Sunday, December 12, 2021

Steve dot TheTutor AT Yahoo dot Com

Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!
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Long Division With 2-Digit Divisors - Remainder Only



Before taking on long division with 2-digit divisors, you should feel comfortable with the following:

- Multiplication tables and patterns of multiplication (Counting by 5s or recognizing patterns for the number 25, for example)

- "Clean" one-step division (no remainders, 1-digit divisor. 15÷3=5 or 72÷9=8, for example)

- One-step division with remainders (15÷2=7R1 or 72÷5=14R2, for example)
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In this lesson, the answers will be in remainder form.
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Example: 706 ÷ 15
If you say this out loud, it sounds like "Seven hundred and six divided by fifteen." In order to make it into a long division problem, you have to re-write it. 

Whether the original problem looks like 706÷15 or 706/15, in order to re-write it, you can think of it like this:
Now you can think of it as "Fifteen goes into seven oh-six how many times?" From this step forward, it is helpful to think of the problem in this fashion. 

The first step sounds like, "Fifteen goes into seven evenly how many times?" The answer is zero. Fifteen cannot go into seven evenly, since seven is too small. 

The following table will take you through the entire process for this problem.
ASK YOURSELF LOOKS LIKE THE ANSWER IS LOOKS LIKE YOU SAY
15 goes into 7 how many times evenly?

15 can go into 7 ZERO times evenly

0 * 15 = 0
LOOKS LIKE SUBTRACT LOOKS LIKE ANY MORE NUMBERS? NEXT STEP
7 - 0 = 7

YES Bring down the zero
ASK YOURSELF LOOKS LIKE THE ANSWER IS LOOKS LIKE YOU SAY
15 goes into 70 how many times evenly?


15 can go into 70 FOUR times evenly


4 * 15 = 60
LOOKS LIKE SUBTRACT LOOKS LIKE ANY MORE NUMBERS? NEXT STEP
70 - 60 = 10

YES Bring down the six
ASK YOURSELF LOOKS LIKE THE ANSWER IS LOOKS LIKE YOU SAY
15 goes into 106 how many times evenly?

15 can go into 106 SEVEN times evenly

7 * 15 = 105
LOOKS LIKE SUBTRACT LOOKS LIKE ANY MORE NUMBERS? NEXT STEP
106 - 105 = 1

NO Put remainder on top
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Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!
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Solving Quadratic Equations Using the QUADRATIC FORMULA


A quadratic equation always equals zero, and always has a squared term. A properly arranged quadratic equation is modeled like this:
ax2 + bx +c = 0
'a', 'b' and 'c' are all coefficients (numbers) and they may be positive OR negative. Here is an example:
4x2 - 17x -15 = 0

HOW TO SOLVE A QUADRATIC EQUATION
There are many ways to solve a quadratic equation. The three main ways are completing the square, using the Quadratic Formula and factoring. This lesson demonstrates how to use the Quadratic Formula to solve a quadratic equation.

USING THE QUADRATIC FORMULA
The Quadratic Formula may seem intimidating, but it is really just a system you use to “plug in” the coefficient values for the variables ‘a’, ‘b’ and ‘c’, just like any other formula. It is important that you learn it well, because out of all three popular methods of solving quadratic equations, the Quadratic Formula is the only one that works every time. Some of the steps in the formula, such as radicals, may mean that you have to elevate your game.
One way to learn how to memorize the Quadratic Formula is by learning a common song that sings the Quadratic Formula to a familiar tune. There are a variety of videos available. HERE IS ONE.

Here is the actual formula:

When I say it out loud, it sounds like this:
“Negative ‘b’ plus or minus the square root of ‘b’ squared, minus four ‘a’ ‘c’, all over two ‘a’ .”
When we plug in the actual coefficient numbers for the variables in the formula, it looks like this:

We start by dealing with the first ‘b’ term. By having a negative in front of the entire coefficient, it is basically saying that we simply change the sign. Now our formula looks like this:

Now we square our next ‘b’ term. The answer will always be a positive number, since a negative number times a negative number equals a positive number. Our changed formula:

At this point, it is time to multiply the ‘a’ and ‘c’ terms, which leaves us with this:

Next, multiply the result of the previous step by (-4):

Add the two numbers left in the discriminant:

Now we find out what the square root of 529 is. I should mention that, at any point in the process, you may solve the bottom terms. Our formula is beginning to look like some easy math by now:

What this means is that we will wind up with two separate answers, as expected. The first answer will be 17 plus 23, over 8, and the other answer will be 17 minus 23, over 8:
solution is x = 5 and -3/4. This means that if you put either number into our original equation, you will make it equal zero. These two points are also where the graphed parabola crosses the x-axis. These are also known as ‘roots’. The ‘solutions’ are the actual ‘x’ values for the coordinates that cross the x-axis: (5, 0) and (-3/4, 0)

Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!
Keywords:algebra,quadratic equation,factoring,tutor,steve,washington,pasco,kennewick,richland,wyzant
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Solving quadratic equations by FACTORING


A quadratic equation always equals zero, and always has a squared term. A properly arranged quadratic equation is modeled like this:
ax2 + bx +c = 0

'a', 'b' and 'c' are all coefficients (numbers) and they may be positive OR negative. Here is an example:
4x2 - 17x -15 = 0

HOW TO SOLVE A QUADRATIC EQUATION
There are many ways to solve a quadratic equation. The three main ways are completing the square, using the Quadratic Formula and factoring. This lesson demonstrates how to use factoring to solve a quadratic equation.

FACTORING
Multiply the 'a' coefficient by the 'c' coefficient. Ignore signs for now.
4 ('a' coefficient) x 15 ('c' coefficient) = 60

Find the factors of 60 ('what' times 'what' equals 60), looking for a pair that equal -17 (the 'b' coefficient) when added together:
Factors of 60: [1, 60] : [2, 30] : [3, 20] : [4, 15] : [5, 12] : [6, 10] 

Out of all of these pairs, only one pair works, and that is only if you have the signs in a specific order:
-20 + 3 = -17

Now you set up two fresh binomials with our 'a' and 'c' terms, which look like this:
(4x2 +      ) and (      - 15)

You then take the two numbers that you factored, and put them, combined with an 'x', in the missing places in the binomial model. It doesn't matter which one you put where.
(4x2 - 20x) and (3x -15)

Now you factor out the common terms, also known as "Reverse Distribution":
4x(x-5) and 3(x-5)

If the new terms in the binomials match EXACTLY, then that is one of your solution sets. In our example, both sets now contain exactly (x-5). Combine the factors on the outside of the binomials. This is your second solution set:

4x(x-5) and 3(x-5)

(x-5) (4x+3)


Use the zero property to solve for 'x' for both sets:
x-5 = 0
x = 5
4x+3 = 0
4x = -3
x = -3/4

Your solution is x = 5 and -3/4. This means that if you put either number into our original equation, you will make it equal zero. These two points are also where the graphed parabola crosses the x-axis. These are also known as ‘roots’. The ‘solutions’ are the actual ‘x’ values for the coordinates that cross the x-axis: (5, 0) and (-3/4, 0)

Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!
Keywords:algebra,quadratic equation,factoring,tutor,steve,washington,pasco,kennewick,richland,wyzant
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Location: Tri-Cities, WA, USA

Trig Water Wheel Word Problem - STUDENT REQUEST

PROBLEM: The top of a bucket 0.5 m high is attached to a waterwheel of diameter 2m, from which the bucket always hangs downward. The wheel sits above the river so that half of the bucket dips below the surface of the water at its lowest position. Write a function for the height of the center of the bucket (in meters) above the river as a function, f, of the angle t as measured counterclockwise from the 3 o'clock position.

The first thing I like to do is draw a picture. We have a water wheel with a diameter of 2, which means the radius is 1. The bucket is .5 tall, but only .25 is between the water wheel and where the bucket meets the water.

As our problem stands, if the water is our x-axis, then the center of our water wheel circle is 1.25 meters above the water. If the radii are all 1, then our water wheel circle coordinates look like this:
The problem is asking about where the center of the bucket is at any given point in its rotation. Well, the bucket is in an entirely different circle than the water wheel. Since the bucket always dangles downward, the positions for the bucket center look like the red marks:
 
Since the center of the bucket is always .25m lower than the water wheel, we can set up the new bucket center circle with the following coordinates:

We have our circle set up, but the instructions say that we need to start at the 3 o' clock position, and calculate our theta angles in a counter-clockwise direction from 3 o' clock. This means that we have to shift our circle so that the origin is at (0, 0), and the x-axis cuts the circle horizontally so that we can start in the 3 o' clock position. Notice that our radius is still 1.
 Well, hmm. Now we have a familiar looking image:
 
At this point we need to reflect on what we know about our sine function. The sine of an angle is the ratio of the opposite side over the hypotenuse (SOH-CAH-TOA). Since our hypotenuse (radius) is 1, then the sine of our angle is actually the measurement of our opposite side. In our model this will give us the distance above the x-axis, but we will need to add 1 to find the actual height from the bottom of the circle, or where the bucket middle meets the water.

To set up an equation, we say "y equals what?" In this case, y = sin(Θ) + 1. We now substitute f(x) for y to create our function f(x) = sin(Θ) + 1. Now we can test the function.
We know that when our bucket is at the bottom of the circle, it is 0 meters above the water. The angle for this point on the circle from the 3 o' clock position, going counter-clockwise, is 270 degrees.When we plug 270 in for theta, we have f(270) = sin(270) + 1. The answer is 0, which is what we expected.
We know that when the bucket is at the top of the circle, it is 2 meters above the water. This puts our theta measurement at 90 degrees. Does sin(90) + 1 equal 2? Yes.
We know that when our radius is even with our x-axis, our bucket should be one meter above the water. There are two places where this occurs - at sin(0) and sin(180). When we plug this into our function, it gives us the expected answer of 1.

Good luck!
Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!
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Steve dot TheTutor AT Yahoo dot Com

Steve Wilson , a.k.a. " Steve the Tutor ," is a tutor in the Tri-Cities ( Kennewick , Pasco , Richland ), WA, and surr...