$We\ are\ given\ that,\\4^4*4^{-8}*4\\We\ know\ that,\ for\ any\ real\ number\ n, n^a*n^b=n^{(a+b)}\\Hence,\\First,\ lets\ write\ all\ the\ terms\ in\ an\ exponential\ form.\\4^4*4^{-8}*4^1\\=4^{(4+(-8)+1)}\\=4^{(5-8)}\\=4^{-3}$

5 0

3 years ago

Answer Question number 5 <br> Plzzzzzz

loris [4]

Answer:

Step-by-step explanation:

a = -4/3

d= -1 +4/3 =1/3

last term = 4 1/3 = 13/3

nth term = a +(n-1)d

$\frac{-4}{3}+(n-1)*\frac{1}{3}=\frac{13}{3}\\\\(n-1)*\frac{1}{3}=\frac{13}{3}-\frac{-4}{3}\\\\(n-1)*\frac{1}{3}=\frac{13+4}{3}\\\\(n-1)*\frac{1}{3}=\frac{17}{3}\\\\n-1=\frac{17}{3}*\frac{3}{1}\\\\n-1=17\\\\n=17+1=18$

Middle terms are 9the term and 10th term

$t_{9}=\frac{-4}{3}+8*\frac{1}{3}=\frac{-4}{3}+\frac{8}{3}=\frac{4}{3}\\\\t_{10}=\frac{-4}{3}+9*\frac{1}{3}=\frac{-4}{3}+\frac{9}{3}=\frac{5}{3}\\\\ t_{9}+t_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3\\\\t_{9}+t_{10}=3$

3 0

3 years ago

I have no idea so just solve it for me

romanna [79]

731/17 is equal to 43

6 0

3 years ago