First find the circumference: $$C = 2\pi r = 2(3.14)(8) = 50.24$$ feet. Border cost: $$50.24 \times 3 = \150.72$$. Then find the area: $$A = \pi r^2 = 3.14(8)^2 = 3.14(64) = 200.96$$ square feet. Mulch cost: $$200.96 \times 2 = \401.92$$. Total: $$150.72 + 401.92 = \552.64$$. Choice B uses diameter instead of radius for circumference. Choice C forgets to double the radius in circumference formula. Choice D adds an extra calculation error.

This question tests calculating both circumference C=2πr and area A=πr² for a given radius, providing exact and approximate values. For r=7 m, C=2π×7=14π m ≈14×3.14=43.96 m, and A=π×49=49π m² ≈49×3.14=153.86 m². For example, if r=5 m, C=10π≈31.4 m, A=25π≈78.5 m². Use the radius directly in both formulas, remembering the factor of 2 for circumference. A common error is swapping the formulas, like using r² for circumference or forgetting the 2 in C. Steps include: (1) for C, 2×π×7=14π, approximate 43.96; (2) for A, π×49=49π, approximate 153.86; (3) units m and m². Circumference is linear in r, area quadratic, so changes affect them differently.

Since Pool A has area 144$$\pi$$, we have $$\pi r^2 = 144\pi$$, so $$r^2 = 144$$ and $$r = 12$$ feet. Pool A's circumference is $$2\pi r = 24\pi$$ feet. Since both pools have the same circumference, Pool B's circumference is also $$24\pi$$ feet. For Pool B: $$\pi d = 24\pi$$, so $$d = 24$$ feet and radius = 12 feet. Pool B's area is $$\pi(12)^2 = 144\pi$$ square feet. Choice A uses radius 6 instead of 12. Choice C doubles the correct area. Choice D uses diameter as radius.

Low pressure radius = 15 feet, area = $$\pi(15)^2 = 225\pi$$ square feet. High pressure radius = $$15 \times 1.4 = 21$$ feet, area = $$\pi(21)^2 = 441\pi$$ square feet. Percentage increase = $$\frac{441\pi - 225\pi}{225\pi} \times 100% = \frac{216\pi}{225\pi} \times 100% = \frac{216}{225} \times 100% = 96%$$. Choice A incorrectly assumes linear relationship. Choice B uses wrong calculation. Choice C doubles the radius increase percentage.

This question tests solving for the diameter from the circumference formula C=πd, rearranging to d=C/π, and handling exact π terms. Circumference C=πd uses diameter, so d=C/π (for C=20π m, d=20 m); since C=2πr, diameter is twice the radius, and circumference doubles if radius doubles. For example, with C=20π m, d=20π/π=20 m, which is the track's diameter. Correctly apply by dividing C by π to get d=20. Common errors include using 2π in the denominator like for radius, or treating it as area and squaring. Steps: (1) identify given circumference, (2) select d=C/π, (3) substitute C=20π, (4) simplify to 20, (5) include units m. Mistakes: confusing with radius formula r=C/(2π) and getting 10 m, not canceling π properly, or using approximate π unnecessarily.

This question tests solving for diameter from the circumference formula C=πd, given C=20π cm, by rearranging to d=C/π. For C=πd=20π cm, divide both sides by π to get d=20 cm. For example, if C=20π, then d=20, or equivalently r=10 using C=2πr. Correct application: use the direct formula d=C/π without extra steps. Common errors include dividing by 2π instead, giving d=10, or confusing with area and squaring. Steps: (1) given C=20π, (2) use d=C/π, (3) compute 20π/π=20 cm. Since C is proportional to d (or r), doubling diameter doubles circumference, but area would quadruple if radius doubles.

This question tests solving for radius from the area formula $A=\pi r^2$, given $A=36\pi \text{ m}^2$, by rearranging to $r=\sqrt{(A/\pi)}$. For area $A=\pi r^2=36\pi \text{ m}^2$, divide both sides by $\pi$ to get $r^2=36$, so $r=6 \text{ m}$ (positive value). For example, if $A=36\pi$, then $r^2=36$, $r=6$, as in a garden of that size. Correct application: rearrange the formula properly without forgetting to take the square root. Common errors include thinking $r=A/\pi=36$, or using circumference formula instead, or taking square root before dividing by $\pi$. Steps: (1) given $A=36\pi$, (2) $r^2=A/\pi=36$, (3) $r=\sqrt{36}=6 \text{ m}$. Remember the relationship: area is quadratic in r, so for $r=6$, $A=36\pi$, and if r doubles to $12$, A quadruples to $144\pi$.

This question tests applying the area formula A=πr² after converting diameter to radius, r=d/2. For d=20 ft, r=10 ft, A=π(10)²=100π ft², and approximating with π≈3.14 gives 100×3.14=314 ft². For example, if d=10 ft, r=5, A=25π≈78.5 ft². First convert d to r, then square r and multiply by π. A common mistake is using diameter in the area formula, like π(20)²=400π, which is four times too large since (2r)²=4r². Steps include: (1) r=20/2=10, (2) r²=100, (3) A=100π exactly, (4) approximate 314, (5) units ft². Area quadruples when radius doubles, reflecting the quadratic relationship.

This question tests applying the circumference formula C=πd or C=2πr, with diameter 12 in given, and providing exact and approximate values using π≈3.14. The circumference C=πd uses the diameter directly, so for d=12 in, C=12π in ≈37.7 in, or equivalently using radius r=6 in, C=2π×6=12π in. For example, a wheel with d=12 in travels 12π in per rotation, approximately 37.7 in. Correct application involves converting diameter to radius if using the 2πr formula, but here πd is straightforward. Common errors include using C=πr without the 2, giving 6π, or squaring the radius like in area, resulting in wrong values. Steps: (1) identify diameter d=12 in, (2) use C=πd, (3) compute 12π exactly, (4) approximate 12×3.14=37.68 rounded to 37.7, (5) include units in. Circumference is linear with radius, so doubling radius doubles circumference, unlike area which quadruples.

This question tests applying the circle area formula A=πr², where you use the given radius to calculate both exact and approximate values. For a radius of 5 cm, the area is A=π(5)²=π×25=25π cm², and approximating with π≈3.14 gives 25×3.14=78.5 cm². For example, if the radius were 3 cm, A=π×9=9π≈28.26 cm². To find the area, identify the radius, square it, multiply by π for the exact value, and then use the approximation if needed. A common mistake is using the diameter instead of the radius or forgetting to square the radius, like calculating π×5=5π instead. Steps include: (1) note the radius r=5 cm, (2) compute r²=25, (3) multiply by π for 25π, (4) approximate 25×3.14=78.5, (5) add units cm². Remember, area scales quadratically with radius, so doubling the radius quadruples the area.