Leigh's Responses in WyzAnt Answers

Hi Rhonda!

To deal with the 1 in this integral, you can break up the original integral as follows:

∫[1+√(9-x^2)]dx = ∫1dx + ∫√(9-x^2)dx

*Note: I have indicated these as indefinite integrals because there is no way to nicely format definite integrals on here.

From there, you take the definite integrals each separately as you normally would, and add them together to find the answer.

Hope this helps! Let me know if you have any further questions.

-Leigh

If the equation is as written and the two equal signs are not typing errors, the equation would be considered unsolvable or false.  If this is a typo, I can show you how to solve for the variable 'k' if you post the corrected form of the equation.

The slope of a line (m) is defined as rise (change in y values) over run (change in x values). Mathematically, the slope can be described as:  m = (y2 - y1)/(x2 - x1)

For this question, you can label the points 1 (6,9) and 2 (-3,9). This means that x1=6, y1=9, x2=-3, and y2=9. From there, you plug the x and y values into the equation for slope.  This gives you:

m = (9 - 9)/(-3 -6) which simplifies to

m = 0/-9 = 0

This tells you that the slope equals zero.  This means that the change in y values for the line is zero, and you have a horizontal line, y = 9 (George is correct).  It might be helpful for you to draw a graph with x and y axes, then plot these two points.  When you connect the two points with a straight line, you will see the horizontal line y = 9.

As a general rule:

A line with a slope of 0 is a horizontal line.

A line with an undefined slope (where the change in x values is zero) is a vertical line.

A line with a positive slope (greater than zero) is a diagonal line pointing up on the right side.

A line with a negative slope (less than zero) is a diagonal line pointing down on the right side.

I hope this helps!

To answer your follow up question:

He took the derivative of both sides of the equation tan y = x.

The derivative of tan y = sec^2(y) y' and the derivative of x = 1 

So, you get sec^2(y) y' = 1, and then you can solve for y'