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NNADVOKAT [17]
3 years ago
14

What is the diameter of a circle with a circumference of 11.27π ​ft?

Mathematics
1 answer:
kiruha [24]3 years ago
3 0

soln,

circumference of circle = 11.27π ft

so,

C = 2πr

or, 11.27π = 2πr

or, 11.27π/2π = r

so, r = 5.635 ft

now,

diameter = 2r

= 2 x 5.635

= 11.27 ft

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A pitcher of fruit punch holds 2 gallons. If 9 people share the entire pitcher equally, how much punch does each person get in a
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2/9 of a gallon per person.
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This is a new version of the question. Make sure you start new workings.
cestrela7 [59]

Answer:

8.5 cm

Step-by-step explanation:

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8 0
2 years ago
create two scenarios of your own that involve relations in real life. Describe the scenarios in detail, including a description
Lorico [155]
Here's one example.

1) Catering Costs

"celebrations" offers catering services. They charge 30 dollars initially, and 1.5 dollars per person.

a) C representing cost and P representing people. The equation of this relation is C = 1.5p + 30

b)  C   P
    32   1
  33.5  2
    35   3
  36.5  4

(The table can be used to make points)
(p,C)

c) You can rewrite this in standard form.
  
   C = 1.5t + 30
   0 = 1.5t - C + 30
-30 = 1.5t - C
 30 = -1.5t + C

Another example:

2) Transportation.

A car is driving from City A to City B at a constant speed of 80km/h The distance away from both cities is 240km.

a) In which D represents distance in km and T represents time in hours, an equation that can represent the distance away from City B is D = -80t + 240

b) D   t
  160  1
   80   2
    0    3
  
(These can be used to make points)

c) You can rewrite in standard form.

D = -80t + 240
0 = -80t - D + 240
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brainliest?

7 0
3 years ago
"You are dating Moon rocks based on their proportions of uranium-238 (half-life of about 4.5 billion years) and its ultimate dec
Olenka [21]

Step-by-step explanation:

a.

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 41% = t

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

N_o = initial mass of isotope  = x

N = mass of the parent isotope left after the time, (t)  = 59% of x = 0.59x

t_{\frac{1}{2}} = half life of the isotope  = 4.5 billion years

\lambda = rate constant

Now put all the given values in this formula, we get

0.59x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}

t = 3.4 billion years

The age a rock is 3.4 billion years.

b.

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 35%= t

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

N_o = initial mass of isotope  = x

N = mass of the parent isotope left after the time, (t)  = 65% of x = 0.65x

t_{\frac{1}{2}} = half life of the isotope  = 4.5 billion years

\lambda = rate constant

Now put all the given values in this formula, we get

0.65x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}

t = 2.8 billion years

The age a rock is 2.8 billion years.

5 0
3 years ago
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