Mar 3, 2017 · -5 < x < 13 When working with absolute value inequalities, we need to remember that the absolute value function will return a positive value regardless of what lies within. For …

Jun 4, 2016 · x^4/x^9=x^4*x^(-9)=x^(4-9)=x^(-5)=1/x^5 x^4/x^9=x^4*x^(-9)=x^(4-9)=x^(-5)=1/x^5

Feb 12, 2015 · 9^(x−4) = 81 implies x = 6 The only value p for which 9^p = 81 is p = 2 so x-4 = 2 rarr x=6

Mar 14, 2016 · First, we must use the rule #alogn = logn^a#. #log_4(x^4) - log_x(262144) = 0# #(logx^4)/(log4) - (log262144)/logx = 0#

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Dec 9, 2016 · # lim_(x rarr 4) (sqrt(x+5)-3)/(x-4) = lim_(x rarr 4) (sqrt(x+5)-3)/(x-4) * (sqrt(x+5)+3)/(sqrt(x+5)+3)# ...

Apr 17, 2018 · A zero of 4 with a multiplicity of 2 tells us that (x-4)(x-4)=(x-4)^2 is another factor of our polynomial. Since the multiplicity is 2, it must show up twice. So, we obtain f(x)=(x+9)(x …

\[9\times 4\] +. > < ...

Apr 16, 2017 · #int dx/sqrt(9x^2+4)# Start with a simple arithmetic adjustment, as clearly we are looking to clean this up first. #=1/2int dx/sqrt(9/4x^2+1)# We then say that: #y^2 = 9/4 x^2 …

Apr 29, 2016 · #9=3^2" and "x^4 = (x^2)^2# Multiply any number by 1 and you do not change that number's value. I am including the value 1 just to demonstrate a process. So write as: #" …