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# MULTIPLYING POLYNOMIALS PRACTICE AND PROBLEM SOLVING A/B LESSON 14-4
If we have a polynomial consisting of only two terms we could instead call it a binomial and a polynomial consisting of three terms can also be called a trinomial. The degree of the polynomial is the greatest degree of its terms. A monomial is a number, a variable or a product of a number and a variable where all exponents are whole numbers. When multiplying two binomial you can use the word FOIL to remember how to multiply the binomials. Algebra 1 Radical expressions Overview The graph of a radical function Simplify radical expressions Radical equations The Pythagorean Theorem The distance and midpoint formulas. Constants have the monomial degree of 0. We just add the like terms to combine the two polynomials into one.
Algebra 1 Rational expressions Overview Simplify rational expression Multiply rational expressions Division of polynomials Add and subtract rational expressions Solving rational expressions. Constants have the monomial degree of 0. Make the two polynomials into one big polynomial by taking away the parenthesis. Algebra 1 Formulating linear equations Overview Writing linear equations using the slope-intercept form Writing linear equations using the point-slope form and the standard form Parallel and perpendicular lines Scatter plots and linear models. Search Pre-Algebra All courses. The degree of the monomial is the sum of the exponents of all included variables.
# Monomials and polynomials (Algebra 1, Factoring and polynomials) – Mathplanet
Make the two polynomials into one big polynomial by taking away the parenthesis. We can add polynomials. A polynomial as oppose to the monomial is a sum of monomials where each monomial is called a term. Just subtract the like terms Or in other words add its multiplynig.
MALACANANG SIGNED THE LAW NO HOMEWORK DURING WEEKENDS
## Monomials and polynomials
If we have a polynomial consisting of only two terms we could instead call it a binomial and a polynomial consisting of three terms can also be called a trinomial. When you multiply polynomials where both polynomials have more than one term you just multiply each of terms in the first polynomial with all problek the terms in the second polynomial.
A polynomial is usually written with the term with the highest exponent of the variable first and then decreasing from left to right. Don’t forget to reverse the signs within the second parenthesis since your multiplying all terms with The degree of the monomial is the sum of the exponents of all included variables.
Algebra 1 Systems of linear equations and inequalities Overview Graphing linear systems The substitution method for solving linear systems The solvinng method for solving linear systems Systems of linear inequalities. Algebra 1 Radical expressions Overview The graph of a radical function Simplify radical expressions Radical equations The Pythagorean Theorem The distance and midpoint formulas.
Algebra 1 Exponents and exponential functions Overview Properties of exponents Scientific notation Exponential growth functions. The same goes for subtracting two polynomials.
We just add the like terms to combine multiplykng two polynomials into one. Algebra 1 Linear inequalitites Overview Solving linear inequalities Solving compound inequalities Solving absolute value equations and inequalities Linear inequalities in two variables. The first term of a polynomial is called the leading coefficient.
# Multiply binomials by polynomials (practice) | Khan Academy
Constants have the monomial degree of 0. Polynomial just means that we’ve got a sum of many monomials. Algebra 1 Rational expressions Overview Simplify rational expression Multiply rational expressions Division of polynomials Add and subtract rational expressions Solving rational expressions.
FELPHAM HOMEWORK WEBSITE
The degree of the polynomial is the greatest degree of its terms. Search Pre-Algebra All courses. Algebra 1 Discovering expressions, equations and functions Overview Expressions and variables Operations in the right order Composing expressions Composing equations and inequalities Representing functions as rules and graphs.
Algebra 1 How to solve linear equations Overview Properties of equalities Fundamentals in solving equations in one or more steps Ratios and proportions and how oplynomials solve them Similar figures Calculating with percents. Algebra 1 Exploring real numbers Overview Integers and rational numbers Calculating with real numbers The Distributive property Square roots.
When multiplying two binomial you can use the word FOIL to remember how to multiply the binomials. Algebra 1 Visualizing linear functions Overview The coordinate plane Linear equations in the coordinate plane The slope of a linear function The slope-intercept form of a linear equation.
A monomial is a number, a pgoblem or a product of a number and a variable where all exponents are whole numbers.
Algebra 1 Formulating linear equations Overview Writing linear equations using the slope-intercept form Writing linear equations using the point-slope form and the standard form Parallel and perpendicular lines Scatter plots and linear models.
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## Precalculus (6th Edition) Blitzer
$\left\{ \left( 2,\ -3,\ 7 \right) \right\}$
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 1 & 2 & 1 \\ 3 & 4 & 2 \\ \end{matrix} \right|\begin{matrix} 0 \\ 3 \\ 8 \\ \end{matrix} \right]$ Now we will solve this matrix as below to get: $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ 3 \\ 1 \\ \end{matrix} \right]$ $By,\ 3{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 0 & 5 & 3 \\ 0 & 2 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ 6 \\ 1 \\ \end{matrix} \right]$ $By,\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 0 & 5 & 3 \\ 0 & 1 & 0 \\ \end{matrix} \right|\begin{matrix} 0 \\ 6 \\ -3 \\ \end{matrix} \right]$ $By,\ 3{{R}_{3}}-{{R}_{2}}\to {{R}_{3}}$ Convert the matrix into equation form: \begin{align} 2x-y-z=0 & \\ 5y+3z=6 & \\ y=-3 & \\ \end{align} Substitute the value of $y$ in $5y+3z=6$ to obtain the value of $z$: \begin{align} & 5\times \left( -3 \right)+3z=6 \\ & 3z=6+15 \\ & 3z=21 \\ & z=7 \end{align} Substitute the value of $y,z$ in $2x-y-z=0$ to obtain the value of $x$: \begin{align} & 2x-\left( -3 \right)-7=0 \\ & 2x=-3+7 \\ & 2x=4 \\ & x=2 \end{align} Hence, the solution is $\left\{ \left( 2,\ -3,\ 7 \right) \right\}$
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So how do you calculate the total resistance for resistors in parallel on your electronic circuit? Put on your thinking cap and follow along. Here are the rules:
• First, the simplest case: Resistors of equal value in parallel. In this case, you can calculate the total resistance by dividing the value of one of the individual resistors by the number of resistors in parallel. For example, the total resistance of two, 1 kΩ resistors in parallel is 500 Ω and the total resistance of four, 1 kΩ resistors is 250 Ω.
Unfortunately, this is the only case that's simple. The math when resistors in parallel have unequal values is more complicated.
• If only two resistors of different values are involved, the calculation isn't too bad:
In this formula, R1 and R2 are the values of the two resistors.
Here's an example, based on a 2 kΩ and a 3 kΩ resistor in parallel:
• For three or more resistors in parallel, the calculation begins to look like rocket science:
The dots at the end of the expression indicate that you keep adding up the reciprocals of the resistances for as many resistors as you have.
In case you're crazy enough to actually want to do this kind of math, here's an example for three resistors whose values are 2 kΩ, 4 kΩ, and 8 kΩ:
As you can see, the final result is 1,142.857 Ω. That's more precision than you could possibly want, so you can probably safely round it off to 1,142 Ω, or maybe even 1,150 Ω.
The parallel resistance formula makes more sense if you think about it in terms of the opposite of resistance, which is called conductance. Resistance is the ability of a conductor to block current; conductance is the ability of a conductor to pass current. Conductance has an inverse relationship with resistance: When you increase resistance, you decrease conductance, and vice versa.
Because the pioneers of electrical theory had a nerdy sense of humor, they named the unit of measure for conductance the mho, which is ohm spelled backward. The mho is the reciprocal (also known as inverse) of the ohm.
To calculate the conductance of any circuit or component (including a single resistor), you just divide the resistance of the circuit or component (in ohms) into 1. Thus, a 100 Ω resistor has 1/100 mho of conductance.
When circuits are connected in parallel, current has multiple pathways it can travel through. It turns out that the total conductance of a parallel network of resistors is simple to calculate: You just add up the conductances of each individual resistor.
For example, suppose you have three resistors in parallel whose conductances are 0.1 mho, 0.02 mho, and 0.005 mho. (These are the conductances of 10 Ω, 50 Ω, and 200 Ω resistors, respectively.) The total conductance of this circuit is 0.125 mho (0.1 + 0.02 + 0.005 = 0.125).
One of the basic rules of doing math with reciprocals is that if one number is the reciprocal of a second number, the second number is also the reciprocal of the first number. Thus, since mhos are the reciprocal of ohms, ohms are the reciprocal of mhos.
To convert conductance to resistance, you just divide the conductance into 1. Thus, the resistance equivalent to 0.125 mho is 8 Ω (1 ÷ 0.125 = 8).
It may help you remember how the parallel resistance formula works when you realize that what you're really doing is converting each individual resistance to conductance, adding them up, and then converting the result back to resistance. In other words, convert the ohms to mhos, add them up, and then convert them back to ohms. That's how — and why — the resistance formula actually works.
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Besides the unit circle method, there is another method for defining (and working with) the basic trigonometric functions. For this method to work, however, the triangle in question must be a right triangle. There is a mneumonic (a memory aid) to help you remember the trig functions: SOHCAHTOA (pronounced "so - cuh - toe - uh"). It stands for Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent. Remember that Opposite is the length of the side opposite the angle in question, Adjacent is the length of the side adjacent to the angle in question, and Hypotenuse is the length of the hypotenuse of the right triangle.
Here's an example of how to apply SOHCAHTOA. Refer to the figure shown below.
We use the Pythagorean Theorem to find the length of the missing side. 122 + x2 = 132 simplifies a little to 144 + x2 = 169 and then to x2 = 25. This says that x = 5 or x = -5. We choose x = 5 (because of the orientation of the triangle, as will be explained later) and add this information to our drawing:
Looking at the angle labeled θ, the side (leg) opposite θ is 5 long while the hypotenuse is 13 long. We remember SOHCAHTOA and SOH stands for Sine is Opposite over Hypotenuse. Therefore, sin(θ) = 5/13. The side (leg) adjacent to θ is 12 long and the hypotenuse is 13 so, from SOHCAHTOA, cos(θ) = 12/13. Finally, we get that tangent is opposite over adjacent so tan(θ) = 5/12. See just how easy SOHCAHTOA is?!
Optional Video Clips
I have found some video clips online (from University of Idaho) that help further explain and give more examples for the trigonometry concepts we've learned in this lesson. While they are optional, many of them are recommended to help you solidify your understanding of the concepts. Note that you will need to have a pretty good internet connection for these to work properly.
To view these clips, you will need to have the Real Player program installed on your computer. If you don't already have Real Player, you can download it to your PC (for free) by visiting this web page: . Just click on the button that says, "Start RealPlayer Download" and follow the instructions. (It will probably offer you the chance to purchase additional programs like Rhapsody, etc., but I don't recommend it because you will have to pay for those out of your own pocket).
Once Real Player is installed on your computer you may view the following video clips. After the name of each video clip is the length of the video in minutes and seconds. Example (5:51) is a video clip that is 5 minutes and 51 seconds long.
Video Clips related to SOHCAHTOA
An Introduction to the First Three Trigonometric Ratios (sin, cos, tan) (8:14) *Strongly Recommended*
Three More Trigonometric Ratios (csc, sec, cot) (5:07) *Strongly Recommended*
Reciprocal and Quotient Identities (6:10) *Strongly Recommended*
Pythagorean Identities (5:20) *Recommended*
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# 6.2: Counting Nails by the Pound
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## Counting by Weighing and Avogadro's number
The size of molecule is so small that it is physically difficult, if not impossible, to directly count out molecules (Figure $$\PageIndex{1}$$). However, we can count them indirectly by using a common trick of "counting by weighing".
Consider the example of counting nails in a big box at a hardware store. You need to estimate the number of nails in a box. The weight of an empty box is $$213 \,g$$ and the weight of the box plus a bunch of big nails is $$1340\, g$$. Assume that we know that the weight of one big nail is 0.450 g. Hopefully it's not necessary to tear open the package and count the nails. We agree that
$\text{mass of big nails} = 1340\, g - 213\, g = 1227 \,g \nonumber$
Therefore
$\text{Number of big nails in box} = \dfrac{1227\, g}{0.450\, g/ \text{big nail}} = 2,726.6\, \text{big nails} = 2,730 \,\text{big nails}. \label{eq2}$
You have just counted the number of big nails in the box by weighing them (rather than by counting them individually).
Now consider if the box of nails weighed the same, but the box were filled with small nails with an individual mass of $$0.23\, g/\text{small nail}$$ instead? You would do the same math, but use a different denominator in Equation \ref{eq2}:
$\text{Number of small nails in box} = \dfrac{1227\, g}{0.230\, g/\text{small nail}} = 5,334.7\, \text{small nails} = 5,335 \, \text{small nails}. \label{eq3}$
The individual mass is the conversion factor used in the calculation and changes, based on the nature of the nail (big or small). Let's ask a different question: how many dozens of nails are there in the same box of small nails described above?
If we know the information from Equation \ref{eq3}, we can just use the conversion of how many nails are in a dozen:
$\dfrac{5,335 \,\text{small nails} }{12\, \text{small nails/dozen}} = 444.6 \,\text{dozen small nails} \label{eq4}$
If we want to get this value from weighing, we use the "dozen mass" instead of individual mass:
$12 \times 0.23 g = 2.76\, g/\text{dozen small nails}. \label{eq5}$
So following Equation \ref{eq3}, we get:
$\text{Number of dozens of small nails} = \dfrac{1227\, g}{2.76\, g/\text{dozen small nails}} = 444.6 \,\text{dozen small nails} \label{eq6}$
and this is the same result as Equation \ref{eq4}. These calculations demonstrate the difference between individual mass (i.e., per individual) and collective mass (e.g., per dozen or per gross). The collective mass of most importance to chemistry is molar mass (i.e., mass per mole or mass per $$6.022 \times 10^{23}$$).
Avogadro's number is an accident of nature. It is the number of particles that delivers a mole of a substance. Avogadro's number = $$6.022 \times 10^{23}$$. The reason why the value is an accident of nature is that the mole is tied to the gram mass unit. The gram is a convenient mass unit because it matches human sizes. If we were a thousand times greater in size (like Paul Bunyan) we would find it handy to use kilogram amounts. This means the kilogram mole would be convenient. The number of particles handled in a kilogram mole is 1000 times greater. The kilo Avogadro number for the count of particles in a kilomole is $$6.022 \times 10^{26}$$.
If humans were tiny creatures (like Lilliputians) only 1/1000 our present size, milligrams would be more convenient. This means the milligram mole would be more useful. The number of particles handled in a milligram mole (millimole) would be 1/1000 times smaller. The milli Avogadro number for the count of particles in a millimole is $$6.022 \times 10^{20}$$.
What do you think would happen to Avogadro's number if the American system was used and amounts were measured in pound moles? Remember 1 pound = 454 grams. Avogadro's number would be larger by a factor of 454. A pound mole of hydrogen would weigh 1 pound, which is 454 grams. A gram mole of hydrogen weighs 1 gram and contains $$6.022 \times 10^{23}$$ H atoms.
## Molar Mass for Elements
You are able to read the periodic table and determine the average atomic mass for an element like carbon. The average mass is 12.01 amu. This mass is a ridiculously tiny number of grams. It is too small to handle normally. The molar mass of carbon is defined as the mass in grams that is numerically equal to the average atomic weight. This means
$1 g/ mole carbon = 12.01 \,g \,carbon \nonumber$
this is commonly written
$1\, mol\, carbon = 12.01\, grams\, carbon. \nonumber$
This is the mass of carbon that contains $$6.022 \times 10^{23}$$ carbon atoms.
• Avogadro's number is $$6.022 \times 10^{23}$$ particles.
This same process gives us the molar mass of any element. For example:
• $$1\, mol\, neon = 20.18\, g\, neon\, Ne$$
• $$1 \,mol\, sodium = 22.99\, g \,sodium\, Na$$
## Molar Mass for Compounds
##### Example $$\PageIndex{1}$$: Molar Mass of Water
The formulas for compounds are familiar to you. You know the formula for water is $$\ce{H2O}$$. It should be reasonable that the weight of a formula unit can be calculated by adding up the weights for the atoms in the formula.
###### Solution
The formula weight for water
weight from hydrogen + weight from oxygen
The formula weight for water
2 H atoms x 1.008 amu + 1 O atom x 16.00 amu = 18.016 amu
The molar mass for water
18.016 grams water or 18 grams to the nearest gram
##### Example $$\PageIndex{2}$$: Molar Mass of Methane
The formula for methane, the major component in natural gas, is $$\ce{CH4}$$.
###### Solution
The formula weight for methane
weight from hydrogen + weight from carbon
The formula weight for methane
4 H atoms x 1.008 amu + 1 C atom x 12.01 amu = 16.04 amu
The molar mass for methane
16.04 grams per mole of methane
##### Example $$\PageIndex{3}$$: Molar Mass of Ethyl Chloride
What is its molar mass for ethyl chloride $$\ce{CH3CH2Cl}$$?
###### Solution
The formula weight
weight from hydrogen + weight from carbon + weight from chlorine
The formula weight
5 H atoms x 1.008 amu + 2 C atom x 12.01 amu + 35.5 amu = 64.5 amu
The molar mass for ethyl chloride
64.5 grams per mole of ethyl chloride
##### Example $$\PageIndex{4}$$: Molar Mass of Sulfur Dioxide
What is the molar mass for sulfur dioxide, $$\ce{SO2 (g)}$$, a gas used in bleaching and disinfection processes?
###### Solution
Look up the atomic weight for each of the elements in the formula.
• 1 sulfur atom = 32.07 amu
• 1 oxygen atom = 16.00 amu
Count the atoms of each element in the formula unit.
• one sulfur atom
• two oxygen atoms
The formula weight
weight from sulfur + weight from oxygen
The formula weight
1 sulfur atom x (32. 07 amu ) + 2 oxygen atoms x (16.00 amu)
The formula weight
$$\ce{SO2}$$ = 32. 07 amu + 32.00 amu = 64.07 amu = 64 amu $$\ce{SO2}$$
The molar mass for $$\ce{SO2}$$ is
64.07 grams of $$\ce{SO2}$$; 1 mol $$\ce{SO2}$$ = 64 grams per mole of $$\ce{SO2}$$
##### Exercise $$\PageIndex{1}$$
What is the formula weight and molar mass for alum, $$\ce{KAl(SO4)2 • 12 H2O}$$ ?
1. Check the periodic table for the atomic masses for each atom in the formula.
2. Count the number of each type of atom in the formula.
3. Multiply the number of atoms by the atomic mass for each element.
4. Add up the masses for all of the elements.
Table $$\PageIndex{1}$$: Masses of each element in alum, $$\ce{KAl(SO4)2 • 12 H2O}$$
element average atomic mass number of atoms in formula rounded to nearest one unit for simplicity
potassium k 39.1 amu 1 39. amu
aluminum 26.98 amu 1 27. amu
sulfur 32.07 amu 2 64. amu
oxygen 16.00 amu 8 + 12 = 20 320. amu
hydrogen 1.008 amu 2 x 12 = 24 24. amu
Molar mass is 474 grams (add up the amu of each element to find the total of 474 amu). This is a mass in grams that is numerically (474) the same as the formula weight.
1 mole alum $$\ce{KAl(SO4)2• 12 H2O}$$ = 474 grams alum $$\ce{KAl(SO4)2• 12 H2O}$$
6.2: Counting Nails by the Pound is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1
• Real Numbers Class 10 Ex 1.1
• Real Numbers Class 10 Ex 1.2
• Real Numbers Class 10 Ex 1.3
• Real Numbers Class 10 Ex 1.4
## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1
Ex1.1 Class 10 Maths Question 1.
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) By Euclid’s Division Algorithm, we have
225 = 135 x 1 + 90 135
= 90 x 1 + 45 90
= 45 x 2 + 0
∴ HCF (135, 225) = 45.
(ii) By Euclid’s Division Algorithm, we have
38220 = 196 x 195 + 0
196 = 196 x 1 + 0
∴ HCF (38220, 196) = 196.
(iii) By Euclid’s Division Algorithm, we have
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
∴ HCF (867, 255) = 51.
Ex1.1 Class 10 Maths Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be a positive odd integer. Also, let q be the quotient and r the remainder after dividing a by 6.
Then, a = 6q + r, where 0 ≤ r < 6.
Putting r = 0, 1, 2, 3, 4, and 5, we get:
a = 6q, a = 6q + 1, a = 6q + 2, a = 6q + 3, a = 6q + 4, a = 6q + 5
But a = 6q, a = 6q + 2 and a = 6q + 4 are even.
Hence, when a is odd, it is of the form 6q + 1, 6q + 3, and 6q + 5 for some integer q.
Hence proved.
Ex1.1 Class 10 Maths Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
Let n be the number of columns such that the value of n be maximum and it must divide both the numbers 616 and 32.
Then, n = HCF (616, 32)
By Euclid’s Division Algorithm, we have:
616 = 32 x 19 + 8 32 = 8 x 4 + 0
∴ HCF (616, 32) = 8
i. e., n = 8
Hence, the maximum number of columns is 8.
Ex1.1 Class 10 Maths Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Lemma,
we have:
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q
⇒ a2 = 9q2
= 3 x 3q2
= 3m (Assuming m = q2)
Then, a = 3q + 1
⇒ a2 = (3q + l)2 = 9q2 + 6q + 1
= 3(3q 2 + 2q) + 1
= 3m + 1 (Assuming m = 3q2 + 2q)
Next, a = 3q + 2
⇒ a2 = (3q + 2)2 =9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1
= 3m + 1. (Assuming m = 3q2 + 4q+l)
Therefore, the square of any positive integer (say, a2) is always of the form 3m or 3m + 1.
Hence, proved.
Ex1.1 Class 10 Maths Question 5.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Algorithm, we have,
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q, a = 3q + 1 and a = 3q + 2
If a = 3q, then a3 = 27q3 = 9(3q3) = 9m. (Assuming m = 3q3.)
If a = 3q + 1, then
a3 = (3q + l)3 = 27q3 + 9q(3q + 1) + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1, (Assuming m = 3q3 + 3q2 + q)
If a = 3q + 2, then a3 = (3q + 2)3
= 27q3 + 18q(3q + 2) + (2)3
= 9(3q3 + 6q2 + 4q) + 8
= 9m + 8, (Assuming m – 3q3 + 6q2 + 4q)
Hence, a3 is of the form 9m, 9m + 1 or 9m + 8.
We hope the NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1, drop a comment below and we will get back to you at the earliest.
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# Speed, Time and Distance Exercise Notes 2021 Download Study Materials BOOK PDF
Every year, they release notifications for several vacant posts, so required people interested in the job can apply for the job. To ace the exam, every candidate needs to prepare for the exam as per the syllabus and information provided in the study materials exclusively for this purpose. To get the study materials that are required, you must download the files provided below. You can click on the links given below to download the files and then study them. We hope you find our resources helpful. Every student who has studied using our study materials finds the exam easy to ace since question banks and solution keys are attached along. Once you’ve mastered the content provided in the study materials we’ve provided for you, you have nothing to worry about when it comes to acing the exam. Use our resources to study well for your exam and get your dream job.
Example 1:
The driver of a maruti car driving at the speed of 68 km/h locates a bus 40 metres ahead of him. After 10 seconds, the bus in 60 metres behind. Find the speed of the bus.
Solution:
Let speed of Bus=sb km/h
Now, in 10 sec, car covers the relative distance
=(60+40)m=100m
ஃ Relative speed of car= 100/10 = 10 m/s
=10 x18/5= 36 km/h
ஃ68-sB=36
=>sB=32 km/h
Example 2:
If a person goes around an equilateral triange shaped field at speed of 10, 20 and 40 kmph on the first, second point, then find his average speed during the journey.
Solution:
Let the measure of each side of triangle is D km. The person travelled the distance from A to B with 10 kmph, B to C with 20 kmph and C to A with 40 kmph.
If TAB = Time taken by the person to travel from A to B,
TBC = Time taken by the person to travel from B to C and
TCA = Time taken by the person to travel from C to A.
Then total time = TAB + TBC + TCA
= D/10 + D/20 + D/40 = D((8+4+2)/80) = 7D/40
Total distance travelled = D + D + D = 3D
Hence, average speed =3D/(7D/40) = 120/7 = 171/7 kmph.
Example 3:
Two guns were fired from the same place at an interval of 15 min, but a person in a bus approaching the place hears the second report 14 min and 30sec after the first. Find the speed of the bus, supposing that sound travels 330 m per sec.
Solution:
Distance travelled by the bus in 14 min 30 sec could be travelled by sound in (15 min – 14 min 30 sec) = 30 sec./ Bus travels = 330 × 30 in 141/2 min./ Speed of the bus per hour
= (330×30×2×60)/(29×1000) = (99×12)/29 = 1188/29 = 4028/29km/hr
Example 4:
A hare sees a dog 100m away from her and scuds off in the opposite direction at a speed of 12 km/h. A minute later the dog perceives her and gives chase at a speed of 16 km/h. How soon will the dog overtake the hare and at what distance from the spot where the hare took flight?
Solution:
Suppose the hare at H sees the dog at D.
D H K/ DH = 100m
Let k be the position of the hare where the dog sees her.
HK = the distance gone by the hare in 1 min = (12×1000)/60×1m = 200m
DK = 100+200 = 300m
The hare thus has a start of 300m.
Now the dog gains (16-12) or 4km/k.
The dog will gain 300m in (60×300)/(4×1000) min or 41/2min.
= (12×1000)/60× 41/2 = 900m
Distance of the place where the hare is caught from the spot H where the hare took flight = 200+900 = 1100m
If two persons(or vehicles or trains) start at the time in opposite directions from two points A and B, and after crossing each other they take x and y hours respectively to complete the journey, then
(Speed of first)/(Speed of second) = √(y/x)
Example 5:
A train starts from A to B and another from B to A at the same time. After crossing each other they complete their journey in 31/2 and 24/7 hours respectively. If the speed of the first is 60 km/h, then find the speed of the second train.
Solution:
(1st train^’ s speed )/(2nd train^’ s speed) = √(y/x) = √((2 4/7)/(3 1/2))
= √(18/7×2/7) = 6/7
60/(2nd train^’ s speed) = 6/7
2nd train’s speed = 70 km/h.
If new speed is a/b of usual speed, then Usual time = (Change in time)/((b/a-1) )
Example 6:
A boy walking at 3/5 of his usual speed, reaches his school 14 min late. Find his usual time to reach the school.
Solution:
Usual time = 14/(5/3-1) = (14×3)/2 = 21 min
Example 7:
A train after travelling 50km, meets with an accident and then proceeds at 4/5 of its former rate and arrives at the terminal 45 minutes late. Had the accident happened 20’km further on, it would have arrived 12 minutes sooner. Find the speed of the train and the distance.
Solution:
Let A be the starting place. B the terminal, C and D the places where the accidents to be placed. A C D B
By travelling at 4/5 of its original rate the train would take 5/4 of its usual time, i.e., 1/4 of its original time more.
1/4 of the usual time taken to travel the distance
CB = 45 min. ….(i)
and 1/4 of the usual time taken to travel the distance
DB = (45-12) min …..(ii)
subtracting (ii) from (i),
1/4 of the usual time taken to travel the distance CD = 12 min.
By Usual time taken on travel 20km = 48 min
By Speed of the train per hour = 20/48×60 = or 25 km/h.
From (i), we have
Time taken to travel CB = 45×4 min = 3 hrs.
∴ The distance CB = 25 × 3 or 75 km
The distance CB = the distance (AC + CB) = 50 + 75 or 25 km.
A man covers a certain distance D. If he moves S1 speed faster, he would have taken t time less and if he moves S2 speed slower, he would have taken t time more. The original speed is given by 2×(S_1×S_2 ))/(S_2- S_1 )
Example 8:
A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 20 min less. If he had moved 2 km/h slower, he would have taken 20 min more. Find the original speed.
Solution:
Speed = (2×(3×2))/(3-2) = 12 km/hr.
If a person with two different speeds U & V cover the same distance, then required distance
= (U×V)/(U-V)×Differnce between arrival time
Also, required distance = Total time taken×(U×V)/(U+V)
Example 9:
A boy walking at a speed of 10km/h reaches is school 12 min late. Next time at a speed of 15 km/h reaches his school 7 min late. Find the distance of his school from his house?
Solution:
Difference between the time = 12 – 7 = 5 min = 5/60 = 1/12hr
Required distance = (15×10)/(15-10)×1/12 = 150/5×1/12 = 2.5km
A man leaves a point A at t1 and reaches the point B at t2. Another man leaves the point B at t3 and reaches the point A at t4 then They will meet at t_(1+) ((t_2-t_1)(t_4-t_1))/((t_2-t_1 )+(t_4-t_3))
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# Question 9.19: Use the inverse matrix to solve the following simultaneous e......
Use the inverse matrix to solve the following simultaneous equations:
x − 2z = 4
2x + 2y + 3z = 15
x + 3y + 2z = 12
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Step 1: Write all the equations in the same order: variables (in order) = RHS. Write down the matrices A and B. Thematrix A is the matrix of coefficients of the three equations, all arranged in the same format; B is the column matrix consisting of the constants from the RHS of the equations.
A = $\left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right)$ B = $\left(\begin{matrix} 4 \\ 15\\ 12 \end{matrix} \right)$
Step 2: Since X = $A^{−1}$ B, determine the inverse of A. However, if you look carefully at A, you will see that this matrix is identical to the matrix D in Worked Examples 9.17 and 9.18 in which the inverse of D was determined; therefore
$A^{−1} =\frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right)$
So proceed straight away to the next step.
Step 3: Premultiply the matrix B by the inverse of A:
$X = A^{−1} B$
$\left(\begin{matrix} x \\ y\\ z \end{matrix} \right) =\frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right) \left(\begin{matrix} 4 \\ 15\\ 12 \end{matrix}\right) = -\frac{1}{13} \left(\begin{matrix} -62 \\ -28\\ -5 \end{matrix}\right) = \left(\begin{matrix} 4.77 \\ 2.15\\ 0.38 \end{matrix}\right)$
$\left(\begin{matrix} x \\ y\\ z \end{matrix} \right) = \left(\begin{matrix} 4.77 \\ 2.15\\ 0.38 \end{matrix}\right)$
The solutions may simply be read off (correct to two decimal places), x = 4.77,
y = 2.15, z = 0.38.
Question: 9.7
## Solve the following equations by Gaussian elimination: x + y − z = 3 (1) 2x + y − z = 4 (2) 2x + 2y + z = 12 (3) ...
All three equations must be written in the same fo...
Question: 9.20
## Given the input/output table for the three-sector economy: ...
Step 1: Use the underlying assumption total input=...
Question: 9.18
## (a) Find the inverse of the matrix D = (1 0 -2 2 2 3 1 3 2) (b) Show that DD^0−1 = I. ...
(a) Use the definition of the inverse given in equ...
Question: 9.17
## (a) Find the inverse of the matrix D = ( 1 0 -2 2 2 3 1 3 2) by Gauss–Jordan elimination. (b) Show that DD^–1 = I. ...
(a) Write out the augmented matrix consisting of t...
Question: 9.12
## Given the supply and demand functions for two related goods, A and B,Good A :{Qda = 30 − 8Pa + 2Pb {Qsa = −15 + 7Pa Good B :{Qdb = 28 + 4Pa − 6Pb {Qsb = 12 + 2Pb (a) Write down the equilibrium condition for each good. Hence, deduce two equations in Pa and Pb . (b) Use Cramer’s rule to find the ...
(a) The equilibrium condition for each good is tha...
Question: 9.9
## Solve the following equations by Gauss–Jordan elimination: 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 ...
Rearrange the equations to have variables on the L...
Question: 9.8
## Solve the following equations by Gaussian elimination: 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 ...
The equations are already written in the required ...
Question: 9.2
## A company manufactures two types of wrought iron gates. The number of labour-hours required to produce each type of gate, along with the maximum number of hours available, are given in Table 9.3. ...
(a) For x type I gates and y type II gates, the nu...
Question: 9.21
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Product of a Fraction and a Whole Number: Problem Type 1
In this lesson, we solve problems where we find the product of a fraction and a whole number.
Rules for finding the product of a fraction and a whole number
• We first write the whole number as a fraction, i.e., we write it divided by one; for example 5 is written as 5/1.
• We then multiply the numerators and then the denominators of both fractions to get the product fraction.
• If any simplification or cross cancelling is required, it is done and final answer is written.
Example
Multiply $\frac{5}{4}$ × 8
Solution
Step 1:
First, we write the whole number 8 as a fraction $\frac{8}{1}$
Step 2:
$\frac{5}{4}$ × 8 = $\frac{5}{4}$ × $\frac{8}{1}$
Step 3:
As 4 and 8 are multiples of 8, cross cancelling 4 and 8, we get
$\frac{5}{4}$ × $\frac{8}{1}$ = $\frac{5}{1}$ × $\frac{2}{1}$
Step 4:
Multiply the numerators and denominators of both fractions as follows.
$\frac{5}{1}$ × $\frac{2}{1}$ = $\frac{(5 × 2)}{(1 × 1)}$ = $\frac{10}{1}$ = 10
Step 5:
So $\frac{5}{4}$ × 8 = 10
Multiply $\frac{4}{5}$ × 15
Solution
Step 1:
First, we write the whole number 15 as a fraction $\frac{15}{1}$
Step 2:
$\frac{4}{5}$ × 15 = $\frac{4}{5}$ × $\frac{15}{1}$
Step 3:
As 5 and 15 are multiples of 5, cross cancelling 5 and 15, we get
$\frac{4}{5}$ × $\frac{15}{1}$ = $\frac{4}{1}$ × $\frac{3}{1}$
Step 4:
We multiply the numerators and denominators of both fractions as follows.
$\frac{4}{1}$ × $\frac{3}{1}$ = $\frac{(4 × 3)}{(1 × 1)}$ = $\frac{12}{1}$ = 12
Step 5:
So $\frac{4}{5}$ × 15 = 12
Multiply $\frac{3}{7}$ × 14
Solution
Step 1:
First, we write the whole number 14 as a fraction $\frac{14}{1}$
Step 2:
$\frac{3}{7}$ × 14 = $\frac{3}{7}$ × $\frac{14}{1}$
Step 3:
As 7 and 14 are multiples of 7, cross cancelling 7 and 14, we get
$\frac{3}{7}$ × $\frac{14}{1}$ = $\frac{3}{1}$ × $\frac{2}{1}$
Step 4:
Multiply the numerators and denominators of both fractions as follows.
$\frac{3}{1}$ × $\frac{2}{1}$ = $\frac{(3 × 2)}{(1 × 1)}$ = $\frac{6}{1}$ = 6
Step 5:
So $\frac{3}{7}$ × 14 = 6
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# 2008 iTest Problems/Problem 34
## Problem
While entertaining his younger sister Alexis, Michael drew two different cards from an ordinary deck of playing cards. Let a be the probability that the cards are of different ranks. Compute $\lfloor 1000a\rfloor$.
## Solutions
### Solution 1
Use complementary counting to count the number of ways one can draw two cards with the same rank. There are $13$ ranks, and each rank has $4$ cards. That means the probability of getting two cards with the same rank is $\tfrac{13 \cdot 4 \cdot 3}{52 \cdot 51} = \tfrac{3}{51}$, so the probability of getting two cards with different ranks is $\tfrac{48}{51}$. That means $\lfloor 1000a\rfloor = \boxed{941}$.
### Solution 2
The first card can be any card, so the probability is $1$. However, of the $51$ cards remaining, only $12 \cdot 4 = 48$ of them have a different rank. Thus, the probability of getting two cards with different ranks is $1 \cdot \tfrac{48}{51} = \tfrac{48}{51}$, so $\lfloor 1000a\rfloor = \boxed{941}$.
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# High School Math : How to find the percentage of a sector from an angle
## Example Questions
### Example Question #1 : Circles
Two pizzas are made to the same dimensions. The only difference is that Pizza 1 is cut into pieces at 30° angles and Pizza 2 is cut at 45° angles. They are sold by the piece, the first for $1.95 per slice and the second for$2.25 per slice. What is the difference in total revenue between Pizza 2 and Pizza 1?
$0 –$5.40
$5.40$2.70
–$2.70 Correct answer: –$5.40
Explanation:
First, let's calculate how many slices there are per pizza. This is done by dividing 360° by the respective slice degrees:
Pizza 1: 360/30 = 12 slices
Pizza 2: 360/45 = 8 slices
Now, the total amount made per pizza is calculated by multiplying the number of slices by the respective cost per slice:
Pizza 1: 12 * 1.95 = $23.40 Pizza 2: 8 * 2.25 =$18.00
The difference between Pizza 2 and Pizza 1 is thus represented by: 18 – 23.40 = –\$5.40
### Example Question #3 : Circles
What percentage of a full circle is the following sector? (Round to the nearest tenth of a percent.)
Note: The figure is not drawn to scale.
Explanation:
In order to find the percentage of a sector from an angle, you need to know that a full circle is .
Therefore, we can find the percentage by dividing the angle of the sector by and then multiplying by 100:
### Example Question #1 : Circles
A sector of a circle contains a center angle that is 36 degrees. What percentage of the circle is the sector?
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## Elementary Algebra
Published by Cengage Learning
# Chapter 8 - Coordinate Geometry and Linear Systems - 8.7 - Graphing Linear Inequalities - Problem Set 8.7 - Page 384: 21
#### Answer
To graph an inequality, we first pretend that the inequality sign is an equal sign, and then we graph the equation. If it is a less than or equal to sign or a greater than or equal to sign, we make the line solid. If not, we make the line dashed. Next, we must shade the region. In order to do this, we pick any point not on the line. If that point satisfies the inequality, we shade the region to the side of the line that the point is in. If the point does not satisfy the inequality, we shade the region on the other side of the line. Once we graph one inequality, we can graph the other inequality doing the same thing. The overlap (the purple area on the attached graph) represents the solution to the system of inequalities.
#### Work Step by Step
To graph an inequality, we first pretend that the inequality sign is an equal sign, and then we graph the equation. If it is a less than or equal to sign or a greater than or equal to sign, we make the line solid. If not, we make the line dashed. Next, we must shade the region. In order to do this, we pick any point not on the line. If that point satisfies the inequality, we shade the region to the side of the line that the point is in. If the point does not satisfy the inequality, we shade the region on the other side of the line. Once we graph one inequality, we can graph the other inequality doing the same thing. The overlap (the purple area on the attached graph) represents the solution to the system of inequalities.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Anyone who wishes to prepare Grade concepts can get a strong foundation by accessing the Eureka Math Book Answer Key. People of highly subject expertise prepared the solutions in a concise manner for easy grasping. Start answering all the questions given in Eureka Math Book Grade 4 Answer Key. Refer to our Eureka Math Answers Grade 4 chapter 11 to enhance your math skills and also to score good marks in the exams.
## Engage NY Eureka Math 4th Grade Module 5 Lesson 11 Answer Key
### Eureka Math Grade 4 Module 5 Lesson 11 Problem Set Answer Key
Question 1.
Label each number line with the fractions shown on the tape diagram. Circle the fraction that labels the point on the number line that also names the shaded part of the tape diagram.
a.
1/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
25 percent of the tape diagram is filled.
25/100 = 1/4.
so 1/4 part is filled.
b.
2/8 = 1/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/8 = 1/4.
25 percent of the tape diagram is filled.
25/100 = 1/4.
so 1/4 part is filled.
c.
3/12 = 1/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
3/12 = 1/4.
25 percent of the tape diagram is filled.
25/100 = 1/4.
so 1/4 part is filled.
Question 2.
Write number sentences using multiplication to show:
a. The fraction represented in 1(a) is equivalent to the fraction represented in 1(b).
2/8 = 1/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/8 = 1/4.
25 percent of the tape diagram is filled.
25/100 = 1/4.
so 1/4 part is filled.
b. The fraction represented in 1(a) is equivalent to the fraction represented in 1(c).
3/12 = 1/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
3/12 = 1/4.
25 percent of the tape diagram is filled.
25/100 = 1/4.
so 1/4 part is filled.
Question 3.
Use each shaded tape diagram below as a ruler to draw a number line. Mark each number line with the fractional units shown on the tape diagram, and circle the fraction that labels the point on the number line that also names the shaded part of the tape diagram.
a.
2/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
so 2/3 part is filled.
b.
4/6 = 2/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/6 = 2/3
so 2/3 part is filled.
c.
8/12 = 2/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/12 = 2/3
so 2/3 part is filled.
Question 4.
Write number sentences using division to show:
a. The fraction represented in 3(a) is equivalent to the fraction represented in 3(b).
4/6 = 2/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/6 = 2/3
so 2/3 part is filled.
b. The fraction represented in 3(a) is equivalent to the fraction represented in 3(c).
8/12= 2/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/12 = 2/3
so 2/3 part is filled.
Question 5.
a. Partition a number line from 0 to 1 into fifths. Decompose $$\frac{2}{5}$$ into 4 equal lengths.
2/5.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
so 2/5 part is filled.
b. Write a number sentence using multiplication to show what fraction represented on the number line is equivalent to $$\frac{2}{5}$$.
8/15 = 2/5.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/15 = 2/5.
1 x 2 = 2.
1 x 5 = 5.
so 2/5 part is filled.
c. Write a number sentence using division to show what fraction represented on the number line is equivalent to $$\frac{2}{5}$$.
8/15 = 2/5.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/15 = 2/5.
8/4 = 2.
15/3 = 5.
so 2/5 part is filled.
### Eureka Math Grade 4 Module 5 Lesson 11 Exit Ticket Answer Key
Question 1.
Partition a number line from 0 to 1 into sixths. Decompose $$\frac{2}{6}$$ into 4 equal lengths.
2/6 = 1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/6 = 1/3.
so the 2/6 part is filled.
Question 2.
Write a number sentence using multiplication to show what fraction represented on the number line is equivalent to $$\frac{2}{6}$$.
8/12 = 2/6.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/12 = 2/6.
so the 2/6 part is filled.
Question 3.
Write a number sentence using division to show what fraction represented on the number line is equivalent to $$\frac{2}{6}$$.
8/15 = 2/5.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
8/15 = 2/5.
8/4 = 2.
15/3 = 5.
so 2/5 part is filled.
### Eureka Math Grade 4 Module 5 Lesson 11 Homework Answer Key
Question 1.
Label each number line with the fractions shown on the tape diagram. Circle the fraction that labels the point on the number line that also names the shaded part of the tape diagram.
a.
1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
33 percent of the tape diagram is filled.
33/100 = 1/3.
so 1/3 part is filled.
b.
2/6 = 1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/6 = 1/3.
33 percent of the tape diagram is filled.
33/100 = 1/3.
so 1/3 part is filled.
c.
4/12 = 1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/12 = 1/3.
33 percent of the tape diagram is filled.
33/100 = 1/3.
so 1/3 part is filled.
Question 2.
Write number sentences using multiplication to show:
a. The fraction represented in 1(a) is equivalent to the fraction represented in 1(b).
2/6 = 1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/6 = 1/3.
33 percent of the tape diagram is filled.
33/100 = 1/3.
so 1/3 part is filled.
b. The fraction represented in 1(a) is equivalent to the fraction represented in 1(c).
4/12 = 1/3.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/12 = 1/3.
33 percent of the tape diagram is filled.
33/100 = 1/3.
so 1/3 part is filled.
Question 3.
Use each shaded tape diagram below as a ruler to draw a number line. Mark each number line with the fractional units shown on the tape diagram, and circle the fraction that labels the point on the number line that also names the shaded part of the tape diagram.
a.
2/4 = 1/2.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
2/4 = 1/2.
50 percent of the tape diagram is filled.
50/100 = 1/2.
so 1/2 part is filled.
b.
4/8 = 1/2.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/8 = 1/2.
50 percent of the tape diagram is filled.
50/100 = 1/2.
so 1/2 part is filled.
c.
5/10 = 1/2.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
5/10 = 1/2.
50 percent of the tape diagram is filled.
50/100 = 1/2.
so 1/2 part is filled.
Question 4.
Write a number sentence using division to show the fraction represented in 3(a) is equivalent to the fraction represented in 3(b).
4/8 = 1/2.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
4/8 = 1/2.
50 percent of the tape diagram is filled.
50/100 = 1/2.
so 1/2 part is filled.
Question 5.
a. Partition a number line from 0 to 1 into fourths. Decompose $$\frac{3}{4}$$ into 6 equal lengths.
3/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
3/4 = 0.75.
75 percent of the tape diagram is filled.
75/100 = 3/4.
so 3/4 part is filled.
b. Write a number sentence using multiplication to show what fraction represented on the number line is equivalent to $$\frac{3}{4}$$.
3/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
3/4 = 0.75.
75 percent of the tape diagram is filled.
75/100 = 3/4.
3 x 1 = 3.
4 x 1 = 4.
so 3/4 part is filled.
c. Write a number sentence using division to show what fraction represented on the number line is equivalent to $$\frac{3}{4}$$.
3/4.
Explanation:
In the above-given question,
given that,
Label each number line with the fractions shown on the tape diagram.
circle the fraction that labels the point on the number line that also names the shaded parts of the tape diagram.
the length of the tape diagram is 1 meter.
3/4 = 0.75.
75 percent of the tape diagram is filled.
75/100 = 3/4.
so 3/4 part is filled.
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# Binomial Theorem and Expansions
## Expansion of binomials raised to a power using combinations.
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Binomial Theorem
The Binomial Theorem tells you how to expand a binomial such as \begin{align*}(2x-3)^5\end{align*} without having to compute the repeated distribution. What is the expanded version of \begin{align*}(2x-3)^5\end{align*}?
### Introduction to the Binomial Theorem
The Binomial Theorem states:
\begin{align*}(a+b)^n=\sum \limits_{i=0}^n \tbinom{n}{i} a^i b^{n-i}\end{align*}
Writing out a few terms of the summation symbol helps you to understand how this theorem works:
\begin{align*}(a+b)^n=\tbinom{n}{0}a^n + \tbinom{n}{1}a^{n-1}b^1 + \tbinom{n}{2}a^{n-2}b^2 + \cdots + \tbinom{n}{n} b^n\end{align*}
Going from one term to the next in the expansion, you should notice that the exponents of \begin{align*}a\end{align*} decrease while the exponents of \begin{align*}b\end{align*} increase. You should also notice that the coefficients of each term are combinations. Recall that \begin{align*}\tbinom{n}{0}\end{align*} is the number of ways to choose objects from a set of \begin{align*}n\end{align*} objects.
Take the following binomial:
\begin{align*}(m-n)^6\end{align*}
It can be expanded using the Binomial Theorem:
\begin{align*}(m-n)^6 &=\tbinom{6}{0}m^6 + \tbinom{6}{1}m^5(-n)^1 + \tbinom{6}{2}m^4(-n)^2 + \tbinom{6}{3}m^3(-n)^3 \\ & \ \ \ + \tbinom{6}{4}m^2(-n)^4 + \tbinom{6}{5}m^1(-n)^5 + \tbinom{6}{6}(-n)^6 \\ &=1m^6-6m^5n+15m^4n^2-20m^3n^3+15m^2n^4-6m^1n^5+1n^6\end{align*}
Be extremely careful when working with binomials of the form \begin{align*}(a-b)^n\end{align*}. You need to remember to capture the negative with the second term as you write out the expansion: \begin{align*}(a-b)^n=(a+(-b))^n\end{align*}.
Another way to think about the coefficients in the Binomial Theorem is that they are the numbers from Pascal’s Triangle. Look at the expansions of \begin{align*}(a+b)^n\end{align*} below and notice how the coefficients of the terms are the numbers in Pascal’s Triangle.
### Examples
#### Example 1
Earlier, you were asked to expand \begin{align*}(2x-3)^5\end{align*}The expanded version of \begin{align*}(2x-3)^5\end{align*} is:
\begin{align*}(2x-3)^5 &=\tbinom{5}{0} (2x)^5+\tbinom{5}{1} (2x)^4(-3)^1+\tbinom{5}{2} (2x)^3(-3)^2\\ & \ \ \ +\tbinom{5}{3} (2x)^2(-3)^3+\tbinom{5}{4} (2x)^1(-3)^4+ \tbinom{5}{5}(-3)^6 \\ &=(2x)^5+5(2x)^4(-3)^1+10(2x)^3(-3)^2\\ & \ \ \ +10(2x)^2(-3)^3+5(2x)^1(-3)^4+(-3)^5 \\ &=32x^5-240x^4+720x^3-1080x^2+810x-243\end{align*}
#### Example 2
What is the coefficient of the term \begin{align*}x^7y^9\end{align*} in the expansion of the binomial \begin{align*}(x+y)^{16}\end{align*}?
The Binomial Theorem allows you to calculate just the coefficient you need.
\begin{align*}\tbinom{16}{9}=\frac{16!}{9!7!}=\frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=11,440\end{align*}
#### Example 3
What is the coefficient of \begin{align*}x^6\end{align*} in the expansion of \begin{align*}(4-3x)^7\end{align*}?
For this problem you should calculate the whole term, since the 3 and the 4 in \begin{align*}(3-4x)\end{align*} will impact the coefficient of \begin{align*}x^6\end{align*} as well. \begin{align*}\tbinom{7}{6}4^1(-3x)^6=7 \cdot 4 \cdot 729x^6=20,412x^6\end{align*}. The coefficient is 20,412.
#### Example 4
Compute the following summation.
\begin{align*}\sum \limits_{i=0}^4 \tbinom{4}{i}\end{align*}
This is asking for \begin{align*}\tbinom{4}{0}+\tbinom{4}{1}+\cdots+\tbinom{4}{4}\end{align*}, which are the sum of all the coefficients of \begin{align*}(a+b)^4\end{align*}.
\begin{align*}1+4+6+4+1=16\end{align*}
#### Example 5
Collapse the following polynomial using the Binomial Theorem.
\begin{align*}32x^5-80x^4+80x^3-40x^2+10x-1\end{align*}
Since the last term is -1 and the power on the first term is a 5 you can conclude that the second half of the binomial is \begin{align*}(? -1)^5\end{align*}. The first term is positive and \begin{align*}(2x)^5=32x^5\end{align*}, so the first term in the binomial must be \begin{align*}2x\end{align*}. The binomial is \begin{align*}(2x-1)^5\end{align*}.
### Review
Expand each of the following binomials using the Binomial Theorem.
1. \begin{align*}(x-y)^4\end{align*}
2. \begin{align*}(x-3y)^5\end{align*}
3. \begin{align*}(2x+4y)^7\end{align*}
4. What is the coefficient of \begin{align*}x^4\end{align*} in \begin{align*}(x-2)^7\end{align*}?
5. What is the coefficient of \begin{align*}x^3 y^5\end{align*} in \begin{align*}(x+y)^8\end{align*}?
6. What is the coefficient of \begin{align*}x^5\end{align*} in \begin{align*}(2x-5)^6\end{align*}?
7. What is the coefficient of \begin{align*}y^2\end{align*} in \begin{align*}(4y-5)^4\end{align*}?
8. What is the coefficient of \begin{align*}x^2y^6\end{align*} in \begin{align*}(2x+y)^8\end{align*}?
9. What is the coefficient of \begin{align*}x^3y^4\end{align*} in \begin{align*}(5x+2y)^7\end{align*}?
Compute the following summations.
10. \begin{align*}\sum \limits_{i=0}^9 \tbinom{9}{i}\end{align*}
11. \begin{align*}\sum \limits_{i=0}^{12} \tbinom{12}{i}\end{align*}
12. \begin{align*}\sum \limits_{i=0}^{8} \tbinom{8}{i}\end{align*}
Collapse the following polynomials using the Binomial Theorem.
13. \begin{align*}243x^5-405x^4+270x^3-90x^2=15x-1\end{align*}
14. \begin{align*}x^7-7x^6y+21x^5y^2-35x^4y^3+35x^3y^4-21x^2y^5+7xy^6-y^7\end{align*}
15. \begin{align*}128x^7-448x^6y+672x^5y^2-560x^4y^3+280x^3y^4-84x^2y^5+14xy^6-y^7\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# Area of a sector of central angle 200°
Question:
Area of a sector of central angle 200° of a circle is 770 cm2. Find the length of the corresponding arc of this sector.
Solution:
Let the radius of the sector AOBA be r.
Given that, Central angle of sector $A O B A=\theta=200^{\circ}$
and area of the sector $A O B A=770 \mathrm{~cm}^{2}$
We know that, area of the sector $=\frac{\pi r^{2}}{360^{\circ}} \times \theta^{\circ}$
$\therefore \quad$ Area of the sector, $770=\frac{\pi r^{2}}{360^{\circ}} \times 200$
$\Rightarrow \quad \frac{77 \times 18}{\pi}=r^{2}$
$\Rightarrow \quad r^{2}=\frac{77 \times 18}{22} \times 7 \Rightarrow r^{2}=9 \times 49$
$\Rightarrow \quad r=3 \times 7$
$\therefore \quad r=21 \mathrm{~cm}$
So, radius of the sector $A O B A=21 \mathrm{~cm}$.
Now, the length of the correspoding arc of this sector $=$ Central angle $\times$ Radius
$=200 \times 21 \times \frac{\pi}{180^{\circ}} \quad\left[\because 1^{\circ}=\frac{\pi}{180} R\right]$
$=\frac{20}{18} \times 21 \times \frac{22}{7}$
$=\frac{220}{3} \mathrm{~cm}=73 \frac{1}{3} \mathrm{~cm}$
Hence, the required length of the corresponding arc is $73 \frac{1}{3} \mathrm{~cm}$.
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## Introduction to Numerical Analysis: Finding Roots of Equations
Let be a function. One of the most basic problems in numerical analysis is to find the value that would render . This value is called a root of the equation . As an example, consider the function . Consider the equation . Clearly, there is only one root to this equation which is . Alternatively, consider defined as with . There are two analytical roots for the equation given by
In this section we are going to present three classes of methods: graphical, bracketing, and open methods for finding roots of equations.
### Graphical Methods
Graphical methods rely on a computational device that calculates the values of the function along an interval at specific steps, and then draws the graph of the function. By visual inspection, the analyst can identify the points at which the function crosses the axis. Graphical methods are very useful for one-dimensional problems, but for multi-dimensional problems it is not possible to visualize a graph of a function of multi-variables.
###### Example
As an example, consider . We wish to find the values of (i.e., the roots) that would satisfy the equation for . The fastest way is to plot the graph (even Microsoft Excel would serve as an appropriate tool here) and then visually inspect the locations where the graph crosses the axis:
As shown in the above graph, the equation has three roots and just by visual inspection, the roots are around , , and . Of course these estimates are not that accurate for , , and . Whether these estimates are acceptable or not, depends on the requirements of the problem. We can further zoom in using the software to try and get better estimates:
Visual inspection of the zoomed-in graphs provides the following estimates: , , and which are much better estimates since: , , and
View Mathematica Code
Clear[x]
f = Sin[5 x] + Cos[2 x]
Plot[f, {x, -1, 1}, AxesLabel -> {"x", "f"}]
f /. x -> -0.525
f /. x -> -0.21
f /. x -> 0.675
Plot[f, {x, -0.53, -0.52}, AxesLabel -> {"x", "f"}]
Plot[f, {x, -0.25, -0.2}, AxesLabel -> {"x", "f"}]
Plot[f, {x, 0.66, 0.675}, AxesLabel -> {"x", "f"}]
f /. x -> -0.5235
f /. x -> -0.225
f /. x -> 0.6731
### Bracketing Methods
An elementary observation from the previous method is that in most cases, the function changes signs around the roots of an equation. The bracketing methods rely on the intermediate value theorem.
#### Intermediate Value Theorem
Statement: Let be continuous and . Then, such that . The same applies if .
The proof for the intermediate value theorem relies on the assumption of continuity of the function . Intuitively, because the function is continuous, then for the values of to change from to it has to pass by every possible value between and . The consequence of the theorem is that if the function is such that , then, there is such that . We can proceed to find iteratively using either the bisection method or the false position method.
#### Bisection Method
In the bisection method, if , an estimate for the root of the equation can be found as the average of and :
Upon evaluating , the next iteration would be to set either or such that for the next iteration the root is between and . The following describes an algorithm for the bisection method given , , , and maximum number of iterations:
Step 1: Evaluate and to ensure that . Otherwise, exit with an error.
Step 2: Calculate the value of the root in iteration as . Check which of the following applies:
1. If , then the root has been found, the value of the error . Exit.
2. If , then for the next iteration, is bracketed between and . The value of .
3. If , then for the next iteration, is bracketed between and . The value of .
Step 3: Set . If reaches the maximum number of iterations or if , then the iterations are stopped. Otherwise, return to step 2 with the new interval and .
###### Example
Setting and applying this process to with and yields the estimate after 9 iterations with as shown below. Similarly, applying this process to with and yields the estimate after 10 iterations with while applying this process to with and yields the estimate after 9 iterations with :
View Mathematica Code
Clear[f, x, ErrorTable, ei]
f[x_] := Sin[5 x] + Cos[2 x]
(*The following function returns x_i and the new interval (a,b) along with an error note. The order of the output is {xi,new a, new b, note}*)
bisec2[f_, a_, b_] := (xi = (a + b)/2; Which[
f[a]*f[b] > 0, {xi, a, b, "Root Not Bracketed"},
f[xi] == 0, {xi, a, b, "Root Found"},
f[xi]*f[a] < 0, {xi, a, xi, "Root between a and xi"},
f[xi]*f[b] < 0, {xi, xi, b, "Root between xi and b"}])
(*Problem Setup*)
MaxIter = 20;
eps = 0.0005;
(*First root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {-0.6};
btable = {-0.5};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = bisec2[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
(*Second Root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {-0.3};
btable = {-0.2};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = bisec2[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
(*Third Root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {0.6};
btable = {0.7};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = bisec2[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
The following code can be used to implement the bisection method in MATLAB. For the sake of demonstration, it finds the roots of the function in above example. The example function is defined in another file
The following tool illustrates this process for and . Use the slider to view the process after each iteration. In the first iteration, the interval . , so, . In the second iteration, the interval becomes and the new estimate . The relative approximate error in the estimate . You can view the process to see how it converges after 12 iterations.
##### Error Estimation
If is the true value of the root and is the estimate, then, the number of iterations needed to ensure that the absolute value of the error is less than or equal to a certain value can be easily obtained. Let be the length of the original interval used. The first estimate and so, in the next iteration, the interval where the root is contained has a length of . As the process evolves, the interval for the iteration number has a length of . Since the true value exists in an interval of length , the absolute value of the error is such that:
Therefore, for a desired estimate of the absolute value of the error, say , the number of iterations required is:
###### Example
As an example, consider , if we wish to find the root of the equation in the interval with an absolute error less than or equal to 0.004, the number of iterations required is 8:
The actual root with 10 significant digits is .
Using the process above, after the first iteration, and so, the root lies between and . So, the length of the interval is equal to 0.5 and the error in the estimate is less than 0.5. The length of the interval after iteration 8 is equal to 0.0039 and so the error in the estimate is less than 0.0039. It should be noted however that the actual error was less than this upper bound after the seventh iteration.
View Mathematica Code
Clear[f, x, tt, ErrorTable, nn];
f[x_] := x^3 + x^2 - 10;
bisec[f_, a_, b_] := If[f[a] f[b] <= 0, If[f[(a + b)/2] f[a] <= 0, {(a + b)/2, a, (a + b)/2}, If[f[(a + b)/2] f[b] <= 0, {(a + b)/2, (a + b)/2, b}]], "Error, root is not bracketed"];
(*Exact Solution*)
t = Solve[x^3 + x^2 - 10 == 0, x]
Vt = N[x /. t[[1]], 10]
(*Problem Setup*)
MaxIter = 20;
eps = 0.0005;
(*First root*)
x = {bisec[f, 1., 2]};
ErrorTable = {1};
ErrorTable2 = {"N/A"};
IntervalLength = {0.5};
i = 2;
While[And[i <= MaxIter, Abs[ErrorTable[[i - 1]]] > eps],
ri = bisec[f, x[[i - 1, 2]], x[[i - 1, 3]]];
If[ri=="Error, root is not bracketed",ErrorTable[[1]]="Error, root is not bracketed";Break[]];
x = Append[x, ri]; ei = (x[[i, 1]] - x[[i - 1, 1]])/x[[i, 1]]; e2i = x[[i, 1]] - Vt; ErrorTable = Append[ErrorTable, ei]; IntervalLength = Append[IntervalLength, (x[[i - 1, 3]] - x[[i - 1, 2]])/2]; ErrorTable2 = Append[ErrorTable2, e2i]; i++]
x // MatrixForm
ErrorTable // MatrixForm
Title = {"Iteration", "x_i", "a", "b", "e_r", "Actual E", "Interval Length (Upper bound for E)"};
T2 = Table[{i, x[[i, 1]], x[[i, 2]], x[[i, 3]], ErrorTable[[i]], ErrorTable2[[i]], IntervalLength[[i]]}, {i, 1, Length[x]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
The procedure for solving the above example in MATLAB is available in the following files. The polynomial function is defined in a separate file.
#### False Position Method
In the false position method, the new estimate at iteration is obtained by considering the linear function passing through the two points and . The point of intersection of this line with the axis can be obtained using one of the following formulas:
Upon evaluating , the next iteration would be to set either or such that for the next iteration the root is between and . The following describes an algorithm for the false position method method given , , “varepsilon_2”, and maximum number of iterations:
Step 1: Evaluate and to ensure that . Otherwise exit with an error.
Step 2: Calculate the value of the root in iteration as . Check which of the following applies:
1. If , then the root has been found, the value of the error . Exit.
2. If , then for the next iteration, is bracketed between and . The value of .
3. If , then for the next iteration, is bracketed between and . The value of .
Step 3: If reaches the maximum number of iterations or if , then the iterations are stopped. Otherwise, return to step 2.
###### Example
Setting and applying this process to with and yields the estimate after 3 iterations with as shown below. Similarly, applying this process to with and yields the estimate after 4 iterations with while applying this process to with and yields the estimate after 3 iterations with :
View Mathematica Code
Clear[f, x, ErrorTable, ei]
f[x_] := Sin[5 x] + Cos[2 x]
(*The following function returns x_i and the new interval (a,b) along with an error note. The order of the output is {xi,new a, new b, note}*)
falseposition[f_, a_, b_] := (xi = (a*f[b]-b*f[a])/(f[b]-f[a]); Which[
f[a]*f[b] > 0, {xi, a, b, "Root Not Bracketed"},
f[xi] == 0, {xi, a, b, "Root Found"},
f[xi]*f[a] < 0, {xi, a, xi, "Root between a and xi"},
f[xi]*f[b] < 0, {xi, xi, b, "Root between xi and b"}])
(*Problem Setup*)
MaxIter = 20;
eps = 0.0005;
(*First root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {-0.6};
btable = {-0.5};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = falseposition[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
(*Second Root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {-0.3};
btable = {-0.2};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = falseposition[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
(*Third Root*)
ErrorTable = {1};
ErrorNoteTable = {};
atable = {0.6};
btable = {0.7};
xtable = {};
i = 1;
Stopcode = "NoStopping";
While[And[i <= MaxIter, Stopcode == "NoStopping"],
ri = falseposition[f, atable[[i]], btable[[i]]];
If[ri[[4]] == "Root Not Bracketed", xtable = Append[xtable, ri[[1]]];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
If[ri[[4]] == "Root Found", xtable = Append[xtable, ri[[1]]]; ErrorTable = Append[ErrorTable, 0];ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]]; Break[]];
atable = Append[atable, ri[[2]]];
btable = Append[btable, ri[[3]]];
xtable = Append[xtable, ri[[1]]];
ErrorNoteTable = Append[ErrorNoteTable, ri[[4]]];
If[i != 1,
ei = (xtable[[i]] - xtable[[i - 1]])/xtable[[i]];
ErrorTable = Append[ErrorTable, ei]];
If[Abs[ErrorTable[[i]]] > eps, Stopcode = "NoStopping", Stopcode = "Stop"];
i++]
Title = {"Iteration", "a", "b", "x_i", "f[a]", "f[b]", "f[x_i]", "er", "ErrorNote"};
T2 = Table[{i, atable[[i]], btable[[i]], xtable[[i]], f[atable[[i]]], f[btable[[i]]], f[xtable[[i]]], ErrorTable[[i]], ErrorNoteTable[[i]]}, {i, Length[xtable]}];
T2 = Prepend[T2, Title];
T2 // MatrixForm
The procedure for implementing the false position root finding algorithm in MATLAB is available in the following files. The example function is defined in a separate file.
The following tool illustrates this process for and . Use the slider to view the process after each iteration. In the first iteration, the interval . , so, . In the second iteration, the interval becomes and the new estimate . The relative approximate error in the estimate . You can view the process to see how it converges after very few iterations.
### Open Methods
Open methods do not rely on having the root squeezed between two values, but rather rely on an initial guess and then apply an iterative process to get better estimates for the root. Open methods are usually faster in convergence if they converge, but they don’t always converge. For multi-dimensions, i.e., for solving multiple nonlinear equations, open methods are easier to implement than bracketing methods which cannot be easily extended to multi-dimensions.
#### Fixed-Point Iteration Method
Let . A fixed point of is defined as such that . If , then a fixed point of is the intersection of the graphs of the two functions and .
The fixed-point iteration method relies on replacing the expression with the expression . Then, an initial guess for the root is assumed and input as an argument for the function . The output is then the estimate . The process is then iterated until the output . The following is the algorithm for the fixed-point iteration method. Assuming , , and maximum number of iterations :
Set , and calculate and compare with . If or if , then stop the procedure, otherwise, repeat.
The Babylonian method for finding roots described in the introduction section is a prime example of the use of this method. If we seek to find the solution for the equation or , then a fixed-point iteration scheme can be implemented by writing this equation in the form:
###### Example
Consider the function . We wish to find the root of the equation , i.e., . The expression can be rearranged to the fixed-point iteration form and an initial guess can be used. The tolerance is set to 0.001. The following is the Microsoft Excel table showing that the tolerance is achieved after 19 iterations:
Mathematica has a built-in algorithm for the fixed-point iteration method. The function “FixedPoint[f,Expr,n]” applies the fixed-point iteration method with the initial guess being “Expr” with a maximum number of iterations “n”. As well, the function “FixedPointList[f,Expr,n]” returns the list of applying the function “n” times. If “n” is omitted, then the software applies the fixed-point iteration method until convergence is achieved. Here is a snapshot of the code and the output for the fixed-point iteration . The software finds the solution . Compare the list below with the Microsoft Excel sheet above.
Alternatively, simple code can be written in Mathematica with the following output
View Mathematica Code
x = {0.1};
er = {1};
es = 0.001;
MaxIter = 100;
i = 1;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = Cos[x[[i]]]; x = Append[x, xnew]; ernew = (xnew - x[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[x];
SolutionTable = Table[{i - 1, x[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm
The following MATLAB code runs the fixed-point iteration method to find the root of a function with initial guess . The value of the estimate and approximate relative error at each iteration is displayed in the command window. Additionally, two plots are produced to visualize how the iterations and the errors progress. Your function should be written in the form . Then call the fixed point iteration function with fixedpointfun2(@(x) g(x), x0). For example, try fixedpointfun2(@(x) cos(x), 0.1)
##### Convergence
The fixed-point iteration method converges easily if in the region of interest we have . Otherwise, it does not converge. Here is an example where the fixed-point iteration method fails to converge.
###### Example
Consider the function . To find the root of the equation , the expression can be converted into the fixed-point iteration form as:
. Implementing the fixed-point iteration procedure shows that this expression almost never converges but oscillates:
View Mathematica Code
Clear[x, g]
x = {-0.9};
er = {1};
es = 0.001;
MaxIter = 100;
i = 1;
g[x_] := (Sin[5 x] + Cos[2 x]) + x
While[And[i <= MaxIter, Abs[er[[i]]] > es],xnew = g[x[[i]]]; x = Append[x, xnew]; ernew = (xnew - x[[i]])/xnew;er = Append[er, ernew]; i++];
T = Length[x];
SolutionTable = Table[{i - 1, x[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm
The following is the output table showing the first 45 iterations. The value of the error oscillates and never decreases:
The expression can be converted to different forms . For example, assuming :
If this expression is used, the fixed-point iteration method does converge depending on the choice of . For example, setting gives the estimate for the root with the required accuracy:
View Mathematica Code
Clear[x, g]
x = {5.};
er = {1};
es = 0.001;
MaxIter = 100;
i = 1;
g[x_] := (Sin[5 x] + Cos[2 x])/x + x
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = g[x[[i]]]; x = Append[x, xnew]; ernew = (xnew - x[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[x];
SolutionTable = Table[{i - 1, x[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm
Obviously, unlike the bracketing methods, this open method cannot find a root in a specific interval. The root is a function of the initial guess and the form , but the user has no other way of forcing the root to be within a specific interval. It is very difficult, for example, to use the fixed-point iteration method to find the roots of the expression in the interval .
###### Analysis of Convergence of the Fixed-Point Method
The objective of the fixed-point iteration method is to find the true value that satisfies . In each iteration we have the estimate . Using the mean value theorem, we can write the following expression:
for some in the interval between and the true value . Replacing and in the above expression yields:
The error after iteration is equal to while that after iteration is equal to . Therefore, the above expression yields:
For the error to reduce after each iteration, the first derivative of , namely , should be bounded by 1 in the region of interest (around the required root):
We can now try to understand why, in the previous example, the expression does not converge. When we plot and we see that oscillates rapidly with values higher than 1:
On the other hand, the expression converges for roots that are away from zero. When we plot and we see that the oscillations in decrease when is away from zero and is bounded by 1 in some regions:
###### Example
In this example, we will visualize the example of finding the root of the expression . There are three different forms for the fixed-point iteration scheme:
To visualize the convergence, notice that if we plot the separate graphs of the function and the function , then, the root is the point of intersection when . For , the slope is not bounded by 1 and so, the scheme diverges no matter what is. Here we start with :
For , the slope is bounded by 1 and so, the scheme converges really fast no matter what is. Here we start with :
For , the slope is bounded by 1 and so, the scheme converges but slowly. Here we start with :
The following is the Mathematica code used to generate one of the tools above:
View Mathematica Code
Manipulate[
x = {5.};
er = {1};
es = 0.001;
MaxIter = 20;
i = 1;
g[x_] := (x + 10)^(1/4);
While[And[i <= MaxIter, Abs[er[[i]]] > es],xnew = g[x[[i]]]; x = Append[x, xnew]; ernew = (xnew - x[[i]])/xnew;er = Append[er, ernew]; i++];
T1 = Length[x];
SolutionTable = Table[{i - 1, x[[i]], er[[i]]}, {i, 1, T1}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
LineTable1 = Table[{{T[[i, 2]], T[[i, 2]]}, {T[[i, 2]], g[T[[i, 2]]]}}, {i, 2, n}];
LineTable2 = Table[{{T[[i + 1, 2]], T[[i + 1, 2]]}, {T[[i, 2]], g[T[[i, 2]]]}}, {i, 2, n - 1}];
Grid[{{Plot[{x, g[x]}, {x, 0, 5}, PlotLegends -> {"x", "g3(x)"},ImageSize -> Medium, Epilog -> {Dashed, Line[LineTable1], Line[LineTable2]}]}, {Row[{"Iteration=", n - 2, " x_n=", T[[n, 2]], " g(x_n)=", g[T[[n, 2]]]}]}}], {n, 2, Length[T], 1}]
The following shows the output if we use the built-in fixed-point iteration function for each of , , and . oscillates and so, it will never converge. converges really fast (3 to 4 iterations). converges really slow, taking up to 120 iterations to converge.
#### Newton-Raphson Method
The Newton-Raphson method is one of the most used methods of all root-finding methods. The reason for its success is that it converges very fast in most cases. In addition, it can be extended quite easily to multi-variable equations. To find the root of the equation , the Newton-Raphson method depends on the Taylor Series Expansion of the function around the estimate to find a better estimate :
where is the estimate of the root after iteration and is the estimate at iteration . Assuming and rearranging:
The procedure is as follows. Setting an initial guess , tolerance , and maximum number of iterations :
At iteration , calculate and . If or if , stop the procedure. Otherwise repeat.
Note: unlike the previous methods, the Newton-Raphson method relies on calculating the first derivative of the function . This makes the procedure very fast, however, it has two disadvantages. The first is that this procedure doesn’t work if the function is not differentiable. Second, the inverse can be slow to calculate when dealing with multi-variable equations.
##### Example
As an example, let’s consider the function . The derivative of is . Setting the maximum number of iterations , , , the following is the Microsoft Excel table produced:
Mathematica has a built-in algorithm for the Newton-Raphson method. The function âFindRoot[lhs==rhs,{x,x0}]â applies the Newton-Raphson method with the initial guess being âx0â. The following is a screenshot of the input and output of the built-in function evaluating the roots based on three initial guesses.
Alternatively, a Mathematica code can be written to implement the Newton-Raphson method with the following output for three different initial guesses:
View Mathematica Code
Clear[x]
f[x_] := Sin[5 x] + Cos[2 x]
fp = D[f[x], x]
xtable = {-0.25};
er = {1};
es = 0.0005;
MaxIter = 100;
i = 1;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = xtable[[i]] - (f[xtable[[i]]]/fp /. x -> xtable[[i]]); xtable = Append[xtable, xnew]; ernew = (xnew - xtable[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[xtable];
SolutionTable = Table[{i - 1, xtable[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm
The following MATLAB code runs the Newton-Raphson method to find the root of a function with derivative and initial guess . The value of the estimate and approximate relative error at each iteration is displayed in the command window. Additionally, two plots are produced to visualize how the iterations and the errors progress. You need to pass the function f(x) and its derivative df(x) to the newton-raphson method along with the initial guess as newtonraphson(@(x) f(x), @(x) df(x), x0). For example, try newtonraphson(@(x) sin(5.*x)+cos(2.*x), @(x) 5.*cos(5.*x)-2.*sin(2.*x),0.4)
###### Analysis of Convergence of the Newton-Raphson Method
The error in the Newton-Raphson Method can be roughly estimated as follows. The estimate is related to the previous estimate using the equation:
Additionally, using Taylor’s theorem, and if is the true root with we have:
for some in the interval between and . Subtracting the above two equations yields:
Since and then:
(1)
If the method is converging, we have and therefore:
Therefore, the error is squared after each iteration, i.e., the number of correct decimal places approximately doubles with each iteration. This behaviour is called quadratic convergence. Look at the tables in the Newton-Raphson example above and compare the relative error after each step!
The following tool can be used to visualize how the Newton-Raphson method works. Using the slope , the difference can be calculated and thus, the value of the new estimate can be computed accordingly. Use the slider to see how fast the method converges to the true solution using , , and solving for the root of .
View Mathematica code
Manipulate[
f[x_] := Sin[5 x] + Cos[2 x];
fp = D[f[x], x];
xtable = {0.4};
er = {1};
es = 0.0005;
MaxIter = 100;
i = 1;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = xtable[[i]] - (f[xtable[[i]]]/fp /. x -> xtable[[i]]); xtable = Append[xtable, xnew]; ernew = (xnew - xtable[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[xtable];
SolutionTable = Table[{i - 1, xtable[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
LineTable1 = Table[{{T[[i, 2]], 0}, {T[[i, 2]], f[T[[i, 2]]]}, {T[[i + 1, 2]],0}}, {i, 2, n - 1}];
Grid[{{Plot[f[x], {x, 0, 1}, PlotLegends -> {"f(x)"}, ImageSize -> Medium, Epilog -> {Dashed, Line[LineTable1]}]}, {Row[{"Iteration=", n - 2, " x_n=", T[[n, 2]], " f(x_n)=", f[T[[n, 2]]]}]}}], {n, 2, 7, 1}]
Depending on the shape of the function and the initial guess, the Newton-Raphson method can get stuck around the locations of oscillations of the function. The tool below visualizes the algorithm when trying to find the root of with an initial guess of . It takes 33 iterations before reaching convergence.
Depending on the shape of the function and the initial guess, the Newton-Raphson method can get stuck in a loop. The tool below visualizes the algorithm when trying to find the root of with an initial guess of . The algorithm goes into an infinite loop.
In general, an initial guess that is close enough to the true root will guarantee quick convergence. The tool below visualizes the algorithm when trying to find the root of with an initial guess of . The algorithm quickly converges to the desired root.
The following is the Mathematica code used to generate one of the tools above:
View Mathematica Code
Manipulate[
f[x_] := x^3 - x + 3;
fp = D[f[x], x];
xtable = {-0.1};
er = {1.};
es = 0.0005;
MaxIter = 100;
i = 1;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = xtable[[i]] - (f[xtable[[i]]]/fp /. x -> xtable[[i]]); xtable = Append[xtable, xnew]; ernew = (xnew - xtable[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[xtable];
SolutionTable = Table[{i - 1, xtable[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
LineTable1 = Table[{{T[[i, 2]], 0}, {T[[i, 2]], f[T[[i, 2]]]}, {T[[i + 1, 2]], 0}}, {i, 2, n - 1}];
Grid[{{Plot[f[x], {x, -3, 2}, PlotLegends -> {"f(x)"}, ImageSize -> Medium, Epilog -> {Dashed, Line[LineTable1]}]}, {Row[{"Iteration=", n - 2, " x_n=", T[[n, 2]], " f(x_n)=", f[T[[n, 2]]]}]}}] , {n, 2, 33, 1}]
#### Secant Method
The secant method is an alternative to the Newton-Raphson method by replacing the derivative with its finite-difference approximation. The secant method thus does not require the use of derivatives especially when is not explicitly defined. In certain situations, the secant method is preferable over the Newton-Raphson method even though its rate of convergence is slightly less than that of the Newton-Raphson method. Consider the problem of finding the root of the function . Starting with the Newton-Raphson equation and utilizing the following approximation for the derivative :
the estimate for iteration can be computed as:
Obviously, the secant method requires two initial guesses and .
##### Example
As an example, let’s consider the function . Setting the maximum number of iterations , , , and , the following is the Microsoft Excel table produced:
The Mathematica code below can be used to program the secant method with the following output:
View Mathematica Code
f[x_] := Sin[5 x] + Cos[2 x];
xtable = {0.5, 0.4};
er = {1,1};
es = 0.0005;
MaxIter = 100;
i = 2;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = xtable[[i]] - (f[xtable[[i]]]) (xtable[[i]] - xtable[[i - 1]])/(f[xtable[[i]]] -f[xtable[[i - 1]]]); xtable = Append[xtable, xnew]; ernew = (xnew - xtable[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[xtable];
SolutionTable = Table[{i - 1, xtable[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm
The following code runs the Secant method to find the root of a function with two initial guesses and . The value of the estimate and approximate relative error at each iteration is displayed in the command window. Additionally, two plots are produced to visualize how the iterations and the errors progress. The program waits for a keypress between each iteration to allow you to visualize the iterations in the figure. Call the function with secant(@(x) f(x), x0, x1). For example try secant(@(x) sin(5.*x)+cos(2.*x),0.5,0.4)
##### Convergence Analysis of the Secant Method
The estimate in the secant method is obtained as follows:
Multiplying both sides by -1 and adding the true value of the root where for both sides yields:
Using algebraic manipulations:
Using the Mean Value Theorem, the denominator on the right-hand side can be replaced with:
for some between and . Therefore,
Using Taylor’s theorem for and around we get:
for some between and and some between and . Using the above expressions we can reach the equation:
and can be assumed to be identical and equal to , therefore:
Therefore:
(2)
Comparing 1 with 2 shows that the convergence in the secant method is not quite quadratic. To find the order of convergence, we need to solve the following equation for a positive and :
Substituting into equation 2 yields:
But we also have:
Therefore: . This equation is called the golden ratio and has the positive solution for :
while
implying that the error convergence is not quadratic but rather:
The following tool visualizes how the secant method converges to the true solution using two initial guesses. Using , , , and solving for the root of yields .
The following Mathematica Code was utilized to produce the above tool:
View Mathematica Code
Manipulate[
f[x_] := x^3 - x + 3;
f[x_] := Sin[5 x] + Cos[2 x];
xtable = {-0.5, -0.6};
xtable = {0.35, 0.4};
er = {1, 1};
es = 0.0005;
MaxIter = 100;
i = 2;
While[And[i <= MaxIter, Abs[er[[i]]] > es], xnew = xtable[[i]] - (f[xtable[[i]]]) (xtable[[i]] - xtable[[i-1]])/(f[xtable[[i]]] -f[xtable[[i - 1]]]); xtable = Append[xtable, xnew]; ernew = (xnew - xtable[[i]])/xnew; er = Append[er, ernew]; i++];
T = Length[xtable];
SolutionTable = Table[{i - 1, xtable[[i]], er[[i]]}, {i, 1, T}];
SolutionTable1 = {"Iteration", "x", "er"};
T = Prepend[SolutionTable, SolutionTable1];
T // MatrixForm;
If[n > 3,
(LineTable1 = {{T[[n - 2, 2]], 0}, {T[[n - 2, 2]], f[T[[n - 2, 2]]]}, {T[[n - 1, 2]], f[T[[n - 1, 2]]]}, {T[[n, 2]], 0}};
LineTable2 = {{T[[n - 1, 2]], 0}, {T[[n - 1, 2]], f[T[[n - 1, 2]]]}}), LineTable1 = {{}}; LineTable2 = {{}}];
Grid[{{Plot[f[x], {x, 0, 1}, PlotLegends -> {"f(x)"}, ImageSize -> Medium, Epilog -> {Dashed, Line[LineTable1], Line[LineTable2]}]}, {Row[{"Iteration=", n - 2, " x_n=", T[[n, 2]], " f(x_n)=", f[T[[n, 2]]]}]}}], {n, 2, 8, 1}]
### Problems
1. Let be such that . Use the graphical method to find the number and estimates of the roots of each of the equations and .
2. Compare the bisection method and the false position method in finding the roots of the two equations and in the interval where is:
Compare with the result using the Solve function and comment on the use of as a measure of the relative error.
3. Locate the first nontrivial root of where is in radians. Use the graphical method and the bisection method with .
4. Water is flowing in a trapezoidal channel at a rate of . The critical depth for such a channel must satisfy the equation:
where , is the cross sectional area (), and is the width of the channel at the surface (). Assuming that the width and the crossâsectional area can be related to the depth by:
Solve for the critical depth using the graphical method, the bisection method, and the false position method to find the critical depth in the interval . Use . Discuss your results (difference in the methods, accuracy, number of iterations, etc.).
(Note: The critical depth is the depth below which the flow of the fluid becomes relatively fast and affected by the upstream conditions (supercritical flow). Above that depth, the flow is subcritical, relatively slow and is controlled by the downstream conditions)
5. Use the graphical method to find all real roots of the equation
You are required to
• Plot the function several times by using smaller ranges for the axis to obtain the roots up to 3 significant digits.
• Explain the steps you have taken to identify the number and estimates of the roots.
• Explain the steps you have taken to obtain the roots to the required accuracy.
• Compare your results to the solution obtained using the “Solve” function in Mathematica.
6. Use the bisection method and the false position method to find all real roots of the equation
Use .
7. Consider the function:
Determine a root for the expression
• using fixed-point iteration
• using the Newton-Raphson method
Use an initial guess of and . Compare your final answer with the Solve function in Mathematica.
8. Consider the function:
Find the lowest positive root for the expression
• graphically
• using the Newton-Raphson method
• using the secant method
9. Use the Newton-Raphson method to find the root of:
Employ initial guesses of 2, 6, and 8. Explain your results.
10. The Manning equation can be written for a rectangular open channel as:
where is the fluid flow in , is the slope in , is the depth in , and is the Manning roughness coefficient. Develop a fixed-point iteration scheme to solve this equation for given , , , and . Consider . Prove that your scheme converges for all initial guesses greater than or equal to zero.
11. Why does the Babylonian method almost always converge?
12. Consider the function
Compare the secant method and the Newton-Raphson method in finding the root of the equation .
13. Solve the previous question using the Fixed-Point Iteration method
14. Use the Fixed-Point Iteration method to find the roots of the following functions:
whats was the error formula in the newton raphson method?
I don’t understand the question
2. aty says:
how can we find the error if we find the root using the graphical method
I don’t think you can quantify the error using the graphical method except visually.
3. Luisa Botero says:
Why does the bisection method not work if the root is exactly in the middle of your intersection?
4. Luisa Botero says:
Why does the bisection method no work when your root is exactly in the middle of the interval.
|
# Solve for: \lim_{x arrow 0} ((sqrt(1+x)-1)/(sqrt(1-x)-1))
## Expression: $\lim_{x \rightarrow 0} \left(\frac{ \sqrt{ 1+x }-1 }{ \sqrt{ 1-x }-1 }\right)$
Evaluate the limits of numerator and denominator separately
$\begin{array} { l }\lim_{x \rightarrow 0} \left(\sqrt{ 1+x }-1\right),\\\lim_{x \rightarrow 0} \left(\sqrt{ 1-x }-1\right)\end{array}$
Evaluate the limit
$\begin{array} { l }0,\\\lim_{x \rightarrow 0} \left(\sqrt{ 1-x }-1\right)\end{array}$
Evaluate the limit
$\begin{array} { l }0,\\0\end{array}$
Since the expression $\frac{ 0 }{ 0 }$ is an indeterminate form, try transforming the expression
$\lim_{x \rightarrow 0} \left(\frac{ \sqrt{ 1+x }-1 }{ \sqrt{ 1-x }-1 }\right)$
Rationalize the denominator
$\lim_{x \rightarrow 0} \left(\frac{ \left( \sqrt{ 1+x }-1 \right) \times \left( \sqrt{ 1-x }+1 \right) }{ -x }\right)$
Simplify the expression
$\lim_{x \rightarrow 0} \left(\frac{ \sqrt{ \left( 1+x \right) \times \left( 1-x \right) }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ -x }\right)$
Use $\left( a-b \right)\left( a+b \right)={a}^{2}-{b}^{2}$ to simplify the product
$\lim_{x \rightarrow 0} \left(\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ -x }\right)$
Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction
$\lim_{x \rightarrow 0} \left(-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }\right)$
Since the function $-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }$ is undefined for $0$, evaluate the left-hand and right-hand limits
$\begin{array} { l }\lim_{x \rightarrow 0^-} \left(-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }\right),\\\lim_{x \rightarrow 0^+} \left(-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }\right)\end{array}$
Evaluate the limit
$\begin{array} { l }-1,\\\lim_{x \rightarrow 0^+} \left(-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }\right)\end{array}$
Evaluate the limit
$\begin{array} { l }-1,\\-1\end{array}$
Since the left-hand and right-hand limits are equal, the limit $\lim_{x \rightarrow 0} \left(-\frac{ \sqrt{ 1-{x}^{2} }+\sqrt{ 1+x }-\sqrt{ 1-x }-1 }{ x }\right)$ equals $-1$
$-1$
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Difference between revisions of "Boolean Algebra"
Boolean algebra is the branch of algebra in which the values of the variables and constants have exactly two values: true and false, usually denoted 1 and 0 respectively.
The basic operators in Boolean algebra are and, or, and not. The secondary operators are exclusive or (often called xor) and equivalence. They are secondary in the sense that they can be composed from the basic operators of and, or, and not.
• The and of two values is true only whenever both values are true. It is written as $xy$ or $x \cdot y$. The values of and for all possible inputs is shown in the following truth table:
$x$ $y$ $x y$
0 0 0
0 1 0
1 0 0
1 1 1
• The or of two values is true whenever either or both values are true. It is written as $x+y$. The values of or for all possible inputs is shown in the following truth table:
$x$ $y$ $x + y$
0 0 0
0 1 1
1 0 1
1 1 1
• The not of a value is its opposite; that is, the not of a true value is false whereas the not of a false value is true. It is written as $\overline{x}$ or $\neg{x}$. The values of not for all possible inputs is shown in the following truth table:
$x$ $\overline{x}$
0 1
1 0
• The xor of two values is true whenever the values are different. It uses the $\oplus$ operator, and can be built from the basic operators: $x \oplus y = x \overline{y} + \overline{x} y$ The values of or for all possible inputs is shown in the following truth table:
$x$ $y$ $x \oplus y$
0 0 0
0 1 1
1 0 1
1 1 0
• The equivalence, or xnor, operator is the opposite of the xor function: The equivalence of two values is true when the values are the same. It uses the $\odot$ or $\equiv$ operator and can be built as follows: $x \odot y = \overline{x \oplus y} = x y + \overline{x} \overline{y}$ The values of equivalence for all possible inputs is shown in the following truth table:
$x$ $y$ $x \odot y$
0 0 1
0 1 0
1 0 0
1 1 1
Just as algebra has basic rules for simplifying and evaluating expressions, so does Boolean algebra.
Why is Boolean Algebra Important for ACSL Students?
Boolean algebra is important to programmers, computer scientists, and the general population.
• For programmers, Boolean expressions are used for conditionals and loops. For example, the following snippet of code sums the even numbers that are not also multiples of 3, stopping when the sum hits 100:
s = 0
x = 1
while (s < 100):
if (x % 2 == 0) and (x % 3 != 0):
s = s + x
x = x + 1
Both s < 100 and (x % 2 == 0) and (x % 3 != 0) are Boolean expressions.
• For computer scientists, Boolean algebra is the basis for digital circuits that make up a computer's hardware. The Digital Electronics category concerns a graphical representation of a circuit. That circuit is typically easiest to understand and evaluate by converting it to its Boolean algebra representation.
• The general population uses Boolean algebra, probably without knowing that they are doing so, when they enter search terms in Internet search engines. For example, the search expression "red sox -yankees" is the Boolean expression "red" and "sox" and not "yankees" that will returns web pages that contain the words "red" and "sox", as long as it does not contain the word "yankees". The search expression "jaguar speed -car" returns pages about the speed of the jaguar animal, not the Jaguar car.
Laws
A law of Boolean algebra is an identity such as $x + (y + z) = (x + y) + z$ between two Boolean terms, where a Boolean term is defined as an expression built up from variables and the constants 0 and 1 using the operations and, or, not, xor, and xnor. LIke ordinary algebra, parentheses are used to group terms.
Monotone laws
Boolean algebra satisfies many of the same laws as ordinary algebra when one matches up or with addition and and with multiplication. In particular the following laws are common to both kinds of algebra:
Associativity of or: $x + (y + z)$ $= (x + y) + z$ Associativity of and: $x \cdot (y \cdot z)$ $= (x \cdot y) \cdot z$ Commutativity of or: $x + y$ $= y + x$ Commutativity of and: $x \cdot y$ $= y \cdot x$ Distributivity of and over or: $x \cdot (y + z)$ $= x \cdot y + x \cdot z$ Identity for or: $x + 0$ $= x$ Identity for and: $x \cdot 1$ $= x$ Annihilator for and: $x \cdot 0$ $= 0$
The following laws hold in Boolean Algebra, but not in ordinary algebra:
Annihilator for or: $x +1$ $= 1$ Idempotence of or: $x +x$ $= x$ Idempotence of and: $x \cdot x$ $= x$ Absorption 1: $x \cdot (x + y)$ $= x$ Absorption 2: $x + (x \cdot y)$ $= x$ Distributivity of or over and: $x + (y \cdot z)$ $=(x + y) \cdot (x +z)$
Nonmonotone laws
Complement of and: $x \cdot \overline{x}$ $= 0$ Complement of or : $x + \overline{x}$ $= 1$ Double Negation: $\overline{ \overline{x}}$ $= x$
De Morgan's Laws
DeMorgan 1: $\overline{ x} \cdot \overline{y}$ $= \overline{x + y}$ DeMorgan 1: $\overline{ x} + \overline{y}$ $= \overline{x \cdot y}$
Sample Problems
Problems in this category are typically of the form "Given a Boolean expression, simplify it as much as possible" or "Given a Boolean expression, find the values of all possible inputs that make the expression true."
Sample Problem 1: Simplify
Problem: Simplify the following expression as much as possible: $\overline{ \overline{A(A+B)} + B\overline{A}}$
Solution:
The simplification proceeds as follows:
$\overline{ \overline{A(A+B)} + B\overline{A}}$
$= \left(\overline{ \overline{A(A+B)}}\right) \cdot \left(\overline{ B\overline{A}}\right)$ (DeMorgan's Law) $= \left(A(A+B)\right) \cdot \left( \overline{B}+\overline{\overline{A}}\right)$ (Double Negation; DeMorgan's Law) $= A \cdot \left( \overline{B}+A\right)$ (Absorption; Double Negation) $= A \overline{B}+AA$ (Distribution) $= A \overline{B}+A$ (Distribution) $=A \left( \overline{B}+1\right)$ $=A \left(1\right)$ $=A$
Sample Problem 2: Find Solutions
Problem: Find all orderd pairs $(A,B)$ that make the following expression true: $\overline{ \overline{(A+B)} + \overline{A}B }$
Solution:
There are typically two approaches to solving this type of problem. One approach is to simplify the expression as much as possible, until it's obvious what the solutions are. The other approach is to create a truth table of all possible inputs, with columns for each subexpression.
The simplification approach is as following:
$\overline{\overline{(A+B)} + \overline{A}B}$
$= \overline{\overline{A+B}} \cdot \overline{\overline{A}B}$
$= (A+B) \cdot (\overline{\overline{A}}+\overline{B} )$
$= (A+B) \cdot (A+\overline{B})$
$= AA + A\overline{B} + BA + B\overline{B}$
$= A + A(\overline{B} + B) + 0$
$= A + A(1)$
$= A + A$
$=A$
This means that all inputs are valid whenever $A$ is true: $(1,0)$ and $(1,1)$
The truth table approach is as following. Each column is the result of a basic operation on two other columns.
#1 #2 #3 #4 #5 #6 #7 #8
OR of Col#1, Col#2 NOT of Col#3 NOT of Col#1 ADD of Col#1, Col#2 OR of Col#4, Col#6 NOT of Col#7
$A$ $B$ $A+B$ $\overline{A+B}$ $\overline{A}$ $\overline{A}B$ $\overline{A+B} + \overline{A}B$ $\overline{\overline{A+B} + \overline{A}B}$
0 0 0 1 1 0 1 0
0 1 1 0 1 1 1 0
1 0 1 0 0 0 0 1
1 1 1 0 0 0 0 1
The rightmost column is the expression we are solving; it is true for the 3rd and 4th rows, where the inputs are $(1,0)$ and $(1,1)$.
Online Resources
Websites
A great online tutorial on Boolean Algebra is part of Ryan's Tutorials.
Videos
The following YouTube videos show ACSL students and advisors working out some previous problems. To access the YouTube page with the video, click on the title of the video in the icon. (You can also play the video directly by clicking on the arrow in the center of the image; however, you'll probably want to have a larger viewing of the video since it contains writing on a whiteboard.) Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
ACSL Prep - Mrs. Gupta - Boolean Algebra (MegaChristian5555) Christian is a student participating in ACSL. This video shows how to solve a half-dozen or so Boolean Algebra problems that have appeared in ACSL contests in recent years in the Intermediate and Senior divisions. ACSL Boolean Algebra Contest 2 Worksheet 1 (misterminich) Mr. Minich is an ACSL advisor. This video was one of two he created to help prepare his students for the ACSL Boolean algebra category. It shows solutions to 5 different problems that have appeared in recent years. ACSL Boolean Algebra Contest 2 Worksheet 2 (misterminich) Mr. Minich is an ACSL advisor. This video was one of two he created to help prepare his students for the ACSL Boolean algebra category. It shows solutions to 5 different problems that have appeared in recent years. ACSL 3 13-14 #1 - AM (Gordon Campbell) This video walks through the solution to finding all ordered triples that make the following Boolean expression true: $(AB+\overline{C})(\overline{A}+BC)(A+\overline{B}+C)$ This problem appeared in 2013-2014 in the Senior Division, Contest #3. A general tutorial on boolean algebra that can be used for American Computer Science League. (Tangerine Code) Walks through the simplification of the following Boolean expression: $\overline{ (\overline{A + \overline{B}})(AB)} + \overline{ (A+B)(\overline{\overline{A}B})}$
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Enter the original rational function and the numerators and denominators of the decomposed fractions into the calculator to determine the fractional decomposition.
## Fractional Decomposition Formula
The following formula is used to calculate the fractional decomposition of a rational function.
A/B = (A1/B1) + (A2/B2) + ... + (An/Bn)
Variables:
• A/B is the original rational function A1, A2, …, An are the numerators of the decomposed fractions B1, B2, …, Bn are the denominators of the decomposed fractions
To calculate the fractional decomposition, divide the original rational function into simpler fractions whose sum is equal to the original function. The numerators and denominators of these simpler fractions are determined by the factors of the original function’s denominator.
## What is a Fractional Decomposition?
Fractional decomposition is a mathematical process used to break down complex fractions or rational expressions into simpler parts, often for the purpose of integration or simplification. It involves expressing the fraction as a sum of simpler fractions with linear or quadratic denominators. This method is particularly useful in calculus and algebra for solving equations and integrating functions.
## How to Calculate Fractional Decomposition?
The following steps outline how to calculate the Fractional Decomposition using the given formula:
1. First, identify the original rational function A/B.
2. Next, decompose the rational function into individual fractions.
3. For each decomposed fraction, identify the numerator (A1, A2, …, An) and the denominator (B1, B2, …, Bn).
4. Write the decomposed fractions in the form (A1/B1) + (A2/B2) + … + (An/Bn).
5. Finally, simplify the expression if possible.
Example Problem:
Use the following variables as an example problem to test your knowledge:
A/B = 3/4
A1 = 1, B1 = 2
A2 = 2, B2 = 3
A3 = 1, B3 = 6
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# Rectangular Hyperbola
## What do we mean by a Rectangular Hyperbola?
The hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The angle between asymptotes of the hyperbola x2/a2 – y2/b2 = 1, is 2 tan–1 (b/a).
This is a right angle if tan–1 b/a = π/4, i.e., if b/a = 1 ⇒ b = a.
The equation of rectangular hyperbola referred to its transverse and conjugate axes as axes of coordinates is therefore x2 – y2 = a2
## How do we compute the Rectangular Hyperbola Equation?
We know that the asymptotes of the hyperbola x2/a2 – y2/b2 = 1 …… (1)
are given by y = + (b/a) x …… (2)
If θ be the angle between the asymptotes, then
θ = tan–1 ((m1–m2)/(1 + m1m2))
= tan–1 [{(b/a)–(–b/a)}/{1+(b/a)(–b/a)}]
= tan–1 [2(b/a)/(1–(b2/a2))]
= 2 tan–1 (b/a)
But if the hyperbola is rectangular, then θ = π/2
i.e., π/2 = 2 tan–1 (b/a) or tan (π/4) = b/a
b = a
Therefore, from (1) the equation of the rectangular hyperbola is x2 – y2 = a2.
In order to obtain the equation of the hyperbola which has asymptotes as coordinate axis we rotate the axes of reference through an angle of -45o
Hence, for this we have to write x/√2 + y/√2 for x and –x/√2 + y/√2 for y.
The equation (i) becomes
(1/2)(x + y)2 – (1/2)(x – y)2 = a2 i.e.
xy = ½ a2 or xy = c2 where c2 = a2/2.
Watch this Video for more reference
### Some Important Points to be noted:
• In a hyperbola b2 = a2 (e2 – 1). In the case of rectangular hyperbola (i.e., when b = a) result becomes a2 = a2(e2 – 1) or e2 = 2 or e = √2
i.e. the eccentricity of a rectangular hyperbola = √2.
• In case of rectangular hyperbola a = b i.e., the length of transverse axis = length of conjugate axis.
• A rectangular hyperbola is also known as an equilateral hyperbola.
• The asymptotes of rectangular hyperbola are y = ± x.
• If the axes of the hyperbola are rotated by an angle of -π/4 about the same origin, then the equation of the rectangular hyperbola x2 – y2 = ais reduced to xy = a2/2 or xy = c2.
• When xy = c2, the asymptotes are the coordinate axis.
• Length of latus rectum of rectangular hyperbola is the same as the transverse or conjugate axis.
• Rectangular Hyperbola with asymptotes as coordinate axis:
• The equation of the hyperbola which has its asymptotes as the coordinate axis is xy = c2 with parametric representation x = ct and y = c/t, t ∈ R-{0}.
• The equations of the directrices of the hyperbola in this case are x + y = ± √2c.
• Since, the transverse and the conjugate axes are the same hence, length of latus rectum = 2√2c = T.A. = C.A.
• Equation of a chord whose middle point is given to be (p, q) is qx + py = 2pq.
• The equation of the tangent at the point P(x1, y1) is x/x1 + y/y1 = 2 and at P(t) is x/t + ty = 2c.
• Equation of normal is y-c/t = t2(x-ct).
• The equation of the chord joining the points P(t1) and Q(t2) is x + t1t2y = c(t1 + t2) and its slope is m = -1/t1t2.
• The vertices of the hyperbola are (c, c) and (-c, -c) and the focus is (√2c, √2c) and (-√2c, -√2c).
• A rectangular hyperbola circumscribing a triangle passes through the orthocentre of this triangle.
• If a circle intersects a rectangular hyperbola at four points, then the mean value of the points of intersection is the mid-point of the line joining the centres of both circle and hyperbola.
### Illustration:
Find the equation of the hyperbola with asymptotes 3x – 4y + 9 = 0 and 4x + 3y + 1 = 0 which passes through the origin.
### Solution:
The asymptotes are given as 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0.
Hence, the joint equation of the asymptotes is (3x – 4y + 9)(4x + 3y + 1) = 0.
Now, we know that the equation of rectangular hyperbola differs from that of the joint equation of asymptotes by a constant only, hence we must have the equation of the hyperbola as
(3x – 4y + 9)(4x + 3y + 1) + r = 0.
It is given that the hyperbola passes through the origin and hence, putting x = y = 0 , we obtain,
9 + r = 0
Hence, r = -9.
Hence, the equation of the hyperbola is (3x – 4y + 9)(4x + 3y + 1) – 9 = 0.
This gives 12x2 + 9xy + 3x -16xy – 12y2 – 4y + 36x + 27y + 9 – 9 = 0.
So, 12x2 – 12y- 7xy + 39x + 23y = 0.
## Related Resources
To read more, Buy study materials of Hyperbola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
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Definitions
# Constant of integration
In calculus, the indefinite integral of a given function (i.e. the set of all antiderivatives of the function) is always written with a constant, the constant of integration. This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function $f\left(x\right)$ is defined on an interval and $F\left(x\right)$ is an antiderivative of $f\left(x\right)$, then the set of all antiderivatives of $f\left(x\right)$ is given by the functions $F\left(x\right) + C$, where C is an arbitrary constant.
## Origin of the constant
The derivative of any constant function is zero. Once one has found one antiderivative $F\left(x\right)$, adding or subtracting a constant C will give us another antiderivative, because $\left(F\left(x\right) + C\right)\text{'} = F,\text{'}\left(x\right) + C,\text{'} = F,\text{'}\left(x\right)$. The constant is a way of expressing that every function has an infinite number of different antiderivatives.
For example, suppose one wants to find antiderivatives of $cos\left(x\right)$. One such antiderivative is $sin\left(x\right)$. Another one is $sin\left(x\right)+1$. A third is $sin\left(x\right)-pi$. Each of these has derivative $cos\left(x\right)$, so they are all antiderivatives of $cos\left(x\right)$.
It turns out that adding and subtracting constants is the only flexibility we have in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for cos(x), we write:
$int cos\left(x\right),dx = sin\left(x\right) + C.$
Replacing C by a number will produce an antiderivative. By writing C instead of a number, however, a compact description of all the possible antiderivatives of cos(x) is obtained. C is called the constant of integration. It is easily determined that all of these functions are indeed antiderivatives of $cos\left(x\right)$:
begin\left\{align\right\}
frac{d}{dx}[sin(x) + C] &= frac{d}{dx}[sin(x)] + frac{d}{dx}[C]
` &= cos(x) + 0 `
` &= cos(x)`
end{align}
## Necessity of the constant
At first glance it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating definite integrals using the fundamental theorem of calculus, the constant will always cancel with itself.
However, trying to set the constant equal to zero doesn't always make sense. For example, $2sin\left(x\right)cos\left(x\right)$ can be integrated in two different ways:
begin\left\{align\right\}
int 2sin(x)cos(x),dx &=& sin^2(x) + C &=& -cos^2(x) + 1 + C int 2sin(x)cos(x),dx &=& -cos^2(x) + C &=& sin^2(x) - 1 + C end{align}
So setting C to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative". By ignoring the constant of integration, one can construct a proof that 1 = 0, which must obviously be invalid.
Another problem with setting C equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an initial value problem). For example, to obtain the antiderivative of $cos\left(x\right)$ that has the value 100 at x = π, then only one value of C will work (in this case C = 100).
This restriction can be rephrased in the language of differential equations. Finding an indefinite integral of a function $f\left(x\right)$ is the same as solving the differential equation $frac\left\{dy\right\}\left\{dx\right\} = f\left(x\right)$. Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed initial value problem. Imposing the condition that our antiderivative takes the value 100 at x = π is an initial condition. Each initial condition corresponds to one and only one value of C, so without C it would be impossible to solve the problem.
There is another justification, coming from abstract algebra. The space of all (suitable) real-valued functions on the real numbers is a vector space, and the differential operator $frac\left\{d\right\}\left\{dx\right\}$ is a linear operator. The operator$frac\left\{d\right\}\left\{dx\right\}$ maps a function to zero if and only if that function is constant. Consequently, the kernel of $frac\left\{d\right\}\left\{dx\right\}$ is the space of all constant functions. The process of indefinite integration amounts to finding a preimage of a given function. There is no canonical preimage for a given function, but the set of all such preimages forms a coset. Choosing a constant is the same as choosing an element of the coset. In this context, solving an initial value problem is interpreted as lying in the hyperplane given by the initial conditions.
## Reason for a constant difference between antiderivatives
This result can be formally stated in this manner: Let $F:mathbb\left\{R\right\}rightarrowmathbb\left\{R\right\}$ and $G:mathbb\left\{R\right\}rightarrowmathbb\left\{R\right\}$ be two everywhere differentiable functions. Suppose that $F,\text{'}\left(x\right) = G,\text{'}\left(x\right)$ for every real number x. Then there exists a real number C such that $F\left(x\right) - G\left(x\right) = C$ for every real number x.
To prove this, notice that $\left[F\left(x\right) - G\left(x\right)\right]\text{'} = 0$. So F can be replaced by F-G and G by the constant function 0, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:
Choose a real number a, and let $C = F\left(a\right)$. For any x, the fundamental theorem of calculus says that
begin\left\{align\right\}
int_a^x 0,dt &= F(x)-F(a)
` &= F(x)-C,`
end{align} which implies that $F\left(x\right)=C$. So F is a constant function.
Two facts are crucial in this proof. First, the real line is connected. If the real line were not connected, we would not always be able to integrate from our fixed a to any given x. For example, if we were to ask for functions defined on the union of intervals [0,1] and [2,3], and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, we can extend this theorem to disconnected domains. For example, there are two constants of integration for $textstyleint dx/x$ and infinitely many for $textstyleint tan x,dx.$
Second, F and G were assumed to be everywhere differentiable. If F and G are not differentiable at even one point, the theorem fails. As an example, let $F\left(x\right)$ be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let $G\left(x\right)=0$. Then the derivative of F is zero where it is defined, and the derivative of G is always zero. Yet it's clear that F and G do not differ by a constant. Even if it is assumed that F and G are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take F to be the Cantor function and again let G = 0.
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## Engage NY Eureka Math 8th Grade Module 6 Mid Module Assessment Answer Key
Question 1.
Many computers come with a Solitaire card game. The player moves cards in certain ways to complete specific patterns. The goal is to finish the game in the shortest number of moves possible, and a player’s score is determined by the number of moves. A statistics teacher played the game 16 times and recorded the number of moves and the final score after each game. The line represents the linear function that is used to determine the score from the number of moves.
a. Was this person’s average score closer to 1130 or 1110? Explain how you decided.
Most of the games had scores between 1125 and 1175. The mean score will be closer to 1130.
b. The first two games she played took 169 moves (1131 points) and 153 moves (1147 points). Based on this information, determine the equation of the linear function used by the computer to calculate the score from the number of moves. Explain your work.
The difference in the scores is 1131 – 1147 or – 16.
The difference in the number of moves is 169 – 153 = 16
The slope is –$$\frac{16}{16}$$ or – 1. This means that 1131 = intercept – 169, So intercept equals 1300
Score = 1300 – moves
c. Based on the linear function, each time the player makes a move, how many points does she lose?
One point last per move.
d. Based on the linear function, how many points does the player start with in this game? Explain your reasoning.
1300, or the score when the number of moves equals 0.
Question 2.
To save money, drivers often try to increase their mileage, which is measured in miles per gallon (mpg). One theory is that speed traveled impacts mileage. Suppose the following data are recorded for five different 300-mile tests, with the car traveling at different speeds in miles per hour (mph) for each test.
a. For the data in this table, is the association positive or negative? Explain how you decided.
As the speed increases in miles per hour, the miles per gallon decrease. This describes a negative association.
b. Construct a scatter plot of these data using the following coordinate grid. The vertical axis represents the mileage, and the horizontal axis represents the speed in miles per hour (mph).
c. Draw a line on your scatter plot that you think is a reasonable model for predicting the mileage from the car speed.
d. Estimate and interpret the slope of the line you found in part (c).
So, slope ≈ $$\frac{20-33}{80-50}$$ ≈ – $$0.43 \overline{3}$$
Each increase of 1 mph in speed predicts a decrease of $$0.43 \overline{3}$$ mpg.
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# Thread: Calculus 1 Questions
1. ## Calculus 1 Questions
1. Let a and b be differentiable.
x^2 x<1
ax+b x> or =1
2. Prove lim as x approaches 3
x^2+x=8
3. A square plot is made at points A,B,C, and D (clockwise). The sides of the square are 100ft long. A pipeline goes from A to a point between B and C and from that point to C. The cost of each side is $20 per ft. and the cost of the pipe inside the square plot is$10 per ft. What is the minimum cost?
2. For #3:
Are you sure the costs aren't reversed. It should maybe be $20 through the square and$10 along the side.
The point between B and C where the pipeline joins the side, call it P.
From the drawing, this length is $\sqrt{100^{2}+x^{2}}$
The distance along the side is 100-x.
Factor in the cost:
$20\sqrt{100^{2}+x^{2}}+10(100-x)$
This is what needs minimized. Therefore, differentiate, set to 0 and solve for x.
3. How did you get 8?
4. Sorry, that was a typo. I fixed it.
5. Originally Posted by Debi
1. Let a and b be differentiable.
x^2 x<1
ax+b x> or =1
You are defining a function:
$
f(x)=\left\{
\begin{array}{cc}
{x^2}& {x<1}\\
{ax+b}& {x \ge 1}
\end{array}
\right{}
$
But what is the question?
At a guess I would say you have been asked to find $a$ and $b$ so the $f(x)$ is differentiable everywhere.
$f(x)$ is piecewise continuous and differentiable, so we need only make sure it is continuous and differentiable at the join between the pieces
First we have to ensure that $f(x)$ is continuous so we need
$\lim_{x \to 1_-} f(x)=\lim_{x \to 1_+}f(x)$,
which is equivalent to:
$1=a+b\ \ \ \dots(1)$.
Similarly we need:
$\lim_{x \to 1_-} f'(x)=\lim_{x \to 1_+}f'(x)$,
which is equivalent to:
$2=a\ \ \ \dots(2)$.
So $s=2$, and substituting back into $(1)$ gives: $b=1$.
RonL
6. your way of solving the prblm isn't possible b/c the square root -99 =x
7. Originally Posted by CaptainBlack
You are defining a function:
$
f(x)=\left\{
\begin{array}{cc}
{x^2}& {x<1}\\
{ax+b}& {x \ge 1}
\end{array}
\right{}
$
But what is the question?
At a guess I would say you have been asked to find $a$ and $b$ so the $f(x)$ is differentiable everywhere.
RonL
YES
8. Originally Posted by Debi
YES
See the rest of that post for the solution, you have only see the first half.
RonL
9. Originally Posted by Debi
2. Prove lim as x approaches 3
x^2+x=8
Not sure what this question should be. The limit as x approaches
3 is not 8
RonL
10. Originally Posted by galactus
For #3:
Are you sure the costs aren't reversed. It should maybe be $20 through the square and$10 along the side.
The point between B and C where the pipeline joins the side, call it P.
From the drawing, this length is $\sqrt{100+x^{2}}$
The distance along the side is 100-x.
Factor in the cost:
$20\sqrt{100+x^{2}}+10(100-x)$
Another typo, should be:
$20\sqrt{100^2+x^{2}}+10(100-x)$
RonL
11. Man, I'm typoin' all over the place. Thanks for the catch, Cap'n.
I fixed my post.
12. Prove as lim x approaches 3 of the equation, x^2+x=12
13. Originally Posted by Debi
Prove as lim x approaches 3 of the equation, x^2+x=12
$\lim_{x\to 3}x=3$
Proof: You need to show for any $\epsilon >0$ there is a $\delta >0$ such as, if
$0<|x-3|<\delta, \, x\in D$ (where $D$ is in the domain of $f(x)=x$) then,
$|x-3|<\epsilon$. If you take $\delta = \epsilon$ you make this condition true.
$\lim_{x\to 3}x+1=4$
Proof: You need to show for any $\epsilon >0$ there is a $\delta >0$ such as, if
$0<|x-3|<\delta, \, x\in D$ (where $D$ is in the domain of $f(x)=x+1$) then,
$|x+1-4|=|x-3|<\epsilon$. If you take $\delta = \epsilon$ you make the condition true.
$\lim_{x\to 3}x^2+x=12$
Proof:Factor,
$\lim_{x\to 3}x(x+1)$
Note that,
$\lim_{x\to 3}x=3$--->First Paragraph
$\lim_{x\to 3}x+1=4$--->Second Paragraph
Therefore,
$\lim_{x\to 3}x(x+1)=3\cdot 4=12$
Proof Complete.
|
Home » Statistics » Geometric Distribution
# Geometric Distribution
## Introduction
Geometric distribution is used to model the situation where we are interested in finding the probability of number failures before first success or number of trials (attempts) to get first success in a repeated mutually independent Beronulli’s trials, each with probability of success $p$.
There are two different definitions of geometric distributions one based on number of failures before first success and other based on number of trials (attempts) to get first success. The choice of the definition is a matter of the context.
In this tutorial we will discuss about various properties of geometric distribution along with their theoretical proofs.
## Geometric Distribution
Consider a series of mutually independent Bernoulli’s trials with constant probability of success $p$ and probability of failure $q =1-p$.
Let random variable $X$ denote the number of failures before first success. Then the random variable $X$ take the values $x=0,1,2,\ldots$.
For getting $x$ failures before first success we required $(x+1)$ Bernoulli trials with outcomes $FF\cdots (x \text{ times}) S$.
$$\begin{eqnarray*} P(\text{x failures and then success} & = & P(FF\cdots (x \text{ times})S)\\ P(X=x) & = & q\cdot q\cdots \text{ (x times) } \cdot p\\ & = & q^x p,\quad x=0,1,2\ldots\\ & & \quad 0 < p < 1, q=1-p. \end{eqnarray*}$$
The name geometric distribution is given because various probabilities for $x=0,1,2,\cdots$ are the terms from geometric progression.
## Definition of geometric distribution
A discrete random variable $X$ is said to have geometric distribution with parameter $p$ if its probability mass function is given by
\begin{align*} P(X=x) &= \begin{cases} q^x p, & x=0,1,2,\ldots; \\ & 0 < p < 1, q = 1-p \\ 0, & Otherwise. \end{cases} \end{align*}
Clearly, $P(X=x)\geq 0$ for all $x$ and
$$\begin{eqnarray*} \sum_{x=0}^\infty pq^x &=& p\sum_{x=0}^\infty q^x \\ &=& p(1-q)^{-1} \\ &=& p\cdot p^{-1}=1. \end{eqnarray*}$$
Hence $P(X=x)$ is a legitimate probability mass function.
## Key features of Geometric Distribution
• A random experiment consists of repeated trials.
• Each trial of an experiment has two possible outcomes, like success and failure.
• The probability of success ($p$) is constants for each trial.
• The trials are independent of each other.
• The random variable $X$ is the number of failures before getting first success $(X = 0,1,2,\cdots)$ OR the number of trials to get first success $(X = 1,2,\cdots)$.
## Distribution Function of Geometric Distribution
The distribution function of geometric distribution is $F(x)=1-q^{x+1}, x=0,1,2,\cdots$.
#### Proof
The distribution function of geometric random variable is given by
\begin{aligned} F(x)&=P(X\leq x)\\ &=1- P(X> x)\\ &=1-\sum_{x=x+1}^\infty pq^x\\ &=1-p(q^{x+1}+q^{x+2}+q^{x+3}+\cdots)\\ &=1-pq^{x+1}(1+q^{1}+q^{2}+\cdots)\\ &=1-pq^{x+1}(1-q)^{-1}\\ &=1-pq^{x+1}p^{-1}\\ &=1-q^{x+1}, x=0,1,2,\cdots \end{aligned}
## Another Form of Geometric Distribution
Sometimes a geometric random variable can be defined as the number of trials (attempts) till the first success, including the trial on which the success occurs. In such situation, the p.m.f. of geometric random variable $X$ is given by
\begin{align*} P(X=x) &= \begin{cases} q^{x-1} p, & x=1,2,\ldots; \\ & 0 < p < 1, q=1-p\\ 0, & Otherwise. \end{cases} \end{align*}
The mean for this form of geometric distribution is $E(X) = \dfrac{1}{p}$ and variance is $\mu_2 = \dfrac{q}{p^2}$. The distribution function of this form of geometric distribution is $F(x) = 1-q^x,x=1,2,\cdots$. The moment generating function for this form is $M_X(t) = pe^t(1-qe^t)^{-1}$.
• The binomial distribution counts the number of successes in a fixed number of trials $(n)$. But geometric distribution counts the number of trials (attempts) required to get a first success.
## Graph of Geometric Distribution with $p=0.5$.
Following graph shows the probability mass function of geometric distribution with parameter $p=0.5$.
## Mean of Geometric Distribution
The mean of Geometric distribution is $E(X)=\dfrac{q}{p}$.
#### Proof
The mean of geometric random variable $X$ is given by
$$\begin{eqnarray*} \mu_1^\prime =E(X) &=& \sum_{x=0}^\infty x\cdot P(X=x) \\ &=& \sum_{x=0}^\infty x\cdot pq^x \\ &=& pq \sum_{x=1}^\infty x\cdot q^{x-1} \\ &=& pq(1-q)^{-2}\\ &=& \frac{q}{p}. \end{eqnarray*}$$
## Variance of Geometric Distribution
The variance of Geometric distribution is $V(X)=\dfrac{q}{p^2}$.
#### Proof
The variance of geometric random variable $X$ is given by
$$\begin{equation*} V(X) = E(X^2) - [E(X)]^2. \end{equation*}$$
Let us find the expected value of $X^2$.
$$\begin{eqnarray*} E(X^2) & = & E[X(X-1)+X]\\ &=& E[X(X-1)] +E(X)\\ &=&\sum_{x=1}^\infty x(x-1) P(X=x) +\frac{q}{p}\\ &=& \sum_{x=2}^\infty x(x-1)pq^x +\frac{q}{p}\\ &=& pq^2 \sum_{x=2}^\infty x(x-1)q^{x-2}+\frac{q}{p}\\ &=& 2pq^2 \sum_{x=2}^\infty \frac{x(x-1)}{2\times 1}q^{x-2} +\frac{q}{p}\\ &=& 2pq^2 (1-q)^{-3}+\frac{q}{p}\\ &=& \frac{2q^2}{p^2} +\frac{q}{p}. \end{eqnarray*}$$
Now,
$$\begin{eqnarray*} V(X) &=& E(X^2)-[E(X)]^2 \\ &=& \frac{2q^2}{p^2}+\frac{q}{p}-\frac{q^2}{p^2}\\ &=& \frac{q^2}{p^2}+\frac{q}{p}\\ &=&\frac{q}{p^2}. \end{eqnarray*}$$
For geometric distribution, $E(X) < V(X)$, i.e., mean < variance.
## MGF of Geometric Distribution
The moment generating function of geometric distribution is $M_X(t) = p(1-qe^t)^{-1}$.
#### Proof
The moment generating function of geometric distribution is
$$\begin{eqnarray*} M_X(t) &=& E(e^{tX})\\ &=& \sum_{x=0}^\infty e^{tx} P(X=x) \\ &=& \sum_{x=0}^\infty e^{tx} q^x p\\ &=& p\sum_{x=0}^\infty (qe^{t})^x\\ &=& p(1-qe^t)^{-1} \qquad \bigg(\text{ \because \sum_{x=0}^\infty q^x = (1-q)^{-1}}\bigg). \end{eqnarray*}$$
## Mean and variance from M.G.F.
The mean and variance of geometric distribution can be obtained using moment generating function as follows
$$\begin{eqnarray*} \text{Mean }=\mu_1^\prime &=& \bigg[\frac{d}{dt} M_X(t)\bigg]_{t=0}\\ &=& \bigg[\frac{d}{dt} p(1-qe^t)^{-1}\bigg]_{t=0} \\ &=& \big[pqe^t(1-qe^t)^{-2}\big]_{t=0} \\ &=& pq(1-q)^{-2}=\frac{q}{p}. \end{eqnarray*}$$
The second raw moment of geometric distribution can be obtained as
$$\begin{eqnarray*} \mu_2^\prime &=& \bigg[\frac{d^2}{dt^2} M_X(t)\bigg]_{t=0}\\ &=&\bigg[\frac{d}{dt} pqe^t(1-qe^t)^{-2}\bigg]_{t=0} \\ &=& pq\bigg[2e^tqe^t(1-qe^t)^{-3}+(1-qe^t)^{-2}e^t\bigg]_{t=0} \\ &=& pq\big[ 2q(1-q)^{-3}+(1-q)^{-2}\big]\\ &=& \frac{2q^2}{p^2}+\frac{q}{p}. \end{eqnarray*}$$
Therefore, the variance of geometric distribution is
$$\begin{eqnarray*} \text{Varaince }=\mu_2 &=& \mu_2^\prime-(\mu_1^\prime)^2 \\ &=& \frac{2q^2}{p^2}+\frac{q}{p} -\frac{q^2}{p^2}\\ &=& \frac{q^2}{p^2}+\frac{q}{p} =\frac{q}{p^2}. \end{eqnarray*}$$
## Cumulant Generating Function
The cumulant generating function of geometric distribution is $K_{X}(t)=\log_e \bigg(\dfrac{p}{1-qe^t}\bigg)$.
#### Proof
The cumulant generating function of geometric distribution is
$$\begin{eqnarray*} K_X(t) &=& \log_e M_X(t) \\ &=& \log_e \bigg( \frac{p}{1-qe^t}\bigg) \\ &=& -\log_e \bigg( \frac{1-qe^t}{p}\bigg)\\ &=& -\log_e \bigg(\frac{1}{p}-\frac{qe^t}{p}\bigg)\\ \end{eqnarray*}$$
$$\begin{eqnarray*} K_X(t)&=& -\log_e\bigg[\frac{1}{p}-\frac{q}{p}\bigg(1+t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg) \bigg]\\ &=& -\log_e\bigg[\frac{1}{p}-\frac{q}{p}-\frac{q}{p}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg) \bigg]\\ &=& -\log_e\bigg[1-\frac{q}{p}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg) \bigg]\\ &=& \bigg[\frac{q}{p}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg)+\bigg(\frac{q}{p}\bigg)^2\frac{1}{2}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg)^2\\ & &+\bigg(\frac{q}{p}\bigg)^3\frac{1}{3}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg)^3\\ & & +\bigg(\frac{q}{p}\bigg)^4\frac{1}{4}\bigg(t +\frac{t^2}{2!}+ \frac{t^3}{3!}+\cdots \bigg)^4+\cdots \bigg] \end{eqnarray*}$$
The cumulant of geometric distribution are
$$\begin{eqnarray*} \kappa_1 &=&\mu_1^\prime \\ &=& \text{coefficient of t in the expansion of K_X(t)}\\ &=& \frac{q}{p}. \end{eqnarray*}$$
$$\begin{eqnarray*} \kappa_2 &= &\mu_2 \\ &=& \text{coefficient of \frac{t^2}{2!} in the expansion of K_X(t)} \\ &=& \frac{q}{p}+ \frac{q^2}{p^2} \\ &=& \frac{q}{p^2}. \end{eqnarray*}$$
$$\begin{eqnarray*} \kappa_3 &=&\mu_3\\ &=& \text{coefficient of \frac{t^3}{3!} in the expansion of K_X(t)} \\ &=& \frac{q}{p}+\frac{3q^2}{p^2}+\frac{2q^3}{p^3} \\ &=& \frac{q}{p}+\frac{q^2}{p^2} +\frac{2q^2}{p^2} +\frac{2q^3}{p^3}\\ &=& \frac{q}{p^2} +\frac{2q^2}{p^3}(p+q)\\ &=& \frac{q}{p^3}(p+2q)= \frac{q}{p^3}(1+q). \end{eqnarray*}$$
$$\begin{eqnarray*} \kappa_4 &=& \mu_4-3k_2^2\\ &=& \text{coefficient of \frac{t^4}{4!} in the expansion of K_X(t)} \\ &=& \frac{q}{p}+(\frac{3q^2}{p^2}+\frac{4q^2}{p^2}) + (\frac{6q^3}{p^3}+\frac{6q^3}{p^3}) + \frac{6q^4}{p^4} \\ &=& \frac{q}{p}+\frac{q^2}{p^2}+\frac{6q^2}{p^2} + \frac{6q^3}{p^3}+\frac{6q^3}{p^3} + \frac{6q^4}{p^4} \\ &=& \frac{q}{p^2}(p+q)+\frac{6q^2}{p^3}(p+q) + \frac{6q^3}{p^4}(p+q)\\ &=& \frac{q}{p^2}+\frac{6q^2}{p^3} + \frac{6q^3}{p^4}\\ &=& \frac{q}{p^2}+\frac{6q^2}{p^4}(p+q)\\ &=& \frac{q}{p^4}(p^2+6q)\\ &=& \frac{q}{p^4}(1+4q+q^2). \end{eqnarray*}$$
Hence
$$\begin{eqnarray*} \mu_4 & = & \kappa_4 + 3\kappa_2^2\\ &=&\frac{q}{p^4}(1+4q+q^2)+ \frac{3q^2}{p^4}\\ &=& \frac{q}{p^4}( 1 + 7q +q^2). \end{eqnarray*}$$
Cumulants can also be determine using the following method:
$$\begin{eqnarray*} \kappa_1 = \mu_1^\prime &=& \bigg[\frac{d}{dt} K_X(t)\bigg]_{t=0} \\ & = & \bigg[\frac{d}{dt} (\log_e p -\log_e(1-qe^t))\bigg]_{t=0} \\ &=& \bigg[-\frac{1}{(1-qe^t)}(-qe^t)\bigg]_{t=0}\\ &=& \frac{q}{1-q}\\ &=&\frac{q}{p}. \end{eqnarray*}$$
$$\begin{eqnarray*} \kappa_2= \mu_2 &=& \bigg[\frac{d^2}{dt^2} K_X(t)\bigg]_{t=0}\\ & = & \bigg[\frac{d}{dt} \frac{qe^t}{(1-qe^t)}\bigg]_{t=0} \\ &=& \bigg[\frac{1}{(1-qe^t)^2}\big[ (1-qe^t) qe^t - qe^t(-qe^t)\big]\bigg]_{t=0} \\ &=& \frac{1}{(1-q)^2}\big[ (1-q)\cdot q+ q^2\big]\\ &=& \frac{q}{(1-q)^2}\\ &=&\frac{q}{p^2}. \end{eqnarray*}$$
## Characteristics function of Geometric Distribution
The characteristics function of geometric distribution is $\phi_X(t)=p(1-qe^{it})^{-1}$.
#### Proof
The characteristics function of geometric distribution is
$$\begin{eqnarray*} \phi_X(t) &=& E(e^{itX})\\ &=& \sum_{x=0}^\infty e^{itx} P(X=x) \\ &=& \sum_{x=0}^\infty e^{itx} q^x p \\ &=& p\sum_{x=0}^\infty (qe^{it})^x\\ &=& p(1-qe^{it})^{-1} \qquad \bigg(\text{ \because \sum_{x=0}^\infty q^x = (1-q)^{-1}}\bigg). \end{eqnarray*}$$
## Recurrence Relation for probabilities
The recurrence relation to calculate probabilities of geometric distribution is
$$\begin{equation*} P(X=x+1) = q\cdot P(X=x). \end{equation*}$$
#### Proof
We have, $P(X=x+1) = pq^{x+1}$ and $P(X=x) = pq^x$. Hence,
$$\begin{equation*} \frac{P(X=x+1)}{P(X=x)} = \frac{pq^{x+1}}{pq^x} = q \end{equation*}$$
$$\begin{equation*} \therefore P(X=x+1) = q\cdot P(X=x),\; x=0,1,2,\cdots. \end{equation*}$$
which is the recurrence relation for probability of geometric
distribution.
## Probability generating function of Geometric Distribution
The probability generating function of geometric distribution is
$P_X(t)=p(1-qt)^{-1}$.
#### Proof
The probability generating function is
$$\begin{eqnarray*} P_X(t) &=& E(t^{X})\\ &=& \sum_{x=0}^\infty t^{x} P(X=x) \\ &=& \sum_{x=0}^\infty t^{x} q^x p\\ &=& p\sum_{x=0}^\infty (qt)^x\\ &=& p(1-qt)^{-1} \qquad \bigg(\text{ \because \sum_{x=0}^\infty q^x = (1-q)^{-1}}\bigg). \end{eqnarray*}$$
Now,
$$\begin{eqnarray*} P_X(t) &=& p(1-qt)^{-1} \\ &=& p\sum_{x=0}^\infty q^xt^x \\ &=& (p+pqt+pq^2t^2+\cdots+pq^xt^x+\cdots) \end{eqnarray*}$$
Hence, the probability mass function of $X$ is
$$\begin{eqnarray*} P(X=x) & = & \text{coefficient of t^x in the expansion of P_X(t)}\\ & = & pq^x, \; x=0,1,2,\cdots,\; 0 < p,q < 1,\; p+q=1. \end{eqnarray*}$$
## Lack of memory property of geometric distribution
Geometric distribution is the only discrete distribution that possesses the lack of memory property.
Suppose a system can fail only at a discrete points of time $0,1,2,\cdots$. Let the life time of a system be denoted by $X$ with the support {$0,1,2,\cdots$}.
Then the conditional probability that a system of age $m$ will survive at least $n$ additional unit of time is the probability that it will survive more than $n$ unit of time. That is,
$$P(X\geq m+n | X\geq m) = P(X\geq n)$$
#### Proof
Let $X\sim Geo(p)$. The p.m.f. of $X$ is
$$\begin{equation*} P(X=x) = pq^x, x=0,1,2,\cdots; 0\leq p\leq 1, q=1-p. \end{equation*}$$
$$\begin{eqnarray*} P(X\geq m+n) &=& pq^{m+n} + p q^{m+n+1} + pq^{m+n+2} +\cdots \\ &=& pq^{m+n}(1 + q + q^2 + \cdots)\\ & = & pq^{m+n} (1-q)^{-1}\\ & = & pq^{m+n} p^{-1} \\ &=& q^{m+n}. \end{eqnarray*}$$
$$\begin{eqnarray*} P(X\geq m) &=& pq^{m} + p q^{m+1} + pq^{m+2} +\cdots \\ &=& pq^{m}(1 + q + q^2 + \cdots)\\ & = & pq^{m} (1-q)^{-1}\\ & = & pq^{m} p^{-1} \\ &=& q^{m}. \end{eqnarray*}$$
$$\begin{eqnarray*} P(X\geq m+n |X\geq m) &=& \dfrac{P(X\geq m + n \cap X\geq m)}{P(X\geq m)}\\ &=& \dfrac{P(X\geq m+n \cap X\geq m)}{P(X\geq m)}\\ &=& \dfrac{P(X\geq m+n)}{P(X \geq m)} \\ &=& \dfrac{q^{m+n}}{q^m}\\ &=& q^n = P(X\geq n). \end{eqnarray*}$$
## Result
Let two independent random variables $X_1$ and $X_2$ have same geometric distribution. Show that the conditional distribution of $X_1/(X_1+X_2)$ is uniform.
#### Proof
Given that $X_1$ and $X_2$ are independent random variable with same geometric distribution. The p.m.f. of $X$ is
\begin{align*} P(X=x) &= \begin{cases} q^x p, & x=0,1,2,\ldots; \\ & 0 < p, q < 1 , p+q=1 \\ 0, & Otherwise. \end{cases} \end{align*}
Now,
$$\begin{eqnarray*} P(X_1 + X_2 = n) &=& \sum_{x=0}^n P(X_1 = x) \cdot P(X_2=n-x)\\ & & \qquad (\because \text{ X_1 and X_2 are independent)} \\ &=& \sum_{x=0}^n q^x p q^{n-x} p = p^2q^n (n+1). \end{eqnarray*}$$
The conditional distribution of $X_1 / (X_1+X_2)$ is
$$\begin{eqnarray*} P(X_1 = x| X_1+X_2= n) &=& \frac{P(X_1 = x, X_1+X_2 =n)}{ P(X_1 + X_2 = n) } \\ &=& \frac{P(X_1 = x, X_2=n-x)}{ P(X_1 + X_2 = n) } \\ &=& \frac{P(X_1 = x)\cdot P(X_2=n-x)}{P(X_1 + X_2 = n)}\\ &=&\frac{q^xp \cdot q^{n-x}p}{p^2q^n (n+1)}\\ &=& \frac{1}{n+1}, \; x=0,1,2, \cdots, n. \end{eqnarray*}$$
which is the p.m.f. of uniform distribution.
## Conclusion
In this tutorial, you learned about theory of geometric distribution like the probability mass function, mean, variance, moment generating function and other properties of geometric distribution.
To read more about the step by step examples and calculator for geometric distribution refer the link Geometric Distribution Calculator with Examples. This tutorial will help you to understand how to calculate mean, variance of geometric distribution and you will learn how to calculate probabilities and cumulative probabilities for geometric distribution with the help of step by step examples.
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# Thread: Find polynomial if n is odd
1. ## Find polynomial if n is odd
Let p(x) be a polynomial of degree n and p(k) = $\frac{k}{k+1}, k = 0,1,2,3,..., n.$ Find p(n+1) if n is odd.
2. ## Re: Find polynomial if n is odd
We claim the degree n polynomial
$p(x) = \sum_{i=1}^n (-1)^{i+1} \frac{1}{i+1} \binom{x}{i}$
where
$\binom{x}{i} = \frac{x(x-1)(x-2) \cdots (x-i+1)}{i!}$
(a generalized binomial coefficient) satisfies
$p(k) = \frac{k}{k+1}$ for $k = 0, 1, 2, \dots n$
and that
$p(n+1) = 1$ for n odd.
The key to these claims is the identity
(*) ... $\sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = \frac{k}{k+1}$
$(1+k)^k = \sum_{i=0}^k \binom{k}{i} x^i$
Integrate both sides from 0 to x and then set x=-1, with the result
$\frac{-1}{k+1} = \sum_{i=0}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = -1 + \sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i}$
so
$\sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1}\binom{k}{i} = 1 - \frac{1}{k+1} = \frac{k}{k+1}$
which shows that $p(k) = \frac{k}{k+1}$ for $k = 0, 1, 2, \dots n$, as claimed.
To show that p(n+1) = 1, start with (*) for k = n+1, i.e.
$\sum_{i=1}^{n+1} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2}$
so
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} + (-1)^{n+2} \frac{1}{n+2}= \frac{n+1}{n+2}$
so for n odd
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} - \frac{1}{n+2}= \frac{n+1}{n+2}$
hence
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2} + \frac{1}{n+2} = 1$
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# Use Concepts of Algebra
In this worksheet, students will use algebraic concepts using mathematical language.
Key stage: KS 3
Curriculum topic: Algebra
Curriculum subtopic: Use and Interpret Algebraic Notation
Difficulty level:
#### Worksheet Overview
We have looked at the command words in algebra in a different activity.
Let's use these along with algebraic language in this activity.
Recap
8n is a term
8n + 4 is an expression
8n + 4 = 20 is an equation
Let's try a question!
Example
Gracie is x years old.
Her brother Max is 3 years older.
Write an expression for their total ages.
We know Gracie's age is x.
If Max is 3 years older we can write his age as x + 3 (that means Gracie's age + 3 yrs).
Their total ages are:
x + x + 3 which we can simplify by collecting like terms:
= 2x + 3 (this is an expression).
Part 2
If we knew that the combined ages of Gracie and Max is 25 years, we could make an equation with the expression we have:
2x + 3 = 25
We could then solve this to find Gracie's age (x).
2x = 22
x = 11
Therefore, Gracie is 11 years old and Max is 3 years older, so he is 11 + 3 = 14 years old.
Let's try some of these!
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# Polynomial Long Division: Definition, Method, Long Division With Monomials, Binomials
7 minutes long
Posted by Osman Gezer, 2/20/24
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Got stuck on homework? Get your step-by-step solutions from real tutors in minutes! 24/7. Unlimited.
Polynomial long division is an essential method in algebra for dividing one polynomial by another. It allows us to simplify complex expressions, find factors, and solve equations involving polynomials. In this article, we will explore the definition, methods, and steps involved in polynomial long division. We will also discuss long division with monomials and binomials, as well as provide solved examples to illustrate the process. So, let’s dive into the world of polynomial long division!
## An Introduction to Polynomial Long Division
Polynomials are algebraic expressions that consist of variables and coefficients combined using addition, subtraction, multiplication, and exponentiation. A polynomial can have one or more terms, and each term may contain multiple variables raised to different powers. Polynomial long division is a method used to divide a polynomial by another polynomial of lower degree.
## What is a Polynomial?
Before we delve into polynomial long division, let’s first understand what a polynomial is. A polynomial is an algebraic expression that includes real numbers, variables, and coefficients. It is composed of various terms, and each term consists of a coefficient multiplied by one or more variables raised to specific powers.
There are different types of polynomials based on the number of terms they have. A monomial is a polynomial with only one term, a binomial has two terms, and a trinomial has three terms.
## What is Polynomial Long Division?
Polynomial long division is an algorithm used to divide one polynomial by another polynomial. It involves the division of the polynomial into smaller parts, resulting in a quotient and a remainder. The process is similar to long division of numbers, where we divide, multiply, subtract, and bring down terms until we reach a remainder of lower degree than the divisor.
The long division process is based on the fact that every polynomial can be expressed as the product of the divisor and the quotient added to the remainder. By dividing polynomials, we can simplify expressions, find factors, and solve equations.
## Polynomial Long Division Methods
There are different methods for performing polynomial long division, depending on the specific scenario. Let’s explore some of the common methods:
### Splitting the Terms Method
In the splitting the terms method, we divide the polynomial by grouping and splitting the terms. This method is useful when the dividend and divisor have similar terms. We divide each term individually and simplify the expression to find the quotient and remainder.
### Factorization Method
The factorization method involves factoring the polynomial to simplify the division process. By factoring the dividend and divisor, we can cancel out common factors and simplify the expression. This method is particularly useful when the divisor is a factor of the dividend.
### Long Division of Polynomial by Missing Terms
In some cases, there may be missing terms in the polynomial expression. When performing long division in such scenarios, we can either leave a gap while dividing or consider the missing terms as having a coefficient of zero. This allows us to divide the polynomial accurately and obtain the quotient and remainder.
### Long Division of Polynomials by Monomials
When dividing a polynomial by a monomial, we divide each term of the polynomial individually by the monomial. This method involves dividing the coefficients and subtracting the exponents of the variables. By simplifying the division process term by term, we can find the quotient and remainder.
### Long Division of Polynomials by Binomials
Long division of polynomials by binomials is similar to long division of numbers. When there are no common factors between the dividend and divisor, or if we cannot identify the factors, we use this method. By following the long division algorithm step by step, we can divide the polynomial by the binomial and obtain the quotient and remainder.
### Long Division of Polynomials with Remainders
In polynomial long division, it is possible to have a remainder. The remainder is the term left over after dividing the dividend by the divisor. If the remainder is zero, it means that the divisor divides evenly into the dividend. If the remainder is nonzero, it indicates that the divisor is not a factor of the dividend.
## Steps For Long Division of Polynomials
The long division of polynomials follows a systematic step-by-step process. Let’s outline the general steps involved:
1. Arrange the terms of the dividend and divisor in descending order of their degrees.
2. Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
3. Multiply the divisor by the quotient term to obtain the product.
4. Subtract the product from the dividend, and bring down the next term if any.
5. Repeat steps 2-4 until you reach a remainder of lower degree than the divisor.
6. Express the remainder as a fraction using the divisor as the denominator, if necessary.
By following these steps, we can perform polynomial long division and obtain the quotient and remainder.
## Long Division Algorithm of Polynomials
The long division algorithm of polynomials provides a systematic approach to dividing polynomials. It ensures that the division is performed correctly and yields accurate results. The algorithm involves arranging the terms, dividing, multiplying, subtracting, and bringing down terms until the division process is complete.
The long division algorithm is based on the concept of dividing the polynomial into smaller parts and simplifying the expression step by step. It allows us to find the quotient and remainder, which is useful for various mathematical computations involving polynomials.
## Types of Polynomials
Polynomials can be classified into different types based on the number of terms they have. Let’s explore some common types of polynomials:
### Monomial
A monomial is a polynomial with only one term. It consists of a coefficient multiplied by one or more variables raised to specific powers. Monomials are the simplest form of polynomials and can be constants or products of constants and variables.
### Binomials
A binomial is a polynomial with two terms. It consists of two monomials separated by an addition or subtraction operator. Binomials can be expressed in the form of “ax + b” or “a – bx,” where “a” and “b” are coefficients and “x” is a variable.
### Trinomials
A trinomial is a polynomial with three terms. It consists of three monomials separated by addition or subtraction operators. Trinomials can be expressed in the form of “ax² + bx + c,” where “a,” “b,” and “c” are coefficients and “x” is a variable.
Understanding the types of polynomials helps us identify their characteristics and perform operations such as addition, subtraction, multiplication, and division more effectively.
## Solved Examples on Polynomial Long Division
To illustrate the process of polynomial long division, let’s work through some solved examples:
Example 1: Divide the polynomial 3x^3 – 5x^2 + 2x + 1 by the polynomial x – 1.
Solution: We start by dividing the highest degree term 3x^3 by x to obtain 3x^2. Then, we multiply (x – 1) by 3x^2 to get 3x^3 – 3x^2, which is subtracted from the original polynomial. We continue this process, dividing the highest degree term of the new dividend by the highest degree term of the divisor at each step. Finally, we obtain the quotient 3x^2 + 2x + 3 and the remainder 4.
Example 2: Divide the polynomial 4x^3 – 10x^2 + 5x by the monomial 2x.
Solution: In this example, we divide each term of the polynomial individually by 2x. Dividing 4x^3 by 2x gives us 2x^2. Dividing -10x^2 by 2x gives us -5x. Dividing 5x by 2x gives us 2.5. So, the quotient is 2x^2 – 5x + 2.5.
These examples demonstrate how polynomial long division is performed step by step to obtain the quotient and remainder.
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# Properties Of Integers
There are a few properties of integers which determine its operations. These principles or properties help us to solve many equations. To recall, integers are any positive or negative numbers, including zero. Properties of these integers will help to simplify and answer a series of operations on integers quickly.
All properties and identities for addition, subtraction, multiplication and division of numbers are also applicable to all the integers. Integers include the set of positive numbers, zero and negative numbers which are denoted with the letter Z.
Z = {……….−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5,………}
## Properties of Integers
Integers have 5 main properties of operation which are:
• Closure Property
• Associative Property
• Commutative Property
• Distributive Property
• Identity Property
Integer Property Addition Multiplication Subtraction Division
Commutative Property x + y = y+ x x × y = y × x x – y ≠ y – x x ÷ y ≠ y ÷ x
Associative Property x + (y + z) = (x + y) +z x × (y × z) = (x × y) × z (x – y) – z ≠ x – (y – z) (x ÷ y) ÷ z ≠ x ÷ (y ÷ z)
Identity Property x + 0 = x =0 + x x × 1 = x = 1 × x x – 0 = x ≠ 0 – x x ÷ 1 = x ≠ 1 ÷ x
Closure Property x + y ∈ Z x × y ∈ Z x – y ∈ Z x ÷ y ∉ Z
Distributive Property x × (y + z) = x × y + x × z
x × (y − z) = x × y − x × z
The explanation of each of the integer properties is given below.
### Property 1: Closure Property
Among the various properties of integers, closure property under addition and subtraction states that the sum or difference of any two integers will always be an integer i.e. if x and y are any two integers, x + y and x − y will also be an integer.
Example 1: 3 – 4 = 3 + (−4) = −1;
(–5) + 8 = 3,
The results are integers.
Closure property under multiplication states that the product of any two integers will be an integer i.e. if x and y are any two integers, xy will also be an integer.
Example 2: 6 × 9 = 54 ; (–5) × (3) = −15, which are integers.
Division of integers doesn’t follow the closure property, i.e. the quotient of any two integers x and y, may or may not be an integer.
Example 3: (−3) ÷ (−6) = ½, is not an integer.
### Property 2: Commutative Property
The commutative property of addition and multiplication states that the order of terms doesn’t matter, the result will be the same. Whether it is addition or multiplication, swapping of terms will not change the sum or product. Suppose, x and y are any two integers, then
⇒ x + y = y + x
⇒ x × y = y × x
Example 4: 4 + (−6) = −2 = (−6) + 4;
10 × (−3) = −30 = (−3) × 10
But, subtraction (x − y ≠ y − x) and division (x ÷ y ≠ y ÷ x) are not commutative for integers and whole numbers.
Example 5: 4 − (−6) = 10 ; (−6) – 4 = −10
⇒ 4 − (−6) ≠ (−6) – 4
Ex: 10 ÷ 2 = 5 ; 2 ÷ 10 = 1/5
⇒ 10 ÷ 2 ≠ 2 ÷ 10
### Property 3: Associative Property
The associative property of addition and multiplication states that the way of grouping of numbers doesn’t matter; the result will be the same. One can group numbers in any way but the answer will remain the same. Parenthesis can be done, irrespective of the order of terms. Let x, y and z be any three integers, then
⇒ x + (y + z) = (x + y) +z
⇒ x × (y × z) = (x × y) × z
Example 6: 1 + (2 + (-3)) = 0 = (1 + 2) + (−3);
1 × (2 × (−3)) =−6 = (1 × 2) × (−3)
Subtraction of integers is not associative in nature i.e. x − (y − z) ≠ (x − y) − z.
Example 7: 1 − (2 − (−3)) = −4; (1 – 2) – (−3) = −2
1 – (2 – (−3)) ≠ (1 − 2) − (−3)
### Property 4: Distributive Property
The distributive property explains the distributing ability of operation over another mathematical operation within a bracket. It can be either distributive property of multiplication over addition or distributive property of multiplication over subtraction. Here, integers are added or subtracted first and then multiplied or multiply first with each number within the bracket and then added or subtracted. This can be represented for any integers x, y and z as:
⇒ x × (y + z) = x × y + x × z
⇒ x × (y − z) = x × y − x × z
Example 8: −5 (2 + 1) = −15 = (−5 × 2) + (−5 × 1)
### Property 5: Identity Property
Among the various properties of integers, additive identity property states that when any integer is added to zero it will give the same number. Zero is called additive identity. For any integer x,
x + 0 = x = 0 + x
The multiplicative identity property for integers says that whenever a number is multiplied by 1 it will give the integer itself as the product. Therefore, 1 is called the multiplicative identity for a number. For any integer x,
x × 1 = x = 1 × x
If any integer multiplied by 0, the product will be zero:
x × 0 = 0 =0 × x
If any integer multiplied by -1, the product will be opposite of the number:
x × (−1) = −x = (−1) × x
### What are the Properties of Integers?
Integers have 5 main properties of operation which are as follows:
• Closure Property
• Associative Property
• Commutative Property
• Distributive Property
• Identity Property
### What is the Difference Between Commutative and Associative Properties of Integers?
In commutative property, the integers can be rearranged in any way and the result will still be the same. In case of associative property, integers can be grouped in any way using parenthesis and the result will still be the same.
• Commutative Property: a + b = b + a
• Associative Property: (a + b) + c = a + (b + c)
### What are the 4 Integer Operations?
The four integer operations are:
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Lesson Objectives
• Learn how to find the common difference of an arithmetic sequence
• Learn how to find a specific term and formula for an arithmetic sequence
• Learn how to evaluate an arithmetic series
## What is an Arithmetic Sequence?
In this lesson, will learn about arithmetic sequences and series. A sequence in which each term after the first is obtained by adding some fixed number to the previous term is known as an arithmetic sequence or an arithmetic progression. The fixed number is known as the common difference and is usually expressed with a lowercase d.
### Finding the Common Difference
To find the common difference for an arithmetic sequence, we can use a simple formula: $$d=a_{n + 1}- a_{n}$$ Let's look at an example.
Example #1: Find the common difference. $$29, 39, 49, 59,...$$ We can choose any two numbers that are next to each other. The one on the right has the higher index value, it will serve as the an + 1, while the one on the left will serve as the a1. Let's choose a1 and a2, which gives us 29 and 39. We plug into our formula: $$d=a_{n + 1}- a_{n}$$ $$d=a_2 - a_1$$ $$d=39 - 29=10$$ Our common difference is 10.
### Finding the nth Term of an Arithmetic Sequence
In some cases, we will be asked to find the nth term of an arithmetic sequence and give the general formula. To accomplish this task, we use the following formula: $$a_{n}=a_{1}+ (n - 1)d$$ Let's look at an example.
Example #2: Find a22 and an. $$28, 31, 34, 37$$ $$d=31 - 28=3$$ Now, let's plug into our formula: $$a_{n}=a_{1}+ d(n - 1)$$ an is what we want to find, here this is a22: $$a_{22}=28 + 3(22 - 1)$$ $$a_{22}=28 + 3 \cdot 21$$ $$a_{22}=28 + 63$$ $$a_{22}=91$$ How do we find the formula for the general term an? We just plug in for a1 and d: $$a_{n}=28 + 3(n - 1)$$ $$a_{n}=28 + 3n - 3$$ $$a_{n}=25 + 3n$$
### Sum of the First n Terms of an Arithmetic Sequence
Recall that a series is sum of the terms of a sequence. When we sum the terms of an arithmetic sequence, this is known as an arithmetic series. We have a very useful formula that allows us to find the sum of the first n terms of an arithmetic sequence. $$S_{n}=\frac{n}{2}(a_{1}+ a_{n})$$ $$S_{n}=\frac{n}{2}[2a_{1}+ (n - 1)d]$$ The first formula will be used when the first and last terms are known, otherwise the second formula is used. Let's look at an example.
Example #3: Evaluate each arithmetic series. $$a_{1}=2, d=7, n=40$$ Let's use the second formula since we don't know the last term. $$S_{n}=\frac{n}{2}[2a_{1}+ (n - 1)d]$$ $$S_{40}=\frac{40}{2}[2(2) + 7(40 - 1)]$$ $$S_{40}=20[4 + 7(39)]$$ $$S_{40}=20(277)$$ $$S_{40}=5540$$
#### Skills Check:
Example #1
Find the common difference $$2, 0, -2, -4$$
Please choose the best answer.
A
$$d=-3$$
B
$$d=-2$$
C
$$d=2$$
D
$$d=\frac{1}{2}$$
E
$$d=4$$
Example #2
Find a20 and an. $$23, 27, 31, 35,...$$
Please choose the best answer.
A
$$a_{20}=94, a_{n}=19 + 2n$$
B
$$a_{20}=100, a_{n}=2 - 5n$$
C
$$a_{20}=9, a_{n}=11 + 5n$$
D
$$a_{20}=99, a_{n}=19 + 4n$$
E
$$a_{20}=16, a_{n}=20 + 3n$$
Example #3
Evaluate each series $$a_{1}=13, d=3, n=12$$
Please choose the best answer.
A
$$177$$
B
$$200$$
C
$$151$$
D
$$354$$
E
$$552$$
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again?
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# Condition of Perpendicularity of Two Lines
We will learn how to find the condition of perpendicularity of two lines.
If two lines AB and CD of slopes m$$_{1}$$ and m$$_{2}$$ are perpendicular, then the angle between the lines θ is of 90°.
Therefore, cot θ = 0
⇒ $$\frac{1 + m_{1}m_{2}}{m_{2} - m_{1}}$$ = 0
⇒ 1 + m$$_{1}$$m$$_{2}$$ = 0
m$$_{1}$$m$$_{2}$$ = -1.
Thus when two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.
Let us assume that the lines y = m$$_{1}$$x + c$$_{1}$$ and y = m$$_{2}$$ x + c$$_{2}$$ make angles α and β respectively with the positive direction of the x-axis and θ be the angle between them.
Therefore, α = θ + β = 90° + β [Since, θ = 90°]
Now taking tan on both sides we get,
tan α = tan (θ + β)
tan α = - cot β
tan α = - $$\frac{1}{tan β}$$
or, m$$_{1}$$ = - $$\frac{1}{m_{1}}$$
or, m$$_{1}$$m$$_{2}$$ = -1
Therefore, the condition of perpendicularity of the lines y = m$$_{1}$$x + c$$_{1}$$, and y = m$$_{2}$$ x + c$$_{2}$$ is m$$_{1}$$m$$_{2}$$ = -1.
Conversely, if m$$_{1}$$m$$_{2}$$ = - 1 then
tan ∙ tan β = - 1
$$\frac{sin α sin β}{cos α cos β}$$ = -1
sin α sin β = - cos α cos β
cos α cos β + sin α sin β = 0
cos (α - β) = 0
Therefore, α - β = 90°
Therefore, θ = α - β = 90°
Thus, the straight lines AB and CD are perpendicular to each other.
Solved examples to find the condition of perpendicularity of two given straight lines:
1. Let P (6, 4) and Q (2, 12) be the two points. Find the slope of a line perpendicular to PQ.
Solution:
Let m be the slope of PQ.
Then m = $$\frac{12 - 4}{2 - 6}$$ = $$\frac{8}{-4}$$ = -2
Therefore the slope of the line perpendicular to PQ = - $$\frac{1}{m}$$ = ½
2. Without using the Pythagoras theorem, show that P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.
Solution:
In ∆ ABC, we have:
m$$_{1}$$ = Slope of the side PQ = $$\frac{4 - 5}{4 - 3}$$ = -1
m$$_{2}$$ = Slope of the side PR = $$\frac{4 - (-1)}{4 - (-1)}$$ = 1
Now clearly we see that m$$_{1}$$m$$_{2}$$ = 1 × -1 = -1
Therefore, the side PQ perpendicular to PR that is ∠RPQ = 90°.
Therefore, the given points P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.
3. Find the ortho-centre of the triangle formed by joining the points P (- 2, -3), Q (6, 1) and R (1, 6).
Solution:
The slope of the side QR of the ∆PQR is $$\frac{6 - 1}{1 - 6}$$ = $$\frac{5}{-5}$$ = -1∙
Let PS be the perpendicular from P on QR; hence, if the slope of the line PS be m then,
m × (- 1) = - 1
or, m = 1.
Therefore, the equation of the straight line PS is
y + 3 = 1 (x + 2)
or, x - y = 1 …………………(1)
Again, the slope of the side RP of the ∆ PQR is $$\frac{6 + 3}{1 + 2}$$ = 3∙
Let QT be the perpendicular from Q on RP; hence, if the slope of the line QT be m1 then,
m$$_{1}$$ × 3 = -1
or, m$$_{1}$$ = -$$\frac{1}{3}$$
Therefore, tile equation of the straight line QT is
y – 1 = -$$\frac{1}{3}$$(x - 6)
or, 3y – 3 = - x + 6
Or, x + 3y = 9 ………………(2)
Now, solving equations (1) and (2) we get, x = 3, y = 2.
Therefore, the co-ordinates of the point of intersection of the lines (1) and (2) are (3, 2).
Therefore, the co-ordinates of the ortho-centre of the ∆PQR = the co-ordinates of the point of intersection of the straight lines PS and QT = (3, 2).
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# 7.12 Central limit theorem: central limit theorem lab i
Page 1 / 1
Class Time:
Names:
## Student learning outcomes:
• The student will demonstrate and compare properties of the Central Limit Theorem.
This lab works best when sampling from several classes and combining data.
## Collect the data
1. Count the change in your pocket. (Do not include bills.)
2. Randomly survey 30 classmates. Record the values of the change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
3. Construct a histogram. Make 5 - 6 intervals. Sketch the graph using a ruler and pencil. Scale the axes.
4. Calculate the following ( $n=1$ ; surveying one person at a time):
• $\overline{x}$ =
• $s$ =
5. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Collecting averages of pairs
Repeat steps 1 - 5 (of the section above titled "Collect the Data") with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs.
1. Randomly survey 30 pairs of classmates. Record the values of the average of their change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collecting the Data". Sketch thegraph using a ruler and a pencil.
3. Calculate the following ( $n=2$ ; surveying two people at a time):
• $\overline{x}$ =
• $s$ =
4. Draw a smooth curve through tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Collecting averages of groups of five
Repeat steps 1 – 5 (of the section titled "Collect the Data") with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of 5.
1. Randomly survey 30 groups of 5 classmates. Record the values of the average of their change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collect the Data". Sketch the graph using a ruler and a pencil.
3. Calculate the following ( $n=5$ ; surveying five people at a time):
• $\overline{x}$ =
• $s$ =
4. Draw a smooth curve through tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Discussion questions
1. As $n$ changed, why did the shape of the distribution of the data change? Use 1 – 2 complete sentences to explain what happened.
2. In the section titled "Collect the Data", what was the approximate distribution of the data? $X$ ~
3. In the section titled "Collecting Averages of Groups of Five", what was the approximate distribution of the averages? $\overline{X}$ ~
4. In 1 – 2 complete sentences, explain any differences in your answers to the previous two questions.
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 7.8 Lab 1: central limit theorem (pocket change)
Page 1 / 1
Class Time:
Names:
## Student learning outcomes:
• The student will demonstrate and compare properties of the Central Limit Theorem.
This lab works best when sampling from several classes and combining data.
## Collect the data
1. Count the change in your pocket. (Do not include bills.)
2. Randomly survey 30 classmates. Record the values of the change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
3. Construct a histogram. Make 5 - 6 intervals. Sketch the graph using a ruler and pencil. Scale the axes.
4. Calculate the following ( $n=1$ ; surveying one person at a time):
• $\overline{x}$ =
• $s$ =
5. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Collecting averages of pairs
Repeat steps 1 - 5 (of the section above titled "Collect the Data") with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs.
1. Randomly survey 30 pairs of classmates. Record the values of the average of their change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collecting the Data". Sketch thegraph using a ruler and a pencil.
3. Calculate the following ( $n=2$ ; surveying two people at a time):
• $\overline{x}$ =
• $s$ =
4. Draw a smooth curve through tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Collecting averages of groups of five
Repeat steps 1 – 5 (of the section titled "Collect the Data") with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of 5.
1. Randomly survey 30 groups of 5 classmates. Record the values of the average of their change.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collect the Data". Sketch the graph using a ruler and a pencil.
3. Calculate the following ( $n=5$ ; surveying five people at a time):
• $\overline{x}$ =
• $s$ =
4. Draw a smooth curve through tops of the bars of the histogram. Use 1 – 2 complete sentences to describe the general shape of the curve.
## Discussion questions
1. As $n$ changed, why did the shape of the distribution of the data change? Use 1 – 2 complete sentences to explain what happened.
2. In the section titled "Collect the Data", what was the approximate distribution of the data? $X$ ~
3. In the section titled "Collecting Averages of Groups of Five", what was the approximate distribution of the averages? $\overline{X}$ ~
4. In 1 – 2 complete sentences, explain any differences in your answers to the previous two questions.
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
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# How do you integrate (x^3 - 2) / (x^4 - 1) using partial fractions?
Sep 19, 2016
$\frac{1}{4} \ln \left({x}^{2} + 1\right) + \arctan \left(x\right) + \frac{3}{4} \ln \left\mid x + 1 \right\mid - \frac{1}{4} \ln \left\mid x - 1 \right\mid + C$
#### Explanation:
We see that:
$\frac{{x}^{3} - 2}{{x}^{4} - 1} = \frac{{x}^{3} - 2}{\left({x}^{2} + 1\right) \left({x}^{2} - 1\right)} = \frac{{x}^{3} - 2}{\left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)}$
Split it up into its partial fractions:
$\frac{{x}^{3} - 2}{\left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)} = \frac{A x + B}{{x}^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$
Multiplying through, we see that:
${x}^{3} - 2 = \left(A x + B\right) \left(x + 1\right) \left(x - 1\right) + C \left({x}^{2} + 1\right) \left(x - 1\right) + D \left({x}^{2} + 1\right) \left(x + 1\right)$
Continue:
${x}^{3} - 2 = \left(A x + B\right) \left({x}^{2} - 1\right) + C \left({x}^{3} - {x}^{2} + x - 1\right) + D \left({x}^{3} + {x}^{2} + x + 1\right)$
${x}^{3} - 2 = A {x}^{3} - A x + B {x}^{2} - B + C {x}^{3} - C {x}^{2} + C x - C + D {x}^{3} + D {x}^{2} + D x + D$
${x}^{3} - 2 = {x}^{3} \left(A + C + D\right) + {x}^{2} \left(B - C + D\right) + x \left(- A + C + D\right) + \left(- B - C + D\right)$
Equating the coefficients on either side of the equation:
$\left\{\begin{matrix}1 = A + C + D \\ 0 = B - C + D \\ 0 = - A + C + D \\ - 2 = - B - C + D\end{matrix}\right.$
Adding the first and third equations, we see that $1 = 2 C + 2 D$. Adding the second and fourth equations, we see that $- 2 = - 2 C + 2 D$. Adding these two equations, this yields $- 1 = 4 D$, so $D = - \frac{1}{4}$.
Substituting $D = - \frac{1}{4}$ into $1 = 2 C + 2 D$, we see that $C = \frac{3}{4}$.
Substituting these values into $1 = A + C + D$, this yields that $A = \frac{1}{2}$.
Furthermore, since $0 = B - C + D$, we see that $B = 1$.
Thus:
$\frac{{x}^{3} - 2}{\left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)} = \frac{\frac{1}{2} x + 1}{{x}^{2} + 1} + \frac{\frac{3}{4}}{x + 1} - \frac{\frac{1}{4}}{x - 1}$
So:
$\int \frac{{x}^{3} - 2}{{x}^{4} - 1} \mathrm{dx} = \frac{1}{2} \int \frac{x + 2}{{x}^{2} + 1} \mathrm{dx} + \frac{3}{4} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{4} \int \frac{1}{x - 1} \mathrm{dx}$
Split up the first integral. The last two integrals can be integrated simply:
$= \frac{1}{2} \int \frac{x}{{x}^{2} + 1} \mathrm{dx} + \frac{1}{2} \int \frac{2}{{x}^{2} + 1} \mathrm{dx} + \frac{3}{4} \ln \left\mid x + 1 \right\mid - \frac{1}{4} \ln \left\mid x - 1 \right\mid$
Rearranging the first term to set up for a natural logarithm substitution, since $2 x$ is the derivative of ${x}^{2} + 1$:
$= \frac{1}{4} \int \frac{2 x}{{x}^{2} + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + 1} \mathrm{dx} + \frac{3}{4} \ln \left\mid x + 1 \right\mid - \frac{1}{4} \ln \left\mid x - 1 \right\mid$
The second integral is a common integral as well:
$= \frac{1}{4} \ln \left({x}^{2} + 1\right) + \arctan \left(x\right) + \frac{3}{4} \ln \left\mid x + 1 \right\mid - \frac{1}{4} \ln \left\mid x - 1 \right\mid + C$
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## Trigonometry 7th Edition
Published by Cengage Learning
# Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 32: 24
#### Answer
The point is $(-1,-1)$ $r = \sqrt{2}$ $\sin{225^{\circ}} = -\dfrac{\sqrt{2}}{2}$ $\cos{225}^{\circ} = -\dfrac{\sqrt{2}}{2}$ $\tan{225}^{\circ} = 1$
#### Work Step by Step
The terminal side of $225^{\circ}$ in standard position is represented by the blue line in the figure. It lies in the 3rd quadrant. The coordinates of points on the terminal side of $225^{\circ}$ can be given by $(-a,-a)$, where $a$ is a positive number. Choosing $a=1$ arbitrarily, the point is $(-1,-1)$. To find the distance from the origin to $(-1,-1)$, we use the distance formula $$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\r=\sqrt{(-1-0)^2+(-1-0)^2}=\sqrt{2}$$ $$\therefore r = \boxed{\sqrt{2}}$$ $\sin{225^{\circ}} = \dfrac{y}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}}$ $\cos{225}^{\circ} = \dfrac{x}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}}$ $\tan{225}^{\circ} = \dfrac{y}{x} = \dfrac{-1}{-1 } = \boxed{1}$
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Illuminations: Rise-Run Triangles
# Rise-Run Triangles
This lesson offers students a method for finding the slope of a line from its graph. The skills from this lesson can be applied as a tool to real-world examples of rate of change and slope.
### Learning Objectives
Students will be able to: Determine if the slope of a line is positive or negative Express the slope of a line as a fraction
### Materials
Activity-Sheet: Counting for Slope Activity Sheet Colored pencils (optional)
### Instructional Plan
To start the lesson, ask students what they already know about slope. They may know terms such as rate of change and rise over run. Often, students have recollection of these terms but don't remember or understand what they mean or how they relate to slope.
Ask students what it means to have positive or negative slope. Encourage a student to come to the front of the room and draw a line with positive slope. Ask classmates if they agree that the line has positive slope, and then ask how they can tell.
A line with positive slope is pointing upward as you look to the right. You always want to see if the line is pointing upward or downward on the right side of the graph, just as we read to the right.
Sketch these two lines with positive slope for students to see.
Ask students to tell you all they can about the two graphs. What's the same? What's different? Emphasize that although both lines have positive slope, there is something different about the direction in which they point. Explain that this description of how slanted a line is can be described by a number called its slope.
Now, draw a third line that has the same slope as the first line, but a different y-intercept. Ask students again for comparisons.
Students should eventually recognize that the third line has the same slope as the first line. Once they do, they are ready to think about the slope number as a description of how slanted a line is.
Use the activity sheet for practice and enforcement. The activity sheet guides students through a process for finding the slope of a given line. Page 1 is meant to be completed as a class, so having an overhead slide of this page will be helpful.
Distribute the activity sheets and make sure each student has 1 or 2 colored pencils. Many students enjoy using a colored pencil to draw and shade the slope triangle, and doing so makes the lesson more memorable. You might ask students to use one color when they're drawing the triangle for a line with positive slope, and another color for triangles representing negative slope.
Shade in the slope triangles with students as shown below.
Encourage students to simplify their fractions on page 1 of the activity sheet. Point out that for each line, the simplified forms of the fractions are equivalent — no matter which two points on the line you student use, or how large the triangle is, you get the correct slope.
On page 2, students are given the slope triangle in the first 3 examples (the top row). In the next 3 examples (middle row), they are given only the points to use to draw the triangle. In the last 3 examples (bottom row), students have to find the points themselves before drawing the triangle and determining the slope. The idea here is to gradually get students comfortable with finding the slope.
While students work on page 2, be sure that they:
• Simplify all fractions
• Determine which lines have negative slope and use a negative fraction to represent the slope of these lines.
This exercise provides students with the skill of finding the slope of a line from a graph. This skill can be applied to less abstract examples using real data from a table or a graph.
### Questions for Students
Which, if any, of the fractions did you have to simplify when you found the slope of a line? How can you avoid the need to simplify a fraction? Suppose you have identified 3 slope triangles for a line to help you find the correct slope. What can you say about the relationship between these triangles? Explain the difference between a line with positive slope and a line with negative slope. Explain the difference between a line with slope 1/2 and a line with slope 2/1.
### Assessment Options
1. Provide students with data in table form and have them graph the points and find the slope of the line connecting the points. Ask students "What does the slope say about the trend of the data?"
2. Can you explain how miles per hour can be seen as a slope? Explain how heart beats per minute can be seen as a slope.
3. Provide a table of data for students to graph, such as the one offered below. Ask the students to graph the data and determine the speed of travel. Ask students, What are the units? How do the words miles per hour relate to what you see in the graph?
Time in Hours
(x)
Distance in Miles
(y)
1 55
2 85
4 145
8 265
### Extensions
Show students a given point on a coordinate grid and give them a value for the slope of a line. Ask them to draw a line that has that slope through the given point. How does the slope triangle method apply to horizontal and vertical lines? What does the slope-triangle look like for a vertical line? for a horizontal? What problems arise, and how do they affect the slope? How is slope formula consistent with the slope triangle method for finding the slope of a line? How does calculating y1 – y2 relate to the height of the slope triangle? How does calculating x1 – x2 relate to the length of the slope triangle?
### Teacher Reflection
Describe what effect using slope triangles had on student understanding of slope. What difficulties did students have with this activity? How could you extend this activity to rate-of-change problems in the curriculum? Describe the effect using colored pencils has on student comprehension of slope.
### NCTM Standards and Expectations
Algebra 6-8Explore relationships between symbolic expressions and graphs of lines, paying particular attention to the meaning of intercept and slope. Use graphs to analyze the nature of changes in quantities in linear relationships. Algebra 9-12Approximate and interpret rates of change from graphical and numerical data.
This lesson was created by Zoe Silver.
1 period
### NCTM Resources
Slope and Y-Intercept Tool
### Lessons
More and Better Mathematics for All Students
© 2000 National Council of Teachers of Mathematics Use of this Web site constitutes acceptance of the Terms of Use The National Council of Teachers of Mathematics is a public voice of mathematics education, providing vision, leadership, and professional development to support teachers in ensuring mathematics learning of the highest quality for all students. The views expressed or implied, unless otherwise noted, should not be interpreted as official positions of the Council.
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# The Pythagorean Theorem
Last Updated on September 11, 2024.
The Pythagorean Theorem states that in a right triangle, the total of the squares of each of the triangle legs lengths is identical to the square of the triangle’s hypotenuse length.
## Online GED Classes
### A simple and easy way of getting your GED diploma.
A long, long time ago, Pythagoras, a Greek mathematician, discovered a fascinating property of right triangles.
1. Determine the missing length in this right triangle using the Pythagorean theorem. Use this formula: $$a^{2} + b^{2} = c^{2}$$
$$b=12$$
$$c=15$$
$$a=?$$
A.
B.
C.
Question 1 of 2
2. Determine the missing length in this right triangle using the Pythagorean theorem. Use this formula: $$a^{2} + b^{2} = c^{2}$$
$$a=12$$
$$b=16$$
$$c=?$$
A.
B.
C.
Question 2 of 2
Next Lesson: Combining like terms
### Video Transcription
If you add the squares of the triangle’s legs lengths, the outcome is equal to the square of the triangle’s hypotenuse length.
This property that has so many applications in architecture, engineering, art, and science is what we call the “Pythagorean Theorem.”
So let’s see how this theorem is helping you to learn more about how triangles are constructed. And the good thing is that you even don’t need to speak the Greek language to apply or understand the discovery of Pythagoras.
## Online GED Classes
### Get a GED Diploma quickly. It doesn’t matter when you left school.
##### Easy Lessons | Practice Tests | Add-ons
The Pythagorean Theorem
The Greek mathematician Pythagoras was studying right triangles and in what way the hypotenuse and the legs of right triangles were related before he derived his theory.
The Pythagorean Theorem
In case a and b symbolize the lengths of a right triangle’s legs, and c symbolizes the hypotenuse’s length, then the sum of the squares of the leg’s lengths is the same as the square of the hypotenuse’s length.
This interesting relationship is represented in this formula: a² + b² = c²
In this box above, you could have noticed the expression “square,” but also the small “2” s to the top of the letters like in a² + b² = c². To square a number, we multiply that number by itself. If we square, for example, the number “5”, we multiply five times 5; If we square “12”, we multiply 12 times 12. A few common squared numbers are shown in the following table.
When we see the following equation: a² + b² = c², we can see this as follows: “the length of the side with “a” times itself (so a²), plus the length of the side with “b” times itself (so b²) equals the length of the side with “c” times itself so c²).”
Let’s take a look at the Pythagorean Theorem using an example of a right triangle.
The Pythagorean theorem holds true for the right triangle above. The sum of both legs’ length squares (a² plus b²) equals the square of the hypotenuse’s length (c²). This actually is holding true for every right triangle!
This Pythagorean Theorem may be represented in terms of “area” as well.
In all right triangles, areas of squares drawn from any hypotenuse equals the sum of areas of squares drawn from the triangles’ two legs. Just take a look at the following illustration in that same 3-4-5 right triangle.
Please note that this Pythagorean Theorem is ONLY working if we have right triangles.
So again, the Pythagorean Theorem states that in right triangles, the total of the squares of each of the triangle legs lengths is identical to the square of the triangle’s hypotenuse length.
|
# What's the probability of rolling a 9 or higher using two fair standard dice?
10/36=5/18=27.bar7%
#### Explanation:
The rolls you can get with 2 six-sided dice is in this table:
$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$
See that there are 10 numbers that are 9 or higher out of a total number of 36 possibilities, giving a probability of 10/36=5/18=27.bar7%
Dec 26, 2017
total no. of outcomes = 6×6
favourable outcomes require that sum should be $\ge 9$ .
favourable outcomes are
for $9 \left(4 , 5\right) , \left(5 , 4\right) , \left(6 , 3\right) , \left(3 , 6\right)$
for $10 \left(5 , 5\right) , \left(6 , 4\right) , \left(4 , 6\right)$
for $11 \left(5 , 6\right) , \left(6 , 5\right)$
for $12 \left(6 , 6\right)$;
probability =10/36=5/18~~27.7777777778%
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3.14
## UNSW Sydney
A parabola translated from the origin
# Scaling and translating quadratic functions
We can get more general quadratic functions by scaling and translating the standard equation $\normalsize{y=x^2}$ . Pleasantly, any quadratic function can be obtained in this way.
In this step we will see how scaling and translation affect the graph of a quadratic function.
## Scaling the standard parabola
We have already seen that the shape of $\normalsize{y=ax^2}$ depends on the value of the constant $\normalsize{a}$. We can consider this graph to be a scaling of the particular standard parabola $\normalsize{y=x^2}$; if $\normalsize{a>0}$ then the parabola opens upwards, and if $% $ then the parabola opens downwards.
For large values of $\normalsize{a}$ the parabola is steep and narrow. The closer to $\normalsize{0}$ the value $\normalsize{a}$ gets, the flatter the parabola. Indeed when $\normalsize{a=0}$ the parabola is so flat that we get the straight line conic $y=0$, which could be considered as a special case of the parabola.
Here for comparison are the graphs of $\normalsize{y=4x^2}$ (green), $\normalsize{y=x^2}$ (blue) and $\normalsize{y=\frac{x^2}{4}}$ (red).
## Translating a standard parabola up or down
Now we investigate what happens if we translate a standard parabola $\normalsize{y=ax^2}$ up or down. This is quite easy. Note that however we translate a parabola, its vertex, and focus, and directrix will move in exactly the same way.
Q1 (E): Where is the vertex of the parabola obtained by translating $\normalsize{y=4x^2}$ by $\normalsize{3}$ in the $\normalsize{x}$ direction and by $\normalsize{5}$ in the $\normalsize{y}$ direction?
Geometrically, if we add a constant $\normalsize{k}$ to the equation $\normalsize{y=ax^2}$ then we translate the parabola by $\normalsize{k}$ in the vertical direction. The algebraic operation that shifts a function vertically by $k$ can be thought of as replacing $y$ with $y-k$.
Recall that the parabola $\normalsize y=\frac{1}{4} x^2$ has vertex $\normalsize V=[0,0]$, focus $\normalsize F=[0,1]$, and directrix $\normalsize l: y=-1$. Here is a graph of the parabola $\normalsize{y=\frac{1}{4}x^2-3}$, obtained by translating $\normalsize{y=\frac{1}{4}x^2}$ down by $\normalsize{3}$. So its vertex will be at $\normalsize{[0,-3]}$, its focus will be at $\normalsize{[0,-2]}$, and its directrix will be the line $\normalsize{ y=-4}$.
## Translating a standard parabola left or right
What if we want to shift $\normalsize{y=\frac{1}{4}x^2}$ over $\normalsize{5}$ units to the right? We apply the same strategy as when we were translating lines: replace $x$ with $x-5$ to get
It is easy to check the old vertex $\normalsize{[0,0]}$ has moved to $\normalsize{[5,0]}$.
We could expand the equation to get
## Combining translations horizontal and vertical
If we take the standard parabola $\normalsize{y=ax^2}$ and translate it by $\normalsize{h}$ in the $\normalsize{x}$ direction and by $\normalsize{k}$ in the $\normalsize{y}$ direction, then we obtain
This is the general form of a standard parabola which has been translated by the vector, or directed line segment, $\normalsize{(h,k)}$. This is an important expression in the theory of quadratic functions.
Q2 (M): What is the vertex of this parabola?
Q3 (M): What are the focus and directrix of $\normalsize{y=1/4(x-5)^2-2}$?
Q4 (C): What is the focus of the parabola $\normalsize{y=ax^2}$?
## A key fact about parabolas
Can we get every parabola this way – just by taking a standard parabola of the form $\normalsize{y=ax^2}$, and then shifting or translating in the $\normalsize{x}$-direction by a certain amount and then in the $\normalsize{y}$-direction by a certain amount? Yes we can!
This is a pleasant and important fact about parabolas. It shows us that all quadratic functions, even ones with complicated formulas like $\normalsize{y=7x^2-11x+8}$ have essentially similar shapes, and that we can find out where they are situated by unravelling how much $\normalsize{x}$ and $\normalsize{y}$ translations are required to obtain them from standard parabolas.
So there is a fundamental question here: how can we translate a standard parabola to get the general parabola $\normalsize{y=ax^2+bx+c}$? Finding the answer will take us to ancient Persia, to a technique called completing the square, and to a remarkable identity that all students of mathematics ought to have seen.
A1. If we translate a parabola, then its vertex translates correspondingly. Since the vertex of $\normalsize{y=4x^2}$ is the origin $\normalsize{[0,0]}$, the vertex of the translated parabola must be $\normalsize{[3,5]}$.
A2. The unshifted parabola has a vertex at $\normalsize{[0,0]}$, so the new vertex will be at $\normalsize{[h,k]}$.
A3. The unshifted parabola $y=\frac14 x^2$ has focus at $[0,1]$ and directrix $y=-1$. We have essentially just moved the entire picture to the right by $5$ and down by $2$. So visually we can see the focus is now located at $[5,-1]$ and the directrix is now $y=-3$.
A4. The focus of the parabola $\normalsize y=ax^2$ is the point $\normalsize F=[0,\frac{1}{4a}]$.
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Students can Download Maths Chapter 4 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions
Additional Questions and Answers
Exercise 4.1
Question 1.
“The sum of any two angles of a triangle is always greater than the third angle”. Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°.
Here sum of two angles 45° + 45° = 90°.
Question 2.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be x, 2x, x.
Using the angle sum property, we have
x + 2x + x = 180°
4x = 180°
x = $$\frac{180^{\circ}}{4}$$
x = 45°
2x = 2 × 45° = 90°
Thus the three angles of the triangle are 45°, 90°, 45°.
Its two angles are equal. It is an isoscales triangle. Its one angle is 90°.
∴ It is a right angled triangle.
Question 3.
Find the values of the unknown x and y in the following figures
Solution:
(i) Since angles y and 120° form a linear pair.
y + 120° = 180°
y = 180° – 120°
y = 60°
Now using the angle sum property of a triangle, we have
x + y + 50° = 180°
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 110°= 70°
x = 70°
y = 60
(ii) Using the angle sum property of triangle, we have
50° + 60° + y = 180°
110° + y = 180°
y = 180° – 110°
y = 70°
Again x and y form a linear pair
∴ x + y = 180°
x + 70° = 180°
x = 180° – 70°= 110°
∴ x = 110°; y = 70°
Question 4.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the third angle be x.
Using the angle sum property of a triangle we have,
30° + 80° + x = 180°
x + 110° = 180°
x = 180° – 110° = 70°
Third angle = 70°.
Exercise 4.2
Question 1.
In an isoscleles ∆ABC, AB = AC. Show that angles opposite to the equal sides are equal.
Solution:
Given: ∆ABC in which $$\overline{A B}$$ = $$\overline{A C}$$.
To Prove: ∠B = ∠K.
Construction: Draw AD ⊥ BC.
Proof: In right ∆ADB and right ∆ADC.
we have side AD = side AD (common)
AB = AC (Hypoteneous) (given)
∆ADB = ADC (RHS criterion]
∴ Their corresponding parts are equal. ∠B = ∠C.
Question 2.
ABC is an isosceles triangle having side $$\overline{A B}$$ = side $$\overline{A C}$$. If AD is perpendicular to BC, prove that D is the mid-point of $$\overline{B C}$$.
Solution:
In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC [∵ AD ⊥ BC]
Side $$\overline{A D}$$ = Side $$\overline{A D}$$ [Common]
Side $$\overline{A B}$$ = Side $$\overline{A C}$$ [Common]
Using RHS congruency, we get
∆ABD ≅ ∆ACDc
Their corresponding parts are equal
∴ BD = CD
∴ O is the mid point of BC.
Question 3.
In the figure PL ⊥ OB and PM ⊥ OA such that PL = PM.
Prove that ∆PLO ≅ ∆PMO.
Solution:
In ∆PLO and ∆PMO, we have
∠PLO = ∠PMO = 90° [Given]
$$\overline{O P}$$ = $$\overline{O P}$$ [Hypotenuse]
PL = PM
Using RHS congruency, we get
∆PLO ≅ ∆PMO
|
## Degree Measurement of Angles
One full revolution makes an angle of $360^{\circ}$, and the angle on a straight line is $180^{\circ}$. Therefore, one degree, $1^{\circ}$, can be defined as $\dfrac{1}{360}$ of one full revolution.
For greater accuracy we define one minute, $1’$, as $\dfrac{1}{60}$ of one degree and one second, $1^{\prime \prime}$, as $\dfrac{1}{60}$ of one minute.
One new area is the concept of a radian. We have already been familiar with measuring angles in degrees ($^{\circ}$) and will recall that there are $360^{\circ}$ in a full circle. An alternative unit for angle measurement is the radian.
An angle is said to have a measure of $1$ radian ($^c$) if it is subtended at the centre of a circle by an equal in length to the radius. The symbol ‘$c$’ is used for radian measure but is usually omitted. By contrast, the degree symbol is always used when the measure of an angle is given in degrees.
It can be seen that $1^c$ is slightly smaller than $60^{\circ}$.
$$1^c \approx 57.2958^{\circ}$$
$$\pi^c \equiv 180^{\circ}$$
To convert from degrees to radians, we multiply by $\dfrac{\pi}{180^{\circ}}$.
For example, $30^{\circ} \equiv 30^{\circ} \times \dfrac{\pi}{180^{\circ}} = \dfrac{\pi}{6}$
To convert from radians to degrees, we multiply by $\dfrac{180^{\circ}}{\pi}$.
For example, $\dfrac{\pi}{3} \equiv \dfrac{\pi}{3} \times \dfrac{180^{\circ}}{\pi} = 60^{\circ}$
$$\text{Degrees} \times \dfrac{\pi}{180^{\circ}} = \text{Radians} \\ \text{Radians} \times \dfrac{180^{\circ}}{\pi} = \text{Degrees}$$
### Example 1
Convert $45^{\circ}$ to radians in terms of $\pi$.
### Example 2
Convert $\dfrac{2 \pi}{3}$ radians to degrees.
### Example 3
Convert $67^{\circ}$ to radians, correcting to three significant figures.
### Example 4
Convert $1.2$ radians to degrees, correcting to three significant figures.
|
SEMATHS.ORG
Updated: 11 August 2021 03:17:00 PM
What is the maximum of a parabola?
Vertical parabolas give an important piece of information: When the parabola opens up, the vertex is the lowest point on the graph - called the minimum, or min. When the parabola opens down, the vertex is the highest point on the graph - called the maximum, or max.
From these considerations, how do you find the maximum of a parabola?
We can identify the minimum or maximum value of a parabola by identifying the y-coordinate of the vertex. You can use a graph to identify the vertex or you can find the minimum or maximum value algebraically by using the formula x = -b / 2a.
In view of this, what is the maximum of a parabola called?
One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value.
In addition how do you find the maximum of a quadratic function?
If you are unable to draw a graph, there are formulas you can use to find the maximum. If you are given the formula y = ax2 + bx + c, then you can find the maximum value using the formula max = c - (b2 / 4a). If you have the equation y = a(x-h)2 + k and the a term is negative, then the maximum value is k.
WHAT IS A in parabola?
The general form of a quadratic is "y = ax2 + bx + c". For graphing, the leading coefficient "a" indicates how "fat" or how "skinny" the parabola will be. Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point, where the parabola changes direction, is called the "vertex".
What is A and B in an ellipse?
Remember the patterns for an ellipse: (h, k) is the center point, a is the distance from the center to the end of the major axis, and b is the distance from the center to the end of the minor axis.
What is the equation of a hyperbola?
A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0) Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2 c 2 = a 2 + b 2 .
What is the maximum and minimum of a parabola called?
The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a parabola.
How do you tell if a parabola is up or down?
Let's look at a few key points about these patterns:
1. If the x is squared, the parabola is vertical (opens up or down). If the y is squared, it is horizontal (opens left or right).
2. If a is positive, the parabola opens up or to the right. If it is negative, it opens down or to the left.
3. The vertex is at (h, k).
What is parabolic equation?
Conic Form of Parabola Equation:
(x - h)2 = 4p(y - k) with the vertex at (h, k), the focus. at (h, k+p) and the directrix. y = k - p. Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same (p = ¼).
What is a parabola in real life?
, When liquid is rotated, the forces of gravity result in the liquid forming a parabola-like shape. The most common example is when you stir up orange juice in a glass by rotating it round its axis. The juice level rises round the edges while falling slightly in the center of the glass (the axis).
How do you solve a parabola problem?
We will learn how to solve different types of problems on parabola.
1. Find the vertex, focus, directrix, axis and latusrectum of the parabola y2 - 4x - 4y = 0.
2. Find the point on the parabola y2 = 12x at which the ordinate is double the abscissa.
3. Write the parametric equation of the parabola (x + 2)2 = - 4(y + 1).
How do you know if its a maximum or minimum?
To see whether it is a maximum or a minimum, in this case we can simply look at the graph. f(x) is a parabola, and we can see that the turning point is a minimum. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).
What does a parabola equation look like?
A quadratic function is a function that can be written in the form f(x)=ax2+bx+c where a,b, and c are real numbers and a≠0. The graph of the quadratic function is a U-shaped curve is called a parabola. The graph of the equation y=x2, shown below, is a parabola.
How do I turn on my notification bar?
The Notification Panel is a place to quickly access alerts, notifications and shortcuts. The Notification Panel is at the top of your mobile device's screen. It is hidden in the screen but can be accessed by swiping your finger from the top of the screen to the bottom. It is accessible from any menu or application.
How do I use notification bar?
Notifications on Android appear in the top bar on your phone. A simple swipe down from the notification bar will pull up the full screen notification drawer, where you can view and interact with your list of notifications.
What are the 3 dots called?
Those three little dots are called an ellipsis (plural: ellipses).
What is a hyperbola in math?
A hyperbola (plural "hyperbolas"; Gray 1997, p. 45) is a conic section defined as the locus of all points in the plane the difference of whose distances and from two fixed points (the foci and ) separated by a distance is a given positive constant , (1)
How do I fix my notification bar?
Solution I. Change your device User.
1. First, reboot your device in safe mode.
2. Once in Safe Mode, go to Android Settings.
3. Here look for the option called Users and switch to Guest Account.
4. Now again switch back to Owner account.
5. Reboot your device and come back to normal mode.
How do you calculate an ellipse?
The standard equation of an ellipse is (x^2/a^2)+(y^2/b^2)=1. If a=b, then we have (x^2/a^2)+(y^2/a^2)=1. Multiply both sides of the equation by a^2 to get x^2+y^2=a^2, which is the standard equation for a circle with a radius of a.
Is a parabola a curve?
In mathematics, a parabola is a plane curve which is mirror-symmetrical and is approximately U-shaped. The point where the parabola intersects its axis of symmetry is called the "vertex" and is the point where the parabola is most sharply curved.
How do you write an equation for a parabola?
To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form y=ax2+bx+c y = a x 2 + b x + c , x is the independent variable and y is the dependent variable. Choose some values for x and then determine the corresponding y-values. Then plot the points and sketch the graph.
What does 5 dots mean in texting?
These dots differ greatly from the previous tattoo – five dots represents time done in prison. Also known as the quincunx, the four dots on the outside represent four walls, with the fifth on the inside representing the prisoner. … The dots are typically found on an inmate's hand, between the thumb and forefinger.
Why is a parabola a strong shape?
The parabola is considered such a strong shape because of its natural oval shape. Both ends are mounted in a fixed bearing while the arch has a uniformly distributed load. The arch itself is evenly shaped overall. It has a natural bend and only uses normal forces, no shear force.
What are the 9 dots called in Google?
The 'Waffle' is the grid of nine small grey boxes that you find in the top right corner of your browser when you are using Gmail, Google Drive, Google Calendar, Google Keep etc.
What does tattoo with 3 triangular dots mean?
The triangular three dots tattoo generally stands for the concept of “mi vida loca”, Spanish for “my crazy life” and is typically associated with the gang community and lengthy prison sentences. The good news is that this meaning only applies when the dots are placed in a triangular fashion.
Is a hyperbola a function?
The hyperbola is not a function because it fails the vertical line test. Regardless of whether the hyperbola is a vertical or horizontal hyperbola
What is ellipse equation?
The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form. (x−h)2b2+(y−k)2a2=1 ( x − h ) 2 b 2 + ( y − k ) 2 a 2 = 1.
|
# Test: Mensuration- 4
## 15 Questions MCQ Test IBPS PO Prelims- Study Material, Online Tests, Previous Year | Test: Mensuration- 4
Description
Attempt Test: Mensuration- 4 | 15 questions in 15 minutes | Mock test for Quant preparation | Free important questions MCQ to study IBPS PO Prelims- Study Material, Online Tests, Previous Year for Quant Exam | Download free PDF with solutions
QUESTION: 1
### Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
Solution:
Explanation:
Length of the cuboid = 8 cm
Breadth of the cuboid = 6 cm
Height of the cuboid = 3.5 cm
Volume of the cuboid = length × breadth × height
= 8 x 6 x 3.5 = 168cm3
Therefore,volume of the cuboid = 168cm3
QUESTION: 2
### Find the area of the given figure.
Solution:
Explanation:
Area of a parallellogram = Base x Height = 4 x 2.5 = 10 cm2
QUESTION: 3
### Find the volume of a cuboid whose length is 8 cm, width is 3 cm and height is 5 cm.
Solution:
:Length of the cuboid = 8 cm
Width of the cuboid = 3 cm
Height of the cuboid = 5 cm
Volume of a cuboid = length × breadth × height
Therefore, Volume of the given cuboid = 8x3x5 = 120 cm3
QUESTION: 4
Find the perimeter of a rectangle whose length is 4 cm and breadth is 3 cm.
Solution:
Explanation:Length of the rectangle,l = 4cmBreadth of the rectangle,b = 3 cmPerimeter of a rectangle = 2(l+b) = 2(4+3) = 2x7 = 14 cm
QUESTION: 5
Area of a rhombus = ______
Solution:
Explanation:Area of a rhombus = Half the product of its diagonals = 1/2 x d1x d2
QUESTION: 6
Area of a trapezium = Half of the sum of the lengths of parallel sides × ______
Solution:
Explanation:Area of a trapezium =( Half of the sum of the length of parallel sides )×(perpendicular distance between them)
QUESTION: 7
Surface area of a cuboid = __________
Solution:
Explanation:Surface area of a cuboid = 2(lb + bh + hl)where, l= lengthb = breadthand h = height
QUESTION: 8
_________ of a solid is the sum of the areas of its faces.
Solution:
Explanation:The surface area of a three-dimensional figure is the sum of the areas of all its faces.
QUESTION: 9
In a right angled triangle, find the hypotenuse if base and perpendicular are respectively 36015 cmand 48020 cm.
Solution:
Let hypotenuse = x cm
Then, by Pythagoras theorem:
x2 = (48020)2 + (36015)2
x fi 60025 cm
QUESTION: 10
The inner circumference of a circular track is 440 cm. The track is 14 cm wide. Find the diameterof the outer circle of the track.
Solution:
Let inner radius = A; then 2pr = 440 \ p = 70
Radius of outer circle = 70 + 14 = 84 cm
Outer diameter = 2 × Radius = 2 × 84 = 168
QUESTION: 11
A pit 7.5 metre long, 6 metre wide and 1.5 metre deep is dug in a field. Find the volume of soilremoved in cubic metres.
Solution:
Volume of soil removed = l × b × h
= 7.5 × 6 × 1.5 = 67.5 m3
QUESTION: 12
The whole surface of a rectangular block is 8788 square cm. If length, breadth and height are inthe ratio of 4 : 3 : 2, find length.
Solution:
Let the common ratio be = x
Then, length = 4x, breadth = 3x and height = 2x
As per question;
2(4x x 3x + 3x x 2x + 2x x 4x) = 8788
2(12x2 + 6x2 + 8x2) = 8788 fi 52x2 = 8788
fi x = 13
Length = 4x = 52 cm
QUESTION: 13
Find the height of a Cuboid , if the volume and area of its base is 1240cm3 and 40 sq.cm respectively.
Solution:
QUESTION: 14
A bicycle wheel makes 5000 revolutions in moving 11 km. What is the radius of the wheel?
Solution:
Let the radius of the wheel be = p
Then 5000 × 2pr = 1100000 cm fi r = 35 cm
QUESTION: 15
The short and the long hands of a clock are 4 cm and 6 cm long respectively. What will be sum ofdistances travelled by their tips in 4 days? (Take p = 3.14)
Solution:
In 4 days, the short hand covers its circumference
4 × 2 = 8 times long hand covers its circumference
4 × 24 = 96 times
Then they will cover a total distance of:-
(2 × p × 4)8 + (2 × p × 6)96 fi 3818.24 cm
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|
# ML Aggarwal Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes
ML Aggarwal Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes are available here, and they contain solutions to the exercise given in the textbook. These solutions are explained in a step-by-step format in simple language by subject experts. We suggest students practise ML Aggarwal Solutions for Class 7 Maths Chapter 15 to obtain excellent marks in the final exam.
Chapter 15 – Visualising Solid Shapes discusses the Introduction to Plane Figures and Solid Figures, Faces, Edges and Vertices, Nets for Building 3D Shapes, Drawing Solids on a Flat Surface, Oblique Sketches, and Isometric Sketches Visualising Solid Objects.
## ML Aggarwal Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes
### Access answers to ML Aggarwal Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes
1. Match the following shapes with their names:
Solution:-
2. Identify the nets which can be folded to form a cube (cut out copies of the nets and try it)
(i)
Solution:-
The given net cannot be folded as a cube.
Because it can be folded as below,
(ii)
Solution:-
The given net can be folded as a cube.
So, it can be folded as below,
(iii)
Solution:-
The given net can be folded as a cube.
So, it can be folded as below,
(iv)
Solution:-
The given net can be folded as a cube.
So, it can be folded as below,
(v)
Solution:-
The given net cannot be folded as a cube.
Because it can be folded as below,
(vi)
Solution:-
The given net can be folded as a cube.
So, it can be folded as below,
3. Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
Solution:-
(i)
By observing the above figure, the sum of the opposite faces is equal to 7.
So, the given net can be folded into a net of dice.
(ii)
By observing the above figure, the sum of the opposite faces is equal to 7.
So, the given net can be folded into a net of dice.
4. Can any of the following be a net for a die? If no, explain your answer.
(i)
Solution:-
Yes
(ii)
Solution:-
The given net is folded as dice, as shown in the figure below
No, this cannot be a net for a die.
By observing the figure, we can say that one pair of the opposite face will have 1 and 4, another pair of the opposite face will have 3 and 6; the sum of these two opposite faces are not equal to 7.
5. Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (You may use a squared paper for easy manipulation).
Solution:-
In the given net, there are 3 faces, and it can be completed as shown below,
6. The dimensions of a cuboid are 5 cm, 3 cm, and 2 cm. Draw three different isometric sketches of this cuboid with height
(i) 2 cm
(ii) 3 cm
(iii) 5 cm
Solution:-
The three different isometric sketches of this cuboid are shown below,
(i) 2 cm
(ii) 3 cm
(iii) 5 cm
7. Give (i) an oblique sketch (ii) an isometric sketch for each of the following:
(a) A cube is with an edge 4 cm long.
(b) A cuboid of length 6 cm, breadth 4 cm, and height 3 cm.
Solution:-
(a)
(i) Oblique sketch
(ii) Isometric sketch
(b) A cuboid of length 6 cm, breadth 4 cm, and height 3 cm.
(i) Oblique sketch
(ii) Isometric sketch
|
# What is the antiderivative of ln(3x)?
Mar 7, 2018
$x \ln \left(3 x\right) - x + C$
#### Explanation:
We can do this in a number of ways.
First, let's get rid of that three using a log property:
$\ln \left(3 x\right) = \ln \left(3\right) + \ln \left(x\right)$
ln(3) is a constant, so its antiderivative will just be $x \ln \left(3\right)$ which is fine to deal with later.
For $\ln \left(x\right)$, we can write the following:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(x\right) \rightarrow \mathrm{dy} = \ln \left(x\right) \mathrm{dx}$
Let's use u substitution with $u = \ln \left(x\right)$, i.e. $\mathrm{du} = \frac{1}{x} \mathrm{dx} = {e}^{- u} \mathrm{dx}$. This yields
$\mathrm{dy} = u {e}^{u} \mathrm{du} \implies y = \int u {e}^{u} \mathrm{du}$
By integration by parts,
$y = u {e}^{u} - \int {e}^{u} \mathrm{du} = u {e}^{u} - {e}^{u} + C$
Putting back $x$,
$y = x \ln x - x + C$
This yields the final antiderivative as
$\int \setminus \ln \left(3 x\right) \mathrm{dx} = x \ln \left(3\right) + x \ln \left(x\right) - x + C = x \ln \left(3 x\right) - x + C$
|
# What is the difference between forces in opposite directions?
## What is the difference between forces in opposite directions?
When the forces acting on an object have equal strength and act in opposite directions, they are balanced. These forces cancel out one another, and the motion of the object they are acting on remains unchanged. When the forces acting on an object are unbalanced, they do not cancel out one another.
What happens when two forces act in the opposite direction?
If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement. equal to the sum of the two forces. The final force and its direction are called a resultant.
### When two forces are moving in opposite directions you should?
The direction of the net force determines the direction of the object’s motion. When two forces act in the same direction, they add together. When forces act in opposite directions, they are combined by subtracting the smaller force from the larger force.
Which two forces are always opposite in direction?
The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs – equal and opposite action-reaction force pairs.
## When two forces act on an object in opposite directions and the object moves it moves in the direction?
When two forces act in the same direction, they add together. When forces act in opposite directions, they are combined by subtracting the smaller force from the larger force. The direction of the resulting force (resultant) is the direction of the larger original force.
When two forces working in opposite directions are not the same strength?
unbalanced forces :When two forces working in opposite directions are not the same strength.
### When two forces are there is a change in position or motion?
When two forces are equal and act on an object in opposite directions, they balance each other out. When this happens, there is no change in an object’s motion. the same direction, the two forces are added together. Adding the forces together cause a change in an object’s motion, speed, or direction.
When you combine forces in two opposite directions one force is and the other force is?
The net force acting on an object is the combination of all of the individual forces acting on it. If two forces act on an object in opposite directions, the net force is the difference between the two forces.
## What are two forces that are equal in magnitude but opposite in direction?
Balanced forces are equal in magnitude but opposite in direction. Explanation: A balanced force is a situation when two forces acting on a body are equal in magnitude but opposite in direction.
What is the reason of the difference between the two forces?
If two forces act on an object in opposite directions, the net force is the difference between the two forces. If two forces act on an object in the same direction, the net force is the sum of the two forces. In this case, the net force is always greater than either of the individual forces.
### When two unequal forces act on an object it causes the object to move these forces are called?
When two forces acting on an object are not equal in size, we say that they are unbalanced forces. The overall force acting on the object is called the resultant force . a moving object changes speed and/or direction in the direction of the resultant force. …
What force is not moving?
Forces that are equal in size but opposite in direction are called balanced forces. Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move.
|
What is a 10 number line?
What is a 10 number line?
A number line that uses the ten frames to visually represent the value of each number 0 through 20. They are in rainbow colors and the rainbow colors repeat at 11 (red, orange, yellow, light green, green, aqua, blue, purple, plum and pink). They’re easy to see and put up above your alphabet line o.
What is number line with example?
The definition of a number line is a straight line with a “zero” point in the middle, with positive and negative numbers listed on either side of zero and going on indefinitely. An example of a number line is what a math student can use to find the answer to addition and subtraction questions. (math.)
How do you teach using a number line?
Basic Number Line Skills
1. Find the dot marked on a number (and name the dot)
2. Place a dot at the desired location.
3. Find the desired number below the number line’s tick mark*
5. Learn the numbers to the right are higher than numbers to left.
6. Determine which number is greater than, less than or equals.
How do you make a number line for kids?
Number Line Jumping Create a number line on the floor by writing the numbers 0 through 9 on cards or dessert-size paper plates. Use yarn or electric tape to create the number line, and place the number cards or plates about 18 to 24 inches apart on the number line. To add, have the child stand on the number 2 or 3.
How do you find a number line?
The numbers on the number line increase as one moves from left to right and decrease on moving from right to left. Here, fractions and decimals have been represented on the number line. A number line can also be used to represent both positive and negative integers as well.
What is a number line for preschool?
Number lines are (in a sense) the address to where a number lives. It helps children to visualize the relationship of one number to another (6 is between 5 and 7; 4 is two more than 2). As children develop in mathematical thinking, using a number line will be a critical part.
What is a number line first grade?
The most common number line used in first grade is a horizontal number line. I begin using this type of number line for counting and identifying missing numbers, but later I use it as a tool to solve subtraction and addition equations. This type of number line helps them see that numbers increase from left to right.
What is a number line to 10?
A number line to 10 is a visual aid used in schools to aid in the teaching of numerical operations. Number lines can show different sequences of numbers, aiding in the teaching of different numerical operations. The above video may be from a third-party source.
What are the benefits of numbernumber line worksheets?
Number line worksheets are carefully prepared to improve learner’s proficiency in recognizing and representing number systems in an effective way! A number line model makes an ideal tool to understand the concept of fundamental math operations.
How do you teach numbers on the number line?
Use sticky notes to line up numbers. Write numbers on sticky notes and ask kids to place them in order on the number line. This simple idea is perfect for little ones first learning to sequence numbers. But number line activities are useful for older students too; try this same exercise with fractions, decimals, or negative numbers.
What is the line to learning in math?
The line to learning starts here. Number lines are a reliable method teachers have been using for ages to teach number sense, arithmetic, and all sorts of other math skills. This collection of number line activities works with littles and not-so littles, as they learn the concepts they need to be successful at math.
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# Adding Fractions with Whole Numbers (Mixed numbers)
Fractions that have whole numbers, also called mixed fractions, can be added following a process similar to adding normal fractions with one additional step. First, we need to convert the mixed fractions to improper fractions. Then, depending on whether the fractions are like fractions or unlike fractions, we will use a different process.
Here, we will learn to add like, and unlike mixed fractions. In addition, we will solve some practice problems to apply these concepts.
##### ARITHMETIC
Relevant for
Learning to add fractions with whole numbers.
See steps
##### ARITHMETIC
Relevant for
Learning to add fractions with whole numbers.
See steps
## Steps to add mixed fractions
Mixed fractions are characterized by having whole numbers. If the denominators of the fractions are the same, we have like fractions and if the denominators are different, we have unlike fractions.
Step 1: Convert mixed fractions to improper fractions. For this, we multiply the whole number by the denominator of the fraction and add the result to the numerator.
Step 2: Determine if the fractions are like or unlike fractions. When we have like fractions, we can skip the next steps and continue to step 6.
Step 3: When we have unlike fractions, we have to find the least common denominator (LCD) of the fractions.
Step 4: Divide the LCD by the denominator of each fraction.
Step 5: Multiply both the numerator and the denominator by the result of step 2. With this, we will obtain like fractions, where the denominator is the LCD.
Step 6: Add the like fractions. For this, we use a single denominator and add the numerators.
Step 7: Simplify the final fraction if possible.
The following examples are solved by applying what has been learned about the addition of mixed fractions. Try to solve the problems yourself before looking at the solution.
### EXAMPLE 1
Solve the addition $latex 1\frac{1}{2}+\frac{1}{2}$.
Step 1: Converting the mixed fractions to improper fractions, we have:
$$1\frac{1}{2}+\frac{1}{2}$$
$$=\frac{3}{2}+\frac{1}{2}$$
Step 2: Both denominators are equal to 2, so we have like fractions. Therefore, we continue to step 6.
Steps 3-5: Not applicable.
Step 6: To add the like fractions, we combine the denominators and add the numerators:
$$=\frac{3}{2}+\frac{1}{2}$$
$$=\frac{3+1}{2}$$
$$=\frac{4}{2}$$
Step 7: Simplifying, we have:
$$=2$$
### EXAMPLE 2
Find the result to the addition $latex 2\frac{2}{3}+1\frac{1}{3}$.
Step 1: Let’s convert the mixed fractions to improper fractions:
$$2\frac{2}{3}+1\frac{1}{3}$$
$$=\frac{8}{3}+\frac{4}{3}$$
Step 2: We have like fractions since both denominators are equal to 3. Therefore, we continue to step 6.
Steps 3-5: Not applicable.
Step 6: We add the like fractions by combining the denominators and adding the numerators:
$$=\frac{8}{3}+\frac{4}{3}$$
$$=\frac{8+4}{3}$$
$$=\frac{12}{3}$$
Step 7: We can simplify the fraction:
$$=4$$
### EXAMPLE 3
Solve the addition of fractions $latex 1\frac{2}{3}+\frac{1}{5}$.
Step 1: Converting the mixed fractions to improper fractions, we have:
$$1\frac{2}{3}+\frac{1}{5}$$
$$=\frac{5}{3}+\frac{1}{5}$$
Step 2: In this case, we have unlike fractions because the denominators are different. Therefore, we continue to step 3.
Step 3: The least common denominator of 3 and 5 is 15.
Step 4: Dividing 15 by 3 (first denominator), we get 5. Dividing 15 by 5 (second denominator), we get 3.
Step 5: We multiply the numerator and denominator by the numbers obtained in step 4, 5 for the first fraction, and 3 for the second:
$$=\frac{5\times 5}{3\times 5}+\frac{1\times 3}{5\times 3}$$
$$=\frac{25}{15}+\frac{3}{15}$$
Step 6: We add the like fractions by combining the denominators and adding the numerators:
$$=\frac{25}{15}+\frac{3}{15}$$
$$=\frac{25+3}{15}$$
$$=\frac{28}{15}$$
Step 7: We can simplify by writing as a mixed number:
$$=1 \frac{13}{15}$$
### EXAMPLE 4
Solve the addition of mixed fractions $latex 2\frac{3}{4}+3\frac{1}{2}$.
Step 1: Converting the mixed fractions to improper fractions, we have:
$$2\frac{3}{4}+4\frac{1}{2}$$
$$=\frac{11}{4}+\frac{9}{2}$$
Step 2: We have unlike fractions because the denominators are different, so we continue to step 3.
Step 3: The least common denominator of 4 and 2 is 4.
Step 4: Dividing 4 by 4 (first denominator), we get 1. Dividing 4 by 2 (second denominator), we get 2.
Step 5: By multiplying the numerator and denominator by the numbers obtained in step 4, we have:
$$=\frac{11\times 1}{4\times 1}+\frac{9\times 2}{2\times 2}$$
$$=\frac{11}{4}+\frac{18}{4}$$
Step 6: Adding the like fractions, we have:
$$=\frac{11+18}{4}$$
$$=\frac{29}{4}$$
Step 7: We can simplify by writing as a mixed number:
$$=7 \frac{1}{4}$$
### EXAMPLE 5
Solve the addition of mixed fractions $latex 1\frac{2}{5}+\frac{3}{5}+3\frac{1}{5}$.
Step 1: Converting the mixed fractions to improper fractions, we have:
$$1\frac{2}{5}+\frac{3}{5}+3\frac{1}{5}$$
$$=\frac{7}{5}+\frac{3}{5}+\frac{16}{5}$$
Step 2: Here, we have like fractions, so we continue to step 6.
Steps 3-5: Not applicable.
Step 6: Combining the denominators and adding the numerators, we have:
$$=\frac{7}{5}+\frac{3}{5}+\frac{16}{5}$$
$$=\frac{7+3+16}{5}$$
$$=\frac{26}{5}$$
Step 7: We can simplify the fraction by converting it to a mixed fraction:
$$=5\frac{1}{5}$$
### EXAMPLE 6
Solve the addition of mixed fractions $latex 2\frac{3}{4}+1\frac{2}{3}+1\frac{4}{5}$.
Step 1: Converting the mixed fractions to improper fractions, we have:
$$2\frac{3}{4}+1\frac{2}{3}+1\frac{4}{5}$$
$$=\frac{11}{4}+\frac{5}{3}+\frac{9}{5}$$
Step 2: We have three unlike fractions, so we continue to step 3.
Step 3: The least common denominator of 4, 3, and 5 is 60.
Step 4: Dividing 60 by 4 (first denominator), we get 15. Dividing 60 by 3 (second denominator), we get 20. Dividing 60 by 5 (third denominator), we get 12.
Step 5: We multiply the numerators and denominators of the fractions by the numbers obtained in step 4, we have:
$$=\frac{11\times 15}{4\times 15}+\frac{5\times 20}{3\times 20}+\frac{9\times 12}{5\times 12}$$
$$=\frac{165}{60}+\frac{100}{60}+\frac{108}{60}$$
Step 6: Adding the like fractions, we have:
$$=\frac{165+100+108}{60}$$
$$=\frac{373}{60}$$
Step 7: We can simplify by writing as a mixed number:
$$=6 \frac{13}{60}$$
## Addition of mixed fractions – Practice problems
Solve the following additions of mixed fractions. If you have trouble with these problems, you can use the solved examples above as a guide.
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# Math: Elementary School: 1st and 2nd Grade Quiz - Place Value - Comparing Numbers (Questions)
This Math quiz is called 'Place Value - Comparing Numbers' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is a fun way to learn if you are in the 1st or 2nd grade - aged 6 to 8.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
Understanding the place value of digits within numbers helps to order and compare numbers up to one hundred. Children should begin to use this knowledge to help them order numbers up to 100 from smallest to largest, place numbers on a number line and compare numbers, knowing which have smaller or larger values than others.
Can you use place value correctly?
1. Which is the largest number?
37, 73, 13, 17
[ ] 73 [ ] 37 [ ] 13 [ ] 17
2. Which number is the smallest?
16, 61, 66, 11
[ ] 61 [ ] 16 [ ] 11 [ ] 66
3. What is the value of the 7 in 27?
[ ] 7 units [ ] 7 tens [ ] 7 hundreds [ ] 7 thousands
4. Put these numbers in order from lowest to highest:
23, 33, 32, 12, 13
[ ] 32, 33, 23, 12, 13 [ ] 33, 32, 23, 13, 12 [ ] 32, 33, 13, 23, 12 [ ] 12, 13, 23, 32, 33
5. Put these numbers in order from highest to lowest:
45, 14, 41, 54, 15
[ ] 54, 45, 41, 15, 14 [ ] 45, 54, 14, 41, 15 [ ] 14, 15, 41, 45, 54 [ ] 15, 45, 54, 14, 41
6. Which numbers are missing?
23, 24, __, 26, __, __
[ ] 27, 28, 29 [ ] 24, 25, 26 [ ] 25, 27, 28 [ ] 28, 38, 48
7. Which numbers are missing?
__, 46, __, __
[ ] 40, 50, 60 [ ] 46, 47, 48 [ ] 55, 56, 57 [ ] 45, 47, 48
8. Which numbers come immediately before and after 79?
[ ] 78 and 80 [ ] 77 and 78 [ ] 69 and 89 [ ] 67 and 76
9. What is the value of the 5 in 57?
[ ] 5 hundreds [ ] 5 units [ ] 5 tens [ ] 5 thousands
10. What is 67 in words?
[ ] Six and seven [ ] Sixty-seven [ ] Sixteen [ ] Seventy
Math: Elementary School: 1st and 2nd Grade Quiz - Place Value - Comparing Numbers (Answers)
1. Which is the largest number?
37, 73, 13, 17
[x] 73 [ ] 37 [ ] 13 [ ] 17
73 has 7 tens and 3 units
2. Which number is the smallest?
16, 61, 66, 11
[ ] 61 [ ] 16 [x] 11 [ ] 66
11 is smaller than the other numbers
3. What is the value of the 7 in 27?
[x] 7 units [ ] 7 tens [ ] 7 hundreds [ ] 7 thousands
The last digit in a whole number represents units or ones
4. Put these numbers in order from lowest to highest:
23, 33, 32, 12, 13
[ ] 32, 33, 23, 12, 13 [ ] 33, 32, 23, 13, 12 [ ] 32, 33, 13, 23, 12 [x] 12, 13, 23, 32, 33
The lowest number is 12 and the highest is 33
5. Put these numbers in order from highest to lowest:
45, 14, 41, 54, 15
[x] 54, 45, 41, 15, 14 [ ] 45, 54, 14, 41, 15 [ ] 14, 15, 41, 45, 54 [ ] 15, 45, 54, 14, 41
54 has the highest value, 14 has the lowest value
6. Which numbers are missing?
23, 24, __, 26, __, __
[ ] 27, 28, 29 [ ] 24, 25, 26 [x] 25, 27, 28 [ ] 28, 38, 48
Counting up aloud might help here
7. Which numbers are missing?
__, 46, __, __
[ ] 40, 50, 60 [ ] 46, 47, 48 [ ] 55, 56, 57 [x] 45, 47, 48
Thinking about the numbers on a number line might be useful here
8. Which numbers come immediately before and after 79?
[x] 78 and 80 [ ] 77 and 78 [ ] 69 and 89 [ ] 67 and 76
One less than 79 is 78 and one more is 80
9. What is the value of the 5 in 57?
[ ] 5 hundreds [ ] 5 units [x] 5 tens [ ] 5 thousands
This column in a number represents the tens
10. What is 67 in words?
[ ] Six and seven [x] Sixty-seven [ ] Sixteen [ ] Seventy
The two words have a hyphen between them
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Paul's Online Notes
Home / Calculus II / Integration Techniques / Integrals Involving Quadratics
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### Section 7.6 : Integrals Involving Quadratics
1. Evaluate the integral $$\displaystyle \int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}}$$.
Show All Steps Hide All Steps
Start Solution
The first thing to do is to complete the square (we’ll leave it to you to verify the completing the square details) on the quadratic in the denominator.
$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \int{{\frac{7}{{{{\left( {w + \frac{3}{2}} \right)}^2} + \frac{3}{4}}}\,dw}}$ Show Step 2
From this we can see that the following substitution should work for us.
$u = w + \frac{3}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dw$
Doing the substitution gives,
$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \int{{\frac{7}{{{u^2} + \frac{3}{4}}}\,du}}$ Show Step 3
This integral can be done with the formula given at the start of this section.
$\int{{\frac{7}{{{w^2} + 3w + 3}}\,dw}} = \frac{{14}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 3 }}} \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{14}}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{{2w + 3}}{{\sqrt 3 }}} \right) + c}}$
Don’t forget to back substitute in for $$u$$!
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Graphical Solution
Custom Search
Graphical Solution If there is a pair of numbers that can be substituted for x and y in two different equations, the pair form the coordinates of a point which lies on the graph of. each equation. The only way in which a point can lie on two lines simultaneously is for the point to be at the intersection of the lines. Therefore, the graphical solution of two simultaneous equations involves drawing their graphs and locating the point at which the graph lines intersect. For example, when we graph the equations x+y=33 and x-y = 5, as in figure 12-5, we see that they intersect in a single point. There is one pair of values comprising coordinates of that point (19, 14), and that pair of values satisfies both equations, as follows: Figure 12-5.-Graph of x + y = 33 and x - y = 5. This pair of numbers satisfies each equation. It is the only pair of numbers that satisfies the two equations simultaneously. The graphical method is a quick and simple means of finding an approximate solution of two simultaneous equations. Each equation is graphed, and the point of intersection of the two lines is read as accurately as possible. A high degree of accuracy can be obtained but this, of course, is dependent on the precision with which the lines are graphed and the amount of accuracy possible in reading the graph. Sometimes the graphical method is quite adequate for the purpose of the problem. Figure 12-6 shows the graphs of x + y = 11 and x-y= -3. The intersection appears to be the point (4, 7). Substituting x = 4 and y = 7 into the equations shows that this is the actual point of intersection, since this pair of numbers satisfies both equations. Figure 12-6 - Graph of x + y = 11 and x - y = -3. The equations 7x - 8y = 2 and 4x + 3y = 5 are graphed in figure 12-7. The lines intersect where y is approximately 1/2 and x is approximately 5/6. Practice problems. Solve the following simultaneous systems graphically: Addition Method The addition method of solving systems of equations is illustrated in the following example: The result in the foregoing example is obtained by adding the left member of the first equation to the left member of the second, and adding the right member of the first equation to the right member of the second. Having found the value of x, we substitute this value in either of the original equations to find the value of y, as follows: Notice that the primary goal in the addition method is the elimination (temporarily) of ,one of the variables. If the coefficient of y is the same in both equations, except for its sign, adding the equations eliminates y as in the foregoing example. On the other hand, suppose that the coefficient of the variable which we desire to eliminate is exactly the same in both equations. In the following example, the coefficient of x is the same in both equations, including its sign:
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ôô
# What Is Modular Arithmetic?
Learn what modular arithmetic is, how to perform it, and how it’s used in the real world.
By
Jason Marshall, PhD
Episode #53
## How to Perform Modular Addition
So that’s the idea of modular arithmetic. Now let’s talk about how to actually do it. It’s helpful here to keep the idea of a clock in mind. For arithmetic with a modulus of 12—also called arithmetic modulo 12—you can think of an actual clock that has 12 numbers on what is essentially a circular number line. For example, what’s 9 + 1 modulo 12? Well, if we think of starting at 9 on a normal 12-hour clock and then moving forward 1 hour, we find that 9 + 1 modulo 12 is just 10. In this case, since 10 is less than the modulus of 12, the answer is the same as with normal math.
But what about something like 9 + 5 modulo 12? Here, in normal non-modular arithmetic, the answer is 9 + 5 = 14. But in modular arithmetic with a modulus of 12, the numbers wrap around after we count up to 12. So to find 9 + 5 modulo 12, we need to count forward 5 hours from 9, being sure to start over once we get to 12. And when we do that we find that 9 + 5 modulo 12 is 2. In other words, using a clock to help visualize this, we start at 9 and then count forward 5 hours: 10, 11, 12, 1, and finally 2.
## More Modulo Arithmetic
What if instead of modulo 12 arithmetic we want to perform modulo 5 arithmetic? For example, what’s 9 + 1 modulo 5? Well, this time you need to imagine a clock that cycles through only 5 numbers—from 0 at the top, then 1, 2, 3, and finally 4, before starting back at 0 again. (By the way, a normal 12-hour clock could just as easily start with 0 at the top instead of 12—we just don’t traditionally do it that way.) To find 9 + 1 modulo 5, we need to start at the top of this new 5 numbered clock, count forward 9 spaces, and then count forward another 1 space. Where do you end up? Well, you end up right back where you started at the top. In other words, 9 + 1 modulo 5 is 0.
[[AdMiddle]So as not to get things confused with normal (meaning non-modular) arithmetic, in modular arithmetic we don’t usually say that 9 + 1 modulo 5 is “equal” to 0. Instead we say that 9 + 1 modulo 5 is “congruent” to 0. And we write this as (9 + 1) (mod 5) ≡ 0. The word congruent here roughly means “the same as,” and we represent this idea of congruence in writing using a symbol that looks like an equals sign but with one extra horizontal line in the middle of it.
## Modular Arithmetic in the Real World
Okay, that’s enough for now to get us started on our way towards understanding the ins-and-outs of this new kind of admittedly strange arithmetic. We’ll return to this topic now and again in future articles to talk about other aspects of modular arithmetic and the other ways in which it’s used in the real world. That’s right, this isn’t all just random math stuff—it actually turns out to be very useful math stuff. But we’ll talk about exactly how it’s useful in later articles.
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+0
# Four-fifths of my current age is greater than three-quarters of my age one year from now.
+6
690
2
+1280
Four-fifths of my current age is greater than three-quarters of my age one year from now.
Four-fifths of my current age is also greater than five-sixths of my age one year ago.
Given that my age is an integer, what are all possible values for my age? Explain how you determined your answer, showing all steps along the way.
AWESOMEEE May 11, 2015
#1
+81090
+15
Let your current age be A .....so your age one year from now is A + 1 and one year ago was A- 1 ....and we have
For the first part, we have
(4/5)(A) > (3/4)(A + 1) multiply through by the common denominator of 4 and 5, i.e., 20
16A > 15(A + 1) simplify
16A > 15A + 15 subtract 15A from both sides
A > 15
And for the second part, we have
(4/5)A > (5/6)(A - 1) multiply through by the common denominator of 5 and 6, i.e., 30
24A > 25(A - 1) simplify
24A > 25A - 25 add 25 to both sides and subtract 24A from both sides
25 > A or, expressed another way.....
A < 25
So.....using both solutions, we have
15 < A < 25 {your possible age is greater than 15 but less than 25 }
CPhill May 11, 2015
Sort:
#1
+81090
+15
Let your current age be A .....so your age one year from now is A + 1 and one year ago was A- 1 ....and we have
For the first part, we have
(4/5)(A) > (3/4)(A + 1) multiply through by the common denominator of 4 and 5, i.e., 20
16A > 15(A + 1) simplify
16A > 15A + 15 subtract 15A from both sides
A > 15
And for the second part, we have
(4/5)A > (5/6)(A - 1) multiply through by the common denominator of 5 and 6, i.e., 30
24A > 25(A - 1) simplify
24A > 25A - 25 add 25 to both sides and subtract 24A from both sides
25 > A or, expressed another way.....
A < 25
So.....using both solutions, we have
15 < A < 25 {your possible age is greater than 15 but less than 25 }
CPhill May 11, 2015
#2
+91510
+10
I always like these questions
Thanks CPhill and Awesomeee
Melody May 12, 2015
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Q:
# How do you calculate percentages?
A:
Percentage, often represented by "%," is a way to say parts per 100. Percentages can be calculated by dividing any fractions that result in a decimal quotient and then multiplying by 100. This can be done by hand or estimated mentally in under a minute in many cases. Use of a calculator is optional.
## Keep Learning
1. ### Determine the whole and the parts
The end result of calculated percentages lets 100 be the whole. Some part-whole relationships begin with a 100-part whole. For example, if you have a quarter and a nickel, you have 30 cents, or 30 parts of a whole that equals 100. However, many wholes are not made of 100 parts. For example, you might have 11 ounces out of the 16-ounce pound, or you could have worked 20 hours of your 32-hour work week.
2. ### Divide the parts by the whole
Divide the parts present by the number of parts required to make up the whole. These expressions look like fractions. For example, write 30 cents as 30/100 or 11 ounces as 11/16. Take the quotient, usually in the form of a decimal. 30/100 = 0.30 11/16 = 0.6875
3. ### Multiply by 100
Multiply the decimal by 100. The product is the percentage or parts per hundred. 0.30 * 100 = 30% 0.6875 * 100 = 68.75% meaning that if the whole were 100 ounces, you would have almost 69 at the present rate.
Sources:
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PEOPLE SEARCH FOR
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# In how many ways can he form a rectangle using all the tiles each time?
A contractor has exactly $1088$ square tiles. In how many ways can he form a rectangle using all the tiles each time?
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Basically there asking how many ways can he multiply two numbers together to get 1088. Because the multiplication of two numbers 'a' and 'b' forms a rectangle with length a, and side b, and area ab. – Ethan Jan 13 '13 at 1:16
Hint: How many $m,n\in \mathbb{N}$ such that $m\times n = n\times m= 1088$ can you find? Note that $1088=2^6\cdot17$
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$$1088 = 2\times2\times2\times2\times2\times2\times17$$ So \begin{align} 1088 & = 1\times1088 \\ & = 2\times544 \\ & = 4 \times272 \\ & = 8 \times136 \\ & = 16\times68 \\ & = 17\times64 \\ & = 32\times 34. \end{align} There's your list of rectangles.
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Basically there asking how many ways can he multiply two numbers together to get 1088. Because the multiplication of two numbers 'a' and 'b' forms a rectangle with length a, and side b, and area ab.
In general the solution to this problem given n tiles will be how many divisors does n have less then or equal to $n^{1/2}$.
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is the rectangle $(a,b)$ equal to the rectangle $(b,a)$ if not then here is a way to solve it: $1088=2^6*17$. since you need to keep the tiles whole and you use all of them the height needs to be a divisor of $1088$. Therefore the number of rectangles $(a,b)$ is the number of divisors of $1088$.(if the number where a perfect square it would be the number of divisors minus 1)
How do we count them? from the fundamental theorem of arithmetic we know every integer number greater than 1 has a single representation as a product of primes.
Therefore how many combinations can we make with six 2's and one 17?? we can chose to add $0,1,2,3...$or $6$ 2's and we can chose to use $0$ or $1$ 17's. Using the fundamental theorem of combinatorics we know the number of combinations is $7*2=14$. Now: if you don't want to count $(a,b)$ and $(b,a)$ the result is $7$ because we are counting all of them twice.
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Question Video: Finding the Absolute Maximum and Minimum Values of a Root Function in a Given Interval Mathematics • 12th Grade
Find, if they exist, the values of the absolute maximum and minimum points for the function π(π₯) = β(3π₯ + 10) where π₯ β [β2, 5].
07:26
Video Transcript
Find, if they exist, the values of the absolute maximum and minimum points for the function π of π₯ is equal to the square root of three π₯ plus 10, where π₯ is in the closed interval from negative two to five.
The question gives us a composite function, and it wants us to find the values of the absolute maximum and minimum if they exist, where π₯ is the closed interval from negative two to five. First, we want to check where our function π of π₯ is continuous. We know that this function is a composite function. We know that if we have two continuous functions π and β, then their composition π composed with β of π₯ is continuous on its domain. We see that our function is composed of two functions: π of π₯ equals the square root of π₯; β of π₯ equals three π₯ plus 10. The square root function is continuous, and all polynomial functions are continuous. Because π of π₯ is a composition of two continuous functions, itβ²s continuous on its domain. This just means π of π₯ is continuous wherever π of π₯ is defined.
We know that the square root of a number will always be defined unless that number is negative. Therefore, π of π₯ is not defined when three π₯ plus 10 is less than zero, which also means it wonβ²t be continuous for these values of π₯. We subtract 10 from both sides of this inequality and then divide by three to see that our function π of π₯ is not defined when π₯ is less than negative 10 over three. Negative 10 over three is equal to negative three and one-third. Negative three and one-third is outside the bounds of our domain. So we must have that our function π of π₯ is continuous for all values of π₯ in the closed interval from negative two to five.
As an aside, itβ²s worth noting at this point since weβ²ve shown that our function π of π₯ is continuous on a closed interval, by applying the extreme value theorem, we know that our function π of π₯ will achieve a maximum and a minimum on the closed interval from negative two to five. So at this point, we know that both of these values must exist. Since our function π of π₯ is continuous on a closed interval, we can find the values of the absolute maximum and absolute minimum by using the following three steps. First, we need to find the critical points of our function π of π₯. Thatβ²s where the derivative is equal to zero or where the derivative does not exist.
Next, we need to evaluate the function π of π₯ at the critical points. Then we need to evaluate the function π of π₯ at the end points. The largest of these values will then be the absolute maximum of our function π of π₯ on the interval, and the lowest will be the minimum of our function π of π₯ on the interval. The first thing we need to do is find the critical points of our function π of π₯. We need to differentiate the square root of three π₯ plus 10. We know that our function π of π₯ is a composite function. To differentiate it, we need to use the chain rule. We set π’ equal to the inner function of three π₯ plus 10. We have π of π’ is equal to the square root of π’, and π’ of π₯ is three π₯ plus 10. π is a function of π’, and π’ in turn is a function of π₯.
The chain rule tells us that if π is a function of π’ and π’ in turn is a function of π₯, then the derivative of π with respect to π₯ is equal to the derivative of π with respect to π’ multiplied by the derivative of π’ with respect to π₯. By applying the chain rule, we have the derivative of our function π with respect to π₯ is equal to the derivative of the square root of π’ with respect to π’ multiplied by the derivative of three π₯ plus 10 with respect to π₯. We can differentiate the square root of π’ with respect to π’ by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. Since the square root of π’ is the same as π’ to the power of a half, this gives us one divided by two root π’.
We can also differentiate three π₯ plus 10 by using the power rule for differentiation. It gives us three. So the derivative of π with respect to π₯ is equal to three divided by two root π’. We want this in terms of π₯. So we substitute π’ is equal to three π₯ plus 10 to show that the derivative of π with respect to π₯ is equal to three divided by two times the square root of three π₯ plus 10. Remember, weβ²re looking for places in our interval where the derivative equals zero or does not exist. We see that the positive square root of three π₯ plus 10, if it exists, will always give a number greater than or equal to zero. This is because weβ²re taking the positive square root of the number. If our denominator were to equal zero, then the derivative wouldnβ²t exist.
However, if the square root of three π₯ plus 10 was positive, then multiplying it by two would give us a positive number. And then three divided by a positive number would give us a positive number. We know that π₯ will be greater than or equal to negative two based on the given interval. This means that the three π₯ plus 10 will never be equal to zero on this interval. Our derivative must be the quotient of two positive numbers and therefore would be positive. On our interval, the derivative is never equal to zero. In fact, if our derivative is the quotient of two positive numbers, on this interval, it will always exist. This shows us that our function has no critical points on this interval. And that means we can skip the second step, as there are no critical points to evaluate the function at.
So finally, we evaluate π of π₯ at the endpoints. π evaluated at negative two is equal to the square root of three times negative two plus 10, which is the square root of four, which equals two. And π evaluated at five is equal to the square root of three times five plus 10, which is the square root of 25 and is equal to five. Since our function π of π₯ is continuous on this closed interval, it must have its minimum at two and its maximum at five on this interval.
In conclusion, we have shown that the function π of π₯ is equal to the square root of three π₯ plus 10, where π₯ is the closed interval from negative two to five, will achieve an absolute minimum value of two and an absolute maximum value of five.
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# NCERT Solutions for Class 8 Maths
Find 100% accurate solutions for NCERT Class VIII Math. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!
1. Given: The side of a square = 60 m
And the length of rectangular field = 80 m According to question,
Perimeter of rectangular field = Perimeter of square field
⇒ 2 (l + b) = 4 × side
⇒ 2 (80 + b) = 4 × 60 ⇒ (160 + 21) = 240
⇒ 2h = 240 – 160 ⇒ 2h =80
⇒ b = 40 m
Now Area of Square field = (Side)2 = (60)2 = 3600 m2
And Area of Rectangular field = length × breadth = 80 × 40 = 3200 m2
Hence, area of square field is larger.
2. Side of a square plot = 25 m
∴ Area of square plot = (Side)2 = (25)2 = 625 m2
Length of the house = 20 m and
Breadth of the house = 15 m
∴ Area of the house = length × breadth = 20 × 15 = 300 m2
Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m2
∵ Cost of developing the garden per sq. m = 55
∴ Cost of developing the garden 325 sq. m = 55 × 325 = 17,875
Hence total cost of developing a garden around is ₹ 17,875.
3. Given: Total length = 20 m
Diameter of semi circle = 7 m
∴ Radius of semi circle = 3.5 m
Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = lxb = 13 x 7 = 91 m2
Area of two semi circles
Area of garden = 91 + 38.5 = 129.5 m2
Now Perimeter of two semi circles
And Perimeter of garden = 22 + 13 + 13 = 48 m
4. Given: Base of flooring tile = 24 cm = 0.24 m
Corresponding height of a flooring tile = 10 cm = 0.10 m
Now Area of flooring tile = Base × Altitude = 0.24 × 0.10 = 0.024 m2
∴ Number of tiles required to cover the floor
Hence 45000 tiles are required to cover the floor.
Circumference of semi circle
Total distance covered by the ant = Circumference of semi circle + Diameter = 4.4 + 2.8 = 7.2 cm
(b) Diameter of semi circle = 2.8 cm
Circumference of semi circle
(c) Diameter of semi circle = 2.8 cm
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.
6. Here one parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and height (h) = 0.8 m
Area of top surface of the table
Hence surface area of the table is 0.88 m2
7. Let the length of the other parallel side be
Length of one parallel side (a) = 10 am and height (h) = 4 cm
Area of trapezium
⇒ 34 = (10 + b) × 2
⇒ 34 = 20+ 2b ⇒ 34 – 20 = 2b
⇒ 14 = 2b ⇒ 7 = b
⇒ b = 7
Hence another required parallel side is 7 cm.
8. Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD = AB + BC + CD + DA
⇒ 120 = AB + 48 + 17 + 40
⇒ 120 = AB = 105
⇒ AB = 120 – 105 = 15 m
Now Area of the field
Hence area of the field ABCD is 660 m2
9. Here h1 = 13 m, h2 = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC
Hence required area of the field is 252 m2
10. Given: d, =7.5 cm and d2 = 12 cm
We know that,
Area of rhombus
Hence area of rhombus is 45 cm2.
11. Since rhombus is also a kind of parallelogram.
Area of rhombus = Base × Altitude = 6 × 4 = 24 cm2
Also Area of rhombus
Hence the length of the other diagonal is 60 cm.
12. Here, d1 = 45 cm and d2 = 30 cm
∵ Area of one tile
∴ Area of 3000 tiles = 675 × 3000 = 2025000 cm2
∵ Cost of polishing the floor per sq. meter = ₹ 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 × 202.50 = ₹ 810
Hence the total cost of polishing the floor is ₹ 810.
13. Given: Perpendicular distance (h) = 100 m
Area of the trapezium shaped field = 10500 m2
Let side along the road be x m and side along the river = 2 × m
Area of the trapezium field
Hence the side along the river = 2x = 2 × 70 = 140 m.
14. Given: Octagon having eight equal sides, each 5 m.
Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.
Now Area of two trapeziums
And Area of rectangle = length × breadth = 11 × 5 = 55 m2
∴ Total area of octagon = 64 + 55 = 119 m2
15. First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
Second way: By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
AM = 30 – 15 = 15 m
Area of pentagon = Area of Δ ABE + Area of square BCDE
Hence total area of pentagon shaped park = 337.5 m2.
16. Here two of given figures (I) and (II) are similar in dimensions.
And also figures (III) and (IV) are similar in dimensions.
Area of figure (I) = Area of trapezium
Also Area of figure (II) = 96 cm2
Now Area of figure (III) = Area of trapezium
Also Area of figure (IV) = 80 cm2
17. (a) Given: Length of cuboidal box (l) = 60 cm
Breadth of cuboidal box (b) = 40 cm Height of cuboidal box (h) = 50 cm
Total surface area of cuboidal box = 2 (lb + bh + hl)
= 2 (60 × 40 + 40 × 50 + 50 × 60)
= 2 (2400 + 2000 + 3000)
= 2 × 7400 = 14800 cm2.
(b) Given: Length of cuboidal box (l) = 50 cm
Breadth of cuboidal box (b) = 50 cm Height of cuboidal box (h) = 50 cm
Total surface area of cuboidal box = 2 (lb + bh + hl)
= 2(50 × 50+ 50 × 50 + 50 × 50)
= 2 (2500 + 2500 + 2500)
= 2 × 7500 = 15000 cm2
Hence cuboidal box (a) requires the lesser amount of materal to make, since surface area of box (a) is less than that of box (b).
18. Given: Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm
And Height of cuboidal box (h) = 24 cm
Total surface area of suitcase box = 2 (lb + bh + hl)
= 2 (80 × 48 + 48 × 24 + 24 × 80)
= 2 (3840 + 1152 + 1920)
= 2 × 6912 = 13824 cm2
Area of Tarpaulin cloth = Surface area of suitcase
Required tarpaulin for 100 suitcases = 144 × 100 = 14400 cm = 144 m
Hence tarpaulin cloth required to cover 100 suitcases is 144 m.
19. Here Surface area of cube = 600 cm2
⇒ 6l2 = 600 ⇒ l2 = 100 ⇒ l = 10 cm
Hence the side of cube is 10 cm
20. Here, Length of cabinet (1) = 2 m, Breadth of cabinet (b) = 1 m
And Height of cabinet (h) = 1.5 m
∴ Surface area of cabinet = lb + 2 (bh + hl)
= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)
= 2 + 2 (1.5 + 3.0)
= 2 + 9.0
= 11 m2
Hence required surface area of cabinet is 11 m2.
21. Here, Length of wall (l) = 15 m, Breadth of wall (b) = 10 m And Height of wall (h) = 7 m
Total Surface area of classroom = lb + 2 (bh + hl)
= 15 × 10 + 2 (10 × 7 + 7 × 15)
= 150 + 2 (70 + 105)
= 150 + 350
= 500 m2
Now Required number of cans
Hence 5 cans are required to paint the room.
22. Given: Diameter of cylinder = 7 cm
Radius of cylinder (r) = 2 – 7 cm
And Height of cylinder (h) = 7 cm
Lateral surface area of cylinder
Now lateral surface area of cube = 4l2 = 4 × (7)2 = 4 × 49 = 196 cm2
Hence the cube has larger lateral surface area.
23. Given: Radius of cylindrical tank (r) = 7 m
Height of cylindrical tank (h) = 3 m
Total surface area of cylindrical tank
Hence 440 m2 metal sheet is required.
24. Given: Lateral surface area of hollow cylinder = 4224 cm2
And Height of hollow cylinder = 33 cm
Curved surface area of hollow cylinder = 2πrh
Now Length of rectangular sheet = 2πr
Perimeter of rectangular sheet = 2 (l + b)
= 2 (128 + 33) = 2 × 161 = 322 cm
Hence perimeter of rectangular sheet is 322 cm.
25. Given: Diameter of road roller = 84 cm
Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller
∵ Area covered by road roller in 750 revolutions = 26400 × 750 = 1,98,00,000 cm2
= 1980 m2 [∵ 1 m2 = 10,000 cm2]
Hence the area of the road is 1980 m2.
26. Given: Diameter of cylindrical container = 14 cm
Height of cylindrical container = 20 cm
Height of the label (h)= 20 – 2 – 2 = 16 cm
Curved surface area of label
Hence the area of the label of 704 cm2.
27. We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.
(a) Volume (b) Surface are (c) Volume
28. Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous).
Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = cm
And Height of cylinder A = 14 cm
Volume of cylinder A
= 539 cm3
Now Diameter of cylinder B = 14 cm
And Height of cylinder B = 7 cm
∴ Volume of cylinder A = = 1078 cm3
Total surface area of cylinder A = πr (2h + r) [∵ It is open from top]
Total surface area of cylinder B = πr (2h + r) [∵ It is open from top]
Yes, cylinder with greater volume also has greater surface area.
29. Given: Base area of cuboid = 180 cm2 and Volume of cuboid = 900 cm3
We know that,
Volume of cuboid = l × b × h [•.• Base area = lxb= 180 (given)]
⇒ 900 = 180 × h
Hence the height of cuboid is 5 m.
30. Given: Length of cuboid (l) = 60 cm, Breadth of cuboid (b) = 54 cm and Height of cuboid (h) = 30 cm
We know that, Volume of cuboid = l × b × h = 60 × 54 × 30 cm3
And Volume of cube = (Side)3 = 6 ×× 6 cm3
∴ Number of small cubes
Hence required cubes are 450.
31. Given: Volume of cylinder = 1.54 m3 and Diameter of cylinder = 140 cm
Volume of cylinder = πr2h
Hence height of the cylinder is 1 m.
32. Given: Radius of cylindrical tank (r) = 5 m and Height of cylindrical tank (h) = 7 m Volume of cylindrical tank = πr2h
= ×1.5 × 1.5 × 7 = 49.5 cm3
= 49.5 × 1000 liters [∵ m3 = 1000 liters]
= 49500 liters
Hence required quantity of milk is 49500 liters.
33. (i) Let the edge of cube be
Since, Surface area of the cube (A) = 6l2 When edge of cube is doubled, then
Surface area of the cube (A’)= 6 (2l)2 = 6 × 412 = 4 × 612
A’ = 4 × A
Hence surface area will increase four times.
(ii) Volume of cube (V) = l3
When edge of cube is doubled, then Volume of cube (V’)= (2l)3 = 8l3
V’ = 8 × V
Hence volume will increase 8 times.
34. Given: volume of reservoir = 108 m3
Rate of pouring water into cuboidal reservoir = 60 liters/minute
∵ mwater filled in reservoir will take = 1 hour
∴ 1 m3 water filled in reservoir will take
∴ 108 mwater filled in reservoir will take hours = 30 hours
It will take 30 hours to fill the reservoir.
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# McGraw Hill My Math Grade 5 Chapter 2 Lesson 5 Answer Key Problem-Solving Investigation: Make a Table
All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 2 Lesson 5 Problem-Solving Investigation: Make a Table will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 5 Answer Key Chapter 2 Lesson 5 Problem-Solving Investigation: Make a Table
1.
Understand
What facts do you know?
Sixty-three out of every ___ households owned at least one pet.
What do you need to find?
how many households out of ten thousand owned at least one pet
Sixty-three out of every 100 households owned at least one pet.
We need to find households owned at least one pet.
households out of ten thousand owned at least one pet.
2. Plan
I will make a ___ to solve the problem.
I will make a table to solve the problem.
3. Solve
So, about ___ out of ten thousand households owned at least one pet.
6300 out of ten thousand households owned at least one pet.
Explanation:
about 6300 out of ten thousand households owned at least one pet.
4. Check
Is my answer reasonable ? Explain.
Use mental math to multiply. 63 × 100 = _____
Explanation:
Given multiplicand is 63 and the multiplier is 100
By multiplying both multiplicand 63 and multiplier 100, we get the value as
The multiplication of 63 × 100 is 6300.
Practice the Strategy
Nestor is saving money to buy a new camping tent. Each week he doubles the amount he saved the previous week If he saves $1 the first week, how much money will Nestor save in 7 weeks? 1. Understand What facts do you know? ________________ Answer: Nestor is saving money to buy a new camping tent. He saves$1 the first week.
What do you need to find?
________________
We need to find out how much money will Nestor save in 7 weeks.
2. Plan
________________
3. Solve
________________
$127 Explanation: 1 + 2 + 4 + 8 + 16 + 32 + 64 =$127
4. Check
________________
Explanation:
He saves $1 the first week Each week he doubles the amount he saved the previous week.$1 + $2 +$4 + $8 +$16 + $32 +$64 = $127 Apply the Strategy Solve each problem by making a table. Question 1. Betsy is saving to buy a bird cage. She saves$1 the first week, $3 the second week,$9 the third week, and so on. How much money will she save in 5 weeks?
81 dollars
Explanation:
Betsy saves $1 for the first week$3 for the second week
$9 for the third week$27 for the fourth week
$81 for the fifth week Question 2. Kendall is planning to buy a laptop for$1,200. Each month she doubles the amount she saved the previous month. If she saves $20 the first mönth, in how many months will Kendall have enough money to buy the laptop? Answer: 6 months Explanation: Kendall is planning to buy a laptop for$1,200 Each month she doubles the amount she saved the previous month If she saves $20 the first month. In 1st month he saved$20
2nd month he saved $20 +$40 = $60 3rd month he saved$80 + $60 =$140
4th month he saved $160 +$140 = $300 5th month he saved$300 + $320 =$620
6th month he saved $620 +$640 = $1280 Hence after 6 months she can save enough to buy the laptop. Question 3. Mrs. Piant’s yearly salary is$42,000 and increases $2,000 per year. Mr. Piant’s yearly salary is$37,000
and increases $3,000 per year. In how many years will Mr. and Mrs. Piant make the same salary? Answer: 5 years Explanation: Let us assume the x years will Mr. and Mrs. Piant make the same salary. 42000 + 2000x = 37000 + 3000x 3000x – 2000x = 42000 – 37000 1000x = 5000 x = 5 Hence in 5 years will Mr. and Mrs. Piant make the same salary. Question 4. Mathematical PRACTICE 8 Look for a Pattern Complete the table that shows the prime factorizations of powers of 10. Use the pattern in the table to mentally find the prime factorization of 109. Write using exponents. Prime factorization of 109 = ____ Answer: The Prime factorization of 109 is 29 × 59 Explanation: Review the Strategies A card shop recorded how many packs of trading cards it sold each day. Use the table to solve Exercises 5—7. Use any strategy to solve each problem. • Make a table. • Use the four-step plan. Question 5. In which week did they sell the most packs of cards? Answer: In week 2 they sold the most packs of cards. Explanation: The week 2 trading cards sold are 48 + 43 + 45 + 41 + 39 = 216 178 216 180 Question 6. In which week did they sell the least amount? Answer: In week 1 they sold the least amount. Explanation: The week 1 trading card sold are 28 + 32 + 38 + 44 + 36 = 178 Question 7. How many more packs did they sell in Week 2 than in Week 3? Answer: There are 36 more packs they sold in week 2 than in week 3. Explanation: In week 2 trading cards sold are 216 In week 3 trading cards sold are 180 The difference between week 2 in week 3 is 216 – 180 = 36 Question 8. A putt-putt course offers a deal that when you purchase 10 rounds of golf, you get 1 round for free. If you played a total of 35 rounds, how many rounds did you purchase? Answer: 32 rounds. Explanation: A putt-putt course offers a deal that when you purchase 10 rounds of golf, you get 1 round for free. 10 + 1 =11 rounds you played a total of 35 rounds. This means you played 35 ÷ 11 3.1 = 3 times of 11 rounds + 2 rounds in 3 times of 11 rounds we purchased 30 rounds and got 3 free rounds plus 2 more rounds. Total round 35 rounds = 3(10 rounds) + 3 free rounds + 2 more rounds You purchased 30 + 2 = 32 rounds and got 3 free rounds. Question 9. Mathematical PRACTICE 1 Plan Your Solution Saketa is saving money to buy a new ferret cage. In the first week, she saved$24. Each week after the first, she saves $6. How much money will Saketa have saved in six weeks? Answer: She will save$60 in 6 weeks.
Explanation:
Given,
The amount she saved in the first week = $24 Each week she saves$6
Let us assume she saved x weeks after the first week
= 24 + 6x
if x = 5
Her total saving in 6 weeks is 24 + 6 (5)
24 + 30 = $54 ### McGraw Hill My Math Grade 5 Chapter 2 Lesson 5 My Homework Answer Key Problem Solving Solve each problem by making a table. Question 1. Mathematical PRACTICE 8 Look for a Pattern In a recent year, one United States dollar was equal to about 82 Japanese yen. How many Japanese yen are equal to$100? $1,000?$10,000?
82 yens, 8200 yens, 82000 yens, 820000 yens.
Explanation:
1 dollar = 82 yens
$100 = 100 × 82 = 8200 yens$1,000 = 1000 × 82 = 82000 yens
$10, 000 = 10,000 × 82 = 820000 yens. Question 2. A local restaurant offers a deal if you purchase 3 medium pizzas, you get 2 side dishes for free. If you get a total of 8 side dishes, how many pizzas did you buy? Answer: 12 medium pizzas. Explanation: He purchased 3 medium pizzas, he got 2 side dishes for free. for buying 6 medium pizzas we will get 4 side dishes free for buying 9 medium pizzas we will get 6 side dishes free for buying 12 medium pizzas we will get 8 side dishes free. Hence to get a total of 8 side dishes, we would buy 12 medium pizzas. Question 3. A recipe for potato salad calls for one teaspoon of vinegar for every 2 teaspoons of mayonnaise. How many teaspoons of vinegar are needed for 16 teaspoons of mayonnaise? Answer: 8 spoons of vinegar. Explanation: 1 teaspoon of vinegar is equal to 2 teaspoons of mayonnaise. Question 4. A package of 4 mechanical pencils comes with 2 free erasers. If you get a total of 1 2 free erasers, how many packages of pencils did you buy? Answer: 6 packages. Explanation: There are 6 packages of pencils. Question 5. Veronica is saving money to buy a saddle for her horse that costs$175. She plans to save $10 the first month and then increase the amount she saves by$5 each month after the first month. How many months will it take her to save $175? Answer: 7 months. Explanation: Veronica is saving money to buy a saddle for her horse that costs$175.
The amount increased every month by $5. Veronica saved in the first month =$10
$15 and$20
175 = n ÷ 2 (2 × 10 + (n-1)5)
350 = 5n2 + 15n
5n2 + 15n – 350 = 0
n2 + 3n – 70 = 0
n = -10, 7
Hence it will take 7 months for veronica to save \$175 for the saddle.
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### Home > CCA > Chapter 11 > Lesson 11.2.2 > Problem11-47
11-47.
Find an equation for each sequence.
1. n
t(n)
$1$
$7$
$2$
$4$
$3$
$1$
To find the equation, you need to know the initial term, when $n = 0$.
When $n = 1$, $t(n) = 7$, and when $n = 2$, $t(n) = 4$, so you know that it decreases by $3$ from one term to the next.
To find the number before $7$, you add $3$, because $3$ was taken away to get $7$. So the initial term is $10$.
To get the number after $10$, you take away $3$ once. To get the next number, you take away $3$ again. To get the next number you take away $3$ again. Since you are always subtracting $3$, you multiply $3$ by $n$ to represent how many times you have subtracted $3$. Now write your equation.
$t(n)=7−3(n−1)$
$t(n)=10−3n$
1. Year
Cost
$2$
$\2000$
$3$
$\6000$
$4$
$\18000$
This sequence is different than the one in part (a). You multiply instead of subtract.
The cost was $2000$ in Year $2$, but you need to know how much it was in Year $0$ to write the equation. Each year it is multiplied by $3$. So the cost in Year $2$ is $3$ times what it was in Year $1$, which is $3$ times what it was in Year $0$. So if you divide the cost in Year $2$ by $9$, you will have the cost from Year $0$.
Similar to part (a), because you are always multiplying by $3$, you can use a variable as the exponent of $3$ to represent how many times you have multiplied by $3$. Now write your equation.
$\textit{t}(\textit{n})=\left(\frac{2000}{3}\right)\cdot3^{\textit{n}-1}$
$t(n)=\left(\frac{2000}{9}\right)\cdot3^n$
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# 103.3.2 Descriptive Statistics
##### Make a look on Basic Descriptive Statistics..
In previous section, we studied about Basic Statistics, Graphs and Reports, now we will be studying about Descriptive Statistics.
As soon as we get some data, we can carry out descriptive statistics on it. Basic descriptive statistics give an idea on the variable and their distribution, we get an overall picture of dataset and it also helps us to create a report on the data. There are 2 types of basic descriptive statistics:
Central tendencies and Dispersion.
Central tendencies deal with the mean, median and mode, whereas the measures of Dispersion are range, variance and standard deviation.
Central tendencies: mean, median
Mean is nothing but the arithmetic mean or the average, i.e, the sum of the values divided by the count of values. It helps us to understand the data, evaluate the data. The mean is a good measure to calculate the average of the variables, but it is not recommended when there are outliers in the data. Outliers are fewer data elements in the dataset which are very much different from the rest of the data elements.
For Example : Let us consider this data
(1.5,1.7,1.9,0.8,0.8,1.2,1.9,1.4,9,0.7,1.1)
Nowhere 90% of the values are below 2, but when we calculate the mean, we get the value as 2. This is because there is a value (i.e.,9), which is very much different from the rest of the values. This is called an outlier. So in such cases, where there are outliers, we need a better approach which gives a more accurate or true middle value. Hence median can be considered in such cases. For calculating the median, the give data is sorted in either ascending or descending order, and then take the middle value which becomes the median, which can be a true average value in such cases
For example, consider the same data;
(1.5,1.7,1.9,0.8,0.8,1.2,1.9,1.4,9,0.7,1.1)
Ascending order:
(0.7,0.8,0.8,1.1,1.2,1.4,1.5,1.7,1.9,1.9,9)
Here the middle value is 1.4, which becomes the median.
Therefore, we can say that even if there are outliers present in the data, we can get a true middle value using median, as the sorting shifts the outliers to the extreme ends.
Let us see how to calculate mean and median in R. We consider the Income data.
`Income<-read.csv("C:\\Amrita\\Datavedi\\Census Income Data\\Income_data.csv")`
From this dataset we calculate the mean and median of the variable “capital.gain”.
```mean(Income\$capital.gain)
## [1] 1077.649
median(Income\$capital.gain)
## [1] 0```
We get mean as 1077.649 and median as 0. As there is a vast difference between the two, we can say that there are outliers in the data. If there are no outliers, there will not be much difference in the mean and median values. So if there are outliers we must always consider the median.
Lab: Mean & Median
Now let us consider the dataset, Online Retail Sales Data.
`Online_Retail<-read.csv("C:\\Amrita\\Datavedi\\Online Retail Sales Data\\Online Retail.csv")`
Calculate the mean and median of the variable “UnitPrice” and let us see if there are any outliers in the data.
```mean(Online_Retail\$UnitPrice)
## [1] 4.611114
median(Online_Retail\$UnitPrice)
## [1] 2.08```
So here the mean is 4.611114 and the median is 2.08, which means mean and meadian are very close. However, we still cannot conclude on the absence of an outlier because if there are balancing outliers on the either side of median, then also the mean and median can be close. Now also find the mean and median of the variable “Quantity”.
```mean(Online_Retail\$Quantity)
## [1] 9.55225
median(Online_Retail\$Quantity)
## [1] 3```
Here we can see that the mean is 9.55225 and the median is 3. In this case, as there is some difference in the mean and the median value, there can be outliers in the data but we cannot be sure. Outliers can be detected using box plot which will be covered in further sessions.
In next section, we will be studying about Percentile and Quartile.
11th October 2018
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# Tracing Straight Lines Worksheets
There are a number of different tracing straight lines worksheets that you can print out for practice, especially in mathematics and geometry. In these areas of study, exactitude is essential. While it’s possible to get a good idea of a problem by thinking about it in your head, it’s a lot easier to see it on paper. This is particularly important when you’re doing geometry problems.
Tracing straight line problems isn’t the same as tracing out a circle or figure. As far as geometry is concerned, you’re only tracing straight lines in relation to an imaginary line. This imaginary line doesn’t exist. In other words, you can never find a single straight line that actually exists from any point on a circle.
However, this concept exists in the real world. You just can’t visualize it directly, unless you make a point of working out a geometry problem, that way you’ll be able to see it clearly on paper. This will be helpful in the future when you are required to solve problems involving angles and measurements.
The next time you’re required to work out a problem involving straight lines, work out an angle. For every point on a circle, draw the side that faces away from you, and the side that faces toward you. For each one, make a simple square with two equal sides, so that you can plot the angle in a vertical line. Then, make the same square but turn it 90 degrees to create a right triangle.
When you have your line of points on the hypotenuse of the triangle, it’s easy to put a simple problem to work in your head. It will be easier if you think of a right triangle as a simple line that points to the right. When you do this, it will also become easier to find the equal sides of the triangle. Now, draw a straight line connecting the opposite side of the equal side.
You should now be able to know where the hypotenuse of the triangle is located, and you can also be confident that the equal sides are pointing in the same direction. At this point, you can use this knowledge to solve a triangle problem. One way to do this is to consider the angle between the two equal sides, and determine the length of the hypotenuse. You should then solve for this angle, or to put it another way, you should find the greatest arc length between these two equal sides.
Now, remember that we used the word “greatest” because the hypotenuse length is the length of the longest side of the triangle. Then, you should find the shortest of these arcs. This may seem like a long way to go, but it’s the most straightforward of all triangle problems.
This type of geometry problem is very useful to work on in mathematics class, especially during your math or physics lab exercises. You may find that when you think about this problem that you suddenly start to see all sorts of things that you didn’t before.
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# 2.11: Synthetic Division of Polynomials
Difficulty Level: At Grade Created by: CK-12
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If you have completed the lesson on oblique asymptotes, you probably know that the problem:
Find the oblique asymptote of \begin{align*}f(x)=\frac{3x^{3}-5x^{2}+2}{x^{2}-3x}\end{align*}
could be an excellent reason to groan and grumble due to the long division that would be required.
Isn't there a better way?
Embedded Video:
### Guidance
Synthetic Division
In this lesson, we explore synthetic division, a derivative of polynomial long division.
To illustrate the value of synthetic division, we will first solve an example with long division and then redo the division by the synthetic division method.
#### Example A
Divide: \begin{align*}f(x)=x^{3}+x^{2}-10x+13\end{align*} by \begin{align*}D(x)=x-2\end{align*}.
Solution A: long division
\begin{align*}& \qquad \qquad \qquad \quad \ \ x^2 + 3x - 4 \qquad \qquad \quad \leftarrow\text{Quotient}\\ & \text{Divisor}\rightarrow x-2 \ \big ) \overline{x^{3} +x^{2} -10x +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \qquad \qquad \quad \ \ x^{3} -2x^{2}\\ & \qquad \qquad \qquad \quad \ \ \searrow\\ & \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-10x\\ & \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-6x\\ & \qquad \qquad \qquad \qquad \qquad \searrow\\ & \qquad \qquad \qquad \qquad \qquad \quad \ -4x+13\\ & \qquad \qquad \qquad \qquad \qquad \quad \ \ -4x+8\\ & \qquad \qquad \qquad \qquad \qquad \qquad \ \searrow\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad +5 \qquad \ \leftarrow \text{Remainder}\end{align*}
Solution B: synthetic division
As you can see, all the math involved relates only to the coefficients of the variables and the constants. We could just as easily complete the division by omitting the variables, as long as we write the coefficients in the proper places.
Doing that, the problem looks like this: \begin{align*}& \qquad \quad \ \ x^2 + 3x - 4 \qquad \quad \leftarrow\text{Quotient}\\ & 1-2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \quad 1-2\\ & \qquad \ \ \searrow\\ & \qquad \qquad +3 -10\\ & \qquad \qquad \quad \ 3 -6\\ & \qquad \qquad \quad \ \searrow\\ & \qquad \qquad \qquad -4+13\\ & \qquad \qquad \qquad \ -4+8\\ & \qquad \qquad \qquad \quad \searrow\\ & \qquad \qquad \qquad \qquad \ +5 \qquad \leftarrow \text{Remainder}\end{align*}
Since the underlined numerals are repetitions of those immediately above them, we can shorten the process by simply deleting them. Further, since these underlined numbers are products of the numbers in the quotient by the 1 in the divisor, we eliminate this 1.
Thus we get the following \begin{align*}& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\ & -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \quad -2\\ & \qquad \ \searrow\\ & \qquad \quad \ +3 -10\\ & \qquad \qquad \quad \ -6\\ & \qquad \qquad \ \searrow\\ & \qquad \qquad \quad \ -4+13\\ & \qquad \qquad \qquad \quad +8\\ & \qquad \qquad \qquad \ \searrow\\ & \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}\end{align*}
It is also unnecessary to bring down the -10 and 13:
\begin{align*}& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\ & -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \quad -2\\ & \qquad \ \searrow\\ & \qquad \quad \ +3\\ & \qquad \qquad \quad \ -6\\ & \qquad \qquad \ \searrow\\ & \qquad \qquad \quad \ -4\\ & \qquad \qquad \qquad \quad +8\\ & \qquad \qquad \qquad \ \searrow\\ & \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}\end{align*}
Moving the numerals upward, we get
\begin{align*}& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\ & -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \quad \ \underline{ -2\ -6 \ +8\;\;}\\ & \qquad \quad \ +3-4 \ +5 \qquad \ \leftarrow \text{Remainder}\end{align*}
When the numeral 1 (the first number in the quotient) is brought down to the last line, it will contain the remainder and the quotient,
\begin{align*}& \qquad \quad -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\ & \qquad \qquad \qquad\ \underline{ -2\ -6 \ +8\;\;}\\ & \text{Quotient} \rightarrow 1 + 3 \ -4 \ +5 \qquad \ \leftarrow \text{Remainder}=+5\end{align*}
Therefore, the quotient is \begin{align*}Q(x)=x^{2}+3x-4\end{align*} and the remainder is \begin{align*}R(x)=5\end{align*}. This is synthetic division'.
#### Example B
Use synthetic division to find the quotient and the remainder of
\begin{align*}\frac{3x^{3}-8x+1}{x+2}\end{align*}
Solution
First we write the divisor \begin{align*}x+2\end{align*} in the form \begin{align*}x-c\end{align*}, as \begin{align*}x-(-2)\end{align*}. Then use -2 as a “divisor” in the synthetic division as follows:
\begin{align*}& \-2 \ \big ) \overline{3 \ \ \ 0 \ -8 \ \ \ \ 1}\\ & \quad \ \underline{\downarrow -6 \ \ \ 12 -8}\\ & \quad \ 3 -6 \ \ \ 4 \ -7\end{align*}
Notice that 0 is used as the coefficient of the “missing” \begin{align*}x^{2}\end{align*} term. Also, we wrote the coefficients of the dividend in descending order. The process of synthetic division is as follows: Bring down the first coefficient 3 and multiply by -2 (the divisor) to get -6, and then add 0 to -6 to get -6. Next, multiply -6 by -2 (the divisor) to get 12, and then add -8 to 12 to get 4. Finally, multiply 4 by -2 to get -8, and then add 1 to -8 to get the remainder, -7. As a result of this process, the quotient is
\begin{align*}Q(x)=3x^{2}-6x+4\end{align*}
and the remainder is
\begin{align*}R(x)=-7\end{align*}
In other words, since
\begin{align*}f(x) & = D(x)Q(x)+R(x)\\ & = (x+2)(3x^2-6x+4)+(-7)\end{align*}
Remember, this method will only work when the divisor is in the form \begin{align*}x-c\end{align*}, that is, when the coefficient of \begin{align*}x\end{align*} in the divisor is 1.
#### Example C
If \begin{align*}h(x)=x^{3}-2x^{2}+5x-3\end{align*}, evaluate
• \begin{align*}h(1)\end{align*}
• \begin{align*}h(-2)\end{align*}
• \begin{align*}h\left ( \frac{1}{2} \right )\end{align*}
Solution
We can simply use the synthetic division to evaluate \begin{align*}h(x)\end{align*} at the given values. By the remainder theorem, the remainder is equal to \begin{align*}h(c)\end{align*}.
• Using synthetic division,
\begin{align*}& \ 1 \ \big ) \overline{1 -2 \ \ \ 5 \ -3}\\ & \quad \ \ \underline{\downarrow \ \ 1 -1 \ \ \ \ 4}\\ & \quad \ \ 1 -1 \ \ \ 4 \ \ \ \ 1\end{align*}
Hence, \begin{align*}h(1)=1\end{align*}.
• By synthetic division,
\begin{align*}& \ -2 \ \big ) \overline{1 -2 \ \ \ 5 \ -3\;}\\ & \qquad \quad \underline{\downarrow \ \ 2 \ \ \ 8 \ \ \ \ 26\;}\\ & \qquad \quad 1 -4 \ 13 -29\end{align*}
Hence, \begin{align*}h(-2)=-29\end{align*}.
• By synthetic division,
\begin{align*}& \ \frac{1}{2} \ \big ) \overline{1 -2 \ \ \ 5 \ -3\;}\\ & \qquad \underline{\downarrow \ \ \frac{1}{2} \ \frac{-3}{4} \ \ \frac{17}{8}\;}\\ & \qquad 1 \ \ \frac{-3}{2} \ \frac{17}{4} \ \frac{-7}{8}\end{align*}
Hence, \begin{align*}h\left ( \frac{1}{2} \right )=\frac{-7}{8}\end{align*}.
### Vocabulary
Synthetic division is a concise 'shortcut' method of dividing polynomials.
Polynomial long division is the standard method of long division, applied to the division of polynomials.
A Dividend is the number (or polynomial) being divided, a divisor is the number (or polynomial) being divided 'into' the dividend, and a quotient is the result, occasionally with a remainder (the amount left over after all the even division has been completed).
### Guided Practice
Problems
Divide using Long Division:
1) \begin{align*}\frac{8x^3 - 7x^2 +10x - 5}{2x + 1}\end{align*}
2) \begin{align*}\frac{x^3 +5x -4}{x^2 - x + 1}\end{align*}
Divide using Synthetic Division:
3) \begin{align*}\frac{2x^4 + 5x^3 -2x -8}{x + 3}\end{align*}
Solutions
1) Solved in the video below:
2) Solved in the video below:
3) Solved in the video below:
### Practice
Divide by using long division:
1. \begin{align*}(20x^2 - 13x + 2)\end{align*} divided by \begin{align*}(4x - 1)\end{align*}
2. \begin{align*}(x^2 - 2x + 3)\end{align*} divided by \begin{align*}(x + 5)\end{align*}
3. \begin{align*}\frac{y^4 - y^2 - 6y}{y^2 - 2}\end{align*}
4. \begin{align*}(x^3 + 2x^2 - x - 2)\end{align*} divided by \begin{align*}(x + 2)\end{align*}
5. \begin{align*}\frac{x^4 - 1}{x^2 +1}\end{align*}
Divide using synthetic division:
1. \begin{align*}(7x^2 - 23x + 6)\end{align*} divided by \begin{align*}(x - 3)\end{align*}
2. \begin{align*}(x^4 - 5x + 10)\end{align*} divided by \begin{align*}(x + 3)\end{align*}
3. \begin{align*}(2x^2 + 13x - 8)\end{align*} divided by \begin{align*}(x - \frac{1}{2})\end{align*}
4. \begin{align*}(x^4 + 6x^3 + 6x^2)\end{align*} divided by \begin{align*}(x + 5)\end{align*}
5. \begin{align*}\frac{x^3 - 7x - 6}{x + 2}\end{align*}
6. \begin{align*}(8y^3 + y^4 + 16 + 32y + 24y^2)\end{align*} divided by \begin{align*}(y + 2)\end{align*}
Use synthetic substitution to evaluate the polynomial function for the given value:
1. \begin{align*}P(x) = 2x^2 - 5x - 3\end{align*} for \begin{align*}x = 4\end{align*}
2. \begin{align*}P(x) = 4x^3 - 5x^2 + 3 \end{align*} for \begin{align*}x = -1\end{align*}
3. \begin{align*}p(x) = 3x^3 - 5x^2 - x =2\end{align*} for \begin{align*}x = -\frac{1}{3}\end{align*}
4. The area of a rectangle is \begin{align*} 3x^3 - 11x^2 - 56x - 48\end{align*} and the length is \begin{align*}3x + 4\end{align*} What is the width?
5. A group of geologists have taken a collection of samples of a substance from a proposed mining site and must identify the substance. Each sample is roughly cylindrical. The volume of each sample as a function of cylinder height (in centimeters) is \begin{align*}V(h) = \frac {1}{4} \pi h^3\end{align*}. The mass (in grams) of each sample in terms of height an be modeled by \begin{align*}M(h) = \frac{1}{4}h^3 - h^2 + 5h\end{align*}. Write an expression that represents the density of the samples. (Hint: \begin{align*}D = \frac{M}{V}\end{align*})
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Dividend In a division problem, the dividend is the number or expression that is being divided.
divisor In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Polynomial long division Polynomial long division is the standard method of long division, applied to the division of polynomials.
Quotient The quotient is the result after two amounts have been divided.
Remainder A remainder is the value left over if the divisor does not divide evenly into the dividend.
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# Problems on Algebraic Fractions
Here we will learn how to simplify the problems on algebraic fractions to its lowest term.
1. Reduce the algebraic fractions to their lowest terms: $$\frac{x^{2} - y^{2}}{x^{3} - x^{2}y}$$
Solution:
$$\frac{x^{2} - y^{2}}{x^{3} - x^{2}y}$$
Factorizing the numerator and denominator separately and cancelling the common factors we get,
= $$\frac{(x + y) (x - y)}{x^{2} (x - y)}$$
= $$\frac{x + y}{x^{2}}$$
2. Reduce to lowest terms $$\frac{x^{2} + x - 6}{x^{2} - 4}$$
Solution:
$$\frac{x^{2} + x - 6}{x^{2} - 4}$$
Step 1: Factorize the numerator x$$^{2}$$ + x – 6
= x$$^{2}$$ + 3x – 2x – 6
= x(x + 3) – 2(x + 3)
= (x + 3) (x – 2)
Step 2: Factorize the denominator: x$$^{2}$$ – 4
= x$$^{2}$$ – 2$$^{2}$$
= (x + 2) (x – 2)
Step 3: From steps 1 and 2: $$\frac{x^{2} + x - 6}{x^{2} - 4}$$
= $$\frac{x^{2} + x - 6}{x^{2} - 2^{2}}$$
= $$\frac{(x + 3) (x - 2)}{(x + 2) (x - 2)}$$
= $$\frac{(x + 3)}{(x + 2)}$$
3. Simplify the algebraic fractions $$\frac{36x^{2} - 4}{9x^{2} + 6x + 1}$$
Solution:
$$\frac{36x^{2} - 4}{9x^{2} + 6x + 1}$$
Step 1: Factorize the numerator: 36x$$^{2}$$ – 4
= 4(9x$$^{2}$$ – 1)
= 4[(3x)$$^{2}$$ – (1)$$^{2}$$]
= 4(3x + 1) (3x – 1)
Step 2: Factorize the denominator: 9x$$^{2}$$ + 6x + 1
= 9x$$^{2}$$ + 3x + 3x + 1
= 3x(3x + 1) + 1(3x + 1)
= (3x + 1) (3x + 1)
Step 3: Simplification of the given expression after factorizing the numerator and the denominator:
$$\frac{36x^{2} - 4}{9x^{2} + 6x + 1}$$
= $$\frac{4(3x + 1)(3x - 1)}{(3x + 1)(3x + 1)}$$
= $$\frac{4(3x - 1)}{(3x + 1)}$$
4. Reduce and simplify: $$\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )$$
Solution:
$$\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )$$
$$\frac{8x^{3}y^{2}z}{2xy^{3}} of \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \times \frac{35x^{2}yz^{3}}{7xy^{2}}$$
$$\frac{4x^{3}y^{2}z}{xy^{3}} \left ( \frac{x^{5}y^{2}z^{2}}{xy^{3}z} \times \frac{x^{2}yz^{3}}{xy^{2}} \right )$$
= 4x$$^{10 - 3}$$ ∙ y$$^{-3}$$ ∙ z$$^{5}$$
$$\frac{4x^{7}\cdot z^{5}}{y^{3}}$$
5. Simplify: $$\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}$$
Solution:
$$\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}$$
Step 1: First factorize each of the polynomials separately:
2x$$^{2}$$ – 3x – 2 = 2x$$^{2}$$ – 4x + x – 2
= 2x(x – 2) + 1 (x – 2)
= (x – 2) (2x + 1)
x$$^{2}$$ + x – 2 = x$$^{2}$$ + 2x - x – 2
= x(x + 2) - 1 (x + 2)
= (x + 2) (x - 1)
2x$$^{2}$$ + 3x + 1 = 2x$$^{2}$$ + 2x + x + 1
= 2x(x + 1) + 1 (x + 1)
= (x + 1) (2x + 1)
3x$$^{2}$$ + 3x – 6 = 3[x$$^{2}$$ + x – 2]
= 3[x$$^{2}$$ + 2x - x – 2]
= 3[x(x + 2) – 1(x + 2)]
= 3[(x + 2) (x - 1)]
= 3[(x + 2) (x - 1)]
= 3(x + 2) (x - 1)
Step 2: Simplify the given expressions by substituting with their factors
$$\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \div \frac{2x^{2} + 3x + 1}{3x^{2} + 3x - 6}$$
$$\frac{2x^{2} - 3x - 2}{x^{2} + x - 2} \times \frac{3x^{2} + 3x - 6}{2x^{2} + 3x + 1}$$
$$\frac{(x - 2) (2x + 1)}{(x + 2) (x - 1)}\times\frac{3(x + 2) (x - 1)}{(x + 1) (2x + 1)}$$
$$\frac{3(x - 2)}{(x + 1)}$$
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# If ${{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}} = {\text{tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x$, then ${\text{y'}}\left( 0 \right)$ can be A) $\dfrac{1}{{3\sqrt \pi }}$B) $- \dfrac{1}{{3\sqrt \pi }}$C) $- \dfrac{1}{{5\sqrt \pi }}$D) $- \dfrac{1}{{3\sqrt {5\pi } }}$
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Hint: This problem comes under implicit function on differentiation. We need to solve separately and want to differentiate on the function and want to find y and then there will be re arranging and substitute to the equation to compare the coordinates for the solving $y'(0) = \dfrac{{dy}}{{dx}}$ which first order differentiation and there will be multiple solvable equation and then complete step by step explanation.
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}$ ${{\text{e}}^{\operatorname{Sin} ({x^2} + {y^2})}} = {\text{ tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x - - - - - - - - (1)$
When $x = 0$, then the inverse function will be zero and the values of $x$ will be so, then we get
$\Rightarrow {{\text{e}}^{\sin {y^2}}} = {\text{ tan }}\dfrac{{{y^2}}}{4}$
Taking $\log$ on both sides,
$\Rightarrow \sin {y^2} = {\text{ log tan }}\dfrac{{{y^2}}}{4}$
Hence we get,
$y = \pm \sqrt x , \pm \sqrt {5x} ,..............$
Now differentiating equation (1) with respect to x, we get
Let us solve separately Left hand side and Right hand side,
Now differentiate Left hand side
$\Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}$
$\Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}\cos \left( {{x^2} + {y^2}} \right)\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right)$
Now differentiate Right hand side
$\Rightarrow {\text{tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x$
Differentiating we get,
$\Rightarrow \dfrac{{2y}}{4}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} + \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Now again substitute $x = 0$ and compare,
$\Rightarrow {{\text{e}}^{\sin {y^2}}}\left[ {\cos {y^2}\left( {2y\dfrac{{dy}}{{dx}}} \right)} \right] = \dfrac{y}{2}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} - - - - (2)$
Now substitute $x$ value and $y$ value in equation (2), we get $y'(0) = \dfrac{{dy}}{{dx}}$
Now, at $x = 0$, $y = \sqrt \pi$ in equation in (2),
$\Rightarrow {{\text{e}}^{\sin {{(\sqrt \pi )}^2}}}[\cos {(\sqrt \pi )^2}2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{{{{\sqrt \pi }^2}}}{4}\dfrac{{dy}}{{dx}}$
Now separate $\dfrac{{dy}}{{dx}}$,
$\Rightarrow {{\text{e}}^{\sin \pi }}[\cos \pi ]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}}$
We know that $\sin \pi = 0$ and $\cos \pi = - 1$,
$\Rightarrow {{\text{e}}^0}[ - 1]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{(\sec \dfrac{\pi }{4})^2}\dfrac{{dy}}{{dx}}$
We know that ${\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}} = 2$ and ${e^0} = 1$,
$\Rightarrow - 2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2} \times 2\dfrac{{dy}}{{dx}}$
Cancelling the term $2$ in denominator and numerator,
$\Rightarrow$$- 2\sqrt \pi \dfrac{{dy}}{{dx}} = \sqrt \pi \dfrac{{dy}}{{dx}}$
Rearranging the terms we get,
$\Rightarrow$$- 2\sqrt \pi \dfrac{{dy}}{{dx}} - \sqrt \pi \dfrac{{dy}}{{dx}} = 0$
Taking common term same as in both terms,
$\Rightarrow$$( - 2\sqrt \pi - \sqrt \pi )\dfrac{{dy}}{{dx}} = 0$
Subtracting the terms we get,
$\Rightarrow$$- 3\sqrt \pi \dfrac{{dy}}{{dx}} = 0$
Hence we get,
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{3\sqrt \pi }}$
Similarly we can find for when $x = 0,y = - \sqrt \pi$ we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt \pi }}$
Similarly we can find for when$x = 0,y = \sqrt {5\pi }$, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = - \dfrac{1}{{3\sqrt {5\pi } }}$
Similarly we can find for when $x = 0,y = - \sqrt {5\pi }$, we get
$\Rightarrow$$\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}$
There will be multiple answers and for this question, these are,
$\Rightarrow$$\dfrac{1}{{3\sqrt \pi }}$, $- \dfrac{1}{{3\sqrt \pi }}$ and $- \dfrac{1}{{3\sqrt {5\pi } }}$
$\therefore$ The correct answers are option A) $\dfrac{1}{{3\sqrt \pi }}$, B) $- \dfrac{1}{{3\sqrt \pi }}$ and D) $- \dfrac{1}{{3\sqrt {5\pi } }}$
Note: This problem needs attention on differentiation and some trigonometric identities, this kind of problem will be able to solve when the when before differentiation and after differentiation for when find x values and y values for finding the first order differential and then simple basic calculation for that arrange and substitute the value in order to find solution.
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## Example Questions
### Example Question #1 : Parallelograms
In the above parallelogram, is acute. Which is the greater quantity?
(A) The area of the parallelogram
(B) 120 square inches
(B) is greater
It is impossible to determine which is greater from the information given
(A) and (B) are equal
(A) is greater
(B) is greater
Since is acute, a right triangle can be constructed with an altitude as one leg and a side as the hypotenuse, as is shown here. The height of the triangle must be less than its sidelength of 8 inches.
The height of the parallelogram must be less than its sidelength of 8 inches.
The area of the parallelogram is the product of the base and the height – which is
Therefore,
(B) is greater.
### Example Question #2 : Parallelograms
Parallelogram A is below:
Parallelogram B is below:
Note: These figures are NOT drawn to scale.
Refer to the parallelograms above. Which is the greater quantity?
(A) The area of parallelogram A
(B) The area of parallelogram B
It is impossible to determine which is greater from the information given
(A) is greater
(B) is greater
(A) and (B) are equal
(A) and (B) are equal
The area of a parallelogram is the product of its height and its base; its slant length is irrelevant. Both parallelograms have the same height (8 inches) and the same base (1 foot, or 12 inches), so they have the same areas.
### Example Question #3 : Parallelograms
Figure NOT drawn to scale
The above figure shows Rhombus ; and are midpoints of their respective sides. Rectangle has area 150.
Give the area of Rhombus .
A rhombus, by definition, has four sides of equal length. Therefore, . Also, since and are the midpoints of their respective sides,
We will assign to the common length of the four half-sides of the rhombus.
Also, both and are altitudes of the rhombus; the are congruent, and we will call their common length (height).
The figure, with the lengths, is below.
Rectangle has dimensions and ; its area, 150, is the product of these dimensions, so
The area of the entire Rhombus is the product of its height and the length of a base , so
.
### Example Question #1 : Parallelograms
In the above parallelogram, is acute. Which is the greater quantity?
(A) The perimeter of the parallelogram
(B) 46 inches
(B) is greater
(A) and (B) are equal
(A) is greater
It is impossible to determine which is greater from the information given
(A) and (B) are equal
The measure of is actually irrelevant. The perimeter of the parallelogram is the sum of its four sides; since opposite sides of a parallelogram have the same length, the perimeter is
inches,
making the quantities equal.
### Example Question #61 : Quadrilaterals
Parallelogram A is below:
Parallelogram B is below:
Note: These figures are NOT drawn to scale.
Refer to the parallelograms above. Which is the greater quantity?
(A) The perimeter of parallelogram A
(B) The perimeter of parallelogram B
(B) is greater
(A) and (B) are equal
(A) is greater
It is impossible to determine which is greater from the information given
(A) is greater
The perimeter of a parallelogram is the sum of its sidelengths; its height is irrelevant. Also, opposite sides of a parallelogram are congruent.
The perimeter of parallelogram A is
inches;
The perimeter of parallelogram B is
inches.
(A) is greater.
### Example Question #62 : Quadrilaterals
Figure NOT drawn to scale.
The above figure depicts Rhombus with and .
Give the perimeter of Rhombus .
All four sides of a rhombus have the same length, so we can find the perimeter of Rhombus by taking the length of one side and multiplying it by four. Since , the perimeter is four times this, or .
Note that the length of is actually irrelevant to the problem.
### Example Question #63 : Quadrilaterals
In Parallelogram , and . Which of the following is greater?
(A)
(B)
It cannot be determined which of (a) and (b) is greater
(a) is the greater quantity
(a) and (b) are equal
(b) is the greater quantity
It cannot be determined which of (a) and (b) is greater
In Parallelogram , and are adjoining sides; there is no specific rule for the relationship between their lengths. Therefore, no conclusion can be drawn of and , and no conclusion can be drawn of the relationship between and .
### Example Question #64 : Quadrilaterals
Which of the following can be the measures of the four angles of a parallelogram?
Opposite angles of a parallelogram must have the same measure, so the correct choice must have two pairs, each of the same angle measure. We can therefore eliminate and as choices.
Also, the sum of the measures of the angles of any quadrilateral must be , so we add the angle measures of the remaining choices:
:
, so we can eliminate this choice.
:
, so we can eliminate this choice.
; this is the correct choice.
### Example Question #65 : Quadrilaterals
Refer to the above figure, which shows a parallelogram. What is equal to?
Not enough information is given to answer this question.
The sum of two consecutive angles of a parallelogram is .
157 is the correct choice.
### Example Question #66 : Quadrilaterals
In Parallelogram , and .
Which is the greater quantity?
(a)
(b)
(b) is the greater quantity
(a) is the greater quantity
It cannot be determined which of (a) and (b) is greater
(a) and (b) are equal
(b) is the greater quantity
In Parallelogram , and are opposite angles and are therefore congruent. This means that
Both are positive, so .
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## Math 160: Study Guide - Chapter 8 & 9
1. Solve a strictly determined game. Show all work.
2. Solve a game. Show all work. Know the formulas from section 8.2.
3. Show that a the square of a probability matrix is also a probability matrix. To do this, you must find the square, and then show that the rows add to one.
4. Show the strategies and find the value of the game if both players randomly select a strategy. Show the strategies and find the value of the game if one player randomly chooses a strategy, but the other player uses their optimal strategy (use the calculator to find the optimal strategy).
5. Setup a game problem and then use the calculator to find the solution.
6. Markov chain problem. Fill in the missing probabilities in the transition diagram, find the transition matrix, find the limiting and steady-state matrices.
7. Markov chain application problem. Write the transition matrix, initial state matrix. Find the second state matrix and the third state matrix. Look at problems 9.1.43 - 53*.
8. Solve a game using the calculator. The column player lets the row player know what her strategy will be, find the best a priori strategy for the row player (this is the best strategy for the row player if he knows what the column player will do) and the value of the game under this strategy. Find the value of the game if the row player uses his a priori strategy, but the column player uses her optimal strategy instead of the one she said she was going to use. Find the best a priori strategy for the column player (what should she play if she knows what the row player will do because he thinks he knows what she is going to do).
9. Write the linear programming problems necessary to solve a game. Then, find the solution using the geometric approach to linear programming.
10. Find the solution to a game using the calculator. Find the value if both players play randomly (all choices are not equally likely - you need to have a concept of what a relative frequency is). Find the payoffs for the row player under the expected value criterion, maximax criterion, and maximin criterion.
11. Absorbing Markov chain application problem. Draw a transition diagram, write the transition matrix, find the fundamental matrix F, the limiting matrix. Find various long-term probabilities and the average duration spent in the matrix. Look at problems 9.3.39 - 9.3.45*.
12. Absorbing Markov chain problem. Create the transition matrix. Find the number of moves before exiting the system. Find the probability of ending up in a particular absorbing state when starting in a transient state.
### Notes
• The solution to a game consists of the optimal row strategy P*, the optimal column strategy Q*, and the value of the game v.
• When solving a game, first check for strictly determined games. Then check for recessive rows or columns.
• The calculator can be used to solve all the problems (except 3), but show work on the first three problems.
• Problems 1 - 7 must be worked individually and turned in before picking up the second half of the test.
• Problems 8 - 12 may be worked in groups of up to three people or alone using notecards. The notecards may contain formulas for absorbing Markov chains and the geometric approach to linear programming.
• Move swiftly through the first portion of the test - the group problems will be lengthy.
# 1 2 3 4 5 6 7 8 9 10 11 12 Total Pts 4 4 4 8 8 9 9 10 8 14 14 8 100
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# Ex.13.2 Q6 Surface Areas and Volumes Solution - NCERT Maths Class 10
Go back to 'Ex.13.2'
## Question
A solid iron pole consists of a cylinder of height $$220\,\rm{cm}$$ and base diameter $$24\,\rm{cm},$$ which is surmounted by another cylinder of height $$60\,\rm{cm}$$ and radius $$8\,\rm{cm}$$.
Find the mass of the pole, given that $$1\,\rm {cm^3}$$ of iron has approximately $$8\,\rm{g}$$ mass.
$$\left( {{\text{Use}}\,\,\pi \,{\text{ = }}\,{\text{3}}{\text{.14}}} \right)$$
## Text Solution
What is known?
A solid iron pole consisting of a cylinder of height $$220\,\rm{cm}$$ and base diameter $$24\,\rm {cm}$$ which is surmounted by another cylinder of height $$60\,\rm {cm}$$ and radius $$8\,\rm {cm}$$
Mass of $$1\;\rm{cm}^3$$ iron $$= 8\rm\,{g}$$
What is unknown?
The mass of the solid iron pole
Reasoning:
Draw the figure to visualize the iron pole
Visually it’s clear that
Volume of the solid iron pole $$=$$ volume of larger cylinder $$+$$ volume of smaller cylinder
Mass of iron in the pole $$=8\,\rm{g}$$ $$\;\times\;$$volume of the solid iron pole in $${\text{c}}{{\text{m}}^3}$$
We will find the volume of the solid by using formula;
Volume of the cylinder$$= \pi {r^2}h$$
where $$r$$ and $$h$$ are the radius and height of the cylinder respectively.
Steps:
Radius of larger cylinder,\begin{align}R = \frac{{24cm}}{2} = 12cm\end{align}
Height of larger cylinder,$$H = 220cm$$
Radius of smaller cylinder,$$r = 8cm$$
Height of smaller cylinder,$$h = 60cm$$
Volume of the solid iron pole $$=$$ volume of larger cylinder $$+$$ volume of smaller cylinder
\begin{align}&= \pi {R^2}H + \pi {r^2}h\\&= \pi \left( {12cm \times 12cm \times 220cm + 8cm \times 8cm \times 60cm} \right)\\&= 3.14 \times \left( {31680c{m^3} + 3840c{m^3}} \right)\\&= 3.14 \times 35520c{m^3}\\&= 111532.8c{m^3}\end{align}
Mass of $$1cm^3$$ iron is $$8g$$
Mass of iron in the pole $$= 8g ×$$ volume of the solid iron pole in $$cm^3$$
\begin{align}&= 8g \times 111532.8\\&= 892262.4g\\&= \frac{{892262.4}}{{1000}}kg\\&= 892.2624kg\end{align}
Mass of iron in the pole is $$892.26\,\rm{ kg}$$
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HomeEducationDê O Valor Na Forma Decimal De
# Dê O Valor Na Forma Decimal De
## Introduction
Decimal notation, often referred to as the base-10 numeral system, is the most commonly used system for representing numbers in everyday life. It’s the foundation for our understanding of quantities, transactions, and measurements. “Dê o valor na forma decimal de” translates to “Give the value in decimal form of” in English, and it’s a phrase often used in mathematical and financial contexts. In this article, we will delve into the significance of decimal notation, its history, and its role in our daily lives.
## What is Decimal Notation?
Decimal notation is a system of representing numbers using ten distinct symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These symbols are called digits, and they can be combined in various ways to represent any positive or negative whole number, fraction, or decimal number. The position of a digit in a number determines its value, and each position is a power of 10. For instance, in the number 437.29, the ‘4’ represents 4 hundreds, the ‘3’ represents 3 tens, the ‘7’ represents 7 ones, the ‘2’ represents 2 tenths, and the ‘9’ represents 9 hundredths.
## History of Decimal Notation
Decimal notation has a long and fascinating history. The earliest known use of decimal notation dates back to ancient civilizations like the Egyptians and the Babylonians, who used base-10 number systems. However, it was the Indian mathematicians who made significant contributions to the development of decimal notation.
Around the 5th century CE, Indian mathematicians invented the decimal system as we know it today. They introduced the concept of a decimal point, which allowed them to represent fractions and decimal numbers efficiently. This innovation revolutionized mathematics and made complex calculations more accessible.
## Decimal Notation in Modern Mathematics
Decimal notation is fundamental in modern mathematics, making it easier to perform arithmetic operations, compare numbers, and understand the relationships between them. It’s especially crucial in fields like science, engineering, economics, and finance, where precision is paramount.
In mathematics, decimal notation plays a vital role in teaching and learning. Students are introduced to it at an early age, and it serves as the foundation for more advanced mathematical concepts. Understanding decimal notation is a stepping stone to mastering concepts like percentages, ratios, and proportions.
## Applications in Everyday Life
Decimal notation is not confined to the realm of mathematics; it has a profound impact on our daily lives. Here are a few examples of how it is used:
1. Money: Decimal notation is crucial in financial transactions. Currencies are divided into smaller units, such as cents in the US dollar or pence in the British pound, with two decimal places. This allows us to make precise calculations when handling money.
2. Measurements: Decimal notation is used in units of measurement, such as meters, liters, and grams. It ensures that measurements are accurate and easily convertible between different units.
3. Cooking and Baking: Recipes often require precise measurements, which are typically expressed in decimal form. This ensures that the dishes turn out as expected.
4. Science and Engineering: Scientists and engineers rely on decimal notation to express measurements and perform calculations in their respective fields. Precision is crucial in experiments and designs.
5. Stock Market: Financial markets use decimal notation to represent stock prices and changes in value. Traders and investors make decisions based on these decimal representations.
## Conclusion
Decimal notation is an essential part of our daily lives, whether we realize it or not. Its history is rich, dating back to ancient civilizations, and its influence on modern mathematics, science, and commerce is undeniable. Understanding decimal notation is not just a mathematical skill; it is a practical tool that empowers us to navigate the complex world of numbers with precision and confidence. So, the next time you encounter the phrase “Dê o valor na forma decimal de,” remember the significance of decimal notation in our world.
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# Cumulative Frequency Histogram
How to make a cumulative frequency histogram from the given data and solve the related problems: examples and their solutions.
## Example 1: Making the Cumulative Frequency Table
To draw the cumulative frequency histogram,
first make a column for the Cumulative Frequency
right next to the given table.
For 50-59 interval,
the frequency is 2.
So the cumulative frequency is 2.
For 60-69 interval,
the frequency is 3.
So the cumulative frequency is
2 + 3 = 5.
For 70-79 interval,
the frequency is 7.
So the cumulative frequency is
5 + 7 = 12.
For 80-89 interval,
the frequency is 6.
So the cumulative frequency is
12 + 6 = 18.
For 90-99 interval,
the frequency is 2.
So the cumulative frequency is
18 + 2 = 20.
Use the Interval column
and the Cumulative Frequency column
to draw a cumulative histogram.
Draw the histogram axes like this.
Set the vertical axis
as Cumulative Frequency:
from 0 to 20.
And set the horizontal axis
as Test Scores:
from 50-59 to 90-99.
Frequency histogram
The cumulative frequency of 50-59 is 2.
So draw a vertical bar in the histogram
whose height is 2.
The cumulative frequency of 60-69 is 5.
So draw a vertical bar
whose height is 5.
(between 4 and 6)
Draw the bar right next to the previous bar
so that there would be no space between the bars.
The cumulative frequency of 70-79 is 12.
So draw a vertical bar
whose height is 12.
Draw the bar right next to the previous bar.
The cumulative frequency of 80-89 is 18.
So draw a vertical bar
whose height is 18.
Draw the bar right next to the previous bar.
The cumulative frequency of 90-99 is 20.
So draw a vertical bar
whose height is 20.
Draw the bar right next to the previous bar.
So this cumulative histogram is the answer.
## Formula: Cumulative Frequency and Percent
By using the cumulative histogram,
it's easy to find the percent of the parts.
If the cumulative frequency is the number of the parts,
use this formula.
(percent) = [x/n]⋅100
x: Cumulative frequency
n: Number of data
## Example 2
The number of students that score less than 80
is the height of the interval 70-79:
12.
This 12 is the number of students
that score from minimum to 79.
x = 12
n = 20
(percent) = [12/20]⋅100
Reduce 12 to 6
and reduce 20 to 10.
Cancel 10
and reduce 100 to 10.
6⋅10 = 60
## Example 3
Recall that
the quartiles divide the data into four parts.
Quartiles
So the percent of the quartiles are
Q1 = 25%, Q2 = 50%, and Q3 = 75%.
Then set the number of students
of 25% as x1, 50% as x2, and 75% as x3.
The goal is to find x1, x2, and x3,
then find the intervals that have these values.
The cumulative frequency of 25% is x1.
n = 20
So [x1/20]⋅100 = 25
Then x1 = 5.
The cumulative frequency of 50% is x2.
n = 20
So [x2/20]⋅100 = 50.
Then x2 = 10.
The cumulative frequency of 75% is x3.
n = 20
So [x3/20]⋅100 = 75
Then x3 = 15.
The cumulative frequency of 25% is 5.
And the cumulative frequency 5
is in 60-69 interval.
So Q1 is in 60-69 interval.
The cumulative frequency of 50% is 10.
And the cumulative frequency 10
is in 70-79 interval.
So Q2 is in 70-79 interval.
The cumulative frequency of 75% is 15.
And the cumulative frequency 15
is in 90-99 interval.
So Q3 is in 90-99 interval.
So Q1 is in 60-69 interval.
Q2 is in 70-79 interval.
And Q3 is in 90-99 interval.
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1.6 Multiplying and Dividing Real Numbers
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# 1.6 Multiplying and Dividing Real Numbers - PowerPoint PPT Presentation
1.6 Multiplying and Dividing Real Numbers. Objective: SWBAT use the properties of multiplication and division. Mini Quiz 6. Simplify -5 + (-2) – (-3) Simplify Evaluate when m = 2 and n = -3. Properties of Multiplication. Identity Property of Multiplication For every real number n,
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### 1.6 Multiplying and Dividing Real Numbers
Objective: SWBAT use the properties of multiplication and division.
Mini Quiz 6
• Simplify -5 + (-2) – (-3)
• Simplify
• Evaluate when m = 2 and n = -3
Properties of Multiplication
• Identity Property of Multiplication
For every real number n,
1 • n = n and n • 1 = n
• Multiplication Property of Zero
For every real number n,
n • 0 = 0 and 0 • n = 0
• Multiplication Property of -1
For every real number n,
-1 • n = -n and n • -1 = -n
• Inverse Property of Multiplication
For every nonzero real number n, there is a multiplicative inverse such that
Multiplying Negatives
+
(+)(+) = ?
(–)(+) = ?
(–)(–)(–)(–)(–) = ?
(+)(–) = ?
(–)(–) = ?
+
*Odd number of negatives = -
Even number of negatives = +
Multiplying Numbers
36
• -9(-4) =
• 4(-6) =
• -4.9(-8) =
-24
39.2
Try These…
Evaluate each expression for c = -8 and d = -7
• -2cd
• (-2)(-1)(cd)
• c(-d)
-112
112
-56
Exponential Expressions
Write in Expanded Form and Simplify:
(-3)(-3)(-3)(-3)
= 81
-3•3•3•3
= -81
(-0.3)(-0.3)
= 0.09
Dividing Numbers
• -42 ÷ 7 =
• -8 ÷ (-2) =
• 8 ÷ (-8) =
• -39 ÷ (-3) =
-6
4
-1
13
Division Using the Reciprocal
17. Evaluate for
and
Wrap Up
• Properties of Multiplication
• Multiply Positive and Negative Numbers
• (+) (+) = +
• (+) (–) = –
• (–) (+) = –
• (–) (–) = +
• Exponential Expressions
• Dividing Positive and Negative Numbers
• Dividing Fractions
Homework: P. 41 #1-61 EOO, 101-109 odd
Write a DLUQ and answer the Question:
If you multiply an even number of negatives, is the answer positive or negative?
|
## Learning Objectives
1. Compute the sum of squares Y
2. Convert raw scores to deviation scores
3. Compute predicted scores from a regression equation
4. Partition sum of squares Y into sum of squares predicted and sum of squares error
5. Define r2 in terms of sum of squares explained and sum of squares Y.
One useful aspect of regression is that it can divide the variation in Y into two parts: the variation of the predicted scores and the variation in the errors of prediction. The variation of Y is called the sum of squares Y and is defined as the sum of the squared deviations of Y from the mean of Y. In the population, the formula is
where SSY is the sum of squares Y, Y is an individual value of Y, and y is the mean of Y. A simple example is given in Table 1. The mean of Y is 2.06 and SSY is the sum of the values in third column and is equal to 4.597. All these points are taken care of by the experts providing Statistic homework help and assignment help at transtutors.com
Table 1. Example of SSY.
Y
Y-y
(Y-y)2
1.00
2.00
1.30
3.75
2.25
-1.06
-0.06
-0.76
1.69
0.19
1.1236
0.0036
0.5776
2.8561
0.0361
When computed in a sample, you should use the sample mean, M, in place of the population mean.
It is sometimes convenient to use formulas that use deviation scores rather than raw scores. Deviation scores are simply deviations from the mean. By convention, small letters rather than capitals are used for deviation scores. Therefore the score, ay, indicates the difference between Y and the mean of Y. Table 2 shows the use of this notation. The numbers are the same as in Table 1.
Table 2. Example of SSY using Deviation Scores.
Y
y
y2
1.00
2.00
1.30
3.75
2.25
-1.06
-0.06
-0.76
1.69
0.19
1.1236
0.0036
0.5776
2.8561
0.0361
The data in Table 3 are reproduced from the introductory section. The column X, has the values of the predictor variable and the column Y has the criterion variable. The third column, y, contains the differences between the column Y and the mean of Y. All such points are covered Statistichomework help and assignment helpat transtutors.com.
Table 3. Example data.
X
Y
y
y2
Y'
y'
y'2
Y-Y'
(Y-Y')2
1.00
2.00
3.00
4.00
5.00
1.00
2.00
1.30
3.75
2.25
-1.06
-0.06
-0.76
1.69
0.19
1.1236
0.0036
0.5776
2.8561
0.0361
1.210
1.635
2.060
2.485
2.910
-0.850
-0.425
0.000
0.425
0.850
0.7225
0.1806
0.0000
0.1806
0.7225
-0.210
0.365
-0.760
1.265
-0.660
0.044
0.133
0.578
1.600
0.436
sum
15.00
10.30
0.00
4.597
10.300
0.000
1.806
0.000
2.791
The fourth column, y2, is simply the square of the ay column. The column Y' contains the predicted values of Y. In the introductory section it was shown that the equation for the regression line for these data is
Y' = 0.425X + 0.785.
The values of Y' were computed according to this equation. The column y' contains deviations of Y' from the mean Y' and y'2 is the square of this column. The next-to-last column, Y-Y' contains the actual scores (Y) minus the predicted scores (Y'). The last column contains the squares of these errors of prediction.
We are now in position to see how the SSY is partitioned. Recall that SSY is the sum of the squared deviations from the mean. It is therefore the sum of the y2 column and is equal to 4.597.
SSY can be partitioned into two parts: the sum of squares predicted (SSY') and the sum of squares error (SSE).
The sum of squares predicted is the sum of the squared deviations of the predicted scores from the mean predicted score. In other words, it is the sum of the y'2 column and is equal to 1.806. The sum of squares error is the sum of the squared errors of prediction. It is there fore the sum of the (Y-Y')2 column and is equal to 2.791. This can be summed up as:
SSY = SSY' + SSE
4.597 = 1.806 + 2.791
There are several other notable features about Table 3. First, notice that the sum of y and the sum of y' are both zero. This will always be the case because these variables were created by subtracting their respective means from each value. Also notice that the mean of Y-Y' is 0. This indicates that although some Y's are higher than there respective Y's and some are lower, the average difference is zero.
The SSY is the total variation, SSY' is the variation explained, and the SSE is the variation unexplained. Therefore, the proportion of variation explained can be computed as:
Proportion explained = SSY'/SSY
Similarly, the proportion not explained is:
Proportion not explained = SSE/SSY
There is an important relationship between the proportion of variation explained and Pearson's correlation: r2 is the proportion of variation explained. Therefore, if r = 1, then, naturally, the proportion of variation explained is 1; if r = 0, then the proportion explained is 0. One last example: for r = 0.4, the proportion of variationexplained is 0.16.
Since the variance is computed by dividing the variation by N (for a population) or N-1 (for a sample), the relationships spelled out above in terms of variation also hold for variance. For example,
where the first term is the variance total, the second term is the variance of Y' and the last term is the variance of the errors of prediction (Y-Y'). Similarly, r2 is the proportion of variance explained as well as the proportion of variation explained.
Our email-based homework help support provides best and intelligent insight and recreation which help make the subject practical and pertinent for any assignment help.
Transtutors.com present timely homework help at logical charges with detailed answers to your Statistic questions so that you get to understand your assignments or homework better apart from having the answers. Our tutors are remarkably qualified and have years of experience providing One-Factor ANOVA-Partitioned Sum of Squares homework help or assignment help.
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# Dice Game: 6-sided die vs 20-sided die
• Master1022
In summary: Both set of dice average 10.5. If the goal is to get the highest sum, then they are equal.However, would the variance come into consideration if we had to choose one set with which to play the game?Not really. It would only indicate if you lost (won) a roll by a lot or a little.
#### Master1022
Homework Statement
There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a sum of numbers wins the game. Which person would you rather be? Is it a fair game?
Relevant Equations
Expectation and variance
Hi,
I was taking a look at the following question.
Question: There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a higher number (for 20-sided die) OR sum of numbers (for three 6-sided dice) wins the game. Which person would you rather be? Is it a fair game? Assume the dice are all fair
Attempt:
I think the general concept is about looking at expectation and variance for both cases and then make a decision from there.
So for the person with the 20-sided die, the expectation and variance are given by:
$$E(\text{20 sided die}) = \frac{n + 1}{2} = \frac{21}{2} = 10.5$$
$$Var(\text{20 sided die}) = \frac{n^2 - 1}{12} = \frac{(20)^2 - 1}{12} = \frac{399}{12}$$
For the person with the three 6-sided dice:
$$E(\text{three 6 sided die}) = E(X_1 + X_2 + X_3) = 3 \cdot E(\text{one 6 sided die}) = 3 \cdot \frac{7}{2} = 10.5$$
$$Var(\text{three 6 sided die}) = Var(X_1 + X_2 + X_3) = 3 \cdot \frac{n^2 - 1}{12} = 3 \cdot \frac{(6)^2 - 1}{12} = \frac{35}{4}$$
Thus, both of the distributions have the same expected value/mean. Therefore, perhaps we would rather be the person who chooses the lower variance side, which is the three 6-sided die. Does that seem reasonable?
Also, what does the 'is the game fair' part mean? Is that basically asking if the mean and variance are the same? If so, then based on the calculations above, I would say the game isn't fair...
Any help would be greatly appreciated.
Last edited:
It kind of depends on the number to win.If the number is 10 or 11, then
- the 1D20 has a 1-in-20 chance of winning
- the 3D6 have a 27-in-216 (1-in-12.5) chance of winning .
If the number is 3 or 18, then
- the 1D20 has a 1-in-20 chance of winning
- the 3D6 have a 1-in-216 chance of winning.
Heck, if the number is 1, the 3D6 has a zero chance of winning.
Or am I misunderstanding the rules?
DaveC426913 said:
It kind of depends on the number to win.
If the number is 3 (or 18), then
- the 1D20 has a 1:20 chance of winning
- the 3D6 have a 1:216 chance of winning.
Heck, if the number is 1, the 3D6 has a zero chance of winning.
Or am I mis-understanding the rules?
Hi @DaveC426913 , thanks for the reply! Sorry, I think there was a word missing from my original question. Basically, the person with the three 6-sided die adds up their 3 numbers and compares the sum to the 1 number from the person who rolled the 20-sided die. Then whichever number is higher wins the game.
Apologies - I hope this is more clear
Master1022 said:
whichever number is higher wins the game.
Ah.
There's a super-duper easy way to answer this question, but your teach might not like it.
Master1022
DaveC426913 said:
Ah.
There's a super-duper easy way to answer this question, but your teach might not like it.
Does it have something to do with the fact that the 20-sided die has two cases which cannot be beaten by the other player (19 and 20)? Or should I continue thinking along the mean and variance line?
For both dice sets, the probability curve is symmetrical.
You could just as easily make the rule "whoever gets the lowest score wins" and you have the exact same problem with the exact same solution - just mirrored.
I could produce a graph of the probabilities and show them to you horizontally flipped or unflipped, and you would not be able to tell me which rule is applied.
Therefore, neither dice set can have an advantage.
Master1022 and jbriggs444
DaveC426913 said:
For both dice sets, the probability curve is symmetrical.
You could just as easily make the rule "whoever gets the lowest score wins" and you have the exact same problem with the exact same solution - just mirrored.
I could produce a graph of the probabilities and show them to you horizontally flipped or unflipped, and you would not be able to tell me which rule is applied.
Therefore, neither dice set can have an advantage.
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?
Master1022 said:
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?
Both set of dice average 10.5. If the goal is to get the highest sum, then they are equal.
Master1022
Master1022 said:
However, would the variance come into consideration if we had to choose one set with which to play the game?
Not really. It would only indicate if you lost (won) a roll by a lot or a little. If the expected values are equal, then the game is fair.
Master1022 and phyzguy
Note that my "solution" is off-topic for the Mathematics Homework forum (though it might be on-topic in a Mathematics or Logic forum).
Master1022 said:
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?
Symmetry, not variance, is the key.
Master1022
DaveC426913 said:
I'm not sure the student is supposed to crunch the numbers here, in terms of probability calculations. If anything, I'd do a computer simulation to check the symmetry argument.
PeroK said:
I'm not sure the student is supposed to crunch the numbers here, in terms of probability calculations.
I guess. My solution bypass any math completely. Not sure if an elegant, intuitive proof is allowed.
PeroK said:
If anything, I'd do a computer simulation to check the symmetry argument.
You mean check that a D20 has a symmetrical distribution, and that 3D6 has a symmetrical distribution?
That's too trivial to warrant a simulation is it not? It's self-evident.
DaveC426913 said:
I guess. My solution bypass any math completely. Not sure if an elegant, intuitive proof is allowed.
You mean check that a D20 has a symmetrical distribution, and that 3D6 has a symmetrical distribution?
That's too trivial to warrant a simulation is it not? It's self-evident.
Your solution is probably what is expected.
There's no harm in running a simulation just to confirm. Even just writing the code may give an insight into why the symmetry argument works.
jedishrfu
When in doubt, write a program to figure it out.
Master1022, Dale and berkeman
jedishrfu said:
When in doubt, write a program to figure it out.
Or use Excel...
jedishrfu
berkeman said:
Or use Excel...
Gloria in excel-sis deo!
SammyS, jedishrfu and berkeman
If you turn it into a three person game, variance becomes a useful heuristic and the d20 has an advantage.
PeroK
jbriggs444 said:
If you turn it into a three person game, variance becomes a useful heuristic and the d20 has an advantage.
That depends on what the third player has, I imagine!
jbriggs444
PeroK said:
That depends on what the third player has, I imagine!
Indeed. We might constrain the third player to a set of dice with an expectation of 10.5. Or to a choice between 3d6 and 1d20.
jbriggs444 said:
Indeed. We might constrain the third player to a set of dice with an expectation of 10.5. Or to a choice between 3d6 and 1d20.
I see what you mean. The D20 player can't be worse, it seems. If the third player has a two-sided die showing 0 and 21, then the D20 and the 3xD6 are still equal.
PeroK said:
I see what you mean. The D20 player can't be worse, it seems. If the third player has a two-sided die showing 0 and 21, then the D20 and the 3xD6 are still equal.
I think that against an intentionally sub-optimal and asymmetric third player with something like a 5, 5, 5, 6, 21, 21 that you can get an advantage with a symmetric low variance (3d6) over a symmetric high variance (1d20).
jbriggs444 said:
I think that against an intentionally sub-optimal and asymmetric third player with something like a 5, 5, 5, 6, 21, 21 that you can get an advantage with a symmetric low variance (3d6) over a symmetric high variance (1d20).
I did a Python simulation and there seems to be a slight edge to D20 there. For simplicity I didn't count draws, only when one player won.
I can't immediately find any third option that gives 3D6 an advantage over D20.
PS I assigned the third player a fixed score of everything from 1-20 and there is no score that creates an advantage for 3D6. The higher the number, the greater the advantage to D20.
Therefore, there is no combination of numbers for the third player that gives 3D6 an advantage.
jbriggs444
PeroK said:
There's no harm in running a simulation just to confirm.
Here's a Python simulation with two players as in the original problem.
Python:
from random import choice
SixDie = [1, 2, 3, 4, 5, 6]
TwentyDie = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
SDTotal = 0
TDTotal = 0
NTrials = 1000000
for i in range(0, NTrials+1):
SDScore = choice(SixDie) + choice(SixDie) + choice(SixDie)
SDTotal += SDScore
TDScore = choice(TwentyDie)
TDTotal += TDScore
print("Six dice average: ", {SDTotal / NTrials})
print("Twenty die average: ", {TDTotal / NTrials})
Output with 1,000,000 trials is
Code:
Six dice average: {10.501267}
Twenty die average: {10.501052}
Running time in Python is about 8 seconds.
The game is who wins the most; not the highest average. The average scores are clearly equal.
Anyway, if a third player is involved, then in general the 20-sided die is better than the three 6-sided dice - or, at least, no worse.
PeroK said:
I did a Python simulation and there seems to be a slight edge to D20 there. For simplicity I didn't count draws, only when one player won.
I had forgotten that the aim was to see which was the winner. I edited my Python code to count wins, and saw a slight advantage for 1D20, likely for the reason already given, that for a 20-sided die, there are possible scores of 19 and 20 that can't be matched by three 6-sided dice.
If we modify the game with four 5-sided dice (no such Platonic solid exists, but never mind) vs. the 20-sided die, the four 5-sided dice have a clear advantage. Each has an expected value of 3, so the expected value of four of them would be 12, which is higher than the 10.5 expected value for the 20-sided die.
Mark44 said:
I had forgotten that the aim was to see which was the winner. I edited my Python code to count wins, and saw a slight advantage for 1D20, likely for the reason already given, that for a 20-sided die, there are possible scores of 19 and 20 that can't be matched by three 6-sided dice.
By a symmetry argument the game is equal. The automatically winning scores of 19 and 20 are counterbalanced by the losing scores of 1 and 2. Likewise scores of 18 and 3 counterbalance etc.
Master1022 and pbuk
PeroK said:
The game is who wins the most; not the highest average. The average scores are clearly equal.
Anyway, if a third player is involved, then in general the 20-sided die is better than the three 6-sided dice - or, at least, no worse.
Can you specify how the third player is involved? I am confused by this.
Master1022
FactChecker said:
Can you specify how the third player is involved? I am confused by this.
You have a third player with a third way of scoring. It could be anything. Absolutely anything. Then, excluding draws between two or more players, you count how often each player wins outright.
The hypothesis is that the 20-sided die is never worse than the three 6-sided dice. And, in general, is almost always better.
PeroK said:
You have a third player with a third way of scoring. It could be anything. Absolutely anything. Then, excluding draws between two or more players, you count how often each player wins outright.
The hypothesis is that the 20-sided die is never worse than the three 6-sided dice. And, in general, is almost always better.
Is the third player rolling another 20-sided die, or another set of three 6-sided die, or doesn't it matter?
Before I think hard about it, I want to know what I am thinking hard about. :-)
FactChecker said:
Is the third player rolling another 20-sided die, or another set of three 6-sided die, or doesn't it matter?
It could be those or anything at all. They might get a fixed score of 15 every time; or scores of 5 or 10 with equal probability; or, absolutely anything.
PeroK said:
It could be those or anything at all. They might get a fixed score of 15 every time; or scores of 5 or 10 with equal probability; or, absolutely anything.
So, as long as the winner has to beat two others, it is an advantage to have a larger variance than at least one other (and not less than others). Is that what you are saying? I'll buy that.
(That is, given equal means.)
Last edited:
PeroK
Mark44 said:
NTrials = 1000000
for i in range(0, NTrials+1):
That loop will execute 1,000,001 times.
|
Notes on Fraction | Grade 7 > Compulsory Maths > Fractions and Decimals | KULLABS.COM
• Note
• Things to remember
• Videos
• Exercise
• Quiz
#### Fraction
A fraction is a number which is usually expressed in the form of $$\frac{a}{b}$$. For examples ( $$\frac{3}{4}$$, $$\frac{5}{3}$$, $$\frac{9}{7}$$, etc.)
Comparison of fraction
A fraction can be compared to their likes and unlike terms. We can compare like fractions by comparing their numerators and to compare unlike fractions, we should convert them into like fractions. By comparing those like fractions, as a result, we can find greater and the smaller fraction. For example, Let's compare $$\frac{5}{3}$$ and $$\frac{7}{4}$$
Here, the L.C.M of denominators 3 and 4 = 3×4 = 12
Now, $$\frac{5}{3}$$ = $$\frac{5×4}{3×4}$$ = $$\frac{20}{12}$$
And, $$\frac{7×3}{4×3}$$ = $$\frac{21}{12}$$
Here, $$\frac{21}{12}$$ > $$\frac{20}{12}$$, So, $$\frac{7}{3}$$ > $$\frac{4}{3}$$
If the like fraction is given, we can add or substract them just by adding or substracting their numenator. For example, $$\frac{5}{6}$$ + $$\frac{7}{6}$$ = $$\frac{12}{6}$$ = 2
$$\frac{10}{7}$$ - $$\frac{9}{7}$$ = $$\frac{10 - 9}{7}$$ = $$\frac{1}{7}$$ ans.
In case of unlike fraction, at first we should convert them into like fractions with the least common denominator, then we add or substract their numenator. For example,
or,$$\frac{4}{5}$$ - $$\frac{3}{7}$$ = $$\frac{4 × 7}{5 × 7}$$ - $$\frac{3 × 5}{7 × 5}$$
= $$\frac{21}{35}$$ - $$\frac{15}{35}$$
= $$\frac{21 -15}{35}$$
= $$\frac{6}{35}$$ ans.
Multiplication of fraction
Multiplication of fraction can be done by the product of a whole number and a fraction as well as product of a two fraction. For example 3× $$\frac{4}{3}$$
$$\frac{3}{1}$$ × $$\frac{4}{3}$$ = $$\frac{12}{3}$$ =4
$$\frac{1}{3}$$ × $$\frac{6}{5}$$ = $$\frac{1×6}{3×5}$$ = $$\frac{6}{15}$$ = $$\frac{2}{5}$$ ans.
Division of fraction
To divide a whole number by a fraction, we should multiply the whole number by the reciprocal of the fraction. For example,
4÷ $$\frac{1}{6}$$ = 4× $$\frac{1}{6}$$ = $$\frac{4}{6}$$ = $$\frac{2}{3}$$ ans.
• To divide a whole number by a fraction, we should multiply the whole number by the reciprocal of the fraction.
• By comparing those like fractions, as a result, we can find greater and the smaller fraction.
.
#### Click on the questions below to reveal the answers
Solution:
Let the required sum be x.
Now, $$\frac{2}{7}$$ of x = Rs. 120
or, $$\frac{2}{7}$$ × x = Rs. 120
or, 2x = 7 × Rs. 120
or, x = $$\frac{7 × Rs. 120}{2}$$
= Rs. 420
So the required sum is Rs. 420.
Solution:
Let, he had x l of milk in the begining.
Remaining parts of the quantity of milk = 1 - $$\frac{3}{4}$$ = $$\frac{1}{4}$$ part.
Now, $$\frac{1}{4}$$ of x = 15 l
or, $$\frac{1}{4}$$ × x = 15 l
or, x = 4 × 15 l = 60 l
So, he had 60 l of milk in the begining.
Solution:
In 1 day, Kamala can do $$\frac{1}{12}$$ part of a piece of work.
In 3 days, Kamala can do 3 × $$\frac{1}{12}$$ part of work = $$\frac{1}{4}$$ parts of a piece of work
Now, the remaining parts of the work done by Reeta = (1-$$\frac{1}{4}$$) = $$\frac{3}{4}$$ parts.
So, $$\frac{3}{4}$$ parts of the work is done by Reeta.
0%
520
400
420
320
450
550
400
400
• ### What is the fraction of decimal 0.45?
(frac{9}{20})
(frac{9}{15})
(frac{8}{15})
(frac{8}{20})
(frac{2}{3})
(frac{3}{2})
(frac{4}{3})
(frac{1}{3})
60 litres
50 litres
65 litrs
55 litres
Rs. 300
Rs. 400
Rs. 500
Rs. 200
40
32.5
35
45
1050
1350
1250
1150
• ### Simplify : (frac{2}{3}) + (frac{3}{4})
(frac{6}{7})
(frac{6}{12})
1(frac{4}{7})
1(frac{5}{12})
• ### Simplify : (frac{5}{6}) - (frac{3}{7})
(frac{8}{13})
(frac{15}{36})
(frac{17}{42})
(frac{2}{-1})
• ### Simplify : 5 - (frac{2}{3})
2(frac{2}{3})
3(frac{2}{3})
1(frac{2}{3})
4(frac{2}{3})
7
6
8
5
• ### Simplify : (frac{7}{9}) × 15
1(frac{2}{3})
2(frac{5}{7})
3(frac{4}{3})
11(frac{2}{3})
• ### Simplify : (frac{6}{5}) ÷ (frac{8}{15})
2(frac{2}{3})
1(frac{1}{2})
2(frac{1}{4})
1(frac{2}{3})
## ASK ANY QUESTION ON Fraction
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Go Math Grade 4 Chapter 13 Answer Key Pdf Algebra: Perimeter and Area
Algebra: Perimeter and Area Go Math Grade 4 Chapter 13 Answer Key Pdf
Students can get the solution for only for the exercises and homework problems but also for the Mid-Chapter checkpoint and review test. So, the students who are practicing seriously for the exams can go through the Go Math 4th Grade Key of Chapter 13 Perimeter and Area. You can test your knowledge by solving the problems in this HMH Go Math Grade 4 Answer Key Chapter Perimeter and Area.
Common Core – New – Page No. 721
Perimeter
Find the perimeter of the rectangle or square.
Question 1.
9+3+9+3=24
24 inches
Explanation:
Length = 9 inches
Width = 3 inches
Perimeter of the rectangle = l + w + l + w
9+3+9+3=24
Therefore the Perimeter of the rectangle = 24 inches.
Question 2.
_____ meters
Explanation:
Side of a square = 8 meters
The perimeter of a square = 4a
= 4 × 8 meters = 32 meters
Thus the perimeter of a square = 32 meters.
Class 4 Maths Chapter 13 Perimeter and Area Question 3.
_____ feet
Explanation:
Length = 10 ft
Width = 12 ft
Perimeter of the rectangle = l + w + l + w
P = 10 + 12 + 10 + 12 = 20 + 24 = 44 feets
Thus the perimeter of the rectangle = 44 feet.
Remember: perimeter is the total distance around the outside, which can be found by adding together the length of each side. In the case of a rectangle, opposite sides are equal in length, so the perimeter is twice its width plus twice its height.
Question 4.
_____ centimeters
Explanation:
Length = 30 cm
Width = 24 cm
Perimeter of the rectangle = l + w + l + w
= 30 + 24 + 30 + 24 = 60 + 48
= 108 centimeters
Therefore the perimeter of the rectangle = 108 centimeters
Question 5.
_____ inches
Explanation:
Length = 25 in.
Width = 83 in.
Perimeter of the rectangle = l + w + l + w
= 25 + 83 + 25 + 83
= 216 inches
Thus the perimeter of the rectangle = 216 inches
Question 6.
_____ meters
Explanation:
The side of a square = 60 meters
The perimeter of the square = 4a
= 4 × 60 meters = 240 meters
Thus the perimeter of the square = 240 meters.
Problem Solving
Question 7.
Troy is making a flag shaped like a square. Each side measures 12 inches. He wants to add ribbon along the edges. He has 36 inches of ribbon. Does he have enough ribbon? Explain.
_____
Answer: No. He needs 48 inches of ribbon.
Explanation:
Troy is making a flag shaped like a square. Each side measures 12 inches.
He wants to add a ribbon along the edges.
He has 36 inches of ribbon.
36 inches + 12 inches = 48 inches
Question 8.
The width of the Ochoa Community Pool is 20 feet. The length is twice as long as its width. What is the perimeter of the pool?
_____ feet
Explanation:
The width of the Ochoa Community Pool is 20 feet.
The length is twice as long as its width.
Length = 2 × 20 feet = 40 feet
Perimeter of the rectangle = l + w + l + w
= 40 + 20 + 40 + 20 = 120 feet
Thus the perimeter of the pool is 120 feet.
Common Core – New – Page No. 722
Lesson Check
Question 1.
What is the perimeter of a square window with sides 36 inches long?
Options:
a. 40 inches
b. 72 inches
c. 144 inches
d. 1,296 inches
Explanation:
Given, Side of a square = 36 inches
The perimeter of the square = 4 × side = 4a
= 4 × 36 inches = 144 inches
Thus the perimeter of the square = 144 inches
The correct answer is option C.
Question 2.
What is the perimeter of the rectangle below?
Options:
a. 11 meters
b. 14 meters
c. 18 meters
d. 400 meters
Explanation:
Length of the rectangle = 5 meter
Width of the rectangle = 4 meters
The perimeter of the rectangle = l + w + l + w
= 5 + 4 + 5 + 4 = 18 meters
Thus the correct answer is option C.
Spiral Review
Question 3.
Which is the most reasonable estimate for the measure of the angle Natalie drew?
Options:
a. 30°
b. 90°
c. 180°
d. 210°
Explanation:
By seeing the above figure we can say that it is the right angle.
The correct answer is option B.
Question 4.
Ethan has 3 pounds of mixed nuts. How many ounces of mixed nuts does Ethan have?
Options:
a. 30 ounces
b. 36 ounces
c. 48 ounces
d. 54 ounces
Explanation:
Given that, Ethan has 3 pounds of mixed nuts.
1 pound = 16 ounces
3 pounds = 3 × 16 ounces = 48 ounces
Therefore the correct answer is option C.
Question 5.
How many lines of symmetry does the shape below appear to have?
Options:
a. 0
b. 1
c. 2
d. more than 2
Explanation:
The above shape has 1 line of symmetry.
The correct answer is option B.
Go Math Grade 4 Chapter 13 Pdf Perimeter for a Square Question 6.
Which of the following comparisons is correct?
Options:
a. 0.70 > 7.0
b. 0.7 = 0.70
c. 0.7 < 0.70
d. 0.70 = 0.07
Explanation:
a. 0.70 > 7.0
7.0 = 7
0.7 is less than 7
b. 0.7 = 0.70
0.7 is nothing but 0.70
So, the comparision is correct.
Page No. 725
Question 1.
Find the area of the rectangle.
A = _____ square cm
Explanation:
Length = 11 cm
Width = 13 cm
Area of the rectangle = l × w
= 11 cm × 13 cm = 143 square cm
Therefore the area of the rectangle = 143 square cm
Find the area of the rectangle or square.
Question 2.
A = _____ square inches
Explanation:
Length = 7 inches
Width = 2 inches
Area of the rectangle = l × w
= 7 inches × 2 inches = 14 inches
Therefore the area of the rectangle = 14 square inches
Question 3.
A = _____ square meters
Explanation:
Side of the square = 9 m
Area of a square = s × s
= 9 m × 9 m = 81 square meters
Thus the area of a square = 81 square meters
Question 4.
A = _____ square feet
Explanation:
Length = 8 feet
Width = 14 feet
Area of the rectangle = l × w
= 8 feet × 14 feet = 112 square feet
Therefore, area of the rectangle = 112 square feet
Find the area of the rectangle or square.
Question 5.
A = _____ square feet
Explanation:
Length of the rectangle = 13 ft
Width of the rectangle = 5 feet
Area of a rectangle = l × w
= 13 feet × 5 feet = 65 square feet
Thus, the area of the rectangle = 65 square feet
Question 6.
A = _____ square yards
Explanation:
Side of the square = 13 yards
Area of a square = s × s
= 13 yards × 13 yards = 169 square yards
Therefore, the area of a square = 169 square yards
Question 7.
A = _____ square centimeters
Explanation:
Length of the rectangle = 20 cm
Width of the rectangle = 2 cm
Area of a rectangle = l × w
= 20 cm × 2 cm = 40 square centimeters
Therefore the area of the rectangle = 40 square centimeters.
Practice: Copy and Solve Find the area of the rectangle.
Question 8.
base: 16 feet
height: 6 feet
A = _____ square feet
Explanation:
base: 16 feet
height: 6 feet
Area of a rectangle = b ×h
= 16 feet × 6 feet = 96 square feet
Thus the area of the rectangle = 96 square feet
Question 9.
base: 9 yards
height: 17 yards
A = _____ square yards
Explanation:
base: 9 yards
height: 17 yards
Area of a rectangle = b × h
9 yards × 17 yards = 153 square yards
The area of the rectangle = 153 square yards
Question 10.
base: 14 centimeters
height: 11 centimeters
A = _____ square centimeters
Explanation:
base: 14 centimeters
height: 11 centimeters
Area of a rectangle = b × h
14 centimeters × 11 centimeters = 154 square centimeters
The area of the rectangle = 154 square centimeters
Question 11.
Terry’s rectangular yard is 15 meters by 18 meters. Todd’s rectangular yard is 20 meters by 9 meters. How much greater is the area of Terry’s yard than Todd’s yard?
_____ square meters
Explanation:
Given,
Terry’s rectangular yard is 15 meters by 18 meters.
Todd’s rectangular yard is 20 meters by 9 meters.
Terry’s rectangular yard:
Area of a rectangle = b × h
= 15 meters × 18 meters = 270 square meters
Todd’s rectangular yard:
Area of a rectangle = b × h
20 meters × 9 meters = 180 square meters
270 square meters – 180 square meters = 90 square meters
Terry’s yard is 90 square meters greater than Todd’s yard.
Question 12.
Reason Quantitatively Carmen sewed a square baby quilt that measures 36 inches on each side. What is the area of the quilt?
A = _____ square inches
Explanation:
Carmen sewed a square baby quilt that measures 36 inches on each side.
Area of a square = s × s
= 36 inches × 36 inches = 1296 square inches
Therefore the area of the quilt is 1296 square inches.
Page No. 726
Question 13.
Nancy and Luke are drawing plans for rectangular flower gardens. In Nancy’s plan, the garden is 18 feet by 12 feet. In Luke’s plan, the garden is 15 feet by 15 feet. Who drew the garden plan with the greater area? What is the area?
a. What do you need to find?
Type below:
__________
Answer: I need to find who drew the garden plan with the greater area.
Question 13.
b. What formula will you use?
Type below:
__________
Answer: I will Area of rectangle and Area of a square formula
Question 13.
c. What units will you use to write the answer?
Type below:
__________
Question 13.
d. Show the steps to solve the problem.
Type below:
__________
First, we need to find the area of Nancy’s plan
Length = 18 feet
Width = 12 feet
Area of a rectangle = l × w
A = 18 feet × 12 feet = 216 square feet
And now we need to find the area of Luke’s plan
A = s × s
A = 15 feet × 15 feet = 225 square feet
Question 13.
e. Complete the sentences.
The area of Nancy’s garden is _______.
The area of Luke’s garden is _______.
_______ garden has the greater area.
Type below:
__________
The area of Nancy’s garden is 216 square feet.
The area of Luke’s garden is 225 square feet.
Luke’s garden has a greater area.
Question 14.
Victor wants to buy fertilizer for his yard. The yard is 35 feet by 55 feet. The directions on the bag of fertilizer say that one bag will cover 1,250 square feet. How many bags of fertilizer should Victor buy to be sure that he covers the entire yard?
______ bags
Explanation:
Given that,
Victor wants to buy fertilizer for his yard. The yard is 35 feet by 55 feet.
The directions on the bag of fertilizer say that one bag will cover 1,250 square feet.
A = b × h
A = 35 feet × 55 feet
A = 1925 square feet
1925 square feet is greater than 1,250 square feet.
So, Victor has to buy 2 bags to be sure that he covers the entire yard.
Question 15.
Tuan is an artist. He is painting on a large canvas that is 45 inches wide. The height of the canvas is 9 inches less than the width. What is the area of Tuan’s canvas?
A = ______ square inches
Explanation:
Tuan is an artist. He is painting on a large canvas that is 45 inches wide.
The height of the canvas is 9 inches less than the width.
So, h = 45 – 9 = 36 inches
A = b × h
A = 45 inches × 36 inches
A = 1,620 square inches
Therefore the area of Tuan’s canvas is 1620 square inches.
Common Core – New – Page No. 727
Area
Find the area of the rectangle or square.
Question 1.
Question 2.
______ square yards
Explanation:
Side of the square = 8 yards
Area of the square = s × s
8 yards × 8 yards = 64 square yards
Therefore, The area of the square is 64 square yards.
Question 3.
_____ square meters
Explanation:
Length of the rectangle = 15 m
Width of the rectangle = 3 m
Area of the rectangle = b × h
= 15 m × 3 m = 45 square meters
Thus the area of the rectangle is 45 square meters.
Question 4.
______ square inches
Explanation:
The base of the rectangle = 13 in.
Height of the rectangle = 6 in.
Area of the rectangle = b × h
13 in. × 6 in. = 78 square inches
Thus the area of the rectangle is 78 square inches.
Question 5.
______ square centimeters
Explanation:
The base of the rectangle = 30 cm
Height of the rectangle = 5 cm
Area of the rectangle = b × h
30 cm × 5 cm = 150 square centimeters
Therefore, the area of the rectangle = 150 square centimeters
Go Math Grade 4 Book My Homework Chapter 13 Perimeter and Area Question 6.
______ square feet
Explanation:
The base of the rectangle = 14 feet
Height of the rectangle = 4 feet
Area of the rectangle = b × h
14 feet × 4 feet = 56 square feet
Therefore, the area of the rectangle = 56 square feet.
Problem Solving
Question 7.
Meghan is putting wallpaper on a wall that measures 8 feet by 12 feet. How much wallpaper does Meghan need to cover the wall?
______ square feet wallpaper
Explanation:
Meghan is putting wallpaper on a wall that measures 8 feet by 12 feet.
The base of the rectangle = 8 feet
Height of the rectangle = 12 feet
Area of the rectangle = b × h
8 feet × 12 feet = 96 square feet
Thus the Area of the rectangle = 96 square feet
Question 8.
Bryson is laying down sod in his yard to grow a new lawn. Each piece of sod is a 1-foot by 1-foot square. How many pieces of sod will Bryson need to cover his yard if his yard measures 30 feet by 14 feet?
______ pieces
Explanation:
Bryson is laying down sod in his yard to grow a new lawn.
Each piece of sod is a 1-foot by 1-foot square.
The base of the rectangle = 30 feet
Height of the rectangle = 14 feet
Area of the rectangle = b × h
= 30 feet × 14 feet = 420 sq. ft.
Therefore Bryson needs 420 pieces of sod to cover his yard.
Common Core – New – Page No. 728
Lesson Check
Question 1.
Ellie and Heather drew floor models of their living rooms. Ellie’s model represented 20 feet by 15 feet. Heather’s model represented 18 feet by 18 feet. Whose floor model represents the greater area? How much greater?
Options:
a. Ellie; 138 square feet
b. Heather; 24 square feet
c. Ellie; 300 square feet
d. Heather; 324 square feet
Explanation:
Given,
Ellie and Heather drew floor models of their living rooms.
Ellie’s model represented 20 feet by 15 feet.
Heather’s model represented 18 feet by 18 feet.
Area of Ellie’s model = 20 feet × 15 feet = 300 square feet
Area of Heather’s model = 18 feet × 18 feet = 324 square feet
Now subtract the area of Ellie’s model from Heather’s model = 324 square feet – 300 square feet = 24 square feet
Thus the area of Heather’s model is greater than Ellie’s model
The correct answer is option B.
Question 2.
Tyra is laying down square carpet pieces in her photography studio. Each square carpet piece is 1 yard by 1 yard. If Tyra’s photography studio is 7 yards long and 4 yards wide, how many pieces of square carpet will Tyra need?
Options:
a. 10
b. 11
c. 22
d. 28
Explanation:
Tyra is laying down square carpet pieces in her photography studio.
Each square carpet piece is 1 yard by 1 yard. Tyra’s photography studio is 7 yards long and 4 yards wide
Area of the rectangle = b × h
= 7 yards × 4 yards
= 28 square yards
Thus the correct answer is option D.
Spiral Review
Question 3.
Typically, blood fully circulates through the human body 8 times each minute. How many times does blood circulate through the body in 1 hour?
Options:
a. 48
b. 240
c. 480
d. 4,800
Explanation:
Blood fully circulates through the human body 8 times each minute.
1 minute = 60 seconds
8 × 60 seconds = 480 seconds
The correct answer is option C.
Question 4.
Each of the 28 students in Romi’s class raised at least $25 during the jump-a-thon. What is the least amount of money the class raised? Options: a.$5,200
b. $700 c.$660
d. $196 Answer:$700
Explanation:
Each of the 28 students in Romi’s class raised at least $25 during the jump-a-thon. Multiply number od students with$25
28 × $25 =$700
The correct answer is option B.
Question 5.
What is the perimeter of the shape below if 1 square is equal to 1 square foot?
Options:
a. 12 feet
b. 14 feet
c. 24 feet
d. 28 feet
Explanation:
Given that 1 square is equal to 1 square foot
There are 14 squares
Length = 14 squares
Width = 2 squares
Area of the rectangle = l × w = 14 × 2 = 28 sq. feets
The correct answer is option D.
Question 6.
Ryan is making small meat loaves. Each small meat loaf uses $$\frac{3}{4}$$ pound of meat. How much meat does Ryan need to make 8 small meat loaves?
Options:
a. 4 pounds
b. 6 pounds
c. 8 pounds
d. 10 $$\frac{2}{3}$$ pounds
Explanation:
Ryan is making small meatloaves.
Each small meatloaf uses $$\frac{3}{4}$$ pound of meat.
Ryan need to make 8 small meatloaves.
$$\frac{3}{4}$$ × 8 = 6 pounds
The correct answer is option B.
Page No. 731
Question 1.
Explain how to find the total area of the figure.
A = ______ square units
Explanation:
Rectangle:
Each square box = 1 unit
There are 7 units
Base = 7 units
Height = 2 units
The area of the figure = b × h
A = 7 units × 2 units = 14 square units
Square:
The side is 3 units
Area of the square = 3 units × 3 units = 9 square units
Add both the areas = 14 square units + 9 square units = 23 square units
Therefore the area of the above figure is 23 square units.
Find the area of the combined rectangles.
Question 2.
A = ______ square mm
Explanation:
Area of top rectangle = b × h
Base = 12 mm
Height = 3 mm
A = 12 mm × 3 mm = 36 square mm
Area of square = s × s
s = 6 mm
A = 6 mm × 6 mm = 36 square mm
Area of the figure = 36 square mm + 36 square mm = 72 square mm
Thus the area of the above figure is 72 square mm.
Question 3.
A = ______ square miles
Explanation:
Area of rectangle = b × h
Area of the first rectangle = 10 mi × 9 mi
A = 90 square miles
Area of the second rectangle = 8 mi × 7 mi
A = 56 square miles
Area of the figure = Area of first rectangle + Area of the second rectangle
Area of the figure = 90 square mi + 56 square miles
Thus the Area of the figure = 146 square miles
Question 4.
A = ______ square feet
Explanation:
There are 2 squares and one rectangle in this figure
Area of the square = s × s
A = 4 ft × 4 ft = 16 square ft
Area of the square = s × s
A = 4 ft × 4 ft = 16 square ft
Area of the rectangle = b × h
A = 16 ft × 4 ft = 64 square ft
Area of the figure = 16 square ft + 16 square ft + 64 square ft
Thus the Area of the figure = 96 square feet.
Find the area of the combined rectangles.
Question 5.
Attend to Precision Jamie’s mom wants to enlarge her rectangular garden by adding a new rectangular section. The garden is now 96 square yards. What will the total area of the garden be after she adds the new section?
A = ______ square yards
Explanation:
There are 2 rectangles in the above figure
Area of rectangle = b × h
A = 12 yard × 8 yards = 96 square yards
Area of rectangle = b × h
A = 6 yards × 14 yards = 84 square yards
Area of the figure = 96 square yards + 84 square yards
Therefore the area of the figure = 180 square yards.
Go Math Workbook Grade 4 Area and Perimeter Answer Key Question 6.
Explain how to find the perimeter and area of the combined rectangles at the right.
P = ______ feet; A = ______ square feet
Answer: A = 92 square feet; P = 52 feet
Explanation:
There are 2 rectangle in the figure
Area of rectangle = b × h
A = 5 ft × 4 ft = 20 square ft
Area of rectangle = b × h
A = 8 ft × 9 ft = 72 square ft
Area of the figure = 20 square ft + 72 square ft = 92 square ft
Perimeter of the rectangle = 2l + 2w
P = 2 × 5 + 2 × 4 = 10 + 8 = 18 feet
Perimeter of the rectangle = 2l + 2w
P = 2 × 8 + 2 × 9 = 16 + 18 = 34 feet
Perimeter of the figure = 52 feet
Page No. 732
Question 7.
The diagram shows the layout of Mandy’s garden. The garden is the shape of combined rectangles. What is the area of the garden?
a. What do you need to find?
Type below:
__________
Answer: I need to find the area of the garden.
Question 7.
b. How can you divide the figure to help you find the total area?
Type below:
__________
Answer: I will divide the figure into 3 parts to find the total area
Question 7.
c. What operations will you use to find the answer?
Type below:
__________
Question 7.
d. Draw a diagram to show how you divided the figure. Then show the steps to solve the problem.
Type below:
__________
There are 2 rectangles and 1 square in this figure.
Area of rectangle = b × h
Base = 1 ft
H = 7 ft
A = 1 ft × 7 ft = 7 square ft
Area of rectangle = b × h
Base = 5 ft
H = 2 ft
A = 5 ft × 2 ft = 10 square ft
Area of the square = s × s
A = 3 ft × 3 ft = 9 square ft
Total area = 7 square ft + 10 square ft + 9 square ft
= 26 square ft
Question 8.
Workers are painting a large letter L for an outdoor sign. The diagram shows the dimensions of the L. For numbers 8a–8c, select Yes or No to tell whether you can add the products to find the area that the workers will paint.
8a. 2 × 8 and 2 × 4
i. yes
ii. no
Explanation:
There are 2 rectangles in the above figure
B = 2 ft
H = 8 ft
A = 2 × 8
B = 4 ft
H = 2 ft
A = 4 × 2
Thus the above statement is correct.
Question 8.
8b. 2 × 6 and 2 × 8
i. yes
ii. no
There are 2 rectangles in the above figure
B = 6 ft
H = 2 ft
A = 2 × 6
Then 2 will be subtracted from 8 = 6
So, the above statement 2 × 6 and 2 × 8 is false.
Question 8.
8c. 2 × 6 and 6 × 2
i. yes
ii. no
Explanation:
There are 2 rectangles in the above figure
B = 6 ft
H = 2 ft
A = 6 × 2
B = 2 ft
H = 6 ft
A = 2 × 6
Thus the above statement is true.
Common Core – New – Page No. 733
Area of Combined Rectangles
Find the area of the combined rectangles.
Question 1.
Question 2.
______ square feet
Explanation:
Area of A = 9 ft × 5 ft = 45 sq. ft.
Area of B = 14 ft. × 7 ft. = 98 sq. ft.
Total Area = Area of A + Area of B
= 45 sq. ft. + 98 sq. ft. = 143 square feet
Therefore the total Area = 143 square feet
Question 3.
______ square inches
Explanation:
Area of A = 9 in. × 5 in. = 45 square inches
Area of B = 6 inches × 3 inches = 18 square inches
Total Area = Area of A + Area of B
Total Area = 45 square inches + 18 square inches
Total Area = 63 square inches
Question 4.
______ square feet
Explanation:
Area of A = 4 feet × 2 feet = 8 square feet
Area of B = 7 feet × 6 feet = 42 square feet
Total Area = Area of A + Area of B
Total Area = 8 square feet + 42 square feet
Total Area = 50 square feet
Question 5.
______ square centimeters
Explanation:
Area of A = 12 cm × 7 cm = 84 square cm
Area of B = 16 cm × 6 cm = 96 square cm
Total Area = Area of A + Area of B
Total Area = 84 square cm + 96 square cm
Total Area = 180 square centimeters
Question 6.
______ square yards
Explanation:
Area of A = 14 yd × 1 yd = 14 square yards
Area of B = 9 yd × 6 yd = 54 square yards
Total Area = Area of A + Area of B
Total Area = 14Â square yards + 54Â square yards
Total Area = 68 square yards
Problem Solving
Use the diagram for 7–8.
Nadia makes the diagram below to represent the counter space she wants to build in her craft room.
Question 7.
What is the area of the space that Nadia has shown for scrapbooking?
______ square feet
Explanation:
Length = 13 feet
Width = 9 feet – 5 feet = 4 feet
Area of scrapbooking = l × w
= 13 feet × 4 feet
= 52 square feet
Therefore the area of the space that Nadia has shown for scrapbooking is 52 square feet.
Question 8.
What is the area of the space she has shown for painting?
______ square feet
Explanation:
The space for painting is a square.
Side of the square is 5 feet
Area of the square = 5 feet × 5 feet
= 25 square feet
Thus the area of the space she has shown for painting is 25 square feet.
Common Core – New – Page No. 734
Lesson Check
Question 1.
What is the area of the combined rectangles below?
Options:
a. 136 square yards
b. 100 square yards
c. 76 square yards
d. 64 square yards
Explanation:
Area of 1st rectangle = 5 yards × 8 yards = 40 square yards
Area of 2nd rectangle = 12 yards × 3 yards = 36 square yards
Area of the figure = Area of 1st rectangle + Area of 2nd rectangle
Area of the figure = 40 square yards + 36 square yards
Therefore, the Area of the figure is 76 square yards.
So, the correct answer is option C.
Question 2.
Marquis is redecorating his bedroom. What could Marquis use the area formula to find?
Options:
a. how much space should be in a storage box
b. what length of wood is needed for a shelf
c. the amount of paint needed to cover a wall
d. how much water will fill up his new aquarium
Answer: the amount of paint needed to cover a wall
Spiral Review
Question 3.
Giraffes are the tallest land animals. A male giraffe can grow as tall as 6 yards. How tall would the giraffe be in feet?
Options:
a. 2 feet
b. 6 feet
c. 12 feet
d. 18 feet
Explanation:
Giraffes are the tallest land animals. A male giraffe can grow as tall as 6 yards.
6 yards + 6 yards + 6 yards = 18 yards
The correct answer is option D.
Question 4.
Drew purchased 3 books for $24. The cost of each book was a multiple of 4. Which of the following could be the prices of the 3 books? Options: a.$4, $10,$10
b. $4,$8, $12 c.$5, $8,$11
d. $3,$7, $14 Answer:$4, $8,$12
Explanation:
Given that,
Drew purchased 3 books for $24. The cost of each book was a multiple of 4. So, the prices of books will be multiple of 4. That means$4 × 1, $4 × 2,$4 × 3
=Â $4,$8, \$12
The correct answer is option B.
Question 5.
Esmeralda has a magnet in the shape of a square. Each side of the magnet is 3 inches long. What is the perimeter of her magnet?
Options:
a. 3 inches
b. 7 inches
c. 9 inches
d. 12 inches
Explanation:
Esmeralda has a magnet in the shape of a square. Each side of the magnet is 3 inches long.
Side = 3 inches
The perimeter of the square = 4s
P = 4 × 3 = 12 inches
The correct answer is option D.
Question 6.
What is the area of the rectangle below?
Options:
a. 63 square feet
b. 32 square feet
c. 18 square feet
d. 16 square feet
Explanation:
Area of the rectangle = base × height
Base = 9 feet
Height = 7 feet
A = 9 feet × 7 feet
A = 63 square feet
Thus the correct answer is option A.
Page No. 735
Choose the best term from the box.
Question 1.
A square that is 1 unit wide and 1 unit long is a ________.
__________
Question 2.
The _______ of a two-dimensional figure can be any side.
__________
Question 3.
A set of symbols that expresses a mathematical rule is called a ______.
__________
Question 4.
The ______ is the distance around a shape.
__________
Find the perimeter and area of the rectangle or square.
Question 5.
Perimeter = ______ cm
Area = ______ square cm
Perimeter = 52 cm
Area = 169 square cm
Explanation:
P = 4s
P = 4 × 13 = 52 cm
A = s × s
A = 13 × 13 = 169 square cm
Go Math 4th Grade Answers Chapter 13 Area and Perimeter Question 6.
Perimeter = ______ ft
Area = ______ square ft
Perimeter: 48 ft
Area = 63 square ft
Explanation:
Base = 21 ft
Height = 3 ft
P = 2l +2w
P = 2 (21 ft + 3 ft)
P = 2 × 24 = 48 feet
A = b × h
A = 21 × 3
A = 63 square ft
Question 7.
Perimeter = ______ in.
Area = ______ square in.
Perimeter = 46 in.
Area = 120 square in.
Explanation:
P = 2l +2w
P = 2 × 15 + 2 × 8
P = 30 + 16 = 46 inches
A = l × w
A = 15 × 8 = 120 square inches
Question 8.
Area = ____ square yd
Area of the rectangle = 20 yards × 5 yards = 100 square yards
Area of the rectangle = 18 yards × 5 yards = 90 square yards
Area of the figure = 100 square yards + 90 square yards = 190Â square yards
Question 9.
Area = ____ square meters
A = b × h
A = 5 m × 2 m = 10 square meters
A = b × h
A = 5 m × 2 m = 10 square meters
A = b × h
A = 4 m × 2 m = 8 square meters
10 square meters + 10 square meters + 8 square meters
= 28 square meters
Therefore the area of the figures is 28 square meters
Question 10.
Area = ____ square feet
Area of the rectangle = b × h
A = 14 ft × 2 ft = 28 square feet
A = s × s
A = 8 ft × 8 ft = 64 square feet
Area of the figures = 64 square feet + 28 square feet
Therefore Area of the figure = 92 square feet
Page No. 736
Question 11.
Which figure has the greatest perimeter?
________
Answer: Figure B has the highest perimeter.
Explanation:
P = 2l +2w
P = 2 × 3 + 2 ×5 = 6 + 10 = 16
P = 2 × 6 + 2 × 3 = 12 + 6 = 18
P = 4a = 4 × 4 = 16
P = 2 × 4 + 2 × 3 = 8+ 6 = 14
Thus the greatest perimeter is figure B.
Question 12.
Which figure has an area of 108 square centimeters?
________
Explanation:
A = 13 cm × 6 cm = 78 square cm.
A = 11 cm × 11 cm = 121 square cm.
A = 12 cm × 9 cm = 108 square cm.
A = 16 cm × 38 cm = 608 square cm.
Thus the area of 108 square centimeters is Figure C.
Question 13.
Which of the combined rectangles has an area of 40 square feet?
________
Explanation:
Area of top rectangle = 6 ft × 2 ft = 12 square feet
Area of bottom rectangle = 6 ft × 2 ft = 12 square feet
Area of square = 4 ft × 4 ft = 16 square feet
Add Area of top rectangle, Area of bottom rectangle and Area of square
= 12 square feet +Â 12 square feet + 16 square feet = 40 square feet.
Thus the correct answer is option A.
Page No. 739
Question 1.
Find the unknown measure. The area of the rectangle is 36 square feet.
A = b × h
The base of the rectangle is ________ .
base = _____ ft
Explanation:
Given,
The area of the rectangle = 36 square feet
Height = 3 feet
Base =?
A = b × h
36 square feet = b × 3 feet
b × 3 feet = 36 square feet
b = 36/3 = 12 feet
The base of the rectangle is 12 feets.
Find the unknown measure of the rectangle.
Question 2.
Perimeter = 44 centimeters
width = _____ cm
Explanation:
Given,
Perimeter = 44 centimeters
Length = 12 cm
width =?
The perimeter of the rectangle = 2 (l + w)
P = 2l + 2w
44 cm = 24 cm + 2w
2w = 44 cm – 24 cm
2w = 20 cm
w = 20/2 = 10
Therefore width = 10 cm
Question 3.
Area = 108 square inches
height = _____ in.
Explanation:
Given,
Area = 108 square inches
Base = 9 inches
height = _____ in.
A = b × h
108 square inches = 9 inches × h
h = 108/9
Height = 12 inches
Therefore the height of the rectangle = 12 inches
Question 4.
Area = 90 square meters
base = _____ cm
Explanation:
Given,
Area = 90 square meters
Height = 5 meters
base = _____ cm
A = b × h
90 square meters = b × 5 meters
b × 5 meters = 90 square meters
b = 90/5 = 18 meters
Therefore the base of the rectangle = 18 meters
Question 5.
Perimeter = 34 yards
length = _____ yd
Explanation:
Given,
Perimeter = 34 yards
Width = 5 yards
Length =?
The perimeter of the rectangle = 2 (l + w)
P = 2l + 2w
34 yards = 2 × l + 2 × 5 yards
34 yards = 2 × l + 10 yards
2 × l + 10 yards = 34 yards
2l = 34 yards – 10 yards
2l = 24 yards
l = 24/2 = 12 yards
Therefore the length of the rectangle = 12 yards.
Question 6.
Area = 96 square feet
base = ______ ft
Explanation:
Given,
Area = 96 square feet
Height = 8 feet
Base =?
A = b × h
96 square feet = b × 8 feet
b × 8 feet = 96 square feet
b = 96/8 = 12 feet
Thus base of the rectangle = 12 feet.
Question 7.
Area = 126 square centimeters
height = _____ centimeters
Explanation:
Given,
Area = 126 square centimeters
Base = 9 cm
height = _____ centimeters
A = b × h
126 square centimeters = 9 cm × h
9 cm × h = 126 square centimeters
h = 126/9 = 14 centimeters
Therefore the Height of the rectangle = 14 centimeters
Question 8.
A square has an area of 49 square inches. Explain how to find the perimeter of the square.
Type below:
________
Explanation:
Given that,
A square has an area of 49 square inches.
A = 49 square inches
s^2 = 49 square inches
The square root of 49 is 7
So, each side of the square is 7 inches
The perimeter of the square = 4 × s
4 × 7 inches = 28 inches.
Therefore the perimeter of the square is 28 inches.
Page No. 740
Question 9.
Identify Relationships The area of a swimming pool is 120 square meters. The width of the pool is 8 meters. What is the length of the pool in centimeters?
length = _____ centimeters
Given that the area of a swimming pool is 120 square meters.
The width of the pool is 8 meters.
We have to find the length of the pool in centimeters.
We know that Area of the rectangle = l × w
A = l × w
120 square meters = l × 8 meters
l × 8 meters = 120 square meters
l = 120/8 = 15 meters
Therefore, the length of the pool = 15 meters
Convert meters to centimeters
1 meter = 100 centimeters
15 meters = 1500 centimeters.
The length of the pool in centimeters = 1500 centimeters
An outdoor deck is 7 feet wide. The perimeter of the deck is 64 feet. What is the length of the deck? Use the numbers to write an equation and solve it. A number may be used more than once.
P=(2 × l) + (2 × w)
So, the length of the deck is _______ feet.
length = _____ ft
An outdoor deck is 7 feet wide.
The perimeter of the deck is 64 feet.
We know that,
P=(2 × l) + (2 × w)
64 feet = (2 × l) + (2 × 7)
64 feet = 2l + 14 feet
2 × l = 64 feet – 14 feet
2 × l = 50 feet
l = 50/2 = 25 feet
Therefore the length of the deck = 25 feet.
Question 11.
A male mountain lion has a rectangular territory with an area of 96 square miles. If his territory is 8 miles wide, what is the length of his territory?
length = _____ miles
A male mountain lion has a rectangular territory with an area of 96 square miles.
Width = 8 miles
Length =?
A = l × w
96 square miles = l × 8 miles
l × 8 miles = 96 square miles
l = 96/8
l = 12 miles
Therefore, length of his territory = 12 miles
Common Core – New – Page No. 741
Find Unknown Measures
Find the unknown measure of the rectangle.
Question 1.
Perimeter = 54 feet
width = 7 feet
Think: P = (2 × l) + (2 × w)
54 = (2 × 20) + (2 × w)
54 = 40 + (2 × w)
Since 54 = 40 + 14, 2 × w = 14, and w = 7.
Question 2.
Perimeter = 42 meters
length = _____ meters
Explanation:
Given, Perimeter = 42 meters
Width = 9 meters
P = (2 × l) + (2 × w)
P = (2 × l) + (2 × 9 m)
42 m = 2l + 18 m
42 m – 18 m = 2l
2l = 24 meters
l = 24 meters/2 = 12 meters
Therefore length = 12 meters
Question 3.
Area = 28 square centimeters
height = _____ centimeters
Explanation:
Given,
Area = 28 square centimeters
Base = 4 cm
A = b × h
28 square centimeters = 4 cm × h
4 × h = 28
h = 28/4 = 7 cm
The height of the rectangle = 7 centimeters
Question 4.
Area = 200 square inches
base = _____ inches
Explanation:
Given,
Area = 200 square inches
Height = 25 inches
Base = ?
Area of the rectangle = b × h
200 square inches = b × 25 inches
b × 25 inches = 200 square inches
b = 200/25 = 8 inches
The base of the rectangle = 8 inches.
Problem Solving
Question 5.
Susie is an organic vegetable grower. The perimeter of her rectangular vegetable garden is 72 yards. The width of the vegetable garden is 9 yards. How long is the vegetable garden?
length = _____ yards
Explanation:
Susie is an organic vegetable grower.
The perimeter of her rectangular vegetable garden is 72 yards.
The width of the vegetable garden is 9 yards.
P = 72 yards
W = 9 yards
L =?
We know that,
P = (2 × l) + (2 × w)
72 yards = (2 × l) + (2 × 9)
72 yards – 18 yards = (2 × l)
(2 × l) = 72 yards – 18 yards
2l = 54 yards
l = 54/2 = 27 yards
Thus the vegetable garden is 27 yards long.
Question 6.
An artist is creating a rectangular mural for the Northfield Community Center. The mural is 7 feet tall and has an area of 84 square feet. What is the length of
the mural?
length = _____ feet
Explanation:
An artist is creating a rectangular mural for the Northfield Community Center.
The mural is 7 feet tall and has an area of 84 square feet.
A = 84 square feet
W = 7 feet
L =?
A = l × w
84 square feet = l × 7 feet
l × 7 feet = 84 square feet
l = 84/7 = 12 feet
Thus the length of Murali is 12 feet.
Common Core – New – Page No. 742
Lesson Check
Question 1.
The area of a rectangular photograph is 35 square inches. If the width of the photo is 5 inches, how tall is the photo?
Options:
a. 5 inches
b. 7 inches
c. 25 inches
d. 30 inches
Explanation:
The area of a rectangular photograph is 35 square inches.
Width = 5 inches
A = l × w
35 square inches = l × 5 inches
Length = 35/5 = inches
Thus the photo is 7 inches tall.
The correct answer is option B.
Question 2.
Natalie used 112 inches of blue yarn as a border around her rectangular bulletin board. If the bulletin board is 36 inches wide, how long is it?
Options:
a. 20 inches
b. 38 inches
c. 40 inches
d. 76 inches
Explanation:
Natalie used 112 inches of blue yarn as a border around her rectangular bulletin board.
Width = 36 inches
A = 112 inches
A = l × w
112 inches = l × 36 inches
l × 36 inches = 112 inches
l = 112/36 = 20 inches
Length = 20 inches
The correct answer is option A.
Spiral Review
Question 3.
A professional basketball court is in the shape of a rectangle. It is 50 feet wide and 94 feet long. A player ran one time around the edge of the court. How far did the player run?
Options:
a. 144 feet
b. 194 feet
c. 238 feet
d. 288 feet
Explanation:
A professional basketball court is in the shape of a rectangle.
It is 50 feet wide and 94 feet long.
A player ran one time around the edge of the court.
P = (2 × l) + (2 × w)
P = (2 × 94 feet) + (2 × 50 feet)
P = 188 feet + 100 feet = 288 feet
Therefore the perimeter of the rectangle is 288 feet.
Question 4.
On a compass, due east is a $$\frac{1}{4}$$ turn clockwise from due north. How many degrees are in a $$\frac{1}{4}$$ turn?
Options:
a. 45°
b. 60°
c. 90°
d. 180°
Explanation:
On a compass, due east is a $$\frac{1}{4}$$ turn clockwise from due north.
$$\frac{1}{4}$$ × 360° = 360°/4 = 90°
The correct answer is option C.
Question 5.
Hakeem’s frog made three quick jumps. The first was 1 meter. The second jump was 85 centimeters. The third jump was 400 millimeters. What was the total length of the frog’s three jumps?
Options:
a. 189 centimeters
b. 225 centimeters
c. 486 centimeters
d. 585 millimeters
Explanation:
Hakeem’s frog made three quick jumps.
The first was 1 meter. The second jump was 85 centimeters. The third jump was 400 millimeters.
Convert other units to centimeters
1 meter = 100 centimeters
400 millimeters = 40 centimeters
100 + 85 + 40 = 225 centimeters
Thus the correct answer is option B.
Question 6.
Karen colors in squares on a grid. She colored $$\frac{1}{8}$$ of the squares blue and $$\frac{5}{8}$$ of the squares red. What fraction of the squares are not colored in?
Options:
a. $$\frac{1}{8}$$
b. $$\frac{1}{4}$$
c. $$\frac{1}{2}$$
d. $$\frac{3}{4}$$
Answer: $$\frac{1}{4}$$
Explanation:
Karen colors in squares on a grid.
She colored $$\frac{1}{8}$$ of the squares blue and $$\frac{5}{8}$$ of the squares red.
$$\frac{1}{8}$$ + $$\frac{5}{8}$$ = $$\frac{6}{8}$$
Total number of fractions = $$\frac{8}{8}$$
$$\frac{8}{8}$$ – $$\frac{6}{8}$$ = $$\frac{2}{8}$$
$$\frac{1}{4}$$ fraction of the squares are not colored.
Page No. 745
Question 1.
Lila is wallpapering one wall of her bedroom, as shown in the diagram. She will cover the whole wall except for the doorway. How many square feet of wall does Lila need to cover?
First, find the area of the wall.
A = b × h
Awall = _____ square feet
Base = 12 feet
Height = 8 feet
A = b × h
Awall = 12 feet × 8 feet
Awall = 96 square feet
Question 1.
Next, find the area of the door.
A = b × h
Base = 3 feet
Height = 7 feet
A = b × h
Adoor = 3 feet × 7 feet
Question 1.
Last, subtract the area of the door from the area of the wall.
_____ – _____ = _____ square feet
So, Lila needs to cover _____ of wall.
Type below:
________
Awall = 96 square feet
Last, subtract the area of the door from the area of the wall.
A = 96 square feet – 21 square feet
A = 75 square feet
So, Lila needs to cover 75 square feet
Question 2.
What if there was a square window on the wall with a side length of 2 feet? How much wall would Lila need to cover then? Explain.
______ square feet
If there is a square window of length 2 feet
Area of square = s × s
Awindow = 2 × 2 = 4 square feet
Now Subtract the area of the door, area of the window from the area of the wall.
A = 96 square feet – 21 square feet – 4 square feet
A = 71 square feet
Therefore Lila need to cover 71 square feet.
Question 3.
Ed is building a model of a house with a flat roof, as shown in the diagram. There is a chimney through the roof. Ed will cover the roof with square tiles. If the area of each tile is 1 square inch, how many tiles will he need? Explain.
_____ tiles
Roof:
Base = 20 inches
Height = 30 inches
Area of the roof = b × h
Aroof = 20 inches × 30 inches
Aroof = 600 inches
Chimney:
Base = 3 inches
Height = 4 inches
Area of the chimney = b × h
Achimney = 3 × 4 = 12 inches
Now subtract Area of Chimney from Area of the roof
A = 600 inches – 12 inches
A = 588 inches
Therefore Ed needs 588 tiles.
Page No. 746
Question 4.
Make Sense of Problems Lia has a dog and a cat. Together, the pets weigh 28 pounds. The dog weighs 3 times as much as the cat. How much does each pet weigh?
cat weight = _____Â pounds dog weight = _____ pounds
Given that, the pets weigh 28 pounds.
28 = 7 + 7 + 7 + 7
The dog weighs 3 times as much as the cat.
= 3 × 7 = 21 pounds
The dog weighs 21 pounds
28 – 21 = 7
The cat weighs = 7 pounds.
Question 5.
Mr. Foster is covering two rectangular pictures with glass. One is 6 inches by 4 inches and the other one is 5 inches by 5 inches. Does he need the same number of square inches of glass for each picture? Explain.
_____
Explanation:
Mr. Foster is covering two rectangular pictures with glass.
One is 6 inches by 4 inches and the other one is 5 inches by 5 inches.
Area of first rectangular picture = 6 × 4 = 24 square inches
Area of second rectangular picture = 5 × 5 = 25 square inches
Area of two rectangular pictures = 25 square inches – 24 square inches
1 square inch.
Therefore, he doesn’t need the same number of square inches of glass for each picture.
Question 6.
Claire says the area of a square with a side length of 100 centimeters is greater than the area of a square with a side length of 1 meter. Is she correct? Explain.
_____
Explanation:
Claire says the area of a square with a side length of 100 centimeters is greater than the area of a square with a side length of 1 meter.
Her statement is not correct because 1 meter = 100 centimeters.
So, the area of a square with a side length of 100 centimeters is equal to the area of a square with a side length of 1 meter.
Question 7.
A rectangular floor is 12 feet long and 11 feet wide. Janine places a rug that is 9 feet long and 7 feet wide and covers part of the floor in the room. Select the word(s) to complete the sentence.
To find the number of square feet of the floor that is NOT covered by the rug,
the the area of the floor.
_____ square feet
Length = 12 feet
Width = 11 feet
Area of the rectangular floor = l × w
= 12 feet × 11 feet = 132 square feet
Room:
Length = 9 feet
Width = 7 feet
Area of the floor in the room = l × w
= 9 feet × 7 feet
= 63 square feet
Subtract the area of the rug from the area of the floor
= 132 square feet – 63 square feet = 69 square feet
The number of square feet of the floor that is NOT covered by the rug is 69 square feet.
Common Core – New – Page No. 747
Problem Solving Find the Area
Solve each problem.
Question 1.
A room has a wooden floor. There is a rug in the center of the floor. The diagram shows the room and the rug. How many square feet of the wood floor still shows?
82 square feet
Area of the floor: 13 × 10 = 130 square feet
Area of the rug: 8 × 6 = 48 square feet
Subtract to find the area of the floor still showing: 130 – 48 = 82 square feet
Question 2.
A rectangular wall has a square window, as shown in the diagram.
What is the area of the wall NOT including the window?
The area of the wall NOT including the window = _____ square feet
Explanation:
Wall:
Base = 14 feet
Height = 8 feet
Area of the wall = b × h
A = 14 feet × 8 feet
A = 112 square feet
Window:
Length = 4 feet
Area of the square = s × s
Area of the window = 4 feet × 4 feet = 16 square feet
Now subtract Area of the window from the area of the rectangular wall
= 112 square feet – 16 square feet
= 96 square feet
Therefore the area of the wall NOT including the window = 96 square feet.
Question 3.
Bob wants to put down new sod in his backyard, except for the part set aside for his flower garden. The diagram shows Bob’s backyard and the flower garden.
How much sod will Bob need?
The area covered with new sod = _____ square yards
Flower Garden:
Base = 20 yards
Height = 14 yards
Area of the rectangular flower garden = b × h
A = 20 yards × 14 yards
A = 280 square yards
Sod:
Base = 5 yards
Height = 9 yards
Area of sod = b × h
= 5 yards × 9 yards = 45 square yards
Now subtract area of sod from area of flower garden
= 280 square yards – 45 square yards
= 235 square yards
Thus the area covered with new sod = 235 square yards
Question 4.
A rectangular painting is 24 inches wide and 20 inches tall without the frame. With the frame, it is 28 inches wide and 24 inches tall. What is the area of the frame not covered by the painting?
The area of the frame = _____ square inches
Explanation:
A rectangular painting is 24 inches wide and 20 inches tall without the frame.
A = b × h
A = 24 inches × 20 inches
A = 480 square inches
With the frame, it is 28 inches wide and 24 inches tall.
A = b × h
A = 28 inches × 24 inches
A = 672 square inches
The area of the frame not covered by the painting
= 672 square inches – 480 square inches
= 192 square inches
Therefore, The area of the frame = 192 square inches
Question 5.
One wall in Jeanne’s bedroom is 13 feet long and 8 feet tall. There is a door 3 feet wide and 6 feet tall. She has a poster on the wall that is 2 feet wide and 3 feet tall. How much of the wall is visible?
The area of the wall visible = _____ square feet
Explanation:
One wall in Jeanne’s bedroom is 13 feet long and 8 feet tall.
Area of Jeanne’s bedroom = 13 feet × 8 feet = 104 square feet
Area of door = 3 feet × 6 feet = 18 square feet
Area of the wall = 2 feet × 3 feet = 6 square feet
To find the area of the wall visible we have to subtract Area of the wall, Area of the door from Area of Jeanne’s bedroom.
104 square feet – 18 square feet – 6 square feet
= 80 square feet
The area of the wall visible = 80 square feet
Common Core – New – Page No. 748
Lesson Check
Question 1.
One wall in Zoe’s bedroom is 5 feet wide and 8 feet tall. Zoe puts up a poster of her favorite athlete. The poster is 2 feet wide and 3 feet tall. How much of the wall is not covered by the poster?
Options:
a. 16 square feet
b. 34 square feet
c. 35 square feet
d. 46 square feet
Explanation:
One wall in Zoe’s bedroom is 5 feet wide and 8 feet tall.
Area of the wall in Zoe’s bedroom = b × h
A = 5 feet × 8 feet
A = 40 square feet
Zoe puts up a poster of her favorite athlete. The poster is 2 feet wide and 3 feet tall.
Area of the poster = b × h
A = 2 feet × 3 feet = 6 square feet
Now subtract Area of the poster from the Area of the wall in Zoe’s bedroom
= 40 square feet – 6 square feet
= 34 square feet
Thus the area of the wall is not covered by the poster = 34 square feet.
The correct answer is option B.
Question 2.
A garage door is 15 feet wide and 6 feet high. It is painted white, except for a rectangular panel 1 foot high and 9 feet wide that is brown. How much of the garage door is white?
Options:
a. 22 square feet
b. 70 square feet
c. 80 square feet
d. 81 square feet
Explanation:
A garage door is 15 feet wide and 6 feet high.
Area of the garage door = b × h
A = 15 feet × 6 feet
A = 90 square feet
It is painted white, except for a rectangular panel 1 foot high and 9 feet wide that is brown.
b = 9 feet
h = 1 foot
A = b × h
A = 9 feet × 1 feet
A = 9 square feet
Area of the garage door is white = 90 square feet – 9 square feet
Area of the garage door is white = 81 square feet
The correct answer is option D.
Spiral Review
Question 3.
Kate baked a rectangular cake for a party. She used 42 inches of frosting around the edges of the cake. If the cake was 9 inches wide, how long was the cake?
Options:
a. 5 inches
b. 12 inches
c. 24 inches
d. 33 inches
Explanation:
Kate baked a rectangular cake for a party. She used 42 inches of frosting around the edges of the cake.
Width = 9 inches
P = (2 × l) + (2 × w)
42 inches = (2 × l) + (2 × 9)
(2 × l) + (2 × 9) = 42 inches
(2 × l) = 42 inches – 18 inches
2l = 24 inches
l = 24/2 = 12 inches
Therefore the cake is 12 inches long.
Thus the correct answer is option B.
Question 4.
Larry, Mary, and Terry each had a full glass of juice. Larry drank $$\frac{3}{4}$$ of his. Mary drank $$\frac{3}{8}$$ of hers. Terry drank $$\frac{7}{10}$$ of his. Who drank less than $$\frac{1}{2}$$ of their juice?
Options:
a. Larry
b. Mary
c. Mary and Terry
d. Larry and Terry
Explanation:
Larry, Mary, and Terry each had a full glass of juice.
Larry drank $$\frac{3}{4}$$, Mary drank $$\frac{3}{8}$$ and Terry drank $$\frac{7}{10}$$ of $$\frac{1}{2}$$
$$\frac{3}{8}$$ is less than $$\frac{1}{2}$$ of their juice.
The correct answer is Option B.
Question 5.
Which of the following statements is NOT true about the numbers 7 and 9?
Options:
a. 7 is a prime number.
b. 9 is a composite number.
c. 7 and 9 have no common factors other than 1.
d. 27 is a common multiple of 7 and 9.
Answer: 27 is a common multiple of 7 and 9
Explanation:
a. 7 is a prime number is true.
b. 9 is a composite number is true
c. 7 and 9 have no common factors other than 1 is true.
d. 27 is a common multiple of 7 and 9 is not true because 7 is not the multiple of 27.
Thus the correct answer is option D.
Question 6.
Tom and some friends went to a movie. The show started at 2:30 P.M. and ended at 4:15 P.M. How long did the movie last?
Options:
a. 1 hour 35 minutes
b. 1 hour 45 minutes
c. 1 hour 55 minutes
d. 2 hours 15 minutes
Explanation:
Tom and some friends went to a movie. The show started at 2:30 P.M. and ended at 4:15 P.M.
Subtract 2:30 P.M. from 4:15 P.M.
4 hour 15 minutes
-2 hour 30 minutes
1 hour 45 minutes
The movie last for 1 hour 45 minutes
Thus the correct answer is option B.
Page No. 749
Question 1.
For numbers 1a–1e, select Yes or No to indicate if a rectangle with the given dimensions would have a perimeter of 50 inches.
a. length: 25 inches; width: 2 inches
i. yes
ii. no
Explanation:
P = (2 × l) + (2 × w)
50 inches = (2 × 25 in.) + (2 × w)
(2 × w) = 50 inches – 50 inches
w = 0
Thus the above statement is false
Question 1.
b. length: 20 inches; width: 5 inches
i. yes
ii. no
Explanation:
P = (2 × l) + (2 × w)
50 inches = (2 × 20 in.) + (2 × 5)
50 inches = 40 in. + 10 in.
Thus the above statement is true.
Question 1.
c. length: 17 inches; width: 8 inches
i. yes
ii. no
Explanation:
P = (2 × l) + (2 × w)
50 inches = (2 × 17 in.) + (2 × 8 in.)
50 inches = 34 in. + 16 in.
Thus the above statement is true.
Question 1.
d. length: 15 inches; width: 5 inches
i. yes
ii. no
Explanation:
P = (2 × l) + (2 × w)
50 inches = (2 × 15 in.) + (2 × 5 in.)
50 inches = 30 in. + 10 in.
50 inches = 40 inches
Thus the above statement is false.
Question 1.
e. length: 15 inches; width: 10 inches
i. yes
ii. no
Explanation:
P = (2 × l) + (2 × w)
50 inches = (2 × 15 in.) + (2 × 10 in.)
50 inches = 30 in. + 20 in.
50 inches = 50 inches
Thus the above statement is true.
Question 2.
The swimming club’s indoor pool is in a rectangular building.
Marco is laying tile around the rectangular pool.
Part A
What is the area of the pool and the area of the pool and the walkway? Show your work.
A(pool) = ____ m2Â Â A(building) = ____ m2
Pool:
Base = 20 m
Height = 16 m
A = b × h
Area of the pool = 20 m × 16 m = 320 square meters
Pool and the walkway:
Area of the pool and the walkway = 26 m × 22 m = 572 square meters
Question 2.
Part B
How many square meters of tile will Marco need for the walkway?
A(walkway) = ____ m2
Explanation:
Area of walkway = Area of the pool and the walkway – Area of pool
Area of the walkway = 572 square meters – 320 square meters
= 252 square meters
Therefore the Area of walkway = 252 square meters
Page No. 750
Question 3.
Match the dimensions of the rectangles in the top row with the correct area or perimeter in the bottom row
Question 4.
Kyleigh put a large rectangular sticker on her notebook. The height of the sticker measures 18 centimeters. The base is half as long as the height. What area of the notebook does the sticker cover?
________ square centimeters
Explanation:
Kyleigh put a large rectangular sticker on her notebook.
The height of the sticker measures 18 centimeters.
The base is half as long as the height.
Base = h/2 = 18/2 = 9 centimeters
Area of the rectangle = b × h
A = 9 cm × 18 cm
A = 162 square centimeters
Thus the area of the notebook the sticker cover is 162 square centimeters.
Question 5.
A rectangular flower garden in Samantha’s backyard has 100 feet around its edge. The width of the garden is 20 feet. What is the length of the garden? Use the numbers to write an equation and solve. A number may be used more than once.
□ = (2 × l) + (2 × □)
□ = 2 × l + □
□ = 2 × l
â–¡ = l
So, the length of the garden _____ feet.
P = (2 × l) + (2 × w)
100 = (2 × l) + (2 × 20)
100 – 40 = 2 × l
2 × l = 60
l = 60/2 = 30 feet
Length = 30 feet
So, the length of the garden 30 feet.
Question 6.
Gary drew a rectangle with a perimeter of 20 inches. Then he tried to draw a square with a perimeter of 20 inches.
Draw 3 different rectangles that Gary could have drawn. Then draw the square, if possible.
Type below:
__________
The possible rectangles with a perimeter of 20 inches are:
The possible square with a perimeter of 20 inches is:
Page No. 751
Question 7.
Ami and Bert are drawing plans for rectangular vegetable gardens. In Ami’s plan, the garden is 13 feet by 10 feet. In Bert’s plan, the garden is 12 feet by 12 feet. For numbers 7a−7d, select True or False for each statement.
a. The area of Ami’s garden is 130 square feet.
i. True
ii. False
Explanation:
A = b × h
Area of Ami’s garden = 13 feet × 10 feet =
Area of Ami’s garden = 130 square feet
The above statement is true.
Question 7.
b. The area of Bert’s garden is 48 square feet.
i. True
ii. False
Explanation:
Area of Bert’s garden = 12 feet × 12 feet = 144 square feet
The above statement is false.
Question 7.
c. Ami’s garden has a greater area than Bert’s garden.
i. True
ii. False
Explanation:
Area of Ami’s garden = 13 feet × 10 feet = 130 square feet
Area of Bert’s garden = 12 feet × 12 feet = 144 square feet
130 square feet is less than 144 square feet
The area of Ami’s garden is less than Area of Bert’s garden.
The above statement is false.
Question 7.
d. The area of Bert’s garden is 14 square feet greater than Ami’s.
i. True
ii. False
Explanation:
Area of Ami’s garden = 13 feet × 10 feet = 130 square feet
Area of Bert’s garden = 12 feet × 12 feet = 144 square feet
144 square feet – 130 square feet = 14 square feet
The above statement is true.
Question 8.
A farmer planted corn in a square field. One side of the field measures 32 yards. What is the area of the cornfield? Show your work.
_______ square yards
Explanation:
A farmer planted corn in a square field. One side of the field measures 32 yards.
Area of the square = 32 yards × 32 yards
A = 1,024 square yards
Therefore the area of the cornfield is 1,024 square yards.
Question 9.
Harvey bought a frame in which he put his family’s picture.
What is the area of the frame not covered by the picture?
_______ square inches
Explanation:
Area of the picture = 12 in. × 18 in.
A = 216 square inches
Area of the frame = 16 in. × 22 in.
A = 352 square inches
The area of the frame not covered by the picture = 352 square inches – 216 square inches
= 136 square inches
Therefore the area of the frame not covered by the picture is 136 square inches.
Question 10.
Kelly has 236 feet of fence to use to enclose a rectangular space for her dog. She wants the width to be 23 feet. Draw a rectangle that could be the space for Kelly’s dog. Label the length and width.
Type below:
________
Kelly has 236 feet of fence to use to enclose a rectangular space for her dog.
She wants the width to be 23 feet.
Perimeter = (2 × l) + (2 × w)
236 = (2 × l) + (2 × w)
236 = (2 × l) + (2 × 23)
236 – 46 = (2 × l)
(2 × l) = 190
l = 190/2
l = 95 feet
Therefore length = 95 feet.
Page No. 752
Question 11.
The diagram shows the dimensions of a new parking lot at Helen’s Health Food store.
Use either addition or subtraction to find the area of the parking lot. Show your work.
_______ square yards
Explanation:
Top:
Base = 40 yards
Height = 20 yards
Area of the top rectangle = b × h
A = 40 yards × 20 yards = 800 square yards
Bottom:
Base = 30 yards
Height = 10 yards
Area of the rectangle = b × h
A = 30 yards × 10 yards = 300 square yards
Area of the parking = Area of top + Area of bottom
A = 800 square yards + 300 square yards
Area of parking = 1100 square yards.
Question 12.
Chad’s bedroom floor is 12 feet long and 10 feet wide. He has an area rug on his floor that is 7 feet long and 5 feet wide. Which statement tells how to find the amount of the floor that is not covered by the rug? Mark all that apply.
Options:
a. Add 12 × 10 and 7 × 5.
b. Subtract 35 from 12 × 10
c. Subtract 10 × 5 from 12 × 7.
d. Add 12 + 10 + 7 + 5.
e. Subtract 7 × 5 from 12 × 10.
f. Subtract 12 × 10 from 7 × 5.
Chad’s bedroom floor is 12 feet long and 10 feet wide.
A = 12 feet × 10 feet = 120 square feet
Area rug on his floor = 7 feet × 5 feet = 35 square feet
To find the amount of the floor that is not covered by the rug we have to subtract 120 square feet from 35 square feet or 35 square feet from 12 × 10.
So, the correct answers are B and F.
Question 13.
A row of plaques covers 120 square feet of space along a wall. If the plaques are 3 feet tall, what length of the wall do they cover?
____ feet
Explanation:
Given that,
A row of plaques covers 120 square feet of space along a wall.
Height = 3 feet
A = b × h
120 square feet = b × 3 feet
b = 120/3 = 40
Therefore the base is 40 feet.
Page No. 753
Question 14.
Ms. Bennett wants to buy carpeting for her living room and dining room.
Explain how she can find the amount of carpet she needs to cover the floor in both rooms. Then find the amount of carpet she will need.
____ square feet
She can find the area of each rectangle and then find the sum. The area of the living room is 20 × 20 = 400 square feet.
The area of the dining room is 15 × 10 = 150 square feet.
The sum of the two rooms = 400 + 150 = 550 square feet.
She needs 550 square feet of carpeting.
Question 15.
Lorenzo built a rectangular brick patio. He is putting a stone border around the edge of the patio. The width of the patio is 12 feet. The length of the patio is two feet longer than the width.
How many feet of stone will Lorenzo need? Explain how you found your answer.
____ feet
Explanation:
Width = 12 feet
Length = 2 × width
Length = 2 + 12 feet = 14 feet
Perimeter = (2 × l) + (2 × w)
P = (2 × 14) + (2 × 12)
P = 28 + 24
P = 52 feet
Page No. 754
Question 16.
Which rectangle has a perimeter of 10 feet? Mark all that apply.
Rectangle: ____
Rectangle: ____
Explanation:
i. Perimeter of A = (2 × l) + (2 × w)
P = (2 × 1) + (2 × 4) = 2 + 8 = 10 feet
ii. Perimeter of B = (2 × l) + (2 × w)
P = (2 × 2) + (2 × 5) = 4 + 10 = 14 feet
iii. Perimeter of C = (2 × l) + (2 × w)
P = (2 × 2) + (2 × 3) = 4 + 6 = 14 feet
iv. Perimeter of D = (2 × l) + (2 × w)
P = (2 × 4) + (2 × 6) = 8 + 12 = 20 feet
The correct answer is option A and C.
Question 17.
A folder is 11 inches long and 8 inches wide. Alyssa places a sticker that is 2 inches long and 1 inch wide on the notebook. Choose the words that correctly complete the sentence.
To find the number of square inches of the folder that is NOT covered by the sticker,
Type below:
________
Answer: Subtract the area of the sticker from the area of the notebook.
Question 18.
Tricia is cutting her initial from a piece of felt. For numbers 18a–18c, select Yes or No to tell whether you can add the products to find the number of square centimeters Tricia needs.
a. 1 × 8 and 5 × 2 _______
b. 3 × 5 and 1 × 8 _______
c. 2 × 5 and 1 × 3 and 1 × 3 _______
a. 1 × 8 and 5 × 2 _______
Yes
b. 3 × 5 and 1 × 8 _______
No
c. 2 × 5 and 1 × 3 and 1 × 3 _______
No
Question 19.
Mr. Butler posts his students’ artwork on a bulletin board.
The width and length of the bulletin board are whole numbers. What could be the dimensions of the bulletin board Mr. Butler uses?
Type below:
________
Answer: 5 feet long by 3 feet wide
Area of the rectangle = l × w
A = 15 square feet
The factor of 15 is 5 and 3.
So, the length = 5 feet long
Width = 3 feet long.
Quick learning is not only important but also understanding is important to learn the concepts. You can’t love maths if you don’t understand the subject. So, to help you guys we have provided the images for your better understanding. Learn the simple techniques to solve the problem in less time in our Go Math Answer Key.
Conclusion:
Hope you are satisfied with the solutions provided in the Go Math Grade 4 Answer Key Chapter 13 Algebra: Perimeter and Area. For unlimited practice check out the questions in the review at the end of the chapter. You can also find all chapter’s solutions in our Go Math Answer Key. Make use of the links and score the highest marks in the exams. Best Of Luck!!!!
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Algebraic equations are mathematical statements where two expressions are separated by an equal sign.
The variables can be on both sides of the equality sign. We use letters to denote variables.
Examples:
$x = 4 + 2$. Here, we observe that the variable is on one side of the equation.
$2x + 3 = x + 4$. Here, we can observe that the variables are on both sides of the equation.
## Writing Algebraic Equations
An algebraic equation contains variables, numbers and an equal sign that can be interpreted into meaningful statements. These variables are also called as literal coefficients.
Any word problem can be changed into a simple and clear algebraic equation, which makes solving it much easier.
Solving a word problem depends on how well one understands it and translates it into a mathematical statement.
So, the main steps that have to be followed while dealing with a word problem are:
• Understand what is given.
• Understand what is to be found
• Understand the relation between what is to be found and what is given.
Word problems get easier if you understand the keywords that are used in the problem.
Lets consider some
Operation Symbol Keyword Addition + Sum Added to Total More than Increased by Subtraction - Minus Less Difference Decreased by Fewer than Multiplication * Multiplied by Product of Times Of Division ÷ Out of Ratio of Per Quotient of Percent Equal = Is Are Will be Gives Were Power of ^ Squared Cubed
Algebraic equations are very useful in solving word problems as they show a symbolic way of solving them.
When reading a word problem, one must be very careful in interpreting the sentences to equations.
The following steps can be followed to convert a word problem into an algebraic equation:
• Understand the problem by reading it thoroughly
• Check what is to be found and assign it a variable, say x.
• Check what is given.
• Form an equation connecting the given values and the variable, x.
Hence, we get an equation which can be solved to get the variable.
### Examples on Writing Algebraic Equations
Given below are some examples that explain how to form an algebraic equation.
Example 1:
Half of a number increased by 5 is 10
Solution:
Let the number be x.
$\frac{1}{2}$ $x + 5 = 10$
Example 2:
Jane gets 20 dollar as pocket money. How much money does she need to buy a toy that costs 92 dollars?
Solution:
Step 1: First consider what is given. Jane has 20 dollar and wants to buy a toy that costs 92 dollars. The question is how many more dollars are needed to buy the toy.
Step2: Let us call the extra amount needed as x.
Step3: So, she already she has 20. We need x more money to get 92 dollars. Hence, an equation can be formed
$20 + x = 92$
Step4: Subtract 20 from both sides
$x = 72$
So, Jane needs 72 dollars more to buy the toy.
Example 3:
The sum of 11 and a quantity is multiplied by 2 to get 8. Find the quantity.
Solution:
Let us consider the quantity as u.
Given, the sum of 11 and a quantity is multiplied by 2 to get 8.
So, the algebraic equation formed will be as follows:
$(u + 11) \times 2 = 8$
Divide both the sides by 2
$(u + 11) = 4$
Subtract 11 from both the sides
$u = 4 – 11$
u = 7 dollars
## Solving Algebraic Equations
Algebraic equations can be solved by isolating the term containing the variable first. This can be done by adding or subtracting terms from both the sides of the equal sign. The value of the variable can be found by then dividing the whole equation by the co-efficient of the variable. The result we thus get is the solution of the algebraic equation.
Let us observe how to solve algebraic equations with the help of the following examples.
Example 1:
$4x = 3x + 1$
Solution:
$4x = 3x + 1$
$4x - 3x = 3x - 3x + 1$ (By subtracting 3x from both the sides, all the variables are on the same side of the equation.)
$x = 1$
Example 2:
$5x = 3x - 16$
Solution:
$5x = 3x - 16$
$5x - 3x = 3x - 3x - 16$ (By subtracting 3x from both the sides, all the variables are on the same side of the equation.)
$2x = 16$
$x = 8$
Example 3:
$5p =$15
Solution:
$5p = 15$
$\frac{5}{5}$ $p =$ $\frac{15}{5}$
$p = 3$
Example 4:
$3a + 5a = 4$
Solution:
$3a + 5a = 4$
$8a = 4$
$\frac{8}{8}$ $a =$ $\frac{4}{8}$
$a =$ $\frac{1}{2}$
Example 5:
$7m = 2m + 18$
Solution:
$7m = 2m + 18$
$7m + 2m = 2m + 2m + 18$ (By adding 2m on both the sides, all the variables are on the same side of the equation)
$9m = 18$
$\frac{9}{9}$ $m =$ $\frac{18}{9}$
$m = 2$
## Solving Algebraic Equations with Fractions
Algebraic equations contain terms in the form of integers, decimals or fractions. When dealing with fractions, the simplest way is to remove the fractional form. This can be done by multiplying the denominator term with the numerator on both sides of the equation. The steps will vary according to the question but the idea is the same - remove the denominator. The process of changing an equation without fractions is called the clearing of fractions.
### Examples on Solving Algebraic Equations with Fractions
Given below are some examples that explain how to solve algebraic equations with fractions.
Example 1:
$3x +$ $\frac{2}{7}$ = $5$
Solution:
Here, there is a fraction $\frac{2}{7}$. The problem would be easier if we can remove the fraction. The only way we can do this is by multiplying both sides by 7.
$7 \times (3x + $$\frac{2}{7}$$)$ = $7 \times 5$
On the left hand side, 7 can be distributed.
$7 \times 3x + 7 \times $$\frac{2}{7} = 7 \times 5 21x + 2 = 35 Subtract 2 from both the sides 21x = 33 Divide both the sides by 21 x = \frac{33}{21} Example 2: (\frac{6}{8})$$ \times (x + 3) = 9$
Solution:
This is an equation that contains a fraction $\frac{6}{8}$.
This can be solved in many ways. But, lets look at the easiest way: Clearing the fractions.
$(\frac{6}{8})$$\times (x + 3) = 9 To remove the fraction from the left hand side, multiply it by 8. Now, this process will alter the equation. So, perform the same process on the right hand side also. 8 \times$$[(\frac{6}{8})$$\times (x + 3)$$]$ $= 8 \times 9$
So, $6 \times (x + 3) = 72$
Now, Distribute 6
$6x + 18 = 72$
Subtract 18 from both the sides
$6x = 72 - 18$
$= 54$
Divide by 6 on both the sides
So, $x = 9$.
The main steps for solving such equations are:
• Choose a common denominator for the equation.
• Multiply the common denominator with every term in the equation
• Remove all the denominators and hence reduce the equation to its simplest form.
Example 3:
$\frac{(x-2)}{2}$$+$$\frac{3}{2}$ = $6$
Solution:
Here, we can see that both the terms have the same denominator 2, on the left hand side. So, that can be taken as common.
$\frac{(( x - 2)+3)}{2}$ = $6$
Multiply both sides by 2
$2 \times $$\frac{(( x - 2)+3)}{2} = 2 \times 6 So, (x - 2) + 3 = 12 Subtract 3 from both sides x - 2 = 9 Add 2 on both the sides So, x = 11 Example 4: \frac{3x}{5}$$ - $$\frac{(x+3)}{10} = -7 Solution: Here, LCM of 5 and 10 is 10. Multiply by 10 on both sides 10$$(\frac{3x}{5}$$-$$\frac{(x+3)}{10})$ = $10(-7)$
$2 \times 3x - (x + 3) = -70$
$6x - x - 3 = -70$
$5x - 3 = -70$
Add 3 on both sides
$5x = -67$
Divide both sides by 5
$x$ = $\frac{-67}{5}$
Example 5:
$\frac{(x – 2)}{3}$$+$$\frac{x}{5}$ = $6$
Solution:
Here, the denominators are different on the left hand side. So, we need a common term.
Find the LCM of 3 and 5, which is 15. Multiply by 15 on either side of the equation
$15 \times $$(\frac{(x – 2)}{3}$$ + $$\frac{x}{5}) = 15 \times 6 15 \times$$(\frac{(x – 2)}{3})$$+ 15 \times$$(\frac{x}{5})$ = $90$
$5 \times (x - 2) + 3 \times x$ = $90$
$5x - 10 + 3x$ = $90$
$8x - 10$ = $90$
Add 10 on both the sides
$8x$ = $100$
Divide 8 on both the sides
$x$ = $\frac{25}{2}$
## Factoring Algebraic Equations
Factoring an algebraic equation is just the reverse of the distribution law. So, by rule
$ab + ac$ = $a(b + c)$
The main steps for factoring the equation are:
• Find the common factor in the expression.
• Factor it out.
In general, factoring is a process by which a sum is converted into a product by removing the common term or factor and putting them outside the parenthesis. So, only if there is a common factor, can any algebraic equation be factored.
### Examples on Factoring Algebraic Equations
Given below are some examples on factoring algebraic equations.
Example 1:
$24x + 9y + 3xy$ = $0$
Solution:
Here, there are three terms $24x$, $9y$ and $3xy$. Lets factor each of them.
$24x$ = $2 \times 2 \times 2 \times 3 \times x$
$9y$ = $3 \times 3 \times y$
$3xy$ = $3 \times x \times y$
The highest common factor is 3.
So,
$24x + 9y + 3xy$ = $0$
$3(12x + 3y + xy)$ = $0$
Example 2:
$2x - 4x^{2}$ = $0$
Solution:
Here, there are two terms $2x$ and $4x^{2}$. Lets factor each of them.
$2x$ = $2 \times x$
$4x^{2}$ = $2 \times 2 \times x \times x$
The highest common factor is $2$ and $x$. Therefore, $2x$ is the highest factor that divides both the terms.
So, $2x$ can be factored out as follows:
$2x - 4x^{2}$ = $0$
$2x(1 - 2x)$ = $0$
Now, it can be solved using the zero property
As $2 \neq 0$, $x = 0$ and $1 - 2x$ = $0$
$x$ = $\frac{1}{2}$
So, the solutions are $x$ = $0$, $\frac{1}{2}$
Example 3:
$-2x^{2}y - 6xy^{2}$
Solution:
Here, the factors are
$-2x^{2}y$ = $-1 \times 2 \times x \times x \times y$
$-6xy^{2}$ = $-1 \times 2 \times 3 \times x \times y \times y$
So, the common term is $-2xy$
So,
$-2x^{2}y - 6xy^{2}$ = $-2xy(x + 2y)$
Example 4:
$x( y + 1 ) + 5(y + 1)$
Solution:
Here, there are two terms $x(y + 1)$ and $5(y + 1)$
The highest common factor is $y+1$
So,
$x(y + 1) + 5(y + 1)$ = $(x + 5)(y + 1)$
Special identities
There are certain special identities that can be used for making the factoring simple
• u2 - v2 = (u + v)(u – v)
• u2 + 2uv + v2 = (u + v)(u + v)
• u2 - 2uv + v2 = (u - v)(u – v)
• u3 + v3 = (u + v)(u2 - uv +v2)
• u3 - v3 = (u - v)(u2 + uv +v2)
• u3 + 3u2v + 3uv2 + v3 = (u +v)(u + v)(u + v)
• u3 - 3u2v + 3uv2 - v3 = (u - v)(u - v)(u - v)
There are many more, but these are the main identities.
Example 4:
$9x^{2} - 64$ = $0$
Solution:
Here, $9x^{2}$ = $(3x)^{2}$
$64$ = $8^{2}$
$9x^{2} - 64$ = $0$
$(3x)^{2} - 8^{2}$ = $0$
Here, the following identity can be used
$u^{2} - v^{2}$ = $(u + v)(u – v)$
So, $(3x)^{2} - 8^{2}$ = $(3x - 8)(3x + 8)$
So, it can be simplified as $(3x - 8) (3x + 8)$ = $0$
### Quadratic Equations:
An algebraic equation in the form $ax^{2} + bx + c$ = $0$ is called a quadratic equation, where $a > 0$. A quadratic equation can be factored in two ways. For factoring, we need two numbers say, $f$ and $g$ such that $f + g$ = $b$ and $fg$ = $c$. If the equation is in the form $x^{2} + bx + c$ = $0$, that is $a$ = $1$, then the method is pretty simple.
### Examples on Factoring Quadratic Equations
Given below are some examples that explain how to factor a quadratic equation.
Example 1:
$x^{2} + 5x + 6$ = $0$
Solution:
Find two numbers so that sum is $5$ and the product is $6$. Checking by factoring we get $2$, $3$ such that, $2 + 3$ = $5$ and $2 \times 3$ = $6$
So, we can factor it as $(x + 2) (x + 3)$
So, $x^{2} + 5x + 6$ = $0$
$(x + 2) (x + 3)$ = $0$
Hence solved.
Example 2:
$9x^{2} + 6x + 1$ = $0$
Solution:
To factor this, we need two numbers so that the sum is $6$ and the product is $9$. Let us consider $3$ and $3$. So that, $3 + 3$ = $6$ and $3 \times 3$ = $9$.
So, just split the middle term as $3x + 3x$
$9x^{2} + 6x + 1$ = $0$
$9x^{2} + 3x + 3x + 1$ = $0$
$3x(3x + 1) + 1(3x + 1)$ = $0$
So, $(3x + 1)(3x + 1)$ = $0$
Hence solved by zero property.
Any quadratic equation can be solved using quadratic formulas.
$x$ = $-b \pm \sqrt{\frac{b^{2} - 4ac}{2a}}$
Example 3:
$x^{2} -7x + 10$
Solution:
Here, $a$ = $1$, $b$ = $-7$, $c$ = $10$
So, $x$ = $-(-7) \pm \sqrt{\frac{((-7)^{2} -4(1) (10))}{(2\times1)}}$
= $7 \pm \sqrt{\frac{(49 – 40)}{2}}$
= $\frac{(7 \pm \sqrt{9})}{2}$
= $\frac{(7 \pm 3)}{2}$
So, $x$ = $\frac{(7-3)}{2}$ and $x$ = $\frac{(7+3)}{2}$
$x$ = $\frac{4}{2}$ and $x$ = $\frac{10}{2}$
$x$ = $2$ and $x$ = $5$
And so, the factors are $(x - 2)$ and $(x - 5)$.
## Conditional Equation
A conditional equation is an equation which is true for the particular value of the variable it contains. The conditional equation represents a problem. We can create as many conditional identities as we want as there is no particular rule to create conditional equations, except that they must not be inconsistent equations which do not have any values of the variable for which the equation is true.
$x^{2} + 3x = 7$ is a conditional equation which asks " What are the values of x for which the square of the value increased by three times the value is seven?" The values that must be assigned to the variable so that the equation is true is called the solution to the conditional equation. If there is more than one value for which the conditional equation is true, then they are the Solution Set for the Conditional Equation.
Consider the equation $3x + 2$ = $8$. This equation is true only when $x$ = $2$. For any other value of $x$, the equation is not true. Hence, $x$ = $2$ satisfies the equation $3x + 2$ = $8$
In many cases, we can predict the existence of a solution to a conditional equation by the concrete conditions that result in the problem. For example, when there is a problem like, " the perimeter of a triangle is 32 and one of its side is 3 cm and the second side is twice the third side. What are the lengths of the sides of the triangle?" This problem clearly says it is a triangle which gives the necessary condition and the clue to solve the conditional equation results from it. In order to find out which values of the variable satisfy a given equation, we must solve it. The values that satisfy a given equation are also referred to as the Roots of the equation.
For example, 3 is a root of the equation $x + 4$ = $7$. If two equations have the same root, they are said to be Equivalent Equations. There are a few theorems that are useful to solve a given conditional equation by transforming them into an equivalent equation.
### Theorems to transform a given conditional equation into an equivalent equation
Given below are some theorems that can help us transform a conditional equation into an equivalent equation.
Theorem 1:
The Addition or Subtraction of a finite quantity to both sides of an equation results in a new equation that is Equivalent to the original equation.
If A = B is an equation and C is any finite quantity or an algebraic expression which will be finite for any finite value assigned to its variables, then A + C = B + C and A - C = B - C are equivalent equations to the equation A = B.
This theorem helps us to transfer a term from one side of the equation to the other so that we can solve it easily.
For example, $x + 4$ = $3$
To solve this equation, we should subtract $4$ from both sides of the equation.
$x + 4 - 4$ = $3 - 4$ which is equivalent to the equation $x + 4$ = $3$
Simplifying we get $x$ = $-1$
Hence, the solution to the $x + 4$ = $3$ is $-1$.
Theorem 2:
An equation which is equivalent to the given equation can be obtained by multiplying or dividing both the sides of the equation by a finite non zero quantity.
Consider the equation A = B. For any finite non zero quantity C, AC = BC and A/C = B/C are equations which are equivalent to the equation A = B. This also helps us to solve the equation.
For example, $2x$ = $8$
This equation can be solved by dividing both the sides of the equation by 2.
$\frac{2x}{2}$ = $\frac{8}{2}$
$x$ = $4$
Hence, the solution to the $2x$ = $8$ is $4$.
### Examples on Conditional Equations
Given below are some examples that explain how to solve conditional equations.
Example 1:
Solve the following conditional equation using the properties of equality
$x - 9$ = $2x - 3$
Solution:
$x - 9$ = $2x - 3$
$x - 9 + 9$ = $2x - 3 + 9$ (Adding 9 to both sides of the equation)
$x$ = $2x + 6$
$x - 2x$ = $2x + 6 -2x$ (Adding -2x to both sides of the equation)
$-x$ = $6$
$-x \times (-1)$ = $6 \times (-1)$ (Multiplying by -1 on both sides of the equation)
$x$ = $-6$
Hence, the solution to the equation $x - 9$ = $2x - 3$ is $x$ = $-6$
Example 2:
Solve $\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$
Solution:
$\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$
$\frac{x}{5}$$+$$\frac{3}{7}$$-$$\frac{3}{7}$ = $\frac{5}{6}$$-$$\frac{3}{7}$ (Subtracting $\frac{3}{7}$ from both the sides of the equation)
$\frac{x}{5}$ = $\frac{17}{42}$
Multiply both sides of the equation by 5 so that we can eliminate the denominator of the variable.
$5 \times $$\frac{x}{5} = 5 \times$$\frac{17}{42}$
$x$ = $\frac{85}{42}$
Hence, the solution to the equation $\frac{x}{5}$$+$$\frac{3}{7}$ = $\frac{5}{6}$ is $x$ = $\frac{85}{42}$
## Infinite Solutions in Algebraic Equations
An equation is said to have infinite solutions if it passes through many points. For example, let us consider y = y. It can be seen that for any real number value, this equation is satisfied. So, it has infinite solutions.
When we consider two lines, to get infinite solutions, they will intersect at many points along the line. If it is a straight line, this will only occur if both of them are on the same line. This kind of solution is called an inconsistent solution. As it represents the same line, it is called an dependent system of equations.
### Infinite solution in a pair of equation
If any of the following conditions are satisfied, then it can be said that there are many solutions.
• If the lines are parallel and overlapping.
• When graphing, all points(x, y) are simultaneously on both lines.
• While solving for a solution, using elimination or substitution method, leads to an identity.
• Slopes are equal with the same y intercepts.
• If A/D = B/E = C/F
### Examples on Infinite Solutions:
Given below are some examples on Infinite Solutions
Example 1:
Solve $3y - x = -3$ and $6y = 2x - 6$
Solution:
$3y - x = -3$
$6y = 2x - 6$
Let us solve this by graphing
$3y - x = -3$
x 0 -3 3 y -1 1 0
$6y = 2x - 6$
x 0 6 3 y -1 1 0
It can be seen that both the lines are parallel and overlapping. So, it has many solutions. And so, it is inconsistent but dependent.
Example 2:
Solve $x + 3y = 1$ and $2x + 12y = 2$
Solution:
$x + 3y = 1$
$2x + 12y = 2$
Using the Substitution method:
$x + 3y = 1$ -------------- (1)
$2x + 6y = 2$ ------------ (2)
Step 1: Isolate x from the first equation
$x = 1 - 2y$
Step 2: Substitute $x = 1 - 3y$ in the second equation
$2(1 - 3y) + 6y = 2$
Step 3: Solve for y
$(2)(1) - (2)(3y) + 6y = 2$
$2 - 6y + 6y = 2$
$2 = 2$
Here, we get an identity.
So, they have many solutions. Hence, the system is inconsistent and dependent.
Example 3:
Solve $x + 5y = 12$ and $4x + 20y = 60$
Solution:
$x + 5y = 12$
$4x + 20y = 60$
While comparing this with the general form of lines Ax + By = C and Dx + Ey = F, we get,
A = 1, B = 5, C = 12, D = 4, E = 20, F = 60
Now, $\frac{A}{D}$ = $\frac{1}{4}$
$\frac{B}{E}$ = $\frac{5}{20}$ = $\frac{1}{4}$
$\frac{C}{F}$ = $\frac{12}{60}$ = $\frac{1}{4}$
So, $\frac{A}{D}$ = $\frac{B}{E}$ = $\frac{C}{F}$
Hence, there are many solutions for the given set of equations. So, the system is inconsistent and dependent.
Example 4:
Solve $y = -2 + 5x$ and $15x - 3y = 6$
Solution:
$y = -2 + 5x$
$15x - 3y = 6$
First, let us write the equation in a slope intercept form y = mx + b where, m and b are the slope and y is the intercept respectively.
So, $y = -2 + 5x$
$y = 5x - 2$
Here, slope = 5
And, y intercept = -2
$15x - 3y = 6$
$y = $$\frac{15x - 6}{3} = \frac{15x}{3} - \frac{6}{3} Here, slope = \frac{15}{3} = 5 And, y intercept = \frac{-6}{3} = -2 So, the slopes are equal with the same y intercepts. Therefore there are many solutions. Hence, the system is inconsistent and dependent. ## No Solution in Algebraic Equations An equation can either be true or false according to the value we assign to the variable. When none of the real numbers makes the equation true, we say that there exists no solution for it. Let us consider a linear equation x - 5 = 0. When we plug in x = 5, it can be seen that the equation is satisfied. Hence, it has one solution x = 5. Now, consider x + 4 = x + 6. Here, whatever value we plug in for x, it can be seen that this equation will never be true. Hence, a contradiction occurs as we assume each as a solution. And so, it has no solution ### No Solution in a Pair of Equation If any of the following condition is satisfied, then it can be said that there is no solution. • If the lines are parallel with no overlapping points. • When graphing, there is no point (x, y) that is simultaneously on both lines. • While solving for a solution, using elimination or substitution method, leads to a contradiction. • Slopes are equal with different y intercepts. • If A/D = B/E ≠ C/F Such equations with no solutions are called inconsistent equations. Such systems are called as an independent system. ### Examples on equations with no solution Given below are some examples on solving equations with no solutions. Example 1: Solve 2x + 4y = 16 and x + 2y = 4 Solution: 2x + 4y = 16 x + 2y = 4 Lets graph each equation. 2x + 4y = 16 x 0 4 8 y 4 2 0 x + 2y = 4 x 0 2 4 y 2 1 0 It can be seen that both the lines are parallel without any overlap. So, it has no solution. Therefore, it is inconsistent. Example 2: Solve x + 3y = 12 and 3x + 9y = 24 Solution: x + 3y = 12 3x + 9y = 24 Using the Substitution method: x + 3y = 12 ------------- (1) 3x + 9y = 24 ----------- (2) Step 1: Isolate x from the first equation x = 12 - 3y Step 2: Substitute x = 12 - 3y in the second equation 3(12 - 3y) + 9y = 24 Step 3: Solve for y (3)(12) - (3)(3y) + 9y = 24 36 - 9y + 9y = 24 36 = 24 Here, the equation becomes a contradiction as 36 and 24 cannot be equal. So, they have no solution. And hence, the system is inconsistent. Example 3: Solve 2x + 3y = 12 and 4x + 6y = 20 Solution: 2x + 3y = 12 4x + 6y = 20 While comparing this with the general form of the lines Ax + By = C and Dx + Ey = F, we get, A = 2, B = 3, C = 12, D = 4, E = 6, F = 20 Now, \frac{A}{D} = \frac{2}{4} = \frac{1}{2} \frac{B}{E} = \frac{3}{6} = \frac{1}{2} \frac{C}{F} = \frac{12}{20} = \frac{3}{5} So, \frac{A}{D} = \frac{B}{E}$$\frac{C}{F}$
Hence, there is no solution for this set of equations. And so, the system is inconsistent.
Example 4:
Solve $x - y = 2$ and $3x - 3y = 5$
Solution:
$x - y = 2$
$3x - 3y = 5$
First, let us write the equation in the slope intercept form y = mx + b where, m and b are slope and y intercept respectively.
So, $x - y = 2$
$y = x - 2$
Here, slope = 1
And, y intercept = -2
$3x - 3y = 6$
$y =$ $\frac{3x - 6}{3}$ = $\frac{3x}{3}$ - $\frac{5}{3}$
Here, slope = $\frac{3}{3}$ = $1$
And, y intercept = $\frac{-5}{3}$
So, the slopes are equal but y intercepts are different. So, there is no solution. Hence, the system is inconsistent.
## Two Step Algebraic Equations
Two-Step Algebraic Equations are defined as solving an equation twice, that is, one step is done two times. The reason it is called so is because there are only two operations that are needed to solve the equation.
1. The opposite of addition operation is subtraction and vice versa.
2. The opposite of multiplication operation is division and vice versa.
There are two types of such equations:
• Two-step Algebraic Multiplication Equations
• Two-step Algebraic Division Equations
### Two-step Algebraic Multiplication Equations:
The processes for solving these types of two way algebraic equations are:
• Add or subtract so that the Coefficient and Variable are on either side of the equal sign, from both sides.
• Divide each side by the coefficient of the corresponding variable.
• Solve the remaining part of the equation.
Example 1:
$7y - 6 = 36$
Solution:
To solve, the variable must be on one side and all the other numbers should be on the other sides.
$7y - 6 = 36$
Add 6 on each side
$7y - 6 + 6 = 36 + 6$
$7y + 0 = 42$
$7y = 42$
Divide by 7
$\frac{7y}{7}$ = $\frac{42}{7}$
$y = 6$
Example 2:
$1 + 10u = 56$
Solution:
$1 + 10u = 56$
Subtract 1 from both the sides
$10u = 55$
Divide by 10
$u$ = $\frac{55}{10}$
= $\frac{11}{2}$
### Two-step Algebraic Division Equations:
The process for solving these types of two way algebraic equations are:
• Add or subtract so that the Coefficient and Variable are in either side of equal sign, from both the sides.
• Multiply both the sides by the coefficient of the variable.
• Solve the remaining part of the equation.
Example 3:
$\frac{y}{5}$ $- 6 = 34$
Solution:
To solve, the variable must be on one side and all other numbers to the other sides.
$\frac{y}{5}$ $- 6 = 34$
Add 6 on either side
$\frac{y}{5}$ $- 6 + 6 = 34 + 6$
$\frac{y}{5}$
$- 0 = 40$
$\frac{y}{5}$ = $40$
Multiply by 5
$\frac{5y}{5}$ = $5 \times 40$
$y = 200$
Example 4:
$\frac{-u}{6}$ $+ 1 = 13$
Solution:
$\frac{-u}{6}$ $+ 1 = 13$
Subtract 1 from both the sides
$\frac{-u}{6}$ = $12$
Multiply by -6 on both the sides
$u = 12 \times (-6) = -72$
## One Step Algebraic Equation
Any algebraic equation that requires only one set to solve can be defined as one step algebraic equation.
To solve an algebraic linear equation means to find an appropriate value for the variable so that the equation is always true. To solve an equation, different operations are performed on either side of the equation according to the question provided. The main idea is to isolate the unknown to the left and all the known values to the right.
Lets consider different forms of solving an algebraic equation requiring only one step. Let C and D be considered as a representative of a real number and let x be a variable.
Form 1: $x + C = D$
Here, subtract C from both sides
$x + C - C = D - C$
$x = D – C$
Hence, if a number is added on one side containing the variable of the equation, then it is subtracted from both the sides.
Example:
$x + 2 = 1$
Solution:
$x + 2 = 1$
Subtract 2 from both the sides
$x + 2 - 2 = 1 - 2$
$x = 1 – 2$
$x = -1$
Form 2: $x - C = D$
Here, add C on both the sides.
$x - C + C = D + C$
$x = D + C$
Hence, if a number is subtracted from one side containing the variable of the equation, it is added on both the sides.
Example:
$x - 5 = 7$
Solution:
$x - 5 = 7$
Subtract 5 on both the sides
$x - 5 + 5 = 7 + 5$
$x = 7 + 5$
$x = 12$
Form 3
: $Cx = D$
Here, divide by C on both the sides.
$\frac{Cx}{C}$ = $\frac{D}{C}$
$x$ = $\frac{D}{C}$
Hence, if a number is multiplied on one side containing the variable of the equation, it is divided on both the sides.
Example:
$2x = 3$
Solution:
$2x = 3$
Subtract 2 on both the sides
$\frac{2x}{2}$ = $\frac{3}{2}$
$x$ = $\frac{3}{2}$
Form 4: $\frac{x}{C}$ = $D$
Here, multiply by C on both the sides.
$\frac{x}{C}$ $\times C$ = $D \times C$
$x = DC$
Hence, if a number is divided from one side containing the variable of the equation, it is multiplied on both the sides.
Example:
$\frac{x}{3}$ = $6$
Solution:
$\frac{x}{3}$ = $6$
Multiply by 3 on both the sides
$\frac{x}{3}$ $\times 3 = 6 \times 3$
$x = 18$
Transforming: Short-cut Method
When shifting from one side to other side, change the signs of each term. So, Form 1 and Form 2 can be written in the form of transforming in order to save even that one step.
Form 1: $x + C = D$
change $+C$ to $–C$ on the other side, hence $x = D – C$
Form 2
: $x - C = D$
change $-C$ to $+C$ on the other side, hence $x = D + C$
Example:
Solve $x + a – b + c = d$
Solution:
To solve x, transform + to – and – to +
So, $x + a – b + c = d$ changes to $x = d – a + b – c$
Hence solved.
In some equations, the unknown variable may fall on the right hand side. So, it’s better to exchange sides so as to bring it to the left hand side. In this process, the sign doesn’t change.
Example:
$1 = x + 6$
This is of Form 1. So, +6 changes to -6
Hence, $1 – 6 = x$
$-5 = x$
Exchanging the sides, we get $x = -5$
## Multi Step Algebraic Equations
Multi-step algebraic equations can be defined as an equation such that solving it will require more than two operations as steps. To solve it, the equation can be simplified by using the distributive property and combining like terms depending on the requirement.
Lets take an example of a simple multi-step algebraic equation and lets check the steps.
$2w + 3 = 6 + w$
The basic idea is to get all the variables on any of the side and all the numbers on the opposite side. Lets take the variables to the left and the numbers to the right. For that, subtract w from both sides
$2w + 3 - w = 6 + w - w$
$w + 3 = 6$
Subtract 3 from both the sides
$w + 3 - 3 = 6 - 3$
$w = 3$
In general, the steps followed for solving multi step algebraic equations are
• Isolate the variable and hence solution of equation can be obtained
• Use addition as inverse of subtraction and vice versa.
• Use division as inverse of multiplication and vice versa.
The numbers/constants used can be integers, fractions or decimals.
### Solving Multi-Step Algebraic Equations with Integers
To solve equations with integers, first we have to isolate the variables and then simplify the equation.
Example:
$3(u + 1) – (2u + 3) = 5$
First use distribution property and open the parenthesis,
$3u + 3 – 2u - 3 = 5$
Combine the like term
$3u – 2u + 3 – 3 = 5$
$u – 0 = 5$
$u = 5$
### Solving Multi-Step Algebraic Equations with Fractions
To solve equations with fraction, first of all simplify the equation without fraction.
Example:
$\frac{5x}{4}$ $+ 2 – 3x =6$
Distribute the terms by opening the parenthesis
$\frac{5x}{4}$
$+ 2 – 3x = 6$
Subtract 2 from both the sides
$\frac{5x}{4}$ $+ 2 – 3x - 2 = 6 - 2$
$\frac{5x}{4}$ $- 3x = 4$
Factor x on the left hand side
$\frac{(5 - 12)x}{4}$ = $4$
$\frac{-7x}{4}$ = $4$
Multiply by 4 on both the sides,
$-7x = 16$
Divide by (-7) on both the sides
$x$ = $\frac{16}{-7}$
$x$ = $\frac{-16}{7}$
### Solving Multi-Step Algebraic Equations with Decimals
To solve equations with decimals, first we have to isolate the variables and then simplify the equation.
Example 1:
$5(x – 2.3) – 2x = 5.3 + x$
Simplify as needed in the equation
$5(x – 2.3) – 2x = 5.3 + x$
Distribute the terms by opening the parenthesis
$5x – 11.5 – 2x = 5.3 + x$
Simplify all the like terms
$3x – 11.5 = 5.3 + x$
Subtract x from both the sides
$3x – 11.5 - x = 5.3 + x - x$
$2x – 11.5 = 5.3$
Add 11.5 on both the sides
$2x – 11.5 + 11.5 = 5.3 + 11.5$
$2x = 9.5$
Divide by 2 on both the sides
$\frac{2x}{2}$ = $\frac{16.8}{2}$
$x = 8.4$
Example 2:
Find three consecutive even integers whose sum is 24
Solution:
To solve lets assume the smallest even integer is $2x$.
Then, the next consecutive even integer = $2x + 2$
Next one = $(2x + 2) + 2 = 2x + 4$
Given sum of all of it is 24
So, $2x + (2x + 2) + (2x + 4) = 24$
Distribute the terms by opening the parenthesis
$2x + 2x + 2 + 2x + 4 = 24$
Simplify all the like terms
$6x + 6 = 24$
Subtract 6 from both the sides
$6x + 6 - 6 = 24 - 6$
$6x = 18$
Divide by 6 on both the sides
$\frac{6x}{6}$ = $\frac{18}{6}$
$x = 3$
So, the smallest number is $2 \times 3$ = $6$
Next is = $6 + 2$ = $8$
Last = $6 + 4$ = $10$
## Rules of Algebraic Equations
Algebra is a branch of mathematics that can help take a real situation and change it in the mathematical form. That is as an equation. As in real number, terms of algebra also follow a set of rules that helps in simplifying the equation and expression. Lets check out certain rules and their corresponding examples:
Basic Rules
Let x, y and z be any variable, then
Rule 1: Any term multiplied by 1 is that term itself.
$x \times 1 = x$
Here, 1 is called as the multiplicative identity.
Example:
$2 \times 1 = 2$
$5 \times 1 = 5$
Rule 2: Any term multiplied by -1 is the negative of that term.
$x \times (-1) = -x$
Example:
$8 \times (-1) = -8$
$2 \times (-1) = -2$
Rule 3: Negative of a negative term is negative.
$-(-x) = x$
Example:
$-(-3) = 3$
Rule 4: The product of a term and its reciprocal is 1.
$x \times ($ $\frac{1}{x}$ $) = 1$
Example:
$6 \times ($ $\frac{1}{6}$ $) = 1$
Rule 5: Sum of any term and zero is that term itself.
$x + 0 = x$
Here, 0 is called additive identity.
Example:
$6 + 0 = 6$
Rule 6: Sum of a positive term and its negative is zero
$x + (-x) = 0$
Hence, –x is the additive inverse of x.
Example:
$4 + (-4) = 0$
Rule 7: Combination of addition and subtraction sign is again a subtraction.
$x + (-y) = x + -y = x – y$
Example:
2 + (-3) = 2 – 3 which is -1
Rule 8: Combination of two subtraction sign is addition.
$x - (-y) = x + y$
Example:
$5 - (-3) = 5 + 3 = 8$
Rule 9: Multiplication of a negative and positive term is negative
$x \times (-y) = -xy$
Example:
$5 \times (-2) = -10$
Rule 10: Multiplication of two negative terms is a positive term
$(-x) \times (-y) = xy$
Example:
$(-2)(-5) = 10$
### Properties Concerning Algebraic Expressions
Addition:
• Commutative: Order of any set of terms doesn’t matter
$x + y = y + x$
Example:
$2 + 1 = 3 = 1 + 2$
• Associative: Consecutive numbers can be grouped accordingly
$(x + y) + z = x + (y + z)$
Example:
$(2 + 4) + 1 = 7 = 2 + (4 + 1)$
Multiplication
:
• Commutative: Order of any set of terms doesn’t matter
$x \times y = y \times x$
Example:
$4 \times 2 = 8 = 2 \times 4$
• Associative: Consecutive numbers can be grouped accordingly
$(x \times y) \times z = x \times (y \times z)$
Example:
$(3 \times 4) \times 2 = 24 = 3 \times (4 \times 2)$
• Distributive Property of Addition over Multiplication:
There are two types: Right and Left
Right: $(x + y)z = xz + yz$
Left: $z(x + y) = zx + zy$
Properties Using Equality
• In any equation, identical terms can be added on both sides, by which the equation remains unaltered.
So, if $x = y$, $x + z = y + z$
Example:
If $x = 7$, $x + 5 = 7 + 5$
$x + 5 = 12$
• In any equation, identical terms can be multiplied on both sides, by which the equation remains unaltered.
So, if $x = y$, $x \times z = y \times z$
Example:
If $u = 2$, $u \times 3 = 2 \times 3$
$3u = 6$
• In any equation, changing the sign on both the sides, makes the equation unaltered
So, if $x = y$, $-x = -y$
Example:
If $x = 5$, $-x = -5$
## Graphing Algebraic Equation
Graphing an algebraic equation is more of a visual way of looking at an equation so it is possible to understand its structure and properties. It is drawn with axes perpendicular to each other. In graphing a linear algebraic equation, two axes, x (horizontal axis) and y (vertical axis), perpendicular to each other, are taken. They intersect a point called as the origin. It is called the coordinate plane or the rectangular coordinate plane. Each point on the plane is represented by an ordered pair (a, b) where a is a point on the x axis and b is a point on the y axis.
There will be infinite points for an algebraic linear equation. But, its not possible to determine all of it. So, we use minimum points to get the whole equation graphed correctly. An algebraic linear equation can be mostly drawn in two ways.
• Using the slope of a line and the y intercept
• Using the T chart
### Using the Slope of a Line and the y-intercept
The standard form of writing any linear equation is y = (m) x + b
where m, being the x coefficient, is called the slope and b, being the constant, is called the y intercept.
The following procedure can to be undertaken:
• First the main point is to write the equation in the slope intercept form.
• Recognize the y intercept and plot it on the graph.
• Recognize the slope and using it determine another point on the xy plane.
• Join the two points and also extend it both to either side of the points to the required line
Example:
Graph $2y - x = 12$
First write it in the slope intercept form
$2y - x = 12$
Add x on both the side
So, $2y = x + 12$
Divide by 2 on both the sides
$y = ($ $\frac{1}{2}$ $) x + 6$
So, slope $m$ = $\frac{1}{2}$ and y intercept = (0, 6)
Plot the point (0, 6) on the graph.
From 6 go 1 units up, so it goes to 7 and 2 units right, so we end up at (2, 7)
Hence draw a line joining the two points.
### Using T chart
When graphing any algebraic linear equations, first form a chart, known as T- chart. It is called so by the shape it takes. It will have two columns. Left column contains x points and right column contains y points. You can choose any reasonable x values, plug it in the equation and solve for y. Grouping the x and corresponding y values makes a pair that can be used for plotting.
Any x values can be chosen such that y value exists.
For any straight line we need at least three points.
Draw a line joining all the three points.
Example :
Graph $y = x - 4$
For x values assume -1, 0, 1. Lets form T chart
x y = x - 4 (x , y) -1 y = (-1) - 4 = -5 (-1, -5) 0 y = 0 - 4 = -4 (0, -4) 1 y = 1 - 4 = -3 (1, -3)
Plot this point on the graph and drawn the line joining the points.
## Types of Algebraic Equations
An algebraic equation can be defied by an equality separated by two algebraic expressions containing variables and constants. The set of algebraic equation can be generally grouped into five sets.
1) Polynomial Equation
A polynomial equation can be defined as an equation that contains variable terms whose power is a whole number. They are classified by the number of terms they contains. The classification is as follows:
Monomial:
It will only contain a single term
Example:
$2x = 0$
Binomial:
It will contain exactly two terms
Example:
$4x +7y = 0$
Trinomial:
It will contain exactly three terms
Example:
$2y + u – 4x = 0$
Polynomial equation can also be classified in terms of degrees
One degree (linear):
It is the simplest form of an algebraic equation. It is commonly known as a linear equation. It is of the form ax + b = c, where a, b, c are constants.
Example:
$3x + 1 = 0$
Two degree (quadratic):
An algebraic equation of degree 2 is called a quadratic equation. The general form is ax2 + bx + c = 0 where a $\neq$ 0 and a, b, c are constants.
Example:
5x2 - x +2
Cubical equation:
An algebraic equation of degree 3 is called a cubic equation. The general form is
ax3 + bx2 + cx + d = 0, where a $\neq$ 0 and a, b, c, d are constants.
Example:
3x3 + 2x2 + x + 3 = 0
2) Rational Equation
Any rational equation will be in the form y= p/q where both p and q are polynomials.
Example:
Y = $\frac{(2x – 5)}{(3x^{2} + x – 1)}$
Its very important the denominator mustn’t be zero.
3) Exponential Equation
Such equation contain variables in the exponent part.
Its form can be abx + c = 0.
If, the independent variable has a positive coefficient, it is exponential growth and if it has negative coefficient, then it is decay.
Example:
y = 3(2x-1) + 5
4) Logarithmic equation
Logarithm is the inverse of exponent.
So if y = ax is in the exponential equation then its logarithmic equation is x = loga y
The number ”e” is the most common logarithmic base and such a logarithm is called the natural logarithm.
Example:
Loga x +loga 4 = loga 3
5) Trigonometric Equations
It will contain trigonometric ratios. The peculiarity of a trigonometric function is that it will have periods so it will be repeat by itself after a particular point. Solving such an equation depends on certain rules and identities of trigonometry.
Example:
sinx + cos x = tan 2x
|
What is a number line? A number line is a straight line which marks numbers at intervals, they are used to help children with various mathematical calculations.
Here is an example:
Question: What is 7 – 5 ? Children can use the number line by placing their finger on the number 7 and then counting down 5 until they get to the answer, number 2.
Learning to add and subtract using a number line is helpful for children, as it enables them to visualise the calculation by counting up or down the line.
Number lines do not have to start at 0, they can be used to mark out any numbers which are relevant to a question.
## When are children taught to use number lines?
From Reception onwards, children will be familiar with number lines, as teachers will use them to show children the order of numbers and how to count up and down.
Children may also use them In Year 1, when they learn how to count in 2s, 5s and 10s. A good understanding of addition will help children when they start learning their times tables in Year 2.
One way that teachers start introducing times tables is by using number lines to demonstrate multiplications. For example:
To work out the answer to 3 x 3, children can use a number line to count up in 3s three times until they get the answer 9.
In a similar way, number lines can also help children learn how to divide. For example:
To work out 16 ÷ 4, children can place their finger over the number 16 and count backwards in sets of 4, marking the number line each time. Once they have got to 0 they can then count how many marks they made on the line, and they will get the answer 4.
For harder subtraction questions, such as 93 – 38, teachers may show children how to use the jump method. The benefit of this method is that it shows children that a subtraction question is just asking you the difference between two numbers.
To work out the calculation above on a number line, the smaller number is put at the start, with the larger number at the end. Then children can jump up in 10s until there is a number less than 10 left to add:
After doing this, children simply need to add up all the numbers from each jump they made:
10 + 10 + 10 + 10 + 10 + 5 = 55 (93 - 38 = 55)
During Years 4 to 6, teachers will use number lines to demonstrate negative numbers, decimal numbers and fractions. Here are some examples of how number lines will be used:
## How does Learning Street help children with number lines?
Learning Street will introduce number lines to the children slowly. This will be combined with regular practice in order that it is understood before then beginning to develop their knowledge and extend it further, as well as constant revision so that it isn’t forgotten.
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# How do you find the limit of cscx-cotx+cosx as x approaches 0?
Dec 16, 2016
#### Explanation:
${\lim}_{x \rightarrow o} \cos x = 1$, so that term is not a problem, it's the other two that challenge us.
In trigonometry, when in doubt, one thing to try is rewriting using just sines and cosines.
$\csc x - \cot x = \frac{1}{\sin} x - \cos \frac{x}{\sin} x = \frac{1 - \cos x}{\sin} x$
I assume that we have learned that
${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = {\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta = 1$ $\text{ }$ and $\text{ }$ ${\lim}_{x \rightarrow \theta} \frac{1 - \cos \theta}{\theta} = 0$
So we'll multiply by $\frac{x}{x}$ to get
$\frac{x \left(1 - \cos x\right)}{x \sin x} = \frac{x}{\sin} x \cdot \frac{1 - \cos x}{x}$
So
${\lim}_{x \rightarrow 0} \left(\csc x - \cot x + \cos x\right) = {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} x \cdot \frac{1 - \cos x}{x} + \cos x\right)$
$= \left(1\right) \cdot \left(0\right) + 1 = 1$
|
## Median to 16
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. You are given a triangle with side lengths of 11 cm, 13 cm, and 16 cm. Determine the length of the median that joins the side of length 16 cm and the opposite vertex.
Source: mathcontest.olemiss.edu 11/2/2009
SOLUTION
Let $AB=16 \textup { cm}, AC=11 \textup{ cm and } BC=13 \textup{ cm}$. Let $\overline{CD}$ be the median joining vertex $C$ to the midpoint $D \textup{ of } \overline{AB}$. From vertex $C$ draw altitude $\overline{CE}$ to side $\overline{AB}$.
Length of $\overline{DE}$
In right triangle $ECB$, by the Pythagorean theorem we have
$EB^2+EC^2=BC^2=13^2=169$
In right triangle $EAC$, we have by the same theorem
$EA^2+EC^2=11^2=121$
Multiplying the second equation by $-1$ then adding the two equations yields
$EB^2+EC^2=169$
$-EA^2-EC^2=-121$
—————————–
$EB^2-EA^2=48$
$\left (EB+EA\right )\left (EB-EA\right )=48$
$AB\left (EB-EA\right )=48$
$16\left (EB-EA\right )=48$
$EB-EA=3$
Since $EB=BD+DE=DA+DE \textup{ and } EA=DA-DE$, rewrite the above equation as follows:
$EB-EA=3$
$DA+DE-\left (DA-DE\right )=3$
$DA+DE-DA+DE=3$
$2DE=3$
$DE=1.5$
Calculate $EC^2$
$EA=DA-DE$
$=8-1.5$
$=6.5$
In right triangle $EAC$:
$EA^2+EC^2=11^2=121$
$6.5^2+EC^2=121$
$42.25+EC^2=121$
Thus, $EC^2=121-42.25=78.75$
Length of median $\overline {CD}$
In right triangle $DEC$:
$DE^2+EC^2=CD^2$
$1.5^2+78.75=CD^2$
$2.25+78.75=CD^2$
$81=CD^2$
Thus, $CD=9$.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Graphs of Square Root Functions | CK-12 Foundation
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11.1: Graphs of Square Root Functions
Created by: CK-12
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Learning Objectives
At the end of this lesson, students will be able to:
• Graph and compare square root functions.
• Shift graphs of square root functions.
• Graph square root functions using a graphing calculator.
• Solve real-world problems using square root functions.
Vocabulary
Terms introduced in this lesson:
increases, decreases
flip
shift
stretch
transform
Teaching Strategies and Tips
Use Example 1 to introduce the basic shape of the square root function.
Use the tables in the examples to show students the square root function’s behavior numerically:
• Why it is undefined on some intervals.
• That it is everywhere increasing.
• That it rises relatively slowly.
• That the square root of a fraction is greater than the fraction. Numbers in the interval $(0,1)$ are smaller than their square roots; $x < \sqrt{x}$ for $x$ in the interval $(0,1)$ and $x > \sqrt{x}$ for $x$ in the interval $(1,\infty)$.
Use the graphs in the examples to make the following observations:
• The square root graph is half a parabola lying sideways.
Emphasize finding the domain of the square root function before making a table.
• When the expression under the square root is negative, table values will be undefined; and the graph corresponding to the interval will be empty.
Use Examples 2-10 to motivate transformations:
• Shifts, stretches, and flips allow graphing without constructing a table of values.
• Teachers are encouraged to use several examples to illustrate the effect of each constant on the graph.
Graph the following functions using transformations of the basic graph $y=\sqrt{x}$.
a. $y=-\sqrt{x}$
Hint: Flip about the $x-$axis.
b. $y=\sqrt{-x}$
Hint: Flip about the $y-$axis.
Remark: Student often claim that the whole function is undefined because of the negative under the radical. Point out that the domain “flips” to negative numbers.
c. $y=2\sqrt{x}$
Hint: Stretch in the vertical direction by a factor of $2$: $y-$values are multiplied by $2$.
d. $y=\sqrt{2x}$
Hint: Point out that $y=\sqrt{2x}=\sqrt{2}\sqrt{x}$ .
e. $y=\sqrt{x}+2$
Hint: Shift the graph up by $2$: $y-$values are increased by $2$.
f. $y=\sqrt{x+2}$
Hint: Shift the graph left by $2$: $x-$values are decreased by $2$.
g. $y=-\sqrt{x}+1$
Hint: Point out that the transformations are in the same direction. $y-$values are reflected across the $x-$axis (parallel to the $y-$axis) and then shifted vertically (parallel to the $y-$axis), in that order. Therefore, the correct sequence is to flip and then shift.
h. $y=\sqrt{-x+1}$
Hint: Have students consider what happens to an input $x$ and do the transformations in the opposite order. Therefore, shift left, then reflect across the $y-$axis.
Graph the following function using shifts, flips, and stretches.
a. $y=4+2\sqrt{2-x}$
Solution: View $y=4+2\sqrt{2-x}$ as a combination of transformations of the basic square root graph $y=\sqrt{x}$.
Start with the simpler equation: $y=\sqrt{2-x}$. If we follow an input $x$, then it first gets multiplied by $-1$ and then increased by $2$. How do the graphs of $y=\sqrt{x}$ and $y=\sqrt{2-x}$ compare? $y=\sqrt{2-x}$ is a shift of $y=\sqrt{x}$ two units LEFT and then flipped across the $y-$axis. Note that the transformations happen in the opposite order in which an input $x$ gets operated on.
To graph $y=2\sqrt{2-x}$ we multiply the $y-$values of $y=\sqrt{2-x}$ by $2$ to obtain a vertically stretched curve. Finally, to obtain the graph of $y=4+2\sqrt{2-x}$, shift the graph of $y=2\sqrt{2-x}$ four units vertically.
Encourage students to keep a list of functions they have studied so far. Include a few examples of each and their graphs. For example:
• Linear: $f(x)=mx+b$
Examples: $f(x)=x, \ f(x)=-x, \ f(x)=x+1, \ f(x)=2$
• Exponential: $f(x)=a \cdot b^x$
Examples: $f(x)=2^x, \ f(x)=2^{-x}, \ f(x)=-2^x$
• Quadratic: $f(x)=ax^2+bx+c$
Examples: $f(x)=x^2, \ f(x)=-x^2, \ f(x)=x^2+1$
• Square root: $f(x)=a\sqrt{bx+c}+d$
Examples: $f(x)=\sqrt{x}, \ f(x)=\sqrt{x+1}, \ f(x)=\sqrt{x}+1, \ f(x)=-\sqrt{x}$
Error Troubleshooting
General Tip: Students may not recognize $y=\sqrt{-x}$ as a valid function at first, stating that the square root of a negative is undefined. Explain that the function’s domain is defined.
General Tip: Have students find the domain of the function they are graphing on a calculator first. This will help find an appropriate window for the graph.
In Example 12 and Review Questions 14-18, state beforehand the number of decimal places required of students when rounding.
Feb 22, 2012
Aug 22, 2014
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# 5.2 Right triangle trigonometry
Page 1 / 12
In this section, you will:
• Use right triangles to evaluate trigonometric functions.
• Find function values for $\text{\hspace{0.17em}}30°\left(\frac{\pi }{6}\right),\text{\hspace{0.17em}}$ $45°\left(\frac{\pi }{4}\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}60°\left(\frac{\pi }{3}\right).$
• Use cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right triangle trigonometry to solve applied problems.
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:
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 $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is its value at $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ radians. First, we need to create our right triangle. [link] shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ to the x -axis, we have a right triangle whose vertical side has length $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and whose horizontal side has length $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ 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
Likewise, we know
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 $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of $\text{\hspace{0.17em}}x,$ we will call the side between the given angle and the right angle the adjacent side to angle $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ (Adjacent means “next to.”) Instead of $\text{\hspace{0.17em}}y,$ we will call the side most distant from the given angle the opposite side from angle $\text{}t.\text{\hspace{0.17em}}$ And instead of $\text{\hspace{0.17em}}1,$ we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in [link] .
## Understanding right triangle relationships
Given a right triangle with an acute angle of $\text{\hspace{0.17em}}t,$
$\begin{array}{l}\mathrm{sin}\left(t\right)=\frac{\text{opposite}}{\text{hypotenuse}}\hfill \\ \mathrm{cos}\left(t\right)=\frac{\text{adjacent}}{\text{hypotenuse}}\hfill \\ \mathrm{tan}\left(t\right)=\frac{\text{opposite}}{\text{adjacent}}\hfill \end{array}$
A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”
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.
## Evaluating a trigonometric function of a right triangle
Given the triangle shown in [link] , find the value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha .$
The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:
$\begin{array}{l}\mathrm{cos}\left(\alpha \right)=\frac{\text{adjacent}}{\text{hypotenuse}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{15}{17}\hfill \end{array}$
Given the triangle shown in [link] , find the value of $\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}t.$
$\frac{7}{25}$
Do somebody tell me a best nano engineering book for beginners?
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s.
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so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Uday
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preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Azam
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Damian
silver nanoparticles could handle the job?
Damian
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I'm interested in Nanotube
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Difference between revisions of "1976 USAMO Problems/Problem 5"
## Problem
If $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ are all polynomials such that $$P(x^5) + xQ(x^5) + x^2 R(x^5) = (x^4 + x^3 + x^2 + x +1) S(x),$$ prove that $x-1$ is a factor of $P(x)$.
## Solutions
### Solution 1
In general we will show that if $m$ is an integer less than $n$ and $P_0, \dotsc, P_{m-1}$ and $S$ are polynomials satisfying $$P_0(x^n) + x P_1(x^n) + \dotsb + x^{m-1} P_{m-1}(x^n) = \sum_{k=0}^{n-1} x^k S(x),$$ then $P_k(1) = 0$, for all integers $0 \le k \le m-1$. For the problem, we may set $n=5$, $m=3$, and then note that since $P(1)= 0$, $(x-1)$ is a factor of $P(x)$.
Indeed, let $\omega_1, \dotsc, \omega_{n-1}$ be the $n$th roots of unity other than 1. Then for all integers $1 \le i \le n-1$, $$\sum_{k=0}^{m-1} \omega_i^k P_k(\omega_i^n) = \sum_{k=0}^{m-1} \omega_i^k P_k(1) = \sum_{k=0}^{n-1} \omega_i^k S(\omega_i) = \frac{\omega_i^{n-1} - 1}{\omega_i - 1} S(\omega_i) = 0 ,$$ for all integers $1 \le i \le n$. This means that the $(m-1)$th degree polynomial $$\sum_{k=0}^{m-1} x^k P_k(1)$$ has $n-1 > m-1$ distinct roots. Therefore all its coefficients must be zero, so $P_k = 0$ for all integers $0 \le k \le m-1$, as desired. $\blacksquare$
### Solution 2
Let $\zeta, \xi, \rho$ be three distinct primitive fifth roots of unity. Setting $x = \zeta, \xi$, we have \begin{align*} P(1) + \zeta Q(1) + \zeta^2 R(1) &= \frac{\zeta^5-1}{\zeta-1} S(\zeta) = 0, \\ P(1) + \xi Q(1) + \xi^2 R(1) &= \frac{\xi^5-1}{\xi-1} S(\xi) = 0 . \end{align*} These equations imply that $$\zeta Q(1) + \zeta^2 R(1) = \xi Q(1) + \xi^2 R(1),$$ or $$Q(1) = - ( \zeta + \xi)R(1).$$ But by symmetry, $$Q(1) = - (\zeta + \rho)R(1) .$$ Since $\xi \neq \rho$, it follows that $Q(1) = R(1) = 0$. Then, as noted above, $$P(1) = P(1) + \zeta Q(1) + \zeta^2 R(1) = 0,$$ so $(x-1)$ is a factor of $P(x)$, as desired. $\blacksquare$
### Solution 3
Let $z, z^2, z^3$ be three of the 5th roots of unity not equal to one that satisfy $1 + z + z^2 + z^3 + z^4 = 0$ as a result. Plugging them into the equation gives the linear system of equations in $(A(1), B(1), C(1))$:
$$A(1) + zB(1) + z^2C(1) = 0$$ $$A(1) + z^2B(1) + z^4C(1) = 0$$ $$A(1) + z^3B(1) + z^6C(1) = 0$$
Direct observation gives $(A(1), B(1), C(1)) = (0,0,0)$ as a solution, and there are no others because the system of equations is linear. Hence, $A(1) = 0$, and so $(x-1)$ is a factor of $A(x)$, as desired. $\blacksquare$
Note: We can generalize this approach to prove the claim in Solution 1 in an equally fast, concise, and readily understandable fashion.
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# Applied Discrete Structures
## Section12.3An Introduction to Vector Spaces
### Subsection12.3.1Motivation for the study of vector spaces
$$\renewcommand{\vec}[1]{\mathbf{#1}}$$
When we encountered various types of matrices in Chapter 5, it became apparent that a particular kind of matrix, the diagonal matrix, was much easier to use in computations. For example, if $$A =\left( \begin{array}{cc} 2 & 1 \\ 2 & 3 \\ \end{array} \right)\text{,}$$ then $$A^5$$ can be found, but its computation is tedious. If $$D =\left( \begin{array}{cc} 1 & 0 \\ 0 & 4 \\ \end{array} \right)$$ then
\begin{equation*} D^5 =\left( \begin{array}{cc} 1 & 0 \\ 0 & 4 \\ \end{array} \right)^5= \left( \begin{array}{cc} 1^5 & 0 \\ 0 & 4^5 \\ \end{array} \right)= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1024 \\ \end{array} \right) \end{equation*}
Even when presented with a non-diagonal matrix, we will see that it is sometimes possible to do a bit of work to be able to work with a diagonal matrix. This process is called diagonalization.
In a variety of applications it is beneficial to be able to diagonalize a matrix. In this section we will investigate what this means and consider a few applications. In order to understand when the diagonalization process can be performed, it is necessary to develop several of the underlying concepts of linear algebra.
### Subsection12.3.2Vector Spaces
By now, you realize that mathematicians tend to generalize. Once we have found a “good thing,” something that is useful, we apply it to as many different concepts as possible. In doing so, we frequently find that the “different concepts” are not really different but only look different. Four sentences in four different languages might look dissimilar, but when they are translated into a common language, they might very well express the exact same idea.
Early in the development of mathematics, the concept of a vector led to a variety of applications in physics and engineering. We can certainly picture vectors, or “arrows,” in the $$x y-\textrm{ plane}$$ and even in the three-dimensional space. Does it make sense to talk about vectors in four-dimensional space, in ten-dimensional space, or in any other mathematical situation? If so, what is the essence of a vector? Is it its shape or the rules it follows? The shape in two- or three-space is just a picture, or geometric interpretation, of a vector. The essence is the rules, or properties, we wish vectors to follow so we can manipulate them algebraically. What follows is a definition of what is called a vector space. It is a list of all the essential properties of vectors, and it is the basic definition of the branch of mathematics called linear algebra.
#### Definition12.3.1.Vector Space.
Let $$V$$ be any nonempty set of objects. Define on $$V$$ an operation, called addition, for any two elements $$\vec{x}, \vec{y} \in V\text{,}$$ and denote this operation by $$\vec{x}+ \vec{y}\text{.}$$ Let scalar multiplication be defined for a real number $$a \in \mathbb{R}$$ and any element $$\vec{x}\in V$$ and denote this operation by $$a \vec{x}\text{.}$$ The set $$V$$ together with operations of addition and scalar multiplication is called a vector space over $$\mathbb{R}$$ if the following hold for all $$\vec{x}, \vec{y}, \vec{z}\in V$$ , and $$a,b \in \mathbb{R}\text{:}$$
• $$\displaystyle \vec{x}+ \vec{y}= \vec{y}+ \vec{x}$$
• $$\displaystyle \left(\vec{x}+ \vec{y}\right)+ \vec{z}= \vec{x}+\left( \vec{y}+\vec{z}\right)$$
• There exists a vector $$\vec{0}\in V\text{,}$$ such that $$\vec{x}+\vec{0} = \vec{x}$$ for all $$x \in V\text{.}$$
• For each vector $$\vec{x}\in V\text{,}$$ there exists a unique vector $$-\vec{x}\in V\text{,}$$ such that $$-\vec{x} +\vec{x}= \vec{0}\text{.}$$
These are the main properties associated with the operation of addition. They can be summarized by saying that $$[V; +]$$ is an abelian group.
The next four properties are associated with the operation of scalar multiplication and how it relates to vector addition.
• $$\displaystyle a\left(\vec{x}+ \vec{y} \right) =a \vec{x}+a \vec{y}$$
• $$\displaystyle (a +b)\vec{x}= a \vec{x} + b \vec{x}$$
• $$\displaystyle a \left(b \vec{x}\right) = (a b)\vec{x}$$
• $$1\vec{x} = \vec{x}\text{.}$$
In a vector space it is common to call the elements of $$V$$ vectors and those from $$\mathbb{R}$$ scalars. Vector spaces over the real numbers are also called real vector spaces.
Let $$V = M_{2\times 3}(\mathbb{R})$$ and let the operations of addition and scalar multiplication be the usual operations of addition and scalar multiplication on matrices. Then $$V$$ together with these operations is a real vector space. The reader is strongly encouraged to verify the definition for this example before proceeding further (see Exercise 3 of this section). Note we can call the elements of $$M_{2\times 3}(\mathbb{R})$$ vectors even though they are not arrows.
Let $$\mathbb{R}^2 = \left\{\left(a_1, a_2 \right) \mid a_1,a_2 \in \mathbb{R}\right\}\text{.}$$ If we define addition and scalar multiplication the natural way, that is, as we would on $$1\times 2$$ matrices, then $$\mathbb{R}^2$$ is a vector space over $$\mathbb{R}\text{.}$$ See Exercise 12.3.3.4 of this section.
In this example, we have the “bonus” that we can illustrate the algebraic concept geometrically. In mathematics, a “geometric bonus” does not always occur and is not necessary for the development or application of the concept. However, geometric illustrations are quite useful in helping us understand concepts and should be utilized whenever available.
Let’s consider some illustrations of the vector space $$\mathbb{R}^2\text{.}$$ Let $$\vec{x}= (1, 4)$$ and $$\vec{y} = (3, 1)\text{.}$$ We illustrate the vector $$\left(a_1, a_2\right)$$ as a directed line segment, or “arrow,” from the point $$(0, 0)$$ to the point$$\left(a_1, a_2\right)\text{.}$$ The vectors $$\vec{x}$$ and $$\vec{y}$$ are as shown in Figure 12.3.4 together with $$\vec{x}+ \vec{y} = (1, 4) + (3, 1) = (4, 5)\text{.}$$ The vector $$2 \vec{x} = 2(1, 4) = (2, 8)$$ is a vector in the same direction as $$\vec{x}\text{,}$$ but with twice its length.
#### Note12.3.5.
1. The common convention is to use that boldface letters toward the end of the alphabet for vectors, while letters early in the alphabet are scalars.
2. A common alternate notation for vectors is to place an arrow about a variable to indicate that it is a vector such as this: $$\overset{\rightharpoonup }{x}\text{.}$$
3. The vector $$\left(a_1,a_2,\ldots ,a_n\right)\in \mathbb{R}^n$$ is referred to as an $$n$$-tuple.
4. For those familiar with vector calculus, we are expressing the vector $$x = a_1 \boldsymbol{\hat{\textbf{i}}}+ a_2 \boldsymbol{\hat{\textbf{j}}} + a_3 \boldsymbol{\hat{\textbf{k}}} \in \mathbb{R}^3$$ as $$\left(a_1,a_2,a_3\right)\text{.}$$ This allows us to discuss vectors in $$\mathbb{R}^n$$ in much simpler notation.
In many situations a vector space $$V$$ is given and we would like to describe the whole vector space by the smallest number of essential reference vectors. An example of this is the description of $$\mathbb{R}^2\text{,}$$ the $$x y$$-plane, via the $$x$$ and $$y$$ axes. Again our concepts must be algebraic in nature so we are not restricted solely to geometric considerations.
#### Definition12.3.6.Linear Combination.
A vector $$\pmb{ y}$$ in vector space $$V$$ (over $$\mathbb{R}$$) is a linear combination of the vectors $$\vec{x}_1\text{,}$$ $$\vec{x}_2, \ldots\text{,}$$ $$\vec{x}_n$$ if there exist scalars $$a_1,a_2,\ldots ,a_n$$ in $$\mathbb{R}$$ such that $$\vec{y} = a_1\vec{x}_1+ a_2\vec{x}_2+\ldots +a_n\vec{x}_n$$
The vector $$(2, 3)$$ in $$\mathbb{R}^2$$ is a linear combination of the vectors $$(1, 0)$$ and $$(0, 1)$$ since $$(2, 3) = 2(1, 0) + 3(0, 1)\text{.}$$
Prove that the vector $$(4,5)$$ is a linear combination of the vectors (3, 1) and (1, 4).
By the definition we must show that there exist scalars $$a_1$$ and $$a_2$$ such that:
\begin{equation*} \begin{array}{ccc} \begin{split} (4,5) &= a_1(3, 1) + a_2 (1, 4)\\ & = \left(3a_1+ a_2 , a_1+4a_2\right) \end{split} &\Rightarrow & \begin{array}{c} 3a_1+ a_2 =4\\ a_1+ 4a_2 =5\\ \end{array}\\ \\ \end{array} \end{equation*}
This system has the solution $$a_1=1\text{,}$$ $$a_2=1\text{.}$$
Hence, if we replace $$a_1$$ and $$a_2$$ both by 1, then the two vectors (3, 1) and (1, 4) produce, or generate, the vector (4,5). Of course, if we replace $$a_1$$ and $$a_2$$ by different scalars, we can generate more vectors from $$\mathbb{R}^2\text{.}$$ If, for example, $$a _1 = 3$$ and $$a_2 = -2\text{,}$$ then
\begin{equation*} a_1(3, 1) + a_2 (1, 4) = 3 (3, 1) +(-2) (1,4) = (9, 3) +(-2,-8) = (7, -5) \end{equation*}
Will the vectors $$(3, 1)$$ and $$(1,4)$$ generate any vector we choose in $$\mathbb{R}^2\text{?}$$ To see if this is so, we let $$\left(b_1,b_2\right)$$ be an arbitrary vector in $$\mathbb{R}^2$$ and see if we can always find scalars $$a_1$$ and $$a_2$$ such that $$a_1(3, 1) + a_2 (1, 4)= \left(b_1,b_2\right)\text{.}$$ This is equivalent to solving the following system of equations:
\begin{equation*} \begin{array}{c} 3a_1+ a_2 =b_1\\ a_1+4a_2 =b_2\\ \end{array} \end{equation*}
which always has solutions for $$a_1$$ and $$a_2$$ , regardless of the values of the real numbers $$b_1$$ and $$b_2\text{.}$$ Why? We formalize this situation in a definition:
#### Definition12.3.9.Generation of a Vector Space.
Let $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ be a set of vectors in a vector space $$V$$ over $$\mathbb{R}\text{.}$$ This set is said to generate, or span, $$V$$ if, for any given vector $$\vec{y} \in V\text{,}$$ we can always find scalars $$a_1\text{,}$$ $$a_2, \ldots\text{,}$$ $$a_n$$ such that $$\vec{y} = a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n\text{.}$$ A set that generates a vector space is called a generating set.
We now give a geometric interpretation of the previous examples.
We know that the standard coordinate system, $$x$$ axis and $$y$$ axis, were introduced in basic algebra in order to describe all points in the $$xy$$-plane algebraically. It is also quite clear that to describe any point in the plane we need exactly two axes.
We can set up a new coordinate system in the following way. Draw the vector $$(3, 1)$$ and an axis from the origin through (3, 1) and label it the $$x'$$ axis. Also draw the vector $$(1,4)$$ and an axis from the origin through $$(1,4)$$ to be labeled the $$y'$$ axis. Draw the coordinate grid for the axis, that is, lines parallel, and let the unit lengths of this “new” plane be the lengths of the respective vectors, $$(3, 1)$$ and $$(1, 4)\text{,}$$ so that we obtain Figure 12.3.10.
From Example 12.3.8 and Figure 12.3.10, we see that any vector on the plane can be described using the standard $$xy$$-axes or our new $$x'y'$$-axes. Hence the position which had the name $$(3,1)$$ in reference to the standard axes has the name $$(1,0)$$ with respect to the $$x'y'$$ axes, or, in the phraseology of linear algebra, the coordinates of the point $$(1,4)$$ with respect to the $$x'y'$$ axes are $$(1, 0)\text{.}$$
From Example 12.3.8 we found that if we choose $$a_1=1$$ and $$a_2=1\text{,}$$ then the two vectors $$(3, 1)$$ and $$(1,4)$$ generate the vector $$(4,5)\text{.}$$ Another geometric interpretation of this problem is that the coordinates of the position $$(4,5)$$ with respect to the $$x'y'$$ axes of Figure 12.3.10 is $$(1, 1)\text{.}$$ In other words, a position in the plane has the name $$(4,5)$$ in reference to the $$xy$$-axes and the same position has the name $$(1, 1)$$ in reference to the $$x'y'$$ axes.
From the above, it is clear that we can use different axes to describe points or vectors in the plane. No matter what choice we use, we want to be able to describe each position in a unique manner. This is not the case in Figure 12.3.12. Any point in the plane could be described via the $$x'y'$$ axes, the $$x'z'$$ axes or the $$y'z'$$ axes. Therefore, in this case, a single point would have three different names, a very confusing situation.
We formalize the our observations in the previous examples in two definitions and a theorem.
#### Definition12.3.13.Linear Independence/Linear Dependence.
A set of vectors $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ from a real vector space $$V$$ is linearly independent if the only solution to the equation $$a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n= \vec{0}$$ is $$a_1 = a_2 = \ldots = a_n = 0\text{.}$$ Otherwise the set is called a linearly dependent set.
#### Definition12.3.14.Basis.
A set of vectors $$B=\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ is a basis for a vector space $$V$$ if:
1. $$B$$ generates $$V\text{,}$$ and
2. $$B$$ is linearly independent.
Assume that $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ is a basis for $$V$$ over $$\mathbb{R}\text{.}$$ We must prove two facts:
1. each vector $$y \in V$$ can be expressed as a linear combination of the $$\vec{x}_i\textrm{'s}\text{,}$$ and
2. each such expression is unique.
Part 1 is trivial since a basis, by its definition, must generate all of $$V\text{.}$$
The proof of part 2 is a bit more difficult. We follow the standard approach for any uniqueness facts. Let $$y$$ be any vector in $$V$$ and assume that there are two different ways of expressing $$y\text{,}$$ namely
\begin{equation*} y = a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n \end{equation*}
and
\begin{equation*} y = b_1 \vec{x}_1+b_2 \vec{x}_2+\ldots +b_n \vec{x}_n \end{equation*}
where at least one $$a_i$$ is different from the corresponding $$b_i\text{.}$$ Then equating these two linear combinations we get
\begin{equation*} a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n=b_1 \vec{x}_1+b_2 \vec{x}_2+\ldots +b_n \vec{x}_n \end{equation*}
so that
\begin{equation*} \left(a_1-b_1\right) \vec{x}_1+\left(a_2-b_2\right) \vec{x}_2+\ldots +\left(a_n-b_n\right) \vec{x}_n=\vec{0} \end{equation*}
Now a crucial observation: since the $$\vec{x}_i's$$ form a linearly independent set, the only solution to the previous equation is that each of the coefficients must equal zero, so $$a_i-b_i=0$$ for $$i = 1, 2, \ldots ,n\text{.}$$ Hence $$a_i=b_i\text{,}$$ for all $$i\text{.}$$ This contradicts our assumption that at least one $$a_i$$ is different from the corresponding $$b_i\text{,}$$ so each vector $$\vec{y} \in V$$ can be expressed in one and only one way.
This theorem, together with the previous examples, gives us a clear insight into the significance of linear independence, namely uniqueness in representing any vector.
Prove that $$\{(1, 1), (-1, 1)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}$$ and explain what this means geometrically.
First we show that the vectors $$(1, 1)$$ and $$(-1, 1)$$ generate all of $$\mathbb{R}^2\text{.}$$ We can do this by imitating Example 12.3.8 and leave it to the reader (see Exercise 12.3.3.10 of this section). Secondly, we must prove that the set is linearly independent.
Let $$a_1$$ and $$a_2$$ be scalars such that $$a_1 (1, 1) + a_2 (-1, 1) = (0, 0)\text{.}$$ We must prove that the only solution to the equation is that $$a_1$$ and $$a_2$$ must both equal zero. The above equation becomes $$\left(a_1- a_2 , a_1 + a_2 \right) = (0, 0)$$ which gives us the system
\begin{equation*} \begin{array}{c} a_1 - a_{2 }=0 \\ a_1 + a_2=0\\ \end{array} \end{equation*}
The augmented matrix of this system reduces in such way that the only solution is the trivial one of all zeros:
\begin{equation*} \left( \begin{array}{cc|c} 1 & -1 & 0 \\ 1 & 1 & 0 \\ \end{array} \right)\longrightarrow \left( \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)\textrm{ }\Rightarrow \textrm{ }a_1 = a_2 =0 \end{equation*}
Therefore, the set is linearly independent.
To explain the results geometrically, note through Exercise 12, part a, that the coordinates of each vector $$\vec{y} \in \mathbb{R}^2$$ can be determined uniquely using the vectors (1,1) and (-1, 1). The concept of dimension is quite obvious for those vector spaces that have an immediate geometric interpretation. For example, the dimension of $$\mathbb{R}^2$$ is two and that of $$\mathbb{R}^3$$ is three. How can we define the concept of dimension algebraically so that the resulting definition correlates with that of $$\mathbb{R}^2$$ and $$\mathbb{R}^3\text{?}$$ First we need a theorem, which we will state without proof.
#### Definition12.3.18.Dimension of a Vector Space.
Let $$V$$ be a vector space over $$\mathbb{R}$$ with basis $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}\text{.}$$ Then the dimension of $$V$$ is $$n\text{.}$$ We use the notation $$\dim V = n$$ to indicate that $$V$$ is $$n$$-dimensional.
### Exercises12.3.3Exercises
#### 1.
If $$a = 2\text{,}$$ $$b = -3\text{,}$$ $$A=\left( \begin{array}{ccc} 1 & 0 & -1 \\ 2 & 3 & 4 \\ \end{array} \right)\text{,}$$ $$B=\left( \begin{array}{ccc} 2 & -2 & 3 \\ 4 & 5 & 8 \\ \end{array} \right)\text{,}$$ and $$C=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 2 & -2 \\ \end{array} \right)$$ verify that all properties of the definition of a vector space are true for $$M_{2\times 3}(\mathbb{R})$$ with these values.
#### 2.
Let $$a = 3\text{,}$$ $$b = 4\text{,}$$ $$\vec{x}\pmb = (-1, 3)\text{,}$$ $$\vec{y} = (2, 3)\text{,}$$and $$\vec{z} = (1, 0)\text{.}$$ Verify that all properties of the definition of a vector space are true for $$\mathbb{R}^2$$ for these values.
#### 3.
1. Verify that $$M_{2\times 3}(\mathbb{R})$$ is a vector space over $$\mathbb{R}\text{.}$$ What is its dimension?
2. Is $$M_{m\times n}(\mathbb{R})$$ a vector space over $$\mathbb{R}\text{?}$$ If so, what is its dimension?
The dimension of $$M_{2\times 3}(\mathbb{R})$$ is 6 and yes, $$M_{m\times n}(\mathbb{R})$$ is also a vector space of dimension $$m \cdot n\text{.}$$ One basis for $$M_{m\times n}(\mathbb{R})$$ is $$\{A_{ij} \mid 1 \leq i \leq m, 1 \leq j \leq n\}$$ where $$A_{ij}$$ is the $$m\times n$$ matrix with entries all equal to zero except for in row $$i\text{,}$$ column $$j$$ where the entry is 1.
#### 4.
1. Verify that $$\mathbb{R}^2$$ is a vector space over $$\mathbb{R}\text{.}$$
2. Is $$\mathbb{R}^n$$ a vector space over $$\mathbb{R}$$ for every positive integer $$n\text{?}$$
#### 5.
Let $$P^3= \left\{a_0 + a_1x + a_2x^2 + a_3x^3 \mid a_0,a_1,a_2,a_3\in \mathbb{R}\right\}\text{;}$$ that is, $$P^3$$ is the set of all polynomials in $$x$$ having real coefficients with degree less than or equal to three. Verify that $$P^3$$ is a vector space over $$\mathbb{R}\text{.}$$ What is its dimension?
#### 6.
For each of the following, express the vector $$\pmb{y}$$ as a linear combination of the vectors $$x_1$$ and $$x_2\text{.}$$
1. $$\vec{y} = (5, 6)\text{,}$$ $$\vec{x}_1 =(1, 0)\text{,}$$ and $$\vec{x}_2 = (0, 1)$$
2. $$\vec{y} = (2, 1)\text{,}$$ $$\vec{x}_1 =(2, 1)\text{,}$$ and $$\vec{x}_2 = (1, 1)$$
3. $$\vec{y} = (3,4)\text{,}$$ $$\vec{x}_1 = (1, 1)\text{,}$$ and $$\vec{x}_2 = (-1, 1)$$
#### 7.
Express the vector $$\left( \begin{array}{cc} 1 & 2 \\ -3 & 3 \\ \end{array} \right)\in M_{2\times 2}(\mathbb{R})\text{,}$$ as a linear combination of $$\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)\text{,}$$ $$\left( \begin{array}{cc} -1 & 5 \\ 2 & 1 \\ \end{array} \right)\text{,}$$ $$\left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right)$$ and $$\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)$$
If the matrices are named $$B\text{,}$$ $$A_1\text{,}$$ $$A_2$$ , $$A_3\text{,}$$ and $$A_4$$ , then
\begin{equation*} B = \frac{8}{3}A_1 + \frac{5}{3}A_2+\frac{-5}{3}A_3+\frac{23}{3}A_4 \end{equation*}
#### 8.
Express the vector $$x^3-4x^2+3\in P^3$$ as a linear combination of the vectors 1, $$x\text{,}$$ $$x^2$$ , and $$x^3\text{.}$$
#### 9.
1. Show that the set $$\left\{\vec{x}_1,\vec{x}_2\right\}$$ generates $$\mathbb{R}^2$$ for each of the parts in Exercise 6 of this section.
2. Show that $$\left\{\vec{x}_1,\vec{x}_2,\vec{x}_3\right\}$$ generates $$\mathbb{R}^2$$ where $$\vec{x}_1= (1, 1)\text{,}$$ $$\vec{x}_2= (3,4)\text{,}$$ and $$\vec{x}_3 = (-1, 5)\text{.}$$
3. Create a set of four or more vectors that generates $$\mathbb{R}^2\text{.}$$
4. What is the smallest number of vectors needed to generate $$\mathbb{R}^2\text{?}$$ $$\mathbb{R}^n\text{?}$$
5. Show that the set
\begin{equation*} \{A_1, A_2 ,A_3, A_4\} =\{ \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)\} \end{equation*}
generates $$M_{2\times 2}(\mathbb{R})$$
6. Show that $$\left\{1,x,x^2 ,x^3\right\}$$ generates $$P^3\text{.}$$
1. If $$x_1 = (1, 0)\text{,}$$ $$x_2= (0, 1)\text{,}$$ and $$y = \left(b_1, b_2\right)\text{,}$$ then $$y = b_1x_1+b_2x_2\text{.}$$ If $$x_1 = (3, 2)\text{,}$$ $$x_2= (2,1)\text{,}$$ and $$y = \left(b_1, b_2\right)\text{,}$$ then $$y =\left(- b_1+2b_2\right)x_1+\left(2b_1-3b_2\right)x_2\text{.}$$
2. If $$y = \left(b_1, b_2\right)$$ is any vector in $$\mathbb{R}^2$$ , then $$y =\left(- 3b_1+4b_2\right)x_1+\left(-b_1+b_2\right)x_2 + (0)x_3$$
3. One solution is to add any vector(s) to $$x_1\text{,}$$ $$x_2\text{,}$$ and $$x_3$$ of part b.
4. 2, $$n$$
5. $$\displaystyle \left( \begin{array}{cc} x & y \\ z & w \\ \end{array} \right)= x A_1+y A_2+ z A_3+ w A_4$$
6. $$a_0+a_1x + a_2x^2+ a_3x^3=a_0(1)+a_1(x) + a_2\left(x^2\right)+ a_3\left(x^3\right)\text{.}$$
#### 10.
Complete Example 12.3.16 by showing that $$\{(1, 1), (-1, 1)\}$$ generates $$\mathbb{R}^2\text{.}$$
#### 11.
1. Prove that $$\{(4, 1), (1, 3)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}\text{.}$$
2. Prove that $$\{(1, 0), (3, 4)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}\text{.}$$
3. Prove that $$\{(1,0, -1), (2, 1, 1), (1, -3, -1)\}$$ is a basis for $$\mathbb{R}^3$$ over $$\mathbb{R}\text{.}$$
4. Prove that the sets in Exercise 9, parts e and f, form bases of the respective vector spaces.
1. The set is linearly independent: let $$a$$ and $$b$$ be scalars such that $$a(4, 1) + b(1, 3) = (0, 0)\text{,}$$ then $$4a + b = 0\textrm{ and } a + 3b= 0$$ which has $$a = b = 0$$ as its only solutions. The set generates all of $$\mathbb{R}^2\text{:}$$ let $$(a, b)$$ be an arbitrary vector in $$\mathbb{R}^2$$ . We want to show that we can always find scalars $$\beta _1$$ and $$\beta _2$$ such that $$\beta _1(4, 1) +\beta _2 (1,3) = (a, b)\text{.}$$ This is equivalent to finding scalars such that $$4\beta _1 +\beta _2 = a$$ and $$\beta _1 + 3\beta _2 = b\text{.}$$ This system has a unique solution $$\beta _1=\frac{3a - b}{11}\text{,}$$ and $$\beta _2= \frac{4b-a}{11}\text{.}$$ Therefore, the set generates $$\mathbb{R}^2\text{.}$$
#### 12.
1. Determine the coordinates of the points or vectors $$(3, 4)\text{,}$$ $$(-1, 1)\text{,}$$ and $$(1, 1)$$ with respect to the basis $$\{(1, 1),(-1, 1)\}$$ of $$\mathbb{R}^2\text{.}$$ Interpret your results geometrically.
2. Determine the coordinates of the points or vector $$(3, 5, 6)$$ with respect to the basis $$\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\text{.}$$ Explain why this basis is called the standard basis for $$\mathbb{R}^3\text{.}$$
#### 13.
1. Let $$\vec{y}_1= (1,3, 5, 9)\text{,}$$ $$\vec{y}_2= (5,7, 6, 3)\text{,}$$ and $$c = 2\text{.}$$ Find $$\vec{y}_1+\vec{y}_2$$ and $$c \vec{y}_1\text{.}$$
2. Let $$f_1(x) = 1 + 3x + 5x^2 + 9x^3$$ , $$f_2(x)=5 + 7x+6x^2+3x^3$$ and $$c = 2\text{.}$$ Find $$f_1(x)+f_2(x)$$ and $$c f_1(x)\text{.}$$
3. Let $$A =\left( \begin{array}{cc} 1 & 3 \\ 5 & 9 \\ \end{array} \right)\text{,}$$ $$B=\left( \begin{array}{cc} 5 & 7 \\ 6 & 3 \\ \end{array} \right)\text{,}$$ and $$c=2\text{.}$$ Find $$A + B$$ and $$c A\text{.}$$
4. Are the vector spaces $$\mathbb{R}^4$$ , $$P^3$$ and $$M_{2\times 2}(\mathbb{R})$$ isomorphic to each other? Discuss with reference to previous parts of this exercise.
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# How would you solve these calculations? 2 + 5 = 2 + 8 = 7 + 7 = 6 + 7 = 15 + 11 = 24 + 9 = 32 + 21 = 45 + 36 = 120 + 11 = Which method did you use? Does.
## Presentation on theme: "How would you solve these calculations? 2 + 5 = 2 + 8 = 7 + 7 = 6 + 7 = 15 + 11 = 24 + 9 = 32 + 21 = 45 + 36 = 120 + 11 = Which method did you use? Does."— Presentation transcript:
How would you solve these calculations? 2 + 5 = 2 + 8 = 7 + 7 = 6 + 7 = 15 + 11 = 24 + 9 = 32 + 21 = 45 + 36 = 120 + 11 = Which method did you use? Does that method work for all of these calculations? What skills were you using?
6 + 5 = Mr Marshall has 6 sweets in one pocket and 5 in the other. How many sweets does he have in total? Counting on using a 0 to 20 number line. 1) Start on 6 2) 5 jumps of 1 3) Answer is 11
Addition across key stage 1 This could also be calculated using partitioning.
National expectations Level 2C
Level 2B
Level 2A
Level 3
What can I do to help?
Rapid recall of number pairs to 10 and then to 20. Doubles and halves
To become familiar with the value and denominations of money Pennies in a pound and different amounts that make £1 Help him/her to add and subtract amount of money perhaps within a pocket money context (workout whether they can afford a particular toy or treat) Shop using money and calculate change.
Learning multiplication facts is a vital part of any child’s mathematical development. Once rapid recall of multiplication facts becomes possible, a whole host of mathematical activities will seem easier. Children need to be able to recall multiplication facts in any order and also to derive associated division facts. The expectations for each year group are set out below: Year 1 Count on or back in ones, twos, fives and tens and use this knowledge to derive the multiples of 2, 5 and 10. Year 2 Derive and recall multiplication facts for the 2, 5 and 10 times-tables and the related division facts.
Year 3 Derive and recall multiplication facts for the 2, 3, 4, 5, 6 and 10 times-tables and the corresponding division facts. Year 4 Derive and recall multiplication facts up to 10 × 10, the corresponding division facts. Year 5 Recall quickly multiplication facts up to 10 x 10 and derive quickly corresponding division facts. Year 6 Use knowledge of place value and multiplication facts to 10 × 10 to derive related multiplication and division facts involving decimals (e.g. If I know 8x7=56, I can use that to workout 0.8 × 7=5.6) Use knowledge of multiplication facts to derive quickly squares of numbers to 12×12. www.teachingtables.co.uk
The aim is that for each times table: The children should be able to say the table in order. E.g. 1 times 3 is 3, 2 times 3 is 6. They should be able to answer questions in any order. E.g. “What is 4 x 5?” “What is 2 x 7?” They should be able to answer – “How many 2’s in 18?” “How many 5’s in 20?” They should also be able to link their tables with division – e.g. 5 x 3 is 15, so 15 ÷ 3 = 5
3) If your child is always on the move try saying them as they go up the stairs of when out walking. They can chant them as they skip or bounce a ball. 4)Make up silly rhymes to help with facts they are struggling to remember e.g. Eight times eight is sixty-four, close your mouth and shut the door! A tree on skates fell on the floor; three times eight is twenty-four. There are lots of ways you can help your child to learn their times tables. Different activities suit different learning styles. Remember it should be fun! 1)Buy a times table CD or tape. Listening to songs and singing can help children learn their tables in a fun way. 2)If your child likes to write or draw they can write out their times tables or copy them from a chart. See how quickly they can do it and can they improve on their time?
Parents evening ‘Maths targets. A booklet for parents’ publication from the Numeracy strategy Ideas for games, key learning skills for that year group.
Useful websites www.topmarks.co.uk www.educationcity.com www.coolmath4kids.com
http://www.bbc.co.uk/schools/parents/
http://www.bbc.co.uk/schools/parents/search/
http://www.bbc.co.uk/schools/numbertime/
http://www.bbc.co.uk/schools/bitesize//
http://www.gridclub.com/ Free trial and membership fee
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# Do diagonals of square bisect each other at 90 degree?
## Do diagonals of square bisect each other at 90 degree?
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Are diagonals of square equal to side?
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## Are diagonals of rhombus angle bisectors?
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What are the properties of the diagonals of a rhombus?
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 6.3, 23 Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.Cone of largest volume inscribed in the sphere of radius R Let OC = x Radius of cone = BC Height of cone = h = OC + OA Finding OC Δ BOC is a right angled triangle Using Pythagoras theorem in ∆BOC 〖𝑂𝐵〗^2=〖𝐵𝐶〗^2+〖𝑂𝐶〗^2 R2 =〖𝐵𝐶〗^2+𝑥^2 BC2 = 𝑅^2 – x2 BC = √(𝑅^2−𝑥^2 ) Thus, Radius of cone = BC = √(𝑅^2−𝑥^2 ) Height of cone = OC + OA = R + x We need to show Maximum volume of cone = 8/27 × Volume of sphere = 8/27 × 4/3 𝜋𝑅^3 = 32/81 𝜋𝑅^3 Let V be the volume of a cone We know that Volume of a cone = 1/3 𝜋(𝑟𝑎𝑑𝑖𝑢𝑠 )^2 (ℎ𝑒𝑖𝑔ℎ𝑡 ) V = 1/3 𝜋(√(𝑅^2−𝑥^2 ))^2 (𝑅+𝑥) V = 1/3 𝜋(𝑅^2−𝑥^2 )(𝑅+𝑥) V = 1/3 𝜋(𝑅^2 (𝑅+𝑥)−𝑥^2 (𝑅+𝑥)) V = 1/3 𝜋(𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 ) Diff w.r.t 𝒙 𝑑𝑉/𝑑𝑥=𝑑/𝑑𝑥 [1/3 𝜋(𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 )] 𝑑𝑉/𝑑𝑥=𝜋/3 [𝑑/𝑑𝑥 (𝑅^(3 )+𝑅^2 𝑥−𝑥^2 𝑅−𝑥^3 )] 𝑑𝑉/𝑑𝑥=𝜋/3 (0+𝑅^2.1−𝑅 ×2𝑥−3𝑥^2 ) 𝑑𝑉/𝑑𝑥=𝜋/3 (𝑅^2−2𝑅𝑥−3𝑥^2 ) Putting 𝒅𝑽/𝒅𝒙=𝟎 1/3 𝜋(𝑅^2−2𝑅𝑥−3𝑥^2 )=0 𝑅^2−2𝑅𝑥−3𝑥^2=0 𝑅^2−3𝑅𝑥+𝑅𝑥−3𝑥^2=0 R(𝑅−3𝑥)+𝑥(𝑅−3𝑥)=0 (𝑅+𝑥)(𝑅−3𝑥)=0 So, x = –R & 𝑥=𝑅/3 Since x cannot be negative 𝑥=𝑅/3 Finding (𝒅^𝟐 𝒗)/(𝒅𝒙^𝟐 ) 𝑑𝑣/𝑑𝑥=1/3 𝜋(𝑅^2−2𝑅𝑥−3𝑥^2 ) (𝑑^2 𝑣)/(𝑑𝑥^2 )=𝜋/3 𝑑/𝑑𝑥 (𝑅^2−2𝑅𝑥−3𝑥^2 ) (𝑑^2 𝑣)/(𝑑𝑥^2 )=𝜋/3 (0−2𝑅−6𝑥) (𝑑^2 𝑣)/(𝑑𝑥^2 )=(−𝜋)/3 (2𝑅+6𝑥) Putting 𝒙=𝑹/𝟑 〖(𝑑^2 𝑣)/(𝑑𝑥^2 )│〗_(𝑥 = 𝑅/3) =(−𝜋)/3 (2𝑅+6(𝑅/3)) =(−𝜋)/3 (2𝑅+2𝑅) =(−𝜋)/3 (4𝑅) =(−4𝜋𝑅)/3 < 0 Thus (𝑑^2 𝑣)/(𝑑𝑥^2 )<0 when 𝑥=𝑅/3 ∴ Volume is Maximum when 𝑥=𝑅/3 Finding maximum volume From (1) Volume of cone = 1/3 𝜋(𝑅^2−𝑥^2 )(𝑅+𝑥) Putting 𝑥 = 𝑅/3 = 1/3 𝜋(𝑅^2−(𝑅/3)^2 )(𝑅+𝑅/3) = 1/3 𝜋(𝑅^2−𝑅^2/9)((3𝑅 + 𝑅)/3) = 1/3 𝜋((9𝑅^2− 𝑅^2)/9)(4𝑅/3) = 𝟑𝟐/𝟖𝟏 𝝅𝑹^𝟑 Hence proved
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# Finding the Area and Perimeter of a Rectangle
Area and Perimeter of a Rectangle
## Area and Perimeter of a Rectangle
The area of a rectangle is found by multiplying the length times the width. The perimeter of the rectangle is found by multiplying 2 times the length, and then adding to that 2 times the width. The reason for that is that the opposite sides of a rectangle are congruent.
If you are trying to find the distance around a rectangle, then you have two sides that have the same length and then two sides that have the same width. If you just multiply 2 times the length and 2 times the width, and add those together, then you’ll find the perimeter of the rectangle. Let’s look at our example.
What is the area and perimeter of a rectangle with a 5ft length and a 3 ft width. We’ve been given a word problem, but we already have this nice picture drawn here of a rectangle. What we can do is we have a 5ft length so we get rid of this l and put that our length is 5ft. Then our width is 3ft, we can get rid of this w and put that our width is 3ft.
Now we’re asked to find area, we should start with our area formula. Area = lw. Area = 5ft × 3ft. The area is 5 times 3 is 15, ft times ft is ft2. Our area is 15ft2. The perimeter. Again, we need to start with our formula. Perimeter is 2 times the length plus 2 times the width. Our perimeter is 2 times, then our length was 5ft. We substitute 5ft for our length plus 2 times, and our width was 3ft. Substitute 3ft for width. P= 2(5ft) + 2(3ft).
Then we need to follow PEMDAS, which means we need to multiply before we add. The perimeter is 2 × 5ft = 10ft, plus 2 × 3ft = 6ft. Finally, we add these together and find the sum. 10ft + 6ft = 16ft. it would be 16ft all the way around our rectangle.
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by Mometrix Test Preparation | Last Updated: June 15, 2020
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# Use Models to Multiply Two Fractions
Sep 20, 2022
## Key Concepts
• Multiply fractions.
• Use models to multiply two fractions.
• Interpret the product a/b x q as a part of a partition of q into b equal parts; equivalently, as the result of a sequence of operations (a q) ÷ b.
• Standard multiplication of two fractions.
• Understand some of the concept distributive property with multiplying fractions.
## Multiply two fractions using models
### Multiplying fractions
Multiplying fractions can be a little tricky to understand.
When adding fractions, you are finding the sum.
When you subtract fractions, you are finding the difference.
When multiplying a fraction by a whole number, you are finding the sum of a
repeaters fraction or a repeated group.
When you multiply two fractions, it means that you are looking for a part of a part. Here is a multiplication problem with two fractions.
Example 1:
There was a pan of lasagna left. Tom ate 1/3 of this amount. What fraction of the whole pan of lasagna did he eat?
Solution: Find 1/3 of 1/4 to solve this problem
One way
Divide one whole part into fourths.
Divide 1/4 into 3 equal parts.
Divide the other 1/4 s into 3 equal parts.
12 parts make one whole, so one part is 1/12
𝟏/𝟑 x 𝟏/𝟒 = 1/12
∴ 1/3 of 1/4 = 𝟏/𝟏𝟐
Another way:
Shade one of 3 columns yellow to represent 1/3 .
Shade 1 of the 4 rows red to represent 1/4.
The orange overlap shows the product.
1 out of 12 parts are shaded orange.
1/3 x 1/4 =1 X 1 / 3 X 4 = 1/12
Tom ate 𝟏/𝟏𝟐
of the pan of lasagna.
Important Note:
You can’t always draw pictures to figure out a problem, so you can multiply fractions using a few simple steps.
To multiply two fractions, multiply the numerator by the numerator and the denominator by the denominator.
a/b × c/d = a × c / b × d
### Multiplying two fractions by using number line
Example 2:
Find
𝟐/𝟑 x 𝟑/𝟒 using a number line.
Solution:
𝟏/𝟑 means 1 of 3 equal parts,
so 𝟏/𝟑 of 𝟑/𝟒 is𝟏/𝟒
𝟐/𝟑 means 2 of 3 equal parts,
so 2/3 of 3/𝟒 is 2 times 1/4
𝟐/𝟑 x 𝟑/𝟒 = 𝟐/𝟒 or 𝟏/𝟐
#### Practice
1. Find 5/𝟔 x 𝟏/𝟐 . Shade the model to help solve.
Solution:
Shade 5 of the 6 columns red to represent
𝟓/𝟔.
Shade 1 of the 2 rows to represent 1/2.
Violet color represents the product. 𝟓/𝟔 x 𝟏/𝟐 = 𝟓/𝟏𝟐
2. Find 3/4 of 4/9.
Solution:
3/4 x 4/9 = 3 X 4/4 X 9
= 12/36
= 1/3
3/4 𝟒/𝟗 = 𝟏/𝟑
3. Find 1/2 of 3/4.
Solution:
= 1/2 x 3/4 = 1 X 3 / 2 X 4
= 3/8
= 1/2 x 3/4 = 3/8
4. A scientist had 3/4 of a bottle of a solution. She used 1/6 of the solution in an experiment. How much of the bottle did she use?
Solution:
Given that,
Total solution that the scientist had = 3/4
Solution used by scientist = 1/6
Then,
1/6 x 3/4 = 1 X 3 / 6 X 4
= 3 / 24
=1/8
∴ She used 1/8 of the solution for the experiment.
### Standard multiplication of two fractions
Example 1:
On dan’s Reader, 2/3 of the books are fiction. Of the fiction 4/5 are mysterious. What fraction of the books on Dan’s eReader are mysterious? Solve this problem any way you choose.
Solution:
Given that,
No. of fiction books = 2/3
No. of books that are mysterious of fiction = 4/5
Then,
2/3 x 4/5 = 2 X 4 / 3 X
= 8/15
= 2 / 3 x 4/5 = 8/15
∴ 8/15 books on Dans eReader are mysterious.
Example 2:
Amelia takes pictures with her smartphone. Of the pictures,5/6 are of animals. 3/4 of her animal photos are of dogs. What fraction of her pictures are of dogs?
Solution:
Step 1
Estimate 3/4 x 5/6. Since both fractions are less than 1,Since both fractions are less than 1, the product will be less than 1the product will be less than 1.
Step 2
Multiply the numerators. Then multiply the denominators.
3/4×5/6 = 3 × 54 × 63 × 54 × 6
=15/24 (∵15/24 < 1 )
= 5/8
The answer is reasonable.
So, 5/8 of all Animal’s pictures have dogs in them.
Example 3:
Is the product of 3/6 x 5/4 equal to the product of 3/4×5/6 ? Explain how you know?
Solution:
Case 1
3/6×5/4 = 3 × 5/6 × 4
= 15/24
∴ 3/6 × 5/4 = 5/8
Case 2
3/4× 5/6 = 3 × 5/4 × 6
= 15/24
∴ 3/4× 5/6 = 5/8
∴ 3/6× 5/4 = 3/4 × 5/6
#### Practice
1. Find 9/10 × 1/2
Solution:
9/10 × 1/2 = 9 × 1/10 × 29
= 9/20
∴ 9/10 ×1/2
=9/20
2. Find 5/6× 1/3
Solution:
5/6×1/3 = 5 × 1/6 × 3
= 5/18
∴ 5/6 × 1/3 = 5/18
3. Find 4/7 of 7/9
Solution:
4/7× 7/9 = 4 × 7/7 × 9
= 28/63
∴ 4/7× 7/9 = 4/9
4. Find ( 1/6 + 1/6 ) × 3/4
Solution:
(1/6+1/6 ) × 3/4 = (1+1 / 6) × 3/4
= (2/6) × 3/4
= 2/6 x 3/4
= 2 x 3/6 x 4
∴ (1/6+1/6 ) x 3/4 = 6/24
5. Find (9/10 – 3/10 ) × 1/4
Solution:
(9/10-3/10 ) × 1/4 = (9 −3/10) × 1/4
= (6/10) × 1/4
= 6/10 × 1/4
= 6 × 1/ 10 × 4
= 6/40
∴ (9/10-3/10 ) x 1/4 = 3/20
6. Edurado runs 6 laps around the track at Lincoln Park school. Then he runs 3/12 miles to get home. How far will he run in all. Show your work.
Solution:
Given that,
Distance covered in one lap = 1/4miles
No of laps around the track = 6
Distance covered to get to home = 3 1/2 miles =7/2
Total distance covered by Edurado = (6 × 1/4) + 7/2
= 6 × 1 + 7 × 2 / 4
= 6 + 14 / 4
= 20/4
= 5 miles
### What have we learned
• Multiply fractions.
• Use models to multiply two fractions.
• Interpret the product ab × q as a part of a partition of q into b equal parts; equivalently as the result of a sequence of operations (a × q) ÷ b.
• Understand standard multiplication of two fractions.
• Understand some of the concepts of distributive property with multiplying fractions.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)
## Section2.3Sage
Many properties of the algebraic objects we will study can be determined from properties of associated integers. And Sage has many powerful functions for analyzing integers.
### Subsection2.3.1Division Algorithm
The code a % b will return the remainder upon division of $$a$$ by $$b\text{.}$$ In other words, the result is the unique integer $$r$$ such that (1) $$0\leq r\lt b\text{,}$$ and (2) $$a=bq+r$$ for some integer $$q$$ (the quotient), as guaranteed by the Division Algorithm (Theorem 2.2.1). Then $$(a-r)/b$$ will equal $$q\text{.}$$ For example,
r = 14 % 3
r
q = (14 - r)/3
q
It is also possible to get both the quotient and remainder at the same time with the .quo_rem() method (quotient and remainder).
a = 14
b = 3
a.quo_rem(b)
A remainder of zero indicates divisibility. So (a % b) == 0 will return True if $$b$$ divides $$a\text{,}$$ and will otherwise return False.
(20 % 5) == 0
(17 % 4) == 0
The .divides() method is another option.
c = 5
c.divides(20)
d = 4
d.divides(17)
### Subsection2.3.2Greatest Common Divisor
The greatest common divisor of $$a$$ and $$b$$ is obtained with the command gcd(a, b), where in our first uses, $$a$$ and $$b$$ are integers. Later, $$a$$ and $$b$$ can be other objects with a notion of divisibility and “greatness,” such as polynomials. For example,
gcd(2776, 2452)
We can use the gcd command to determine if a pair of integers are relatively prime.
a = 31049
b = 2105
gcd(a, b) == 1
a = 3563
b = 2947
gcd(a, b) == 1
The command xgcd(a,b) (“eXtended GCD”) returns a triple where the first element is the greatest common divisor of $$a$$ and $$b$$ (as with the gcd(a,b) command above), but the next two elements are values of $$r$$ and $$s$$ such that $$ra+sb=\gcd(a,b)\text{.}$$
xgcd(633,331)
Portions of the triple can be extracted using [ ] (“indexing”) to access the entries of the triple, starting with the first as number 0. For example, the following should always return the result True, even if you change the values of a and b. Try changing the values of a and b below, to see that the result is always True.
a = 633
b = 331
extended = xgcd(a, b)
g = extended[0]
r = extended[1]
s = extended[2]
g == r*a + s*b
Studying this block of code will go a long way towards helping you get the most out of Sage's output. Note that = is how a value is assigned to a variable, while as in the last line, == is how we compare two items for equality.
### Subsection2.3.3Primes and Factoring
The method .is_prime() will determine if an integer is prime or not.
a = 117371
a.is_prime()
b = 14547073
b.is_prime()
b == 1597 * 9109
The command random_prime(a, proof=True) will generate a random prime number between $$2$$ and $$a\text{.}$$ Experiment by executing the following two compute cells several times. (Replacing proof=True by proof=False will speed up the search, but there will be a very, very, very small probability the result will not be prime.)
a = random_prime(10^21, proof=True)
a
a.is_prime()
The command prime_range(a, b) returns an ordered list of all the primes from $$a$$ to $$b-1\text{,}$$ inclusive. For example,
prime_range(500, 550)
The commands next_prime(a) and previous_prime(a) are other ways to get a single prime number of a desired size. Give them a try below if you have an empty compute cell there (as you will if you are reading in the Sage Notebook, or are reading the online version). (The hash symbol, #, is used to indicate a “comment” line, which will not be evaluated by Sage. So erase this line, or start on the one below it.)
# Practice area (not linked for Sage Cell use)
In addition to checking if integers are prime or not, or generating prime numbers, Sage can also decompose any integer into its prime factors, as described by the Fundamental Theorem of Arithmetic (Theorem 2.2.7).
a = 2600
a.factor()
So $$2600 = 2^3\times 5^2\times 13$$ and this is the unique way to write $$2600$$ as a product of prime numbers (other than rearranging the order of the primes themselves in the product).
While Sage will print a factorization nicely, it is carried internally as a list of pairs of integers, with each pair being a base (a prime number) and an exponent (a positive integer). Study the following carefully, as it is another good exercise in working with Sage output in the form of lists.
a = 2600
factored = a.factor()
first_term = factored[0]
first_term
second_term = factored[1]
second_term
third_term = factored[2]
third_term
first_prime = first_term[0]
first_prime
first_exponent = first_term[1]
first_exponent
The next compute cell reveals the internal version of the factorization by asking for the actual list. And we show how you could determine exactly how many terms the factorization has by using the length command, len().
list(factored)
len(factored)
Can you extract the next two primes, and their exponents, from a?
# Practice area (not linked for Sage Cell use)
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# Coordinate Geometry
A coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates.
### 3. Conic Sections
Parabola
Ellipse
Hyperbola
Parametric Form of Conics
Asymptotes
Rectangular Hyperbola
### Cartesian Plane
The horizontal line is called the x-axis, and the vertical line is called the y-axis. The coordinate axes divide the plane into four parts called quadrants. The point of intersection of the axes is called the origin.
The distance of a point from the y-axis is called its x-coordinate, or abscissa, and the distance of the point from the x-axis is called its y-coordinate, or ordinate.
Coordinates of Point
If the abscissa of a point is x and the ordinate is y, then (x, y) are called the coordinates of the point.
The coordinates of a point on the x-axis are of the form (x, 0) and that of the point on the y-axis are (0, y). The coordinates of the origin are (0, 0).
The coordinates of a point are of the form (+,+) in the first quadrant, (-,+) in the second quadrant, (-,-) in the third quadrant and (+,-) in the fourth quadrant.
Distance Formula
The distance between P(x1, y1) and Q(x2, y2) is
Mid-Point Formula
The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is
Area of Triangle
If A(x1, y1), B(x2 , y2) and C(x3, y3) be three vertices of a triangle ABC, then its area is given by
Colinear Points
Three points A, B and C are said to be collinear (lying on the same straight line) if:
• AB + BC = AC or AC + CB = AB or AB + AC = BC
• The area of the triangle formed by A, B and C is zero
Section Formula
If A and B are two points in a plane, then the coordinates of the point P which divides the line joining AB internally in the ratio m : n are
### Straight Lines
A linear equation of the first degree in two variables x and y represents a straight line. The equation ax + by + c = 0 is the general form of equation of a line. Examples:
• 2x + 3y + 6 = 0
• 3y + 5 = 0 [a = 0]. The line is parallel to x-axis
• 2x + 7 = 0 [b = 0]. The lines is parallel to y-axis
• 2x + 3y = 0 [c = 0]. The line passes through the origin
The equation of x-axis is y = 0 and equation of y-axis is x = 0.
Slope of Line (m)
The slope is the tangent of the angle which the line makes with the x-axis, in the positive direction. You measure the angle from the x-axis towards the line, in the anti-clockwise direction.
The slope m is given as m = tan θ°. If θ is acute, the slope is positive and if θ is obtuse the slope is negative.
The slope of a line parallel to x-axis is m = tan 0° = 0. The slope of a line perpendicular to x-axis is m = tan 90° = ∞.
Parallel Lines
Two lines are parallel to each other, if their slopes are equal. Thus, if the slope of line is m, the slope of a line parallel to it is also m. The equation of a line parallel to ax + by + c = 0 is ax + by = k.
Perpendicular Lines
Two lines are perpendicular to each other, if and only if the product of their slopes is -1. Thus, if the slope of a line is m, the slope of a line perpendicular to it is -1/m. The equation of a line perpendicular to ax + by + c = 0 is bx – ay = k.
Angle Between Two Lines
If the angle between two lines of slopes m1 and m2 be θ, then
tan θ = (m1 - m2)/(1 + m1m2)
### Standard Forms of the Equation of a Line
Slope Intercept Form
The equation of a straight line having slope m and making an intercept c on y-axis is y = mx + c.
Two Point Form
The equation of a straight line passing through the points (x1, y1) and (x2, y2) is
Point Slope Form
The equation of a straight line passing through the point (x1, y1) and having slope m is
(y – y1) = m(x – x1).
Double Intercept Form
The equation of a line making intercepts a and b on the x and y axes respectively is
x/a + y/b = 1.
The area of a triangle formed by the coordinate axes and the lines having intercepts a and b is ab/2.
### Circles
The equation of circle whose centre is (h, k) and radius r is given by
(x - h)2 + (y - k)2 = r2
The equation of circle whose centre is origin and radius r is
x2 + y2 = r2
General Equation
The equation x2 + y2 + 2gx + 2fy + c = 0 represents a circle of radius of √(g2 + f2 - c)
• If g2 + f2 - c > 0, then the radius of the circle is real and hence the circle is also real.
• If g2 + f2 - c = 0, then the radius of the circle is zero. Such a circle is known a point circle.
• If g2 + f2 - c < 0, then the radius of circle is imaginary and such a circle is called an imaginary circle.
### Conic Sections
A conic section is the locus of a point which moves in such a way that its distance from a fixed point always bears a constant ratio to its distance from a fixed line. If S is a fixed point in the plane and ZZ´ is a fixed line in the same plane, then the locus of a point P which moves in the same plane in such a way that
SP/PM = Constant
Focus, Directrix, Ecentricity
The fixed point is called the focus of the conic section and the fixed line is known as its directrix. The constant ratio e is called the eccentricity of the conic section.
• If e = 1, the conic is called a parabola
• If e < 1, the conic is called an ellipse
• If e > 1, the conic is called a hyperbola
General Equation of a Conic
Let S(α, β) be the focus, ax + by + c = 0 be the directrix and e be the eccentricity of a conic. Let P(h, k) be any point on the conic such that PM is the perpendicular from P on the directrix. Then, general equation is
The general equation of the second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will represent
• circle, if a = b and h = 0
• parabola, if h2 = ab
• ellipse, if h2 < ab
• hyperbola, if h2 > ab
• rectangular hyperbola, if h2 > ab and a + b = 0
Useful Terms
Axis: The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section.
Vertex: The point of intersection of the conic section and the axis is called the vertex of the conic section.
Centre: The point which bisects every chord of the conic passing through it, is called the centre of the conic section.
Latus-rectum: The latus rectum of a conic is the chord passing through the focus and perpendicular to the axis.
Focal Chord: Any chord passing through the focus is called a focal chord of the conic section.
Double Ordinate: A chord perpendicular to the axis of a conic is known as its double ordinate.
### Parabola
The standard equation of parabola is
y2 = 4ax
(a, 0) is the focus of the parabola.
Parabola Types
### Ellipse
An ellipse is the focus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is always less than unity.
The equation of ellipse whose focus is (ae, 0) is
b2 = a2 (1 - e2)
Vertices: The point A and A', where the curve meets the line joining the foci S and S', are called the vertices of the ellipse. The coordinates of A and A' are (a, 0) and (–a, 0) respectively.
Major and Minor Axes
The distance AA' = 2a and BB' = 2b are called the major and minor axes of the ellipse respectively. Since, e < 1 and b2 = a2(1-e2). Therefore, a > b.
Foci
The points S(ae, 0) and S'(–ae, 0) are the foci of the ellipse.
Directrices
ZK and Z'K' are two directrices of the ellipse and their equations are x = a/e and x = - a/e respectively.
### Hyperbola
A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from fixed line (called directrix) is always constant which is always greater than unity.
Standard equation of hyperbola is:
b2 = a2 (e2 - 1)
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# Grid Method Multiplication Explained For Parents
Although grid method is fairly simple once you get to grips with it, it can be a bit of a challenge on first sight, so we've written a handy guide to get you through.
This step-by-step breakdown shows how to use the grid method to solve a variety of multiplication problems your children are likely to encounter in school. This might be simple money questions in Year Three all the way up to multiplying 4-digit numbers in Year Six.
## What Is The Grid Method?
The grid method of multiplication, also known as the box method, is a way of doing long multiplication by breaking numbers down into place values and writing them out in a grid. A school usually starts introducing the multiplication grid method in maths at the start of Key Stage 2, when children go into Year Three, although some introduce it as early as Year Two.
Using the grid method to do long multiplication gets children to break the numbers down into hundreds, tens and ones before multiplying them. This helps the child to understand what each digit in a number represents and what's actually happening to the numbers when they are multiplied. This helps children who are struggling by letting them visualise the process more easily.
In this article, we will walk you through solving various types of multiplication problems using grid method.
## Grid Method: Multiplying A Two-Digit Number By A Two-Digit Number
The problem: 23x15 = ?
The first number, 23, is made up of the number 20 and the number 3. That means we need to write 20 and 3 in the boxes to the right of the X.
Next, add the other number down the side:
Now, we do the actual multiplication. It doesn't really matter what order you multiply the boxes in, but we suggest starting on the right because it makes it easier for children to adjust to column method later.
Multiply the ones column by the tens row:
Now do the tens times the tens:
Now the ones times the ones:
And finally, the tens row times the ones column:
Now, we just need to add all the numbers together. Take all four of the answers you have just found, and write them out as a column addition (or whichever addition method the child you are helping is most comfortable with):
## Grid Method: Multiplying A Three-Digit Number By A One-Digit Number
In some ways, this is even easier than the grid method example above with a 2-digit number, since using a 1-digit number means there's only one row to deal with. We just need an extra column in for the hundreds. Then, follow the same method as above, multiplying each number in the top row by the one in the left-hand column:
Once all the numbers have been multiplied, write out a column addition to find the sum of all three.
Sorted!
## Grid Method: Multiplying A Three- (Or More)-Digit Number By A Two-Digit Number
In Year 6, children will have to use the maths grid to multiply a three- or four-digit number by a two-digit one.
Put the numbers into the grid as before:
Next, multiply the top row:
Once you've multiplied all the numbers in the top row, it's time for the second row:
Some less confident children might find adding so many numbers at once intimidating, so it's fine to do this step in two parts.
## Grid Method: Multiplying A Decimal
Many children get intimidated at the thought of working with decimals. The good part about grid method is that it really isn't much different to using it without the decimal.
In the following example, we're working out 12.5 x 2.2.
Put the numbers into the grid as usual. This time we have a column called "tenths" for the decimal place.
Multiply the top row:
Multiply the bottom row:
Find the total of the answers:
## Grid Method: Multiplying Money
As long as you're clear about whether you're working in pounds or pence, multiplying money using the grid method is very similar to any other multiplication in grid method.
Here is an example of a question your child may come across in Year 4:
Anna buys two packs of muffins from the bakery. Each pack costs £1.25. How much did she pay altogether?
Once your child has worked out that they need to multiply £1.25 by two, put the numbers into the grid as usual.
Most children won't have covered decimal multiplication in Lower KS2, so convert the numbers to pence before you start. Then, work through the grid as if you were multiplying a three-digit number by a one-digit number.
Finally, find the total as usual, then convert back to pence at the end:
## Grid Method: Troubleshooting
If your child is struggling with the grid method despite your best efforts, here are our top three suggestions to help.
1) Times tables knowledge. Make sure your child's times tables knowledge is solid. Often, children who struggle with long multiplication understand the method - they just don't have instant recall of their times' tables facts, so while they know they need to multiply 3 by 12, they don't know 3 x 12 = 36. Practice is key - get them repeating their tables in the car, while they help with washing up, or any other time they can.
2) Addition struggles. If your child handles the grid well but gets the wrong answer at the end, they may need a quick refresher on column addition, or to break the addition down into smaller steps.
3) Confidence. Especially for children who have struggled with maths in the past, it's easy to lose confidence. The grid method can look intimidating at first, and some children get so nervous they simply stop being able to work through the steps logically. Reassurance will do wonders here, as will working through a couple of examples together slowly.
If you've tried all these and your child is still finding grid method difficult, check-in with their teacher, especially if you've noticed them struggling in school more generally.
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## Volume of Three-Dimensional Shapes
This chapter gives an overview of volume formulas for more complicated three-dimensional objects: triangular prisms and pyramids, cylinders, cones, and spheres.
The volume of a solid is the measure of how much space an object takes up. It is measured by the number of unit cubes it takes to fill up the solid.
Counting the unit cubes in the solid, we have 30 unit cubes, so the volume is: 2 units \begin{align*}\cdot\end{align*}3 units \begin{align*}\cdot\end{align*}5 units = 30 cubic units.
#### Volume of a Prism
A prism is a solid with two congruent polygon bases that are parallel and connected by rectangles. Prisms are named by their base shape.
To find the volume of a prism, find the area of its base and multiply by its height.
\begin{align*}V_{\text{prism}}\end{align*} \begin{align*} = A_{\text{base}}\cdot h\end{align*}
#### Volume of a Cylinder
A cylinder is a three-dimensional solid consisting of two congruent, parallel, circular sides (the bases), joined by a curved surface.
To find the volume of a cylinder, find the area of its circular base and multiply by its height.
\begin{align*}V_{\text{cylinder}}\end{align*} \begin{align*} = \pi r^2 h\end{align*}
#### Volume of a Pyramid
A pyramid is a three dimensional solid with a polygonal base. Each corner of a polygon is attached to a singular vertex, which gives the pyramid its distinctive shape. Each base edge and the vertex form a triangle. Pyramids are named by their base shape.
To find the volume of a pyramid, find the volume of the prism with the same base and divide by three.
\begin{align*}V_{\text{pyramid}} = \frac{A_{\text{base}} \cdot h}{3}\end{align*}
#### Volume of a Cone
A cone is a three-dimensional solid with a circular base whose lateral surface meets at a point called the vertex.
To find the volume of a cone, find the volume of the cylinder with the same base and divide by three.
\begin{align*}V_{\text{cone}}=\frac{\pi r^2 h}{3}\end{align*}
##### Consider:
What is the ratio between the volume of a cylinder and of a cone, having the same radius and height?
The ratio between the volume of a cylinder and a cone with the same radius and height is (1:1 / 3:1 / 1:3 / 2:1).
#### Volume of a Sphere
A sphere is the set of all points in space equidistant from a center point. The distance from the center point to the sphere is called the radius.
The volume of a sphere relies on its radius.
\begin{align*}V_{\text{sphere}}= \frac{4}{3} \pi r^3\end{align*}
#### Volume of a Composite Solid
A composite solid is a solid made up of common geometric solids. The solids that it is made up of are generally prisms, pyramids, cones, cylinders, and spheres.
The volume of a composite solid is the sum of the volumes of the individual solids that make up the composite.
Let's look at some problems where we find the volume.
1. Find the volume of the rectangular prism below.
To find the volume of the prism, you need to find the area of the base and multiply by the height. Note that for a rectangular prism, any face can be the "base", not just the face that appears to be on the bottom.
$Volume = Area \, { }_{\text {base }} \cdot \, height$
$Volume = (length \, \cdot \, width) \cdot \, height$
$Volume = (4) \quad (4) \quad (5)$
$Volume =80 \, in ^{3}$
2. Find the volume of the cone below.
To find the volume of the cone, you need to find the area of the circular base, multiply by the height, and divide by three.
$Volume =\frac{1}{3} \quad Area \, { }_{\text {base }} \, \cdot \, height$
$Volume =\frac{1}{3} \, \pi \, r^{2} \, h$
$Volume =\frac{1}{3} \, \pi \, (7)^{2}\left(1 \frac{4}{2}\right)$
$Volume =196 \, \pi \, cm ^{3}$
3. Find the volume of a sphere with radius 4 cm.
\begin{align*}\text{Volume of Sphere} & = \frac{4}{3} \pi r^3 \\ & = \frac{4}{3} \pi (4)^3 \\ & = \frac{256 \pi}{3} \ \text{cm}^3\end{align*}
\begin{aligned} \text { Volume of Sphere } &=\frac{4}{3} \pi r^{3} \\ &=\frac{4}{3} \pi(4)^{3} \\ &=\frac{256 \pi}{3} cm ^{3} \end{aligned}
#### Examples
##### Example 1
The composite solid below is made of a cube and a square pyramid. The length of each edge of the cube is 12 feet and the overall height of the solid is 22 feet. What is the volume of the solid? Why might you want to know the volume of the solid?
To find the volume of the solid, find the sum of the volumes of the prism (the cube) and the pyramid. Note that since the overall height is 22 feet and the height of the cube is 12 feet, the height of the pyramid must be 10 feet.
\begin{align*} \text{Volume of Prism (Cube)}&=A_{\text{base}} \cdot h \\ &=(12 \cdot 12) (12) \\ &=1728 \ {\text{ft}}^3 \\ \text{Volume of Pyramid} &=\frac{A_{\text{base}} \cdot h}{3}\\ & =\frac{(12 \cdot 12) (10)}{3}\\ & =480 \ \text{ft}^3 \\ \text{Total Volume} & = \text{Volume of Prism (Cube)}+ \text{Volume of Pyramid} \\ &=1728+480 \\ & =2208 \ \text{ft}^3\end{align*}
The volume helps you to know how much the solid will hold. One cubic foot holds about 7.48 gallons of liquid, so
\begin{align*}\text{Gallons of liquid the solid can hold } & = \text{ Volume of solid } \cdot \text{ Number of gallons/cubic foot}\\ & = (2208)(7.48) \\ & = 16,515.84 \ \text{gallons} \\\end{align*}
##### Example 2
The area of the base of the pyramid below is \begin{align*}100 \ \text{cm}^2\end{align*}. The height is 5 cm. What is the volume of the pyramid?
\begin{align*}V_{\text{prism}} & = A_{\text{base}} \cdot h \\ & =(100 \ {\text{cm}}^2) \cdot (5 \ \text{cm}) \\ & =500 \ {\text{cm}}^3\end{align*}
##### Example 3
The volume of a sphere is \begin{align*}\frac{500 \pi}{3} \ \text{in}^3\end{align*}. What is the radius of the sphere?
To find the radius, use the formula: Volume of Sphere = \begin{align*}\frac{4}{3} \pi r^3\end{align*}
\begin{align*}\frac{500 \pi}{3} & = \frac{4}{3} \pi r^3 \\ 4r^3 &=500 \\ r^3 &=125 \\ r &=5 \ \text{in}\end{align*}
##### Example 4
The volume of a square pyramid is \begin{align*}64 \ {\text{in}}^3\end{align*}. The height of the pyramid is three times the length of a side of the base. What is the height of the pyramid?
\begin{aligned} V &=\frac{A_{\text {base }} \cdot h}{3} \\ 64 &=\frac{\left(s^{2}\right)(\not{3} s)}{\not{3}} \\ 64 &=s^{3} \\ \sqrt[3]{64} &=s \\ 4 &=s \end{aligned}
Side \begin{align*}s=4 \ \text{in}\end{align*} and height \begin{align*}h=3(4)=12 \ \text{in}.\end{align*}
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# A-level Mathematics/CIE/Pure Mathematics 1/Trigonometry
## The Trigonometric Functions
A right-angled triangle consists of three sides: the hypotenuse, the adjacent, and the opposite.
### Sine
The sine of an angle is defined as the ratio between the opposite and the hypotenuse. For a given angle, this ratio will always be the same, even if the triangle is scaled up or down.
${\displaystyle \sin(\theta )={\frac {opposite}{hypotenuse}}}$
### Cosine
The cosine of an angle is defined as the ratio between the adjacent and the hypotenuse.
${\displaystyle \cos(\theta )={\frac {adjacent}{hypotenuse}}}$
### Tangent
The tangent of an angle is defined as the ratio between the opposite and the adjacent.
${\displaystyle \tan(\theta )={\frac {opposite}{adjacent}}}$
### The Unit Circle
The unit circle is a circle of radius 1. It can be used to provide an alternate way of looking at trigonometric functions.
In the unit circle, a right-angled triangle can be drawn with the radius as its hypotenuse. Thus, the hypotenuse is 1 and the sine and cosine functions refer to the coordinates of a point on the unit circle.
Sine and cosine correspond to the coordinates of a point on the unit circle
## Graphing Trigonometric Functions
A sine wave
A sine graph starts at ${\displaystyle (0,0)}$, then oscillates with a period of ${\displaystyle 2\pi }$ and an amplitude of ${\displaystyle 1}$.
A cosine wave
A cosine graph is like a sine graph in that it oscillates with a period of ${\displaystyle 2\pi }$ and an amplitude of ${\displaystyle 1}$, but it starts at ${\displaystyle (1,0)}$
A tangent wave
A tangent graph starts at ${\displaystyle (0,0)}$, goes to infinity as it approaches ${\displaystyle x={\frac {\pi }{2}}}$, emerges from negative infinity after ${\displaystyle x={\frac {\pi }{2}}}$, then repeats this at ${\displaystyle (\pi ,0)}$. The tangent graph has a period of ${\displaystyle \pi }$.
## Exact Values
It is useful to know the following exact values of trigonometric functions:
Exact Values
x/° x/rad sin x cos x tan x
0 0 0 1 0
30 π/6 1/2 √3/2 1/√3
45 π/4 1/√2 1/√2 1
60 π/3 √3/2 1/2 √3
90 π/2 1 0 undefined
## Inverse Trigonometric Functions
A Note on Notation
Some sources may use ${\displaystyle \arcsin }$, ${\displaystyle \arccos }$, and ${\displaystyle \arctan }$ to represent the inverse trigonometric functions, but this notation is not endorsed by Cambridge International
The inverse trigonometric functions are functions that reverse the trigonometric functions, just like any other inverse function. The inverse trigonometric functions are: ${\displaystyle \sin ^{-1}}$, which is the inverse of ${\displaystyle \sin }$; ${\displaystyle \cos ^{-1}}$, which is the inverse of ${\displaystyle \cos }$; and ${\displaystyle \tan ^{-1}}$, which is the inverse of ${\displaystyle \tan }$.
## Trigonometric Identities
An identity is a statement that is always true, such as ${\displaystyle a+b\equiv b+a}$. A trigonometric identity, therefore, is a trigonometric statement that is always true.
It is helpful to know the following identities:
• ${\displaystyle \sin(x)\equiv \cos({\frac {\pi }{2}}-x)}$
• ${\displaystyle \cos(x)\equiv \sin({\frac {\pi }{2}}-x)}$
• ${\displaystyle {\frac {\sin(x)}{\cos {x}}}\equiv \tan(x)}$
• ${\displaystyle \sin ^{2}(x)+\cos ^{2}(x)\equiv 1}$
These identities can be used to prove other identities.
e.g. Prove that ${\displaystyle {\frac {\sin(x)\cos ^{2}(x)}{\cos(x)}}\equiv \tan(x)-\sin ^{2}(x)\tan(x)}$
{\displaystyle {\begin{aligned}{\frac {\sin(x)\cos ^{2}(x)}{\cos(x)}}&\equiv {\frac {\sin(x)(1-\sin ^{2}(x))}{\cos(x)}}\\&\equiv \tan(x)(1-\sin ^{2}(x))\\&\equiv \tan(x)-\sin ^{2}(x)\tan(x)\end{aligned}}}
## Solving Trigonometric Equations
When solving a trigonometric equation, it is important to keep the interval in mind.
e.g. Solve ${\displaystyle {\sqrt {3}}\tan 2x-1=0}$ for ${\displaystyle -{\frac {\pi }{2}}.
{\displaystyle {\begin{aligned}{\sqrt {3}}\tan 2x-1&=0\\{\sqrt {3}}\tan 2x&=1\\\tan 2x&={\frac {1}{\sqrt {3}}}\\&{\text{The interval is }}-{\frac {\pi }{2}}
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## Trigonometry (11th Edition) Clone
There is one value of $\theta$ satisfying the equation: $$\{180^\circ\}$$
$$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $\frac{\theta}{2}$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $\frac{\theta}{2}$, the inequality would be $$0^\circ\le\frac{\theta}{2}\lt180^\circ$$ Thus, the corresponding interval for $\frac{\theta}{2}$ is $[0^\circ,180^\circ)$. 2) Now we examine the equation: $$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$ Here we have both sine and cosecant functions. It would be beneficial if we can change $\csc\frac{\theta}{2}$ into a sine function, using the identity $\csc x=\frac{1}{\sin x}$ Thus, $$\sin\frac{\theta}{2}=\frac{1}{\sin\frac{\theta}{2}}$$ ($\sin\frac{\theta}{2}\ne0$) Multiply both sides with $\sin\frac{\theta}{2}$: $$\sin^2\frac{\theta}{2}=1$$ $$\sin\frac{\theta}{2}=\pm1$$ With $\sin\frac{\theta}{2}=1$, over interval $[0^\circ,180^\circ)$, there is 1 value whose sine equals $1$, which are $\{90^\circ\}$ With $\sin\frac{\theta}{2}=-1$, over interval $[0^\circ,180^\circ)$, there is no value whose sine equals $-1$. Combining 2 cases, only 1 value has been found out, meaning $$\frac{\theta}{2}=\{90^\circ\}$$ It follows that $$\theta=\{180^\circ\}$$ This is the solution set of the equation.
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Virtual University of Pakistan Lecture No. 36 of the course on Statistics and Probability by Miss Saleha Naghmi Habibullah
IN THE LAST LECTURE, YOU LEARNT •Confidence Interval for µ (continued) •Confidence Interval for µ1-µ2
TOPICS FOR TODAY •Determination of Sample Size (with reference to Interval Estimation) •Large Sample Confidence Intervals for p and p1-p2 •Estimation of sample size •Hypothesis-Testing (An Introduction)
In the last lecture, we discussed the construction and the interpretation of the confidence intervals for µ and µ1 - µ2. We begin today’s lecture by focusing on the confidence intervals for p and p1-p2.
First, we consider the confidence interval for p, the proportion of successes in a binomial population:
Confidence Interval for a Population Proportion (p): For a large sample drawn from a binomial population, the C.I. for p is given by where pˆ = n = zα/2 = =
pˆ ± z α / 2
pˆ(1 − pˆ ) n
proportion of “successes” in the sample sample size 1.96 for 95% confidence 2.58 for 99% confidence
(In
a practical situation, the criterion for deciding whether or not n is sufficiently large is that if both np and nq are greater than or equal to 5, then we say that n is sufficiently large.)
We illustrate this concept with the help of a few examples:
EXAMPLE-1 As a practical illustration, let us look at a survey of teenagers who have appeared in a juvenile court three times or more. A survey of 634 of these shows that 291 are orphans (one or both parents dead).
What proportion of all teenagers with three or more appearances in court are orphans? The estimate is to be made with 99% confidence.
SOLUTION In this problem, we have n = 634, and
pˆ = 291/634 = 0.459, qˆ = 1 − pˆ = 0.541, Hence, the 99% confidence limits for p are: 0.459 ± 2.58 0.459 × 0.541 634 = 0.459 ± 0.051 = 0.408 and 0.510
Hence, we estimate that the percentage of teenagers of this type who are orphans lies between 40.8 per cent and 51.0 per cent.
It should be noted that, in this problem, happily, the confidence interval has come out to be pretty narrow, and this is happening in spite of the fact that the level of confidence is very high ! This very desirable situation can be ascribed to the fact that the sample size of 634 is pretty large.
EXAMPLE-2 After a long career as a member of the City Council, Mr. Scott decided to run for Mayor. The campaign against the present Mayor has been strong with large sums of money spent by each candidate on advertisements.
In the final weeks, Mr. Scott has pulled ahead according to polls published in a leading daily newspaper. To check the results, Mr. Scott’s staff conducts their own poll over the weekend prior to the election.
The results show that for a random sample of 500 voters 290 will vote for Mr. Scott. Develop a 95 percent confidence interval for the population proportion who will vote for Mr. Scott. Can he conclude that he will win the election?
SOLUTION We begin by estimating the proportion of voters who will vote for Mr. Scott. The sample included 500 voters and 290 favored Mr. Scott. Hence, the sample proportion is 290/500 = 0.58.
The value 0.58 is a point estimate of the unknown population proportion p.
The 95% Confidence Interval for p is:
pˆ ± z α / 2
pˆ(1 − pˆ ) n
= 0.58 ± 1.96
0.58(1 − 0.58) 500
= 0.58 ± 0.043 = ( 0.537, 0.623)
The end points of the confidence interval are 0.537 and 0.623. The lower point of the confidence interval is greater than 0.50. So we conclude that the proportion of voters in the population supporting Mr. Scott is greater than 50 percent. He will win the election, based on the polling results.
EXAMPLE-3 A group of statistical researchers surveyed 210 chief executives of fastgrowing small companies. Only 51% of these executives had a management-succession plan in place.
A spokesman for the group made the statement that many companies do not worry about management succession unless it is an immediate problem. However, the unexpected exit of a corporate leader can disrupt and unfocus a company for long enough to cause it to lose its momentum.
Use the survey-figure to compute a 92% confidence interval to estimate the proportion of all fast-growing small companies that have a managementsuccession plan.
SOLUTION The point estimate of the proportion of all fast-growing small companies that have a management-succession plan is the sample proportion found to be 0.51 for that particular sample of size 210 which was surveyed by the group of researchers. Realizing that the point estimate might change with another sample selection, we calculate a confidence interval, as follows:
The value of n is 210
;
pˆ is 0.51 and qˆ = 1 − pˆ = 0 . 49 .
Because the level of confidence is 92%, the value of Z .04= 1.75.
0.04
-zÎą/2= -1.75
0.92 0
0.04
zÎą/2= 1.75
Z
The confidence interval is computed as:
0.51 −1.75
(0.51)(0.49) 210
≤0.51 +1.75
≤p
(0.51)(0.49)
210 0.51 −0.06 ≤ p ≤0.51 +0.06 0.45 ≤ p ≤0.57
P ( 0.45 ≤ p ≤0.57 ) =0.92 .
Conclusion: It is estimated with 92% confidence that the proportion of the population of fastgrowing small companies that have a management-succession plan is between 0.45 and 0.57.
Next, we consider the Confidence Interval for the difference in the population proportions (p1 – p2):
Confidence Interval for p1-p2: For large samples drawn independently from two binomial populations, the C.I. for p1-p2 is given by
( pˆ1 − pˆ 2 ) ± z α / 2
pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) + n1 n2
where subscript 1 denotes the first population, and subscript 2 denotes the second population.
We illustrate this concept with the help of an example:
EXAMPLE In a poll of college students in a large university, 300 of 400 students living in students’ residences (hostels) approved a certain course of action, whereas 200 of 300 students not living in students’ residences approved it. Estimate the difference in the proportions favouring the course of action, and compute the 90% confidence interval for this difference.
SOLUTION Let pˆ1 be the proportion of students favouring the course of action in the first sample (i.e. the sample of resident students).
pˆ 2
And, let be the proportion of students favouring the course of action in the second sample (i.e. the sample of students not residing in students’ residences). 300 200 = 0.75,and pˆ 2 = = 0.67. Then pˆ1 = 400 300 ∴ Difference in proportions = pˆ1 − pˆ 2 = 0.75 – 0.67 = 0.08
The required level of confidence is 0.90. Therefore z0.05 = 1.645, and hence, the 90% confidence interval for p1 – p2 is:
90% C.I. for p1-p2:
( pˆ1 − pˆ 2 ) ± (1.645)
pˆ1 qˆ1 pˆ 2 qˆ 2 + n1 n2
or
( 0.75)( 0.25) ( 0.67 )( 0.33) 0.08 ± (1.645) +
or or or or
0.000469 + 0.000737 0.08 ± (1.645) 0.08 ± (1.645) (0.0347) 0.08 ± 0.057 0.023 to 0.137
400
300
Hence the 90 per cent confidence interval for p1 – p2 is (0.023, 0.137). In other words, on the basis of 90% confidence, we can say that the difference between the proportions of resident students and non-resident students who favour this particular course of action lies somewhere between 2.3% and 13.7%.
Evidently, this seems to be a rather wide interval, even though the level of confidence is not extremely high. Hence, it is obvious that, in this example, sample sizes of 400 and 300 respectively, although apparently quite large, are not large enough to yield a desirably narrow confidence interval.
In the last lecture, we discussed the construction and interpretation of confidence intervals.
Next, we consider the determination of sample size:
In this regard, the first point to be noted is that, in any statistical study based on primary data, the first question is: What is going to be the size of the sample that is to be drawn from the population of interest?
We present below a method of finding the sample size in such a way that we obtain a desired level of precision with a desired level of confidence:
First, we consider the determination of sample size in that situation when we are trying to estimate Âľ, the population mean:
SAMPLE SIZE FOR ESTIMATING POPULATIONMEAN.
α) per cent confidence In deriving the 100(1– Interval for µ, we have the expression σ σ = 1 − α P − z α / 2 ≤ X − µ ≤ zα / 2 n n
which implies that the maximum X and allowable difference between µ is: σ x − µ = zα , /2 n
Ďƒ where
n
is the
standard
X when sampling is error of performed with replacement of population is very large (infinite).
The quantity
x − µ is also called
X and is the error of the estimator denoted by e. α) per cent Thus a 100(1– error bound µ is for estimating σ . given by z α / 2 n
In other words, in order to have a 100(1–α) per cent confidence that the error is estimating µ withX to be less than e, we need n such that
:
e = z
or
or
n
σ α/2 = z
n σ α/2 e
σ 2 z n = α/2 e
Hence the desired sample size for being 100(1–α)% confident that the error in estimating µ will be less than e, when sampling is with replacement or the population is very large, is given by 2
zα / 2 σ n = . e
It is important to note that the population standard deviation Ďƒ is generally not known, and hence, its estimate is found either from past experience or from a pilot sample of size n > 30.
In case of fractional result, it is always to be rounded to the next higher integer for the sample size.
EXAMPLE A research worker wishes to estimate the mean of a population using a sample sufficiently large that the probability will be 0.95 that the sample mean will not differ from the true mean by more than 25 percent of the standard deviation. How large a sample should be taken?
SOLUTION: If the sample mean is not being allowed to differ from the true mean by more than 25% of σ with a probability of 0.95, then
25σ σ e = x −µ = = , and z α / 2 = 1.96. 100 4
Substituting these values in the formula σ 2 z n = α/ 2 , we get e 1 .96 × σ 2 = 61 .4656 . n= σ/4 Hence the required sample size is 62, (the next higher integer), as the sample size cannot be fractional.
Next, we consider the determination of sample size in that situation when we are trying to estimate p, the proportion of successes in the population:
SAMPLE SIZE FOR ESTIMATING POPULATION PROPORTION. The large sample confidence interval for p is given by pˆ qˆ pˆ = z α/ 2 n This implies that
e=z
α/ 2
pˆ qˆ n
Therefore, solving for n, we obtain (z )2pˆ qˆ n = α/ 2 e2 Since the values of pˆ and qˆ are not nown as the sample has not yet been selected, we therefore use an estimate pˆ obtained from pilot sample information .
EXAMPLE In a random sample of 75 axle shafts, 12 have a surface finish that is rougher than the specification will allow. How large a sample is required if we want to be 95% confident that the error in using to estimate p is less than 0.05?
pˆ
Solution: e = pˆ − p = 0 . 05 , Here
pˆ =
12 75
= 0 . 16 ,
= 1 . 96 qˆ = 1 − pˆ − 0 . 84 and z 0.025 ( α / 2 = 0 . 025 )
Substituting these values in the formula z 2 n = α/ 2 pˆqˆ , we obtain e 1.96 2 ( )( ) = × n 0.16 0.84 = 206 .52 0.05 which, upon rounding upward, yields 207 as the desired sample size.
As stated earlier, Inferential Statistics can be divided into two parts, estimation and hypothesis-testing. Having discussed the concepts of point and interval estimation in considerable detail, we now begin the discussion of Hypothesis-Testing:
Hypothesis-testing is a very important area of statistical inference. It is a procedure which enables us to decide on the basis of information obtained from sample data whether to accept or reject a statement or an assumption about the value of a population parameter.
Such a statement or assumption which may or may not be true, is called a statistical hypothesis. We accept the hypothesis as being true, when it is supported by the sample data. We reject the hypothesis when the sample data fail to support it.
It is important to understand what we mean by the terms ‘reject’ and ‘accept’ in hypothesis-testing. The rejection of a hypothesis is to declare it false. The acceptance of a hypothesis is to conclude that there is insufficient evidence to reject it. Acceptance does not necessarily mean that the hypothesis is actually true.
The basic concepts associated with hypothesis testing are discussed below:
NULL AND ALTERNATIVE HYPOTHESES: Null Hypothesis: A null hypothesis, generally denoted by the symbol H0, is any hypothesis which is to be tested for possible rejection or nullification under the assumption that it is true.
A null hypothesis should always be precise such as ‘the given coin is unbiased’ or ‘a drug is ineffective in curing a particular disease’ or ‘there is no difference between the two teaching methods’.
The hypothesis is usually assigned a numerical value. For example, suppose we think that the average height of students in all colleges is 62″. This statement is taken as a hypothesis and is written symbolically as H0 : µ = 62″. In other words, we hypothesize that µ = 62″.
Alternative Hypothesis: An alternative hypothesis is any other hypothesis which we are willing to accept when the null hypothesis H0 is rejected. It is customarily denoted by H1 or HA.
A null hypothesis H0 is thus tested against an alternative hypothesis H1. For example, if our null hypothesis is H0 : µ = 62″, then our alternative hypothesis may be H1 : µ ≠ 62″ or H1 : µ < 62″.
LEVEL OF SIGNIFICANCE The probability of committing Type-I error can also be called the level of significance of a test.
Now, what do we mean by Type-I error? In order to obtain an answer to this question, consider the fact that, as far as the actual reality is concerned, H0 is either actually true, or it is false. Also, as far as our decision regarding H0 is concerned, there are two possibilities --either we will accept H0, or we will reject H0. The above facts lead to the following table:
Decision
True Situation
Accept H 0
Reject H 0 (or accept H 1)
H 0 is true
Correct decision (No error)
Wrong decision (Type-I error)
H 0 is false
Wrong decision (Type-II error)
Correct decision (No error)
A close look at the four cells in the body of the above table reveals that the situations depicted by the top-left corner and the bottom right-hand corner are the ones where we are taking a correct decision. On the other hand, the situation depicted by the top-right corner and the bottom lefthand corner are the ones where we are taking an incorrect decision.
The situation depicted by the top-right corner of the above table is called an error of the first kind or a Type I-error, while the situation depicted by the bottom left-hand corner is called an error of the second kind or a Type II-error. In other words:
TYPE-I AND TYPE-II ERRORS: On the basis of sample information, we may reject a null hypothesis H0, when it is, in fact, true or we may accept a null hypothesis H0, when it is actually false.
The probability of making a Type I error is conventionally denoted by α and that of committing a Type II error is indicated by β. In symbols, we may write α = P (Type I error) = P (reject H0|H0 is true), β = P (Type II error) = P (accept H0|Ho is false).
IN TODAY’S LECTURE, YOU LEARNT •Large Sample Confidence Intervals for p and p1-p2 •Determination of Sample Size (with reference to Interval Estimation) •Hypothesis-Testing (An Introduction)
IN THE NEXT LECTURE, YOU WILL LEARN •Hypothesis-Testing (continuation of basic concepts) •Hypothesis-Testing regarding µ • p-value •Hypothesis-Testing regarding µ1 - µ2 •Hypothesis Testing regarding p
STA301_LEC36
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# How to factor polynomials
## How to factor polynomials
We have studied about polynomial addition,subtraction ,multiplication and division. Now it time to learn How to factor polynomials. As you are already aware factoring is the process by which we go about determining what we multiplied to get the given quantity. So factoring polynomials means finding the terms which when multiplied together produced the polynomials
The various methods to perform factoring of polynomails are
### Greatest Common Factor
We can look at each of the term in the polynomials ,factorize each term and then find common factors to factorize the expression. This should be the first steps always when you are factoring the polynomials
In these problem, we need to factorize each term and then find common factors to factorize the expression
i) 7x-42
= (7×x) – (7×6)
=7(x-6)
ii) 6p-12
= (6×p) – (6×2)
=6(p-2)
iii) 7a2+14a
=(7×a×a) + (2×7×a)
=7a(a+2)
iv) -16z+ 20z3
=4z(5z2-4)
v) 20 l² m + 30 a l m
=(2×2×5×l×m×l) + (2×3×5×a×l×m)
=2lm(10l+15a)
vi) 5 x²y -15 xy2
=(5×x×x×y) + (5×3×x×y×y)
=5x(xy-3y2)
vii) 10a² -15 b2+20c2
=(2×5×a×a) - (5×3×b×b) + (5×2×2×c×c)
=5(2a2-3b2+4c2)
viii) -4a² +4a b-4ca
=(-4×a×a) +(4×b×a) - (4×c×a)
=-4a(a-b+c)
ix) x²yz + xy2z +xyz2
=(x×x×y×z) +(x×y×y×z) +(x×y×z×z)
=xyz(x+y+z)
x) ax²yz + bxy2z +cxyz2
=(x×x×y×z×a) +(x×y×y×z×b) +(x×y×z×z×c)
=xyz(ax+by+cz)
### factor polynomials by grouping
When we dont see common factor across all the terms, we may look at grouping the terms and check if we find binomial factor from both the groups .Let us take example to take close look at the method
1) x2 + 8x - x - 8
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x2 + 8x = x(x+8)
What about the last two terms? Observe them. If you take -1 out
-x-8 = -1(x+8)
Step 3 Putting both part togethes together,
x2 + 8x - x - 8= x(x+8) -1(x+8)
= (x+8)(x-1)
2) x5 -2x3 -2x2 +4
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x5 -2x3 = x3(x2-2)
What about the last two terms? Observe them. If you take -2 out
-2x2 +4 = -2(x2-2)
Step 3 Putting both part togethes together,
x5 -2x3 -2x2 +4= x3(x2-2) -2(x2-2)
= (x2-2)(x3-2)
### factor Quadratic polynomials by splitting the middle term
P(x) =ax2 +bx +c
Let its factors be (px + q) and (rx + s). Then
ax2 +bx +c= (px + q) (rx + s) = pr x2 + (ps + qr) x + qs
Comparing the coefficients of x2, we get a = pr.
Similarly, comparing the coefficients of x, we get b = ps + qr.
And, on comparing the constant terms, we get c = qs
This shows us that b is the sum of two numbers ps and qr, whose product is
(ps)(qr) = (pr)(qs) = ac.
Therefore, to factorise ax2 +bx +c, we have to write b as the sum of two numbers whose product is ac
1) 4x2 + 10x - 6
Step 1 Here b=10 and ac=-24
Step 2 We need to find factor of -24,whose sum is 10, the number which satisfy this is 12 and -2
4x2 + 10x - 6
=4x2 +(-2+12)x -6
=4x2 -2x + 12x -6
= 2x(2x-1)+ 6(2x-1)
= (2x-1)(2x+6)
2) x5 -2x3 -2x2 +4
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x5 -2x3 = x3(x2-2)
What about the last two terms? Observe them. If you take -2 out
-2x2 +4 = -2(x2-2)
Step 3 Putting both part togethes together,
x5 -2x3 -2x2 +4= x3(x2-2) -2(x2-2)
= (x2-2)(x3-2)
### Factor's Theorem's
If x-a is a factor of polynomial p(x) then p(a)=0 or if p(a) =0,x-a is the factor the polynomial p(x)
### 4) Factorising a Polynomial by Factor Theorem
We know by factor theorem if (x-a) is the factor of the polynomial ,then P(a)=0. We can use this theorem to find factors of the polynomials
Lets first take a look at quadratic polynomial
1) 4x2 + 10x - 6
Step 1 Take 4 out of the expression 4( x2 + 5x/2 -3/2)
Step 2 This can be further expressed as
4( x2 + 5x/2 -3/2) = 4(x-a) (x-b)
Here so ab= -3/2
We need to look for all the factor of the term -3/2 i.e +1/2,-1/2,-3,+3,1/4,-1/4,3/2,-3/2
Whereever it satisfies the factor theorem, we are good
In this particular case
P(1/2)=P(-3)=0, we can write like this
4( x2 + 5x/2 -3/2) = 4(x-1/2) (x+3)
= (2x-1)(2x+6)
For the last example, it may seems splitting the terms seems more efficient
Now let us take a look at cubic polynomial
Suppose the Polynomial is the form
P(x)= x3 +6x2+11x+6
Step 1 We need to look at the constant 6 and factorise it
The factor of 6 will be 1,2,3
Now we can try the polynomial for all the values -3,-2,-1,1,2,3
Whereever it satisfies the factor theorem, we are good
In this particular case
P(-1)=P(-2)=P(-3)=0, we can write like this
Step 2 x3 +6x2+11x+6=K(x+1)(x+2)(x+3)
We can put any value of x in this identity and get the value of x
In this particular case putting x=0, we get K=1
So the final identity becomes
x3 +6x2+11x+6=(x+1)(x+2)(x+3)
Or we can just find the one factor (x+1) and try the splitting or long division method to get the remaining quadratic polynomial and solve that by splitting method
x3 +6x2+11x+6
= x3 +x2+5x2+5x+6x +6
Splitting cubic polynomial to get (x+1)
=x2(x+1) + 5x(x+1) + 6(x+1)
=(x+1)(x2+5x+6)
Now solving the quadratic polynomial by splitting method
=(x+1)(x2+2x+3x+6)
=(x+1)[x(x+2)+3(x+2)]
=(x+1)(x+2)(X+3)
In General Term,
S(x)=anxn+a(n-1)x(n-1)+a(n-2)x(n-2)+....+ax+a0
Look for the factors in a0/an, Take both the positive and negative values and find out which suites your polynomial and then find the value of k
### Factoring polynomial using Algebraic Identities
Identity I : (x + y)2 = x2 + 2xy + y2
Identity II : (x - y)2 = x2 - 2xy + y2
Identity III : x2 - y2 = (x + y) (x - y)
Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab
Identity V :(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Identity VI : (x + y)3 = x3 + y3 + 3xy (x + y)
Identity VII :(x - y)3 = x3 - y3 - 3xy(x - y)
Identity VIII : x3 + y3 + z3 -3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
Identity IX : x3 + y3 = (x + y )(x2 + y2 - xy )
Identity X : x3 - y3 = (x - y )(x2 + y2 + xy )
1) 27x3 +1
Step 1 Both th terms are exact cubes (3x) and 1
Step 2 We can the algebric identity IX
27x3 +1
=(3x+1)(9x2+1-3x)
2)8x3 + y3 + 27z3 -18xyz
= (2x)3 + y3 + (3z)3 - 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)2 + y2 + (3z)2 - (2x)(y) - (y)(3z) - (2x)(3z)]
=
(2x + y + 3z) (4x2 + y2 + 9z2 - 2xy - 3yz - 6xz)
## Solved Examples Polynomials
1) If x + 1/x = 11, evaluate x2 + 1/x2
Solution x + 1/x = 12
(x + 1/x)2 = (x)2 +1/x2 + 2(x)(1/x) = 144
x2 + 1/x2 + 2 = 144
x2+ 1/x2 = 144 - 2 = 142
2) If 9x2+ 25y2 = 181 and xy = -6. Find the value of 3x + 5y
Solution 9x2+ 25y2 = 181
(3x)2 + (5y)2 = 181
(3x)2 + (5y)2 + 30xy - 30xy = 181
(3x)2 + (5y)2 + 2(5x)(6y) = 181 + 30xy
(3x + y)2 = 181 + 30(-6) = 181 - 180
(3x + y)2 =1
3x + y = +1 or -1
3) Prove that x2 + y2 + x2 -xy -yz -zx is always non-negative for all values of x, y and y.
Solution To prove x2 + y2 + x2 -xy -yz -zx = 0.
We know that square of any number (+ve or -ve) is always +ve.
x2 + y2 + z2 -xy -yz -zx
= (1/2)[2x2 + 2y2 + 2z2 -2xy -2yz -2zx]
=(1/2)[x2 + y2 -2xy + y2 + z2 -2yz + x2 + z2 -2zx]
=(1/2)[(x2 + y2 -2ab) + (y2 + z2 -2yz) + (x2 + z2 -2zx)]
=(1/2)[(x - y)2 + (y - z)2 + (z - x)2 ]
Here, all the terms are always be positive,
x2 + y2 + x2 -xy -yz -zx = 0
## Practice Questions
1) If ( x - 4) is a factor of the polynomial 2x2 + Ax + 12 and ( x - 5) is a factor of the polynomial x3 - 7x2+ 11 x + B , then what is the value of ( A - 2B )?
2) if x -1/x =3; then find the value of x3 -1/x3
3) if a+b+c=7 and ab+bc+ca=20 find the value of (a+b+c)2
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# Factors of 4
Top
FAQ
## What are Factors of 4?
View Notes
Factors of 4 can be defined as the numbers that divide the original number uniformly or completely. The two numbers which when gets multiplied and gives the result as the original number, can be termed as pair factors.
We can use the factorization method to find the factors of 4 as well as multiples of 4.
In this factorization method we need to first consider the given number itself and one as the factors, so 4 and 1 are the factors of the number 4 and then we need to keep finding the factors of number 4 which shall give us the number 4.
### What are Pair Factors of 4?
To find the pair factors of the number 4, multiply the two numbers in a pair to have the original number, such numbers are as follows:
(1, 4) and (2, 2)
The above are the positive pair factors for 4.
### The Factors of 4 in Pairs
(1, 4) are the factors of 4 as 1 x 4 is equal to 4.
(2, 2) are the factors of 4 as 2 x 2 is equal to 4.
So the factors of the number 4 are 1, 2 and 4.
### Prime Factors of 4 (Prime Factorization)
The prime factorisation of 4 gives:
4 equals 1 x 2 x 2Â
### Factors of 4 in Pairs:
(1 x 4)Â
(2 x 2)Â
(4 x 1)
Therefore, the prime factors of 4 : 2 * 2
### What are the Factors of 4?
These are the integers that can be evenly divided into 4; they can be expressed as either individual factors or as factor pairs. In this case, we present them both ways. This is the mathematical decomposition of a particular number.
### What is the Prime Factorization of 4?
A prime factorization is a result of factoring any given number into a set of components where every member is a prime number. This is generally written by showing 4 as a product of its prime factors. For 4, this result would be:
4 equals 2 x 2
(this is also known as the prime factorization method; the smallest prime number in this series is described as the smallest prime factor among all of them.)
### Is 4 a Composite Number?
No! The number 4 is not a composite number. None of the two-factor pairs was the same number. A composite number must be formed as the product of any smaller positive integer.
### Is 4 a Square Number?
Yes! 4 is a square number. Its square root is an integer, specifically the number 2.
### How Many Factors Does 4 Have?
This number has three factors: 1, 2, 4
More specifically, shown as pairs...
(1*4) (2*2) (4*1)
### What is the Greatest Common Factor of 4 and Another Number?
The greatest common factor (GCF) of two numbers can be determined by comparing the prime factorization or factorization of the two numbers and taking the highest common prime factor. If there is no common factor, then the GCF is equal to 1. This is also known as the highest common factor and is part of the common prime factors of two given numbers. It is the largest factor or the largest number the two numbers share as a prime factor. The least common factor or the smallest number in common of any pair of integers is the number 1.
### What is a Factor Tree?
A factor tree can be defined as a graphic representation of the possible factors of numbers and their sub-factors. It is designed to simplify factorization. A factor tree is created by finding the factors of any number, then finding the factors of the number. The process continues recursively until you've derived a bunch of prime factors, this is the prime factorization of the original number. In constructing the factor tree, be sure to remember that the second item in a factor pair.
How Do You Find the Factors of Any Negative Number? (eg. -4)
To find the factors of the number, let’s suppose -4, find all the positive factors (see above) and then duplicate them by adding a minus sign before each one (effectively multiplying them by -1). This addresses negative factors. (handling negative integers)
### Is 4 a Whole Number?
Yes, 4 is a whole number.
### What are the Divisibility Rules?
Divisibility can be referred to a given integer number being divisible for a given divisor. The divisibility rule can be defined as a shorthand system to determined what is or isn't divisible. This includes rules about even number and odd number factors.
1. What are All the Factors of 4?
Ans. Factors of 4: 1, 2, 4. Prime factorization: 4 equal 2 x 2, which can also be written 4 equals 2². Since √4 equals 2, a whole number, 4 is a perfect square.
2. What is a Factor of 4 and 6?
Ans. The greatest common factor (GCF) of 4 and 6 is 2. We will now calculate the prime factors of the numbers 4 and 6 then find the greatest common factor (greatest common divisor (that is gcd)) of the numbers by matching the biggest common factor of the numbers 4 and 6.
3. What Does a Factor of 4 Mean?
Ans. "By a factor of " can commonly mean the same as "multiplied by" or "divided by." If the variable x is increased by a factor of 4, it further becomes 4x. If the variable x is decreased by a factor of 4, then it becomes x/4. The keyword is the direction of change that is, (increased/decreased) by a factor of.
4. What Does by a Factor of 10 Mean?
Ans. "By a factor of " commonly means the same as "multiplied by" or "divided by." "Costs were reduced by a factor of ten" would mean that the costs are 1/10th what they used to be. "Costs increased by a factor of ten" , this means that costs are ten times what they actually used to be.
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# The Pressure P and Volume V of a Gas Are Connected by the Relation Pv1/4 = Constant. the Percentage Increase in the Pressure Corresponding to a Deminition of 1/2 % in the Volume is - CBSE (Science) Class 12 - Mathematics
#### Question
The pressure P and volume V of a gas are connected by the relation PV1/4 = constant. The percentage increase in the pressure corresponding to a deminition of 1/2 % in the volume is
• $\frac{1}{2} \%$
• $\frac{1}{4} \%$
• $\frac{1}{8} \%$
• none of these
#### Solution
$\frac{1} {8}$%
We have
$\frac{\bigtriangleup V}{V} = \frac{- 1}{2} %$
$P V^\frac{1}{4} = \text { constant }= k \left( \text { say } \right)$
$\text { Taking log on both sides, we get }$
$\log \left( P V^\frac{1}{4} \right) = \log k$
$\Rightarrow \log P + \frac{1}{4}\log V = \log k$
$\text { Differentiating both sides w . r . t . x, we get }$
$\frac{1}{P}\frac{dP}{dV} + \frac{1}{4V} = 0$
$\Rightarrow \frac{dP}{P} = - \frac{dV}{4V} = - \frac{1}{4} \times \frac{- 1}{2} = \frac{1}{8}$
$\text { Hence, the increase in the pressure is } \frac{1}{8} \% .$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution The Pressure P and Volume V of a Gas Are Connected by the Relation Pv1/4 = Constant. the Percentage Increase in the Pressure Corresponding to a Deminition of 1/2 % in the Volume is Concept: Approximations.
S
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# numerator of a fraction
The numerator function allows the calculation of the numerator of a fraction.
numerator(4/5+3/7), returns 43 numerator(0.5), returns 1
### Numerator, online calculus
#### Summary :
The numerator function allows the calculation of the numerator of a fraction.
#### Description :
A fraction is a number that is written as follows : a/b with a and b integers and b not zero. It's the quotient of a by b, that is to say : a/b = a:b. a is the numerator of the fraction and b is the denominator of the fraction. In other words, in a fraction, the numerator term is located above the stroke, the denominator is the end located below the stroke. The calculator with the numerator function used to find the numerator of a fraction or an expression with fractions.
# Fraction numerator
The calculator is able to determine the numerator of a fraction : to find the fraction numerator of 4/5, enter numerator(4/5), after calculation, the result 4 is returned.
The calculator also applies to fractions that contain letters : to pour calculate the fraction numerator of a/b, enter numerator(a/b) , after calculation, the result a is returned.
The numerator function is used on fractions but also to expressions containing fractions, after calculation, the simplified result is returned.
# Numerator of an expression with fraction
When the numerator function is applied to an algebraic expression, the expression is first converted to a fraction, then the numerator of the fraction obtained is returned. So to find the numerator of the following expression 4/5+3/7, enter numerator(4/5+3/7), after calculation, he fraction obtained is 43/35, then the function returns the numerator 43.
The numerator function also applies to literal expressions.
#### Syntax :
numerator(expression), where expression is an algebraic expression or a fraction.
#### Examples :
numerator(4/5+3/7), returns 43 numerator(0.5), returns 1
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# If Paisa : Rupee :: ? : Kilometre
1. Metre
2. Hectometre
3. Quintal
4. Decametre
Option 4 : Decametre
## Detailed Solution
As, 100 paisa equals to 1 rupee. Similarly, 100 decametre equals to 1 Kilometre.
Hence, ‘Decametre’ is the correct answer.
# Directions: Select the set in which the numbers are related in the same way as are the numbers of the following set. (2, 7, 57)
1. (4, 21, 463)
2. (3, 20, 81)
3. (5, 21, 121)
4. (3, 27, 729)
Option 1 : (4, 21, 463)
## Detailed Solution
The logic followed here is:
Second number = [First number × (First number + 1)] + 1
Third number = [Second number × (Second number + 1)] + 1
The given set is (2, 7, 57)
Second number = [First number × (First number + 1)] + 1
= [2 × (2 + 1)] + 1
= [2 × 3] + 1
= 6 + 1 = 7
Third number = [Second number × (Second number + 1)] + 1
= [7 × (7 + 1)] + 1
= [7 × 8] + 1
= 56 + 1 = 57
Checking the options:-
1) (4, 21, 463)
Second number = [First number × (First number + 1)] + 1
= [4 × (4 + 1)] + 1
= [4 × 5] + 1
= 20 + 1 = 21
Third number = [Second number × (Second number + 1)] + 1
= [21 × (21 + 1)] + 1
= [21 × 22] + 1
= 462 + 1 = 463
2) (3, 20, 81)
Second number = [First number × (First number + 1)] + 1
= [3 × (3 + 1)] + 1
= [3 × 4] + 1
= 12 + 1 = 13 ≠ 20
3) (5, 21, 121)
Second number = [First number × (First number + 1)] + 1
= [5 × (5 + 1)] + 1
= [5 × 6] + 1
= 30 + 1 = 31 ≠ 21
4) (3, 27, 729)
Second number = [First number × (First number + 1)] + 1
= [3 × (3 + 1)] + 1
= [3 × 4] + 1
= 12 + 1 = 13 ≠ 27
Hence, "(4, 21, 463)" is the correct answer.
# Select the option that is related to the third term in the same way as the second term is related to the first term. Singapore : SIMBEX :: France : ?
1. VARUNA
2. INDRA
3. KONKAN
4. SLINEX
Option 1 : VARUNA
## Detailed Solution
The pattern followed here is:
Singapore India Maritime Bilateral Exercise (SIMBEX) is an annual bilateral naval exercise conducted by the Indian Navy and the Republic of Singapore Navy (RSN). The exercise has been held annually since 1994.
Similarly,
The annually held VARUNA naval exercise is an integral part of France–India strategic relationship in the 21st century and consists of naval cooperation drills between the French Navy and the Indian Navy.
JOINT MILITARY EXERCISE - NAVY Name of Exercise Participating Nations SIMBEX India and Singapore VARUNA India and France INDRA India and Russia KONKAN India and UK SLINEX India and Sri Lanka
# In the following question, select the related letters/numbers from the given alternatives.666 : 15 :: 888 : ?
1. 18
2. 13
3. 17
4. 19
Option 3 : 17
## Detailed Solution
According to the given information-
666 → (6)3 + (6)2 + 6 → 216 + 36 + 6 → 258 → 2 + 5 + 8 → 15
In the same way-
888 → (8)3 + (8)2 + 8 → 512 + 64 + 8 → 584 → 5 + 8 + 4 → 17
Hence, the correct answer is 17
# In the following question, select the related word pair from the given alternatives.Book : Pages ∷ ? : ?
1. Writer : Write
2. Pen : Write
3. Car : Tyre
4. Tyre : Round
Option 3 : Car : Tyre
## Detailed Solution
As books contain pages,
Pages are part of book.
Similarly,
Tyres are the part of a car.
Hence, ‘Car : Tyre’ is the right answer.
# Select the related word from the given alternatives.Bhangra ∶ Punjab ∶∶ ? ∶ Uttar Pradesh
1. Kathak
2. Bihu
3. Garba
4. Lavani
Option 1 : Kathak
## Detailed Solution
Bhangra is the dance form of Punjab.
Similarly,
The dance form of Uttar Pradesh is Kathak.
Hence, ‘kathak’ is the correct answer.
The differences between kathak or kathakali.
Both styles are parts of Indian classical dance but both come from different states of India.
Kathak had evolved from north Indian states Jaipur, Banaras, or Lucknow, which are also known as gharanas and the kathakali has come from south Indian state Kerala which is known as home several of traditional classical dance.
So the conclusion is all about kathak or kathakali dance are both are different dance style not are same but most people think that both styles are same because of pronouncing sound of styles.
# In the following question select the related term from the given alternatives. 11 : GO :: 14.5 : ?
1. CF
2. KM
3. IT
4. LP
Option 3 : IT
## Detailed Solution
Consider, A =1, B = 2, C = 3,..........., Y = 25, Z = 26
G is 7th and O is 15th in the alphabetical order.
GO = 7 + 15 = 22 → 22/2 = 11
Similarly,
CF = 3 + 6 = 9 → 9/2 = 4.5 ≠ 14.5
KM = 11 + 13 = 24 → 24/2 = 12 ≠ 14.5
IT = 9 + 20 = 29 → 29/2 = 14.5 = 14.5
LP = 12 + 16 = 28 → 28/2 = 14 ≠ 14.5
Hence, 'IT' is the correct alternative.
# In the following question, select the related word from the given alternatives. Eloquent : Inarticulate ∷ Obsolete : ?
1. Current
2. Bereave
3. Divest
4. Bleak
Option 1 : Current
## Detailed Solution
Eloquent and Inarticulate are exactly opposite in meaning. Eloquent means fluent or persuasive in speaking or writing, whereas Inarticulate means unable to express one's ideas or feelings clearly or easily, opposite of Inarticulate.
Similarly, Obsolete and Current are opposite to each other in meaning. Obsolete means no longer produced or used; out of date, whereas Current is up to date, exactly opposite to Obsolete.
# Select the related letters from the given alternative. ACFJ : KMPT ∷ DIBE : ?
1. MSLO
2. MLOS
3. NLSO
4. NSLO
Option 4 : NSLO
## Detailed Solution
In ACFJ : KMPT
Similarly;
Hence, “NSLO” is the correct answer.
# Select the option that is related to the third term in the same way as the second term is related to the first term. Dispur : Assam :: Kohima : ?
1. Capital
2. Sikkim
3. Meghalaya
4. Nagaland
Option 4 : Nagaland
## Detailed Solution
logic: State and Capital combination is given.
Dispur : Assam → Dispur is the capital of Assam.
Similarly,
Kohima : ? → Kohima is the capital of Nagaland.
Hence, ‘Nagaland’ is the correct answer.
# Select the option that is related to the third letter-cluster in the same way as the second letter-cluster is related to the first letter-cluster. ROAST : UTBPS :: HONEY : ?
1. ZFOPI
2. XBNTE
3. RPKLR
4. ZFKPY
Option 1 : ZFOPI
## Detailed Solution
Similarly,
Hence, ZFOPI is the correct answer.
# Select the related word from the given alternatives. Pratibha Patil : First Woman President of India :: C.D. Deshmukh : ?
1. First Chief Justice of India.
2. First Railway Minister of India.
3. First Indian Governor of RBI.
4. First Indian Governor General of Independent India.
Option 3 : First Indian Governor of RBI.
## Detailed Solution
Pratibha Patil was the first woman president of India.
Similarly,
C.D. Deshmukh was the first Indian Governor of RBI.
Hence, the correct answer is "First Indian Governor of RBI".
IN India First Present Chief justice of India Harilal Jekisundas Kania N V Ramana Railway Minister of India. John Mathai Ashwini Vaishnaw
1. Chakravarti Rajagopalachari, who was a fierce Mahatma Gandhi loyalist, made Hindi a compulsory language in Tamil schools when he was CM. New Delhi: Chakravarti Rajagopalachari, popularly known as Rajaji, was independent India's first Indian Governor-General.
2. Louis Mountbatten, Earl Mountbatten of Burma became governor-general and oversaw the transition of British India to independence. Chakravarti Rajagopalachari (1878-1972) became the only Indian and last governor-general after independence.
# Select the relative word for the second pair from the given options which will replace the question mark (?) following the same relation as in the first pair.SUGARCANE : ETHANOL :: JATROPHA : ?
1. BIODIESEL
2. MANURE
3. PETROL
4. NATURAL GAS
Option 1 : BIODIESEL
## Detailed Solution
Ethanol is obtained by fermentation of sugar cane.
Similarly,
Biodiesel is obtained from Jatropha plants.
Hence, “Biodiesel” is the correct answer.
# Select the related word from the given alternatives.Cat : Feline :: Dog : ?
1. Canine
2. Cunning
3. Cunine
4. Bovine
Option 1 : Canine
## Detailed Solution
The logic is -
Cat : Feline → Cat belongs to Feline family.
Similarly,
Dog : ? → Dog belongs to Canine family.
Hence, 'Canine' is the correct answer.
# ‘E’ is related to “BH” in the same way as “N” is related to?
1. IP
2. KQ
3. KP
4. KR
Option 2 : KQ
## Detailed Solution
‘E’ is in the middle of series ‘BH’. Similarly ‘N’ is in middle of series ‘KQ’.
# Select the related number from the given alternatives.21 : 11 :: 19 : ?
1. 10
2. 25
3. 22
4. 41
Option 1 : 10
## Detailed Solution
The logic is:
11 × 2 = 22 -1 = 21
Similarly,
10 × 2 = 20 -1 = 19
Hence the correct answer is 10.
# Select the option that is most like the given set. (9, 29, 67)
1. 2, 10, 45
2. 3, 18, 35
3. 2, 10, 30
4. 3, 15, 26
Option 3 : 2, 10, 30
## Detailed Solution
The pattern followed here is,
n3 + 1, (n+ 1)3 + 2 ; (n+2)3 + 3
So,
(9, 29, 67) → (23 + 1 = 9; 33 + 2 = 29; 4+ 3 = 67)
Similarly,
(2, 10, 30) → (13 + 1 = 2, 23 + 2 = 10; 33 + 3 = 30)
Hence, (2, 10, 30) is the correct answer.
# In the following question, select the related word from the given alternatives. Trivial : Essential ∷ Outrage : ?
1. Agile
2. Bereft
3. Tranquil
4. Berate
Option 3 : Tranquil
## Detailed Solution
Trivial and Essential are exactly opposite in meaning. Trivial means 'of little value or importance', whereas Essential means 'important'.
Similarly,
Outrage and Tranquil are opposite to each other in meaning. Outrage means 'an extremely strong reaction of anger, shock', whereas Tranquil is the opposite of outrage meaning 'in a peaceful state'.
# Choose the pair which is related in the same way as the words in the first pair from the given choicesSurgeon : Scalpel :: Gardener : ______A. BurrowB. HarrowC. PloughD. Shoes
1. C
2. D
3. B
4. A
Option 3 : B
## Detailed Solution
The scalpel is the tool used by a surgeon.
Similarly,
Harrow is used by a gardener.
Hence, the harrow is the correct answer.
# Arrange the given numbers in the order they are arranged in first part and choose the one that comes in place of question mark(?).1024 : 18 ∷64 : ?
1. 10
2. 13
3. 12
4. 9
Option 1 : 10
## Detailed Solution
The pattern followed here is;
1024 = 210; 2 × 10 = 20; 20 - 2 = 18
In the same way;
64 = 26; 2 × 6 = 12; 12 - 2 = 10
Hence, 10 will replace the question mark.
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# How do you find a unit vector perpendicular to both (2,1,1) and the x-axis?
Nov 23, 2016
Please see the explanation for steps leading to: $\hat{C} = \frac{\sqrt{2}}{2} \hat{j} - \frac{\sqrt{2}}{2} \hat{k}$
#### Explanation:
Given: $\overline{A} = 2 \hat{i} + \hat{j} + \hat{k}$
Let $\overline{B} = \text{a vector along the x axis} = \hat{i}$
The vector $\overline{C} = \overline{A} \times \overline{B}$ will be perpendicular to both but is will not be a unit vector.
barC =barA xx barB = | (hati, hatj, hatk, hati, hatj), (2,1,1,2,1), (1,0,0,1,0) |=
$\left\{1 \left(0\right) - 1 \left(0\right)\right\} \hat{i} + \left\{1 \left(1\right) - 2 \left(0\right)\right\} \hat{j} + \left\{2 \left(0\right) - \left(1\right) \left(1\right)\right\} \hat{k}$
$\overline{C} = \hat{j} - \hat{k}$
Compute the magnitude of $\overline{C}$
$| \overline{C} | = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$
The unit vector perpendicular to both is:
$\hat{C} = \frac{\overline{C}}{|} \overline{C} |$
$\hat{C} = \frac{\hat{j} - \hat{k}}{\sqrt{2}}$
$\hat{C} = \frac{\sqrt{2}}{2} \hat{j} - \frac{\sqrt{2}}{2} \hat{k}$
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# Feet to Miles
Instructions: Use this step-by-step Feet to Miles Calculator, to compute amount of miles corresponding to a given number of feet $$F$$. Please provide the required information in the form below:
Number of Feet $$F$$ (number or fraction) =
## More about the conversion of feet to miles
The idea of converting feet to miles may appear to be strange, considering that a foot and a mile are fairly different in sizes, a mile being much bigger.
Typically, there is always a reason for a conversion of magnitude of distances: either we are working with metric systems of different countries, or sometimes just for the pleasure in making things harder (really).
### Feet to Miles conversion
When stating how to do a feet to miles conversion, it is typically easier to start with miles to feet. Indeed, there is a round number for that:
$\large 1 \text{ mile} = 5280 \text{ feet}$
So you need lots of feet to make one mile. Now, taking the above equation and dividing both sides by 5280, we get that
$\large \frac{1}{5280} \text{ mile} = 1 \text{ foot}$
and calculating the numeric form of the fraction we get:
$\large 1 \text{ foot} = 0.000189394 \text{ miles}$
So then, you may be asking yourself now: How many feet are in a mile? . The answer is, there are 5280 feet in a mile.
And you may ask yourself: " How many miles are there in a foot? ". The answer is that there are 0.000189394 miles in a foot.
Hence, if we have $$F$$ feet, the corresponding amount of miles is
$\large F \text{ feet} = 0.000189394 \times F \text{ miles}$
### So based on the above, how do you convert miles into feet?
The formula above is telling to get the number of feet, and multiply that by 0.000189394, to get the corresponding number of miles.
### Examples
Convert 41 feet into miles
Solution: In this case we get $$F = 41$$ feet, so we can apply directly the formula above to say that
$\large F \text{ feet} = 0.000189394 \times 41 \text{ miles} = 0.007765154 \text{ miles}$
### Other uses of the feet-to-miles conversion
Sometimes, the conversion of feet-to-miles is done as an intermediary step for making a conversion for a velocity unit. For example, we may be interested in converting speed given in feet per second to miles per hour.
We know that 1 mile is equal to 5280 feet. Also, we know that 1 hour is the same as $$60 \times 60 = 3,600$$ seconds. Therefore,
$\large \displaystyle 1 \text{ } \frac{\text{mile}}{\text{hour}} = \frac{5280 \text{ feet}}{3600 \text{ seconds}}$ $\large \displaystyle = 1.466666667 \text{ } \frac{\text{ feet}}{\text{ seconds}}$
The above expression also can be written as
$\large \displaystyle 1 \text{ } \frac{\text{ feet}}{\text{ seconds}} =\frac{1}{1.466666667} \frac{\text{miles}}{\text{hour}} = 0.6818182 \text{ } \frac{\text{miles}}{\text{hour}}$
In other words, 1 feet per second is equivalent to 0.6818182 miles per hour.
Therefore, if have $$v$$ feet per second, that is equivalent
$\large \displaystyle v \text{ } \frac{\text{ feet}}{\text{ seconds}} = 0.6818182 \times v \text{ } \frac{\text{miles}}{\text{hour}}$
Granted, the above formulas are not easy to memorize, so you will probably use a calculator or make the full deduction to derive the constants.
### Example
What is 20 feet per seconds in miles per hour?
Solution: Using the above formula, we get directly that 20 feet per seconds is
$\large \displaystyle 0.6818182 \times 20 \text{ } \frac{\text{miles}}{\text{hour}} = 13.63636364 \text{ } \frac{\text{miles}}{\text{hour}}$
### Other related conversion tools
For other types of conversion, please check out our conversion calculator tool , where you will find many types of conversions. For example, you can compute meters per second when you are given the distance in miles instead.
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