Recent content by skeeterrr
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Is f(x) = g(x+ root(x^2+1)) an even, odd, or neither function?
Yes, moments after I posted this, I realized I did not have to replace g with a logarithm. Thanks for your input!- skeeterrr
- Post #5
- Forum: Precalculus Mathematics Homework Help
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Is f(x) = g(x+ root(x^2+1)) an even, odd, or neither function?
Ok so I took time to examine this problem again. g is a function defined for x>0, and satisfies the following properties for a,b > 0: g(1) = 0, g(a/b) = g(a) - g(b) Having seen that, I can assume that g is a logarithmic function because: let h be log x h is a function defined for x>0, and...- skeeterrr
- Post #3
- Forum: Precalculus Mathematics Homework Help
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Is f(x) = g(x+ root(x^2+1)) an even, odd, or neither function?
Homework Statement Suppose g is a function defined for all x > 0 which satisfies the following properties for all a, b > 0: g(1) = 0 g(a/b) = g(a) - g(b) Determine whether the function f(x) = g(x+ root(x^2+1)) is even, odd or neither, and justify your answer. Homework...- skeeterrr
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- even Function
- Replies: 4
- Forum: Precalculus Mathematics Homework Help
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Sketching Graphs: Homework Answers
Does the graph look like a bunch of triangles on the x-axis?- skeeterrr
- Post #7
- Forum: Precalculus Mathematics Homework Help
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Sketching Graphs: Homework Answers
Homework Statement Suppose we define {x} to be the distance from x to the nearest integer. a) Sketch the graph of f(x) = {x} b) " g(x) = {2x} c) " h(x) = {x} + 1/2{x} d) " all points (x,y) which satisfy {x} + {y} =...- skeeterrr
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- Graphs
- Replies: 7
- Forum: Precalculus Mathematics Homework Help
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Parallelogram is formed by joining midpoints of a quadrilateral
Yes, just do the strategy that you thought out. As for explaining, read the guideline on the MAT137 site, I am working on this problem set as well.- skeeterrr
- Post #8
- Forum: Calculus and Beyond Homework Help
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Parallelogram is formed by joining midpoints of a quadrilateral
You don't need Pythagorean theorem. And the question says show, not prove.- skeeterrr
- Post #4
- Forum: Calculus and Beyond Homework Help
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Area of Window w/ Perimeter 15ft: Function of Side Length
Cool, thanks for your help!- skeeterrr
- Post #5
- Forum: Precalculus Mathematics Homework Help
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Area of Window w/ Perimeter 15ft: Function of Side Length
Thanks for the reply Can I go further with this equation or anything? I'm a little skeptical...- skeeterrr
- Post #3
- Forum: Precalculus Mathematics Homework Help
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Proving Inequality: x1, x2, y1, y2 Non-zero Constants
Homework Statement Let x1, x2, y1, y2 be arbitrary non-zero constants. Let x = x1 / root((x1)^2+(x2)^2) y = y1 / root((y1)^2+(y2)^2) Show that (2(x1)(y1))/(root(x1)^2+(x2)^2)root(y1)^2+(y2)^2)) =< (x1)^2/(x1)^2+(x2)^2 + (y1)^2/(y1)^2+(y2)^2 Homework Equations The...- skeeterrr
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- Inequality
- Replies: 1
- Forum: Precalculus Mathematics Homework Help
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Area of Window w/ Perimeter 15ft: Function of Side Length
Homework Statement A window has the shape of a rectangle surmounted by an equilateral triangle. Given that the perimeter of the window is 15 feet, express the area as a function of the length of one side of the equilateral triangle. Homework Equations A = lw A = 1/2(bh) The...- skeeterrr
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- Area Function
- Replies: 4
- Forum: Precalculus Mathematics Homework Help
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Proving 2xy =< x^2 + y^2 for all real numbers x and y: A Contradictory Approach
Homework Statement Prove for all real numbers x and y that 2xy =< x^2 + y^2 Homework Equations The Attempt at a Solution Well, since this is a problem regarding proof, I thought I would start with a contradictory statement like: 2xy >= x^2+y^2 0 >= x^2-2xy+y^2 0 >=...- skeeterrr
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- Inequality Proof
- Replies: 2
- Forum: Precalculus Mathematics Homework Help
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Parallelogram is formed by joining midpoints of a quadrilateral
"I'm guessing I need to show that the line formed by the midpoints of say, P1P4 and P1P2, is parallel to the line formed by joining midpoints P2P3 and P3P4?" You already figured it out.- skeeterrr
- Post #2
- Forum: Calculus and Beyond Homework Help
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Proving the Geometric Mean Inequality
There are 3 things you need to show: a < root ab root ab < (a+b)/2 (a+b)/2 < b So you have the given condition 0 < a < b (Let's assume that < or > also includes the equal to, since I don't know how to use the notations) since 0 < a < b, a < b as well try working around a < b PS. is this...- skeeterrr
- Post #2
- Forum: Calculus and Beyond Homework Help