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DedPeter [7]
3 years ago
9

the price of 9-volt batteries is increasing according to the function below, where t is years after January 1, 1980. During what

year will the price reach $4? use formula P(t)=1.1*e^0.047t A.)2007 B.)2005 C.)2003 D.)2009
Mathematics
2 answers:
dangina [55]3 years ago
4 0
4=1.1e^(0.047t)  divide both sides by 1.1

40/11=e^(0.047t)  take the natural log of both sides

ln(40/11)=0.047t  divide both sides by 0.047

t=ln(40/11)/0.047

t≈24.5

1980+24.5

2007.5

So during 2007 the price will reach $4.

OverLord2011 [107]3 years ago
3 0

Answer:

The option A.) 2007 is correct

Step-by-step explanation:

The formula which is to be used is given :

P(t) = 1.1\cdot e^{0.047t}

where P(t) is the function of time t and t is the time in years after January 1 , 1980

Now, we need to find the year when the price will reach $4

So, substituting P(t) = 4 and finding the value of t from the given equation.

\implies 4=1.1\cdot e^{0.047t}\\\\\implies 3.64=e^{0.047t}\\\\\text{Taking natural log ln on both the sides}\\\\\implies \ln 3.64=\ln e^{0.047t}\\\\\implies 1.29=0.047\cdot t\\\\\implies t = 27.49

So, t = 27.49 which is approximately equals to 27.5 years

So, 27.5 years after January 1, 1980 is the year 2007

Hence, The price will reach $4 in the year 2007

Therefore, The option A.) 2007 is correct

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Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

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⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

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Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

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