Answer:
That is true!
Let me know if you need more help with multiplication. I may be able to help.
Have a great day.
Use the graph to evaluate the function for the given input value. fl-1) = f(1) =
2 answers:
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Remark
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer