Question

1. ignoring (as the real clt does on questions like this!) the fact that these quantities are measured in different units, suppose the following proposition is offered to you:

\for every circle, the magnitude of the circumference is always greater than or equal to twice the magnitude of its area.\
which of the following would qualify as a counterexample that disproves that statement?
a) a circle with a radius of 1.0.
b) a circle with an area of 0.64π.
c) a circle with a circumference of 1.5π.
d) a circle with a radius of 4.0.

1. if the operation is defined such that ab = 3a + 4b, then which of the following is the value of 2*(-3)?

a) -6
b) 0
c) 18
d) 30

1. suppose the line y = x + 4 is reflected across the line x = 2. which of the following points is on the line, after reflection?

a) (1,-5)
b) (4,4)
c) (7,5)
d) (10,0)

1. consider a list of all the odd numbers between 1 and 60, inclusive. suppose one number in that list must be replaced with an even number between 2 and 60, inclusive. which of the following properties of the list must be different from the original?

i. the sum of all the integers
ii. the mean of all the integers
iii. the median of all the integers
a) i and ii only
b) ii and iii only
c) i and iii only
d) i, ii, and iii

Explanation:

Question 26

Step1: Recall circle formulas

Circumference \( C = 2\pi r \), Area \( A=\pi r^2 \). The statement is \( C\geq 2A \), i.e., \( 2\pi r\geq 2\pi r^2 \) or \( r\leq 1 \) (dividing both sides by \( 2\pi r \), \( r>0 \)). A counterexample has \( r > 1 \).

Step2: Analyze options

• A: \( r = 1 \), \( C = 2\pi \), \( 2A=2\pi \), \( C = 2A \), not a counterexample.
• B: \( A = 0.64\pi=\pi r^2\Rightarrow r = 0.8 \leq 1 \), \( C = 1.6\pi \), \( 2A = 1.28\pi \), \( C>2A \), not a counterexample.
• C: \( C = 1.5\pi=2\pi r\Rightarrow r = 0.75 \leq 1 \), \( 2A = 2\pi(0.75)^2=1.125\pi \), \( C>2A \), not a counterexample.
• D: \( r = 4>1 \), \( C = 8\pi \), \( 2A = 2\pi(16)=32\pi \), \( 8\pi<32\pi \), so \( C<2A \), counterexample.

Step1: Apply the operation

Given \( a\bullet b = 3a + 4b \), for \( 2\bullet(-3) \), \( a = 2 \), \( b=-3 \).

Step2: Substitute values

\( 3(2)+4(-3)=6 - 12=-6 \).

Step1: Find reflection of a point on \( y=x + 4 \) over \( x = 2 \)

The reflection of a point \( (x,y) \) over \( x = 2 \) is \( (4 - x,y) \). First, find two points on \( y=x + 4 \): when \( x = 2 \), \( y = 6 \) (point \( (2,6) \), on the axis of reflection, so it stays). When \( x = 0 \), \( y = 4 \), its reflection is \( (4,4) \).

Step2: Check options

• A: \( (1,-5) \): Reflect \( x = 1 \) over \( x = 2 \) is \( x = 3 \), \( y=-5 \) not on \( y=x + 4 \) (reflected line: let's find the reflected line. The original line has slope 1, reflected line also has slope -1? Wait, no: reflection over vertical line \( x = 2 \). The original line \( y=x + 4 \), slope 1. The reflected line: for a point \( (x,y) \) on original, \( (4 - x,y) \) on reflected. So \( y=(4 - x)+4=8 - x \). Check \( (4,4) \): \( 4=8 - 4 \), yes. \( (1,-5) \): \( -5

eq8 - 1 = 7 \). \( (7,5) \): \( 5
eq8 - 7 = 1 \). \( (10,0) \): \( 0
eq8 - 10=-2 \). So \( (4,4) \) is on the reflected line.

Answer:

D. A circle with a radius of 4.0.

Question 27