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3 years ago

What are the domain and range of the logarithmic function f(x) = log7x?

Mathematics

1 answer:

kap26 [50]3 years ago

6 0

The domain is (0, ∞)
The range is (-∞, ∞)

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8 cups lemon juice
8 cups vinegar
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3 years ago

Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by p = 130e0.0205t,wh

svetoff [14.1K]

Answer: a) 236,000.

b) 2021

Step-by-step explanation:

Given : Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by $p = 130e^{0.0205t}$ (1)

where t = 0 corresponds to 1980.

Then , for 2009

t= 2009-1980=29

a. ⇒The population of Phoenix in 2009 = $p = 130e^{0.0205(29)}$

$p = 130e^{0.5945}$

$p =130(1.81212465608)=235.57620529\approx236$

Hence, the population of Orlando in 2009 was about 236,000.

b) Substitute p= 300 in (1) , we get

$300 = 130e^{0.0205t}$

$2.3077= e^{0.0205t}$ (Divide both sides by 130)

Taking Natural log on both sides,

$\ln(2.3077) = 0.0205t\\\\\Rightarrow\ t=\dfrac{\ln(2.3077)}{0.0205}$

$t=\dfrac{0.836251357528}{0.0205}=40.7927491477\approx41$

Hence, The year Orlando will have a population of 300,000 =1980+41 =2021

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3 years ago

I need help on my quiz!

Fantom [35]

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3 years ago

Equation y=0.25x+10 explain in words what each letter and number on your equation means

enot [183]

The 0.25 is the rate of change per each time. the 10 is what it started it

8 0

3 years ago

If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}$

therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

$a=-\frac{1}{4c^{-6}}$

6 0

3 years ago