Tag: midtermreview

Week 10 – Pre Calc 11

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This week in Pre-Calculus 11 we prepared for the midterm. Although we didn’t learn anything new, we did review many things from the previous units that I forgot about.

Sequences and Series 

Sequences and Series was the very first unit of the year. One of the concepts that I found confusing was dealing with percentages in word problems.

Ex. Billy makes $1240 a month. His boss recently told him that he will receive a 3.02% raise at the end of every month. How much will he make at the end of the first month? How much in total will he make at the end of the ninth month?

First Month:

 a= $1240

r=1.00+0.0302=1.0302 

1240\times1.0302

1277.448

$1277.45

Ninth Month:

a= $1277.45

r=1.00+0.0302=1.0302

n= 9

S_n=\frac{a(1-r^n)}{1-r}

S_n=\frac{1277.45(1-1.0302^9}{1-1.0302})

S_n=\frac{1277.45(1-1.307055142)}{-.0302})

S_n=\frac{1277.45(-.307055142)}{-.0302})

S_n={-392.2475909}\div{-.0302}

S_n=12988.33082

$12,988.33

In the example, the first thing I did was figure out what the common ratio was. I knew that the common ratio was above 1.0 because in the example it says that Billy will receive a 3.02% raise in his cheque, which means that his pay is increasing. Then, I found the amount he made at the end of the first month by multiplying 1240\times1.0302 because that would be the “a” value in the second part of the question.

Infinite Sum: An infinite sum can be calculated when a geometric series converges because the series decreases a sum can be calculated. The sum can only be calculated if -1< r< 1. This means that r is a fractional value and cannot be an improper fraction. The formula to calculate the infinite sum is S_{\infty}=\frac{1}{1-r}

Ex. 3, 9, 27, 81… 

r=\frac{t_n}{t_{n-1}} 

r=\frac{9}{3}

r=3 

INFINITE SUM CANNOT BE CALCULATED 

The infinite sum for the example above cannot be calculated because the common ratio is 3.

Ex. 10, -5, \frac{5}{2} …

r=\frac{t_n}{t_{n-1}} 

r=\frac{-5}{10}

r=\frac{-1}{2} 

S_{\infty}=\frac{a}{1-r}

S_{\infty}=\frac{10}{1-\frac{-1}{2}}

S_{\infty}=\frac{10}{1+\frac{1}{2}}

S_{\infty}=\frac{10}{\frac{3}{2}}

S_{\infty}=\frac{10}{1}\times\frac{2}{3}

S_{\infty}=\frac{20}{3}

Absolute Value and Radicals 

The absolute value and radicals unit was the second unit of the year. In this unit we learned about how the absolute value of a number is really the number of spaces it is away from 0.

Ex. \mid-7x+x^2-10\mid , x=6

\mid-7(6)+6^2-10\mid

\mid-42+36-10\mid

\mid-16\mid

16

In the example above, I was given a value for the variable so I just had to plug in the value to find the absolute value of the expression.

Analyzing Quadratic Functions

This was the most latest unit we finished. In this unit I struggled with some of the word problem questions involving substitution.

Ex. A graph of a quadratic equation passes through (6,2) and (0,9), the axis of symmetry is x= 7. Determine an equation of the function.

STEP 1: y=a(x-x_1)^2+q

y=a(x-7)^2+q

2=a(6-7)^2+q

2=a(-1)^2+q

2=a+q

STEP 2: y=a(x-x_1)^2+q

y=a(x-7)^2+q

9=a(0-7)^2+q

9=49a+q

9-49a=q

STEP 3: 2=a+q

2=a+9-49a

2=-48a+9

2-9=-48a

-7=-48a

\frac{7}{48}=a

STEP 4: q=9-49a

q=9-49(\frac{7}{48})

q=9-\frac{343}{48}

q=\frac{89}{48}

STEP 5: y=a(x-x_1)^2+q

y=\frac{7}{48}(x-7)^2+\frac{89}{48}

In the first step, I began by using standard form because I had the axis of symmetry. Then, I plugged in the point (6,2) into the x and y, this created an equation with two unknown variables. In step two, I figured out the value of “q” by using the other point given in the question (0, 9). In step three, I plugged in the value I got for “q” into the first equation I developed to find a value for “a” because we found a value for “q” we just needed to isolate “a” to find the numerical value of it. In step 4, I took the value I got for “a” and plugged it into the equation q=9-49a to find the numerical value of q. Then, in step 5 I took all the information we figured out in the 4 steps and made it into an equation.