Monthly Archives: September 2018

Math 11 Week #4

Posted on by under Grade 11, Math 11, Pre-calc 11

This week we learned about how to multiply and divide radicals. It is very similar to the protocol with multiplying and dividing fractions, which was a very helpful tip for me since radicals can look a little confusing and hard to do.

For multiplying, you multiply the coefficients together and then you multiply the radicals together to find the answer:
ex. 4\sqrt{2} (2\sqrt{3})

= 4(2) and 2(3)

= 8\sqrt{6}

and also, a good trick to remember with multiplying AND dividing is that you should always simplify the question as much as you can before starting to multiply/divide so that you don’t end up having big numbers at the end.

For dividing, the rule is that you cannot ever have a irrational number (a radical) as the denominator

ex.

\frac{2}{\sqrt{5}}

In order to solve this question we would have to rationalize denominator so that it is not a radical anymore. In order to do that we mulitiply the top and the bottom by square root.

ex.

\frac{2}{\sqrt{5}} x \frac{\sqrt{5}}{\sqrt{5}}

= \frac{2\sqrt{5}}{\sqrt{25}}

Now that the denominator is a rational number, our process is over

= \frac{2\sqrt{5}}{5}

 

 

 

Math 11 Week #3

Posted on by under Grade 11, Math 11, Pre-calc 11

This week in math we learned about finding the absolute value of a real number.

We determined the absolute value of a real number is defined as the principal square root of a number 

(principal square root of a number = the positive square root)

The absolute value can be explained as the distance of the number from zero on a number line.

For example, the absolute value of 109 or |109| = 109 because it is 109 numbers away from the number line. Another example is the absolute value of -63 or |-63| is 63 because even though it is a negative number, it is still 63 numbers away from zero, therefore the absolute value is positive 63

*absolute value symbols (|x|) are grouping symbols and are very similar to brackets, so you solve the problem as you would any other.*

ex. -4|5-2(-2)|

-4|5+4|

 

-4|9|

 

-36

 

Math 11 Week #2

Posted on by under Grade 11, Math 11, Pre-calc 11

This week, I learned how to tell if a geometric sequence or series is converging or diverging. Converging is when you can find an infinite sum, whereas diverging is when there is no infinite sum.

The rule to know if the sequence is diverging is if:
1<r

or

-1>r

The rule to know if the sequence is converging 

-1<r<1

ex. If we are given a geometric sequence such as: 1, \frac{1}{3},\frac{1}{9},\frac{1}{27} and r= \frac{1}{3}

We know that this sequence is converging and will have an infinite sum because -1<r<1

To find the infinite sum we use the equation:

S∞= \frac{a}{1-r}

Now insert the numbers :

S∞= \frac{1}{1-\frac{1}{3}}

S∞= \frac{3}{2}

 

Math 11 Week #1

Posted on by under Grade 11, Math 11, Pre-calc 11

For the first week of Pre-calc 11, we learned all about arithmetic sequences and series. In simpler terms, patterns. Today I am going to focus on how we learned to find a term when given only 2 terms out of a arithmetic sequence.

Find t_1 : t_5 = 20, t_9 = 36

So first off, to find t_1 you need to find the common difference or “d” in the sequence. To do that, you must find the difference between the two terms (t_5 and t_9) In this equation the difference between term 5 and term 9 is 4 terms, therefore the equation is:

t_5 + 4d = t_9

Then, you replace the variables with your known numbers and find “d”:

20 + 4d = 36

4d = 36-20

4d = 16

4d/4 = 16/4

d = 4

Now we have found that d = 4, so the common difference in the sequence is +4.

Since we know d, we can finally use an equation to find t_1.

The equation for this is t_n = t_1 + d(n – 1)

The next step is to insert our known numbers into their variables and find t_1 :
(In this case, t_n would = 20 or 36 but for this example I will be using 20)

20 = t_1 + 4(5-1)

20 = t_1 + 20 – 4

20 = t_1 + 16

20 – 16 = t_1

4 =t_1

We have found t_1!

Answer:  t_1 = 4

 

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