# Breaking Down a Story Problem

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Story problems can become frustrating if you do not understand what the problem is asking. I had a story problem in my math text-book for homework that I had to walk through step by step in order to fully understand it. The problem was broken into four parts: a, b, c, and d. Part a involved converting the units into liters of water; part b involved dividing the volume; part c involved division of a given volume; part d involved finding perimeter. Looking back on the problem gives me a different perspective because now I realized it is a great problem. It has many different parts which makes you think critically about the problem and how to find the answer. I have provided the story problem below and how I worked through it!

The story problem stated: the number of fish that can be put in an aquarium depends on the amount of water the tank holds, the size of the fish, and the capacity of the pump and filter system (Mathematics for Elementary Teachers Text book ninth edition). The picture from the story problem is above.

Part a states: How many liters of water will this tank hold? I found the volume of the tank with the formula V = Area of base x h. (50cm x 25 cm) x 30 cm = 37500 cm cubed. Once I found the volume in cubic centimeters, I converted it to liters. Since 1 cubic centimeter is equivalent to an mL, I did that first. Once I had the problem in mL it was easy to find liters because there are 1000 mL in a liter; therefore, the answer is 37.5 liters.

Part b states: The recommended number of tropical fish for this tank is 30. How many cubic centimeters of space would each fish have? This was pretty easy because all I did was divide the volume (37500) by 30 fish which equals 1250 cubic centimeters.

Part c states: Goldfish need more space and oxygen than tropical fish. Goldfish that are about 5 centimeters long require 3000 cubic centimeters of water. How many goldfish could live in this tank? For this I divided the volume (37500 cubic centimeters) by the required 3000 cubic centimeters of water; this comes out to about 12 goldfish.

Part d states: How many square centimeters of glass are needed for this tank if there is glass on all sides except the top? This question is simply asking for the surface area; therefore I found the area of all the shapes and added them together. (2 x 25 x 30) + (50 x 25) + (2 x 30 x 50) = 5750 cm squared.

With patience and effort any story problem can be solved!

# Discovering Volume

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Understanding volume can be a tricky topic because it is easy to get it confused with area and perimeter. There are some main things that can help you understand volume. First, you want to make sure that all of the dimensions are labeled in cubic units. For example, if you have a triangular prism with sides measured in centimeters, you want the answer to be in cubic centimeters. Secondly, there are separate formulas to find the volume of a prism and the volume of a pyramid. Thirdly, you want to know the shape or 3 dimensional object you are dealing with because that will help you to determine the formula you need to use. Let’s look at some examples!

The formula for a prism is V = Area of base x height of prism. A prism is a three-dimensional shape with two parallel bases that are the same. In the picture below I have drawn a right prism. The base in this case is the square; therefore, area = length x width (3cm x 3cm = 9cm squared). Once we find the area of the base, we need to multiply it by the height of the prism (5 cm). The answer to the example is 45cm cubed because 45 cm cubed = 9 cm squared x 5 cm.

The formula for a pyramid is V = 1/3 x Area of base x height of pyramid. A pyramid is a three-dimensional shape with one base of any shape with sides all meeting at one point. In the picture above I have drawn a triangular pyramid. The one base in the pyramid is a triangle; therefore, area = 1/2 length x height (1/2 x 4 cm x 8 cm= 16 cm squared). Once we find the area of the base, we need to multiply it by the height of the pyramid which in the example is 10 cm. Lastly, we multiply the whole answer by 1/3. The answer to the example is about 53.3 cm cubed. The descriptions above can help clear up any confusion, and help to explain the two different formulas for volume of a prism and the volume of a pyramid.

# Having Fun in Math Class

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Today in math class I presented an activity. I had the class divide into groups of two, and each group was given a shape; there were squares, triangles, and rectangles. The measurements of the sides were labeled to help find the area and perimeter of each shape. One person in each group was in charge of finding the perimeter, and the other person had to find the area. I did have a formula sheet under the projector to help if the students got stuck on a problem. The class did a great job with area and perimeter and understood how to find the solutions. Before the activity, we discussed circumference and diameter.

We went on a circle scavenger hunt! This was fun because we grouped into partners and went around the college measuring circles. Each group had a ruler, pencil, piece of string, and a worksheet. To measure circumference we wrapped the string around the circle and measured the length of the string with the ruler. After we found the circumference, we found the diameter. This was done by placing the string on one side of the circle’s face and measuring the length across. Some objects that my partner and I found were a master lock, door knob, and elevator button. The scavenger hunt was fun and productive because it helped me understand how to find the circumference and diameter of a circle.

Although the scavenger hunt was productive and helped me understand the meaning of circumference and diameter of a circle, there is a formula for circumference. Circumference = ∏ x diameter or ∏ x 2radius. (∏=pie) Radius can be found by taking half of the diameter. If you have a circle with a diameter of 2, you can multiply it by ∏ and get 6.28; therefore, 6.28 is the circumference of your circle. If you have a circle with a radius of 3, you can multiply it by 2 then multiply it by ∏ and get 18.85; therefore, 18.85 is the circumference of your circle. I have included a picture of these two problems below to clear up any confusion you may have.

# Drawing Shapes

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Drawing shapes may sound simple, but it can be challenging if you are trying to draw a regular shape. The difference between a shape and a regular shape is that a regular shape has equal sides and equal angles. For example, a pentagon may have 5 different side lengths and 5 different angle degrees, and a regular pentagon would have 5 equal side lengths and 5 equal angles.

In my math class we had to draw a regular hexagon and a regular decagon. In order to do this we had to use a protractor, which allowed us to make equal angles and straight sides. First, we determined the central angle of the shape. Central angle = 360/number of sides; therefore, to find the central angle of a hexagon we divided 360 by 6. 60 degrees is the central angle. This means that every line we draw in the hexagon must be drawn at a 60 degree angle. The picture below is the worksheet that includes the drawings of a regular hexagon and a regular decagon.

As well as finding the central angle, it is helpful if you find the vertex angle. Vertex angle = 180 – central angle. I have highlighted the vertex angle in the worksheet above. This helps to determine the degrees of every angle in the shape. If you were to draw a regular decagon you must find the central angle first (360/10 = 36 degrees). Once you have the central angle you can find the vertex angle (180-36 = 144 degrees). Understanding how to find the central angle and vertex angle can help you determine any angle of a shape; I think these two formulas are great to know!

# Tricky Significant Digits

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Significant digits can be a frustrating concept if you do not understand the rules. Significant digits are the digits needed to have a precise answer. If you were to take a chemistry class, then you will be familiar with them. Recently, I began working with significant digits in my math class. I think the more you work on a concept the more you understand it. There are different rules when dealing with significant digits. Adding and subtracting numbers have different rules compared to multiplying and dividing numbers.

Adding and subtracting numbers focuses on the accuracy of the answer. The number with the least amount of decimal places is the “winning” number. For example if we are adding 5.36+2.2, the “winning” number is 2.2 because it has the least amount of decimal places. The answer to this problem is 7.56, but since we need to have 1 decimal place the answer is 7.6. In the picture below I have provided an example of a subtraction problem.

Multiplying and dividing numbers focuses on the least number of significant digits. Instead of looking at the amount of decimal places, we look at the whole number and count the amount of significant numbers; therefore, the “winning” number is the number with the least amount of significant digits. The solution to the problem should be rounded to the same amount of numbers as the “winning” number. For example if we are multiplying 82.1×3.2, the “winning” number is 3.2 because it only has 2 significant digits. In the picture above I have included a division problem and some tricky problems dealing with zeroes. I think with effort and determination anyone can understand the rules and differences between adding and subtracting numbers compared to multiplying and dividing!

# What is the difference?

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What is the difference between rational and irrational numbers? A rational number can be written in a simple fraction, and an irrational number cannot. I think rational numbers are easier to understand because they can be written in a ratio. I have provided you with an example of a rational number below. On the other hand, irrational numbers can be tricky.

Irrational numbers cannot be written in a ratio; therefore, it can be hard to determine what a number is equal to. In the picture above I have given an example of an irrational number. Nonrepeating decimals are an example of irrational numbers. 8.08008000800008… is an example because there are no repeats in the numbers.

Rational and irrational numbers are both real numbers. There are also non-real numbers; these numbers are usually not learned until higher grade levels, but they do exist. An example of a non-real number is the square root of -4. Having a negative number under a square root is a sign that the number you are dealing with is non-real. This is a good tip to know if you have to categorize numbers into different groups. I experienced this while I was working on a worksheet in my math class. Understanding the difference between real and non-real numbers helped me on my worksheet and can help you too!

# Picking Apart Percents

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There are multiple ways to look at percents. One way to look at percents is by using the “easy 10%” method. The “easy 10%” method simply means to move the decimal to the left one spot. If you are teaching this to a group of young students it may be easier to understand this method by saying you are making the number smaller. A couple of  examples are 10% of 53 = 5.3 and 10% of 249 = 24.9. In both of these cases we move the decimal to the left one spot which makes the number smaller.

Understanding the “easy 10%” method can help you figure out other percentages. If something in 30% off, you need to find 10% off of the item, then multiply it by 3. An example of this could be if an item costs \$25.00 and it is 30% off. We can start by finding 10% of the item which is \$2.50, then multiply that by 3; this will equal \$7.50. I think percentages are very important to learn because they are common in our lives. Everytime we shop we can use percentages to figure out our discounts, coupons, and the total of our purchase!

Another way to look at percentages is by using percent charts. If we are taking 50% of 200, we can set up a 10×10 grid with each unit equalling 2. By setting the grid up like this we are essentially multiplying 2×100 which makes 200. In order to get 50% of this grid we would fill in 50 units. In this case 50 units would equal 100 because each unit is equal to 2. This example is drawn in the picture provided.