This week we review the final exam.

It’s about GCF(greatest common factor), LCM(least common muitples), slope, Pythagorean Theorem, and relations.

First of all, GCF. For example, 90 and 225, 90=2*3*3*5. 225=3*3*5*5. GCF=3*3*5, GCF=45.

LCM. For example, 30 and 36. 30=2*3*5, 36=2*2*3*3. LCM=2*2*3*3*5, LCM=180.

The next one is pythagoren Theorem. 

Hypontenuse (longest side) = a2+b2=c2.

We have three ratio: Sine ratio=o/h, Cos ratio=a/h, Tan=o/a. 

And then we learned the relation part.

The cost of jujubes is related to the weight.

Cost=dependent, y, output, range, Weight=independent, x, input, domain. 

And other important thing is x-intercept=(x,0). y-intercept=(0,y).

Domain is x scope, Range is y scope.

This week we leared about arithmetic sequences. 

For example: 8, 14, 20… They add six every time, so t4=26, t5=32, t6=38.

And then there’s another type of question. ( ), ( ), 0, ( ), ( ). d=3.

So we can know they add by 3 every time. So the anwser is -6, -3, 0, 3, 6.

There are another two questions here.

1: If t2=10 and t6=34, determine t1 and the general term(tn).

We can use the step before, we can write t1-t6 down. Then we can know, t2 to t6 need 4d.

So we can write: t2+4d=t6. And turn them into Numbers. 

10+4d=34, 4d=24, d=6.

t1=t2-d, t1=4.

tn=t1+(n-1)(6). This is a formula: tn=t1+(n-1)d. 

tn=4+6n-6, tn=6n-2. So this is the anwser.

Other example is: Given the sequence: x+2, 3x-1, 2x++1… determine what x would be to make this an arithmetic sequence..

   Find d= tn-tn+1. 

d=(3x-1)-(x+2）       d=(2x+1)-(3x-1)

=3x-1-x-2                   =2x=1-3x+1

=2x-3                          =-x+2

And then we need: 2x-3=-x+2

3x-3=2

3x=5, x=5/3.

https://sd43bcca-my.sharepoint.com/personal/132-wzhou_sd43_bc_ca/_layouts/15/doc.aspx?sourcedoc={90a3f6ab-b44e-47da-a56b-2ca8ebee189f}&action=edit

This week we learned about systems of linear equations.

There are four ways: 1. If you have a separate x or y, write the separate x or y as: x=.Y =.

2. I have minus 3y in both, and 3y is zero pair.I can just cancel out

3. If you don’t have the same x and y, and you don’t have minus 3y and 3y, you have to turn an x or a y into a positive or negative number.

4. Find the least common multiple.

And we also learned elimination and substitution.

Elimination: 4x+2y-31=0, -4x+6y-13=0. 4x and -4x can Directly to cancel.

Substitution: Let’s get an x or a y out.

This week we learned verifying a solution.

1 solution: m1 Is not equal to m2. no solution: m1=m2, b1=b2. coincident solution: m1=m2, b1=b2.

And then we learned substitution.

An algebraic method of finding solution for systems.

For example, x+4y=-3 ->x=-4y-3

3x -y=29.

First, we nned to find the x or y who is It’s independent. It’s easy to calculate.

3(-4y-3)-7y=29.

-12y-9-7y=29.

-12y-7y=29+9=38.

-19y=38, y=-2.

We know y=-2, We can substitute y into the original formula. So the answer is (5, -2).

Steps: 1 pick one equation and rearrarrge so it is x or y.

2 substitute this into the other equation. (use brackets)

3 solve equation.

4 substitute the number just found, back into the rearranged equation.

5 verify solution.

This week we did the wonky initials project.

First, we need to draw the first letter of our name on the graph. It has to be horizontal, vertical, and diagonal. My three letter are: Z, Y, and W.

We need to write down the coordinates. (Z) X: -14, -12, -5, -5. Y: 11, 3, 11, 4. And then we need to look at the slope of Z. m=8/7. And find x and y. I calculated it by the slash at the bottom, so I want to find the coordinates of the slash. (-12, 3). Finally, we need to use the formula, m(x-x1)=y-y1. m=slope. 8/7(x+12)=y-3.

Based on this rule, let’s figure out Y and W again. And then you can connect the letters.

https://www.desmos.com/calculator/7e9uzxgymz

We learned distance, midpoint and slope.

Line segment have: horizontal, vertical, distance.

Here are a few formulas, midpoint: (x1+x2/2, y1+y2/2). distance: a2+b2=c2. slope: rise/run. rise=y, run=x= y2-y1/x2-x1. (rise=y2-y1, run=x2-x1)

For example, how to find A(2,7) to B (5,7). We can see 7 are same. So we can use 5-2, the answer is 3. How to determine whether each line segment is horizontal or vertical? We can draw a picture for it, then we can know. And determine the cooridinates of the midpoint of the line segment with the given pair of endpoints.

This week we review the chapter 5 chapter 6, and have a exam.

The first coordinate of an ordered pair is called the x coordinate and the second coordinate is called y coordinate. And the numbers of an ordered pair are called the corrdinates of a point on the grid. And then we learned that the top right is quadrant top left is quadrant two, the bottom left is quadrant hree, and the last one is quadrant four.

Then we learend: The mathematical relationship between two quantities is called a relation, and we learned independent, dependent, inputs and outputs. X=value, y=name. x=inputs, y=outputs. x=independent, y=dependent. Learned to substitute coordinate Numbers into the problem to get the answer.

The we leared domain and range. x=domain, y=range.

R=real number, W=whole number, I=integer: -2,-1,0,1,2. N=natural number.

Finally we learned relations and functons. If you want to determine whether the coordinate or the coordinate graph is a function, if the coordinate graph has more than two points, it is not functions; if there is only one point or no point, it is functions.

We learned function this week.

For example, determine the value of the y-intercept of the graph of each equation.

2y+3x-12=0, y-int=(0,y), 2y-12=0, -12=2y, 6=y.

For example, f(x)=5-3x, what is f(1)? x=1, So we can substitute that in. 5-3(1)=2. So the answer is 2.

Solve for a if h(3)=8 and h(x)=3×2-ax-1. We can know x=3. 3×2-ax-1=8, because h(3)=8. then 27-3a-1=8, 3a=8+1-27=-18, a=-18/-3, a=6.