Category Archives: Math 11

Totem response

It has made them much more reliant on Canadian traditions and in some words much more “whitewashed” than they might have been if uninfluenced. Still, some keep with there traditions and old ways but at the same time some also keep up with the times. I believe that even if left to their traditions and techniques they would still have adapted to the rest of the world’s technology. I think that right now everyone is worked up over something that’s not there. Why should I have to pay taxes and give up land that should be shared because some great-great relative was a racist. I think that every person should have equal rights, but I do believe that if they were left to their ways they would end up still adapting to modern day society.

week 13 – Math 11 – Advanced graphing of radicals

this week in math Precalc I learned about radicals and how to graph them
for example y=\frac{1}{2x+5}

so first you want to graph the parent function 2x+5

now you want to look for your invariant points (-1 and +1) then you want to draw two L shapes not touching the x-intercept or the x=x int asymptotes making this

the red line doesn’t actually touch the x int even if it looks like it does because a radical is just flipping the numbers 2 is now 1/2.

Week 12 – Math 11 – Graphing absolute values

in this week we learned how to graph absolute values

for the first example a linear line y=|x+3|

so first you just graph the parent function y=x+3

when it touches the x-intercept it has to go back up because it always has to stay positive

the right side stays the same because it’s positive and the left side just flips making a v shape

we also learned the absolute value of parabolas, for example, y=|x^2 -3|

so you want to draw the parent function again and just flip the negative parts so they are positive making a w shape

WEEK 10 – MATH 11 – GRAPHING INEQUALITYS

In this week we learned how to graph an inequality.
Say we have 2x + 3y > 6

First, we must make the Y a 0
2x > 6
Then we must divide both sides by 2
x > 3
Then change the > to a =
x = 3
we then repeat the same steps for the Y axis to end up with
y = 2

Once you have these two points plot both on the grid and connect them. after this follow the line to reach each side of your graph.

Next, we must figure out which side is true. to do this we must replace the 2x + 3y with a 0 so we end up with
0 > 6
Now we ask our self, is 0 greater than 6?
since its no, the shaded side will be the side without 0 on it.

WEEK 8 – MATH 11 – GRAPHING TRINOMIALS

In this week we learned how to graph trinomials.

if the graph is Y=X^2
notice how this graph has its axis at (0,0) and how it is going up 1,3,5

if the graph is Y=X^2 +2
The + 2 makes the parabola just go up to spaces so the axis is now (0,2) and still has the same pattern 1,3,5

if the graph is y= (x-2)^2
the -2 means to go to the right 2 so its the opposite of the symbol to which side you go so if it was +2 you would have gone 2 to the left

if the graph is y=2x^2
the 2 in front of the x makes it stretched out upwards because before there is an imaginary 1 in front of the x and now its 2 so it goes from 1,3,5 to 2,6,10 which makes the graph look skinner

WEEK 7 – MATH 11 – FACTORING TRINOMIALS

In this week we learned how to factor Trinomials

For example 16X^2 + 36X + 8

First, multiply 16 * 8 and find the factors of numbers that multiply towards it until you find the factors that add together to create 36.

So 16 * 8 = 128
128 = 4 * 32
4 + 32 = 36

so once you have done this you will end up with 4X and 32X in your empty squares.

and finally, you will end with (4 + 32X) and (4X + 8).

Week 6 – Math 11 – Solving Trinomials

In this week we learned how to balance trinomials, to do this we use this formula,

X= \frac{-B \sqrt{(b)^{2}-4(a)(c)}}{2(a)}

so lets say we have this as our equation,
3X^{2} + 8X + 24=0

first we must identify A, B and C,
3 is A
8 is B
24 is C

So using these values we must replace them in the spots of the formula with the matching letters,
X= \frac{-8 \sqrt{(8)^{2}-4(3)(24)}}{2(3)}

Now we will evaluate the equation,
X= \frac{-8 \sqrt{64-288}}{6}
then,
X= \frac{-8 \sqrt{-224}}{6}

after you get to this point we must find a common factor to divide everything with,
X= \frac{-8 \sqrt{-224}}{6} / \frac{2}{2}

Giving us,
X= \frac{-4 \sqrt{-56}}{3}

simplified again,
X= \frac{-4 -2\sqrt{14}}{3}

and that’s the final answer!