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SpyIntel [72]
2 years ago
5

What’s the difference between numbers that are to the power of with parentheses and no parentheses. For example (-2/5)^3 and -2/

5^3. Please explain this rule and thank you.
Mathematics
1 answer:
Verdich [7]2 years ago
8 0

Answer:

See below

Step-by-step explanation:

The first one 'cubes' the denominator AND the numerator:

(-2/5)^3  = -2/5  * -2/5  * -2/5 =  -8/125

The second one 'cubes' only the denominator:

-2/5^3  = -2 / (5 * 5 * 5) =  - 2 / 125

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Answer:

210

Step-by-step explanation:

The answer for 7p3 is : 210. This can be solved in the following way: 7p3 is an expression for permutation which means the number of ways of arranging 3 items from a total of 7 items.

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Evaluate 6ab when a=1/2<br> and b = 7.
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6*1/2*7

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Step-by-step explanation:

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Let f(x) = 5x + 12. Find f−^1(x)
soldi70 [24.7K]

{f}^{ - 1}(x)  =  \frac{x - 12}{5}

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3 years ago
Convert 58 to a percent.<br><br> Do not include the "%" sign in your answer.
irga5000 [103]

Answer:

5800

Step-by-step explanation:

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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20
vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of y=f(x) is a vertical has an asymptote at x=a if at least one of the following statements is true:

1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty

The function is:

f(x)=\frac{x^2+7x+10}{x^2+9x+20}

First of all, let't factor out:

f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq  5

From here:

\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty

\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty

Accordingly:

x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

5 0
4 years ago
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