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## Sunday, November 29, 2015
### Two methods for finding outliers
There are two methods for finding outliers, numbers that are far away from "the middle". If we count the middle as the median, we use the five number summary to find the threshold for high and low outliers. If we have the average, we will need to calculate the standard deviation for a sample sx, and find the z-score. With a calculator, this is very simple, but it can also be done by hand with small sets.
The five number summary
Here is a list of the number of win for the Pac-12 teams as of November 29, 2015. Obviously, the length of the list is 12.
8, 7, 6, 4, 4, 0, 6, 6, 5, 4, 3, 1
We need to put the list in order, either top-to-bottom or bottom-to-top. Since the 8 is the first number on the list, let's go top-to-bottom.
8, 7, 6, 6, 6, 5, 4, 4, 4, 3, 1, 0
The five number summary are the high and low values - very easy, and the Quartiles, Q3, Q2 and Q1. We already know how to get Q2, because it is the median. Q3 is the median of the top half of the data and Q1 is the median of the bottom half of the data. Because there are 12 items on the list, it splits into the top six and the bottom six, and the median is the average of the two middle values.
8, 7, 6, 6, 6, 5 || 4, 4, 4, 3, 1, 0
The median Q2 is (5+4)/2 = 4.5
Q3: For the top half, the median is between the first 6 and the second 6, so it is 6.
Q1: For the bottom half, the median is between the 4 and the 3, so the median is (4+3)/2 = 3.5
High = 8
Q3 = 6
Q2 = 4.5
Q1 = 3.5
Low = 0
Next we get the IQR = Q3 - Q1, which in our instance is 6 - 3.5 = 2.5
The high threshold for outliers is Q3 + 1.5*IQR, or 6 + 1.5*2.5 = 6 + 3.75 = 9.75. This threshold is above 8, so 8 is not an outlier.
The low threshold for outliers is Q1 - 1.5*IQR, or 3.5 - 1.5*2.5 = 3.5 - 3.75 = -0.25. This threshold is just barely below 0, so 0 is not an outlier.
The z-score method
We know how to take z-scores if we have the average and standard deviation, but here we are going to have to compute the average and standard deviation instead of them being given. Average isn't hard by hand with smallish data sets, and if you have a calculator that is set up for statistics, both the standard deviation and average are given to you as quickly as you can input the set. If you don't have a calculator. Here is what we need to do.
1. Find the sum of the list, which we will call sum(x).
In our case, it's 8+7+6+6+6+5+4+4+4+3+1+0 = 54
2. Find the sum of the squares of the list, which we will call sum(x²)
In our case, it's 64+49+36+36+36+25+16+16+16+9+1+0 = 304
3. Then we get sum(x²) - [sum(x)]²/n
This is 304 - 54²/12 = 304 - 243 = 61
4. The standard deviation is the square root of the value from step 3 divided by n-1.
sqrt(61/11) ~= 2.35487881..., which we can round to 2.35.
The average is 54/12 = 4.5
To be a high outlier, we need a z-score over 2. To be a low outlier, we need a z-score under -2.
z(8) = (8-4.5)/2.35 ~= 1.48936..., which isn't above 2, so it's not an outlier.
z(0) = (0-4.5)/2.35 ~= =1.91489..., which isn't below -2, so it's not a low outlier, but it was close.
With this particular set, our two methods agreed there were no outliers. The methods sometimes disagree. We can have sets with just high outliers, just low outliers, outliers in both directions or no outliers at all.
## Sunday, November 15, 2015
### Notes on the metric system
The United States uses a measurement system originated in Great Britain, which is sometimes called the Imperial system or the customary system. The relationships between measurements aren't very user friendly, and even Americans aren't very good at remembering all of them, according to tests I gave students at the beginning of classes years ago.
Here are some examples of the stuff you have to remember.
Length
12 inches = 1 foot
3 feet = 1 yard
5,280 feet = 1 mile
There are also odd measurements that very few people use any more, like rods, fathoms, furlong and nautical miles.
Volume
Of all the parts of the system This one makes some sense because the next biggest named measurement is almost always a factor of 2 away, either 2, 4 or 8 times bigger. Let's start with the fluid ounce (fl. oz.) as the basic unit.
8 fl. oz. = 1 cup (8 fl. oz.)
2 cups = 1 pint (16 fl. oz.)
2 pints = 1 quart (32. fl. oz.)
4 quarts = 1 gallon (128 fl. oz.)
Yay, powers of two! Unfortunately it breaks down as we get smaller than a fluid ounce.
2 tablespoons = 1 fl. oz.
3 teaspoons = 1 tablespoon
Another way in the system to measure volume is the cubic inch. This is not related nicely to the fluid ounce, as 1 cubic inch = 0.554413... fluid ounces.
Weight
16 ounces = 1 pound or lb.
2,000 lbs. = 1 ton
There are also little used units like a hundredweight (100 lbs.), a stone (14 lbs.) and a long hundredweight (8 stone or 112 lbs.). When we get smaller than an ounce, the standard is a grain. 7,000 grains is a pound, which means 437.5 grains is equal to an ounce.
The metric system
Unlike the customary system, which was thrown together over time, the metric system was made all at the same time, so length is related to volume and volume is related to weight using water as the standard thing we will weigh. All you need to learn for most measuring are three words and three prefixes for most stuff.
Length
The standard length unit is the meter or m. 10,000,000 meters is the distance from the equator to the North Pole.
The important prefixes here. For long distances that Americans would measure in miles, the metric system uses kilometers or km. Kilo means "1,000", so this is 1,000 meters.
For short distances Americans would measure in inches, the metric system uses centimeters or cm, which is 1/100 of a meter.
For even smaller lengths where Americans would use fractions of inches, the metric standard is a millimeter, which is 1/1000 meters.
Volume
The standard measurement for volume is the liter or L. It is based on a cube 1/10 or a meter (or 10 cm) on each side. The liter is just slightly larger than a quart.
When dealing with very large volumes, the unit is still the liter.
For smaller volumes, the standard unit is the milliliter or mL. This is a cube one centimeter on each side, so it is also also called a cubic centimeter or cc.
Weight
The basic unit of weight is a gram, which is the weight of a cc of water at about 4 degrees Celsius, or about 39.2 degrees Fahrenheit. The temperature was chosen as the lowest point where at normal pressure, water shows no sign of freezing.
The gram is very small, so when measuring things the customary system would measure in pounds, the metric system uses kilograms or kg. A kilogram is slightly more than 2 pounds, and for very heavy weights that Americans would measure in tons, the metric system users resort to metric tons, which are 1,000 kg. (technically, this would be a megagram, but this word is almost never used.)
As small as the gram is, some small measurements like medicine doses are doled out in milligrams or mg. When medicine is measured, not even Americans use ounces or grains, though when I was a kid, grains were used on aspirin packages. One gram is about 15 grains and sometimes there was confusion, leading to massive overdoses or underdoses.
Conversion values
We only need a single conversion number, and depending which direction the conversion is in, we either multiply by the conversion number or divide by it. Here is an example.
1 inch = 2.54 centimeters (cm)
If we have 40 inches, the 40 is related to the 1, so 40/1 = x/2.54. With fractions, we cross-multiply to get 40*2.54 = x*1, so x = 101.6 cm. If we are asked to round to the nearest whole number, it would be 102 cm.
If instead we have 40 cm, the fraction would be x/1 = 40/2.54 = 15.7480315... inches, which could round to 16 inches, 15.7 inches or 15.75 inches, depending on the level of rounding.
Here are some of the standard numbers used for conversion.
Length, middle distances: 1 meter = 3.2808 feet or 39.37 inches
Length, long distances: 1 kilometer = 0.6214 miles
Weight, small weights: 1 gram = 0.03527 ounces
Weight, medium weights: 1 kilogram = 35.27 ounces = 2.2046 pounds
Weight, large weights: 1 metric ton = 2,204.6 pounds = 1.1023 metric tons
Volume, small measures: 1 milliliter = 0.033814 fluid ounces
Volume, medium to large measures: 1 liter = 33.814 fluid ounces = 1.0567 quarts
Volume, liters to cubic inches or cubic feet: 1 liter = 61.0237 cubic inches = 0.0353 cubic feet
## Tuesday, November 10, 2015
### Compound interestEffective interest ratesDoubling periods and half lifesThe Consumer Price IndexMonthly payments on loans
A lot of the math of "the real world" is linear, but some of it involves exponents. The best known interest rate situations are those of a savings account. Here is the equation; instead of using superscripts for exponents, I use the symbol ^, which is the symbol for exponentiation on most of the TI calculators, though some have an "x rasied to the y" button.
Compound interest rate equation
A(t) = P(1+r)^t
A(t) is the amount in the bank at time t
P is the original principle invested
r is the rate of interest, usually given as a percent
Effective rate compound interest rate equation
Some saving institutions will give the rate as "4% compounded quarterly". This is slightly different from 4% a year. If we use n as the number of compounding periods in a year, the equation changes to
A(t) = P(1+r/n)^nt
4% compounded quarterly
In the stated example (1+.04/4)^4 = (1.01)^4 = 1.04060401, so the rate r = .04060401, usually rounded to the nearest hundredth of a percent at 4.06%. The difference is small but noticeable over long stretches of time.
\$1000 invested at 4% over 20 years is \$2,191.12
\$1000 invested at 4% compounded quarterly over 20 years is \$2,216.72
Doubling periods and half lifes
How long will it take at 4% to turn \$1000 into \$2000. This amount of time is called the doubling period and is found by the formula log(2)/log(1 + r)
In our case, that would be log(2)/(1 + .04) = 17.6729.....
If the bank is actually compounding annually, the amount only changes on the anniversary of your putting the money in.
After 17 years you have (1.04)^17 = \$1,947.90
After 18 years you have (1.04)^18 = \$2,025.82
The Consumer Price Index
This week, students got a green handout sheet with the consumer price index number for the years 1950 to 2015. The simplest way to use it goes as follows.
How did prices change from 1960 to 1980? We use the two CPI values.
CPI(1960) = 29.6
CPI(1980) = 82.4
82.4/29.6 = 2.78378.....
29.6/82.4 = .35922...
What these numbers mean is that on average, a \$10 item in 1960 sold for 10*2.78378... = \$27.84 in 1980, while a \$10 item in 1980 would have sold for 10*.35922... = \$3.59.
We can use this to figure out the cost of living increase in any given year by dividing the CPI for that year by the CPI for the previous year. For example, the rate in 1975 would be
CPI(1975)/CPI(1974) = 53.8/49.3 = 1.09127789... This is 1 + rate, so rounded to the nearest tenth of a percent we would have 9.1% and to the nearest hundredth of a percent it would be 9.13%.
Let's look at the number CPI(1980)/CPI(1960) = 2.783783... This gives us how much prices increased in the 20 year period from 1961 to 1980. To get the average increase over those 20 years, we take (2.783783)^(1/20) = 1.05252... This is 1 + rate, so the average rate = .05252... or 5.3% if rounded to the nearest tenth of a percent and 5.25% rounded to the nearest hundredth.
Monthly payments on loans
If you know the interest rate r, the number of years y you will take to pay off a loan and the amount you wish to borrow A. The formula for the monthly payment is
Monthly payment = A(1 + ry)/(12y)
We can use this to figure out monthly payments or, if we know how much we are willing to pay a month, the rate and the numbers of years we can get the loan for, conversely we can get the amount A with the formula
A = 12y(Monthly payment)/(1 + ry)
Practice problems
a) If you put \$1,000 in the bank invested at 4.5%, how much will you have in 10 years?
b) If you put \$1,000 in the bank invested at 4.5%, how much will you have in 15 years?
c) what is the effective rate of 5.5% compounded monthly?
d) what is the effective rate of 5.5% compounded daily?
e) A radioactive isotope loses 0.1% of its radioactivity every year. What is the half-life, rounded to the nearest tenth of a year?
f) The worst of inflation in last century in the United States happened in the 1970s and early 1980s. Find the average inflation rates for the following presidential terms. (Remember to start with the year prior to the start of the term.
1. Nixon (1969-1974)
2. Ford (1974-1976)
3. Carter(1977-1980)
4. Reagan(1981-1988)
g) If you borrow \$20,000 at 6% and have a ten year loan, what is the monthly payment?
Answers in the comments. |
# Lettering With Math
1,223
6
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This instructable was inspired by the old "All you need is MATH (LOVE)" picture, which is old as math itself :) OK, if I remember correctly, I have seen it first more then ten years ago, so in IT world, it is old.
I have used the knowledge of equations to form the solution flag into one of a CTF (capture the flag) game, and now I show you how you can do this as well.
My pictures were made with the online math tool from Wolfram - the Wolfram Alpha.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Know Your Basic Forms - Straight Line
Here I will show you some of the basics, they are the bricks we are building from. Sometimes we search for "x", but sometimes the "y" value in the Descartes coordinate system. So let the fun begin!
Let's draw a line...
We have some options here (a and b are constants):
• For a horizontal line: y=a
• For a vertical line: x=a
• For a slope line y=a*x+b
The constants will tell you where and how to draw your line. On the pictures you can see a few examples. For drawing purposes you have to know which constant in the equation does what. The first two are simple, you draw the line at the given position. But the third…
The "a" tells you the angle of your line and "b" tells you how far you have to move your line at the vertical direction (or if you want to think a bit more - that moves your line at the horizontal direction with -b/a). This piece of information will be helpful when you want to draw your letters into a strict position.
## Step 2: Basic Forms - Curved Lines
If you plot some functions, they become nice curved line(s).
Just to mention a few:
• Logarithmic function: log(x)
• Power function: x^n or x*x*...*x n times
• Inverse function: 1/x
• Sine or cosine function: sin(x)
• Tangent or cotangent function: tan(x)
These can be used to form our letters as well, just think about where does they get a value. For example if you want to get an "U" shape, just plot the y=x^6 function (and move it to the right position as you can see in later steps).
## Step 3: Basic Modificators - Mirroring
Mirroring is easy to the horizontal axis. You have to multiply by -1 your values. However, if you want to mirror your shape to the vertical axis, you can do that as well. It is a multiplication with -1 as well, but this time you have to move that multiplication inside your equation.
Of course there are a lot of examples, when these two are the same...
## Step 4: Basic Modificators - Non Negative Values
The absolute function. It simply mirrors everything below the horizontal zero line to the zero line. If you have a line with negative values, they will be all positive. This forms the most simple "V" letter. You can mark the absolute function with two vertical lines: |all these things will be positive|. (Or you can use the abs() function as well.)
## Step 5: Basic Modificators - Moving
To move your lines, you have to calculate a little bit. There are two directions (staying at the 2D system) which you can use. The horizontal and the vertical. To move your image at the vertical direction, simply add a constant value to the very end of the calculation.
But what if you want to move in the horizontal direction? That is a bit tricky - you have to put your constant inside the equation! And that is not all… You have to put it in the right place and everywhere. Oh, and almost forget to mention - this constant will move it in the opposite direction you would think :)
To be clear - let's use a new variant instead of "x", call it "z". z=x+constant value. If we put this inside for example the power function, the whole thing will move into the left with the constant value we choose.
## Step 6: Basic Modificators - Rotate by 90 Degree
This modification is not so basic as would you think. So far we plot our function by searching for "y" based the "x" value. To "rotate" a function, we don't rotate the function's plot itself, but we search for the "x" now.
Let me explain:
Lets plot the function y=sin(x). This gives us a nice sine wave plot. Ok, but if we want this wave "rotated", we have to plot x=sin(y) on the same coordinate system.
## Step 7: Advanced Shapes - Sphere/Ellipse
If you want to draw a sphere, you can do it, it has a nice equation:
r^2=x^2+y^2, where the "r" means the radius of the sphere.
This equation is the direct meaning of the sphere, which is the points equidistant from the centre point. You can check this by selecting a point on the circle line and calculate with Pythagorean theorem.
But we now just using these things for drawing, so let's play with our sphere a little bit.
Let's stretch it along the "x" axis and make an ellipse, but how? Just modify the equation's "x" part a little: r^2=(x/a)^2+y^2
The same works on the other axis as well.
## Step 8: Combine Them - Let's Form a "W"
We already know that a simple "V" letter can be drawn with the absolute equation. Now try to form a "W" with the same procedure.
1. Draw a simple "V" in the desired position y=|a*x+b|
2. Move our "V" a bit lower y=|a*x+b|-c (now the breakpoint is at the negative "c" value)
3. Make a "W" from the "V" with another absolute equation y=||a*x+b|-c|
Of course this method looks fine only in a specific range (just try looking the y=||2x+2|-2| at the -100..100 range, you will not see the middle rippling, but in the -3..3 range it looks great).
## Step 9: Make It Harder for a Competition...
With the basic knowledge we can draw almost anything we want, but when there is a competition where we want our contestants to think and our solution should not be obvious - we have to make it hard. How can we use some facts to complicate our math?
Let me show you with the simplest formula.
1. 1=1
2. 1=sin^2(x)+cos^2(x)
3. -i^2=sin^2(x)+cos^2(x)
4. -i^2+log(1)=sin^2(x)+cos^2(x)
5. ...
So when we write down to our contestants the formula y=5*ln^x(e), they won't be happy and they have to think a little bit to see this is just a horizontal line :)
This is an entry in the
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## 6 Discussions
Thanks ! Great instructable. It inspirés me to generate waveforms for signal processing testing. |
# A metal cylindrical can is to have a volume of 3.456pi cubic feet. How do you find the radius and height of the can which uses the smallest amount of metal?
Nov 29, 2016
The smallest area occurs when we have a radius of $1.2$ feet, leading to an area of $8.64 \pi \approx 27.143$ square feet
#### Explanation:
Let us set up the following variables:
$\left\{\left(r , \text{Radius (feet"), (y, "Height of can (feet)"), (A, "Surface Area of can (sq feet)}\right)\right.$
We want to vary the radius $r$ such that we minimise $A$, ie find a critical point of $\frac{\mathrm{dA}}{\mathrm{dr}}$ that is a minimum, so we to find a function $A \left(r\right)$
Then the volume is fixed:
$\pi {r}^{2} h = 3.456 \pi$
$\therefore {r}^{2} h = 3.456$
$\therefore h = \frac{3.456}{r} ^ 2$
And, The Surface Area is given by:
$A = 2 \pi r h + 2 \pi {r}^{2}$
$\therefore A = 2 \pi r \left(\frac{3.456}{r} ^ 2\right) + 2 \pi {r}^{2}$
$\therefore A = \frac{6.912 \pi}{r} + 2 \pi {r}^{2}$
Differentiating wrt $r$ gives us;
$\therefore \frac{\mathrm{dA}}{\mathrm{dr}} = \left(6.912 \pi\right) \left(- \frac{1}{r} ^ 2\right) + 4 \pi r$
$\therefore \frac{\mathrm{dA}}{\mathrm{dr}} = \frac{- 6.912 \pi}{r} ^ 2 + 4 \pi r$
At a critical point, $\frac{\mathrm{dA}}{\mathrm{dr}} = 0$
$\therefore \frac{- 6.912 \pi}{r} ^ 2 + 4 \pi r = 0$
$\therefore - 6.912 \pi + 4 \pi {r}^{3} = 0$
$\therefore {r}^{3} = 1.728$
$\therefore r = 1.2$
With $r = 1.2$ we have:
$A = \frac{6.912 \pi}{1.2} + 2 \pi {\left(1.2\right)}^{2}$
$\therefore A = 5.76 \pi + 2.88 \pi$
$\therefore A = 8.64 \pi \approx 27.143$
We should check that this value leads to a minimum (rather than a maximum) area. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that $\frac{{d}^{2} A}{\mathrm{dr}} ^ 2 > 0$ when $r = 1.2$ Instead I will just use the graph $A = \frac{6.912 \pi}{r} + 2 \pi {r}^{2}$
graph{(6.912pi)/x + 2pix^2 [-2, 15, -500, 500]}
Hopefully you can visually confirm that a minimum does indeed occur when $r = 1.2$ |
# Converse of Pythagorean Theorem
Converse of Pythagorean Theorem states that:
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.
Given: A ∆PQR in which PR2 = PQ2 + QR2
To prove: ∠Q = 90°
Construction: Draw a ∆XYZ such that XY = PQ, YZ = QR and ∠Y = 90°
So, by Pythagora’s theorem we get,
XZ2 = XY2 + YZ2
⇒ XZ2 = PQ2 + QR2 ……….. (i), [since XY = PQ and YZ = QR]
But, PR2 = PQ2 + QR2 ………… (ii), [given]
From (i) and (ii) we get,
PR2 = XZ2 ⇒ PR = XZ
Now, in ∆PQR and ∆XYZ, we get
PQ = XY,
QR = YZ and
PR = XZ
Therefore ∆PQR ≅ ∆XYZ
Hence ∠Q = ∠Y = 90°
Word problems using the Converse of Pythagorean Theorem:
1. The side of a triangle are of length 4.5 cm, 7.5 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 4.5 cm, 7.5 cm and 6 cm are the lengths of angled triangle, then 7.5 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(7.5)2 = (6)2 + (4.5)2
56.25 = 36 + 20.25
56.25 = 56.25
Since, both the sides are equal therefore, 4.5 cm, 7.5 cm and 6 cm are the side of the right angled triangle having hypotenuse 7.5 cm.
2. The side of a triangle are of length 8 cm, 15 cm and 17 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 8 cm, 15 cm and 17 cm are the lengths of angled triangle, then 17 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(17)2 = (15)2 + (8)2
289 = 225 + 64
289 = 289
Since, both the sides are equal therefore, 8 cm, 15 cm and 17 cm are the side of the right angled triangle having hypotenuse 17 cm.
3. The side of a triangle are of length 9 cm, 11 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(11)2 = (9)2 + (6)2
121 = 81 + 36
121 ≠ 117
Since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle.
The above examples of the converse of Pythagorean Theorem will help us to determine the right triangle when the sides of the triangles will be given in the questions.
Congruent Line-segments
Congruent Angles
Congruent Triangles
Conditions for the Congruence of Triangles
Side Side Side Congruence
Side Angle Side Congruence
Angle Side Angle Congruence
Angle Angle Side Congruence
Right Angle Hypotenuse Side congruence
Pythagorean Theorem
Proof of Pythagorean Theorem
Converse of Pythagorean Theorem
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# How do you express x^(4/3) in simplest radical form?
Jul 24, 2015
You raise $x$ to the ${4}^{\text{th}}$ power, then take the cube root.
#### Explanation:
When dealing with fractional exponents, it's always useful to remember that the exponent can be written as a product of an integer and of a fraction that has the numerator equal to 1.
In general, this looks like this
$\frac{a}{b} = a \cdot \frac{1}{b}$
This is important when dealing with fractional exponents because an exponent that takes the form $\frac{1}{b}$, like in the above example, is equivalent to taking the ${b}^{\text{th}}$ root.
${x}^{\frac{1}{b}} = \sqrt[b]{x}$
Since, for any $x > 0$, you have ${\left({x}^{a}\right)}^{b} = {x}^{a \cdot b}$, you can write
${x}^{\frac{4}{3}} = {x}^{4 \cdot \frac{1}{3}} = {\left({x}^{4}\right)}^{\frac{1}{3}} = \textcolor{g r e e n}{\sqrt[3]{{x}^{4}}}$
SImply put, you need to take the cube root from $x$ raised to the ${4}^{\text{th}}$ power.
Of course, you can also write
x^(4/3) = x^(1/3*4) = (x^(1/3))^4 = color(green)((root(3)(x))^4 |
# General term of Binomial Expansion
I assume that if you are reading this post you are already familiar of binomial expansion. You can read more about binomial theorem here. This topic will focus on how to determine the rth term of binomial expansion and other problems similar to it but will not expand it. General formula of expansion (a+b)n
rth term = nC(r-1)(a)n-r+1(b)r-1 where C is combination operator
Sample problem 1: What the 2nd term of the expansion of (2x-y)5 ? Solution: r=2, n=5,a=2x,b=-y
By using the formula above and substitution
2nd term = 5C(2-1)(2x)5-2+1(-y)2-1
=5C1(2x)4(-y)1
= -80x4y
Sample problem 2: Find the middle term in the expansion of (x2+3y2)6. Solution: In the expansion of (a+b)n there are n+1 terms.Therefore, there are 7 terms and the middle term is the 4th term.
r=4,n=6,a= x2, b=3y2 By substitution,
4th term=6C(4-1)( x2)6-4+1(3y2)4-1
=6C(3) ( x2)3(3y2)3
=540x6y6
Sample problem 3: Find the term involving x5 in the expansion of (2x-3y2)8.
Solution: It is easy to recognize such term since a and b are contain different coefficients. Since the power of x is decreasing, that is the power of first term is 8, second term is 7 and so on.
We can conclude that the term involving x5 is the 4th term. n=8, r=4,a=2x,b=-3y2
4th term=8C(4-1)(2x)8-4+1(-3y)4-1
=8C3(2x)5(-3y)3
=-48384x5y3
Sample problem 4:Find the term involving x5 in the expansion of (x2-1/x)7
Solution: Since a and b has common literal coefficient (x) we can’t do what we did in problem number 3. To solve this we will focus first to get the value of r.
From the literal coefficient part of our formula we have, (x)n-r+1(y)r-1 By substitution, we have n=7, we need to look for r. (x2)7-r+1(-1/x)r-1=x5, now we have an exponential equation, we can solve for r.
(x2)8-r(-x-1)r-1=x5 (x16-2r)(-x-r+1)=x5, the sign of x in the left side for now is irrelevant
By law of exponents,
16-2r+(-r+1)=5
17-3r=5
3r=12
r=4
therefore, term involving x5 is the 4th term, Going back to our formula above and by substitution, n=7, r=4, a=x2,y=-1/x
4th term= nC(r-1)(a)n-r+1(b)r-1
4th term= 7C(4-1)(x2)7-4+1(-1/x)4-1
4th term= 7C3(x2)4(-1/x)3
4th term= 7C3 x8(-1/x3)
4th term=-35x5
Sample problem 5: Find the sum of all numerical coefficients of (2x-3y)4.
Solution: You can expand the equation and add each numerical coefficient to get the answer for this question however that is so impractical especially if you are only given 15 seconds to answer the question.
The shortest way to get the answer is to let x=1 and let y=1 and evaluate the expression.
Sum=(2x-3y)4,x=1 , y=1
Sum=(2(1)-3(1))4
Sum=(-1)4
Sum=1
Practice problems
Note: If the answer is too large you can keep your answer to the product of its prime factor
1. Find the 8th term of (3x+2y)12
2. What is the middle term in the expansion of (x-4y)6?
3. Find the sum of the of all numerical coefficients of (2x+5y)5
4. What is the term involving x3 in the expansion of (x+2)7
5. Find the term involving x2 in the expansion of (x2-1/x)10 Answer key here
### Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
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# Surface Areas and Volumes Formulas for Class 9 Maths Chapter 13
## Surface Areas and Volumes Formulas for Class 9 Maths Chapter 13
Are you looking for Surface Areas and Volumes formulas or important points that are required to understand Surface Areas and Volumes for class 9 maths Chapter 13? You are the right place to get all information about Surface Areas and Volumes Class 9 maths chapters 13. Surface Areas and Volumes formulas play a vital role in preparing you for the class 9 exam as well as higher studies. Surface Areas and Volumes formulas are very helpful for better scores in the exam. Check Surface Areas and Volumes formulas according to class 9:
$\ Surface\;area\;of\;Cube=6a^{2}$ $\ Volume\;of\;a\;cube=a^{3}$ Where, a is the side length of the cube. $\ Surface\;area\;of\;Cuboid = 2(lb + bh + hl)$ $\ Volume\;of\;a\;Cuboid = h \times l \times w$ Where, l: Height, h: Legth, w: Depth $\ Diameter\;of\;a\;sphere=2r$ $\ Circumference\;of\;a\;sphere=2\pi r$ $\ Surface\;area\;of\;a\;sphere=4\pi r^{2}$ $\ Volume\;of\;a\;sphere=\frac{4}{3}\: \pi r^{3}$ $\ Curved\;Surface\;area\;of\;a\;Hemisphere =4\pi r^{2}$ $\ Total\;Surface\;area\;of\;a\;Hemisphere =3\pi r^{2}$ $\ Volume\;of\;a\;Hemisphere =\frac{2}{3}\: \pi r^{3}$ Where, r: Radius $\ Curved\;Surface\;area\;of\;a\;Cylinder =2\pi rh$ $\ Total\;Surface\;area\;of\;a\;Cylinder =2\pi r(r+h)$ $\ Volume\;of\;a\;Cylinder = \pi r^{2} h$ Where, r: Radius, h: Height $\ Total\;Surface\;Area\;of\;cone=\pi r \left (s+r \right )$ Where, r: Radius $\ Vomule\;of\;cone=\frac {1}{3}\pi r^{2}h$ $\ Curved\;Surface\;Area\;of\;cone=\pi rs$ Where, r: radius of cone. h:height of cone. s: slant height of the cone.
#### Summary of Surface Areas and Volumes
We have shared very important formulas for Surface Areas and Volumes which helps to score in the class 9 exams. If you have any questions and doubts related to Surface Areas and Volumes please let me know through comment or mail as well as social media. When you understand the formulas behind each Surface Areas and Volumes topics then it would be easier to solve the most complex problems related to Surface Areas and Volumes too.
#### NCERT Solutions For Class 9 Maths by Chapters
• Chapter 1 – Number Systems
• Chapter 2 – Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 – Linear Equations in Two Variables
• Chapter 5 – Introduction to Euclids Geometry
• Chapter 6 – Lines and Angles
• Chapter 7 – Triangles |
# Study Guide for Intermediate Algebra
Lesson 1
Properties of Real Numbers and Linear Equations
Lesson Objectives
After completing this lesson, students will be able to do the following:
1. Use the distributive property and other properties that are presented
2. Solve linear equations including those with fractions and decimals
3. Solve and apply formulas
4. Solve percent problems
Important Terms
distributive property identity property associative property algebraic expressions linear equation addition and multiplication properties of equality inverse property commutative property multiplication property of 0 algebraic equations formula first-degree equations fractional equations
Lecture
Note: This is a good time to become familiar with your textbook. You will notice that the important
words and/or concepts are highlighted in bold text and in blue boxes. You may find it helpful to put this
information on 3 x 5 index cards and make yourself a set of “flash cards” as a study tool. Also, when I
assign practice or written problems, I will just give you the first page number even though the problems
may be spread out over two or three pages.
1.4 Properties of Real Numbers
♦ Real numbers are the numbers we work with every day. These include integers, fractions and
decimals
. Please don’t let the language of mathematics scare you off!
♦ The distributive property is incredibly important. Be sure you understand how to use it. While the
other properties listed are important, I don’t really care if you know their names. It’s more important
that you know how and when to use them, and that will come with practice.
♦ Be sure you can recognize like terms .
♦ Although the book hasn’t made a big deal about it, I think it is time to point out the difference
between terms and factors .
1. Factors are the numbers and/or variables being multiplied together.
Example:
–5abc has factors of –5, a, b, and c.
2. A term is an indicated product that may have any number of factors.
Example:
In 7xy + 5abc, 7xy is a term with factors 7, x, and y. 5abc is a term with factors 5, a, b, and c.
♦ As a general rule, terms are separated by a + or – sign.
Practice Problems
Work page 41, #1–10, 11, 15, 19, 23, 27.
(Answers to even-numbered problems are given toward the back of this study guide.)
2.1 Linear Equations in One Variable
♦ An expression does not contain an equal sign. It can only be simplified. It is considered an algebraic
expression when it contains variables such as x or y.
♦ An equation does contain an equal sign . Equations must be solved. An algebraic equation contains
at least one variable .
♦ You will see the term linear equation used frequently. What makes an equation linear is the fact that
the variable is never raised to a power other than one. When the power (or exponent) of a number or
variable is one, we generally don’t write it. For example: x1 is the same as x. For this reason, these
equations are sometimes called first-degree equations.
♦ As you read through this material, take your time! These are the building blocks for much of what we
will do during the course. It is essential that you understand the concepts.
♦ You can do anything you want to an equation as long as you do it to both sides.
♦ I really like the way this text explains solving equations involving fractions (I call these fractional
equations) and decimals.
♦ Here is another example of a fractional equation:
15(3a – 1) + 20(a – 2) – 12(a – 1) = 3(21)
45a – 15 + 20a – 40 – 12a + 12 = 63
53a – 43 = 63
53a = 106
a = 2
Practice Problems
Work page 61, #9–41 (every other odd), 43, 49, 55, 59.
2.2 Formulas
♦ Solving formulas for a specific variable is not only fun, it is also very useful. I like to call these
manipulation problems because you are manipulating the formula to make it useful for you!
♦ Percent problems always give people trouble. Percent is from Latin: per means “by,” centum, “one
hundred.” This means, for us, “by hundreds.”
♦ I teach my students that percent is the part divided by the whole. In this text, the authors say the same
thing, but they phrase it differently, saying that percent is the partial amount (a) divided by the whole
amount (b). What you need to do is figure out what makes sense to you and use that.
Practice Problems
Work page 72, #7–17 odd, 21, 27, 35, 37.
Written Assignment
Please read the Appendix in the back of this study guide for essential Independent Study in
Idaho policies and procedures, and forms you will need to successfully complete this course.
You are responsible for understanding and following ISI policies and procedures. If there is
anything on these pages you do not understand, contact the ISI office for clarification. Before
starting the written assignment for Lesson 1, see the letter in your registration packet for your
instructor’s requirements: how to format and submit lessons, number of lessons you may
submit at one time, and lesson guidelines.
Work the following pages and problems: (10 points)
42 (#34, 36)
62 (#24, 32, 54, 62)
72 (#16, 22, 28, 36)
Prev Next |
# How Do You Determine Where To Place The Decimal Point In The Product?
Multiplying a decimal by 10 moves the decimal point one place to the right. Dividing a decimal by 10 moves the decimal point one place to the left. Multiplying or dividing by a power of 10 will move the decimal point equal to the number of 0s.
## Which way do decimals go?
Move the decimal point to the left when dividing a decimal number by a power of 10 number. As with multiplying decimals by power of 10 numbers, count the number of zeros to know how many spaces to move the decimal, but move the decimal in the opposite direction.
## Why do you move the decimal point when multiplying?
It’s simply a matter of counting how many factors of 10 appear in the denominator after the multiplication. Each factor of 10 in the denominator moves the decimal point one place to the left.
### When multiplying a number by 10 Where is the decimal point moved?
When you multiply a decimal by a power of 10, all you need to do is move the decimal point! as an example. For each zero in the power of 10, move the decimal point one place to the right. Since there are 3 zeros in 1,000, move the decimal point in 12.7 to the right 3 places.
### What are the four rules of decimals?
You should become efficient in using the four basic operations involving decimals—addition, subtraction, multiplication, and division.
### How do you find the quotient with decimals?
Divide the dividend by the whole-number divisor to find the quotient.
1. Multiply the divisor by a power of 10 to make it a whole number.
2. Multiply the dividend by the same power of 10. Place the decimal point in the quotient.
3. Divide the dividend by the whole-number divisor to find the quotient.
### What are 6 steps to multiplying decimals?
You can multiply numbers with decimals just as you would multiply whole numbers, as long as you remember to factor in the decimals at the end of the problem.
Multiply the numbers while ignoring the decimal points.
1. Start by multiplying the 6 in . …
2. Multiply the 6 by the 4 in . …
3. When you start multiplying .
### What is the product when you multiply decimals by 1?
So, the product will have 1 + 1 = 2 digits after the decimal point.
### Are numbers whose decimal form is non terminating and do not have a pattern?
Solution: The number which has an infinite number of digits after the decimal point without any pattern of repetition of digits is known as non-terminating non-recurring decimal. Non-terminating non-recurring decimals are also known as irrational numbers.
### What does 3 decimal places mean?
“Three decimal places” is the same as “the nearest thousandth.” So, for example, if you are asked to round 3.264 to two decimal places it means the same as if your are asked to round 3.264 to the nearest hundredth.
### How many is 3 decimal places?
However many decimal places we have, that’s the number of zeroes for the number matching the “place” we’re at. For instance, the number 83.295 has three digits after the decimal place (the digits 2, 9, and 5), so the number has three decimal places. The number with “1” followed by three zeroes is 1,000: one thousand.
### What is rounded to three decimal places?
As the digits 759458 are more than half i.e. 500000 , rounding them means the digit in thousandth place would have to increased by one. Hence, rounding 435.738759458 to 3 decimal places means it becomes 435.739 .
### How many times do you move the decimal when multiplying?
The difference is that the decimal point moves to the right when you multiply; it moves to the left when you divide.
### What does divided by look like?
The division sign (÷) is a symbol consisting of a short horizontal line with a dot above and another dot below, used to indicate mathematical division.
### How do you move decimals on a calculator?
To change to four decimal places, press the 4 key followed by the ENTER key. To change the number of decimal places with the HP 10BII, press the Shift key (the key with the yellow square on it), followed by the DISP key (the = key), followed by the 4 key. Set your calculator to one (1) payment per year.
### How do you divide a decimal by 100?
To divide a whole or a decimal number by 100, move the decimal point two places to the left. Note that although a whole number does not have a decimal point, we can always add it at the end of the number. (For example, 35 and 35. are the same numbers.)
### How do you start teaching decimals?
Start by explaining whole number place values, like tens and hundreds. Describe decimals as “in-between” numbers with their own set of place values, including tenths and hundredths. Mention that decimals are related to fractions, and show your students how to convert one to the other. |
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# Lecture #4: Eigenvalues and eigenvectors
Definition:
Let A be a square matrix, then a non-zero vector v is called an eigenvector of A if and only if
there exists a number (real or complex) such that
v v A =
If such a number exists, it is called an eigenvalue of A. The vector v is called eigenvector associated to
the eigenvalue .
Remark:
The eigenvector v must be non-zero since we have
0 0 0 = = A
for any number . For example
let
(
=
0 1
2 3
A ,
(
=
1
2
1
v and
(
=
2
3
2
v find if v
1
and v
2
are an eigenvectors of A.
Solution:
v
1
is and eigenvector of A with an associate eigenvalue =2.
v
2
is not an eigenvector of A because Av
2
is not a multiple of v
2
.
(
=
(
=
(
=
2
3
3
13
2
3
0 1
2 3
2
Av
(
=
(
=
(
=
1
2
2
2
4
1
2
0 1
2 3
2
Av
Calculating Eigenvalues and Eigenvectors of a matrix
The eigenvalue equation
v v A =
for a matrix A can be express as
0 = I v v A
Where
I
is the identity matrix, and can be rearrange to
0 ) ( = v I A
Because the eigenvector must not be zero, the determinant of ) ( I A must be zero. In Linear Algebra
this is called the characteristic equation of a square matrix A.
0 ) det( = I A
The solutions of the characteristic equation are precisely the eigenvalues of the matrix A. The
polynomial which results from evaluating the determinant is the characteristic polynomial of the matrix.
The eigenvectors associated to a specific eigenvalue () can be found by solving
0 ) ( = v I A .
Example #1:
Let
(
=
2 2
2 5
A compute the eigenvalues and the associated eigenvectors
1) To calculate the eigenvalues ( ) we must solve determinant of ( A- I )=0, where I is the
identity matrix.
(
=
(
2 2
2 5
1 0
0 1
2 2
2 5
0 6 7
0 4 ) 2 )( 5 (
0
2 2
2 5
2
= +
= =
=
Solving for we have =6 and =1.
2) To compute the eigenvectors v we solve (A- I )v = 0.
Case =6
(
=
(
=
(
0
0
4 2
2 1
0
0
6 2 2
2 6 5
2
1
2
1
v
v
v
v
Solving for v
1
and v
2
we have
2 1 2 1
2 0 2 v v v v = = +
If 2 1
1 2
= = v then v
(
=
(
1
2
2
1
v
v
Case =1
(
=
(
=
(
0
0
1 2
2 4
0
0
1 2 2
2 1 5
2
1
2
1
v
v
v
v
1 2 2 1
2 0 2 4 v v v v = = +
If 2 1
2 1
= = v then v
Solving for v1 and v
2
(
=
(
2
1
2
1
v
v
Example #2:
Let
(
(
(
=
1 1 0
1 2 1
0 1 1
A compute the eigenvalues and the associated eigenvectors
We solve det ( A- I )=0
| | | |
| |
| |
3 0 1
0 ) 3 )( )( 1 ( 3 ) 1 (
0 2 ) 1 )( 2 ( ) 1 (
0 0 ) 1 ( 1 1 ) 1 )( 2 ( ) 1 (
0
1 0
2 1
0
1 0
1 1
1
1 1
1 2
) 1 (
0
1 1 0
1 2 1
0 1 1
2
= = =
= =
=
= + +
=
Solving for we have =1, =0 and =3.
Case =1:
(
(
(
=
(
(
(
(
(
(
(
(
(
=
(
(
(
(
(
(
0
0
0
0 1 0
1 1 1
0 1 0
0
0
0
) 1 1 ( 1 0
1 ) 1 2 ( 1
0 1 ) 1 1 (
3
2
1
3
2
1
v
v
v
v
v
v
3 1 3 2 1 3 2 1
2
0
0
v v v v v v v v
v
= = = +
=
(
(
(
=
1
0
1
v
Case =0:
(
(
(
=
(
(
(
(
(
(
(
(
(
=
(
(
(
(
(
(
0
0
0
1 1 0
1 2 1
0 1 1
0
0
0
) 0 1 ( 1 0
1 ) 0 2 ( 1
0 1 ) 0 1 (
3
2
1
3
2
1
v
v
v
v
v
v
(
(
(
=
1
1
1
v
Case =3:
(
(
(
=
(
(
(
(
(
(
(
(
(
=
(
(
(
(
(
(
0
0
0
2 1 0
1 1 1
0 1 2
0
0
0
) 3 1 ( 1 0
1 ) 3 2 ( 1
0 1 ) 3 1 (
3
2
1
3
2
1
v
v
v
v
v
v
(
(
(
=
1
2
1
v
2 1 3 3 2 1
2 1 2 1
2 0 2
1 0
v v v v v v
v v v v
+ = = +
= = =
2 1 3 3 2 1
1 2 2 1
0
2 0 2
v v v v v v
v v v v
= =
= = + |
# Thread: Finding inverse of a 3X3 Matrix
1. ## Finding inverse of a 3X3 Matrix
I've found inverses for 2X2 matrices in the past but have never done a 3X3. I looked at a few sites and they performed things such as row operations which I've never seen before. I tried to use the method myself but got more confused . Does anyone know a way that is easy to understand?
2. Have you learned how to reduce a matrix to row echelon form before? If so, then take your matrix and reduce it to row echelon form. For each step you do, you do it to the identity matrix. What is left after row reducing is your inverse matrix.
For example, say that your first step to reducing B to the identity matrix is dividing the entire first row by 4, then you do the same step to the identity matrix:
$\left[ \begin{array}{ccc} 4 & 5 & 8 \\ 7 & 6 & -1 \\ 0 & 9 & 2 \end{array} \Bigg| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \: \: \Rightarrow$ ${\color{white}.} \: \: \left[ \begin{array}{ccc} 1 & \frac{5}{4} & 2 \\ 7 & 6 & -1 \\ 0 & 9 & 2 \end{array} \Bigg| \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$
See how what I did to the left side, I did to the right side? Now continue to row reduce the left matrix to the identity matrix and perform the same steps to the right matrix. Again, whatever your result is on the right side after row reducing the left matrix is your inverse.
-----------------------
If not, you can always use this ghastly formula: $B^{-1} = \frac{adj(B)}{det(B)}$
$adj(B)$ is found by finding the cofactors of each entry of B and then taking its transpose.
Here's a resource that you may find helpful: Paul's Online Notes: Linear Algebra
3. oo I see what you did, but how did you know to divide by 4? Is there some way of telling what to multiply or divide by? Are we able to subtract and add too? |
# 6th Grade Math Ratio Word Problems
The following are some examples of 6th Grade Math Word Problems that deals with ratio and proportions. The word problems are solved with the help of tape diagrams, block diagrams or bar model (Singapore Math)
You may also want to check out how to solve Ratio Word Problems using Fractions.
Example:
Mark and Fred had some money in the ratio 6:1. Mark gave half of his money to Fred. Find the ratio of the amount of money Mark had left to the amount of money Fred had in the end.
Solution:
Before
After
The ratio of the amount of money Mark had left to the amount of money Fred had in the end is 3:4.
The following diagram gives another example of a ratio word problem solved using modeling.
How to solve ratio word problems?
Examples:
1. A board was cut into two pieces whose lengths are in the ratio 2:5. The longest piece was 85 inches. How long was the shortest piece?
2. At a track meet, 5/8 of the students were boys. There were 48 more boys. How many students were there altogether?
3. Theresa read two and one-fourth times as many pages on Saturday as on Sunday. She read 120 pages more on Saturday. How many pages did she read altogether?
How to solve more difficult ratio word problems?
Examples:
1. The ratio of pigs to cows to sheep on the farm is 2:4:7. There are 65 more sheep than pigs. How many cows are there on the farm?
2. Archimedes, Hypatia and Zeno shared a sum of money. Archimedes received 1/6 of the money. Hypatia and Zeno shared the rest of the money in the ratio 2:3. Zeno got \$7 more than Hypatia. How much money did they share altogether?
3. One-third of Elmer’s age is the same as two-fifths of Sam’s age.
a) Draw a bar model representing the situation
b) Elmer is 5 years older than Sam. How old is Sam?
How to use proportions to solve ratio problems?
1. At Al’s Catering Service, they often make a large batch of fruit salad. The ratio of cups of apples to cups of oranges in the recipe is 8 : 5.
a) If they use 15 cups of oranges, how many cups of apples should they use?
b) If they only use 2 cups of apples, how many cups of oranges should they use?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Right Triangle Applications
Discussion in 'Algebra' started by nycmathguy, Jul 6, 2022.
1. ### nycmathguy
Joined:
Jun 27, 2021
Messages:
5,386
422
College Algebra
Chapter 1/Section 3
Radius of Earth = 3960 miles
Feet in one mile = 5280
For 53
Let d = hypotenuse
(d)^2 + (radius of Earth)^2 = [(radius of Earth) plus (distance above sea level)/(feet in one mile)]^2
(d)^2 + (3960)^2 = [(3960) + (20/5280)]^2
Is this the correct set up?
Can you set up 54? I will show the math work?
nycmathguy, Jul 6, 2022
2. ### MathLover1
Joined:
Jun 27, 2021
Messages:
2,989
2,884
53. set up is correct
54.
Use the facts that the radius of Earth is 3960 miles and 1 mile =5280 feet.
Solution. The Earth is a sphere, so we cut this sphere with a plane
passing through the person, the ship, and the center of the Earth. This
gives the following cross section (certainly not to scale!):
The radius of the Earth in feet is (3960 miles)(5280 feet/mile) = (3960)(5280) feet = 20,908,800 feet.
So the distance from the center of the Earth to the eyes of the observer (technically, to the top of the observer’s head ) is 20,908,806 feet.
We draw a line tangent to the circle passing through the observer’s eyes. Lines tangent to a circle are perpendicular to a radius of the circle containing the point of tangency, so we get the pictured right triangle and we want to find d.
Since we have a right triangle, then the Pythagorean Theorem gives
(20,908,800 ft)^2 + d^2 = (20,908,806 ft)^2
d^2 = (20,908,806^2 − 20,908,800)^2ft^2
d = sqrt(250905636ft^2 )
Since d is a distance it is positive and so
d =15840 ft
Now (15840 feet feet)(1/5280miles/feet) = 3 miles.
So the ship is 3 miles away .
MathLover1, Jul 6, 2022
nycmathguy likes this.
Joined:
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Messages:
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# Multiplication Chart Year 4
Understanding multiplication right after counting, addition, as well as subtraction is ideal. Children find out arithmetic by way of a normal progression. This progress of studying arithmetic is truly the subsequent: counting, addition, subtraction, multiplication, lastly department. This document results in the concern why learn arithmetic in this particular pattern? Furthermore, why learn multiplication right after counting, addition, and subtraction just before section?
## The next information respond to these concerns:
1. Youngsters find out counting very first by associating aesthetic physical objects making use of their hands. A real case in point: Just how many apples are there any in the basket? A lot more abstract illustration is just how older are you currently?
2. From counting amounts, the subsequent logical phase is addition followed by subtraction. Addition and subtraction tables are often very beneficial instructing tools for the kids as they are aesthetic instruments generating the move from counting much easier.
3. Which should be discovered after that, multiplication or division? Multiplication is shorthand for addition. At this point, children use a firm knowledge of addition. For that reason, multiplication may be the after that rational form of arithmetic to find out.
## Overview fundamentals of multiplication. Also, assess the basics using a multiplication table.
Allow us to overview a multiplication case in point. By using a Multiplication Table, multiply 4 times 3 and have a solution twelve: 4 by 3 = 12. The intersection of row a few and line 4 of the Multiplication Table is a dozen; 12 will be the solution. For kids beginning to find out multiplication, this is certainly straightforward. They may use addition to solve the issue therefore affirming that multiplication is shorthand for addition. Case in point: 4 x 3 = 4 4 4 = 12. It is an exceptional overview of the Multiplication Table. A further benefit, the Multiplication Table is visible and displays to discovering addition.
## In which do we commence understanding multiplication using the Multiplication Table?
1. Very first, get knowledgeable about the table.
2. Begin with multiplying by 1. Start off at row number one. Go on to column primary. The intersection of row 1 and column one is the solution: one particular.
3. Repeat these actions for multiplying by one. Grow row 1 by columns 1 through a dozen. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Repeat these techniques for multiplying by two. Multiply row two by posts one by way of five. The solutions are 2, 4, 6, 8, and 10 respectively.
5. We will hop in advance. Recurring these methods for multiplying by five. Flourish row 5 various by columns 1 through twelve. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now allow us to raise the level of trouble. Replicate these actions for multiplying by a few. Multiply row about three by columns one particular by way of 12. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be comfortable with multiplication up to now, try out a check. Resolve the following multiplication issues in your mind after which assess your answers towards the Multiplication Table: multiply 6 as well as two, increase nine and three, flourish one and eleven, grow 4 and a number of, and grow 7 and two. The problem replies are 12, 27, 11, 16, and 14 respectively.
In the event you received four out of five troubles right, create your very own multiplication exams. Estimate the replies in your mind, and look them using the Multiplication Table.
</ |
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# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.1
• Difficulty Level : Expert
• Last Updated : 24 Nov, 2020
### Question 1. Identify the terms, their coefficients for each of the following expressions:
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii) 3x2y2 – 5x2y2z2 + z2
(iv) 9 – ab + bc – ca
(v) a/2 + b/2 – ab
(vi) 0.2x – 0.3xy + 0.5y
Solution:
(i) 7x2yz – 5xy
In this expression there are two terms:
7x2yz and – 5xy
The coefficient of 7x2yz is 7
The coefficient of –5xy is –5
(ii) x2 + x + 1
In this expression there are three terms:
x2, x and 1
The coefficient of x2 is 1
The coefficient of x is 1
The coefficient of 1 is 1
(iii) 3x2y2 – 5x2y2z2 + z2
In this expression there are three terms:
3x2y, -5x2y2z2 and z2
The coefficient of 3x2y is 3
The coefficient of -5x2y2z2 is -5
The coefficient of z2 is 1
(iv) 9 – ab + bc – ca
In this expression there are four terms:
9, -ab, bc and -ca
The coefficient of 9 is 9
The coefficient of -ab is -1
The coefficient of bc is 1
The coefficient of -ca is -1
(v) a/2 + b/2 – ab
In this expression there are three terms:
a/2, b/2, and -ab
The coefficient of a/2 is 1/2
The coefficient of b/2 is 1/2
The coefficient of -ab is -1
(vi) 0.2x – 0.3xy + 0.5y
In this expression there are three terms:
0.2x, -0.3xy and 0.5y
The coefficient of 0.2x is 0.2
The coefficient of -0.3xy is -0.3
The coefficient of 0.5y is 0.5
### Question 2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x + y
(ii) 1000
(iii) x + x2 + x3 + x4
(iv) 7 + a + 5b
(v) 2b – 3b2
(vi) 2y – 3y2 + 4y3
(vii) 5x – 4y + 3x
(viii) 4a – 15a2
(ix) xy + yz + zt + tx
(x) pqr
(xi) p2q + pq2
(xii) 2p + 2q
Solution:
(i) x + y
The given expression contains two terms x and y
∴ It is Binomial
(ii) 1000
The given expression contains one term 1000
∴ It is Monomial
(iii) x + x2 + x3 + x4
The given expression contains four terms
∴ It belongs to none of the categories
(iv) 7 + a + 5b
The given expression contains three terms
∴ It is Trinomial
(v) 2b – 3b2
The given expression contains two terms
∴ It is Binomial
(vi) 2y – 3y2 + 4y3
The given expression contains three terms
∴ It is Trinomial
(vii) 5x – 4y + 3x
The given expression contains three terms
∴ It is Trinomial
(viii) 4a – 15a2
The given expression contains two terms
∴ It is Binomial
(ix) xy + yz + zt + tx
The given expression contains four terms
∴ It belongs to none of the categories
(x) pqr
The given expression contains one term
∴ It is Monomial
(xi) p2q + pq2
The given expression contains two terms
∴ It is Binomial
(xii) 2p + 2q
The given expression contains two terms
∴ It is Binomial
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Home » Uncategorized » Inequality 55 (Unknown Author)
# Inequality 55 (Unknown Author)
Problem:
If $x,y$ are positive real numbers then prove that
$\displaystyle x^y+y^x>1.$
Solution:
The Inequality is obvious for $x,y\geq 1$. So, let’s consider the case where $x,y\in (0,1)$. We will now divide the Inequality into two cases.
• Case 1st: $x^{y-1}>y^{x-1}.$ Then it holds that
$\displaystyle (x^y+y^x)^{1/x}=\left[y^x\left(\frac{x^y}{y^x}+1\right)\right]^{1/x}=y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}$
therefore, according to Bernulli’s Inequality it will hold that
\displaystyle \begin{aligned}y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}&\geq y\left(1+\frac{x^{y-1}}{y^x}\right)\\&\geq y\left(1+\frac{y^{x-1}}{y^x}\right)\\&=y+1>1.\end{aligned}
• Case 2nd: $y^{x-1}>x^{y-1}.$ Then it is
$\displaystyle (x^y+y^x)^{1/y}=\left[x^y\left(\frac{y^x}{x^y}+1\right)\right]^{1/y}=x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}$
therefore, once again from Bernulli’s Inequality we get
\displaystyle \begin{aligned}x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}&\geq x\left(1+\frac{y^{x-1}}{x^y}\right)\\&\geq x\left(1+\frac{x^{y-1}}{x^y}\right)\\&=x+1>1\end{aligned}
and the proof is completed Q.E.D. |
1. Class 10
2. Important Questions for Exam - Class 10
3. Chapter 11 Class 10 Constructions
Transcript
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Given sides other than hypotenuse in a right angled triangle. ∴ Both sides will be perpendicular to each other Steps of construction Draw a line segment AB = 4 cm. Draw a ray RA making 90° with it. Taking A as center and 3 cm as radius, draw an arc intersecting the ray RA at C. Join BC. ΔABC is the required triangle. Locate 5 points (as 5 is greater in 5 and 3), A_1, A_2, A_3, A_4, A_5, on line segment AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3 = A_3 A_4= A_4 A_5. Join A_3B. Draw a line through A_5 parallel to A_3B intersecting extended line segment AB at B'. Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle. Justification Here, (AB^′)/AB=(AA_5)/(AA_3 ) = 5/3 Also, A’C’ is parallel to AC So, the will make the same angle with line BC ∴ ∠ AB’C’ = ∠ ABC Now, In Δ AB’C’ and Δ ABC ∠ A = ∠ A ∠ AB’C’ = ∠ ABC Δ A’BC’ ∼ Δ ABC Since corresponding sides of similar triangles are in the same ratio AB/AB′=BC/(B^′ C^′ )=AC/(AC^′ ) So, AB/AB′=BC/(B^′ C^′ )=AC/(AC^′ )=5/3 This justifies the construction.
Chapter 11 Class 10 Constructions
Class 10
Important Questions for Exam - Class 10 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.1: Tree Diagrams
Difficulty Level: Basic Created by: CK-12
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Suppose you flip a coin three times in succession. How can you illustrate all your possible outcomes? What is the probability that your coin will come up heads three times in a row?
### Watch This
First watch this video to learn about tree diagrams.
Then watch this video to see some examples.
Watch this video for more help.
### Guidance
In previous Concepts, we studied independent and dependent events, as well as mutually inclusive and mutually exclusive events. We used the Addition Rule for dependent events, as well as mutually inclusive and mutually exclusive events. The Addition Rule, or Addition Principle, is used to find P(A or B)\begin{align*}P(A \ or \ B)\end{align*}, while the Multiplication Rule is used for independent events.
Addition Rule – For 2 events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, the probability of selecting one event or another is given by: P(A or B)=P(A)+P(B)P(A and B)\begin{align*}P(A \ \text{or} \ B) = P(A) + P(B) - P(A \ \text{and} \ B)\end{align*}.
Multiplication Rule – For 2 independent events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, where the outcome of A\begin{align*}A\end{align*} does not change the probability of B\begin{align*}B\end{align*}, the probability of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} is given by: P(A and B)=P(A)×P(B)\begin{align*}P(A \ \text{and} \ B) = P(A) \times P(B)\end{align*}.
Tree diagrams are another way to show the outcomes of simple probability events. In a tree diagram, each outcome is represented as a branch on a tree.
#### Example A
If you toss a coin 2 times, what is the probability of getting 2 heads? Use a tree diagram to find your answer.
This is an example of independent events, because the outcome of one event does not affect the outcome of the second event. What does this mean? Well, when you flip the coin once, you have an equal chance of getting a head (H) or a tail (T). On the second flip, you also have an equal chance of getting a a head or a tail. In other words, whether the first flip was heads or tails, the second flip could just as likely be heads as tails. You can represent the outcomes of these events on a tree diagram.
From the tree diagram, you can see that the probability of getting a head on the first flip is 12\begin{align*}\frac{1}{2}\end{align*}. Starting with heads, the probability of getting a second head will again be 12\begin{align*}\frac{1}{2}\end{align*}. But how do we calculate the probability of getting 2 heads? These are independent events, since the outcome of tossing the first coin in no way affects the outcome of tossing the second coin. Therefore, we can calculate the probability as follows:
P(A and B)P(A and B)=12×12=14\begin{align*}P(A \ \text{and} \ B) &= \frac{1}{2} \times \frac{1}{2}\\ P(A \ \text{and} \ B) &= \frac{1}{4}\end{align*}
Therefore, we can conclude that the probability of getting 2 heads when tossing a coin twice is 14\begin{align*}\frac{1}{4}\end{align*}, or 25%. Let’s try an example that is a little more challenging.
#### Example B
Irvin opens up his sock drawer to get a pair socks to wear to school. He looks in the sock drawer and sees 4 red socks, 8 white socks, and 6 brown socks. Irvin reaches in the drawer and pulls out a red sock. He is wearing blue shorts, so he replaces it. He then draws out a white sock. What is the probability that Irvin pulls out a red sock, replaces it, and then pulls out a white sock?
First let’s draw a tree diagram.
There are 18 socks in Irvin’s sock drawer. The probability of getting a red sock when he pulls out the first sock is:
P(red)P(red)=418=29\begin{align*}P(\text{red}) &= \frac{4}{18}\\ P(\text{red}) &= \frac{2}{9}\end{align*}
Irvin puts the sock back in the drawer and pulls out the second sock. The probability of getting a white sock on the second draw is:
P(white)P(white)=818=49\begin{align*}P(\text{white}) &= \frac{8}{18}\\ P(\text{white}) &= \frac{4}{9}\end{align*}
Therefore, the probability of getting a red sock and then a white sock when the first sock is replaced is:
P(red and white)P(red and white)=29×49=881\begin{align*}P(\text{red and white}) &= \frac{2}{9} \times \frac{4}{9}\\ P(\text{red and white}) &= \frac{8}{81}\end{align*}
One important part of these types of problems is that order is not important.
Let’s say Irvin picked out a white sock, replaced it, and then picked out a red sock. Calculate this probability.
P(white and red)P(white and red)=49×29=881\begin{align*}P(\text{white and red}) &= \frac{4}{9} \times \frac{2}{9}\\ P(\text{white and red}) &= \frac{8}{81}\end{align*}
So regardless of the order in which he takes the socks out, the probability is the same. In other words, P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*}.
#### Example C
In Example B, what happens if the first sock is not replaced?
The probability that the first sock is red is:
P(red)P(red)=418=29\begin{align*}P(\text{red}) &= \frac{4}{18}\\ P(\text{red}) &= \frac{2}{9}\end{align*}
The probability of picking a white sock on the second pick is now:
So now, the probability of selecting a red sock and then a white sock, without replacement, is:
P(red and white)P(red and white)P(red and white)=29×617=12153=451\begin{align*}P(\text{red and white}) &= \frac{2}{9} \times \frac{6}{17}\\ P(\text{red and white}) &= \frac{12}{153}\\ P(\text{red and white}) &= \frac{4}{51}\end{align*}
If the first sock is white, will P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*} as we found in Example 1? Let's find out.
P(white)P(white)=618=13\begin{align*}P(\text{white}) &= \frac{6}{18}\\ P(\text{white}) &= \frac{1}{3}\end{align*}
The probability of picking a red sock on the second pick is now:
As with the last example, P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*}. So when does order really matter? We'll find out in the next concept.
Points to Consider
• How are tree diagrams helpful for determining probabilities?
### Guided Practice
In a survey, baseball fans were asked who they would like to win the National League playoffs. 54% responded that they would like the Phillies to win, and 46% responded that they would like the Giants to win. The fans were then asked who they would like to win the American League playoffs if the Phillies win the National League playoffs, and who they would like to win the American League playoffs if the Giants win the National League playoffs. If the Phillies win the National League playoffs, 42% of the fans responded that they want the Rangers to win the American League playoffs, and 58% said that they want the Yankees to win. If the Giants win the National League playoffs, 48% of the fans responded that they want the Rangers to to win the American League Playoffs, and 52% said that they want the Yankees to win. The results of the survey are shown in the following tree diagram:
According to the survey, what percentage of the fans want each of the possible World Series match-ups? Do all of the probabilities add up to 100%?
The percentage of the fans wanting each of the possible World Series match-ups can be calculated as follows:
P(Phillies and Rangers)P(Phillies and Rangers)P(Phillies and Rangers)=0.54×0.42=0.2268=22.68%\begin{align*}P(\text{Phillies and Rangers}) &= 0.54 \times 0.42\\ P(\text{Phillies and Rangers}) &= 0.2268\\ P(\text{Phillies and Rangers}) &= 22.68\%\end{align*}
P(Phillies and Yankees)P(Phillies and Yankees)P(Phillies and Yankees)=0.54×0.58=0.3132=31.32%\begin{align*}P(\text{Phillies and Yankees}) &= 0.54 \times 0.58\\ P(\text{Phillies and Yankees}) &= 0.3132\\ P(\text{Phillies and Yankees}) &= 31.32\%\end{align*}
P(Giants and Rangers)P(Giants and Rangers)P(Giants and Rangers)=0.46×0.48=0.2208=22.08%\begin{align*}P(\text{Giants and Rangers}) &= 0.46 \times 0.48\\ P(\text{Giants and Rangers}) &= 0.2208\\ P(\text{Giants and Rangers}) &= 22.08\%\end{align*}
P(Giants and Yankees)P(Giants and Yankees)P(Giants and Yankees)=0.46×0.52=0.2392=23.92%\begin{align*}P(\text{Giants and Yankees}) &= 0.46 \times 0.52\\ P(\text{Giants and Yankees}) &= 0.2392\\ P(\text{Giants and Yankees}) &= 23.92\%\end{align*}
Now let's add up all the probabilities.
22.68%+31.32%+22.08%+23.92%=100%\begin{align*}22.68\%+31.32\%+22.08\%+23.92\% = 100\%\end{align*}
All of the probabilities do, in fact, add up to 100%.
### Practice
1. A bag contains 3 red balls and 4 blue balls. Thomas reaches in the bag and picks a ball out at random from the bag. He places it back into the bag. Thomas then reaches in the bag and picks another ball at random.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Thomas picks:
1. 2 red balls
2. a red ball in his second draw
2. A teacher has a prize box on her front desk for when students do exceptional work in math class. Inside the box there are 20 math pencils and 10 very cool erasers. Janet completed a challenge problem for Ms. Cameron, and Ms. Cameron rewarded Janet’s innovative problem-solving approach with a trip to the prize box. Janet reaches into the box and picks out a prize and then drops it back in. Then she reaches in again and picks out a prize a second time.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Janet reaches into the box and picks out an eraser on the second pick?
3. Students in BDF High School were asked about their preference regarding the new school colors. They were given a choice between green and blue as the primary color and red and yellow as the secondary color. The results of the survey are shown in the tree diagram below. You can see that 75% of the students choose green as the primary color. Of this 75%, 45% chose yellow as the secondary color. What is the probability that a student in BDF High School selected red as the secondary color if he or she chose blue as the primary color?
4. In question 3, what color combination do the students in BDF High School want the most?
5. According to the following tree diagram, what is the probability of getting 1 head and 1 tail when flipping a coin 2 times? Note that 1 head and 1 tail can mean a head and then a tail or a tail and then a head.
6. If a coin is flipped 4 times, how many branches will the tree diagram have that represents this situation?
7. If a die is rolled 2 times, how many branches will the tree diagram have that represents this situation?
8. Suppose a tree diagram has 4 branches, and the probabilities of the outcomes representing 3 of the branches are 0.12, 0.53, and 0.28, respectively. What is the probability of the outcome representing the remaining branch?
9. The following tree diagram represents the responses to 2 yes/no questions:
10. If the probability of answering yes to both questions is 0.1947, what is the probability of answering yes to the second question if the answer to the first question was yes?
11. For question 9, if the probability of answering yes to both questions is 0.1947, what is the probability of answering no to the second question if the answer to the first question was yes?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
tree diagrams
Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree.
If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)
Multiplication Rule
States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
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# How do you graph f(x)=(x-3)/(-2x-8) using holes, vertical and horizontal asymptotes, x and y intercepts?
Jan 12, 2018
see explanation
#### Explanation:
$\textcolor{b r o w n}{\text{y-intercept}}$ at $x = 0$ giving: $y = \frac{- 3}{- 8} = + \frac{3}{8}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{x-intercept}}$ at $y = 0$ giving:
$0 = \frac{x - 3}{- 2 x - 8}$
Multiply both sides by $- 2 x - 8$ giving
$0 = x - 3 \to x = 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Hole}}$
Not 'allowed' to divide by 0 (undefined)
So $- 2 x - 8 \ne 0$
$2 x \ne - 8$
$x \ne - \frac{8}{2}$
$x \ne - 4 \leftarrow \text{ Hole at } x = 4$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Horizontal asymptotes}}$
These occur as $x$ tends to infinity.
Set $y = \frac{x - 3}{- 2 x - 8}$
As $x$ becomes 'larger' in both positive and negative directions the -3 and -8 become insignificant. Consequently the equation is 'tending' towards:
$y = - \frac{x}{2 x} \to - \frac{1}{2}$
${\lim}_{x \to \pm \infty} \frac{x - 3}{- 2 x - 8} = - \frac{1}{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Vertical asymptotes - positive}}$
These will occur as the denominator tend to 0. That is, tends towards any holes. This has to be looked at as you approach it from both the left and right. So lets investigate.
Set $x = - 4 + \delta$ where $0 < \delta < 1$
y=(-4+ delta -3)/((-2(-4+delta)-8)
$y = \frac{- 7 + \delta}{- 2 \delta}$
Negative $\div$ negative is positive
$7 \div \text{something minute} \to \infty$
So from the right
$y = {\lim}_{\textcolor{b l u e}{x \to {0}^{+}}} \frac{x - 3}{- 2 x - 8} = + \infty$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Vertical asymptotes - negative}}$
Using the same approach as above but approaxhing $x = - 4$ from the left we end up with:
$y = {\lim}_{\textcolor{b l u e}{x \to {0}^{-}}} \frac{x - 3}{- 2 x - 8} = - \infty$ |
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Quadratics are transformations of the function $$f(x)=x^2$$. Quadratics commonly arise from problems involving area and projectile motion, providing some interesting applications.
#### Example 1
A backyard farmer wants to enclose a rectangular space for a new garden. She has purchased 80 feet of wire fencing to enclose three sides, and will put the fourth side against the backyard fence. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length $$L$$.
In a scenario like this involving geometry, it is often helpful to draw a picture. It might also be helpful to introduce a temporary variable, $$W$$, to represent the side of fencing parallel to the fourth side or backyard fence.
Since we know we only have 80 feet of fence available, we know that $$L+W+L=80$$, or more simply, $$2L+W=80$$. This allows us to represent the width, $$W$$, in terms of $$L$$: $$W=80-2L$$
Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so $$A=LW=L(80-2l)$$, so $A(L)=80L-2L^2.$ This formula represents the area of the fence in terms of the variable length $$L$$.
The standard form of a quadratic function is $$f(x)=ax^2+bx+c$$.
The transformation form of a quadratic function is $$f(x)=a(x-h)^2+k$$.
The vertex of the quadratic function is located at $$(h, k)$$, where $$h$$ and $$k$$ are the numbers in the transformation form of the function. Because the vertex appears in the transformation form, it is often called the vertex form.
#### Example 2
Write an equation for the quadratic graphed below as a transformation of $$f(x)=x^2$$.
We can see the graph is the basic quadratic shifted to the left 2 and down 3, putting the vertex at $$(-2, -3)$$, giving a formula in the form $$g(x)=a(x+2)^2-3$$. By plugging in a point that falls on the grid, such as $$(0,-1)$$, we can solve for the stretch factor: \begin{align*} -1=&a(0+2)^2-3 \\ 2=& 4a \\ a=& \frac{1}{2} \end{align*}
The equation for this formula is $g(x)=\frac{1}{2}(x+2)^2-3$
### Short run Behavior: Intercepts
As with any function, we can find the vertical intercepts of a quadratic by evaluating the function at an input of zero, and we can find the horizontal intercepts by solving for when the output will be zero. Notice that depending upon the location of the graph, we might have zero, one, or two horizontal intercepts.
Notice that in the standard form of a quadratic, the constant term c reveals the vertical intercept of the graph, since $$f(0)=a(0)^2+b(0)+c=c$$.
#### Example 3
Find the vertical and horizontal intercepts of the quadratic $$f(x)=3x^2+5x-2$$.
We can find the vertical intercept by evaluating the function at an input of zero: $f(0)=3(0)^2+5(0)-2=-2$ So the vertical intercept is at (0,-2)
For the horizontal intercepts, we solve for when the output will be zero: $0=3x^2+5x-2.$ In this case, the quadratic can be factored easily, providing the simplest method for solution.: $0=(3x-1)(x+2),$ so either \begin{align*} 0=& 3x-1\\ x=& \frac{1}{3} \end{align*} or \begin{align*} 0=& x+2\\ x=& -2 \end{align*} So the Horizontal intercepts are at $$\left(\frac{1}{3},0\right)$$ and $$(-2,0)$$.
When a quadratic is not factorable or is hard to factor, we can turn to the quadratic formula.
For a quadratic function given in standard form $$f(x)=ax^2+bx+c$$, the quadratic formula gives the horizontal intercepts of the graph of this function:$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
#### Example 4
A ball is thrown upwards from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation $H(t)=-16t^2+80t+40 .$ When does the ball hit the ground?
To find when the ball hits the ground, we need to determine when the height is zero, i.e., when $$H(t) = 0$$. While we could do this using the transformation form of the quadratic, we can also use the quadratic formula: $t=\frac{-80\pm \sqrt{80^2-4(-16)(40)}}{2(-16)}=\frac{-80\pm\sqrt{8960}}{-32}$
Since the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions:$t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458 \quad\text{or}\quad t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458$
The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. |
## Why does the commutative property of multiplication work?
The commutative property of multiplication states that when two numbers are being multiplied, their order can be changed without affecting the product.
For example, has the same product as .
These properties apply to all real numbers.
Let’s take a look at a few addition examples..
## What is commutative property of whole numbers?
Commutative Property When we add two or more whole numbers, their sum is the same regardless of the order of the addends. The sum of both 2 + 4 and 4 + 2 is 6. That means, we can add whole numbers in any order.
## What is associative and commutative property of addition?
In math, the associative and commutative properties are laws applied to addition and multiplication that always exist. The associative property states that you can re-group numbers and you will get the same answer and the commutative property states that you can move numbers around and still arrive at the same answer.
## What is an example of distributive property?
The distributive property of multiplication over addition can be used when you multiply a number by a sum. For example, suppose you want to multiply 3 by the sum of 10 + 2. … According to this property, you can add the numbers and then multiply by 3. 3(10 + 2) = 3(12) = 36.
## What do you mean by commutative property?
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
## What is the difference between closure property and commutative property?
In summary, the Closure Property simply states that if we add or multiply any two real numbers together, we will get only one unique answer and that answer will also be a real number. The Commutative Property states that for addition or multiplication of real numbers, the order of the numbers does not matter.
## What are the three properties?
There are four basic properties of numbers: commutative, associative, distributive, and identity. You should be familiar with each of these. It is especially important to understand these properties once you reach advanced math such as algebra and calculus.
## What are the 3 properties of multiplication?
There are three properties of multiplication: commutative, associative, and distributive.Commutative Property.Associative Property.Distributive Property.
## What is commutative and distributive property?
When you rewrite an expression by a commutative property, you change the order of the numbers being added or multiplied. … The distributive property can be used to rewrite expressions for a variety of purposes. When you are multiplying a number by a sum, you can add and then multiply.
## How do you explain the commutative property of addition?
The word “commutative” comes from “commute” or “move around”, so the Commutative Property is the one that refers to moving stuff around. For addition, the rule is “a + b = b + a”; in numbers, this means 2 + 3 = 3 + 2. For multiplication, the rule is “ab = ba”; in numbers, this means 2×3 = 3×2.
Commutative Property The commutative property of addition says that when the order of the addends is changed, the answer still stays the same. Tip: The word commutative is kind of like the word commute, which means to move around.
## What’s the associative property of addition?
Basically the Associative Law of Addition means that the order that the numbers are added does not matter.
## What is the distributive property of addition?
According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.
## What is the commutative law of addition?
Commutative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: a + b = b + a and ab = ba. … From these laws it follows that any finite sum or product is unaltered by reordering its terms or factors.
## What do you mean by commutative?
Mathematics. (of a binary operation) having the property that one term operating on a second is equal to the second operating on the first, as a × b = b × a. having reference to this property: commutative law for multiplication.
## What is the difference between associative and commutative property?
For that reason, it is important to understand the difference between the two. The commutative property concerns the order of certain mathematical operations. … The associative property, on the other hand, concerns the grouping of elements in an operation. This can be shown by the equation (a + b) + c = a + (b + c).
## What is an example of commutative property of multiplication?
Commutative property of multiplication: Changing the order of factors does not change the product. For example, 4 × 3 = 3 × 4 4 \times 3 = 3 \times 4 4×3=3×44, times, 3, equals, 3, times, 4.
## Why is the commutative property of addition important?
The Commutative Property is a great strategy to use when adding multi-digit numbers. When I taught first grade, I taught counting up as an addition and subtraction strategy. But if students know that they can switch the order of the addends and start adding with the greater number FIRST, it makes counting up easier.
## What is an example of commutative property?
Commutative property of addition: Changing the order of addends does not change the sum. For example, 4 + 2 = 2 + 4 4 + 2 = 2 + 4 4+2=2+44, plus, 2, equals, 2, plus, 4. Associative property of addition: Changing the grouping of addends does not change the sum. |
# How to Divide Fractions (like ½ ÷ ¼) with Free Online Tutoring in Math
###### Unlock the mystery of dividing fractions with our latest blog post, the final installment in our comprehensive eleven-part series!
Designed to follow a cyclical learning approach, this article provides a complete lesson on fractions, guiding your child step-by-step through the complexities of dividing fractions like ½ ÷ ¼.
Dividing fractions can be perplexing. Why does ½ ÷ ¼ result in an answer larger than both ½ and ¼? Shouldn’t division make numbers smaller? The key to understanding this lies in illustrating mathematical models. When children visualize the division process, they can clearly see why ½ ÷ ¼ yields a larger answer. Without this visual aid, many children struggle to grasp the concept, leading to confusion and frustration in higher levels of math.
In our blog post, your child will:
• Learn to Illustrate Division: Visual models help demystify why dividing fractions often results in larger numbers.
• Build a Strong Foundation: Step-by-step lessons ensure your child thoroughly understands each concept.
• Achieve Long-term Success: A solid grasp of fractions is crucial for future mathematical success.
Visit our website (https://www.teachersdungeon.com/) for a comprehensive educational program designed to help kids become proficient in mathematics. By mastering these concepts, your child will gain a deeper, more concrete understanding of dividing fractions, paving the way for a successful educational journey. Don’t miss out on this valuable resource—empower your child’s learning today!
Articles within this series on Fractions:
Solving problems that deal with fractions is simple when you develop a concrete understanding. I have had incredible results with with the students in my class! The strategies taught within this article work with children who have ADHD, Dyslexia, and other learning disabilities. Virtually every one of my students who has learned the strategies within this HOW TO DO FRACTIONS article has passed the standards based assessment for adding, subtracting, multiplying and dividing fractions.
###### I have scaffold the problems in each lesson.
The first problem in this article is a “Watch Me” problem. The second is a “Work with Me” problem. All the rest are “On Your Own” problems.
*If your child needs a bit more support, they should complete the “On Your Own” problems as a “Work with Me” problem. I have a number of students with gaps in their learning and others with a variety of learning disabilities. I have had incredible success, by having those students complete 5 to 7 problems within each lesson as a “Work with Me” problem. They play a bit of the video, then pause it and copy, then watch a bit more, pause it and copy. My students Play – Pause – and Copy until the entire problem is solved. This is like having a personal tutor working through each and every problem with your child. Every one of my students who has used this strategy has passed the Common Core Proficiency Exam.
### How to Divide Fractions Online Tutoring in Math: Challenge 1 Watch Me Terrance the Ticklish Mule
Terrance loves his alfalfa, but his owner keeps tickling his nose. His owner loves Terrance, but he can’t help tickling his favorite mule. Each day, Terrance eats 2/3 pounds of alfalfa. His owner divides the alfalfa into portions that are 2/7 pounds, how many portions will Terrance get to eat?
Hint: How many 2/7’s are in 2/3?
Watch this Free Tutoring for Math Video!
### How to Divide Fractions Online Tutoring in Math: Challenge 2 Work With Me Singing Camels
Kami & Connie Camel love to sing. They walk around their paddock and sing to all the on lookers. Kami & Connie’s paddock is 8/9 miles long.
If Kami& Connie break the paddock into portions that are 3/4 miles, how many portions will our “Singing Camels” walk and sing?
Hint: How many 3/4’s are in 8/9?
Watch this Free Tutoring for Math Video!
## Do your children get frustrated when they make a mistake?
We all make mistakes. As a matter of fact, making mistakes is an essential part of the learning process. This is why at the end of each of the following “On Your Own” challenges I encourage children to fix their mistakes. Finding and fixing your own mistake is the fastest way to learn.
### How to Divide Fractions Online Tutoring in Math: Challenge 3 On Your Own Honovi the Honey Loving Grizzly Bear
Honovi is always on the lookout for bees. She follows them back to the hive and collects as much honey as possible. Yesterday, Hovoni found a hive. While the bees crawled under her fur and stung again and again, Honovi stole 4/5 pounds of honey. If Hovoni divides the honey into portions that are each 1/8 pound, how many portions can she make?
Hint: How many 1/8’s are in 4/5?
Watch this Free Tutoring for Math Video!
### How to Divide Fractions Online Tutoring in Math: Challenge 4 On Your Own Shelby & Sherman Seahorse
Shelby & Sherwood are a couple of hungry seahorses. They love nothing more than a big old mouthful of plankton. The marine biologist that feeds them has a 2/5 bucket of plankton.
A serving of the plankton is 3/8 of the bucket.
How many servings are in the bucket?
Watch this Free Tutoring for Math Video!
### How to Divide Fractions Online Tutoring in Math: Challenge 5 On Your Own Elyssa the Thirsty Elephant
Elyssa Elephant is thirsty. There is very little water on the African Plains at this time of the year. Luckily for Elyssa and her herd, a local farmer gives a bucket of cool water to each elephant in Elyssa’s family. Each bucket holds 5/6 gallons of water.
If Elyssa’s trunk holds 2/7 of a gallon, how many trunkfuls of water will Elyssa get to drink?
Watch this Free Tutoring for Math Video!
### Want More Tutorials?
Discover the transformative power of learning with TeachersDungeon today. Dive into a world where education meets adventure, empowering students from grades 3 to 6 with personalized math instruction that adapts to their needs. Whether you’re an educator looking to enrich classroom learning or a parent seeking to support your child’s academic journey, The Teacher’s Dungeon offers interactive gameplay, instant help with video tutorials, and comprehensive progress tracking through its Stats Page. Visit The Teacher’s Dungeon’s website now to explore how our innovative approach can elevate your child’s math education. Embark on this exciting educational journey with us and watch your students thrive!
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# Areas related to Circles Class 10 Notes Maths Chapter 12
On this page, you will find Areas related to Circles Class 10 Notes Maths Chapter 12 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 12 Areas related to Circles will seemingly help them to revise the important concepts in less time.
## CBSE Class 10 Maths Chapter 12 Notes Areas related to Circles
### Areas related to Circles Class 10 Notes Understanding the Lesson
1. Circle: Circle is set of all points in a plane which are at the fixed distance from a fixed point i.e., centre. Centre: Mid-point of a circle is called centre of a circle.
2. Radius: The distance between the centre of a circle to the circumference of the circle.
It is denoted by r or R.
3. Chord: A line segment which joins the two points in the circumference of the circle.
4. Diameter: It is the longest chord which passes through the centre of a circle. It is denoted by d or D.
$$\text { Radius }=\frac{\text { Diameter }}{2}$$
5. Circumference of a Circle (or Perimeter)
A perimeter is a path that surrounds a two dimensional shape.
The perimeter of a circle is called its circumference. Circumference
$$\frac{\text { Circumference }}{\text { Diameter }}=\pi$$
Circumference = n x diameter = π x 2r = 2πr
or
The value of π is $$\frac{22}{7}$$ or 3.14 (approximately)
6. Arc: An arc is the part of the circumference of a circle.
An arc AB is denoted as $$\widehat{\mathrm{AB}}$$
Length of arc AB is l $$\widehat{\mathrm{AB}}$$ or l.
Length of an arc of a sector of angle
$$\theta=\frac{2 \pi r \theta}{360^{\circ}}$$
7. Sector: The portion (or part) of circular region enclosed by two radii and the corresponding arc is called a sector of the circle.
8. Minor and Major sector of the circle: Shaded region OAPB is called sector or minor sector of a circle with centre O.
∠AOB is called angle of the sector.
And OAQB is called major sector.
Angle of major sector = 360° – ∠AOB.
9. Area of Circle
Let r be the radius of circle. If we cut the circle in sectors and arrange then we see this figure like a rectangle whose length is $$\frac{1}{2}$$ 2πr = πr and breadth is r.
Hence,
Area of circle = Area of rectangle
= l x b = $$\frac{1}{2}$$ 2πr = r
Area of circle = πr2
10. Segment: The portion or part of a circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
The shaded region APB is the minor segment.
And the region AQB is the major segment.
Area of circle = πr2
Area of the sector of an angle $$\theta=\frac{\pi r^{2} \theta}{360^{\circ}}$$
Area of major sector = Area of circle – Area of minor sector
Area of minor segment = Area of sector – Area of ΔOAB$$\frac{\pi r^{2} \theta}{360^{\circ}}$$– area of ΔOAB
Area of major segment = Area of circle – Area of minor segment
11. Area of Combinations of Plane Figures
In our daily life we have observed various plane figures which are combinations of two or more figures and i also in the form of various interesting designs like flower beds, curtains, drain covers, window designs, ] designs on table covers. To calculate areas of such figures, we field the area of shapes used and then and/subtract as per need. |
Interior Angles of Polygons
Another example:
Triangles
The Interior Angles of a Triangle add up to 180°
Let's try a triangle:
90° + 60° + 30° = 180°
It works for this triangle
Now tilt a line by 10°:
80° + 70° + 30° = 180°
It still works!
One angle went up by 10°,
and the other went down by 10°
(A Quadrilateral has 4 straight sides)
Let's try a square:
90° + 90° + 90° + 90° = 360°
A Square adds up to 360°
Now tilt a line by 10°:
80° + 100° + 90° + 90° = 360°
It still adds up to 360°
Because there are 2 triangles in a square ...
The interior angles in a triangle add up to 180° ...
... and for the square they add up to 360° ...
... because the square can be made from two triangles!
Pentagon
A pentagon has 5 sides, and can be made from three triangles, so you know what ...
... its interior angles add up to 3 × 180° = 540°
And when it is regular (all angles the same), then each angle is 540° / 5 = 108°
(Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°)
The Interior Angles of a Pentagon add up to 540°
The General Rule
Shape Sides Sum of Shape Each Angle Interior Angles If it is a Regular Polygon (all sides are equal, all angles are equal) Triangle 3 180° 60° Quadrilateral 4 360° 90° Pentagon 5 540° 108° Hexagon 6 720° 120° Heptagon (or Septagon) 7 900° 128.57...° Octagon 8 1080° 135° Nonagon 9 1260° 140° ... ... .. ... ... Any Polygon n (n-2) × 180° (n-2) × 180° / n
So the general rule is:
Sum of Interior Angles = (n-2) × 180°
Each Angle (of a Regular Polygon) = (n-2) × 180° / n
Perhaps an example will help:
Example: What about a Regular Decagon (10 sides) ?
Sum of Interior Angles = (n-2) × 180°
= (10-2)×180° = 8×180° = 1440°
And it is a Regular Decagon so:
Each interior angle = 1440°/10 = 144°
Note: Interior Angles are sometimes called "Internal Angles" |
# Calculate the product using complex numbers. p=cos20*cos40*cos80
giorgiana1976 | Student
We'll write the polar form of z:
z = cos 20 + i*sin 20 (1)
We'll write the conjugate of z:
z^-1 = cos 20 - i*sin 20
1/z = cos 20 - i*sin 20 (2)
z + 1/z = cos 20 + i*sin 20 + cos 20 - i*sin 20
We'll combine and eliminate like terms:
z + 1/z = 2*cos 20
We'll multiply z by z:
(z^2 + 1)/z = 2*cos 20
We'll use symmetric property:
2*cos 20 = (z^2 + 1)/z
We'll divide by 2:
cos 20 = (z^2 + 1)/2z (3)
cos 40 = cos 2*(20) = 2 (cos 20)^2 - 1
We'll substitute cos 20 by (3):
cos 40 = 2 (z^2 + 1)^2/4z^2 - 1
cos 40 = (z^2 + 1)^2/2z^2 - 1
We'll raise to square the binomial z^2 + 1:
cos 40 = (z^4 + 2z^2 + 1)/2z^2 - 1
We'll multiply -1 by 2z^2:
cos 40 = (z^4 + 2z^2 + 1 - 2z^2)/2z^2
We'll eliminate like terms:
cos 40 = (z^4 + 1)/2z^2 (4)
cos 80 = cos 2*(40) = 2 (cos 40)^2 - 1 (5)
We'll substitute (4) in (5):
cos 80 = 2 (z^4 + 1)^2/4z^4 - 1
cos 80 = (z^4 + 1)^2/2z^4 - 1
We'll raise to square the binomial z^4 + 1:
cos 80 = (z^8 + 2z^4 + 1)/2z^4 - 1
We'll multiply -1 by 2z^4:
cos 80 = (z^8 + 2z^4 + 1 - 2z^4)/2z^4
We'll eliminate like terms:
cos 80 = (z^8+ 1)/2z^4
Now, we'll calculate the product:
P = [(z^2 + 1)/2z]*[(z^4 + 1)/2z^2]*[(z^8+ 1)/2z^4]
P = (z^2 + 1)*(z^4 + 1)*(z^8+ 1)/8z^7
If z^2 + z + 1 = 0 and z^3 = 1
P = (-z)(-z^2)(-z)/8z
P = -z/8z
P = -1/8 |
# Common Core: 3rd Grade Math : Solve Real World and Mathematical Problems Involving Area: CCSS.Math.Content.3.MD.C.7b
## Example Questions
### Example Question #321 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #322 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #321 : Quadrilaterals
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #331 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #31 : Solve Real World And Mathematical Problems Involving Area: Ccss.Math.Content.3.Md.C.7b
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #333 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #334 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #335 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #336 : How To Find The Area Of A Rectangle
What is the area of the rectangle?
Explanation:
The formula to find area is . We are given the length and the width from the problem, so we can plug those values into our equation and solve.
*Area is the number of square units inside a shape, which is why area is always written with square units.
### Example Question #337 : How To Find The Area Of A Rectangle
What is the area of the rectangle? |
# 2.3: Right Triangle Trigonometry
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Learning Objectives
• Use right triangles to evaluate trigonometric functions.
• Find function values for 30°($$\dfrac{\pi}{6}$$),45°($$\dfrac{\pi}{4}$$),and 60°($$\dfrac{\pi}{3}$$).
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.
Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.
We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle:
\begin{align*} \cos t &= x \\ \sin t &=y \end{align*}
In this section, we will see another way to define trigonometric functions using properties of right triangles.
## Using Right Triangles to Evaluate Trigonometric Functions
In earlier sections, we used a unit circle to define the trigonometric functions. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of $$t$$ is its value at $$t$$ radians. First, we need to create our right triangle. Figure $$\PageIndex{1}$$ shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point $$(x,y)$$ to the x-axis, we have a right triangle whose vertical side has length $$y$$ and whose horizontal side has length $$x$$. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.
We know
$\cos t= \frac{x}{1}=x$
Likewise, we know
$\sin t= \frac{y}{1}=y$
These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using $$(x,y)$$ coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of $$x$$,we will call the side between the given angle and the right angle the adjacent side to angle $$t$$. (Adjacent means “next to.”) Instead of $$y$$,we will call the side most distant from the given angle the opposite side from angle $$t$$. And instead of $$1$$,we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in Figure $$\PageIndex{2}$$.
## Understanding Right Triangle Relationships
Given a right triangle with an acute angle of $$t$$,
\begin{align} \sin (t) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (t) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (t) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef}\end{align}
A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”
how to: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle
1. Find the sine as the ratio of the opposite side to the hypotenuse.
2. Find the cosine as the ratio of the adjacent side to the hypotenuse.
3. Find the tangent is the ratio of the opposite side to the adjacent side.
Example $$\PageIndex{1}$$: Evaluating a Trigonometric Function of a Right Triangle
Given the triangle shown in Figure $$\PageIndex{3}$$, find the value of $$\cos α$$.
Solution
The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so via Equation \ref{cosdef}:
\begin{align*} \cos (α) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[4pt] &= \dfrac{15}{17} \end{align*}
Exercise $$\PageIndex{1}$$
Given the triangle shown in Figure $$\PageIndex{4}$$, find the value of $$\sin t$$.
$$\frac{7}{25}$$
## Relating Angles and Their Functions
When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure $$\PageIndex{5}$$. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.
We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.
how to: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles
1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. Find the required function:
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side
Example $$\PageIndex{2}$$: Evaluating Trigonometric Functions of Angles Not in Standard Position
Using the triangle shown in Figure $$\PageIndex{6}$$, evaluate $$\sin α, \cos α, \tan α, \sec α, \csc α,$$ and $$\cot α$$.
Solution
\begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}
Exercise $$\PageIndex{2}$$
Using the triangle shown in Figure $$\PageIndex{7}$$, evaluate $$\sin t, \cos t,\tan t, \sec t, \csc t,$$ and $$\cot t$$.
\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}
## Finding Trigonometric Functions of Special Angles Using Side Lengths
We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of $$30°, 60°,$$ and $$45°$$, however, remember that when dealing with right triangles, we are limited to angles between $$0° \text{ and } 90°$$.
Suppose we have a $$30°,60°,90°$$ triangle, which can also be described as a $$\frac{π}{6}, \frac{π}{3},\frac{π}{2}$$ triangle. The sides have lengths in the relation $$s,\sqrt{3}s,2s.$$ The sides of a $$45°,45°,90°$$triangle, which can also be described as a $$\frac{π}{4},\frac{π}{4},\frac{π}{2}$$ triangle, have lengths in the relation $$s,s,\sqrt{2}s.$$ These relations are shown in Figure $$\PageIndex{8}$$.
We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.
Given trigonometric functions of a special angle, evaluate using side lengths.
1. Use the side lengths shown in Figure $$\PageIndex{8}$$ for the special angle you wish to evaluate.
2. Use the ratio of side lengths appropriate to the function you wish to evaluate.
Example $$\PageIndex{3}$$: Evaluating Trigonometric Functions of Special Angles Using Side Lengths
Find the exact value of the trigonometric functions of $$\frac{π}{3}$$, using side lengths.
Solution
\begin{align*} \sin (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{hyp}}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2} \\ \cos (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{hyp}}=\dfrac{s}{2s}=\dfrac{1}{2} \\ \tan (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{adj}} =\dfrac{\sqrt{3}s}{s}=\sqrt{3} \\ \sec (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{adj}} = \dfrac{2s}{s}=2 \\ \csc (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{opp}} =\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\ \cot (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{opp}}=\dfrac{s}{\sqrt{3}s}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3} \end{align*}
Exercise $$\PageIndex{3}$$
Find the exact value of the trigonometric functions of $$\frac{π}{4}$$ using side lengths.
$$\sin (\frac{π}{4})=\frac{\sqrt{2}}{2}, \cos (\frac{π}{4})=\frac{\sqrt{2}}{2}, \tan (\frac{π}{4})=1,$$
$$\sec (\frac{π}{4})=\sqrt{2}, \csc (\frac{π}{4})=\sqrt{2}, \cot (\frac{π}{4}) =1$$
## Using Equal Cofunction of Complements
If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of $$\frac{π}{6}$$ and $$\frac{π}{3}$$, we see that the sine of $$\frac{π}{3}$$, namely $$\frac{\sqrt{3}}{2}$$, is also the cosine of $$\frac{π}{6}$$, while the sine of $$\frac{π}{6}$$, namely $$\frac{1}{2},$$ is also the cosine of $$\frac{π}{3}$$ (Figure $$\PageIndex{9}$$).
\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}
This result should not be surprising because, as we see from Figure $$\PageIndex{9}$$, the side opposite the angle of $$\frac{π}{3}$$ is also the side adjacent to $$\frac{π}{6}$$, so $$\sin (\frac{π}{3})$$ and $$\cos (\frac{π}{6})$$ are exactly the same ratio of the same two sides, $$\sqrt{3} s$$ and $$2s.$$ Similarly, $$\cos (\frac{π}{3})$$ and $$\sin (\frac{π}{6})$$ are also the same ratio using the same two sides, $$s$$ and $$2s$$.
The interrelationship between the sines and cosines of $$\frac{π}{6}$$ and $$\frac{π}{3}$$ also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, and the right angle is $$\frac{π}{2}$$, the remaining two angles must also add up to $$\frac{π}{2}$$. That means that a right triangle can be formed with any two angles that add to $$\frac{π}{2}$$—in other words, any two complementary angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure $$\PageIndex{10}$$.
Using this identity, we can state without calculating, for instance, that the sine of $$\frac{π}{12}$$ equals the cosine of $$\frac{5π}{12}$$, and that the sine of $$\frac{5π}{12}$$ equals the cosine of $$\frac{π}{12}$$. We can also state that if, for a certain angle $$t, \cos t= \frac{5}{13},$$ then $$\sin (\frac{π}{2}−t)=\frac{5}{13}$$ as well.
COFUNCTION IDENTITIES
The cofunction identities in radians are listed in Table $$\PageIndex{1}$$.
$$\cos t= \sin (\frac{π}{2}−t)$$ $$\sin t= \cos (\dfrac{π}{2}−t)$$ $$\tan t= \cot (\dfrac{π}{2}−t)$$ $$\cot t= \tan (\dfrac{π}{2}−t)$$ $$\sec t= \csc (\dfrac{π}{2}−t)$$ $$\csc t= \sec (\dfrac{π}{2}−t)$$
how to: Given the sine and cosine of an angle, find the sine or cosine of its complement.
1. To find the sine of the complementary angle, find the cosine of the original angle.
2. To find the cosine of the complementary angle, find the sine of the original angle.
Example $$\PageIndex{4}$$: Using Cofunction Identities
If $$\sin t = \frac{5}{12},$$ find $$( \cos \frac{π}{2}−t)$$.
Solution
According to the cofunction identities for sine and cosine,
$\sin t= \cos (\dfrac{π}{2}−t). \nonumber$
So
$\cos (\dfrac{π}{2}−t)= \dfrac{5}{12}. \nonumber$
Exercise $$\PageIndex{4}$$
If $$\csc (\frac{π}{6})=2,$$ find $$\sec (\frac{π}{3}).$$
Solution
2
## Using Trigonometric Functions
In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.
how to: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides
1. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
2. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
3. Using the value of the trigonometric function and the known side length, solve for the missing side length.
Example $$\PageIndex{5}$$: Finding Missing Side Lengths Using Trigonometric Ratios
Find the unknown sides of the triangle in Figure $$\PageIndex{11}$$.
Solution
We know the angle and the opposite side, so we can use the tangent to find the adjacent side.
$\tan (30°)= \dfrac{7}{a} \nonumber$
We rearrange to solve for $$a$$.
\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber
We can use the sine to find the hypotenuse.
$\sin (30°)= \dfrac{7}{c} \nonumber$
Again, we rearrange to solve for $$c$$.
\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}
Exercise $$\PageIndex{5}$$:
A right triangle has one angle of $$\frac{π}{3}$$ and a hypotenuse of 20. Find the unknown sides and angle of the triangle.
$$\mathrm{adjacent=10; opposite=10 \sqrt{3}; }$$ missing angle is $$\frac{π}{6}$$
## Using Right Triangle Trigonometry to Solve Applied Problems
Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure $$\PageIndex{12}$$.
how to: Given a tall object, measure its height indirectly
1. Make a sketch of the problem situation to keep track of known and unknown information.
2. Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
3. At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
4. Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
5. Solve the equation for the unknown height.
Example $$\PageIndex{6}$$: Measuring a Distance Indirectly
To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure $$\PageIndex{13}$$. Find the height of the tree.
Solution
We know that the angle of elevation is $$57°$$ and the adjacent side is 30 ft long. The opposite side is the unknown height.
The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of $$57°$$, letting $$h$$ be the unknown height.
$\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array}$
The tree is approximately 46 feet tall.
Exercise $$\PageIndex{6}$$:
How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of $$\frac{5π}{12}$$ with the ground? Round to the nearest foot.
media:
Access these online resources for additional instruction and practice with right triangle trigonometry.
Visit this website for additional practice questions from Learningpod.
## Key Equations
Cofunction Identities
\begin{align*} \cos t &= \sin ( \frac{π}{2}−t) \\ \sin t &= \cos (\frac{π}{2}−t) \\ \tan t &= \cot (\frac{π}{2}−t) \\ \cot t &= \tan (\frac{π}{2}−t) \\ \sec t &= \csc (\frac{π}{2}−t) \\ \csc t &= \sec (\frac{π}{2}−t) \end{align*}
## Key Concepts
• We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example.
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example.
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example.
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example.
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side. See Example.
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example. |
Top
# Ratio & Proportion
1.
In a mixture $$60 litres,$$ the ratio of milk and water$$2 : 1.$$ If this ratio is to be $$1 : 2,$$ then the quanity of water to be further added is:
Quantity of milk $$= (60 \times \frac{2}{3}) littres = 40 littres.$$
Quantity of water in it $$= (60 - 40) littres = 20 littres.$$
New ratio $$= 1:2$$
Let quantity of water to be added further be $$x$$ $$littres.$$
Then, milk : water $$= (\frac{40}{20 + x})$$
Now, $$(\frac{40}{20 + x}) = \frac{1}{2}$$
$$\Rightarrow 20 + x = 80$$
$$\Rightarrow x = 60.$$
$$\therefore$$ Quantity of water to be added $$= 60$$ $$littres.$$
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2.
In a bag, there are coins of $$25 p, 10 p$$ and $$5 p$$ in the ratio of $$1 : 2 : 3.$$ If there is $$Rs. 30$$ in all, how many $$5 p$$ coins are there?
Let the number of $$25 p, 10 p$$ and $$5 p$$ coins be $$x, 2x, 3x$$ respectively.
Then, sum of there value $$= Rs. (\frac{25x}{100} + \frac{10 \times 2x}{100} + \frac{5 \times 3x}{100}) = Rs. \frac{60x}{100}$$
$$\therefore \frac{60x}{100} = 30 \Leftrightarrow x = \frac{300 \times 100}{60} = 50.$$
Heance, the number of $$5 p$$ coins $$= (3 \times 50) = 150.$$
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3.
If $$40$$% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
Let $$40$$% of $$A = \frac{2}{3} B$$
Then, $$\frac{40A}{100} = \frac{2B}{3}$$
$$\Rightarrow \frac{2A}{5} = \frac{2B}{3}$$
$$\Rightarrow \frac{A}{B} = (\frac{2}{3} \times \frac{5}{2}) = \frac{5}{3}$$
$$\therefore A : B = 5 : 3.$$
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4.
If $$0.75 : x :: 5 : 8,$$ then $$x$$ is equal to:
$$(x \times 5) = (0.75 \times 8 )$$
$$\Rightarrow x = (\frac{6}{5}) = 1.20$$
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5.
$$A$$ and $$B$$ together have $$RS. 1210.$$ If $$\frac{4}{15}$$ of $$A,s$$ amount is equal to $$\frac{2}{5}$$ of $$B's$$ amount, how much amount does $$B$$ have?
$$\frac{4}{12}A = \frac{2}{5}B$$
$$\Rightarrow A = (\frac{2}{5} \times \frac{15}{4})B$$
$$\Rightarrow A = \frac{3}{2} B$$
$$\Rightarrow \frac{A}{B} = \frac{3}{2}$$
$$\Rightarrow A : B = 3 : 2.$$
$$\therefore B's$$ share $$= RS. (1210 \times \frac{2}{5}) = Rs. 484$$
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6.
Two number are in the ratio $$3 : 5.$$ If $$9$$ is subtracted from each, the new numbers are in the ratio $$12 : 23.$$ The smaller number is:
Let the numbers be $$3x$$ and $$5x.$$
Then, $$\frac{3x - 9}{5x - 9} = \frac{12}{23}$$
$$\Rightarrow 23(3x - 9) = 12(5x - 9)$$
$$\Rightarrow 9x = 99$$
$$\Rightarrow x = 11.$$
$$\therefore$$ The smaller number $$= (3 \times 11) = 33.$$
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7.
The fourth proportional to $$5, 8, 15$$ is:
Let the fourth proportional to $$5, 8 , 15$$ be $$x.$$
Then, $$5 : 8 : 15 : x$$
$$\Rightarrow 5x = (8 \times 15)$$
$$x = \frac{(8 \times 15)}{5} = 24.$$
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8.
Two numbers are respectively $$20$$% and $$50$$% more than a $$3rd$$ number. The ratio of the two numbers is:
Let the $$3rd$$ number be $$x.$$
Then, first number $$= 120$$% of $$x = \frac{120x}{100} = \frac{6x}{5}$$
Second number $$= 150$$% of $$x = \frac{150x}{100} = \frac{3x}{2}$$
$$\therefore$$ Ratio of first two numbers $$= (\frac{6x}{5} : \frac{3x}{2}) = 12x : 15x = 4:5$$
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9.
The ratio of the number of boys and girls in a college is $$7 : 8.$$ If the percentage increase in the number of boys and girls be $$20$$% and $$10$$% respectively, what will be the new ratio?
Let the number of boyes and girls in the collage be $$7x$$ and $$8x$$ respectively. Their increased number is $$(120$$% of $$7x)$$ and $$(110$$% of $$8x)$$.
$$\Rightarrow (\frac{120}{100} \times 7x)$$ and $$(\frac{110}{100} \times 8x)$$
$$\Rightarrow \frac{42x}{5}$$ and $$\frac{44x}{5}$$
$$\therefore$$ The required ratio $$= (\frac{42x}{5} : \frac{44x}{5}) = 21 :22.$$
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10.
Seats for Mathematics, Physics and Biology in a school are in the ratio $$5 : 7 : 8.$$ There is a proposal to increase these seats by $$40$$%, $$50$$% and $$75$$% respectively. What will be the ratio of increased seats?
Let the number of seats for Mathematics, Physics and Biology be $$5x, 7x$$ and $$8x$$
Number of increased seats are $$(140$$% of $$5x),$$ $$(150$$% of $$7x)$$ and $$(175$$% of $$8x).$$
$$\Rightarrow (\frac{140}{100} \times 5x), (\frac{150}{100} \times 7x)$$ and $$(\frac{175}{100} \times 8x)$$
$$\Rightarrow 7x, \frac{21x}{2}$$ and $$14x.$$
$$\therefore$$ The required ratio $$= 7x : \frac{21x}{2} : 14x$$
$$\Rightarrow 14x : 21x : 28x$$
$$\Rightarrow 2 : 3 : 4.$$
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# Rate of Change Formula
• Last Updated : 15 May, 2022
The rate of change of a variable quantity is defined as the rate at which it changes in relation to another quantity. To put it another way, the rate of change divides the amount of change in one variable by the equivalent amount of change in another variable. It describes the concept of how one quantity changes in proportion to the other. It has formulas defined for coordinates and linear functions as well.
### Rate of Change Formula
The rate of change of coordinates (x1, y1) with respect to the other coordinates (x2, y2) is given by the ratio of the difference between the y-coordinates to that of the x-coordinates. In other words, the rate of change is equal to the slope of the line joining these points. It can also be denoted by the symbol m.
R = Δy/Δx = (y2 – y1) / (x2 – x1)
Where,
R is the rate of change of variable y with respect to variable x,
(x1, y1) and (x2, y2) are the coordinates.
For a linear function y = f(x), the rate of change is calculated for the interval (a, b) where b > a. Its formula is given by the ratio of the difference between the function values for each interval to the difference between interval values.
R = (f(b) – f (a)) / (b – a)
Here, R is the rate of change of variable y with respect to variable x for the interval (a, b).
### Sample problems
Problem 1: Calculate the rate of change for the coordinates (4, 11) and (2, 5).
Solution:
We have,
(x1, y1) = (4, 11)
(x2, y2) = (2, 5)
Using the formula, we have
R = (y2 – y1)/(x2 – x1)
= (5 – 11)/(2 – 4)
= -6/(-2)
= 3
Problem 2: The rate of change for two points (x, 3) and (1, 6) is 7. Find the value of x.
Solution:
(x1, y1) = (x, 3)
(x2, y2) = (1, 6)
R = 7
Using the formula, we have
R = (y2 – y1)/(x2 – x1)
=> 7 = (6 – 3)/(1 – x)
=> 7 = 3/(1 – x)
=> 7 – 7x = 3
=> 7x = 4
=> x = 4/7
Problem 3: The rate of change for two points (5, 4) and (3, y) is 4. Find the value of y.
Solution:
(x1, y1) = (5, 4)
(x2, y2) = (3, y)
R = 4
Using the formula, we have
R = (y2 – y1)/(x2 – x1)
=> 4 = (y – 4)/(3 – 5)
=> 4 = (y – 4)/(-2)
=> -8 = y – 4
=> y = -4
Problem 4: Calculate the rate of change for the function f(x) = x2 if the interval is (3, 5).
Solution:
We have, f(x) = x2
Calculate the value of f(3) and f(5).
f(3) = 32 = 9
f(5) = 52 = 25
Using the formula, we have
R = (f(b) – f(a))/(b – a)
= (f(5) – f(3))/ (5 – 3)
= (25 – 9)/2
= 16/2
= 8
Problem 5: Calculate the rate of change for the function f(x) = 4 – 3x3 if the interval is (1, 2).
Solution:
We have, f(x) = 4 – 3x3
Calculate the value of f(1) and f(2).
f(1) = 4 – 3(1)3 = 4 – 3 = 1
f(2) = 4 – 3(2)3 = 4 – 24 = -20
Using the formula,
R = (f(b) – f(a))/(b – a)
= (f(2) – f(1))/ (2 – 1)
= -20 – 1
= -21
Problem 6: The rate of change for two points (x, 7) and (9, 2) is 5. Find the value of x.
Solution:
(x1, y1) = (x, 7)
(x2, y2) = (9, 2)
R = 5
Using the formula,
R = (y2 – y1)/(x2 – x1)
=> 5 = (2 – 7)/(9 – x)
=> 5 = -5/(9 – x)
=> 45 – 5x = -5
=> 5x = 50
=> x = 10
Problem 7: The rate of change for two points (1, 5) and (8, y) is 9. Find the value of y.
Solution:
(x1, y1) = (1, 5)
(x2, y2) = (8, y)
R = 9
Using the formula,
R = (y2 – y1)/(x2 – x1)
=> 9 = (y – 5)/(8 – 1)
=> 9 = (y – 5)/7
=> y – 5 = 63
=> y = 68
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# Roman numeration system
The Roman numeration system evolved around 500 BCE. Just like other anscient numeration systems, it uses special symbols to represent numbers.
The basic Roman numerals are the followings. Study them and memorise them if you can. It could become handy one day
Any other roman numerals are found by combining these basic numerals
Examples:
1) 154 is equivalent to CLIIII in Roman numerals
2) 1492 is equivalent to MCCCCLXXXXII in Roman numerals
3) 3495 is equivalent to MMMCCCCLXXXXV in Roman numerals
Over time, two useful attributes were introduced that made the Roman numeration system very useful and efficient
The first one is the subractive principle
With the subtractive principle, Roman numerals can be combined or paired so that when reading from left to right, the values of the symbols in any pair increase.
The value of the new pair is:
bigger number in the pair − smaller number in the pair
For instance, I can pair I and V to make IV and the value of this pair will be V − I = 5 − 1 = 4
I can pair C and D to make CD and the value of this pair will be D − C = 500 − 100 = 400
I can pair X and L to make XL and the value of this pair will L − X = 50 − 10 = 40
This subtractive principle will make the writing of examples 1), 2), and 3) a lot simpler
1) CLIIII = CLIV
2) MCCCCLXXXII
Instead of CCCC, we can pair C and D to get CD and CD = 400 as demonstrated above
Also, instead of LXXXX, we can pair X and C to make XC since XC still equal to 90
Replacing CCCC (in bold) by CD, we get:
MCCCCLXXXII = MCDLXXXXII
Replacing LXXXX by XC (in blue), we get:
MCDLXXXXII= MCDXCII
So, instead of using 11 symbols, we can just use 7 to represent the same number
3) MMMCCCCLXXXXV = MMMCDXCV
The second one is the multiplicative principle
Basically, a horizontal bar above any number means 1000 times the number
Examples:
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# 2007 iTest Problems/Problem 58
The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.
## Problem
For natural numbers $k,n\geq 2$, we define $$S(k,n)=\left\lfloor\frac{2^{n+1}+1}{2^{n-1}+1}\right\rfloor+\left\lfloor\frac{3^{n+1}+1}{3^{n-1}+1}\right\rfloor+\cdots+\left\lfloor\frac{k^{n+1}+1}{k^{n-1}+1}\right\rfloor$$ Compute the value of $S(10,112)-S(10,55)+S(10,2)$.
## Solution
The function $S(k,n)$ can be rewritten as $$\sum_{j=2}^{k} \left\lfloor\frac{j^{n+1}+1}{j^{n-1}+1}\right\rfloor$$
Let $x = j^{n-1}$. With the substitution, each similar part becomes $$\sum_{j=2}^{k} \left\lfloor\frac{j^2 \cdot x+1}{x+1}\right\rfloor$$ Performing polynomial division results in $$\sum_{j=2}^{k} \left\lfloor j^2 + \frac{1 - j^2}{x+1}\right\rfloor$$ $$\sum_{j=2}^{k} \left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor$$ When $n = 112$ or $n = 55$, then $\frac{j^2 - 1}{j^{n-1}+1}$ is close to zero, which means that $\left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor$ would be the same for a given $j$ when $n = 112$ or $n = 55$. Thus, $S(10,112) - S(10,55) = 0$.
That means $S(10,112) - S(10,55) + S(10,2) = S(10,2)$, and that equals $$\sum_{j=2}^{10} \left\lfloor j^2 - \frac{j^2 - 1}{j+1}\right\rfloor$$ $$\sum_{j=2}^{10} \left\lfloor j^2 - (j-1) \right\rfloor$$ $$\sum_{j=2}^{10} j^2 - \sum_{j=2}^{10} (j-1)$$ $$(4+9 \cdots 100) - (1+2 \cdots 9)$$ $$(\frac{10 \cdot 11 \cdot 21}{6} - 1) - (\frac{10 \cdot 9}{2})$$ $$384-45$$ $$\boxed{339}$$
## See Also
2007 iTest (Problems, Answer Key) Preceded by:Problem 57 Followed by:Problem 59 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 |
# Factors of 20 in pair
Factors of 20 in pair are (1, 20) , (2, 10) and (4, 5)
#### How to find factors of a number in pair
1. Steps to find factors of 20 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
### Example: Find factors of 20 in pair
Factor Pair Pair Factorization
1 and 20 1 x 20 = 20
2 and 10 2 x 10 = 20
4 and 5 4 x 5 = 20
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 20. They are called negative pair factors.
Hence, the negative pairs of 20 would be ( -1 , -20 ) .
#### Definition of factor pairs?
In mathematics, factor pair are often given as pair of numbers which when multiplied together give the original number. Every natural number is a product of atleast one factor pair. Eg- Factors of 20 are 1 , 2 , 4 , 5 , 10 , 20. So, factors of 20 in pair are (1,20), (2,10), (4,5).
#### What are factors?
In mathematics, a factor is that number which divides into another number exactly, without leaving a remainder. A factor of a number can be positive or negative.
#### Properties of Factors
• Each number is a factor of itself. Eg. 20 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 20.
• Every number is a factor of zero (0), since 20 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 4, 5, 10, 20 are exact divisors of 20.
• Factors of 20 are 1, 2, 4, 5, 10, 20. Each factor divides 20 without leaving a remainder.
• Every factor of a number is less than or equal to the number, eg. 1, 2, 4, 5, 10, 20 are all less than or equal to 20.
#### Steps to find Factors of 20
• Step 1. Find all the numbers that would divide 20 without leaving any remainder. Starting with the number 1 upto 10 (half of 20). The number 1 and the number itself are always factors of the given number.
20 ÷ 1 : Remainder = 0
20 ÷ 2 : Remainder = 0
20 ÷ 4 : Remainder = 0
20 ÷ 5 : Remainder = 0
20 ÷ 10 : Remainder = 0
20 ÷ 20 : Remainder = 0
Hence, Factors of 20 are 1, 2, 4, 5, 10, and 20
• Is 20 a composite number?
Yes 20 is a composite number.
• Is 20 a prime number?
No 20 is not a prime number.
• Is 20 a perfect square?
No 20 is not a perfect square.
• Write five multiples of 20.
Five multiples of 20 are 40, 60, 80, 100, 120.
• Write all odd factors of 20?
The factors of 20 are 1, 2, 4, 5, 10, 20.
Odd factors of 20 are 1 , 5.
#### Examples of Factors
Rustom has been assigned the following tasks by the teacher:
- Finding out all positive factors of 20.
- Writing all prime factors of 20.
- Writing all the possible factors of 20 in pair.
Help him in writing all these.
Positive factors of 20 are 1, 2, 4, 5, 10, 20.
Prime factors of 20 are 2, 2, 5.
Factors of 20 in pair are (1,20), (2,10), (4,5).
What are the pair factors of 20?
Factors of 20 are 1, 2, 4, 5, 10, 20. Hence, the factors of 20 in pair are (1,20), (2,10), (4,5).
Can you help Sammy list the factors of 20 and also find the factor pairs?
Factors of 20 are 1, 2, 4, 5, 10, 20.
Factors of 20 in pair are (1,20), (2,10), (4,5).
The area of a rectangle is 20 square meters. List all the possible combinations possible for length and breadth in which a designer can design out all the combinations.
For the possible combinations possible for length and breadth in which a designer can design out all the rectangles can be calculated by calculating the factors of 20 in pair. So, the possible combinations (1,20), (2,10), (4,5).
Help Diji in finding factors of 20 by Prime Factorization method and then sorting factors of 20 in pairs.
Prime factorization of 20 is 2 x 2 x 5. Factors of 20 in pair can be written as (1,20), (2,10), (4,5).
Write the smallest prime factor of 20.
Smallest prime factor of 20 is 2.
Write the largest prime factor of 20.
Largest prime factor of 20 is 5.
Diji wants to write all the negative factors of 20. Can you help Diji in doing the same?
Negative factors of 20 are -1, -2, -4, -5, -10, -20. |
# RELATIONS AND FUNCTIONS for Class 11 MATHS
## Introduction
In mathematics, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, we notice that a relation involves pairs of objects in certain order. We will learn how to link pairs of objects from two sets and then introduce relations between the two objects in the pair. Finally, we will learn about special relations which will qualify to be functions.
## Introduction
Function is the most important part of calculus as it deals with dependency of one physical quantity on the other. For example volume of a sphere is dependent on its radius. This dependency of volume ‘v’ on radius ‘r’ is shown by v==43πr3 or v=f(r) Here volume ‘v’ is said to be a function of radius ‘r’ and ‘f’ is a symbol used to show dependency of one variable on the other. To Understand the Concept of Function first we will introduce the concept of relation.
## Index
RELATIONS|TYPES OF RELATION|FUNCTIONS|DOMAIN AND RANGE OF A FUNCTION|METHODS OF FINDING DOMAIN AND RANGE OF A FUNCTION|RANGE OF A FUNCTION|FEW IMPORTANT FUNCTIONS|VARIOUS TYPE OF FUNCTIONS|PERIODIC FUNCTION|EQUAL OR IDENTICAL FUNCTIONS|HOMOGENEOUS FUNCTION|CLASSIFICATION OF FUNCTIONS|METHODS OF FINDING WHETHER FUNCTION IS ONE-ONE OR MANY-ONE|COMPOSITION OF FUNCTIONS|INVERSE OF A FUNCTION|SOME IMPORTANT POINTS|ELEMENTRY TRANSFORMATIONS OF GRAPHS
## Cartesian Product of Sets
So guys jaise hum aaj tak numbers ka multiplication karte aaye hai ab hum sets ke product ki baat karnge but sets ka multpication normal multiplication process se alag hai ,Let's understand .Consider A is set of 2 colours and B is set of 3 object.
A={r,g} and B={b,c,h}
where r,g,b,candh represent a particular red, green, bag, chair and hat, respectively.
We can select ordered pair in 6 ways:
(r,b),(r,c),(r,h),(g,b),(g,c),(g,h).
Each pair can be visualized by intersection of lines.
Given two non-empty sets P and Q. The Cartesian product P×Q is the set of all ordered pairs of elements from P and Q, i.e., P×Q={(p,q):pP,qQ}
If either P or Q is the null set, then P×Q will also be empty set, i.e., P×Q=ϕ. From above example we can write A×B={(r,b),(r,c),(r,h),(g,b),(g,c),(g,h)}.
Note: 1. Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal. i.e. if (a,b)=(c,d)a=c,b=d
2. If there are p elements in A and q elements in B, then there will be pq elements in A×B, i.e., if n(A)=p and n(B)=q, then n(A×B)=pq.
3. If A and B are non-empty sets and either A or B is an infinite set, then A×B is also infinite set.
4. A×A×A={(a,b,c):a,b,cA}. Here (a,b,c) is called an ordered triplet.
5. n(A×B×C)=n(A)×n(B)×n(C)
Some useful results:
1. A×(BC)=(A×B)(A×C)
2. A×(BC)=(A×B)(A×C)
3. A×(BC)=(A×B)(A×C)
4. if A and B have n elements common then A×B and B×A have n2 elements common.
## 1. RELATIONS
A relation between two sets A and B is a subset of AxB. R is a relation from A to B iff RA×B, we write aRb iff (a,b)R and say that a is R−related to b or b is R−relative of a. We also write a (~R)b if a is not R−related to b.
If A consists of m elements and B consists of n elements, then the total no. of different relations from ABis2^(mn).
Any subset of A×A is said to be a relation on A.
If R is relation from A to B, then domain and range are defined as
Domain = {x:(x,y)R}
Range = {y:(x,y)R}
## Example 1
Find A×B , if A ={1,2,3} and B= {x,y}.Also find total number of elements in set A×B.
So guys we will pair each element of set A with each element of B in such a way that all elements of set A will come up at first place and all elements of set B will come at second place in the ordered pair.
So we have ,A×B = {(1,x) , (1,y) , (2,x) , (2,y) , (3,x), (3,y)}
Now as we have the formula to determine the total number of elements in set A×B = mn
where m= total elements in set A
n =total elements in set B
So Total elements in A×B=3×2=6
## Example 1:
Let A={1,2,3} & B={2,4,6,8}. Let R1={(1,2),(2,4),(3,6)} and R2={(2,4),(2,6),(3,8),(1,6)} then find domains and range of relation R1 and R2.
DomainR1={1,2,3}
RangeR1={2,4,6}
Domain R2={2,3,1}
RangeR2={4,6,8}
## Example 2
If (x3+1, y23)=(53, 13), find the values of x and y.
Since hum jaante hai jab 2 ordered pairs equal hote hai tab corresponding elements of those pairs are equal, iska matlab hai agar 2 ordered pairs
(a,b) = (c,d) then a= b and c=d , so we have
x3+1=53x3=(531)=23x3=23x=2
and y23=13y=(13+23)=33=1y=1.
Hence, x=2andy=1
## Types of Relation
( i ) Reflexive relation
A relation R on a set A is said to be a reflexive relation on A if
xRxi.e.(x,x)R;xA
( ii ) Symmetric relation
A relation R on a set A is said to be a symmetric relation on A if xRyyRx i.e. (x,y)R(y,x)Rx,yA
( iii ) Transitive relation
A relation R on a set A is said to be a transitive relation on A if (x,y)R and (y,z)R(x,z)R;x,y,zA i.e. xRy and yRzxRz
( iv ) Identity relation
A relation R on a set A is said to be an identity relation on A if R={(x,y):xA,yA,x=y}
This is denoted by IAIA={(x,x):xA}
(v) A×A is said to be the universal relation on A.
(vi) ϕA×A. So ϕ is a relation on A, called void relation on A
Identity relation is always reflexive but a reflexive relation need not to be identity relation.
( vii ) Equivalence Relation
Let A be a non empty set, then a relation R on A is said to be equivalence relation if
(i) R is reflexive
(ii) R is symmetric
(iii) R is transitive
## Example 3
Express A={(a,b):2a+b=5,a,bW} as the set of ordered pairs.
Here, W denotes the set of whole numbers (non-negative integers).
We have, 2a+b=5, where a,bW
So pehle hum a=0 ko 2a+b=5 me put karenge so we will get , 2(0)+ b= 5
this gives b=5
similarly a=1b=3 and a=2b=1
But ab agar hum 'a' ki koi 3 se badi value put karenge then we will get negative values but question me given hai ke 'b'should be whole number
so , For a>3, the values of 'b' given by the above relation are not whole numbers.
A={(0,5),(1,3),(2,1)}
## Example 2:
Let T be the set of triangles in a plane and a relation r be defined by xryx is similar to y;x,yT. Show that r is an equivalence relation on T.
1. Every triangle is similar to itself.
x is similar to x,xT i.e. xrx. So r is reflexive on T
2.xryx is similar to y
y is similar to x
yrx
r is symmetric relation on T.
3.xry and yrzx is similar to y and y is similar to z
x is similar to zxrz
r is transitive relation. Thus r is an equivalence relation on T.
## Example 4
If A×B={(p,q),(p,r),(m,q),(m,r)}, find A and B.
So we know if (a,b) is any ordered pair which belong to the relationA×B then elements placed first belong to first set and the other belong to second set B.
So collect all elements placed on first place {p , p, m, m} but we know in a set elements do not repeat
So , A = {p, m} and B ={ r,q}
## Example 3:
Let N be the set of all natural numbers. A relation R be defined on N×N by (a,b)R(c,d)a+d=b+c. Show that R is an equivalence relation.
1. (a,b)R(a,b). For a+b=b+a
R is reflexive
2. (a,b)R(c,d)a+d=b+cc+b=d+a
(c,d)R(a,b)
R is symmetric
3. (a,b)R(c,d) and (c,d)R(e,f)a+d=b+c and c+f=d+e
a+d+c+f=b+c+d+e
a+f=b+e (a,b)R(e,f)
R is transitive.
Thus R is an equivalence relation on N×N.
## Example 5
Let R be the set of all real numbers. What does (R×R×R) represent?
(R×R×R)={(a,b,c):a,b,cR}.
Thus, (R×R×R) represents the set of all points in three-dimensional space.
Remark : For any nonempty sets A, B and C, we always have
(A×B)×C=A×(B×C)=A×B×C.
## FUNCTIONS
Function can be easily defined with the help of the concept of mapping. Let X and Y be any two non-empty sets. A function from X to Y is a rule or correspondence that assigns to each element of set X, one and only one element of set Y". Let the correspondence be f then mathematically we write f:XY where y=f(x),xεX and yεY. We say that 'y' is the image of 'x' under ‘f ’ (or x is the pre image of y).
A mapping f:XY is said to be a function if each element in the set X has it's image in set Y. It is possible that a few elements in the set Y are present which are not the images of any element in set X.
Every element in set X should have one and only one image. That means it is impossible to have more than one image for a specific element in set X. Functions can't be multi-valued (A mapping that is multi-valued is called a relation from X to Y)
Graphical method: to check whether a relation between x and y is a function or not, draw a line parallel to y-axis and, if it interest the graph at one and only one point then the given relation represent a function.
## Relations
So guys Relation ka matlab hai 2 ya 2 se jayda concepts, objects, or people kis tareeke se apaas me connected hai.
Mathematically you can say ; A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A×B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A×B. Second element ko hum first element ki image kehte hai.
Kisi relation R from a set A to a set B ke saare first element ke collection ko hum domain of the relation R kehte hai.
Relation R from a set A to a set B ke saare second element ke collection ko hum of the range relation R kehte hai.
Poore set B ko co domain of the relation R kehte hai. Note that range co domain.
Note : (i) A relation may be represented algebraically either by the Roster method or by the Set-builder method.
(ii) An arrow diagram is a visual representation of a relation.
Consider the two sets P={a,b,c} and Q={Ali, Bhanu, Binoy, Chandra, Divya}. The Cartesian product of P and Q has 15 ordered pairs which can be listed as P×Q={(a,Ali),(a,Bhanu),(a,Binoy),...,(c,Divya)}. Let there be some relation R between the first element x and the second element y of each ordered pair (x,y) as R={(x,y):x is the first letter of the name y,xP,yQ}. Then R={(a,Ali),(b,Bhanu),(b,Binoy),(c,Chandra)}. This can be visualized by arrow diagram
Note: (i) The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A×B. If n(A)=p and n(B)=q, then n(A×B)=pq and the total number of relations is 2pq.
(ii) A relation R from A to A is also stated as a relation on A.
Representation of relation:-
(i) Roster Form: In this form a relation is represented by the set of all ordered pairs belonging to R .
For example, if R is a relation from set A={2,1,0,1,2} to set B={0,1,4,9,10} by the rule aRba2=b. Then,0R0,2R4,1R1,1R1 and 2R4.
So,R can be described in roster form as R={(0,0),(1,1),(2,4),(1,1),(2,4)}
(ii) Set-Builder Form: In this form the relation R from set A to set B is represented as R={(a,b):aA,bB and a,b satisfy the rule which associates a and b}.
For example, if A={1,2,3,4,5},B={1,12,13,14,15,16,....} and R is a relation from A to B, given by R={(1,1),(2,12),(3,13),(4,14),(5,15)}
Then, R in set-builder form can be described as: R={(a,b):aA,bB and b=1a}
(iii) By Arrow Diagram: In order to represent a relation from set A to a set B by an arrow diagram, we draw arrows from first components to the second components of all ordered pairs belonging to R.
For example, relation R={(1,2),(2,4),(3,2),(1,3),(3,4)} from set A={1,2,3,4,5} to set B={2,3,4,5,6,7} can be represented by the following arrow diagram:
(iv) By Lattice : In this form, the relation R from set A to set B is represented by darking the dots in the lattice for A×B which represent the ordered pairs in R.
For example, if R={(3,9),(2,4),(1,1),(0,0),(1,1),(2,4),(3,9)} is a relation from set A={3,2,1,0,1,2,3} to set B={0,1,2,3,4,5,6,7,8,9}, then R can be represented by the following lattice.
Inverse Relation: Let A,B be two sets and let R be a relation from a set A to a set B. Then the inverse of R, denoted by R1, is a relation from B to A and is defined by R1={(b,a):(a,b)R}.
Clearly, (a,b)R(b,a)R1
Also, Dom(R)=Range(R1) and, Range (R)=Dom(R1).
## DOMAIN AND RANGE OF A FUNCTION
The set X is called the domain of the function. ƒ and set Y is called the co-domain. The set of the image of all elements of X under the function ƒ is called the range of ƒ and is denoted by ƒ(x). It is obvious that range could be a subset of co-domain as we may have few elements in co-domain which are not the images of any element of the set X.
Thus range of ‘ƒ’ i.e. ƒ(x)={ƒ(x):xX}. Clearly ƒ(x)Y
## Example 6
Define a relation R on the set N of natural numbers by R={(x,y):y=x+5, x is a natural number less than 4;x,yN}. Depict this relationship using roster form. Write down the domain , codomain and the range.Also depicit the relation using arrow diagram.
So guys question ke according do elements x, y jo ki natural numbers hai , tab relation me honge jab y = x +5 hoga provided hume x is value natural numbers ke set se pick karni hai and x should be less than 4 , so for x we have the following choices:
x can be 1, 2, 3
Now we will put these values in y = x+ 5 to get the second element (y) of the ordered pair (x, y), for that put all above values of x one by one in y = x + 5
We will get y = 6 for x =1 ; y = 7 for x = 2 and y = 8 for x = 3
so , R = { (1, 6) , (2, 7) , (3, 8)}
This is the required roaster form of the given relation.
Now , Domain = Collection of first element in the ordered pair = {1, 2, 3}
Range = Collection of second element in ordered pair of relation R = { 6, 7, 8}
Codomain = The set from which y can be chosen = Set of natural number.
## Example 4
Find the domain of f(x)=x2+1x24
x240x±2
Hence domain is R{2,2}
## Example 7
Let A = {1, 2} and B = { 3, 4} , Find the number of relations from A to B .
So we know total number of relations from set A to B = Total number of subsets of A×B = 2mn
Where m= number of elements of set A = n(A) and n = number of elements of set B = n(B)
Now n(A) = m = 2 and n(B) = n = 2
So , total number of relations from set A to B = 22×2=24=16
## Example 5
The domain of function f(x)=log12(x2+4x+3) is
f(x)=log12(x2+4x+3)
f is defined if x2+4x+3>0
i .e., if (x+1)(x+3)>0
i.e., if x>1orx<3
Domain of f is : (,3)(1,)
## Functions
So guys function means a connection or a rule defined between two sets .Mathematically you can define it as , A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.
In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element.
If f is a function from A to B and (a,b)f, then f(a)=b, where b is called the image of a under f and a is called the pre-image of b under f.
The function f from A to B is denoted by f:AB.
Note: For every value of x in relation (x,y) if it gives unique y then that relation is called function.
A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function.
Two functions are said to be equal if for all value of x,f(x)=g(x), then f=g.
The value of function can be determined by replacing variable x with some constant or variable k then function value will be f(k). i.e. if we want to find f(x2) replace x with x2.
Some functions and their graphs:
(i)Identity function: Man lo we have a set A then identity function is a function jisme each element of set A has an image jo ki wo element khud hoga or a real valued function f:RR by y=f(x)=x for each xR. Such a function is called the identity function. Here the domain and range of f are R. The graph is a straight line. It passes through the origin.
Note: The function f defined by f(x)=mx+c,xR, is called linear function, where m and c are constants.
(ii) Constant function: Ek constant function aisa linear function hai jisme domain me koi bhi object le lo range will be same. Mathematically , The function f:RR by y=f(x)=c, xR where c is a constant and each xR. Here domain of f is R and its range is {c}.
The graph is a line parallel to x-axis.
(iii) Polynomial function: A function f:RR is said to be polynomial function if for each x in R , y=f(x)=a0+a1x+a2x2+...+anxn , where n is a non-negative integer and a0,a1,a2,...,anR. The functions defined by f(x)=x32x2+x+2, f(x)=4x2+4x8 are example of polynomial.
(iv) Rational functions: They are functions of the type f(x)g(x), where f(x) and g(x) are polynomial functions of x defined in a domain, where g(x)0.
(v)The Modulus function: The function f:RR defined by f(x)=|x| for each xR is called modulus function. For each non-negative value of x,f(x) is equal to x . But for negative values of x, the value of f(x) is the negative of the value of x, i.e.,
f(x)={xx0xx<0}
(vi) Signum function: The function f:RR defined by
f(x)=1x>00x=01x<0
is called the signum function. The domain of the signum function is R and the range is the set {–1, 0, 1}. It can also be given by |x|x , x0
and 0 for x=0.
(vii) Greatest integer function: The function f:RR defined by f(x)=[x], xR assumes the value of the greatest integer, less than or equal to x. Such a function is called the greatest integer function.
[x]=1 for 1x<0
[x]=0 for 0x<1
[x]=1 for 1x<2
[x]=2 for 2x<3 and so on.
Algebra of real functions :-
Now, we will learn how to add two real functions, subtract a real function from another, multiply a real function by a scalar (real number), multiply two real functions and divide one real function by another.
(i) Addition of two real functions: Let f:XR and g:XR be any two real functions, where XR. Then, we define (f+g):XR by (f+g)(x)=f(x)+g(x), for all xX.
For Example, f(x)=x2 and g(x)=x+1,
(f+g)(x)=f(x)+g(x)=x2+x+1
(ii) Subtraction of a real function from another: Let f:XR and g:XR be any two real functions, where XR. Then, we define (fg):XR by (fg)(x)=f(x)g(x), for all xX.
For Example, f(x)=x2 and g(x)=x+1,
(fg)(x)=f(x)g(x)=x2(x+1)=x2x1
(iii) Multiplication by a scalar: Let f:XR be a real valued function and α be a scalar. Here by scalar, we mean a real number. Then the product αf is a function from X to R defined by (αf)(x)=αf(x),xX.
For Example, f(x)=x2andα=5,
(αf)(x)=αf(x)=5x2
(iv) Multiplication of two real functions: The product (or multiplication) of two real functions f:XR and g:XR is a function fg:XR defined by (fg)(x)=f(x)g(x), for all xX. This is also called point wise multiplication.
For Example, f(x)=x2 and g(x)=x+1,
(fg)(x)=f(x)g(x)=x2×(x+1)=x3+x2
(v) Quotient of two real functions: Let f and g be two real functions defined from XR, where XR. The quotient of f by g denoted by fg is a function defined by, (fg)(x)=f(x)g(x) provided g(x)0, xX. For Example, f(x)=x2 and g(x)=x+1,
(fg)(x)=f(x)g(x)=x2x+1, x+10
Important point to remember :
1. To find Domain exclude the point at which function will become not define i.e. where denominator =0, square root becoming negative, etc.
2. To find range put f(x)=y and express x in terms of y i.e. x=f(y) and follow same step for y that we follow for domain exclude those point from range.
For example, f(x)=x2x3
here we have x30 which gives x3
Therefore, Domain is R{3}
Now put y=x2x3
(x2)=y(x3)=xy3y
xxy=23y
x(1y)=23y
x=23y1y
here we have 1y0 which gives y1
Therefore, Range= R –{1}
## Example 6
The domain of function f(x)=1|x|x is
f is defined if |x|x>0
If x0, then |x|=x
If x<0, then |x|=x>0 and |x|>x
∴ Domain of f={x:x<0}
## Example 8
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11 , 1), (14 , 1), (17 , 1)}
(ii) {(2, 1), (4, 6), (6, 3), (8, 4), (2,10)}
So guys we know ek relation ko hum function tab he kehte hai jab domain ke har ek elements ki sirf ek unique image ho in codomain , let see which given relation satisfy this definition.
(i) As yaha domain ke har ek element {2 , 5 , 8 , 11 , 14 , 17} ke pas unique image hain in codomain , so we can surely say it is a function.
Domain = {2 , 5 , 8 , 11 , 14 , 17}
Range = {1}
(ii)Since in this relation an element 2 has two different image 1 & 10 such that (2 ,1) and (2, 10) , so it is not a function.
## METHODS OF FINDING DOMAIN AND RANGE OF A FUNCTION
By definition, domain of a function are those values of 'x' (Independent variable) for which f(x) exists or is defined.
If a function is expressed as sum, difference, product or division of two or more functions then the domain of function can be defined by the following method:
Let f:D1R and g:D2R. Let D1D2=D. Then we describe functions f+g,fg,gf and fg as follows
(1) f+g:DR
is a function defined by
(f+g)(x)=f(x)+g(x)
(2) fg:DR
is a function defined by
(fg)(x)=f(x)g(x)
(3) fg:DR
is a function defined by
(fg)(x)=f(x)g(x)
(4) fg:CR
is a function defined by
(fg)(x)=f(x)g(x) , g(x)0
where C={xD:g(x)0}.
## Example 9
A function f is defined by f(x)=2x5. Write down the values of (i) f(0), (ii) f(7), (iii) f(3)
(i)So guys this is the most easiest question among all , we simply need to put value of unknown variable x accordingly.
for finding f(0) put x = 0 in f(x)=2x5
this gives , f(0) = 2(0) - 5 = -5
Similarly f(7) = 2(7) - 5 = 14 - 5 = 9
f(-3) = 2(-3) -5 =- 6- 5 = -11
## Example 7
Find the domain of definition of the function
f(x)=x25x+4+ln2ln3x+sin1(x+1)2
LetD1 is domain of x25x+4
D2 is domain of ln2ln3x and
D3 is domain of sin1(x+1)2
Now Domain D of f(x) is D1D2D3
(i) Now D1x25x+40 (since square root function operates on non-negative values only)
D1x1x4 (1)
(ii) Similarly ln2M is defined when M>0(M is ln3x)
(ln3)x>0 or, x>1
D2x>1 (2)
(iii)sin1(x+12) is defined whenever 1x1
3x1
D33x1 (3)
Clearly DD1D2D3 is ϕ. Hence Domain of f(x) is null set which means that f(x) doesnot exists for any value of x.
## Example 10
Find the domain and range of the following real functions:(i) f(x)=|x| (ii) f(x)=9x2
So guys domain is a set where the given function has some defined value and range is set of values which we get after putting all values of domain in given function f(x), so according to this
(i) As |x| is defined for each real number so Domain = Set of real number
Toh agar aap saare real numbers ko f(x) = -|x| me put karenge toh apko saare negative real numbers milenge
Range = Set of all negative real numbers or (,0]
(ii)Similarly hum jaante hai ke square root is not defined (jab function real valued ho) jab iske andar negative value aajaye it means we have to exclude the values jaha 9x2 me negative values aaye
So we can conclude ke jaise he hum x ki koi value 3 se badi ya -3 se choti put karenge f(x) not defined ho jayega kyuki square root ke andar negative aajayega .
so , Domain ={x:3x3} .
Now to get the range just put all values of domain in f(x) , so we get all values between x = 0 and x = 3 including these two as well
So , Range = {x;0x3}
## RANGE OF A FUNCTION
Range of a function are the values which f(x) takes when 'x' takes the all values which are in the domain. Although there is no particular method to find the Range, in most of the cases, range can be found out by one of the following methods:
(a) Transform and find x in terms of y, then find domain of y.
(b) If f(x) is in the form of , where p(x) and Q(x) are polynomial function of second degree we use the concept of quadratic equation.
(c) If f(x) is continuous function, then range of f(x) is [Min f(x), Max f(x)] i.e. from Minimum value of f(x) to maximum value of f(x).
## Summary
(i) Ordered pair: A pair of elements grouped together in a particular order.
(ii)Cartesian product: A × B of two sets A and B is given by A×B={(a,b):aA,bB}
In particular R×R={(x,y):x,yR} and R×R×R=(x,y,z):x,y,zR}
(iii) If (a, b) = (x, y), then a = x and b = y.
(iv) If n(A) = p and n(B) = q, then n(A×B) = pq.
(v) A×ϕ=ϕ
(vi) In general, A×BB×A.
(vii)Relation: A relation R from a set A to a set B is a subset of the cartesian product A×B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A×B.
(viii) The image of an element x under a relation R is given by y, where (x,y)R
(ix) The domain of R is the set of all first elements of the ordered pairs in a relation R.
(x) The range of the relation R is the set of all second elements of the ordered pairs in a relation R.
(xi) Function: A function f from a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B. We write f:AB, where f(x) = y.
(xii) A is the domain and B is the codomain of f.
## Example 8
Iff(x)=x2+2x+3x . Find the range of f(x).
Let x2+2x+3x=y
x2+(2y)x+3=0
xεRD0(2y)24x1x30
yε(,223]U[2+23,)
## Let's Revise
Cartesian Product of sets: Let 'A' and 'B' be two non empty sets then A×B is defined as Cartesian product of 'A' and 'B' which is a set of ordered pair of elements of set A andB.
A×B= { (a,b): aA and bB}
Where (a,b) = ordered pair, and 'a' always belongs to set A and 'b' always belong to set B.
Relation:If we have two sets suppose 'A' and 'B', then Relation is nothing but the sub sets of Cartesian product of set A and set B .
Which can be further expressed as
R = { (a,b): a related to b, a A and b B}
Here a related to b means some elements of A is Related to some elements of B where elements of A is represented as a and elements of B is represented as b.
If (a,b)R, then we can say 'a' is related to 'b' with relation R. Where 'b' is called Image of 'a'
Domain of Relation: Relation ke elements ordered pair hote hai to agar hum in pairs ke first elements ko le kr ek new set bna de to us new set ko Domain or R kehte hai.
For example: Let we have a Relation R from Ato B:
R={(1,3),(2,4),(3,5),(4,6),(5,7)}
Then Domain of R ={ a: (a,b)R}
Domain of R={1,2,3,4,5}
Range of Relation: Jaisa ki humne dekha ki Relation ke elements ordered pair hote hai to agar hum in pairs ke second elements ko le kr ek new set bna de to us new set ko Range or R kehte hai. For ex:
R={(1,3),(2,4),(3,5),(4,6),(5,7)}
Then Range of R ={ b: (a,b)R}
Range of R={3,4,5,6,7}
Note: Jab kabhi hum set A and set B ke beach relation bnate hai then elements of set B will always form a new set which is called Codomain of R.
Function: A function ffrom a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B. We write f: A→B, where f(x) = y.A is the domain and B is the codomain of f.
The function f from A to B is denoted by f: A B.
If f is a function from A to B and (a, b) ∈ f, then f(a)=b, where 'b' is called the image of a under f and 'a' is called the preimage of b under f.
Some Important functions and their graphs.
(i)Identity function:Man lo we have a set A then identity function is a function jisme each element of set A has an image jo ki wo element khud hoga.
For identity function f(x)=x
(ii) Constant function: Ek constant function aisa linear function hai jisme domain me koi bhi object le lo range will be same.
The graph is a line parallel to x-axis.
(iii) Polynomial function: A function f:RR is said to be polynomial function if for each x in R , y=f(x)=a0+a1x+a2x2+...+anxn , where n is a non-negative integer and a0,a1,a2,...,anR.
(iv) Rational functions: They are functions of the type f(x)g(x), where f(x) and g(x) are polynomial functions of x defined in a domain, where g(x)0.
(v)The Modulus function: The function f:RR defined by f(x)=|x| for each xR is called modulus function i.e.,
f(x)={xx0xx<0}
(vi) Signum function: The function f:RR defined by
f(x)=1x>00x=01x<0
is called the signum function. The domain of the signum function is R and the range is the set {–1, 0, 1}. It can also be given by |x|x , x0 and 0 for x=0.
(vii) Greatest integer function: It is denoted by f(x)=y=[x]. For all real numbers, ⌊x⌋=the largest integer that is less than or equal to x. In other words, it rounds down any real number to the nearest lower integer. For example: [2] = 2 [2.5] = 2, [3.7] = 3 [4.3] = 4.
Algebra of functions
Let a function f:XR and function g:XR, we have
(f+g)(x)=f(x)+g(x),xX
(fg)(x)=f(x)g(x),xX
(f.g)(x)=f(x).g(x),xX
(kf)(x)=k(f(x)),xX,where k is a real number.
(fg)(x)=f(x)g(x), xX,g(x)0
## Example 9
f(x)=x2+2x+3x . Find the range of f(x).
y=x2x2+1x2=y1y
We know for any real x,x20
y1y0yy10
Range ε[0,1)
## Example 10
Find the range of the function x2+x+1.
Since x2+x+1 is a polynomial function, it is continuous over all xR.
Also f(x)=x2+x+1 is a quadratic expression is a parabola with mouth opening upwards.
We know from our knowledge on Quadratic Equation that maximum value off(x) tends to infinity as x do so. However minimum value of f(x) occurs at x=b2a=12
The value of f(12)=34
Hence range of function is [34,)
## Example 11
Find the range of the function, f(x)=3ex.
y=3ex
ex=3y
or, x=ln(3y)............(1)
Domain of above function (1) is 3y>0 or, y<3. This domain becomes range of the original function f(x).
Note that f(x) is a bijective.
Note : This technique can be used even if function is not bijective, but case should be taken to include relevant values only.
## Example 12
Find the domain and range of the function, f(x)=x2.
Domain x[2,) Here y2=x2;y0.
x=y2+2 and domain of this is y0.
Thus, Range is y[0,).
## FEW IMPORTANT FUNCTIONS
(a) Absolute Value Function
f:RR defined by
f(x)=|x|={x x0x x<0
is called absolute value function.
Its domain is R and its range is [0,).
(b) Constant Function
f:RR defined by f(x)=cxR, where c is a constant, is called a constant function. Its domain is R and range is {c}. Graph of a constant function is a straight line parallel to x − axis.
(c) The Identity Function
f:RR defined by f(x)=x is called the identity function. Its domain is R and range is also R.
(d) The Exponential Functions
Let a1 be a positive real number. Then
f:RR defined by f(x)=ax is called exponential function. Its domain is R and range is (0,)
(e) The Logarthmic Functions
Let a1 be a positive real number. Then f:(0,)R defined by f(x)=logax is called logarthmic function. Its domain is (0,) and range is R.
(f) The Greatest Integer Function or Step function
f:RZ defined by f(x)=[x], where [x] denotes the greatest integer among all the integers less than or equal to x is called the greatest integer function.
i.e. f(x)=n, where nx<n+1,nI (set of integers)
Its domain is R and range is I.
Example [2.5]=2 as 22.5<3
[π2]=9 as 9π2<10
[4.5]=5 as 54.5<4
Properties of greatest integer function
• [x]x<[x]+1
• [x+y]=[x]+[y] if {x}+{y}<1
• [x+y]=[x]+[y]+1 if {x}+{y}1 where {x} denotes fraction part of x
• (g) Fractional-Part Function {x} denotes fractional part of 'x'. It is equal to x[x].
e.g {2.7}=0.7 {3}=0
{3.2}=0.8
(h) Polynomial Functions
f:RR defined by f(x)=a0+a1x+a2x2++anxn, where a0,a1,a2,,anR, is called a polynomial function. If an0, then the degree of f (x) is n.
Remarks
If n is odd, then its range is R.
(i) Trigonometric Functions
(j) Inverse Circular Functions
(k) Rational Functions
Let g(x) and h(x) be two polynomials over the set of real numbers where h(x)0.
Then is called a rational function. Its domain is R{xh(x)=0}.
(l) Signum Function
A function y=f(x)=Sgn(x) is defined as follows:
y=f(x)=1forx>00forx=01forx<0
It is also written as Sgnx=|x|x;x0;f(0)=0
## VARIOUS TYPE OF FUNCTIONS
Even and Odd Functions
A function y=f(x) is even if f(x)=f(x) for every number x in the domain of f.
e.g. f(x)=x2 is even because f(x)=(x)2=x2=f(x).
A function y=f(x) is odd if f(x |
Problems on Trigonometric Ratio of Standard Angle
How to solve the problems on Trigonometric Ratio of Standard Angle?
We know the standard angles are 0°, 30°, 45°, 60° and 90°. The questions are based on these standard angles. Here we will learn how to solve the standard angle of trigonometry related question.
Standard angles in trigonometry generally mean those angles whose trigonometric ratios can determine without using calculators. To find the values of trigonometric ratios of these standard angles we need to follow the trigonometric table.
Worked-out problems on trigonometric ratio of standard angle:
1. If β = 30°, prove that 3 sin β - 4 sin$$^{3}$$ β = sin 3β.
Solution:
L.H.S = 3 sin β - 4 sin$$^{3}$$ β
= 3 sin 30° – 4. sin$$^{3}$$ 30°
= 3 ∙ (1/2) - 4 ∙ (1/2)$$^{3}$$
= 3/2 – 4 ∙ 1/8
3/2 – ½
= 1
R.H.S. = sin 3A
= sin 3 ∙ 30°
= sin 90°
= 1
Therefore, L.H.S. = R.H.S. (Proved)
2. Find the value of 4/3 tan$$^{2}$$ 60° + 3 cos$$^{2}$$ 30° - 2 sec$$^{2}$$ 30° - 3/4 cot$$^{2}$$ 60°
Solution:
The given expression
$$\frac{4}{3} \cdot (\sqrt{3})^{2} + 3 \cdot (\frac{\sqrt{3}}{2})^{2} - 2 \cdot (\frac{2\sqrt{3}}{3})^{2} - \frac{3}{4} \cdot (\frac{\sqrt{3}}{3})^{2}$$
= $$\frac{4}{3} \cdot 3 + 3 \cdot \frac{3}{4} - 2 \cdot \frac{12}{9} - \frac{3}{4} \cdot \frac{3}{9}$$
= 4 + 9/4 - 8/3 – 1/4
= 10/3
= $$3\tfrac{1}{3}$$
3. If θ = 30°, prove that cos 2θ = cos$$^{2}$$ θ - sin$$^{2}$$ θ
Solution:
L. H. S. = cos 2θ
= cos 2 ∙ 30°
= cos 60°
= 1/2
And R. H. S. = cos$$^{2}$$ θ - sin$$^{2}$$ θ
= cos$$^{2}$$ 30° - sin$$^{2}$$ 30°
= (√3/2)$$^{2}$$ – (1/2)$$^{2}$$
= ¾ - ¼
= 1/2
Therefore, L.H.S = R.H.S. (Proved)
4. If A = 60° and B = 30°, verify that sin (A - B) = sin A cos B - cos A sin B
Solution:
L.H.S. = sin (A - B)
= sin (60° - 30°)
= sin 30°
= ½
R.H.S. = sin A cos B - cos A sin B
= sin 60° cos 30° - cos 60° sin 30°
= $$\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}$$
= ¾ - ¼
= 2/4
= ½
Therefore, L.H.S. = R.H.S. (Proved)
5. If sin (x + y) = 1 and cos (x - y) = $$\frac{\sqrt{3}}{2}$$, find x and y.
Solution:
sin (x + y) = 1
sin (x + y) = sin 90°, [since sin 90° = 1]
⇒ x + y = 90° .........................(A)
cos (x - y) = $$\frac{\sqrt{3}}{2}$$
⇒ cos (x - y) = cos 30°
⇒ x - y = 30° .........................(B)
Adding, (A) and (B), we get
x + y = 90°
x - y = 30°
2x = 120°
x = 60°, [Dividing both sides by 2]
Putting the value of x = 60° in (A) we get,
60° + y = 90°
Subtract 60° from both sides
60° + y = 90°
-60° -60°
y = 30°
Therefore, x = 60° and y = 30°.
Trigonometric Functions
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# Fractions of Shapes
This Math quiz is called 'Fractions of Shapes' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.
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Finding fractions of shapes means being able to understand that a shape split into two equal parts has been halved, that a shape split into four equal parts (perhaps by halving and then halving again) has been quartered and a shape split into three equal parts is in thirds. Children should recognize the equivalence of two quarters equaling one half, and also that three quarters is greater than a half, but less than the whole. Using division is key when working out and finding fractions of whole numbers.
Use this quiz to practice finding fractions of shapes.
Question 1
12
34
25
14
There are four quarters. If I shaded two of them, it would be half of the shape
Question 2
23
14
13
12
One of the three equal parts is called a third
Question 3
24
14
34
12
If you shaded in 1 quarter, and then 2 more, you would have shaded 3 of the 4 quarters
Question 4
24
14
44
12
The whole shape is shaded – this is 4 quarters
Question 5
24
34
44
14
3 of the 4 quarters are not shaded, only one quarter is shaded
Question 6
24
They are both the same
12
34
12 is the same as 24 so 34 is greater
Question 7
24
14
12
34
The shape is divided into four equal sections – each one is a quarter. One is shaded
Question 8
34, 14, 12
14, 34, 12
12, 14, 34
14, 12, 34
12 is the same as 24
Question 9
8
6
5
10
Each shape has 4 quarters so there are 8 altogether
Question 10
5
3
6
9
Each shape has 3 thirds. There are 3 shapes so 3 x 3 = 9
Author: Angela Smith |
# How do you find the average value of f(x)=(x-3)^2 as x varies between [2,5]?
Feb 14, 2018
The average value is $= 1$
#### Explanation:
The average value of a function $f \left(x\right)$ over an interval $\left[a , b\right]$ is
$\overline{x} = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$
Here,
$f \left(x\right) = {\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$
and $\left[a , b\right] = \left[2 , 5\right]$
Therefore,
$\overline{x} = \frac{1}{5 - 2} {\int}_{2}^{5} {\left(x - 3\right)}^{2} \mathrm{dx}$
$= \frac{1}{3} {\int}_{2}^{5} \left({x}^{2} - 6 x + 9\right) \mathrm{dx}$
$= \frac{1}{3} {\left[{x}^{3} / 3 - 6 {x}^{2} / 2 + 9 x\right]}_{2}^{5}$
$= \frac{1}{3} \left(\left(\frac{125}{3} - 75 + 45\right) - \left(\frac{8}{3} - 12 + 18\right)\right)$
$= \frac{1}{3} \left(\frac{117}{3} - 36\right)$
$= \frac{1}{3} \cdot 3$
$= 1$ |
# Bernoulli Trials: Definition, Examples
## What are Bernoulli Trials?
A Bernoulli trial is an experiment with two possible outcomes: Success or Failure. “Success” in one of these trials means that you’re getting the result you’re measuring. For example:
• If you flip a coin 100 times to see how many heads you get, then the Success is getting heads and a Failure is getting tails.
• You might want to find out how many girls are born each day, so a girl birth is a Success and a boy birth is a Failure.
• You want to find the probability of rolling a double six in a dice game. A double six dice roll = Success and everything else = Failure.
Note that “Success” doesn’t have the traditional meaning of triumph or prosperity. In the context of Bernoulli trials, it’s merely a way of counting the result you’re interested in. For example, you might want to know how many students get the last question on a test wrong. As you’re measuring the number of incorrect answers, the “Success” is answered incorrectly and a “Failure” is answered correctly. Of course, this might get confusing so there’s nothing stopping you tweaking your hypothesis so that you’re measuring the number of correct answers instead of incorrect ones.
## Probability Distribution for Bernoulli Trials
Bernoulli trials are a special case of i.i.d. trials; Trials are i.i.d. if all the random variables in the trials have the same probability distribution.
The probability distribution for a Bernoulli trial is given by the binomial probability distribution:
Where:
• ! is a factorial,
• x is the number of successes,
• n is the number of trials.
Bernoulli Distribution Probability & PDF
## Assumptions for Bernoulli Trials
The three assumptions for Bernoulli trials are:
1. Each trial has two possible outcomes: Success or Failure. We are interested in the number of Successes X (X = 0, 1, 2, 3,…).
2. The probability of Success (and of Failure) is constant for each trial; a “Success” is denoted by the letter p and “Failure” is q = 1 − p.
3. Each trial is independent; The outcome of previous trials has no influence on any subsequent trials.
Sampling in Statistics > Bernoulli Sampling
## What is Bernoulli Sampling?
Bernoulli sampling is an equal probability, without replacement sampling design. In this method, independent Bernoulli trials on population members determines which members become part of a sample. All members have an equal chance of being part of the sample. The sample sizes in Bernoulli sampling are not fixed, because each member is considered separately for the sample. The method was first introduced by statistician Leo Goodman in 1949, as “binomial sampling”.
The sample size follows a binomial distribution and can take on any value between 0 and N (where N is the size of the sample). If π is the probability of a member being chosen then the expected value (EV) for the sample size is πN. for example, let’s say you had a sample size of 100 and the probability of choosing any one item is 0.1, then the EV would be 0.1 * 100 = 10. However, the sample could theoretically be anywhere from 0 to 100.
Example of Bernoulli Sampling: A researcher has a list of 1,000 candidates for a clinical trials. He wants to get an overview of the candidates and so decides to take a Bernoulli sample to narrow the field. For each candidate, he tosses a die: if it’s a 1, the candidate goes into a pile for further analysis. If it’s any other number, it goes into another pile that isn’t looked at. The EV for the sample size is 1/6 * 1,000 = 167.
An advantage to Bernoulli sampling is that it is one of the simplest types of sampling methods. One disadvantage is that it’s not known how large the sample is at the outset.
In SAS: Bernoulli sampling is specified with METHOD=BERNOULLI. The sampling rate is specified with the SAMPRATE= option.
In R: S.BE(N, prob) will choose a sample from population of size N with a probability of prob. For example (UPenn):
# Vector U contains the label of a population of size N=5
U <- c("Yves", "Ken", "Erik", "Sharon", "Leslie") # Draws a Bernoulli sample without replacement of expected size n=3 # The inclusion probability is 0.6 for each unit in the population sam <- S.BE(5,0.6) sam # The selected sample is U[sam]
## The Bernoulli Distribution
A Bernoulli Distribution is the probability an experiment produces a particular outcome. It is a binomial distribution with a single event (n = 1).
There are two variables in a Bernoulli Distribution: n and p.
• “n” represents how many times an experiment is repeated. In a Bernoulli, n = 1.
• “p” is the probability of a specific outcome happening. For example, rolling a die to get a six gives a probability of 1/6. The Bernoulli Distribution for a die landing on an odd number would be p= 1/2.
The Bernoulli and binomial distribution are often confused with each other. However, the difference between the two is slim enough for both to be used interchangeably. Technically, the Bernoulli distribution is the Binomial distribution with n=1.
### Bernoulli Trial
A Bernoulli distribution is a Bernoulli trial. Each Bernoulli trial has a single outcome, chosen from S, which stands for success, or F, which stands for failure. For example, you might try to find a parking space. You are either going to be successful, or you are going to fail. Many real-life situations can be simplified to either success, or failure, which can be represented by Bernoulli Distributions.
## References
Evans, M.; Hastings, N.; and Peacock, B. “Bernoulli Distribution.” Ch. 4 in Statistical Distributions, 3rd ed. New York: Wiley, pp. 31-33, 2000.
UPenn. Retrieved April 1, 2020 from: http://finzi.psych.upenn.edu/library/TeachingSampling/html/S.BE.html
Governors State University. General PPT.
CITE THIS AS:
Stephanie Glen. "Bernoulli Trials: Definition, Examples" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/bernoulli-trials/ |
Question 245529
<font face="Garamond" size="+2">
The Linear Combinations Method is also known as the Addition Method which is also known as the Elimination Method. I like to call it the Elimination Method because that name most aptly describes what you are trying to accomplish, namely the elimination of one of the variables so that you can easily solve for the other.
The first step of the method is to multiply one or the other (or sometimes both) of the equations by a number or numbers so that the coefficients on one of the variables becomes the additive inverse of the coefficient on that same variable in the other equation.
In your very simple example, you don't have to do the first step -- it is already done for you with respect to the variable *[tex \Large y]. Note that the coefficient on *[tex \Large y] in the first equation is 1 and the coefficient on *[tex \Large y] in the second equation is -1. 1 and -1 are additive inverses because 1 + (-1) = 0.
The next step is to add the two equations together (that's where the Addition Method name comes from). Just like any other simplification, you combine like terms.
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ y\ =\ 4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,x\ -\ y\ =\ 2]
*[tex \Large 2x\ +\ x\ =\ 3x], *[tex \Large y\ +\ (-y)\ =\ 0y], and *[tex \Large 4\ +\ 2\ = 6], that is:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 0y\ =\ 6]
Notice that we now have a zero coefficient on *[tex \Large y], so we can just eliminate (there's that name again) the *[tex \Large y] variable and simply write:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ =\ 6]
Which solves to:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,x\ =\ 2]
Now that you have a value for *[tex \Large x], it is a simple matter to substitute that value back into one of the original equations and solve for the other variable.
However, I'm going to do this the hard way as a demonstration of the step 1 that you didn't have to do. Let's eliminate the *[tex \Large x] variable instead.
Begin with the given equations:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ y\ =\ 4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,x\ -\ y\ =\ 2]
Multiply the second equation by -2:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 2x\ +\ y\ \ \,=\ \ \,4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x\ +\ 2y\ =\ -4]
Now the coefficients on *[tex \Large x] are additive inverses, that is 2 + (-2) = 0. Add the equations:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0x\ +\ 3y\ =\ 0]
and finally
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 0]
That's all there is to it.
John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font> |
# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/6, the angle between sides B and C is (pi)/2, and the length of B is 12, what is the area of the triangle?
Jan 21, 2017
The area of the triangle is 41.5692
#### Explanation:
$\frac{\Pi}{2}$ indicates a right angled triangle
Calculating the length of C will allow us to use
$\frac{1}{2} \cdot B a s e \cdot H e i g h t$ to calculate the area
The length of C can be calculated by using
$T a n \left(\frac{\Pi}{6}\right) = 0.5774 = \frac{C}{12}$
Rearranging
$C = 12 \cdot 0.57735 = 6.9282$
Using $\frac{1}{2} \cdot B a s e \cdot H e i g h t$
$\frac{1}{2} \cdot 12 \cdot 6.9288 = 41.56928$ |
# Triangle Identification as Similar, Congruent, or Neither
## Understand the terms similar and congruent and apply them to triangles.
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Triangle Identification as Similar, Congruent, or Neither
### Source: https://commons.wikimedia.org/wiki/File%3AReglas.svg License: CC BY-NC 3.0 [Figure1]
Gabriel is in geometry. He uses different tools to measure different items. One day he sits a couple of his tools together and he notices that two of his tools are triangles. His class has recently compared triangles and identified them as similar, congruent or neither. The class is supposed to be using their measuring tools to measure items around the classroom and determine if they are similar, congruent or neither. Gabriel takes a good look at the two shapes then decides to identify them so that he can add them to his list. Are Gabriel's triangles similar, congruent or neither?
In this concept, you will learn to distinguish between triangles.
### Identifying Triangles as Similar, Congruent, or Neither
Congruent means being exactly the same. When two line segments have the same length, they are congruent. When two figures have the same shape and size, they are congruent.
These two triangles are congruent. They are exactly the same in every way. They are the same size and the same shape. Their side lengths are the same and that their angle measures are the same.
Sometimes, two figures will be similar. Similar means that the figures have the same shape, but not the same size. Similar figures are not congruent.
These two triangles are similar. They are the same shape, but they are not the same size.
### Examples
#### Example 1
Earlier, you were given a problem about Gabriel and his measuring tools.
He notices that two of his tools are triangles. Are his triangles similar, congruent or neither?
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
No
Then, identify the triangles.
The answer is that the triangles are neither similar, nor congruent. Gabriel will list the triangles as neither.
#### Example 2
Identify the following triangles as similar, congruent or neither.
First, check to see if the triangles have the same size.
Yes
Next, check to see if the triangles have the same shape.
Yes
Then, identify the triangles.
Congruent
The answer is that the triangles are congruent.
#### Example 3
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
No
Then, identify the triangles.
Neither congruent, nor similar
The answer is that the triangles are neither congruent, nor similar.
#### Example 4
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
Yes
Then, identify the triangles.
Similar
The answer is that the triangles are similar.
#### Example 5
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
Yes
Next, check to see if the triangles are the same shape.
Yes
Then, identify the triangles.
Congruent
The answer is that the triangles are congruent.
### Review
Identify the given triangles as visually similar, congruent or neither.
Answer each of the following questions.
1. Triangles and are congruent. Does this mean that their angle measures are the same? Why?
2. True or false. If triangles and are similar, then the side lengths are different but the angle measures are the same.
3. True or false. Similar figures have exactly the same size and shape.
4. True or false. Congruent figures are exactly the same in every way.
5. Triangles and are similar. If this is true, then the side lengths are the same, true or false.
6. True or false. To figure out if two figures are similar, then their side lengths form a proportion.
7. Define similar figures
8. Define congruent figures.
9. Use a ruler to draw a congruent pair of triangles.
10. Use a ruler to draw a pair of triangles that is similar.
To see the Review answers, open this PDF file and look for section 9.15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Congruent Congruent figures are identical in size, shape and measure.
Similar Two figures are similar if they have the same shape, but not necessarily the same size. |
Counting
2018-05-30
this post is still a draft as I'm working on the examples design and ux
We don't often think (or know) about it, but our everyday use of numbers is only one representation of values amongst many.
If I ask you for 12 dollars, you'll understand what quantity twelve represents and how 1 and 2 work together to represent 12.
This is because we learnt to use the digits 0 to 9 and combine them to form numbers.
We could decide to use only 0 and 1 instead, in which case twelve would be represented by 1100. You've probably heard of this system - it's called binary.
Just like we use different alphabets, words and languages to communicate ideas, there is an infinite amount of digits, systems and ways to represent values.
I just mentioned the decimal and binary systems. You probably heard of hexadecimal, often used for html color codes. You can even build your own.
All those systems work the same way and you can go from one to another. Let's see how all this works!
Here you can have some fun sliding the numbers and see how binary, decimal and hexadecimal relate to each others!
{{ formatCount }}
2 3 10 16
Decimal
Since we are used to count using decimals, let's start with it. This system uses 10 digits, from 0 to 9. We call it its base. So decimal is base-10.
Counting until 9 is pretty straight forward, since it uses one single digit. You probably know how to count beyond 9, and how to write 10, 11, 123, 456, etc. But how does it actually work?
Imagine that we have a 3-digit rolodex, [0][0][0].
Start from the right-most roll and switch the different ones until you reach the number you want to represent.
So to represent 3, you'd turn the right-most roll from 0 to 3, ending with [0][0][3].
To represent 10, you'd turn the right-most roll from 0 to 9. Then as you turn it once more, it would get back to 0, at which point you'd add one turn to the roll on its left, switching it from 0 to 1, ending with [0][1][0] would be shown. Continuing like that allows you to go up to [9][9][9] at which point our rolodex would need additional rolls. You get the point!
Now that we can write numbers, how do we read them?
Looking at 456, someone who doesn't know the decimal system might think that the value represented is 4 + 5 + 6 = 15. How is 1 representing 100? and 2, 20?
If you look at digit positions from right to left, and starting from 0 (like arrays), we can count the value from 456 the following way:
• 6 is in position 0: 6 * 100 = 6
• 5 is in position 1: 5 * 101 = 50
• 4 is in position 2: 4 * 102 = 400
• 400 + 50 + 6 = 456
Binary
Now that we got the basics for decimals, let's see how we can apply the same things with only two digits, 0 and 1, using the binary system.
Let's try to use our three digits rolodex and try to write five in binary.
We start with [0][0][0] and roll once. We get [0][0][1]. Then we add one and have [0][1][0]. We can see we are moving positions much faster, but it works exactly the same way as decimals.
If we do this 5 times, we'll end up with [1][0][1].
Let's now try to read binary.
We have 101, how do we count this?
• 1 is in position 0: 1 * 2*0 = 1
• 0 is in position 1: 0 * 21 = 0
• 1 is in position 2: 1 * 22 = 4
• 1 + 0 + 4 = 5
Counting any base
As we've seen with decimals and binary, we were able to use the same formula to count. We get the sum of each `d * b^p` with `d` a digit, `b` the base and `p` the digit's position.
Here is a table showing a preview of the value of each position. Multiply it by the actual digit in it.
3 2 1 0
Binary 8 4 2 1
Decimal 1000 100 10 1
Hexa 4096 256 16 1
Let's use this table to write a few numbers in hexadecimal, since we haven't worked with this system yet!
3 2 1 0
4096 256 16 1
fifteen 0 0 0 F
one hundred sixty 0 0 A 0
seven thousand fifty eight 1 B F 2
An example with web colors
We started with an example showing the relation between binary, decimal and hexadecimal. Let's see another example here, helping you converting colors between hexa and decimal.
Hopefully now you'll understand a bit better how they relate!
{{ red }}, {{ green }}, {{ blue }}
#{{ hexa.red }}{{ hexa.green }}{{ hexa.blue }} |
### Solving Problems Using the Principle of `Awl
Posted on Updated on
In this post, insha’Allah, we will learn how to apply `awl by solving example problems.
It should be understood that `awl can only be applied when the sum of the shares is greater than 1. When the sum of the shares is greater than 1, this will result in the sum of the portions being greater than the base number.
The application of `awl is actually very simple: As discussed in the previous post, after calculating the base number and the resulting portions we find that the sum of the portions is actually greater than the base number. To apply `awl, all we have to do is make this sum our new base number, while keeping the portions the same. This will result in reducing the shares of the categories proportionally.
Lets demonstrate with a few examples, insha’Allah.
Example 1: A man dies leaving behind
• 1 Wife
• Father
• Mother
• 2 Daughters
All of them inherit. Note that this is the exact same problem which was presented in the previous post. This time, insha’Allah, we will apply `awl and solve it.
For now, we do everything as usual: Make the chart, list the shares, calculate the base number and the portions. We end up with the following:
Notice that the share of the father is “1/6 + Residue”, as the father (and paternal grandfather, or higher) are both zaawil furoodh and `asabaat simultaneously and therefore also receive the residue (if there is any) along with their set share of 1/6. However, we already know that there will be no residue in this case. The shares already add up to more than 1, so the question of any residue being left over does not even arise.
Now, as explained, we apply `awl by adding up all the portions. The sum of all portions gives us our new base number.
3 + 4 + 4 + 16 ＝ 27 <—— New base number.
Now all we have to do is add another row to our chart where we write our new base number.
Note that there is no “New Portions” row, because the portions of each heir stay the same. When applying `awl, its only the base number that will change.
Final Answer: The estate will be divided into 27 equal portions:
The Wife gets 3 portions.
The Father gets 4 portions.
The Mother gets 4 portions.
Each Daughter gets 8 portions.
This problem we just solved is actually a famous case of inheritance called Mimbariyya for the reason that Caliph `Ali solved it while delivering a sermon on the mimbar (i.e. pulpit) in a mosque at Kufa, in present day Iraq. He was asked what the wife’s share will be if the surviving heirs of a deceased are wife, both parents and 2 daughters. There and then, he answered, “The wife’s 1/8 becomes 1/9.”
Example 2: A woman dies leaving behind
• Husband
• 2 Allaati sisters
• 1 Paternal aunt
The paternal aunt is a non-heir. Only the husband and allaati sisters inherit.
Our initial chart is the following:
Again, shares add up to greater than 1, so we immediately know this is an `awl problem. As per the rule, the sum of the portions becomes our new base number.
3 + 4 ＝ 7 <——- New base number.
Final chart:
Final Answer: The estate will be divided into 7 equal portions:
Each Allaati sister receives 2 portions.
Example 3: A woman dies leaving behind
• Husband
• 3 Granddaughters
• 1 Great-Granddaughter
• Mother
• Paternal grandmother
• 2 Akhyaafi brothers
• 1 Akhyaafi sister
• 2 Haqeeqi sisters
• 3 Nieces
• 1 Daughter of haqeeqi paternal uncle
There’s a lot going on here, so lets break it down insha’Allah:
Firstly, the daughter of haqeeqi paternal uncle (i.e. female cousin) and the 3 nieces are non-heirs.
The great-granddaughter is excluded by the 3 granddaughters. This is the application of Rule #6 of hujub hirmaan, except on lower generations. The paternal grandmother is excluded by the mother; the akhyaafi siblings are excluded by the granddaughters.
So we are left with this short list:
• Husband
• 3 Granddaughters
• Mother
• 2 Haqeeqi sisters
This gives us our initial chart:
The 2 haqeeqi sisters are `asabaat due to the presence of the granddaughters. See the post How Some Zaawil Furoodh Relatives Become `Asabaat. However, they will not be receiving anything in this case because there is never any residue left over in `awl problems. As explained in example 1 of this post, the shares add up to more than 1 so the question of residue being left over does not even arise. In fact, just to illustrate this point, watch what happens when we try to calculate the sisters’ portions:
12 – 3 – 8 – 2 ＝ -1
We end up with a negative number. Obviously an heir cannot have negative portions, so it follows that the sisters will not be receiving anything.
Before we apply `awl, we must first resolve the granddaughters category. 8 portions cannot be divided evenly among 3 heads. Once we resolve this category (by calculating a new base number and new portions), then we can apply `awl.
By now, you should be familiar with all the rules for adjusting the base number. If not, please take a second to review them in the post Summary of All Rules for Adjusting the Base Number.
3 and 8 are tabayun. Therefore, we are to multiply 3 by the current base number to get our new base number. We find that our new base number is 36. Now that we know our new base number, we can calculate the new portions and finally apply `awl.
See the chart below:
We used the adjusted base number (36) to calculate our new portions, which resolved the granddaughters category. After this we applied `awl by adding up all the portions, which gave us our final base number of 39.
9 + 24 + 6 ＝ 39
So essentially, we went through three base numbers. Our original base number (12), which we then adjusted to get our new base number (36). We then applied `awl to get the final base number (39).
Note: `Awl should always be applied after resolving all categories. This is the easier way to solve these types of problems.
Final Answer: The estate will be divided into 39 equal portions:
The Husband gets 9 portions.
Each Granddaughter gets 8 portions.
The Mother gets 6 portions.
The Haqeeqi sisters are deprived.
Example 4: A woman dies leaving behind
• 1 Haqeeqi sister
• 5 Allaati sisters
• 3 Akhyaafi brothers
• 1 Akhyaafi sister
• Paternal grandmother
• Maternal grandmother
All will inherit. Below is our starting chart:
For those who may be confused, the allaati sisters get 1/6 due to the presence of the haqeeqi sister. There share and the share of the haqeeqi sister has to sum up to 2/3. This is parallel to the rule of daughter and granddaughter(s) inheriting together. See item #8, rule #3 in the post Inheritance of Siblings.
Before applying `awl, there are 3 categories that need resolving: The allaati sisters, the akhyaafi siblings, and the grandmothers.
You should be familiar with how to resolve three categories, but nonetheless, we will do a quick walk-through.
Let us resolve the akhyaafi siblings and grandmothers categories first.
The heads-portions of the grandmothers category are tabayun.
The heads-portions of the akhyaafi siblings category are tawafuq. The wafq of their heads is 2, which is tamathul to the heads of the grandmothers category.
X ＝ 2.
The heads-portions of the remaining category – the allaati sisters – is tabayun. Therefore Y ＝ number of heads ＝ 5.
Now we resolve X and Y.
X and Y are tabayun.
5 x 2 ＝ 10 ——-> 10 x 6 ＝ 60 <——- New base number.
As shown above, after calculating the new portions based on the adjusted base number, we applied `awl to get our final base number:
30 + 10 + 20 + 10 ＝ 70
Final Answer: The estate will be divided into 70 equal portions:
The Haqeeqi sister will get 30 portions.
Each Allaati sister will get 2 portions.
Each Akhyaafi sibling will get 5 portions.
Paternal grandmother will get 5 portions.
Maternal grandmother will get 5 portions.
And now, we leave you with a final example to do on your own. Its solution will be given in the next post, insha’Allah.
A woman dies leaving behind:
• Husband
• 3 Haqeeqi sisters
• 2 Allaati sisters
• 3 Akhyaafi brothers
• Mother
• Maternal grandfather
How will the estate be divided and distributed? |
# How do you multiply sqrt2(3sqrt14 - sqrt7)?
Mar 27, 2018
$6 \sqrt{7} - \sqrt{14}$
#### Explanation:
Multiply $\sqrt{2}$ by each term in the parentheses
$\sqrt{2} \left(3 \sqrt{14} - \sqrt{7}\right)$
$\sqrt{2} \cdot 3 \sqrt{14} - \sqrt{2} \cdot \sqrt{7}$
$3 \sqrt{28} - \sqrt{14} \rightarrow 3 \sqrt{28}$ can be simplified
$3 \sqrt{28} = 3 \sqrt{4} \cdot \sqrt{7} \rightarrow$ $4$ is a perfect square
$6 \cdot \sqrt{7} = 6 \sqrt{7} = 3 \sqrt{28}$
$6 \sqrt{7} - \sqrt{14}$ |
Chapter 2:
Transformations and Symmetry
The word "transformation" refers to a movement or change in a geometric shape. If we slide a geometric shape to the right or left, this is called a "translation". In the example below, the original triangle is triangle ABC, and the second triangle A'B'C' is a translation of the original triangle. In moving the first triangle two inches to the right, we say that we translated triangle ABC two inches horizontally to create triangle A'B'C'.
In the second example of translation below, triangle DEF has been translated one inch vertically to create triangle D'E'F':
The second kind of transformation is called a "reflection". When you look in a mirror, you see a reflection of yourself, but left and right are reversed. As you see in the diagram below, when we reflect triangle ABC, the reflected triangle A'B'C' is "flipped" or reversed. Line "m" is the mirror.
A third kind of transformation is called a "rotation". If you want to rotate an object, you must first choose the Center of Rotation. This point is the point that you will rotate the object around, just as the center of a wheel is the center about which the wheel rotates. In the diagram below, triangle DEF has been rotated 90 degrees counterclockwise to its new position, triangle D'E'F'. P is the center of rotation.
In the diagram below, a triangle has been rotated 6 times about point P, at 60 degrees each time. Since 6 times 60 is 360 degrees, we see that this completes a full circle:
The fourth transformation is called "dilation". When the eye doctor "dilates your eyes" during an eye examination, the pupils of your eyes become enlarged so that the doctor can see the veins of your eyes more clearly. Dilation, mathematically, means to take a shape and make it larger or smaller but keep it the same proportion. In the geometric example below, the triangle on the left is "dilated" by a factor of 2 (it is "doubled") to create the dilated triangle on the right. Each side of the new triangle is twice as long as the corresponding side of the original triangle. Point "P" is the center of dilation. We would therefore say that the "scale factor is 1 to 2" comparing the original to the new figure. Another way to compare the sizes of the figures is to say that each side of the smaller triangle is one-half the size of the corresponding side of the larger figure.
What do you think is true of the angles of the new triangle compared to the angles of the original triangle? If you answered "they are the same", or "the new angles are the same measure, or number of degrees, as the original triangle" then you are correct! Since the angles are the same, the two triangles are the same shape.
http://mathforum.com/sum95/suzanne/symsusan.html
Test Question #1: What is special about the word ATOYOTA, as explained on the web site linked above?
To create an interesting geometric, you can combine the two or three of the transformations (translation, reflection, rotation) and use a variety of shapes and colors as in the example below. Which of the transformations were used?
If you answered "reflection and rotation", then you are correct!
Translation and dilation were used to construct the "frieze pattern" below. A "frieze pattern" is a design made from horizontal or vertical repetitions of the same shape.
Frieze patterns are often used in fabric designs for clothing, draperies, and wallpaper, and even in architecture, such as the example below, a Tile Frieze from the Palacio de Velazquez in Madrid, Spain
(photo from http://www.fandm.edu/Departments/Mathematics/a.crannell/hm/Math/seven.html)
You will find a mathematical discussion about these and other transformations at the following web page, on the Math Forum website. The Math Forum is a fascinating web site, with thousands of pages about math.
http://forum.swarthmore.edu/sum95/suzanne/symsusan.html
Test Question #2: What type of transformation is discussed on the Math Forum web page that was NOT discussed on this MathArt page, what is the definition of that type of transformation, and how does it differ from the 3 types of transformation discussed in MathArt?
Symmetry
We say that an object has symmetry, or is symmetrical, if the two sides are "of similar measure". Many animals and insects are symmetrical; a beautiful example of this is the Butterfly below.
We say that a butterfly, and the human body, are symmetrical because the left and right sides appear to be very much alike. This type of symmetry is called "reflection symmetry" because the left side, if it were reflected across the middle line, would match the right side. You can see this general symmetry in the following famous drawing by Leonardo Da Vinci. In this drawing, Da Vinci illustrates the form with two positions of both the arms and legs, depicting the symmetry and proportion of the human figure.
Leonardo Da Vinci is often called the "Renaissance Man", because he combined imagination, curiosity and creativity. He was a scientist, an inventor, and an artist. His most famous painting is The Mona Lisa. The web page below will give you an introduction to the artistic side of Leonardo:
http://www.mos.org/leonardo/artist.html
I would like you to read the following web page from the Leonardo web site, but you do not have to do any of the activities (although you will probably find them intriguing):
http://www.mos.org/sln/Leonardo/LeonardoRighttoLeft.html
You will find some interesting information about symmetry at the following web page, on the Da Vinci web site:
http://www.fi.edu/qa97/me3/
Now here's a challenge for you. Using the method explained in the web page above, "de-code" the following question and answer:
(photo from http://members.aol.com/alohaquilt/index.html)
Many pieces of art work are symmetrical, and so are many graphic designs. Some of the most beautiful examples of symmetry can be found in Hawaiian quilts, such as the ones above. You will find out how to create your own geometric quilt pattern at the web page below:
MACsw.html
You will find out more information about symmetry by reading the one page linked below, from "The Geometry Pages". You do not have to go to any other pages besides this one page; please come back to the MathArt pages when you have carefully read the page below. Click on this link now:
http://mathforum.org/~sanders/geometry/
Test Question #4: Name the two types of symmetry, and give an example (from the above web page) for each type.
From earliest recorded history, we find examples of artists and architects designing their paintings and buildings with a strong emphasis on symmetry. For example, most buildings built before the 20th century had very strong symmetry: the left side of the structure is exactly the same as the right side, as you can see in the images below:
(Photos from http://www.GreatBuildings.com/)
From left to right:
Torii of Itsukushima on Miyajima Island in Japan 10th century AD, Norwegian storehouse 13th century AD, Piazza of St. Peter's in Rome 17th century, S. Maria Novella Church in Florence Italy 15th century AD, The Taj Mahal in Agra India 17th century, United States Capitol in Washington D.C. 19th century.
The following two images are from http://homer.reed.edu/Parthenon.html and the text is from http://www.johnco.cc.ks.us/~jjackson/part.html
"The supreme achievement of the Greeks was their claim that everything--from the human body to the entire cosmos--was governed by an order accessible to human reason. That claim has been the basis for western civilization ever since. The Temple of Athena Parthenos, the Parthenon, now a ruin lying atop the Acropolis of Athens, is the testament in stone to that claim."
And this " order accessible to human reason" is mathematical symmetry! The Parthenon is one of the classical examples of ancient Greek civilization, built in 447 BC. The structure, with beautiful proportions and perfect symmetry, is also an example of the importance of symmetry and proportion, and therefore mathematics, in architecture.
The floor plan of the Parthenon below shows the very strong order and symmetry, and emphasis on geometric shapes that is found in most architecture:
You will find further information at the following web page, called "The Golden Section in Architecture".
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibInArt.html#arch
This page is rich with information on both mathematics and architecture and you should spend quite a lot of time reading and looking through it. As you scroll down you will also find "The Golden Section in Art", which you should also read. I would like you to go to some of the links from this page, and read what you find there. You will find many good ideas for your project!
So what is the "Golden Section"? You will find the answer to this question, and some good project ideas, at the following web page:
http://mathforum.org/~sanders/geometry/GP21MoreCurves.html
As you have seen, many architects design their buildings so that they are symmetrical, and this has been true throughout history. You will also find many examples of artistic use of symmetry in much of the ornamental designs on the walls of many buildings. One very good example is the Alhambra Palace, in Spain. You will find some interesting information about this beautiful building, and examples of the symmetry in the ornamental tiles of the Alhambra by clicking on the link below:
http://en.wikipedia.org/wiki/Alhambra
The transformation called "dilation" produces figures that are the same shape, but not necessarily the same size, as explained earlier on this page. In the example above, the original figure was one-half the size of the new figure. This meant that each side of the original triangle was one-half as long as the corresponding ("matching") side of the "dilated" triangle.
This mathematical concept is used extensively in building model airplanes, dollhouses, and in architectural design as well. Certainly an architect cannot possible draw a building at its full size - it would take an enormous sheet of paper! So an architect (or an engineer, or an airplane designer, etc.) draws the building "to scale". The phrase "scale drawing" refers to a drawing that is smaller (or occasionally larger) than the real thing. The floor plan of the house shown below has been drawn at to scale.
The original scale at which this floor plan was created was 1/4 inch represents 1 foot, which means that every 1/4" inch on the drawing represents 1 foot in the real house. (Note: to fit the drawing onto a web page, it has been reproduced, so is not measurable as you see it now.)
Now, some mathematics: if 1/4 inch represents 1 foot, then it would also be true that 1/4 inch represents 12 inches. We can simplify this equation by multiplying both sides by 4 as shown below:
Therefore, the ratio is 1 to 48, and the actual drawing was one forty-eighth the size of the real house.
Here's another example: The length of a stadium is 200 yards and its width is 120 yards. If 1 inch represents 10 yards, what would be the dimensions of the stadium drawn on a sheet of paper?
Answer: 20 inches by 12 inches.
http://regentsprep.org/Regents/math/scale/Lscale.htm
In searching on in the internet for information about architects and architecture, I found the following website. It is fascinating! If you are interested in possibly becoming an architect, please visit this website.
http://www.abc.net.au/architecture/
As you have seen, architects, artists, and designers use the mathematical concepts of transformations - translation, reflection, rotation and dilation - in creating their work. You, too, can create beautiful designs using the mathematical concepts of transformations and symmetry. In your project for this chapter, you will design a beautiful geometric graphic. You will find out more about how to create these geometric graphics at the following web page:
http://mathforum.org/~sanders/geometry/GP05Constructions.html
Chapter 2 Project
The project for this chapter is to create an artistic symmetrical design, using geometric shapes and transformations. First do some research on the Internet, and look at geometric artwork (net search: geometry art OR geometric art). Read some information about the geometric artwork that you find, and explore any aspects that interest you. Then design and create a symmetrical geometric design yourself. If you would like to see some examples of graphics created by students like yourselves, please go to the Student Work section from the Table of Contents.
• You must use only geometric shapes in your design (triangles, pentagons, hexagons, circles, half-circles etc.)
• Create the graphic using compass, ruler, and tracing paper, in pencil on unlined white paper. If you are using pencil and compass and ruler, draw a geometric figure (an equilateral triangle, for example) then trace it onto a small piece of tracing paper. Draw a point on the original paper, and place the tracing on top so that it lines up with the original triangle. Put the point of your compass at the center, and rotate the top sheet about this point 30 degrees, 45 degrees, or 60 degrees (or any number that divides evenly into 360 degrees).For angle measures, you will need to use a protractor - or you will need to construct the angles you need as follows: construct a hexagon with compass and that construction will give you a circle divided into 6 equal parts, each of which is 60 degrees. Constructing a square and drawing both diagonals will give you 45 degree angles at each corner, etc.). If you are using The Geometer's Sketchpad, follow the instructions on the GSP pages - you will find a link to those pages in the Introduction to MathArt Connections.
• Make your design more interesting by constructing a second geometric shape (a square or diamond, for example) and then use tracing paper to rotate and/or reflect copies of the shape, around that same central point, OR use The Geometer's Sketchpad.
• When you are satisfied with your design, trace it in ink (or go over the pencil lines in ink and then erase those that still show). Then color it using colored pens or pencils OR if you are using The Geometer's Sketchpad, hide all the points, print it out and then color it with colored pens or pencils.
• If you use pencil, paper and compass, lay out your design with light pencil lines, then complete the graphic using compass and ruler, in pen, then color it using colored pens
• If you use Sketchpad, hide all the points before you print it. (Click on one point. Then use the Edit menu to Select All Points. Then use the Display menu to Hide the points.)
• For both of these options, complete your geometric graphic using a black pen for the outlines, and colored pens or colored pencils for shading.
• Write at least 30 words about the transformations that you used, the kind of symmetry that your design has, and the names of the geometric figures in your design (for example: "I drew three connected lines on the left side of the rectangle and translated the lines to the right. Then I drew a half-circle and a triangle on the top of the rectangle ..." etc.)
I certainly hope you have fun doing this project!
Go To Chapter 3: Tessellations |
## Which statement is true about the equation (x – 4)(x + 2) = 16? The equation x – 4 = 16 can be used to solve for a solution of the given e
Question
Which statement is true about the equation (x – 4)(x + 2) = 16? The equation x – 4 = 16 can be used to solve for a solution of the given equation. The standard form of the equation is x2 – 2x – 8 = 0. The factored form of the equation is (x + 4)(x – 6) = 0. One solution of the equation is x = –6.
in progress 0
2 weeks 2021-09-10T22:47:05+00:00 2 Answers 0
Factored form…
Step-by-step explanation:
Foil out (x-4)(x+2)=16
First: x*x=
Outer: 2*x=2x
Inner: -4*x = -4x
Last: -4*2 = -8
Combine them all:
Simplify:
What multiplies together to make -24 but adds together to make -2?
Lets list the factors of -24 to decide:
1 x -24
2 x -12
3 x -8
4 x -6
-6+4 = -2
Therefore…
3rd statement
Step-by-step explanation:
Lets go through the choices and see.
The first one says:
The equation x – 4 = 16 can be used to solve for a solution of the given equation.
If we solve this we get x=20. I just added 4 on both sides.
Is 20 a solution thr original equation? Let’s check. We need to replace x with 20 in
(x – 4)(x + 2) = 16 to check.
(20-4)(20+2)=16
(16)(22)=16
16 times 22 is definitely not equal to 16 so the first statement is false.
Lets check option 2:
The standard form of the equation is
x2 – 2x – 8 = 0.
So lets put our equation in standard form and see:
(x – 4)(x + 2) = 16
Foil is what we will use:
First: x(x)=x^2
Inner: (-4)x=-4x
Outer: x(2)=2x
Last: -4(2)=-8
Add together to get: x^2-2x-8. We still have the equal 16 part.
So the equation is now x^2-2x-8=16. Subtracting 16 on both sides will put the equation in standard form. This gives us
x^2-2x-24=0. This is not the same as the standard form suggested by option 2 in our choices.
Checking option 3:
This says:
The factored form of the equation is
(x + 4)(x – 6) = 0.
So we already put our original equation in standard form. Lets factor our standard form and see if is the same as option 3 suggests.
To factor x^2-2x-24, we need to find two numbers that multiply to be -24 and add to be -2. These numbers are 4 and -6 because 4(-6)=-24 and 4+(-6)=-2. So the factored form of our equation is (x+4)(x-6)=0 which is what option 3 says. So option 3 is true.
Let’s go ahead and check option 4: It says: One solution of the equation is x = –6. This is false because solving (x+4)(x-6)=0 gives us the solutions x=-4 and x=6. Neither one of those is -6. *
* I solved (x+4)(x-6)=0 by setting both factors equal to zero and solving them for x. Like so,
x+4=0 or x-6=0. |
## Success Criteria: Learning Goals
• To Complete the Square in order to find the maximum/minimum value of the quadratic and its vertex
• To use the Quadratic Formula in order to find the x-intercepts along with the vertex of a quadratic function
• To find how much x-intercepts a quadratic function has by finding the Discriminant
## Summary
• Standard Form Equation: y = ax² + bx + c
• a value gives you the shape and direction of opening of the quadratic
• c value gives you the y-intercept of the quadratic
• To get x-intercepts, SOLVE using the "Quadratic Formula"
• Quadratic Formula: x =[ -b + √(b² - 4ac)] / 2a *on picture*
• To identify if the quadratic has a maximum or minimum value, complete the square to transform standard form equation in vertex form
• Vertex Form Equation: y = a(x - h)² + k
• To identify how much x-intercepts a quadratic has, use the discriminant formula
• Discriminant Formula : x = b² - 4ac
## Completing the Square
• Involves transforming a standard form equation into vertex form in order to get the vertex and maximum or minimum value of the quadratic
Below are steps in order of how to change a standard form equation into vertex form
Ex. 1
y = x² - 8x + 10
y = (x² - 8x )+ 10 Step One: Isolate ax² and bx terms, also (-8/2)² basically (b/2)²
y = (x² - 8x + 16 - 16 ) + 10 Step Two: Add and subtract the value from above
y = (x² - 8x + 16) + 10 - 16 Step 3: Bring the -16 outside the bracket
y = (√ x² - 8x + √ 16) - 6 Step 4: Square root both ax² and c terms)
y = (x - 4)² Step Five: Write as a perfect square.
Ex. 2
y = 2x² + 20x - 8
y = 2(x² + 10x) - 8 *If there is a coefficient, factor it out (applies to ax² and bx)
y = 2(x² + 10x + 25 - 25) - 8 *Remember, multiply coefficient with subtracted value (2)(-25)
y = 2(√ x² + 10x + √ 25) - 50 - 8
y = 2(x + 5)² - 58
Ex. 3
y = -x² + 10x + 20
y = -(x² - 10x) + 20 *Even if there is a negative sign, factor it out*
y = -(x² - 10x +25 - 25) + 20
y = -(√ x² - 10x + √ 25) + 20 + 25
y = -(x - 5)² + 45
*The video below will help further your understanding of Completing the Square*
❖ Completing the Square - Solving Quadratic Equations ❖
• Using the quadratic formula is a direct way of calculating roots (also known as zeroes, the solution or the x-intercepts)
• A quadratic equation can have 2, 1 or 0 solutions
• If the number under the square root is negative, the quadratic has no solution
• Quadratic Formula: x =[ -b + √(b² - 4ac)] / 2a
Ex. 1
9x² - 24x + 16 = 0
x =[ -b + √(b² - 4ac)] / 2a
x =[ -(-24) + √((-24)² - 4(9)(16))] / 2(9)
x =[ 24 + √(576 - 576)] / 18
x =[ 24 + √0)] / 18
x = (24 + 0) / 18
x = 24 / 18
x = 1.3 or 4/3
(1.3, 0) or (4/3, 0)
Ex. 2
2x² + 4x = -1
2x² + 4x - 1 = 0
x =[ -b + √(b² - 4ac)] / 2a
x =[ -4 + √((4)² - 4(2)(1))] / 2(2)
x =[ -4 + √(16 - 8)] / 4
x =[ -4 + √8] / 4
x = (-4 + 2.8) / 4
x = -1.2/4
x = -0.3
x = (-4 - 2.8) / 4
x = -6.8/4
x = -1.7
(-0.3, 0) (-1.7, 0)
*MY OWN VIDEO BELOW will help further your understanding of using the Quadratic Formula*
## Graphing Standard Form
In order to graph a standard form equation follow these steps:
• Step One: Find the x-intercepts using "The Quadratic Formula"
• Step 2: Find the vertex (Use AOS = (r + s)/2 to find x-value and sub in x-value into original equation to find y-value of vertex
• Plot the vertex along with the two x-intercepts to form parabola
Ex. 1
y = x² + 2x - 3
Step One: x-intercepts
x =[ -b + √(b² - 4ac)] / 2a
x =[ -2 + √((2)² - 4(1)(-3))] / 2(1)
x =[ -2 + √(4 + 12)] / 2
x =[ -2 + √(16)] / 2
x = (-2 + 4) / 2
x = 2/2
x = 1
x = (-2 - 4) / 2
x = -6/2
x = -3
x- intercepts = (1, 0) (-3, 0)
Step Two: Vertex
AOS = (r + s)/2 = (-3 + 1) / = -2/2 = -1
y = x² + 2x - 3
y = (-1)² + 2(-1) - 3
y = 1 - 2 - 3
y = - 4
Vertex (-1, -4)
Step 3: Plots x-Intercepts and Vertex on Graph *graph on side*
## The Discriminant
• The Discriminant Formula is used to identify the number of x-intercepts in a quadratic function
• Given that a, b and c are rational numbers, they are needed to determine the Discriminant value.
• Discriminant Formula = b² - 4ac
The following will tell you how to determine how much x - intercepts a quadratic function has:
• D>0 = 2 Solutions - If D is greater than 0 the quadratic will have two x-intercepts.
• D<0 = 0 Solutions - If D is less than 0 the quadratic will have no x-intercepts.
• D = 0 = 1 Solution - If D is equal to 0, the quadratic will have 1 x-intercept.
Ex. 1
y = 2x² + 2x - 3
D = b² - 4ac
= (2)² - 4(2)(-3)
= 4 + 24
= 28
Therefore, since 28 is greater than O, this quadratic will have two x-intercepts.
Ex. 2
y = x² - 2x + 5
D = b² - 4ac
= (-2)² - 4(1)(5)
= 4-20
= -16
Therefore, since -16 is less than 0, this means the quadratic will have no x-intercepts.
Ex. 3
y = x² + 2x + 1
D = b² - 4ac
= (2)² - 4(1)(1)
= 4 - 4
= 0
Therefore, since D = 0, this means the quadratic will have only 1 x-intercept.
## Word Problem #1 - Complete the Square
Sherri sells photos of athletes to baseball, basketball and hockey fans after their games. Her regular price is \$10 per photo, and she usually sells about 30 photographs. Sherri finds that, for each \$0.50 reduction in price, she can sell two more photographs.
a) Write an equation to represent Sherri's total sales revenue.
[Hint: R = (price)(quantity)]
Let x represent the number of price decreases.
EQUATION:
R = (10 - 0.50x)(30 + 2x)
b) What price must Sherri charge in order to maximize her revenue.
R = (10 - 0.50x)(30 + 2x)
= 300 + 20x - 15x - x²
= - x² + 5x + 300
= - (x² - 5x) + 300
= - (x² - 5x + 6.25 - 6.25) + 300
= - (x² - 5x + 6.25) + 300 + 6.25
= - (√x² - 5x + √6.25) + 306.25
= - (√x² - 2.5)² + 306.25
Vertex (2.5, 206.25)
Price = 10 - 0.50x
= 10 - 0.50(2.5)
= 10 - 1.25
= 8.75
Therefore, to maximize her revenue, Sherri must charge \$8.75 per photo.
## Word Problem #2 - The Quadratic Formula
The product of two consecutive numbers is 3306. What are the numbers?
Let x represent the first number.
Let x + 1 represent the second number.
x(x + 1) = 3306
x² + x = 3306
x² + x - 3306 = 0
x = [-b + √(b² - 4ac)] / 2a
x = [-1 + √((1)² - 4(1)(- 3306))] / 2(1)
x = [-1 + √( 1 + 13224)] / 2
x = [-1 + √13225] / 2
x = (-1 +115) / 2
x = 114 / 2
x = 57
x = (-1 -115) / 2
x = -116 / 2
x = -58
x = 57
x + 1 = 57 + 1 = 58
Check: (57)(58) = 3306
x = -58
x + 1 = -58 + 1 = -57
Check: (-58)(-57) = 3306
Therefore, the numbers are 57 and 58 or -57 and -58.
## Reflection
• Overall the Quadratics Unit has been a fairly interesting unit to learn about where it connects to a whole lot of applications throughout the world
• With the exception of a few silly mistakes, I was able to exceed throughout this whole unit in all the 3 components and understand as well as apply the key concepts of the quadratics unit
Connections:
1. Vertex Form Connects to Standard Form
• Standard Form can be obtained by expanding Vertex Form
• y = a(x – h)² + k -------> y = ax² + bx + c = 0
• Vertex Form can be obtained by "Completing the Square" from Standard Form
• y = ax² + bx + c = 0 -------> y = a(x – h)² + k
2. Vertex Form Connects to Graphing
• Once your quadratic function is put into vertex form you are able to identify the vertex of the function and plot it on your graph
• Once vertex is plotted, use the step pattern (a value included) to continue graphing from both sides
3. Standard Form Connects to Graphing
• In a standard form equation, the c value gives you the y-intercept, begin with plotting the y-intercept on the graph
• Transforming your Standard Form Equation into Vertex Form by Completing the Square, will give you your vertex as well as your a value
• Once vertex is plotted use the a value as well as the step pattern to continue to plot your quadratic
4. Factored Form Connects to Graphing
• In a factored form equation the two x-intercepts as well your a value is easily obtainable by factoring your equation. Plot your two x-intercepts
• Use x-intercepts to find the vertex of the equation (x-value - AOS, y - value subbing x into original equation
• Once vertex is plotted use a value along with the step pattern to graph quadratic
5. Factored Form Connects to Standard Form
• In order to find the vertex, this two forms involve similar methods involving finding the vertex
• In Factored Form: Once two x-intercepts are found by factoring, find the vertex by using AOS formula (x), and subbing x into the original equation for (y).
• In Standard Form: Once two x-intercepts are found by using the "Quadratic Formula", find the vertex by using AOS formula (x), and subbing x into the original equation for (y).
6. All Forms Connect to Each Other (Vertex, Factored, Standard)
• In order to find the y-intercept for all of these forms, when x is subbed in for 0 (x = 0) the y-intercept is easily obtainable in all forms
• The y intercept is written as (0, y) - for all forms
## Assessment
• This assessment strengthened my knowledge to understand that their are multiple ways to come to a solution
• In terms of finding the vertex, I originally did it by using the Quadratic Formula to find the x-intercepts then went along to find the vertex
• I ensured that I was right by Completing the Square to the the equation in Vertex Form to obtain the vertex of this quadratic relation
• This made me realize there's more than one way of doing a problem involving quadratics and doing it a second time and arriving at the same answer will ensure that your answer is correct |
# Geometry: Polygons
### Properties of Polygons
#### Diagonals
One property of all convex polygons has to do with the number of diagonals that it has:
Every convex polygon with n sides has n(n-3)/2 diagonals.
With this formula, if you are given either the number of diagonals or the number of sides, you can figure out the unknown quantity. Diagonals become useful in geometric proofs when you may need to draw in extra lines or segments, such as diagonals.
Figure %: Diagonals of polygons
The figure with 4 sides, above, has 2 diagonals, which accords to the formula, since 4(4-3)/2 = 2. The figure with 8 sides has twenty diagonals, since 8(8- 3)/2 = 20.
#### Interior Angles
The interior angles of polygons follow certain patterns based on the number of sides, too. First of all, a polygon with n sides has n vertices, and therefore has n interior angles. The sum of these interior angles is equal to 180(n-2) degrees. Knowing this, given all the interior angle measures but one, you can always figure out the measure of the unknown angle.
#### Exterior Angles
An exterior angle on a polygon is formed by extending one of the sides of the polygon outside of the polygon, thus creating an angle supplementary to the interior angle at that vertex. Because of the congruence of vertical angles, it doesn't matter which side is extended; the exterior angle will be the same.
The sum of the exterior angles of any polygon (remember only convex polygons are being discussed here) is 360 degrees. This is a result of the interior angles summing to 180(n-2) degrees and each exterior angle being, by definition, supplementary to its interior angle. Take, for example, a triangle with three vertices of 50 degrees, 70 degrees, and 60 degrees. The interior angles sum to 180 degrees, which equals 180(3-2). Because the exterior angles are supplementary to the interior angles, they measure, 130, 110, and 120 degrees, respectively. Summed, the exterior angles equal 360 degreEs.
A special rule exists for regular polygons: because they are equiangular, the exterior angles are also congruent, so the measure of any given exterior angle is 360/n degrees. As a result, the interior angles of a regular polygon are all equal to 180 degrees minus the measure of the exterior angle(s).
Notice that the definition of an exterior angle of a polygon differs from that of an exterior angle in a plane. A polygon's exterior angle is not equal to 360 degrees minus the measure of the interior angle. A polygon's interior and exterior angles at a given vertex don't span the entire plane, they only span half the plane. That is why they are supplementary--because their measures sum to 180 degrees instead of 360.
## Take a Study Break
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##### Common Core Math For Parents For Dummies with Videos Online
Sixth grade is the first time that Common Core math students start to solve problems using variables and the property of equivalence (specifically, the distributive property).
They typically use variables that closely match the values they represent, so s for side length, v for volume, and so on. They also notice that some expressions — even though they may look quite different — always have the same values as each other, and they explore this property, called equivalence.
## Solve problems using variables
Sixth graders are introduced to variables (letters that stand for numbers) for the first time. Mostly, they're concrete applications of variables in which students represent simple numerical or measurement relationships.
An example of using variables to represent numerical relationships might be expressing the number of wheels on x cars as 4x or an unknown number of boys in a class as b, an unknown number of girls in a class as g, and the number of students in the class as b + g.
An example of using variables to represent measurement relationships is showing the perimeter of a square as 4s and the area of a square as
In these two cases, s represents the side length of the square, and students show how to use the side length to compute the other measure — either perimeter or area.
Beginning the use of variables by expressing these relationships in familiar scenarios allows students to get a feeling for the way algebra shows the structure of a situation, whereas arithmetic only shows individual values.
## Solve problems using the property of equivalence
Students come to sixth grade knowing that there are many ways to write the same value. The number 459 can be written as 400 + 50 + 9, for example, or 460 – 1,
and so on. Similarly, the fractions
all have the same value. Students are accustomed to calling fractions equivalent when they have the same value.
This idea of same value is extended in sixth grade to include equivalent expressions. Two equivalent expressions have the same value whenever their variable has the same value. The expressions 4(x + 2) and 4x + 8 have the same value as each other for all values of x, so they're equivalent. The expressions x + 2 and 2x only have the same value when x is equal to 2, so they aren't equivalent.
In sixth grade, students use simple examples of the distributive property to write and to justify that two expressions are equivalent. The distributive property states that
If you think of
as 70 + 14, then you're using the distributive property because
The example of 4(x + 2) and 4x + 8 is an example of the distributive property. |
Math Calculators, Lessons and Formulas
It is time to solve your math problem
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# Prime factorization calculator
Calculator finds prime factorization of a given number and creates a factor tree. The calculator generates a step by step explanation on how the work has been done.
Prime Factorization Calculator
that shows a step by step explanation
show help ↓↓ examples ↓↓
Find prime factorization and factor tree
Find all factors
Find all factor pairs
Check if the number is prime or not
working...
examples
example 1:ex 1:
Create factor tree for 154 ?
example 2:ex 2:
Find prime factorization of 70 and create factor tree?
example 3:ex 3:
What are the prime factors of 18240 ?
example 4:ex 4:
Find all factor pairs of 548.
example 5:ex 5:
Is the number 1361 prime or composite?
### Finding prime factorization and factor tree
Example: Find prime factorization of 60.
Step 1: Start with any number that divides 60, in this we will use 10. So, $\color{blue}{60 = 6 \cdot 10}$.
• 60
• 10
• 6
Step 2: Write 10 as a product of 2 and 5. $\color{blue}{10 = 2 \cdot 5}$.
• 60
• 10
• 5
• 2
• 6
Step 3: Write 6 as a product of 2 and 3. $6 = 2 \cdot 3$.
• 60
• 10
• 5
• 2
• 6
• 2
• 3
Step 4: The end nodes are the prime factors of 60.
• 60
• 10
• 5
• 2
• 6
• 2
• 3
$$60 = 2 \cdot 2 \cdot 3 \cdot 5$$
### Finding all factors
Example: Find all factors of 54.
Step 1: Begin the list with 1 and end it with 54.
$$\color{blue}{\boxed{1}} ~ , ~ . ~ . ~ . ~ , ~ \color{blue}{\boxed{54}}$$
Step 2: Since $54 = 2 \cdot 27$ we put in 2 and 27 into an array.
$$1 ~ , ~ \color{blue}{\boxed{2}} ~ , ~ . ~ . ~ . ~ , ~ \color{blue}{\boxed{27}} ~ , ~ 54$$
Step 3: Since $54 = 3 \cdot 18$ so we will put in 3 and 18 into an array.
$$1 ~ , ~ 2, ~ \color{blue}{\boxed{3}} ~ , ~ . ~ . ~ . ~ , ~ \color{blue}{\boxed{18}} ~ , ~ 27 ~ , ~ 54$$
Step 4:At the end we have $54 = 6 \cdot 9$
$$1 ~ , ~ 2 ~ , ~ 3 ~ , ~ \color{blue}{\boxed{6}} ~ , ~ \color{blue}{\boxed{9}} ~ , ~ 18 ~ , ~ 27 ~ , ~ 54$$
### Check if the number is prime
Example: Check whether a number 581 is prime or not.
Step 1: Find the square root of 581.
$$\sqrt{581} = 24.1 \approx = 25$$
Step 2: Try to divide 581 by all prime numbers less or equal to 25.
In this example we will try to divide 581 with :
2 , 3, 5, 7, 11, 13, 17, 19 and 23.
\begin{aligned} 581 : 2 & = 290.5 \\ 581 : 3 & = 484 \\ 581 : 5 &= 116.2 \\ 581 : 7 &= 83 \end{aligned}
581 is divisible by 3 so is is not prime.
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# Prime Numbers and Euclids Proof
Prime Numbers and Euclids Proof
Prime numbers have been fascinating mathematicians for thousands of years.
A prime number is always greater than 1 and can only be divided by itself and 1 – no other number will divide into it. 2 is the first prime number (also the only even prime number), followed by 3, 5, 7, and so on.
Non-prime numbers are defined as composite numbers.
Prime numbers are distributed randomly throughout all the other numbers. Also, there is no simple and quick way to find a specific (new) prime number. As a result, large prime numbers are used in encryption to make the online platform protected for safe communication and financial transaction.
Did you know the number of primes is infinite? Explore Euclid’s Theorem of infinite primes with proof, solved examples and interactive questions, the Cuemath way.
## Lesson PLan
1 2 What is Euclid’s Proof for Infinite Primes? 3 Solved Examples on Infinite Prime Numbers 4 Interactive Questions on Infinite Primes 5 Important Notes on Infinite Prime Numbers
## How Do You Identify Prime Numbers?
### Definition
In numbers, we have learnt that prime numbers have two factors - 1 and the number itself. To find a prime number, we need to find the numbers which consist of only two factors. This is possible by using a simple method, which is called prime factorization.
### Steps to Finding Prime Numbers Using Factorization
Step 1. Divide the number into factors
Step 2. Check the number of factors of that number. If the number of factors is more than 2 then it is composite.
Example: $$8$$ has four factors $$1,\, 2,\, 4,\, 8$$. So 8 and therefore is not prime
Step 3. All prime numbers greater than 3 can be represented by the formula $$6n+ 1$$ and $$6n -1) for n greater than equal to 1. Exception if the number ends with 5 or higher multiples of 7 such as 49, 77, 91 Examples: \(5=6\times 1-1$$
$$7=6\times 1+1$$
$$11=6\times 2-1$$
$$13=6\times 2-1$$
etc.
### Steps to Identify A Large Number Is Prime Number
Lets us take a number $$1249$$
Step 1. If the number ends with $$0,\, 2,\, 4,\,5,\, 6\, and\, 8,$$ then it is not a prime number.
Step 2. Add the digits of your number if the number is divisible by $$3$$ then we can say that, it is not a prime number.
$$1249=1+2+4+9=16$$
Step 3. If the condition for steps 1 and 2 are not true, then find the square root of the number
$$\sqrt(1249)=35.3$$
Step 4. Divide the number by all prime numbers less than its square root value
Prime number less than $$35:\,\, 1,\,2,\,3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29,\,31$$
Step 5. If the number is divisible by any of the prime numbers less than its square root, it is not a prime number; otherwise, it is prime.
Hence, after following all steps we can say that $$1249$$ is a prime number
## What Is Euclid’s Proof for Infinite Primes?
Euclid, in 4th century B.C, pointed out that there has been an infinite number of primes. The concept of infinity is not known at that time. He said ”prime numbers are quite any fixed multitude of them” meaning if you imagine 100, there are more, and if you imagine a million there are still more.
### Euclid's Proof for an Infinite Number of Primes
let us assume that, there are a finite number of primes, $$n$$
the largest prime number is $$p_n$$.
Consider the number, $$N$$. that is the product of these primes number plus one
By construction, N is not divisible by any of the $$p_i$$.
Hence it's either prime itself, or divisible by another prime greater than $$p_n$$ , contradicting the idea.
For example:
$$2 + 1 = 3$$, is prime
$$2\times 3 + 1 = 7$$, is prime
$$2\times 3 \times 5 + 1 = 31$$, is prime
$$2\times 3 \times 5\times 7 + 1 = 211$$, is prime
$$2\times 3 \times 5\times 7\times 11 + 1 = 2311$$, is prime
etc.
Thus, we can conclude this the number of primes is infinite.
## Solved Examples
Example 1
Is $$233$$ a prime number?
Solution
If $$233$$ is a prime number, then it must not be divisible by a prime that is less than or equal to $$\sqrt{233}$$.$$\sqrt{213}$$ is between $$15$$ and $$16$$, so the largest prime number that is less than $$\sqrt{213}$$ is $$13$$. therefore we test $$2,\, 3,\, 5,\, 7,\, 11,\, and\, 13\,$$ for divisibility.
233 is not divisible by any of those numbers, so it must be prime
$$233$$ is a prime number
Example 2
<
Is $$1001$$ a prime number?
Solution
If $$1001$$ is a prime number, then it must not be divisible by a prime that is less than or equal to $$\sqrt{1001}$$.$$\sqrt{1001}$$ is between $$31$$ and $$32$$, so the largest prime number that is less than $$\sqrt{1001}$$ is $$31$$. It is therefore sufficient to test $$2,\, 3,\, 5,\, 7,\, 11,\,13,\,17,\,19,\, 23,\, 29,\, and\, 31$$ for divisibility.
$$1001=7 \times 11\times 13$$. $$1001$$ is divisible by $$7,\,11,\, and \, 13$$ so it is not a prime number
$$1001$$ is not a prime number
## Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Important Notes
• $$1$$ is the only whole number that we don't consider as prime nor composite.
• The lowest even prime number is $$2$$ and is also the only even prime number.
• The lowest odd prime number is $$3$$.
• Most common pattern in all prime numbers is except $$2$$ and $$5$$ ends in $$1,\, 3,\, 7,\, or\, 9$$.
• If $$N$$ is composite, then it must be divisible by any prime number $$p$$ such that $$p<=\sqrt{N}$$.
## Let's Summarize
The mini-lesson targeted the fascinating concept of "Euclid’s Proof for Infinite Primes." The math journey around "Euclid’s Proof for Infinite Primes" starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
At Cuemath, our team of math experts is dedicated to making learning fun for our favourite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
## FAQs on Prime Numbers And Euclid's Proof
### 1. What did Euclid prove about prime numbers in 300 BC?
Euclid, in 4th century B.C, points out that there have been an infinite Primes. The concept of infinity is not known at that time. He said ”prime numbers are quite any fixed multitude of them” meaning if you imagine 100, there are more, and if you imagine a million there are still more.
### 2. Are twin primes infinite?
"Twin primes" are primes that are two steps apart from each other on that line: 3 and 5, 5 and 7, 29 and 31, 137 and 139, and so on. The twin prime conjecture states that there are infinitely many twin primes, and that you'll keep encountering them no matter how far down the number line you go.
### 3. Is 1 a Prime Number?
There are people that believe so because they say that 1 can only be divided by 1 and itself but in mathematics, the number 1 has been discarded as a prime number because it only has one divisor. In fact, the criteria of ”a positive integer is prime if it has exactly two positive divisors” is used to exclude the number one from the prime number list. It is not because we are being picky about it, but if the number one was considered prime then many mathematical properties would have to be said differently.
### 4. Is 0 finite or infinite?
The concept of zero and that of infinity are linked, but, obviously, zero is not infinity. Rather, if we have N / Z, with any positive N, the quotient grows without limit as Z approaches 0. Hence we readily say that N / 0 is infinite.
### 5. What is meant by the Euclidean Geometry
Euclidean geometry, the study of plane and solid figures on the basis of axioms and theorems employed by the Greek mathematician Euclid (c. 300 BCE). In its rough outline, Euclidean geometry is the plane and solid geometry commonly taught in secondary schools
### 6. Is there a pattern to prime numbers?
A clear rule determines exactly what makes a prime: it's a whole number that can't be exactly divided by anything except 1 and itself. But there's no discernable pattern in the occurrence of the primes. ... That's because after the number 5, there are only four possibilities — 1, 3, 7 and 9 — for prime last digits.
Arithmetic Integers |
# How do you solve \frac{8}{1}\times \frac{3}{1}?
Oct 18, 2016
24
#### Explanation:
$\textcolor{red}{\frac{8}{1}} \cdot \textcolor{b l u e}{\frac{3}{1}}$
This is simple multiplication with fractions.
When multiplying fractions, you multiply across the top and bottom. Then simplify.
$\textcolor{red}{\frac{8}{1}} \cdot \textcolor{b l u e}{\frac{3}{1}} = \frac{\textcolor{red}{8} \cdot \textcolor{b l u e}{3}}{\textcolor{red}{1} \cdot \textcolor{b l u e}{1}} = \textcolor{p u r p \le}{\frac{24}{1}} = \textcolor{p u r p \le}{24}$
Another thing you could do is simplify the original fractions. $\textcolor{red}{\frac{8}{1}}$ is $\textcolor{red}{8}$ and $\textcolor{b l u e}{\frac{3}{1}}$ is $\textcolor{b l u e}{3}$. After simplifying, multiply like you normally would.
$\textcolor{red}{\frac{8}{1}} \cdot \textcolor{b l u e}{\frac{3}{1}} = \textcolor{red}{8} \cdot \textcolor{b l u e}{3} = \textcolor{p u r p \le}{24}$ |
# Difference between revisions of "2018 AMC 10B Problems/Problem 4"
## Problem
A three-dimensional rectangular box with dimensions $X$, $Y$, and $Z$ has faces whose surface areas are $24$, $24$, $48$, $48$, $72$, and $72$ square units. What is $X$ + $Y$ + $Z$?
$\textbf{(A) }18 \qquad \textbf{(B) }22 \qquad \textbf{(C) }24 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36 \qquad$
## Solution 1
Let $X$ be the length of the shortest dimension and $Z$ be the length of the longest dimension. Thus, $XY = 24$, $YZ = 72$, and $XZ = 48$. Divide the first two equations to get $\frac{Z}{X} = 3$. Then, multiply by the last equation to get $Z^2 = 144$, giving $Z = 12$. Following, $X = 4$ and $Y = 6$.
The final answer is $4 + 6 + 12 = 22$. $\boxed{B}$
## Solution 2
Simply use guess and check to find that the dimensions are $4$ by $6$ by $12$. Therefore, the answer is $4 + 6 + 12 = 22$. $\boxed{B}$
## Solution 3
If you find the GCD of $24$, $48$, and $72$ you get your first number, $12$. After this, do $48 \div 12$ and $72 \div 12$ to get $4$ and $6$, the other 2 numbers. When you add up your $3$ numbers, you get $22$ which is $\boxed{B}$. |
## Exercises - The GCD and Euclidean Algorithm
1. Use the Euclidean algorithm to compute each of the following gcd's.
1. gcd(12345,67890)
2. gcd(54321,9876)
1. \displaystyle{\begin{align*} 67890 &= 5 \cdot 12345 + 6165\\ 12345 &= 2 \cdot 6165 + \fbox{15} \leftarrow \textrm{gcd}\\ 6165 &= 411 \cdot 15 + 0 \end{align*}}
2. \displaystyle{\begin{align*} 54321 &= 5 \cdot 9876 + 4941\\ 9876 &= 1 \cdot 4941 + 4935\\ 4941 &= 1 \cdot 4935 + 6\\ 4935 &= 822 \cdot 6 + \fbox{3} \leftarrow \textrm{gcd}\\ 6 &= 2 \cdot 3 + 0 \end{align*}}
2. A number $L$ is called a common multiple of $m$ and $n$ if both $m$ and $n$ divide $L$. The smallest such $L$ is called the least common multiple of m and n and is denoted by $\textrm{lcm} (m,n)$
1. Find the following:
1. lcm (8, 12)
2. lcm (20, 30)
3. lcm (51, 68)
4. lcm (23, 18)
2. Compare the value of $\textrm{lcm} (m,n)$ with the values of $m$, $n$, and gcd($m,n$). In what way are they related?
3. Prove the relationship you found in part (b) always holds.
4. Compute $\textrm{lcm} (301337,307829)$.
5. Find all $m$ and $n$ where $\gcd (m,n) = 18$ and $\textrm{lcm} (m,n) = 720$.
1. 24
2. 60
3. 204
4. 414
1. It appears that the product of the lcm and the gcd always equals $mn$.
2. Let $p_1,p_2,p_3, ...$ be the primes: 2, 3, 5, ..., in order of magnitude. Let $p_r$ be the largest prime that divides either $m$ or $n$.
Then we can find write a prime factorization for both $m$ and $n$ in the following way: $m = p^{m_1}_1 p^{m_2}_2 \cdots p^{m_r}_r$ and $n = p^{n_1}_1 p^{n_2}_2 \cdots p^{n_r}_r$. Note, that some primes may not divide $m$ or $n$, so some of the exponents may be zero.
It should be clear that $$\gcd(m,n) = p^{\min(m_1,n_1)}_1 p^{\min(m_2,n_2)}_2 \cdots$$ $$\textrm{lcm}(m,n) = p^{\max(m_1,n_1)}_1 p^{\max(m_2,n_2)}_2 \cdots$$
As such, the product of the gcd and lcm is \begin{align} \gcd \cdot \textrm{lcm} &= p^{\min(m_1,n_1)+\max(m_1,n_1)}_1 p^{\min(m_2,n_2)+\max(m_2,n_2)}_2 \cdots\\ &= p^{m_1+n_1}_1 p^{m_2+n_2}_2 \cdots\\ &= (p^{m_1}_1 p^{m_2}_2 \cdots) (p^{n_1}_1 p^{n_2}_2 \cdots)\\ &= mn \end{align}
3. 171460753
4. (144,90) or (720,18)
3. Prove for all positive integers $a$ and $b$, it must be the case that $\textrm{gcd}(a+b,a-b) \ge \textrm{gcd}(a,b)$.
Let $d = \textrm{gcd}(a,b)$, can you argue $d | a+b$ and $d | a-b$?
4. 🔎 In number theory we are often curious about the greatest common divisors of two integers of interest. In this spirit, let us ask the question "Can anything interesting be said about the gcd of any two Fibonacci numbers?" Recall the $n^{th}$ Fibonacci number is defined by $f_1 = 1$, $f_2 = 1$, and $f_n = f_{n-1} + f_{n-2}$ for $n \gt 2$. Experimenting with several pairs of Fibonacci numbers should lead one to make the following startling conjecture:
$$\textrm{gcd}(f_m,f_n) = f_{\textrm{gcd}(m,n)}$$
If one knows $\textrm{gcd}(a,b)$, what is $\textrm{gcd}(a-b,b)$? |
# PERCENTAGES TO DECIMALS
Converting percentages to decimals you must remember
1=100%
0.5=50%
0.25=25%
0.75=75%
0.3dot3=33\1/3%
0.6dot6=66\2/3%
0.1=10%
0.2=20%
But knowing 1=100% will also help you solve any inbetween percentages.
Example 1
Express 31% as a decimal
to do this remember 1=100%
and also read this in conjunction with the section on percentages.
Which states
1. Write out what 100% is
2. Write out what we know
3. Keep % on one side
1. Write out what 100% is 100%=1
2. Write out what we know 31%=x
(Keep the % sign on the same side)
(100%)/(31%)=1/x
Now work out what x is
Multiply by x on both sides to get x on its own.
(100%)/(31%)timesx=(1timescancelx)/cancelx
100/31timesx=1
Multiply both sides by 31% to get x on its own.
100/cancel31timesxtimescancel31=1times31
100%x=31
Divide both sides by 100 to get x on its own.
(cancel100\x)/cancel100=31/100
x=31/100
x=0.31
Example 2
Convert 12.5% to a decimal
1. Write out what 100% is 100%=1
2. Write out what we know 12½%=x
(Keep the % sign on the same side)
(100%)/(12.5%)=1/x
Now work out what x is
Multiply by x on both sides to get x on its own.
(100%)/(12.5%)timesx=(1timescancelx)/cancelx
100/12.5timesx=1
Multiply both sides by 12.5 to get x on its own.
100/cancel12.5timesxtimescancel12.5=1times12.5
100x=12.5
Divide both sides by 100 to get x on its own.
(cancel100\x)/cancel100=12.5/100
x=12.5/100
x=0.125
Example 3
Convert 104% to a decimal
1. Write out what 100% is 100%=1
2. Write out what we know 104%=x
(Keep the % sign on the same side)
(100%)/(104%)=1/x
Now work out what x is
Multiply by x on both sides to get x on its own.
(100%)/(104%)timesx=(1timescancelx)/cancelx
100/104timesx=1
Multiply both sides by 104 to get x on its own.
100/cancel104timesxtimescancel104=1times104
100x=104
Divide both sides by 100 to get x on its own.
(cancel100\x)/cancel100=104/100
x=104/100
x=1.04
Example 4
Convert 0.5% to a decimal
1. Write out what 100% is 100%=1
2. Write out what we know 0.5%=x
(Keep the % sign on the same side)
(100%)/(0.5%)=1/x
Now work out what x is
Multiply by x on both sides to get x on its own.
(100%)/(0.5%)timesx=(1timescancelx)/cancelx
100/0.5timesx=1
Multiply both sides by 0.5 to get x on its own.
100/cancel0.5timesxtimescancel0.5=1times0.5
100x=0.5
Divide both sides by 100 to get x on its own.
(cancel100\x)/cancel100=0.5/100
x=0.5/100
x=0.005 |
proving this summation
• Sep 30th 2009, 04:04 AM
thereddevils
proving this summation
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$
Thanks a lot .
• Sep 30th 2009, 05:36 AM
Amer
Quote:
Originally Posted by thereddevils
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$
Thanks a lot .
let prove it by induction okay
let $P_n : \sum_{r=1}^{n} [(r+2)(r+1) - 3(r+1) +2]r! = n(n+1)!$
first see if n=1 and 2 true
$(1+2)(1+1)-3(2)+2 = 2$ and
$(1)(2)! = 2$ at 1 the statement is true try when n=2
you will find it is true now suppose that n is true try n+1
$P_{n+1}:
\sum_{r=1}^{n+1} [(r+2)(r+1)-3(r+1)+2]r! = (n+1)(n+2)!$
......(*)
$\sum_{r=1}^{n+1}[(r+2)(r+1)-3(r+1)+2]r! = \underbrace{\sum_{r=1}^{n} [(r+2)(r+1)-3(r+1)+2]r! }_{n(n+1)!}+$ $[(n+3)(n+2)-3(n+2)+2](n+1)!$
$\sum_{r=1}^{n+1}[(r+2)(r+1)-3(r+1)+2]r! = n(n+1)!+[(n+3)(n+2)-3(n+2)+2](n+1)!$
$=(n+1)![n+(n+3)(n+2)-3(n+2)+2]$
$=(n+1)![n+n^2+5n+6-3n-6+2]$
$=(n+1)!(n^2+3n+2)$
$=(n+1)!(n+1)(n+2)=(n+1)(n+2)!$ you begin with the RHS in (*) and finished with LHS in the (*) formula
so P_{n+1} is true the proof end
• Sep 30th 2009, 07:38 AM
Hello thereddevils
Quote:
Originally Posted by thereddevils
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$
Thanks a lot .
It's much easier if you simplify the expression first!
$(r+2)(r+1)-3(r+1)+2 = r^2+1$
So the induction hypothesis is:
$S_n=\sum^{n}_{r=1}(r^2+1)r!=n(n+1)!$
$\Rightarrow S_{n+1} = S_n+\Big((n+1)^2+1\Big)(n+1)!$
$=n(n+1)! + (n^2+2n+2)(n+1)!$
$=(n+1)!(n^2+3n+2)$
$=(n+1)!(n+2)(n+1)$
$=(n+2)!(n+1)$
...etc |
# How do you solve x(x-7)=0?
Nov 1, 2015
$x = 0$
or
$x = 7$
#### Explanation:
The format of this queston is already in factored form. This makes it quite easy to solve for $x$. They key to this is to know that anything multiplied by $0$ is $0$. We have two things multiplied by each other:
$x$ multiplied by $\left(x - 7\right)$ = $0$
So, this means, that if either term is equal to zero, the whole equation will equal to zero.
The first term is just $x$, so $x = 0$ is the first solution.
The second term is $\left(x - 7\right)$. We need to set that equal to zero and solve for $x$.
$x - 7 = 0$
$x = 7$ which is our second solution.
To check, plugging back into $x \left(x - 7\right) = 0$:
For $x = 0$
$0 \left(- 7\right) = 0$
$0 = 0$
so it works.
For $x = 7$
$7 \left(7 - 7\right) = 7 \left(0\right) = 0$
$0 = 0$
which also works. |
FutureStarr
22 Is What Percent of 50.
## 22 Is What Percent of 50
If 22 is what percent of 50, then 50 is what percent of 22. The percent change calculator can help with these types of problems.
### Use
In calculating 22% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,22% off, 22% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.percent dollar to pound = 0 pound
I've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below). (Source: visualfractions.com)
### Percent
It isn’t correct to divide it on 100 and to use a Percent sign, at the same time. (For example, 15% = 15/100 = 0,15, instead of 15% / 100, which actually (15/100)/100 = 0,0015. The term such as (100/100) % also will be wrong, it can be read as (1) Percent even if the purpose consists to tell 100%. )
Okay, let’s take this problem step-by-step. 50% of 100 is equivalent to half of 100, so 50. But how do we calculate 25% of 200? To calculate that, you can use our percentage calculator, or you can read through the section below containing formulas and explanations to gain a better grasp on these concepts and avoid technical help in the future. (Source: www.blitzresults.com)
### Value
Let’s take this one step at a time. The percentage value (W) is 50, the basic value (G) is 200, and our starting formula is as follows: 100 x W = G x P. Dividing both sides of this initial equation by G leaves P alone on one side of the equation, which is what we want, as it is the value we are looking for.
However, you can also make it very easy for yourself by breaking down the problem. For the question “50 is 25% of what value?”, we know that the basic value (G), is the value we are looking for. The percentage value (W) is 50, the percentage (P) is 25%, and we can recall that our starting formula is 100 x W = G x P. Since we are looking for G, we can divide by P, so G is isolated on one side of the equation, resulting in the following rearranged formula: Base value (G) = Percentage value (W)/ Percentage (P) × 100 %. (Source: www.blitzresults.com)
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# Selina Solutions Concise Maths Class 10 Chapter 4 Linear Inequations (In one variable) Exercise 4(A)
Solving linear inequations in one variable algebraically is done by following certain rules. The understanding of the replacement set and the solution set is made clear when solving problems of this exercise. The Selina Solutions for Class 10 Maths is a powerful tool for students to do any quick reference or doubt clearance anytime. Students can now get the Concise Selina Solutions for Class 10 Maths Chapter 4 Linear Inequations (In one variable) Exercise 4(A) PDF in the links given below.
## Selina Solutions Concise Maths Class 10 Chapter 4 Linear Inequations (In one variable) Exercise 4(A) Download PDF
### Access Selina Solutions Concise Maths Class 10 Chapter 4 Linear Inequations (In one variable) Exercise 4(A)
1. State, true or false:
(i) x < -y ⇒ -x > y
(ii) -5x ≥ 15 ⇒ x ≥ -3
(iii) 2x ≤ -7 ⇒ 2x/-4 ≥ -7/-4
(iv) 7 > 4 ⇒ 1/7 < 1/5
Solution:
(i) Given statement is true. (according to Rule 5)
(ii) Given statement is false. (according to Rule 4)
(iii) Given statement is true. (according to Rule 4)
(iv) Given statement is true. (according to Rule 6)
2. State whether the following statements are true or false.
(i) a < b, then a – c < b – c
(ii) If a > b, then a + c > b + c
(iii) If a < b, then ac > bc
(iv) If a > b, then a/c < b/cÂ
(v) If a – c > b – d, then a + d > b + c
(vi) If a < b, and c > 0, then a – c > b – c
Where a, b, c and d are real numbers and c ≠ 0.
Solution:
(i) Given statement is true. (Subtracting equals on both sides will not change the inequality)
(ii) Given statement is true. (Adding equals on both sides will not change the inequality)
(iii) Given statement is false. (According to rule 3)
(iv) Given statement is false. (According to rule 3)
(v) Given statement is true. As a – c > b – d ⇒ a + d > b + c
(vi) Given statement is false. As a < b, a – c < b – c (since c > 0)
3. If x ∈ N, find the solution set of inequations.
(i) 5x + 3 ≤ 2x + 18
(ii) 3x – 2 < 19 – 4x
Solution:
(i) 5x + 3 ≤ 2x + 18
5x – 2x ≤ 18 – 3
3x ≤ 15
x ≤ 5
As, x ∈ N, thus the solution set is {1, 2, 3, 4, 5}.
(ii) 3x – 2 < 19 – 4x
3x + 4x < 19 + 2
7x < 21
x < 3
As, x ∈ N, thus the solution set is {1, 2}.
4. If the replacement set is the set of whole numbers, solve:
(i) x + 7 ≤ 11
(ii) 3x – 1 > 8
(iii) 8 – x > 5
(iv) 7 – 3x ≥ -1/2
(v) x – 3/2 < 3/2 – xÂ
(vi) 18 ≤ 3x – 2
Solution:
(i) x + 7 ≤ 11
x ≤ 11 – 7
x ≤ 4
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {0, 1, 2, 3, 4}
(ii) 3x – 1 > 8
3x > 8 + 1
x > 3
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {4, 5, 6, …}
(iii) 8 – x > 5
– x > 5 – 8
– x > -3
x < 3
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {0, 1, 2}
(iv) 7 – 3x ≥ -1/2
-3x ≥  -1/2 – 7
-3x ≥ -15/2
x ≥ 5/2
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {0, 1, 2}
(v) x – 3/2 < 3/2 – xÂ
x + x < 3/2 + 3/2
2x < 3
x < 3/2
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {0, 1}
(vi) 18 ≤ 3x – 2
18 + 2 ≤ 3x
20 ≤ 3x
x ≥ 20/3
As the replacement set = W (set of whole numbers)
Therefore, the solution set = {7, 8, 9, …}
5. Solve the inequation:
3 – 2x ≥ x – 12 given that x ∈  N.
Solution:
3 – 2x ≥ x – 12
-2x – x ≥ -12 – 3
-3x ≥ -15
x ≤ 5
As, x ∈ N,
Thus, the solution set = {1, 2, 3, 4, 5}
6. If 25 – 4x ≤ 16, find:
(i) the smallest value of x, when x is a real number,
(ii) the smallest value of x, when x is an integer.
Solution:
25 – 4x ≤ 16
-4x ≤ 16 – 25
-4x ≤ -9
x ≥ 9/4Â
x ≥ 2.25
Now,
(i) The smallest value of x, when x is a real number is 2.25.
(ii) The smallest value of x, when x is an integer is 3. |
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# How do you solve for x in $5x + 3 > 2x - 9$ ?
Last updated date: 10th Aug 2024
Total views: 388.5k
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Hint:In this question, we are given that $5x + 3$ is greater than $2x - 9$ , it is an algebraic expression containing one unknown variable quantity (x). We know that we need an “n” number of equations to find the value of “n” unknown variables. In the given algebraic expression; we have exactly one equation and 1 unknown quantity, so we can easily find the value of x by rearranging the equation such that one side of the equation contains the terms containing x and all other terms lie on the other side. Then by applying the given arithmetic operations, we can find the value of x.
We are given that $5x + 3 > 2x - 9$
$5x - 2x > - 9 - 3 \\ \Rightarrow 3x > - 12 \\$
$x > \dfrac{{ - 12}}{3}$
$x > \dfrac{{ - 2 \times 2 \times 3}}{3}$
$\Rightarrow x > - 4$
Hence, when $5x + 3 > 2x - 9$ , we get $x > - 4$ . |
Find a Perpendicular line through a Point Video Demonstration
Students are often asked to find the equation of a line that is perpendicular to another line and that passes through a point. Watch the video tutorial below to understand how to do these problems and, if you want, download this free worksheet if you want some extra practice.
Practice Problems
Problem 1
Step 1
Find negative reciprocal of the slope:
Slope = -1 or $$-\frac{1}{1}$$
Negative reciprocal = $$\frac{1}{1} = 1$$
Step 2
Plug the x and y given in the equation into the slope interecept formula.
$$0 = 1\left(0\right) + b$$
Step 3
Solve for b.
$$0 = 0 + b$$
$$b = 0$$
Step 4
Substitute b = 0 into slope-intercept equation.
$$y = x$$
Problem 2
Step 1
Find negative reciprocal of the slope:
Slope = 5 or $$\frac{5}{1}$$
Negative reciprocal = $$-\frac{1}{5}$$
Step 2
Plug the x and y given in the equation into the slope interecept formula.
$$1= -\frac{1}{5} \left(0\right) + b$$
Step 3
Solve for b.
$$1 = 0 + b$$
$$b = 1$$
Step 4
Substitute b = 1 into slope-intercept equation.
$$y = -\frac{1}{5} x + 1$$
Problem 3
Step 1
Find the negative reciprocal of the slope.
Slope = $$\frac{1}{4}$$
Negative reciprocal = $$-\frac{4}{1} = -4$$
Step 2
Plug the x and y given in the question into the slope interecept formula.
$$-4 = -4(-4) + b$$
Step 3
Solve for b.
$$-4 = 16 + b$$
$$-16 -16$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$b = -20$$
Step 4
Substitute b into slope-intercept formula.
$$y = -4x - 20$$
Problem 4
Step 1
Line $$y = -8$$ is parallel to the x-axis. So, the perpendicular line to this line must be parallel to the y-axis.
Step 2
The equation of the perpendicular line is in the form $$x = n$$.
Step 3
This line is perpendicular to the first line at $$y = -8$$; therefore, the second line must be perpendicular at a point with the x-coordinate of -8. That is, $$x = -8$$.
Problem 5
Step 1
Find the negative reciprocal of the slope.
Slope = -3 or $$-\frac{3}{1}$$
Negative reciprocal = $$\frac{3}{1} = 3$$
Step 2
Plug the x and y given in the question into the slope interecept formula.
$$-5 = 3(0) + b$$
Step 3
Solve for b.
$$-5 = 0 + b$$
$$b = -5$$
Step 4
Substitute b into slope-intercept formula.
$$y = 3x - 5$$
Practice Problems II
Problem 6
Step 1
Line $$x = 32$$ is a vertical line. Any line to be perpendicular to this line must be horizontal.
Step 2
We can draw infinite number of perpendicular lines to this line. Some of them are, for example.
$$y = 21$$
$$y = 22$$
$$y = 23$$
$$y = 24$$
$$\dots \dots$$
$$\dots \dots$$
$$\dots \dots$$
$$y = 32$$
Step 3
Out of all the infinite number of perpendicular lines only one passes through (-32, 32).
Therefore, $$y = -32$$ is the only perpendicular line the problem asked for.
Problem 7
Step 1
Multiply both sides of the equation by 3.
$$3\left(\frac{x+y}{3} \right) = 3$$
Step 2
Simplify the product on the left.
$$x + y = 3$$
Step 3
Subtract x from each side.
$$x + y - x = 3 - x$$
Step 4
Cancel out the x' on the left.
$$y = -x + 3$$
Step 5
Find the negative reciprocal of the slope.
Slope = -1 or $$-\frac{1}{1}$$
Negative reciprocal = $$\frac{1}{1} =1$$
Step 6
Plug the x and y given in the question into the slope interecept formula.
$$6 = 1(-3) + b$$
Step 7
Solve for b.
$$6 = -3 + b$$
$$+3 +3$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$9 = b$$
Step 8
Substitute b into slope-intercept formula.
$$y = x + 9$$
Problem 8
Step 1
Find the slope of the given line first. The line is not in slope-Intercept Form. Add -3x + 5 to both sides.
Step 2
Find the negative reciprocal of the slope.
Slope = -3 or $$-\frac{3}{1}$$
Negative reciprocal = $$\frac{3}{1} =3$$
Step 3
Plug the x and y given in the question into the slope interecept formula.
$$5 = 3(0) + b$$
Step 4
Solve for b.
$$5 = 0 + b$$
$$5 = b$$
Step 5
Substitute b into slope-intercept formula.
$$y = 3x + 5$$
Problem 9
Step 1
The slopes of the lines are as follows:
A: -7
B: $$\frac{1}{7}$$
C: $$-\frac{1}{7}$$
D: $$\frac{1}{7}$$
Step 2
Only the products of the slopes of the following pairs give -1.
$$(Slope of A)(Slope of B) = -1$$
$$(Slope of A)(Slope of D) = -1$$
Step 3
Check the given point to see which lines it fits. Replace (-2, 4) in A.
$$4 = -7(-2) - 10$$
$$4 = 14 - 10$$
$$4 = 4$$
Step 4
So, A is acceptable as one of the perpendicular lines.
Replace (-2, 4) in B.
$$4=\frac{1}{7} (-2)+12$$
$$4=\frac{-2}{7} +12$$
$$4=\frac{-2+84}{7}$$
$$4=\frac{82}{7}$$
Step 5
Both sides are not equal. Thus, line B is not accepted.
Replace (4, -2) in D.
$$4=\frac{1}{7} \left(-2\right)+\frac{30}{7}$$
$$4=\frac{-2}{7} +\frac{30}{7}$$
$$4=\frac{-2+30}{7}$$
$$4=\frac{28}{7}$$
$$4=4$$
Both sides are equal. So, the given point is on D.
Step 6
Point (4, -2) fits two perpendicular lines A and D. Therefore, A and D only are perpendicular lines at (2, -4).
Problem 10
Step 1
All the equations are not in the form of $$y = mx + b$$. So, we must change their forms to y = mx + b first.
Step 2
Equation K: Add 2x to both sides.
$$-2x + 8y + 2x = 11 + 2x$$
Step 3
Cancel out the similar terms on the left side.
$$8y = 2x + 11$$
Step 4
Divide both sides by 8.
$$\frac{8y}{8} =\frac{2x}{8} +\frac{11}{8}$$
Step 5
Simplify the fractions.
$$y=\frac{1}{4} x+\frac{11}{8}$$
Thus, the slope of K is $$\frac{1}{4}$$.
Step 6
L: $$y = 5x - 6$$
This line is already in Slope-Intercept form. The coefficient of x is 5.
L: slope is 5.
Step 7
M: $$-x = 4y + 12$$
This line is not in slope-intercept form. Subtract 12 from each side.
$$-x - 12 = 4y + 12 - 12$$
Step 8
Cancel out numbers on the right side.
$$-x - 12 = 4y$$
Step 9
Divide both sides by 4.
$$\frac{x}{4} -\frac{12}{4} =\frac{4y}{4}$$
Step 10
Simplify all the fractions.
$$y=-\frac{1}{4} x-3$$
That is, the slope of M is $$-\frac{1}{4}$$.
Step 11
N: $$y = -4x + 9$$ is in slope -- intercept form.
The coefficient of x is the slope of the line. Thus, -4 is the slope of N.
Step 12
Now we have four slopes:
Slope of $$K = \frac{1}{4}$$
Slope of $$L = 5$$
Slope of $$M = -\frac{1}{4}$$
Slope of $$N = -4$$
The product of the slopes of K and N is $$\left(\frac{1}{4} \right)\left(-4\right)=-1$$. So K and N are perpendicular. |
# Multisets of a given set
A multiset is an unordered collection of elements where elements may repeat any number of times. The size of a multiset is the number of elements in it counting repetitions.
(a) What is the number of multisets of size $4$ that can be constructed from $n$ distinct elements so that at least one element occurs exactly twice?
(b) How many multisets can be constructed from $n$ distinct elements?
For part b, infinite is correct.
For part a, taking $n=3$ and elements $\{1,2,3\}$ we have multisets as: $\{1,1,2,2\}, \{1,1,3,3\}, \{1,1,2,3\}, \{2,2,3,3\}, \{2,2,1,3\}, \{3,3,1,2\}$, for a total of $6$.
Similarly for $n=4$ and using elements $\{1,2,3,4\}$, we have $18$ multisets. There must be some formula, or we have to develop one!
I am in particular looking for a formula when there is a restriction on the number occurrences in the multiset.
-
There are four places to be filled in the multiset using the $n$ distinct elements. Atleast one element has to occur exactly twice. That would leave 2 more places in the multiset. This means, atmost two elements can occur exactly twice. We can thus divide this into 2 mutually exclusive cases as follows:
1. Exactly one element occurs exactly twice: Select this element in ${n\choose{1}} = n$ ways. Fill up the remaining two spots using 2 distinct elements from the remaining $n-1$ elements in ${{n-1}\choose{2}}$ ways. Overall: $n \cdot {{n-1}\choose{2}} = \frac{n(n-1)(n-2)}{2}$ ways.
2. Exactly two elements that occur twice each: These two will fill up the multiset, so you only have to select two elements out of $n$ in ${n\choose 2} = \frac{n(n-1)}{2}$ ways.
Since these are mutually exclusive, the total number of ways to form the multiset is: $$\frac{n(n-1)(n-2)}{2} + \frac{n(n-1)}{2}$$$$= \frac{n(n-1)^2}{2}$$
Note, that $n \ge 2$ otherwise no element can be present twice. This is also obvious from the formula (when $n = 0, 1$).
You can check that the formula tallies with your counts.
-
one more thing suppose we are given {a,b,c,d} and {1*a,2*b,3*c,5*d} than way to distribute 10 vertices among a,b,c,d ? – user1771809 Dec 25 '12 at 13:46
@user1771809 What do you mean by {1*a,2*b,3*c,5*d}? What do you mean by distribute 10 vertices among a,b,c,d? Please be clearer. And if this is a reasonably different question, you might want to post a separate question too. – Paresh Dec 25 '12 at 17:53 |
## Developing an Algorithm to Divide Polynomials
Let us explore polynomial division through an example, hoping it might lead to a general process/algorithm for dividing any two given polynomials.
In doing so, it will be useful to have some verbiage to quickly identify different terms of the polynomials in question. As these can be distinguished by the exponents with which they are associated, let us define the degree of a term in a polynomial of one variable to be the exponent on the variable in that term. Further, let us call the term of highest degree in such a polynomial the leading term. We also call the degree of the leading term the degree of the polynomial.
Additionally, let us refer to the term of a polynomial with degree zero as the constant term (since its value doesn't change, even when the value of the variable changes). We call the term with degree one the linear term (for reasons that will be forthcoming), the term with degree two the quadratic term (from the Latin quadratus which literally means "made square"), the term with degree three the cubic term (from the greek kubikos whose root kobos means "cube"), and the term with degree four the quartic term (in Latin quartus means "fourth"), and the term with degree five the quintic term (similar to the last one, quintus means "fifth" in Latin).
With those definitions aside, let us now suppose we wish to divide $x^3 - 2x^2 - 4$ (the dividend) by $x-3$ (the divisor).
Presuming that the divisor divides the dividend evenly for the moment (recall this need not be the case), let us assume that $(x-3)(\cdots) = x^3 - 2x^2 - 4$. We hope the as-yet-unknown factor on the left will be a polynomial. If it is, it seems clear the leading term of the unknown factor must be $x^2$ so that the product of the two leading terms on the left can produce the leading term of $x^3$ seen on the right. This $x^2$ was of course the result of a quick division of the $x^3$ of the dividend and the $x$ in the divisor.
Of course if we have an $x^2$ as a term in the unknown factor, we can find a partial product $$(x-3)(x^2 + \cdots) = x^3 - 3x^2 + \cdots$$ As such, some of the dividend has been accounted for. We can subtract this $x^3 - 3x^2$ from the dividend to find what remains. As $(x^3 - 2x^2 - 4) - (x^3 - 3x^2) = x^2 - 4$, we can write $$(x-3)(x^2 + \cdots) = (x-3)x^2 + (x-3)(\cdots) = (x^3 - 3x^2) + (x^2 - 4)$$ Note, this allows us to equate $(x-3)(\cdots) = x^2 - 4$ as seen above.
This gives us a new "smaller" problem of the same form (note the degree of the polynomial on the right has been reduced). As such, we can apply the same strategy again (i.e., determine the leading term of what remains unknown is $x$, find the product of this $x$ and $x-3$ and subtract it off to see what still remains of the quotient) -- and again and again, as necessary.
Seeing that we may have to solve a number of such "smaller problems", we will want to keep our work as organized and efficient as possible.
To this end, let us restart the problem using a notation reminiscent of that used for long division of integers. It will also help if we use zero coefficients for any missing powers in our dividend: $$\require{color}\begin{array}{l} x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ \end{array}$$
Recall we earlier found the first part of the quotient, $x^2$, by dividing the leading terms of the dividend and divisor (i.e., $x^3/x$). We do so again, only this time, we place this $x^2$ above the bar but slightly to the right, to align with the term in the dividend of the same degree. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}{\color{blue}x^2} \\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ \end{array}$$ Again, we now find what the inclusion of this $x^2$ contributes to the product of $(x-3)$ and the unknown factor/quotient by multiplying it by $(x-3)$. Only this time, we write our previously found product $x^2(x-3) = x^3 - 3x^2$, underneath the dividend -- taking care to vertically align terms of common degree so that we easily subtract things to reveal what remains unaccounted for.. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 \\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ {\color{white}x-3 \ )\ }\underline{{\color{blue}x^3 - 3x^2}}\\ \end{array}$$ Apart from the visual difference of seeing a couple of $0x$ terms, the subtraction results in the same result: $(x^3-2x^2+0x-4) - (x^3-3x^2) = x^2 + 0x -4$. Note however, how we only write the $x^2$ part of this difference on the next line. This is solely to keep our work from getting too cluttered -- we must remember that all three of these terms (shown in red) are still unaccounted for.. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 \\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2 \ {\color{red}+ \ 0x-4}}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} {\color{red}x^2} \\ \end{array}$$ Now, we simply repeat the process by asking what must we multiply the $x$ in $x-3$ (i.e., the highest degree term in the divisor) by to get $x^2$ (the highest degree term in what is still unaccounted for from the dividend). A quick division of these two terms reveals this to be $x^2/x = x$, which we again place above the bar. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 + {\color{white}0}{\color{blue}x}\\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} x^2\\ \end{array}$$ Again, let us find out what including this new $x$ in the quotient adds to the mix by multiplying it by the divisor (i.e., $x(x-3) = x^2-3x$), writing this again on a new line below our existing work, aligning same powers to make the coming subtraction easy... $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 + {\color{white}0}x\\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} x^2 {\color{white}+ 0x}\\ {\color{white}x-3 \ )\ x^3 - 3} \underline{{\color{blue}x^2 - 3x}}\\ \end{array}$$ We again subtract this from what we previously had left to discover what still remains unaccounted for. This time the subtraction performed is $(x^2 + 0x - 4) - (x^2-3x) = 3x - 4$. We again leave these terms (shown in red) a bit "separated" in the interest of not cluttering up things too badly. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 + {\color{white}0}x\\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x \ {\color{red}{}-4}}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} x^2 {\color{white}+ 0x}\\ {\color{white}x-3 \ )\ x^3 - 3} \underline{x^2 - 3x}\\ {\color{white}x-3 \ )\ x^3 - 3x^2 - {}} {\color{red}3x} \\ \end{array}$$ Close to being done, we ask what we must multiply the $x$ in $(x-3)$ (i.e., the leading term of the divisor) by to get $3x$, the leading term in what remains unaccounted for. This is of course simply $3$, which we write above the bar $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 + {\color{white}0}x + {\color{blue}3}\\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} x^2 {\color{white}+ 0x}\\ {\color{white}x-3 \ )\ x^3 - 3} \underline{x^2 - 3x}\\ {\color{white}x-3 \ )\ x^3 - 3x^2 - {}} 3x \\ \end{array}$$ We again find the contribution of this $3$'s addition by multiplying it by the divisor -- subtracting this product $3(x-3) = 3x-9$ from what remains as before. Notice that this time, the entirety of the difference $(3x-4)-(3x-9) = 5$ ends up in the bottom-most line -- it is no longer "split" in two different locations any more. $$\require{color}\begin{array}{l} {\color{white}x-3 \ )\ x^3-2}x^2 + {\color{white}0}x + 3\\ x-3 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3-2x^2+0x-4}\\ {\color{white}x-3 \ )\ }\underline{x^3 - 3x^2}\\ {\color{white}x-3 \ )\ x^3 - 3} x^2 {\color{white}+ 0x}\\ {\color{white}x-3 \ )\ x^3 - 3} \underline{x^2 - 3x}\\ {\color{white}x-3 \ )\ x^3 - 3x^2 - {}} 3x \\ {\color{white}x-3 \ )\ x^3 - 3x^2 - {}} \underline{{\color{blue}3x - 9}}\\ {\color{white}x-3 \ )\ x^3 - 3x^2 - 3x - {}} {\color{red}5}\\ \end{array}$$ As asking the question "What can we multiply the $x$ in $x-3$ by to get the $5$ that remains?" would result in an answer involving a negative exponent, we see that our divisor did not evenly go into the dividend, and we consequently have a remainder of $5$.
Our process has thus come to an end, allowing us to say: $$(x^3 - 2x^2 - 4) = (x^2+x+3)(x-3) + 5$$ In this case, we call the $(x^2+x+3)$ the quotient, and $5$ the remainder, and may also write $$(x^3 - 2x^2 - 4) \div (x-3) = (x^2 + x + 3) \textrm{ with a remainder of } 5$$ While above the remainder was a constant value, it need not be. Indeed, when dividing any two polynomials, stopping the process above before any negative exponents appear may leave a polynomial as a remainder -- although its degree will always be less than the divisor polynomial.
When performing the division of polynomials $f(x)$ and $g(x)$, if we find a quotient $q(x)$ and remainder $r(x)$, then we can say $$f(x) = q(x)g(x) + r(x)$$ Dividing both sides by $g(x)$ gives us an alternate way to express this fact. Namely, $$\cfrac{f(x)}{g(x)} = q(x) + \cfrac{r(x)}{g(x)}$$ In our example above, we might then also say:
$$\frac{x^3 - 2x^2 - 4}{x-3} = x^2 + x + 3 + \frac{5}{x-3}$$
This process is easily extended to divisors with higher degree. Consider the work shown below to divide $(x^3 - 8)$ by $(x^2 - 5x + 6)$: $$\begin{array}{l} \hphantom{x^2 - 5x + 6 \ )\ x^3 + 0x^2 + \ 0}x \ + \ 5\\ x^2 - 5x + 6 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3 + 0x^2 + \ 0x \ - \ 8}\\ \hphantom{x^2 - 5x + 6 \ )\ } \underline{x^3 - 5x^2 + \ 6x}\\ \hphantom{x^2 - 5x + 6 \ )\ x^3 - {}} 5x^2 - \ 6x\\ \hphantom{x^2 - 5x + 6 \ )\ x^3 - {}} \underline{5x^2 - 25x + 30}\\ \hphantom{x^2 - 5x + 6 \ )\ x^3 - 5x^2 - {}} 19x - 38 \end{array}$$
Note, here we again stop when the degree of what remains (i.e., $19x-38$) is less than the degree of the divisor (i.e., $x^2 - 5x + 6$).
We again can state the result in two ways: $$(x^3-8) \div (x^2 - 5x + 6) = (x+5) \textrm{ with a remainder of } (19x - 38)$$ or equivalently, $$\cfrac{x^3 - 8}{x^2 - 5x + 6} = x + 5 + \cfrac{19x-38}{x^2 - 5x + 6}$$
As a matter of verbiage, we call a rational expression where the degree of the numerator equals or exceeds that of the denominator an improper rational expression, and rational expressions where the degree of the numerator is less than that of the denominator a proper rational expression. As such, expressing the quotient as we have immediately above suggest we can always write an improper rational expression as the sum of a polynomial and a proper rational expression -- much in the same way that an improper fraction can be written as the sum of an integer and a proper fraction (i.e., "mixed numeral" form).
Importantly, and perhaps unexpectedly -- the long division algorithm described above forces us to generalize what we mean by a "polynomial".
Heretofore, we had only considered polynomials with integer coefficients, but the many divisions of leading terms that appear in the process described above can produce expressions of the form $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0$ where the coefficients $a_k$ are rational.
Consider the case below, where we divide $(x^2-5x+6)$ by $(19x-38)$. Certainly, the leading term $x^2$ of the dividend must be the product of the leading term of the divisor, $19x$, and the leading term of what we seek. This tells us that we need $\frac{1}{19}x$ as the first term of our answer. An additional rational coefficient, $-\frac{3}{19}$, shows up in a later term as well: $$\begin{array}{l} \hphantom{19x - 38 \ )\ x^2 + {}} \frac{1}{19}x - \frac{3}{19}\\ 19x - 38 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^2 \ - 5x \ + \ 6}\\ \hphantom{19x - 38 \ )\ } \underline{x^2 \ -2x}\\ \hphantom{19x - 38 \ )\ x^2 } \ -3x\\ \hphantom{19x - 38 \ )\ x^2 } \ \underline{{}- 3x \ + \ 6}\\ \hphantom{19x - 38 \ )\ x^2 \ -7x \ + {}}\ 0 \end{array}$$ Note the coefficients of the quotient are no longer integers! Naturally, we call expressions like this a polynomial with rational coefficients. |
# Assume y varies inversely as x, if y=8 when x=4, how do you find x when y=2?
Jul 6, 2017
$x = 16$
#### Explanation:
$y \text{ varies inversely as } x \implies y \propto \frac{1}{x}$
$\implies y = \frac{k}{x} \text{ where "k" is the constant of proportionality}$
$y = 8 , x = 4$
we have
$8 = \frac{k}{4} \implies k = 8 \times 4 = 32$
$\therefore y = \frac{32}{x}$
$y = 2$
$2 = \frac{32}{x} \implies x = \frac{32}{2} = 16$
$x = 16$
Jul 6, 2017
$x = 16$
#### Explanation:
$\text{the initial statement is } y \propto \frac{1}{x}$
$\text{to convert to an equation multiply by k, the constant }$
$\text{of variation}$
$\Rightarrow y = k \times \frac{1}{x} = \frac{k}{x}$
$\text{to find k, use the condition given for x and y}$
$y = 8 \text{ when } x = 4$
$y = \frac{k}{x} \Rightarrow k = x y = 4 \times 8 = 32$
$\text{the equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{32}{x}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{when y = 2}$
$x y = 32 \Rightarrow x = \frac{32}{y} = \frac{32}{2} = 16$ |
# How to find the coefficient of a power in a power series??
How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$
Is there a general method to this?
We can use the fact that \begin{align} 1+x+x^2+\dotsb&=\frac{1}{1-x}\\ x^2+x^4+x^6+\dotsb&=\frac{x^2}{1-x^2} \end{align} so the expression is $$\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}$$ and we can try doing partial fraction decomposition: $$\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}= x^4+x^3+3x^2+3x+5+\frac{6x^2+2x-5}{(1-x)^2(1+x)}$$ so we want to decompose the fraction into $$\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}$$ which gives $A=3/2$, $B=-25/4$, $C=-1/4$. Therefore your product can be written as $$x^4+x^3+3x^2+3x+5 +\frac{3}{2}\sum_{k\ge0}(k+1)x^k -\frac{25}{4}\sum_{k\ge0}x^k -\frac{1}{4}\sum_{k\ge0}(-1)^kx^k$$ and the coefficient of $x^{80}$ can be read directly as $$\frac{3}{2}\cdot 81-\frac{25}{4}-\frac{1}{4}=115$$
Note that this method gives all coefficients.
I've used the formula $$\frac{1}{(1-x)^2}=\sum_{k\ge0}(k+1)x^k$$ that can be deduced by differentiating $$\frac{1}{1-x}=\sum_{k\ge0}x^k$$
We have $(1+x+x^2+\cdots)(x^2+x^4+\cdots)(1+x^3+x^5)$ Then we want The coeficient of $x^{80}$, note that $x^nx^m=x^{n+m}$ then searching The ways of write $80$ using the sets $\{0,1,\cdots\},\{2,4,\cdots\},\{0,3,5\}$. \begin{align} 80&=0+80+0\\ &=2+78+0\\ &=4+76+0\\ &\vdots\\ &=78+2+0\\ &=1+76+3\\ &=3+74+3\\ &=5+72+3\\ &\vdots\\ &=75+2+3\\ &=1+74+5\\ &=3+72+5\\ &\vdots\\ &=73+2+5 \end{align} Then The coeficiente will be \begin{align} C_{80}&=\stackrel{2n}{(40-1+1)}+\stackrel{2n+1}{(37-0+1)}+\stackrel{2n+1}{(36-0+1)}\\ &=40+38+37\\ &=115 \end{align} Then if nothing as wrong The answer must be $115$
• Unfortunately when I inputted it in, the answer was wrong. I did however find out that the "power series representation" of all three products is x^5+x^3+1 / (1-x)(1-x^2). – Mathy Person Dec 9 '14 at 3:41
• sure you have not forgotten any data from the given problem? i will revise and fix the answer later, thx. – cand Dec 9 '14 at 3:58
• There was a typo in my original problem statement....it should have been *(1+x^2+x^5), rather than (1+x^3+x^5). Sorry for the confusion!! – Mathy Person Dec 9 '14 at 4:35
• You sure that The middle product is $(x^2+x^4+\cdots)$ instead of $(1+x^2+x^4+\cdots)$? – cand Dec 9 '14 at 13:10
• Yes, I'm sure. [[15 characters in length left]] – Mathy Person Dec 9 '14 at 21:08
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\bracks{x^{80}}\bracks{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots} \pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots} \pars{1 + x^{3} + x^{5}}}:\ {\large ?}}$
$${\rm Lets}\quad \dsc{\fermi\pars{x}}\equiv \dsc{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots} \pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots}}$$
Then, $$\bracks{x^{80}}\bracks{\fermi\pars{x}\pars{1 + x^{3} + x^{5}}} =\bracks{x^{80}}\fermi\pars{x} + \bracks{x^{77}}\fermi\pars{x} + \bracks{x^{75}}\fermi\pars{x}$$
and \begin{align} \fermi\pars{x}&=\sum_{i\ =\ 0}^{\infty}x^{i}\sum_{j\ =\ 0}^{\infty}x^{2j + 2} =\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty}x^{i + 2j + 2} =\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty} \sum_{n\ =\ 2}^{\infty}x^{n}\delta_{n,i + 2j + 2} \\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n}\pars{% \sum_{j\ =\ 0}^{\infty}\sum_{i\ =\ 0}^{\infty}\delta_{i,n - 2j - 2}} =\sum_{n\ =\ 2}^{\infty}x^{n} \pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{n - 2j - 2\ \geq\ 0}} \\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n} \pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{j\ \leq\ n/2 - 1}} =\sum_{n\ =\ 2}^{\infty}x^{n}\pars{\sum_{j\ =\ 0}^{\floor{n/2 - 1}}1} =\sum_{n\ =\ 2}^{\infty}\pars{\floor{{n \over 2} - 1} + 1}x^{n} \end{align}
$$\bracks{x^{n}}\fermi\pars{x} = \floor{n - 2 \over 2} + 1\,,\qquad n \geq 2$$
The $\color{#66f}{\large\mbox{final result}}$ is given by: $$\pars{\floor{78 \over 2} + 1} + \pars{\floor{75 \over 2} + 1} +\pars{\floor{73 \over 2} + 1}=40 + 38 + 37=\color{#66f}{\Large 115}$$ |
How do you find all the zeros of f(x) = x^2 + 6x + 18?
Mar 10, 2016
Complete the square and use the difference of squares identity to find zeros:
$x = - 3 \pm 3 i$
Explanation:
Complete the square then use the difference of squares identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
with $a = \left(x + 3\right)$ and $b = 3 i$ as follows:
$f \left(x\right) = {x}^{2} + 6 x + 18$
$= {x}^{2} + 6 x + 9 + 9$
$= {\left(x + 3\right)}^{2} + {3}^{2}$
$= {\left(x + 3\right)}^{2} - {\left(3 i\right)}^{2}$
$= \left(\left(x + 3\right) - 3 i\right) \left(\left(x + 3\right) + 3 i\right)$
$= \left(x + 3 - 3 i\right) \left(x + 3 + 3 i\right)$
Hence $f \left(x\right) = 0$ when $x = - 3 \pm 3 i$ |
In this chapter you will work with whole numbers smaller than 0. These numbers are called negative numbers. The whole numbers larger than 0, 0 itself and the negative whole numbers together are called the integers. Mathematicians have agreed that negative numbers should have certain properties that would make them useful for various purposes. You will learn about these properties and how they make it possible to do calculations with negative numbers.
## What is beyond
### Why people decided to have negative numbers
Below, you can see how Jimmy prefers to work when doing calculations such as 542 + 253.
$500 + 200 = 700 \\ 40 + 50 = 90 \\ 2 + 3 = 5 \\ 700 + 90 + 5 = 795$
He tries to calculate 542 - 253 in a similar way:
\begin{align} 500 - 200 &= 300 \\ 40 - 50 &= ? \end{align}
Jimmy clearly has a problem. He reasons as follows:
I can subtract 40 from 40; that gives 0. But then there is still 10 that I have to subtract.
He decides to deal with the 10 that he still has to subtract later, and continues:
\begin{align} 500 - 200 &= 300 \\ 40 - 50 &= 0 \textit{, but there is still 10 that I have to subtract.} \\ 2 - 3 &= 0 \textit{, but there is still 1 that I have to subtract.} \end{align}
1. What must Jimmy still subtract, and what will his final answer be?
2. When Jimmy did another subtraction problem, he ended up with this writing at one stage:
600 and (-)50 and (-)7
What do you think is Jimmy's final answer for this subtraction problem?
About 500 years ago, some mathematicians proposed that a "negative number" may be used to describe the result in a situation such as in Jimmy's subtraction problem above, where a number is subtracted from a number smaller than itself.
For example, we may say $10 - 20 = (-10)$
This proposal was soon accepted by other mathematicians, and it is now used all over the world.
Mathematicians are people who do mathematics for a living. Mathematics is their profession, like health care is the profession of nurses and medical doctors.
1. Calculate each of the following:
1. $16 - 20$
2. $16 - 30$
3. $16 - 40$
4. $16 - 60$
5. $16 - 200$
6. $5 - 1 000$
2. Some numbers are shown on the lines below. Fill in the missing numbers.
The numbers 1; 2; 3; 4 etc. are called the natural numbers. The natural numbers, 0 and the negative whole numbers together are called the integers.
The following statement is true if the number is 5:
15 - (a certain number) = 10
A few centuries ago, some mathematicians decided they wanted to have numbers that will also make sentences like the following true:
15 + (a certain number) = 10
But to go from 15 to 10 you have to subtract 5.
The number we need to make the sentence 15 + (a certain number) = 10 true must have the following strange property:
If you add this number, it should have the same effect as to subtract 5.
Now the mathematicians of a few centuries ago really wanted to have numbers for which such strange sentences would be true. So they thought:
Let us decide, and agree amongst ourselves, that the number we call negative 5 will have the property that if you add it to another number, the effect will be the same as when you subtract the natural number 5.
This means that the mathematicians agreed that $15 + (-5)$ is equal to $15 - 5$.
Stated differently, instead of adding negative 5 to a number, you may subtract 5.
Adding a negative number has the same effect as subtracting a natural number.
For example: $20 + (-15) = 20 - 15 = 5$
1. Calculate each of the following:
1. $500 + (-300)$
2. $100 + (-20) + (-40)$
3. $500 + (-200) + (-100)$
4. $100 + (-60)$
2. Make a suggestion of what the answer for $(-20) + (-40)$ should be. Give reasons for your suggestion.
3. Continue the lists of numbers below to complete the table.
(a) (b) (c) (d) (e) (f) (g) 10 100 3 -3 -20 150 0 9 90 6 -6 -18 125 -5 8 80 9 -9 -16 100 -10 7 70 12 -12 -14 75 -15 6 60 15 -15 50 -20 5 50 -25 4 40 3 30 2 20 1 10 -1 -10
The following statement is true if the number is 5:
15 + (a certain number) = 20
What properties should a number have so that it makes the following statement true?
15 - (a certain number) = 20
To go from 15 to 20 you have to add 5. The number we need to make the sentence 15 - (a certain number) = 20 true must have the following property:
If you subtract this number, it should have the same effect as to add 5.
Let us agree that $15 - (-5)$ is equal to $15 + 5$.
Stated differently, instead of subtracting negative 5 from a number, you may add 5.
Subtracting a negative number has the same effect as adding a natural number.
For example: $20 - (-15) = 20 + 15 = 35$
1. Calculate.
1. $30 - (-10)$
2. $30 + 10$
3. $30 + (-10)$
4. $30 - 10$
5. $30 - (-30)$
6. $30 + 30$
7. $30 + (-30)$
8. $30 - 30$
You probably agree that
\begin{align} 5 + (-5) &= 0 \\ 10 + (-10) &= 0 \\ \text{and} \\ 20 + (-20) &= 0 \end{align}
We may say that for each "positive" number there is a corresponding or opposite negative number. Two positive and negative numbers that correspond, for example 3 and (-3), are called additive inverses. They wipe each other out when you add them.
What may each of the following be equal to?
$(-8) + 5$
$(-5) + (-8)$
When you add any number to its additive inverse, the answer is 0 (the additive property of 0). For example, $120 + (-120) = 0$.
1. Write the additive inverse of each of the following numbers:
1. 24
2. -24
3. -103
4. 2 348
The idea of additive inverses may be used to explain why $8 + (-5)$ is equal to 3:
$8 + (-5) = 3 + \boxed{5 + (-5)} = 3 + 0 = 3$
1. Use the idea of additive inverses to explain why each of these statements is true:
1. $43 + (-30) = 13$
2. $150 + (-80) = 70$
### Statements that are true for many different numbers
For how many different pairs of numbers can the following statement be true, if only natural (positive) numbers are allowed?
a number + another number = 10
For how many different pairs of numbers can the statement be true if negative numbers are also allowed?
## Adding and subtracting with integers
### Adding can make less and subtraction can make more
1. Calculate each of the following:
1. $10 + 4 + (-4)$
2. $10 + (-4) + 4$
3. $3 + 8 + (-8)$
4. $3 + (-8) + 8$
The numbers 1; 2; 3; 4; etc. that we use to count, are called natural numbers.
Natural numbers can be arranged in any order to add and subtract them. This is also the case for integers.
1. Calculate each of the following:
1. $18 + 12$
2. $12 + 18$
3. $2 + 4 + 6$
4. $6 + 4 + 2$
5. $2 + 6 + 4$
6. $4 + 2 + 6$
7. $4 + 6 + 2$
8. $6 + 2 + 4$
9. $6 + (-2) + 4$
10. $4 + 6 + (-2)$
11. $4 + (-2) + 6$
12. $(-2) + 4 + 6$
13. $6 + 4 + (-2)$
14. $(-2) + 6 + 4$
15. $(-6) + 4 + 2$
2. Calculate each of the following:
1. $(-5) + 10$
2. $10 + (-5)$
3. $(-8) + 20$
4. $20 - 8$
5. $30 + (-10)$
6. $30 + (-20)$
7. $30 + (-30)$
8. $10 + (-5) + (-3)$
9. $(-5) + 7 + (-3) + 5$
10. $(-5) + 2 + (-7) + 4$
3. In each case, find the number that makes the statement true. Give your answer by writing a closed number sentence.
1. 20 + (an unknown number) = 50
2. 50 + (an unknown number) = 20
3. 20 + (an unknown number) = 10
4. (an unknown number) + (-25) = 50
5. (an unknown number) + (-25) = -50
Statements like these are also called number sentences.
An incomplete number sentence, where some numbers are not known at first, is sometimes called an open number sentence:
8 - (a number) = 10
A closed number sentence is where all the numbers are known:
$8 + 2 = 10$
1. Use the idea of additive inverses to explain why each of the following statements is true:
1. $43 + (-50) = -7$
2. $60+ (-85) = -25$
2. Complete the table as far as you can.
(a) (b) (c) $5 - 8 =$ $5 + 8 =$ $8 - 3 =$ $5 - 7 =$ $5 + 7 =$ $7 - 3 =$ $5 - 6 =$ $5 + 6 =$ $6 - 3 =$ $5 - 5 =$ $5 + 5 =$ $5 - 3 =$ $5 - 4 =$ $5 + 4 =$ $4 - 3 =$ $5 - 3 =$ $5 + 3 =$ $3 - 3 =$ $5 - 2 =$ $5 + 2 =$ $2 - 3 =$ $5 - 1 =$ $5 + 1 =$ $1 - 3 =$ $5 - 0 =$ $5 + 0 =$ $0 - 3 =$ $5 - (-1) =$ $5 + (-1) =$ $(-1) - 3 =$ $5 - (-2) =$ $5 + (-2) =$ $(-2) - 3 =$ $5 - (-3) =$ $5 + (-3) =$ $(-3) - 3 =$ $5 - (-4) =$ $5 + (-4) =$ $(-4) - 3 =$ $5 - (-5) =$ $5 + (-5) =$ $(-5) - 3 =$ $5 - (-6) =$ $5 + (-6) =$ $(-6) - 3 =$
3. Calculate.
1. $80 + (-60)$
2. $500 + (-200) + (-200)$
1. Is $100 + (-20) + (-20) = 60$, or does it equal something else?
2. What do you think $(-20) + (-20)$ is equal to?
4. Calculate.
1. $20 - 20$
2. $50 - 20$
3. $(-20) - (-20)$
4. $(-50) - (-20)$
5. Calculate.
1. $20 - (-10)$
2. $100 - (-100)$
3. $20 + (-10)$
4. $100 + (-100)$
5. $(-20) - (-10)$
6. $(-100) - (-100)$
7. $(-20) + (-10)$
8. $(-100) + (-100)$
6. Complete the table as far as you can.
(a) (b) (c) $5 - (-8)$ = $(-5) + 8 =$ $8 - (-3) =$ $5 - (-7) =$ $(-5) + 7 =$ $7 - (-3) =$ $5 - (-6) =$ $(-5) + 6 =$ $6 - (-3) =$ $5 - (-5) =$ $(-5) + 5 =$ $5 - (-3) =$ $5 - (-4) =$ $(-5) + 4 =$ $4 - (-3) =$ $5 - (-3) =$ $(-5) + 3 =$ $3 - (-3) =$ $5 - (-2) =$ $(-5) + 2 =$ $2 - (-3) =$ $5 - (-1) =$ $(-5) + 1 =$ $1 - (-3) =$ $5 - 0 =$ $(-5) + 0 =$ $0 - (-3) =$ $5 - 1 =$ $(-5) + (-1) =$ $(-1) - (-3) =$ $5 - 2 =$ $(-5) + (-2) =$ $(-2) - (-3) =$ $5 - 3 =$ $(-5) + (-3) =$ $(-3) - (-3) =$ $5 - 4 =$ $(-5) + (-4) =$ $(-4) - (-3) =$ $5 - 5 =$ $(-5) + (-5) = -$ $(-5) - (-3) =$
7. In each case, state whether the statement is true or false and give a numerical example to demonstrate your answer.
1. Subtracting a positive number from a negative number has the same effect as dding the additive inverse of the positive number.
2. Adding a negative number to a positive number has the same effect as adding the additive inverse of the negative number.
3. Subtracting a negative number from a positive number has the same effect as subtracting the additive inverse of the negative number.
4. Adding a negative number to a positive number has the same effect as subtracting the additive inverse of the negative number.
5. Adding a positive number to a negative number has the same effect as adding the additive inverse of the positive number.
6. Adding a positive number to a negative number has the same effect as subtracting the additive inverse of the positive number.
7. Subtracting a positive number from a negative number has the same effect as subtracting the additive inverse of the positive number.
8. Subtracting a negative number from a positive number has the same effect as adding the additive inverse of the negative number.
### Comparing integers and solving problems
1. Fill <, > or = into the block to make the relationship between the numbers true:
1. -103 ☐ -99
2. -699 ☐ -701
3. 30 ☐ -30
4. 10-7 ☐ -(10-7)
5. -121 ☐ -200
6. -12 - 5 ☐ -(12 + 5)
7. -199 ☐ -110
2. At 5 a.m. in Bloemfontein the temperature was -5 °C. At 1 p.m., it was 19 °C. By how many degrees did the temperature rise?
3. A diver swims 150 m below the surface of the sea. She moves 75 m towards the surface. How far below the surface is she now?
4. One trench in the ocean is 800 m deep and another is 2 200 m deep. What is the difference in their depths?
5. An island has a mountain which is 1 200 m high. The surrounding ocean has a depth of 860 m. What is the difference in height?
6. On a winter's day in Upington the temperature rose by 19 °C. If the minimum temperature was -4 °C, what was the maximum temperature?
## Multiplying and dividing with integers
### Multiplication with integers
1. Calculate.
1. $-5 + -5 + -5 + -5 + -5 + -5 + -5 + -5 + -5 + -5$
2. $-10 + -10 + -10 + -10 + -10$
3. $-6 + -6 + -6 + -6 + -6 + -6 + -6 + -6$
4. $-8 + -8 + -8 + -8 + -8 + -8$
5. $-20 + -20 + -20 + -20 + -20 + -20 + -20$
2. In each case, show whether you agree (✓) or disagree (✗) with the given statement.
1. $10 \times (-5) = 50$
2. $8 \times (-6) = (-8) \times 6$
3. $(-5) \times 10 = 5 \times (-10)$
4. $6 \times (-8) = -48$
5. $(-5) \times 10 = 10 \times (-5)$
6. $8 \times (-6) = 48$
7. $4 \times 12 = -48$
8. $(-4) \times 12 = -48$
Multiplication of integers is commutative:
$(-20) \times 5 = 5 \times (-20)$
2. Calculate.
1. $20 \times (-10)$
2. $(-5) \times 4$
3. $(-20) \times 10$
4. $4 \times (-25)$
5. $29 \times (-20)$
6. $(-29) \times (-2)$
3. Calculate.
1. $10 \times 50 + 10 \times (-30)$
2. $50 + (-30)$
3. $10 \times {\bf(}50 + (-30){\bf)}$
4. $(-50) + (-30)$
5. $10 \times (-50) + 10 \times (-30)$
6. $10 \times {\bf(}(-50) + (-30){\bf)}$
The product of two positive numbers is a positive number, for example $5 \times 6 = 30$.
The product of a positive number and a negative number is a negative number, for example $5 \times (-6) = -30$.
The product of a negative number and a positive number is a negative number, for example $(-5) \times 6 = -30$.
1. Four numerical expressions are given below. Underline the expressions that you would expect to have the same answers. Do not do the calculations.
$14 \times (23 + 58) \\ 23 \times (14 + 58) \\ 14 \times 23 + 14 \times 58 \\ 14 \times 23 + 58$
2. What property of operations is demonstrated by the fact that two of the above expressions have the same value?
1. Does multiplication distribute over addition in the case of integers?
2. Three numerical expressions are given below. Underline the expressions that you would expect to have the same answers. Do not do the calculations.
$10 \times {\bf(}(-50) - (-30){\bf)} \\ 10 \times (-50) - (-30) \\ 10 \times (-50) - 10 \times (-30)$
3. Do the three sets ofcalculations given in question 8.
Your work in questions 5, 8 and 9 demonstrates that multiplication with a positive number distributes over addition and subtraction of integers. For example:
$10 \times {\bf(}5 + (-3){\bf)} = 10 \times 2 = {\bf20} \text{ and } 10 \times 5 + 10 \times (-3) = 50 + (-30) = {\bf 20}$
$10 \times {\bf(}5 - (-3){\bf)} = 10 \times 8 = {\bf 80} \text{ and } 10 \times 5 - 10 \times (-3) = 50 - (-30) = {\bf 80}$
1. Calculate: $(-10) \times {\bf(}5 + (-3){\bf)}$
Now consider the question of whether multiplication with a negative number distributes over addition and subtraction of integers. For example, would $(-10) \times 5 + (-10) \times (-3)$ also have the answer -20, as does $(-10) \times {\bf(}5 + (-3){\bf)}$?
1. What must $(-10) \times (-3)$ be equal to, if we want $(-10) \times 5 + (-10) \times (-3)$ to be equal to -20?
In order to ensure that multiplication distributes over addition and subtraction in the system of integers, we have to agree that
(a negative number) $\times$ (a negative number) is a positive number,
for example $(-10) \times (-3) = 30$.
1. Calculate.
1. $(-10) \times (-5)$
2. $(-10) \times 5$
3. $10 \times 5$
4. $10 \times (-5)$
5. $(-20) \times (-10) + (-20) \times (-6)$
6. $(-20) \times {\bf(}(-10) + (-6){\bf)}]$
7. $(-20) \times (-10) - (-20) \times (-6)$
8. $(-20) \times {\bf(}(-10) - (-6){\bf)}$
Here is a summary of the properties of integers that make it possible to do calculations with integers:
• When a number is added to its additive inverse, the result is 0, for example $(+12) + (-12) = 0$.
• Adding an integer has the same effect as subtracting its additive inverse. For example, $3 + (-10)$ can be calculated by doing $3 - 10$, and the answer is -7.
• Subtracting an integer has the same effect as adding its additive inverse. For example, $3 - (-10)$ can be calculated by doing $3 + 10$, and the answer is 13.
• The product of a positive and a negative integer is negative, for example $(-15) \times 6 = -90$.
• The product of a negative and a negative integer is positive, for example $(-15) \times (-6) = 90$.
### Division with integers
1. Calculate $25 \times 8$.
2. How much is $200 \div 25?$
3. How much is $200 \div 8?$
Division is the inverse of multiplication. Hence, if two numbers and the value of their product are known, the answers to two division problems are also known.
1. Calculate.
1. $25 \times\ (-8) )$
2. $(-125) \times 8$
2. Use the work you have done for question 2 to write the answers for the following division questions:
1. $(-1 000) \div (-125)$
2. $(-1 000) \div 8$
3. $(-200) \div 25$
4. $(-200) \div 8$
3. Can you also work out the answers for the following division questions by using the work you have done for question 2?
1. $1 000 \div (-125)$
2. $(-1 000) \div (-8)$
3. $(-100) \div (-25)$
4. $100 \div (-25)$
When two numbers are multiplied, for example $30 \times 4 = 120$, the word "product" can be used in various ways to describe the situation:
• An expression that specifies multiplication only, such as $30 \times 4$, is called a product or a product expression.
• The answer obtained is also called the product of the two numbers. For example, 120 is called the product of 30 and 4.
An expression that specifies division only, such as $30 \div 5$, is called a quotient or a quotient expression. The answer obtained is also called the quotient of the two numbers. For example, 6 is called the quotient of 30 and 5.
1. In each case, state whether you agree or disagree with the statement, and give an example to illustrate your answer.
1. The quotient of a positive and a negative integer is negative.
2. The quotient of a positive and a positive integer is negative.
3. The quotient of a negative and a negative integer is negative.
4. The quotient of a negative and a negative integer is positive.
2. Do the necessary calculations to enable you to provide the values of the quotients.
1. $(-500) \div (-20)$
2. $(-144) \div 6$
3. $1 440 \div (-60)$
4. $(-1 440) \div (-6)$
5. $-14 400 \div 600$
6. $500 \div (-20)$
### The associative properties of operations with integers
Multiplication of whole numbers is associative. This means that in a product with several factors, the factors can be placed in any sequence, and the calculations can be performed in any sequence. For example, the following sequences of calculations will all produce the same answer:
1. $2 \times 3$, the answer of $2 \times 3$ multiplied by 5, the new answer multiplied by 10
2. $2 \times 5$, the answer of $2 \times 5$ multiplied by 10, the new answer multiplied by 3
3. $10 \times 5$, the answer of $10 \times 5$ multiplied by 3, the new answer multiplied by 2
4. $3 \times 5$, the answer of $3 \times 5$ multiplied by 2, the new answer multiplied by 10
1. Do the four sets of calculations given in A to D to check whether they really produce the same answers.
1. If the numbers 3 and 10 in the calculation sequences A, B, C and D are replaced with -3 and -10, do you think the four answers will still be the same?
2. Investigate, to check your expectation.
Multiplication with integers is associative.
The calculation sequence A can be represented in symbols in only two ways:
• $2 \times 3 \times 5 \times 10$. The convention to work from left to right unless otherwise indicated with brackets ensures that this representation corresponds to A.
• $5 \times (2 \times 3) \times 10$, where brackets are used to indicate that $2 \times 3$ should be calculated first. When brackets are used, there are different possibilities to describe the same sequence.
1. Express the calculation sequences B, C and D given above symbolically, without using brackets.
2. Investigate, in the same way that you did for multiplication in question 2, whether addition with integers is associative. Use sequences of four integers.
1. Calculate: $80 - 30 + 40 - 20$
2. Calculate: $80 + (-30) + 40 + (-20)$
3. Calculate: $30 - 80 + 20 - 40$
4. Calculate: $(-30) + 80 + (- 20) + 40$
5. Calculate: $20 + 30 - 40 - 80$
### Mixed calculations with integers
1. Calculate.
1. $-3 \times 4 + (-7) \times 9$
2. $-20(-4 - 7)$
3. $20 \times (-5) - 30 \times 7$
4. $-9(20 - 15)$
5. $-8 \times (-6) - 8 \times 3$
6. $(-26 - 13) \div (-3)$
7. $-15 \times (-2) + (-15) \div (-3)$
8. $-15(2 - 3)$
9. $(-5 + -3) \times 7$
10. $-5 \times (-3 + 7) + 20 \div (-4)$
2. Calculate.
1. $20 \times (-15 + 6) - 5 \times (-2 - 8) - 3 \times (-3 - 8)$
2. $40 \times (7 + 12 - 9) + 25 \div (-5) - 5 \div 5$
3. $-50(20 - 25) + 30(-10 + 7) - 20(-16 + 12)$
4. $-5 \times (-3 + 12 - 9)$
5. $-4 \times (30 - 50) + 7 \times (40 - 70) - 10 \times (60 - 100)$
6. $-3 \times (-14 + 6) \times (-13 + 7) \times (-20 + 5)$
7. $20 \times (-5) + 10 \times (-3) + (-5) \times (-6) - (3 \times 5)$
8. $-5(-20 - 5) + 10(-7 - 3) - 20(-15 - 5) + 30(-40 - 35)$
9. $(-50 + 15 - 75) \div (-11) + (6 - 30 + 12) \div (-6)$
## Squares, cubes and roots with integers
### Squares and cubes of integers
1. Calculate.
1. $20 \times 20$
2. $20 \times (-20)$
2. Write the answers for each of the following:
1. $(-20) \times 20$
2. $(-20) \times (-20)$
3. Complete the table.
$x$ 1 -1 2 -2 5 -5 10 -10 $x^2$ which is $x \times x$ $x^3$
4. In each case, state for which values of $x$, in the table in question 3, the given statement is true.
1. $x^3$ is a negative number
2. $x^2$ is a negative number
3. $x^2$ > $x^3$
4. $x^2$ < $x^3$
5. Complete the table.
$x$ 3 -3 4 -4 6 -6 7 -7 $x^2$ $x^3$
6. Ben thinks of a number. He adds 5 to it, and his answer is 12.
1. What number did he think of?
2. Is there another number that would also give 12 when 5 is added to it?
7. Lebo also thinks of a number. She multiplies the number by itself and gets 25.
1. What number did she think of?
2. Is there more than one number that will give 25 when multiplied by itself?
8. Mary thinks of a number and calculates (the number) $\times$ (the number) $\times$ (the number). Her answer is 27.
What number did Mary think of?
$10^2$ is 100 and $(-10)^2$ is also 100.
Both 10 and (-10) are called square roots of 100. 10 maybe called the posistive square root of 100> and (-10) may be called the neagtive square root of 100
1. Write the positive square root and the negative square root of each number.
1. 64
2. 9
2. Complete the table.
Number 1 4 9 16 25 36 49 64 Positive square root 3 8 Negative square root -3 -8
3. Complete the tables.
1. $x$ 1 2 3 4 5 6 7 8 $x^3$
2. $x$ -1 -2 -3 -4 -5 -6 -7 -8 $x^3$
$3^3$ is 27 and $(-5)^3$ is -125.
3 is called the cube root of 27, because $3^3 = 27$.
-5 is called the cube root of -125 because $(-5)^3 = -125$.
1. Complete the table.
Number -1 8 -27 -64 -125 -216 1 000 Cube root -3 10
The symbol $\sqrt{}$ is used to indicate "root".
$\sqrt[3]{-125}$ represents the cube root of -125. That means $\sqrt[3]{-125} = -5$.
$\sqrt[2]{36}$ represents the positive square root of 36, and $- \sqrt[2]{36}$ represents the negative square root. The "2" that indicates "square" is normally omitted, so $\sqrt{36} = 6$ and $- \sqrt{36} = - 6$.
1. Complete the table.
$\sqrt[3]{-8}$ $\sqrt{121}$ $\sqrt[3]{-64}$ $- \sqrt{64}$ $\sqrt{64}$ $\sqrt[3]{-1}$ $-\sqrt{1}$ $\sqrt[3]{-216}$
1. Use the numbers -8, -5 and -3 to demonstrate each of the following:
1. Multiplication with integers distributes over addition.
2. Multiplication with integers distributes over subtraction.
3. Multiplication with integers is associative.
4. Addition with integers is associative.
2. Calculate each of the following without using a calculator:
1. $5 \times (-2)^3$
2. $3 \times (-5)^2$
3. $2 \times (-5)^3$
4. $10 \times (-3)^2$
3. Use a calculator to calculate each of the following:
1. $24 \times (-53) + (-27) \times (-34) - (-55) \times 76$
2. $64 \times (27 - 85) - 29 \times (-47 + 12)$
4. Use a calculator to calculate each of the following:
1. $-24 \times 53 + 27 \times 34 + 55 \times 76$
2. $64 \times (-58) + 29 \times (47 - 12)$
If you don't get the same answers in questions 3 and 4, you have made mistakes. |
# algebraic expression examples
Document Sample
``` Section 1.2 – Algebraic Expressions and Models
Numerical expression – numbers, operations, and grouping
symbols
25 = 22222 2 to the 5th power
5 facotrs of 2
Base – 2
Exponent – 5
Power – the expression 25
Example 1/ Evaluating Powers
a. (-3)4 =
b. -34 =
ORDER OF OPERATIONS
1. ______________________________________________
2. ______________________________________________
3. ______________________________________________
4. ______________________________________________
Example 2/ Using Order of Operations
1.) -4 + 2(-2 + 5)2 2.) -8 + 5(1 – (-3))3
Variable – a letter that is used to represent one or more numbers
Algebraic Expression – an expression involving variables
Example 3/ Evaluating an Algebraic Expression
Evaluate –3x2 – 5x + 7 when x = -2
Example 4/ Writing and Evaluating a Real-Life Model
You have \$75 and are buying some CDs that cost \$15 each.
Write an expression that shows how much money you have left
after buying n movies. Evaluate the expression when n = 2 and n =
3.
SIMPLIFYING ALGEBRAIC EXPRESSIONS
- To simplify algebraic expressions you must combine like terms.
Like terms are parts that have the same variable with the
same exponent. When combining like terms you add the
coefficients of the like terms together.
Like Terms Unlike Terms
2 2
6x , 2x 6x, 2x2
Example 5/ Simplifying by Combining Like Terms
a. 7x + 4x
b. 3n2 + n – n2
c. 2(x + 1) – 3(x – 4)
Example 6/ Using a Real-Life Model
You want to buy either a CD or a cassette as a gift for each
of 10 people. CDs cost \$13 each and cassettes cost \$8
each. Write an expression for the total amount you must
spend. Then evaluate the expression when 4 of the people
get CDs.
Example 7/
Write an expression for the total monthly cost of phone
service if you pay a \$5 fee and \$0.08 per minute. Then,
find the cost if you talk 6 hours during the month.
HOMEWORK – ASSIGNMENT #2
```
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# 4.3 Extending Operations to Fractions
## Unit Goals
• Students learn that a fraction $\frac{a}{b}$ is a product of a whole number $a$ and a unit fraction $\frac{1}{b}$, or $\frac{a}{b} = a \times \frac{1}{b}$, and that $n \times \frac{a}{b} = \frac{(n \space \times \space a)}{b}$. Students learn to add and subtract fractions with like denominators, and to add and subtract tenths and hundredths.
### Section A Goals
• Recognize that $n \times \frac{a}{b} = \frac{(n \space \times \space a)}{b}$.
• Represent and explain that a fraction $\frac{a}{b}$ is a multiple of $\frac{1}{b}$, namely $a \times \frac{1}{b}$.
• Represent and solve problems involving multiplication of a fraction by a whole number.
### Section B Goals
• Create and analyze line plots that display measurement data in fractions of a unit ($\frac18, \frac14, \frac12$).
• Represent and solve problems that involve the addition and subtraction of fractions and mixed numbers, including measurements presented in line plots.
• Use various strategies to add and subtract fractions and mixed numbers with like denominators.
### Section C Goals
• Reason about equivalence to solve problems involving addition and subtraction of fractions and mixed numbers.
### Problem 1
#### Pre-unit
Practicing Standards: 3.NF.A.1
What fraction of the rectangle is shaded? Explain how you know.
### Solution
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### Problem 2
#### Pre-unit
Practicing Standards: 3.NF.A.2
1. Locate and label $$\frac{3}{4}$$ and $$\frac{6}{4}$$ on the number line.
2. Explain why your points represent $$\frac{3}{4}$$ and $$\frac{6}{4}$$.
### Solution
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### Problem 3
#### Pre-unit
Practicing Standards: 3.OA.A.1
Write a multiplication expression for each image. Explain your reasoning.
1.
2.
### Solution
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### Problem 4
#### Pre-unit
Practicing Standards: 3.MD.B.4
Here are the lengths of some lizards in inches. Use the lengths to complete the line plot.
• $$2\frac{1}{4}$$
• $$1\frac{1}{2}$$
• $$2\frac{2}{4}$$
• 3
• $$3\frac{2}{4}$$
• 2
• $$2\frac{1}{4}$$
• $$2 \frac{1}{4}$$
• $$2\frac{3}{4}$$
• 2
• $$2\frac{1}{4}$$
• 3
### Solution
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### Problem 5
Write an expression that matches each diagram. Then, find the value of each expression.
### Solution
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### Problem 6
Five friends go on a hike. They each bring $$\frac{1}{4}$$ cup of nuts.
1. If the shaded parts represent the amount of nuts the friends bring on their hike, which diagram matches the story? Explain your reasoning.
2. How many cups of nuts do the friends bring on the hike?
### Solution
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### Problem 7
Kiran’s cat eats $$\frac{1}{2}$$ cup of food each day.
1. How much food does Kiran’s cat eat in a week?
2. Draw a diagram to represent the situation.
### Solution
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### Problem 8
1. Draw a diagram to show $$3 \times \frac{7}{8}$$.
2. How does the diagram help you find the value of the expression $$3 \times \frac{7}{8}$$?
### Solution
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### Problem 9
Find the number that makes each equation true. Draw a diagram if it is helpful.
1. $$\frac{10}{3} = \underline{\hspace{0.7cm}} \times \frac{1}{3}$$
2. $$\frac{10}{3} = \underline{\hspace{0.7cm}} \times \frac{2}{3}$$
3. $$\frac{10}{3} = \underline{\hspace{0.7cm}} \times \frac{5}{3}$$
### Solution
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### Problem 10
Each bead weighs $$\frac{5}{8}$$ gram. How much do 7 beads weigh? Explain or show your reasoning.
### Solution
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### Problem 11
#### Exploration
1. Measure how thick your workbook is to the nearest $$\frac{1}{8}$$ inch.
2. If all of your classmates stacked their workbooks together, how tall would the stack be? Explain or show your reasoning.
### Solution
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### Problem 12
#### Exploration
Diego walked the same number of miles to school each day. He says that he walked $$\frac{48}{5}$$ miles in total, but does not say how many days that distance includes.
What are some possible number of days Diego counted and the distance he walked each of those days?
### Solution
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### Problem 1
1. Write $$\frac{4}{3}$$ in as many ways as you can as a sum of fractions.
2. Write $$\frac{9}{8}$$ in at least 3 different ways as a sum of fractions.
### Solution
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### Problem 2
1. Draw “jumps” on the number lines to show two ways to use fourths to make a sum of $$\frac{7}{4}$$.
2. Represent each combination of jumps as an equation.
### Solution
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### Problem 3
1. Explain how the diagram represents $$\frac{13}{5} - \frac{4}{5}$$.
Use the diagram to find the value of $$\frac{13}{5}- \frac{4}{5}$$.
2. Use a number line to represent and find the difference $$\frac{9}{4} - \frac{3}{4}$$.
### Solution
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### Problem 4
Show two different ways to find the difference: $$2 - \frac{3}{4}$$
### Solution
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### Problem 5
Elena is making friendship necklaces and wants the chain and clasp to be a total of $$18\frac{1}{4}$$ inches long. She is going to use a clasp that is $$2\frac{3}{4}$$ inches long. How long does her chain need to be? Explain or show your reasoning.
### Solution
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### Problem 6
For each of the expressions, explain whether you think it would be helpful to decompose one or more numbers to find the value of the expression.
1. $$\frac{4}{3} + \frac{5}{3}$$
2. $$5\frac{1}{5} - 2\frac{2}{5}$$
3. $$9\frac{5}{6} - 6\frac{1}{6}$$
### Solution
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### Problem 7
The lengths of the shoes of a dad and his two daughters are shown.
For each question, show your reasoning.
1. How much longer is the older daughter’s shoes than her sister’s?
2. Which is longer, the dad’s shoes or the combined lengths of his daughters’ shoes?
### Solution
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### Problem 8
#### Exploration
A chocolate chip cookie recipe calls for $$2\frac{3}{4}$$ cups of flour. You only have a $$\frac{1}{4}$$-cup measuring cup and a $$\frac{3}{4}$$-cup measuring cup that you can use.
1. What are different combinations of the measuring cups that you can use to get a total of $$2\frac{3}{4}$$ cups of flour?
2. Write each of the combinations as an addition equation.
### Solution
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### Problem 9
#### Exploration
The table shows some lengths of different shoe sizes in inches.
U.S. shoe size insole length
1 $$7\frac{6}{8}$$
1.5 8
2 $$8\frac{1}{8}$$
2.5 $$8\frac{2}{8}$$
3 $$8\frac{4}{8}$$
3.5 $$8\frac{5}{8}$$
4 $$8\frac{6}{8}$$
4.5 9
5 $$9\frac{1}{8}$$
5.5 $$9\frac{2}{8}$$
6 $$9\frac{4}{8}$$
6.5 $$9\frac{5}{8}$$
7 $$9\frac{6}{8}$$
1. What do you notice about the insole lengths as the size increases?
2. What will the insole length increase be from size 7 to 7.5? What is the insole length of a size 7.5 shoe?
3. Predict the insole length for sizes 9, 10, and 12. Explain your prediction. Then solve to find out if your prediction is true.
### Solution
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### Problem 1
Andre is building a tower out of different foam blocks. These blocks come in three different thicknesses: $$\frac{1}{2}$$-foot, $$\frac{1}{4}$$-foot, and $$\frac{1}{8}$$-foot.
Andre stacks two $$\frac{1}{2}$$-foot blocks, two $$\frac{1}{4}$$-foot blocks, and two $$\frac{1}{8}$$-foot blocks to create a tower. What will the height of the tower be in feet? Explain or show how you know.
### Solution
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### Problem 2
Find the value of each of the following sums. Show your reasoning. Use number lines if you find them helpful.
1. $$\frac{1}{10} + \frac{3}{100}$$
2. $$\frac{24}{100} + \frac{4}{10}$$
3. $$\frac{7}{10} + \frac{13}{100}$$
### Solution
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### Problem 3
Is the value of each expression greater than, less than or equal to 1? Explain how you know.
1. $$\frac{3}{10} + \frac{7}{100}$$
2. $$\frac{13}{10} + \frac{7}{100}$$
3. $$\frac{30}{100} + \frac{7}{10}$$
### Solution
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### Problem 4
Diego and Lin continued to play with their coins.
Diego said that he has exactly 3 coins whose thickness adds up to $$\frac{50}{100}$$ cm. What coins does Diego have? Explain or show your reasoning.
coin thickness in cm
1 centavo $$\frac{12}{100}$$
10 centavos $$\frac{22}{100}$$
1 peso $$\frac{16}{100}$$
2 pesos $$\frac{14}{100}$$
5 pesos $$\frac{2}{10}$$
20 pesos $$\frac{25}{100}$$
### Solution
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### Problem 5
#### Exploration
A chocolate cake recipe calls for 2 cups of flour. You gather your measuring cups and notice you have these sizes: $$\frac{1}{2}$$ cup, $$\frac{1}{3}$$ cup, $$\frac{1}{4}$$ cup, and $$\frac{1}{6}$$ cup.
1. What are the different ways you could use all 4 measuring cups to measure 2 cups of flour?
2. What are other ways you could use just some of the 4 measuring cups to measure exactly 2 cups of flour?
### Solution
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### Problem 6
#### Exploration
A dime is worth $$\frac{1}{10}$$ of a dollar and a penny is worth $$\frac{1}{100}$$ of a dollar.
1. If I have $$\frac{89}{100}$$ of a dollar, how many different combinations of dimes and pennies could I have? Use equations to show your reasoning.
2. A nickel is worth $$\frac{5}{100}$$ of a dollar. How many different combinations of dimes, nickels and pennies could I have if I still have $$\frac{89}{100}$$ of a dollar? Use equations to show your reasoning.
### Solution
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# Angle Between Two Straight Lines
THEOREM 1:
To find the angle between two given straight lines.
Solution:
Let the two straight lines be AT1 and AT2 which meet at x-axis at T1 and T2 points respectively.
Let the equations of the two lines AT1 and AT2 be
y = m1x + c1 and y2 = m2x + c2 (1)
Hence tan AT1X = m1 and tan AT2X = m2
Now ∠T1AT2 = ∠AT1X - ∠AT2X
Hence tan T1AT2 = tan [AT1X - AT2X]
Hence the required angle = ∠T1AT2
(2)
If equation (2) is a positive quantity then it is the tangent of the acute angle between the two lines, if (2) is a negative quantity then it is the tangent of the obtuse angle.
Example:
Find the angle between two straight lines y = 6x - 11 and
Solution:
The equation of the first line is y = 6x - 11 which implies m1 = 6.
The equation of the second line is hence
THEOREM 2:
To find the condition that two straight lines can be parallel.
Two straight lines are parallel when the angle between them is zero and hence the tangent of this angle is zero. Hence in theorem 1 the equation (2) becomes
m1 = m2 which is the required result.
Example:
Find the equation of a straight line which passes through the point (4, -5) and which is parallel to the straight line 3x + 4y + 5 = 0.
Solution:
Any straight line which is parallel to 3x + 4y + 5 = 0 will be 3x + 4y + c2 = 0 because m1 = m2.
Now given that this line 3x + 4y + c2 = 0 passes through the point (4, -5) hence we get,
3. 4 + 4. (-5) + c2 = 0 or c2 = 8
Putting the value of c2 = 8 we get 3x + 4y + 8 = 0. (Answer)
THEOREM 3:
To find the condition that two straight lines can be perpendicular.
Let the two straight lines be y = m1x + c1 and y2 = m 2x + c2
If the angle between them is θ then by theorem 1,
(1)
If the two lines are perpendicular then θ = 90 ο and so tan θ = ∞.
Hence in equation (1) which is only possible if the denominator is zero
Therefore the condition of perpendicularity is
1 + m1m2 = 0 or m1m2 = -1.
Hence the straight line y2= m2x + c2 is perpendicular to y = m1x + c1 if
Example:
Find the equation of a straight line which passes through the point (4, -5) and is perpendicular to the straight line 3x + 4y + 5 = 0.
Solution:
Let the equation of a straight line be y = m1x + c
Now given that this line passes through the point (4, -5) hence it becomes, -5 = 4m1 + c (1)
Given that the above line is perpendicular to the straight line 3x + 4y + 5 = 0 (2)
From (2) we get where
Then by the condition of perpendicularity m1m2 = -1.
Hence
Or putting which in eq. (1) we get,
Hence the required equation will be Or 4x - 3y = 31 (Answer)
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# Matrix Solution of Equations
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1 Contents 8 Matrix Solution of Equations 8.1 Solution by Cramer s Rule Solution by Inverse Matrix Method Solution by Gauss Elimination 22 Learning outcomes In this Workbook you will learn to apply your knowledge of matrices to solve systems of linear equations. Such systems of equations arise very often in mathematics, science and engineering. Three basic techniques are outlined, Cramer's method, the inverse matrix approach and the Gauss elimination method. The Gauss elimination method is, by far, the most widely used (since it can be applied to all systems of linear equations). However, you will learn that, for certain (usually small) systems of linear equations the other two techniques may be better.
2 Solution by Cramer s Rule 8.1 Introduction The need to solve systems of linear equations arises frequently in engineering. The analysis of electric circuits and the control of systems are two examples. Cramer s rule for solving such systems involves the calculation of determinants and their ratio. For systems containing only a few equations it is a useful method of solution. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be able to evaluate 2 2 and 3 3 determinants state and apply Cramer s rule to find the solution of two simultaneous linear equations state and apply Cramer s rule to find the solution of three simultaneous linear equations recognise cases where the solution is not unique or a solution does not exist 2 HELM (2008): Workbook 8: Matrix Solution of Equations
3 1. Solving two equations in two unknowns If we have one linear equation ax = b in which the unknown is x and a and b are constants then there are just three possibilities: a 0 then x = b a = a 1 b. In this case the equation ax = b has a unique solution for x. a = 0, b = 0 then the equation ax = b becomes 0 = 0 and any value of x will do. In this case the equation ax = b has infinitely many solutions. a = 0 and b 0 then ax = b becomes 0 = b which is a contradiction. In this case the equation ax = b has no solution for x. What happens if we have more than one equation and more than one unknown? We shall find that the solutions to such systems can be characterised in a manner similar to that occurring for a single equation; that is, a system may have a unique solution, an infinity of solutions or no solution at all. In this Section we examine a method, known as Cramer s rule and employing determinants, for solving systems of linear equations. Consider the equations ax + by = e cx + dy = f (1) (2) where a, b, c, d, e, f are given numbers. The variables x and y are unknowns we wish to find. The pairs of values of x and y which simultaneously satisfy both equations are called solutions. Simple algebra will eliminate the variable y between these equations. We multiply equation (1) by d, equation (2) by b and subtract: first, (1) d adx + bdy = ed then, (2) b bcx + bdy = bf (we multiplied in this way to make the coefficients of y equal.) Now subtract to obtain (ad bc)x = ed bf (3) Task Starting with equations (1) and (2) above, eliminate x. HELM (2008): Section 8.1: Solution by Cramer s Rule 3
4 Answer Multiply equation (1) by c and equation (2) by a to obtain acx + bcy = ec and acx + ady = af. Now subtract to obtain (bc ad)y = ec af If we multiply this last equation in the Task above by 1 we obtain (ad bc)y = af ec (4) Dividing equations (3) and (4) by ad bc we obtain the solutions x = ed bf ad bc, af ec y = ad bc (5) There is of course one proviso: if ad bc = 0 then neither x nor y has a defined value. If we choose to express these solutions in terms of determinants we have the formulation for the solution of simultaneous equations known as Cramer s rule. If we define as the determinant equations a c b d and provided 0 then the unique solution of the ax + by = e cx + dy = f is by (5) given by x = x, y = y where x = e f b d, y = a c e f Now is the determinant of coefficients ( ) on the left-hand sides of the equations. In the expression a x the coefficients of x (i.e. which is column 1 of ) are replaced by the terms on the c ( ) e right-hand sides of the equations (i.e. by ). Similarly in f y the coefficients of y (column 2 of ) are replaced by the terms on the right-hand sides of the equations. 4 HELM (2008): Workbook 8: Matrix Solution of Equations
5 The unique solution to the equations: is given by: in which = a c Key Point 1 Cramer s Rule for Two Equations ax + by = e cx + dy = f x = x, b d x = e f If = 0 this method of solution cannot be used. y = y b d, y = a c e f Task Use Cramer s rule to solve the simultaneous equations 2x + y = 7 3x 4y = 5 Answer Calculating = = 11. Since 0 we can proceed with Cramer s solution. = = 11 x = , y = ( 28 5) (10 21) i.e. x =, y = implying: x = 33 ( 11) ( 11) 11 = 3, y = = 1. You can check by direct substitution that these are the exact solutions to the equations. HELM (2008): Section 8.1: Solution by Cramer s Rule 5
6 Task Use Cramer s rule to solve the equations (a) 2x 3y = 6 4x 6y = 12 (b) 2x 3y = 6 4x 6y = 10 Answer You should have checked first, since = 12 ( 12) = 0. Hence there is no unique solution in either case. In the system (a) the second equation is twice the first so there are infinitely many solutions. (Here we can give y any value we wish, t say; but then the x value is always (6 + 3t)/2. So for each value of t there are values for x and y which simultaneously satisfy both equations. There is an infinite number of possible solutions). In (b) the equations are inconsistent (since the first is 2x 3y = 6 and the second is 2x 3y = 5 which is not possible). Hence there are no solutions. Notation For ease of generalisation to larger systems we write the two-equation system in a different notation: a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 Here the unknowns are x 1 and x 2, the right-hand sides are b 1 and b 2 and the coefficients are a ij where, for example, a 21 is the coefficient of x 1 in equation two. In general, a ij is the coefficient of x j in equation i. Cramer s rule can then be stated as follows: If a 11 a 12 a 21 a 22 0, then the equations a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 have solution b 1 a 12 b 2 a 22 x 1 = a 11 a 12 a 21 a 22, x 2 = a 11 b 1 a 21 b 2 a 11 a 12. a 21 a 22 6 HELM (2008): Workbook 8: Matrix Solution of Equations
7 2. Solving three equations in three unknowns Cramer s rule can be extended to larger systems of simultaneous equations but the calculational effort increases rapidly as the size of the system increases. We quote Cramer s rule for a system of three equations. Key Point 2 Cramer s Rule for Three Equations The unique solution to the system of equations: a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 is in which and x1 = b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 x 1 = x 1, x 2 = x 2, x 3 = x 3 = x2 = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 x3 = a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 If = 0 this method of solution cannot be used. Notice that the structure of the fractions is similar to that for the two-equation case. For example, the determinant forming the numerator of x 1 is obtained from the determinant of coefficients,, by replacing the first column by the right-hand sides of the equations. Notice too the increase in calculation: in the two-equation case we had to evaluate three 2 2 determinants, whereas in the three-equation case we have to evaluate four 3 3 determinants. Hence Cramer s rule is not really practicable for larger systems. HELM (2008): Section 8.1: Solution by Cramer s Rule 7
8 Task Use Cramer s rule to solve the system x 1 2x 2 + x 3 = 3 2x 1 + x 2 x 3 = 5 3x 1 x 2 + 2x 3 = 12. First check that 0: Answer = Expanding along the top row, = ( 2) = 1 (2 1) + 2 (4 + 3) + 1 ( 2 3) = = 10 Now find the value of x 1. First write down the expression for x 1 in terms of determinants: Answer x 1 = Now calculate x 1 explicitly: 8 HELM (2008): Workbook 8: Matrix Solution of Equations
9 Answer The numerator is found by expanding along the top row to be ( 2) = ( 17) = 30 Hence x 1 = = 3 In a similar way find the values of x 2 and x 3 : Answer x 2 = 1 10 = { } = 1 { } = 1 10 x 3 = 1 { ( 2) = 1 { ( 5)} = } HELM (2008): Section 8.1: Solution by Cramer s Rule 9
10 Engineering Example 1 Stresses and strains on a section of material Introduction An important engineering problem is to determine the effect on materials of different types of loading. One way of measuring the effects is through the strain or fractional change in dimensions in the material which can be measured using a strain gauge. Problem in words In a homogeneous, isotropic and linearly elastic material, the strains (i.e. fractional displacements) on a section of the material, represented by ε x, ε y, ε z for the x-, y-, z-directions respectively, can be related to the stresses (i.e. force per unit area), σ x, σ y, σ z by the following system of equations. ε x = 1 E ( σ x vσ y vσ z ) ε y = 1 E ( vσ x +σ y vσ z ) ε z = 1 E ( vσ x vσ y +σ z ) where E is the modulus of elasticity (also called Young s modulus) and v is Poisson s ratio which relates the lateral strain to the axial strain. Find expressions for the stresses σ x, σ y, σ z, in terms of the strains ε x, ε y, and ε z. Mathematical statement of problem The given system of equations can be written as a matrix equation: ε x ε y = 1 1 v v σ x v 1 v σ y E ε z v v 1 σ z We can write this equation as ε = 1 E Aσ ε x where ε = ε y, A = ε z 1 v v v 1 v v v 1 and σ = This matrix equation must be solved to find the vector σ in terms of the vector ε and the inverse of the matrix A. σ x σ y σ z 10 HELM (2008): Workbook 8: Matrix Solution of Equations
11 Mathematical analysis ε = 1 E Aσ Multiplying both sides of the expression by E we get Eε = Aσ Multiplying both sides by A 1 we find that: A 1 Eε = A 1 Aσ But A 1 A = I so this becomes σ = EA 1 ε To find expressions for the stresses σ x, σ y, σ z, in terms of the strains ε x, ε y and ε z, we must find the inverse of the matrix A. 1 v v To find the inverse of v 1 v we first find the matrix of minors which is: v v 1 1 v v 1 v v v 1 v v 1 v v v v 1 1 v v 1 1 v v v We then apply the pattern of signs: to obtain the matrix of cofactors 1 v 2 v + v 2 v 2 + v v + v 2 1 v 2 v + v 2. v 2 + v v + v 2 1 v 2 v 1 v v 1 v v v = 1 v v 1 1 v 2 v v 2 v 2 + v v v 2 1 v 2 v v 2 v 2 + v v v 2 1 v 2 To find the adjoint we take the transpose of the above, (which is the same as the original matrix since the matrix is symmetric) 1 v 2 v + v 2 v 2 + v v + v 2 1 v 2 v + v 2. v 2 + v v + v 2 1 v 2 The determinant of the original matrix is 1 (1 v 2 ) v(v + v 2 ) v(v 2 + v) = 1 3v 2 2v 3.. HELM (2008): Section 8.1: Solution by Cramer s Rule 11
12 Finally we divide the adjoint by the determinant to find the inverse, giving 1 v 1 2 v + v 2 v + v 2 v + v 2 1 v 2 v + v v 2 2v 3 v + v 2 v + v 2 1 v 2 σ x 1 v Now we found that σ = EA 1 ε so σ y E 2 v + v 2 v + v 2 ε x = v + v 2 1 v 2 v + v 2 ε 1 3v σ 2 2v 3 y z v + v 2 v + v 2 1 v 2 ε z We can write this matrix equation as 3 equations relating the stresses σ x, σ y, σ z, in terms of the strains ε x, ε y and ε z, by multiplying out this matrix expression, giving: σ x = σ y = σ z = E ( ) (1 v 2 )ε 1 3v 2 2v 3 x + (v + v 2 )ε y + (v + v 2 )ε z E ( ) (v + v 2 )ε 1 3v 2 2v 3 x + (1 v 2 )ε y + (v + v 2 )ε z E ( ) (v + v 2 )ε 1 3v 2 2v 3 x + (v + v 2 )ε y + (1 v 2 )ε z Interpretation Matrix manipulation has been used to transform three simultaneous equations relating strain to stress into simultaneous equations relating stress to strain in terms of the elastic constants. These would be useful for deducing the applied stress if the strains are known. The original equations enable calculation of strains if the applied stresses are known. 1. Solve the following using Cramer s rule: (a) 2x 3y = 1 4x + 4y = 2 (b) Exercises 2x 5y = 2 4x + 10y = 1 2. Using Cramer s rule obtain the solutions to the following sets of equations: (c) 6x y = 0 2x 4y = 1 (a) 2x 1 + x 2 x 3 = 0 x 1 + x 3 = 4 x 1 + x 2 + x 3 = 0 (b) x 1 x 2 + x 3 = 1 x 1 + x 3 = 1 x 1 + x 2 x 3 = 0 Answers 1. (a) x = 1 2, y = 0 (b) = 0, no solution (c) x = 1 22, y = (a) x 1 = 8 3, x 2 = 4, x 3 = 4 3 (b) x 1 = 1 2, x 2 = 1, x 3 = HELM (2008): Workbook 8: Matrix Solution of Equations
### 8.1. Cramer s Rule for Solving Simultaneous Linear Equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Cramer s Rule for Solving Simultaneous Linear Equations 8.1 Introduction The need to solve systems of linear equations arises frequently in engineering. The analysis of electric circuits and the control
### Solving simultaneous equations using the inverse matrix
Solving simultaneous equations using the inverse matrix 8. Introduction The power of matrix algebra is seen in the representation of a system of simultaneous linear equations as a matrix equation. Matrix
### 8.2. Solution by Inverse Matrix Method. Introduction. Prerequisites. Learning Outcomes
Solution by Inverse Matrix Method 8.2 Introduction The power of matrix algebra is seen in the representation of a system of simultaneous linear equations as a matrix equation. Matrix algebra allows us
### 3.4. Solving Simultaneous Linear Equations. Introduction. Prerequisites. Learning Outcomes
Solving Simultaneous Linear Equations 3.4 Introduction Equations often arise in which there is more than one unknown quantity. When this is the case there will usually be more than one equation involved.
### Solving simultaneous equations using the inverse matrix
Solving simultaneous equations using the inverse matrix 8.2 Introduction The power of matrix algebra is seen in the representation of a system of simultaneous linear equations as a matrix equation. Matrix
### 8.3. Gauss elimination. Introduction. Prerequisites. Learning Outcomes
Gauss elimination 8.3 Introduction Engineers often need to solve large systems of linear equations; for example in determining the forces in a large framework or finding currents in a complicated electrical
### 3.4. Solving simultaneous linear equations. Introduction. Prerequisites. Learning Outcomes
Solving simultaneous linear equations 3.4 Introduction Equations often arise in which there is more than one unknown quantity. When this is the case there will usually be more than one equation involved.
### UNIT 2 MATRICES - I 2.0 INTRODUCTION. Structure
UNIT 2 MATRICES - I Matrices - I Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices 2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress
### 8.3. Solution by Gauss Elimination. Introduction. Prerequisites. Learning Outcomes
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# Trigonometric Equations
## Presentation on theme: "Trigonometric Equations"— Presentation transcript:
Trigonometric Equations
In quadratic form, using identities or linear in sine and cosine
Solving a Trig Equation in Quadratic Form
Solve the equation: 2sin2 θ – 3 sin θ + 1 = 0, 0 ≤ θ ≤ 2p Let sin θ equal some variable sin θ = a Factor this equation (2a – 1) (a – 1) = 0 Therefore a = ½ a = 1
Solving a Trig Equation in Quadratic Form
Now substitute sin θ back in for a sin θ = ½ sin θ = 1 Now do the inverse sin to find what θ equals θ = sin-1 (½) θ = sin-1 1 θ = p/6 and 5p/6 θ = p/2
Solving a Trig Equation in Quadratic Form
Solve the equation: (tan θ – 1)(sec θ – 1) = 0 tan θ – 1 = 0 sec θ – 1 = 0 tan θ = 1 sec θ = 1 θ = tan θ = sec-1 1 θ = p/4 and 5p/4 θ = 0
Solving a Trig Equation Using Identities
In order to solve trig equations, we want to have a single trig word in the equation. We can use trig identities to accomplish this goal. Solve the equation 3 cos θ + 3 = 2 sin2 θ Use the pythagorean identities to change sin2 θ to cos θ
Solving a Trig Equation Using Identities
sin2 = 1 – cos2 θ Substituting into the equation 3 cos θ + 3 = 2(1 – cos2 θ) To solve a quadratic equation it must be equal to 0 2cos2 θ + 3 cos θ + 1 = 0 Let cos θ = b
Solving a Trig Equation using Identities
2b2 + 3b + 1 = 0 (2b + 1) (b + 1) = 0 (2b + 1) = 0 b + 1 = 0 b = -½ b = -1 cos θ = -½ cos θ = -1 θ = 2p/3, 4p/3 θ = p
Solving a Trig Equation Using Identities
cos2 θ – sin2 θ + sin θ = 0 1 – sin2 θ – sin2 θ + sin θ = 0 -2sin2 θ + sin θ + 1 = 0 2 sin2 θ – sin θ – 1 = 0 Let c = sin θ 2c2 – c – 1 = 0 (2c + 1) (c – 1) = 0
Solving a Trig Equation Using Identities
(2c + 1) = 0 c – 1 = 0 c = -½ c = 1 sin θ = -½ sin θ = 1 θ = p/3 + p q=2p-p/3 θ = p/2 θ = 4p/3, q = 7p/3
Solving a Trig Equation Using Identities
Solve the equation sin (2θ) sin θ = cos θ Substitute in the formula for sin 2θ (2sin θ cos θ)sin θ=cos θ 2sin2 θ cos θ – cos θ = 0 cos θ(2sin2 – 1) = 0 cos θ = 0 2sin2 θ=1
Solving a Trig Equation Using Identities
cos θ = 0 θ = 0, p θ = p/4, 3p/4, 5p/4, 7p/4
Solving a Trig Equation Using Identities
sin θ cos θ = -½ This looks very much like the sin double angle formula. The only thing missing is the two in front of it. So multiply both sides by 2 2 sin θ cos θ = -1 sin 2θ = -1 2 θ = sin-1 -1
Solving a Trig Equation Using Identities
2θ = 3p/2 θ = 3p/4 2θ = 3p/2 + 2p 2q = 7p/2 q = 7p/4
Solving a Trig Equation Linear in sin θ and cos θ
sin θ + cos θ = 1 There is nothing I can substitute in for in this problem. The best way to solve this equation is to force a pythagorean identity by squaring both sides. (sin θ + cos θ)2 = 12
Solving a Trig Equation Linear in sin θ and cos θ
sin2 θ + 2sin θ cos θ + cos2 θ = 1 2sin θ cos θ + 1 = 1 2sin θ cos θ = 0 sin 2θ = 0 2θ = 0 2θ = p θ = 0 θ = p/2 θ = p θ = 3p/2
Solving a Trig Equation Linear in sin θ and cos θ
Since we squared both sides, these answers may not all be correct (when you square a negative number it becomes positive). In the original equation, there were no terms that were squared
Solving a Trig Equation Linear in sin θ and cos θ
Check: Does sin 0 + cos 0 = 1? Does sin p/2 + cos p/2 = 1? Does sin p + cos p = 1? Does sin 3p/2 + cos 3p/2 = 1?
Solving a Trig Equation Linear in sin θ and cos θ
sec θ = tan θ + cot θ sec2 θ = (tan θ + cot θ)2 sec2 θ = tan2 θ + 2 tan θ cot θ + cot2 θ sec2 θ = tan2 θ cot2 θ sec2 θ – tan2 θ = 2 + cot2 θ 1 = 2 + cot2 θ -1 = cot2 θ
Solving a Trig Equation Linear in sin θ and cos θ
q is undefined (can’t take the square root of a negative number).
Solving Trig Equations Using a Graphing Utility
Solve 5 sin x + x = 3. Express the solution(s) rounded to two decimal places. Put 5 sin x + x on y1 Put 3 on y2 Graph using the window 0 ≤ θ ≤2p Find the intersection point(s)
Word Problems Page 519 problem 58
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# How do you determine the limit of ((x+1)/(x-2) - x/(x-2))^2 as x approaches 2+?
Aug 12, 2016
${\lim}_{x \to {2}^{+}} {\left(\frac{x + 1}{x - 2} - \frac{x}{x - 2}\right)}^{2} = + \infty$
#### Explanation:
firstly
${\left(\frac{x + 1}{x - 2} - \frac{x}{x - 2}\right)}^{2}$
$= {\left(\frac{1}{x - 2}\right)}^{2}$
$= \frac{1}{x - 2} ^ 2$
So
${\lim}_{x \to 2} {\left(\frac{x + 1}{x - 2} - \frac{x}{x - 2}\right)}^{2}$
$= {\lim}_{x \to 2} \frac{1}{x - 2} ^ 2 = + \infty$
because the denominator is a square (even) term, the right and left side limits are the same
so ${\lim}_{x \to {2}^{\textcolor{red}{+}}} \frac{1}{x - 2} ^ 2 = + \infty$
and ${\lim}_{x \to {2}^{\textcolor{red}{-}}} \frac{1}{x - 2} ^ 2 = + \infty$ |
1. ## fractions & derivatives
I have forgotten some of the simplest things from my school days.
How would I go about getting rid of the denominator in a problem such as 2/t.
How would I solve a problem when there is a fraction in a fraction? I am doing derivatives and the actual problem is 2/t if t=1/2.
Using the formula f(x) lim h->0 f(x+h) - f(x)/h.
2. This belongs in the Calculus section, not the elementary math section.
Ok. So using the definition of a derivative $f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{f(t+\Del ta{t})-f(t)}{\Delta{t}}$.
Thus for finding the derivative of $\frac{2}{t}$ we find that:
$f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{\frac{2} {t+\Delta{t}}-\frac{2}{t}}{\Delta{t}}$
Multiply the numerator and denominator by $(t+\Delta{t})$
$f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{2-\left(\frac{2t+2\Delta{t}}{t}\right)}{\Delta{t}(t+ \Delta{t})}$
Split up the numerator to cancel some things:
$f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{2-2-\frac{2\Delta{t}}{t}}{\Delta{t}(t+\Delta{t})}$
And from here it's just algebra.
$f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{-2\Delta{t}}{t}*\frac{1}{\Delta{t}(t+\Delta{t})}$
Distribute and apply the limit:
$f'(t)=\lim_{\Delta{t}\rightarrow{0}}\frac{-2}{t^2+t\Delta{t}}$
And once you get your answer, just plug in for $t=\frac{1}{2}$
EDIT: For some reason, the Latex on the definition of the limit isn't working, but I think you're familiar with it. |
# Lesson 2.3 Real Zeros of Polynomials. The Division Algorithm.
## Presentation on theme: "Lesson 2.3 Real Zeros of Polynomials. The Division Algorithm."— Presentation transcript:
Lesson 2.3 Real Zeros of Polynomials
The Division Algorithm
Dividing by a polynomial Set up in long division 2 terms in divisor (x + 1). How does this go into 1 st two terms in order to eliminate the 1 st term of the dividend. 2x 2x 2 + 2x Multiply by the divisor Write product under dividend Subtract Carry down next term Repeat process -- x + 5 + 1 x + 1 -- 4 Answer:
HINTS: If a term is missing in the dividend – add a “0” term. If there is a remainder, put it over the divisor and add it to the quotient (answer) Example 1 (x 4 – x 2 + x) ÷ (x 2 - x + 1)
Synthetic Division Less writing Uses addition Setting Up Divisor must be of the form: x – a Use only “a” and coefficients of dividend Write in “zero terms” 450-25 x – 2: a = 2 x + 3: a = -3
450-25
Steps Bring down Multiply diagonally Add Remainder = last addition Answer Numbers at bottom are coefficients Start with 1 degree less than dividend REPEAT
Example 2: (2x 3 – 7x 2 – 11x – 20) (x – 5)
Example 3: (2x 4 – 30x 2 – 2x – 1) (x – 4) Problem Set 2.3 (1 – 21 EOO)
The Factor Theorem If f(x) has a factor (x – a) then f(a) = 0 The Remainder Theorem If f(x) is divided by x – a, the remainder is r = f(a)
Example 4 Show that (x – 2) and (x + 3) are factors of
Rational Zero Test Every rational zero = Factors of constant term Factors of leading coefficient =
Descartes’ Rule Number of positive real roots is: ► the number of variations in the signs, or ► less than that by a positive even integer 5x 4 – 3x 3 + 2x 2 – 7x + 1 variations: possible positive real roots:
Example 5 List possible zeros, verify with your calculator which are zeros, and check results with Descartes’ Rule Problems Set 2.3 |
## Algebra Solutions: Hungerford I.1
1) Give examples of semigroups and monoids which are not groups
Solution- $(\mathbb{N},+)$ is a semigroup, but neither a monoid nor a group. This is because the number 0 is not in the natural numbers, thus there is no additive identity.
$(\mathbb{N},\cdot)$ is a monoid because 1 is the identity, however it has no inverses.
2) Let G be a group, S a nonempty set, and M(S,G) the set of all functions $f:S \rightarrow G$. Define addition in M (S,G) as follows: $(f+g):S \rightarrow G$ is given by $s\mapsto f(s) + g(s) \in G$. Prove that M (S,G) is a group, which is abelian if G is.
Proof-
We must check that the operation is associative. We have that $(f+(g+h))(s) = f(s) + (g+h)(s) = f(s) + g(s) + h(s) = (f + g)(s) + h(s) = ((f+g)+h)(s)$
Now, G is a group so it has an identity element by definition, call it 0. Define a function $e:S \rightarrow G$ by $e(s) = 0$ $\forall s \in S$. This is by definition a function so it belongs to M (S,G). Then we have $\forall f \in M (S,G), (e+f)(s) = e(s) + f(s) = 0 + f(s) = f(s)$ so that e is the identity of M (S,G).
Lastly, define for any function $f \in M (S,G)$ the function $g:S \rightarrow G$ defined by $g(s)= -f(s) \forall s \in S$. Then we have $(g+f)(s) = g(s) + f(s) = -f(s) + f(s) = 0 = e(s)$.
If G is abelian then we have $(f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s) \forall f,g \in M(S,G)$ Therefore M(S,G) is abelian.
7) If p is prime, then the nonzero elements of $\mathbb{Z}_p$ form a group of order p-1 under multiplication (mod p).
Proof-
$\mathbb{Z}_p \backslash \{0\}$ inherits associativity from multiplication of integers, as well as the identity element 1. Since p is prime, for any $a \in \mathbb{Z}_p \backslash \{0\}$ we have that gcd(a,p) = 1. Thus, there exists $k_1, k_2 \in \mathbb{Z}$ such that $k_1a+k_2p = 1$. Then by doing mod p on both sides we have $k_1a = 1$ (mod p), so that $a^{-1} = k_1$. Thus we have an inverse for each element and hence a group of order p-1 elements.
8) a) The relation given by $a \sim b \leftrightarrow a - b \in \mathbb{Z}$ is a congruence relation on the additive group $\mathbb{Q}$
Proof-
Let $a,b,c,d \in \mathbb{Q}$. First, since $a-a=0 \in \mathbb{Z}$ we have that $\sim$ is reflexive. If $a \sim b$ then $a-b = k$ for some $k \in \mathbb{Z}$ so then $b - a = -k \in \mathbb{Z}$ so that the relation is symmetric. If $a \sim b$ and $b \sim c$ then we have $a-b = m$ and $b - c = n$ for some integers, m and n. Then putting them together we have $a-c = b+m+n-b = m+n \in \mathbb{Z}$ so the relation is transitive. Lastly, suppose that $a \sim b$ and $c \sim d$. Then $a-b = r$ and $c-d=s$ for some integers r and s. To show congruence we must show that $(a+c) \sim (b+d)$. Consider $(a+c) - (b+d) = ((b+r)+(d+s))-((a-r)+(c-s))= 2r+2s+b-a+d-c=2r+2s-r-s = r+s \in \mathbb{Z}$ Thus it is a congruence relation.
b) The set $\mathbb{Q} \backslash \mathbb{Z}$ of equivalence classes is an infinite abelian group.
Proof-
From part a) we know that the relation is a congruence relation. Theorem 1.5 in Hungerford gives us that the set of equivalence classes of an abelian group under a congruence relation is an abelian group.
9) Let p be a fixed prime. Let $R_p$ be the set of all those rational numbers whose denominator is relatively prime to p. Let $R^p$ b the set of rationals whose denominator is a power of p. Prove both $R_p$ and $R^p$ are groups under ordinary addition.
Proof-
First we consider $R_p$ and we show closure. Let $\frac{a}{b}, \frac{c}{d} \in R_p$. Then $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ We have that $gcd(b,p)=1=gcd(d,p)$ so that $p \nmid b$ and $p \nmid d$ which gives that $p \nmid bd$. Thus $gcd(bd,p) = 1$ so the operation is closed.
Associativity is inherited from the ordinary addition of rationals.
The identity 0, can be written as $\frac{0}{p}$ where p is some other prime so that $gcd(p,p`)= 1$.
The inverse of an element $\frac{a}{b}$ is $\frac{-a}{b}$
$R^p$ is very similar except easier.
11) The following conditions on a group G are equivalent:
(i) G is abelian
(ii)$(ab)^2 = a^2b^2 \forall a,b \in G$
(iii)$(ab)^{-1} = a^{-1}b^{-1} \forall a,b \in G$
(iv)$(ab)^n = a^nb^n \forall n \in \mathbb{Z} , \forall a,b \in G$
(v) $(ab)^n = a^nb^n$ for 3 consecutive integers n and $\forall a,b \in G$
Proof-
$(i) \Rightarrow (iv)$ Let G be abelian. The case when n is 0 is trivial. Let $n \geq 1$ Then $(ab)^1 =a^1b^1$ Assume it is true for some natural number k and then use induction. If $n \leq 1$ then we have $-n \geq 1$ so we can use the previous case to show it is true.
$(iv) \Rightarrow (v)$ follows immediately.
$(v) \Rightarrow (i)$ Let $a,b \in G$. Assuming for some n the hypothesis is true, we have $(ab)^n = a^nb^n$, $(ab)^{n+1} = a^{n+1}b^{n+1}$, $(ab)^{n+2} = a^{n+2}b^{n+2}$. We use these facts to say first that
$(ab)^{n+1} = a^{n+1}b^{n+1}$
$(ab)^n(ab) = a^{n}ab^{n}b$
$a^nb^nab = a^{n}ab^{n}b$
$b^na = ab^{n}$
We also have
$(ab)^{n+2} = a^{n+2}b^{n+2}$
$(ab)^n(ab)^2 = a^{n}a^2b^{n}b^2$
$a^nb^nabab = a^{n}aab^{n}bb$
$b^naba = aab^{n}b$
Then using the fact that $b^na = ab^{n}$, we get
$ab^nba = ab^{n}ab$
$ba = ab$ So that G is abelian.
$(i) \Rightarrow (ii)$ Trivial
$(ii) \Rightarrow (iii)$ Assume that $(ab)^2 = a^2b^2 \forall a,b \in G$. Then we have:
$(a^{-1}b^{-1})^2 = (a^{-1})^2(b^{-1})^2$
$a^{-1}b^{-1}a^{-1}b^{-1} = a^{-1}a^{-1}b^{-1}b^{-1}$
$b^{-1}a^{-1} = a^{-1}b^{-1}$
$(ab)^{-1} = a^{-1}b^{-1}$
$(iii) \Rightarrow (i)$ Assume that $(ab)^{-1} = a^{-1}b^{-1}$. Then
$(ab)^{-1} = a^{-1}b^{-1}$
$(ab)^{-1} = (ba)^{-1}$
$e = (ba)^{-1}ab$
$ba = ab$ so that G is abelian.
13) If $a^2 = e \forall a \in G$, then G is abelian.
Proof-
The fact that $a^2 =e$ gives us that $a = a^{-1} \forall a \in G$. Let $a,b \in G$. Then we know $(ab)^2 = e = abab = a^{-1}b^{-1}ab$
$\Rightarrow e =a^{-1}b^{-1}ab$
$\Rightarrow ba =ab$
This is true for any arbitrary elements a and b, so G is abelian.
14) If G is a finite group of even order then G contains an element $a \neq e$ such that $a^2 = e$.
Proof-
Suppose that no element satisfies the given condition. So that there is no element a such that $a = a^{-1}$. Then each element can be paired with it’s inverse to give 2n elements in G for some n. All of these pairs together with the identity element e, gives 2n + 1 elements in G, which is a contradiction to G being of even order. Therefore such an element exists. |
What is the Area of this Figure? Enter Your Answer in the Box. M²
What is the Area of this Figure? Enter Your Answer in the Box M²
You may often find the question about calculating area on Brainly. The complete question like this; What is the area of this figure? Enter your answer in the box m² A parallelogram with a right triangle made inside it with a short leg length of 5 m and a long leg length of 9 m. Two triangles attached to the top of the parallelogram are touching and share a long leg length of 8 m.
Calculating Area
In this paragraph, we are going to talk about calculating the area. For your information, area is a measure of how much space inside a shape. Calculating the area of shape or surface is able to be useful in everyday life. For instance, you may need to know how much paint to purchase for covering a wall, or how much grass seed you require to sow a lawn. In addition, you can also learn how to calculate the areas of common shapes including squares, rectangles, triangles and circles.
Calculating Area Using the Grid Method
When a shape is drawn on a scaled grid, you are able to find the area by counting the number of grid squares inside the shape. For example: There are 10 grid squares inside the rectangle. To find an area value by using the grid method, you need to know the size that grid square represents. This example needs to use centimetres, but the same method applies for any unit of length or distance. You could be using inches, miles, metres, feet etc.
In this example, each grid square has a width of 1cm and a height of 1cm. In other words, each grid square is one square centimetre. Please count the grid squares inside the large square to discover its area.
There are 15 little squares, so the area of the square is 15 square centimetres. In mathematics, we are able to abbreviate square centimetres to cm2. For note: 2 means squared. Each grid square is 1cm2. The area of the large square is 15cm2. However, this method becomes challenging when shapes do not suit the grid, or when you require to count fractions of the grid squares. In this example, the square does not suit exactly onto the grid.
You are still able to calculate the area by counting grid squares.
• There are 25 full grid squares.
• 10 half grid squares – 10 half squares is the same as 5 full squares.
• Also, there is 1 quarter square – (¼ or 0.25 of a whole square).
• Now, you are able to add the entire squares and fractions: 25 + 5 + 0.25 = 30.25.
The area of this square is 30.25cm2. Also, you are able to write this as 30¼cm2.
Although using a grid and count the squares within a shape is an easy way to learn the area concepts, but it is less useful to find the exact areas with more complex shapes, when there can be many fractions of the grid squares to add. Area is able to be calculated by using simple formulae, depending on the kind of shape you are learning. The remainder of this paragraph will explain and give the examples of how to calculate the area of a shape without using the grid system. |
# NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area
### NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1
Ex 11.1 Class 7 Maths Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m2 of the land costs ₹ 10,000
Solution:
Given: l = 500 m, b = 300 m
(i) Area = l × b
= 500 m × 300 m = 150000 m2
(ii) Cost of land = ₹ 10,000 × 150000 = ₹ 15,00,000,000
Ex 11.1 Class 7 Maths Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
Given: Perimeter = 320 m
Area of the square = Side × Side
= 80 m × 80 m = 6400 m2
Ex 11.1 Class 7 Maths Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also, find its perimeter.
Solution:
Given: Area = 440 m2
Length = 22 m
Perimeter = 2[l + b] = 2 [22 m + 20 m]
= 2 × 42 m = 84 m
Ex 11.1 Class 7 Maths Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:
Given: Perimeter = 100 cm
Length = 35 cm
Perimeter = 2(l + b)
100 = 2(35 + b)
100/2=35+b
⇒ 50 = 35 + b
⇒ b = 50 – 35 = 15 cm
Area = l × b = 35 cm × 15 cm
= 525 cm2
Ex 11.1 Class 7 Maths Question 5.
The area of a square park is same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
Given: Side of the square park = 60 m Length of the rectangular park = 90 m Area of the rectangular park = Area of the square park
90 m × 6 = 60 m × 60 m
⇒ b = 40m
Hence, the required breadth = 40 m.
Ex 11.1 Class 7 Maths Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Solution:
Given: Length = 40 cm, Breadth = 22 cm Perimeter of the rectangle
= Length of the wire
= 2(l + b) = 2(40 cm + 22 cm)
= 2 × 62 cm = 124 cm
Now, the wire is rebent into a square.
Perimeter = 124 cm
⇒ 4 × side = 124
∴ side = 124/4 cm = 31 cm
So, the measure of each side = 31 cm
Area of rectangular shape = l × b
= 40 cm x 22 cm
= 880 cm2
Area of square shape = (Side)2
= (31)2 = 961 cm2
Since 961 cm2 > 880 cm2
Hence, the square encloses more area.
Ex 11.1 Class 7 Maths Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:
Given: Perimeter = 130 cm
Perimeter = 2 (l + b)
130 cm = 2(l + 30 cm)
⇒ 130/2 cm = l + 30 cm
⇒ 65 cm = l + 30 cm
⇒ 65 cm – 30 cm = l
∴ l = 35 cm
Area of the rectangle = l × b = 35 cm × 30 cm
= 1050 cm2
Ex 11.1 Class 7 Maths Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2.
Solution:
Given: Length of wall = 4.5 m
Breadth of the wall = 3.6 m
Length of the door = 2 m
Breadth of the door = 1 m
Area of the wall = l × b = 4.5 m × 3.6 m = 16.20 m2
= 16.20 m2
Area of the door = l × b = 2 m × 1 m = 2 m2
∴ Area of the wall to be white washed = Area of the wall – Area of the door
= 16.20 m2 – 2 m2= 14.20 m2
Cost of white washing
= ₹ 14.20 × 20 = ₹ 284.00
Hence, the required area = 14.20 m2 and the required cost = ₹ 284
Exercise 11.2
Exercise 11.3
Exercise 11.4
## SabDekho
The Complete Educational Website |
Smith's Magic Division Squares
A magic square is a square of numbers with each row or each column adding up to the same number.
There is another 3x3 square of numbers, with some entries missing, such that the first number in each row or column divided by the second equals the third entry in that row or column. It so happens that we can write four numbers in the square in almost any way so that the rest of the square can be completed.
$\begin{array}{ccc} 2 & 3 & ? \\ 4 & 6 & ? \\ ? & ? & ? \end{array}$
The square is completed for the first two rows with
$2 \div 3=6$
$4 \div 6=2/3$
We have then
$\begin{array}{ccc} 2 & 3 & 2/3 \\ 4 & 6 & 2/3 \\ ? & ? & ? \end{array}$
Now the columns.
$\begin{array}{ccc} 2 & 3 & 2/3 \\ 4 & 6 & 2/3 \\ 1/2 & 1/2 & 1 \end{array}$
.
Now notice for the last row
$1/2 \div 1/2 =1$
.
This always happens.
If we fill out the square differently, say
$\begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ ? & ? & ? \end{array}$
.
Completing the columns first gives
$\begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ 1 & ? & 7/8 \end{array}$
.
Completing the rows gives
$\begin{array}{ccc} 1 & 1/7 & 7 \\ 1 & 1/8 & 8 \\ 1 & 8/7 & 7/8 \end{array}$
.
Again the first enty in a row or column divided by the second entry in that row or column equals the third entry. |
Just the Factors, Ma am
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1 1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a divisor of N and write d N if N/d is a positive integer Thus, for example, 2 6 Denote by D N the set of all positive integer divisors of N For example D 6 = {1, 2, 3, 6} There are four parts to this note In the first part, we count the divisors of a given positive integer N based on its prime factorization In the second part, we construct all the divisors, and in the third part we discuss the geometry of the D N In part four, we discuss applications to contest problems Study the figure above There are two things to work out One is how does the set of dots and segments represent a number, and the other is how do the digit strings just above the digits 1 through 9 represent those digits Take some time now before reading more to figure this out 2 Counting the divisors of N First consider the example N = 72 To find the number of divisors of 72, note that the prime factorization of 72 is given by 72 = Each divisor d of 72 must be of the form d = 2 i 3 j where 0 i 3 and 0 j 2 Otherwise, /d could not be an integer, by the Fundamental Theorem of Arithmetic (the theorem that guarantees the unique factorization into primes of each positive integer) So there Saylor URL : wwwsaylororg/courses/ma111 1 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
2 are 4 choices for the exponent i and 3 choices for j Hence there are 4 3 = 12 divisors of 72 Reasoning similarly, we can see that for any integer the number of divisors is N = p e 1 1 p e 2 2 p e k k, Π k i=1(e i + 1) = (e 1 + 1)(e 2 + 1) (e k + 1) 3 Constructing the divisors of N In part 1 we found the number of members of D N for any positive integer N In this part, we seek the list of divisors themselves Again we start with the prime factorization of N Suppose N = p e 1 1 p e 2 2 p e k k If k = 2 the listing is straightforward In this case, we build a table by first listing the powers of p 1 across the top of the table and the powers of p 2 down the side, thus obtaining an (e 1 + 1) (e 2 + 1) matrix of divisors Again we use N = 72 as an example Notice that each entry in the table is the product of its row label and its column label What do we do when the number of prime factors of N is more than 2? If k = 3 we can construct e matrices of divisors, one for each power of p 3 For example, if N = 360 = , we construct one 3 4 matrix for 5 0 and one for 5 1 The result is a matrix of divisors The two 3 4 matrices are shown below and For larger values of k we can create multiple copies of the matrix associated with the number N = p e 1 1 p e 2 2 p e k 1 k 1 4 The geometry of D N To investigate the geometry of D N, we first explore the relation divides Recall that a b means that a and b are positive integers for which b/a is an integer The relation has several important properties, three of which are crucial to our discussion Saylor URL : wwwsaylororg/courses/ma111 2 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
3 1 Reflexivity For any positive integer a, a a 2 Antisymmetry For any pair of positive integers a, b, if a b and b a, then a = b 3 Transitivity For any three positive integers, a, b, c, if a b and b c, then a c These properties are easy to prove The first says that each integer is a divisor of itself; that is, a/a is an integer The second says that no two different integers can be divisors of one another This is true since a larger integer can never be a divisor of a smaller one The third property follows from the arithmetic b/a c/b = c/a together with the property that the product of two positive integers is a positive integer Any set S with a relation defined on it that satisfies all three of the properties above is called a partially ordered set, or a poset A branch of discrete mathematics studies the properties of posets, (S, ) Each finite poset has a unique directed graph representation This pictorial representation is what we mean by the geometry of D N To construct the directed graph of a poset (S, ), draw a vertex (dot) for each member of S Then connect two vertices a and b with a directed edge (an arrow) if a b Of course, in our case D N this means we connect a to b if a b The case D 6 is easy to draw: Fig 1 The digraph of D 6 The circles at each of the four vertices are called loops They are included as directed edges because each number is a divisor of itself The reader should imagine that all the non-loop edges are upwardly directed The directed edge from 1 to 6 indicates that 1 6 But since we know that the vertices of D 6 satisfy all three properties required of a poset, we can leave off both (a) the loops, which are implied Saylor URL : wwwsaylororg/courses/ma111 3 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
4 by the reflexive property, and (b) the edges that are implied by the transitivity condition The slimmed down representation, called the Hasse diagram, is much easier to understand It captures all the essential information without cluttering up the scene The Hasse diagram of D 6 is shown below Fig 2, the Hasse diagram of D 6 In general, the Hasse diagram for D N has only those non-loop edges which are not implied by transitivity, that is, those edges from a to b for which b is a prime number multiple of a The Hasse diagrams of D 72, D 30, and D 60 are shown below Fig 3 The Hasse diagram of D 72 Saylor URL : wwwsaylororg/courses/ma111 4 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
5 Notice that each prime divisor of 30 can be considered a direction, multiplication by 2 moves us to the left ( ), by 3 moves us upward ( ) and, by 5 moves us to the right ( ) Also note that if a and b are divisors of 30 then a b if and only if there is a sequence of upwardly directed edges starting at a and ending at b For example, 1 30 and (1, 3), (3, 15), (15, 30) are all directed edges in the digraph of D 30 On the other hand, we say 2 and 15 are incomparable because neither divides the other, and indeed there is no upwardly directed sequence of edges from either one to the other Fig 4 The Hasse diagram of D 30 What would the divisors of 60 look like if we build such a diagram for them? Try to construct it before you look at D Fig 5 The Hasse diagram of D 60 Saylor URL : wwwsaylororg/courses/ma111 5 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
6 Consider the same (lattice/hasse) diagram for the divisors of 210 We can draw this in several ways The first one (Fig 6a) places each divisor of 210 at a level determined by its number of prime divisors The second one (Fig 6b) emphasizes the degrees of freedom These two diagrams are representations of a four dimensional cube, not surprising since the Hasse diagram for D 30 is a three-dimensional cube A mathematical way to say the two digraphs are the same is to say they are isomorphic This means that they have the same number of vertices and the same number of edges and that a correspondence between the vertices also serves as a correspondence between the edges Note that the digraphs in 6a and 6b have the required number of vertices (16) and the required number of edges (32) Can you find an N such that the Hasse diagram of D N is a representation of a five-dimensional cube? Such a digraph must have 2 5 = 32 vertices, and = 80 edges Fig 6a The Hasse diagram of D 210 Saylor URL : wwwsaylororg/courses/ma111 6 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
7 Fig 6b The Hasse diagram of D 210 How can the geometry help us do number theory? One way to use the geometry is in the calculation of the GCF and LCM of two members of D N Note that each element d of D N generates a downward cone of divisors and an upward cone of multiples We can denote these cones by F (d) and M(d) respectively Then the GCF (d, e) = max{f (d) F (e)} and LCM(d, e) = min{m(d) M(e)} Saylor URL : wwwsaylororg/courses/ma111 7 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
8 5 Problems On Divisors Some of the following problems come from MathCounts and the American Mathematics Competitions 1 Find the number of three digit divisors of 3600 Solution: 15 Construct three 3 5 matrices of divisors The first matrix has all the divisors of the form 2 i 3 j, the second all divisors of the form 2 i 3 j 5 1 and the third all the divisors of the form 2 i 3 j 5 2 There is one three digit divisor in the first matrix, 144, five in the second, and nine in the third for a total of = 15 2 How many positive integers less than 50 have an odd number of positive integer divisors? Solution: A number has an odd number of divisors if and only if it is a perfect square Therefore, there are exactly seven such numbers, 1, 4, 9, 16, 25, 36, and 49 3 (The Locker Problem) A high school with 1000 lockers and 1000 students tries the following experiment All lockers are initially closed Then student number 1 opens all the lockers Then student number 2 closes the even numbered lockers Then student number 3 changes the status of all the lockers numbered with multiples of 3 This continues with each student changing the status of all the lockers which are numbered by multiples of his or her number Which lockers are closed after all the 1000 students have done their jobs? Solution: Build a table for the first 20 lockers, and notice that the lockers that end up open are those numbered 1, 4, 9, and 16 This looks like the squares Think about what it takes to make a locker end up open It takes an odd number of changes, which means an odd number of divisors We know how to count the number of divisors of a number N = p e 1 1 p e 2 2 p en n N has D N = (e 1 + 1)(e 2 + 1) (e n + 1) divisors So the issue is how can this last number be odd It s odd if each factor e i + 1 is odd, which means all the e i have to be even This is true precisely when N is a perfect square So the lockers that are closed at the end are just the non-square numbered lockers 4 If N is the cube of a positive integer, which of the following could be the number of positive integer divisors of N? (A) 200 (B) 201 (C) 202 (D) 203 (E) 204 Solution: C The number N must be of the form N = p 3e q 3f So the number of divisors must be of the form D = (3e+1)(3f +1) Such a number Saylor URL : wwwsaylororg/courses/ma111 8 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
9 must be one greater than a multiple of 3 The number N = (2 67 ) 3 = has exactly 202 divisors 5 Let N = How many positive integers are factors of N? (A) 3 (B) 5 (C) 69 (D) 125 (E) 216 Solution: The expression given is the binomial expansion of N = (69 + 1) 5 = 70 5 = So N has (5 + 1)(5 + 1)(5 + 1) = 216 divisors 6 A teacher rolls four dice and announces both the sum S and the product P Students then try to determine the four dice values a, b, c, and d Find an ordered pair (S, P ) for which there is more than one set of possible values Solution: If (S, P ) = (12, 48), then {a, b, c, d} could be either {1, 3, 4, 4} or {2, 2, 2, 6} Note: strictly speaking, these are not sets, but multisets because multiple membership is allowed 7 How many of the positive integer divisors of N = have exactly 12 positive integer divisors? Solution: Let n be a divisor of N with 12 divisors Then n = p 3 q 2 where p and q belong to {2, 3, 5, 7, 11} or n = p 2 qr where p, q, r {2, 3, 5, 7, 11} There are 5 4 = 20 of the former and 5 (4) 2 = 30 of the later So the answer is = 50 8 How many ordered pairs (x, y) of positive integers satisfy xy + x + y = 199? Solution: Add one to both sides and factor to get (x+1)(y+1) = 200 = which has 4 3 factors, one of which is 1 Neither x + 1 not y + 1 can be 1, so there are 10 good choices for x + 1 Saylor URL : wwwsaylororg/courses/ma111 9 Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
10 9 How many positive integers less than 400 have exactly 6 positive integer divisors? Solution: Having exactly 6 divisors means that the number N is of the form p 5 or p 2 q where p and q are different prime numbers Only 2 5 = 32 and 3 5 = 243 are of the first type If p = 2, then q could be any prime in the range 3, 5, 97 of which there are 24 primes If p = 3, then q could be 2 or any prime in the range 5, 43 and there are 13 of these If p = 5, then q could be 2 or 3 or any prime in the range 7, 13 If p = 7 then q = 2, 3, or 5 There are two values of q for p = 11 and one value of q for p = 13 Tally these to get = The product of four distinct positive integers, a, b, c, and d is 8! The numbers also satisfy What is d? Solution: Add 1 to both sides and factor to get Build a factor table for 392 ab + a + b = 391 (1) bc + b + c = 199 (2) (a + 1)(b + 1) = 392 = (3) (b + 1)(c + 1) = 200 = (4) Thus it follows that (a+1, b+1) is one of the following pairs: (1, 392), (2, 196), (4, 98), (8, 49), (7, 56), (14, 28), (392, 1), (196, 2), (98, 4), (49, 8), (56, 7), (28, 14) Each of these lead to dead ends except (49, 8) and (8, 49) For example, a+1 = 7, b + 1 = 56 leads to a = 6 and b = 55 = 5 11, which is not a factor of 8! Next note that b = 48 cannot work because b + 1 = 49 is not a factor of 200 Therefore a = 48 and b = 7 This implies that c + 1 = 200/(b + 1) = 25 and c = 24 Finally, d = 5 Thus, a = 48, b = 7, c = 24 and d = 5 11 How many multiples of 30 have exactly 30 divisors? Solution: I am grateful to Howard Groves (Uniter Kingdom) for this problem A number N = 2 i 3 j 5 k M is a multiple of 30 if each of i, j, k are at least 1 Saylor URL : wwwsaylororg/courses/ma Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
11 Since 30 has exactly three different prime factors, it follows that M = 1, and that (i + 1)(j + 1)(k + 1) = 30 Thus, {i, j, k} = {1, 2, 4}, and there are exactly 6 ways to assign the values to the exponents The numbers are 720, 1200, 7500, 1620, 4050, and Find the number of odd divisors of 13! Solution: The prime factorization of 13! is Each odd factor of 13! is a product of odd prime factors, and there are = 144 ways to choose the five exponents 13 Find the smallest positive integer with exactly 60 divisors 14 (2008 Purple Comet) Find the smallest positive integer the product of whose digits is 9! 15 (1996 AHSME) If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have? Solution: 35 Let 2 e 1 3 e 2 5 e3 be the prime factorization of n Then the number of positive divisors of n is (e 1 + 1)(e 2 + 1)(e 3 + 1) In view of the given information, we have and 28 = (e 1 + 2)(e 2 + 1)P 30 = (e 1 + 1)(e 2 + 2)P, where P = (e 3 +1)(e 4 +1) Subtracting the first equation from the second, we obtain 2 = (e 1 e 2 )P, so either e 1 e 2 = 1 and P = 2, or e 1 e 2 = 2 and P = 1 The first case yields 14 = (e 1 + 2)e 1 and (e 1 + 1) 2 = 15; since e 1 is a nonnegative integer, this is impossible In the second case, e 2 = e 1 2 and 30 = (e 1 + 1)e 1, from which we find e 1 = 5 and e 2 = 3 Thus n = , so 6n = has (6 + 1)(4 + 1) = 35 positive divisors 16 Find the sum of all the divisors of N = Solution: Consider the product ( )( )( ) Notice that each divisor of N appears exactly once in this expanded product This sum can be thought of as the volume of a ( ) ( ) ( ) box The volume of the box is = (2010 Mathcounts Competition, Target 7) Find the product P of all the divisors of 6 3, and express your answer in the form 6 t, for some integer t Saylor URL : wwwsaylororg/courses/ma Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
12 Solution: Build the 4 4 matrix of divisors of 6 3, and notice that they can be paired up so that each has 6 3 as their product There are 8 such pairs, so P = (6 3 ) 8 = 6 24 So t = Find the largest 10-digit number with distinct digits that is divisible by 11 Solution: We want to keep the digits in decreasing order as much as possible Trying 98765xyuvw works because we can arrange for y + v = 28, while x + u + w = 17 works More later 19 (This was problem 25 on the Sprint Round of the National MATHCOUNTS competition in 2011) Each number in the set {5, 9, 10, 13, 14, 18, 20, 21, 25, 29} was obtained by adding two number from the set {a, b, c, d, e} where a < b < c < d < e What is the value of c? Solution: c = 7 The sum of all 10 numbers is 4 times the sum a+b+c+d+e Why? Now a + b = 5 and d + e = 29, so c = (164 4) 29 5 = = 7 20 (This was problem 8 on the Target Round of the National MATHCOUNTS competition in 2011) How many positive integers less than 2011 cannot be expressed as the difference of the squares of two positive integers? Solution: The answer is 505 If n = a 2 b 2 = (a b)(a + b), then a b and a + b have the same parity, so n is either odd or a multiple of 4 So we want to count all the numbers of the form 4k + 1, k = 0, 1,, 502 together with two values 2 and 4, for a total of Consider the number N = = (a) How many divisors does N have? Solution: Use the formula developed here: τ = (3 + 1)(2 + 1)(3 + 1)(1 + 1) = 96 divisors (b) How many single digit divisors does N have? (c) How many two-digit divisors does N have? (d) How many three-digit divisors does N have? (e) How many four-digit divisors does N have? (f) How many five-digit divisors does N have? Solution: The answers are respectively 9, 25, 33, 23 and 6 You can get these numbers by building the six 4 4 matrices of divisors Use the basic matrix Saylor URL : wwwsaylororg/courses/ma Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
13 of divisors of 1000 a model The other five matrices are obtained from this one by multiplying by 3, by 9, by 7, by 21, and by The integer N has both 12 and 15 as divisors Also, N has exactly 16 divisors What is N? Solution: In order to have both 12 and 15 as divisors, N must be of the form d for some positive integer d Since the number of divisors of N is 16, d must be even But has exactly 16 divisors, so d = 2 and N = How many multiples of 30 have exactly 3 prime factors and exactly 60 factors Solution: This requires some casework We re considering numbers of the form N = 2 i 3 j 5 k which satisfy (i + 1)(j + 1)(k + 1) = 60, i, j, k 1 There are 21 such numbers 24 How many whole numbers n satisfying 100 n 1000 have the same number of odd divisors as even divisors? Solution: Such a number is even but not a multiple of 4, so it follows that we want all the numbers that are congruent to 2 modulo 4, and there are 900/4 = 225 of them 25 What is the sum of the digits in the decimal representation of N = ? Solution: The number is N = = = , the sum of whose digits is What is the fewest number of factors in the product 27! that have to be removed so that the product of the remaining numbers is a perfect square? Solution: Factor 27! to get 27! = The largest square divisor of 27! is , so each of the primes 2, 3, 7, 17, 19, 23 must be removed It takes five factors to do this We must remove each of 17, 19, 23, and then we have 2, 3, 7 to remove We cannot do this with just one factor, so it takes two, either 6 and 7 or 21 and 2 Saylor URL : wwwsaylororg/courses/ma Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
14 27 Euclid asks his friends to guess the value of a positive integer n that he has chosen Archimedes guesses that n is a multiple of 10 Euler guesses that n is a multiple of 12 Fermat guesses that n is a multiple of 15 Gauss guesses that n is a multiple of 18 Hilbert guesses that n is a multiple of 30 Exactly two of the guesses are correct Which persons guessed correctly? Solution: Hilbert must be wrong because if 30 n then so do 10 and 15, which would mean that three of them were right Now 2 n because all but on of the numbers are multiples of 2, and 3 n for the same reason But 5 does not divide n because, if it did, then 30 n which contradicts our first observation Thus Euler and Gauss are the two correct mathematicians Saylor URL : wwwsaylororg/courses/ma Saylororg This material may be copied, stored and distributed for non-commercial purposes only Any commercial use of this material is strictly forbidden
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Solve Lesson 36 Page 59 Math 10 – Kite >
Topic
Which of the following statements is correct?
A. The solution set of the equation $$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)}$$ is the solution set of the equation $$f\left( x$$ right) = g\left( x \right)\)
B. The solution set of the equation $$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)}$$ is the solution set of the equation $${\left)[ {f\left( x \right)} \right]^2} = {\left[ {g\left( x \right)} \right]^2}$$
C. The solution set of the equation $$f\left( x \right) = g\left( x \right)$$ is the solution set of the equation $$\sqrt {f\left( x \right)} = \ sqrt {g\left( x \right)}$$
D. The solution set of the equation $$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)}$$ is the solution set of the equation $$f\left( x$$ right) = g\left( x \right)\) satisfying the inequality $$f\left( x \right) \ge 0$$ (or $$g\left( x \right) \ge 0$$ )
Solution method – See details
$$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)} \Leftrightarrow \left\{ \begin{array}{l}f(x) \ge 0\\ f\left( x \right) = g\left( x \right)\end{array} \right.$$or $$\left\{ \begin{array}{l}g(x) \ge 0\ \f\left( x \right) = g\left( x \right)\end{array} \right.$$
Detailed explanation
We have: $$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)} \Leftrightarrow \left\{ \begin{array}{l}f(x) \ge 0\\f\left( x \right) = g\left( x \right)\end{array} \right.$$
Thus the solution set of the equation $$\sqrt {f\left( x \right)} = \sqrt {g\left( x \right)}$$ is the set of solutions of the equation $$f\left( x$$ right) = g\left( x \right)\) satisfying the inequality $$f\left( x \right) \ge 0$$ (or $$g\left( x \right) \ge 0$$ )
Choose D. |
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### Section 5-5 : Solutions to Systems
Now that we’ve got some of the basics out of the way for systems of differential equations it’s time to start thinking about how to solve a system of differential equations. We will start with the homogeneous system written in matrix form,
$$$\vec x' = A\,\vec x \label{eq:eq1}$$$
where, $$A$$ is an $$n \times n$$ matrix and $$\vec x$$ is a vector whose components are the unknown functions in the system.
Now, if we start with $$n = 1$$then the system reduces to a fairly simple linear (or separable) first order differential equation.
$x' = ax$
and this has the following solution,
$x\left( t \right) = c{{\bf{e}}^{a\,t}}$
So, let’s use this as a guide and for a general $$n$$ let’s see if
$$$\vec x\left( t \right) = \vec \eta \,{{\bf{e}}^{r\,t}}\label{eq:eq2}$$$
will be a solution. Note that the only real difference here is that we let the constant in front of the exponential be a vector. All we need to do then is plug this into the differential equation and see what we get. First notice that the derivative is,
$\vec x'\left( t \right) = r\vec \eta \,{{\bf{e}}^{r\,t}}$
So, upon plugging the guess into the differential equation we get,
\begin{align*}r\vec \eta {{\bf{e}}^{r\,t}} & = A\vec \eta {{\bf{e}}^{r\,t}}\\ \left( {A\vec \eta - r\vec \eta } \right){{\bf{e}}^{r\,t}} & = \vec 0\\ \left( {A - rI} \right)\vec \eta {{\bf{e}}^{r\,t}} & = \vec 0\end{align*}
Now, since we know that exponentials are not zero we can drop that portion and we then see that in order for $$\eqref{eq:eq2}$$ to be a solution to $$\eqref{eq:eq1}$$ then we must have
$\left( {A - rI} \right)\vec \eta = \vec 0$
Or, in order for $$\eqref{eq:eq2}$$ to be a solution to $$\eqref{eq:eq1}$$, $$r$$ and $$\vec \eta$$ must be an eigenvalue and eigenvector for the matrix $$A$$.
Therefore, in order to solve $$\eqref{eq:eq1}$$ we first find the eigenvalues and eigenvectors of the matrix $$A$$ and then we can form solutions using $$\eqref{eq:eq2}$$. There are going to be three cases that we’ll need to look at. The cases are real, distinct eigenvalues, complex eigenvalues and repeated eigenvalues.
None of this tells us how to completely solve a system of differential equations. We’ll need the following couple of facts to do this.
#### Fact
1. If $${\vec x_1}\left( t \right)$$ and $${\vec x_2}\left( t \right)$$ are two solutions to a homogeneous system, $$\eqref{eq:eq1}$$, then ${c_1}{\vec x_1}\left( t \right) + {c_2}{\vec x_2}\left( t \right)$
is also a solution to the system.
2. Suppose that $$A$$ is an $$n$$ x $$n$$ matrix and suppose that $${\vec x_1}\left( t \right)$$, $${\vec x_2}\left( t \right)$$, …, $${\vec x_n}\left( t \right)$$ are solutions to a homogeneous system, $$\eqref{eq:eq1}$$. Define, $X = \left( {{{\vec x}_1}\,\,{{\vec x}_2}\,\, \cdots \,\,{{\vec x}_n}} \right)$
In other words, $$X$$ is a matrix whose ith column is the $$i^{\text{th}}$$ solution. Now define,
$W = \det \left( X \right)$
We call $$W$$ the Wronskian. If $$W \ne 0$$ then the solutions form a fundamental set of solutions and the general solution to the system is,
$\vec x\left( t \right) = {c_1}{\vec x_1}\left( t \right) + {c_2}{\vec x_2}\left( t \right) + \cdots + {c_n}{\vec x_n}\left( t \right)$
Note that if we have a fundamental set of solutions then the solutions are also going to be linearly independent. Likewise, if we have a set of linearly independent solutions then they will also be a fundamental set of solutions since the Wronskian will not be zero.
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# Lesson 10
Representing Large Numbers on the Number Line
## 10.1: Labeling Tick Marks on a Number Line (5 minutes)
### Warm-up
This warm-up prompts students to reason about values on a number line that end in a power of 10. It enables them to visualize and make sense of numbers expressed as a product of a single digit and a power of 10, which prepares them to begin working with scientific notation.
Expect student responses to include a variety of incorrect or partially-correct ideas. It is not important that all students understand the correct notation at this point, so it is not necessary to extend the time for this reason.
During the partner discussions, identify and select students who have partially-correct responses to share during the whole-class discussion.
### Launch
Arrange students in groups of 2. Give students 1 minute of quiet work time and then 2 minutes to compare their number line with their partner. Tell the partners to try and come to an agreement on the values on the number line. Follow with a whole-class discussion.
### Student Facing
Label the tick marks on the number line. Be prepared to explain your reasoning.
### Anticipated Misconceptions
• Some students may count the tick marks instead of the segments and think $$10^7$$ is being divided into 9 or 11 parts. Consider asking these students to mark each segment with a highlighter to count them.
• Some students may try to label the tick marks $$10^1$$, $$10^2$$, $$10^3$$, etc., and as a result, they may say that the $$10^7$$ is shown in the wrong place. Ask these students how many equal parts $$10^7$$ is divided into, and how they would write that as a division problem or with a fraction.
• Some students may label the tick marks as $$10^6$$, $$20^6$$, $$30^6$$, etc. Ask these students to expand $$20^6$$ and $$10^6$$ into their repeated factors and compare them so they see that $$20^6$$ is not twice as much as $$10^6$$.
• Some students may work with the idea of $$\frac{1}{2}$$ and label the middle tick mark as $$10^{3.5}$$. Explain that this tick mark would have a value of $$\frac12 \boldcdot 10^7$$ and ask these students to use what they know about powers of 10 to find the value of the first tick mark after 0.
### Activity Synthesis
Ask selected students to explain how they labeled the number line. Record and display their responses on the number line for all to see. As students share, use their responses, correct or incorrect, to guide students to the idea that the first tick mark is $$1 \boldcdot 10^6$$, the second is $$2 \boldcdot 10^6$$, etc. This gives students the opportunity to connect the number line representation with the computational rules they developed in previous lessons. For example, if a student claims that the second tick mark is $$20^6$$, they can check whether $$20^6$$ is equal to $$2 \boldcdot 10^6$$ by expanding both expressions.
If not uncovered in students' explanations, ask the following questions to make sure students see how to label the number line correctly:
• “How many equal parts is $$10^7$$ being divided into?” (10)
• “If the number at the end of this number line were 20, how would we find the value of each tick mark?” (Divide 20 by 10)
• “Can we use the same reasoning with $$10^7$$ at the end?” (Yes)
• “What is $$10^7 \div 10$$?” ($$10^6$$) “What does this number represent?” (The distance between two tick marks)
• “Can we write $$10^6$$ as $$1 \boldcdot 10^6$$?” (Yes).
• “If the first tick mark is $$1 \boldcdot 10^6$$, then what is the second tick mark?” ($$2 \boldcdot 10^6$$) “
## 10.2: Comparing Large Numbers with a Number Line (10 minutes)
### Activity
This activity encourages students to use the number line to make sense of powers of 10 and think about how to rewrite expressions in the form $$b \boldcdot 10^n$$, where $$b$$ is between 1 and 10 (as in the case of scientific notation). It prompts students to use the structure of the number line to compare numbers, and to extend their use to estimate relative sizes of other numbers when no number lines are given.
As students work, notice the ways in which they compare expressions that are not written as multiples of $$10^6$$. Highlight some of these methods in the discussion.
### Launch
Arrange students in groups of 2. Give students 4 minutes of quiet time to work on the first problem, followed by 1–2 minutes to exchange and discuss their work with their partner (second problem).
Then, tell students that representing numbers as a single digit times a power of 10 is useful for making rough comparisons. Give an example: $$9 \boldcdot 10^{11}$$ is roughly 200 times as large as $$4\boldcdot 10^9$$, because $$10^{11}$$ is 100 times as much as $$10^9$$, and 9 is roughly twice as much as 4. Give students the remaining time to answer the last question. Follow with a brief whole-class discussion.
Classes using the digital version have an interactive applet. Students need to drag the points, marked with open circles and their coordinates, to their proper places on the number line. When all five points are on the line, feedback is available. Note: labels are placed above or below the points only to avoid crowding on the number line.
Representation: Internalize Comprehension. Activate or supply background knowledge about number lines. Keep the display from the warm-up visible for students to reference.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
1. Drag the points to their proper places on the number line. Be prepared to explain your reasoning.
2. Discuss with a partner how you decided where each point should go.
3. Which is larger, 4,000,000 or $$75 \boldcdot 10^5$$? Estimate how many times larger.
### Launch
Arrange students in groups of 2. Give students 4 minutes of quiet time to work on the first problem, followed by 1–2 minutes to exchange and discuss their work with their partner (second problem).
Then, tell students that representing numbers as a single digit times a power of 10 is useful for making rough comparisons. Give an example: $$9 \boldcdot 10^{11}$$ is roughly 200 times as large as $$4\boldcdot 10^9$$, because $$10^{11}$$ is 100 times as much as $$10^9$$, and 9 is roughly twice as much as 4. Give students the remaining time to answer the last question. Follow with a brief whole-class discussion.
Classes using the digital version have an interactive applet. Students need to drag the points, marked with open circles and their coordinates, to their proper places on the number line. When all five points are on the line, feedback is available. Note: labels are placed above or below the points only to avoid crowding on the number line.
Representation: Internalize Comprehension. Activate or supply background knowledge about number lines. Keep the display from the warm-up visible for students to reference.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
1. Place the numbers on the number line. Be prepared to explain your reasoning.
1. 4,000,000
2. $$5 \boldcdot 10^6$$
3. $$5 \boldcdot 10^5$$
4. $$75 \boldcdot 10^5$$
5. $$(0.6) \boldcdot 10^7$$
2. Trade number lines with a partner, and check each other’s work. How did your partner decide how to place the numbers? If you disagree about a placement, work to reach an agreement.
3. Which is larger, 4,000,000 or $$75 \boldcdot 10^5$$? Estimate how many times larger.
### Anticipated Misconceptions
Some students may misplace expressions like $$(0.6) \boldcdot 10^7$$ or $$75 \boldcdot 10^5$$ on the number line. For $$75 \boldcdot 10^5$$, point out that 75 is the same as $$(7.5) \boldcdot 10$$, so $$75 \boldcdot 10^5$$ is equivalent to $$(7.5) \boldcdot 10 \boldcdot 10^5$$. For $$(0.6) \boldcdot 10^7$$, point out that 0.6 is the same as $$6 \boldcdot 10^{\text-1}$$, so $$(0.6) \boldcdot 10^7$$ is equivalent to $$6 \boldcdot 10^{\text-1} \boldcdot 10^7$$. Alternatively, tell the student to think of $$(0.6) \boldcdot 10^7$$ as between 0 and $$10^7$$ in the same way that 0.6 is between 0 and 1 to guide them to the correct placement on the number line.
### Activity Synthesis
Ask students, “How could you change 4,000,000, $$75 \boldcdot 10^5$$, and $$(0.6) \boldcdot 10^7$$ so that all the expressions have the same power of 10?” Highlight the main idea that it’s always possible to rewrite an expression that is a multiple of a power of 10 so that the leading factor is between 1 and 10. For example, 75 can be written as $$(7.5) \boldcdot 10$$, and 0.6 can be written as $$6 \boldcdot 10^{\text-1}$$.
If time allows, consider presenting a problem that allows students to use powers of 10 to estimate the relative sizes of large numbers and use them to answer a question in context:
• The population of the United States is roughly $$3 \boldcdot 10^8$$ people. The global population is roughly $$7 \boldcdot 10^9$$ people. Estimate how many times larger the global population is than the U.S. population. ($$7 \boldcdot 10^9$$ is roughly 20 times as large as $$3 \boldcdot 10^8$$, because 7 is roughly twice as large as 3, and $$10^9$$ is 10 times as large as $$10^8$$.)
Writing, Speaking: MLR1 Stronger and Clearer Each Time. Use this routine to give students a structured opportunity to revise and refine their response to the last question. Ask each student to meet with 2–3 other partners in a row for feedback. Provide students with prompts for feedback that will help them strengthen their ideas and clarify their language (e.g., “Why do you think _____ is larger?”, “How did you compare the two terms?”, “Can you represent your thinking in another way?”, etc.). Students can borrow ideas and language from each partner to strengthen their final version.
Design Principle(s): Optimize output (for explanation)
## 10.3: The Speeds of Light (20 minutes)
### Activity
This activity guides students to thinking in terms of scientific notation while investigating the properties of light. A number line that shows a power of 10 partitioned into 10 equal intervals is again used to illustrate the base-ten structure. Plotting numbers along it gives a clearer meaning to expressions that are a product of a single digit and a power of 10.
To distinguish more easily between the different speeds of light through various materials, the interval between $$2 \boldcdot 10^8$$ and $$3 \boldcdot 10^8$$ is magnified on the number line. This illustrates numbers with more decimal places and allows students to see how they are expressed in scientific notation.
Once a number line is labeled with powers of 10 and its structure is understood, numbers given in scientific notation can be placed on the number line fairly straightforwardly. This encourages students to look for ways to write the other numbers in scientific notation.
### Launch
Display the following number line for all to see. Explain to students that as light moves through different materials, it slows down. The speed of light through empty space, with nothing in its way, is roughly 300,000,000 meters per second. The speed of light through olive oil is much slower at roughly 200,000,000 meters per second.
Ask students to decide what the power of 10 to use for the label of the rightmost tick mark on the number line so that the speed of light through space and through olive oil can be plotted. Give 1 minute of quiet think time before asking 1–2 students to share their responses. Make sure students see that $$10^9$$ is appropriate because for 200,000,000 (which is $$2 \boldcdot 10^8$$) to be plotted between 0 and the last tick mark, the last power of 10 needs to be greater than $$10^8$$).
Next, give students 10–12 minutes to work followed by a whole-class discussion.
Students using the digital materials can use the applet to plot the numbers. The magnifying glass allows them to zoom into any interval between two tick marks and plot numbers to an additional decimal place.
Writing, Conversing: MLR5 Co-Craft Questions. Display only the table and ask pairs of students to write possible questions that could be answered by the data in the table. Select 2–3 groups to share their questions with the class. Highlight questions that ask students to compare quantities. Next, reveal the questions of the activity. This routine allows students to produce the language of mathematical questions and talk about the quantities in this task that are represented in different ways (i.e., powers of 10, expanded form) prior to being asked to solve questions based on the values.
Design Principle(s): Maximize meta-awareness; Support sense-making
### Student Facing
The table shows how fast light waves or electricity can travel through different materials:
material speed of light (meters per second)
space 300,000,000
water $$(2.25) \boldcdot 10^8$$
copper wire (electricity) 280,000,000
diamond $$124 \boldcdot 10^6$$
ice $$(2.3) \boldcdot 10^8$$
olive oil 200,000,000
1. Which is faster, light through diamond or light through ice? How can you tell from the expressions for speed?
Let’s zoom in to highlight the values between $$(2.0) \boldcdot 10^8$$ and $$(3.0) \boldcdot 10^8$$.
2. Plot a point for each speed on both number lines, and label it with the corresponding material.
3. There is one speed that you cannot plot on the bottom number line. Which is it? Plot it on the top number line instead.
4. Which is faster, light through ice or light through diamond? How can you tell from the number line?
### Launch
Display the following number line for all to see. Explain to students that as light moves through different materials, it slows down. The speed of light through empty space, with nothing in its way, is roughly 300,000,000 meters per second. The speed of light through olive oil is much slower at roughly 200,000,000 meters per second.
Ask students to decide what the power of 10 to use for the label of the rightmost tick mark on the number line so that the speed of light through space and through olive oil can be plotted. Give 1 minute of quiet think time before asking 1–2 students to share their responses. Make sure students see that $$10^9$$ is appropriate because for 200,000,000 (which is $$2 \boldcdot 10^8$$) to be plotted between 0 and the last tick mark, the last power of 10 needs to be greater than $$10^8$$).
Next, give students 10–12 minutes to work followed by a whole-class discussion.
Students using the digital materials can use the applet to plot the numbers. The magnifying glass allows them to zoom into any interval between two tick marks and plot numbers to an additional decimal place.
Writing, Conversing: MLR5 Co-Craft Questions. Display only the table and ask pairs of students to write possible questions that could be answered by the data in the table. Select 2–3 groups to share their questions with the class. Highlight questions that ask students to compare quantities. Next, reveal the questions of the activity. This routine allows students to produce the language of mathematical questions and talk about the quantities in this task that are represented in different ways (i.e., powers of 10, expanded form) prior to being asked to solve questions based on the values.
Design Principle(s): Maximize meta-awareness; Support sense-making
### Student Facing
The table shows how fast light waves or electricity can travel through different materials.
material speed (meters per second)
space 300,000,000
water $$(2.25) \boldcdot 10^8$$
copper wire (electricity) 280,000,000
diamond $$124 \boldcdot 10^6$$
ice $$(2.3) \boldcdot 10^8$$
olive oil 200,000,000
1. Which is faster, light through diamond or light through ice? How can you tell from the expressions for speed?
Let’s zoom in to highlight the values between $$(2.0) \boldcdot 10^8$$ and $$(3.0) \boldcdot 10^8$$.
2. Label the tick marks between $$(2.0) \boldcdot 10^8$$ and $$(3.0) \boldcdot 10^8$$.
3. Plot a point for each speed on both number lines, and label it with the corresponding material.
4. There is one speed that you cannot plot on the bottom number line. Which is it? Plot it on the top number line instead.
5. Which is faster, light through ice or light through diamond? How can you tell from the number line?
### Student Facing
#### Are you ready for more?
Find a four-digit number using only the digits 0, 1, 2, or 3 where:
• the first digit tells you how many zeros are in the number,
• the second digit tells you how many ones are in the number,
• the third digit tells you how many twos are in the number, and
• the fourth digit tells you how many threes are in the number.
The number 2,100 is close, but doesn’t quite work. The first digit is 2, and there are 2 zeros. The second digit is 1, and there is 1 one. The fourth digit is 0, and there are no threes. But the third digit, which is supposed to count the number of 2’s, is zero.
1. Can you find more than one number like this?
2. How many solutions are there to this problem? Explain or show your reasoning.
### Anticipated Misconceptions
Students may struggle to rewrite a number written using one power of 10 as a number with a different power, for example writing $$125 \boldcdot 10^6$$ using $$10^8$$. Some may know the relationship between the powers of 10 (say, between $$10^6$$ and $$10^8$$), but may not know how expressing one in terms of the other affects the other factor. Help students make sense of the rewriting process with a series of questions such as these:
• “What do we need to multiply $$10^6$$ by to get $$10^8$$?” (100 or $$10^2$$
• “If we multiply $$10^6$$ by $$10^2$$, what must we also do to maintain the value of the expression? (Divide the other factor in the expression by $$10^2$$.)
• “What is the resulting expression?” ($$(125 \div 10^2) \boldcdot (10^6 \boldcdot 10^2) = (1.25) \boldcdot 10^8$$)
• “How do we know that the two expressions are equivalent?” (We multiplied the expression by $$10^2$$ and then divided it by $$10^2$$, which is equal to multiplying it by 1.)
### Activity Synthesis
Ask students to explain how they were able to compare the speeds of light. Students do not yet know the definition of scientific notation, but this activity should help them see that expressing values in this format allows us to more easily compare them. Consider using the applet to further illustrate how each number could be plotted on the number line.
Tell students, “We saw that the speed of light through ice can be written as $$(2.3) \boldcdot 10^8$$ meters per second, and the speed of electricity can be written as $$(2.8) \boldcdot 10^8$$ meters per second. When you write them both the same way like this, it makes it much easier to compare them.”
## Lesson Synthesis
### Lesson Synthesis
The purpose of the discussion is to check that students understand how to express a large number as a multiple of a power of 10, find the value of a given multiple of a power of 10, and compare different large numbers by expressing them as multiples of the same power of 10.
It is important students understand that “multiple of a power of 10” does not mean integer multiple, necessarily. Tell students that they will be asked to express numbers as “multiples of a power of 10,” which might mean writing 52,000 as $$(5.2) \boldcdot 10^4$$, for example.
Here are some questions for discussion:
• “What are some ways you came up with to write 230,000,000 using powers of 10?” (Since the value of 230,000,000 will stay the same if it is multiplied by 10 and then divided by 10, we can think of it as: $$230,\!000,\!000 = 23,\!000,\!000 \boldcdot 10 = 2,\!300,\!000 \boldcdot 10^2 = \ldots = 23 \boldcdot 10^7 = (2.3) \boldcdot 10^8$$.)
• “What are some of the ways you came up with to find the value of $$(5.4) \boldcdot 10^5$$?” (The value of $$(5.4) \boldcdot 10^5$$ is 540,000 because $$(5.4) \boldcdot 10^5 = 54 \boldcdot 10^4 = 540 \boldcdot 10^3 = \ldots = 540,\!000$$.)
• “How did you compare which was faster—the speed of light through diamond and the speed of light through ice?” (The speed of light through diamond was given as $$124 \boldcdot 10^6$$ meters per second, and the speed of light through ice was given as $$(2.3) \boldcdot 10^8$$ meters per second. The speed through diamond could be rewritten as $$(1.24) \boldcdot 10^8$$ meters per second, which makes it clear that it is slower than the speed of light through ice because $$1.24 < 2.3$$.)
## Student Lesson Summary
### Student Facing
There are many ways to compare two quantities. Suppose we want to compare the world population, about 7.4 billion
There are many ways to do this. We could write 7.4 billion as a decimal, 7,400,000,000, and then we can tell that there were more pennies made in 2015 than there are people in the world! Or we could use powers of 10 to write these numbers: $$\displaystyle 7.4 \boldcdot 10^9$$ for people in the world and $$\displaystyle 8.9 \boldcdot 10^9$$ for the number of pennies.
For a visual representation, we could plot these two numbers on a number line. We need to carefully choose our end points to make sure that the numbers can both be plotted. Since they both lie between $$10^9$$ and $$10^{10}$$, if we make a number line with tick marks that increase by one billion, or $$10^9$$, we start the number line with 0 and end it with $$10 \boldcdot 10^{9}$$, or $$10^{10}$$. Here is a number line with the number of pennies and world population plotted: |
# Difference between revisions of "1962 AHSME Problems/Problem 6"
## Problem
A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:
$\textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these}$
## Solution
To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is $\dfrac{x^2\sqrt{3}}{4}$. This has to be equal to $9 \sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It follows that each side of the square is $\dfrac{18}{4}=4.5$. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is $\boxed{\text{(D)} \dfrac{9\sqrt{2}}{2}}$. |
# How do you graph, find the intercepts and state the domain and range of f(x)=6^x+3?
May 30, 2017
y-intercept $\to \left(x , y\right) = \left(0 , 4\right)$
No x-intercept
domain $\textcolor{w h i t e}{.} \to \left\{x : x \in \left(- \infty , + \infty\right)\right\}$
range $\text{ } \to \left\{y : y \in \left(+ 3 , + \infty\right)\right\}$
#### Explanation:
Set $\text{ } y = {6}^{x} + 3$
$\textcolor{b l u e}{\text{Determine the y-intercept}}$
Set $x = 0$ giving
$y = {6}^{0} + 3 = 4$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the x-intercept}}$
Set $y = 0 = {6}^{x} + 3$
${6}^{x} = - 3$
Take logs of both sides
$x \ln \left(6\right) = \ln \left(- 3\right)$
But ln(-3) is undefined thus there is no x-intercept
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine end behaviour } x \to - \infty}$
${\lim}_{x \to - \infty} {6}^{x} + 3 = {\lim}_{z \to + \infty} \frac{1}{6} ^ z + 3 \to k = 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine end behaviour } x \to + \infty}$
${\lim}_{x \to + \infty} {6}^{x} + 3 \text{ " ->" } k + 3 = \infty + 3 = \infty$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
## Equivalence
At the heart of our approach to mathematics is the technique of calculation. When we calculate, we rearrange expressions to form new expressions in accordance with certain rules. These expressions may have many different values. One way we cope with this complexity is to identify values which have properties in common and collect them into groups. Such groups are called types. If a value $v$ belongs to a type $T$, we say that $v$ is of type $T$.
***
The first type we will explore is a very simple one – so simple it consists of only two values. The type is called boolean and its values are $\top$, pronounced “top” and $\bot$, pronounced “bottom”.
Now we can talk about the value of $X = Y$ :
“The value of $X = Y$ is of type boolean.”
or, equivalently,
“The value of $X = Y$ is either $\top$ or $\bot$.”
***
Equality between booleans is special, and so when we express equality between $X$ and $Y$, where $X$ and $Y$ are boolean we write
$X\equiv Y$
pronounced “$X$ equivales $Y$ “.
What is so special about boolean equality – called equivalence – that we have special symbol for it? The answer gives us our first rule for calculating:
Associativity:
$(X \equiv (Y \equiv Z)) \equiv ((X \equiv Y ) \equiv Z)$
We give the rule a name to help us remember what it is. Here is how we would use the rule in calculations:
$X \equiv (Y \equiv Z)$ $\equiv\quad$ $\{\textrm{associativity}\}$ $(X \equiv Y ) \equiv Z$
An important consequence of $\eqv$’s associativity is that when we have a series of expressions punctuated by $\eqv$ signs it does not matter where we put the brackets. Indeed, we need not write the brackets at all, as for instance in our next rule:\\
Symmetry: $X\equiv Y\equiv Y\equiv X$
We may parse this rule in several ways:
$latex (X\equiv Y)\equiv (Y\equiv X)$
$X\equiv (Y\equiv Y\equiv X)$
$(X\equiv Y\equiv Y)\equiv X$
The last two expressions tell us that `equivaling’ $Y\equiv Y$ with any boolean $X$ is $X$. We say that $Y\equiv Y$ is the identity of $\equiv$. Our next rule gives a name to the identity:
(2) $X\equiv\top\equiv X$
Up to this point we have simply postulated rules i.e. we have taken it as given that they hold. Our next rule, however, we will prove. How do we prove? When we postulate a rule we are saying that it is an expression which is equivalent to $\top$. So to show that an expression is a rule, we present a calculation showing that the said expression is also equivalent to $\top$.
Reflexivity: $X\equiv X$
Proof
$X\equiv X$
$\equiv$ {{(2), parsed as $X\equiv(\top\equiv X)$
$X\equiv\top\equiv X$
$\equiv$ {(2)}
$\top$
Such is the equivalence.
Challenge: Does anyone know how to get the latex \quad and \begin{tabular} commands to work in WordPress?
### One Response to “Equivalence”
1. Science After Sunclipse Says:
Carnival of Mathematics
Mark Chu-Carroll hosts the latest Carnival of Mathematics with a theme dear to my heart, the way cholesterol is: spam!
Among the notable posts are My Tiny Kingdom’s report on helping with long-division homework. This reminds me: can any of the … |
# Triangles and TrigonometryPythagoras’ Theorem
We have now reached an important point in geometry – being able to state and understand one of the most famous theorems in all of mathematics: Pythagoras’ Theorem. It is named after the ancient Greek mathematician Pythagoras of Samos.
Pythagoras’ Theorem
In any right-angled triangle, the square of the length of the hypotenuse (the side that lies opposite the right angle) is equal to the sum of the squares of the other two sides. In other words,
a2 + b2 = c2
The converse is also true: if the three sides in a triangle satisfy a2+b2=c2, then it must be .
Right angles are everywhere, and that’s why Pythagoras’ Theorem is so useful.
Here you can see a 6m long ladder leaning on a wall. The bottom of the ladder is 1m away from the wall. How far does it reach up the wall?
Notice that there is a right-angled triangle formed by the ladder, the wall and the ground. Using Pythagoras’ theorem, we get
h2+12 =62 ⇒h2 = ⇒h =35=5.92m
Whenever you’ve got a right-angled triangle and know two of its sides, Pythagoras can help you find the third one.
## Proving Pythagoras’ Theorem
Pythagoras’ theorem was known to ancient Babylonians, Mesopotamians, Indians and Chinese – but Pythagoras may have been the first to find a formal, mathematical proof.
There are actually many different ways to prove Pythagoras’ theorem. Here you can see three different examples that each use a different strategy:
### Rearrangement
Have a look at the figure on the right. The square has side length a+b, and contains four right-angled triangles, as well as a smaller square of area .
Now let’s rearrange the triangles in the square. The result still contains the four right-angles triangles, as well as two squares of size .
Comparing the area of the red area and the rearrangement, we see that
a2+b2=c2.
This is the original proof that Pythagoras came up with.
### Algebra
Here we have the same figure as before, but this time we’ll use algebra rather than rearrangement to prove Pythagoras’ theorem.
The large square has side length a+b and area .
It consists of four triangles, each with an area of , and one square of area .
If we combine all of that information, we have
a+b2 =4×12ab+c2 a2+2ab+b2 =2ab+c2 a2+b2 =c2
And, once again, we get Pythagoras’ theorem.
### Similar Triangles
Here you can see another right-angled triangle. If we draw one of the altitudes, it splits the triangle into two smaller triangle. It also divides the hypotenuse c into two smaller parts which we’ll call x and y. Continue
Let’s separate out the two smaller triangles, so that it’s clearer to see how they are related… Continue
Both smaller triangles share one angle with the original triangle. They also all have one right angle. By the AA condition, all three triangles must be .
Now we can use the equations we already know about similar polygons:
xa=acx=a2c yb=bcy=b2c
But remember that c=x+y. Therefore
c=a2c+b2c
c2=a2+b2
Once more, we’ve proven Pythagoras’ theorem!
Much about Pythagoras’ life is unknown, and no original copies of his work have survived. He founded a religious cult, the Pythagoreans, that practiced a kind of “number worship”. They believed that all numbers have their own character, and followed a variety of other bizarre customs.
The Pythagoreans are credited with many mathematical discoveries, including finding the first irrational number, 2. Irrational numbers cannot be expressed as a simple fraction – a concept the Pythagoreans found deeply troubling and (unsuccessfully) tried to cover up!
“Pythagoreans celebrate sunrise” by Fyodor Bronnikov
## Calculating Distances
One of the most important application of Pythagoras’ Theorem is for calculating distances.
On the right you can see two points in a coordinate system. We could measure their distance using a ruler, but that is not particularly accurate. Instead, let’s try using Pythagoras.
We can easily count the horizontal distance along the x-axis, and the vertical distance along the y-axis. If we draw those two lines, we get a right-angled triangle.
Using Pythagoras,
d2 =\${b.x-a.x}2+\${b.y-a.y}2 d2 =\${(b.x-a.x)**2 + (b.y-a.y)**2} d =\${(b.x-a.x)**2 + (b.y-a.y)**2}=\${round(distance(a,b),4)}
This method works for any two points:
The Distance Formula
If you are given two points with coordinates (x1,y1) and (x2,y2), the distance between them is
d2=x2x12+y2y12
d=x2x12+y2y12
## Pythagorean Triples
As you moved the vertices of the triangle in the previous step, you might have noticed that in most cases, the length of the hypothenuse d ended up being a . However there are a few examples of right-angled triangles where the lengths of all three sides happens to be whole numbers.
One famous example is the 3-4-5 triangle. Since 32+42=52, any triangle with sides of length 3, 4 and 5 must be right-angled.
The ancient Egyptians didn’t know about Pythagoras’ theorem, but they did know about the 3-4-5 triangle. When building the pyramids, they used knotted ropes of lengths 3, 4 and 5 to measure perfect right angles.
Three integers like this are called Pythagorean Triples. (3, 4, 5) is one example of a Pythagorean triple. If we multiply every number by 2, we get another Pythagorean triple: (6, 8, ).
We can think of these triples as grid points in a coordinate systems. For a valid Pythagorean triples, the distance from the origin to the grid point has to be a whole number. Using the coordinate system below, can you find any other Pythagorean triples?
Do you notice any pattern in the distribution of these points?
Archie |
# What is standard algorithm multiplication?
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Listen
## Introduction
Standard algorithm multiplication is a widely used method for multiplying multi-digit numbers. It provides a systematic approach to perform multiplication by breaking down the process into smaller, more manageable steps. This article will explore the concept of standard algorithm multiplication, its step-by-step procedure, and its importance in mathematics education.
## Understanding Standard Algorithm Multiplication
Standard algorithm multiplication, also known as long multiplication, is a method used to multiply two or more multi-digit numbers. It is based on the principles of place value and requires a solid understanding of basic multiplication facts.
The process involves breaking down the multiplication problem into smaller, more manageable steps, multiplying digits in each place value position, and then summing up the partial products to obtain the final result.
## The Step-by-Step Procedure
The standard algorithm multiplication can be best understood through an example. Let’s consider multiplying a two-digit number, 23, by a three-digit number, 456.
Step 1: Write the numbers vertically, aligning them by their place values:
“`
456
x 23
——
“`
Step 2: Starting from the rightmost digit of the bottom number (3 in this case), multiply it by each digit of the top number (456). Write the partial products below the line, aligning them by their place values:
### What is primavera software?
“`
456
x 23
——
1368
+ 9120
——
“`
Step 3: Add up the partial products to obtain the final result:
“`
456
x 23
——
10488
——
“`
In this example, the result of multiplying 456 by 23 using the standard algorithm is 10,488.
## Importance in Mathematics Education
Standard algorithm multiplication is an essential skill for students to develop in their mathematical education. It helps them understand the concept of place value, reinforces basic multiplication facts, and improves their problem-solving abilities.
By breaking down the multiplication process into smaller steps, students can better grasp the underlying principles and develop a deeper understanding of multiplication. It also provides a systematic approach that can be easily applied to multiply larger numbers.
Moreover, standard algorithm multiplication lays the foundation for more advanced mathematical concepts, such as long division and algebraic equations. Mastering this method early on helps students build confidence in their mathematical abilities and prepares them for more complex mathematical tasks.
## Conclusion
Standard algorithm multiplication is a fundamental method for multiplying multi-digit numbers. By breaking down the process into smaller steps, it provides a systematic approach that enhances students’ understanding of multiplication and lays the groundwork for more advanced mathematical concepts. Developing proficiency in standard algorithm multiplication is crucial for students’ mathematical education.
## References
– MathisFun: mathisfun.com/long_multiplication.html |
## Combinations
### Learning Outcomes
• Find the number of combinations of n distinct choices.
So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations. A selection of $r$ objects from a set of $n$ objects where the order does not matter can be written as $C\left(n,r\right)$. Just as with permutations, $\text{C}\left(n,r\right)$ can also be written as ${}_{n}{C}_{r}$. In this case, the general formula is as follows.
$\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}$
An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are $3!=3\cdot 2\cdot 1=6$ ways to order 3 paintings. There are $\frac{24}{6}$, or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings.
### A General Note: Formula for Combinations of n Distinct Objects
Given $n$ distinct objects, the number of ways to select $r$ objects from the set is
$\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}$
### How To: Given a number of options, determine the possible number of combinations.
1. Identify $n$ from the given information.
2. Identify $r$ from the given information.
3. Replace $n$ and $r$ in the formula with the given values.
4. Evaluate.
### Example: Finding the Number of Combinations Using the Formula
A fast food restaurant offers five side dish options. Your meal comes with two side dishes.
1. How many ways can you select your side dishes?
2. How many ways can you select 3 side dishes?
### Q & A
#### Is it a coincidence that parts (a) and (b) in Example 4 have the same answers?
No. When we choose r objects from n objects, we are not choosing $\left(n-r\right)$ objects. Therefore, $C\left(n,r\right)=C\left(n,n-r\right)$.
### Try It
An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?
## Finding the Number of Subsets of a Set
We have looked only at combination problems in which we chose exactly $r$ objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible?
To answer this question, we need to consider pizzas with any number of toppings. There is $C\left(5,0\right)=1$ way to order a pizza with no toppings. There are $C\left(5,1\right)=5$ ways to order a pizza with exactly one topping. If we continue this process, we get
$C\left(5,0\right)+C\left(5,1\right)+C\left(5,2\right)+C\left(5,3\right)+C\left(5,4\right)+C\left(5,5\right)=32$
There are 32 possible pizzas. This result is equal to ${2}^{5}$.
We are presented with a sequence of choices. For each of the $n$ objects we have two choices: include it in the subset or not. So for the whole subset we have made $n$ choices, each with two options. So there are a total of $2\cdot 2\cdot 2\cdot \dots \cdot 2$ possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time.
### A General Note: Formula for the Number of Subsets of a Set
A set containing n distinct objects has ${2}^{n}$ subsets.
### Example: Finding the Number of Subsets of a Set
A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?
### Try It
A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
You will need
• Calculator, pen
Instruction
1
Own speed is the speed of the boat or another vehicle in still water. Her mark - V own.
The water in the river is moving. So it has its speed, called the speedu of the flow (V tech.)
The speed of boats on the river mark - V via tech., and the speed upstream is V PR. tech.
2
Now remember the formulas required to solve problems on the motion:
V PR. tech.= V own. - V tech.
V on tech.= V own. + V tech.
3
So, based on these formulas, we can draw the following conclusions.
If the boat moves against the river's current, V own. = V PR. tech. + V tech.
If the boat moves with the current, then V own. = V on tech. - V tech.
4
Solve some of the problems on the motion on the river.
Task 1. The speed of the boat against the river's current 12.1 km/h. Find your own speed boats, knowing that the speed of river is 2 km/h.
Solution: 12,1 + 2 = 14, 1 (km/h) - private speed boat.
Task 2. The speed of boats on the river 16.3 km/h, the speed of the river is 1.9 km/HR How many metres would have passed this boat for 1 min., if it was placed in still water?
The decision of 16.3 and 1.9 = 14.4 V (km/h) - private speed boat. Translate km/h to m/min: 14.4 V / 0,06 = 240 (m/min). So in 1 minute the boat was 240 m.
Task 3. Two boats went at the same time towards each other from two points. The first boat moved up the river, and the second - against the current. They met in three hours. During this time the first boat was 42 km, and the second is 39 km away and Find your own speed of each boat if it is known that the speed of river is 2 km/h.
Solution: 1) 42 / 3 = 14 (km/h) - speed of traffic on the river the first boat.
2) 39 / 3 = 13 (km/h) - the speed of the movement against the current of the river a second boat.
3) 14 - 2 = 12 (km/h) - speed of the first boat.
4) 13 + 2 = 15 (km/h) - speed of a second boat. |
Absolute value facts for kids
Kids Encyclopedia Facts
The graph of the absolute value function for real numbers.
In mathematics, the absolute value or modulus of a real number x, written as |x| or $\text{abs}(x)$, is the non-negative value of x when the sign is dropped. That is, |x| = x for a positive x, |x| = −x for a negative x (in which case x is positive), and |0| = 0.
For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a real number may be thought of as its distance from zero. It can be defined as follows:
$\begin{cases} \ \;\,\ \ x &\mathrm{if}\ x \ge 0\\ \ \;\, - x &\mathrm{otherwise} \end{cases}$
Similarly, the absolute value (or modulus) of a complex number may be thought of as its distance from the origin. It is defined by the equation
$|x + iy| = \sqrt{x^2+y^2}\$
Properties
Real numbers
For any real number x, the absolute value of x is denoted by | x | (a vertical bar on each side of the quantity), and is defined as
$|x| = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x < 0. \end{cases}$
The absolute value of x is always either positive or zero, but never negative. From an analytic geometry point of view, the absolute value of a real number is that number's distance from zero along the real number line. The absolute value of the difference of two real numbers is the distance between them.
In calculus, the absolute value function is differentiable except at 0. It is everywhere continuous.
In linear algebra, the norm of a vector is defined similarly as the distance from the tip of the vector to the origin. This is similar to the way the absolute value of a complex number is defined.
The square-root notation without sign represents the positive square root. So, it follows that
$|a| = \sqrt{a^2}$ (1)
which is sometimes used as a definition of absolute value.
The absolute value has the following four main properties:
$|a| \ge 0$ (2) Non-negativity $|a| = 0 \iff a = 0$ (3) Positive-definiteness $|ab| = |a||b|\,$ (4) Multiplicativeness $|a+b| \le |a| + |b|$ (5) Subadditivity
Other important properties of the absolute value include:
$||a|| = |a|\,$ (6) Idempotence (the absolute value of the absolute value is the absolute value) $|-a| = |a|\,$ (7) Symmetry $|a - b| = 0 \iff a = b$ (8) Identity of indiscernibles (equivalent to positive-definiteness) $|a - b| \le |a - c| +|c - b|$ (9) Triangle inequality (equivalent to subadditivity) $|a/b| = |a| / |b| \mbox{ (if } b \ne 0) \,$ (10) Preservation of division (equivalent to multiplicativeness) $|a-b| \ge ||a| - |b||$ (11) (equivalent to subadditivity)
Two other useful properties related to inequalities are:
$|a| \le b \iff -b \le a \le b$
$|a| \ge b \iff a \le -b \mbox{ or } b \le a$
These relations may be used to solve inequalities involving absolute values. For example:
$|x-3| \le 9$ $\iff -9 \le x-3 \le 9$ $\iff -6 \le x \le 12$
Diagram showing $|z|$, the modulus of z.
Complex numbers
For a complex number $z = x + iy$, where x is the real part of z and y is the imaginary part of z,
$|z| = |x + iy| = \sqrt{x^2+y^2}\$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Sample Spaces and Events
## Words or diagrams that detail favorable outcomes and intersections, complements, and unions of events.
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Sample Spaces and Events
Jacqueline prepares the Halloween bowl for trick-or-treaters with equal numbers of Reese's, Kit Kat, Twix and M&Ms. Her first trick-or-treater randomly pulls a candy bar from the bowl, puts it back, then randomly pulls another candy bar from the bowl. Jacqueline wonders how many outcomes are possible from the sample space for the two candy bar pulls from the bowl.
In this concept, you will learn about sample spaces.
### Guidance
When you conduct an experiment, there are many possible outcomes. If you are doing an experiment with a coin, there are two possible outcomes because there are two sides of the coin. You can either have heads or tails. If you experiment with a number cube, there are six possible outcomes because there are six sides of the number cube and the sides are numbered one to six. We can think of all of these possible outcomes as the sample space.
A sample space is the set of all possible outcomes for a probability experiment or activity. For example, on the spinner there are 5 different colors on which the arrow can land. The sample space, , for one spin of the spinner is then:
These are the only outcomes that result from a single spin of the spinner.
Changing the spinner changes the sample space. This second spinner still has 5 equal-sized sections. But its sample space now has only 3 outcomes:
Let’s look at an example with sample spaces.
A small jar contains 1 white, 1 black, and 1 red marble. If one marble is randomly chosen, how many possible outcomes are there in the sample space?
Since only a single marble is being chosen, the total number of possible outcomes, or sample space matches the marble colors.
Sometimes, the sample space can change if an experiment is performed more than once. If a marble is selected from a jar and then replaced and if the experiment is conducted again, then the sample space can change. The number of outcomes is altered. When this happens, we can use tree diagrams and multiplication to help us figure out the number of outcomes in the sample space.
A jar contains 1 white and 1 black marble. If one marble is randomly chosen, returned to the jar, then a second marble is chosen, how many possible outcomes are there?
This is a situation where a tree diagram is very useful. Consider the marbles one at a time. After the first marble is chosen, it is returned to the jar so now there are again two choices for the second marble.
Use a tree diagram to list the outcomes.
From the tree diagram, you can see that the sample space is:
We could also multiply by multiplying the two options and the number of selections. Two marble colors and two selections:
There are four outcomes in the sample space.
### Guided Practice
June flipped a coin three times. How many outcomes are in the sample space?
First, make a list of the possible outcomes for each flip.
Tails
Next, count the number of the possible outcomes for each flip.
There are two outcomes for each flip of a coin: heads or tails.
Then, multiply the number of outcomes by the number of flips.
June flipped the coin three times.
The answer is there are 6 outcomes in the sample space.
### Examples
#### Example 1
How many outcomes are in the sample space of three spins of a spinner with red, blue, yellow and green?
First, make a list of the possible outcomes for each spin.
Red
Blue
Yellow
Green
Next, count the number of possible outcomes for each spin.
There are four possible outcomes for each spin: red, blue, yellow, green.
Then, multiply the number of outcomes by the number of spins.
June flipped the coin three times.
The answer is there are 12 outcomes in the sample space.
#### Example 2
What is the sample space for the roll of a number cube numbered 1–6 and how many possible outcomes are there from one roll?
First, determine all possible outcomes:
1, 2, 3, 4, 5, 6
Next, count the number of possible outcomes for one roll:
There are 6 possible outcomes: 1, 2, 3, 4, 5, 6
Then, multiply the number of outcomes by the number of rolls.
Since we are only rolling once, the number of possible outcomes is 6.
The answer is the sample space is 1, 2, 3, 4, 5, 6 and the number of possible outcomes is 6.
#### Example 3
A bag with a blue and a red marble. One marble is drawn and then replaced. What is the sample space?
First, list the possible outcomes for each draw:
blue
red
Next, list the possible outcomes for Draw 1 and Draw 2:
Draw 1: red blue
Draw 2: red blue red blue
Then, list all possible outcomes from the two draws:
red-red, red-blue, blue-red, blue-blue
The answer is the sample space is red-red, red-blue, blue-red, blue-blue
Remember Jacqueline's curiosity about the candy from the Halloween bowl?
The trick-or-treater pulled one time randomly, returned the candy bar to the bowl, then pulled randomly once more. How many possible outcomes are there from the two pulls?
First, list the different candy choices:
Twix, Reese's, M&Ms, Kit Kat
Next, list the possible outcomes of Pull 1 and Pull 2:
Pull 1: Twix M&Ms
Pull 2: Twix Reese's M&Ms Kit Kat Twix Reese's M&Ms Kit Kat
Pull 1: Reese's Kit Kat
Pull 2: Twix Reese's M&Ms Kit Kat Twix Reese's M&Ms Kit Kat
Then, list the sample space, or all possible outcomes of the two pulls from the candy bowl:
Twix-Twix, Twix-Reese's, Twix-M&Ms, Twix-Kit Kat, M&Ms-Twix, M&Ms-Reese's, M&Ms-M&Ms, M&Ms-Kit Kat, Reese's-Twix, Reese's-Reese's, Reese's-M&Ms, Reese's-Kit Kat, Kit Kat-Twix, Kit Kat-Reese's, Kit Kat-M&Ms, Kit Kat-Kit Kat
The answer is there are 16 possible outcomes.
### Explore More
1. What is the sample space for a single toss of a number cube?
2. What is the sample space for a single flip of a coin?
3. A coin is flipped two times. List all possible outcomes for the two flips.
4. A coin is flipped three times in a row. List all possible outcomes for the two flips.
5. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball from the bag?
6. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball, returning the ball to the bag, then drawing a second ball?
7. What is the sample space for a single spin of the with red, blue, yellow and green sections spinner?
8. What is the sample space for 2 spins of the first spinner?
9. What is the sample space for three spins of the spinner?
10. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing a single sock, NOT returning the sock to the box, then drawing a second sock?
11. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box?
12. A box contains 3 black socks. What is the sample space for drawing all 2 socks from the box, one at a time, without returning any of the socks to the box?
13. A box contains 2 black socks and 1 white sock. What is the sample space for drawing all 2 socks from the box, one at a time, without returning any of the socks to the box?
14. True or false. A sample space is the total possible outcomes.
15. True or false. A sample space is a percentage.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 12.8.
### Vocabulary Language: English Spanish
experiment
experiment
An experiment is the process of taking a measurement or making an observation.
Favorable Outcome
Favorable Outcome
A favorable outcome is the outcome that you are looking for in an experiment.
Outcome
Outcome
An outcome of a probability experiment is one possible end result.
Sample Space
Sample Space
In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.
simple events
simple events
A simple event is the simplest outcome of an experiment.
tree diagrams
tree diagrams
Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree. |
The Application of Differential Equations in Biology
Differential equations are frequently used in solving mathematics and physics problems. In the following example we shall discuss the application of a simple differential equation in biology.
Example:
In a culture, bacteria increases at the rate proportional to the number of bacteria present. If there are 400 bacteria initially and are doubled in 3 hours, find the number of bacteria present 7 hours later.
Let $x$ be the number of bacteria, and the rate is $\frac{{dx}}{{dt}}$. Since the number of bacteria is proportional to the rate, so
$\frac{{dx}}{{dt}} \propto x$
If $k\,\left( {k > 0} \right)$ is the proportionality constant, then
$\frac{{dx}}{{dt}} = kx$
Separating the variables, we have
$\frac{{dx}}{x} = kdt\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$
Since there are 400 bacteria initially and they are doubled in 3 hours, we integrate the left side of equation (i) from 400 to 800 and integrate its right side from 0 to 3 to find the value of $k$ as follows:
$\begin{gathered} \int\limits_{400}^{800} {\frac{{dx}}{x} = k\int\limits_0^3 {dt} } \\ \Rightarrow \left| {\ln x} \right|_{400}^{800} = k\left| t \right|_0^3 \\ \Rightarrow \ln 800 – \ln 400 = k\left( {3 – 0} \right) \\ \Rightarrow 3k = \ln \frac{{800}}{{400}} = \ln 2 \\ \Rightarrow k = \frac{1}{3}\ln 2 \\ \end{gathered}$
Putting the value of $k$ in (i), we have
$\frac{{dx}}{x} = \left( {\frac{1}{3}\ln 2} \right)dt\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$
Next, to find the number of bacteria present 7 hours later, we integrate the left side of (ii) from 400 to $x$ and its right side from 0 to 7 as follows:
$\begin{gathered} \int_{400}^x {\frac{{dx}}{x} = \frac{1}{3}\ln 2\int_0^7 {dt} } \\ \Rightarrow \left| {\ln x} \right|_{400}^x = \frac{1}{3}\ln 2\left| t \right|_0^7 \\ \Rightarrow \ln x – \ln 400 = \frac{1}{3}\ln 2\left( {7 – 0} \right) \\ \Rightarrow \ln x = \ln 400 + \frac{7}{3}\ln 2 \\ \Rightarrow \ln x = \ln 400 + \ln {2^{\frac{7}{3}}} \\ \Rightarrow \ln x = \ln \left( {400} \right){2^{\frac{7}{3}}} \\ \Rightarrow x = \left( {400} \right)\left( {5.04} \right) = 2016 \\ \end{gathered}$
Thus, there are 2016 bacteria after 7 hours. |
# volume and surface area
1. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
A. 12 cm3 B. 15 cm3 C. 16 cm3 D. 20 cm3
Explanation:
Clearly, we have r = 3 cm and h = 4 cm.
Volume = 1 r2h = 1 x x 32 x 4 cm3 = 12 cm3. 3 3
2. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m B. 750 cu. m C. 7500 cu. m D. 75000 cu. m
Explanation:
1 hectare = 10,000 m2
So, Area = (1.5 x 10000) m2 = 15000 m2.
Depth = 5 m = 1 m. 100 20
Volume = (Area x Depth) = 15000 x 1 m3 = 750 m3. 20
3. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
A. 720 B. 900 C. 1200 D. 1800
Explanation:
2(15 + 12) x h = 2(15 x 12)
h = 180 m = 20 m. 27 3
Volume = 15 x 12 x 20 m3 = 1200 m3. 3
4. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:
A. 84 B. 90 C. 168 D. 336
Explanation:
Let the length of the wire be h.
Radius = 1 mm = 1 cm. Then, 2 20
22 x 1 x 1 x h = 66. 7 20 20
h = 66 x 20 x 20 x 7 = 8400 cm = 84 m. 22
5. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:
A. 3.6 kg B. 3.696 kg C. 36 kg D. 36.9 kg
Explanation: |
# How do you evaluate the integral int cos(root3x)?
Apr 2, 2017
$\int \cos \left(\sqrt[3]{x}\right) \mathrm{dx} = 3 \sqrt[3]{{x}^{2}} \sin \left(\sqrt[3]{x}\right) + 6 \sqrt[3]{x} \cos \left(\sqrt[3]{x}\right) - 6 \sin \left(\sqrt[3]{x}\right) + C$
#### Explanation:
$I = \int \cos \left(\sqrt[3]{x}\right) \mathrm{dx}$
First, let $t = \sqrt[3]{x}$. This implies that ${t}^{3} = x$, which tells us that $3 {t}^{2} \mathrm{dt} = \mathrm{dx}$. We can substitute these in:
$I = \int \cos \left(t\right) \left(3 {t}^{2}\right) \mathrm{dt} = \int 3 {t}^{2} \cos \left(t\right) \mathrm{dt}$
Now we perform integration by parts. Let:
$\left\{\begin{matrix}u = 3 {t}^{2} & \implies & \mathrm{du} = 6 t \mathrm{dt} \\ \mathrm{dv} = \cos \left(t\right) \mathrm{dt} & \implies & v = \sin \left(t\right)\end{matrix}\right.$
Then:
$I = 3 {t}^{2} \sin \left(t\right) - \int 6 t \sin \left(t\right) \mathrm{dt}$
Integration by parts again:
$\left\{\begin{matrix}u = 6 t & \implies & \mathrm{du} = 6 \mathrm{dt} \\ \mathrm{dv} = \sin \left(t\right) \mathrm{dt} & \implies & v = - \cos \left(t\right)\end{matrix}\right.$
Paying attention to sign:
$I = 3 {t}^{2} \sin \left(t\right) - \left(6 t \left(- \cos \left(t\right)\right) - \int \left(- 6 \cos \left(t\right)\right) \mathrm{dt}\right)$
$I = 3 {t}^{2} \sin \left(t\right) + 6 t \cos \left(t\right) - \int 6 \cos \left(t\right) \mathrm{dt}$
$I = 3 {t}^{2} \sin \left(t\right) + 6 t \cos \left(t\right) - 6 \sin \left(t\right) + C$
Using $t = \sqrt[3]{x}$:
$I = 3 \sqrt[3]{{x}^{2}} \sin \left(\sqrt[3]{x}\right) + 6 \sqrt[3]{x} \cos \left(\sqrt[3]{x}\right) - 6 \sin \left(\sqrt[3]{x}\right) + C$ |
# Solving Linear Inequalities
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The following math revision questions are provided in support of the math tutorial on Solving Linear Inequalities. In addition to this tutorial, we also provide revision notes, a video tutorial, revision questions on this page (which allow you to check your understanding of the topic) and calculators which provide full, step by step calculations for each of the formula in the Solving Linear Inequalities tutorials. The Solving Linear Inequalities calculators are particularly useful for ensuring your step-by-step calculations are correct as well as ensuring your final result is accurate.
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Inequalities Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
10.1Solving Linear Inequalities
## Solving Linear Inequalities Revision Questions
1. . Which of the following numbers is a solution for the inequality
1 - 2x ≥ 7
1. -3
2. -2
3. -1
4. 0
2. . Which of the following numbers is NOT a solution for the inequality
11 > 3 - 2x
1. 7
2. 6
3. 5
4. 4
3. . What is the simplest form of the inequality
5 - 2x ≥ 3x + 15
5x ≥ -10
1. 5x ≤ -10
2. x ≤ -2
3. x ≥ -2
4. x ≤ -10
4. . The inequality
x - 3 < 4x - 15
is equivalent to
1. x > 12/5
2. x > 4
3. x < 4
4. x < -4
5. . Which of the following is the solution set of the inequality
3x - 2 ≤ 16 + 6x
1. [-6, +∞)
2. (-∞, 6]
3. (-∞, -6)
4. (6, +∞)
6. . Which of the following is the solution set of the double inequality
-5 < 2x + 1 ≤ 11
1. x ϵ (-3, 5)
2. x ϵ (-3, 5]
3. x ϵ [-3, 5)
4. x ϵ (-5, 3]
7. . Which of the following pairs of inequalities is equivalent to the double inequality
2x - 1 < 3x - 3 ≤ 2x + 9
1. x > 4 and x ≤ 8
2. x < 2 and x ≥ -12
3. x > 2 and x ≤ 12
4. x < 2 and x ≥ 12
8. . The inequalities 3 - x ≥ 0 and 2x + 4 > 0 can be expressed as a double inequality as
1. 2 < x < 3
2. -2 < x ≤ 3
3. 2 < x ≤ 3
4. -3 ≤ x < 2
9. . Which of the following number pairs is a solution for the linear inequality in two variables
y < 4x - 1
1. (3, 12)
2. (-1, -5)
3. (2, 9)
4. (1, 2)
10. . Which of the following number pairs is NOT a solution for the linear equation in two variables
3x - y - 5 ≤ 0
1. (0, -4)
2. (2, 2)
3. (-1, -3)
4. (3, 3)
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## Help others Learning Math just like you
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# Mathematics
### L'Hospital Rule
Guillaume de l'Hôpital, famously known as L'Hospital, came up with a method to deal with fractions, when they take the indeterminate forms, approaching certain limits.
The indeterminate form can be as follows:
• 0/0 - as in the case of sin(x)/x, when x approaches 0
• ∞ / ∞ - as in the case of ex2 / x2, when x approaches ∞
L'Hospital Rule help us deal with situations of this kind. It is as follows:
#### L'Hospital Rule
If Limit[f(x)/g(x)] as x approaches a is 0/0 or ∞ / ∞, then Limit[f(x)/g(x)] as x approaches a is [f'(x)/g'(x)].
In all the following animations, [f(x)/g(x)] is drawn in red and [f'(x)/g'(x)] in purple.
Please note the convergence pf the two curves / lines to the same point, as the limit approaches.
E.g.1
Find Limit [(4x-3)/(5x-6)] as x approaches ∞.
If x = ∞, then [(4x-3)/(5x-6)] = ∞/∞ - indeterminate
Let's use L'Hospital rule for this:
f'(x)/g'(x) = 4/5 = 0.8, as x approaches ∞
E.g.2
Find Limit [(x-4)/ln(x-3)] as x approaches 4.
If x = 4;, then [(x-4)/ln(x-3)] = 0/0 - indeterminate
Let's use L'Hospital rule for this:
f'(x)/g'(x) = 1/(1/x-3) = 1, as x approaches 4.
E.g.3
Find Limit [ln(x)/√x] as x approaches ∞.
If x = ∞, then [ln(x)/√x] = ∞/∞- indeterminate
Let's use L'Hospital rule for this:
f'(x)/g'(x) = (1/x)/(1/2)x-1/2 = 2/√x = 0, as x approaches ∞.
The behaviour of the curve will be clearer when x is really large.
E.g.4
Find Limit [(x2 -x - 6)/(x2 -3x)] as x approaches 3;.
If x = 3, then [(x2 -x - 6)/(x2 -3x)] = 0/0 - indeterminate
Let's use L'Hospital rule for this:
f'(x)/g'(x) = (2x-1)/(2x-3) = 5/3, as x approaches 3.
E.g.5
Find Limit sin(x) /x as x approaches 0 and hence sketch y = sin(x)/x.
sin(x) /x when x approaches 0 = sin(0)/0 = 0/0 - indeterminate
Let's use L'Hospital rule for this:
f'(x) = cos(x); g'(x) = 1
So, f'(x)/g'(x) = cos(x)/1
When x approaches 0, f'(x)/g'(x) = 1/1 = 1
Therefore, sin(x)/x, when x approaches 0 = 1.
Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.
### Recommended - GCSE & iGCSE
This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.
### Recommended for A Level
This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented. |
This is an option extra page summarizing the method of Proof By Induction, applied to a particular Fibonacci number formula. We show how not to use the method also as well as the complete proof of the formula.
The icon means there is a Things to do investigation at the end of the section.
# The Powers of Phi Formula to prove
The formula we will prove is
Phin = Fib(n+1) + Fib(n) phi
We assume the standard definition of Fib(n):
Fib(n+2)=Fib(n+1)+Fib(n) for all n>0;
Fib(0)=0 and Fib(1)=1
and for Phi and phi we use the definitions
Phi = √5 + 1 phi = √5 – 1 2 2
## What is a Proof By Induction?
A proof by induction involves two steps:
1. Proving that IF the above formula is true for any particular value of n, let's say n=k, then it must automatically follow that it isrue for k+1 too.
Since (k+1) is another particular value, the same argument shows the formula is therefore true for k+2. "By induction" we can therefore reason that it will be true for all following values of n from k onwards.
Note that we assume the formula is true for n=k and, on that basis, show that it must inevitably be true for the next larger value k+1.
This assumption f(that the formula is true for n=k in particular) is called the Inductive Assumption. It is not the same as assuming the formula is true for all n, since we are only assuming it is true for one particular value of n, namely n=k.
The proof that "IF the formula is true for n=k THEN it must also be true for n=k+1 also" must be sufficiently general that it applies to all values of k, that is, it does not rely on specific values of k to work.
2. There must be a value of n for which the formula can be shown to be true.
This value will start off the "induction" process. So, for the formula to be true for all n we need the starting value to be n=1 or perhaps n=0.
## A Faulty Proof By Induction
Failure to observe these two conditions fully gives rise to faulty induction "proofs".
For instance, here is a "proof" that All cars are red. We show a faulty application of proof by induciton to the statement All sets of n cars are red cars. and show it is true for all values of n:
1. Suppose that we have a specific instance where it is true: n=k.
Our assumption is that ALL sets of k cars contain only red cars.
So let's take any other car and add it in to make a set of k+1 cars.
We need to show it is red too.
To show this new set contains only red cars means we have to show the new car is red too.
We do this by considering a subset of the new set formed by taking out any other car from the set, leaving a set of size n containing the original assumption that all sets of size k are of red cars then this set is too.
But that set contains the new car and so we have shown, on the basis if the assumption that the new car is red too.
2. My car is red so I can find a set of size 1 for which the statement is true.
So where has this "proof" gone wrong since clearly there are cars of other colours than red?
Well, although there is no fault in the reasoning of the inductive part of the argument (stage 1 above), the second part is not always true. Whilst my car is red, my neighbour's car is white, so it wouldn't work for him! We cannot verify the second condition, that the statement "All sets of 1 car are red cars" is always true.
So we have to be particularly careful as to what the statement is (about n) that we are trying to prove.
## Proving this Formula by Induction
First, assume it is true for n=k, that is that
Phik = Fib(k+1) + Fib(k) phi -- our starting assumption
and we want to show that
Phik+1 = Fib(k+2) + Fib(k+1) phi
Since the left hand sides differ by a factor of Phi, let's start by multiplying our assumed result by Phi on both sides:
Phi × Phik = Phi × ( Fib(k+1) + Fib(k) phi ) Phik+1 = Fib(k+1) Phi + Fib(k) Phi phi Phik+1 = Fib(k+1) Phi + Fib(k) using Phi phi = 1 Phik+1 = Fib(k+1) (1 + phi) + Fib(k) since Phi = 1 + phi Phik+1 = Fib(k+1) + Fib(k) + Fib(k+1) phi by rearranging Phik+1 = Fib(k+2) +Fib(k+1) phi by the Fibonacci Rule
Notice that the above argument would work no matter what the value of k is.
Secondly, we need to find a starting value for n for which the formula is true.
When n=0, we have
Phi0 = Fib(0+1) + Fib(0) phi
which simplifies to
Ph0 = 1 + 0 phi = 1
which we know is true.
So the two parts of our proof by induction are now complete. Things to do |
Home | | Maths 10th Std | Cartesian Product
# Cartesian Product
If A and B are two non-empty sets, then the set of all ordered pairs (a, b) such that a ∈ A, b ∈ B is called the Cartesian Product of A and B, and is denoted by A×B . Thus, A×B = {(a,b) |a ∈ A,b ∈ B}.
Cartesian Product
### Illustration 1
Let us consider the following two sets.
A is the set of 3 vegetables and B is the set of 4 fruits. That is,
A = {carrot, brinjal, ladies finger} and B = {apple, orange, grapes, strawberry} What are the possible ways of choosing a vegetable with a fruit? (Fig.1.2)
We can select them in 12 distinct pairs as given below.
(c, a), (c, o), (c, g), (c, s), (b, a), (b, o), (b, g), (b, s), (l, a), (l, o), (l,g), (l, s)
This collection represents the cartesian product of the set of vegetables and set of fruits.
### Definition
If A and B are two non-empty sets, then the set of all ordered pairs (a, b) such that a Ab is called the Cartesian Product of A and B, and is denoted by A×B . Thus, A×B = {(a,b) |a A,b B}.
### Note
· A × B is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of A and the second coordinate is an element of B.
· B × A is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of B and the second coordinate is an element of A.
· If a = b, then (a, b) = (b, a).
· The “cartesian product” is also referred as “cross product”.
### Illustration 2
Let A = {1, 2, 3} and B = {a, b}. Write A ×B and B ×A ?
A ×B = {1,2,3}×{a,b} = {(1, a ),(1, b ),(2, a ),(2, b ),(3, a ),(3, b )} (as shown in Fig.1.3)
B×A = {a,b} × {1,2,3} = {(a,1), (a,2), (a,3),(b,1), (b,2), (b,3)} (as shown in Fig.1.3)
Recall of standard infinite sets
Natural Numbers N = {1, 2, 3, 4…} ;
Whole Numbers W = {0,1,2,3, ...};
Integers Z ={..., –2,–1,0,1,2, ...} ;
Rational Numbers
Real Numbers R = Q Q , where Q’ is the set of all irrational numbers.
### Illustration 3
For example, let A be the set of numbers in the interval [3, 5] and B be the set of numbers in the interval [2,3]. Then the Cartesian product A×B corresponds to the rectangular region shown in the Fig. 1.4. It consists of all points (x, y) within the region
Progress check
1. For any two non-empty sets A and B, A×is called as ______.
2. If n(A× B) = 20 and n(A) = 5 then n(B) is ______.
3. If A = {-1,1} and B = {−1,1} then geometrically describe the set of points of A×B .
4. If A, B are the line segments given by the intervals (–4, 3) and (–2, 3) respectively, represent the cartesian product of A and B.
Note:
The set of all points in the cartesian plane can be viewed as the set of all ordered pairs (x, y) where x, y are real numbers. In fact, × is the set of all points which we call as the cartesian plane.
Example 1.1 If A = {1,3,5} and B = {2,3} then (i) find A ×B and B ×A.
(ii) Is A × B = B ×A? If not why? (iii) Show that n(A×B) = n(B×A) = n(An(B)
Solution Given that A = {1,3,5} and B = {2,3}
(i) A×B = {1,3,5} × {2,3} = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)} ...(1)
B× A = {2,3} × {1,3,5} = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)} ...(2)
(ii) From (1) and (2) we conclude that A × B B ×A as (1, 2) (2, 1) and (1, 3) (3, 1), etc.
(iii) n(A)=3; n (B) = 2.
From (1) and (2) we observe that, n (A×B) = n (B×A) = 6;
we see that, n (A) ×n (B) = 3 × 2 = 6 and n (B) × n (A) = 2×3 = 6
Hence, n (A×B) =n (B×A) = n(A) × n (B) = 6.
Thus, n (A×B) =n (B×A) = n(A) × n (B).
Example 1.2 If A×B = {(3,2), (3,4), (5,2), (5,4)} then find A and B.
Solution A×B ={(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A × B }. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A × B }. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.
Example 1.3 Let A = {x N | 1 < x <4} , B = {x W| 0 x < 2} and C = {x N | x<3} .
Then verify that
(i) A × (B U C) = (A × B) U (A × C)
(ii) A × (B Ո C) = (A × B) + (A × C)
Solution A = {x N| 1 < x < 4} = {2, 3} , B = {x W | 0 x < 2} = {0, 1},
C = {x N | x < 3}= {1,2}
(i) A × (B C) = (A × B ) , (A × C)
B = {0, 1} {1, 2} = {0, 1, 2}
A × (B C) = {2, 3} × {0, 1, 2} = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ...(1)
A × B = {2, 3} × {0, 1} = {(2,0),(2,1),(3,0),(3,1)}
A × C = {2, 3} × {1, 2} = {(2, 1), (2, 2), (3, 1), (3, 2)}
(A × B) (A ×C) = {(2, 0), (2, 1), (3, 0), (3, 1)} , {(2, 1), (2, 2), (3, 1), (3, 2)}
= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ...(2)
From (1) and (2), A × (B C ) = (A×B ) (A×C) is verified.
(ii) A × (B C) = (A ×B) (A ×C)
B Ո C = {0, 1}{1, 2} = {1}
A × (B C) = {2, 3} ×{1} = {(2,1),(3,1)} ... (3)
A × B= {2, 3} ×{0, 1} = {(2, 0),(2, 1),(3, 0),(3, 1)}
A × C = {2, 3} ×{1, 2} = {(2, 1),(2, 2),(3, 1),(3, 2)}
(A × B) (A ×C) = {(2, 0),(2, 1),(3, 0),(3, 1)}{(2, 1),(2, 2),(3, 1),(3, 2)}
= {(2, 1),(3, 1)} ... (4)
From (3) and (4), A × (B C ) = (A×B ) (A×C) is verified.
### Note
The above two verified properties are called distributive property of cartesian product over union and intersection respectively. In fact, for any three sets A, B, C we have
(i) A × (B C) = (A ×B) (A ×C) (ii) A × (B C ) = (A×B) (A×C).
## Cartesian Product of three Set
If A, B, C are three non-empty sets then the cartesian product of three sets is the set of all possible ordered triplets given by
A× B ×C= {(a,b,c) for all a A,b B,c C }
### Illustration for Geometrical understanding of cartesian product of two and three sets
Let A = {0,1}, B = {0,1}, = {0,1}
A×B = {0,1}×{0,1} = {(0, 0),(0,1),(1, 0),(1,1)}
Representing A×B in the xy - plane we get a picture shown in Fig. 1.5.
(A×BC= {(0, 0),(0,1),(1, 0),(1,1)} ×{0,1}
= {(0, 0, 0),(0, 0,1),(0,1, 0),(0,1,1),(1, 0, 0),(1, 0,1)(1,1, 0),(1,1,1)}
Representing A×B ×C in the xyz - plane we get a picture as shown in Fig. 1.6
Thus, A×B represent vertices of a square in two dimensions and A×B ×C represent vertices of a cube in three dimensions.
NOTES
In general, cartesian product of two non-empty sets provides a shape in two dimensions and cartesian product of three non-empty sets provide an object in three dimensions.
Tags : Definition, Illustration, Example, Solution , 10th Mathematics : UNIT 1 : Relation and Function
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
10th Mathematics : UNIT 1 : Relation and Function : Cartesian Product | Definition, Illustration, Example, Solution |
FutureStarr
A Percent Minus Percent Calculator
# Percent Minus Percent Calculator
via GIPHY
The Percent Minus Percent Calculator is a number-formatting app for calculating the difference between two numbers.
### Calculator
For this calculator, a "stackable additional discount" means getting a further percent off of a product after a discount is applied. Using the same example, assume that the 20% discount is a discount applied by the store to the product. If you have a coupon for another 15% off, the 15% off would then be applied to the discounted price of \$223.20. It is not a total of 35% off of the original price. It is less: Let’s enter a thought bubble and attempt to mentally process this with ideas and concepts we may already be familiar with. If we think about the meaning of the word percent, which we touched on earlier, we can use the fact that they are parts of a whole to help us understand what the calculators might be doing. For instance, 25% is 25 (the “part”) out of 100 (the “whole” in this case). We can also take advantage of the fact that we know 200 is 2 times 100. So, if 25% of 100 is 25, then 25% of 200, in theory should be 2 time that value or 2 x 25, which is 5.
Okay, let’s take this problem step-by-step. 50% of 100 is equivalent to half of 100, so 50. But how do we calculate 25% of 200? To calculate that, you can use our percentage calculator, or you can read through the section below containing formulas and explanations to gain a better grasp on these concepts and avoid technical help in the future. (Source: www.blitzresults.com)
### Formula
However, you can also make it very easy for yourself by breaking down the problem. For the question “50 is 25% of what value?”, we know that the basic value (G), is the value we are looking for. The percentage value (W) is 50, the percentage (P) is 25%, and we can recall that our starting formula is 100 x W = G x P. Since we are looking for G, we can divide by P, so G is isolated on one side of the equation, resulting in the following rearranged formula: Base value (G) = Percentage value (W)/ Percentage (P) × 100 %.
Putting our known values into this gives us: Base value (G) = 50/ 25 x 100. Converting the fraction into its integer equivalent results in our formula looking as follows: Base value (G) = 2 x 100. Now, that it is in its simplest form, directly doing the math leaves us with a base value of 200. So, we can now confidently say that 50 (the percentage value) is 25% (percentage) of 200 (base value)! (Source: www.blitzresults.com)
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# Lesson 6
Comparemos números decimales en la recta numérica
## Warm-up: Observa y pregúntate: Rectas anidadas (10 minutes)
### Narrative
The purpose of this Notice and Wonder is for students to look at different number lines that all start at 0 but show different decimals. The number lines are nested, that is each successive one is contained in the previous one. The key points for students to notice are that the number lines all have decimals on them and that the size of those decimals is getting smaller. In the lesson, they will plot decimals on number lines like these.
### Launch
• Groups of 2
• Display the image.
• “¿Qué observan? ¿Qué se preguntan?” // “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• “Discutan con su compañero lo que pensaron” // “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
¿Qué observas? ¿Qué te preguntas?
### Activity Synthesis
• “¿Qué representan las marcas de la recta numérica de arriba?” // “What are the tick marks on the top number line?” (one tenth, two tenths, three tenths, and so on)
• “Hoy vamos a usar rectas numéricas como estas para ubicar varios números decimales” // “Today we are going to use number lines like these to locate different decimals.”
## Activity 1: Ubiquemos 1 milésima (15 minutes)
### Narrative
The purpose of this activity is for students to plot the same number on different number lines and recognize that the location of the number on the number line can only be accurately determined when it lies on a tick mark. As they work on locating the number, students reinforce their understanding of place value as the tick marks on the number lines are tenths, hundredths, and thousandths (MP7). Students may struggle to locate 0.001 on the first two number lines. The important take-away is that when a decimal does not lie on a tick mark estimation is needed to locate the number and it can be difficult or impossible to locate it precisely (MP6).
MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations. Monitor and clarify any questions about the context.
Engagement: Provide Access by Recruiting Interest. Provide choice. Invite students to decide which number line to label first.
Supports accessibility for: Attention, Organization
• Groups of 2
### Activity
• 5 minutes: independent work time
• 5 minutes: partner discussion
• Monitor for students who reason about place value to:
• label each tick mark
• locate 0.001 on each number line
### Student Facing
• Escribe el número que corresponde debajo de cada marca.
• Ubica y marca el número 0.001.
### Activity Synthesis
• Ask previously identified students to share.
• “¿Cómo decidieron qué escribir debajo de las marcas de la primera recta numérica?” // “How did you decide what to label the tick marks on the first number line?” (There are ten tick marks and there are ten tenths in one whole so I counted by tenths.)
• “¿En qué lugar de la primera recta numérica está ubicado diez décimas?” // “Where is ten tenths on the first number line?” (The number 1 is ten tenths.)
• “¿Cómo decidieron qué escribir debajo de las marcas de la segunda y de la tercera recta numérica?” // “How did you decide what to label the tick marks on the second and third number lines?” (There are 10 hundredths in 1 tenth. There are ten thousandths in 1 hundredth.)
• “¿En qué fue diferente ubicar 0.001 en cada una de las rectas numéricas?” // “How was locating 0.001 different for each of the number lines?” (It was so close to 0 on the first number line that I could not plot it. I could estimate its location on the second number line and then it was the first tick mark on the last number line.)
## Activity 2: Marquemos rectas y comparemos decimales (10 minutes)
### Narrative
The purpose of this activity is for students to label number lines where the end tick marks are tenths or hundredths written as decimals. They will use their understanding of place value when they label the tick marks (MP7). Students also choose one of the number lines to compare two numbers, preparing them for the comparison work in the next activity and in future lessons.
• Groups of 2
### Activity
• 4–5 minutes: independent work
• 2–3 minutes minutes: partner discussion
• Monitor for students who accurately label the number lines with hundredths and thousandths.
### Student Facing
1. En cada recta numérica, escribe el número que corresponde debajo de cada marca.
2. ¿Cuál de las rectas numéricas usarías para comparar 0.534 y 0.537? Explica o muestra tu razonamiento.
### Activity Synthesis
• Ask previously identified students to share their solutions and reasoning.
• “¿Qué recta numérica escogerían para comparar 0.534 y 0.537?” // “Which number line would you choose to compare 0.534 and 0.537?” (I liked the middle one because those numbers were labeled tick marks and I could see which one was further to the right.)
• “¿Cuál número es mayor: 0.534 o 0.537? ¿Por qué?” // “Which number is greater, 0.534 or 0.537? Why?” (0.537 because it is further to the right on the number line.)
• Display inequality: $$0.534 < 0.537$$
• “También podemos usar símbolos para decir que 0.534 es menor que 0.537” // “We can also say that 0.534 is less than 0.537 with symbols.”
## Activity 3: Ubiquemos y comparemos usando símbolos (10 minutes)
### Narrative
The purpose of this activity is for students to compare decimal numbers using the number line for support. All of the numbers lie on tick marks and students will use their understanding of place value to accurately place the decimals. They will also use their understanding that one number is greater than another when it lies farther to the right on the number line.
• Groups of 2
### Activity
• 5 minutes: independent work
• 2 minutes: partner discussion
### Student Facing
1. Usa el símbolo < o > para comparar los decimales 0.2 y 0.02. Usa la recta numérica para explicar o mostrar cómo pensaste.
2. Usa el símbolo < o > para comparar los decimales 0.3 y 0.14. Usa la recta numérica para explicar o mostrar cómo pensaste.
3. Usa el símbolo < o > para comparar los decimales 0.23 y 0.216. Usa la recta numérica para explicar o mostrar cómo pensaste.
### Activity Synthesis
• “¿Cómo nos ayudan las rectas numéricas a comparar números decimales?” // “How do number lines help compare decimals?” (We can put the decimals exactly on tick marks and then see which number is farther to the right.)
## Lesson Synthesis
### Lesson Synthesis
“Hoy razonamos sobre el valor posicional para ubicar y comparar números decimales hasta la posición de las milésimas usando rectas numéricas” // “Today we used place value reasoning to locate and compare decimals to the thousandths place using number lines.”
Display the last number line from the last activity.
“¿Qué número está ubicado en la primera marca que está después de 0.23?” // ”What number is located at the first tick mark after 0.23?” (0.231)
“¿Qué número está ubicado en la última marca que está antes de 0.22?” // “What number is located at the last tick mark before 0.22?” (0.219)
Label the numbers as students respond.
“¿Cuál número es mayor? ¿Cómo lo saben?” // “Which number is greater? How do you know?” (0.231 because it is farther to the right on the number line) |
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