CPhill

√(27) =√(9*3)= √9* √3 = 3√3

First, if x = 7 and y = 3, you couldn't have a resultant of 8. The resultant would be √(7² + 3²) = √ (49 + 9) = √(58)

To find theta, use the tangent inverse. So we have tan^-1 (3/7)  ≈ 23.2°

In order, they would be:

-4.45 min, -1.8 min, -(1 + 2/3) min, -1.375 min

A fuction....call it  y = f(x)..... reflected about the y axis has the form, y = f(-x).

So, if y= f(x) = (1/2)x + 8, then the reflected function is  y = f(-x) = (-1/2)x + 8.

Ah...Melody beat me to it.....rats!!!!!!

There is a "formula" for this. But notice when we add the first number, 1, and the last number, 100, we get 101. And addnig the next-to-the-last number, 99, and the second number, 2, we also get 101. And, continuing in this manner, we have 50 pairs of numbers that each add to 101. Therefore, 50 * 101 = 5050.

Distance = √[(5 - 1)² + (8 - 3)²] =  √ (4² + 5² ) = √(16 + 25) = √41

The Slope is   [8 - 3 ] / [5 - 1 ] = 5 / 4

3x^2(4x^3)^3 =

(3x^2) * (4x^3)^3 =

(3x²) * 4³ * (x³)³  =

(3x²) * 64 * x⁹ =

3*64* x² * x⁹ =

192 x¹¹

I'll do the first one, the others are similar....(only the values have changed to protect the innocent...)

For, 1.5, we have the point on the curve

f(1.5) = 1 /(1 - 1.5) = 1 / (-.5) = -2

So we have two points  (2, -1) and (1.5 , -2)   and the slope of the secant line is  [-2 - (-1)] / [1.5 - 2] =         -1 / -.5    = 2

Here's the graph, here......note that the secant line drawn with these two enpoints would have a positive slope.

The point of this exercise is that is to show that the closer we get to x = 2, the more the secant line models the tangent line to the function at that point (i.e., the "derivative" at that point)

Hope this helps...

I do this one a little differently from Melody.......but the answers are the same.

First...note that S30°W = 240°    .....and S20°E = 290°

So we have two vectors here with the following components....

<300cos240°, 300sin240° >   and < 50cos290°, 50sin290° >

And adding the first to components of the two vectors we get.....300cos240° + 50cos240° = -132.8989928337

And adding the second components of the two vectors we get.....300cos290° + 50cos290° = -306.7922521745

And to find the magnitude of the resultant vector, we have.......

√[( -132.8989928337)²  + ( -306.7922521745)² ] = √(111783.6282905138439343) ≈ 334.34 km/h

And to find the resultant heading, we have

tan-1 [(-306.7922521745) / ( -132.8989928337)] = tan-1 (2.308461829792775) = 66.578275555105°

And to find the resulting heading, we have.... S(90 - 66.578275555105)°W  ≈ S(23.42)° W

I actually like this method better because I always have a little trouble visualizing the "parallelogram law."  In effect, I just let the math "do the work" for me and I don't have to worry about drawing any pictures.

But, it's purely a matter of preference....!!!!