackrel{\checkmark}{=}3 \end{align*}

   So the solution set is \(\{7\}\)

7. Since the equation \((3-2x)^2-9=9\) is a quadratic equation, we again have several options. Since the variable only appears once in this equation we will use the the square root method to solve.

   \begin{align*} (3-2x)^2-9\amp=9\\ (3-2x)^2\amp=18 \end{align*}
   \begin{align*} 3-2x\amp=\sqrt{18}\amp\amp\text{or}\amp 3-2x\amp=-\sqrt{18}\\ 3-2x\amp=3\sqrt{2}\amp\amp\text{or}\amp 3-2x\amp=-3\sqrt{2}\\ -2x\amp=-3+3\sqrt{2}\amp\amp\text{or}\amp -2x\amp=-3-3\sqrt{2}\\ x\amp=\frac{-3+3\sqrt{2}}{-2}\amp\amp\text{or}\amp x\amp=\frac{-3-3\sqrt{2}}{-2}\\ x\amp=\frac{3-3\sqrt{2}}{2}\amp\amp\text{or}\amp x\amp=\frac{3+3\sqrt{2}}{2} \end{align*}

   The solution set is \(\left\{\frac{3-3\sqrt{2}}{2},\frac{3+3\sqrt{2}}{2}\right\}\text{.}\)

8. Since the equation \(3=\sqrt[3]{5-2x}\) is a radical equation, we will isolate the radical (which is already done) and then raise both sides to the third power to cancel the cube root.

   \begin{align*} 3\amp=\sqrt[3]{5-2x}\\ \firsthighlight{(\unhighlight{3})^3}\amp=\firsthighlight{\left(\unhighlight{\sqrt[3]{6x-5}}\right)^3}\\ 27\amp=6x-5\\ 32\amp=6x\\ x\amp=\frac{32}{6}\\ x\amp=\frac{16}{3} \end{align*}

   The solution set is \(\left\{\frac{16}{3}\right\}\text{.}\)

9. Since

   \begin{align*} \left\{ \begin{alignedat}{4} 5x\amp-{}\amp y \amp={}\amp -6 \\ 2x\amp-{}\amp 3y \amp={}\amp 8 \\ \end{alignedat} \right. \end{align*}

   is a system of linear equations, we can either use substitution or elimination to solve. Here we will use substitution. To use substitution, we need to solve one of the equations for one of the variables. We will solve the top equation for \(y\text{.}\)

   \begin{align*} 5x-y\amp=-6\\ -y\amp=-5x-6\\ y\amp=5x+6 \end{align*}

   Now, we substitute \(\substitute{5x+6}\) where ever we see \(\substitute{y}\) in the other equation.

   \begin{align*} 2x-3\substitute{y}\amp=8\\ 2x-3\substitute{(5x+6)}\amp=8\\ 2x-15x-18\amp=8\\ -13x-18\amp=8\\ -13x\amp=26\\ x\amp=-2 \end{align*}

   Now that we have found \(x\text{,}\) we can substitute that back into one of the equations to find \(y\text{.}\) We will substitute into the first equation.

   \begin{align*} 5(\substitute{-2})-y\amp=-6\\ -10-y\amp=-6\\ -y\amp=4\\ y\amp=-4 \end{align*}

   So, the solution must be the point \((-2,-4)\text{.}\)

10. Since the equation \(x^2+10x=12\) is quadratic and we are instructed to solve by using completing the square, we should recall that Fact 13.3.3 tells us how to complete the square, after we have sufficiently simplified. Since our equation is already in a simplified state, we need to add \(\left(\frac{b}{2}\right)^2=\left(\frac{10}{2}\right)^2=\substitute{25}\) to both sides of the equation.

    \begin{align*} x^2+10x\amp=12\\ x^2+10x+\substitute{25}\amp=12+\substitute{25}\\ (x+5)^2\amp=37\\ x+5\amp=\pm\sqrt{37}\\ x\amp=-5\pm\sqrt{37} \end{align*}

    So, our solution set is \(\left\{-5+\sqrt{37},-5-\sqrt{37}\right\}\)

First we make a number line with both intervals drawn to understand what both sets mean.

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The two intervals should be viewed as a single object when stating the union, so here is the picture of the union. It looks the same, but now it is a graph of a single set.

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First we will solve each inequality for \(z\text{.}\)

The solution set to the compound inequality is:

This is a three-part inequality. The goal is to isolate \(x\) in the middle and whatever you do to one “side,” you have to do to the other two “sides.”

The solutions to the three-part inequality \(4\ge x\gt -2\) are those numbers that are trapped between \(-2\) and \(4\text{,}\) including \(4\) but not \(-2\text{.}\) The solution set in interval notation is \(\left(-2,4\right]\text{.}\)

This is an “and” inequality. We will solve each part of the inequality and then combine the two solutions sets with an intersection.

The solution set to \(t\gt\frac{2}{3}\) is \(\left(\frac{2}{3},\infty\right)\) and the solution set to \(t\le\frac{5}{4}\) is \(\left(-\infty,\frac{5}{4}\right]\text{.}\) Shown is a graph of these solution sets.

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Recall that an “and” problem finds the intersection of the solution sets. Intersection finds the \(t\)-values where the two lines overlap, so the solution to the compound inequality must be

Since the mulch costs \(\$27\) per cubic yard and delivery is \(\$40\text{,}\) the formula for the cost of \(x\) yards of mulch is \(27x+40\text{.}\) Since Mishel wants to spend between \(\$200\) and \(\$300\text{,}\) we just trap their cost between these two values.

Most companies will only sell whole number cubic yards of mulch, so we have to round appropriately. Since Mishel wants to spend more than \(\$200\text{,}\) we have to round our lower value from \(5.93\) up to \(6\) cubic yards.

If we round the \(9.63\) up to \(10\text{,}\) then the total cost will be \(27\cdot10+40=310\) (which represents \(\$310\)), which is more than Mishel wanted to spend. So we actually have to round down to \(9\)cubic yards to stay below the \(\$300\) maximum.

In conclusion, Mishel could buy \(6\text{,}\) \(7\text{,}\) \(8\text{,}\) or \(9\) cubic yards of mulch to stay between \(\$200\) and \(\$300\text{.}\)

To solve the inequality \(\abs{4-2x} \lt 10\text{,}\) we will start by making a graph with both \(y=\abs{4-2x}\) and \(y=10\text{.}\)

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The portion of the graph of \(y=\abs{4-2x}\) that is below \(y=10\) is highlighted and the \(x\)-values of that highlighted region are trapped between \(-3\) and \(7\text{:}\) \(-3 \lt x \lt 7\text{.}\) That means that the solution set is \((-3,7)\text{.}\) Note that we shouldn't include the endpoints of the interval because at those values, the two graphs are equal whereas the original inequality was only less than and not equal.

1. To solve \(G(x)\lt-2\text{,}\) we first draw a dotted line (since it's a less-than, not a less-than-or-equal) at \(y=-2\text{.}\) Then we examine the graph to find out where the graph of \(y=G(x)\) is underneath the line \(y=-2\text{.}\) Our graph is below the line \(y=-2\) for \(x\)-values less than \(-5\text{.}\) So the solution set is \((-\infty,-5)\text{.}\)

   

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2. To solve \(G(x)\ge1\text{,}\) we first draw a solid line (since it's a greater-than-or-equal) at \(y=1\text{.}\) Then we examine the graph to find out what parts of the graph of \(y=G(x)\) are above the line \(y=1\text{.}\) Our graph is above (or on) the line \(y=1\) for \(x\)-values between \(-2\) and \(0\) as well as \(x\)-values bigger than \(4\text{.}\) So the solution set is \([-2,0]\cup[4,\infty)\text{.}\)

   

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3. To solve \(-1\lt G(x)\le1\text{,}\) we first draw a solid line at \(y=1\) and dotted line at \(y=-1\text{.}\) Then we examine the graph to find out what parts of the graph of \(y=G(x)\) are trapped between the two lines we just drew. Our graph is between those values for \(x\)-values between \(-4\) and \(-2\) as well as \(x\)-values between \(0\) and \(2\) as well as as well as \(x\)-values between \(2\) and \(4\text{.}\) We use the solid and hollow dots on the graph to decide whether or not to include those values. So the solution set is \((-4,-2]\cup[0,2)\cup(2,4]\text{.}\)

   

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