Question

1. proof use the diagram and the steps below to prove that the distance from the restaurant to the movie theater is the same as the distance from the café to the dry cleaners.

restaurant shoe store movie theater café florist dry cleaners
a. state what is given and what is to be proven for the situation.
b. write a two - column proof.

1. thought provoking the distance from springfield to lake city is equal to the distance from springfield to bettsville. janisburg is 50 miles farther from springfield than bettsville is. moon valley is 50 miles farther from springfield than lake city is. use line segments to draw a diagram that represents this situation. then prove that the distance from springfield to janisburg is equal to the distance from moon valley to springfield.

review & refresh
in exercises 20 and 21, solve the equation using any method. explain your choice of method.

1. 4x² - 87 = 109
2. 3x² - 2x - 7 = 0
3. does the table represent a linear or nonlinear function? explain.

x	2	4	6	8	10
y	1/2	1	2	4	8

in exercises 23 and 24, find the angle measure.

1. ∠1 is a complement of ∠4, and m∠1 = 33°. find m∠4.
2. ∠3 is a supplement of ∠2, and m∠2 = 147°. find m∠3.

Explanation:

Step1: Solve the equation \(4x^{2}-87 = 109\)

Add 87 to both sides: \(4x^{2}=109 + 87=196\). Then divide both sides by 4: \(x^{2}=\frac{196}{4}=49\). Take the square - root of both sides: \(x=\pm\sqrt{49}=\pm7\).

Step2: Solve the equation \(3x^{2}-2x - 7 = 0\)

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 3\), \(b=-2\), \(c=-7\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times3\times(-7)=4 + 84 = 88\). Then \(x=\frac{2\pm\sqrt{88}}{6}=\frac{2\pm2\sqrt{22}}{6}=\frac{1\pm\sqrt{22}}{3}\).

Step3: Determine if the table represents a linear or non - linear function

For a linear function, the rate of change \(\frac{\Delta y}{\Delta x}\) is constant. Calculate \(\frac{\Delta y}{\Delta x}\) for consecutive points:

• From \((2,\frac{1}{2})\) to \((4,1)\): \(\frac{1-\frac{1}{2}}{4 - 2}=\frac{\frac{1}{2}}{2}=\frac{1}{4}\).
• From \((4,1)\) to \((6,2)\): \(\frac{2 - 1}{6 - 4}=\frac{1}{2}\).

Since the rate of change is not constant, it is a non - linear function.

Step4: Find the measure of \(\angle4\)

If \(\angle1\) is a complement of \(\angle4\) and \(m\angle1 = 33^{\circ}\), then \(m\angle4=90^{\circ}-m\angle1=90^{\circ}-33^{\circ}=57^{\circ}\).

Step5: Find the measure of \(\angle3\)

If \(\angle3\) is a supplement of \(\angle2\) and \(m\angle2 = 147^{\circ}\), then \(m\angle3=180^{\circ}-m\angle2=180^{\circ}-147^{\circ}=33^{\circ}\).

Answer:

1. For \(4x^{2}-87 = 109\), \(x = 7\) or \(x=-7\).
2. For \(3x^{2}-2x - 7 = 0\), \(x=\frac{1+\sqrt{22}}{3}\) or \(x=\frac{1-\sqrt{22}}{3}\).
3. The table represents a non - linear function.
4. \(m\angle4 = 57^{\circ}\).
5. \(m\angle3 = 33^{\circ}\).