Category: Grade 11
Week 17-The sin , cos law
This week we learned about the sin law
which is :sinA/a = sinB/b = sinC/c
we can use this formula to solve the angle or the line
so we have to solve what is the AC
we already know what is the angle c is 80 degree and the c line is 5 cm , and we also know the angle b is 30 degree
so we can use this function that just use two of them
sin C/c = sin C /b
sin80 degree/5cm=sin30 degree/AC
AC=(sin30*5 )/sin80 degree
AC=2.5cm(use the calculator)
ok , and we also learn how to solve the cos line
thanks for reading
Week 16-Angles in standard position in all quadrants
Rotational :is a angle that from the 0 line
Reference : is the angle that always close to the middle linelike this
Coterminal angle: is a angle that always negative and is a angle that are shared same arm with the rotational.
example:rotational angle is 243 degree
so reference is 243-180=63 degree
and we use the way
look at this picture:
so we use this graph to solve
The angle in quadrant 1 :63 degree
The angle in quadrant 2: 180-63=117degree
The angle in quadrant 3 :180+63=243degree
The angle in quadrant 4:360-63=297degree
because we already know the reference angle
so it is very easy to solve
And we also learned the angles with triangle
an isosceles right triangle and an acute right triangle
remember we know the triangle in 4 different quadrant
use the formula y/r , x/r , y/x
and use the way that Mrs. Burton taught
“All Student Take Calculus”
In the quadrant 1 the sin,cos,tan are all positive
in quadrant 2 just sin is the positive
in quadrant 3 just tan is the positive
in quadrant 4 just cos is the positive
so it is a good way to solve the degree
for example:sin 135 degree
we use the picture we find this is a 45 degree triangle
so we use o/H (y/r)=1/√2=√2/2
Thank you for reading
Week15-Applications of Rational Equations
This week we learned how to solve the unknown value from the score formula
For example:
–=3
Then we make all the term time common denominator
so we get the common denominator is 4v
8-5=12v
12v=3
v=0.25
Then we learn how to solve the unknow number in the life things
This is the one of the example
we get Mary’s apprentice need 9 hours long to build a deck than it takes Marcy
So we make Mary need x hours to finish , her apprentice need x+9 hours to finish
And if they do together , so they will finish in 20 hours
so we know after 1 hour Mary finished , her apprentice finished
so we got a function that are after 20 hours
+=1(this ‘1’ is mean 20 over 20)
And you use the way that times common denominator
we got
then we factor
we get (x-36)(x+5)=0
And we get 36 ,-5
but -5 is not belong to the question
so we just get 36 is mary
than we get 36+9=45 is her apprentice
Week 14 -Rational Expressions and Equations
This week we learned how to make a function simply
For example:
we need to factor first
so
we can see the same thing from the top and the bottom, so we can get rid of them
so it is (x+3)
After that we can get
Next step we have to make sure the range of values of “x” and what number can’t it become,
First we can see this formula: for values of bottom can’t be “0”
“x “can’t be -3 and -5
second we can see this formula : ,in the bottom x can’t be -5
WARNING : you have to check the all function , even the first one ,one special is start by diving by a fraction , and then you have to multiply by the inverse of that fraction , and you have to look at the denominator of that fraction to make sure that the unknow numbers before and after reciprocal and which number can’t be.(“0″ CAN’T BE THE BOTTOM”)
Week 13-Reciprocal Quadratic Functions
This week we learned reciprocal quadratic functions graph
So when we get the function, we have to graph the (parent function)quadratic formula first ,and then we have to find the invariant points ,which is when y=1 . y=-1 ,the points that on the parent function graph ,we can see this picture don’t have find the vertical asymptotes ,so there is no vertical asymptotes, because the x=0 is the asymptotes already so we don’t have to draw another one ,just have the horizontal asymptotes y=0 (in this term we just earn y=0)
First we also we have draw the parent formula, and there is no invariant point.we can find the reciprocal of number in the end (2) to find a point that the reciprocal of the quadratic function touch in the y line(x=0)
Ok so we draw the parent function again , find the x-intercept to find the vertical asymptotes (two) ,and there have the four invariant numbers ,and for the third graph line that function of the reciprocal of the quadratic function ,you also can find the point that reciprocal of the end number (four to be one over four:0.25) that where it touch at the y line(x=0)
Week 12-Solving Absolute value Equations
This we learn how to solve the solution of the absolute value Equation
First we learned how to use the graph to solve
this question is asking that what’s the x solution are by looking the graph
we can see the equation
|-3x+3|=9
The y intercept is 3
So there two possible inside the absolute
First:-3x+3=9
-3x=6
x=-2
Second:-(-3x+3)=9
3x-3=9
3x=12
x=4
And because of when the y=9 touch the linear the x=-2,4
so the solution are -2,4.
EX:Another example is about the x square
Formula:
First possible
x is positive
(x+2)(x+4)=0
x=-2,x=-4
Second posible
x is negative
(x+2)(x+8)=0
x=-2,x=-8
Because absolute can’t be the negative.so when x=-4 ,x=-8
4+2x is negative , so they can’t be the solution of this,
so the solution is x=-2
Week 11-Graphing Linear inequalities in two Variables
PART 1:If you want to slove the graph is in which part of, you can use the test point
Ex:
3x-2y>(=)-16
Is (-3, 4) can be in this part?
so 3(-3)-2(4)>(=)-16
-9-8>(=)-16
-17>(=)-16
From there we can see this point can be the formula used
so if this point is outside, so you can draw the part inside(can be use), so if this point is inside, you can draw outside(can be use).
WARN:And don’t forgot if the formula have (=) the line should be the soild line, if there is no (
=),so the line should be the broken line!
PART 2:SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY
we learn from grade 10
How much solution can linear system have?
There possible:1 , countless ,no
And we leaned the possible of two quadratic functions
one, two and none
two solution is the max
same with the linear function with the quadratic function
(use the desmos if you can’t see the solutions(the point or points that they meet)
Week 10-solving Quadratic inequalities in one variable
The formula that we learned
where a, b, and c are constants and a not =0
Ex:
The solution of the quadratic inequality
is the value of x for which y>0 ; that is ,the values of x for which the graph is above the x -axis.Visualize the shadow of the graph on the x-axis.
so we first have to factor this formula
(x-3)(x+1)=0
ok, so we can get the value of x for equal to zero
x=3,x=-1
and because y>0
so the part of y>0 are
-1>x>3
Week 9-Modelling and Solving Problems with Quadratic Functions
This week we learned three formula form
General Form :
Standard Form(Vertex form:Ms.Burton’s Favorite):
X point form(x intercept form):y=(x-x_1)(x-x_2)
Ex:the rectangular have the area is to be bounded by 120 m of fencing(determine max area)
so we can get two formula
2L+2W=120
L*W=mix
so we can use the first formula get this form
L+W=60
L=60-W
SO PUT THIS FORMULA INTO THE SECOND FORMULA
(60-W)*W=MIX
We can get
So the vertex is (30,900):the 30 m is the w , and the is the MIX area
so we can use the formula L=60-W ,to get the L=30
so it is a square .the special rectangular