# The tutor shows how to easily solve a two-variable linear system with the amazing Sharp el-520w.

Let’s imagine you’re posed the following problem:

Solve the linear system:

(2/3)x – (1/5)y = 2

-4x + 3y = 6

With the Sharp el-520w, solving it couldn’t be easier:

1. MODE 2 0
2. It asks for a1: key in 2÷3, then =. Next, it asks for b1: key in -1÷5, then =. For c1, key in 2, then =. (Note these are the numbers from the first equation.)
3. Now it starts asking for the second row numbers. Key them in the same manner as you just did for the first row.
4. After you key in c2 and press =, the screen should say x=6. Press = again to reveal y=10.
5. To review the inputs, and the answers, you can press = repeatedly.

Source:

Sharp el-520w Operation Manual.

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

# The tutor talks a bit about “auto”: is it the default setting for layout?

I’ve read that “auto” is default: unless you set the property to something else in particular, the element will be rendered as though that property was “auto”. Not always, in my experience; I’ll show you now:

Here’s an image without the “margin:auto” setting:

<img src=”source0.abc” style=”display:block” />

Here’s the same image with “margin:auto” included:

<img src=”source0.abc” style=”display:block;margin:auto” />

I’ve viewed this post in Chrome, ie11, and Firefox. In each case, the top image is left-aligned, while the bottom one is centered.

So many times I’ve keyed “margin:auto” wondering “Is this redundant?” Certainly not always:)

Source:

w3schools.com

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

# The tutor shows absolute value on the Sharp el-520w.

As I mentioned in yesterday’s post, absolute value of x, commonly written |x|, can be defined as follows:

I can’t find a dedicated absolute value function on the Sharp el-520w; however, it certainly does have square root and square functions. Consider the following example:

Solve the equation |x| = -2x + 7

Solution: We’ll use the Sharp el-520w equation solver. First, we can rewrite the equation as

(x2)0.5 +2x -7 = 0

Next, we enter the equation, pressing = but not 0. Then, press MATH, then 0 (for SOLV). It asks for a start point (0 is fine, then ENTER), then a step size (you can just go with what it suggests, so just press ENTER again).

Hopefully you’ll be rewarded with the answer 2.3333.

There we have it: absolute value handled with the Sharp el-520w:)

Source:

Sharp el-520w Operation Manual.

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

# The tutor defines absolute value in two ways.

The absolute value of x is normally written as |x|; a calculator might call it abs(x) but show it either way. It can be defined as follows:

 x = { x, x≥0 -x, x<0

Then we have |6| = 6, since 6≥0; |-3| = -(-3) = 3, since -3<0

Note: |x| ≥ 0 for all x.

Another definition of absolute value is

From this definition, we have |-4| = ((-4)2)0.5 = (16)0.5 = 4

By definition, square root is positive only, unless ± is written in front. Therefore, (9)0.5 = 3, not -3. Hence, by the second definition as well, |x|≥0.

Each definition of absolute value has its advantage; for example, the first is probably easier to understand. The second definition has the advantage that it always does the same operations, regardless of input. Therefore, it’s more convenient to use in combination with other formulas.

I’ll be showing an application of the second definition in a coming post.

Source:

Roland, Larson E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: D C Heath and Company, 1989.

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

# The tutor continues about polynomial long division.

I put across the main idea of polynomial long division in my Sept 24 post.

What if the polynomial is missing a power of x: for example, like so

In such a case, the question must be rewritten with a zero for the missing power of x:

Then, the question can be done as shown in my Sept 24 post.

HTH:)

Source:

Travers, Kenneth et al. Using Advanced Algebra. Toronto: Doubleday Canada Limited, 1977.

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

# The tutor shares information about the hazard of green potatoes.

Potatoes are meant to grow covered beneath the soil. When a tuber (the edible part of a potato) gets exposed to the sun, its exposed cells develop chlorophyll, so turn green: they become photosynthetic so they can produce carbohydrates, rather than just storing them.

To protect themselves from being eaten, the exposed, newly-photosynthetic cells also produce dangerous toxins related to strychnine. Hence, the wisdom that any green parts on a potato must be discarded.

A study reported in 2006 looked at the toxicity of green potatoes. Results showed that the green skin can host the toxins at dangerous concentration; however, the flesh of the green potato was not found to – in that study, anyway. Yet, the researchers urged caution: the toxins can persist in the human body for more than a day, and even low-level concentrations of them may cause symptoms difficult to detect.

To be as safe as possible, one can obey the old wisdom and discard all green parts of a potato.

Source:

curiouscook.com

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

# The tutor shows an example of how to divide a polynomial by a binomial.

Long division with polynomials is appreciated by math teachers, but few others. What follows is an illustrated example of how to do it.

A simple polynomial long division might begin like this:

The process starts with the question, “How many times does x (from the divisor) go into x2?” The answer is x, and is written on top, above x2:

Next, the x you just wrote on top is multiplied by the divisor; the product is written below the polynomial:

Subsequently, subtraction is done, then the next term (in this case, 7) is brought down:

Now, the process restarts, with the question, “How many times does x go into -x (the result of the subtraction)?” The answer, -1, is written on top, above the column that contains the -x. Afterwards, the -1 is multiplied to the divisor, and the result is written below:

Subtraction is the next step:

The quotient is x-1, the remainder, 5. If the division is correct,

(x-1)(x-2) + 5 should equal x2 -3x + 7

Let’s check:

(x-1)(x-2) + 5 = (using FOIL) x2 -2x -x + 2 + 5

Simplifying gives

x2 -3x + 7

Apparently, this long division was successful.

There is yet more to discuss about polynomial long division. Look for more in coming posts:)

Source:

Travers, Kenneth, et al. Using Advanced Algebra. Toronto: Doubleday Canada Limited, 1977.

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

# The tutor identifies a London planetree.

A London planetree is a hybrid between American sycamore and Oriental planetree. (The Oriental planetree is a sycamore as well.)

London planetree’s leaves are similar to American sycamore, but its fruits come in strings of 2 to 4, like Oriental planetree. There’s a tree like that a few blocks away.

HTH:)

Source:

Brockman, Frank and Rebecca Merrilees. Trees of North America: A Guide to Field Identification. New York: Golden Press, 1968.

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

# The tutor tells where the comma is on the Staples BD-6120G.

With the Staples BD-6120G, finding the comma took me a few minutes. It’s a blue speck just above the left corner of the hyp key. You would enter a comma by pressing SHIFT hyp.

HTH:)

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

# The tutor shows the integral function on the TI-83 Plus.

To evaluate an integral on the TI-83 Plus, go to the Math menu, then arrow down to fnInt(, which is choice 9. It accepts the parameters as follows:

fnInt(function, variable of integration, lower bound, upper bound)

Therefore, to integrate xsinx from 1 to 2 (in rads), one would put

fnInt(xsin(x), x, 1, 2)