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Maurinko [17]
3 years ago
10

Write a rule in function notation for each situation:

Mathematics
1 answer:
algol [13]3 years ago
5 0

Answer:

1. \bold{S=f(P) = 0.07P}

2. x = 16

3. Part 1: P is the independent variable and S is the dependent variable.

Part 2: x is the independent variable and y is the dependent variable.

Step-by-step explanation:

1. To write the function notation for:

Sales tax is 7% of the total price.

Let the total price be P.

And sales tax be S.

As per the given statement:

S = 7\% \ of\ P\\\Rightarrow S =\dfrac{7}{100}P\\\Rightarrow S=0.07P

Writing it in the function notation:

\bold{S=f(P) = 0.07P}

2. To find the value of x such that f(x) = 14 and

f(x) = \dfrac{1}4x + 10

Putting the value of f(x) = 14

14 = \dfrac{1}4x + 10\\\Rightarrow \dfrac{1}4x =14-10\\\Rightarrow \dfrac{1}4x =4\\\Rightarrow x =4\times 4\\\Rightarrow \bold{x =16}

3. To find the dependent and independent variable.

Independent variables are those whose value is not dependent on the other variable's values.

Dependent variables are dependent on the value of other variables.

In question 1:

\bold{S=f(P) = 0.07P}

P is the independent variable.

S is the dependent variable.

In question 2:

If we write it as follows:

y=f(x) = \dfrac{1}4x + 10

x is the independent variable and y is the dependent variable.

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So, for example if your dog needed 7 vaccinations, it would be: 45+10(7) = y -- answer being $115.

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Answer:

Step-by-step explanation:

Part 1:

Let 

     Q₁ = Amount of the drug in the body after the first dose.

     Q₂ =  250 mg

As we know that after 12 hours about 4% of the drug is still present in the body.

For Q₂,

we get:

            Q₂ = 4% of Q₁ + 250

                  = (0.04 × 250) + 250

                  = 10 + 250

                  = 260 mg

Therefore, after the second dose, 260 mg of the drug is present in the body.

Now, for Q₃ :

We get;

          Q₃ = 4% of Q2 + 250

               = 0.04 × 260 + 250

               = 10.4 + 250

               = 260.4

For Q₄,

We get;

          Q₄ = 4% of Q₃ + 250 

                = 0.04 × 260.4 + 250

                = 10.416 + 250 

                = 260.416

Part 2:

To find out how large that amount is, we have to find Q₄₀.

Using the similar pattern

for Q₄₀,

We get;

           Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹

Taking 250 as common;

           Q₄₀ = 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)

                 = 2501 − 0.04401 − 0.04

           Q₄₀ = 260.4167

Hence, The greatest amount of antibiotics in Susan’s body is 260.4167 mg.

Part 3:

From the previous 2 components of the matter, we all know that the best quantity of the antibiotic in Susan's body is regarding 260.4167 mg and it'll occur right once she has taken the last dose. However, we have a tendency to see that already once the fourth dose she had 260.416 mg of the drug in her system, that is simply insignificantly smaller. thus we will say that beginning on the second day of treatment, double every day there'll be regarding 260.416 mg of the antibiotic in her body. Over the course of the subsequent twelve hours {the quantity|the quantity|the number} of the drug can decrease to 4% of the most amount, that is 10.4166 mg. Then the cycle can repeat.

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