Recent content by adillhoff

Homework Statement (a) The quadratic Chebyshev approximation of a function on [-1, 1] can be obtained by finding the coefficients of an arbitrary quadratic y = ax^2 + bx + c which fit the function exactly at the points (-sqrt(3)/2), 0, (sqrt(3)/2). Find the quadratic Chebyshev approximation of...

Wow that seems so obvious now. I completely overlooked it. Thank you so much for the help.

Homework Statement Define the function at a so as to make it continuous at a. f(x)=\frac{4-x}{2-\sqrt{x}}; a = 4 Homework Equations \lim_{x \rightarrow 4} \frac{4-x}{2-\sqrt{x}} The Attempt at a Solution I cannot think of how to manipulate the denominator to achieve f(4), so I...

Of course. I knew I missed something simple. Thanks for the reply.

Homework Statement Solve the compound interest formula for t by using natural logarithms. Homework Equations A=P(1+\frac{r}{n})^{nt} The Attempt at a Solution I start by dividing both sides by P. I then take the natural log of both sides and end up with ln(\frac{A}{P})=nt *...

Absolutely. Thank you so much for the help. I ended up with x=\frac{ln\frac{y+1}{y-1}}{20}

Homework Statement Use Natural Logarithms to solve for x in terms of y y = \frac{e^{10x}+e^{-10x}}{e^{10x}-e^{-10x}} Homework Equations I am not too sure. The Attempt at a Solution I multiplied both sides by the denominator first. Then I multiply by an LCD of e^{10x} I end up...

Yes that was a typo. I also mean to say that the midpoint of PQ = (-13/2, 1). I completely over-analyzed this problem as I do with many problems. I took your advice and found point R(-11, -23) by taking the difference between the x- and y-coordinates of P & Q. Thanks for your help!

Homework Statement Given P(-5, 9) and Q(-8, -7), find a point R such that Q is the midpoint of PR Homework Equations d = \sqrt{(x+8)^2+(y+7)^2} The Attempt at a Solution Because Q is the midpoint of PR, I know that d(P, Q) = d(Q, R). I also know that d(P, R) = d(Q, R), which is...

I wasn't quite sure if I could solve it that way, but it makes a lot of sense. After solving for a using the quadratic formula I get a = 1 - 2 and a = 1 + 2. The problem stated that the point is in the third quadrant which means a < 0. So I am left with a = 1 - 2 = -1. The answer ends up...

Homework Statement Find the point with coordinates of the form (2a, a) that is in the third quadrant and is a distance 5 from P(1, 3) Homework Equations \begin{distance} d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \end{distance} The Attempt at a Solution To be quite...

That is a great reply. I've gotten a little further, but am stuck again. I used the Quadratic Formula to solve this and came up with y = (-1/m +- sqrt((1/m^2) - (4b - m))) / 2. I know that the problem has only one solution so sqrt((1/m^2) - (4b - m)) = 0. I am not sure what to do from this...

I do have the figure. It is simply a graph of x = y^2 and another line intersecting x = y^2 at (4, 2). I am given no other information on the line that is intersected x = y^2. I am instructed to find the slope of the line so that it only intersects x = y^2 at (4, 2). I have the answer to this...

Homework Statement Shown in the figure is the graph of x = y^2 and a line of slope m that passes through the point (4, 2). Find the value of m such that the line intersects the graph only at (4, 2) and interpret graphically.Homework Equations x = y^2 y = mx + bThe Attempt at a Solution Since...

Thank you for the quick reply. I did run into one more issue with this problem. After plugging in the values into the quadratic formula, I run into this step: (4y +– sqrt(32y^2 - 16)) / 8 There are numbers inside the sqrt that I could pull out: 32 and -16. My next planned step it to re-arrange...