Week four, we learned how to solve for domain, range, find f of x, g of x, how to sketch f of x, g of x, and inverses. the thing that i caught on to very quickly is f of x and g of x. here are some examples of how to f and g of x problems:
f(x)=x^2+1 g(x)=x-3
1:(g+f)(x)
=x^2+1+x-3
=x^2+x-2
when solving for this equation all you do is solve by adding f(x) and g(x)
2:(f-g)(0)
=x^2+1-(x-3)
=x^2+1-x+3
=x^2-x+4
=4
when solving for this equation all you do is solve by adding f(x) and g(x)
something i still don't get is the domain and range i still don't know when to use brackets and parenthesis and when to put the negative and positive infinity. just some practice might do the trick if you know anything that might help let me know..
Saturday, September 12, 2009
REFLECTION #4
What a long week this has been!! I don't know about you guys, but to me this week went on forever even though we were off Monday. Anyway, this week was a little tough and I'm still catching on, but I know I'll get it soon. And I have to say that I'm glad we had those extra domain and range worksheets to do while Ms Robinson was out. It helped me to better understand the concepts of domain and range. The problems are much easier for me to do now since I had a little more extra practice.
What we learned this week was about functions, how to find symmetries, and inverses. I basically understand the concept for all of them, but all the information is still kinda processing in my head. It's a little hard, but the steps for solving are really easy to follow.
What I thought was the easiest thing that we learned this week is solving functions. (I don't know what the proper name for this is at the moment, but I'm talking about the things with the f(x) and f(g(x)). I think you know what I mean.) I think this is the easiest thing for me to explain so here it goes..
Okay, for functions you'll always get one, two, or more equations>>f(x)>> that you will use to solve individual problems like this for example:
Example 1
f(x)=x+4 g(x)=2x-3
(these are the two equations that you are given^)
a.) find (f + g) (x) >>By the plus sign, this notation tells you that you're going to add your two equations together like this:
x + 4 + 2x - 3
= 3x - 1
b.) f(g(x)) >>This is a composite function. What this function is telling you to do is plug in the "g" equation every time you see an x in the "f" equation. like this:
(2x-3) + 4
(*you always want to put what you plugged in in parenthesis in case you have to distribute a negative.)
then you get >> 2x-3+4
= 2x + 1
c.) find g(6) >> this notation has a number in parenthesis. Whenever you see a number in parenthesis you know that your answer should be a number.
So the function g(6) tells you to plug in the number 6 every time you see an x in the "g" equation.
so you would get >> 2(6) - 3 = 9
**Now for what I didn't understand this week. First off, I had a little trouble understanding the "pictures" part when you have to reflect on the x-axis and y-axis and you have to switch which way the picture is going and whatnot. (sorry it's hard to explain) And also, I get confused a lot with inverses. I mean I understand the basics and how to do simple things with them. But for the directions, "Find y^-1 and prove that it's an inverse" for a problem like this>> f(x)=4x-2 I get stuck and I don't know where to start without staring at my notes for ten minutes. If anyone would like to explain these things to me that would be great :)
Well, overall I thought this week wasn't too bad. I just have to work on the new things we just learned.
What we learned this week was about functions, how to find symmetries, and inverses. I basically understand the concept for all of them, but all the information is still kinda processing in my head. It's a little hard, but the steps for solving are really easy to follow.
What I thought was the easiest thing that we learned this week is solving functions. (I don't know what the proper name for this is at the moment, but I'm talking about the things with the f(x) and f(g(x)). I think you know what I mean.) I think this is the easiest thing for me to explain so here it goes..
Okay, for functions you'll always get one, two, or more equations>>f(x)>> that you will use to solve individual problems like this for example:
Example 1
f(x)=x+4 g(x)=2x-3
(these are the two equations that you are given^)
a.) find (f + g) (x) >>By the plus sign, this notation tells you that you're going to add your two equations together like this:
x + 4 + 2x - 3
= 3x - 1
b.) f(g(x)) >>This is a composite function. What this function is telling you to do is plug in the "g" equation every time you see an x in the "f" equation. like this:
(2x-3) + 4
(*you always want to put what you plugged in in parenthesis in case you have to distribute a negative.)
then you get >> 2x-3+4
= 2x + 1
c.) find g(6) >> this notation has a number in parenthesis. Whenever you see a number in parenthesis you know that your answer should be a number.
So the function g(6) tells you to plug in the number 6 every time you see an x in the "g" equation.
so you would get >> 2(6) - 3 = 9
**Now for what I didn't understand this week. First off, I had a little trouble understanding the "pictures" part when you have to reflect on the x-axis and y-axis and you have to switch which way the picture is going and whatnot. (sorry it's hard to explain) And also, I get confused a lot with inverses. I mean I understand the basics and how to do simple things with them. But for the directions, "Find y^-1 and prove that it's an inverse" for a problem like this>> f(x)=4x-2 I get stuck and I don't know where to start without staring at my notes for ten minutes. If anyone would like to explain these things to me that would be great :)
Well, overall I thought this week wasn't too bad. I just have to work on the new things we just learned.
Friday, September 11, 2009
Reflection 4
WOW! What a week. I found that this week was sorta difficult. Although we didn't learn as much as normal, you had to try and understand more steps. This week we learned about domain, range, functions, finding any symmetry, and inverses. What I found was the easiest to learn was finding the symmetry.
EXAMPLE: y=x^3=4x
a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No
b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No
c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No
d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes
I also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti
EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)
EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)
What I am having the most trouble understanding is the inverses. I get most of the concept, but I'm still getting confused. Can anyone help me with that??
EXAMPLE: y=x^3=4x
a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No
b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No
c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No
d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes
I also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti
EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)
EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)
What I am having the most trouble understanding is the inverses. I get most of the concept, but I'm still getting confused. Can anyone help me with that??
Wednesday, September 9, 2009
umm.....
anybody wanna help me figure out whether the thing goes up or down on a absolute value domain and range
Monday, September 7, 2009
This week was alright because we took the chapter 2 test over which helped me a lot. Now I understand chapter 2 way better. We also took a quiz that I did the best on. I realized that today is not sunday and reflections are due on sunday. So im hoping brob takes this under the circumstances of a three day weekend. One thing that I did get right away was inequalities from chapter 3.
1 .2x - 3> 3
+3 +3
2. 2x>6
2 2
3. the answer x>3
The one thing i still dont get though is the word problems and how you break them down into knowing what to do with all the numbers they give you. Hope this week will be good.
1 .2x - 3> 3
+3 +3
2. 2x>6
2 2
3. the answer x>3
The one thing i still dont get though is the word problems and how you break them down into knowing what to do with all the numbers they give you. Hope this week will be good.
reflection 3
This week was a review of Chapter 2 and some of chapter 3, such as solving inequalities:
1.) x + 4 >/= 12
x > or = 8
2.) -2x - 6 < 20
-2x < 26
x > -13
3.) x + 7 < or = 24
-24 < x + 7 < 24
-31 < x < 17
im having some trouble with some word problems, and problems like #s 5 and 11 on the packet
5.) y=2x^3 -7x + 2
11.) y=3x^4 + 13x^3 + 15x^2 -4
some help would really be usefull, thank you...
1.) x + 4 >/= 12
x > or = 8
2.) -2x - 6 < 20
-2x < 26
x > -13
3.) x + 7 < or = 24
-24 < x + 7 < 24
-31 < x < 17
im having some trouble with some word problems, and problems like #s 5 and 11 on the packet
5.) y=2x^3 -7x + 2
11.) y=3x^4 + 13x^3 + 15x^2 -4
some help would really be usefull, thank you...
Reflection 3
Well, aint it a surprise, another week of school is gone and another week of math is over with. That was a quickie, and its gonna be a retarded crazy week. I mean really, crazyness is known alot around riverside, but last week was kind of chill for once. The only thing that happened that was new last week was the equations and inequalities, but we learned that last year, and this was just another review of it. And THANK YOU Brob for lettin us retake the chapter 2 test. The only thing that was wrong with that test was the length of time it took for some of us, and that horrible typo on one of the problems. But luckily, since you found the typo, you let us get 10 extra minutes on the test on friday, and for some of us, it saved us big time. Like me, i couldnt finish the test because of the typo, but you found it and made up for your mistake, and it benefitted indirectly :D. But wow, this week's gonna be retarded. We got so many tests, assignments, etc. due and to do this week, its gonna be really hectic, so you better get ready for some busy ppl to teach. But back to math, unfortunately, all of the problems on the test were on stuff like the homework, and those of which ive needed help to understand how to do them, which is why i use skype (advertisement agian XD). Yea, i used skype to understand how to do many of the problems, stephen, sparky, chase, and justin help out alot on there.
Well, one thing i do understand is the equations and inequalities.
Its only 10x>3
then, x>3/10
its too simple
but i didnt understand how to do stuff that you had to use multiple ways of solving to solve problems, like 11 on the packet
Well, one thing i do understand is the equations and inequalities.
Its only 10x>3
then, x>3/10
its too simple
but i didnt understand how to do stuff that you had to use multiple ways of solving to solve problems, like 11 on the packet
reflection #3
this week we worked inequalities in chapter 3 and had a review of chapter 2. i found inequalities to be easy and the chapter 2 review helped a lot. inequalities are what i found easiest.
|x-2|<6
-6
-4
inequalities arent that bad, but what im still having some issues with are word problems from chapter 2.
Reflection #3
This week was mainly just a review but we were taught something semi new which was not on our first chapter 2 test..
-Sloving inequalities
x-9>-18
-first you would want to get rid of the -9, so you would +9 to both sides..
then you would end up with x>-9
Remember when solving inequalities is that if you multiply or divide by a negative number, you must flip the sign!!
The only thing I'm mainly having trouble with still are those darn word problems :/
reflection 3
So this week we had a test on 2 and 3 that i think i didnt do too bad on. I thought the week went by pretty fast (as usual) and I thought that generally everything was easy. I think i might need a little more practice on all the stuff overall though. We went over stuff on Domain and Range. Now this stuff was pretty easy but it took me a little getting used to. Really the only trick in it is learning and memorizing the notes that Brandi gave us. We had to learn stuff like finding the Domain and Range of fractions. To do this we have to
a. set the bottom equal to zero
b. solve the problem for x
c. set up all of the intervals
(-inf. , ____) (___,___) (____,-inf.)
All this is actually very easy and you should be able to pick up on it very quickly.
questions on our packet!
From:
Ashleigh, Ashton, and Connor....
#'s:
5. y=2x^3-7x+2
-we're stuck after we find negative 2; our second equation is unfactorable.
11.y= 3x^4+13x^3+15^2-4
17.y=2x^4-x^3-7x^2+x+2
Sunday, September 6, 2009
Reflection #3
Well for starters, Chapter 3 wasn't that difficult! :)
Also, using this week as a review of Chapter 2 helped me out tremendously! :)
I understood practically everything but i'll just explain solving inequalities:
First, isolate the x, if the leading coefficient is negative(which in this case, it is) you would have to change the signs.
ex: -4x<12
x>-3
In more complex inequalities, or absolute value inequalities, such as AND and OR equations, you're solution would be two answers.
you'd solve like this:
If the equation is greater than or equal to, it is an OR equation and vice versa would be an AND equation.
ex of an OR:
|4x-12|(gt=)7
FIRST, FORM YOUR TWO EQUATIONS:
4x-12(gt=)7 OR 4x-12(lt=)-7
NOTICE THAT YOUR SECOND EQUATION IS THE OPPOSITE OF THE VALUE AND OPPOSITE OF THE CONSTANT.
second, solve for x:
x(gt=)19/4 OR x(lt=)5/4
ex of an AND:
|4x-12|(lt=)7
4x-12(lt=)7 AND 4x-12(gt=)-7
x(lt=)19/4 x(gt=)5/4
------------------------------------------------------------------------------------------------
Now, i understand this is a part of section 2, but this is still a little iffy in my mind...
i wasn't here when we got the notes and for some reason i can't find them anywhere!
does anyone mind if i ask them to please explain?! :)
thankssss!
Also, using this week as a review of Chapter 2 helped me out tremendously! :)
I understood practically everything but i'll just explain solving inequalities:
First, isolate the x, if the leading coefficient is negative(which in this case, it is) you would have to change the signs.
ex: -4x<12
x>-3
In more complex inequalities, or absolute value inequalities, such as AND and OR equations, you're solution would be two answers.
you'd solve like this:
If the equation is greater than or equal to, it is an OR equation and vice versa would be an AND equation.
ex of an OR:
|4x-12|(gt=)7
FIRST, FORM YOUR TWO EQUATIONS:
4x-12(gt=)7 OR 4x-12(lt=)-7
NOTICE THAT YOUR SECOND EQUATION IS THE OPPOSITE OF THE VALUE AND OPPOSITE OF THE CONSTANT.
second, solve for x:
x(gt=)19/4 OR x(lt=)5/4
ex of an AND:
|4x-12|(lt=)7
4x-12(lt=)7 AND 4x-12(gt=)-7
x(lt=)19/4 x(gt=)5/4
------------------------------------------------------------------------------------------------
Now, i understand this is a part of section 2, but this is still a little iffy in my mind...
i wasn't here when we got the notes and for some reason i can't find them anywhere!
does anyone mind if i ask them to please explain?! :)
thankssss!
Reflection#3
This week we had a test on the combined materials in chapter 2 and in chapter 3. I thought the week went by pretty fast and i also thought that the stuff we went over was pretty much easy. Ever now and then i would forget a step or two but after working on that graded project and alot of homework i believe i can solve any problem she gives me quite easily. Although the first test i took wasnt to good u were nice enough to give us a retake and now i understand everything pretty good. This week we reviewed over synthetic division, finding the sum and product of the roots, solve anything bigger than a quadratic. To do that u have to solve by graphing, quadratic form, make g=x, or the rational root theorem, or grouping. Doing a problem by grouping is definitly the easiest.
Example: 3x^3-16x^2-12x+64, now take out like terms and get x^2(3x-16)-4(3x-16)
now you will have (x^2-4)(3x-16), then solve for x and you should get three answers which are (2,0)( -2,0)(16/3,0), remember to have answers in point form. I really didnt like doing rational root theorem because it takes way to long, and i can easily make a mistake on my calculater. Now solving inequalities was super easy, its like solving an equation. If you cant do that get out of advance math. The new stuff that we started in chapter 4 seems a little tricky but all u got to do is study and remember the steps. As of right now i'm not having trouble with any problem. Hope this week goes by goo.
Example: 3x^3-16x^2-12x+64, now take out like terms and get x^2(3x-16)-4(3x-16)
now you will have (x^2-4)(3x-16), then solve for x and you should get three answers which are (2,0)( -2,0)(16/3,0), remember to have answers in point form. I really didnt like doing rational root theorem because it takes way to long, and i can easily make a mistake on my calculater. Now solving inequalities was super easy, its like solving an equation. If you cant do that get out of advance math. The new stuff that we started in chapter 4 seems a little tricky but all u got to do is study and remember the steps. As of right now i'm not having trouble with any problem. Hope this week goes by goo.
Reflection #3
This week was really just a review of chapter 2 but we did learn solving inequalities, and I know how to do those pretty well:
-Sloving inequalities is very similar to solving regular equations when it comes to the algebraic part of it.
- For ex: x+4>5.......you would solve by subtracting 4 from each side giving you x>1
If you had an equation x+4=5 you would do the same and get x=1
- The main difference to remember when solving inequalities is that if you multiply or divide by a negative number, you must flip the sign:
Ex: -4x>8....divide by -4 and get x<-2, note that the sign was flipped, and it works the same when multiplying by a negative number.
- In the case of absolute value inequalities, set it up like this:
Equation: /x+2/>4............set it up x+2>4 an x+2<-4, then solve both inequlities.
_________________________________________________________
The only thing I had trouble with is the inequalities that like:
3/5x +3/10>11/12
I dont know where to start and these are killing me. Please and thanks.
-Sloving inequalities is very similar to solving regular equations when it comes to the algebraic part of it.
- For ex: x+4>5.......you would solve by subtracting 4 from each side giving you x>1
If you had an equation x+4=5 you would do the same and get x=1
- The main difference to remember when solving inequalities is that if you multiply or divide by a negative number, you must flip the sign:
Ex: -4x>8....divide by -4 and get x<-2, note that the sign was flipped, and it works the same when multiplying by a negative number.
- In the case of absolute value inequalities, set it up like this:
Equation: /x+2/>4............set it up x+2>4 an x+2<-4, then solve both inequlities.
_________________________________________________________
The only thing I had trouble with is the inequalities that like:
3/5x +3/10>11/12
I dont know where to start and these are killing me. Please and thanks.
Reflection 3
Well, this week did seem a lot easier than last week. Inequalities are very simple. Here is an example: 2x-3<3 Move all numbers to one side; therefore you add three to both sides. 2x<6 Next divide both sides by two to solve for x. Your answer is x<3. That is very simple. Now absolute value inequalities are a little more difficult. If the sign is greater than/ greater than or equal to, it will be an or problem (meaning your solution will be x or x) and if the sign is less than/less than or equal to then it will be an and problem (meaning your solution will be x and x). Those are the two types of absolute inequalities. Here is an example: |2x + 3| < 6 Step 1: 2x + 3 < 6 and 2x + 3 > -6 2x < 3 and 2x > -9 x < 3/2 and x > -9/2. The test was not so bad even though most of it was the same material from last week. I found that solving polynomials seemed to be a breeze. The three methods are 1: factor by grouping, 2: Quadratic form, 3: Rational root theorem which we have gone over in class multiple times. One small thing that may have gotten anyone who did not study is the product and the sum of the roots. Sum of the roots = -2nd coefficient/leading coefficient. Product of the roots= constant/leading coefficient (if the degree is even) or –constant/leading coefficient if the degree is odd.
Reflection #3
Well, for Ch. 3 I understood everything so I'll just explain domain and range.
For domain and range, you must remember the rules:
the only time you don't find the range is when the problem is a fraction. When working with a polynomial, the domain will always be (-infinity,infinity) but for range, it depends: if the leading exponent is an odd number, the range is the same as the domain, but if the leading exponent is 2, then it's [vertex,infinity) if it's positive and if it's negative, it's (-infinity,vertex]. (The vertex is -b/2a). For square roots:
1)set the inside = 0
2)set up a # line
3)try values on either side (like graphing in Ch.2)
4)eliminate anything -ve
5)set up intervals (infinity, )( , )...
( ,infinity)
and for range, just graph
Now, for fractions:
1)set bottom = 0
2)solve for x
3)set up intervals
For absolute value: the domain is always (-infinity,infinity) and the range is [shift,infinity) if positive and if negative its (-infinity,shift). (Shift is when the entire graph moves to however many spaces the problem implies such as {x+1}-1, the last -1 indicates that the graph moves on the y-axis to -1, so the range would be (-1,infinity).)
Like I said, there wasn't anything I didn't understand in Ch.3, but on the adv. math packet, I don't understand how to do number 18 or 22.
#18) y=x^3-6x^2
#22) y=x^5-3x^4+3x^3-5x^2+12
for 22 I know you have to work the rational root theorem, but I tried it and it didn't work for some reason.
Anyone out there want to help?
For domain and range, you must remember the rules:
the only time you don't find the range is when the problem is a fraction. When working with a polynomial, the domain will always be (-infinity,infinity) but for range, it depends: if the leading exponent is an odd number, the range is the same as the domain, but if the leading exponent is 2, then it's [vertex,infinity) if it's positive and if it's negative, it's (-infinity,vertex]. (The vertex is -b/2a). For square roots:
1)set the inside = 0
2)set up a # line
3)try values on either side (like graphing in Ch.2)
4)eliminate anything -ve
5)set up intervals (infinity, )( , )...
( ,infinity)
and for range, just graph
Now, for fractions:
1)set bottom = 0
2)solve for x
3)set up intervals
For absolute value: the domain is always (-infinity,infinity) and the range is [shift,infinity) if positive and if negative its (-infinity,shift). (Shift is when the entire graph moves to however many spaces the problem implies such as {x+1}-1, the last -1 indicates that the graph moves on the y-axis to -1, so the range would be (-1,infinity).)
Like I said, there wasn't anything I didn't understand in Ch.3, but on the adv. math packet, I don't understand how to do number 18 or 22.
#18) y=x^3-6x^2
#22) y=x^5-3x^4+3x^3-5x^2+12
for 22 I know you have to work the rational root theorem, but I tried it and it didn't work for some reason.
Anyone out there want to help?
Reflection # 3
This week went by very quickly. I'm glad B-Rob decided to redo the Chapter 2 test and combine it with the Chapter 3 test. The test was easier in that there were a lot of repeat problems on it, and I understand all of the material now. After completing that packet for homework, I pretty much perfected solving for x. Even though we didn't learn that much this week, I remember the inequalities and finding the domain and range. I'm going into detail about inequalities.
To solve an inequality, you treat it as a regular equation, but when dividing by a -ve you must switch the sign.
Example:
-2x + 5 lt 7
-2x lt 2
x gt -1 (You switch the sign because you have to divide by -2)
Absolute value inequalities, on the other hand, are a little bit different. You have to come out with two answers. If the initial sign is greater than or = to, or just greater than, it is called an or problem. If the initial sign is less than or = to, or just less than, it is called an and problem.
First, you isolate the absolute value, then you set up two different problems out of the initial problem. One problem is the original. For the second problem, you switch the sign of the value, and switch the sign of the constant on the other side.
Example:
|2x + 3| lt 6 (the absolute value is already isolated, so form two equations)
2x + 3 lt 6 and 2x + 3 gt -6
2x lt 3 and 2x gt -9
x lt 3/2 and x gt -9/2
Now before you make that your final answer, you have to plug back in and check. Whichever answers work, you circle or box.
3/2 does not work because if you plug it in, 6 lt 6.
-9/2 does work because -6 lt 6.
So, your answer would simply be -9/2.
Reflectinon 3
So i think this week was my favorite week of adv. math because I actually understood almost everything we learned. For our Chapter 3 test, we combinded the Chapter 2 test with Chapter 3, so we could review some more on Chapter 2. I'm glad reviewed Chapter 2 some more because I'm starting to get better at the word problems and sketching polynomial functions. What I understood the most this week was the Inequalities, it had an easy concept to grasp.
Example 1.) 7x-12<9
FIRST: Add 12 to each side. 7x-12<9
+12+12
This then gives you 7x<21
SECOND: Divide each sid by 7. 7x/7<21/7
This gives you the final answer of x<3
Example 2.) Absolute Value Inequalities abs. value of 3x-9>4
FIRST: This is an "or" equation because it is greater than (if it was less than it would be
"and") 3x-9<-4 or 3x-9>4
SECOND: Add nine to each side of the equations.
3x-9<-4 or 3x-9>4
+9 +9 +9+9
This gives you 3x<5>or3x>13
THIRD: Divide by 3 on each side of equations.
3x/3<5/3 or 3x/3>13/3
This gives you the answer of x<5/3 or x>13/3
So basically this week I understood mostly everthing. I did have some trouble understanding word problems still, but I'll keep working on it.
Example 1.) 7x-12<9
FIRST: Add 12 to each side. 7x-12<9
+12+12
This then gives you 7x<21
SECOND: Divide each sid by 7. 7x/7<21/7
This gives you the final answer of x<3
Example 2.) Absolute Value Inequalities abs. value of 3x-9>4
FIRST: This is an "or" equation because it is greater than (if it was less than it would be
"and") 3x-9<-4 or 3x-9>4
SECOND: Add nine to each side of the equations.
3x-9<-4 or 3x-9>4
+9 +9 +9+9
This gives you 3x<5>or3x>13
THIRD: Divide by 3 on each side of equations.
3x/3<5/3 or 3x/3>13/3
This gives you the answer of x<5/3 or x>13/3
So basically this week I understood mostly everthing. I did have some trouble understanding word problems still, but I'll keep working on it.
Reflection #3
Another week gone by this week was much more smoother than the last 2 weeks. I guess its because im gettn use to the way mrs. robinson teach and im getting a feel of what is expected from her. But i chapter 3, one thing i really understood was solving inequalities. basically its like solving for x just a few different steps/changes...
2x-6>12
First you would have to get all your like terms to one side so it would be 2x-6+6>+6+12. Then it should look like this 2x>18. after you get all your X's to one side then you solve for x. and you would end up with x>9 as your final answer.
When you have a negative for your x you have to switch the sign around so it would be like this.
-2x-6>12
First you would do the same thing for the last problem. get all like terms to one side and Xs on the other. you would have -2x>18 then you divide by -2 but when you do that you have to change the sign because your dividing by a neg. x #. so your final answer would be x<-9
2x-6>12
First you would have to get all your like terms to one side so it would be 2x-6+6>+6+12. Then it should look like this 2x>18. after you get all your X's to one side then you solve for x. and you would end up with x>9 as your final answer.
When you have a negative for your x you have to switch the sign around so it would be like this.
-2x-6>12
First you would do the same thing for the last problem. get all like terms to one side and Xs on the other. you would have -2x>18 then you divide by -2 but when you do that you have to change the sign because your dividing by a neg. x #. so your final answer would be x<-9
REFLECTION #3
I think this week was a pretty good week. It deffinitely went by faster than I thought it would. It was basically like a review week for me since we were going over material from Chapter 2 again. Now that I can easily recognize which solving methods to use on the problems, it's made that part of advanced math much simpler.
What I thought was really easy that we learned this past week was inequalities. Inequalities are not something totally new to me because I learned it last year, so it was more like a review for me.
There are two types of absolute value inequalities--AND & OR. You use AND when you see a less than sign (<) in the problem, and you use OR when you see a greater than sign (>) in the problem.
Here are some examples:
Example 1:
(by the way, there's supposed to be an absolute value symbol around 6-3x)
6-3x < 12
1.) First you have to ask yourself, "Is this an AND problem or an OR problem?" It's an AND problem because the sign is less than.
2.) Next, since this is an absolute value equation, you know you will have to get two answers, so you would change the problem to look like this:
-12 < 6-3x < 12
3.) (*When you set the problem up like this, ^^^ the absolute value symbol goes away.) Now you just solve the equation from the inside out like this:
-first you subtract 6 from all sides
-then you divide by -3 on all sides
-12 < 6-3x < 12
-6 -6 -6
-18 < -3x < 6
(divide by -3)
6 > x > -2
**Your final answer can be written 2 ways:
1.) 6 > x > -2
2.) x > -2 and x < 6
Here's another example:
(supposed to have absolute value symbol around 3x-9)
3x-9 = 9
Okay this problem is a little different because it doesn't have a greater than or less than sign. But because it still has the absolute value symbol, you will still get 2 answers. You would have to set up 2 problems like this:
3x-9 = 9 3x-9 = -9
(again the absolute value symbol goes away. Then you just solve each equation for x.)
3x-9 = 9 3x-9 = -9
+9 +9 +9 +9
3x = 18 3x = 0
x = 6 x = 0
Your final answers would be 6 and 0.
And in point form that would be (6,0) (0,0)
***Alright, now for what I didn't understand this week. I had a little trouble with the inequality equations that had a whole bunch of fractions in them. They're not too hard, I'm just really really slow at them. And also, I don't understand a lot of this weekend's homework on domain and range. (Page 122 #'s 1-16) I mean I caught on to it easy in class, but when I tried to do the homework I got really lost. I don't really have a specific problem (because I'm stuck on all of them), but if someone could try and help me out that would be greatly appreciated :)
Overall, I didn't have too many problems this week. It was pretty easy.
What I thought was really easy that we learned this past week was inequalities. Inequalities are not something totally new to me because I learned it last year, so it was more like a review for me.
There are two types of absolute value inequalities--AND & OR. You use AND when you see a less than sign (<) in the problem, and you use OR when you see a greater than sign (>) in the problem.
Here are some examples:
Example 1:
(by the way, there's supposed to be an absolute value symbol around 6-3x)
6-3x < 12
1.) First you have to ask yourself, "Is this an AND problem or an OR problem?" It's an AND problem because the sign is less than.
2.) Next, since this is an absolute value equation, you know you will have to get two answers, so you would change the problem to look like this:
-12 < 6-3x < 12
3.) (*When you set the problem up like this, ^^^ the absolute value symbol goes away.) Now you just solve the equation from the inside out like this:
-first you subtract 6 from all sides
-then you divide by -3 on all sides
-12 < 6-3x < 12
-6 -6 -6
-18 < -3x < 6
(divide by -3)
6 > x > -2
**Your final answer can be written 2 ways:
1.) 6 > x > -2
2.) x > -2 and x < 6
Here's another example:
(supposed to have absolute value symbol around 3x-9)
3x-9 = 9
Okay this problem is a little different because it doesn't have a greater than or less than sign. But because it still has the absolute value symbol, you will still get 2 answers. You would have to set up 2 problems like this:
3x-9 = 9 3x-9 = -9
(again the absolute value symbol goes away. Then you just solve each equation for x.)
3x-9 = 9 3x-9 = -9
+9 +9 +9 +9
3x = 18 3x = 0
x = 6 x = 0
Your final answers would be 6 and 0.
And in point form that would be (6,0) (0,0)
***Alright, now for what I didn't understand this week. I had a little trouble with the inequality equations that had a whole bunch of fractions in them. They're not too hard, I'm just really really slow at them. And also, I don't understand a lot of this weekend's homework on domain and range. (Page 122 #'s 1-16) I mean I caught on to it easy in class, but when I tried to do the homework I got really lost. I don't really have a specific problem (because I'm stuck on all of them), but if someone could try and help me out that would be greatly appreciated :)
Overall, I didn't have too many problems this week. It was pretty easy.
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