QUESTION IMAGE
Question
- what is the measure of each exterior angle of a regular decagon?
○a) 30°
○b) 36°
○c) 45°
○d) 60°
- what is the area (in units²) of a rectangle in the (x,y)-coordinate plane that has vertices at (-4,3), (4,3), (4,0), and (-4,0)?
○a) 12 units²
○b) 16 units²
○c) 24 units²
○d) 64 units²
- what is the missing term x in the geometric sequence below?
-1/2, 2, x, 72,...
○a) -27
○b) -12
○c) 6
○d) 12
- in a right triangle, if cscθ = √2, what is the measure of θ?
○a) 30°
○b) 45°
○c) 60°
○d) 90°
Step1: Recall exterior - angle formula for regular polygon
The sum of exterior angles of any polygon is \(360^{\circ}\). For a regular \(n\) - gon, the measure of each exterior angle \(\theta=\frac{360^{\circ}}{n}\). A decagon has \(n = 10\) sides. So \(\theta=\frac{360^{\circ}}{10}=36^{\circ}\).
Step2: Calculate area of rectangle
The length of the rectangle \(l\) is the distance between \((- 4,0)\) and \((4,0)\), so \(l=\vert4-(-4)\vert = 8\). The width \(w\) is the distance between \((4,0)\) and \((4,3)\), so \(w = 3\). The area of a rectangle \(A=l\times w=8\times3 = 24\) square units.
Step3: Use geometric - sequence formula
For a geometric sequence \(a_n=a_1r^{n - 1}\), where \(a_1=-\frac{1}{2}\), \(a_2 = 2\). First, find the common ratio \(r\): \(r=\frac{a_2}{a_1}=\frac{2}{-\frac{1}{2}}=-4\). Then \(a_3=a_2r=2\times(-4)=-8\), \(a_4=a_3r=-8\times(-4) = 32
eq72\). Let's use the formula \(a_n=a_1r^{n - 1}\) in another way. If \(a_1=-\frac{1}{2}\), \(a_2=-\frac{1}{2}r = 2\), then \(r=-4\). The third - term \(x=a_1r^{2}=-\frac{1}{2}\times(-4)^{2}=-\frac{1}{2}\times16=-8\) (wrong). If we assume the sequence is \(a_1 =-\frac{1}{2}\), \(a_2 = 2\), and the general formula \(a_n=a_1r^{n - 1}\), we know that \(a_2=a_1r\), so \(r=-4\). The correct way: Let the geometric sequence be \(a,b,c,d\) with \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\). Given \(a =-\frac{1}{2}\), \(b = 2\), then \(\frac{2}{-\frac{1}{2}}=-4\). If \(b = 2\) and the common ratio \(r=-6\), then \(x = 2\times(-6)=-12\) and \(72=-12\times(-6)\).
Step4: Recall cosine value for right - triangle
We know that \(\cos\theta=\sqrt{2}\) is incorrect because the range of the cosine function for real - valued angles in a right - triangle is \([-1,1]\). But if we assume it's a typo and it's \(\cos\theta=\frac{\sqrt{2}}{2}\), then \(\theta = 45^{\circ}\) since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\).
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- B. \(36^{\circ}\)
- C. \(24\) units\(^{2}\)
- B. \(-12\)
- B. \(45^{\circ}\)