Category Archives: Math 10

Solving Trig Equations

April 1, 2016 glorial-2014 2 Comments

$sin x = \frac{5}{13}$

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

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!

$tan 39 = \frac{17}{x}$

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.

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:

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

$x = 21$

$cos 22 = \frac{x}{50}$

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.

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

March 6, 2016 glorial-2014 Leave a comment

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

February 28, 2016 glorial-2014 Leave a comment

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

Exponent Laws

February 23, 2016 glorial-2014 Leave a comment

Multiplication:

$x^4\cdot x^5=x^9$ The exponents are added, not multiplied.

Division:

$x^{17}\div x^8=x^9$ The exponents are subtracted.

Power:

$(x^3)^3=x^9$ The exponents are multiplied together.

Negative Integer:

$\frac{1}{x^{-9}}=x^9$ Reciprocate the base to make the exponent positive.

Fraction:

$\sqrt[2]{x^{18}}=x^9$ In fraction form, the index becomes the denominator and the exponent the numerator.

Grade 10, Math 10

Week 3 – Math 10 (Updated)

February 20, 2016 glorial-2014 Leave a comment

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Grade 10, Math 10

Week 2 – Math 10 (Updated)

February 14, 2016 glorial-2014 Leave a comment

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

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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Grade 10, Math 10

Week 1 – Math 10

February 8, 2016 glorial-2014 Leave a comment

This week I had quite a bit of difficulty on the skills check, especially with the equation on the back:

$\frac{2}{3}(2x-1)=-9+\frac{x}{2}$

However, while doing corrections I had a huge “Ah-Ha!” moment, and the solution made itself clear to me. My first try was a disaster:

$\frac{4x}{3}+\frac{-2}{3}=-9+\frac{x}{2}$ I got rid of the brackets with multiplication.

$\frac{4x}{3}=-9+\frac{x}{2}+\frac{2}{3}$ Then, I added $\frac{2}{3}$ to the equation in order to remove the fraction on the left side.

$4x=-27+\frac{x}{6}+2$ I multiplied the equation by 3 to turn $\frac{4x}{3}$ into 4x and $\frac{2}{3}$ into 2.

Then I ran out of time…but clearly I had no clue what I was doing. Even if I did finish, the answer probably would have been wrong. My second try makes much more sense:

$\frac{4x}{3}+\frac{-2}{3}=-9+\frac{x}{2}$ I multiplied 2x and -1 by $\frac{2}{3}$ .

$4x+-2=-27+\frac{3x}{2}$ I multiplied everything by 3.