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just olya [345]
2 years ago
13

Write as a single fraction in its simplest form.

Mathematics
1 answer:
yaroslaw [1]2 years ago
5 0

\dfrac{4}{3x-5} + \dfrac{x+2}{x-1}\\\\\\=\dfrac{4(x-1) + (x+2)(3x-5)}{(3x-5)(x-1)}\\\\\\=\dfrac{4x-4 + 3x^2 -5x +6x -10}{(3x-5)(x-1)}\\\\\\=\dfrac{3x^2 +5x-14}{(3x-5)(x-1)}

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The function f(x) = 25(4)^x represents the growth of a bird population every year in a national park. Kyle wants to manipulate t
Natasha_Volkova [10]

f(x)=25(4)^{\frac{x}{2}} will be the function to calculate population every half year.

Further explanation:

The given formula is:

f(x) = 25(4)^x

In the given formula, x represents the amount of time in years. So, in order to convert the given function for yearly calculation of population, to find the population every half year the time will be converted into half. This means that instead of x, x/2 will be used.

So the new function will be:

f(x) = 25(4)^{\frac{x}{2}}

Keywords: Population growth, Population growth function

Learn more about population growth at:

  • brainly.com/question/10689103
  • brainly.com/question/118412

#LearnwithBrainly

6 0
3 years ago
Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.
Mkey [24]

Answer:

1.~e^{-2} \\2.~e^{\frac{7}{2}}\\3.~e^8\\4.~e^{\frac{-11}{2}}

Step-by-step explanation:

We have to simplify the given exponential exponents.

Exponential Properties:

e^0 =1\\e^a.e^b = e^{a+b}\\\\\displaystyle\frac{e^a}{e^b} = e^{a-b}\\\\(e^a)^b = e^{ab}\\\\e^{-a} = \frac{1}{e^a}

Simplification takes place in the following manner:

a)

(e^{-3})^\frac{2}{3}\\(e^a)^b = e^{ab}\\=e^{-3\times \frac{2}{3}}\\=e^{-2}

b)

(e^4)(e^{\frac{-1}{2}})\\e^a.e^b = e^{a+b}\\=e^{(4+\frac{-1}{2})} \\= e^{\frac{7}{2}}

c)

(e^{-2})^{-4}\\(e^a)^b = e^{ab}\\= e^{-2\times -4}\\=e^8

d)

(e^{-4})(e^{\frac{-3}{2}})\\e^a.e^b = e^{a+b}\\=e^{(-4+\frac{-3}{2})} \\= e^{\frac{-11}{2}}

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3 years ago
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krok68 [10]
2.6 lb

Hope this helps!

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4 years ago
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Paha777 [63]

form an idea of the amount, number, or value of; assess.

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