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Friday, December 6, 2013

Student Problem #6: Unit K Concept 10: Writing a repeating decimal as a rational number using geometric series


The trickiest part of this concept is people think 29/99 is the answer but the forget they have a whole number in the front and they have to add that as well to find it all. They also forget that you can simplify if you can but since its a big number people are too lazy to check if you can so you just leave it like that and then you get the answer wrong but always check if you can simplify. Always check your math go back to if if you think you got the fractions wrong or the math wrong. The HARDEST part of this concept is that you CANT USE A CALCULATOR!!!   


Sunday, November 17, 2013

SP#5 Unit J Concept 6 Partial Fraction Decomposition with repeated factors



                    The tricky part of this was when you get your answer when you decompose it since we got a zero in out matrix we dont have to put it as your answer there is no need to. make sure to add subtract and multiply right or your whole answer will be wrong. this second picture is how you can check if your decompose answer was correct if not you probably messed up on your matrx by not plugging in the numbers right in your calculator to get your rref (reduced row echolen form) hope this helped.

Thursday, November 14, 2013

SP#4 Unit J Concept 5 Partial Fraction decomposition with distinct factors



                  The trickiest part is composing it because some people might think that all you got to do is distribute the factors and they only do it in ones thinking that's the answer but you have to do it in all 3 and after that combine the like terms to get the answer. For the second pick you have to be very careful when you distribute and put the like terms where they have to be. For the last pic don't be in a rush when plugging in the numbers because people usually get it wrong because they put the numbers all quick without checking or if its not that you made a mistake in the (second picture) that might of gave you the wrong answer.



Monday, November 11, 2013

SV#5 Unit J Concept 3-4: Solving three- variable systems with Gaussian Elimination and Solving non- square systems


     The trickiest part for doing matrx is the math. As in the multiplying  and adding or subtraction to get a zero on where they have to be. Also as you can see here I got 0001 but I got 00011 and that is because in the calculator I put rref as the reduced form thats why I got confused but my answer is still right. Well hope this helped thanks for watching. :D

Sunday, October 27, 2013

SV#4 Unit I Concept 2 - Graphing Logarithmic Functions

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     The trickiest part of this concept is putting it in the calculator right because parenthesis matter a lot if you forget your answer will be totally different. The y intercept is tricky too because when it comes with the log people think its a times not an exponent so they get it wrong and when they see that its an exponent they get stuck there. When that happens you have to use the natural log its in your calculator and you divide them. So at the numerator it should be the exponent and at the denominator it is the base.


Wednesday, October 23, 2013

SP#3 Unit I Concept 1:Graphing exponential functions and identifying x-intercept, y-intercept, asymptotes, domain, range



      You first write out what a, b, h, and k is. When you do that you find the asymptote which is y=k. you then find the x-intercept which is y=0. you add the coefficient first then you divide the first number to leave the exponent alone. in this case if you have a negative ln ( which i didnt put there sorry) it should be ln of 2 divided by ln of -8 but whenever theres a negative the calculator will tell its a imaginary number.so it will be undefined and your answer is you dont have an x-intercept.
      Then you move on to the y-intercept and that you equal x to 0. then you just use PEMDAS pretty much. When you go to your graph you plug in the asymptote which is -2 and put it as a horizontal asymptote you plug in the y-intercept you got. You go back to your key points and your third key point is always h even if you found it on your x- intercept. you look for the numbers you dont have in there and put it on the key points. to see how it will look like you put the original equation dont forget its an exponent so you put a ^ before you put the (x+1) and the put 2nd table and you will find your points then you plug them in and it should look like that.




Tuesday, October 15, 2013

SV #3 Unit H Concept 7- Finding Logs Given Approximation

           

      So first you have the clues to find your treasure which is the color green. You first multiply to see if you can reduce that to be able to get some of your clues. When you do find your clues you put them in log form. The tricky part of this is the log base 9 to 7 to the 3rd power you can change that to make it look easier to you and make you understand it as well.
      You would put the exponent in the front which is 3 so it would be 3 log base 9 to 7. Then you look at the logs and see where should go where so you can put the right letter where it needs to be. Don't forget when you have 3 log base to 7 in this case log base 9 to 7 is R BUT don't forget the 3 in the front so it should be 3R plus your other answers that you had. ALSO DON'T FORGET THAT SOME ARE - BECAUSE OF THE DIVISION SIGN. Thanks for Watching!!!! :D

Sunday, October 6, 2013

SV#2 Unit G Concept 1-7- Finding all parts and graphing a rational function

   

          Hi my name is Derik and this is Unit G Concept 1-7. This Video will show you how to Figure out the Vertical and Slant Asymptotes and as well as the holes if there is something canceled out in the equation if not then there will not be a hole at all. You will also learn how to graph them as well and how they are  suppose to look like.
          The tricky part for me was the calculator because in this problem it looked the right side was a positive infinity but at the same time it also looked like a negative infinity. It just got very confusing for me at that point now. I go know though that the extra numbers you have to to plot more to make a line in the calculator i know how to do that. Other than that if you follow these steps and listen you will get it. Thanks for watching..

Sunday, September 29, 2013

SV#1 Unit F Concept 10- Finding all real and imaginary zeroes of a polynomial




       First you find the p's and q's then divide p/q and find the possible zeroes. right after that you use the decartes rule of sign to which how many possible zeroes you will get. Use synthetic division until you find a zero hero you keep going until your able to distribute it or you cant anymore.
      Then use the quadratic formula -b+or- the square root of b^2 -4(a)(c)/ 2(A). you will either get a perfect square or a imaginary number but you usually always get an imaginary number. You find the zeroes which are the numbers that equaled to zero in your synthetic division and quadratic formula you put and x-then your quadratic answer your suppose to get two different answers. Then the factors is the equation you reduced while doing the synthetic division and the quadratic but you dont put x- on it. Thanks for watching!!!!!!!

Sunday, September 15, 2013

SP#2 Unit E Concept 7- Graphing a polynomial and identifying all key parts




       This is my own equation that i made up. what you do is you factor x^2-4x+5. You should of gotten (x-5) and (x+1) if you didn't go back and try it again and see what you did wrong. So those two points should be where it says Factored Equation. To find your End Behavior you go back to the original equation and see the x^2 only the exponent is to see if it is even or odd in this case it is even and the x to see if it is positive or negative in this case it is positive so you sing the song and it should be the first one you say which both ends go up.Then you write the limited notation since both go up the f(x) of both should be positive infinity.
      To find the x-intercepts you go back to the factored equations and you equal them to 0. You should have (5,0)M1 because there is only one of those and (-1,0)M1 because it also only has one there's no other one that is the same as that one. Then you graph it you put the 5,0 and since it is multiplicity of 1 it is going thru the graph then you put the other one -1,0 and since it is also the multiplicity of one it is also going thru. To find the y-intercept you go back to the original equation and substitute the x's to 0's and you should of gotten 5 but to put it as a y-intercept it should be (0,5) then you put that one in as well to see how low its going to go. and that's how the graph should look like.

Sunday, September 8, 2013

SP#1 Unit E Concept 1-Graphing a quadratic and identifying all key parts



       You first write the equation of course. you equal it to zero. then subtract 12 on both sides. you find what they have in common then you take it out and try and find the third number that you are able to factor. then multiply on the other side by the number you can take out from the two numbers to make it smaller. that will be your parent function equation that's at the top right corner. then you use (h,k) to find the vertex. for the y-intercept you plug in for x to the original equation and you find the y-intercept.to find the x-axis you look at all the points you put and draw a line to the middle of it and that would be the x-axis.
      Then you put the parent function equation and you divide it after that you square root it to delete the squared  and whatever is left should be your x-intercept and if you have an i then you are going to have an imaginary number number which is pretty much nothing so you will have to have 3 points in your graph not 5 that's only if you have an imaginary number but in this cause i don't.