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NikAS [45]
3 years ago
13

A) R is inversely proprtional to A                                 ,when R=12, A =1.5

Mathematics
1 answer:
guapka [62]3 years ago
8 0
12=k1÷1•5
k=18

9=18÷A
A=2
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natta225 [31]

Answer:

2x -y ≥ 4

Step-by-step explanation:

The intercepts of the boundary line are given, so it is convenient to start with the equation of that line in intercept form:

... x/(x-intercept) + y/(y-intercept) = 1

... x/2 + y/(-4) = 1

Multiplying by 4 gives the equation of the line.

... 2x -y = 4

This line divides the plane into two half-planes. The half-plane that is shaded is the one for larger values of x and/or smaller values of y than the ones on the line. So, for some given y, if we increase x we will get a number from our equation above that is greater than 4. Hence, the inequality we want is ...

... 2x -y ≥ 4

We use the ≥ symbol because the line is solid, so part of the solution space.

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Anna [14]

Answer:

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2 years ago
The half life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
Veronika [31]

Using an exponential function, it is found that:

a) N(t) = 75(0.5)^{\frac{t}{3.8}}

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

A(t) = A(0)(1 - r)^t

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

N(t) = 75(0.5)^{\frac{t}{3.8}}

Item b:

This is N when t = 3.8, hence:

N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

N(t) = 75(0.5)^{\frac{t}{3.8}}

10 = 75(0.5)^{\frac{t}{3.8}}

(0.5)^{\frac{t}{3.8}} = \frac{10}{75}

\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}

\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}

t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}

t \approx 11

The amount remaining will be of 10 grams after approximately 11 days.

More can be learned about exponential functions at brainly.com/question/25537936

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