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Grade 10, Math 10, Uncategorized

Week 7 – Math 10

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This question stumped me for a very long time, since it had multiple parts that weren’t all trigonometry. For example, I was not sure if I had to use the distance over time triangle from physics, or if I should use unit analysis from the measurement unit. I tried many different things, and I finally found out how to solve the problem!

At first I used sine to figure out the length of the hypotenuse, but later on as I thought about it, I realized it would not make sense. The biker is going along a highway, it’s impossible to travel along the hypotenuse side! So I changed the equations:

tan 8 = \frac{60}{x}

\frac{60}{tan 8} = 426.92

\frac{426.92 km}{20 s}={x km}{1 hr}

I multiplied 20s by 3 into 60s, which is 1 minute. Then I multiplied it by 60 to convert it into 1 hour. You must do the same to the top numbers, so I got an answer of:

426.92\cdot 30\cdot 60 = 76.84

So the average speed the biker travelled at is 77km/hr!

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Grade 10, Math 10

Solving Trig Equations

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sin x = \frac{5}{13}

For this triangle, the degree is the variable. To find the value of x, first we will need to label the sides:

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Sine is being used to solve the equation because the measurements of the hypotenuse and opposite sides are provided.

To isolate x, use algebra to cancel out the sine on the left side. What you do to one side, you must do to the other, so the equation becomes:

x = sin^{-1}\cdot \frac{5}{13}

Plug those numbers in on your calculator and you get x = 23 , which is the answer!

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tan 39 = \frac{17}{x}

This time, a degree and only one side length is given. In this case, x is the denominator of the fraction, which makes the problem a bit more complicated.

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We know that when we switch two fractions around, like so:

\frac{1}{2} = \frac{3}{6} —> \frac{6}{2} = \frac{3}{1}

They are equivalent. So, we can use this trick to make the equation easier to solve:

\frac{tan 39}{1} = \frac{17}{x} —> \frac{x}{1} = \frac{17}{tan 39}

x = 21

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cos 22 = \frac{x}{50}

The x is the numerator in this equation. So you will have to solve this equation  in a different way compared to the last two.

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Since the x is being divided by 50, if we multiply it by 50 the 50 will be canceled out. This means we also have to multiply the other side by 50:

50\cdot cos 22 = x

x = 46

Grade 10, Math 10

Week 5 – Math 10

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This week I learnt a simple way to convert squared and cubed units. I had trouble converting 0.37 km^3 to m^3 , until I used this strategy:

1 km = 1000 m , or  10^3m

0.37\cdot 10^3 = 370000000

Written in scientific notation: 3.7\cdot 10^8

The same works for any exponent, all you need to know is the relationship between the two units:

60 yds^2  to  in^2

1 yd = 36 in

60\cdot 36^2 = 77760

Written in scientific notation and rounded: 7.78\cdot 10^4

 

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Grade 10, Math 10

Week 4 – Math 10

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While doing measurement homework, I discovered that I needed to use scientific notation. However, I never learnt it before so I had to teach myself. Being the lazy individual I am, I didn’t bother reading the notes and dove right into some practice questions, such as:

2. Express each number in scientific notation.

a) 2 300          b) 7 580 000          c) 41 000 000 000

All I knew was that I had to write it into x\cdot 10^y (x and y represent integers). After struggling and repeatedly getting the answers wrong, I started to notice a pattern. I found that if I counted the numbers after the first one, the number I received became the exponent. Also, the numbers that aren’t 0 would be turned into a decimal less than 10 and more than 0. For example:

2 300

  There are 3 numbers after the 2, so the exponent is 3. 23 becomes 2.3. Written in scientific notation: 2.3\cdot 10^3

I tried this strategy for the rest of the questions and it worked! Until I ran into these sneaky questions:

0.000 023

Here, I figured out that I can count the number of digits after the decimal point, then subtract the number of digits that aren’t 0s. However, I need to leave out one digit when subtracting. Example:

000 023 = 6 digits

There are 2 digits that aren’t 0, but I need to leave one out, so:

6-1=5 ,so 0.000 023 = 2.3\cdot 10^{-5}

Through trial and error I also discovered that the exponent becomes negative when dealing with these tiny numbers. I was delighted to find out that my made-up strategy was similar to the one my math teacher taught, when we went over our homework.

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Grade 10, Math 10

Week 2 – Math 10 (Updated)

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This week I had a bit of trouble with a couple of questions, including this one:

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I had to arrange them in order. Since entire radicals are easier to organize than mixed radicals, I tried converting them. However, that did not go so well. I converted them by multiplying the coefficient by itself x times (x = index). For example, on the first radical, 7x7x7x7x7x7x1. I got a ridiculously huge number, and since I was not allowed to use a calculator, I stopped multiplying after the 4th or 5th 7. I did the same for the remaining radicals, except for the last one which had me confused. As you can see, I received huge, huge numbers which aren’t very efficient for calculations…

After I ordered the first three radicals, I checked the answers, only to find that they were all incorrect. My strategy had worked in the past before, but for some reason I got the wrong answers with this conversion method.

I turned to my friend for help. She asked me what times itself 6 times equals 1, and I said 1. She then asked me to multiply 1 by 7, which is 7. And that’s the answer! Trying the same with the others, I converted them into simpler entire radicals without having to do complicated multiplication.

The last radical: 3\sqrt[2]{\sqrt[3]{64}}

4\cdot 4\cdot 4=64 ——–> 3\sqrt[2]{4}

2\cdot 2=4 , so 3\sqrt[2]{\sqrt[3]{64}}=6 !

 

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