Math 10 summary

Unit 1: Numbers

Key points:

 

The entire basis of chapter 1 is knowing how to find the prime factor of numbers. This allows you to find the gcf and lcm of numbers, and help solve some entire and mixed radical questions.

to find the prime factorization of a number, you first need to know the list of prime numbers up to about 50. Here is a list up to 100,

download

now what you do is take the number you are factoring, and try dividing by a prime number. Here are some tricks to save time looking for a number, if the number is even, it will go into two. If it ends in 0 or 5, it will go into 5. If it is a 2 digit number and both digits are the same, it will go into 11. If the number does not fit into any of the guidelines, start with 3 and work your way up. keep trying numbers until you are left with 1. That set of prime numbers you are left with is the prime factor of that number!

here is an example,

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Unit 2: Exponents 

In my opinion, the most key part of the exponents unit is the rule “Flower Power”. The mistake I most commonly find myself making out of all of math 10 is when I am a number with a fraction as an exponent. To do this you must first draw a square root sign with the base going to the radicand in the square root, the denominator going to the index in the square root and the numerator going to the exponent of the radicand in the square root. This may seem a bit wordy, so here is a visual to help,

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Also, if the denominator is 2, there is no need to write that in the index spot since all regular square root have the index of 2 already there.

Where the flower power part comes in is when you are looking at the fraction and cant remember what goes where in the square root, think of the denominator as the bottom part or root of the flower, and put the denominator in the root part of the square root.

 

Unit 3: Measurement 

An important part of measurement that people often make mistakes of in converting inits that are cubed or squared. It is a fairly simple switch from doing it normally, but can often be forgotten. When converting, you MUST do the multiplication of that certain measurement the number of the dimensions of the unit. ex. squared = 2x, and cubed = 3x. A simple way to remember this is if you see an exponent, do whatever number it is that many times for the conversion. If you were converting 53cm to cm, you would multiply 53 by 100/1 or 100 to get 5300cm³. If you were converting 53m³ to cm³, you would  multiply 53 by 100/1 or 100 3 times to get 53000000cm³. See how big a difference 1 small change can make?

 

Unit 4: Trigonometry 

A big mistake in unit 4 is regarding the angle of depression. Here is the triangle we will be going off of,

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The angle of elevation is angle bac or cab, which is on the inside of the triangle and facing upwards. Because of this, you would assume that the angle of depression is angle abc or cba which is also on the inside of the triangle but facing downwards. Howevery the angle of depression is actually the angle on the outside of point abc or cba facing downwards. Here is a visual, x = angle of depression

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If we are given the measurements ca = 5m, bc = 4m, and angle cab = 34°, we can find the angle of depression. Here is an equation solving for cba or y,

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Now, since the angle of a straight line = 180°, we can find the angle of depression by 90 – 51.3 = 38.7

this makes all the angles add up to 180°, so the angle of depression = 38.7°

 

 

Unit 5: Polynomial operations

When simplifying polynomials, I find the most used step is FOIL. It is used when distributing a bracket with 2 or more terms into another bracket with 2 or more terms. The letters FOIL stand for First, Outer, Inner, Last. First means multiply the terms which occur first in each binomial. Then Outer means multiply the outermost terms in the product. Inner means multiply the innermost two terms. Last means multiply the terms which occur last in each binomial. Then simplify the products and combine any like terms which may occur.

Here is an example of me going through the steps

  1. (x + 5) (x + 5) Multiply the first terms in each bracket (x),  → (x + 5) (x + 5) = x²
  2.  (x + 5) (x + 5) = x² Now take x and multiply it with 5, the outer most terms in each bracket (x and 5) →(x + 5) (x + 5) = x² + 5x
  3. (x + 5) (x + 5) = x² + 5x  Next multiply the two innermost terms together (x and 5) →(x + 5) (x + 5) = x² + 5x + 5x
  4. (x + 5) (x + 5) = x² + 5x + 5x Now multiply the two last terms in the brackets together (5 and 5)  →(x + 5) (x + 5) = x² + 5x + 5x + 25
  5. The last step is to collect any like terms. The only two like terms in this question is 5x and 5x which when collected will get you the final answer of  x² + 10x + 25

 

Unit 6: Factoring polynomial expressions 

When factoring an polynomial in the x² + bx + c, it is important to remember that the factors of the 2 last terms in the equation must multiply to = c, and must add to equal bx. For example if I am factoring x² + 3x + 2, the x² will be the first terms in each bracket (x +  ) and (x +  ). Now, we list all the factors of the outermost term. For 2 it is -2 x -1, and 2 x 1. Now we pick the set that also adds to = 3x. -2 + -1 + -3, and 2 + 1 = 3, so the right set of numbers to use is 2 and 1. Now we plug them into the bracket to get (x + 2) (x + 1). You can check your answer by re-expanding the equation.

 

Unit 7: Relations and functions

An important part of relations and functions is finding the x and y intercepts of equations. If we were to find the x intercept of the equation, we know that the coordinates for an x intercept must be (?,0). The y is always zero, same for the y intercept, except the x is zero. Now you plug the 0 into whatever equation you are doing and solve for the variable. Here is an example,

y = 4x + 8

to find the x intercept we plug zero into then y spot

0 = 4x + 8

next we begin to isolate x by removing 8. Minus 8 on both sides

-8 = 4x

now divide 4x by 4 to completely isolate x. Do it on the other side as well

-8/4 = x

finally, simplify the equation

-2 = x-int

 

Unit 8: Characteristics of linear relations

The main point of unit 7 is finding the slope with 2 sets of 2 given coordinates. Slope is used to find the equation of a line, and finding things like x and y intercepts of graphs. The basic formula for slope is rise over run, which translates to m = (y2 — y1) / (x2 — x1)

Slope_formula_2

it is a common mistake to put y as the numerator and x as the denominator, so to remember this think of the word run. If you were to run on the graph, you would be running straight across, or on the x axis. So x would go where run is, which is on the bottom. Here is a written example of me finding the slope using a given pair of coordinates,

IMG_20160622_114500

 

When plugging in the y and x into the equation it does not matter which is which. However, if the y in one set of coordinates is y2, then the x must be x2 as well, not x1.

 

Unit 9: Equations of linear relations

An important skill learned in unit 9 is being able to convert general form into slope y-int form and vice versa. Most questions will either start off in general form, or start in slope y-int form and answer in general form. To solve an equation in general from, you must first change it to slope y-int form. Here is a video I found useful while doing this,

 

when converting to general form, it is important to note that the first number or Ax, must be positive. If it is negative you must divide it by -1, along with all the other numbers.

Week 17 – Math 10

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Here is one of the questions I had trouble with. The mistake I made is that I tried subtracting the equations (x+3y=10 – 5x-3y=14) rather that adding them (x+3y=10 + 5x-3y=14). This did not create a zero pair when subtracted so I had nowhere to go in the equation

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When I tried adding the equation, a zero pair was made with 3y and -3y, so I was able to isolate x, then plug x into one of the equations to find y.

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Week 16 – Math 10

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This is one of the questions I had trouble with. When converting from slope y int from to general form, I would get the numbers correct, but half the time the negatives and positives would be reversed. this is because when writing an equation in general form (Ax+By+C=0), I did not know the ”A” had to be positive.

here is the wrong answer,

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here is the correct answer,

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What I do now when converting to general form, is if the Ax part of the equation is negative, I move it to the other side and inverse it along with the C, If it is positive, I leave it where it is and move the Y to the other side of the equation and inverse that.

Week 13 – Math 10

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(To view mindmap, right click and press open link in new tab)

Expressing Relations: There are 4 ways we learned to express relations; An equation, graph, ordered pairs, and table of values. 

Equations:download (2)

graph: images

ordered pairs: download (3)

table of values: download (4)

cartasien graph: download (5)

Discrete data: points in the graph are not connected. The data is counted rather than calculated, so the data is often put as whole numbers.

Continuous data: points in the graph are connected. The data is calculated, so the points between points could be any real number.

Linear: the line in the graph is straight.

Nonlinear: the line in the graph is bent or curved.

 

Input: The input in a relation, or x is the independent variable because its value is not effected by another value. The input is always placed on the horizontal axis on any graph

Domain: the values of x on a graph. If the domain is being determined using discrete data, you line up the points on the graph using the x axis, ex. domain:  {2, 3, 4, 6}. If using continuous data, you must include all real numbers between the points, ex. download (6)

x intercept: to find the x intercept in a equation, leave x as it is and replace y with the value of 0, then isolate x, ex. download (7)

 

Output: The output in a relation, or y is the dependent variable because its value is dependent on another value. The out put is always placed on the vertical axis on any graph

Range: the values of y on a graph. If the range is being determined using discrete data, you line up the points on the graph using the y axis, ex. range:  {–3, –1, 3, 6}

If using continuous data, you must include all real numbers between the points, ex. download (6)

y intercept: to find the y intercept in a equation, leave y as it is and replace x with the value of 0, then isolate y, ex.  download (8)

 

Functions: A function is a relation, but each input only has 1 output. For example, if you plug 2 in a function, you can only get the output of 3, not 5 or 7.

download (9)

A quick way to determine whether a graph is a function or not, is placing your pencil vertically on the graph and sliding it from one end to the other. if the pencil and line on the graph intersect more than once on any point on the graph, it is not a function and if it does not, it is a function.

There are 5 similar, but different ways we learned to write functions as equations. The different ways you can write them are called notation.

 

Function notation: f(x)=3+x

mapping notation: f:x—>5x

set notation: {x>8}

Interval notation: [3, 8)

two variables: y=3x+2

 

 

Week 12 – Math 10

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This is one of the questions I had trouble with. I had to find the y-intercept of this equation. When finding the y-intercept, you are supposed to input x as 0, because in a y-intercept on an xy graph the x would have to be zero. Here is a picture that shows this,

download (1)

However, I made the mistake of imputing y as 0, instead of x. Here is the incorrect equation,

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The 3x should have cancelled out, instead of the 2x. Here is the correct equation,

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To not make this mistake again, I wrote down this multiple times,

X-int = (x,0)

Y-int = (0,y)

Week 9 – Math 10

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One of the questions in the homework I had trouble with was this one. I was making the mistake of subtracting 2 from 8, then distributing the product into the bracket and finally adding like terms. When I did this, the incorrect answer I was getting was this,

IMG_20160417_174542

 

The correct way of doing this question is distributing the nearest term to the bracket, then adding like terms, rather than taking the nearest term to the bracket and multiplying, dividing, adding, or subtracting it by whatever is beside it, then distributing and adding like terms. When I did the question properly, the answer was,

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In this unit it is important to recognize that even the smallest mistake can drastically effect your final answer, making attention to detail very important if you want to get a high mark.

Week 7 – Math 10

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This is one of the questions that I had difficulty with.

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I set up he equation correctly, but I calculated 4/6 sin, instead of inversing sin, which was 4/6 sin. The incorrect answer I was getting was 0.01 degrees, which I knew did not seem to fit the triangles proportions. If I did not catch that, I would have kept making the same mistake.

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Here is the correct equation where I inversed sin. The answer was 41.8 degrees.

 

Solving Trigonometry Equations

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To get the length of AB, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse. I then chose cosine as the trigonomic ratio because the adjacent length was given, and I was trying to find the length of the hypotenuse. I then wrote cos 37 = 38/x because cos = adj/hyp and in this case, 38 = adj and x = hyp. I then isolated x by dividing both sides by 38. I then wrote each fraction as reciprocals, and calculated 38/cos 37, which gave me x =48.

IMG_20160403_205821

To get the length of AC, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse. I chose tangent as the trigonomic ratio because the adjacent length was given, and I was trying to find the length of the opposite side. I then wrote  tan 37 = x/38 because tan = opp/adj and in this case, adj = 38 and x = opp. I then isolated x by multiplying both sides by 38, then calculated 38 multiplied by tan 37, which gave me the answer x = 29.

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To get the angle of CAB, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse.  I chose the trigonomic ratio tan because tan = opp/adj, and I knew that for this triangle that opp = 29 and adj = 38. I then isolated x by moving tan to the other side of the equation, which inversed tan and changed it to tan-1.  I then calculated 29/38 tan-1  , which gave me 37.

 

 

Week 5 – Math 10

 

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This question stumped me because I was only multiplying one set of a conversion fraction. When dealing with 3-D objects, you need to multiply with 3 sets of conversion fractions, and when dealing with 2-D objects, you need to multiply with 2 sets of conversion fractions. In this case, I multiplied with 3 sets of conversion fractions.

Here is the correct answer,

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