Asked by carry on
For all x in the domain of the function x+1 over x^3x this function is equivalent to
\[a)\frac{1}{x^2}-\frac{1}{x^3}\\
b)\frac{1}{x^3}-\frac{1}{x}\\
c)\frac{1}{x^2-1}\\
d)\frac{1}{x^2-x}\\
e) \frac{1}{x^3}\]
This is one of the questions to my math test can someone tell if im right?
Is it D im very confused on this one
Which of the following is a rational number
All numbers are squared
2
Pie
7
5/25
64/49
Is it 5/25
\[a)\frac{1}{x^2}-\frac{1}{x^3}\\
b)\frac{1}{x^3}-\frac{1}{x}\\
c)\frac{1}{x^2-1}\\
d)\frac{1}{x^2-x}\\
e) \frac{1}{x^3}\]
This is one of the questions to my math test can someone tell if im right?
Is it D im very confused on this one
Which of the following is a rational number
All numbers are squared
2
Pie
7
5/25
64/49
Is it 5/25
Answers
Answered by
Reiny
For #1, please retype.
On this forum we write fractions in the form a/b ,
or something like (4x+6)/(x^2 - 5)
is your original function (x+1)/(x^3)*x ?
that trailing x looks like a typo
for your #2
did you mean "All numbers are under a squareroot" ?
In that case √(64/49) = 8/7 , all others are irrational
On this forum we write fractions in the form a/b ,
or something like (4x+6)/(x^2 - 5)
is your original function (x+1)/(x^3)*x ?
that trailing x looks like a typo
for your #2
did you mean "All numbers are under a squareroot" ?
In that case √(64/49) = 8/7 , all others are irrational
Answered by
Steve
Just massage things to see. For example,
\frac{1}{x^2}-\frac{1}{x^3} = 1/x^2 - 1/x^3
= x/x^3 - 1/x^3
= (x-1)/x^3
Play around with the others, and it should become clear.
D is clearly wrong. 1/(x^3-x) = 1/ [x(x^2-1)]
there's no way you're going to get that to a fraction with only x^3 in the bottom.
\frac{1}{x^2}-\frac{1}{x^3} = 1/x^2 - 1/x^3
= x/x^3 - 1/x^3
= (x-1)/x^3
Play around with the others, and it should become clear.
D is clearly wrong. 1/(x^3-x) = 1/ [x(x^2-1)]
there's no way you're going to get that to a fraction with only x^3 in the bottom.
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