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Arturiano [62]
2 years ago
14

A CD has a diameter of 12 cm. The hole in the middle of the CD has a diameter of 1.5 cm. Find the area of the CD to the nearest

tenth. Use 3.14 for ​π.
A. 111.3 cm2
b. 113.0 cm2
c. 349.4 cm2
d. 445.1 cm2
Mathematics
1 answer:
Aleks [24]2 years ago
8 0
C. because to work out the area it is pi radius squared. the area of the CD as a whole is 354.94. The smaller circle inside is 5.55. so 354.94 take away 5.55 is 349.39 rounded up so 349.4 cm^2
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Step-by-step explanation:

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Question 9 (3 points)
Mila [183]

From the dot plot, we have that 90.91% of the students have 5 or more vowels in their first and last  names.

---------------------

The dot plot states that:

  • 2 students have 4 vowels in their first and last name.
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  • 8 students have 6 vowels.
  • 9 students have 7 vowels.

Then:

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  • 3 + 8 + 9 = 20 have or more vowels.

p = \frac{20}{22} \times 100\% = 90.91\%

90.91% of the students have 5 or more vowels in their first and last  names.

A similar problem is given at brainly.com/question/14354536

5 0
2 years ago
In a lecture demonstration, an object is suspended from a spring scale which reads 8N when the object is in air. The object is t
katovenus [111]

Answer:

a) density of the object is 3995.01, b) the weight scale reads 22N c) the sum individually will be the same with when added together.

Step-by-step explanation:

The weight of the object in air is 8N,

and weight = Mass * acceleration due to gravity = m * 9.81

8/9.81 = 0.815,

upthrust( force acting on the body from the liquid impeding the immersion) on the body when fully submerged = weight in air - weight in water = 8N - 6N =2N

Upthrust = weight of water displaced = 2N = mass * acceleration

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volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)

volume of water displaced = volume of the solid

density of solid = mass/ volume = 0.815/0.000204 = 3995.01kg/m^3

b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N

c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.

6 0
2 years ago
PLEASE HELP ME!
algol13

Step-by-step explanation:

1.\sum_{i=1}^{5}3i

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∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)

∑ = 3 + 6 + 9 + 12 + 15

∑ = 45

2.\sum_{k=1}^{4}(2k)^{2}

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

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3.\sum_{k=3}^{6}(2k-10)

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∑ = -4 + -2 + 0 + 2

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a = 1 (1/4)ⁿ⁻¹

So the series is:

\sum_{j=1}^{7}(\frac{1}{4})^{j-1}

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a = -5 + 4(n−1)

a = -5 + 4n − 4

a = 4n − 9

So the series is:

\sum_{j=1}^{5}(4j-9)

5 0
2 years ago
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