√

Question

Which of the following is an irrational number?

A.
0.333

B.
√8
C.
√49
D.
5/22

Answer 1

To determine which of the options is an irrational number, let's evaluate each one:

A. \(0.333\) - This is a repeating decimal that can be expressed as the fraction \(\frac{1}{3}\). It is a rational number.

B. \(\sqrt{8}\) - This can be simplified to \(2\sqrt{2}\). Since \(\sqrt{2}\) is an irrational number, \(\sqrt{8}\) is also irrational.

C. \(\sqrt{49}\) - This simplifies to \(7\), which is a rational number.

D. \(5/22\) - This is a fraction and can be expressed as a decimal, which is approximately \(0.227\), making it a rational number.

Therefore, the only irrational number from the options provided is:

B. \(\sqrt{8}\)

Answer 2

An irrational number, d, has a value such that 6 < d < 7 . Which of the following is a possible value for d?

A.
√6.2
B.
2π
C.
√13
D.
6π

Answer 3

To determine which of the given options could be an irrational number \(d\) such that \(6 < d < 7\), we will evaluate each option:

A. \(\sqrt{6.2}\)
Calculating \(\sqrt{6.2} \approx 2.49\). This value is not within the range of \(6 < d < 7\).

B. \(2\pi\)
Calculating \(2\pi \approx 6.28\). This value is within the range of \(6 < d < 7\).

C. \(\sqrt{13}\)
Calculating \(\sqrt{13} \approx 3.61\). This value is not within the range of \(6 < d < 7\).

D. \(6\pi\)
Calculating \(6\pi \approx 18.85\). This value is not within the range of \(6 < d < 7\).

Therefore, the option that satisfies \(6 < d < 7\) is:

B. \(2\pi\)

Answer 4

The relationship between Amanda’s distance from school and time since she left school is linear, as shown in the table below.

The table shows time in hours as 0, 1, 3, and distance in miles as 3, 4.25, 6.75

Based on the table, determine whether each statement is true or false.

True
False
Amanda's initial distance from her school is 1.25 miles.
Amanda's distance increases by 1.25 miles each hour.
Amanda's distance from school after 4 hours is 9.75 miles.

Answer 5

Let's analyze the data provided in the table and each of the statements step by step.

Time and Distance Data:

• Time (hours): 0, 1, 3
• Distance (miles): 3, 4.25, 6.75

First, let's interpret the points:

• When time = 0, distance = 3 miles.
• When time = 1, distance = 4.25 miles.
• When time = 3, distance = 6.75 miles.

1. Amanda's initial distance from her school is 1.25 miles.

• False. The initial distance (when time = 0) is 3 miles, not 1.25 miles.

2. Amanda's distance increases by 1.25 miles each hour.

To find out if this is true, let's calculate the change in distance for the given times:

• From 0 to 1 hour: \(4.25 - 3 = 1.25\) miles.
• From 1 to 3 hours: \(6.75 - 4.25 = 2.5\) miles over 2 hours (which is \(2.5/2 = 1.25\) miles per hour).

Since the increase is consistently 1.25 miles per hour:

• True.

3. Amanda's distance from school after 4 hours is 9.75 miles.

To find the distance at 4 hours, we can extrapolate based on the linear relationship. We already determined that the slope (rate of change) is 1.25 miles per hour.

Starting from the known distance at 3 hours (6.75 miles) and adding the increase for the next hour:

• After 4 hours: \(6.75 + 1.25 = 8\) miles.

Thus, she is not at 9.75 miles after 4 hours:

• False.

Summary of Statements:

1. False
2. True
3. False

Answer 6

Peter solved an equation as shown below.

Given: minus left parenthesis 3x minus 2 right parenthesis plus 2 left parenthesis x minus 3 right parenthesis equals minus 7. Step 1 shows, minus 3x plus 2 plus 2x minus 6 equals minus 7. Step 2 shows, minus 3x plus 2x plus 2 minus 6 equals minus 7. Step 3 shows, minus x minus 4 equals minus 7. Step 4 shows, minus x minus 4 plus 4 equals -7 plus 4. Step 5 shows, minus x over -1 equals -3 over -1. Answer: x equals 3.

When did Peter apply the distributive property?

A.
from the given to step 1

B.
from step 1 to step 2

C.
from step 3 to step 4

D.
from step 4 to step 5

Answer 7

To determine when Peter applied the distributive property in solving the equation, let's analyze each step:

1. From the given to step 1: \[

   • (3x - 2) + 2(x - 3) \rightarrow -3x + 2 + 2x - 6 \] Here, the distributive property is used twice:
   • For \(- (3x - 2)\), which becomes \(-3x + 2\).
   • For \(2(x - 3)\), which becomes \(2x - 6\).
     This is where the distributive property is applied.

2. From step 1 to step 2: \[ -3x + 2 + 2x - 6 \rightarrow -3x + 2x + 2 - 6 \] This step involves rearranging terms, not applying the distributive property.

3. From step 3 to step 4: \[ -x - 4 \rightarrow -x - 4 + 4 = -7 + 4 \] This step is focused on isolating \(x\) and does not use the distributive property.

4. From step 4 to step 5: \[ -x = -3 \] This step simplifies to \(x\) but does not involve the distributive property.

Based on this analysis, Peter applied the distributive property:

A. from the given to step 1.