2018年5月24日 · #"the factors of + 12 which sum to + 13 are + 12 and + 1"# #x^2+13x+12=(x+12)(x+1)# Answer link. Jacobi J.

2017年7月7日 · What we can do first is complete the square under the radical to give the equivalent expression. #intsqrt((x+2)^2 + 9) dx#

x^11 Expression = x^13/x^2 Here we will use two of the laws of exponents as follows: (i) 1/a^n = a^-n (ii) a^n xx a^m = a^(n+m) We can rewrite the expression as: Expression = 1/x^2 xx x^13 Apply (i) Expression = x^-2 xx x^13 Apply (ii) Expression = x^(-2+13) = x^11

2017年4月21日 · The solution is x in (-oo,-13) uu (2,+oo) We solve this inequality with a sign chart. Let f(x)=(x+13)/(x-2 ...

2016年7月4日 · #= x [x^2 + 13x + 42]# Now getting the factors of 42 and seeing if the sum of its factors equals to 13 ( middle value). By getting its factors ( #7*6# = 42) we have come to know that 7 + 6 equals to 13.

x^4-13x^2-30 = (x^2-15)(x^2+2) = (x-sqrt(15))(x+sqrt(15))(x^2+2) x^4-13x^2-30 is quadratic in x^2. To factor it, first find a pair of factors of 30 whose difference is 13. Such a pair is 15 and 2. That gives us: x^4-13x^2-30 = (x^2-15)(x^2+2) The first of these quadratic factors can be considered a difference of squares: x^2-15 = x^2 - (sqrt(15))^2 = (x-sqrt(15))(x+sqrt(15)) using the ...

2018年5月7日 · By arranging the equation 25x^2 = 13 x^2 = 13/25 x = (13/25)^0.5 There are two answers remember: 1^2=1 and (-1)^2=1. x=0.72 or x=-0.72

2017年10月21日 · First, standard form is ax+by+c=0 In this question, we can expend the formula and rearrange it. You can try it first, and the steps are as follow: Step 1: Expend it 2y+3=-1/3x+2/3 As there are fraction in the formula and it doesn't look good, we can eliminate it by multiplyng 3. Step 2: Multiply by 3 6y+9=-1x+2 Step 3: Rearrange the terms x+6y+7=0 And we can get the …

2017年8月20日 · x=sqrt(13/20)+1/2 x=-sqrt(13/20)+1/2 Given - 6x^2-5x-13=x^2-11 Take all the terms to the left-hand side 6x^2-5x-13-x^2+11=0 Simplify it. 5x^2-5x-2=0 Take the constant term to the right-hand side 5x^2-5x=2 Divide all the terms by the coefficient of x^2 (5x^2)/5-(5x)/5=2/5 x^2-x=2/5 Take half the coefficient of x, square it and add it to both ...

2016年2月8日 · If #(12-k^2)x^2 = -13x^2# then we find #k^2=25# so #k=+-5# and our original quartic factors as: #x^4-13x^2+36 = (x^2-5x+6)(x^2+5x+6)# In our example, each of the remaining quadratic factors is reducible to linear factors with Real coefficients: #(x^2-5x+6)(x^2+5x+6) = (x-3)(x-2)(x+3)(x+2)#