Adding fractions and evaluating functions

AI Thread Summary
The discussion focuses on solving algebraic expressions and evaluating functions. The least common multiple (LCM) for the first problem is confirmed as 15a²b², while the correct LCM for the second problem is m²n, not m²(mn). For the third problem, the correct approach involves using the conjugate, specifically multiplying by (sqrt(x + h + 9) + sqrt(x + 9)). Regarding the fourth problem, the relation x = y² is not a function because it yields two values for y (positive and negative square roots) for each x, which violates the definition of a function. The participants clarify their misunderstandings and confirm the correct methods for solving these problems.
Jimmy84
Messages
190
Reaction score
0

Homework Statement


1.) (a + 3b) /3ab + (a^2 b - 4ab^2) / 5a^2b^2

2.) n/(m^2) + 3/mn + 2/m


3.) Given F(x) = square root of x + 9 determine F(x+h) - F(x)/h

4.) say whether or not {(x,y) l x= y^2} is a function.

Homework Equations





The Attempt at a Solution



im having problems to find the least common multiple of the algebric expresions in the first two problems.

I think that on problem one the LCM is 15 a^2 b^2

I think that on the second problem the LCM is m^2 (mn) Is this right?


On problem 3 I don't know how to solve (square root of(x+h) +9 - square root of x+9)/ h

How can I solve that? should I use the square root's conjugate? and if so which is it?

Thanks a lot.
 
Physics news on Phys.org
You're correct about the LCM for problem 1. Recheck what you got for problem 2. You are very close. You have an extra term in it.
 
Jimmy84 said:

Homework Statement


1.) (a + 3b) /3ab + (a^2 b - 4ab^2) / 5a^2b^2

2.) n/(m^2) + 3/mn + 2/m


3.) Given F(x) = square root of x + 9 determine F(x+h) - F(x)/h

4.) say whether or not {(x,y) l x= y^2} is a function.

Homework Equations





The Attempt at a Solution



im having problems to find the least common multiple of the algebric expresions in the first two problems.

I think that on problem one the LCM is 15 a^2 b^2
Right
Jimmy84 said:
I think that on the second problem the LCM is m^2 (mn) Is this right?
That's not the least common multiple. The LCM is m2n. Notice that all three denominators divide this evenly.
Jimmy84 said:
On problem 3 I don't know how to solve (square root of(x+h) +9 - square root of x+9)/ h
Multiply by 1 in the form of the conjugate over itself. I can't tell you any more because what you wrote is ambiguous. Is it f(x) = sqrt(x + 9) or sqrt(x) + 9?
Jimmy84 said:
How can I solve that? should I use the square root's conjugate? and if so which is it?

Thanks a lot.
You didn't ask about 4, but you included it. Graph the relation. If it's a function, no two points will be on the same vertical line.
 
Mark44 said:
RightThat's not the least common multiple. The LCM is m2n. Notice that all three denominators divide this evenly.
Multiply by 1 in the form of the conjugate over itself. I can't tell you any more because what you wrote is ambiguous. Is it f(x) = sqrt(x + 9) or sqrt(x) + 9?

You didn't ask about 4, but you included it. Graph the relation. If it's a function, no two points will be on the same vertical line.

I ve done the first 2 problems.

On problem 3 I meant sqrt(x + 9) sorry.
Is the conjugate of problem 3 sqrt(x + h + 9) + sqrt(x+9) ?

on problem 4 I considered that x = y^2 could be seen as sqrt x = y which is a function however in the book it said that problem 4 wasent a function. So I am a bit confused about it.
 
For 3, multiply your expression by 1 in the form of [sqrt(x + h + 9) + sqrt(x+9)] over itself.

For 4, x = y2 <==> y = +/-sqrt(x). Now do you understand your book's answer?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top