Estimating Irrational Roots

Since $7^2=49$ and $8^2=64$, we have $49<54<64$, so $7<\sqrt{54}<8$. The largest whole number below the side length is $\boxed{{\displaystyle 7}}$. A bigger number always has a bigger square root, so trapping $54$ between two perfect squares traps the root between the two whole numbers.

From $82^2=6724$ and $83^2=6889$ we get $6724<6800<6889$. Taking roots keeps that order, so $82<\sqrt{6800}<83$ and the largest whole number below the side is $\boxed{{\displaystyle 82}}$. The numbers are bigger but the method is the same, since the two perfect squares around the radicand are all you ever need.

The halfway point between $9$ and $10$ is $9.5$, and $9.5^2=90.25$. Since $83<90.25$, the root is below $9.5$, so it rounds down. To the nearest inch, $\sqrt{83}\approx \boxed{{\displaystyle 9}}$. Squaring the midpoint is the whole trick, since it turns a question about an endless decimal into one comparison of plain numbers.

The midpoint between $4$ and $5$ is $4.5$, and $4.5^2=20.25$. Since $21>20.25$, the root is past $4.5$, so it rounds up. To the nearest inch, $\sqrt{21}\approx \boxed{{\displaystyle 5}}$. Rounding down to $4$ is the usual slip here, because $21$ looks close to $16$. The midpoint square settles it instead of eyeballing.

The midpoint between $4.7$ and $4.8$ is $4.75$, and $4.75^2=22.5625$. Since $23>22.5625$, the root is past the midpoint and rounds up. To the nearest tenth, $\sqrt{23}=\boxed{{\displaystyle 4.8}}$. If the long decimal feels shaky, scale by $10,000$ and compare $475^2=225,625$ with $230,000$, which is exact whole-number work.

The midpoint between $6.4$ and $6.5$ is $6.45$. Comparing exactly, $645^2=416,025$ while $41\times 10,000=410,000$, so $41<6.45^2$ and $\sqrt{41}<6.45$. The root sits below the midpoint, so it rounds down and $\sqrt{41}\approx \boxed{{\displaystyle 6.4}}$. Scaling both sides by $10,000$ keeps every comparison in whole numbers, so no rounding error can creep in.

Squaring both bounds turns $7<\sqrt{n}<11$ into $49<n<121$, so $n$ runs from $50$ through $120$. That is $120-50+1=\boxed{{\displaystyle 71}}$ whole numbers. Both ends drop out because $\sqrt{49}=7$ and $\sqrt{121}=11$ exactly, and the inequalities are strict.

A whole number $k$ fits exactly when $130<k^2<320$. Since $11^2=121$ is too small and $12^2=144$ works, the smallest is $12$. Since $17^2=289$ works and $18^2=324$ is too big, the largest is $17$. Counting $12$ through $17$ gives $17-12+1=\boxed{{\displaystyle 6}}$. Squaring $k$ instead of estimating the two roots keeps the whole problem in exact arithmetic.

Square each length. Cable A gives $(6\sqrt{7}{)}^2=6^2\times 7=252$, and Cable B gives $(4\sqrt{15}{)}^2=4^2\times 15=240$. Since $252>240$ and both lengths are positive, Cable A is longer, option $\boxed{{\displaystyle 1}}$. The number out front squares too, which is the step people skip when they try to compare $6\sqrt{7}$ and $4\sqrt{15}$ by eye.

Pair the factors. Each pair gives ${\left(\sqrt{6}\right)}^2=6$, so ${\left(\sqrt{6}\right)}^4=6^2=\boxed{{\displaystyle 36}}$. Any even power of a root collapses this way, since the copies pair off and each pair leaves just the number underneath.

Sharpen each root to tenths. Since $4.2^2=17.64<18$, we know $\sqrt{18}>4.2$, and since $5.7^2=32.49<33$, we know $\sqrt{33}>5.7$, so the sum is more than $4.2+5.7=9.9$. Going the other way, $4.25^2=18.0625>18$ and $5.75^2=33.0625>33$ make the sum less than $4.25+5.75=10$. The total is caught in $9.9<\sqrt{18}+\sqrt{33}<10$, so the largest whole number below it is $\boxed{{\displaystyle 9}}$. Whole-number brackets alone only narrow the sum to between $9$ and $11$, which is why the tenths are needed.

The sum already beats $14$, since $\sqrt{40}>6$ and $\sqrt{75}>8$. So the real test is $15$. Ask whether $\sqrt{40}<15-\sqrt{75}$, which is safe to square because $\sqrt{75}<9$ keeps both sides positive. Squaring gives $40<300-30\sqrt{75}$, which rearranges to $30\sqrt{75}<260$. Square again and compare $30^2\cdot 75=67,500$ with $260^2=67,600$. Since $67,500<67,600$, the sum really is under $15$, so it sits between $14$ and $15$ and the answer is $\boxed{{\displaystyle 14}}$. Every squaring step reverses cleanly here because both sides stay nonnegative throughout.

Since $5^2=25$ and $6^2=36$, we have $25<34<36$, so $5<\sqrt{34}<6$. The largest whole number less than $\sqrt{34}$ is $\boxed{{\displaystyle 5}}$. Taking roots keeps the order, so bracketing $34$ between perfect squares brackets the root between whole numbers.

Squaring each length gives $63$ and $57$. Since $63>57$, we get $\sqrt{63}>\sqrt{57}$, so Rope 1 is longer, option $\boxed{{\displaystyle 1}}$. Both roots have floor $7$, so the floors tie and only the numbers under the radicals tell them apart.

The midpoint between $8$ and $9$ is $8.5$, and $8.5^2=72.25$. Since $75>72.25$, the root is past the midpoint and rounds up, so $\sqrt{75}\approx \boxed{{\displaystyle 9}}$ inches. To skip the decimal entirely, double both sides and compare $4\times 75=300$ with $17^2=289$, which gives the same verdict.

The midpoint between $6$ and $7$ is $6.5$, and $6.5^2=42.25$. Since $44>42.25$, the root is past the midpoint, so $\sqrt{44}$ rounds up and the height is $\boxed{{\displaystyle 7}}$ feet. Doubling first keeps it all in whole numbers, comparing $4\times 44=176$ with $13^2=169$.

The midpoint between $5.5$ and $5.6$ is $5.55$, and $5.55^2=30.8025$. Since $31>30.8025$, the root is above $5.55$ and rounds up, so $\sqrt{31}=\boxed{{\displaystyle 5.6}}$ to the nearest tenth. Rounding to tenths works exactly like rounding to whole numbers, only the midpoint you square is a halfway tenth.

The midpoint between $9.3$ and $9.4$ is $9.35$, and $9.35^2=87.4225$. Since $88>87.4225$, the root is past the midpoint and rounds up, so $\sqrt{88}\approx \boxed{{\displaystyle 9.4}}$. Comparing $935^2=874,225$ with $880,000$ reaches the same answer using only whole numbers.

Since $15^2=225$ and $16^2=256$, we have $225<250<256$, so $15<\sqrt{250}<16$. The largest whole number below $\sqrt{250}$ is $\boxed{{\displaystyle 15}}$.

Testing squares in the nineties, $94^2=8836$ and $95^2=9025$. Since $8836<9000<9025$, the root sits between $94$ and $95$, so the floor is $\boxed{{\displaystyle 94}}$. When no perfect square is handed to you, size it up first. Here $90^2=8100$ and $100^2=10,000$ show the answer is in the nineties.

Testing $1,732^2=2,999,824$, which is under $3,000,000$, and $1,733^2=3,003,289$, which is over, traps the root between the two. So the largest whole number below $\sqrt{3,000,000}$ is $\boxed{{\displaystyle 1732}}$. Only the two squares straddling the target matter, no matter how large the numbers get.

Squaring both bounds turns the question into $50<n<130$, so $n$ runs from $51$ through $129$. That is $129-51+1=\boxed{{\displaystyle 79}}$ whole numbers. Both ends are strict, so $50$ and $130$ themselves are left out of the count.

Since $6^2=36<40<49=7^2$, the smallest whole number above $\sqrt{40}$ is $7$. Since $14^2=196<210<225=15^2$, the largest below $\sqrt{210}$ is $14$. Counting $7$ through $14$ gives $14-7+1=\boxed{{\displaystyle 8}}$. Both $7$ and $14$ do count here, which is why the run is $8$ long and not $7$.

Square each. Beam 1 gives ${\left(7\sqrt{5}\right)}^2=49\times 5=245$, and Beam 2 gives ${\left(4\sqrt{15}\right)}^2=16\times 15=240$. Since $245>240$ and both lengths are positive, Beam 1 is longer, option $\boxed{{\displaystyle 1}}$.

Pair the six factors. Each ${\sqrt{3}}^2$ is $3$, so ${\sqrt{3}}^6={\left({\sqrt{3}}^2\right)}^3=3^3=\boxed{{\displaystyle 27}}$. Any even power of a root comes out a whole number this way, since the copies pair off and each pair leaves the number underneath.

Bracket each root with tenths. Since $5.2^2=27.04<28<28.09=5.3^2$, we get $5.2<\sqrt{28}<5.3$, and since $7.2^2=51.84<52<53.29=7.3^2$, we get $7.2<\sqrt{52}<7.3$. Adding the low ends and the high ends separately, the sum lies between $5.2+7.2=12.4$ and $5.3+7.3=12.6$. The largest whole number below the total is $\boxed{{\displaystyle 12}}$. Whole-number brackets would only give $12$ to $14$, so the tenths are what pin the floor down.