The equation y+mx+b is slope. Y is equal to the equation. Mx is the fraction of how many times you move up, down, right, or left. B is the number or fraction that you start with. First, in an equation of y=2/3+5 you start with five at the y-axis and plot your first point. Then since the fraction is a positive you go up three and right three. There you plot your second point. There you have a positive slope. Another example, is y=5/7=12, since it is not a fraction change 12 into a fraction which will be 12/1. Start at 12 and go over one then plot your point. Next go up 5 and over 7. Therefore you have a positive slope.
You follow the steps and if b is a negative just start from the negative of the y-axis. This method is used in all the negative equations. The number you got is what y equals. As you can see, slope isn't very hard if you know the process.
There are many questions to lines such as, where does the numerator of a fraction go? Where do you start? Where is the y intercept found? Which number is to plot first? Which way do we go if there is a negative denominator? Where do we start when there is a positive? What do you do to the fraction? How many spaces is two to the right? What do you do to find the answer? Where do you plot the coordinates if they are whole numbers?
Well, all the questions will be answered. First of all you start in the y-axis or the vertical line of the graph. Pretend the problem is seven, negative two thirds (7, -2/3). Let's look at the y-axis, in slope we start with the y-axis first. Since it is positive seven we go up seven and plot our point there. Then do rise over run with the fraction. Negative two will be the rise, however since it is negative go down two from seven, which will be six. Then since three is a positive it will go three spaces to the right. Now plot your second point and connect them. That's your slope. It is a negative slope for it you read it left to right and it is up to down.
Another different example is fifteenth, twenty-seven(15, 27). If both of the numbers are whole number in slope then the slope is always zero, a straight line. To solve this start at fifteeen in the y-axis and go up twenty-seven more for the rise. Next connect the dots and there's your answer. As you can see slopes are really simple. Just use the rise over run to help you with the fractions. There are lots more ways that the problem is written out so this is just a couple of examples.
Which is a better deal, buying a 12 pack Mountain Dew or one liter Mountain Dew. Well, a 12 pack cost 12.99 and a liter cost 5.39. To find the better deal you find the cost of one of the 12 pack cans ounces. Divide 12 by 12.99, which is a long number, so I round to the nearest hundredths. The amount per ounce is 0.92. Next divide one by 5.39, which also happens to be a long number, so I round to the nearest hundredths and got 0.19. As you can see buying the liter cost you less and is a better deal.
Some other options you can use this for is when buying meats, fruits, and sending a package. All these things you need to choose which is a better deal by the weight of cost.
However, sometimes it could be something else. Like sleeping in a hotel where sleeping over in 1 night cost $50 sleeping over in three nights cost $175.
A topic from last semester that reminds me most of math is arithmetic. It is a math problem that is mixed with lots of other lesson to solve it. A problem may have more than three lesson combined together. Such as this problem: /2-(-3)/+8. Firstly, the first step you do is the inside of what's in the absolute value sign, which was 2-(-3). If two negatives are together like this problem, change the two negatives to a positive or a plus. Now it is 2+3, which equals five. And the absolute value sign change it to a positive, but it is already positive, so 5 stays the same. Next is the simple step, you just add 5+8=13. After the problem we have already done three step to solve this problem. So as you can see, arithmetic has almost all the steps of math to solve it.
All the step that involves in the problem I have already learned, so it isn't really hard to solve the problem. Sometimes the problem could turn out to be real easy when you simplify it down.
Some of my main problems were with division. In third grade I never get the point of division. It was a confusing step in math for me. I was taught, dad, mom, sister, and brother or divide, multiply, subtract, and bring down. Even with these step I still do not get it if someone didn't guide me through. The mistake I've made in division is that I do not no the steps in order even with the d, m, s, and b. It was a big step from what I was use to learning, and that was one of the first hardest challenge I had in math.
My confusing thought of division was the step to divide into the first of second number. The first step! I was suppose to divide the divisor into the number. Then I did not know my multiplications until a long study in one night! The second step was to multiply how many times the divisor into, number then write it on the top of the line or on top of the number. The third step was to subtract the two numbers then bring down the next number and do the step all over again. When you reach the end of the problem if there is an extra number, that is called the remainder. It is the left over the number and to write the answer out put the answer then lower case or upper case ''r'' and write the remaining number. You may not get it because it is a confusing thing to describe, however you'll understand better with the problem written down.
61R1
6/367
-36
07
-6
1
There are many uses of Pythagorean, especially for triangles. The formula for Pythagorean is A^2+B^2=C^2. The ''A'' is usually the shorter line of the right triangle, while the B and C are longer. We use this method to find one of the side. The name for side ''A'' and ''B'' are ''Leg A'' and ''Leg B.'' However the name for side ''C'' is different; it is hypotenuse. There are two ways of doing the Pythagorean. To solve for sides ''A'' and ''B'' is to do it like equations. It doesn't really matter what order you put side ''A'' and ''B'', anyways if side ''C'' does matter. The number for side ''A'' and ''B'' should be given. Then write it out: ex.8^2+6^2=C^2. 8^2 equals 64 and 6^2 equals 36. Next add 64 and 36, it should equal 100. Afterwards find the base of 100 and C^2, which equals 10 and ''C." Therefore''C=10."
However if you get side ''A'' and ''C'' or side "B" and "C", there's a different way of solving it. For example if "B" equals 3^2 and "C" equals 5^2, then side "B" equals 9 and side "C" equals 25. Next subtract 9 from itself and cross it off, then subtract it from 25, which equals 16. Then find the base of A^2 and 16. It is "A" and 4. Which meant "A" equals 4.
A funny situation I thought of was that a hunter was hunting for deers in the woods. He had went 3^2 miles in the woods. Then he heard a rustling sound that startled him, so he ran 4^2 miles away from the sound. However he got lost and needed to track how many miles he covered. Luckily his compass was recording the miles, so he didn't have to guess. Dependent on the compass, he forgot it was set to go back where ever he left. During a long walk he thought to do Pythagorean and figured how many miles he needs to go. He ended up with 25 miles left. So concentrated on his compass he didn't notice that he had reached his starting point. Again he heard the loud rustle,but now he did not know where to go.
A personal experience I have, was racing with my siblings. We ran 2 yards, turn left, and ran 3 yards back. To solve this I convert yards to inches. First we ran 72 inches, then turn, and 108 inches. I find 72^2 and 108^2, which equals 5184 and 11664. Then I subtract 5184 from 11664, equals 6480. Afterward, I find the base of 6480 and the turn. The turn equals a number in the hundredths.
Square roots are called square roots because they are the squared of the the original numbers. In other words the original number multiply by itself is is called a square root. Imagine a square with the number 36 in it. That's the square and to find the root or square root you have to divide it by a number and see if the quotient turns out the same as the divisor. If it does, then that's the answer. Afterwards each of the two sixes would be written on the bottom and left side of the square.
Usually the square roots needs the product of the same number times itself. But if there are other cases, such as the root is 14, what could you do. Well first you will find square roots that are divisible and choose the two roots that the products lay closest with the number. The answer could be written between 3 and 4, which are the roots of 9 and 16. If you want to find the exact whole number subtract the products from the number.(Notice that if the product or number is bigger than each other it doesn't matter the point is to find the difference so there should be no integers...) 16-14=2 and 14-9=5 so the answer is 4 because there is far less difference from 16 and 14 than 14 and 9. So that is how you solve square roots problems. A name for the square root could be a number exponent two because two is the number that tells how many times the number needs to multiply itself. Another name could be a number two to the second. It is also the area of a square.
5^-2 equals 1/25 because five to the second power equals 25, it is five times five. Most people forget or don't get this part, but it is actually really easy. Since the exponent is a negative you put a one on top of the 25, and that makes 1/25. You always put a one on top of the answer when the exponent is negative. 1 represents the negative, so you do not put the negative after solving the problem.
Another example could be 9^-5. Nine times nine equals eighty-one, and nine times nine again equals eighty-one. Eighty-one times eighty one equals six thousand, five hundred fifty-one. Then five times six thousand, five hundred fifty-one times five again equals thirty-two thousand, eight hundred five. Then change the negative into a one and the answer will be one thirty-two thousand, eight hundred five(32,805). So as you can see this is a very simple way of solving the answer. All you have to do is to do the exponent out and add the one. (REMEMBER it is always okay to show out your work or the problem.)
I don't see much of exponents anywhere, but only in math. Exponents is used for multiplying how many times number a number multiply itself together. It looks like a smaller version of the number and is up on the right hand corner of the number. When you solve exponents you multiply the number how many times the exponent tells you. Like if you have four and an exponent that says five, don't multiply the it together. Since the exponent is five do four times four times four times four times four. Four times four is sixteen, sixteen times four is sixty-four, sixty-four times four is two hundred fifty-six, and two hundred fifty-six times four is one thousand twenty-four. So the answer is one thousand twenty-four. Exponents are shorter ways to write out the whole repeating problem. However they are a little complicated to solve because sometimes people forget and multiply the exponent with the number.
To solve exponet is to know the process too. If you do not know it well sometimes it looks like another type of problem you've never seen before. So examine the problem closely because there are also several kinds of exponent problems. Such as exponent that must turn to a fraction when solved or exponent used in solving for (x).
I learned that from playing this math game online it helps me think faster. It is labeled in circles and lines. The circles are across, vertical, and center in a squared way. The lines connect all the circles together. I played the three games, decimals, money, and integers three times each. I think the hardest one is money because it involves more steps. I try solving all the problems in my head and completed the three games.
My method in some integer problems were to follow the direction because it was just adding and subtracting. However I did used a method I learn from math class to solve problems with negatives in it. The problem was a positive minus a negative. Our teacher taught us how to use ''Keep, Change, Change'' and I keep the positive, change the minus to a plus, and change the negative to a positive. Then I only had to add to solve the problem. My method I used for decimals is remove the decimal, add/subtract the the number that in the farthest right and if I need to regroup in the farthest right I subtract the second to last number by one. Instead of regroup I also add when needed. Like add the front number of how much the two farthest right hand number equal to the second to last number. Then solve the problem with adding back the decimal where it is suppose to be or keep on repeating the method to solve the problem if I need to. I use the same method in decimals with money, except I only go up to the hundredth place. I think this a great game to improve my speed and skills of math.
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