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How do you simplify #5(x-3)#? - Socratic
2018年6月30日 · #5(x-3)# To simplify this, use the distributive property: Following this image, we know that: #color(blue)(5(x-3) = (5 * x) + (5 * -3) = 5x - 15)#
How do you find the domain and range of #y=5/(x-3)#? - Socratic
2018年6月6日 · The domain is x in (-oo,3)uu(3,+oo). The range is y in (-oo,0)uu(0, oo). The denominator must be !=0. Therefore, x-3!=0 =>, x!=3 The domain is x in (-oo,3)uu(3,+oo ...
How do you find the inverse of y=x^3 +5? - Socratic
2016年1月4日 · Reformulate the equation with x isolated on one side to find inverse function: f^(-1)(y) = root(3)(y-5) Let f(x) = x^3+5 Let y = f(x) = x^3+5 Subtract 5 from both ends to get: y - 5 = x^3 Take the cube root of both sides and transpose to get: x = root(3)(y - …
5-x - Respuesta | Solucionador de Problemas Matemáticos - Cymath
\[5-x\] +. > < ...
How do you integrate #int 5^x-3^xdx# from #[0,1]#? - Socratic
2017年5月5日 · Let #y=5^x#. Taking log on both sides. #lny=xln5# #y=e^(xln5)# Therefore, #int5^xdx=inte^(xln5)dx=e^(xln5)/ln5=5^x/ln5#
How do you find the inverse of f(x)= x^5+x^3+x? - Socratic
2015年7月3日 · The inverse of f(x) = x^3+x is tricky enough: Using Cardano's method, we can find that f^-1(y) = root(3)((9y+sqrt(3y^2+12))/18) + root(3)((9y-sqrt(3y^2+12))/18) We can show that g(x) = x^5+x^3+x does have an inverse, in that it is continuous, monotonically increasing, with derivative always >= 1. d/(dx)g(x) = 5x^4+3x^2+1 >= 1 for all x in RR ...
How do you multiply #(x+5)(x-3)#? - Socratic
2017年1月11日 · How do you find the volume of a prism if the width is x, height is #2x-1# and the length if #3x+4#? How do you multiply #(a^2+2)(3a^2-4)#? How do you simplify #(x – 8)(x + 5)#?
Solve the Equation x-3=5 - Answer | Math Problem Solver - Cymath
Solve the Equation x-3=5 - Answer | Math Problem Solver - Cymath ... \\"Get
If 5^(-x) = 3, what does 5^(3x) equal? - Socratic
2017年1月14日 · So, if we currently know #5^-x# and want to determine #5^(3x)#, we see their powers are off by a factor of #-3#. Thus, we can write: #5^(3x)=(5^-x)^(-3)# Which is useful because we know #5^-x=3#, so: #5^(3x)=3^-3# Using the rule #a^-b=1/a^b# this becomes: #5^(3x)=1/3^3=1/27#