We started with a 30 minute pre-test just to see what skills we need to focus on.
Then we briefly reviewed the rules of algebra/arithmetic.
\(\dfrac{4}{5}+\dfrac{5}{7}\)
\(\dfrac{4}{5} \cdot \dfrac{5}{7}\)
and
For more videos, see this unit in Khan Academy.
Today we talked about solving simple equations.
\(5x - 7\)
\(7 - \dfrac{3}{x^2} = 4\)
\(2x + 7 = 3x - 5\)
\(-\tfrac{1}{5}x + 7 = -\tfrac{2}{5}\)
Don’t forget:
You can do anything you want to an equation, as long as you do it to both sides.
For more videos, see this unit in Khan Academy.
Today we talked about factoring and expanding expressions. Expanding is when you apply the distributive law to convert factors into terms. For example we expanded the expressions:
Factoring is when you use the distributive law in reverse to un-expand an expression. The easiest kind of factoring is when you can factor out a common factor. For example, we factored out the \((x+3)\) from both terms in the expression:
There is also factoring of quadratic polynomials (of the form \(ax^2 + bx + c\)) which is a little more complicated. I recommend using factoring by grouping (also known as the AC-method). A similar method is the box method). We factored these examples in class:
Today we talked about solving polynomial equations.
Key Idea: Factor to solve for when a polynomial equals zero.
The places where a polynomial equals zero are called the roots. We solved the following examples.
\(x^2-4x = 5\) (see also: https://youtu.be/BI_jmI4xRus)
\(3x^4+12x^3+9x^2 = 0\) (see also: https://youtu.be/KbFwLvCOBUI)
\(\dfrac{x-7}{x-4} = \dfrac{1}{x-8}\) (see also: https://youtu.be/bRwJ-QCz9XU)
\(\dfrac{x^2+6x+5}{x^2+4x+3} = 0\) (see also: https://youtu.be/h9MoF-ZA6n0)
We talked about function notation and how graphs. We did the following examples.
We also defined the domain and range of a function. Then we looked at different ways to express functions. For example, we had these two examples:
The volume of a sphere: \(V = \frac{4}{3} \pi r^3\).
The function to convert Celsius to Fahrenheit: \(f(x) = \tfrac{9}{5}x + 32\).
We also talked about inverse functions
Find the inverse of \(V(r) = \tfrac{4}{3}\pi r^3\) and explain what it does in words.
Find the inverse of \(f(x) = 2x +1\). (see: https://youtu.be/W84lObmOp8M)
Finally we looked at using the graph of a function to answer questions like the following:
We spent a lot of the first part of the class discussing Homework 5.
After that we talked briefly about word problems. I recommend reviewing these tips for word problems.
Finally we talked about dimensional analysis (also known as factor-label method) for converting units. We did this problem:
After that we did one example of a word problem that needs algebra to solve.
In the Wednesday section we did this example of translating words into math instead of #2 above:
20% is the same as \(\tfrac{20}{100}\). “Of” means multiply. The word “what” means that there is an unknown quantity which we can call \(x\). And “is” means \(=\). So translating into math, the sentence above is the same as the equation \[\dfrac{20}{100} \cdot x = 36\] which you can solve by multiplying both sides by the reciprocal \(\tfrac{100}{20}.\)
We spent the first part of class talking about Homework 6 and word problems.
After that, we started the review problems for the midterm exam.
After you try all of the problems on the midterm review, you can check your answers here:
This week we talked about using algebra to solve inequalities.
For two functions defined using the usual operations (\(+\),\(-\),\(\cdot\),\(/\)), the function which is greater can only change at an \(x\)-value where the graphs cross (are equal) or where one is undefined (a bad point).
Therefore you can solve complicated inequalities with these two steps:
We used this method to solve these examples.
\(-5c \le 15\) (Video solution: https://youtu.be/D1cKk48kz-E)
\(x^2 \ge 7x - 12\) (Example video: https://youtu.be/a10EuIFcphk)
\(\displaystyle\frac{6}{x-4} < 2\)
We also reviewed how to solve equations and inequalities with absolute values like this:
An equation with an absolute value is really two equations: one where the inside is positive (and the absolute value works like parentheses), and another where the inside is negative (so you can replace the absolute value symbol with an expression in parentheses multiplied by \(-1\)).
For any two numbers \(a\) and \(b\), the expression \(|a-b|\) is the distance between \(a\) and \(b\) on a number line.
This week was about working with exponents.
Here are the rules you need to know.
We simplified these examples in class:
\(9^{-1/2}\) (https://youtu.be/tn53EdOr6Rw)
\((64)^{2/3}\) (https://youtu.be/S34NM0Po0eA)
\(\left( \dfrac{a^3 b^4}{a^2 b} \right)^3\) (https://youtu.be/AR1uqNbjM5s?t=332)
\(\sqrt{20x^6}\) (https://youtu.be/kCP-ptqeNMQ)
\(\dfrac{8 a^6 b^{-4}}{24 a^{-8} b^9}\) (https://youtu.be/7gTfmXhFOVo)
This week we introduced logarithms. We started by talking about how base-10 logarithms tell you the order of magnitude of a number. We also drew pictures of logarithmic scales. Then we settled on this definition of a logarithm:
The base-b logarithm is the function \(\log_b(x)\) that gives the power \(y\) such that \(b^y = x\).
We calculated the following examples in class:
\(\log_2(16)\) (https://youtu.be/Z5myJ8dg_rM)
\(\log_2(\frac{1}{8})\)
\(\ln(\sqrt{e})\)
We also talked about the
\(\log(xy) = \log(x) + \log(y)\)
\(\log\left(\dfrac{x}{y}\right) = \log(x)-\log(y)\)
\(\log(x^p) = p \log(x)\)
We used these properties to find the solution to
Today we talked more about exponential and logarithmic functions.
For any positive base \(b\), the exponential function \(b^x\) is the inverse of the logarithmic function \(\log_b(x)\). In particular, \(e^x\) is the inverse of \(\ln(x)\). Therefore the cancellation rules apply:
If \(f\) and \(f^{-1}\) are inverse functions, then \[f(f^{-1}(x)) = x ~~~~\text{ and }~~~~ f^{-1}(f(x)) = x.\]
We did these problems in class:
Solve \(10^{2t-3} = 7\) (https://youtu.be/R443Db-wJ5o)
The milligrams of medication in someone’s bloodstream is modeled by \[M(t) = 20 \cdot e^{-0.8t}\] where \(t\) is the time in hours since the medicine was taken. Solve for the time until there is only 1 mg of medicine left in the bloodstream. (https://youtu.be/HnHCQ2X_meg)
What is the inverse of the function \(M(t)\) from the last problem?
This week is about solving Systems of Equations. Typically, if you have two unknowns, then you need two equations. Then you can use the idea of substitution to reduce the number of variables. We did these examples to illustrate the idea:
\(\begin{aligned} -3x &- 4y = -2 \\ y &= 2x-5 \end{aligned}\) (https://youtu.be/GWZKz4F9hWM)
\(\begin{aligned} x^2 &+y^2 = 25 \\ y &= x+1 \end{aligned}\) (https://youtu.be/swFohliPgmQ)
\(\begin{aligned} y &= -x^2 + 6 \\ y &= -2x -2 \end{aligned}\) (https://youtu.be/hjigR_rHKDI)
A population of bacteria in a petri dish is growing exponentially, so the population is \[P(t) = C e^{kt}\] where \(t\) is time in days and \(C\) and \(k\) are constants. If \(P(1) = 5,000\) and \(P(2) = 10,000\), then what are \(C\) and \(k\)?