Category Archives: Grade 10

Math 10 – Week 13

Last week in math we looked at functions. We learned that they are a very special relationship. For example; “a function is like a marriage because you have one spouse and only one, and a friendship is like a relation because you can have more than one friend.” Like how an input has one output.

FUNCTIONS;

In the first example on the table of values there is an example of a function, due to the input having one individual output if the 2 for example had a partner of 2 and 5 than that would make the expression a relation. In the second example the 1 (input) had an output of 4 & 5 which made the expression a relation. In the last example graphs are shown on how relations and functions are graphed. The function has a perfectly diagonal line which is not necessary, but 2 points that are equal to each cannot be perfectly vertical.

Math 10 – Week 12

The introduction week to the Relations & Functions unit was reviewing concepts from grade 9. For example linear equations, working with a table of values, and inputs and outputs. But this week intercepts were introduced.

SOLVING AN EQUATION;

 

What ever intercept you are trying to solve that will mean that the other variable will be zero

ex; x – intercept = 0

then you work through the steps with the expression (BFSD) in your mind

Brackets

Fractions

Sorting (Zero pairs)

Divide

Once you remove the brackets and Fractions you are on sorting will means having numbers on each side.

For example;

4x -8 = 8

+8 = 8

then you divide with the coefficient.

20 dollar potato day

May 9th 2017

Cameron Yeung,Emilio Deras, Runa Nagai, Alan Hu

Foods 10

Mrs. Keating

Potato for a day

Ingredients:

– X14 baking potatoes ($9.10/$0.65 per)

– X1 no name egg carton ($2.83/$2.83 per)

– X1 no name paprika ($2.00/$2.00 per)

– X1 Sour Cream ($1.98/$1.98 per)

– X1 laughing cow Cheddar cheese ($3.47/$3.47 per)

– All adds up to $19.38

Recipes:

– Breakfast: Baked hash browns (2 potatoes cut into a large bowl of hash browns spiced with paprika)

– Lunch: Baked Potatoes (1 potato per person, whipped in sour cream and splattered with bits of green onion)

– Dinner: Potato Cheddar Soup (Rich soup with 500 mls of milk, 5 mls of sugar, two sliced potatoes all covered in cheddar cheese)

Links:

– https://www.realcanadiansuperstore.ca/?cid=bs_google_cpc_Branded-Core-Vancouver-EXA_superstore_e&gclid=CjsKDwjw0cXIBRCxjqnE3K3sHhIkAL1LezTY-ODyY88gDx60IRXOc8gManVPYkvpxB7Y0JhtlDjzGgLbUPD_BwE

– http://www.thekitchn.com/10-breakfast-potato-recipes-to-start-your-morning-right-224645

– https://finger-licking-recipies.com/

Math 10 – Week 10

This week in math we looked at factoring ugly trinomials. At first I struggled with the concept, but I found a very simple strategy to figure it out.

Box Method

The Box Method:

The way you start is by putting the variable with the highest degree of the top right and the constant (The term with no variables) in the bottom left, once you have those two numbers you put cross multiply. In the example the answer is 6. The way you find the top left and the bottom is you list the factors and you pick the factors that add up to the middle value which in the case is 5. Once you put the terms in the box. Then you horizontally find the things (factors, variables) and then do the same thing vertically. After that you group the like terms into 2 different brackets. Then you have your answer.

Math 10 -Week 8

This week we learned about simplifying polynomial equations, which for me in grade was a major struggle. Espically the equations that had 2 terms in a bracket for example; (5x+4)(3x+7), but now I have learned a way to deal with the following types of equations

 

Taking it one step at a time made it easy to complete. Slowly doing the steps of distributive property, was made the questions easy. Using simple math like; 24x+6x was easy.

Math 10 – Week 7

This week the class learned about trigonometry we learned how to solve the angles and the side lengths. We are currently with right angle triangles where one of the angles is  90° and the two other angles add up to  90° making the whole triangle add up to 180°. This following week we learned how to find the missing angles or side lengths when one is not apparent. The formulas that help figure out the missing side length or angle are as stated S = \frac{O}{H} or Sine = \frac{Opposite}{Hypotenuse} (SOH), C = \frac{A}{H} or Cosine = \frac{Adjacent}{Hypotenuse} (CAH), and  T = \frac{O}{A} or Tangent = \frac{Opposite}{Adjacent} (TOA). SOH CAH TOA is a very useful expression that helps us remember the needed formula. Labelling the triangles sides (Opposite, Adjacent, Hypotnuse) is the most important and helpful thing to do. The another essential thing is to know the reference angle as it makes up what sides are the opposite or the adjacent, because the hypotenuse is directly across from the right angle.

Finding the side length 

To find the side length you need to take it one step at a time. FIRST you need to label the sides (Opposite, cosine, sine) THEN locate the needed sides. One will be the side you need and the other will be the side have. In this case I had the hypotenuse, and I needed the the opposite side. Then I put sin because the first letter was S in SOH and the reference, and in this case “X” was the opposite angle and the hypotenuse was 64. To isolate the variable you multiply the hypotenuse by itself then plug that in your calculator.

When the side length is the denominater

 

To find the side length when the missing length is the denominater is actually not difficult. You do everything in the beginning like the regular side length. Like label the sides, find the reference point, etc. Though you now have to put the length you have up top as the numerator. So you divide the known side length by the (Sin, cosine or Tangent) angle.

 

How to find the angle 

 

To find the angle is simple as well as you just again label the sides, find the reference angle, etc. But this time. For example, cos x is put as the reciprocal.

Math 10 – Week 5

This week in Math we learned how to find surface area and volume in 3D shapes, Prisms and shapes like a pyramid or a cone

Formualas for pyramids & cones

The formulas are  very easy to apply once you have the numbers right for example;

 

The formula for volume of a cone is very easy to apply, you first plug the radius in, which in this case is 5 then square it, then multiply it by the height which is 10. Last you divide by 3, because the volume of 3 cones is equal to a cube, though they have to have the same dimensions. Then punch it in to your calculator.

 

Math 10 – Week 4

A couple weeks ago, we started our exponents unit. From using from using exponent laws to figure out the new simplified expression, I caught to the pretty early. One concept I struggled with, was the intergral exponents concept.

An example of a question which I made a mistake on:

I always messed up, whether it was not put the exponents in the wrong spots, to not calculating the expression properly.

Property simplified equation:

I learned to always deal with the outside exponent first. So I factor in exponent to get the proper exponents. Then you group the like terms, and most importantly get rid of the brackets. Then you get rid of any intergral(negative) exponents by doing the reciprocal, negative exponents on the top go down to the bottom and vice versa, you then subtract or add exponents and divide or multiply the actual number.

Math 10 – Week 3

This week in Math 10 we learned exponents. The exponent laws, integral exponents, and rational exponents. I found the exponent laws the easiest to understand

Multiplication Law

• What you do to simplify is take the exponents and add them while keeping the base the same

Division Law

• What you do to simplify is take the exponents and subtract them while keeping the base the same

Power of a Power Law

• What you do to simplify is take the exponents and multiply them while keeping the base the same

Zero Exponent Law

• Anytime there is something to the power of 0 it is equal to 1

Math 10 – Week 2

This week in Math we learned about entire and mixed radicals, we learned on how to convert and simplify both types. I easily identified how to convert entire radicals to mixed radicals and vice versa.

Converting a entire radical into a mixed radical is very easy. First step was to multiply by multiples that equals the radicand. You simplify all the multiples and then the number that isn’t a square root becomes the coefficient .

Converting a mixed radical to an entire radical is very simple as you simply take the coefficient then you square it take the radicals within the radical sign and just multiply them for example, \sqrt{4} \cdot \sqrt{7}=\sqrt{28}