StartRoot 53 EndRoot is between 7.2 and 7.3. Estimate further to the hundredths place. Which two consecutive values does StartRoot 53 EndRoot fall between?

7.26 and 7.27
7.27 and 7.28
7.28 and 7.29
7.29 and 7.30

1 answer

To estimate \( \sqrt{53} \) more accurately between the suggested values, we can calculate a few close squares and then narrow it down.

We know:

  • \( 7.2^2 = 51.84 \)
  • \( 7.3^2 = 53.29 \)

Since \( 51.84 < 53 < 53.29 \), \( \sqrt{53} \) is indeed between 7.2 and 7.3.

Next, we can further estimate \( \sqrt{53} \) by checking \((7.25^2)\):

\[ 7.25^2 = 52.5625 \]

Now we see:

  • \( 7.2^2 = 51.84 \)
  • \( 7.25^2 = 52.5625 \)
  • \( 7.3^2 = 53.29 \)

Since \( 52.5625 < 53 < 53.29 \), \( \sqrt{53} \) is between 7.25 and 7.3.

Next, we check \( (7.27^2) \):

\[ 7.27^2 = 52.8129 \]

Now we see:

  • \( 7.25^2 = 52.5625 \)
  • \( 7.27^2 = 52.8129 \)
  • \( 7.3^2 = 53.29 \)

Since \( 52.8129 < 53 < 53.29 \), \( \sqrt{53} \) is between 7.27 and 7.3.

Continuing, we check \( (7.28^2) \):

\[ 7.28^2 = 52.9984 \]

Now we see:

  • \( 7.27^2 = 52.8129 \)
  • \( 7.28^2 = 52.9984 \)
  • \( 7.3^2 = 53.29 \)

Since \( 52.9984 < 53 < 53.29 \), \( \sqrt{53} \) is between 7.28 and 7.3.

Finally, we check \( (7.29^2) \):

\[ 7.29^2 = 53.0641 \]

Now we see:

  • \( 7.28^2 = 52.9984 \)
  • \( 7.29^2 = 53.0641 \)

Since \( 52.9984 < 53 < 53.0641 \), we find that \( \sqrt{53} \) lies between \( 7.28 \) and \( 7.29 \).

Thus, the correct answer is:

7.28 and 7.29.