Monthly Archives: April 2016

Week 10 – Math 10

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IMG_20160424_185041

This is one of the questions I had trouble with. The mistake I was making was that when I separated the coefficient from the x2 to make the polynomial easy and not ugly, I did not divide the rest of the terms by the coefficient.

The incorrect factoring of this expression is,

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The correct factoring of the expression is,

IMG_20160424_185014

After I divided the other terms by the coefficient, I put x in both brackets, then listed the factor of 3; 3 x 1, -3 x -1, -3 x 1, and 3 x -1. I knew the product also had to equal -4, so I chose -3 x -1 because -3 x -1 = 3, and -3 + -1 = -4.

 

Week 9 – Math 10

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IMG_20160417_174519

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,

IMG_20160417_174634

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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IMG_20160404_204550

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

IMG_20160404_205821

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.

IMG_20160404_205732

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

 

Solving Trigonometry Equations

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IMG_20160403_205811

 

 

 

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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.

IMG_20160403_205831

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.