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2 years ago

Classify the real numbers as rational or irritational numbers

Mathematics

1 answer:

Andre45 [30]2 years ago

3 0

Answer:

god loves you <3 .

Step-by-step explanation:

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Consider two circular swimming pools. Pool A has a radius of 22 feet, and Pool B has a diameter of 13.6 meters. Complete the des

maw [93]

Answers:

13 42 m; B; 0.57 m

Step-by-step explanation:

Data:

Pool A: r = 22 ft

Pool B: D = 13.6 m

Calculations:

1. Radius of Pool A

r = 22 ft × (0.305 m/1 ft) = 6.71 m

2. Diameter of Pool A

D =2r = 2 × 6.71 = 13.42 m

The diameter of Pool A is 13.42 m.

3. Compare pool diameters

The diameter of Pool B is 13.6 m.

So, the diameter of Pool B is greater.

4. Compare circumferences

The formula for the circumference of a circle is

C = 2πr or C = πD

Pool A: C = 2π × 6.71 = 42.16 m

Pool B: r = π × 13.6 = 42.73 m

Pool B - Pool A = 42.73 - 42.16 = 0.57 m

The circumference is greater by 0.57 m.

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2 years ago

A linear function, f, has an x-intercept at (5,0) and a y-intercept at (0,3). Which best describes the shape of the graph of thi

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As x approaches positive infinity, f(x) approaches negative infinity, since the graph goes down

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2 years ago

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Show a way to count 170 to 410 using tens and hundreds circle at least 1 benchmark number

Lostsunrise [7]

What is a benchmark number?

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3 years ago

which statement below is represented by the correct ratio?Choose all that apply. for ever 50 shoppers 25 watermelons are sold

ASHA 777 [7]

For every 50 shoppers, 25 watermelons are sold, so 50/25

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I'm trying to move up a rank and it would help!

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3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago