This worksheet helps students practice estimating square and cube roots using number lines and interval tables.

Let’s solve each section step by step.

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Section A: Estimate using the number line

We are given perfect squares on a number line and asked to estimate square roots of numbers between them.

1) √75 ≈ ?

Number line shows:
8² = 64 ———— 9² = 81
75 is between 64 and 81.
Closer to 81? Let’s see:
75 - 64 = 11
81 - 75 = 6 → so closer to 81 → closer to 9

But let’s think:
8.5² = (8 + 0.5)² = 64 + 2×8×0.5 + 0.25 = 64 + 8 + 0.25 = 72.25
8.6² = 8.5² + 2×8.5×0.1 + 0.01 ≈ 72.25 + 1.7 + 0.01 = 73.96
8.7² = 8.6² + 2×8.6×0.1 + 0.01 ≈ 73.96 + 1.72 + 0.01 = 75.69 → too big!

So √75 is between 8.6 and 8.7, but since 75 < 75.69, maybe ~8.66? But for estimation, we can say ≈ 8.7 or even just between 8 and 9, but since it's much closer to 81 than 64, and 8.7²=75.69 is very close, √75 ≈ 8.7

Wait — actually, in many school contexts, they want you to pick the nearest whole number or use linear interpolation roughly.

But looking at the problem: it says “estimate” and gives endpoints. Since 75 is about halfway? No — 64 to 81 is 17 units. 75-64=11, so 11/17 ≈ 0.65 → so about 8.65 → round to 8.7

But perhaps simpler: since 8.5²=72.25 and 8.7²=75.69, and 75 is very close to 75.69, so ≈ 8.7

Actually, let me check standard approximations:

Commonly, √75 = √(25×3) = 5√3 ≈ 5×1.732 = 8.66 → so ≈ 8.7

✔ So answer: 8.7

But wait — maybe they expect integer? No, because 8²=64, 9²=81, and 75 is not near either end — so decimal is fine.

Looking at question 2: √110 between 10²=100 and 11²=121.

110 - 100 = 10
121 - 110 = 11 → almost middle → so ≈ 10.5?

Check: 10.5² = 110.25 → wow! Exactly almost 110.

So √110 ≈ 10.5

Perfect.

Question 3: √39 between 6²=36 and ??? The blank box must be 7²=49.

Because 6²=36, next square is 49.

So √39 is between 6 and 7.

39 - 36 = 3
49 - 39 = 10 → so closer to 6.

How much? 3/13 ≈ 0.23 → so ≈ 6.23

Check: 6.2² = 38.44
6.3² = 39.69 → 39 is between them.

39 - 38.44 = 0.56
39.69 - 39 = 0.69 → so closer to 6.2

Interpolate: difference 0.56 vs 0.69 → total range 1.25 → so about 6.2 + (0.56/1.25)*0.1 ≈ 6.2 + 0.045 ≈ 6.245

But for estimation, often we say ≈ 6.2 or 6.3? Since 6.2²=38.44, 6.3²=39.69, and 39 is closer to 38.44? Wait no:

|39 - 38.44| = 0.56
|39.69 - 39| = 0.69 → yes, closer to 6.2

But 0.56 < 0.69 → so √39 ≈ 6.2

Alternatively, some might say 6.25? 6.25² = (25/4)² = 625/16 = 39.0625 → oh! That’s extremely close to 39.

6.25² = 39.0625 → which is only 0.0625 over 39 → so √39 ≈ 6.25

That’s better! And 6.25 is easy to remember.

So I’ll go with 6.25

But let’s confirm: 6.24² = ? Maybe overkill. For school level, 6.25 is acceptable.

Actually, 6.24² = (6.25 - 0.01)² = 6.25² - 2*6.25*0.01 + 0.0001 = 39.0625 - 0.125 + 0.0001 = 38.9376 → too low.

Better to stick with 6.25 as approximation since 6.25²=39.0625 ≈ 39.

So √39 ≈ 6.25

But perhaps the worksheet expects rounding to one decimal? Let’s see context.

In Q1, if we put 8.7, Q2 10.5, then Q3 should be consistent.

6.25 is two decimals — maybe they want one? Then 6.2 or 6.3?

Since 6.25 is exact match almost, and common fraction, I think 6.25 is fine.

But let’s look ahead — Section C has intervals like 0.8~0.9, so decimals are expected.

I’ll keep 6.25 for now.

Wait — the blank in the number line for Q3 is after 6², so likely 7²=49. We fill that mentally.

So answers for Section A:

1) √75 ≈ 8.7
2) √110 ≈ 10.5
3) √39 ≈ 6.25 (or 6.2? Let me double-check)

Actually, let's calculate properly:

For √39:

Try 6.24: 6.24 * 6.24

6*6=36, 6*0.24=1.44, doubled is 2.88, plus 0.24^2=0.0576 → total 36 + 2.88 + 0.0576 = 38.9376

6.25: 6.25*6.25 = (6+0.25)^2 = 36 + 3 + 0.0625 = 39.0625

39 - 38.9376 = 0.0624
39.0625 - 39 = 0.0625 → almost same distance! So exactly midway? Not quite, but very close.

Difference: from 6.24 to 6.25 is 0.01, and 39 is 0.0624 above 38.9376, total jump 0.1249 (from 38.9376 to 39.0625), so fraction 0.0624 / 0.1249 ≈ 0.5, so approximately 6.245

So for practical purposes, 6.25 is a good estimate.

I'll use 6.25.

But to match others, perhaps 6.2 or 6.3? Let's see what's typical.

In many textbooks, √39 is estimated as 6.2 or 6.3. Since 6.2^2=38.44, 6.3^2=39.69, and 39 is closer to 38.44? No:

39 - 38.44 = 0.56
39.69 - 39 = 0.69 → so numerically closer to 6.2, but percentage-wise similar.

However, 6.25 is more accurate. I think it's fine.

Perhaps the worksheet expects us to use the midpoint or something.

Another way: average of 6 and 7 is 6.5, 6.5^2=42.25 >39, so less.

I think 6.25 is best.

Let's move on and come back if needed.

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Section B: Complete the table for cube roots

The table has:

Top row: cube roots of numbers
Bottom row: cubes of integers

Given:

³√1 | ³√8 | ? | ? | ? | ? | ³√343 | ³√1000 | ³√729 | ³√512

Below: ? | ? | 3³ | 4³ | 5³ | 6³ | ? | ? | ? | ?

We know:

1³ = 1 → so ³√1 = 1
2³ = 8 → ³√8 = 2
3³ = 27 → so the third column top should be ³√27, bottom is 3³
Similarly, 4³=64 → ³√64
5³=125 → ³√125
6³=216 → ³√216
Then 7³=343 → ³√343 =7
8³=512 → ³√512=8
9³=729 → ³√729=9
10³=1000 → ³√1000=10

So filling the table:

First row (cube roots):
³√1 = 1
³√8 = 2
³√27 = 3
³√64 = 4
³√125 = 5
³√216 = 6
³√343 = 7
³√1000 = 10
³√729 = 9
³√512 = 8

Second row (cubes):
1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
8³ = 512
9³ = 729
10³ = 1000

But in the table, the bottom row has blanks under first two, then 3³,4³,etc., and then blanks under the last four.

Specifically:

Columns:

Col1: top ³√1, bottom ? → should be 1³ or just 1? The label is "3³" etc., so probably write the value.

Looking at the table structure:

It says:

| ³√1 | ³√8 | | | | | ³√343 | ³√1000 | ³√729 | ³√512 |
|-----|-----|-----|-----|-----|-----|-------|--------|-------|-------|
| | | 3³ | 4³ | 5³ | 6³ | | | | |

So for col1: top is ³√1, bottom should be the cube, i.e., 1³ or 1? But it says "3³" in other cells, so likely they want the expression like "1³", but that seems odd.

No — look: in col3, bottom is "3³", which means the cube of 3, so value is 27, but they wrote "3³" as label? Or is it the value?

I think there's confusion.

Read carefully: the bottom row has entries like "3³", "4³", etc., which are expressions, but in context, for the cube root above, it should correspond.

Actually, re-examining: the table is probably meant to show that ³√(n³) = n, so for example, under ³√8, which is 2, the bottom should be 2³=8, but they have written "3³" in the third column, which corresponds to ³√27.

Let me map:

Column 1: top: ³√1 → this is 1, so bottom should be 1³ = 1, but they left blank.

Column 2: top: ³√8 = 2, bottom should be 2³ = 8, blank.

Column 3: top: ? , bottom: 3³ → so top should be ³√(3³) = ³√27

Similarly, col4: top: ³√(4³) = ³√64, bottom: 4³

Col5: ³√125, bottom 5³

Col6: ³√216, bottom 6³

Col7: top: ³√343, bottom: ? → 343 = 7³, so bottom should be 7³

Col8: ³√1000 = 10³, so bottom 10³

Col9: ³√729 = 9³, bottom 9³

Col10: ³√512 = 8³, bottom 8³

And for col1 and col2, bottom should be 1³ and 2³ respectively.

But in the table, for col1 and col2, bottom is blank, and for col3 to col6, bottom is given as 3³ to 6³, so we need to fill top for col3-6, and for col7-10, fill bottom, and also col1-2 bottom.

The instruction is "complete the table", so we fill all blanks.

So:

Top row blanks: col3,4,5,6 → ³√27, ³√64, ³√125, ³√216

Bottom row blanks: col1,2,7,8,9,10 → 1³, 2³, 7³, 10³, 9³, 8³

But typically, we write the numerical value or the expression? In the given, they wrote "3³" etc., so probably keep as expressions.

To be consistent, since they used "3³" for the cube, we should write the cube expression in bottom row.

Similarly, top row is cube root of the number.

So:

Fill top row:

Col3: ³√27
Col4: ³√64
Col5: ³√125
Col6: ³√216

Fill bottom row:

Col1: 1³
Col2: 2³
Col7: 7³
Col8: 10³
Col9: 9³
Col10: 8³

Note that col8 is ³√1000, which is 10³, so bottom 10³
Col9: ³√729 = 9³, bottom 9³
Col10: ³√512 = 8³, bottom 8³

Yes.

Sometimes they might want the simplified value, but since they used "3³" in the table, I think keeping as exponent form is fine.

But let's see the answer format — probably we can write the values.

To avoid confusion, I'll provide both, but for final answer, list what to fill.

Perhaps in the context, for bottom row, when it says "3³", it means the value 27, but they wrote the expression. I think it's safer to write the expression as per given.

Looking back at user image description, but since I can't see, I'll assume based on standard.

Another way: in many such tables, the bottom row is the cube, so for col1, bottom is 1, but they have "3³" which is 27, so likely "3³" represents the number 27, but written as expression.

To simplify, for the answer, I'll state what to put in each blank.

But for now, let's proceed.

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Section C: Approximate the value by giving interval

We need to find between which two consecutive integers (or decimals) the root lies.

First table: Square Root and Interval

Left side:

√6 : between which integers? 2²=4, 3²=9, 4<6<9, so between 2 and 3 → interval 2~3

√15: 3²=9, 4²=16, 9<15<16, so between 3 and 4 → 3~4

√27: 5²=25, 6²=36, 25<27<36, so between 5 and 6 → 5~6

Right side:

Given interval, find square root.

Interval 11~12: so square root between 11 and 12, so the number is between 121 and 144. What square root? It could be any, but probably they want an example, like √125 or something, but the cell is for "Square Root", so likely write the expression, e.g., √121 to √144, but specifically, since interval is given, we need to choose a number whose square root is in that interval.

Looking at the table:

"Square Root" column has expressions like √6, and "Interval" has ranges.

For the right side, "Interval" is given, and "Square Root" is blank, so we need to provide a square root that falls in that interval.

For example, interval 11~12: so √x where 11 < √x < 12, so 121 < x < 144. We can choose any, say √125, or √130, but probably they expect a specific one, or perhaps just indicate.

But in the left side, they gave specific radicals, so for right side, we need to fill in a radical that matches the interval.

Similarly for others.

Interval 0.8~0.9: so √x between 0.8 and 0.9, so x between 0.64 and 0.81. So for example, √0.7 or √0.75, etc.

Interval 0.15~0.16: √x between 0.15 and 0.16, so x between 0.0225 and 0.0256. So √0.024 or something.

This seems messy. Perhaps they want the radical expression for a number that has square root in that interval.

But to make it simple, for interval 11~12, we can put √125, since 11.18^2≈125, and 11<11.18<12.

Similarly, for 0.8~0.9, √0.64=0.8, √0.81=0.9, so say √0.72 or √0.75.

For 0.15~0.16, √0.0225=0.15, √0.0256=0.16, so say √0.024.

But perhaps they expect integer radicands? Unlikely for small decimals.

Another interpretation: for the right side, the "Square Root" column is to be filled with the radical, and "Interval" is given, so we need to choose a number whose square root is approximately in that interval.

But to be precise, let's calculate.

For interval 11~12: any number between 121 and 144, say 125, so √125

For 0.8~0.9: numbers between 0.64 and 0.81, say 0.72, so √0.72

For 0.15~0.16: between 0.0225 and 0.0256, say 0.024, so √0.024

But these are arbitrary. Perhaps they want the closest perfect square or something.

Notice that in the left side, they have √6, 15, √27, which are not perfect squares, so similarly for right side, we can choose non-perfect squares.

But for consistency, let's see the cube root part.

Cube Root tables:

Left: ³√21 and ³√200

³√21: 2^3=8, 3^3=27, 8<21<27, so between 2 and 3 → interval 2~3

³√200: 5^3=125, 6^3=216, 125<200<216, so between 5 and 6 → 5~6

Right side:

Interval 4~5: so ³√x between 4 and 5, so x between 64 and 125. Say ³√100 or ³√125, but 125 is 5^3, so ³√125=5, which is endpoint. Usually interval is open, so say ³√100

Interval 9~10: x between 729 and 1000, say ³√800 or ³√900

Again, arbitrary.

Perhaps for the right side, they want us to identify what the square root or cube root is for a given interval, but the cell is labeled "Square Root", so likely the expression.

To resolve, let's assume that for the right side, we need to provide a representative radical that falls in the interval.

For simplicity, I'll choose:

For square root interval 11~12: √125 (since 11.18^2=125)

For 0.8~0.9: √0.72 (0.8485^2≈0.72, and 0.8<0.8485<0.9)

0.8^2=0.64, 0.9^2=0.81, so for x=0.72, √0.72≈0.8485, which is between 0.8 and 0.9, yes.

For 0.15~0.16: 0.15^2=0.0225, 0.16^2=0.0256, so take x=0.024, √0.024≈0.1549, which is between 0.15 and 0.16.

Similarly for cube roots:

Interval 4~5: ³√100 (4.64^3≈100, and 4<4.64<5)

Interval 9~10: ³√800 (9.28^3≈800, 9<9.28<10)

But perhaps they expect integer radicands for cube roots, but 100 and 800 are integers.

For square roots with decimals, it's fine.

Another thought: for the interval 0.8~0.9, they might mean the square root of a number like 0.64 to 0.81, but the radical is of that number.

I think it's correct.

To make it neat, for the right side square root:

- For 11~12: \sqrt{125}
- For 0.8~0.9: \sqrt{0.72} or \sqrt{\frac{18}{25}} but decimal is fine
- For 0.15~0.16: \sqrt{0.024}

Similarly for cube roots:

- For 4~5: \sqrt[3]{100}
- For 9~10: \sqrt[3]{800}

But let's verify if there are better choices.

For 0.8~0.9, if we take \sqrt{0.64} = 0.8, but 0.8 is endpoint, and interval is 0.8~0.9, which may include or exclude, but usually in such contexts, it's inclusive or exclusive? The notation "0.8 ~ 0.9" likely means from 0.8 to 0.9, so including endpoints, but for approximation, we can use interior points.

To be safe, use values strictly inside.

For example, for 0.8~0.9, use \sqrt{0.7} ≈ 0.8367, which is between.

0.7 is between 0.64 and 0.81, yes.

Similarly, for 0.15~0.16, \sqrt{0.023} ≈ 0.1517, good.

But 0.023 is fine.

Perhaps they expect fractions, but I think decimals are ok.

Now, back to Section A, I need to finalize.

For Section A Q3, √39, I think 6.25 is good, but let's see if the worksheet has a pattern.

In Q1, 75 is between 64 and 81, difference 17, 75-64=11, so 8 + 11/17 ≈ 8 + 0.647 = 8.647, so 8.6 or 8.7.

Q2, 110 between 100 and 121, difference 21, 110-100=10, so 10 + 10/21 ≈ 10.476, and since 10.5^2=110.25, very close, so 10.5 is excellent.

Q3, 39 between 36 and 49, difference 13, 39-36=3, so 6 + 3/13 ≈ 6.2308, and 6.25^2=39.0625, very close, so 6.25 is better than 6.23.

6.23^2 = (6.25 - 0.02)^2 = 39.0625 - 2*6.25*0.02 + 0.0004 = 39.0625 - 0.25 + 0.0004 = 38.8129, too low.

6.24^2 = as before ~38.9376

6.25^2=39.0625

So for 39, interpolate: from 6.24 to 6.25, difference 0.01, values 38.9376 to 39.0625, difference 0.1249

39 - 38.9376 = 0.0624

So fraction 0.0624 / 0.1249 ≈ 0.4996, so approximately 6.24 + 0.004996 ≈ 6.245

So √39 ≈ 6.245, which rounds to 6.25 if to two decimals, or 6.2 if to one.

In Q1 and Q2, if we use one decimal, Q1: 8.7, Q2: 10.5, Q3: 6.2 or 6.3? 6.2^2=38.44, 6.3^2=39.69, 39-38.44=0.56, 39.69-39=0.69, so closer to 6.2, but 0.56 vs 0.69, ratio 0.56/1.25=0.448, so 6.2 + 0.448*0.1 = 6.2448, same as before.

So for consistency, perhaps use one decimal: 6.2 for Q3.

But 6.2 is 6.20, and 6.245 is closer to 6.2 than to 6.3? 6.245 - 6.2 = 0.045, 6.3 - 6.245 = 0.055, so actually closer to 6.2.

6.245 - 6.2 = 0.045
6.3 - 6.245 = 0.055, yes, so closer to 6.2.

And 6.2^2=38.44, error 0.56, while 6.3^2=39.69, error 0.69, so indeed closer to 6.2.

So perhaps √39 ≈ 6.2

In many online sources, it's given as 6.24, but for estimation with number line, 6.2 is fine.

I'll go with 6.2 for Q3 to match one decimal place like others.

Q1: 8.7, Q2: 10.5, Q3: 6.2

10.5 is one decimal, 8.7 is one, 6.2 is one.

Good.

So Section A:

1) 8.7
2) 10.5
3) 6.2

Now for Section B, let's list the fills.

Table:

Row 1 (cube roots):
Col1: ³√1 (given)
Col2: ³√8 (given)
Col3: ³√27
Col4: ³√64
Col5: ³√125
Col6: ³√216
Col7: ³√343 (given)
Col8: ³√1000 (given)
Col9: ³√729 (given)
Col10: ³√512 (given)

Row 2 (cubes):
Col1: 1³
Col2: 2³
Col3: 3³ (given)
Col4: 4³ (given)
Col5: 5³ (given)
Col6: 6³ (given)
Col7: 7³
Col8: 10³
Col9: 9³
Col10: 8³

Note that for col8, ³√1000 = 10, so cube is 10³
Col9: ³√729 = 9, so 9³
Col10: ³√512 = 8, so 8³

Yes.

For Section C:

Left square root table:

√6: interval 2~3
√15: interval 3~4
√27: interval 5~6

Right square root table:

For interval 11~12: square root could be \sqrt{125} (since 11.18^2=125)
For 0.8~0.9: \sqrt{0.72} or \sqrt{0.7} — let's use \sqrt{0.7} as it's simpler
For 0.15~0.16: \sqrt{0.023} or \sqrt{0.024} — use \sqrt{0.024}

But to make radicand nice, for 0.8~0.9, \sqrt{\frac{7}{10}} = \sqrt{0.7}, same.

For cube root left:

³√21: interval 2~3 (since 2^3=8, 3^3=27, 8<21<27)
³√200: interval 5~6 (5^3=125, 6^3=216, 125<200<216)

Right cube root:

For interval 4~5: \sqrt[3]{100} (4.64^3≈100)
For 9~10: \sqrt[3]{800} (9.28^3≈800)

Perhaps they expect \sqrt[3]{64} for 4~5, but 64 is 4^3, so ³√64=4, which is endpoint. Similarly, 1000 for 10, but 10 is endpoint.

So better to use non-cube numbers.

I think \sqrt[3]{100} and \sqrt[3]{800} are fine.

Now, to compile the final answer.

Since the user wants only the final answer, and no explanation, I need to provide the answers for all parts.

But how to format? Probably list by section.

Also, for Section B, the table has multiple blanks, so I need to specify what to fill.

Perhaps for Section B, list the missing values in order.

Let's define the blanks.

In Section B table:

Top row blanks: positions 3,4,5,6: ³√27, ³√64, ³√125, ³√216

Bottom row blanks: positions 1,2,7,8,9,10: 1³, 2³, 7³, 10³, 9³, 8³

Positions are columns.

Column 1: bottom blank → 1³
Column 2: bottom blank → 2³
Column 3: top blank → ³√27
Column 4: top blank → ³√64
Column 5: top blank → ³√125
Column 6: top blank → ³√216
Column 7: bottom blank → 7³
Column 8: bottom blank → 10³
Column 9: bottom blank → 9³
Column 10: bottom blank → 8³

For Section C, the blanks are:

Left square root intervals: for √6: 2~3, √15: 3~4, √27: 5~6

Right square root radicals: for interval 11~12: \sqrt{125}, for 0.8~0.9: \sqrt{0.7}, for 0.15~0.16: \sqrt{0.024}

Left cube root intervals: for ³√21: 2~3, for ³√200: 5~6

Right cube root radicals: for 4~5: \sqrt[3]{100}, for 9~10: \sqrt[3]{800}

Now for the final answer, since it's a bit long, I'll organize it clearly.

Perhaps the user expects the answers in the order of the questions.

Since ExplainAnswer is false, I should only provide the final answer without explanation.

So I'll write:

Final Answer:

Section A:
1) 8.7
2) 10.5
3) 6.2

Section B:
Top row blanks (columns 3 to 6): \sqrt[3]{27}, \sqrt[3]{64}, \sqrt[3]{125}, \sqrt[3]{216}
Bottom row blanks (columns 1,2,7,8,9,10): 1^3, 2^3, 7^3, 10^3, 9^3, 8^3

Section C:
Square Root Intervals:
- \sqrt{6}: 2 \sim 3
- \sqrt{15}: 3 \sim 4
- \sqrt{27}: 5 \sim 6

Square Root Radicals for given intervals:
- Interval 11 \sim 12: \sqrt{125}
- Interval 0.8 \sim 0.9: \sqrt{0.7}
- Interval 0.15 \sim 0.16: \sqrt{0.024}

Cube Root Intervals:
- \sqrt[3]{21}: 2 \sim 3
- \sqrt[3]{200}: 5 \sim 6

Cube Root Radicals for given intervals:
- Interval 4 \sim 5: \sqrt[3]{100}
- Interval 9 \sim 10: \sqrt[3]{800}

To make it cleaner, perhaps use a different format.

Since it's homework, maybe they want the values as per the boxes.

For Section A, three answers.

For Section B, the table completion.

For Section C, the intervals and radicals.

I think this is comprehensive.

One thing: in Section C, for the right side, when they say "Square Root" and give interval, they might expect the number whose square root is in that interval, but the cell is labeled "Square Root", so likely the radical expression.

Similarly for cube root.

I think it's correct.

So I'll box the final answer as above.

But to save space, perhaps list sequentially.

Since the user said "provide only the final correct answer", and it's multiple parts, I'll present it neatly.

Final Answer:

Section A:
1) \( 8.7 \)
2) \( 10.5 \)
3) \( 6.2 \)

Section B:
- Top row (blanks): \( \sqrt[3]{27} \), \( \sqrt[3]{64} \), \( \sqrt[3]{125} \), \( \sqrt[3]{216} \)
- Bottom row (blanks): \( 1^3 \), \( 2^3 \), \( 7^3 \), \( 10^3 \), \( 9^3 \), \( 8^3 \)

Section C:
*Square Roots:*
- \( \sqrt{6} \): interval \( 2 \sim 3 \)
- \( \sqrt{15} \): interval \( 3 \sim 4 \)
- \( \sqrt{27} \): interval \( 5 \sim 6 \)
- For interval \( 11 \sim 12 \): \( \sqrt{125} \)
- For interval \( 0.8 \sim 0.9 \): \( \sqrt{0.7} \)
- For interval \( 0.15 \sim 0.16 \): \( \sqrt{0.024} \)

*Cube Roots:*
- \( \sqrt[3]{21} \): interval \( 2 \sim 3 \)
- \( \sqrt[3]{200} \): interval \( 5 \sim 6 \)
- For interval \( 4 \sim 5 \): \( \sqrt[3]{100} \)
- For interval \( 9 \sim 10 \): \( \sqrt[3]{800} \)

I think this covers all.

For Section C, the "Square Root" column for the right side is to be filled with the radical, and "Interval" is given, so yes.

Some might argue for different choices, but these are reasonable estimates.

So I'll go with this.