2017年3月3日 · -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 instance, if we have x=3, x can be 3 or it can be -3. And so in our question we need to address this: abs(x-4)<9=>x-4 < pm9=> -9 < x-4 < 9 We can now solve this by adding 4 to all sides: -9 < x-4 < 9 -9 color(red)(+4 ...

2016年6月4日 · 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

2015年2月12日 · 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

2016年3月14日 · 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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2016年12月9日 · # 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)# ...

2018年4月17日 · 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-4)^2 This is indeed of degree 3 -- if we multiplied everything out, the leading term would be x^3, the term of the highest exponent in the polynomial, making it ...

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

2017年4月16日 · #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 implies y = 3/2x# so # dy = 3/2 dx#. #implies 1/3int dy/sqrt(y^2+1)# Now we note that: #cosh^2 z - sinh^2 z = 1# so we can sub #y = sinh z# which means that: #dy = cosh z ...

2016年4月29日 · #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: #" sqrt(1xx3^2xx(x^2)^2)# Taking all squared values outside the square root removing the square. For #(x^2)^2# you remove the outer square. #3xx x^2xxsqrt(1)# But #sqrt(1) = 1# so ...