How do you simplify #(ab)^-2 - Socratic
2017年3月16日 · See the entire simplification process below: First, remove the negative exponent by using this rule for exponents: x^color(red)(a) = 1/x^color(red)(-a) (ab)^color(red)(-2) = 1/(ab)^color(red)(- -2) = 1/(ab)^2 This may be enough simplification depending on what has been requested. However, if you want to remove the terms from parenthesis you can use these two …
What is the square root of ab^2? - Socratic
2015年9月16日 · It is sqrt(a*b^2)=abs(b)*sqrta. 8827 views around the world You can reuse this answer
How do you factor #a^2 - 2ab +b^2# - Socratic
2018年3月4日 · (a - b)^2 Break down the expression into a simpler form by regrouping and putting brackets. a^2 - 2ab +b^2 a^2 - ab - ab +b^2 a(a - b) - b(a - b) (a - b) (a - b) (a - b)^2
What is the length of AB, given #A(5, -2)# and #B(-3, -4)#? - Socratic
2018年1月17日 · AB=sqrt68~~8.25" to 2 dec. places" >"calculate the length using the "color(blue)"distance formula" color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2 ...
How do you factor completely a^2 - ab - a + b? | Socratic
2016年4月6日 · =color(green)( (a -1 ) ( a -b) a^2 - ab - a + b Factorising by grouping: We can factor this expression by making groups of two terms: (a^2 - ab) + ( - a + b) a is common to both the terms in the first group, and -1 is common to both the terms in the second group: = a( a -b) - 1(a-b) a-b is common to both the terms now: =color(green)( (a -1 ) ( a -b)
How do you factor the expression -2a^2-ab+3b^2? - Socratic
2016年3月24日 · -(a-b)(2a+3b) −2a^2−ab+3b^2 => Factor out -1: =-(2a^2+ab-3b^2) Write the 2nd term as factors of 1st and last term i.e: ab=-2ab+3ab => We get: =-(2a^2-2ab+3ab-3b^2 ...
How do you expand log AB^2? - Socratic
2018年6月19日 · logA+2logB Using the product property, we have log(AB^2) = logA+logB^2 The power is moved in the front: logB^2=2logB :. log(AB^2) = logA+2logB
The graph y=ab^x passes through (2, 400) and (5,50). Find
2017年12月17日 · Please see below. As y=ab^x and it passes though (2,400) and (5,50), we have 400=ab^2 and 50=ab^5 Dividing latter by former, we get b^3=1/8 or b=1/2 As such 400=axx(1/2)^2 or 400=axx1/4 and hence a=400xx4=1600 Let for some x, we have ab^x>k, where k>0 then log_n1600+xlog_n(1/2)>log_nk or xlog_n(1/2)>log_nk-log_n1600 or 0 …
How do you prove that a group G is abelian if and only if a
2017年9月2日 · If a group G is Abelian, then for any two elements a and b, a*b = b*a Now, Define c=a*b which is also a member of G on virtue of closure property. Then c^2 = c*c implies (a*b)^2=(a*b)*(a*b) By associative axiom, (a*b)^2 = a*(b*a)*b But, by property of commutativity (for an Abelian group), implies (a*b)^2=a*(a*b)*b implies (a*b)^2=(a*a)*(b*b) Thus we get our …
How do you simplify #(ab^4)(ab^2)#? - Socratic
2016年12月26日 · a^2b^6 In simplifying the given expression, Law of addition of exponents would apply. The exponent of both 'a' is one, hence on multiplication exponents would add to become 2. Likewise the exponents of 'b' would add to become 6. Thus the final result would be a^2b^6.