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klemol [59]
3 years ago
15

Scott wants to buy fencing to place around a circular rock garden. The diameter of the circle is 16 feet. Using 3.14 for π, Calc

ulate the amount of fencing Scott will need to buy?
Explain your answer and how you found it in a complete sentence.
Mathematics
1 answer:
Nadya [2.5K]3 years ago
6 0

Amount of fencing Scott will need to buy for the circular rock garden we need to find circumference of circular garden i.e. 50.24feet .

<u>Step-by-step explanation:</u>

Here we have, Scott wants to buy fencing to place around a circular rock garden. The diameter of the circle is 16 feet. Using 3.14 for π. In order to calculate amount of fencing Scott will need to buy for the circular rock garden we need to find circumference of circular garden .

Circumference of circle = 2\pi r

⇒ Circumference of circle = 2\pi r

⇒ Circumference of circle = 2(3.14)(\frac{Diameter}{2} )

⇒ Circumference of circle = 2(3.14)(\frac{16}{2} )

⇒ Circumference of circle = 3.14(16)

⇒ Circumference of circle = 50.24feet

Therefore, Amount of fencing Scott will need to buy for the circular rock garden we need to find circumference of circular garden i.e. 50.24feet .

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Answer:

1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻

Step-by-step explanation:

1. Let b be the number and B be the set, so mathematically, it is written as

b ∈ B.

2. Let  a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write

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3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.

Since each set is a subset of the latter set, we write

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4 0
3 years ago
Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
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Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

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We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

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So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

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Step-by-step explanation:


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Answer:

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Step-by-step explanation:

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Then convert  3^{\frac{1}{10} } to \sqrt[10]{3^{1} } = \sqrt[10]{3}             (here a = 3, n = 10, m = 1)

Now we get  2^{\frac{2}{5} } + 3^{\frac{1}{10} } = \sqrt[5]{4}  + \sqrt[10]{3} . This is the simplified form, which is the answer.

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