




a) 2w + s
b) w stands for the number of wins, s stands for the number of shootout losses
c)20-12-4=4 losses
d) 2w+s
2(12)+4
24+4
28
*Just replace the variables with the given numbers.
e) 2x5+18=28
2x9+10=28
*The combinations are for team B, and has to equal 28.
Then, we had a review about polynomials, their coefficients, variables, and exponents.
Here's a chart about that (we had to fill in the coefficients, variables, and exponents):
After that, Ms. Yoo, started us on unit 5.2. Which is equivalent expressions. But we talked about like terms first. What are like terms? Like terms are terms that differ only by their numerical coefficients.
*Variables, and it's power stay the same. Coefficients changes however.
examples:
3x and -2x
6y² and 3y²
-5xy and -2 xy
17 and 12
Here's are other examples featuring the variable x:
-1x=-x
+x= x
-x²=-x²
+1x²=x²
*Mathematicians are lazy, they don't like writing 1 in front of a variable.
Next, we had to identify like terms:
(like terms are colour coded, dark green, and blue)
a) 5b² 3bc -2b 7c 6b
b) 3x² 4xy -2x² 7x² 1/2 y
c) 3pq 11 -4q² -3 pq
Then, we were combining like terms. (the variables and coefficients are bolded, the constants are "slanted", while variables with higher power than 1 are left with regular font)
a) 4x-2x+3-6
2x-3
*4x-2x= 2x, +3-6= -3
b) 2x²+3x -1+x²-4x-2
3x²-x-3
c) k-2k²+3+5k²-3k-4
-3k²-2k-1
-Always start with the highest degree (power)
-Try to gather like terms first
After, we had to simplify like terms.
1) x²-(-3x) +4+7x²-8x-6
x²+7x²-8x+3x+4-6 (<= work should look like that)
8x²-5x-2
explanation: 7x²+x² is 8x², while -8x+3x is -5x, and 4+(-6) is -2
2) -x-5x+(-3x²)-9-2x+7
-3x²-8x-2
3) 4x³+6x²+6x-1+5x³-x²-(-9)
9x³+5x²+6x+8
That's all we did in class. Hope I covered everything. Here's the homework:
-5.2 Extra Practice
-5.2 Workbook
-2 sheets (double sided)
-journal
-textbook work (since we don't have class tomorrow because of the Tec Voc tour)
-5.2 Mathlink (Mr. B didn't assign this, so do it if you have time)
-COMMENT ON MY POST!
*By the way, Friday's scribe is Melanie (since there's no math class tomorrow)
To do this first question, I changed the mixed numbers into improper fractions. Then, I found out the common denominator, for just the first two fractions. With those two fractions, I tried different operations, until I got to the pattern. Then you would have to see if the pattern actually works. so it's basically trial and error, for me.
On the second fraction, I did the same thing too. Except I only used the absolute value on the first fraction, to find out the pattern. This question, I had a longer time figuring out than the first one. Since the changing of signs (positive then to negative) caught me off guard. Then I remembered that a negative number, divided by a negative number is positive. Also a positive number divided by a negative number is negative. So the pattern must have been the number divided by a negative number. That gave me a clue. Sooner or later, I finally got it.
absolute value: only how far a number is from zero (eg. the absolute value of 6 is 6, and the absolute value of -6 is 6 also.
I guess that's it. I tried my best explaining this. I think that the easier question you get, the harder it is explaining it. Since there's nothing much to explain.
Have a good long weekend everyone! :D
In this particular shape, I would use the formula for finding the surface area of a rectangular prism. I would have to use that twice (top and bottom). Add them together. Then, find the area of the overlap, then subtract.
S.A. (bottom)= 2(lw)+2 (hw)+ 2(hl)
= 2 (5)(2) + 2 (2)(2)+ 2 (2)(5)
= 2 (10)+2 (4)+ 2 (10)
= 20+8+20
= 48
S.A. (top)= 2(lw)+2(hw)+2(hl)
= 2(5)(1)+2 (2)(1)+2(2)(5)
= 2(5)+2(2)+2(10)
= 10+ 4+ 20
= 34
Area of overlap = (1)(5)(1)
= 5
S.A.=area (bottom)+ area (top) -area of overlap
=48+34-5
S.A. =77 u²
How does symmetry help us solve some of these surface area problems?
- Well, that's an interesting question. It's also an easy question to answer. Most solid shapes (eg. rectangular prism) has planes of symmetry in it. Planes of symmetry helps us how many faces are equal. Which helps us to know that when we see one side of that shape, the opposite side is the same. Planes of symmetry only applies to some shapes. So, it can't be like that for every shape.
What happens when a part of any of these shapes is missing? How do I find the surface area then?
-Simple! Most shapes have planes of symmetry. So if one side is missing, you will know that it's the same as the given part. That helps on finding the missing part of a shape, then calculating the surface area with the missing shape.
Hey! Feel free to comment on how great this post is! Also feel free to criticize this post! Or just comment! :D
The angle of rotation is: 60 degrees.
Those are two of the questions that was on the test.
We had a few homework for today
1) Provide an algebraic formula for finding the surface area of a rectangular prism.
2) Provide a formula for finding the surface area of a cylinder.
*Then, when we come to class, we should be able to explain, make a net, and be able to talk about the symmetry of the shapes. (for rectangular prisms and cylinders)
3) Write in our journals.
4) Last but not least, comment on my post! :)
The next scribe is Aleiah. The homework is due. next class. So you better get going!