2017年12月22日 · 1 + 6x + 12x^2 + 8x^3 We must use our knowledge of the binomial expansion: Method 1: We can use: (x+1)^n = 1 + nx + (n(n-1))/(2!) x^2 + (n(n-1)(n-2))/(3!) x^3 + ...

dy/dx=6(2x-1)^2 Let y=(2x-1)^3 dy/dx=3(2x-1)^2(2) dy/dx=6(2x-1)^2. 24517 views around the world . You can reuse this answer Creative Commons License

2018年5月28日 · #2x - 1= 3; 2x-1 >=0 # and #-(2x - 1)= 3; 2x-1 <0 # The inequalities that describe the domain restrictions probably do not make much sense to you, in their current form, so let's take a moment to simplify them:

What is the antiderivative of #(2x-1)^3#? Calculus Introduction to Integration Integrals of Polynomial ...

x> -2 or x <1 |2x +1| > -3 Solve for the absolute value We know either: 2x+1> -3 or 2x + 1 < - (-3) Now let's solve the first possibility which is 2x + 1 > -3 2x + 1 > -3 Add -1 on both sides 2x + 1 -1 > -3 - 1 2x > -4 Divide both sides by 2 (2x)/2 > (-4)/2 x > -2 Now let's solve the second possibility 2x + 1 < - (-3) 2x + 1< 3 Add -1 on both sides 2x + 1 - 1 < 3 -1 2x < 2 (2x)/2 < 2/2 x < 1 ...

2018年4月10日 · (x+3)(2x+1) >"using the a-c method of factorising" "the factors of "2xx3=6 "which sum to + 7 are + 6 and + 1" "split the middle term using these factors" rArr2x^2+6x ...

2015年10月13日 · This is really a calculus problem. The expansion is an infinite series because the power is a fraction: (1-2x)^(1/3)=1-2/3 x-4/9 x^2-40/81 x^3-160/243 x^4-704/729 x^5-cdots for |x|<1/2. This could be used as an approximation for small x, such as (1-2x)^(1/3) approx 1-2/3 x-4/9 x^2. The general form of the binomial theorem discovered by Newton, that works for any number p, can be written: (1+y ...

2016年12月3日 · How do you evaluate the definite integral #int (2x-1) dx# from [1,3]? Calculus Introduction to Integration Definite and indefinite integrals

2018年4月13日 · #x^2-2x-1=x+3# Subtract #x+3# from both sides (Or bring all terms to the left-hand side) #x^2-2x-1-x-3=x+3-x-3# Thus, #x^2-2x-1-x-3=0# Add like terms #x^2-3x-4=0# Split middle term so as to obtain #(-4)# as product and #(-3)# as sum #x^2+x-4x-4=0# Take common factors out from paired terms #x(x+1)-4(x+1)=0# Then, #(x+1)(x-4)=0# #x+1=0# or #x-4=0 ...

2017年10月13日 · #2^(2x+1) = 3(color(red)(2^x)) - 1# #2^(2x+1) = 3color(blue)w - 1# However, the other one is a little harder to spot. Let's look back at our rules of exponents: #a^(b+c) = a^b*a^c#. #a^(bc) = (a^b)^c#. Using these rules, we can manipulate the left side of the equation a bit: #2^(2x+1) = 3color(blue)w - 1# #2^1 * 2^(2x) = 3color(blue)w - 1#