Using a number line, what whole number and rational number to the nearest half is V 40 between? (1 point)

Question

V40 is between and

Answer 1

To find the whole number and the rational number to the nearest half that \( V 40 \) is between, let's first calculate \( V 40 \):

\[ V 40 = \sqrt{40} \approx 6.3246 \]

Now, examining the whole number:

• The whole numbers surrounding \( 6.3246 \) are \( 6 \) and \( 7 \).
• Therefore, \( V 40 \) is between \( 6 \) and \( 7 \).

Next, we determine the rational number to the nearest half:

• The nearest halves around \( 6.3246 \) are \( 6.0 \) and \( 6.5 \).
• Since \( 6.3246 \) is closer to \( 6.5 \) than to \( 6.0 \), we round \( 6.3246 \) to \( 6.5 \).

Thus, we can conclude that:

\[ V 40 \text{ is between } 6 \text{ and } 7 \text{, and between } 6 \text{ and } 6.5. \]

Final answer:

V 40 is between 6 and 7, and 6 and 6.5.

Answer 2

Using a number line, what whole number and rational number to the nearest half is V 58 between? (1 point)
• between 7 and 7.5
• between 7.5 and 8
• between 8 and 8.5
• between 8.5 and 9

Answer 3

To determine the whole number and rational numbers to the nearest half that \( V 58 \) is between, we first need to calculate \( V 58 \):

\[ V 58 = \sqrt{58} \approx 7.6158 \]

Now, let's identify the whole numbers surrounding \( 7.6158 \):

• The whole numbers around \( 7.6158 \) are \( 7 \) and \( 8 \).

Next, we need to find the nearest half:

• The nearest halves are \( 7.5 \) and \( 8.0 \).
• Since \( 7.6158 \) is closer to \( 7.5 \) than to \( 8.0 \), we round to \( 7.5 \).

Thus, the closest range for \( V 58 \) is:

V 58 is between 7.5 and 8.

So the correct choice is:

between 7.5 and 8.

Answer 4

Using rational approximations, what is an inequality statement comparing v 81
and 88? Enter <, >, or = (1 point)
The inequality is: v 81
188.

Answer 5

To compare \( V 81 \) and \( 88 \), we first calculate \( V 81 \):

\[ V 81 = \sqrt{81} = 9. \]

Now, we compare \( 9 \) and \( 88 \):

\[ 9 < 88. \]

Thus, the inequality statement is:

The inequality is: \( V 81 < 88 \).

Answer 6

Write 6,739 in expanded form. (1 point)
• (6 × 1,000) + (7 × 100) + (3 × 10
• (7 × 100) + (3 × 10) + (9 × 1)
O (6 × 1,000) + (7×100) + (3 x10
• (9 × 1, 000) + (3 × 100) + (7×10.

Answer 7

To write the number \( 6,739 \) in expanded form, we break it down according to the place value of each digit:

• The digit \( 6 \) is in the thousands place, so it represents \( 6 \times 1,000 \).
• The digit \( 7 \) is in the hundreds place, so it represents \( 7 \times 100 \).
• The digit \( 3 \) is in the tens place, so it represents \( 3 \times 10 \).
• The digit \( 9 \) is in the ones place, so it represents \( 9 \times 1 \).

Putting that all together, the expanded form of \( 6,739 \) is:

\[ (6 \times 1,000) + (7 \times 100) + (3 \times 10) + (9 \times 1). \]

Among the options you provided, the correct expression in expanded form is:

(6 × 1,000) + (7 × 100) + (3 × 10) + (9 × 1).

Since \( (9 \times 1) \) wasn't included in your options, the correct part of your expanded form choices from what you referenced would be:

(6 × 1,000) + (7 × 100) + (3 × 10) (assuming it implies that the \( 9 \) is accounted for as its own term in the complete context of expanded notation).

Answer 8

Express II as a repeating decimal.
Mark the cycle of repeating digits with a bar.

Answer 9

To express the fraction \( \frac{1}{11} \) as a repeating decimal, we can perform the division:

1. Divide \( 1.000000 \) by \( 11 \):
   • \( 11 \) goes into \( 100 \) nine times (i.e., \( 11 \times 9 = 99 \)), leaving a remainder of \( 1 \).
   • Bring down a zero to make it \( 10 \).
   • \( 11 \) goes into \( 10 \) zero times.
   • Bring down a zero to make it \( 100 \) again.
   • This process repeats, cycling through the same steps.

From this division, we find:

\[ \frac{1}{11} = 0.090909\ldots \]

In repeating decimal notation, we can express this as:

\[ 0.\overline{09}. \]

So, the final answer is:

\( 0.\overline{09} \) (the bar denotes that "09" repeats indefinitely).

Answer 10

Express 8/11 as a repeating decimal.
Mark the cycle of repeating digits with a bar.

Answer 11

To express \( \frac{8}{11} \) as a repeating decimal, we perform the division of \( 8 \) by \( 11 \):

1. Divide \( 8.000000 \) by \( 11 \):
   • \( 11 \) goes into \( 80 \) seven times (since \( 11 \times 7 = 77 \)), leaving a remainder of \( 3 \).
   • Bring down a zero to make it \( 30 \).
   • \( 11 \) goes into \( 30 \) two times (since \( 11 \times 2 = 22 \)), leaving a remainder of \( 8 \).
   • Bring down another zero to make it \( 80 \) again.
   • We notice that it returns to \( 80 \), which starts the cycle over.

From this division, we find:

\[ \frac{8}{11} = 0.727272\ldots \]

This means:

\[ 0.\overline{72} \]

The final answer is:

\( 0.\overline{72} \) (the bar indicates that "72" repeats indefinitely).