Question

1. a section of highway is represented by a line - segment with endpoints (30,50) and (90,100) in the standard (x,y) coordinate plane. exactly halfway along this highway section is a road sign. what are the coordinates of the road sign? e. (30,25) g. (60,50) h. (60,75) j. (70,65) k. (75,60)
2. the student council is preparing a budget for an upcoming fund - raising dance. they have decided to spend $150 for a light and sound show, $400 for refreshments, and $50 for decorations. these are the only expenses. given that the student council estimates 500 students will attend the dance, what should be the price, per student, for admission to the dance if the student council wants to raise as close as possible to $300 after paying expenses? a. $0.60 b. $1.20 c. $1.67 d. $1.80 e. $3.33
3. the following shapes are a circle and some regular polygons that have their centers marked. all units are given in feet. which of these shapes has the greatest area? f. j.

g. k.

Explanation:

14.

Step1: Find mid - point formula

The mid - point formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). Here \(x_1=30,y_1 = 50,x_2=90,y_2 = 100\).

Step2: Calculate x - coordinate of mid - point

\(x=\frac{30 + 90}{2}=\frac{120}{2}=60\).

Step3: Calculate y - coordinate of mid - point

\(y=\frac{50+100}{2}=\frac{150}{2}=75\).

Step1: Calculate total expenses

The total expenses for the dance are \(150 + 400+50=600\) dollars.

Step2: Let the price per student be \(x\).

The number of students is \(n = 500\). The money raised from students is \(500x\). The council wants to raise as close as possible to \(300\) after paying expenses. So the equation is \(500x-600\approx300\).

Step3: Solve for \(x\)

\(500x=300 + 600=900\), then \(x=\frac{900}{500}=1.8\).

Step1: Recall area formulas

The area of a circle with radius \(r = 1\) is \(A_{circle}=\pi r^{2}=\pi\times1^{2}=\pi\approx3.14\).
The area of a square with apothem \(a = 1\) (distance from center to side) has side - length \(s = 2\), and area \(A_{square}=s^{2}=4\).
For a regular octagon with apothem \(a = 1\), the central angle \(\theta=\frac{360^{\circ}}{n}\), where \(n = 8\), \(\theta = 45^{\circ}\). The area of a regular polygon is \(A=\frac{1}{2}aP\), where \(a\) is the apothem and \(P\) is the perimeter. For an octagon with apothem \(a = 1\), the side - length \(s = 2\tan22.5^{\circ}\), and perimeter \(P = 8s=8\times2\tan22.5^{\circ}\). \(A_{octagon}=\frac{1}{2}\times1\times8\times2\tan22.5^{\circ}\approx2.83\).
For a non - standard polygon in option J with central angle \(67.5^{\circ}\), \(n=\frac{360^{\circ}}{67.5^{\circ}} = 5.33\) (not a regular polygon in the traditional sense). But if we consider a regular octagon - like figure in option K with apothem \(a = 1\), its area is less than that of the square.

Step2: Compare areas

Since \(4>\pi>2.83\), the square has the greatest area.

Answer:

H. \((60,75)\)

15.