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Example: Solve following.
| 10 x - 3 |= 0
Solution
Let's approach this one through a geometric standpoint. It is saying that the quantity in the absolute value bars has a distance of zero from the origin. There is just one number that has the property and i.e. zero itself. Thus, we must have,
10x - 3 = 0 ⇒ x = 3/10
In this case we acquire a single solution.
Thus, if b is zero then we can just drop the absolute value bars and solve the equation. Likewise, if b is negative then there will be no solution to the equation.
To this point we've only looked at equations which involve an absolute value being equivalent to a number, however there is no cause to think that there ought to only be a number on the other side of the equal sign. Similarly, there is no cause to think that we can only have one absolute value in the problem. Thus, we have to take a look at a couple of these kinds of equations.
g^2-6g-55/g
simplifications
ac + xc + aw^2 +xw^2 i need to factor the problem above
(3b+1)^2=16
if A is an ideal and phi is onto S,then phi(A)is an ideal.
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using T for term- p T-1,2,3,4,5,6 = 8,16,24,32,40. Formula is T x _ +_=8, T x_+_=16 etc same 2 numbers must be used. how do i figure this out? an brackets be used or minus?
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