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Practice EQAO today

January 11, 2019

Grade 9s are trying a practice EQAO test today to get the feel of things before the real test on Tuesday/Thursday next week.

For Tues/Thurs bring a calculator and be sure to have eaten and gone to the bathroom first!  Leave backpacks in lockers.

For more practice over the weekend check out 5 years of past tests on the eqao site in English and here in French.  (Our test will be written in English).

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Equations of lines

January 9, 2019

We are finding linear equations using all sorts of information.  We can find an equation if we are given 2 points.

We can graph the points and sketch our line.  Our graph can help us find the slope by using a triangle and comparing the rise and run.  We can also use a table and look at the difference in y values and x values and show the slope as the change in y value divided by the change in x value.


The next calculation is to figure out the “b” value or the constant.

Here’s a different question, where a table was used to find the constant (the value of y when x=0).  The slope was found (-4/7 or -0.57) and then the table was filled in where for each increase in 1 for x, the y value decreased by 0.57.


Another way to solve for the b value is to substitute the value of one given point for x and y in the equation y=mx+b.  

Y=-0.57x+b

Sub in the given point (4,-9)

-9=-0.57(4)+b

-9=-2.28+b

-9+2.28=-2.28+2.28+b

-6.72=b

(Note, the difference in value from above is due to rounding)

We looked at some complex problems where we are looking for a line that is parallel to a given line, and has the same y intercept as another given line.  We needed to isolate y in the equations and then extract the useful information (either m or b) and use that to create our new line.


We’re working hard and helping each other

Trigonometry

January 9, 2019

Grade 10s are working with similar triangles and trigonometry. 


We started with a 90 degree angle and the sides 10cm and 9cm in one triangle, and 27cm and 30cm in the other.  Using our knowledge of the pythagorean theorem we calculated the hypotenuse.  We also used our learning from Monday to calculate the value of the angles using our trig tables.  We had the option to use the ratio for sine, cosine or tangent since we knew all 3 sides.  

We noticed that the angles are the same in both triangles, and the side lengths are proportional.  This is always the case!  We have 2 similar triangles.  We also looked at the areas and compare them.  Our big triangle has sides that are 3 times the length of the small triangle sides.  The are of the big triangle is 9 times the area of the small one. Thats 3^2 (three squared).  This relationship holds true for similar triangles.  If the proportionality constant is “k” between the sides of the triangle, it will mean that the areas have a proportionality constant of “k^2”.

We looked at a situation where we can use similar triangles.


We can determine the height of an object by using shadows.  

Parallel and Perpendicular

January 8, 2019

We’ve been working a lot with lines since coming back from the break.  We already looked at parallel lines a while back.  We know they don’t cross each other, and they run like train tracks always separated by the same distance.  We now can identify that they are increasing or decreasing at the same rate, or they have the same slope (“même pente”)….in fact we have been doing lots of chanting out loud… when someone say “parallèle” the response is “même pente”, and most of us have this word association stuck in our minds.


We know that the slope is the coefficient of x when the equation is written with y isolated.  This is the y=mx+b form.  The b value is the constant, the initial value, the y intercept (“ordonnée à l’origine”).  

We learned that perpendicular lines intersect, and always form a 90 degree angle.  If one slope is positive the other is negative, if one slope is 0 the other is undefined (indéfinie).  If one slope is steep, the other is not.  That’s the only way they’ll have a 90 degree angle between them.


The slopes are related.  This example has one slope of 1/2 and the other of -2/1.   The fraction is inverted, and one is positive and the other is negative.  We determined a process for finding a perpendicular slope is to invert the fraction and to multiply by negative 1.  Our chanting and word association continued, and when someone says “perpendiculaire” the response is “inverse négative”, the rhythm is a bit catchy.  

We see the slopes as the x coefficients again, when y is isolated.

Solving equations

December 12, 2018

Today we started to work with equations (they are like two expressions on either side of an equal sign). We used algebra and our visual representations help us solve the equation.  We do the same things on both sides, and our goal is to isolate the variable.


We worked through several different types of questions.  Some of us had different approaches to solving…. like here we divided every term by 2 (the coefficient) first.


Some groups tried to add 2 to both sides to start….but at the end it didn’t verify.  (We couldn’t substitute our answer into the variable and have both sides remain equal).  Their next step was to restart with a new first step.  In this case they divided by 3 on both sides, and then have f-2=6 so f=8.


We are learning lots, and getting more confident with our new skills.

The next step was to manage equations with variables  on each side of the equal sign.  We need to choose a side, and use algebra (opposite operations) to make sure that the variables are on one side, and constants on the other.


Our final type of equation had fractions.  We can eliminate fractions first by multiplying by the denominator on both sides.  This is one strategy of many that we used to manage fractions.


We’re practicing a lot this week, and making good progress.

Modelling quadratics

December 10, 2018

Grade 10s are working on solving problems involving quadratics.  This example shows how we can determine dimensions of a rectangle given the area and perimeter.  We can create equations, substitute, expand to make a trinomial, use algebra to get a trinomial that equals zero, and then use the quadratic formula to solve for the roots (if they exist).

We also saw that sometimes there are no roots.  In this case on desmos we can see that there are no intersection points between our two equations, so we have no possible dimensions that will work.  It corresponds to the fact that the discriminant found is a negative number.


Our work this week is mostly based on modelling (creating equations) and using them to solve problems.  

Grade 9 Review

December 5, 2018




Some good review questions