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babunello [35]
3 years ago
11

Circumference of the circle

Mathematics
2 answers:
lutik1710 [3]3 years ago
8 0

Step-by-step explanation:

The formula for calculating a circle's circumference is πd, where π is an irrational number (3.14....) and d is the diameter of the circle.

In the circle with diameter 19.5cm, the circumference is around 3.14 * 19.5cm = 61.23cm. (A)

stealth61 [152]3 years ago
8 0

Answer:

61.2

Step-by-step explanation:

circumference= 2πR

r=d/2

r= 19.5/2

r=9.75

circumference= 2×3.14×9.75

circumference= 61.2

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mario62 [17]

Answer:

what the heck

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Step-by-step explanation:

4 0
3 years ago
Solve for x<br><br> 15x^2+16x= -4
Klio2033 [76]
15x²+16x+4 =0      (ax² +bx +c=0)

Δ = b²-4ac =256 - 4×15×4 =16

x1 = (-b+√Δ) / 2a = (-16+√16) / 30 =( -16+4) / 30 = -12/30 = - 2/5
x2 = (-b -√Δ) / 2a = (-16 -√16) / 30 = (-16 -4) /30 = -20/30 = -2/3
4 0
3 years ago
Read 2 more answers
Please help i wanna move to the next grade,10 points
AfilCa [17]

Answer:

C -1

Step-by-step explanation:

- 3x + 2 < 5

5 - 2 = 3

- 3x  \div  - 3

3 \div  - 3 =  - 1

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4 0
3 years ago
3.53 divided by 51<br> 24.2 divided by 42<br> 9.13 divided by 23<br> 79.2 divided by 39
patriot [66]

Answer:

1. 0.0692156863

2. 0.5761904762

3. 0.3969565217

4. 2.030769231

Step-by-step explanation:

5 0
3 years ago
An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0
3 years ago
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