Category Archives: Math 11
Week 6 Foundations 11
Week 7 foundations 11
This week in Foundations 11 I learned the difference between linear scale factor, area scale factor and volume scale factor and how to be able to measure things with those different scale factors. To be able to find the linear scale factor from having the volume from each shape.
the volume given for the original shape is and the new shapes volume is
To be able to find the volume scale factor, you take the two volumes and divide them from each other, . 128 divided by 40 is 3.2, meaning that the volume scale factor is 3.2. Once you have the volume scale factor, you take the number you got a cube root it to be able to find the linear scale factor. The linear scale factor is 1.47cm.
the area of the original square is and the width of the larger square is
To be able to find the linear scale factor, you take the two numbers and divide them by each other with the new square over the old square, 48 divided by 8 which is 6, meaning that the area scale factor is . Once you have the area scale factor, you take the number you got and you square root it to be able to find the linear scale factor. The linear scale factor is 2.4cm.
*i was not able to upload my photos on here*
Week 4 – Foundations 11
This week I learned how vertically opposite angles work.
This is a vertically opposite angle. These two diagonal lines intersect each other. Each line adds up to 180, meaning that when they intersect it adds to 360. At the top, we know that one angle is 140. Since the top is 140, we now know that the bottom is 140 as well. In order to find B, we can take 140 + 140 = 280 then subtract 365 from 280 which is 85. You must divide 85 by two in order to get both sides. You would end up getting 42.5 for B. to double check that the answers are correct, you can take all the numbers and add them all together to make sure they add up to 360.
Week 3 – Foundations 11
In this week of foundations 11, i have learned the two other forms of quadratic equations, the vertex form and the factored form. I have learned how to take a vertex equation and be able to input a specific point the parabola goes through to be able to figure out the spacing is for the parabola.
You have the equation y=a(x-6)^2 + 3 and a point that the parabola goes through is (8,4). you would replace the 8 for where the x is and the 4 where the y is, to be able to find A. Once you do that, the equation would look like this… 4= a(8-6)^2 +3. you then do the brackets, 8-6. 4= a(2)^2+3. You then square the 2 in the bracket making it 4. the equation would then look like this… 4=4a+3. You would then subtract 3 from each side making the equation 1=4a. You then have to divide 4 from each side to isolate A. making it 1/4=A. The equation would then look like y=1/4(x-6)^2+3.
I could not upload a photo so i wrote it out.
Example: Finding A
Equation: y=a(x+7)^2 +6
Point that parabola crosses: (3,8)
Replace (3,8) with y and x in the equation
8=a(3+7)^2 +6
then you do the brackets
8=a(10)^2 +6
Then you square 10
8=100a+6
you then subtract 6 from each side
2=100a
then divide both sides by 100 to isolate A
leaving 1/50=A
then you replace A with the number you got from the original equation
y = 1/50 (x+7)^2 +6
Week 2 – Foundations 11
This week in foundations, we learned about parabolas and what they are. A parabola is a plane curve which is mirror-symmetrical meaning that one side looks exactly the same on the other side. First, we learned the proper vocabulary and then we learned how to graph them. The highest or lowest point of the parabola is called the vertex. the “line” that goes down the center of the parabola is known as the line of symmetry or the axis of symmetry. By using the line of symmetry, and technically cutting the parabola in half, both sides will look the exact same but mirrored. The best way to find out where the line of symmetry is, is by using the vertex, since it is the highest/lowest point of the parabola. In all equations there are exponents if the exponent is it is automatically a parabola.
To be able to graph a simple parabola, we can use the 1-23-5 rule. these numbers represent the slope that we need to graph the parabola. For the equation , we would start at 3 on the Y axis, since it is the y-intercept. Then we would use the 1-3-5 rule by going up 1 from 3 and over 1, then 3 up and over 1, and finally 5 up and over 1.
Week 1 Foundation’s 11
This week in Foundations 11 we learned about graphing linear inequalities. I believe that the most important concept that we learned was how to figure out what side to shade when graphing the line on the graph. To be able to figure out what side to shade, you must pick any point on the graph that is not on the line. You then replace X and Y in your equation with the X and Y numbers from the point you chose. After you have inserted the new values in the equation, you solve it, and if it is true, you shade the side with the point you chose. If the equation is incorrect, you shade the opposite side of the line that does not include the point you chose.
Example:
In my example I used y<6x+4 and I plotted it on the graph. Since the equation does not go through (0.0), I used that point and replaced it in the equation. When I inserted (0,0) into the inequality, it was correct, meaning I shaded the side that included (0,0).