if you add 'O', you're talking blood types.
is this an electrical question or an algebra??
Ab+ universal receiver o- universal donor blood types: can donate to: can receive from: ab+: ab+: ab+ ab- a+ a- b+ b- o+ o- ab-: ab+ ab-: ab- b- a- o- a+: a+ ab+: a+ a- o+ o- a-: a+ a- ab+ ab-: a- o- b+: b+ ab+: b+ b- o+ o- b-: b- b+ ab- ab+: b- o- 0+: o+ a+ b+ ab+: o- o+ o-: o+ o- a+ a- b+ b- ab+ ab-: o-
(a x b)^b =ab x b^2 =ab^3
a+b(a+B)=ab
b*ab = ab2 Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then ab2 = ab + b2 b = 1 + b/a would have to be true for all b. Counter-example: b = 2; a = 3 b(ab) = 2(3)(2) = 12 = ab2 = (4)(3) ab + b2 = (2)(3) + (2) = 10 but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
ab * -b = -ab^-2
No. A times B = AB
a=7 b=9 ab=? ab is the multiplication of a & b there fore the value of ab=7*9=63
No, A+B is left as A+B AB would be A x B
Here is a proof. Let a and b be any two real numbers. Consider the number x defined as x = ab + (-a)(b) + (-a)(-b). We can write this out differently as x = ab + (-a)[ (b) + (-b) ] Then, by factoring out -a , we find that x= ab + (-a)(0) = ab + 0 = ab. Also, x = [ a + (-a) ]b + (-a)(-b) And by factoring out b, we find that x=0 * b + (-a)(-b) = 0 + (-a)(-b) = (-a)(-b). Therefore x = ab and x = (-a)(-b) Then, by the transitivity of equality, we have ab = (-a)(-b).
Barack is type AB, so there are several possibilities: Mom A, Dad B Mom B, Dad A Mom AB, Dad A Mom AB, Dad B Mom A, Dad AB Mom B, Dad AB Mom AB, Dad AB
The types of blood include: A, B, AB, and O. (A-, A+, B-, B+, AB-, AB+, O-, O+)
a+, a-, b+, b-, o+, o-, ab+, ab-