Category Archives: Mathematics

Graphing Linear Functions Practice Worksheet

Linear functions are functions where the solution forms a straight line. This worksheet will give you a chance to practice graphing these functions.

Graphing Linear Functions WorksheetThe equations included in this worksheet are all simple equations where the graph will fit on the included 10×10 grid. Download the PDF and sketch the graph of the functions.

This can be done a couple of ways. One way is to form a T- table of a few values of x. Plot those points and connect the line it forms. Another way is to solve the equation to the slope-intercept form y = mx + b. Plot the y-intercept and use the slope to count off other points to connect.

Once you finish, check your work with this completed answer sheet.

Graphing Linear Functions Worksheet Answer SheetIf you’d like, you could download the PDF of the completed worksheet.

Slope Intercept Form – Equation of a Line Example

Slope Intercept Example Line 1Slope intercept form is a function of a straight line  with the form

y = mx + b
where
m is the slope of the line and
b is the y-intercept.

Slope refers to how steep the line changes as the values of x change where the y-intercept is where the line crosses the y-axis.

Here are a pair of worked examples to show how to find the equation of a line from different initial conditions.

Slope Intercept Form Example 1

This example shows how to find the equation of the line when given the slope and one point on the line.

Q: What is the Slope intercept form of the line with slope 23 and passes through the point (-3, -4)

Slope Intercept Example 1 GraphHere’s the graph of that line. A slope of 23 means the line will rise 2 points for every 3 points of x to the right. If we start at the point (-3, -4) we can count 2 up and three over to see the way the line will progress.

The slope intercept form is

y = mx + b

To solve this, we need to know both the slope and the y-intercept. We know the slope, so we need to find the value of b. We know from the given:
m = 23
x = -3
y = -4

Plug these into the equation

-4 = 23 (-3) + b
-4 = -2 + b
-2 = b

Now we have all we need for the equation of the line.

y = 23 x + (-2)
y = 23 x – 2

We can see from the graph the line crosses the y-axis at -2 but it nice to have a second opinion.

Slope Intercept Form Example 1 – Alternate Method

There is an alternate method to solve this problem where you can memorize an equation to eliminate the first couple steps. That formula is

(y – y1) = m(x – x1)

where x1 and yare the coordinates of the given point.

Let’s plug in the point (-3, -4) from above.

(y – y1) = m(x – x1)
(y – (-4)) = 23(x – (-3))
y + 4 = 23(x + 3)
y + 4 = 23x + 23(3)
y + 4 = 23x + 2
y = 23x + 2 – 4
y = 23x – 2

Slope Intercept Form Example 2

The second example, two points of the line are given in the initial conditions.
Let’s find the same line as before. The red line passes through points (-6, -6) and (9, 4). Find the equation of the line.

First we need to find the slope.

The formula to find slope between two points (x1, y1) and (x2, y2) is

Slope formula

For our two points:
x1 = -6
y1 = -6
x2 = 9
y2 = 4

Plug these into the formula

slope intercept example step 1 math
slope intercept example step 2
m = 1015
reduce the fraction by factoring out a 5
m = 23

Now we can find the y-intercept in the same way we did above. Choose one of the points for x and y. Let’s try the (-6, -6) point.

y = mx + b
-6 = 23(-6) + b
-6 = –123 + b
-6 = -4 + b
-2 = b

Just to show it doesn’t matter which point you choose, let’s use the (9, 4) point.

y = mx + b
4 = 2(9) + b
4 = 183 + b
4 = 6 + b
-2 = b

Plug into the slope intercept formula

y = 23 x + (-2)
y = 23 x – 2

Which matches what we expected to see from the first example.

The key to this type of problem is to find the slope of the line. Once you have the slope, finding the y-intercept is easy. For more examples of finding the slope, check out the What is Slope? page and examples

What Is Slope? How to Find the Slope of a Line

Graph of a Pair of Straight LinesWhat is slope?

Simply put, slope refers to the steepness of a line. The larger the slope, the steeper the line.

Slope is often referred to as ‘rise over run’ because it is calculated by the change in the vertical (rise) divided by the change in the horizontal (run).

When calculated, the value of slope can tell you how steep the line is or its general direction. For example, a high value of slope means a very steep line. A positive value of slope means the line is rising as it moves along the x-axis. A negative slope means the line is falling as it travels along. A flat line is said to have no slope. In this picture, the red line has positive slope. The values of y are increasing as you move along the x-axis. The green line has negative slope since the values of y are decreasing as x-increases.

The formula to calculate slope is

Formula for Slope

where
m is the slope
Δy is the change in y values and
Δx is the change in x values.

Let’s use this formula to find the slopes of the two lines above.

What Is the Slope of the Red Line?

Graph of a LineTo find the slope, we need to know two points on the line. I’m going to choose two obvious points: (-2,2) and (6,6).

Graph of a Line 2

Formula for Slope
or
slope formula
from the points I chose:
x1 = -2
y1 = 2
x2 = 6
y2 = 6

Plug these into the formula:
Step 1 of what is slope red line
2nd step of What is slope red line
3rd step of what is slope red line
m = ½

The slope of the red line is ½. This means for every two units of x, the line will rise one unit. Two over, one up. Follow the path of the line and see that it is true. Now let’s try the green line.

What Is Slope of the Green Line?

slopeexample2

This line decreases as it moves towards the right. This means we should expect the slope to be negative. Let’s check.  First, pick two points on the line. I’m going to choose (-3, 5) and (1, -7).

slope formula
x1 = -3
y1 = 5
x2 = 1
y2 = -7

Plug these into the formula:
What is Slope Green Line step 1
What is slope green line example step 2
What is slope green line example step 3
m = -3

The slope is negative like we expected. As x increases one point, the value of y will decrease by three points.

Just to show it doesn’t make any difference which point you choose, let’s switch the two points around: (1, -7) and (-3, 5). Plug in these values:

x1 = 1
y1 = -7
x2 = -3
y2 = 5

Slope formula
What is Slope Green Line step 4
What is Slope Green Line step 5
What is Slope Green Line step 6
m = -3

Notice how we got the same value and it didn’t matter which points we called (x1, y1) and (x2, y2). The important thing to keep track of is once you choose, maintain that choice through the entire problem.

 

SOHCAHTOA Example Problem – Trigonometry Help

SOHCAHTOA is the mnemonic used to remember which sides of a right triangle are used to find the ratios needed to determine the sine, cosine or tangent of an angle. Here are a pair of SOHCAHTOA example problems to help show how to use these relationships. If you have no idea what SOHCAHTOA means, check out this introduction to SOHCAHTOA.

SOHCAHTOA Example Problem 1

A mathematically inclined squirrel sits atop a 14-foot tall tree. He spies a nut on the ground some distance away. After careful measurements, he determines the nut is 74º from the base of his tree.
How far away is the nut from:
A) The base of the tree?
B) The math squirrel?

Here is a layout of the problem.

Squirrel and Nut SOHCAHTOA Example Problem

Part A: How far is the nut from the base of the tree?

Looking at our triangle, we see we know the angle and the length of the adjacent side. We want to know the length of the opposite side of the triangle. The part of SOHCAHTOA with these three parts is TOA, or tan θ = opposite / adjacent.

tan θ = opposite / adjacent
tan ( 74º ) = opposite / 14 ft

solve for ‘opposite’

opposite = 14 ft ⋅ tan ( 74º )
opposite = 14 ft ⋅ 3.487
opposite = 48.824 ft

The nut is 48.824 ft from the base of the tree.

Part B: How far is the nut from the math squirrel?

This time, the needed distance is the hypotenuse of the triangle. The part of SOHCAHTOA with  both adjacent and hypotenuse is CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos ( 74º ) = 14 ft / hypotenuse
hypotenuse = 14 ft / cos ( 74º )
hypotenuse = 14 ft / 0.276
hypotenuse = 50.791 ft

We can use the Pythagorean Theorem to check our work.

a2 + b2 = c2
(14 ft)2 + (48.824 ft)2 = c2
196 ft2+ 2383.783 ft2= c2
2579.783 ft2 = c2
50.792 ft = c

The two numbers are close enough to account for rounding errors.

SOHCAHTOA Example Problem 2

A prince has arrived to rescue a princess standing in a tower balcony 15 meters up across a moat 12 meters wide. He knew the princess was high up and there was a moat, so he brought the longest ladder in the kingdom, a 20-meter monster.

A) Is the ladder long enough?
B) If the top of the 20 m ladder touches the balcony, how far away from the tower wall is the bottom of the ladder?
C) What is the angle between the ground and the ladder?

Here is an illustration of our situation.

Princess rescue trigonometry examplePart A: Is the ladder long enough?  To know if the ladder is long enough, we need to know how long it needs to be. At the very least, the prince’s ladder must reach the 15-meter balcony from the edge of the 12-meter moat. Use the Pythagorean theorem to find out how far up the tower wall a 20-meter ladder will reach.

a2 + b2 = c2
height2 + moat2 = ladder2
height2 = ladder2 – moat2
height2 = (20 m)2 – (12 m)2
height2 = 400 m2 – 144 m2
height2 = 256 m2
height = 16 m

A 20-meter ladder can reach 16 meters up the side of the tower. This is a meter longer than the 15 meters needed to reach the princess balcony.

Part B: When the ladder touches the edge of the balcony, how far away from the tower wall is the bottom of the ladder?

The first part measured the height the 20-meter ladder reached when the ladder was placed at the edge of the moat. We found we had more ladder than needed. If the ladder touches the balcony, we know the height it reaches is 15 meters. The ladder is still 20 meters long. We just need to find out how far from the tower to stick the bottom of the ladder. Use the Pythagorean theorem again.

a2 + b2 = c2
(balcony height)2 + (ground distance)2 = ladder2
(ground distance)2 = ladder2 – (balcony height)2

(ground distance)2 = (20 m)2 – (15 m)2
(ground distance)2 = 400 m2 – 225 m2
(ground distance)2 = 175 m2
ground distance = 13.23 m

The ladder’s base is 13.23 m away from the tower.

Part C: What is the angle between the ground and the ladder?

We were given the height of the balcony, which in this case is the ‘opposite’ side of the triangle from the needed angle. We also know the length of the ladder that forms the hypotenuse of the triangle. The part of SOHCAHTOA that has both of these parts is SOH, or sin θ = opposite/hypotenuse. Use this to solve for the angle.

sin θ = opposite/hypotenuse
sin θ = 15 m / 20 m
sin θ = 0.75
θ = sin-1(0.75)
θ = 48.59°

You can check your work in Part B if you use the answer we got as the ‘adjacent’ side of the triangle.

If you use the ladder as the hypotenuse, use CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos θ = 13.23 m / 20 m
cos θ = 0.662
θ = cos-1(0.662)
θ = 48.59°

The angle matches the value above.

If you choose the balcony height as the opposite side, use TOA, or tan θ = opposite / adjacent

tan θ = opposite / adjacent
tan θ = 15 m / 13.23 m
tan θ = 1.134
θ = tan-1(1.134)
θ = 48.59°

Pretty neat how it doesn’t really matter which sides of the right triangle you use, as long as you use the correct part of SOHCAHTOA.

 

Right Triangle Trigonometry and SOHCAHTOA

Sohcahtoa isn't actually an Egyptian god, but if it helps to remember him that way, you'll have an easier time recalling right angle trig relationships.

Sohcahtoa isn’t actually an Egyptian god, but if it helps to remember him that way, you’ll have an easier time recalling right angle trig relationships.

Right triangles are extremely common in science homework. Even though they are common, they can be confusing to new students. That is why we have the Egyptian god SOHCAHTOA.

SOHCAHTOA is a handy mnemonic trigonometry students learn to remember which sides of a triangle are used for the three main trig functions: sine, cosine, and tangent.

These functions are defined by the ratios of various lengths of the sides of a right triangle. Let’s look at this right triangle.

Right TriangleThis triangle is made up of three sides of lengths a, b and c. Note the angle marked θ. This angle is formed by the intersection of b and c. The hypotenuse is always the longest of the three sides and opposite of the right angle. The side b is ‘adjacent’ to the angle, so this side is known as the adjacent side. It follows the side ‘opposite’ of the angle is known as the opposite side. Now that we have all our sides labelled, we can use SOHCAHTOA.

SOHCAHTOA

S – Sine
O – Opposite
H – Hypotenuse

C – Cosine
A – Adjacent
H – Hypotenuse

T – Tangent
O – Opposite
A – Adjacent

SOH = sin θ = opposite over hypotenuse = ac
CAH = cos θ = adjacent over hypotenuse = bc
TOA = tan θ = opposite over adjacent = ab

Easy to remember. Now let’s see how easy it is to apply.

Example Problem

Consider this triangle.

trig example for SOHCAHTOAThe hypotenuse has a length of 10 and one angle of the triangle is 40º. Find the lengths of the other two sides.

Let’s start with the side with length a. This side is opposite the angle and we know the length of the hypotenuse. The part of SOHCAHTOA with both hypotenuse and opposite is SOH or sine.

sin 40º = opposite / hypotenuse
sin 40º = a / 10

solve for a by multiplying both sides by 10.

10 sin 40º = a

Punch 40 into your calculator and hit the sin key to find the sine of 40º.

sin 40º = 0.643

a = 10 sin 40º
a = 10 (0.643)
a = 6.43

Now let’s do side b. This side is adjacent to the angle, so we should use CAH or cosine.

cos 40º = adjacent / hypotenuse
cos 40º = b / 10

solve for b

b = 10 cos 40º

Enter 40 and hit the cos button on your calculator to find:

cos 40º = 0.766

b = 10 cos 40º
b = 10 (0.766)
b = 7.66

The sides of our triangle are 6.43 and 7.66. We can use the Pythagorean equation to check our answer.

a2 + b2 = c2
(6.43)2 + (7.66)2 = c2
41.35 + 58.68 = c2
100.03 = c2
10.00 = c

10 is the length of the triangle’s hypotenuse and matches our calculation above.

As you can see, our friend SOHCAHTOA can help us calculate the angles and lengths of the sides of right triangles with very little information. Make him your friend too.

 

What a Fractal Is and Why You Should Care

Since I’ve started making fractal art, I’ve been asked many times, “What is a fractal?” and “Yes, they look pretty, but what good are they?” Here are the basics.

What Is a Fractal?

A fractal is a mathematical equation that displays a repeating pattern, no matter what scale you examine it. It can also be described as a pattern of chaos. Fractals can be described using mathematical sets, but you also see them all the time in nature. Basically, anything that can be described using mathematical equations may be considered a form of fractal. The difference between natural fractals and pure equations is that the repeating scale in nature tend to be (or at least appear) finite. Examples of natural fractal features include many familiar patterns:

  • fern fronds
  • snowflakes
  • the rings of Saturn
  • Lichtenberg figures and lightning
  • DNA
  • heart beats
  • trees
  • river systems
  • mountain ranges
  • Brownian motion
  • coastlines
  • the stock market
  • blood vessels
  • nautilus shells
  • ocean waves
The spiral shape of fern fronds is a natural approximations of a fractal. (Wingchi Poon)

The spiral shape of fern fronds is a natural approximations of a fractal. (Wingchi Poon)

Take fern fronds, for example. The spiral shape of the frond may be described mathematically. If you then view the unfurling of the smaller leaves of the frond, the spiral pattern repeats. The difference between the frond shape and the fractal equation is that you could keep “zooming in” in a graphical representation of the equation, while the natural phenomenon only covers a few iterations.

Here’s an example of a spiral-shaped fractal. See the resemblance?

Sea Slug Animated Fractal

Sea Slug Animated Fractal

Uses of Fractals

Fractals are aesthetically-pleasing art, but they have practical applications, too. In many cases, using fractals is much more efficient and accurate than physically measuring phenomena. One of the first papers linking fractals to useful analysis was Benoit Mandelbrot’s “How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension”, which he published in the 1960s and illustrated using computer-generated visualizations. (Before computers, only a few iterations of an equation could be drawn, so it was difficult to visualize the math.)

Here is the now-famous Mandelbrot Set, a recursive set of equations, so that a modern computer can zoom in to see infinite detail from the initial image:

Mandelbrot Fractal

Mandelbrot Fractal

Today, various types of fractals are used in real-life to:

  • map topology
  • model fluid transport (like human blood flow or petroleum flow)
  • to produce more efficient cooling systems for computer chips
  • to model turbulent mixing
  • to compress digital images (fractal image compression is used by most programs)
  • to predict the structure of galaxies and the universe
  • to model crystals
  • to calculate the amount of carbon in a tree based on carbon content of a single leaf
  • for analysis of earthquakes and seismic patterns
  • Fractal-shaped antennae reduce the size and weight of antennas.
  • To model drug interactions and describe the functioning of biosensors.
  • Fractals are used to describe how rough or smooth a surface is.
  • Fractals are used to help predict circulation patterns to make longterm weather forecasts.
  • to predict stock market fluctuations

And, of course, fractals make cool art:

Copper City Fractal Animated Gif (Anne Helmenstine)

Copper City Fractal Animated Gif (Anne Helmenstine)

Cool Math Tricks To Amaze Your Friends

Cool Math Trick Using 1

Cool Math Trick Using 1

Math is like magic, except better! Here is a collection of cool math tricks you can learn to amaze your friends and learn more about how numbers work.

The 11 Rule

Everyone knows the “10 Rule” where you multiply by 10 simply by adding a 0 to the end of a number, but you might not know the 11 rule. This rule works for any two digit number to multiply it by 11:

For an example, let’s use the number 62

  • Separate the two digits in your mind (6 __ 2).
  • Add together the two digits of the numbers. (6 + 2 = 8).
  • Place this number in the space or hole between the two digits (6 8 2).
  • That’s it! 11 x 62 = 682

The only tricky part to remember is that if adding the two digits results in a number greater than 9, then you put the “ones” digit in the space and carry the “tens” digit. For example:

11 x 57 … 5 __ 7 … 5 + 7 = 12

so you put the 2 in the space and add the 1 to the 5, giving you the number 627

11 x 57 = 627

Countdown Math Trick

Countdown Math Trick

Single Digit Numbers Math Trick

  1. Think of 2 single digit numbers.
  2. Take either of the numbers and double it.
  3. Add 5 to the result.
  4. Add the second number to your answer.
  5. Subtract 25 from the answer.
  6. You’ll get the 2 single digit numbers in the answer.

Using Shoe Size To Tell Your Age – Algebra Trick

There are many math tricks that ask you to supply a number to get a “hidden” number that you actually supply in a different form during the trick. The first number isn’t too important, since it gets removed during the trick, so you can change the wording of this math trick.

  1. Use your shoe size to tell your age. Take your shoe size (whole number, so round up if it’s a half size).
  2. Multiply it by 5.
  3. Add 50.
  4. Multiply it by 20.
  5. Add 1016. (if you’re doing the trick in the year 2016… if it’s 2017 use 1017, in 2018 use 1018, etc.)
  6. Subtract the year you were born.
  7. The first digit is your shoe size and the last 2 digits are your age.

As you might guess, this trick is meant to reveal a number that is less than 100. The trick uses algebra to solve for the answer. Let’s do the trick again using s for shoe size and b for birth year:

Multiply s x 5: 5s

Add 50: 5s + 50

Multiply by 20: 20(5s + 50) = 100s + 1000

Add 1016 (depending on current year): 100s + 1000 + 1016 = 100s + 2016

Subtract birth year: 100s + 2016 – b

Why does it work (and why does it sometimes fail)? No matter what your shoes size is, it will be the first two digits of the answer. If your shoe size is 9, the 100s is 900. If you use a European size chart and wear a size 36, then 100s = 3600.

The age part takes the current year minus your birth year. The trick does not take into account your birth month, so if your birthday this year has not arrived, the answer will be a year off!

1, 2, 4, 5, 7, 8 Math Trick

  1. Choose a number between 1 and 6.
  2. Multiply the number by 9.
  3. Multiply the result by 111.
  4. Multiply the answer by 1001.
  5. Divide the number by 7.
  6. The answer will contains all the numbers 1, 2, 4, 5, 7, 8.

The 1089 Math Trick

  1. Think of a 3 digit number.
  2. Arrange the digits in descending order.
  3. Reverse the order and subtract it from the number in step 2.
  4. Reverse the order of the answer.
  5. Add it to the result from step 3. You get 1089!
  • For example, let’s say I chose 423.
  • Arrange in descending order: 432
  • Reverse the order and subtract it from the previous number: 234… 432 – 234 = 198
  • Reverse the order: 891
  • Add the numbers together: 198 + 891 = 1089

The Answer Is 5

  1. Think of a number.
  2. Double it.
  3. Add 10.
  4. Divide it by 2.
  5. Subtract your original number.
  6. Your answer is 5!

Three Digits the Same Trick

  1. Think of a 3 digit number where all the digits are the same  (e.g., 333, 777).
  2. Add up the digits.
  3. Divide your 3-digit number by the added value.
  4. Your answer is 37.
Cheryl's Birthday Math Riddle

Cheryl’s Birthday Math Riddle

 

How To Solve Cheryl’s Birthday Math Riddle

“Cheryl’s Birthday” is more of a math riddle or logic puzzle that is solved using the process of elimination or deductive reasoning. Albert can’t know the birthday because he only has the month and all months have multiple dates, but he has enough information to know Bernard does not have the date. If Cheryl had told Bernard 19 or 18, then he would know the whole birthday because there is only one month with each number. This rules out May 19 and June 18.

Since Albert knows Bernard doesn’t know, Albert must have been told July or August, as this rules out any possibility of Bernard being told 18 or 19. This excludes any dates in May or June.

When Bernard says he did not know the answer, but now he does, this means Bernard has the one remaining unique number in the list. If Bernard had a 14, he wouldn’t know whether it was in July or August. If he had a 15 or a 17, he wouldn’t know which date in August was correct. Thus Cheryl’s birthday must be July 16!

If you want to get really tricky, you can rephrase the riddle to say Bernard starts out saying he doesn’t know when the birthday is, with Albert replying he doesn’t know either. If Bernard then says he didn’t know, but now he does, and Albert replies he now knows too, then the answer is August 17. Can you see why?

Do you have other cool math tricks to add? Post a reply and share them!