I think that the most successful parts of my projects were the creativity, perspectives and diversity that I had throughout my work. I think that the most successful projects, where the creative zone, elements and principles of photography, and the aquarium power point. Throughout these projects, I found a consistent theme of what I found successful – the time that I took to go out and shoot the images. I think that in order to take images to the next level you have to be able to take time to go out and choose a location, and plan a pose, to be able to produce more powerful images. However, I think that the least successful parts of my projects where my studio portraits and my avatar project. I think that one of the main reasons that I didn’t have as much success with these projects, is because I didn’t have much time to complete them, didn’t plan them out well, and wasn’t as motivated to create when completing them. I think if I had more time I would like to redo the studio portrait project, because I feel that it could have been better planed out and of a better quality had I given myself more time to complete it.

I think that throughout this photography course, I learned a lot about my style of photography and who I am and want to be as a photographer. I also discovered more about other people’s styles and strategies in photography, which I think is very interesting. Beyond that, I also discovered more about what photography is, in the elements and principles of photography. I really enjoyed this class, as it pushed me to explore different corners of photography (like macro), that I likely would not have tried otherwise. I am very happy with the outcome of this class, and would take photography 12 to continue growing and expanding my understanding of photography, and who I am as a photographer.

]]>There are two special triangles the first having a 45 degree angle, a root 2 of the hypotenous, and 1 on the other lengths. The only option for this angle is to have45 degrees. The second special triangle is 60 degrees and 30 degrees, 2 on the hypotenous, root 3 on the adjascent and 1 on the opposite length. The two degrees for this triangle, are 30 and 60. For the circle chart, you can apply it to triangles that are 1, -1, 0 or undefined. The top and bottom quadrants are (0, 1) and (0, -1), while both sides are (-1, 0) and (1, 0). The degrees for the circle are 0, 90, 180, and 270. The CAST rule, begins in the fourth quadrant, with “C, then the first with “A”, then the second with “S”, then the third with “T”. In the first quadrant, all are positive (Cos, Sin, Tan). In the second quadrant, Sin is positive, and Tan and Cos are negative. In the third quadrant, Tan is positive, and Sin and Cos are negative. In the fourth quadrant, Cos is positive, and Tan and Sin are negative. You would decide to use th Cosine Law when there are two sides and one angle, which surround each other, making “C shape.

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To solve for piecewise notation, first, you will put the equation in absoulute value form. Next, you will will find the zero of the equation (isolate for “x” if it is a linear equation, or factor if it is a quadratic equation). After you have done this, you will put the number that you got from solving for “x” on a number line. You will choose two (if there is one “x” value) or three (if there are two “x” values) test points on your number line. Once youo have tested the points (by putting them back into the original equation), and you have determined which points are negative and positive, you can write the equation in piecewise form. To do this, you write “y” and then “{” and multiply in a negative for the negative value, writing whether it is < or > where you “x” value or values are on the number line. And you will leave the positive points as the original equation, and write < or > to represent where on the numberline “x” is positive.

To solve a reciprocal function, first you will put it into a fraction. Next, you will take the original equation (not in a fraction), and find the NVP. To do this, you will find the zero (the equation cannot be equal to zero) of the equation, and do this by isolating “x” (linear) or by factoring (quadratic). Once you have found what “x” cannot equal, this will be your asymptote. After, you go one to the left from your asymptote and up or down (positive, up or negative, down) and go one to the right from the asymptote and one up or down. You will then draw the shapes (two if it is linear, or three, one is a parabola, if it is quadratic).

]]>When you are putting an equation into piecewise form, you first take the equation out of the absolute value signs. Once you have done this, you make the equation equal to zero. After this, you put all the numbers (without variables) on the other side of the equation. Next, you isolate “x” by diving by the number that is next to it (coefficient) on both sides, and then you will have your answer for what “x’ equals. Once you have solved for “x”, you put the point or points on a number line, and choose two (one point) or three (two points) on the number line. Based on which point is negative, you will then write the formula in piecewise form. To do this, you will write the original formula as it is, and then write “x” is < or > depending on where your point is positive and negative for all your points (one or two). Next, you will write out the original equation again, this time for the negative point, and since it is a negative, you will have to multiply in a negitive sign in the front of the original equaiton to make it positive. Once you have writtent that, you will then write “x” is < or > the point or points that are negative on your number line.

To solve an equation, you seperate it into two different equations. One will be the original equation (without the absolute value symbols) and the other will be the equation with the negative symbole multiplying in (to make the negative part of the number line positive). To solve your first side (the original one), you will move any numbers without variables to the other side. Once you have done this, you divide both sides by the number in front of “x” (coefficient), and this will give you your answer. For the second side (the original one with the negative multiplied in), you will start by multiplying the negative symbole in. Once you have done this, you move all the numbers without variables to the other side. Next, you will divide both sides by the number in front of “x” (coefficient), and this will give you the other answer.

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First, you put the first equation into the second equation (where the “y” is in the second equation, since the first equation is equal to “y”). After you have done this, you simplify the equation, first by foiling the equation, and then by making it equal to zero. After you have done this, you factor the equation. Once you have the factor of the equation, you are able to put those points into the the first equation where “x” was. After you have put both points into the first equation (seperatly), you now have the second points for both numbers. Your first point will be the first number you got from factoring the equation, and the first nunber you got from putting that number as “x” into the first equation. The same will be for your second point of the equation.

]]>First, when you are solving a linear equation, you want to isolate for “x”, and when it is a quadratic equation you want to isolate for “y”. To solve a linear equation, the first step is to seperate the numbers from the numbers with variables (“x”). After you have put both numbers on the other side of the less than or greater than sign, you find what the answer is to both of the numbers on the other side. Once you have done this, you divide both sides of the equation by the number in front of the variable. Once you have done this, you have the point or points that you will use. After, you put the point or points on a number line, and choose two ( one point) or three (two points) test points to determine which way works for the equation. If you are solving a quadratic equation, you could first find the factors for the equation, and write (x + 3) (x – 2). Once you have done this, you take the first number from the original equation, and would divide both numbers in the brackets by this number. Once you have done this, you take the answer that you get from dividing and flip the symbols of both numbers and use these as the points for the number line. Once you have put them on a number line, you choose three test points, and you can determine this way which parts of the number line work for the equation.

]]>First, you factor the equation. Next, you take the two numbers that will be used to factor and write them in two separate brackets next to “x”. Next, you take the first number from the original equation, and use that as the bottom number of a fraction of each number next to “x” in each bracket. After, you simplify the fractions, and if the number is still in fraction form, then you take the bottom number and put it in front of the “x” in the bracket, because you cannot have it in fraction form. Next you take the numbers next to “x” and give them the opposite sign that they originally had (positive or negative) and then you use those numbers, however if one or both of the numbers was a fraction, then you would put it back into it’s simplified (if possible) fraction form, and still reverse the original sign that it had, as you did with the other number. After, you put both numbers on a number line. You would use a closed circle if it is not = to zero (greater or lesser), or an open circle if it is = to zero (greater or lesser). Once both numbers are placed on the number line, you choose one number in each section (left, middle and right) of the numbers, and test them by plugging them into the equation that you created, and then you are able to determine which sections work with the equation, and therefore determine which numbers are greater, lesser or equal to “x”.

]]>First, every word problem wants you to create two equations from it’s poblem. These two equations will be what you use to solve the problem. From these equations, you will create a quadratic equation. Then, you will use that equation to solve the problem. Next, if the equation is asking you about people and money (or two different things) you would put people with people and money with money in your equation. Then, like the last one you will turn this equation into a quadratic equation by using FOIL and add all the like terms together to give you a quadratic equation that you will use to solve the question.

]]>If you had y = 2x^2 + 8x + 6 you would begin by finding the vertex. You would do that by factoring out 2 from the equation (because that is the value of the first number). So the equation would now become y = 2 (x^2 + 4x + 3) The next step would be to take the 4 divide it by two and then square it. Then, you would take that number (4) and create a zero pair after the 4x in the equation. The equation would now becomre y = 2 (x^2 + 4x + 4 – 4 +3). Then you put brackets around the first three numbers of the equation, y = 2 (x^2 + 4x + 4)(-4 +3). After this, you would go back to the number that you had to divide by two and then square (4), and root it, which gives you two. So the equation would now become, 2 (x + 4) (x + 4 ) (-1), which would become 2 (x + 4)^2 (-1). This would give you your vertex of (-4, -1). Once you have found your vertex, you are able to put it into the middle of the table, and move up by two, and down by two. The top number on the “y” values of your table will be the same on the top and bottom. These will give you your points for your graph. Once you have these you will be able to plot them on the graph, which you can draw based on where you points on (move the graph higher or lower). That is how you find the vertex, points, and graph an equation.

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