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By February 14, 2019.  No Comments on My Post  ELD 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

ex: 

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

see the example

thanks for reading

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@gachspeech

This is my favorite speech because this speech is very funny.his actions and his voice are actually petty good for the watcher to understand what he(they) want to show the audience.And also that in the beginning of this speech, they will make some scenes to let the audience in to this story, I think this is petty good to make this speech not very boring .So i hope you like it!

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:

\frac{2}{v}\frac{5}{4v}=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\frac{1}{x} , her apprentice finished \frac{ 1 }{x+9}

so we got a function that are after 20 hours

\frac{20}{x}+\frac{20}{x+9}=1(this ‘1’ is mean 20 over 20)

And you use the way that times common denominator

we got x^2 -31x-180=0

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:

\frac{x^2-9}{x^2+8x+15}

we need to factor first

so \frac{(x+3)(x-3)}{(x+3)(x+5)}

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 \frac{(x-3)}{(x+5)}

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:\frac{(x+3)(x-3)}{(x+3)(x+5)} for values of bottom can’t be “0”

“x “can’t be -3 and -5

second we can see this formula :\frac{(x-3)}{(x+5)} ,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

Ex

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:4+2x=|x^2 +8x+12|

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

ax^2 +bx+c<0 ax^2 +bx+c>0 ax^2 +bx+c<(=)0 ax^2 +bx+c>(=)0

where a, b, and c are constants and a not =0

Ex:

The solution of the quadratic inequality

x^2 -2x-3>0 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^2 -2x-3=0

(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