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mario62 [17]
3 years ago
8

Is a positive integer raised to a negative power always positive

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

Yes

Step-by-step explanation:

Many students gets confused with these exponential problems, they often get misguided but understand that there is nothing to be confused of.

When you have a negative number, you just take the reciprocal of the whole exponent number.

What is a reciprocal?

Let's take an integer for example, let's take the number 3.

The reciprocal for 3 is (1/3)

Let's take an other number, let's take 2/3

The reciprocal for 2/3 is (3/2)

In conclusion, we just reverse the denominator and the numerator or just switch it.

We take 3 as (3/1) and that is the reason, the reciprocal would be (1/3)

Now, coming to the negative integers. Taking an example:

(2)⁻¹

This would be become (1/2¹) = (1/2)

Hence, the result of a negative integer is positive but would be a fraction.

Hope I helped!


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Nadia compares the weights, in grams, of some apples and oranges she find the median abs interquartile range of the weights.
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Answer:

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3 years ago
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kow [346]
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hope this helps:)
7 0
2 years ago
Read 2 more answers
Solve the equation on the interval [0,2π]
WARRIOR [948]
\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------

\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ -  b \pm \sqrt {  b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}

\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}
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3 years ago
Help plz its confusing
DedPeter [7]
Nothing ;) Bc i don’t know hahahaha banana
4 0
3 years ago
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