# Tutoring math, you realize that arithmetic is the foundation.  The math tutor brings forward a quick method for determining divisibility by 11, with a couple of applications.

Consider the following problem:

Reduce 55/528 to lowest terms.

Of course, the key is knowing the number(s) that divide into both the top and the bottom. In this case, 11 does:

55/528=(5×11)/(48×11) = 5/48

Since knowing that 11 divides into a number is obviously useful, how do we do so?

Multiply the ones digit (8, in the case of 528) by 1, then the tens digit (2, in 528) by -1, then the hundreds digit (5) by 1. Add up the three products: 8(1) +2(-1) + 5(1) = 11. If the sum is divisible by 11, the number is. Since, for 528, the sum is 11, 528 is divisible by 11.

231 is also divisible by 11: 1(1) +3(-1) + 2(1) = 0. Since 0 is divisible by 11 (0÷11=0), 231 also is. Indeed, 231=11(21).

The pattern continues for numbers in the thousands. Consider 1936. 6(1) + 3(-1) + 9(1) + 1(-1) = 11. Therefore, 1936 is divisible by 11. In particular, 1936=11(176).

Now consider a polynomial factoring example:

Factor x2 – 4x – 165

Solution: From my post on easy trinomials, you know that we must find two numbers that multiply to make -165, but add to make -4. Knowing by sight that 11 divides into 165, we easily recognize our numbers as 11 and -15.

x2 – 4x – 165 = (x+11)(x-15).

I’ll have more to say about divisibility in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring high school math, this is a celebrated topic.  It even gets revisited at various levels of calculus. The math tutor introduces it, having just come in from shoveling the drive….

One of many situations in which you might need to complete the square is the following:

x^2 – 6x =17

It’s a quadratic; naturally, you want to subtract 17 from both sides and hopefully factor.

x^2 – 6x -17 = 0

However, you already know that x2 – 6x -17 doesn’t factor. What might you do instead?

Completing the square is one option. To do so, leave 17 on the right side:

x^2 -6x =17

Step 1:

Focus on the coefficient of the x term: in this case, the -6. Specifically, take half of it, then square that.

-6÷2 = -3→(-3)^2=9

Step 2:

Add the result from Step 1 to both sides of the equation:

x^2 – 6x + 9 = 17 + 9

Step 3:

Notice that now, the left side does indeed factor. In this case, it factors to (x-3)(x-3), aka (x-3)2. We rewrite the equation correspondingly:

(x-3)^2=26

Step 4:

With the left side in square form, we can square root both sides, to get

x-3=±√(26)

We add three to both sides, yielding

x=3+√(26) or x=3-√(26)

This is a light introduction to a topic that can be more complicated. However, it shows the usefulness and elegance of completing the square. More uses and complications of it will be visited in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, you visit this topic with your grade 10 & 11 students.  The math tutor shows one method.

A complex trinomial is of the form

ax2 + bx + c,

where a≠0,1.  An example is 2x2 -3x -35. How do you factor such a trinomial?

Well, first of all, the easy trinomial method won’t work for a complex trinomial. Let’s see, now, what will:

Example: Factor 2x2 – 3x – 35

Step 1: Multiply the lead coefficient (2 in this case) by the constant term (-35 in this case) to get -70.

Step 2: Find two numbers that multiply to make the product from step 1, but add to make the middle term coefficient (-3, in this case). Therefore, for our example, we need find the two numbers that multiply to make -70 but add to make -3. Of course, the numbers are -10 and 7.

Step 3: Rewrite the original trinomial, replacing the middle term with two terms whose coefficients are the numbers from step 2.

In other words,

2x2 -3x – 35 becomes

2x2 -10x +7x -35.

Step 4

Common factor the first two terms from step 3. Then, common factor the last two. Do the pairs separately; it won’t be the same common factor for the first two as for the last two.

2x(x-5) + 7(x-5)

Step 5

Notice from Step 4 that, although the common factors you took out front don’t match, the brackets do match. Put the common factors in their own bracket, then rewrite:

(2x+7)(x-5)

Step 6 (optional): Foil out your answer from Step 5 to check it.

First: 2x(x)=2x2

Outer: 2x(-5)=-10x

Inner: 7(x)=7x

Last: 7(-5)=-35

Add the four terms:

2x2 -10x +7x -35 = 2x2 -3x -35

That’s one way to factor complex trinomials. Trial and error is faster; I’ll explore it in a future post:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring physics or chemistry, solutions to problems are often given in scientific notation.  The tutor asks, “Is your calculator set conveniently?”

Let’s imagine you have a physics problem whose solution is given by the following calculation:

v = √(1.55×10-9)

Of course, the answer is 3.94×10-5. However, your calculator might give 0.0000394, which is the same number but in normal (aka, float) notation. Likely, the answer key is in scientific notation. If your calculator gives the float version, you need to decide if your answer matches with the key. You might “count back” the decimal places in order to compare the two answers.

Among my students, the Sharp scientific is the most popular. If you want it to automatically give all answers in scientific notation, you can, by doing the following:

First, press the SET UP key. You will see DRG FSE TAB. Under DRG will be a flashing number. Press the right arrow key to get to FSE. Now, the number under it will be flashing. Press=, then right arrow to get to SCI. Press =. The calculator returns to the regular screen, but it’s in scientific display now. Even the calculation 8×3=24 will display the answer 2.4000000×101, which you may not want. However, with Physics 12, the scientific display is probably more convenient overall.

Let’s suppose you want to switch back to regular display. Press SETUP, then right arrow over to FSE, then press =. Next, arrow over to NORM1 (which you can’t see at first, but it’s three steps over), then press=. You’re back in normal display.

In future posts I’ll explore the scientific display on other calculators:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring biology, the cell is fundamental.  Heading towards another weekend Biology 12 workshop, the biology tutor recalls Bruce Willis’s advice in Last Boy Scout: “Be prepared.”

A cell, like anyone, needs storage.  More than that, it needs storage for different purposes.  While you keep your food in the pantry, you store your clothes in the bedroom closet.  The cell faces similar storage challenges.

The three main types of storage vessels a cell uses are vesicles, vacuoles, and lysosomes.  Each has its own particular use and features:

Vesicles are often used for transport.   For example, they are used to store molecules or food arriving from outside the cell.  Vesicles are also used to hold partially completed molecules the cell is making as they are moved to different “work sites”. Then, when a molecule is to be secreted from the cell, it is shipped to the outer membrane in a vesicle, then released outside.

A vacuole is larger than a vesicle.  Vacuoles are meant for storage until use – or even permanent storage.  Water, sugars, and even pigments are stored in vacuoles.  In the case of pigment, it will remain in the vacuole for the life of the cell, giving the cell color. A water vacuole holds a large amount of water in order to “fill out” the cell, giving it the proper shape.  Plants derive their rigid shape partially from the water in their cells’ vacuoles.  Vacuoles can also hold toxic by-products until the cell gets around to destroying them.  Although both plant and animal cells contain vacuoles, plants use them more.

A lysosome is a special type of vesicle that stores digestive enzymes (see my previous post). A vesicle containing food will be fused with a lysosome for digestion to take place.

A cell can be a busy place. Any such place needs ample storage. Fortunately for the cell, it can make new vesicles, vacuoles, or lysosomes as needed:)

Source: Inquiry into Life, 11th Edition. Mader, Sylvia. New York: McGraw-Hill, 2006.

Jack or Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring Biology 12, you mention enzymes often.  The biology tutor defines the term enzyme, having used it in previous posts.

Recalling elementary school, your teacher likely led holiday crafting.  For example, she might have handed out pages with snowmen traced on them for the kids to cut out and decorate.  Likely, the finished snowmen were affixed to the window or wall.

Back then, photocopying was still pretty new; not all schools had a photocopier.  If they did, they didn’t use them much; likely, photocopies were expensive.  They still are, by the way:  going “over-budget” on photocopies can be a real problem in bureaucracies.

Therefore, the teacher didn’t use a photocopier.  She traced each snowman by hand.  However, she didn’t do it free-hand; she used a pattern.  Likely, she drew one “good one” by hand on cardboard, then cut it out.  To produce a snowman sheet for a child, she laid the cardboard pattern on a sheet of blank paper and traced around.  Tracing around the cardboard was quick: likely, she could produce twenty-five snowmen sheets just as quickly that way as by going down the hall to the photocopier, running them off there, then bringing them back.  What’s more, she could afterwards put the cardboard pattern away amongst her other supplies.  Next year, it would be there, waiting to be put to use again.  From one careful snowman drawing done on cardboard in her early days, she could produce unlimited snowman sheets over her teaching career.

Let’s consider a cell in the human body.  Like the teacher, it needs to repeat the same job(s).  To a cell, a job is a chemical reaction.

Does the cell do the reaction “free-hand”?  No.  Like the teacher, it makes a physical pattern that can be reused as many times as needed.  That pattern, to a cell, is an enzyme.

Most enzymes are proteins.  Like the teacher’s snowman pattern, the enzyme’s shape defines which job it helps with.  To the teacher, a different art project would require a different cardboard pattern.  To the cell, each specific reaction requires its own specific enzyme.

If the teacher could not use patterns, preparing for the art project might take prohibitively long; similarly, the cell’s life processes can not occur quickly enough without enzymes.  Some poisons specifically target enzymes and shut them down.

The veteran teacher found her job easy, since she’d accumulated such a helpful toolkit over her years.  Similarly, a cell whose enzymes are all available and functional is likely very prosperous:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring English, you get asked how to make writing more interesting. The English tutor offers a way to add variety to your sentences:  occasional use of the dash.

The dash is not the most commonly used punctuation mark; however, some people use it frequently.  It can mark a sudden change in thought, which is the way many people really do think.  Consider the following examples:

I bought him some wine, afterwards recalling he is a recovering alcoholic – what was I thinking?

I bought my bed at Robinson’s Furniture during Saturday Morning Markdown – I don’t know if they still do that.

We discussed the exam afterwards – none of us felt confident.

Dashes can give the flavour of how people truly talk and think.  They can also be used to enclose an idea, rather like parentheses:

If we invite Dave – we’ve invited him every year so far – we’ll need a vegetarian dish.

While we’re in Berlin – we’ll be there for two weeks – I hope we can attend the opera.

I might overuse dashes.  If you don’t use them already, start slowly:  put one or two in an assignment, then wait for feedback.  If your teacher doesn’t complain, you can start experimenting.  If, on the other hand, they do complain, try to comply with their advice the next time you use a dash.  You needn’t use them often in order to benefit from the variety they can add to your writing:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, you realize that to most people, math’s main importance is its applications.  The tutor presents a business example.

Any small business owner knows the break-even point is when you’ve paid your expenses, but haven’t “made” any money.  At the break-even point, you haven’t lost or gained.

In math, you can notice the break-even point on a graph.  (For a refresher in graphing, see my post here.) Consider the following example:

The Chess Club, in order to fund its overseas tournaments, holds a bake sale each year.  The items at the sale cost \$1.50 each.  To rent the hall, the club pays \$240.  Find the break-even point.

Solution:

Let P=profit
Let n=number of baked goods sold

Then P=1.50n – 240

Notice, of course, that profit doesn’t mean income; rather, it means income less expenses.  The only expense we are considering is the hall rental; the baked goods, we can assume, are donated.

At the break-even point, P=0.

Let’s look at a graph that models the situation.  To make some points to plot, we choose a few values of n, then plug each into our equation P=1.50n-240 to get the corresponding values for P.  For instance, let’s find P when n=100:

P=1.50(100) – 240

P=150-240

P=-90

Therefore, when n=100, P=-90.  By repeating that process for various values of n we arrive at the following table:

 n (number of baked goods sold) P (Profit) 0 -240 100 -90 200 60

Each (n,P) from the table means a point on the graph. For instance, when n=100, P=-90; therefore, (100,-90) will be on our line. We plot the points from our table as follows:

You can see the break-even point: it’s where our line cuts the horizontal axis. Note that, in this case, the x axis has been renamed the “n” axis; similarly, the y axis is called the “P” axis. In word problems, textbooks often rename the variables, calling them letters that stand for elements in the specific problem.

In our case, of course, the break-even point is 160; when 160 baked goods are sold, the expenses are paid. Any additional sales are profit.

You can also find the break-even point by setting P to 0:

0=1.50n – 240

Add 240 to both sides:

240=1.50n

Now, divide both sides by 1.50:

160=n

If desired, we could add (160,0) to our table of values.

Finding the break-even point can be a common question in math 10 and math 11. Its attractiveness is its real-world meaning:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, your grade 12 students ask about logs.  The math tutor continues the discussion from last post, showing how to use logs to solve an exponential equation.

In my previous post, I showed how you might solve an exponential equation without logs when the sides can be brought to a common base.

Common base or not, you can always use logs to solve an exponential equation (provided it has a defined answer). Consider the following example:

Solve

13^(x+3)=7^(2.1x+5)

To tackle this equation, the rule of logs that you need to know is

loga^b=bloga

In other words, after you take the log, you can “bring down” the exponent to the front, where it becomes a multiple.

We take the log of each side of our equation:

log13^(x+3)=log7^(2.1x+5)

We invoke the “exponent-to-multiple” rule for logs:

(x+3)log13=(2.1x+5)log7

Apply the distributive law to each side:

xlog13+3log13=2.1xlog7+5log7

Now, we use algebra to get all the x terms to one side, then all the constants to the other. Specifically, we subtract 2.1xlog7 from both sides, then subtract 3log13 from both sides, to get

xlog13-2.1xlog7=5log7-3log13

Next, we factor out x from the left side:

x(log13-2.1log7)=5log7-3log13

Finally, we divide both sides by (log13 – 2.1log7) to isolate x on the left. We arrive at the solution

x=(5log7-3log13)/(log13-2.1log7)

If you carefully key the solution into a calculator, you’ll hopefully get -1.337334 (rounded). In order to ascertain that it truly is the solution, we substitute it for x in the original equation as follows:

13^(-1.337334+3)=7^(2.1(-1.337334)+5)

You’ll get, rounded to four decimal places,

71.1403=71.1403

Since the left side equals the right side, our solution is correct.

The logs method works whether or not the sides have a common base. Some people prefer it for every exponential equation they face.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, you might be asked about these by your grade 12 precalculus students.  The math tutor opens the discussion….

There are two types of exponential equations:  those that don’t need logs to be solved, and those that do.  We’ll cover the ones that don’t need logs this time; next time, we’ll cover those that do.

Consider the following example:

Solve

25^(2x-3)=(1/125)^(-x+7)

Solution:

First of all, we notice that both sides are base 5 numbers: 25=52, while 1/125=5^(-3). (You may want to brush up on negative exponents; see my post here.) The fact that both sides are in the same base means we don’t need logs; otherwise, we would.

We rewrite both sides in the common base (in this case, 5):

(5^2)^(2x-3)=(5^(-3))^(-x+7)

Recall the exponent-to-an-exponent law: (xa)b=xab. For example, (x2)5=x10. We apply that law to each side:

5^(4x-6)=5^(3x-21)

Next, we invoke another exponent law: Ap=Aq→p=q. That is, if the bases are the same, and the left side equals the right side, the exponents must be equal. (Of course, there must be no other terms, nor any arithmetic operations pending, on either side.)

Therefore,

4x-6=3x-21

We subtract 3x from both sides, then add 6 to both sides, yielding

x=-15

If you have trouble believing -15 is the answer, here’s an option: Let’s take an impartial calculator and, substituting -15 for x, carefully enter the original left side:

25^(2(-15)-3)=7.37869763×10^(-47)

Let’s do the same for the right side:

(1/125)^(-(-15)+7)=7.37869763×10^(-47)

Not necessarily everyday numbers; still, they’re equal:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.