This week in Math 11, we learned how to use the Cosine Law. To use this method, you need atleast 2 side lengths, an angle or 3 side lengths. The cosine law is used to find an angle, for the first step you have the givens, a, b, and c is what i use. The equation we use for the first method is, c2 = a2 + b2 – 2ab cos C. One thing to keep in mind is that you can have information from the opposite side. For the example below, we used A= 10 cm, B= 18 cm, C= 20cm. Since we’re given three side angles, i used the second method which requires 3 side angles.
This week in math, we learned how to multiply and divide binomials. The first step is always to factor if possible, we always need to take a look at what we’re doing; either multiplication or dividing. From previous math, we know that if it’s dividing then we need to reciprocate the fraction. The next step is to simplify all like terms, you always need to find what X can’t equal to, because the denominator cannot equal 0.
In the examples below, i forgot my non-permissible values, for the first one it’s b cannot equal 0, and same for the second one, except B and A cannot=0
My first step was to examine what equation i was dealing with, and i applied all the steps above to help me solve the solution
This week in math, we learning how to simplify radical expressions. If the top the same term as the bottom they can cancel each other out. Our first step is to simplify the expression and then you find what the variable cannot equal to, sometimes there are more than 1 solution. In the example below:
I first found what the X value cannot equal to, so it was 0,9. Next i found all like terms and cancelled each other out X, (X-3). This left me with 24/18x which i could then simplify to 4/3x.
This week is math, we learned how to graph absolute values, meaning the graph cannot cross the X axis to become negative. How this works, is the point it normally crosses to the negative side, reflects going the opposite direction using the same pattern, let’s say we were using 3x+2, when it reaches the point where it’s supposed to cross it goes up 3 then to the other way 2. If we were going up 3 over 2 to the right the reflected side would be up 3 over 2 to the left. We also learned how to do this with parabolas but instead of one line, it will be the vertex that will be reflected, the same rules also apply, the X intercept is like an invisible wall which will not let the line/ parabola cross to become a negative.
In the first picture, this graph represents the graph of x2+3x-5 and |x2+3x-5|. The green line represents the parabola if it were to be just graphed, no absolute values; we can see that it crosses the X intercept making it negative. In the orange lines, we can see the vertex is reflected, because the invisible wall reflected it back to a positive, having the same vertex but positive.
In the second picture, the orange line represents the normal value where it goes into negative 3x+2, and the blue line represents the absolute value version, where it’s reflected, which the equation is|3x+2|. This represents the absolute value, where it reflects right before it crosses the X intercept using the same pattern as the normal version but going up 3 to the left 2…
This week in math week 12, we re-learned how to substitute y=mx+b but with a quadratic, to solve the quadratic version, you must have 2 quadratic equations. The first step in this is to make sure Y or X is isolated generally i like to isolate Y, and make sure they have no coefficients, and that you have an x2 to make sure you have a quadratic equation. Now that you have that you have one of the variables isolated, you use that solution and put it into the other solution wherever you see the variable you isolated on the other solution. Next you want to subtract one end to another making one side equal to 0 (Generally you subtract the soltion without x2)
Now you want to find the 0’s, because this will allow you to find the y axis. Finding the 0’s will help you complete the equation. Once you find the 2 zero’s, you replace it in any of the equations finding the y axis but it has to be the same equation.
This week in math, we learned how to graph inequalities and systems of equations. This week, we combined our last 2 units and made it into one (Solving Quadratics, and Graphing inequalities and systems). We learned that if the equal sign has a line on the bottom which means equal that it will have a solid line when graphing. $slatex y\le4x+5$. Knowing this we can add a point to check which part to shade in, in this case i used (-2,1)
-2>4(1)+5 = -2> 8+5
This shows us that we need to shade the bottom side of the graph, because the test points for this solution was not true.
This week, i went over 3 things that i forgot how to do. The first one i did was rationalizing radicals. I knew the general idea of this, but i forgot how you have to multiply the conjugate making it a zero pair rather than multiplying by the same thing. For my second one, i did arithmetic sequences, i didn’t forget how to do it, it was just a memory refresher. And for the last one, i did solving discriminant which tells me how many roots i have and this was confusing for me on the test, because i would divide it by 2a.
This week in math, I learned how to find the X intercepts in the standard form. I learned this late, but i still learned it in week 9. So to find the X intercepts we have to go back to our last chapter Solving for quadradics. First you would isolate to make X by itself so i subtracted by 1, then after that i sqaure rooted everything to get rid of the brackets, the from there i added 2 to the other side make it 2+/- square root of 1.
This week in Pre calc 11, I learned how to graph the vertex. I learned that when you are given the standard form, you can graph the vertex. In the brackets, you have the X value but a cool thing is that whatever number is in there, you change the sybol. So if it is -3 it really means +3. and the number right outside the brackets is the Y value; so given y=a(x-3)+3, we can tell that we need to graph (3,3) and that is our vertex.