This week I learned that you can find the equation of a line without a graph only needing two order pairs or 1 ordered pair and the slope to do this you with the two pairs you simply find the slope of theses two lines then reduce or increase one of the order pairs by  the rise and run of the slope until the x variable is 0 and then now we know that the y number is the y-intercept to show the equation we use this where the m= the slope and the b= y-intercept and get this y=mx+b. the other way to do this is with 1 order pair and its slope we first start of by showing these number with this 1/2(x-3)=y-4. this is with the example slope 1/2 and ordered pair 3,4 we can then type this into desmos to show the line:

if we want to show this in slope intercept for we simply solve it to this: 1/2x-1.5=y-4. then we move the numbers on the y side to the x side which is this y=1/2x+2.5.

You can find slope of a line through two different ways first is by first find the y coordinate difference between one coordinate and another on the line to the right of it. so for example the y coordinate difference between (2,3) and (3,5) is 2 and then find the x difference which is 1 after that we first divide the y difference by the x difference which is 2/1 or just 2 then if needed we simplify it. Another example of this is if we have two order pair (4,2) and (5, 3,) which are points close to 1 after the other on a linear relation line the y difference is 1 and the x difference 1 so the slope of the line is 1/1 or just 1. Another way to do this to same as we do in the first way up until find the difference but instead of subtracting we instead look on a graph of the line and find how many numbers we go up by to reach the other point and the rest is the same. Good things to remember is if there is no difference in x or it is just a vertical line the slop is undefined if the line is horizontal the slop is 0.

This week in math 10 I learned how to show a function with function notation, mapping notation. To first start turning a simple equation into a function notation You write f: x as Y or in words F of x  so for example we see the function y=3x+2 in function notation we write as f(x)= 3x+2. To write a mapping notation we go from what we learned previously and change the “=” into an Arrow “->” which is shown as  f(x)->3x+2 You use mapping notation when there are a multiple amount of relation between input and output variables. Also another good way to use function notation would be to switch the F into other letters so that we can tell the difference. Another way of using function notion is in equation between different functions and showing the input an example of this is in the equation f(2)+f(8)-f(2)=8+26-8=26.

This week in math we learned how to find the X and Y intercepts of a linear equation which are where the linear relations line touches the X or Y lines this can be done in an equation to find the X you should set Y to 0 and find a number for X that works with that. an example of the X and Y intercept in a linear relationship would be with y = 2x + 1 to find the x-intercept this can be done by setting y to 0 = 2x + 1 so x would be -1/2 or -0.5 and the Y-intercept can be found by solving this y=2×0+1 which equals 1 so the X and Y intercepts are (-0.5,0) (0, 1). Another example of this being used is In this equation 3x+7y=21 so for the x-intercept we set y to 0 then we have to find out what times 3 equals 21 is 7 so the ordered pair for the x-intercept is (7,0). Now for the Y-intercept, we set x to 0 leaving us with the relation 7y=21 and we know that 7 times 3 = 21 from the other intercept so the y-intercept is (0,3).

This week I learned how to find the linear relation between two sets of numbers this can be done by first identifying the independent variable or x to input this is 1,2,3 and seeing the difference between each number in the dependent variable which is 6,8,10 which is 2 and multiplying it by the independent variable plus a number that will make the number the matching variable like 4 this can be shown like this y=2x+4. so if 1 were x that y would be 6 this is the relation between these two numbers. another example is between 1,2,3,4 and 8,4,0,-4,-8 the linear relation is -4x+12 this relation can also be plotted in a calculated to show that it Is in a straight line which means it is a linear relation.

This week I learned how to factor polynomials with three terms and determine if a trinomial is factorable. this can be done by first finding variable a which is the coefficient of the squared variable. the variable b is the variable with a power of 1 coefficient lastly the c variable is the constant. After this find, two numbers that add to b and multiply to c if there aren’t trinomial is not factorable anymore. you can show factors by having two polynomials each with one of the numbers we found and multiply those two polynomials together to check again.

An example of this is if we factor the trinomial x^2+6x+8 we find that 6 is b and 8 is C next we see the two numbers that add two 6 and multiply to 8 which are 2 and 4 because 2+4=6 and 2×4=8. we then write the factored polynomial like (x+2)(x+4). Another example of factoring trinomial can be shown with x^2+9x+8 we find that 9 is b and 8 is C next we see the two numbers that add two 1 and multiply to 8 which are 1 and 8 because 1+8=9 and 1×8=8. we then write the factored polynomial like (x+8)(x+1) we can also check to make sure that this is the right answer by solving the expression with the distributive law.

This week we learned how to multiple binomials and higher using the distributive property this can be done by first looking at the smaller polynomial and multiplying each term by each of the other polynomials’ terms individually not just the like terms and doing that with every term in the polynomial. After multiplying the terms we then combine and arrange the like terms like how we did in adding polynomials to get the simplified form of the product of both of the polynomials. I also learned a way to remember the order of the multiply the binomials with the acronym FOIL which stands for First: first terms of each binomial, Outer the outer terms of each binomial, Inner the two terms closest to the middle, Last the two last terms of each binomial. An example of the distributive property we can show with the binomials (2x-2)(4x+2) we would first start with the F which is 2x times 4x which equals 8×2 next is the O terms 2x times 2 is 4x next the I which are -2 and 4x equals -8x. Lastly, we have the L terms which are -2, and 2 which equals -4. After simplifying this polynomial we get a trinomial that is 8×2-4x-4.

This week in math 10 we learned how to add and subtract polynomials and monomials this could be done by grouping up the like terms from both polynomials a like term is a term whose variables and exponents are the same and then simplifying the answer by combining the like terms or subtracting them based on the question. After that arrange the remaining terms in order of decreasing degree. An example of this would be in the equation (2×2 + 6x +5) + ( 3×2 – 2x – 1) we would start by grouping them which equals 2×2+3×2+6x-2x+5-1 we then add the like terms together this can be done by adding the amount of each like term variable together or just plain adding the constant terms together. This equals 5×2+4x+4 and this is the answer to the question. Another example of subtracting polynomials is (5×2+3x-2)-(x2+2x-4) we first start off doing the same as adding polynomials and group up the like terms this equals 5x+x2+3x+2x-2-4 then we subtract them instead of adding to get the answer in order of decreasing degree this equals (4x-x+2)

I learned how to find the other sides of a right triangle using the Pythagoras theory with only one side and angle this can be done by first finding the trigonometric ratio that the side we have and the other one we don’t are both in. Then cross multiply the ratio and the angle with the side we don’t have as a variable and the side in their ratio form. An example of this being done was if we had the adjacent of angle 30 and the angle is 40 to find the hypotenuse we know that they are both in cosine ratio with the adjacent divided by the hypotenuse we write it as 30/A then we can cross multiple by the angle 40 cosine this equation equals (Cos 40 = 30/h) then the hypotenuse is 39.2(nearest tenth). Another example of this being done is if we had the hypotenuse is 20 and the angle we have is 16 then we want to find the adjacent side we can notice that they are both in cosine ratio which we write as x/20 then we cross multiply that by cos 16 which equals 19.2 so 19.2 is the adjacent side.

This week at school we learned how to find the trigonometric ratios of a right-angle triangle which can be found by first determining where the angle is which is the acute angle given this is the variable θ then we find out what type of side they are from this. The longest side in the triangle is the hypotenuse the side next to the angle is the adjacent side and the opposite side is opposite to the angle. we use those sides in the formulas  sin(θ) = Opposite / Hypotenuse, cos(θ) = Adjacent / Hypotenuse and  tan(θ) = Opposite / Adjacent. An example of this is if a right triangle has a hypotenuse side of 6 in length and an opposite side of 3 based on the given angle the sine ratio is 3/6= 2. Another example is if a triangle’s adjacent side length to the given acute angle is 4 and the longest length side of the hypotenuse is 5 then the cosine ratio of this right triangle is 4/5. to find the ratio angle we use the second function of a scientific calculator and press the ratio we have these functions are cos-1, sin-1, and tan-1.