Order these number in order to least to greatest using less than or greater than signs
1 answer:
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Easy, it says "for ANY positive integer" so just test any positive integer
remember that n! means times all integers from 1 to that number n
lets try 1
(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1
if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1
te answer is 1, B
and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=
![\frac{r^{2} \sqrt[3]{r^{2}} }{t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Br%5E%7B2%7D%20%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D%20%7D%7Bt%7D%20)
remember that n! means times all integers from 1 to that number n
lets try 1
(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1
if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1
te answer is 1, B
and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=
Answer:
ln(5/3)
Step-by-step explanation:
The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.
<h3>Limit</h3>We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.