Explain your reasoning as you compare the values of a^n and a^-n when n<0 brainly

Mathematics

1 answer:

mojhsa [17]3 years ago

4 0

n < 0, is another way to say "n is negative", so let's check

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad
a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
a^n~\hspace{10.5em}\stackrel{n = -n}{a^{-n}}\implies \cfrac{1}{a^n}
\\\\\\
a^{-n}~\hspace{10em}\stackrel{n=-n}{a^{-(-n)}}\implies a^{+n}\implies a^n$

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determine whether the lines intersect and if so find the point of intersection and the cosine of the angle of intersection x=4t+

Fittoniya [83]

Answer:

As the lines are not intersecting nor parallel, they must be skew.

Step-by-step explanation:

Question is incomplete, we consider the nearest match available online.

Parametric equations of two lines are:

L₁ : x=4t+2 , y = 3 , z =-t+1

L₂: x=2s+2 , y= 2s+5 , z = s+1

If lines are parallel then parametric coordinates must be equal scalar multiple of each other which s not true here.

$4t+2=2s+2 ---(1)\\ \\3=2s+5---(2) \\\\-t+1=s+1---(3)$

If lines are intersecting then parametric coordinates must be equal for some value of t and s.

$From (3)\\-t=s\\From (2)\\\\s=\frac{3-5}{2}\\\\s=-1\\\implies t=1\\\\To\,\,check\,\, the \,\,values \,\,of \,\,s\,\, and \,\,t \,\,for\,\,intersection\,\,put \,\,them\,\, in\,\, (1)\\\\4(1)+2= 2(-1)+2\\ 6\ne0$