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3 years ago

Use properties of exponents to rewrite the following expressions as a number or an exponential expression with

Mathematics

1 answer:

Lerok [7]3 years ago

7 0

Answer:

[(2)^√3]^√3 = 8

Step-by-step explanation:

Hi there!

Let´s write the expression:

[(2)^√3]^√3

Now, let´s write the square roots as fractional exponents (√3 = 3^1/2):

[(2)^(3^1/2)]^(3^1/2)

Let´s apply the following exponents property: (xᵃ)ᵇ = xᵃᵇ and multiply the exponents:

(2)^(3^1/2 · 3^1/2)

Apply the following property of exponents: xᵃ · xᵇ = xᵃ⁺ᵇ

(2)^(3^(1/2 + 1/2)) =2^3¹ = 2³ = 8

Then the expression can be written as:

[(2)^√3]^√3 = 8

Have a nice day!

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In 2009, there were 1570 bears in a wildlife refuge. In 2010, the population had increased to

natita [175]

Answer:

$8,101\ bears$

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

x ----> is the number of years since 2009

y ----> is the population of bears

a ----> is the initial value

b ---> is the base

step 1

Find the value of a

For x=0 (year 2009)

y=1,570 bears

substitute

$1.570=a(b)^{0}$

$a=1.570\ bears$

$y=1.570(b)^{x}$

step 2

Find the value of b

For x=1 (year 2010)

y=1,884 bears

substitute

$1,884=1.570(b)^{1}$

$b=1,884/1.570$

$b=1.2$

The exponential function is equal to

$y=1.570(1.2)^{x}$

step 3

How many bears will there be in 2018?

2018-2009=9 years

For x=9 years

substitute in the equation

$y=1.570(1.2)^{9}$

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7 0

3 years ago

SHOW YOUR WORK, <br><br>PLEASE HELP!!!!!!!!!

Shkiper50 [21]

To find the inverse, we swap the variables y and x, then solve for the new y.

3a. $y=\frac{3}{x-1}$

Swapping the variables: $x=\frac{3}{y-1}$
Solving for y: $x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}$
The domain of this inverse is $x ≠ 0$ .
3b. $y=x^2-1$

Swapping: $x = y^2 - 1$
Solving for y: $y^2 = x + 1 \\ y = \sqrt{x+1}$
The domain of this inverse is $x ≥ -1$ .
3c. $y=\sqrt[3]{\frac{x-7}{3}}$
Swapping: $x=\sqrt[3]{\frac{y-7}{3}}$
Solving for y: $x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7$
The domain of this inverse is all real numbers.
4a. $y=\frac{3}{x-1}$ , $y=1+\frac{3}{x}$
$y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x$
$y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x$

4c. $y=\sqrt[3]{\frac{x-7}{3}}$ , $y=3x^3+7$
$y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x$
$y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x$

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3 years ago

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Answer:

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35* left to the right angle

x above right angle

Step-by-step explanation:

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2 years ago