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hawaiidude
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foueri series
how would you solve f(t)=1+t with -pi<x<pi?
we know that a0/2+sigma n=1 (an cos( nt) + bn(sin( nt) ?
how would you solve f(t)=1+t with -pi<x<pi?
we know that a0/2+sigma n=1 (an cos( nt) + bn(sin( nt) ?
The first step would be to replace f(t) with 1+t, since they are equivalent. This will give us the equation 1+t=1+t with -pi Since the variable is not isolated, we can use algebraic manipulation to isolate it. In this case, we can subtract 1 from both sides of the equation to get t=0. Yes, you can solve this equation using a graphing calculator by graphing both sides of the equation and finding the point of intersection, which will give you the solution for t. Since the equation is a straight line, there will only be one solution. In this case, the solution is t=0. Yes, you can use substitution to solve this equation. For example, you can substitute t=0 into the original equation to check if it satisfies the equation.2. How can I solve this equation when the variable is not isolated?
3. Can I solve this equation using a graphing calculator?
4. Is there more than one solution to this equation?
5. Can I use substitution to solve this equation?
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