Number 6 Explain how to use two simultaneous linear equations, obtained from two given pointsm to obtain the equation of the line through the two points?

Can you explain this one to me?

I believe I explained this before, so be sure to double-check. Let me know if I didn't.

In a nutshell, this means that you substitute the coordinates of each point into the form y = m x + b. This gives you two simultaneous equations, which you can then solve for m and b.

number 4 At clock times 13.4, 20.1, 26.8, 33.5 seconds, we observe water depths of 28, 20.9, 16, and 13.4cm. At what average rate does the depth change during each time interval? Sketch a graph of depth vs. clock time, which i can do.

Then it says use a sketch to explain why the slope of this graph between 13.4 and 20.1 seconds represents the average rate at which depth changes during this time interval. Why would it be between 13.4 and 20.1 seconds instead of 20.1 and 26.8 which is the middle set of times?

This was your answer

The slope corresponding to a given time interval is rise / run. Since rise is change in depth and run is change in clock time, the slope is equal to (change in depth) / (change in clock time), and this is the definition of the average rate of change of depth with respect to clock time.

The same question could have been asked for the middle interval, or it could have been asked for the last interval. This question could be asked of any interval.

I still do not understand your answer

After I graph all the points on a graph of depth vs. clock time, how would I use that sketch to explain why the slope of the graph between 13.4 and 20.1 represents the average rate at which depth changes during this time interval.

What is the rise between the two points (13.4, 28) and (20.1, 20.9)? What is the meaning of the rise?

What is the run between these two points? What is the meaning of this run?

What therefore is the slope, and what is the meaning of the slope?

Evaulate both the linear and the quadratic depth function at four equally spaced points between t=7 and t=18 seconds. HOw closely does the linear function approximate the quadratic function at each of these times.

If there is no slope for the linear model the line will not change vertically at all, it will just increase as time increases no matter what time you'd use.

that is correct but you should get the linear function, which is found by any method to be y = 0 x + 19.4, or just plain y = 19.4.

For the quadratic depth function the depth would remain at a constant 32 because there is no slope

The quadratic function is depth(t)=.1t^2+-2.5+32. This will not be the same at all 4 points. I am still confused on the very last part of this, what exactly would I do, would I use four evenly spaced points

7, 10.6, 14.3, and 18 and then plug them in for the time?

You don't show all the information on this problem, but for this part of the question, yes, you would evaluate the depth function for these t values.

If a(n)=a(n-1)+b, a(0)=5, then if a(300)=0 what is the value of b?

I'm not sure where to plug in what on this one. Would it be 300=(5-1)+b then just solve to get b to be equal to 296?

What are the laws of exponents?

The laws of exponents are stated at the beginning of the worksheet Final Assignment Prior to Test 1.

The following is copied from that page:

(x ^ a) ^ b = x ^ (ab) e.g., (x ^ 2) ^ 3 = x^2 * x^2 * x^2 = x*x * x*x * x*x = x^6

(x ^ a) (x ^ b) = x ^ (a + b) e.g., (x^2)(x^3) = x*x * x*x*x = x^5

x^0 = 1 (x^a) * (x^0) = x^(a+0) = x^a. Since (x^a) * (x^0) = x^a, x^0 gotta be 1.

x ^ -a = 1 / (x^a) since (x^-a) (x^a) = x^(-a + a) = x^0 = 1, (x^-a) (x^a) = 1 and x^-a = 1 / x^a

Find the first 6 terms of the sequence defined by a(n)=a(n-1)+-1n, a(1)=3 Show that the second difference sequence is constant.

I am just totally lost on this question.

Here is a response I posted last night to a very similar question. See if you can adapt the methods of this solution to the question you ask here. You will get a(2) = 1, a(3) = -2, a(4) = -6, a(5) = -11, a(6) = -17. So the first 6 members of the sequence are

3, 1, -2, -6, -11, -17. The first difference of this sequence is -2, -3, -4, -5, -6. The second difference is therefore -1, -1, -1, -1.

The second difference is unchanging, so it is constant.

The response I posted last night, which goes through the mechanics of finding a(2), a(3), etc., is as follows:

recurrence relation

If a(n+1)=a(n)+.5n with a(0)=2then ....Find a(1) a(2) ect. I am not sure what a need to do to solve this problem....in order for me to find a(1) should I say...

a(1+1)=a(1+1)+.5(1)...

You wouldn't say this because the a to the left of the = sign has argument n + 1, whereas the a to the right has argument n. The two expressions won't have the same argument, since n + 1 is different than n. See more below.

and then try to solve? I dont understand the difference in a(n) and n .....what is the difference? Is it 1+1 and 1?..Thanks.

a(n) means a as a function of n, just like f(x) means f as a function of x.

You are given a(0), meaning the value of the a function at n = 0.

If you let n = 0 you get

a(0 + 1) = a(0) + .5 * 0 = 2 + 0 = 2, so since 0 + 1 = 1 we have a(1) = 2.

Then letting n = 1 you get

a(1 + 1) = a(1) + .5 * 1 = 2 + .5 = 2.5, so since 1 + 1 = 2 we have a(2) = 2.5.

Then letting n = 2 you get

a(2 + 1) = a(2) + .5 * 2 = 2.5 + 1 = 3.5, so since 2 + 1 = 3 we have a(3) = 3.5.

Then letting n = 3 you get

a(3 + 1) = a(3) + .5 * 3 = 3.5 + 1.5 = 5, so since 3 + 1 = 4 we have a(4) = 5.

You continue the process as long as necessary.