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Karolina [17]
2 years ago
3

What is the best approximation for the perimeter of a semicircle with a diameter of 64 meters?

Mathematics
1 answer:
eduard2 years ago
5 0

Correct option A) 100.48m .

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

Here we have , Diameter of a semicircle is 64 meters . We need to find the perimeter of semicircle . Let's find out:

We know that perimeter of a circle = 2\pi r , where r is radius of circle .

⇒ Perimeter = 2\pi r

⇒ Perimeter = 2\pi \frac{D}{2} , where D is diameter of a circle .

Since, a semicircle is half of a circle so Perimeter of semicircle will also be half of circle i.e.

⇒ Perimeter = \frac{2\pi \frac{D}{2}}{2}

⇒ Perimeter = \pi \frac{D}{2}}

Diameter = 64 meters

⇒ Perimeter = (3.14) \frac{64}{2}}

⇒ Perimeter = (3.14) 32

⇒ Perimeter = 100.48 m

Therefore , Perimeter of semicircle is 100.48 m . Correct option A) 100.48m .

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe
Natalka [10]

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}

Circumference in meters:

40,076km\times \frac{1000m}{1km}=40,076,000 m

Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

Hence, they encircle the planet 3.76\times 10^{11} times.

8 0
2 years ago
Sharon and Jacob started at the same place. Jacob walked 3 m north and then 4 m west. Sharon walked 5 m south and 12 m east. How
Nesterboy [21]

Answer:

d

Step-by-step explanation:

4 0
3 years ago
You are buying boxes of cookies at a bakery. Each box of cookies costs \$4$4dollar sign, 4. In the equation below, ccc represent
Basile [38]

Answer:

The required equation is

d = 4c

where d represents the cost of cookies in dollar and c represents the number of boxes of cookies.

Step-by-step explanation:

Given that, the cost of each box of cookies is $4.

It means,

The cost of 1 box of cookies is $4.

The cost of 2 boxes of cookies is $(4+4)  =$(2×4)

The cost of 3 boxes of cookies is $(4+4+4)  =$(3×4)

The cost of cookies

=(Number of boxes × 4)

The required equation is

d = 4c

where d represents the cost of cookies in dollar and c represents the number of boxes of cookies.                                                    

   

8 0
2 years ago
Find a single expression that represents the area of the outer ring of the circle if the area of the whole circle is represented
sp2606 [1]

Answer:  The answer is A_o=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


Step-by-step explanation: Given that the area of the whole circle is represented by the expression

A_c=25x^2-12x-9.

We are to find the area of the outer ring of the circle, i.e., to find the circumference of the circle.

Now, if 'r' represents the radius of the circle, then we have

A_c=\pi r^2\\\\\Rightarrow \dfrac{22}{7}r^2=25x^2-12x-9\\\\ \Rightarrow r^2=\dfrac{7}{22}(25x^2-12x-9)\\\\\Rightarrow r=\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.

Thus, the area of the outer ring is

A_o=2\pi r=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


5 0
2 years ago
The surface area of an oil spill gets 105% larger every day, represented by the function s(x) = (1.05)x − 1. On the first day, i
ioda
The given scenario above is like a geometric sequence with r or common ratio being 1.05 and the first term being 31 square meters. The nth term of the sequence is calculated through the equation,
                        A15 = (A1)(r)^(n - 1)
Substituting,
                        A15 = (31)(1.05)^(15 - 1) = 61.3778 square meter
Therefore, the area covered by the spill after the 15th day is approximately 61.4 square meters. 
6 0
2 years ago
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