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3 years ago

Rewrite the rational exponent as a radical by extending the properties of integer exponents.

Mathematics

2 answers:

agasfer [191]3 years ago

8 0

The correct answer is:

the eighth root of 2 to the fifth power .

Explanation:

What we have is:

$\frac{2^{\frac{7}{8}}}{2^{\frac{1}{4}}}$

Using the rules of exponents, we know that when we divide powers with the same base, we subtract the exponents. The base of each exponent is 2, so we subtract:

7/8 - 1/4

We find a common denominator. The smallest thing that both 8 and 2 will evenly divide into is 8:

7/8 - 2/8 = 5/8

This gives us:

$2^{\frac{5}{8}}$

When rewriting rational exponents as radicals, the denominator is the root and the numerator is the power. This means that 8 is the root and 5 is the power, which gives us:

$\sqrt[8]{2^5}$ ,

or in words, the eighth root of 2 to the fifth power.

Vitek1552 [10]3 years ago

6 0

A. 2^(7/8) / 2^(1/4) = 2^(7/8) / 2^(2/8) = 2^(5/8)B. ( 2^(1/8) ) 5 = 2^(5/8)C. 2^(8/5)D. 2^(5/8)E. ( 2^(5/8) )^(1/2) = 2^(5/16)F. ( 2^6 )^(1/4) = 2^(6/4) = 2^(1/3)

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Which is a possible number of distinct real roots for a cubic function? Select all that apply.

Neko [114]

Answer:

Step-by-step explanation:

3 is a possible number of distinct real roots for a cubic function.

The maximum possible number of distinct roots are equal to the degree of any polynomial function.

Hence quadratic function has 2 roots

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Linear has 1

4 0

3 years ago

jake makes 9 loaves of olive bread. he uses 30 grams of olives in each loaf.he started with 1 kilogram of olives.how many grams

8090 [49]

Considering that 1 kilogram is equal to 1000 grams and he made 9 loaves of bread using 30 grams each time, that equates to 270 grams of olives. 1000 - 270 = 730. Your answer is 730 grams of olives remain. Hope this helped.

6 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$