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Yanka [14]
3 years ago
14

The fountain is made up of two semicircles and a quarter circle. Find the perimeter and the area of the fountain. Round the peri

meter to the nearest tenth of a foot and the area to the nearest square foot.

Mathematics
1 answer:
Andrej [43]3 years ago
5 0
FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE.
 For this case what we have is the following:
 For the two semicircles we can model it as a complete circle.
 We have to then:
 
 Perimeter: 
 P = 2 * pi * r
 or
 P = pi * d
 Where,
 r = radius
 d = diameter
 Therefore the perimeter is:
 P = 10 * pi
 For the largest circle we have:
 radius = 10
 Perimeter:
 P '= 2pi10
 P '= 20pi
 1/4 since 1/4 circle:
 P '' = 20pi / 4 = 5pi
 Then, the total perimeter of the source is:
 Pt = P + P '' = 10pi + 5pi = 15pi
 Pt = 15 * (3.141592)
 Pt = 47.1239
 round
 Pt = 47.1 ft

 Area:
 The total area will be:
 A = A (two semicircles) + A (quarter big circle)
 A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
 A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
 A = 98.17477042 feet ^ 2
 Round:
 A = 98.2 feet ^ 2

 Answer: 
 Perimeter of the source: 
 Pt = 47.1 ft 
 Area of the source: 
 A = 98.2 feet ^ 2

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What is the decibel level of the noise from a firecracker with intensity 10^-4 watts per square inch? Use a logarithmic model
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Answer:

\large \boxed{\text{110 db}}

Step-by-step explanation:

The intensity β of a sound wave in decibels is given by the formula

\beta = 10 \log \left (\dfrac{I}{I_{0}} \right)

and I₀ = the reference intensity (10⁻¹² W/m²)

Data:

I = 10⁻⁴ W·in⁻²

Calculations:

1. Convert watts per square inch to watts per square metre

\text{Sound level} = \dfrac{10^{-4}\text{ W}}{\text{1 in}^{2}} \times \left(\dfrac{\text{39.37 in}}{\text{1 m}}\right )^{2} = 1.5 \times 10^{-1} \text{ W$\cdot$m}^{-2}

2. Convert the intensity to decibels.

\beta = 10 \log \left (\dfrac{1.5 \times 10^{-1}}{1\times 10^{-12}} \right) = 10\log(1.5 \times 10^{11}) = 10 \times 11 = \textbf{110 db}\\\text{The sound intensity of the firecracker is $\large \boxed{\textbf{110 db}}$}

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Answer:

The frame will be 4 inches in width

Step-by-step explanation:

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The width of the outside is 20+2x

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6 0
2 years ago
I recently (yesterday) took the AMC 10A, can anyone help me with this problem?
lianna [129]
First, we claim that \frac{10^n-1}{9} is the n-digit number with all digits equal to one.

Note that 10^n is a one followed by n zeroes, so subtracting one gives n nines.  Divide that number by nine and you get n ones, completing the proof.

Therefore, we have that A_n = a \frac{10^n - 1}{9}, B_n = b \frac{10^n-1}{9}, and C_n = c \frac{10^{2n} - 1}{9}.

Let x = 10^n.  Then, we have:

c \frac{x^2-1}{9} - b \frac{x-1}{9} = (a \frac{10^n - 1}{9})^2 = a^2 \frac{x^2 - 2x + 1}{81}.

Multiplying by 81 gives:

9c(x^2-1) - 9b(x-1) = a^2 (x^2-2x+1)

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Since 9c = a^2, we must have a = 3 \sqrt{c}.  Since a must be a multiple of three, we plug in values.  If a = 9, we get c = 9 and thus 9b - 162 = 0, which is impossible.  So a = 9 doesn't work.

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With a = 3, c = 1, and thus 9b - 18 = 0, so b = 2.  This obviously is worse.

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