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# 2010 AMC 12B Problems/Problem 20
## Problem
A geometric sequence $(a_n)$ has $a_1=\sin x$, $a_2=\cos x$, and $a_3= \tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\cos x$?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
## Solution
By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2x$.
The common ratio of the sequence is $\frac{\cos x}{\sin x}$, so we can write
$$a_1= \sin x$$ $$a_2= \cos x$$ $$a_3= \frac{\cos^2x}{\sin x}$$ $$a_4=\frac{\cos^3x}{\sin^2x}=1$$ $$a_5=\frac{\cos x}{\sin x}$$ $$a_6=\frac{\cos^2x}{\sin^2x}$$ $$a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}$$ $$a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}$$
Since $\cos^3x=\sin^2x=1-\cos^2x$, we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$, which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$.
## Solution 2
Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$. Then, working backwards we have $a_3=\frac{1}{r}$, $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$. Now notice that since $a_1=\sin(x)$ and $\a_2=cos(x)$ (Error compiling LaTeX. ! LaTeX Error: Command \_ unavailable in encoding OT1.), we need $a_1^2+a_2^2=1$, so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$. Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$, which the left side is equal to $1+\cos(x)$; we see as well that the right hand side is equal to $a_8$ given $a_4=1$, so the answer is $\boxed{E}$. - mathleticguyyy |
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Module 9 Assessment Answer Key.
Vocabulary
Choose the best term from the box.
Vocabulary
cubic units
formula
volume
Question 1.
_Volume____ is the amount of space occupied by a solid figure. (p. 355)
Volume,
Explanation:
As we know in math, volume is defined as the 3 dimensional space
enclosed by a boundary or occupied by an object.
A ___formula______ is an equation that expresses a mathematical rule. (p. 349)
formula,
Explanation:
A formula can also be called an equation that expresses
a relationship between two or more variables.
Concepts and Skill
Use formulas to find the perimeter and area. (TEKS 5.4.H)
Question 3.
Perimeter: _______________
Area: ______________
Question 4.
Perimeter: ____20 ft___________
Area: _____25 square ft_________
Perimeter: 20 ft,
Area: 25 square ft,
Explanation:
Given square with side 5 feet as we know perimeter of square is
4 X side, so perimeter = 4 X 5 feet = 20 ft and area of square is
side X side = 5 fet X 5 ft = 25 square ft.
Use formulas to find the volume. (TEKS 5.4.G, 5.4.H)
Question 5.
Volume: _____512 cubic m________
Volume is 512 cubic m,
Explanation:
Given rectangular cube with edge 8 m,
so volume is edge X edge X edge =
8 m X 8 m X 8 m = 512 cubic m.
Question 6.
Volume: _____________
Volume is 1287 cubic cm,
Explanation:
Given rectangular cuboid with length 11 cm, width 9 cm and
height 13 cm, so volume is length X width X height =
11 cm X 9 cm X 13 cm = 1287 cubic cm.
Find the area of the polygon. Show how you find your answer. (TEKS 5.4.H)
Area: _____52 square cm________
Area of the polygon is 52 square cm,
Explanation:
Given to find area of the polygon but is irregular polygon we divide into
regular shapes of two rectangles as shown above one with 8 cm length and
width 5 cm other with length 6 cm and width 2 cm then we add to get
the area of total polygon, so first rectangle area is 8 cm X 5 cm = 40 square cm,
other rectangle area is 6 cm X 2 cm = 12 square cm,
total area is 40 square cm + 12 square cm = 52 square cm.
Question 8.
Which equation shows how to find the volume of the model? (TEKS 5.4.G, 5.4.H)
(A) 8 × 5 = 40
(B) (2 × 8) + (2 × 5) = 26
(C) 8 × 5 × 4 = 160
(D) 8 + 5 + 4 = 17
(C) 8 × 5 × 4 = 160
Explanation:
Given model shape is regular cuboid with length 8 cm,
width 5 cm and height 4 cm, so equation showing
to find the volume of the model as V = length X width X height is
V = 8 X 5 X 4 = 160 which matches with (C).
Question 9.
Sanaya has a rectangular bulletin board that is 3 feet long and 2 feet wide.
She wants to cover the board with felt and add a satin ribbon around
the border of the bulletin board.
How much felt and ribbon does Sanaya need? (TEKS 5.4.H)
(A) 6 feet of felt, 10 feet of ribbon
(B) 6 sq feet of felt, 10 feet of ribbon
(C) 10 sq feet of felt, 6 feet of ribbon
(D) 6 sq feet of felt, 5 feet of ribbon
(B) 6 sq feet of felt, 10 feet of ribbon,
Explanation:
Given Sanaya has a rectangular bulletin board that is 3 feet long and 2 feet wide.
She wants to cover the board with felt and add a satin ribbon around
the border of the bulletin board. So felt and ribbon does Sanaya need is
as felt is base area A = long X wide, A= 3 feet X 2 feet = 6 sq feet,
and ribbon is perimeter = 2 (long + wide) = 2 X (3feet + 2 feet) = 3 X 5 feet = 10 feet,
therefore Sanaya needs 6 sq feet of felt, 10 feet of ribbon which matches with (B).
Hicham buys a rectangular sheet of plywood that measures 18 square meters.
The sheet of plywood is 3 meters long.
What is the perimeter of the sheet of plywood? (TEKS 5.4.H)
(A) 9 meters
(B) 6 meters
(C) 18 meters
(D) 18 square meters
(C) 18 meters,
Explanation:
Given Hicham buys a rectangular sheet of plywood that measures 18 square meters.
The sheet of plywood is 3 meters long. To find the perimeter of the sheet of plywood
we need first width as Area = long X wide, 18 square meters = 3 meters X w,
w = 18 ÷ 3 meters = 6 meters, Now the perimeter of the sheet of plywood is
2( long + wide) = 2 X (3 meters + 6 meters) = 2 X (9 meters) = 18 meters
which matches with (C).
Question 11.
Melinda is making a tin box shaped like a rectangular prism.
For the base, Melinda cuts out a square with a side length of 12 inches.
If Melinda wants the volume of the tin box to be 1,440 cubic inches,
how tall should the box be? (TEKS 5.4.H )
(A) 12 inches
(B) 10 inches
(C) 6 inches
(D) 8 inches
(B) 10 inches,
Explanation:
Given Melinda is making a tin box shaped like a rectangular prism.
For the base, Melinda cuts out a square with a side length of 12 inches.
If Melinda wants the volume of the tin box to be 1,440 cubic inches,
the height of the box must be given length is 12 inches and
width also 12 inches(square) as we know volume = length X width X height ,
1,440 cubic inches = 12 inches X 12 inches X height/tall,
1,440 cubic inches = 144 square inches X height/tall,
therefore height/tall = 1,440 cubic inches ÷ 144 square inches,
height/tall = 10 inches which matches with (B).
Deval has a rectangular patio that is 15 feet long and 5 feet wide.
If Deval uses square tiles with a side length of 1 foot,
how many tiles will he need to cover the patio? (TEKS 5.4.H) |
# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2014 | June | Q#4
Question
a.
i. Express in the form where p and q are integers.
ii. Hence write down the maximum value of .
b.
i. Factorise .
ii. Sketch the curve with equation , stating the values of x where the curve crosses the x-axis and the value of the y-intercept.
Solution
a.
i.
We have the expression;
We use method of “completing square” to obtain the desired form.
Next we complete the square for the terms which involve .
We have the algebraic formula;
For the given case we can compare the given terms with the formula as below;
Therefore we can deduce that;
Hence we can write;
To complete the square we can add and subtract the deduced value of ;
ii.
We are given;
We are required to find the maximum value of .
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph.
If
(‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
We recognize that given curve , is a parabola opening downwards.
Vertex form of a quadratic equation is;
The given curve , as demonstrated in (a:i) can be written in vertex form as;
Coordinates of the vertex are .Since this is a parabola opening downwards the vertex is the maximum point on the graph.
Here y-coordinate of vertex represents maximum value of
and x-coordinate of vertex represents corresponding value of .
For the given curve coordinates of vertex are .
Therefore, maximum value of is 25.
b.
i.
ii.
We are required to sketch the given equation;
It is evident that it is quadratic equation which represents a parabola.
First we find the coordinates of x-intercepts of the parabola.
The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore;
As demonstrated in (b:i), we know that;
Therefore;
Now we have two options.
Two values of x indicate that there are two intersection points.
Next we find the y-intercept of the parabola.
The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore;
Now we can sketch the parabola represented by as follows. |
## Solution #018 - circle of hats
40
This post contains my proposed solution to Problem #018 - circle of hats. Please do not read this solution before making a serious attempt at the problem.
### Solution
The answer is really interesting and can be formulated in mathematical terms in a rather simple way; after that I will walk you through what the mathematical formulation means with some doodles I drew.
Let $$a_i$$ be the number in the hat of the $$i$$th mathematician and $$g_i$$ be the guess the $$i$$th mathematician writes down. Then $$g_i$$ satisfies the equation
$$g_i + \sum_{j \neq i} a_j \equiv i \hspace{0.5cm} \text{mod } n$$
What follows from here is that if $$k \equiv \sum_i a_i \text{ mod } n$$, then the $$k$$th mathematician will get its guess right and all other mathematicians will fail.
#### Why it works (mathematically)
Let $$a_i, g_i$$ be as above and let $$k = \sum_i a_i \text{ mod } n$$. Notice how
$$g_k + \sum_{j \neq k} a_j \equiv k \iff g_k \equiv k - \sum_{j \neq k} a_j.$$
Of course, we defined $$k = \sum_i a_i$$ so the above becomes
$$g_k = \sum_{j} a_j - \sum_{j \neq k} a_j = a_k \hspace{0.5cm} \text{mod } n,$$
hence the $$k$$th mathematician will guess correctly.
#### Explanation of the solution
Let us take $$n = 5$$ and number the mathematicians from $$0$$ to $$4$$, starting from the top and going in the clockwise direction. The numbers inside the circles represent the numbers in the hats, that I distributed randomly.
Now we pretend we are the mathematician number $$0$$, and so we see the numbers $$0$$, $$3$$, $$0$$ and $$1$$, which give $$4$$ when added up. Now, we are the mathematician $$0$$ so our guess $$g_0$$ is how much is left to go from $$4$$ to $$0$$ which is... $$1$$, so that is our guess.
In case you are not familiar with modular arithmetics, adding and subtracting is just like you are used to, but numbers wrap around the modulus that in this case is $$5$$. This is like the hours in a day wrapping around $$24$$! To do additions and subtractions modulo $$5$$, just look at the small numbers next to the mathematicians and count clockwise when adding, counter-clockwise when subtracting. For example, $$3 + 3$$ is $$1$$ because if you start at mathematician $$3$$ and you count $$3$$ mathematicians starting from that one, you end up at mathematician $$1$$.
To recap, mathematician $$0$$ guessed $$1$$ because the other mathematicians' hats added up to $$4$$ and $$4 + 1 \equiv 0$$ modulo $$5$$. Unfortunately, the mathematician got it wrong because its hat had a $$3$$ on it...
Mathematician $$1$$ sees $$3$$, $$0$$, $$1$$ and $$3$$, which add up to $$3 + 0 + 1 + 3 \equiv 2$$ mudulo $$5$$ and to get to $$1$$ we must add $$4$$, so mathematician $$1$$ guesses $$4$$, which is also wrong...
Then came mathematician $$2$$ who sees $$0$$, $$1$$, $$3$$ and $$0$$, adding up to $$4$$. Now, to get from $$4$$ to $$2$$ modulo $$5$$ we need to add $$3$$, which is what our mathematician guesses... and it is correct!
Just for the sake of completeness, can you tell what mathematicians $$3$$ and $$4$$ will write down as their guesses..? Hint: none of them will get it right.
#### A piece of code for you to play with
I wrote a small piece of APL code so you can check for yourself which mathematician writes down its number correctly, just follow this link and change the numbers in the last line to whatever you like, then hit the arrow on the top to run the code:
HatSolver ← {
⎕IO ← 0 ⍝ IO delenda est
n ← ≢⍵
s ← ⍵ - ⍨ +/⍵
⍵ = n| s -⍨ ⍳n
}
⎕← HatSolver 3 0 3 0 1
I included the example from this blog post in the code, the 3 0 3 0 1 you can see above. The numbers that get printed show a $$0$$ for every mathematician that fails and shows a $$1$$ for the mathematician that guesses correctly. |
## Eight Knights
Eight dashing knights are in love with [the] kingdom’s princess. Starting on January $1$, the first knight visits the princess every day. The second night visits the princess on January $2$ and every second day thereafter. The third knight visits the princess on January $3$ and every third day thereafter. The pattern continues for each of the $8$ knights. What is the total number of knight visits up to and including the first day on which all $8$ knights visit the princess?
Source: mathcontest.olemiss.edu 9/3/2012
SOLUTION
Let’s simplify the problem by assuming there are only $4$ knights. We make a little spreadsheet to detect any pattern.
The first time all $4$ knights visit the princess is on day $12$ which is the least common multiple (LCM) of $1,2,3,4$.
During those $12$ days, the number of visits by each knight is as follows
Knight $1\!: 12$ times $\left ( 12\div 1=12\right )$
Knight $2\!: 6$ times $\left ( 12\div 2=6\right )$
Knight $3\!: 4$ times $\left ( 12\div 3=4\right )$
Knight $4\!: 3$ times $\left ( 12\div 4=3\right )$
Total number of visits
$12+6+4+3=25$
Applying the pattern to 8 knights
LCM of $1,2,3,4,5,6,7,8 = 840$
During those $840$ days, the number of visits by each knight is as follows
Knight $1\!: 840$
Knight $2\!: 420$
Knight $3\!: 280$
Knight $4\!: 210$
Knight $5\!: 168$
Knight $6\!: 140$
Knight $7\!: 120$
Knight $8\!: 105$
Total number of visits
$840+420+280+210+168+140+120+105=2283$
Answer: $2283$. |
# Review Day for Equations and Inequalities
27 teachers like this lesson
Print Lesson
## Objective
SWBAT apply an understanding of how to solve equations and inequalities using tables and diagrams.
#### Big Idea
Practicing for tests helps build good study habits.
## DO NOW
15 minutes
Students will be working on a problem that requires them to do multiple steps to solve. I chose this problem to help stimulate their minds and get them thinking about math.
Possible solution: Students will use multiplication to try and solve the problem. They will do 25 x 4 to get to 100 and then add \$50 to get to \$150. Once they get this, the students can do a guess and check to estimate their solutions.
For example:
Number of watches Amount she earns 25 \$150 50 \$150 + \$150 = \$300 75 \$300 + \$150 = \$450 100 \$450 + \$150 = \$600
Students can estimate that she will need to sell more than 75 watches to earn \$506.
Then, they could say that Sarah had to sell 75 watches to earn \$450. Next, they will subtract \$506 - \$450(amount sold for 75 watches) to get \$56(Remaining dollar amount) . Students should divide to 56 by 4(How much she earns for each watch) to get 14.
Finally, they can use the guess and check estimation of 75 and add it to the 14 to get 89 watches sold.
To help students out with this problem, you could ask them about the important information:
How much does she earn per watch?
How much does she earn for a bonus?
What earns her the bonus?
Is there a tool that can be use to help us organize our data?
This problem attends to SMP 2 by understanding what the numbers mean and SMP 5 by using an appropriate tool to organize our information.
Tools: Do now problem
60 minutes
I chose Numbered Heads Together because it is my favorite and also because it allows students time to work independently and check solutions with a group. Each problem represents a similar problem for their assessment. Students will be writing and solving inequalities and using the coordinate grid to graph their solutions.
For questions 1 – 3 the students will be writing equations. If needed, remind students to circle key words and identify both expressions. Question 2: the students may want to use 2 variables. “If both coefficients relate to firefighters, do we need to have 2 variables?” Once they see that they do not need two variables, they can combine like terms to write a simplified equation.
Questions 4 and 5 deal with inequalities. Students will need to write and graph the inequality. For the word problem, if students get stuck with what symbol to use say to them “if you have \$20 to spend, think about the amounts you can spend?”
Questions 6 and 7 are asking the students to identify the independent and dependent variable. If students get stuck ask them to draw a table. Remember that what gets substituted “in” for the variable is the independent variable, and what comes out is the dependent variable. Students could use made up information and a table to model this mathematics (SMP4 and 5).
Question 8 is asking the students to write an equation to represent the information in the table. For students to use precision in their equation, they should use the recommended variables from the problem. (SMP 6)
Question 9 is asking the students to use the equation to complete the information from table. They are finding both the independent and dependent variable given one of them. Students may have trouble when the dependent variable is given. If this happens, ask the student to say to themselves “what number times 3 makes 24”?
Questions 10 and 11 are asking students to write an equation, complete a table and graph the information in a grid. Students will need to create their own table and grid. If students have trouble with creating a grid, it might be a good idea to have some available for them. In question 11, the word function is used. Students may be confused by this as we have yet to cover this word. Explain to students that a function is an algebraic term to describe the table and the equation.
Question 12 is asking the students to write the equation given the points on a line. If students are having difficulty, encourage them to write the ordered pairs in a table so they can see the pattern. Students will understand the pattern and write it as an equation. (SMP 7 and 8)
Numbered Heads Together supports mathematical practices:
SMP1: making sense of problems
SMP2: understanding what the numbers mean.
SMP3: constructing viable arguments to prove their solution is correct.
Tools: NHT power point, whiteboards, markers, coordinate grids.
## Closure + Homework
10 minutes
The students will receive their study guides for the test. The problems on the study guide are similar to NHT and are helpful for practicing problems for the assessment. Allow students time to look over their study guide and ask any questions they may have. We will go over the study guide before the assessment the next day. Any extra time can be used to start the study guide.
Tools: Study Guide for Equations. |
College Physics for AP® Courses
# 10.2Kinematics of Rotational Motion
College Physics for AP® Courses10.2 Kinematics of Rotational Motion
## Learning Objectives
By the end of this section, you will be able to:
• Observe the kinematics of rotational motion.
• Derive rotational kinematic equations.
• Evaluate problem solving strategies for rotational kinematics.
Just by using our intuition, we can begin to see how rotational quantities like $θθ size 12{θ} {}$, $ωω size 12{ω} {}$, and $αα size 12{α} {}$ are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration $αα size 12{α} {}$ is large for a long period of time $tt size 12{α} {}$, then the final angular velocity $ωω size 12{ω} {}$ and angle of rotation $θθ size 12{θ} {}$ are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large.
Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating $ωω size 12{ω} {}$, $αα size 12{α} {}$, and $tt size 12{t} {}$. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:
10.15
Note that in rotational motion $a=ata=at size 12{a=a rSub { size 8{t} } } {}$, and we shall use the symbol $aa size 12{a} {}$ for tangential or linear acceleration from now on. As in linear kinematics, we assume $aa size 12{a} {}$ is constant, which means that angular acceleration $αα size 12{α} {}$ is also a constant, because $a=rαa=rα size 12{a=rα} {}$. Now, let us substitute $v=rωv=rω size 12{v=rω} {}$ and $a=rαa=rα size 12{a=rα} {}$ into the linear equation above:
$rω = rω 0 + rαt . rω = rω 0 + rαt . size 12{rω=rω rSub { size 8{0} } +rαt} {}$
10.16
The radius $rr size 12{r} {}$ cancels in the equation, yielding
10.17
where $ω0ω0 size 12{ω rSub { size 8{0} } } {}$ is the initial angular velocity. This last equation is a kinematic relationship among $ωω size 12{ω} {}$, $αα size 12{α} {}$, and $tt size 12{t} {}$ —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart.
## Making Connections
Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion.
Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts):
Rotational Translational
$θ = ω ¯ t θ = ω ¯ t size 12{θ= {overline {ωt}} } {}$ $x = v - t x = v - t size 12{x= { bar {v}}t} {}$
$ω = ω 0 + αt ω = ω 0 + αt size 12{ω=ω rSub { size 8{0} } +αt} {}$ $v = v 0 + at v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}$ (constant $αα size 12{α} {}$, $aa size 12{a} {}$)
$θ = ω 0 t + 1 2 αt 2 θ = ω 0 t + 1 2 αt 2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {}$ $x = v 0 t + 1 2 at 2 x = v 0 t + 1 2 at 2 size 12{x=v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}$ (constant $αα size 12{α} {}$, $aa size 12{a} {}$)
$ω 2 = ω 0 2 + 2 αθ ω 2 = ω 0 2 + 2 αθ size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {}$ $v 2 = v 0 2 + 2 ax v 2 = v 0 2 + 2 ax$ (constant $αα$, $aa$)
Table 10.2 Rotational Kinematic Equations
In these equations, the subscript 0 denotes initial values ($θ0θ0 size 12{θ rSub { size 8{0} } } {}$, $x0x0 size 12{x rSub { size 8{0} } } {}$, and $t0t0 size 12{t rSub { size 8{0} } } {}$ are initial values), and the average angular velocity $ω-ω- size 12{ { bar {ω}}} {}$ and average velocity $v-v- size 12{ { bar {v}}} {}$ are defined as follows:
10.18
The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which $aa size 12{a} {}$ and $αα size 12{α} {}$ are constant.
## Problem-Solving Strategy for Rotational Kinematics
1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion.
2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion.
5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
## Example 10.3
### Calculating the Acceleration of a Fishing Reel
A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of $110rad/s2110rad/s2 size 12{"110""rad/s" rSup { size 8{2} } } {}$ for 2.00 s as seen in Figure 10.8.
(a) What is the final angular velocity of the reel?
(b) At what speed is fishing line leaving the reel after 2.00 s elapses?
(c) How many revolutions does the reel make?
(d) How many meters of fishing line come off the reel in this time?
### Strategy
In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown.
### Solution for (a)
Here $αα size 12{α} {}$ and $tt size 12{α} {}$ are given and $ωω size 12{ω} {}$ needs to be determined. The most straightforward equation to use is $ω=ω0+αtω=ω0+αt size 12{ω=ω rSub { size 8{0} } +αt} {}$ because the unknown is already on one side and all other terms are known. That equation states that
$ω=ω0+αt .ω=ω0+αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}$
10.19
We are also given that $ω0=0ω0=0 size 12{ω rSub { size 8{0} } =0} {}$ (it starts from rest), so that
$ω=0+110 rad/s22.00 s=220rad/s .ω=0+110 rad/s22.00 s=220rad/s . size 12{ω=0+ left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right )="220 rad/s."} {}$
10.20
### Solution for (b)
Now that $ωω size 12{ω} {}$ is known, the speed $vv size 12{v} {}$ can most easily be found using the relationship
$v=rω ,v=rω , size 12{v=rω","} {}$
10.21
where the radius $rr size 12{α} {}$ of the reel is given to be 4.50 cm; thus,
$v=0.0450 m220 rad/s=9.90 m/s.v=0.0450 m220 rad/s=9.90 m/s. size 12{v= left (0 "." "0450"" m" right ) left ("220"" rad/s" right )=9 "." "90"" m/s."} {}$
10.22
Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have $m × rad = m m × rad = m size 12{m times "rad"=m} {}$.
### Solution for (c)
Here, we are asked to find the number of revolutions. Because $1 rev=2π rad1 rev=2π rad size 12{1" rev"=2π" rad"} {}$, we can find the number of revolutions by finding $θθ size 12{θ} {}$ in radians. We are given $αα size 12{α} {}$ and $tt size 12{t} {}$, and we know $ω0ω0 size 12{ω rSub { size 8{ {} rSub { size 6{0} } } } } {}$ is zero, so that $θθ size 12{θ} {}$ can be obtained using $θ=ω0t+12αt2θ=ω0t+12αt2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {}$.
θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . alignl { stack { size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} # " "=0+ left (0 "." "500" right ) left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right ) rSup { size 8{2} } ="220"" rad" {} } } {}
10.23
$θ=220 rad1 rev2π rad=35.0 rev.θ=220 rad1 rev2π rad=35.0 rev. size 12{θ= left ("220"" rad" right ) { {1" rev"} over {2π" rad"} } ="35" "." 0" rev."} {}$
10.24
### Solution for (d)
The number of meters of fishing line is $xx size 12{x} {}$, which can be obtained through its relationship with $θθ size 12{θ} {}$:
$x = rθ = 0.0450 m 220 rad = 9.90 m . x = rθ = 0.0450 m 220 rad = 9.90 m . size 12{x=rθ= left (0 "." "0450"" m" right ) left ("220"" rad" right )=9 "." "90"" m"} {}$
10.25
### Discussion
This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites.
Figure 10.8 Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel.
## Example 10.4
### Calculating the Duration When the Fishing Reel Slows Down and Stops
Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of $–300rad/s2–300rad/s2 size 12{"300""rad/s" rSup { size 8{2} } } {}$. How long does it take the reel to come to a stop?
### Strategy
We are asked to find the time $tt size 12{α} {}$ for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is $ω0=220 rad/sω0=220 rad/s size 12{ω rSub { size 8{0} } ="220"" rad/s"} {}$ and the final angular velocity $ωω size 12{ω} {}$ is zero. The angular acceleration is given to be $α=−300rad/s2α=−300rad/s2 size 12{α= - "300" "rad/s" rSup { size 8{2} } } {}$. Examining the available equations, we see all quantities but t are known in $ω=ω0+αt,ω=ω0+αt, size 12{ω=ω rSub { size 8{0} } +αt} {}$ making it easiest to use this equation.
### Solution
The equation states
$ω=ω0+αt .ω=ω0+αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}$
10.26
We solve the equation algebraically for t, and then substitute the known values as usual, yielding
$t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s.t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s. size 12{t= { {ω - ω rSub { size 8{0} } } over {α} } = { {0 - "220"" rad/s"} over { - "300""rad/s" rSup { size 8{2} } } } =0 "." "733"" s."} {}$
10.27
### Discussion
Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration.
## Example 10.5
### Calculating the Slow Acceleration of Trains and Their Wheels
Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of $0.250rad/s20.250rad/s2 size 12{0 "." "250""rad/s" rSup { size 8{2} } } {}$. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?
### Strategy
In part (a), we are asked to find $xx size 12{x} {}$, and in (b) we are asked to find $ωω size 12{ω} {}$ and $vv size 12{v} {}$. We are given the number of revolutions, the radius of the wheels $rr size 12{r} {}$, and the angular acceleration $αα size 12{α} {}$.
### Solution for (a)
The distance $xx size 12{x} {}$ is very easily found from the relationship between distance and rotation angle:
$θ = x r . θ = x r . size 12{θ= { {x} over {r} } } {}$
10.28
Solving this equation for $xx size 12{x} {}$ yields
$x=rθ.x=rθ. size 12{x=rθ.} {}$
10.29
Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:
$θ = 200 rev 2π rad 1 rev = 1257 rad . θ = 200 rev 2π rad 1 rev = 1257 rad . size 12{θ= left ("200"" rev" right ) { {2π" rad"} over {"1 rev"} } ="1257"" rad"} {}$
10.30
Now we can substitute the known values into $x=rθx=rθ size 12{x=rθ} {}$ to find the distance the train moved down the track:
$x = rθ = 0.350 m 1257 rad = 440 m . x = rθ = 0.350 m 1257 rad = 440 m . size 12{x=rθ= left (0 "." "350"`m right ) left ("1257"" rad" right )="440"" m"} {}$
10.31
### Solution for (b)
We cannot use any equation that incorporates $tt$ to find $ωω$, because the equation would have at least two unknown values. The equation $ω2= ω02+2αθω2= ω02+2αθ$ will work, because we know the values for all variables except $ωω$:
$ω2= ω02+2αθω2= ω02+2αθ$
10.32
Taking the square root of this equation and entering the known values gives
10.33
We can find the linear velocity of the train, $vv size 12{v} {}$, through its relationship to $ωω size 12{ω} {}$:
$v = rω = 0.350 m 25.1 rad/s = 8.77 m/s . v = rω = 0.350 m 25.1 rad/s = 8.77 m/s . size 12{v=rω= left (0 "." "350"" m" right ) left ("25" "." 1" rad/s" right )=8 "." "77"" m/s"} {}$
10.34
### Discussion
The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h).
There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.
Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly).
## Example 10.6
### Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate
A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)
### Strategy
First, find the total number of revolutions, and then the linear distance $xx size 12{x} {}$ traveled.
### Solution
The number of revolutions is the product of rpm and time:
$6.0 rpm2.0 min=12 rev.6.0 rpm2.0 min=12 rev.$
10.35
$θ=12 rev2π rad1 rev=75.4 rad.θ=12 rev2π rad1 rev=75.4 rad. size 12{θ= left ("12"" rev" right ) left ( { {2π" rad"} over {"1 rev"} } right )="75" "." 4" rad"} {}$
10.36
Now, using the relationship between $xx size 12{x} {}$ and $θθ size 12{θ} {}$, we can determine the distance traveled:
10.37
### Discussion
Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics.
Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.) |
# How do you use the sum and difference identities to find the exact value of cos 75?
Dec 6, 2014
The sum and difference identities are given as follows:
sum:
$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$
difference:
$\cos \left(\alpha - \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\alpha\right) \sin \left(\beta\right)$
If you have an angle whose trig values you don't know, but that can be made by two angles you do know, you will use one of these identities. ${75}^{o}$ can be made of the sum of ${45}^{o}$ and ${30}^{o}$, both of which are on the unit circle!
$\cos \left({75}^{o}\right) = \cos \left({45}^{o} + {30}^{o}\right)$
Using our sum identity, where $\alpha = {45}^{o}$ and $\beta = {30}^{o}$;
$= \cos \left({45}^{o}\right) \cos \left({30}^{o}\right) - \sin \left({45}^{o}\right) \sin \left({30}^{o}\right)$
Now just find these values on your unit circle and you get;
$= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2}$
and the rest is arithmetic.
To use the difference identity, I would recommend using the angles;
$\cos \left({75}^{o}\right) = \cos \left({120}^{o} - {45}^{o}\right)$
Then, using difference identity;
$\cos \left({120}^{o}\right) \cos \left({45}^{o}\right) + \sin \left({120}^{o}\right) \sin \left({45}^{o}\right)$
Which gives;
$- \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}$
Both expressions simplify to;
$\frac{\sqrt{2} \left(\sqrt{3} - 1\right)}{4}$ |
# What Is 7/44 as a Decimal + Solution With Free Steps
The fraction 7/44 as a decimal is equal to 0.159.
We can express the division of two numbers p and q in the form of a fraction p/q, where p (the dividend) is now called the numerator and q (the divisor) is called the denominator. Fractions, like any division, produce either an integer or decimal result.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 7/44.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 7
Divisor = 44
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 7 $\div$ 44
This is when we go through the Long Division solution to our problem.
Figure 1
## 7/44 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 7 and 44, we can see how 7 is Smaller than 44, and to solve this division, we require that 7 be Bigger than 44.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 7, which after getting multiplied by 10 becomes 70.
We take this 70 and divide it by 44; this can be done as follows:
 70 $\div$ 44 $\approx$ 1
Where:
44 x 1 = 44
This will lead to the generation of a Remainder equal to 70 – 44 = 26. Now this means we have to repeat the process by Converting the 26 into 260 and solving for that:
260 $\div$ 44 $\approx$ 5Â
Where:
44 x 5 = 220
This, therefore, produces another Remainder which is equal to 260 – 220 = 40. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 400.
400 $\div$ 44 $\approx$ 9Â
Where:
44 x 9 = 396
Finally, we have a Quotient generated after combining the three pieces of it as 0.159, with a Remainder equal to 4.
Images/mathematical drawings are created with GeoGebra. |
# Fascinated by Fibonacci
Purpose
This unit looks at Fibonacci numbers and how they occur in nature. Fibonacci numbers provide a rich context in which to apply algebra at level 4. It is recommended that this unit be used with students that already have some experience with level 4 algebra.
Achievement Objectives
GM4-3: Use side or edge lengths to find the perimeters and areas of rectangles, parallelograms, and triangles and the volumes of cuboids.
NA4-1: Use a range of multiplicative strategies when operating on whole numbers.
NA4-8: Generalise properties of multiplication and division with whole numbers.
Specific Learning Outcomes
• Use a recursive rule to generate the sequence of Fibonacci numbers.
• Create a Fibonacci spiral using squares with Fibonacci side lengths.
• Find a pattern of odd and even numbers in the sequence.
• Identify and represent patterns we find for consecutive numbers in the sequence.
Description of Mathematics
In this unit we look at Fibonacci numbers and how they occur in nature. We also look at situations where the next number in a sequence is created by adding the previous two numbers. We also explore some intriguing patterns that occur with the Fibonacci sequence and try to explain why some of those patterns occur.
Fibonacci numbers are a specific example of sequences that are generated by rules that tell how to operate on the previous term to get the next term. Such rules are called recurrence relations because they can be applied infinitely in a ‘recurring’ way. The Fibonacci sequence is created using the recursive rule Fn = Fn-2 + Fn-1. That rule means that the nth Fibonacci number is the sum of the previous two Fibonacci numbers.
The sequence progresses as 1, 1, 2, 3, 5, 8, 13,.. Other sequences can be created using the same rule but different starting numbers, e.g. 2, 1, 3, 4, 7, 11, … (Lucas numbers).
Required Resource Materials
Activity
#### Lesson One
In this lesson we explore how to generate the Fibonacci sequence. Show the students Slide One of PowerPoint One. After the first sequence show the tree growing until it has 13 branches.
Watch carefully how the tree grows new branches each spring.
Is there a pattern to the growth?
How many branches will the tree have after next spring?
Click on the mouse to reveal that the tree has 21 branches next spring.
Do students notice that each new branch does not sprout another branch until its second spring?
Do they notice that each new number of branches is the sum of the two springs before it?
Progress to Slide Two which shows the growing family tree of a male bee. Male bees have only one parent, a mother (female). Female bees have two parents, a father (male) and a mother (female). Watch the animation until three generations are past.Comment that some plants sprout in this manner, though often conditions like weather and animals mean that branches get lost.
How many ancestors are in the family tree after five generations?
Ask students to draw the family tree to five generations. Slide Three has a model answer. Create a table of values for the numbers of ancestors and their genders.
Generation Male ancestors Female ancestors Total number 1 0 1 1 2 1 1 2 3 1 2 3 4 2 3 5 5 3 5 8
Ask the students to consider what will happen in the next two generations:
• Each male ancestor will have only a mother.
• Each female ancestor will have both a mother and father.
• There will be female ancestor for each bee in the layer below but a male ancestor only for each female bee.
The next layers of the table are:
Generation Male ancestors Female ancestors Total number 6 5 8 13 7 8 13 21
Students should note that the same sequence of numbers 1, 2, 3, 5, 8, 13, 21,…, occurs in the tree branching pattern and the bee’s ancestor patterns.
What is similar about both situations?
Students might note that extra branches or ancestors are added by splitting into two. They might also observe that there is a delay of one spring or generation. Branches must wait one spring before sprouting other branches, and male bees must wait one generation before they have another female male ancestor.
Slides 5 and 6 show Fibonacci numbers appearing in the growth spirals of plants. The pinecone has 13 rows of bracts spiralling outwards. Similarly, the aloe vera plant has 5 spirals of leaves. The pineapple has 8 rows of scales spirally from the base to the top. 13, 5, and 8 are Fibonacci numbers.
Given the frequency of Fibonacci numbers appearing as spirals in nature it is no surprise that these same numbers can generate a beautiful spiral. Slide 9 shows how to create the spiral using arrangements of squares with sides lengths that are consecutive Fibonacci numbers. Copymaster One is provided for students to create their own spiral pattern. Slides 10-12 show examples of where the spiral occurs in nature.
Slide 13 shows the areas of the squares in the spiral design appearing. Click once for the areas to appear in sequence.
What do these numbers represent?
You may need to remind students about the connection between the side length of a square and its area. Slides 14 and 15 provide two examples. Ask your students to complete the areas on their version of Copymaster One.
Why do the areas increase so rapidly?
Two things contribute to the rapid growth in area. The Fibonacci numbers themselves grow non-linearly (greater and greater differences) and those numbers are being multiplied by themselves.
For early finishers, pose this problem (see Slide 16):
Here are the first few Fibonacci numbers:
n 1 2 3 4 5 6 … Fibonacci Number 1 1 2 3 5 8 …
There are two Fibonacci numbers that are equal to the square of their place (n). Ask students to identify them.
You may need to illustrate some non-examples, such as:
The 6th Fibonacci number is 8. 6 x 6 = 36 so the sixth Fibonacci number is not six squared.
Both the first and twelfth Fibonacci numbers, 1 and 144, are the square of their place (n). 144 is the 12th Fibonacci number, and 12 x 12 = 144 (122 = 144).
#### Lesson Two
In the next two lessons students explore Fibonacci numbers in context and some of the patterns that exist with numbers in the sequence.
Introduce the Property Developers Problem using PowerPoint Two. For small numbers of sections in a row, it is easy to find all the possible layouts. For example, there are five layouts for a row of four sections (see Slide Five). Look for students to be systematic about finding all the possible layouts, so none are omitted. Copymaster Two can be used so students can cut and paste the layouts in their workbooks or on a poster.
Let’s organise the data we have in a table. Call on students to fill in the cells for numbers of layouts.
Number of sections 1 2 3 4 5 6 7 Number of layouts 1 2 3 5 ? ? ?
Does anything look familiar in the way the number of layouts is growing?
Students should notice that the pattern is the Fibonacci sequence.
Is there a way to check that the sequence works for any number of sections?
Students might suggest testing to see if the numbers of layouts for 5, 6, and 7 sections are Fibonacci numbers. This will give evidence that the pattern continues but not conclusive proof that it holds for any number of sections. You might use code to list all the layouts for 5, 6, and 7 sections.
5 sections (Begin with all that has S first, then move to D first)
• S, S, S, S, S.
• S, D, S, S.
• S, S, D, S.
• S, S, S, D.
• S, D, D.
• D, S, S, S.
• D, D, S.
• D, S, D.
6 Sections
• S, S, S, S, S, S.
• S, S, D, S, S.
• S, S, S, D, S.
• S, S, S, S, D.
• S, S, D, D.
• S, D, S, S, S.
• S, D, D, S.
• S, D, S, D.
• D, S, S, S, S.
• D, D, S, S.
• D, S, S, D, S.
• D, S, S, S, D.
• D, S, D, D.
Listing gets increasingly cumbersome with more sections. Support the attendance of your students on why the Fibonacci numbers work in this case.
Slide Six has a systematic listing of layouts to four sections.
Can you use the listings for three, and four sections to work out how many different layouts are possible for five sections?
Let students discuss how they might predict the number of layouts for five sections. Slide Six animates how a systematic listing may be done.
What two ways can a layout start? (S or D)
Suppose we start a six-section layout with S. How can we find all the layouts that start with S? (Go to all the five section layouts since there are five sections to fill)
How can we find all the layouts that start with D? (Go to all the four section layout since there are four sections to fill)
How many six-section layouts are possible? (8 + 5 = 13)
Challenge your students to write down a convincing argument about how to find the number of different layouts for any number of sections. Look for the following:
• Do they say anything about the importance for the first section layouts?
• Do they specify how the two previous numbers of layouts help to create the number of layouts for a given number of sections? (e.g. To find the number of 10 section layouts go to the numbers of layouts for 8 and 9 sections)
• Do they explain why the two previous numbers of layouts are helpful?
Finish the lesson with the Assembly Problem which is similar but subtly different to the housing layout problem. If allocating 1 to a talker and 0 to a listener means the problem can be represented using strings of 1s and 0s. For three seats there are eight possible arrangements:
111, (three talkers)
110, 101, 011, (two talkers, one listener)
100, 010, 001, one talker, two listeners)
000 (three listeners)
The underlined arrangements are not allowed as there are two talkers together. There are five possible arrangements. Five is a Fibonacci number.
If there are four seats, then either a 1 or 0 can be used as the first seat. If 0 is used, then all five three seat arrangements can be ‘tacked on.’ If 1 is used, then only the three-seat arrangement starting with 0 can be used. There are three of those arrangements, 010, 001, and 000. In total 5 + 3 = 8 arrangements are possible with four seats. 3, 5 and 8 are Fibonacci numbers.
#### Lesson Three
In this lesson students explore some number patterns in the Fibonacci sequence. Use PowerPoint Three to drive the lesson. Begin with displaying the sequence using Slide One. Use mouse clicks to ask students to anticipate the next numbers in sequence.
Which of these Fibonacci numbers are even?
Is there a pattern to where they occur?
Continue the sequence to see if your predictions are correct.
A second row of Fibonacci numbers appear when you click the mouse. Students can check their sequence and confirm whether their prediction about when even numbers occur.
Can you explain why even numbers occur every third Fibonacci number?
Look for students to notice that the sequence goes:
Odd, odd, even, odd, odd, even, odd, odd, even, …
If two odd numbers occur consecutively the third number must be even since two odd numbers add to an even number. The next two numbers must be odd since an odd and even number make an odd sum.
Work through the second slide that shows a pattern with sets of four consecutive Fibonacci numbers. The example uses the numbers, 2, 3, 5, and 8.
2 + 8 = 10 and 10 ÷ 2 = 5. The sum is five, the third number in the set of four.
Is that just lucky, or does it happen all the time?
Let students investigate the pattern with other sets of four Fibonacci numbers. Here are some examples.
Fibonacci numbers Sum Sum divided by two 2, 3, 5, 8 2 + 8 = 10 10 ÷ 2 = 5 3, 5, 8, 13 3 + 13 = 16 16 ÷ 2 = 8 5, 8, 13, 21 5 + 21 = 26 26 ÷ 2 = 13 8, 13, 21, 34 8 + 34 = 42 42 ÷ 2 = 21
Challenge your students to explain why the pattern occurs. An informal proof is possible if students assign arbitrary names like a, and b to the first two numbers in the set of four.
If the set of four numbers start with a and b, what will the third and fourth numbers be? (a + b, and a + 2b, since the sequence is produced by adding consecutive terms to get the next)
What do we get if we add the first and last terms, a + (a + 2b)? The sum is 2a + 2b, or two lots of a + b, 2 (a + b).
If 2a + 2b is divided by two, what do you get? (a + b)
Which number is the set of four is that?
For student who find the use of variables difficult, you might try examples where the numbers are known. Cuisenaire rods or stack of connecting cubes can act as a model. Slides 3 and 4 show how specific examples can be modelled to help students to see the structure of what occurs. They might then try to describe that structure.
Slide Five introduces a different pattern with four consecutive Fibonacci numbers. The example uses the numbers, 2, 3, 5, 8. 2 x 8 = 16 and 3 x 5 = 15. There is a difference of one.
Is there always a difference of one, no matter which four consecutive Fibonacci numbers you choose?
Let your students investigate whether the pattern works for other sets of our consecutive numbers. It does though the differences oscillate between positive and negative. Here are some examples:
Fibonacci numbers Products Difference 2, 3, 5, 8 2 x 8 = 16, 3 x 5 = 15 -1 3, 5, 8, 13 3 x 13 = 39, 5 x 8 = 40 +1 5, 8, 13, 21 5 x 21 = 105, 8 x 13 = 104 -1 8, 13, 21, 34 8 x 34 = 272, 13 x 21 = 273 +1
An explanation of why the pattern occurs is beyond students at this level. Using a and b as the first two Fibonacci numbers and a + b, and a + 2b as the third and fourth requires algebra that is normally accessible at Level 6.
#### Lesson Four
Pattern One
In lesson four students investigate patterns that occur when Fibonacci numbers are squared. They may not be aware that they did this in Lesson One after constructing the Fibonacci spiral. Slide One of PowerPoint Four reminds them of this. Slides One and Two ask students to create a table of squares. You may need to discuss the notation, n is the ordinal number, Fn is the nth Fibonacci number, and (Fn)2 is the square of Fn. Copymaster Three can be used to speed this process up.
Look at the table. Try adding consecutive pairs of squares. For example, 4 and 9 are consecutive. 4 + 9 = 13.
What do you notice?
After a suitable period bring the students back together to share their findings. They should notice that each sum is a Fibonacci number. Slide Three shows those results.
That’s interesting but can you be more specific. Let’s look more carefully.
Can you predict where the Fibonacci can be found?
Let’s take the sum of two squared and three squared. What Fibonacci number is thirteen? (Thirteen is the 7th Fibonacci number so can be written as F7.)
It is the sum of squares of what two Fibonacci numbers? (22 + 32 = 13, and 2 is F3 and 3 is F4).
Write:
(F3)2 + (F4)2 = F7
What do you notice? Does this happen all the time?
Check some other examples.
Students should find several examples and conclude that “If you know the first Fibonacci number (Fn) then the sum of the two squares is Fn + Fn+1 (or F2n+1).”
You might write this as (Fn)2 + (Fn+1)2 = Fn+n+1. Proving the theorem takes algebra that is too complex for this level.
Pattern Two
Slides Four to Seven reintroduce the students to the arrangement of rectangles they used to create the Fibonacci spiral. They are asked to find the area of given rectangles. The area of the rectangles can be found by summing the areas of the squares within them.
Record the results in sequential order like this:
1 + 1 + 4 + 9 + 25 = 40
1 + 1 + 4 + 9 + 25 + 64 = 104
1 + 1 + 4 + 9 + 25 + 64 + 169 = 273
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 = 714
Return to Slides Eight and Nine of PowerPoint Four to pose the problem of finding side lengths. To do so students will have work backwards from the areas to find the side lengths. Using the square root function on the calculator is one way to do this. For example, a square with area of 169 has side lengths of 13 since 169= 13. The side lengths of both rectangles are Fibonacci numbers since they are the sum of consecutive Fibonacci numbers.
For example, to find the side lengths of this rectangle students must first find the side lengths of the internal squares. Using 64= 8 and 25= 5 gives side lengths of 8 and 8 + 5 = 13. The area of the rectangle equals 8 x 13 = 104 square units.
The side lengths for the larger rectangle on Slide are 34 and 34 + 21 = 55, using 1156= 34 and 441= 21. Both 34 and 55 are Fibonacci numbers and the total area is 34 x 55 = 1870. Introduce this new information to the sequence you started previously:
1 + 1 + 4 + 9 + 25 = 40 = 5 x 8
1 + 1 + 4 + 9 + 25 + 64 = 104
1 + 1 + 4 + 9 + 25 + 64 + 169 = 273
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 = 714
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 + 1 156 = 1870 = 34 x 55
Ask students what they notice. Expect attention to both the addends and factors:
The squares add up to multiplication of Fibonacci numbers.
Copymaster Three requires students to complete the middle equations and add to the set above and below. The aim is for students to get more specific about what they notice. Completed patterns might look like this:
1 + 1 + 4 + 9 = 13 = 3 x 5
1 + 1 + 4 + 9 + 25 = 40 = 5 x 8
1 + 1 + 4 + 9 + 25 + 64 = 104 = 8 x 13
1 + 1 + 4 + 9 + 25 + 64 + 169 = 273 = 13 x 21
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 = 714 = 21 x 34
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 + 1 156 = 1 870 = 34 x 55
1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 + 1 156 + 3025 = 4 895 = 55 x 89
Ask the students to attempt a rule for the pattern. You make need to support them with introducing series notation, such as:
(F1)2 + (F2)2 + (F3)2 +… + (Fn)2 to represent the sum of squares up to the square of the nth Fibonacci number. Specific examples might help the student to define the factors that combine to the sum.
12 + 1+ 22 + 32 + 52 + 82 = 104 = 8 x 13
12 + 12 + 22 + 32 + 52 + 82 + 132 + 212 = 714 = 21 x 34
The first factor is always Fn and the second factor is Fn+1, the next Fibonacci number after it. The theorem can be written as:
(F1)2 + (F2)2 + (F3)2 +… + (Fn)2 = Fn × F(n+1)
There are many other number patterns arising from the Fibonacci sequence. The units titled develop conjectures around such patterns.
#### Lesson Five
PowerPoint Five has a final problem set for the students to attempt independently or in small groups. The context of a telephone tree will be unfamiliar, but they should notice the Fibonacci sequence occur as the number of new people are added with each minute. Continuing the tree will be very cumbersome so students should be encouraged to use number pattern.
Minutes 0 1 2 3 4 5 6 7 8 … New People 1 1 2 3 5 8 13 21 34 …
Note that the sequence begins at minute zero, so the Fibonacci numbers are Fn+1, that is, one term along. Slide Four of PowerPoint Five has a continuation of the tree to support students who find drawing it difficult.
The answer is found by summing the Fibonacci numbers until 143 is reached or exceeded. That occurs when nine minutes elapse.
Discuss why the Fibonacci sequence is involved. Through drawing the tree students will see that the conditions for the sequence apply:
1. It begins with 1, 1.
2. Callers from both the previous layers of tree contribute one new person.
Students might be interested that the sum of consecutive Fibonacci numbers to any n is always Fn+2 – 1. They might also be interested to note that all the rectangles in the spiral diagram are very similar, that is, they are scaled copies of one another. Slide Five shows some rectangles in sequence of size. Note that the sides are always consecutive Fibonacci numbers. The rectangles have similar side ratios, that is the long side divided by the short side. Slide Six has the beginnings of a table that students might continue. The ratio converges on a special number 1.61803398875, which is known as the golden ratio, and represented by the Greek letter phi, φ .
Slide Seven shows the Parthenon, in Athens, with a superimposed Fibonacci spiral. Some students might like to investigate the connection between Fibonacci numbers and the golden ratio further.
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# 2. STATISTICS: REGRESSION AND CORRELATION
Introduction
The work conducted by fishery biologists generally requires a fair amount of statistical analysis and most courses in fishery biology therefore include elementary statistics, at least.
Most often, however, lack of practice causes one to forget what was learnt, which results in a very valuable tool remaining underutilized.
This note aims at briefly recalling two very powerful statistical techniques - regression and correlation analysis - and to indicate some of their most common fields of application by fishery biologists.
Linear Regression
Put simply, linear regression is a technique for quantifying the relationship that can be seen when a scatter diagram involving two variables is drawn (Figure la), which relationship being summarized by a “best fitting” equation of the form:
y = a + bx (1)
In this equation, y represents the coordinate values along the vertical axis of the graph (ordinate), while x represents the coordinate values along the horizontal axis (absissa). The value of “a” (which can be negative, positive or equal to zero) is called the intercept, while the value of b (which can be negative or positive) is called the slope or regression coefficient.
Table 1
Data set for calculating a regression (a and b) and correlation coefficient (r)
Numberx-valuesy-valuesNumberx-valuesy-values
19.00.50 7 6.71.00
29.40.50 8 8.40.50
37.41.23 9 8.00.50
49.71.001010.00.50
510.40.3011 9.20.50
65.01.5012 6.21.00
13 7.70.50
The procedure to obtain values of a and b for a given set of y and x data pairs (such as in Figure 1 and/or Table 1) is as follows:
Step 1 Compute, for each pair of y, x values the quantities x², y² and x.y. Step 2 Compute the sums (Σ) of these quantities for all x, y data pairs, along with the sums of the x and y values. The results of Steps 1 and 2 should look similar to this:
Number of data pairsxyx.y
1
2
3
.
.
.
n
Names of sums
ΣxΣx²ΣyΣy²Σx·y
Step 3 Estimate the slope (b) by means of the relationship
Step 4 Estimate the intercept (a) by means of the relationship
Using values of “a” and “b” obtained by means of Equations 2 and 3, one then can draw through the points of a scatter diagram the best fitting straight line and visually assess if the points are well “explained” by the line (Figure 1b).
Correlation
Correlation analysis is closely related to regression analysis and both can be viewed, in fact as two aspects of the same thing.
The correlation between two variables, is, again put in the simplest terms, the degree of association between two variables. This degree of association is expressed by a single value called a correlation coefficient (r), which can take values ranging between -1 and +1. When r is negative, it means that one variable (either x or y) tends to decrease as the other increases - there is a “negative correlation” (corresponding to a negative value of b in regression analysis). When r is positive, on the other hand, it means that the one variable increases with the other (which corresponds to a positive value of b in regression analysis).
Values of r are easily computed for a set of x, y data pairs, using the same table and sums as shown in Step 2 of the “regression” section of this note. Thus r can then be obtained - indirectly - from the relationship
Figure 1a A scatter diagram (or scattergram) of x, y values. Note that y generally decreases as x increases, suggesting negative regression and correlation coefficients (based in Table 1)
Figure 1b Same data as in la, but fitted with the regression y = 2.16 - 0.173, with r = -0.756
which provides a value of the “coefficient of determination” (= r²). All we need is then to compute
that is to take the square root of the coefficient of determination to obtain the (absolute) value of r, and then to add the sign (+ or -) depending on whether the correlation is positive or negative (which can be assessed by visual inspection of a scattergram or by computing the b value of the corresponding regression and using for r the sign of b).
When we compute values of r, we would also like to know, however, whether the correlation that was identified could have arisen by chance alone. This can be established by testing whether the computed value of r is “significant” that is whether the (absolute) value of r is higher than, or equal to a “critical” value of r as given in a statistical table (see table of critical values of r in Appendix 1).
Exercise: Compute a, b and r for the data given in Table 1 and test, by means of the table in Appendix 1 whether the computed value of r is significant at P = 0.01 and P = 0.05.
Linearizing Transformation in Regression Analysis
Both the regression and correlation analysis, as outlined above are based on the assumption of a “linear” relationship between the two variables involved (meaning that the best fitting line is straight). There are many cases in fishery biology, however, where the relationship between two variables is non-linear, and a well known example for this is the length-weight relationship, where
W = α · Lb (6)
where the weight (W) is proportional to a certain power (b) of the length (L) (see Figure 2a).
Length-weight data can, however, be fitted with a (linear) regression if logarithms are taken of both sides, resulting in
log10 W = a + b log10L (7)
As may be seen from Figure 2b, the logarithm of the length and weight are fitted extremely well by a linear regression, where
y = log10W (8a)
and
x = log10L (8b)
Thus, fitting a length weight relationship of the form given in Expression 6 to a set of length/weight data (such as given in Table 2) consists of the following:
Table 2
Data for the estimation of a length-weight relationship in the threadfin bream Nemipterus marginatus1
NumberTL (cm)W (g)Log10 L (=x)Log10 W (=y)
1 8.1 6.30.9080.799
2 9.1 9.60.9590.982
310.211.61.0091.064
411.918.51.0761.267
512.226.21.0861.425
613.836.11.1401.558
714.840.11.1701.603
815.747.31.1961.675
916.665.61.2201.817
1017.769.41.2481.841
1118.776.41.2721.883
1219.082.51.2791.916
1320.6106.61.3142.028
1421.9119.81.3402.078
1522.9169.21.3602.228
1623.5173.31.3712.239
Step 1 Take the logarithm of the length and weight values. Step 2 Compute the sums given in the regression section, with x and y values as defined in 8a and 8b. Step 3 Compute a and b using Equations 2 and 3. Step 4 Take the antilogarithm of a to obtain α in Equation 6. Step 5 Write your version of Equation 6. Step 6 Using the sums computed in Step 2, compute the value of r² and r, and check significance. Exercise: (a) Perform Steps 1 to 6 (with P = 0.01) for the length-weight data given in Table 2. (b) List other linearizing transformations, and give examples of their use in fishery biology.
Figure 2a Length-weight relationship of Nemipterus marginatus in the South China Sea (based on data in Table 2)
Figure 2b The same data converted to base 10 logarithms |
# 6.7 Using the Fundamental Theorem of Algebra
## Presentation on theme: "6.7 Using the Fundamental Theorem of Algebra"— Presentation transcript:
6.7 Using the Fundamental Theorem of Algebra
What is the fundamental theorem of Algebra? What methods do you use to find the zeros of a polynomial function? How do you use zeros to write a polynomial function?
German mathematician Carl Friedrich Gauss ( ) first proved this theorem. It is the Fundamental Theorem of Algebra. If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one root in the set of complex numbers.
Solve each polynomial equation
Solve each polynomial equation. State how many solutions the equation has and classify each as rational, irrational or imaginary. x = ½, 1 sol, rational 2x −1 = 0 x2 −2 = 0 x3 − 1 = 0 (x −1)(x2 + x + 1), x = 1 and use Quadratic formula for
Solve the Polynomial Equation.
x3 + x2 −x − 1 = 0 Notice that −1 is a solution two times. This is called a repeated solution, repeated zero, or a double root. 1 −1 −1 1 1 2 1 1 2 1 x2 + 2x + 1 (x + 1)(x + 1) x = −1, x = −1, x = 1
Finding the Number of Solutions or Zeros
x +3 x3 + 3x2 + 16x + 48 = 0 (x + 3)(x2 + 16)= 0 x + 3 = 0, x = 0 x = −3, x2 = −16 x = − 3, x = ± 4i x2 x3 3x2 16x 48 +16
Finding the Number of Solutions or Zeros
f(x) = x4 + 6x3 + 12x2 + 8x f(x)= x(x3 + 6x2 +12x + 8) 8/1= ±8/1, ±4/1, ±2/1, ±1/1 Synthetic division x3 + 6x2 +12x + 8 Zeros: −2,−2,−2, 0
Finding the Zeros of a Polynomial Function
Find all the zeros of f(x) = x5 − 2x4 + 8x2 − 13x + 6 Possible rational zeros: ±6, ±3, ±2, ±1 1 − − 1 1 −1 −1 7 −6 1 −1 −1 7 −6 −2 −2 6 −10 6 1 −3 5 −3 1 1 −2 3 1 −2 3 x2 −2x + 3 Use quadratic formula
Graph of polynomial function
Turn to page 367 in your book. Real zero: where the graph crosses the x-axis. Repeated zero: where graph touches x-axis.
Using Zeros to Write Polynomial Functions
Write a polynomial function f of least degree that has real coefficients, a leading coefficient of 1, and 2 and 1 + i as zeros. x = 2, x = 1 + i, AND x = 1 − i. Complex conjugates always travel in pairs. f(x) = (x − 2)[x − (1 + i )][x − (1 − i )] f(x) = (x − 2)[(x − 1) − i ][(x − 1) + i ] f(x) = (x − 2)[(x − 1)2 − i2 ] f(x) = (x − 2)[(x2 − 2x + 1 −(−1)] f(x) = (x − 2)[x2 − 2x + 2] f(x) = x3 − 2x2 +2x − 2x2 +4x − 4 f(x) = x3 − 4x2 +6x − 4
What is the fundamental theorem of Algebra?
If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one root in the set of complex numbers. What methods do you use to find the zeros of a polynomial function? Rational zero theorem (6.6) and synthetic division. How do you use zeros to write a polynomial function? If x = #, it becomes a factor (x ± #). Multiply factors together to find the equation.
Assignment is p. 369, 15-29 odd, odd Show your work
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Using Mathematics to Win the Lottery
Lottery Basics
Many of you are probably familiar how lottery works. A lottery is a game where a smaller group of numbers is chosen from a larger group. If you bet on the right combination, you win the jackpot prize, which is usually staggering.
Although there is a common concept about lottery, there are variations in different places or countries. In this post, I will use ours as an example. In the Philippines, as of this writing, we have three types of lottery: 6/42, 6/45 and 6/49. Yes, you guessed it right, 6/42 means 6 numbers are randomly chosen from a set of numbers from 1 through 42. We use the 6/42 lottery in the following discussion.
In our country, the process of choosing numbers in a lottery is shown on television. First, balls of the same weight numbered from 1 through 42 are placed in a machine-operated transparent container. Next, the balls are mixed using a blower located at the bottom of the container. These balls are light enough to be floated by the air from the blower. Finally to determine the winning combination, a button is pushed to eject 6 balls, one after the other. See video below.
Now, suppose we bought a ticket of the 6/42 lottery, what is our chance of winning? Is there a way that we can win for sure?
We will answer the first question later. For the second question, yes, there is a way and, in principle, it’s simple. Just bet on all the possible number combinations.
Okay, so you’re not a math major and you do not know what I am talking about. Well, let me explain it in plain language. Suppose a mini-lottery has 5 numbers, say 1 through 5, and two numbers are drawn to determine the winner. The pair, {1,2} is a possible bet as well as {4,3}. It is clear that the order of the numbers chosen does not matter, so betting on {5,2} is the same as betting on {2,5}.
Finding all the possible pairs or combinations in the 2/5 mini-lottery is a bit easy. As we can see from the table, if we choose two numbers from 5 choices, we have 10 possible combinations or pairs (see red text). The pairs in black texts are just the reverse of those in red. We did not also include the pair having the same number, say (2,2), because in a lottery draw, drawn balls are not returned back into the container before drawing the next ball.
If we buy 10 tickets and bet on all the 10 pairs on the table, surely, we will win. We will use the same principle in winning the 6/42 lottery draw.
Systematic Betting
It is very hard to use the strategy above if we have six numbers taken from 42. There is no easy way to tally our bets. However, we can use a computer in finding all the possible number combinations. The method below maybe done:
1. We can start with 1,2,3,4,5,6, then 1,2,3,4,5, 7, and then 1,2,3,4,5,6,8 all the way up to 1, 2, 3, 4, 5, 42.
2. After that, we can increase the fifth digit by 1 and we start with 1, 2, 3, 4, 6, 7, and then 1, 2, 3, 4, 6, 8, and so on. We can do this all the way to 1, 2, 3, 4, 6, 42.
3. When our sixth digit reached 42, we now increase the fifth digit by 1; that is, 1, 2, 3, 4, 7, 8; 1, 2, 3, 4, 7, 9 all the way up to 1, 2, 3, 4, 5, 7, 42.
4. Every time we increase the fifth digit by 1, we get the sixth digit by adding the fifth digit by 1, and then keep adding 1 until the sixth digit reached 42. For instance, 1, 2, 3, 4, 8, 9; 1, 2, 3, 4, 8, 10 all the way up to 1, 2, 3, 4, 8, 42. In varying the fifth digit, our last number would be 1, 2, 3, 4, 41, 42.
5. After the last combination is finished, we change the third digit: 1, 2, 4, 5, 6, 7 and so on…
Exercise:
Another mini-lottery has 5 balls and 3 balls to be drawn. List all the possible combinations. How many tickets should we buy for a sure win?
Winning the 6/42 lottery
In the 2/5 mini-lottery example above, we have two numbers chosen from a group of 5 numbers. This is equivalent to the combination of 5 numbers taken 2 at time. We have learned from the Introduction to Combination post that 5 taken 2 at a time is equal to $\frac{5!}{(5-2)!2!} = 10$, and that can be verified from the table above. In the 6/42 lottery, 42 taken 6 is equal to $\frac{42!}{(42-6!)6!} =5,245,786$. This means if we want a sure win, we must buy 5,245,786 tickets.
The Difficulties of Winning
There are, however, several difficulties in getting our fortune and becoming a millionaire, although it is definitely possible. Some of them are enumerated below.
1. Ticket Price. One 6/42 lottery ticket is 10 pesos as of this writing (note that it may increase in the future). This means that today we will need 52,457,860 pesos (about 1,165,000 US dollars) to buy all the tickets. Now you really have to be quite rich to have that kind of money.
2. Buying the Tickets. Assuming that we have the money, we also have a problem where to buy 5 million tickets. Suppose each lottery outlet has 5,000 tickets, we need to go to 1000 lottery outlets to buy all the tickets we need.
3. Time Filling Out the Tickets. If you are going to fill out 5,245,786 tickets alone, then you will probably not be able to do it. Assuming that you can fill out 1 ticket in 1 minute, then 5,245,786 minutes is equal to 3642.9 days or nearly 10 years: that is, without doing anything. If you are going to spend 8 hours a day filling out the ticket, like a regular working employee, you will triple the time above. You will finish filling the tickets out in 30 years!
4. Hiring People to Fill out the Tickets. Suppose we hire 100 people to solve problem 3, then, the 100 people will share the 3642.9 days x 24 hours/day = 87,429.6 hours. This means that each of the 100 person will work for about 874.3 hours, or about, 110 days in an 8-hour work day. So, if the salary of 1 person is Php450 a day, then, the money you need to pay them is 450/day x 110 days/person x 100 persons = Php4.9 million pesos. Well, it is not as expensive as the tickets, so we can afford it.
5. Number Combinations. We should also have a way of organizing our number combinations. Note that in the five million plus tickets, we must fill out our numbers accurately and without duplicates. The mini-lottery above can be easily listed, but we will need a computer program to enumerate the 5,245,786 combinations. Well, we can definitely afford a programmer since we are ready to spend more than 50 million pesos.
6. Prize money. Needless to say, if we want to spend 52 million pesos, the jackpot prize should be more than that.
7. More than 1 Jackpot prize winners. Even if the prize money is more than 50 million, we still have one problem. If another person wins, we are in big trouble. If the jackpot prize, for instance, is 100 million pesos, and there are three winners, each of the winners will only bag 33 million pesos, and we are 19 million pesos short.
Now, you probably understand why many mathematicians do not bother to buy lottery tickets, despite the fact that they know how to win. The answer to our second question above is that the chance of winning is 1 in 5,245,786.
What does that mean?
That means that if you buy a single 6-42 ticket and bet randomly 5,245,786 times, it is likely that you will win only once (well technically, that is a bit wrong, but I cannot use a better analogy). And, if the chance is so small, why is there that there is always a winner in almost every draw?
The Chance of Winning
Millions of people bet on lottery in each draw, so it is likely, that one or more would win. If 3 million people bet different combinations in one draw, that is equivalent to one person betting 3 million times which results to about 60% probability of winning. That is the reason that most of the time, there are lottery winners.
It has been done
Despite the numerous constraints, in 1995(?), two computer scientists from United States flew to Australia just to bet on a lottery (I saw the documentary several years ago). They used a computer to systematically bet on all combinations. One employee from a lottery outlet was surprised when they told him that they were going to buy 100,000 tickets. The problem was they didn’t have enough tickets, so they had not bet on all possible number combinations.
They won anyway.
There is no secret to win a lottery.
There are websites that lure and convince people that they have apps or formulas that increase the chance of winning a lottery. These websites are definitely FAKE. Any high school student with good understanding of basic probability will tell you the same thing.
The only secret to increase your chance of winning is to buy more tickets. The more tickets you buy, the higher your chance of winning. In the 6-42 lottery above, if you are going to buy 2, 622, 893 tickets, then you have a 50% chance of winning. That means that in two bets, you are likely to win one jackpot prize.
***
Notes:
1. I am not really sure the process of lottery betting done in other countries, but here, first we buy tickets, then fill up our numbers of choice, then return it to the lottery outlet for processing. They place the ticket in a machine, and the machine ejects the receipt containing our number combinations. The receipt will be our official ticket.
2. Prices and exchange rate discussed in this article are based on the time this article was written. They may change in the near future.
3. Photos: Money by AMagill, The Thinker by Andrew Horne |
# What are Consecutive Numbers – Consecutive Numbers Example
## What are Consecutive Numbers ?
What are Consecutive Numbers ? This question often comes to the mind of Maths students in early classes. Consecutive numbers are numbers, that follow each other , in the order of smallest to largest.
In other words, we can say that consecutive number of any given number is obtained by adding 1 to that number.
### Consecutive numbers Examples
Write the three consecutive numbers of the following numbers:
#### Example 1
25367
Explanation : –
The next three consecutive number of 25367 can be obtained by adding 1, 2 and 3 to 25367
that is,
25367 + 1 = 25368
25367 + 2 = 25369
25367 + 3 = 25370
Hence the three consecutive number occurring just after 25367 are 25368, 25369 and 25370 respectively.
#### Example 2
36135
Explanation:-
The next three consecutive number of 36135 can be obtained by adding 1, 2 and 3 to 36135
that is,
36135 + 1 = 36136
36135 + 2 = 36137
36135 + 3 = 36138
Hence the three consecutive number occuring just after 36135 are 36136 , 36137 and 36138 respectively.
#### Example 3
46903
Explanation:-
The next three consecutive number of 46903 can be obtained by adding 1, 2 and 3 to 46903
that is,
46903 + 1 = 46904
46903 + 2 = 46905
46903 + 3 = 46906
Hence the three consecutive number occuring just after 46903 are 46904 , 46905 and 46906 respectively.
#### Example 4
57671
Explanation:-
The next three consecutive number of 57671 can be obtained by adding 1, 2 and 3 to 57671
that is,
57671 + 1 = 57672
57671 + 2 = 57673
57671 + 3 = 57674
Hence the three consecutive number occuring just after 57671 are 57672 , 57673 and 57674 respectively.
#### Example 5
68439
Explanation:-
The next three consecutive number of 68439 can be obtained by adding 1, 2 and 3 to 68439
that is,
68439 + 1 = 68440
68439 + 2 = 68441
68439 + 3 = 68442
Hence the three consecutive number occuring just after 68439 are 68440 , 68441 and 68442 respectively. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Equations with Decimals
## Solve equations including decimal values
Estimated8 minsto complete
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Progress
Practice Equations with Decimals
Progress
Estimated8 minsto complete
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Equations with Decimals
Karen wants to design a garden for her back yard. She knows she only has space for a rectangular garden of a perimeter equal to 46.5 feet. She needs to know the dimensions. The width will be half the length. What will be the dimensions of the garden?
### Guidance
You can solve equations that contain decimals in the same way that you solve any equation. You can also multiply both sides of the equation by a multiple of 10 in order to get rid of the decimals if you prefer. Consider the equation . At first this looks difficult because of the decimals. But multiply all of the numbers by 10 and see what happens:
Now you can see that the answer is .
When you have decimals in an equation, you can get rid of them by multiplying by 10 (if one decimal place), 100 (if two decimal places), or 1000 (if three decimal places). Then, solve the equation as usual. Remember that if you feel confident with adding, subtracting, multiplying, and dividing decimals, you can also solve equations with decimals by isolating the variable and solving without first removing the decimals.
#### Example A
Solution: You can think about this problem with the balance method. You know that the two sides are equal so the balance has to stay horizontal. You can place each side of the equation on each side of the balance.
Like always, in order to solve the equation, you have to get the all by itself. Always remember that you need to keep the balance horizontal. This means that whatever you do to one side of the equation, you have to do to the other side.
Subtract 2.3 from both sides to get rid of the 2.3 on the left and isolate the variable.
If you simplify this expression, you get:
Therefore .
You can, as always, check your answer to see if you are correct.
#### Example B
Solution: This time try solving the problem using algebra tiles. Multiply this problem by 10 to get rid of the decimals.
Now you can use algebra tiles to solve for the variable.
Your first step is to add 16 to both sides of the equal sign.
Simplify and rearrange and you get:
Therefore .
Note: since the numbers were so large with this problem, the balance method is more efficient. This is true with most problems involving decimals. In fact, algebra tiles are rarely more efficient when solving problems with variables involving decimals.
#### Example C
Solution: You can again use the balance method to solve this problem.
Let’s first add 2.1 to both sides to get rid of the 2.1 on the left.
Simplifying you get:
Since 6.4 is multiplied by , you can get a by itself (or isolate it) by dividing by 6.4. Remember that whatever you do to one side, you have to do to the other.
If you simplify this expression, you get:
Therefore .
#### Concept Problem Revisited
Karen wants to design a garden for her back yard. She knows she only has space for a rectangular garden of a perimeter equal to 46.5 feet. She needs to know the dimensions. The width will be half the length. What will be the dimensions of the garden?
Therefore:
or
Since you know that Karen has 60 feet for her perimeter, you can substitute 46.5 in for .
Now you can solve for (the length).
You now know that the length is 15.5 feet. You also know that the width is the length. So you can solve for the width.
Therefore the dimensions of Karen’s garden are .
### Vocabulary
Equation
An equation is a mathematical statement with expressions separated by an equals sign.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is sometimes referred to as the literal coefficient.
### Guided Practice
1. Use a model to solve for the variable in the equation .
2. Use a model to solve for the variable in the equation .
3. Solve for x in the equation .
1.
You can use either method to solve this problem but let’s use algebra tiles for this one. Remember since there are decimals (to one decimal place to be exact), you first must multiply the equation by 10 to get rid of the decimals.
You now can add 14 to each side to isolate the variable.
Simplifying and rearranging leaves you with:
Therefore, .
2.
First you have to subtract 3.9 from both sides of the equation in order to start to isolate the variable.
Now, in order to get all by itself, you have to divide both sides by 1.2. This will isolate the variable
Therefore, rounded to the nearest hundredth, .
3.
You can use any method to solve this equation. Remember to isolate the variable. You will notice here that there are two values, one on each side of the equation. First combine these terms by adding to both sides of the equation.
Simplifying you get:
Now you can use any method to solve the equation. You now should just have to subtract 1.75 from both sides to isolate the variable.
Simplifying you get:
Now to solve for the variable, you need to divide both sides by 1.75.
You can now solve for .
### Practice
Use the balance model to solve for each of the following variables.
Use algebra tiles to solve for each of the following variables.
Use the rules that you have learned to solve for the variables in the following problems.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 2.4.
### Vocabulary Language: English
Equation
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
1. [1]^ License: CC BY-NC 3.0
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5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
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### Explore More
Sign in to explore more, including practice questions and solutions for Equations with Decimals. |
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# 1.5: Solving Equations
Difficulty Level: At Grade Created by: CK-12
## Introduction
The cookie sale was a much bigger success than either Cameron or Tracy had even thought. Within the first month, the students were selling out the inventory every single day. They were having such a difficult time keeping up, that they hired the home economics classes to help them with the baking. So some of the money went to the students who agreed to bake, but the rest of it went into the student council.
By midterms, the students calculated that after paying the home economics students, that they were still averaging $60.00 profit per week. By midterms, they had collected$540.00 total.
“How many weeks did it take us to make that much?” Jesse asked at lunch one day.
“I don’t know, but I am sure that we can make $1000.00 by end of the semester,” Tracy said smiling. “How can you be so sure?” “You just do the math. First, we can write an equation and solve it to figure out how long it took us to make the$540.00. Then double it for the end of the semester,” Tracy explained.
In the last lesson you worked with evaluating expressions. Now let’s look at writing and solving simple equations. These are equations that you can use mental math to solve. Pay close attention and you will see this problem again at the end of the lesson.
What You Will Learn
In this lesson you will learn how to complete the following skills.
• Solve single-variable equations using mental math.
• Check suggested solutions to equations by substituting for the variable and simplifying.
• Write and solve equations using verbal models for given problem situations.
• Model and solve real-world problems using simple whole-number equations.
Teaching Time
I. Solve Single-Variable Equations Using Mental Math
Like an expression, an equation includes groups of numbers, symbols, and variables. However, equations also include equal signs. The key thing to remember about an equation is that the quantity on one side of the equals must be the same as the quantity on the other side of the equals. There are different ways to solve an equation. In this lesson, you will learn to solve equations using mental math.
When you solve an equation, you are solving to determine the value of the variable. If you choose the correct value for the variable, then the equation will be a true statement. Let’s look at an example of an equation without a variable.
Example
\begin{align*}4+16=20\end{align*}
We can look at the quantity on the left side of the equation first. It is equal to 20. The right side of the equation is also 20. This is a true statement.
An equation must always make a true statement. We can say that this is a balanced equation.
What if this equation had a variable in place of one of the numbers?
Example
\begin{align*}x+16=20\end{align*}
Now we have a puzzle to solve. We can start by thinking about what number plus sixteen is equal to 20. We know that four plus sixteen is equal to 20. So, the value of \begin{align*}x\end{align*} must be four.
We write the answer to an equation in a particular way.
The answer is that \begin{align*}x=4\end{align*}.
Think a little deeper about how you solved this. If you think about it you probably subtracted 20 – 16 in your head. This is called using an inverse operation. The inverse operation is the opposite operation. We can use inverse operations to solve equations.
Example
\begin{align*}4x=12\end{align*}
Here we have a multiplication problem. We can ask ourselves, what number times four is equal to 12? The answer is 3.
\begin{align*}x=3\end{align*}
We could also use the inverse operation to solve this. Twelve divided by four is three. Our answer is the same and both methods can be completed using mental math.
Example
\begin{align*}\frac{x}{2}= 7\end{align*}
Now we have a division problem. We have a number divided by two is equal to seven. To solve this one, we can use the inverse operation. Two times seven is fourteen.
\begin{align*}x=14\end{align*}
Example
\begin{align*}\frac{22}{x}=11\end{align*}
This one is tricky because the inverse operation won’t work. We can’t multiply 22 times 11 to get an answer that makes any sense. We want to figure out what number times 11 is 22 or 22 divided by what number is 11?
We can use the 11 times table to figure this out.
11, 22, 33, 44
\begin{align*}x=2\end{align*}
Sometimes, you will have problems that are a bit more challenging. You can still complete these using mental math. You will just have to think of two operations and not one.
Example
\begin{align*}5x+3=18\end{align*}
Let’s break down this equation by using saying it to ourselves.
“Five times some number plus three is equal to eighteen.” Now you can think through the five times table for an answer that makes sense.
5, 10, 15, 20
15 makes sense so that would make the variable equal to three since five times three is fifteen.
\begin{align*}x=3\end{align*}
Sometimes looking at a problem like this one can be hard to check to see if your answer is correct. This is when checking your answer is so important.
II. Check Suggested Solutions to Equations by Substituting for the Variable and Simplifying
A solution of an equation is the value for a variable that makes an equation true. For example, five is the solution in the equation \begin{align*}2 + x = 7\end{align*} because \begin{align*}2 + 5 = 7\end{align*}. You can check to see if the solution given for an equation is correct by substituting it for the variable. Let’s look at the last example from the last section.
Example
\begin{align*}5x+3=18\end{align*}
After solving this equation using mental math, we figure out the value of the variable is three. We can check this answer by substituting the value of the variable back into the original equation. Then we simplify it. If the equation makes a true statement, then we know that we have the correct answer.
\begin{align*}5(3) + 3 &= 18\\ 15 + 3 &= 18\\ 18 &= 18\end{align*}
This is a true statement so our work is accurate.
Sometimes you will be given a value and you will need to determine if it makes a true statement.
Example
Is 40 a solution of \begin{align*}15x + 15 = 615\end{align*}?
Step 1: Substitute 40 for the variable “\begin{align*}x\end{align*}.”
\begin{align*}15x + 15 &= 615\\ 15(40) + 15 &= 615\end{align*}
Step 2: Solve the equation using the standard order of operations.
\begin{align*}15(40) + 15 &= 615\\ 600 + 15 &= 615\\ 615 &= 615\end{align*}
In this case, when 40 is substituted for the variable, both sides of the equation equal 615. Therefore, 40 is a solution of \begin{align*}15x + 15 = 615\end{align*}.
Let’s look at an example where you will also need to use the order of operations.
Example
Is 6 a solution for \begin{align*}7m + 7 - 2m = 37\end{align*}?
Recall that 6 should be substituted for the variable in the equation. Then follow the order of operations to solve.
\begin{align*}7m + 7 - 2m &= 37\\ 7(6) + 7 - 2(6) &= 37 \ (\text{Substitute the variable and multiply})\\ 42 + 7 - 12 &= 37 \ (\text{Add})\\ 49 - 12 &= 37 \ (\text{Subtract})\\ 37 &= 37\end{align*}
When 6 is substituted for the variable, both sides of the equation equal 37. Therefore, 6 is a solution for the equation \begin{align*}7m + 7 - 2m = 37\end{align*}.
6 is a solution for this equation.
III. Write and Solve Equations using Verbal Models for Given Problem Situations
Sometimes if you think of a problem in terms of words and parts it will be easier to write an equation and solve it. Writing a verbal model is similar to making a plan for solving a problem. When you write a verbal model, you are paraphrasing the information stated in the problem. After writing a verbal model, insert the values from the problem to write an equation. Then, use mental math or an inverse operation to solve it.
Let’s look at an example.
Example
Monica purchased a pair of tennis shoes on sale for $65.99. The shoes were originally$99.00. Use a verbal model to write and solve an equation to determine the amount of money Monica saved by purchasing the shoes on sale
First write a verbal model to represent the problem.
Verbal Model: \begin{align*}\text{Sale Price} + \text{Amount Saved} = \text{Original Price}\end{align*}
Let “\begin{align*}s\end{align*}” represent the amount saved.
Equation: \begin{align*}65.99 + s = 99.00\end{align*}
Solution: Recall that to solve for “\begin{align*}s\end{align*},” complete the inverse operation. Since addition is used in the equation, use subtraction to solve.
It makes sense to subtract 65.99 from 99.00.
\begin{align*}99.00 - 65.99 = \33.01\end{align*}
Example
The cost to run a thirty second commercial on prime time television is seven hundred fifty-thousand dollars. Use a verbal model to write and solve an equation to determine the cost per second.
Verbal Model: \begin{align*}\frac{\text{Total Cost}}{\text{Number of Seconds}} = \text{Cost per Second}\end{align*}
Let “\begin{align*}x\end{align*}” represent the unknown cost per second.
Equation: \begin{align*}\frac{\750,000}{30} = x\end{align*}
Solution: To solve, divide 750,000 by 30.
\begin{align*}\frac{750,000}{30} &= x\\ 25,000 &= x\end{align*}
Now remember that we were talking about money in this problem. So our answer needs to be written as a money amount.
The answer is that is costs 25,000 per second for a thirty second commercial. IV. Model and Solve Real-World Problems using Simple Whole-Number Equations You can construct models to help you solve a problem. You have already learned to use a verbal model to write and solve equations to solve problems. Creating a model for a problem may also include methods such as drawing a diagram or picture or making a table or chart. Example The triangles below were constructed using toothpicks. Determine the number of toothpicks needed to construct twenty triangles. As you can see, three toothpicks were needed to construct one triangle. Two more were needed to construct the second triangle. Therefore, five toothpicks were used to make two triangles. Continue to make more triangles along the row. Each time you construct a new triangle, record the number of toothpicks used on a chart. Triangle #: Toothpick # 1 3 2 5 3 7 4 9 5 11 6 13 7 15 8 17 9 19 10 21 Looking at the table, you can identify a pattern. You can see that two toothpicks are needed each time a new triangle is constructed. You can write a verbal model to express this amount. Total Number of Toothpicks Needed = Two Times the Number of Triangles + One Toothpick Let \begin{align*}n =\end{align*} number of triangles Total Number of Toothpicks Needed \begin{align*}= 2n + 1\end{align*} To determine the number of toothpicks needed to construct twenty triangles, substitute twenty for the variable. \begin{align*}& 2n + 1\\ & 2(20) + 1\\ & 40 + 1\\ & 41\end{align*} 41 toothpicks are needed to construct twenty triangles. Now let’s go back to the problem from the introduction and work on applying what we have learned in this lesson. With our new knowledge, we can work on solving the introductory problem. ## Real-Life Example Completed Cookie Sale Success Here is the original problem once again. Reread it first. Then write an equation to solve for the number of weeks that it took the students to earn540.00. There are two parts to this problem, an equation and the number of weeks.
The cookie sale was a much bigger success than either Cameron or Tracy had even thought. Within the first month, the students were selling out the inventory every single day. They were having such a difficult time keeping up, that they hired the home economics classes to help them with the baking. So some of the money went to the students who agreed to bake, but the rest of it went into the student council.
By midterms, the students calculated that after paying the home economics students, that they were still averaging $60.00 profit per week. By midterms, they had collected$540.00 total.
“How many weeks did it take us to make that much?” Jesse asked at lunch one day.
“I don’t know, but I am sure that we can make $1000.00 by end of the semester,” Tracy said smiling. “How can you be so sure?” “You just do the math. First, we can write an equation and solve it to figure out how long it took us to make the$540.00. Then double it for the length of the semester.” Tracy explained.
Now write an equation and solve it using mental math.
Solution to Real – Life Example
To work on this problem, first we need to write an equation. Let’s look at what we know.
We know that the students averaged $60.00 profit per week. We know that their gross profit was$540.00.
We need to know how many weeks it took them to earn that. Our variable is the number of weeks, \begin{align*}w\end{align*}.
Here is our equation.
\begin{align*}60w=540\end{align*}
We can solve this using mental math.
It took the students 9 weeks to earn the money.
## Vocabulary
Here are the vocabulary words found in this lesson.
Equation
a group of numbers, operations and variables where the quantity on one side of the equal sign is the same as the quantity on the other side of the equal sign.
Inverse Operation
the opposite operation. Equation can often be solved by using an inverse operation.
Verbal Model
using words to decipher the mathematical information in a problem. An equation can often be written from a verbal model.
## Time to Practice
Directions: Solve each equation using mental math. Be sure to check each answer by substituting your solution back into the original problem. Then simplify to be sure that your equation is balanced.
1. \begin{align*}x+4=22\end{align*}
2. \begin{align*}y+8=30\end{align*}
3. \begin{align*}x-19=40\end{align*}
4. \begin{align*}12-x=9\end{align*}
5. \begin{align*}4x=24\end{align*}
6. \begin{align*}6x=36\end{align*}
7. \begin{align*}9x=81\end{align*}
8. \begin{align*}\frac{y}{5}=2\end{align*}
9. \begin{align*}\frac{a}{8} = 5\end{align*}
10. \begin{align*}\frac{12}{b}=6\end{align*}
11. \begin{align*}6x+3=27\end{align*}
12. \begin{align*}8y-2=54\end{align*}
13. \begin{align*}3b+12=30\end{align*}
14. \begin{align*}9y-7=65\end{align*}
15. \begin{align*}12a-5=31\end{align*}
16. \begin{align*}\frac{x}{2} + 4 = 8\end{align*}
17. \begin{align*}\frac{x}{4}+3=7\end{align*}
18. \begin{align*}\frac{10}{x}+9=14\end{align*}
19. \begin{align*}5a-12=33\end{align*}
20. \begin{align*}7b-9=33\end{align*}
Directions: Use what you have learned about writing verbal models to solve each equation. There are three parts to each problem.
1. Alexander’s resting heart rate is 75 beats per minute. This is thirty-beats less than his heart rate after exercising. Use a verbal model. Write an equation. Then solve the equation to determine Alexander’s heart rate after exercising.
2. The rectangles below were constructed using toothpicks. Four toothpicks were used to make one rectangle. Three more toothpicks were added to make a second rectangle. Determine the number of toothpicks needed to make fifteen rectangles in a row. Write a verbal model. Write an equation. Solve the equation for the solution.
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# INEQUALITIES IN ONE TRIANGLE
Theorem 1 :
If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.
Theorem 2 :
If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.
## Exterior Angle Inequality
The measure of an exterior angle of a triangle is greater than the measure of either of the two nonadjacent interior angles.
In the triangle above, according to theorem 3, we have
m∠1 > m∠A
m∠1 > m∠B
## Triangle Inequality
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In the triangle ABC above, according to theorem 4, we have
AB + BC > AC
AC + BC > AB
AB + AC > BC
## Examples
Example 1 :
Write the sides of the triangle shown below in order from least to greatest.
Solution :
In the triangle GHJ above, we have
m∠G < m∠H < m∠J
So, the smallest angle is m∠G and the largest angle m∠J.
In any triangle, the smallest angle is always across from the shortest side and the largest angle is always across from the longest side.
Hence, the order of sides of the triangle from least to greatest is
JH < JG < GH
Example 2 :
Write the angles of the triangle shown below in order from smallest to largest.
Solution :
In the triangle PQR above, we have
PQ < PR < QR
So, the shortest side is PQ and the longest side is QR.
In any triangle, the smallest angle is always across from the shortest side and the largest angle is always across from the longest side.
Hence, the order of angles of the triangle from smallest to largest is
m∠R < m∠Q < m∠P
Example 3 :
In the triangle shown below AB ≅ AC and BC > AB. What can we conclude about the angles in triangle ABC ?
Because AB ≅ AC, triangle ABC is isosceles.
So, we have
m∠B ≅ m∠C
Therefore, m∠B = m∠C.
Because BC > AB, m∠A > m∠C by Theorem 1.
By substitution, m∠A > m∠B.
In addition, you can conclude that m∠A > 60°. m∠B < 60° and m∠C < 60°.
Example 4 :
Construct a triangle with the given group of side lengths, if possible.
2 cm, 2 cm, 5 cm
Solution :
We can not construct a triangle with the given side lengths.
Because, sum of the lengths of any two sides of a triangle must be greater than the third side by Theorem 4.
In the given side lengths, we have
2 + 2 < 5
(Does not satisfy the theorem)
The diagram given below illustrates that a triangle can not be constructed with the given side lengths.
Example 5 :
Construct a triangle with the given group of side lengths, if possible.
3 cm, 2 cm, 5 cm
Solution :
We can not construct a triangle with the given side lengths.
Because, sum of the lengths of any two sides of a triangle must be greater than the third side by Theorem 4.
In the given side lengths, we have
3 + 2 = 5
(Does not satisfy the theorem)
The diagram given below illustrates that a triangle can not be constructed with the given side lengths.
Example 6 :
Construct a triangle with the given group of side lengths, if possible.
4 cm, 2 cm, 5 cm
Solution :
In the given side lengths, we have
4 + 2 > 5
4 + 5 > 2
2 + 5 > 4
In the given side lengths, it is clear that the sum of any two sides is greater than the third side.
Because the given side lengths satisfy Theorem 4, we can construct a triangle with the given side lengths.
The diagram given below illustrates that a triangle can be constructed with the given side lengths.
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WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Why to know how to multiply numbers does not mean to know the result?
Contents
## Why do we need to know how do you multiply numbers?
Builds Confidence in Mathematics
The ability to fully understand multiplication and have fluency and instant recall will boost your child’s confidence in the subject. Many of the tasks they are required to do both at school, and in the home requires this basic skill.
## Is the result you get when you multiply?
The result of a multiplication is called a product.
## What is the identity law of multiplication?
The multiplicative identity property means that when something is multiplied by 1 then it will remain the same as the other number.
## When we multiply any number with one we always get result?
According to the multiplicative identity property of 1, any number multiplied by 1, gives the same result as the number itself. It is also called the Identity property of multiplication, because the identity of the number remains the same.
## Is there an identity element for multiplication in the whole numbers explain your answer?
Answer and Explanation: The multiplicative identity of whole numbers is 1. This means that any whole number multiplied by 1 will equal that same whole number.
## How do you show identity property of multiplication?
Multiplicative Identity Property Formula
The multiplicative identity formula is expressed as a × 1 = a, where ‘a’ is any real number. This shows that when any number is multiplied by 1, the product is the number itself. For example, if we multiply 65 with 1 we get 65 as the product. 65 × 1 = 65.
## When we multiply a whole number and the multiplicative identity of whole numbers then we get?
The product of the multiplicative identity of whole number i.e., 1 and a whole number we get the number itself.
## What number is always used in the identity property of multiplication?
The identity property of 1 says that any number multiplied by 1 keeps its identity. In other words, any number multiplied by 1 stays the same. The reason the number stays the same is because multiplying by 1 means we have 1 copy of the number. For example, 32×1=32.
## Is the multiplicative identity?
Which is the multiplicative identity, 0 or 1? Multiplicative identity of a real number is 1. When we multiply 1 to any real number then we get the same number.
## What is multiplicative math?
Definition of multiplicative
1 : tending or having the power to multiply. 2 : of, relating to, or associated with a mathematical operation of multiplication the multiplicative property of 0 requires that a × 0 = 0 and 0 × a = 0.
## Where do we use multiplication in real life?
It is also important in many everyday situations. For children, it could be helpful as well, in calculations like dividing the money among siblings. In daily life, there are many situations, such as cooking, gardening, and collecting data about the number of people that can benefit from having a multiplication table.
## Does multiply mean times?
Often a student will say the multiplication symbol means “times.” But when pushed further, they can only define it as a synonym for multiplication. (An informal canvas of friends at dinner revealed the same level of awareness.) “Times” is one of those words we use without thinking. |
# Quadratic Inequalities - Examples, Exercises and Solutions
The quadratic inequality shows us in which interval the function is positive and in which it is negative - according to the inequality symbol. To solve quadratic inequalities correctly, it is convenient to remember two things:
1. Set of positivity and negativity of the function:
Set of positivity - represents the $X$s in which the graph of the parabola is above the $X$ axis, with $Y$ value positive.
Set of negativity - represents the $X$s in which the graph of the parabola is below the $X$ axis, with $Y$ value negative.
2. Dividing by a negative term - reverses the sign of the inequality.
#### Method to solve the quadratic inequality:
1. We will carry out the transposition of members and isolate the quadratic equation until one side equals 0. Remember that when we divide by a negative term, the inequality is reversed.
2. Let's draw a diagram of the parabola - placing points of intersection with the $X$ axis and identifying the maximum and minimum of the parabola.
3. Let's calculate the corresponding interval according to the exercise and the diagram.
Quadratic equation $>0∶$ Set of positivity
Quadratic equation $<0∶$ Set of negativity
### Suggested Topics to Practice in Advance
1. Solution of a system of equations - one of them is linear and the other quadratic
## Examples with solutions for Quadratic Inequalities
### Exercise #1
Solve the following equation:
x^2+4>0
### Video Solution
All values of $x$
### Exercise #2
Solve the following equation:
-x^2+2x>0
0 < x < 2
### Exercise #3
Solve the following equation:
-x^2-9>0
### Video Solution
There is no solution.
### Exercise #4
Solve the following equation:
x^2+9>0
### Video Solution
All values of $x$
### Exercise #5
Solve the following equation:
x^2-9<0
-3 < x < 3
### Exercise #6
Solve the following equation:
-x^2+3x+4>0
-1 < x < 4
### Exercise #7
Solve the following equation:
x^2-3x+4<0
### Video Solution
There is no solution.
### Exercise #8
Solve the following equation:
x^2-6x+8<0
2 < x < 4
### Exercise #9
Solve the following equation:
x^2+4>0
### Video Solution
All values of $x$
### Exercise #10
Solve the following equation:
x^2-8x+12>0
x < 2,6 < x
### Exercise #11
Solve the following equation:
x^2+4x>0
x < -4,0 < x
### Exercise #12
Solve the following equation:
x^2+4x>0
x < -4,0 < x
### Exercise #13
Solve the following equation:
x^2-2x-8>0
### Exercise #14
Solve the following equation:
x^2-25<0
-5 < x < 5
### Exercise #15
Solve the following equation:
-x^2-25<0
### Video Solution
All values of $x$ |
The formula for variance and standard deviation for grouped data is very similar to the formula for ungrouped data. Below we show the formula for ungrouped data and grouped data.
## Variance for grouped data
deviation population sample Ungrouped data \$\$ Variance = frac {Σ (x – µ) ^ {2}} {N} \$\$ \$\$ Variance = frac {Σ (x – bar x) ^ {2}} {n – 1} \$\$ Grouped data \$\$ Variance = frac {Σf (m – µ) ^ {2}} {N} \$\$ \$\$ Variance = frac {Σf (m – bar x) ^ {2}} {n – 1} \$\$
For grouped data, m is the midpoint of a class and f is the frequency of a class.
Did you notice the resemblance? Remember that a class is a group of values like 1-3 that includes 1, 2, and 3.
For grouped data, we’ll use the midpoint of a class instead of x or the exact value. Then, just like with the mean, we multiply the numerator by f or the frequency before we add the sum.
To get the standard deviation, just take the square root of the variance. To get the variance, simply increase the standard deviation to the power of 2.
## Standard deviation for grouped data
Standard deviation population sample Ungrouped data \$\$ σ = sqrt { frac {Σ (x – µ) ^ {2}} {N}} \$\$ \$\$ s = sqrt { frac {Σ (x – bar x) ^ {2}} {n – 1}} \$\$ Grouped data \$\$ σ = sqrt { frac {Σf (m – µ) ^ {2}} {N}} \$\$ \$\$ s = sqrt { frac {Σf (m – bar x) ^ {2}} {n – 1}} \$\$
Let s be the standard deviation of the sample, then s² is the sample variance.
\$\$ s = sqrt {s ^ {2}} \$\$
Let σ be the population standard deviation, then σ² is the population variance.
\$\$ σ = sqrt {σ ^ {2}} \$\$
1. ### Chebyshev’s theorem
May 31, 21:32 a.m.
What is Chebyshev’s theorem? Definition and simple examples to help you understand quickly. |
Categories: MCQ Questions
# MCQ Questions for Class 10 Maths Areas Related to Circles with Answers
Free PDF Download of CBSE Class 10 Maths Chapter 12 Areas Related to Circles Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Areas Related to Circles MCQs with Answers to know their preparation level.
## Class 10 Maths MCQs Chapter 12 Areas Related to Circles
Circle Multiple Choice Question 1. The area of the circle is 154 cm2. The radius of the circle is
(a) 7 cm
(b) 14 cm
(c) 3.5 cm
d) 17.5 cm
Areas Related To Circles Class 10 MCQ With Answer: dExplaination: Reason: Area of circle = 154 cm²
⇒ nr² = 154 cm2
⇒ $$\frac{22}{7}$$ × r² = 154
⇒ r² = 154 × $$\frac{22}{7}$$
⇒ r² = 7 × 7 = 49
∴ r = √49 = 7
2. If angle of sector is 60°, radius is 3.5 cm then length of the arc is
(a) 3 cm
(b) 3.5 cm
(c) 3.66 cm
(d) 3.8 cm
Explaination: Reason: Here r = 3.5 cm = $$\frac{35}{10}$$ = $$\frac{7}{2}$$ cm θ = 60°
Length of arc = $$\frac{θ}{360}$$ × 2πr = $$\frac{60}{360}$$ × 2 × $$\frac{22}{7}$$ = × $$\frac{7}{2}$$ × $$\frac{1}{6}$$ × 22 = $$\frac{11}{6\3}$$ = 3.66 cm
Areas Related To Circles MCQs Question 3. The area of a quadrant of a circle whose circumference is 22 cm, is
Explaination: Reason: Here 2πr = 22 cm
2 × $$\frac{22}{7}$$ × r = 22
⇒ r = 22 × $$\frac{7}{22}$$ × $$\frac{1}{2}$$ = $$\frac{7}{2}$$ cm
∴ Area of quadrant of circle = $$\frac{1}{4}$$πr² = $$\frac{1}{4}$$ × $$\frac{22}{7}$$ × $$\frac{7}{2}$$ × $$\frac{7}{2}$$ = $$\frac{77}{8}$$ cm²
4. If 0 is the angle in degrees of a sector of a circle of radius V, then area of the sector is
5. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 7 m long rope. The area of that part of the field in which the horse can graze, is
(a) 77 cm²
(b) $$\frac{77}{2}$$ cm²
(c) 154 cm²
(d) $$\frac{77}{4}$$ cm²
MCQ on Areas Related To Circles With Answer: b
Explaination:
6. The area of the circle whose diameter is 21 cm is
(a) 346.5 cm²
(b) 37.68 cm²
(c) 18.84 cm²
(d) 19.84 cm²
Explaination: Reason: Here diameter = 21 cm
∴ Radius r = $$\frac{21}{2}$$ cm
Area of the circle, A = πr²
∴ $$A=\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}=11 \times 3 \times \frac{21}{2}=\frac{693}{2}=346.5 \mathrm{cm}^{2}$$
7. The area of the sector of a circle with radius 6 cm and of angle 60° is
(a) 9.42 cm²
(b) 37.68 cm²
(c) 18.84 cm²
(d) 19.84 cm²
Explaination: Reason: Here r = 6 cm, θ = 60°
Area of the sector = $$\frac{θ}{360}$$
∴ Area = $$\frac{60}{360}$$ × 3.14 × 6 × 6 = $$\frac{1}{6}$$ × 3.14 × 6 × 6 = 3.14 × 6 = 18.84 cm²
8. The area of a circle whose circumference is 22 cm, is
(a) 11 cm²
(b) 38.5 cm²
(c) 22 cm²
(d) 77 cm²
Explaination:
9. The area of a circle is 154 cm2. Its diameter is
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 28 cm
Explaination: Reason: Here area of the circle, A = 154 cm², Radius, r = ?
Area of the circle = 154 cm² …(Given)
∴ πr² = 154
⇒ $$\frac{22}{7}$$ × r² = 154
⇒ r² = 154 × latex]\frac{7}{22}[/latex] = 7 × 7
⇒ r = 7 cm
∴ Diameter of the circle = 2 × r = 2 × 7 = 14 cm
10. The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is
(a) 153.9 cm²
(b) 102.6 cm²
(c) 51.3 cm²
(d) 205.2 cm²
Explaination: Reason: Angle swept by the minute hand in 1 minute = (360° ÷ 60) = 6°
∴ θ = 30°
∴ Angle swept by the minute hand in 5 minutes = 6° × 5 = 30°
Length of minute hand (r) = 14 cm
∴ Area swept = $$\frac{θ}{360}$$πr² = $$\frac{30}{360}$$ × $$\frac{22}{7}$$ × 14 × 14 = $$\frac{154}{3}$$ = 51.3 cm²
MCQ Questions For Class 10 Maths Areas Related To Circles Question 11. The radii of two circles are 19 cm and 9 cm respectively. The radius of the circle which has circumference equal to the sum of the circumference of two circles is
(a) 35 cm
(b) 10 cm
(c) 21 cm
(d) 28 cm
Explaination: Reason: Let the radii of two circles be r1 and r2 and the radius of large circle be r.
∴ r1 = 19 cm, r2 = 9 cm
Circumference of two circles = C1+ C2 …(where C = circle)
= 2πr1 + 2πr2 = 2π × 19 + 2π × 9 = 38π + 18π = 56π
∴ Circumference of large circle = 56π
⇒ 2πr = 56π
⇒ r = 28
∴ Radius of large circle = 28 cm
12. The area of the circle that can be inscribed in a square of side 6 cm, is
(a) 18π cm²
(b) 12π cm²
(c) 9π cm²
(d) 14π cm²
Explaination: Reason: Size of square = 6 cm, radius = $$\frac{6}{2}$$ = 3 cm;
Area of the circle = πr² = π × 3 × 3 = 9π cm²
13. The radii of two circles are 4 cm and 3 cm respectively. The diameter of the circle having area equal to the sum of the areas of the two circles (in cm) is [Delhi 2011]
(a) 5
(b) 7
(c) 10
(d) 14
Explaination:
MCQ on Area Related To Circles Class 10 Question 14. The perimeter (in cm) of a square circumscribing a circle of radius a cm, is [AI2011]
(a) 8 a
(b) 4 a
(c) 2 a
(d) 16 a
Explaination:
(a) Side of a square circumscribing a circle of radius a cm = diameter of circle = 2 a cm
∴ Perimeter of the square
= 4 × 2a = 8a cm
15. If the area of a circle is numerically equal to twice its circumference, then the diameter of the circle is
(a) 4 units
(b) n units
(c) 8 units
(d) 2 units
Explaination:
(c) πr² = 2πr × 2
⇒ r = 4
⇒ 2r = 8 units
16. If the circumference of a circle is 352 metres, then its area in square metres is
(a) 5986
(b) 6589
(c) 7952
(d) 9856
Explaination:
Area Related To Circle Class 10 MCQ Question 17. The diameter of a wheel is 1.26 m. The distance travelled in 500 revolutions is
(a) 2670 m
(b) 2880 m
(c) 1980 m
(d) 1596 m
Explaination:
(c) Radius of the wheel = $$\frac{1.26}{2}$$ = 0.63 m
Distance travelled in one revolution
= 2πr = 2 × $$\frac{22}{7}$$ × 0.63
= 3.96 m
∴ Distance travelled in 500 revolutions
= 500 × 3.96
= 1980 m.
18. If the sum of the circumferences of two circles with radii Rj and R2 is equal to the circumference of a circle of radius R, then [NCERT Exemplar Problems]
(a) R1 + R2 = R
(b) R1 + R2 > R
(C) R1 + R2 < R
(d) nothing definite can be said about the relation among Rp1, R2 and R.
Explaination:
(a) 2πR1 + 2πR2 = 2πR
⇒ R1 + R2 = R.
19. If the circumference of a circle and the perimeter of a square are equal, then [NCERT Exemplar Problems]
(a) area of the circle = area of the square
(b) area of the circle > area of the square
(c) area of the circle < area of the square
(d) nothing definite can be said about the relation between the areas of the circle and square.
Explaination:
(b) Let circumference of a circle = C
2πr = C C
⇒ r = $$\frac{C}{2π}$$
Perimeter of a square = C
⇒ 4a = C
⇒ a = $$\frac{C}{4}$$
MCQ on Area Related To Circles Question 20. Area of the largest triangle that can be inscribed in a semi-circle of radius r units is [NCERT Exemplar Problems]
(a) r² sq. units
(b) $$\frac{1}{2}$$ r² sq. units
(c) 2 r² sq. units
(d) √2 r² sq. units
Explaination:
21. Match the columns
(a) 1 → A, 2 → C, 3 → D, 4 → E
(b) 1 → B, 2 → C, 3 → F, 4 → E
(c) 1 → D, 2 → B, 3 → A, 4 → F
(d) 1 → D, 2 → B, 3 → E, 4 → F
Explaination: (c) Formulae
22. In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. The area of the shaded region (in cm2) is [Using π = $$\frac{22}{7}$$] [Foreign 2012]
(a) 77
(b) 154
(c) 44
(d) 22
Explaination:
(a) Area of shaded region = area of sector with angle
(60° + 80° + 40°) = $$\frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7 \times 7$$
= 77 cm²
Areas Related To Circles Question 23. If the difference between the and the radius of of a circle is 37 cm, then 22 using π = $$\frac{22}{7}$$ the circumference (in cm) of the circle is: [Delhi 2013]
(a) 154
(b) 44
(c) 14
(d) 7
Explaination:
24. If 7i is taken as $$\frac{22}{7}$$, the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is [AI2013]
(a) 2.2
(b) 1.1
(c) 9.625
(d) 96.25
Explaination:
(b) Distance covered by a wheel in one
revolution = 2 πr = 2 × $$\frac{22}{7}$$ × $$\frac{35}{2}$$
= 110 cm = 1.1 m
Circle MCQ Pdf Question 25. If the circumferences of two circles are in the ratio 4 : 9, then the ratio in their area is
(a) 9 : 4
(b) 4 : 9
(c) 2 : 3
(d) 16 : 81
Explaination:
26. The ratio of the areas of the incircle and circumcircle of a square is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : √2
Explaination:
(a) Let side of square = x units
∴ Diagonal of the square = √2 x units
Diameter of the incircle = x units
Diameter of the circumcircle = √2 x units
27. A circular wire of radius 42 cm is cut and bent into the form of a rectangle whose sides are in the ratio of 6 : 5. The smaller side of the rectangle is
(a) 30 cm
(b) 60 cm
(c) 70 cm
(d) 80 cm
Explaination:
(b) Length of wire = 2πrr = 2πr × 42
= 2 × $$\frac{22}{7}$$ × 42 = 264 cm
Let sides of rectangle are 6x and 5x
⇒ 2(6x + 5x) = 264
⇒ 11x = 132
⇒ x = 12
∴ Smaller side = 12 × 5 = 60 cm
28. Match the columns.
(a) 1 → B, 2 → C, 3 → D, E → 4
(b) 1 → B, 2 → D, 3 → C, E → 4
(c) 1 → A, 2 → C, 3 → D, E → 4
(d) 1 → B, 2 → C, 3 → D, E → 4
Explaination: (b) Formulae
29. ABC is an equilateral triangle. The area of the shaded region if the radius of each of the circle is 1 cm, is
Explaination:
30. ABCDEF is any hexagon with different vertices A, B, C, D, E and F as the centres of circles with same radius r are drawn. The area of the shaded portion is
(a) πr²
(b) 2πr²
(c) 3πr²
(d) 4πr²
Explaination:
31. In the figure, PQRS is a square and O is centre of the circle. If RS = 10 √2, then area of shaded region is
(a) 90 π – 90
(b) 80 π – 80
(c) 50 π-100
(d) 100 π – 100
Explaination:
(c) Diagonal of square = √2 × (10√2) = 20 units
∴ Diameter of circle = 20 units
Area of circle = π × (10)² = 100π sq.units
Area of square = (10√2)² = 200 sq. units
Area of circle not included in the square = (100π – 200) sq.units
∴ Area of shaded portion = $$\frac{1}{2}$$(100π – 200)
= 50π – 100.
32. The diameter of the wheel of a bus is 1.4 m. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle (in kmph) is ______ .
Explaination:
33. The area of a quadrant of a circle whose circumference is 44 cm is ______ .
Explaination:
34. If the wheel of an engine of a train is 4$$\frac{2}{7}$$m in circumference makes seven revolutions in 4 seconds, then the speed of the train is _____ km/h.
Explaination:
35. The area of the largest possible square inscribed in a circle of unit radius (in sq. units) is ______.
Explaination:
2 units
Hints: Diameter of circle = 2 units
∴ Diagonal of the square = 2 units
Side of the square = $$\frac{2}{\sqrt{2}}$$ = √2 units
∴ Area of the square = (√2)² = 2 sq. units
36. In the fig., O is the centre of a circle. The area of sector OAPB is $$\frac{5}{18}$$ of the area of the circle. Find x.
Explaination:
37. Find the perimeter of the given figure, where $$\widehat{\mathrm{AED}}$$ is a semicircle and ABCD is a rectangle.
Explaination:
Length of $$\widehat{\mathrm{AED}}$$ = πr
= π x 7 cm = 7π cm
AB + BC + DC = 20 + 14 + 20 = 54 cm
∴ Perimeter of figure = (7π + 54) cm
38. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel. (use π = $$\frac{22}{7}$$)
Explaination:
39. A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of pendulum, (use π = $$\frac{22}{7}$$)
Explaination:
40. An arc of a circle is of length 5TI cm and the sector it bounds has an area of 20π cm². Find the radius of the circle.
Explaination:
41. If the diameter of a semicircular protractor is 14 cm, then find its perimeter. [π = $$\frac{22}{7}$$]
Explaination:
42. Find the perimeter of the shaded region in figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [Use π = $$\frac{22}{7}$$]
Explaination:
Perimeter = AD + BC + length of DPC + length of APB
= 14 + 14 + πr + πr
= 28 + 2 × $$\frac{22}{7}$$ × $$\frac{14}{2}$$= 72 cm
We hope the given MCQ Questions for Class 10 Maths Areas Related to Circles with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 12 Areas Related to Circles Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.
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# Lesson 13Tables and Double Number Line Diagrams
Let’s contrast double number lines and tables.
### Learning Targets:
• I can create a table that represents a set of equivalent ratios.
• I can explain why sometimes a table is easier to use than a double number line to solve problems involving equivalent ratios.
• I include column labels when I create a table, so that the meaning of the numbers is clear.
## 13.1Number Talk: Constant Dividend
1. Find the quotients mentally.
2. Locate and label the quotients on the number line.
## 13.2Moving 3,000 Meters
The other day, we saw that Han can run 100 meters in 20 seconds.
Han wonders how long it would take him to run 3,000 meters at this rate. He made a table of equivalent ratios.
1. Do you agree that this table represents the situation? Explain your reasoning.
20 100 10 50 1 5 3,000
1. Complete the last row with the missing number.
1. What could Han do to improve his table?
1. Priya can bike 150 meters in 20 seconds. At this rate, how long would it take her to bike 3,000 meters?
1. Priya’s neighbor has a dirt bike that can go 360 meters in 15 seconds. At this rate, how long would it take them to ride 3,000 meters?
## 13.3The International Space Station
The International Space Station orbits around the Earth at a constant speed. Your teacher will give you either a double number line or a table that represents this situation. Your partner will get the other representation.
1. Complete the parts of your representation that you can figure out for sure.
2. Share information with your partner, and use the information that your partner shares to complete your representation.
1. What is the speed of the International Space Station?
2. Place the two completed representations side by side. Discuss with your partner some ways in which they are the same and some ways in which they are different.
3. Record at least one way that they are the same and one way they are different.
### Are you ready for more?
The Earth’s circumference is about 40,000 kilometers and the orbit of the International Space Station is just a bit more than this. About how long does it take for the International Space Station to orbit the Earth?
## Lesson 13 Summary
On a double number line diagram, we put labels in front of each line to tell what the numbers represent. On a table, we put labels at the top of each column to tell what the numbers represent.
Here are two different ways we can represent the situation: “A snail is moving at a constant speed down a sidewalk, traveling 6 centimeters per minute.”
Both double number lines and tables can help us use multiplication to make equivalent ratios, but there is an important difference between the two representations.
On a double number line, the numbers on each line are listed in order. With a table, you can write the ratios in any order. For this reason, sometimes a table is easier to use to solve a problem.
For example, what if we wanted to know how far the snail travels in 10 minutes? Notice that 60 centimeters in 10 minutes is shown on the table, but there is not enough room for this information on the double number line.
## Lesson 13 Practice Problems
1. The double number line shows how much water and how much lemonade powder to mix to make different amounts of lemonade.
Make a table that represents the same situation.
2. A bread recipe uses 3 tablespoons of olive oil for every 2 cloves of crushed garlic.
1. Complete the table to show different-sized batches of bread that taste the same as the recipe.
2. Draw a double number line that represents the same situation.
3. Which representation do you think works better in this situation? Explain why.
olive oil (tablespoons) crushed garlic (cloves)
3 2
1
2
5
10
3. Clare travels at a constant speed, as shown on the double number line.
At this rate, how far does she travel in each of these intervals of time? Explain or show your reasoning. If you get stuck, consider using a table.
1. 1 hour
2. 3 hours
3. 6.5 hours
4. Lin and Diego travel in cars on the highway at constant speeds. In each case, decide who was traveling faster and explain how you know.
1. During the first half hour, Lin travels 23 miles while Diego travels 25 miles.
2. After stopping for lunch, they travel at different speeds. To travel the next 60 miles, it takes Lin 65 minutes and it takes Diego 70 minutes.
5. A sports drink recipe calls for tablespoons of powdered drink mix for every 12 ounces of water. How many batches can you make with 5 tablespoons of drink mix and 36 ounces of water? Explain your reasoning.
6. In this cube, each small square has side length 1 unit.
1. What is the surface area of this cube?
2. What is the volume of this cube? |
Calculus 2 : Parametric Calculations
Example Questions
← Previous 1 3
Example Question #1 : Parametric Calculations
Calculate the length of the curve drawn out by the vector function from .
Explanation:
The formula for arc length of a parametric curve in space is for .
Taking derivatives of each of the vector function components and substituting the values into this formula gives
We need to recognize that underneath the square root we have a perfect square, and we can write it as
Solving this we get
Example Question #2 : Parametric Calculations
Calculate at the point on the curve defined by the parametric equations
Explanation:
We use the equation
But we need a value for to substitute into our derivative. We can obtain such a by setting as our given point suggests.
Since our values of match, is our correcct value. Substituting this into the derivative and simplifying gives us our answer of
Example Question #3 : Parametric Calculations
Which of the answers below is the equation obtained by eliminating the parametric from the following set of parametric equations?
Explanation:
When the problem asks us to eliminate the parametric, that means we want to somehow get rid of our variable t and be left with an equation that is only in terms of x and y. While the equation for x is a polynomial, making it more difficult to solve for t, we can see that the equation for y can easily be solved for t:
Now that we have an expression for t that is only in terms of y, we can plug this into our equation for x and simplify, and we will be left with an equation that is only in terms of x and y:
Example Question #4 : Parametric Calculations
Suppose and . Find the arc length from .
Explanation:
Write the arc length formula for parametric curves.
Find the derivatives. The bounds are given in the problem statement.
Example Question #5 : Parametric Calculations
Solve for if and .
None of the above
Explanation:
Given equations for and in terms of , we can find the derivative of parametric equations as follows:
, as the terms will cancel out.
Using the Power Rule
for all and given and :
.
Example Question #6 : Parametric Calculations
Solve for if and .
None of the above
Explanation:
Given equations for and in terms of , we can find the derivative of parametric equations as follows:
, as the terms will cancel out.
Using the Power Rule
for all and given and :
Example Question #7 : Parametric Calculations
Solve for if and .
None of the above
Explanation:
Since we have two equations and , we can find by dividing the derivatives of the two equations - thus:
since the terms cancel out by standard rules of division of fractions.
In order to find the derivatives of and , let's use the Power Rule
for all :
Therefore,
.
Example Question #8 : Parametric Calculations
Solve for if and .
Explanation:
Since we have two equations and , we can find by dividing the derivatives of the two equations - thus:
since the terms cancel out by standard rules of division of fractions.
In order to find the derivatives of and , let's use the Power Rule
for all :
Therefore,
.
Example Question #9 : Parametric Calculations
Solve for if and .
None of the above
Explanation:
Since we have two equations and , we can find by dividing the derivatives of the two equations - thus:
(since the terms cancel out by standard rules of division of fractions).
In order to find the derivatives of and , let's use the Power Rule
for all :
Therefore, .
Example Question #10 : Parametric Calculations
Given and , what is the length of the arc from ?
Explanation:
In order to find the arc length, we must use the arc length formula for parametric curves:
.
Given and , we can use using the Power Rule
for all , to derive
and .
Plugging these values and our boundary values for into the arc length equation, we get:
Now, using the Power Rule for Integrals for all , we can determine that:
← Previous 1 3 |
# Show $\,1897\mid 2903^n - 803^n - 464^n + 261^n\,$ by induction
Show that: $2903^n - 803^n - 464^n + 261^n$ is divisible by $1897$ for all integers $n\geq1$ using induction.
(I missed out the "by induction" in the question when I wrote this answer, but I will leave it here anyway. Do feel free to have it removed.)
I remember this as a classic Olympiad problem for factorization.
We can rewrite the given expression as:
$$(2903^n - 803^n) - (464^n - 261^n)$$
Then, factorizing (using the difference of powers factorization) we get:
$$(2903 - 803)(\dots) - (464 - 261)(\dots)$$ $$= 2100\cdot(\dots) - 203(\dots)$$
which is divisible by $7$. (I've omitted the expression in the bracket, but the important thing is that they are integers)
Next, we can also rewrite the given expression as:
$$(2903^n - 464^n) - (803^n - 261^n)$$
Factorizing, we get:
$$(2903 - 464)(\dots) - (803 - 261)(\dots)$$ $$=2439\cdot(\dots) - 542\cdot(\dots)$$
which is divisible by $271$ (because $271$ divides both $2439$ and $542$).
Since both $7$ and $271$ divides the given expression, it follows that $7\cdot271 = 1897$ must also divide it.
Hint $\$ Exploit innate $\rm\color{#c00}{symmetry}$! Consider a simpler analogous example
$\qquad\phantom{\Rightarrow}\ \ \{ 52,\ \ \ \ 23\}\ \ \equiv\, \{2,\ \ \ 3\}\ \ \ \ {\rm mod}\,\ 10,7,\ \$
$\qquad\Rightarrow\ \{52^n,\ \ 23^n\} \equiv \{2^n,\ \,3^n\}\,\ {\rm mod}\,\ 10,7,\ \$ by the Congruence Power Rule
$\qquad\Rightarrow\ \ \ 52^n\!+\! 23^n\ \ \equiv \ \ 2^n\!+3^n\ \ \,{\rm mod}\,\ 10,7,\ \$ so also $\,{\rm mod}\ 70 = {\rm lcm}(10,7)$
since addition $\,f(x,y)\, =\, x + y\$ is $\rm\color{#c00}{symmetric}$ $\,f(x,y)= f(y,x),\$ therefore its value depends only upon the (multi-)set $\,\{x,\,y\}.\$ Your problem is completely analogous since
$\qquad\!\phantom{\Rightarrow} \{2903,\, 261\}\ \equiv\ \{803,\, 464\}\,\ {\rm mod}\,\ 271,7,\ \,$ where $\,\ 271\cdot 7 = 1897$
Generally if a polynomial $\,f\in\Bbb Z[x,y]\,$ is $\rm\color{#c00}{symmetric}$ then
$\qquad\qquad\quad \{A, B\}\, \equiv\, \{a,b\}\,\ {\rm mod}\,\ m,\, n\,\ \Rightarrow\,\ f(A,B)\equiv f(a,b)\, \pmod{\!{\rm lcm}(m,n)}\qquad\quad$
a generalization of the constant-case optimization of CRT = Chinese Remainder, combined with a generalization of the Polynomial Congruence Rule to (symmetric) bivariate polynomials.
• See this answer for a significant generalization. – Bill Dubuque Jun 22 '14 at 17:12
Let
$$S_n = 2903^n - 803^n - 464^n + 261^n$$
$S_1 = 1897$ so the statement is true for $n=1$.
Assume true for n, and evaluate for (n+1):
$$S_{n+1} = 2903^{n+1} - 803^{n+1} - 464^{n+1} + 261^{n+1}=$$ $$=261\cdot S_n + 2642\cdot 2903^n - 542\cdot803^n - 203\cdot464^n$$
The first term divides by $1897$ by assumption. Let
$$T_n = 2642\cdot2903^n - 542\cdot803^n - 203\cdot464^n$$
If we can prove that $T_n$ divides by 1897 then we are done. Proceed by induction (again). $T_1 = 7140308 = 3764\cdot1897$, i.e. $T_1$ divides by 1897. Assume true for n and evaluate for (n+1).
$$T_{n+1} = 2642\cdot2903\cdot2903^n -542\cdot803\cdot803^n - 203\cdot464\cdot464^n =$$ $$=464\cdot T_n + (2903-464)\cdot2642\cdot2903^n - (803-464)\cdot542\cdot803^n$$
Again the first term by assumption divides by 1897 and let $$U_n = 6443838\cdot2903^n - 183738\cdot803^n$$
Now we just need to prove that $U_n$ divides by 1897. Proceed (yet again) by induction. $U_1 = 18558920100 = 9783300\cdot1897$. So $U_1$ divides by 1897. Assume true for n and evaluate for (n+1).
$$U_{n+1} = 6443838\cdot2903\cdot2903^n - 183738\cdot803\cdot803^n =$$ $$=803\cdot U_n +6443838\cdot(2903-803)\cdot2903^n$$
again the first term divides by 1897 (by assumption). The second term = $6443838\cdot2100\cdot2903^n = 13532059800\cdot 2903^n$ and $13532059800 = 7133400\cdot 1897$, so $U_{n+1}$ divides by 1897.
(phew !! ).
• This amounts to constructing a constant coeff order $4$ recurrence for $\,S_n,\,$ see the explanation here. – Bill Dubuque Apr 26 '15 at 21:48 |
#### Solving Multi-Step Equations
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##### Sections
Exercise name Free?
###### Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
Monitoring Progress 10
Monitoring Progress 11
###### Exercises
Exercise name Free?
Exercises 1 We are able to combine terms when they are like terms. (-7)+1=-62x+3x=5x5m−4m=m Like terms can be combined because they have the same exact variable, or because they are constants and have no variable.
Exercises 2 There are two ways to approach this problem, you can use the Distributive Property or you can treat (4x−11) as a single quantity. Let's try both and compare our answers.Distributive Property We can solve this problem by first distributing the 2 to the terms inside the parentheses and simplifying. 2(4x−11)=10Distribute 22⋅4x−2⋅11=10Multiply8x−22=10 Now we can continue solving by using the Addition Property of Equality to isolate the 8x. 8x−22=10LHS+22=RHS+228x−22+22=10+22Add terms8x=32 Finally, we can find our answer by using the Division Property of Equality. 8x=32LHS/8=RHS/888x=832Calculate quotientx=4 Using this method, we found that x=4.Treating the parentheses as a single quantity This time, let's try dividing both sides of the equation by 2 first, instead of distributing. We are allowed to do this by the Division Property of Equality. 2(4x−11)=10LHS/2=RHS/222(4x−11)=210Calculate quotient4x−11=5 Next, we can use the Addition Property of Equality to isolate the 4x. 4x−11=5LHS+11=RHS+114x−11+11=5+11Add terms4x=16 And finally, we will once again use the Division Property of Equality to solve for x. 4x=16LHS/4=RHS/444x=416Calculate quotientx=4 Using this method, we also found that x=4.Conclusion Both methods came to the conclusion that x=4. Both methods are equally valid and will find the correct answer when used properly!
Exercises 3 To solve the equation, we first have to isolate 3w by subtracting 7 from both sides. 3w+7=19LHS−7=RHS−73w+7−7=19−7Subtract term3w=12 The next step is to isolate w on the left-hand side by dividing both sides of the equation by 3. 3w=12LHS/3=RHS/333w=312Calculate quotientw=4 Therefore, w=4 is the solution to the equation. We can check our solution by substituting it into the original equation. 3w+7=19w=43(4)+7=?19Multiply12+7=?19Add terms19=19 Since the left-hand side and right-hand side are equal, we have the correct solution.
Exercises 4 To solve the equation, we first isolate 2g by adding 13 to both sides. 2g−13=3LHS+13=RHS+132g−13+13=3+13Add terms2g=16 The next step is to isolate g on the left-hand side. By dividing both sides of the equation by 2, we can isolate g. 2g=16LHS/2=RHS/222g=216Calculate quotientg=8 Therefore, g=8 is the solution to the equation. We can check our solution by substituting it into the original equation and simplifying. 2g−13=3g=82(8)−13=?3Multiply16−13=?3Subtract term13=13 Since the left-hand side and right-hand side are equal, g=8 is the correct solution.
Exercises 5 To solve the equation, we first have to isolate -q. We can do this by subtracting 12 from both sides. 11=12−qLHS−12=RHS−1211−12=12−q−12Subtract terms-1=-q Now, we can continue isolating q by multiplying both sides of the equation by -1. -1=-qLHS⋅-1=RHS⋅-11=qRearrange equationq=1 Therefore, q=1 is the solution to the equation. We can check our solution by substituting it into the original equation. 11=12−qq=111=?12−1Subtract term11=11 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 6 To solve the equation, we first have to isolate -m. We can do this by subtracting 7 from both sides. 10=7−mLHS−7=RHS−710−7=7−m−7Subtract terms3=-m Now, we can continue isolating m by multiplying both sides of the equation by -1. 3=-mChange signs-3=mRearrange equationm=-3 Therefore, m=-3 is the solution to the equation. We can check our solution by substituting it into the original equation. 10=7−mm=-310=?7−(-3)a−(-b)=a+b10=?7+3Add terms10=10 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 7 To solve the equation, we first have to isolate the fraction by removing -3 from the right-hand side. Adding 3 to both sides of the equation will cancel out -3. 5=-4z−3LHS+3=RHS+35+3=-4z−3+3Add terms8=-4z The next step is to isolate z on the right-hand side. Division can be canceled out by multiplying both sides of the equation by the fraction's denominator. 8=-4zLHS⋅(-4)=RHS⋅(-4)8(-4)=-4z(-4)a(-b)=-a⋅b-32=-4z(-4)-4a⋅-4=a-32=zRearrange equationz=-32 Therefore, z=-32 is the solution to the equation. Finally, let's check our solution by substituting it into the original equation. 5=-4z−3z=-325=?-4-32−3-b-a=ba5=?432−3Calculate quotient5=?8−3Subtract term5=5 Because the left-hand side and right-hand side are equal, we know that our solution is correct.
Exercises 8 To solve the equation, we first have to isolate 3a by subtracting 4 from both sides. 3a+4=6LHS−4=RHS−43a+4−4=6−4Subtract term3a=2 The next step is to isolate a by multiplying both sides by 3. This will allow us to cancel out the fraction. 3a=2LHS⋅3=RHS⋅33a⋅3=2⋅33a⋅3=aa=2⋅3Multiplya=6 Therefore, a=6 is the solution to the equation. We can check our solution by substituting it into the original equation. 3a+4=6a=636+4=?6Calculate quotient2+4=?6Add terms6=6 Since the left-hand side and right-hand side are equal, the solution is correct.
Exercises 9 To solve the equation, we first have to isolate h+6. We can do this by multiplying both sides by 5. 5h+6=2LHS⋅5=RHS⋅55h+6⋅5=2⋅55a⋅5=ah+6=2⋅5Multiplyh+6=10 Now we continue isolating h by subtracting 6 from both sides. h+6=10LHS−6=RHS−6h+6−6=10−6Subtract termh=4 Therefore, h=4 is the solution to the equation. We can check our solution by substituting it into the original equation. 5h+6=2h=454+6=?2Add terms510=?2Calculate quotient2=2 Since the left-hand side and right-hand side are equal, we know the solution is correct.
Exercises 10 To begin solving this equation, let's first isolate d−8 by multiplying both sides by -2. -2d−8=12LHS⋅(-2)=RHS⋅(-2)-2d−8(-2)=12(-2)-2a⋅-2=ad−8=12(-2)a(-b)=-a⋅bd−8=-24LHS+8=RHS+8d−8+8=-24+8Add termsd=-16 Therefore, d=-16 is the solution to the equation. We can check if this solution is correct by substituting it into the original equation. -2d−8=12d=-16-2(-16)−8=?12Subtract term-2-24=?12-b-a=ba224=?12Calculate quotient12=12 Since both sides are equal, the solution is correct.
Exercises 11 We can solve the equation by combining like terms and then isolating y. 8y+3y=44Add terms11y=44LHS/11=RHS/111111y=1144Calculate quotienty=4 Let's check our answer by substituting y=4 into the original equation. 8y+3y=44y=48(4)+3(4)=?44 Simplify LHS Multiply32+12=?44Add terms 44=44 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 12 We can solve the equation by combining like terms and then isolating n. 36=13n−4nSubtract terms36=9nLHS/9=RHS/9936=99nCalculate quotient4=nRearrange equationn=4 Let's check our answer by substituting n=4 into the original equation. 36=13n−4nn=436=?13(4)−4(4) Simplify LHS Multiply36=?52−16Subtract term 36=36 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 13 We can solve the equation by combining like terms and then isolating v. 12v+10v+14=80Add terms22v+14=80LHS−14=RHS−1422v+14−14=80−14Subtract terms22v=66LHS/22=RHS/222222v=2266Calculate quotientv=3 Let's check our answer by substituting v=3 into the original equation. 12v+10v+14=80v=312(3)+10(3)+14=?80 Simplify LHS Multiply36+30+14=?80Add terms 80=80 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 14 We can solve the equation by combining like terms and then isolating c. 6c−8−2c=-16Subtract terms4c−8=-16LHS+8=RHS+84c−8+8=-16+8Add terms4c=-8LHS/4=RHS/444c=4-8Put minus sign in front of fraction44c=-48Calculate quotientc=-2 Let's check our answer by substituting c=-2 into the original equation. 6c−8−2c=-16c=-26(-2)−8−2(-2)=?-16 Simplify LHS a(-b)=-a⋅b-12−8−2(-2)=?-16(-a)(-b)=a⋅b-12−8+4=?-16Subtract term-20+4=?-16Add terms -16=-16 Since the left-hand side and right-hand side are equal, we know our solution is correct.
Exercises 15 We are given a formula for the altitude (in feet) of a plane with two variables, a and t. In this formula, a is the altitude of the plane and t is the time after liftoff in minutes. To find how many minutes after liftoff the plane is at an altitude of 21000 feet, we substitute this altitude into the given formula and solve for t. a=3400t+600a=2100021000=3400t+600LHS−600=RHS−60021000−600=3400t+600−600Subtract terms20400=3400tLHS/3400=RHS/3400340020400=34003400tCalculate quotient6=tRearrange equationt=6 The plane reaches an altitude of 21,000 feet 6 minutes after liftoff.
Exercises 16 To write an equation for the number of hours of labor spent repairing the car, let's break down the given information into individual expressions and then combine these parts to form the equation. First, we know that the total bill came to \$533. …=533 This cost represents the sum of the cost of parts and the cost of labor. We are given that the parts cost \$265. labor cost+265=533 We are also told that the labor cost is \$48 per hour, but not the number of hours spent on the car. If we call this unknown variable t, the expression for the labor cost is the product of t and the hourly cost. Finally, our equation becomes: 48t+265=553. Now we can solve for t using the Properties of Equality to isolate the variable. 48t+265=553LHS−265=RHS−26548t+265−265=553−265Subtract terms48t=288LHS/48=RHS/484848t=48288Use a calculatort=6 6 hours of labor were spent repairing the car.
Exercises 17 To begin solving this equation, we distribute the 4 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 4(z+5)=32Distribute 44z+20=32LHS−20=RHS−204z+20−20=32−20Subtract terms4z=12LHS/4=RHS/444z=412Calculate quotientz=3 Therefore, z=3 is the solution to the equation. We can check our answer by substituting it back into the original equation. 4(z+5)=32z=34(3+5)=?32Add terms4(8)=?32Multiply32=32 Because both sides are equal, we know that our answer is correct.
Exercises 18 To begin solving this equation, we distribute the -2 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. -2(4g−3)=30Distribute -2-2(4g)−(-2)(3)=30Multiply-8g+6=30LHS−6=RHS−6-8g+6−6=30−6Subtract term-8g=24LHS/-8=RHS/-8-8-8g=-824Put minus sign in front of fraction-8-8g=-824Calculate quotientg=-3 Therefore, g=-3 is the solution to the equation. We can check our answer by substituting it back into the original equation. -2(4g−3)=30g=-3-2(4(-3)−3)=?30a(-b)=-a⋅b-2(-12−3)=?30Subtract terms-2(-15)=?30-a(-b)=a⋅b30=30 Because both sides are equal, we know that our answer is correct.
Exercises 19 To begin solving this equation, we can distribute the 5 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 6+5(m+1)=26Distribute 56+5m+5=26Add terms11+5m=26LHS−11=RHS−1111+5m−11=26−11Subtract terms5m=15LHS/5=RHS/555m=515Calculate quotientm=3 Therefore, m=3 is the solution to the equation. We can check our answer by substituting it back into the original equation. 6+5(m+1)=26m=36+5(3+1)=?26Add terms6+5(4)=?26Multiply6+20=?26Add terms26=26 Because both sides are equal, we know that our answer is correct.
Exercises 20 To begin solving this equation, we can distribute the 2 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 5h+2(11−h)=-5Distribute 25h+11(2)−h(2)=-5Multiply5h+22−2h=-5Subtract term3h+22=-5LHS−22=RHS−223h+22−22=-5−22Subtract term3h=-27LHS/3=RHS/333h=3-27Calculate quotienth=-9 Therefore, h=-9 is the solution to the equation. We can check our answer by substituting it back into the original equation. 5h+2(11−h)=-5h=-95(-9)+2(11−(-9))=?-5a(-b)=-a⋅b-45+2(11−(-9))=?-5a−(-b)=a+b-45+2(11+9)=?-5Add terms-45+2(20)=?-5Multiply-45+40=?-5Add terms-5=-5 Because both sides are equal, we know that our answer is correct.
Exercises 21 To begin solving this equation, we can distribute the -3 to each term in the parentheses on the right-hand side. Then we can continue solving by using the Properties of Equality. 27=3c−3(6−2c)Distribute -327=3c+6(-3)−2c(-3)a(-b)=-a⋅b27=3c−18−2c(-3)-a(-b)=a⋅b27=3c−18+6cAdd terms27=9c−18LHS+18=RHS+1827+18=9c−18+18Add terms45=9cLHS/9=RHS/9945=99cCalculate quotient5=cRearrange equationc=5 Therefore, c=5 is the solution to the equation. We can check our answer by substituting it back into the original equation. 27=3c−3(6−2c)c=527=?3(5)−3(6−2(5))Multiply27=?15−3(6−10)Subtract term27=?15−3(-4)-a(-b)=a⋅b27=?15+12Add terms27=27 Because both sides are equal, we know that our answer is correct.
Exercises 22 To begin solving this equation, we begin by distributing the -5 to each term in the parentheses on the right-hand side and simplifying. -3=12y−5(2y−7)Distribute -5-3=12y+2y(-5)−7(-5)a(-b)=-a⋅b-3=12y−10y−7(-5)-a(-b)=a⋅b-3=12y−10y+35Subtract term-3=2y+35 We continue solving by using the Properties of Equality to isolate y. -3=2y+35LHS−35=RHS−35-3−35=2y+35−35Subtract term-38=2yLHS/2=RHS/22-38=22yCalculate quotient-19=yRearrange equationy=-19 Therefore, y=-19 is the solution to the equation. We can check our answer by substituting it back into the original equation. -3=12y−5(2y−7)y=-19-3=?12(-19)−5(2(-19)−7)a(-b)=-a⋅b-3=?-228−5(-38−7)Subtract terms-3=?-228−5(-45)-a(-b)=a⋅b-3=?-228+225Add terms-3=-3 Because both sides are equal, we know that our answer is correct.
Exercises 23 On the left-hand side of the equation, we have products that can be simplified using the Distributive Property. -3(3+x)+4(x−6)=-4Distribute -3 & 43(-3)+x(-3)+x(4)−6(4)=-4Multiply-9−3x+4x−24=-4Add and subtract terms-33+x=-4 Now, we can continue to solve using the Properties of Equality. -33+x=-4LHS+33=RHS+33-33+x+33=-4+33Add termsx=29 The solution to the equation is x=29. We can check our solution by substituting it into the original equation. -3(3+x)+4(x−6)=-4x=29-3(3+29)+4(29−6)=?-4Add and subtract terms-3(32)+4(23)=?-4Multiply-96+92=?-4Add terms-4=-4 Since the left-hand side is equal to the right-hand side, our solution is correct.
Exercises 24 On the left-hand side of the equation, we have products that can be simplified using the Distributive Property. 5(r+9)−2(1−r)=1Distribute 5 & -2r(5)+9(5)+1(-2)−r(-2)=1Multiply5r+45−2+2r=1Add and subtract terms7r+43=1 Now, we can continue to solve using the Properties of Equality. 7r+43=1LHS−43=RHS−437r+43−43=1−43Subtract term7r=-42LHS/7=RHS/777r=7-42Put minus sign in front of fraction77r=-742Calculate quotientr=-6 The solution to the equation is r=-6. We can check our solution by substituting it into the original equation. 5(r+9)−2(1−r)=1r=-65(-6+9)−2(1−(-6))=?1a−(-b)=a+b5(-6+9)−2(1+6)=?1Add terms5(3)−2(7)=?1Multiply15−14=?1Subtract term1=1 Since the left-hand side is equal to the right-hand side, our solution is correct.
Exercises 25 It is given that the sum of all interior angles of the triangle is 180∘. We can thus write an expression for the sum of the interior angles and set it equal to 180∘ to form an equation. 45+2k+k=180 Solving this equation for k will give us the measure of the unknown angle. 45+2k+k=180Add terms45+3k=180LHS−45=RHS−4545+3k−45=180−45Subtract terms3k=135LHS/3=RHS/333k=3135Calculate quotientk=45 The unknown variable k is equal to 45∘. The angle measures are therefore: k=45∘,2⋅k=90∘ and 45∘.
Exercises 26 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘: 2a+a+2a+a=360 Solving this equation for a will give us the measure of the unknown angle. 2a+a+2a+a=360Add terms6a=360LHS/6=RHS/666a=6360Calculate quotienta=60 The unknown variable is 60∘ so the angle measures are a=60∘,2⋅a=120∘,60∘ and 120∘.
Exercises 27 It is given that the sum of all interior angles in the given pentagon is 540∘. Therefore, we can write an equation that adds all of the angles together and is equal to 540∘. (2b−90)+23b+b+(b+45)+90=540 Solving this equation for b will give us the measure of the unknown angle. (2b−90)+23b+b+(b+45)+90=540Remove parentheses2b−90+23b+b+b+45+90=540 Simplify left-hand side Add terms4b+23b+45=540a=22⋅a28b+23b+45=540ca⋅b=ca⋅b28b+23b+45=540Add fractions28b+3b+45=540Add terms211b+45=540ca⋅b=ca⋅b 211b+45=540 Isolate b LHS−45=RHS−45211b+45−45=540−45Subtract terms211b=495LHS⋅2=RHS⋅2211b⋅2=495⋅2Multiply11b=990LHS/11=RHS/111111b=11990Calculate quotient b=90 The unknown variable is 90∘, so we can calculate the angle measures.Given AngleSubstitute b=90Simplify b9090∘ 23b23(90)135∘ b+4590+45135∘ 2b−902(90)−9090∘ 90—90∘
Exercises 28 It is given that the sum of all interior angles in the given hexagon is 720∘. We can thus write an equation that adds all of the angles together and is equal to 720∘: x+120+100+120+(x+10)+120=720 Solving this equation for x will give us the measure of the unknown angle. x+120+100+120+(x+10)+120=720Remove parenthesesx+120+100+120+x+10+120=720Add terms2x+470=720LHS−470=RHS−4702x+470−470=720−470Subtract terms2x=250LHS/2=RHS/222x=2250Calculate quotientx=125 The unknown variable is 125∘ so the angle measures are x=125∘,120∘,100∘,120∘,x+10=135∘ and 120∘.
Exercises 29 The part of the sentence that says "is 75" tells us that something should be equal to 75. …=75 On the other side of the equation, the key phrase is "the sum of," indicating that we are adding two terms. The two terms being added are "twice a number" and "13." The phrase "a number" indicates a variable, which we will call n. 2n +13=… Finally, we can bring both sides of the equation together. The sum of twice a number and 13 is 75.2n +13=75 The next step is to find the number n. 2n+13=75LHS−13=RHS−132n+13−13=75−13Subtract terms2n=62LHS/2=RHS/222n=262Calculate quotientn=31
Exercises 30 We are asked to write an equation and solve it.Writing the equation The last part of the sentence says "is -19", which tells us that something should be equal to -19. The equation so far is shown below. …=-19 The first part of the sentence tells us what should be written on the left-hand side of the equation. The phrasing "the difference of three times a number and 4" means that we should subtract 4 from three multiplied by a number, which can be called x. We can now write the full equation. The difference of three times a number and 4 is -193x − 4=-19Finding the number To find the number, we need to solve the equation we've just written. 3x−4=-19LHS+4=RHS+43x−4+4=-19+4Add terms3x=-15LHS/3=RHS/333x=3-15ca⋅b=ca⋅b33x=3-15Calculate quotientx=-5
Exercises 31 Before we can solve for anything, we need to write an algebraic equation from the given sentence. The word "is" indicates where the equal sign is placed. In this case, we have "is -2." …=-2 The left-hand side of the equation can be broken down into two key phrases, "eight plus" and "the quotient of a number and 3." We can add the first of these to our equation. 8+…=-2 If we let the mystery number be n, then we have n÷3 which can also be written as 3n. We can now form our final equation. Eight plus the quotient of a number and 3 is -28+3n=-2 Finally, let's solve for n. 8+3n=-2LHS−8=RHS−88+3n−8=-2−8Subtract terms3n=-10LHS⋅3=RHS⋅33n⋅3=-10⋅33a⋅3=an=-10⋅3Multiplyn=-30
Exercises 32 The part of the sentence that says "is ten," tells us that something should be equal to 10. So far we have …=10. The first part of the sentence tells us what should be on the left-hand side of the equation. The phrasing "the sum of twice a number and half the number" means that we should add a number multiplied by 2 and the same number multiplied by 21. Let's name the number n. We can now write the full equation. The sum of twice n and half of n is ten 2n + 21n=10 Now, let's solve the equation for n! 2n+21n=10LHS⋅2=RHS⋅22(2n+21n)=2⋅10Distribute 22⋅2n+2⋅21n=2⋅102a⋅2=a2⋅2n+n=2⋅10Multiply4n+n=20Add terms5n=20LHS/5=RHS/5n=4 The solution to the equation is n=4.
Exercises 33 The last part of the sentence, "...is -42," tells us that something should be equal to -42. …=-42 The part that comes before "is -42" tells us what should be on the left-hand side of the equation. "Six times the sum" indicates that two numbers should be added together before being multiplied by 6. 6(the sum of two numbers)=-42 The two numbers being added together are "a number and 15." If we call this unknown number n, we can form our final expression: Six times the sum of a number and 15 is -426(n+15)=-42 We can solve the equation by isolating n on the left-hand side. 6(n+15)=-42LHS/6=RHS/666(n+15)=6-42 Calculate Quotient 6⋅6a=an+15=6-42Put minus sign in front of fractionn+15=-642Calculate quotient n+15=-7LHS−15=RHS−15n+15−15=-7−15Subtract termsn=-22 The solution to the equation is n=-22.
Exercises 34 The last part of the sentence, "...is 12," tells us that something should be equal to 12. …=12 Before the "is 12," we are given "four times the difference." This indicates that a subtraction should be calculated before being multiplied by 4. 4(the difference of two numbers)=12 The numbers being subtracted are "a number and 7." If we call this unknown number n, we can form our final expression. Four times the difference of a number and 7 is 124(n−7)=12 We can solve the equation by isolating n on the left-hand side. 4(n−7)=12LHS/4=RHS/444(n−7)=412 Calculate Quotient 4⋅4a=an−7=412Calculate quotient n−7=3LHS+7=RHS+7n−7+7=3+7Add termsn=10 The solution to the equation is n=10.
Exercises 35 To calculate the total amount earned in a week at one job, we need to calculate the product of earnings per hour and the number of hours worked. At the gas station job, we earn \$8.75 per hour and work 30 hours per week. 30⋅8.75 ⇔ 8.75(30)=gas station earnings We will need to apply this operation to the wage and number of hours worked at the landscaping job as well. We make \$11 per hour at this job, but, we don't yet know the number of hours we have to work to accomplish our goal. Let's call this unknown value t. t⋅11 ⇔ 11t=landscaping earnings By adding these two expressions, we can form one master expression for calculating the total amount earned at our summer jobs each week. Our given goal is \$400, so the only remaining unknown is the number of hours we need to work landscaping t. 11t+8.75(30)=400 Using the Properties of Equality, we will isolate t to find our solution. 11t+8.75(30)=400Multiply11t+262.5=400LHS−262.5=RHS−262.511t+262.5−262.5=400−262.5Subtract terms11t=137.5LHS/11=RHS/111111t=11137.5Use a calculatort=12.5 We must work 12.5 hours at the landscaping job each week to meet our weekly goal of \$400.
Exercises 36 Using the formula for the area of rectangle, we can write an equation to solve for our missing length d. A=ℓ⋅w We are given that the area is A=210 square feet and the width is w=10 feet. The length is found by adding the length of the deep and shallow ends together, ℓ=d+9 feet. A=ℓ⋅w⇒210=(d+9)⋅10 Now we can solve for d. 210=(d+9)⋅10LHS/10=RHS/1021=d+9LHS−9=RHS−912=dRearrange equationd=12 A solution of d=12, in the context of this word problem, means that the deep end of the pool is 12 feet long. To check that our units are correct, we can think back to the formula for area of rectangle. 10 feet ×(12+9) feet =210 square feet
Exercises 37 To write an equation for the given situation, we really need to think about each factor that goes into the calculation of your total cost.Total Cost and Tip The total cost is \$13.80 The \$3 tip is a constant added after the tax and ordered items are already taken into consideration. …+3=13.80Food Purchased Let the unknown cost of a taco be t. We can write an expression for the ordered items, two tacos and a salad, using the fact that the salad is known to be \$2.50. 2t+2.50Sales Tax To account for sales tax, we need to use the percent change in the total cost. The sales tax is 8% so the total cost of the meal is going to be 8% higher. 100%+8%=108% We can multiply the cost of the meal by this percent change. However, for simplicity, let's think about this in the decimal format rather than as a percentage, 108%=1.08.Final Equation Now, we can combine all of the components of the bill to have an equation that we can solve for the cost of tacos. sales tax ×( food cost )+ tip = total cost1.08(2t+2.50)+3=13.80 Now we can solve for t. 1.08(2t+2.50)+3=13.80LHS−3=RHS−31.08(2t+2.50)=10.80LHS/1.08=RHS/1.082t+2.50=10LHS−2.50=RHS−2.502t=7.50LHS/2=RHS/2t=3.75 Each taco costs \$3.75.
Exercises 39 We are shown the solving of an equation and asked to justify each step. To do this, we need to think about what is changing on both sides of the equation with each step.Line 1 to Line 2 Between the first and second lines, 2 is distributed according to the Distributive Property to each term inside the parentheses. 2(x+3)+x=-9Distribute 22(x)+2(3)+x=-9Line 2 to Line 3 Between the second and third lines, the multiplication in 2(x) and 2(3) is simplified. 2(x)+2(3)+x=-9Multiply2x+6+x=-9Line 3 to Line 4 Between the third and fourth lines, the x-terms are combined. 2x+6+x=-9Add terms3x+6=-9Line 4 to Line 5 In the next step, 6 is subtracted from both sides of the equation. 3x+6=-9LHS−6=RHS−63x=-15Line 5 to Line 6 Finally, both sides of the equation are divided by 3. 3x=-15 LHS/3=RHS/3 LHS/3=RHS/333x=3-15Calculate quotientx=3-15Put minus sign in front of fractionx=-315Calculate quotient x=-5Conclusion The steps that were taken to solve the equation are:Distributive Property Simplify. Combine like terms. Subtract 6 from each side. Divide each side by 3.
Exercises 40 When solving an equation, we need to remember to watch our positives and negatives. The error in the given solution can be found in the very first step. When using the Distributive Property, the solution shows that a negative times a negative is a negative, but it should be positive. Shown: Correct: -2(-y)=-2y×-2(-y)=2y✓ With that mistake corrected, let's solve for y. -2(7−y)+4=-4Distribute -2-14+2y+4=-4Add terms-10+2y=-4LHS+10=RHS+102y=6LHS/2=RHS/2y=3
Exercises 41 When solving an equation, we need to remember which operations are inverse. The error in the given solution can be found between the second and third lines. When removing the 41 from the left-hand side, the solution shows a division by 4. To remove the fraction, we should instead be multiplying by its reciprocal, 4. With that mistake corrected, let's solve for x. 41(x−2)+4=12LHS−4=RHS−441(x−2)=8LHS⋅4=RHS⋅4x−2=32LHS+2=RHS+2x=34
Exercises 42 The perimeter P of a rectangle is found by adding the lengths of all sides of the figure. P=ℓ+ℓ+w+w⇒P=2ℓ+2w In the above formula, ℓ is the length and w is the width. In this exercise, we are told that ℓ=2w+6 and that P=228 feet. We can substitute these values into our formula and solve for w. Remember, we want to keep our units in mind as well. P=2ℓ+2wP=228, ℓ=2w+6228=2(2w+6)+2wDistribute 2228=4w+12+2w Solve for w Add terms228=6w+12LHS−12=RHS−12216=6wLHS/6=RHS/636=wRearrange equation w=36 A solution of w=36, in the context of this word problem, tells us that the width of the tennis court is 36 feet. Now we can solve for the length of the tennis court using the expression provided in the diagram. ℓ=2w+6w=36ℓ=2(36)+6Multiplyℓ=72+6Add termsℓ=78 The length of the tennis court is 78 feet.
Exercises 43 The perimeter P of a rectangle is found by adding together the lengths of all of its sides: P=ℓ+ℓ+w+w⇒P=2ℓ+2w, where ℓ is the length and w is the width. In this exercise, we are told that w=y, ℓ=811y, and P=190 inches. We can substitute these values into our formula and solve for y. P=2ℓ+2wSubstitute P=190,ℓ=811y,w=y190=2(811y)+2y Simplify terms a⋅cb=ca⋅b190=82⋅11y+2yba=b/2a/2 190=411y+2y Combine Like Terms Factor out y190=(411+2)ya=44⋅a190=(411+48)yAdd fractions 190=419yLHS⋅194=RHS⋅194190⋅194=y Multiply a⋅cb=ca⋅b19190⋅4=yba=b/19a/19110⋅4=y1a=a10⋅4=yMultiply 40=yRearrange equationy=40 A solution of y=40 tells us that the width of the flag is 40 inches. Now we can solve for the length of the flag using the expression provided in the diagram. ℓ=811yy=40ℓ=811⋅40 Multiply Multiply fractionsℓ=811⋅40ba=b/8a/8ℓ=111⋅5Multiplyℓ=1551a=a ℓ=55 The length of the flag is 55 inches.
Exercises 44 The perimeter P of a pentagon is found by adding together the lengths of all sides of the figure. This crossing sign has two sides equal to s, two equal to s+6 and one side 2s. P=s+s+(s+6)+(s+6)+2s In this exercise, we are told that P=102 inches. By solving for s we will be able to calculate all of the side lengths of the sign. P=s+s+(s+6)+(s+6)+2sP=102102=s+s+(s+6)+(s+6)+2s Simplify Terms Remove parentheses102=s+s+s+6+s+6+2sAdd terms 102=6s+12 Subtract terms LHS−12=RHS−12102−12=6s+12−12Subtract terms 90=6s LHS/6=RHS/6 LHS/6=RHS/6690=66sCalculate quotient 15=sRearrange equations=15 We found that s=15. Let's use this information to find the other side lengths.Side LengthSubstituteSimplify s 15 15 s 15 15 s+6 15+6 21 s+6 15+6 21 2s 2⋅15 21
Exercises 45 There is often more than one way to solve an equation. Here we will look at two methods to solving the same equation and then look at the benefits of each one.Method 1 For the first method, we will use the Distributive Property to solve the equation. 2(4−8x)+6=-1LHS−6=RHS−62(4−8x)=-7Distribute 28−16x=-7 Solve for x LHS−8=RHS−8-16x=-15Change signs16x=15LHS/16=RHS/16 x=1615Method 2 For the second method, we will treat the parentheses as a single quantity. 2(4−8x)+6=-1LHS−6=RHS−62(4−8x)=-7LHS/2=RHS/24−8x=2-7 Solve for x LHS−4=RHS−4-8x=2-7−4Multiply by 22-8x=2-7−4⋅22a⋅cb=ca⋅b-8x=2-7−28Subtract fractions-8x=-215Change signs8x=215LHS/8=RHS/8x=215÷8ba/c=b⋅ca x=1615Benefits of each method In general, there are benefits to both methods. If the numbers are very large, or if the number to be distributed is a fraction, it may be easier to use the second method. In this exercise, the second method proved to be very difficult because 2 does not divide evenly into 7 which caused us to have to work with fractions the entire time. For this exercise, the easier method was definitely Method 1. We were able to avoid fractions until the very end and the numbers remained simple.
Exercises 46 To solve for the number of tickets purchased, which we can call t, we should first write an equation to represent the situation. We know that the total cost of the order is \$220.70 and that the order is charged a flat fee of \$5.90. Let's use these values to begin writing our equation. 220.70=5.90+… Next, let's think about the cost of the tickets themselves. For each ticket purchased, we need to pay the price of the ticket, \$32.50, and the convenience charge \$3.30. (32.50+3.30)t We can combine the total cost, flat fee, and ticket prices to have our equation. 220.70=5.90+(32.50+3.30)t Finally, let's solve for t. 220.70=5.90+(32.50+3.30)tAdd terms220.70=5.90+35.80tLHS−5.90=RHS−5.90214.80=35.80tLHS/35.80=RHS/35.806=tRearrange equationt=6 6 tickets were purchased through the online ticketing agency.
Exercises 47 We must have our units matching throughout, so before we can start solving this exercise, we need to rewrite the total amount of change into cents. \$2.80⇒280 cents Since dimes d are worth 10 cents and quarters q are worth 25 cents, we can multiply the number of dimes by 10 and the number of quarters by 25 to know the total value of having that many of each coin. 280=10d+25q Now, if we have 280 cents made up of quarters and dimes, is it possible to have 8 more quarters than dimes? We can substitute q=d+8 into our equation to find out! 280=10d+25qq=d+8280=10d+25(d+8)Distribute 25280=10d+25d+200 Solve for d Add terms280=35d+200LHS−200=RHS−20080=35dLHS/35=RHS/352.2857=dRearrange equation d=2.2857 A solution of d=2.2857 means that, if we have 8 more quarters than dimes and a total of 280 cents, we would need to have approximately 2.3 dimes. It is not possible to have a partial number of dimes so our friend cannot be correct. Alternative solution info Alternate Equation Another way to write the equation would be to keep the total in dollars, \$2.80, and write the values of the coins in decimal format. Dimes are worth 0.10 dollars and quarters are worth 0.25 dollars. This is how the book chooses to solve the exercise. However, decimals are not as easy to work with so we chose to convert the dollars into cents instead.
Exercises 48 Before we begin, please know that there are many possible solutions to this exercise. The first thing we need to do is assign weights to the three remaining columns.Homework weight =0.20 Midterm Exam weight =0.30 Final Exam weight =0.30 Any combination would be fine, as long as the total adds to 1. 0.20+0.20+0.30+0.30=1 Our table now looks as shown below.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.20Midterm Exam88%0.30Final Exam0.30Total1The next step is to calculate the values in the "Score × Weight" column for the known exam scores. To find these values, we need to multiply the scores by the weight. This column tells us the portion of the final grade awarded by each component of grading.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Exam0.30Total1If the student wants to earn a 90% in the class, we can use this to solve for the portion of the final grade required from the Final Exam score. Right now the student has an overall 63.8% in the class. 18.4+19+26.4=63.8% To earn a 90%, they need 90−63.8=26.2% in the "Score × Weight" column.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Examp0.30p%×0.30=26.2% Total190%Finally, we can solve for the score required on the Final Exam to earn the 90% in the course. p%×0.30=26.2%LHS/0.30=RHS/0.30p%=87.3333% The students needs to earn at least 87.3ˉ% on the Final Exam.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Exam87.3ˉ%0.3087.3ˉ%×0.30=26.2% Total190%
Exercises 49 Even and odd integers alternate. If we begin at 0 and label the next few integers as even or odd, we have the following. 0even,1odd,2even,3odd,4even,5odd,6even Therefore, any integer can be represented by the expression 2n. To get to the next even integer we have to add 2. If the first of the three consecutive even integers is 2n, the other integers must be: First Integer: Second Integer: Third Integer: 2n2n+22n+4 To find three consecutive even integers whose sum is 54, we need to add together these three expressions and isolate n. 2n+(2n+2)+(2n+4)=54Remove parentheses2n+2n+2+2n+4=54Add terms6n+6=64 LHS−6=RHS−6 LHS−6=RHS−66n+6−6=54−6Subtract term 6n=48 LHS/6=RHS/6 LHS/6=RHS/666n=648Calculate quotient n=8 Since the first integer in our set of consecutive even integers is 2n, the first number is 2⋅8=16. Therefore, the second number is 18 and the third number is 20.
Exercises 50
Exercises 51
Exercises 52
Exercises 53
Exercises 54
Exercises 55
Exercises 56
Exercises 57
Exercises 58
Exercises 59
Exercises 60
Exercises 61
Exercises 62
Exercises 63
Exercises 64
Exercises 65
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# 分數的介紹
下載 嵌入 關閉 嵌入啓動模擬教學之圖樣
返回 HTML5 版本
### 標題
• Fractions
• Equivalent Fractions
• Improper Fraction
• Number Line
### 描述
Amber Chang (2012)
HTML5 版本之翻譯：
### 學習目標
• Predict and explain how changing the numerator of a fraction affects the fraction's value
• Predict and explain how changing the denominator of a fraction affects the fraction's value
• Convert between a picture of a fraction, a numeric fraction, and a point on a number line
• Build matching fractions using numbers and pictures
• Compare fractions using numbers and patterns
### 標準對齊
#### 共用核心 - 數學
1.G.A.3
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
2.G.A.2
Partition a rectangle into rows and columns of same-size squares and count to find the total number of them.
2.G.A.3
Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
3.NF.A.1
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
3.NF.A.2
Understand a fraction as a number on the number line; represent fractions on a number line diagram.
3.NF.A.2a
Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line.
3.NF.A.2b
Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line.
3.NF.A.3
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
3.NF.A.3a
Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
3.NF.A.3b
Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3. Explain why the fractions are equivalent, e.g., by using a visual fraction model.
3.NF.A.3d
Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
### 教學提示
模擬教學的控制概觀、模型簡化與引導學生思考 ( PDF ).
### 老師提供的活動
What is a Fraction? Anonymous K-5 其它 數學
Comparing and Ordering Fractins Day 2 of 3 Meaghan Hixson K-5 指引 數學
What is a Fraction? Day 1 of 3 Meaghan Hixson K-5 指引 數學
Equivalent Fractions Day 3 of 3 Meaghan Hixson K-5 指引 數學
Fractions 3 Day Unit Meaghan Hixson K-5 指引 數學
PRIMARIA: Alineación con programas de la SEP México (2011 y 2017) Diana López 國中
K-5
How do PhET simulations fit in my middle school program? Sarah Borenstein 國中 其它 物理
MS and HS TEK to Sim Alignment Elyse Zimmer 國中
분수 소개 SIM 사용설명서 이화국(Wha Kuk Lee) 國中 作業
NGSS Simulation Alignment/Correlation Matthew Huffine 大學-簡介
K-5
Comparing Fractions with Like Numerators and Denominators Margaret McCabe K-5
Building Fractions Michellle Fiumefreddo K-5 概念問題 數學
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NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3
# NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3
## NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3.
### Ex 6.3 Class 6 Maths Question 1.
Find:
(a) 35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (-20) – (13)
(e) 23 – (-12)
(f) (-32) – (-40)
Solution:
(a) 35 – (20) = 35 – 20 = 15
(b) 72 – 90 = –18
(c) (-15) – (-18) = -15 + 18 = 3
(d) (-20) – (13) = -20 – 13 = –[20 + 13] = –33
(e) 23 – (-12) = 23 + 12 = 35
(f) (-32) – (-40) = -32 + 40 = 8
### Ex 6.3 Class 6 Maths Question 2.
Fill in the blanks with >, < or = sign.
(a) (-3) + (-6) _______ (-3) – (-6)
(b) (-21) – (-10) ______ (-31) + (-11)
(c) 45 – (-11) ______ 57 + (-4)
(d) (-25) – (-42) ______ (-42) – (-25)
Solution:
(a) (-3) + (-6) = – [3 + 6] = – 9 and (-3) – (-6) = (-3) + 6 = 3
Here, – 9 < 3
(-3) + (-6) < (-3) – (-6)
(b) (-21) – (-10) = (-21) + 10 = -11 and (-31) + (-11) = –(31 + 11) = –42
Here, -11 > -42
(-21) – (-10) > (-31) + (-11)
(c) 45 – (-11) = 45 + 11 = 56 and 57 + (-4) = 57 – 4 = 53
Here, 56 > 53
45 – (-11) > 57 + (-4)
(d) (-25) – (-42) = -25 + 42 = 17
and (-42) – (-25) = -42 + 25 = -17
Here, 17 > -17
(-25) – (-42) > (-42) – (-25)
### Ex 6.3 Class 6 Maths Question 3.
Fill in the blanks.
(a) (-8) + _____ = 0
(b) 13 + _____ = 0
(c) 12 + (-12) = _____
(d) (-4) + _____ = – 12
(e) _____ -15 = –10
Solution:
(a) (-8) + (additive inverse of -8) = 0
= (-8) + (8) = 0
Value in the blank space is 8.
(b) 13 + (additive inverse of 13) = 0
= 13 + (-13) = 0
Value in the blank space is –13.
(c) 12 + (-12) = 0 [ -12 is additive inverse of 12]
Value in the blank space is 0.
(d) (-4) + (-8) = -[4 + 8] = -12
Value in the blank space is -8.
(e) (+5) – 15 = -10
Value in the blank space is +5.
### Ex 6.3 Class 6 Maths Question 4.
Find:
(a) (-7) – 8 – (-25)
(b) (-13) + 32 – 8 – 1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Solution:
(a) (-7) – 8 – (-25)
= (-7) – 8 + 25
= -15 + 25 = 10
(b) (-13) + 32 – 8 – 1
= (-13) + 32 – (8 + 1)
= 19 – 9
= 10
(c) (-7) + (-8) + (-90) = –(7 + 8) + (-90)
= -15 + (-90)
= -(15 + 90)
= -105
(d) 50 – (-40) – (-2)
= 50 – [-40 – 2]
= 50 – (-42)
= 50 + 42
= 92
You can also like these:
NCERT Solutions for Maths Class 7
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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Think about a delicious hot dog for a snack 🌭. A familiar problem for people who prepare it is knowing how many packs of hot dogs and buns are needed to have no leftovers. Although it sounds challenging, the greatest common factor and least common multiple are properties that are helpful to solve this and other similar situations. These properties will be discussed in this lesson.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Comparing Factors of and
The following applet creates arrangements using the factors of and as the width of an arrangement of blocks. It considers vertical and horizontal arrangements as different arrangements.
Now, think about the following questions.
• How many factors do these two numbers have in common?
• What is the greatest factor of both and that arranges the blocks in rectangles with the same width?
Discussion
Breaking a Number Into Prime Factors
An essential property of any whole number greater than is that it can be expressed as a product of prime numbers. This process is called prime factorization.
Concept
Prime Factorization
Prime factorization, also called complete factorization, is the decomposition of a whole number into a product of its prime factors.
The prime factors are found by dividing the number by the smallest prime number that is a factor of that number. This process is repeated with the quotient until the resulting quotient is a prime number. The following table shows the prime factorization of
Number Smallest Prime Factor Quotient
Prime Factorization
Any whole number greater than can be factored into primes, and its factorization is unique. It should be noted that the prime factorization of a prime number is the number itself. Factor trees are also used to find prime factorizations.
Discussion
Factor Tree
A factor tree is a diagram that shows the prime factors of a number. The tree begins with a root node that contains the number whose prime factorization is needed. Two branches extend from the root node and connect to a factor pair of the number. This process continues by breaking each factor into its factors until only prime factors appear at the end of the branches.
Factor trees help express the prime factorization of a number by using the prime factors at the end of the branches.
Example
Use Factor Trees to Find the Missing Values in a Math Puzzle
Magdalena loves puzzles. She is currently solving a special edition of a Sudoku, a game that consists of a grid with some cells containing numbers and others blank. The purpose of the game is to find the missing values by using the numbers only once in each row and column and in each of the nine boxes.
In this special version of the game, each cell contains a particular math challenge whose solution helps find the corresponding missing number. Magdalena is now focusing on filling in the red and yellow cells.
a The number that goes in the red cell is the most repeated prime factor of the prime factorization of What is the missing value for the red cell?
b The yellow cell must contain the most repeated factor of the prime factorization of What is the missing value for the yellow cell?
Hint
a Make a factor tree of
b Find the prime factorization of by using a factor tree.
Solution
a To find the missing value on the red cell, begin by finding the prime factorization of This can done with a factor tree by following these steps.
1. Create a root node that contains the number
2. From the root node, draw two branches that a factor pair of
3. Break each factor of into its own factor pair.
4. Continue this process until only there are only prime factors on each branch.
With this information in mind, the factor tree of can now be created.
Now, consider that the missing value on the red cell is the most repeated prime factor of the prime factorization of
Here, the number is repeated three times, while the number appears only once. This means that the missing value on the red cell is
b Follow a similar procedure to find the prime factorization of
In this case, the prime factorization of contains only the number six times.
Therefore, the missing value on the yellow cell of the Sudoku game is Magdalena can now fill the yellow and red cells in the Sudoku board!
Discussion
Greatest Common Factor
Factors are used to divide a set of items into equal amounts. However, it may be difficult when two or more different sets are to be divided into a certain number of groups with equal amounts of each item. The common factors of the sets, and the greatest common factor in particular, can help find a solution to this type of problem.
Concept
Greatest Common Factor
Factors that are shared by two or more numbers are called common factors. The greatest of these common factors is called the greatest common factor (GCF). Consider, for example, the factors of and
The common factors of and are and The The GCF can also be determined by multiplying the prime factors shared in the prime factorization of the numbers in question.
It should be noted that the greatest common factor is also called the greatest common divisor because a factor of a number divides that number evenly.
Discussion
Finding the Greatest Common Factor
The greatest common factor (GCF) of two numbers can be determined by finding the prime factorization of the numbers. Next, the common prime factors of the prime factorizations are identified. The GCF is then given by the product of the common prime factors. Consider the following pair of numbers.
The will be found by following these four steps.
1
Find the Prime Factorization of the First Number
expand_more
Use a factor tree to determine the prime factorization of the first number. In this case, the factor tree of must be found.
2
Find the Prime Factorization of the Second Number
expand_more
Now, follow the same process to find the prime factorization of the second number. Consider the factor tree of
3
Identify the Common Prime Factors
expand_more
Write the two prime factorizations together and circle the common prime factors. In this example, the prime factorizations of and will be written together.
4
Multiply the Prime Common Factors
expand_more
The greatest common factor of the numbers is given by the product of the common prime factors of the prime factorization. In this case, the common prime factors of and are and
Therefore, the greatest common factor of and is
Example
Find the GCF to Fill in a Set of Sudoku Cells
Magdalena continues with the process of solving this special version of the Sudoku game. This time, she wants to know which number goes in the purple cells.
The clue is that this purple cells must be filled with the greatest common factor (GCF) of and What number goes into the purple cells?
Hint
Find the prime factorization of Next, find the prime factorization of Identify the common prime factor between the two prime factorizations. Multiply the common prime factors to get the GCF.
Solution
The clue must be solved to help Magdalena figure out the number that goes in the purple cells. The clue claims that the number in the cells is given by the greatest common factor of and To find the GCF, these steps can be followed.
1. Find the prime factorization of
2. Find the prime factorization of
3. Identify the common prime factors between the factorizations of and
4. Multiply the common prime factors.
Each of these steps will now be applied.
Prime Factorization of
The prime factorization of can be found by using a factor tree. The root node of the tree is and then factor pairs of will extend from the root node.
The prime factorization of is
Prime Factorization of
A similar process can be followed to find the prime factorization of
The prime factorization of is
Identify Common Prime Factors
Now that the two prime factorizations are stated, find the common prime factor shared by these factorizations.
The only common prime factors shared by these factorization is the number
Multiply
In this case, only one factor is shared by the prime factorizations of and so the greatest common factor of and is This number goes into the purple cells.
What a great achievement! Magdalena has made great progress on her puzzle.
Explore
Comparing Multiples of and
Consider the following applet that creates different arrangements of blocks using multiples of and
Now, consider the following questions and think about the possible answers to each.
• Is there any multiples in common?
• What is the least of the common multiples?
Discussion
Least Common Multiple
Similar to factors, it may be of interest to find the smallest multiple of two or more different numbers. This number is called the least common multiple.
Concept
Least Common Multiple
The least common multiple (LCM) of two whole numbers and is the smallest whole number that is a multiple of both and It is denoted as The least common multiple of and is the smallest whole number that is divisible by both and Some examples can be seen in the table below.
Numbers Multiples of Numbers Common Multiples Least Common Multiple
and
and
A special procedure exists for finding the of a pair of numeric expressions.
Discussion
Finding the Least Common Multiple
To determine the least common multiple (LCM) of two or more numbers, begin by finding the prime factorization of each number. Then, highlight all the instances of each prime factor in the prime factorization that is repeated most. Finally, the LCM is given by the product of the highlighted factors. This process will be illustrated with this pair of numbers.
The will be found by following these four steps.
1
Find the Prime Factorization of the First Number
expand_more
Use a factor tree to determine the prime factorization of each number. For this situation, the factor tree of will be drawn.
2
Find the Prime Factorization of the Second Number
expand_more
Follow the same process to find the prime factorization of the next number. Consider the factor tree of
3
Match Prime Factors Vertically
expand_more
It can be helpful to write the prime factorizations of the numbers in a table. Place each prime factorization in a row of the table. Use columns for each factor and match common factors vertically when possible. The table for the prime factorization of and is shown here.
4
Bring Down the Primes in Each Column and Multiply
expand_more
Lastly, bring down the prime factors in each column of the table. This process is done with the table containing the prime factorizations of and below.
The product of these prime factors is the least common multiple of the numbers.
Prime Factors Multiplication
and
The least common multiple of and which can also be written as is
Example
Cracking a Pair of Mystery Numbers With the LCM
Magdalena is having a great time solving her special Sudoku puzzle but the next clue looks scary.
The clue claims that to find the numbers to fill in the orange and pink cells, first find the least common multiple (LCM) of and Next, these conditions must be met.
• The number for the orange cells multiplied by is the
• The number for the pink cells multiplied by is the
What are the numbers that Magdalena will write in the orange and pink cells? Write the number for the orange cells first.
Hint
Begin by determining the prime factorization of each number. Label each factor in the prime factorizations that are repeated most. Multiply the labeled factors.
Solution
Before finding the numbers to fill in the blue and pink cells, Magdalena first needs to find the least common multiple (LCM) of and Consider these steps to find the LCM.
1. Determine the prime factorization of
2. Determine the prime factorization of
3. Write the prime factorization in a table to match prime factors vertically.
4. Bring down the primes in each column and multiply them to get the LCM.
Next, each of these steps will be performed.
Prime Factorization of and
Both prime factorizations will be found using factor trees. One tree will have as its root, while the root of the second tree will be Two branches extend from each root node to connect a factor pair. The process is repeated with the factors until only prime factors are on each branch.
Math Prime Factors Vertically
Now that the prime factorizations of and are found write them in a table. In the first row, place the prime factorization of and the factorization of in the second row. The table's columns will contain each factor while matching them vertically when possible.
Bring Down the Primes in Each Column and Multiply
Lastly, bring down the factors of and in each column of the table created previously.
The product of these prime factors is the least common multiple of and
Factors Multiplication
and
The least common multiple of and is
Finding the Numbers to Fill in the Orange and Pink Cells
Magdalena found that the is but she is not ready to fill in the cells in the puzzle yet. For the orange cells, she needs to find the number that gives when multiplied by
The number that goes in the pink cells can be found in a similar way.
This means that the number that goes in the orange cells is and the number that goes in the pink cells is What a big step Magdalena has made now in her solution process!
Pop Quiz
The GCF and the LCM of Random Numbers
Find the greatest common factor (GCF) or the least common multiple (LCM) of the given numbers, as requested.
Discussion
Distributive Property
An important use of the greatest common factor is that it can help simplify numeric expressions. Consider, for example, the following sum.
The greatest common factor of these numbers is This means that each number can be rewritten using this common factor.
The greatest common factor can be pulled out of each addend. This factor will be multiplied by the sum of the numbers left after pulling it out.
The property applied to pull the greatest common factor out a sum is called the Distributive Property. This property states that multiplying a number by the sum of two or more addends produces the same result as multiplying the number by each addend individually and then adding all the products together.
The factor outside the parentheses is multiplied by, or distributed to, every term inside the parentheses. It should be noted that pulling out a common factor can make calculations more straightforward in some cases, but distributing that factor may be better in other calculations. Which way is better for the calculations will be dictated by the situation being studied.
Example
Creating Different Bouquets With the Same Amount of Flowers
Exhausted from solving math problems to solve the puzzle, Magdalena goes to the kitchen to have something to eat. She chooses a delicious slice of Italian pizza.
She is just about to return to her puzzle when her mother asks her for some help. Magdalena's mother works as a volunteer in a retirement home. She plans to gift some bouquets to the ladies in the retirement home next weekend.
She has roses and tulips. Each bouquet will have the same number of flowers and contain only roses or only tulips.
a What is the greatest number of flowers that Magdalena and her mother can use in each bouquet?
b How many bouquets can Magdalena and her mother create?
Hint
a Find the greatest common factor of and
b Write an expression for the total number of flowers. Use the Distributive Property to rewrite the expression for the total number of flowers.
Solution
a Magdalena and her mother want each of the bouquets to have an equal number of flowers. This number must be the greatest possible number of flowers. Also, each bouquet must contain only one type of flower. In other words, they need to find the greatest common factor of and
First, find the prime factorizations of these numbers using a factor tree.
Now that the prime factorizations have been found, look for the common prime factors between them. |
My Math Forum quick q
Algebra Pre-Algebra and Basic Algebra Math Forum
June 15th, 2010, 09:44 AM #1 Newbie Joined: Jun 2010 Posts: 24 Thanks: 0 quick q given 2x+3y=12, what is the slope, y intercept, and what would the graph look like is the slope??? m= -3 y int: b= 6 any help? thanks
June 15th, 2010, 10:15 AM #2 Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: quick q The slope and y intercept are most easily identified when the equation is in "slope intercept form" y=mx + b
June 15th, 2010, 10:25 AM #3
Member
Joined: Feb 2010
Posts: 53
Thanks: 0
Re: quick q
Quote:
Originally Posted by bignick79 given 2x+3y=12, what is the slope, y intercept, and what would the graph look like is the slope??? m= -3 y int: b= 6 any help? thanks
Remeber, if you can get a linear equation like this one into the following form:
$y= mx + b$
Then you know the y-intercept and the slope.
$SLOPE= m$
$Y-INTERCEPT= b$
So, we have the equation:
$2x+3y=12$
Subtracting 2x from both sides gives:
$3y= 12 - 2x$
Then dividing both sides (meaning the whole of both sides) by 3 gives:
$y= \frac{12 - 2x}{3}$
this becomes, by basic rules of fractions, the following:
$y= \frac{-2}{3}x + 4$
This is in the form:
$y= mx + b$
So,
$m= \frac{-2}{3}$
and
$b= 4$
This graph would look like a line, slanted from top left to bottom right, intersecting the y-axis at four. Basically draw a point at (0, 4) and move down two [-2 in the slope, m] and to the right 3 [+3 in the slope, m] to find your next point. Continue this to get a picture of the graph, in both directions. Hope this helps, anymore questions post them here and I'll try to clarify.
June 15th, 2010, 10:41 AM #4
Math Team
Joined: Dec 2006
From: Lexington, MA
Posts: 3,267
Thanks: 408
Re: quick q
Hello, bignick79!
Quote:
Given:[color=beige] .[/color]$2x\,+\,3y \:=\:12$ What is the slope, y-intercept, and what would the graph look like?
Get the equation into the form:[color=beige] .[/color]$y \:=\:mx\,+\,b$
[color=beige]. . [/color]That is, solve for $y.$
$\text{We have: }\:2x\,+\,3y \:=\:12$
[color=beige]. . . [/color]$\text{Then: }\;\;\;\;\;\;\;\;\;3y \:=\:-2x\,+\,12$
[color=beige]. .[/color]$\text{Finally: }\;\;\;\;\;\;\;\;\;y \:=\;-\frac{2}{3}x \,+\,4$
[color=beige]. . . . . . . . . . . . . . . . . . [/color]$\uparrow$[color=beige] - - [/color]$\uparrow$
[color=beige]. . . . . . . . . . . . . . . . [/color]slope[color=beige] . [/color]y-intercept
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This line has a y-intercept at $(0,\,4)$
$\text{The slope is: }\:\frac{-2}{+3} \:=\:\frac{\text{rise}}{\text{run}}$
$\text{It has a "run" of +3 . . . and a "rise" of -2.}$
$\text{So, from (0, 4), we move: right 3 units and down 2 units.}$
[color=beige]. . [/color]$\text{This locates a second point and we can draw the line.}$
Code:
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| * : -2
| * :
| * :
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- - - + - - - - - - - - - - - * - - -
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2022-01-10
The top view of a circular table shown on the right has a radius of 120cm.find the area of the smaller segment of the table (shaded region) determined by 60° arc
star233
Area of segment (the shaded region) = Area of sector - area of triangle
Where:
Area of sector $=\left(\frac{\theta \pi }{360}\right){r}^{2}$
Area of segment $=\left(\frac{\mathrm{sin}\theta }{2}\right){r}^{2}$
Derive the equation:
Area of segment $=\left(\frac{\theta \pi }{360}\right){r}^{2}$$\left(\frac{\mathrm{sin}\theta }{2}\right){r}^{2}$
Area of segment $={r}^{2}\left(\frac{\theta \pi }{360}-\frac{\mathrm{sin}\theta }{2}\right)$
Given:
Central angle, $\theta =60°$
Radius, $r=120$ cm
pi, $\pi \approx 3.14$
Solve for the area of segment or shaded region:
Area of segment $={r}^{2}\left(\frac{\theta \pi }{360}-\frac{\mathrm{sin}\theta }{2}\right)$
Area = ($120$ cm)${}^{2}$ $\left[60×\frac{3.14}{360}-\frac{\mathrm{sin}60}{2}\right]$
Area = $14,400$ cm${}^{2}$ $\left[0.523-\frac{0.866}{2}\right]$
Area = $14,400$ cm${}^{2}$ $\left[0.523-0.433\right]$
Area = $14,400$ cm${}^{2}$ $\left(0.09\right)$
Area of segment or shaded region = $1,296$ cm${}^{2}$
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# Table of 26
Tables are the basics of Mathematics based on which many multiplication calculations could be simplified quickly. Table of 26 gives the repeated addition of number 26, repeated for a certain number of times. For example, 4 baskets of 26 apples each, gives the sum of 26 + 26 + 26 + 26 = 104. Therefore, the total number of apples in all the 4 baskets are 104, or we can also write it as, 4 x 26 = 104, where 4 is the number of baskets and 26 is the number of apples in each basket.
The multiplication tables are useful to all the level of students studying from Classes 1st to 12th.
By memorizing the tables 1 to 20, we can spare our time from consuming in typical multiplication problems. These tables can be used by candidates appearing for competitive examinations or common entrance test, which includes maths aptitude questions. The table of 26 is provided below which can be used for maths aptitude calculations.
## Maths Table of 26
Let us draw the table of 26 from 1 to 50 times.
26 x 1 = 26 26 x 2 = 52 26 x 3 = 78 26 x 4 = 104 26 x 5 = 130 26 x 6 = 156 26 x 7 = 182 26 x 8 = 208 26 x 9 = 234 26 x 10 = 260 26 x 11 = 286 26 x 12 = 312 26 x 13 = 338 26 x 14 = 364 26 x 15 = 390 26 x 16 = 416 26 x 17 = 442 26 x 18 = 468 26 x 19 = 494 26 x 20 = 520 26 x 21 = 546 26 x 22 = 572 26 x 23 = 598 26 x 24 = 624 26 x 25 = 650 26 x 26 = 676 26 x 27 = 702 26 x 28 = 728 26 x 29 = 754 26 x 30 = 780 26 x 31 = 806 26 x 32 = 832 26 x 33 = 858 26 x 34 = 884 26 x 35 = 910 26 x 36 = 936 26 x 37 = 962 26 x 38 = 988 26 x 39 = 1014 26 x 40 = 1040 26 x 41 = 1066 26 x 42 = 1092 26 x 43 = 1118 26 x 44 = 1144 26 x 45 = 1170 26 x 46 = 1196 26 x 47 = 1222 26 x 48 = 1248 26 x 49 = 1274 26 x 50 = 1300 |
# Chapter 7 - Newton's Third Law - Exercises and Problems - Page 188: 40
The acceleration of the 2.0-kg block is $2.29~m/s^2$
#### Work Step by Step
Note that the three blocks will all have the same magnitude of acceleration and the 3.0-kg will move downward while the 1.0-kg moves upward. We can set up a force equation for the 1.0-kg block. Let $T_1$ be the tension in the rope attached to this block. Let $m_1$ be the mass of this block. $\sum F = m_1~a$ $T_1- m_1~g= m_1~a$ $T_1= m_1~(g+a)$ We can set up a force equation for the 3.0-kg block. Let $T_2$ be the tension in the rope attached to this block. Let $m_3$ be the mass of this block. $\sum F = m_3~a$ $m_3~g - T_2 = m_3~a$ $T_2 = m_3~(g-a)$ To find the acceleration of the 2.0-kg block, we can set up a force equation for the 2.0-kg block. Let $m_2$ be the mass of this block. $\sum F = m_2~a$ $T_2 - T_1 - F_f = m_2~a$ $m_3~(g-a) - m_1~(g+a) - m_2~g~\mu_k = m_2~a$ $m_3~g - m_1~g - m_2~g~\mu_k = m_1~a+m_2~a+m_3~a$ $a = \frac{m_3~g - m_1~g - m_2~g~\mu_k}{ m_1+m_2+m_3}$ $a = \frac{(3.0~kg)(9.80~m/s^2) - (1.0~kg)(9.80~m/s^2) - (2.0~kg)(9.80~m/s^2)(0.30)}{1.0~kg+2.0~kg+3.0~kg}$ $a = 2.29~m/s^2$ The acceleration of the 2.0-kg block is $2.29~m/s^2$
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. |
# Introduction to Complex Numbers
Hint 1; Complex numbers are not more complicated than any other numbers, just different.
Complex number = real number + imaginary number
Hint 2; imaginary numbers are not more ethereal than real numbers, just different.
Imaginary numbers were invented to solve problems involving square roots of negative numbers.
Hint 3; we lied to you when we told you that you cannot take the square root of a negative number. We just waited until now to tell you how to do it.
• i = √(-1)
The letter i is used to signify the square root of -1. Any number multiplied by i is an imaginary number.
Thus the square root of any negative number equals the square root of its positive value multiplied by i. Yes, it really is that easy. Examples
• √(-25) = 5i
• √(-7) = i√(7)
• √(-12) = i√(12) = 2i√(3)
The powers of i follow a repeating pattern that is illustrated below.
• i0 = 1 because any number (even imaginary ones) raised to the power of zero equals 1
• i1 = i because any number raised to the power of 1 equals that number
• i2 = -1 because just like [√(3)]2 = 3, so does [√(-1)]2 = -1
• i3 = -i because i3 = (i)(i2) = (i)(-1) = -i
Ok, that was harder but go over it a couple times to convince yourself and the next part is easy.
The value of in can be found by dividing n by 4 and matching the remainder with the power of i in the pattern above.
For example;
• i53 = i because the remainder of 53 divided by 4 is 1 and i1 = i
• i18 = -1 because the remainder of 18 divided by 4 is 2 and i2 = -1
• i31 = -i because the remainder of 31 divided by 4 is 3 and i3 = -i
• i48 = 1 because the remainder of 50 divided by 4 is 0 and i0 = 1
The conjugate of a complex number is another complex number except the imaginary part have opposite signs.
• 6 + 3i has a complex conjugate of 6 - 3i
• 6 - 3i has a complex conjugate of 6 + 3i
• -6 + 3i has a complex conjugate of -6 - 3i
• -6 - 3i has a complex conjugate of -6 + 3i
The conjugate is important because the product of a complex number and its conjugate is always a real number because the sum of the inner products is always zero. For example
(4 - 3i)(4 + 3i) = 16 - 12i + 12i - 9i^2) = 16 - (9)(-1) = 16 + 9 = 25
Hint 4: If you have gotten to here then you have a basic introduction to complex numbers.
Carry on!
Steven G. |
# 11.2: Simplification of Radical Expressions
Difficulty Level: Basic Created by: CK-12
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Suppose that a shoemaker has determined that the optimal weight in ounces of a pair of running shoes is 200004\begin{align*}\sqrt[4]{20000}\end{align*}. How many ounces would this be? Is there a way that you could rewrite this expression to make it easier to grasp? In this Concept, you'll learn how to simplify radical expressions like this one so that you can write them in multiple ways.
### Guidance
Radicals are the roots of values. In fact, the word radical comes from the Latin word “radix,” meaning “root.” You are most comfortable with the square root symbol x\begin{align*}\sqrt{x}\end{align*}; however, there are many more radical symbols.
A radical is a mathematical expression involving a root by means of a radical sign.
y3=xy4=xyn=xbecause x3=ybecause x4=ybecause xn=y273=3, because 33=27164=2 because 24=16\begin{align*}\sqrt[3]{y}=x && \text{because} \ x^3=y && \sqrt[3]{27}=3, \ because \ 3^3=27\\ \sqrt[4]{y}=x && \text{because} \ x^4=y && \sqrt[4]{16}=2 \ because \ 2^4=16\\ \sqrt[n]{y}=x && \text{because} \ x^n=y && \end{align*}
Some roots do not have real values; in this case, they are called undefined.
Even roots of negative numbers are undefined.
xn\begin{align*}\sqrt[n]{x}\end{align*} is undefined when n\begin{align*}n\end{align*} is an even whole number and x<0\begin{align*}x<0\end{align*}.
#### Example A
• 643\begin{align*}\sqrt[3]{64}\end{align*}
• 814\begin{align*}\sqrt[4]{-81}\end{align*}
Solution:
643=4\begin{align*}\sqrt[3]{64} = 4\end{align*} because 43=64\begin{align*}4^3=64\end{align*}
814\begin{align*}\sqrt[4]{-81}\end{align*} is undefined because n\begin{align*}n\end{align*} is an even whole number and 81<0\begin{align*}-81<0\end{align*}.
In a previous Concept, you learned how to evaluate rational exponents:
axy where x=power and y=root\begin{align*}a^{\frac{x}{y}} \ where \ x=power \ and \ y=root\end{align*}
This can be written in radical notation using the following property.
Rational Exponent Property: For integer values of x\begin{align*}x\end{align*} and whole values of y\begin{align*}y\end{align*}:
axy=axy\begin{align*}a^{\frac{x}{y}}= \sqrt[y]{a^x}\end{align*}
#### Example B
Rewrite x56\begin{align*}x^{\frac{5}{6}}\end{align*} using radical notation.
Solution:
This is correctly read as the sixth root of x\begin{align*}x\end{align*} to the fifth power. Writing in radical notation, x56=x56\begin{align*}x^{\frac{5}{6}}=\sqrt[6]{x^5}\end{align*}, where x5>0\begin{align*}x^5>0\end{align*}.
You can also simplify other radicals, like cube roots and fourth roots.
#### Example C
Simplify 1353\begin{align*}\sqrt[3]{135}\end{align*}.
Solution:
Begin by finding the prime factorization of 135. This is easily done by using a factor tree.
1353=33353=33353353\begin{align*}&\sqrt[3]{135}= \sqrt[3]{3 \cdot 3 \cdot 3 \cdot 5} = \sqrt[3]{3^3} \cdot \sqrt[3]{5}\\ & 3 \sqrt[3]{5}\end{align*}
### Guided Practice
Evaluate \begin{align*}\sqrt[4]{4^2}\end{align*}.
Solution: This is read, “The fourth root of four to the second power.”
\begin{align*}4^2=16\end{align*}
The fourth root of 16 is 2; therefore,
\begin{align*}\sqrt[4]{4^2}=2\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Radical Expressions with Higher Roots (8:46)
1. For which values of \begin{align*}n\end{align*} is \begin{align*}\sqrt[n]{-16}\end{align*} undefined?
1. \begin{align*}\sqrt{169}\end{align*}
2. \begin{align*}\sqrt[4]{81}\end{align*}
3. \begin{align*}\sqrt[3]{-125}\end{align*}
4. \begin{align*}\sqrt[5]{1024}\end{align*}
Write each expression as a rational exponent.
1. \begin{align*}\sqrt[3]{14}\end{align*}
2. \begin{align*}\sqrt[4]{zw}\end{align*}
3. \begin{align*}\sqrt{a}\end{align*}
4. \begin{align*}\sqrt[9]{y^3}\end{align*}
Write the following expressions in simplest radical form.
1. \begin{align*}\sqrt{24}\end{align*}
2. \begin{align*}\sqrt{300}\end{align*}
3. \begin{align*}\sqrt[5]{96}\end{align*}
4. \begin{align*}\sqrt{\frac{240}{567}}\end{align*}
5. \begin{align*}\sqrt[3]{500}\end{align*}
6. \begin{align*}\sqrt[6]{64x^8}\end{align*}
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### Vocabulary Language: English Spanish
TermDefinition
radical A mathematical expression involving a root by means of a radical sign. The word radical comes from the Latin word radix, meaning root.
Rational Exponent Property For integer values of $x$ and whole values of $y$: $a^{\frac{x}{y}}= \sqrt[y]{a^x}$
Rationalize the denominator To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
Variable Expression A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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# 6.7: Area of Regular Polygons
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Recognize and use the terms involved in developing formulas for regular polygons.
• Calculate the area and perimeter of a regular polygon.
You already know how to find areas and perimeters of some figures – triangles, parallelograms, and other quadrilaterals. Not surprisingly, the new formulas in this lesson will build on those basic figures – in particular, the triangle.
## Parts and Terms for Regular Polygons
Do you remember the names of different polygons from Chapter 1?
First of all, “poly-” means “many” and “-gon” refers to the sides of a shape.
A regular polygon is a shape whose many sides are all congruent.
Regular polygons also have congruent interior angles and congruent central angles (which you will learn about on the next page.)
Polygons are classified by how many sides they have.
Here are a few names to review:
• A pentagon has 5 sides.
• A hexagon has 6 sides.
• A heptagon has 7 sides.
• An octagon has ______ sides. (Hint: how many legs does an octopus have?)
• A nonagon has 9 sides.
• A decagon has ______ sides. (Hint: how many years are in a decade?)
Here is a general regular polygon with \begin{align*}n\end{align*} sides, where \begin{align*}n\end{align*} stands for some number. Some of its sides are shown in the diagram:
In the diagram, here is what each variable represents:
• \begin{align*}s\end{align*} is the length of each side of the polygon.
• \begin{align*}r\end{align*} is the length of a “radius” of the polygon, which is a segment from the center of the polygon to a vertex (or corner).
• \begin{align*}x\end{align*} is the length of one-half of a side of the polygon (so \begin{align*}x = \frac{1}{2} \ s\end{align*} or \begin{align*}2x = s\end{align*}).
• \begin{align*}a\end{align*} is the length of a segment called the apothem — a segment from the center to a side of the polygon, perpendicular to the side. (Notice that \begin{align*}a\end{align*} is the altitude of each of the triangles formed by two radii and a side.)
A triangle would have \begin{align*}n = \underline{\;\;\;\;\;\;}\end{align*} sides.
A square would have \begin{align*}n = \underline{\;\;\;\;\;\;}\end{align*} sides.
An octagon would have \begin{align*}n = \underline{\;\;\;\;\;\;}\end{align*} sides.
The angle between two consecutive radii measures \begin{align*}\frac{360^\circ}{n}\end{align*} because \begin{align*}n\end{align*} congruent central angles are formed by the radii from the center to each of the \begin{align*}n\end{align*} vertices of the polygon.
We can figure this out because an entire circle is \begin{align*}360^\circ\end{align*}, and you can think of the center of the polygon as having a circle of angles around it. If there are \begin{align*}n\end{align*} central angles (all equivalent), each central angle between each radius is \begin{align*}\frac{360^\circ}{n}\end{align*}.
An apothem divides each of these central angles into two congruent halves; each of these half angles measures \begin{align*}\frac{1}{2} \cdot \frac{360^\circ}{n} = \frac{360^\circ}{2n} = \frac{180^\circ}{n}\end{align*}.
## Perimeter of a Regular Polygon
We continue with the regular polygon diagrammed on the previous page. Let \begin{align*}P\end{align*} be the perimeter. Remember that _____ is the number of sides in the polygon and _____ is the length of each side. In simplest terms,
\begin{align*}P = ns\end{align*}
We know this because the perimeter of a shape is the sum of ______________________. Another way to express perimeter is the number of sides times the length of each side.
Example 1
A square has a radius of 6 inches. What is the perimeter of the square?
Notice that a side and two radii make an isosceles right triangle:
• The triangle is isosceles because the legs of the triangle are each a radius of the square. Each radius is _____ inches long and both are the same length, so the triangle is isosceles because its legs are congruent.
• The triangle has a right angle because the central angle is \begin{align*}\frac{360^\circ}{n}\end{align*} and the square has 4 sides (which means \begin{align*}n = 4\end{align*}) so each central angle is \begin{align*}\frac{360^\circ}{4} = 90^\circ\end{align*}.
Not only is this an isosceles right triangle, but it is also a 45–45–90 triangle!
You may remember that if the legs are each _____ inches long, then the hypotenuse of the triangle is \begin{align*}6 \sqrt{2}\end{align*} inches long. Notice that the hypotenuse is also a side of the square.
To find the perimeter, use the formula \begin{align*}P = ns\end{align*}. We know \begin{align*}n = 4\end{align*} and \begin{align*}s = 6 \sqrt{2}\end{align*}.
\begin{align*}P & = ns\\ & = 4 \cdot 6 \sqrt{2} = 24 \sqrt{2} \ \text{inches \ (on a calculator, this length} \approx 33.9 \ \text{inches})\end{align*}
The perimeter of the square is \begin{align*}24 \sqrt{2}\end{align*} inches.
## Area of a Regular Polygon
The next logical step is to complete our study of regular polygons by developing area formulas.
Take another look at the regular polygon figure below (it is the same one you saw earlier in this lesson.) Here’s how we can find its area, \begin{align*}A\end{align*}.
Two radii and a side make a triangle with base \begin{align*}s\end{align*} and altitude \begin{align*}a\end{align*}:
There are \begin{align*}n\end{align*} of these triangles in the polygon.
The area of each triangle is: \begin{align*}\frac{1}{2} base \cdot height = \frac{1}{2} sa\end{align*}
The entire area of the polygon is:
\begin{align*}A &= \text{number of triangles} \cdot \text{area of each triangle}\\ A &= n \left( \frac{1}{2} sa \right ) = \frac{1}{2} (ns)a = \frac{1}{2}(Pa) \ \text{because perimeter} \ P = ns\end{align*}
Therefore, the Area of a regular polygon with perimeter \begin{align*}P\end{align*} and apothem \begin{align*}a\end{align*}:
\begin{align*}A = \frac{1}{2} \ Pa\end{align*}
1. How many sides does a pentagon have?
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
2. True or false: All sides of a regular polygon are the same length.
3. If you know the length of one side of a regular pentagon, can you find its perimeter? How? Explain the steps you would use.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
4. True or false: A regular polygon that has \begin{align*}n\end{align*} sides also has \begin{align*}n\end{align*} vertices.
5. In the figure below (you have seen it a few times already!),
a. What does \begin{align*}a\end{align*} stand for? ___________________________
Describe what this is:
b. What does \begin{align*}r\end{align*} stand for? ___________________________
c. What does \begin{align*}s\end{align*} stand for? ___________________________
## Graphic Organizer for Lessons 2 – 6: Area
Shape Draw a Picture Area Formula What does each letter in the Area Formula stand for?
Parallelogram
Triangle
Trapezoid
Rhombus
Kite
Regular
Polygon
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# How to solve for y
It’s important to keep them in mind when trying to figure out How to solve for y. We can solve math word problems.
## How can we solve for y
College algebra students learn How to solve for y, and manipulate different types of functions. If you're struggling to remember all of the trigonometric identities, there's now an app for that! With a trig identities solver, all you need to do is input the equation you're trying to solve and it will give you the answer. No more memorizing dozens of identities!
A triangle solver is a useful tool for finding the area of a triangle. It works by taking into account the size of each side and then comparing them to each other to find the average size of each side. The calculation can be done in one of two ways: either treating the sides as equal, or by calculating the difference between the three measurements. The latter method is more accurate and less prone to rounding error, but it’s also more complex. In most cases, calculating the difference is not necessary and just treating both sides as equal will suffice. However, if you have very small sides that are difficult to measure accurately, you may want to consider using this option. • Solving triangles by area: This method requires determining the area of each triangle’s base. To do this, multiply each side’s length (in centimeters) by its corresponding value from the table below (to convert values into inches, divide by 25.4). Subtract these results from 100. The result is the total base area (in square centimeters). Next, use a calculator to find the area of the triangle’s height (in square centimeters). Finally, use a formula to find the total area of all three triangles (in square centimeters). • Solving triangles by height: This method involves finding the difference between each side’s height (in centimeters),
Factorization is a process that involves breaking down a large number into smaller pieces. The key to factorization is being able to break down large numbers into their prime factors. If you are having trouble doing this, check out some resources on the internet that will walk you through this process step by step. Once you are comfortable with factorization, it will be much easier for you to solve quadratic equations. If you have any questions or comments about this article, please feel free to leave a message in the comment section below.
It's also a good tool for students who want to learn how to solve equations on their own, without having to rely on someone else. The steps can be simplified or complex depending on your needs. You can also save the steps you've solved so you can refer back to them later.
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# Difference between revisions of "2008 AMC 10B Problems/Problem 24"
## Problem
Quadrilateral $ABCD$ has $AB = BC = CD$, angle $ABC = 70$ and angle $BCD = 170$. What is the measure of angle $BAD$?
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$
## Solution
### Solution 1
Draw the angle bisectors of the angles $ABC$ and $BCD$. These two bisectors obviously intersect. Let their intersection be $P$. We will now prove that $P$ lies on the segment $AD$.
Note that the triangles $ABP$ and $CBP$ are congruent, as they share the side $BP$, and we have $AB=BC$ and $\angle ABP = \angle CBP$.
Also note that for similar reasons the triangles $CBP$ and $CDP$ are congruent.
Now we can compute their inner angles. $BP$ is the bisector of the angle $ABC$, hence $\angle ABP = \angle CBP = 35^\circ$, and thus also $\angle CDP = 35^\circ$. (Faster Solution picks up here) $CP$ is the bisector of the angle $BCD$, hence $\angle BCP = \angle DCP = 85^\circ$, and thus also $\angle BAP = 85^\circ$.
It follows that $\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ$. Thus the angle $APD$ has $180^\circ$, and hence $P$ does indeed lie on $AD$. Then obviously $\angle BAD = \angle BAP = \boxed{ 85^\circ }$.
$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,N); label("P",P,W); label("35^\circ",B + dir(180-17.5)); label("35^\circ",B + dir(180-35-17.5)); label("85^\circ",C + .5*dir(120+42.5)); label("85^\circ",C + .5*dir(120+85+42.5)); [/asy]$
Faster Solution: Because we now know three angles, we can subtract to get $360 - 35 - 85 - 85 - 35 - 35$, or $\boxed{85}$.
### Solution 2
Draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. Then, $\triangle ABC$ and $\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\angle BAC = 55^{\circ}$, $\angle CBD = 5^{\circ}$, and $\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}$. Draw $E'$ such that $EE'B = 60^{\circ}$ and so that $E'$ is on $\overline{AE}$, and draw $E''$ such that $\angle EE''C = 60^{\circ}$ and $E''$ is on $\overline{DE}$. It follows that $\triangle BEE'$ and $\triangle CEE''$ are both equilateral. Also, it is easy to see that $\triangle BEC \cong \triangle DE''C$ and $\triangle BCE \cong \triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\triangle ADE$ is isosceles. Since $\angle AED = 120^{\circ}$, then $\angle DAC = \frac{180 - 120}{2} = 30^{\circ}$, and $\angle BAD = 30 + 55 = 85^{\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label("A",(-0.096,0.005),NE*lsf); dot((1,0),ds); label("B",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label("C",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label("D",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label("E",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label("E'",(0.1,0.23),NE*lsf); label("60^\circ",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label("E''",(0.423,0.957),NE*lsf); label("60^\circ",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
### Solution 3
Again, draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\angle ABE = 65^\circ$ and $m\angle DCE = 115^\circ$. Applying the Law of Sines to $\triangle AEB$ and $\triangle EDC$, it follows that $DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA$, so $\triangle AED$ is isosceles. We finish as we did in solution 2.
$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,N); label("E",P,W); [/asy]$
### Solution 4
Start off with the same diagram as solution 1. Now draw $\overline{CA}$ which creates isosceles $\triangle CAB$. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is $\boxed{85}.$
### Solution 5 (Cheap solution)
Draw the diagram accurately with a protractor and ruler. $\angle BAD$ comes out to be $\boxed{85}$, or $\boxed{\text{C}}$. (You should always carry a protractor and ruler to a competition.) |
# How do you differentiate f(x)=sqrt(cote^(4x) using the chain rule.?
Mar 5, 2018
$f ' \left(x\right) = \frac{- 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right) {\left(\cot \left({e}^{4 x}\right)\right)}^{- \frac{1}{2}}}{2}$
color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x))
#### Explanation:
$f \left(x\right) = \sqrt{\cot \left({e}^{4 x}\right)}$
$\textcolor{w h i t e}{f \left(x\right)} = \sqrt{g \left(x\right)}$
$f ' \left(x\right) = \frac{1}{2} \cdot {\left(g \left(x\right)\right)}^{- \frac{1}{2}} \cdot g ' \left(x\right)$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{g ' \left(x\right) {\left(g \left(x\right)\right)}^{- \frac{1}{2}}}{2}$
$g \left(x\right) = \cot \left({e}^{4 x}\right)$
$\textcolor{w h i t e}{g \left(x\right)} = \cot \left(h \left(x\right)\right)$
$g ' \left(x\right) = - h ' \left(x\right) {\csc}^{2} \left(h \left(x\right)\right)$
$h \left(x\right) = {e}^{4 x}$
$\textcolor{w h i t e}{h \left(x\right)} = {e}^{j \left(x\right)}$
$h ' \left(x\right) = j ' \left(x\right) {e}^{j \left(x\right)}$
$j \left(x\right) = 4 x$
$j ' \left(x\right) = 4$
$h ' \left(x\right) = 4 {e}^{4 x}$
$g ' \left(x\right) = - 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right)$
$f ' \left(x\right) = \frac{- 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right) {\left(\cot \left({e}^{4 x}\right)\right)}^{- \frac{1}{2}}}{2}$
color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x)) |
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Exponents with variables can be quite confusing. When you multiply two numbers whose exponents are both variable, you get a result that is also variable. For example, let's say you have the variable x, and the number y = 6x + 5. In this case, the exponent of y is variable because x is a variable. Now let's say you want to solve for y because you know that the exponent of y is 4. How do you solve this problem? You would factor out the variable x from both sides of the equation and find 4y = 4x + 1. This gives you the answer for y because now you know that 4y = 4(x + 1) = 4x –1. When this happens, we say that there is an "intractable" relationship between the variables on one side of an equation when they cannot be separated.
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# In the figure above, point O is the center of the circle and OC = AC = AB. What is the value of x?
In the figure above, point O is the center of the circle and OC = AC = AB. What is the value of x?
(A) 40
(B) 36
(C) 34
(0) 32
(E) 3
## Explanation + a simple rule that will help you on challenging GMAT triangle questions (scroll down for video solution)
This GMAT geometry question from the official guide (quantitative review) hinges on a simple rule that is overrepresented on challenging GMAT geometry questions involving triangles.
What is it?: the exterior angle theorem!
Eh? Sounds complicated but it’s actually pretty basic one you see it in action.
Angle ACB above is an “exterior angle” of triangle AOC. The exterior angle theorem states that the exterior angle ACB equals the sum of the “remote interior angles” OAC and AOC.
Basically, the exterior angle equals the sum of the angles that it’s not connected to, the other two angles of the triangle known as the “remote interior angles”.
If triangles share sides it’s not unlikely that you’ll have an exterior angle and will be able to use (and will probably need to use) the exterior angle theorem. So, keep EAT (exterior angle theorem) in mind.
### Here are the steps that I’d follow to solve:
1. Draw a diagram and make easy inferences. The first one to label is that OC = AC = AB.
2. Once you have those equal sides you can also label angles that are equal as sides that are equal within the same triangle have angles opposite them that are also equal.
3. Now you should be thinking, hmmm, triangles sharing sides maybe there’s an exterior angle. Yes, of course there is! Angle ACB. ACB is equal to AOC + OAC.
4. Now you should also be wondering why this triangle is in a circle. How can the circle help you make an inference? For GMAT circle questions always think about “helpful radii”. There’s usually a radius whether it’s drawn or you have to draw it that will help. In this case OA and OB are both radii so they are equal boom!
5. That’s it. You should have all angles defined by x. Go ahead and sum all of the x’s and set that equal to 180 to solve. |
# RD Sharma Solutions Chapter 11 Constructions Exercise 11.3 Class 10 Maths
Chapter Name RD Sharma Chapter 11 Constructions Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 11.1Exercise 11.2Exercise 11.3 Related Study NCERT Solutions for Class 10 Maths
### Exercise 11.3 Solutions
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution
Steps of construction :
1. Draw a circle with O centre and 6 cm radius.
2. Take a point P, 10 cm away from the centre O.
3. Join PO and bisect it at M.
4. With centre M and diameter PO, draw a circle intersecting the given circle at T and S.
5. Join PT and PS.
Then PT and PS are the required tangents.
2. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution
Steps of construction :
1. Draw a circle with centre O and radius 3 cm.
2. Draw a diameter and produce it to both sides.
3. Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.
4. Bisect PO at M and QO at N
5. With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S', T' respectively.
6. Join PS, PT, QS' and QT'.
Then PS, PT, QS' and QT' are the required tangents to the given circle.
3. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution
Steps of construction :
1. Draw a line segment AB = 8 cm.
2. With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.
3. Bisect AB at M.
4. With centre M and diameter AB, draw a circle which intersects the two circles at S', T' and S, T respectively.
5. Join AS, AT, BS' and BT'.
Then AS, AT, BS' and BT' are the required tangent.
4. Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre.
Solution
Steps of construction :
1. Draw a circle with centre O and radius 3.5 cm
2. Take a point P which is 6.2 cm from O.
3. Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.
4. Join PT and PS.
PT and PS are the required tangents to circle.
5. Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.
Solution
Steps of construction :
Angle at the centre 180° – 45° = 135°
(i) Draw a circle with centre O and radius 4.5 cm.
(ii) At O, draw an angle ∠TOS = 135°
(iii) At T and S draw perpendicular which meet each other at P.
PT and PS are the tangents which inclined each other 45°.
6. Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Solution
Steps of Construction :
1. Draw a line segment BC = 8 cm
2. From B draw an angle of 90°
3. Draw an arc BA = 6cm cutting the angle at A.
4. Join AC.
5. ΔABC is the required A.
6. Draw ⊥ bisector of BC cutting BC at M.
7. Take M as centre and BM as radius, draw a circle.
8. Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification :
∠ABC = 90° (Given)
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also, AE is a tangent to the circle.
7. Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.
Solution
Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.
Steps of construction :
1. Draw two concentric circles with centre O and radii 3 cm and 5 cm.
2. Taking any point P on outer circle. Join OP.
3. Bisect OP, let M' be the mid-point of OP.
4. Taking M' as centre and OM' as radius draw a circle dotted which cuts the inner circle as M and P'.
5. Join PM and PP’. Thus, PM and PP’ are the required tangents.
6. On measuring PM and PP', we find that PM = PP' = 4 cm.
Actual calculation:
In right angle ΔOMP, ∠PMO = 90°
∴ PM2 = OP2 – OM2
[by Pythagoras theorem i.e. (hypotenuse)2 = (base)2 + (perpendicular)2]
⇒ PM2 = (5)2 – (3)2 = 25 – 9 = 16
⇒ PM = 4 cm
Hence, the length of both tangents is 4 cm. |
# Pre-Algebra : Area of a Triangle
## Example Questions
← Previous 1 3 4
### Example Question #1 : Area Of A Triangle
Find the area of the triangle:
Explanation:
The area of the triangle can be determined using the following equation:
The base is the side of the triangle that is intersected by the height.
### Example Question #1 : Area Of A Triangle
The length of the base of a triangle is inches. The height of the triangle is inches. Find the area of the triangle.
Explanation:
The formula for the area of a triangle is .
To solve the equation, plug in the base and height:
Once you multiply these three numbers, the answer you find is .
The units for area are always squared, so the unit is .
### Example Question #1 : Area Of A Triangle
If a right triangle has dimensions of inches by inches by inches, what is the area?
Explanation:
The question is asking you to find the area of a right triangle.
First you must know the equation to find the area of a triangle,
.
A right triangle is special because the height and base are always the two smallest dimensions.
This makes the equation
### Example Question #1 : Area Of A Triangle
Please use the following shape for the question.
What is the area of this shape?
Explanation:
From this shape we are able to see that we have a square and a triangle, so lets split it into the two shapes to solve the problem. We know we have a square based on the 90 degree angles placed in the four corners of our quadrilateral.
Since we know the first part of our shape is a square, to find the area of the square we just need to take the length and multiply it by the width. Squares have equilateral sides so we just take 5 times 5, which gives us 25 inches squared.
We now know the area of the square portion of our shape. Next we need to find the area of our right triangle. Since we know that the shape below the triangle is square, we are able to know the base of the triangle as being 5 inches, because that base is a part of the square's side.
To find the area of the triangle we must take the base, which in this case is 5 inches, and multipy it by the height, then divide by 2. The height is 3 inches, so 5 times 3 is 15. Then, 15 divided by 2 is 7.5.
We now know both the area of the square and the triangle portions of our shape. The square is 25 inches squared and the triangle is 7.5 inches squared. All that is remaining is to added the areas to find the total area. Doing this gives us 32.5 inches squared.
### Example Question #1 : Area Of A Triangle
What is the area of the triangle?
Explanation:
Area of a triangle can be determined using the equation:
### Example Question #1 : Area Of A Triangle
Bill paints a triangle on his wall that has a base parallel to the ground that runs from one end of the wall to the other. If the base of the wall is 8 feet, and the triangle covers 40 square feet of wall, what is the height of the triangle?
Explanation:
In order to find the area of a triangle, we multiply the base by the height, and then divide by 2.
In this problem we are given the base and the area, which allows us to write an equation using as our variable.
Multiply both sides by two, which allows us to eliminate the two from the left side of our fraction.
The left-hand side simplifies to:
The right-hand side simplifies to:
Now our equation can be rewritten as:
Next we divide by 8 on both sides to isolate the variable:
Therefore, the height of the triangle is .
### Example Question #1 : How To Find The Area Of A Triangle
A triangle has a height of 9 inches and a base that is one third as long as the height. What is the area of the triangle, in square inches?
None of these
Explanation:
The area of a triangle is found by multiplying the base times the height, divided by 2.
Given that the height is 9 inches, and the base is one third of the height, the base will be 3 inches.
We now have both the base (3) and height (9) of the triangle. We can use the equation to solve for the area.
The fraction cannot be simplified.
### Example Question #6 : Area Of A Triangle
Find the area of this triangle:
Explanation:
The formula for the area of a triangle is . In this case, the base is 11 and the height is 9. So, we're multiplying
### Example Question #1 : Area Of A Triangle
Given the following measurements of a triangle: base (b) and height (h), find the area.
Explanation:
The area of triangle is found using the formula
.
Provided with the base and the height, all we need to do is plug in the values and solve for A.
.
Since this is asking for the area of a shape, the units are squared.
### Example Question #1 : Area Of A Triangle
Given the following measurements of a triangle: base (b) and height (h), find the area.
Explanation:
The area of triangle is found using the formula
.
Provided with the base and the height, all we need to do is plug in the values and solve for A.
.
Since this is asking for the area of a shape, the units are squared. |
Abraham Lincoln once famously said, “Everybody loves a compliment.” I suspect that if he had been a mathematician he would have loved complements, too. We’ve already seen what complements are and talked about the two most prolific: the radix complement and the diminished radix complement. Now it’s time to explore how we can leverage complements to do some really interesting integer arithmetic. Using complements we can subtract one positive integer from another or add a negative integer to a positive one by simply performing addition with two positive integers. The algorithm behind this black magic is called the Method of Complements.
The Method of Complements, it turns out, is a slight misnomer. It really should be called the Methods of Complements since there are two different ways we can use complements to achieve the same goal.
Method 1
Supposed we have two positive n-digit, radix b integers x and y, and we would like to subtract y from x. If we actually want to do subtraction, the answer is straightforward, simply calculate x – y. As I stated above, however, the method of complements allows us to calculate the same answer using addition alone.
First, we calculate the diminished radix complement of x:
Now, instead of subtracting y from x, add y to the diminished radix complement of x:
Using the the associative property of addition and subtraction we can rearrange our parenthesis:
In this form, we now have the diminished radix complement of the value we trying to solve for: x – y.
One important issue I haven’t mentioned previously about complements is the fact that taking a compliment or diminished radix compliment of a number is an involutory function, a function that is in an inverse of itself. Put another way the radix compliment of the radix compliment of a number x is x itself.
So, if is the diminished radix complement of , then to find the value of , we need to calculate the diminished radix complement of .
To summarize: in order to calculate the difference using this method:
1. Calculate the diminished radix complement of x.
2. Add the diminished radix complement of x to y.
3. Calculate the diminished radix complement of the sum from #2.
Method 2
Again, we have two positive n-digit, radix b integers x and y, and we want to calculate the difference . This time, we first calculate the radix complement of y:
Next add the radix complement of y to x:
We can rearrange this to resemble the value we are looking to solve for:
So we see this is our difference, , plus . It is actually very easy to solve this for if we think about the value of and how it relates to x, y, and their difference.
As initially stated, x and y are n-digit numbers with radix b. For such an n-digit number, the place values from left to right are: . , therefore, is one more than the largest number that can be represented by an n-digit number. Because of this, is a 1 followed by n 0s – regardless of the radix, the smallest (n + 1)-digit number. This is important.
Back to x and y. x and y are both positive, n-digit numbers. Assuming that , and since y is positive, the difference has to be at most an n-digit number. Even it’s got fewer digits, we can consider it an n-digit number by adding leading 0s.
If we add to this difference, ala , what we get is an (n + 1)-digit number where the first digit is 1 and the remaining n digits are the value of the difference . Remember, is n + 1 digits beginning with a 1 followed by n 0s. Adding the n-digit value of to the first n digits of b^n, 0s, just gives us again.
So, to extract the value of from the value of all you need to do is drop the leading 1 from your solution. It is that simple.
Looking at these steps without the exposition reveals that this method is not nearly as confusing as it seems. To calculate the value of using Method 2:
1. Calculate the radix complement of y.
2. Add the radix complement of y to x.
3. Drop the leading 1 and any leading 0s from the result.
Since we know the easy method for calculating the radix complement, the steps above should really be written:
1. Calculate the diminished radix complement of y and add 1.
2. Add the radix complement of y to x.
3. Drop the leading 1 and any leading 0s from the result.
Decimal Example
Lets’s look at the decimal subtraction problem using both methods.
Method 1
We first calculate the diminished radix complement of the minuend, 737, by subtracting each digit from 9, ie., to get 262.
Next, add the diminished radix complement of the minuend, 262, to the subtrahend, 234:
Calculate the diminished radix complement of the sum, 496, to get the answer to the original subtraction problem, 503.
Method 2
First calculate the radix complement of the subtrahend, 234. The diminshed radix complement is calculated by subtracting each digit of 234 from 9 and is found to be 765. Add 1 to the dimished radix complement to find the radix complement, 766.
Next, add the radix complement of the subtrahend, 766, to the minuend, 737:
Drop the leading 1 from this sum to get the answer to original subtraction problem, 503.
Leading Zeros
Leading zeros can come into play in one of two places: one of which requires adding them while the other requires dropping them.
Scenario 1
If you are using method 2, and the subtrahend has fewer digits than the minuend, assuming you are calculating the radix complement via the diminished radix complement, then you need to right pad the subtrahend with 0s so that it has the same number of digits as the minuend before calculating the diminished radix complement.
Consider :
Before we can find the diminished radix complement of the subtrahend 92 we need to right pad it with two 0s so that it is four digits like the the minuend 1221. The diminished radix complement of 0092 is 9907 to which we add 1 to get the radix complement 9908. Now we can proceed with our addition:
Drop the leading 1 from the sum, and 1129 is the answer to the subtraction problem .
Note that if you are finding the radix complement of the subtrahend 92 the hard way, by calculating , then you don’t need to bother with leading 0s.
Scenario 2
The other place where leading zeros are an issue, regardless of the method you choose, are in the sum. Using the first method, it is possible that when you calculate the dimished radix complement of the sum that you end up with leading 0s. What do with them? Drop them. The same goes for the sum you get using method 2. We are already dropping the leading 1 as per the method, but the same goes for any 0s following the leading one. Drop them, too.
Consider calculating the difference using method 1. First we calculate the diminished radix complement for the minuend 138, which turns out to be 861. Then we add that value to our subtrahend 131:
The diminished radix complement of the sum, 992, is 007. (Cool, but that wasn’t on purpose.) Drop the leading 0s, and 7 is the answer.
Binary Example
The method of complements is a natural fit for working with binary numbers, and is ultimately responsible for why the twos complement representation is the representation of choice for binary signed integers.
In my post about the ones complement representation, I showed, but did not go into the detail that I do here, how easy it is to calculate the ones complement, or diminished radix complement of a binary number. All you need to do is flip every bit in your number. 1s become 0s and 0s become 1s. If you want the twos complement, or radix complement, just add 1 to this value.
Let’s look at the subtraction problem .
Method 1
First we need to calculate the ones complement of the minuend, 10101101. To do this, we just flip its individual bits and get 01010010. Now we add the ones complement of the minuend to the subtrahend:
The ones complement of our sum, 10110001, will be our difference. Flipping the bits gives us 01001110, or , the correct answer.
Method 2
The first step using this method is to calculate the twos complement of the subtrahend, 01011111. First we calculate the ones complement by flipping the bits to get 10100000. To that we add 1 to get the twos complement 10100001.
Now we add our minuend to the twos complement:
Dropping the leading 1 (and subsequent 0 if you are so inclined) gives us our answer 01001110. |
## In a triangle ABC right angled at B, AB = 24 cm, BC = 7 cm. then sinC = ?
Question
In a triangle ABC right angled at B, AB = 24 cm, BC = 7 cm. then sinC = ?
in progress 0
2 months 2021-12-03T06:43:03+00:00 2 Answers 0 views 0
In Δ ABC, right-angled at B
Using Pythagoras theorem
AC² = AB² +BC²
AC² = 576 + 49 = 625
AC = √635
AC = +25
Now
AC = 25 CM, AB = 24cm , BC = 7cm
sinC=
side opposite to angle c / hypotenuse =>
AB / AC => 24/25
In ∆ ABC, ∠B = 90°
AB = 24 cm BC = 7 cm
( by pythyorous therom )
We have ,
➠ AC = 25 cm
Now , sin A =
Cos A =
ANOTHER METHOD IS : –
➠ In a given triangle ABC, right-angled at B = ∠B = 90°
➠ Given: AB = 24 cm and BC = 7 cm
➠ That means, AC = Hypotenuse
➠ According to the Pythagoras Theorem,
➠In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.
➠ By applying Pythagoras theorem, we get
➠ AC2 = AB2 + BC2
AC2 = (24)2 + 72
➠ AC2 = (576 + 49)
➠ AC2 = 625 cm2
➠ Therefore, AC = 25 cm
(i) We need to find Sin A and Cos A.
➠ As we know, sine of the angle is equal to the ratio of the length of the opposite side and hypotenuse side. Therefore,
➠ Sin A = BC/AC = 7/25
➠Again, the cosine of an angle is equal to the ratio of the adjacent side and hypotenuse side. Therefore,
➠ cos A = AB/AC = 24/25
(ii) We need to find Sin C and Cos C.
➠ Sin C = AB/AC = 24/25
➠Cos C = BC/AC = 7/25
hence proved , |
Back
# Do Cubes and Squares Have the Same Properties as Spheres and Circles?
Calculus students are often intrigued when they realize that the derivative of the volume of a sphere formula, with respect to the radius of the sphere, is the surface of the sphere formula: . They are also fascinated by the fact that the derivative of the area of a circle formula is the formula for its circumference: . These results don't seem to apply to cubes and squares because , the surface area of the cube, and , the perimeter of the square. Why do spheres and circles behave one way and cubes and squares another?
Consider the derivative of the area of a circle. According to the definition of the derivative:
Geometrically, this result is easy to see because the region between two concentric circles, one with a radius of r and another with a radius of r + h, is essentially a band of width h and length 2πr, as shown in Figure 1.
A similar calculation is true for the derivative of the volume of a sphere. According to the definition of the derivative:
Geometrically, this result is easy to understand because the region between two concentric spheres, one with a radius of r and another with a radius of r + h, is essentially a hollow ball of thickness h and surface area 4πr 2.
Similar results don't hold for the standard formula for the volume of a cube and the area of a square, as shown in the first paragraph. Let us consider different formulas: Let s be the distance from the center of a square perpendicular to the opposite side, as shown in Figure 2. Then, since s is half the length of the edge of the square, we have the formula A = (2s)2 = 4s2, and P = 8s, for the area of a square and the perimeter of a square, respectively.
Consider the derivatives of this new formula for the area of a square. Since A = 4s2, , which is our formula for the perimeter of the square. Similarly, let s be the distance from the center of a cube perpendicular to the opposite side. Then, since s is half the length of the edge of the cube, we have the formula V = (2s)3 = 8s3 for the volume of the cube, and (2s)2 = 4s2 for the area of each face. Since V = 8s3, then , which is 6 times the area of a face of the cube, and thus this expression is equal to the surface area of the cube. By writing the formulas in terms of s, half the length of the edge, we now have formulas that have properties consistent with those of spheres and circles.
We can extend this result to equilateral triangles. Let e be the length of the edge of the triangle, as shown in Figure 3.
The area of the triangle is , and its perimeter is 3e. The derivative of the area is , which doesn't remotely look like the perimeter of the triangle. Let us now change the variable and let s represent the perpendicular distance from the center of the triangle to one of the sides, as shown in Figure 4.
Using the Pythagorean relationship, or . In terms of s, the area of the triangle is , and the perimeter is . Since , the derivative of the area of the equilateral triangle is the perimeter of the triangle. Figure 5 shows that the difference between the triangular regions, using s and s + h, is essentially a three-sided strip of width h and length equal to the perimeter of the triangle, which geometrically confirms our result.
The case of the equilateral triangle suggests a way of generalizing this result to other regular polygons. The segment s, from the center of the polygon perpendicular to the side, is called the apothem of the polygon. The length of the side of the polygon then is , where n is the number of sides, and therefore the perimeter of the polygon is . The area of the polygon is the length of the apothem multiplied by half the length of the perimeter. In symbols:
We can now check that , the perimeter of the regular polygon.
Challenge your students to apply this method to a regular hexagon. They should find formulas for the area and perimeter of the hexagon in terms of its apothem. They then should show that one is the derivative of the other.
We can also extend our results to three-dimensional figures. For regular polyhedra, the apothem is the radius of the inscribed sphere. Consider a tetrahedron.
The standard formulas for the volume and surface area of the tetrahedron are normally based on the length of the edge, a, of the tetrahedron. It isn't difficult, however, to rewrite them in terms of the length, r, of the apothem—the radius of the inscribed sphere.
r = radius of the inscribed sphere = apothem of the polyhedron
R = radius of the circumscribed sphere
a = length of the edge of the tetrahedron
S = surface area of the tetrahedron
V = volume of the tetrahedron
In terms of r, the inscribed radius:
Thus, we can easily see that, in terms of the apothem, the derivative of the formula for the volume of a tetrahedron is the formula for its surface area.
Similarly, using the standard formulas for the volume and surface area of the octahedron based on the length of the edge, a, of the octahedron. We can rewrite them in terms of the length, r, of the apothem.
Again, it is clear that the derivative of the volume formula, with respect to r, is the formula for surface area.
Here are the formulas for the volume and surface area of the other two regular polyhedra in terms of the length of the inscribed radius, r.
### Icosahedron:
You may wish to assign students the challenging problem of verifying that the derivative of the above formula for the volume of a dodecahedron is, in fact, the given formula for its surface area.
What is the value of this result? Is it just an academic exercise? In fact, these formulas provide an easy way to find the formula for surface area of regular solids if you know the formula for its volume—or vice versa. Just write the formula in terms of the apothem of the solid and differentiate, or antidifferentiate, accordingly to find the other formula.
#### References
http://mathforum.org/dr.math/faq/formulas/faq.polyhedron.html#octahedron
http://mathforum.org/dr.math/faq/formulas/faq.polyhedron.html
John F. Mahoney introduced AP Statistics to Banneker High School in 2002-03 and taught 59 students that year in the course. Each of the students took the AP Exam, and Mahoney believes that they may have had the highest percentage of seniors taking AP Statistics in the country. This year he is also teaching both AB level and BC level AP Calculus. He is an AP Consultant and a longtime participant in the AP Calculus Reading. Currently, he is one of the Exam Leaders. He also chairs the editorial panel of NCTM's ON-Math: www.nctm.org/onmath. At Banneker High School, he is one of the mentors for the robotics team and helps students design gear-based drive trains—just as Banneker himself did more than 200 years ago. Many engineers help them, including those from Howard University, located across Georgia Avenue from the high school. He is one of the coaches of the school's award-winning It's Academic team. When he decided to teach at Banneker High School three years ago, after a long career in private schools, he explored Banneker's mathematics, and this paper is the result of that work.
John F. Mahoney |
# Power Rule for Differentiation
In this section we learn how to differentiate, find the derivative of, any power of $$x$$.
That's any function that can be written: $f(x)=ax^n$ We'll see that any function that can be written as a power of $$x$$ can be differentiated using the power rule for differentiation.
In particular we learn how to differentiate when:
• the power is a positive integer like $$f(x) = 3x^5$$.
• the power is a negative number, this means that the function will have a "simple" power of $$x$$ on the denominator like $$f(x) = \frac{2}{x^7}$$.
• the power is a fraction, this means that the function will have an $$x$$ under a root like $$f(x) = 5\sqrt{x}$$.
We start by learning the formula for the power rule.
## Power Rule
Given a function which is a power of $$x$$, $$f(x)=ax^n$$, its derivative can be calculated with the power rule: $\text{if} \quad f(x)=ax^n \quad \text{then} \quad f'(x)=n\times ax^{n-1}$ We can also write this: $\text{if} \quad y=ax^n \quad \text{then} \quad \frac{dy}{dx}=n\times ax^{n-1}$
## Tutorial 1: Power Rule for Differentiation
In the following tutorial we illustrate how the power rule can be used to find the derivative function (gradient function) of a function that can be written $$f(x)=ax^n$$, when $$n$$ is a positive integer.
## Example 1
Find the derivative of the function defined by: $f(x) = 2x^4$
### Detailed Solution
Comparing the function $$f(x) = 2x^4$$ to the generic "power function" $$f(x) = ax^n$$, we can see that: $a = 2 \quad \text{and} \quad n = 4$ The power rule for differentiation: $f'(x) = n\times ax^{n-1}$ therfore leads to: \begin{aligned} f'(x) & =4\times 2x^{3-1} f'(x) & = 8x^2 \end{aligned} The derivative is therefore: $f'(x) = 8x^2$
## Exercise 1
Use the power rule for differentiation to find the derivative function of each of the following:
1. $$f(x) = 6x^3$$
2. $$y = x^4$$
3. $$f(x) = -2x^6$$
4. $$y = \frac{x^2}{2}$$
5. $$f(x) = 3x$$
6. $$y = \frac{2}{5}x^{10}$$
7. $$f(x) = -6x^3$$
8. $$y = 4x^4$$
## Answers Without Working
1. For $$f(x) = 6x^3$$ we find: $f'(x) = 18x^2$
2. For $$y = x^4$$ we find: $\frac{dy}{dx} = 4x^3$
3. For $$f(x) = -2x^6$$ we find: $f'(x) = -12x^5$
4. For $$y = \frac{x^2}{2}$$ we find: $\frac{dy}{dx} = x$
5. For $$f(x) = 3x$$ we find: $f'(x) = 3$
6. For $$y = \frac{2}{5}x^{10}$$ we find: $\frac{dy}{dx} = 4x^9$
7. For $$f(x) = -6x^3$$ we find: $f'(x) = -18x^2$
8. For $$y = 4x^4$$ we find: $\frac{dy}{dx}= 16x^3$
## Negative Exponents
The power rule also works for negative exponents.
Remember that: $\frac{a}{x^m} = ax^{-m}$ this allows us to use the power rule to differentiate any function that can be written: $f(x)=\frac{a}{x^m}$
## Tutorial 2: Negative Exponents
In the following tutorial we illustrate how the power rule can be used to find the derivative function (gradient function) of a function that can be written $$f(x)=\frac{a}{x^m}$$, when $$m$$ is a positive integer.
We use the fact that $$\frac{a}{x^m} = a.x^{-m}$$ to then use the power rule.
## Exercise 2
Differentiate each of the following:
1. $$f(x) = \frac{3}{x^2}$$
2. $$f(x) = \frac{1}{x}$$
3. $$y = \frac{5}{x^3}$$
4. $$f(x) = -\frac{5}{x^3}$$
5. $$y = \frac{6}{x^4}$$
6. $$f(x) = - \frac{2}{x}$$
7. $$y = \frac{3}{4x^2}$$
8. $$f(x) = -\frac{2}{3x^3}$$
### Solution
1. For $$f(x) = \frac{3}{x^2}$$, we find: $f'(x) = -6.x^{-3}$ Which we can also write: $f'(x) = -\frac{6}{x^3}$
2. For $$f(x) = \frac{1}{x}$$ we find: $f'(x) = -1.x^{-2}$ Which can/should be written: $f'(x) = -\frac{1}{x^2}$
3. For $$y = \frac{5}{x^3}$$ we find: $\frac{dy}{dx} = -15.x^{-4}$ which can/should be written: $\frac{dy}{dx} = - \frac{15}{x^4}$
4. For $$f(x) = -\frac{5}{x^3}$$ we find: $f'(x) = 15.x^{-4}$ which can/should be written: $f'(x) = \frac{15}{x^4}$
5. For $$y = \frac{6}{x^4}$$ we find: $\frac{dy}{dx} = -24.x^{-4}$ which can/should be written: $\frac{dy}{dx} = -\frac{24}{x^5}$
6. For $$f(x) = - \frac{2}{x}$$ we find: $f'(x) = 2.x^{-2}$ which can/should be written: $f'(x) = \frac{2}{x^2}$
7. For $$y = \frac{3}{4x^2}$$ we find: $\frac{dy}{dx} = -\frac{3}{2}.x^{-3}$ which can/should be written: $\frac{dy}{dx} = -\frac{3}{2x^3}$
8. For $$f(x) = -\frac{2}{3x^3}$$ we find: $f'(x) = 2.x^{-4}$ which can/should be written: $f'(x) = \frac{2}{x^4}$
## Fractional Exponents
The power rule for differentiation also works for any fraction.
Remembering that: $\sqrt[n]{x^m}=x^{\frac{m}{n}} \quad \text{and} \quad \frac{1}{\sqrt[n]{x^m}} = x^{-\frac{m}{n}}$ we can differentiate any function that can be written: $f(x)=a.\sqrt[n]{x^m} \quad \text{and} \quad f(x)=\frac{a}{\sqrt[n]{x^m}}$
## Exercise 3
Differentiate each of the following:
1. $$f(x)=3.\sqrt{x}$$
2. $$y=5.\sqrt[3]{x^4}$$
3. $$y = 12.\sqrt[6]{x}$$
4. $$f(x) = 10.\sqrt[5]{x^3}$$
5. $$f(x) = -4.\sqrt{x^5}$$
6. $$y = 9.\sqrt[3]{x^2}$$
7. $$f(x)=\frac{4}{\sqrt{x}}$$
8. $$y = - \frac{8}{\sqrt{x^5}}$$
### Solution
Each of these can be differentiated using the power rule.
1. For $$f(x)=3\sqrt{x}$$ we find: $f'(x) = \frac{3}{2}.x^{-\frac{1}{2}}$ which we can write: $f'(x) = \frac{3}{2\sqrt{x}}$
2. For $$y=5\sqrt[3]{x^4}$$ we find: $\frac{dy}{dx} = \frac{20}{3}x^{\frac{1}{3}}$ which we can write: $\frac{dy}{dx} = \frac{20}{3}\sqrt[3]{x}$
3. For $$f(x)=\frac{4}{\sqrt{x}}$$ we find: $f'(x) = -2.x^{-\frac{3}{2}}$ which we can write: $f'(x) = -\frac{2}{\sqrt{x^3} }$
4. For $$f(x) = 10.\sqrt[5]{x^3}$$ we find: $f'(x) = 6.x^{-\frac{2}{5}}$ which we can write: $f'(x) = \frac{6}{\sqrt[5]{x^2}}$
5. For $$f(x) = -4.\sqrt{x^5}$$ we find: $f'(x) = -10.x^{\frac{3}{2}}$ which we can also write: $f'(x) = - 10 \sqrt{x^3}$
6. For $$y = 9.\sqrt[3]{x^2}$$ we find: $\frac{dy}{dx} = 6.x^{-\frac{1}{3}}$ which we can also write: $\frac{dy}{dx} = \frac{6}{\sqrt[3]{x}}$
7. For $$f(x)=\frac{4}{\sqrt{x}}$$ we find: $f'(x) = -2.x^{-\frac{3}{2}}$ which we can also write: $f'(x) = - \frac{2}{\sqrt{x^3}}$
8. For $$y = - \frac{8}{\sqrt{x^5}}$$ we find: $\frac{dy}{dx} = 20.x^{-\frac{7}{2}}$ which we can also write: $\frac{dy}{dx} = \frac{20}{\sqrt{x^7}}$ |
RD Sharma Solutions Class 8 Mathematics Solutions for Compound Interest Exercise 14.2 in Chapter 14 - Compound Interest
Question 33 Compound Interest Exercise 14.2
What will Rs. 125000 amount to at the rate of 6%, if the interest is calculatedafter every four months?
Given details are,
Principal (p) = Rs 125000
Rate (r) = 6% per annum
Time (t) = 1 year
Since interest is compounded after 4months, interest will be counted as 6/3 = 2% andTime will be 12/4 = 3quarters
By using the formula,
A = P (1 + R/100)^n
= 125000 (1 + 2/100)^3
= 125000 (102/100)^3
= Rs 132651
∴ Amount is Rs 132651
Video transcript
hello students welcome to elitos india's best online classroom my name is shai safirozi class and today we are going to find out the amount but before we go ahead let's quickly see what the question says the question says what will rupees 1 lakh 25 000 amount to at the rate of 6 percent if the interest is calculated after every 4 months okay so here we have got as one lakh 25 000 as our principal and the rate is six percent per annum but we want to calculate after every four months so let us find out for four months first let's jot down the details whatever is given and accordingly we will be calculating it for the four months so my principle is rupees one lakh twenty 25 000 my rate of interest is rupees six percent do remember this is for per annum okay and time period since it is calculating after every four four months this means that is it is calculating for one year because it is per annum so it is for one year okay now since interest you can say here that interest is compounded annually always interest is compounded annually okay so interest will be counted annually but here it is not counted annually so what we are going to do is in this will be compounded as 6 upon 3 so this 6 upon 3 becomes 2 percent correct and our time we are dividing it by 3 because we are calculating after four four months so four four months means it becomes the whole year gets divided into three parts of four months each therefore we are dividing by three so to get two percent as a rate of interest and time since it is of one years one years means it is of 12 months so 12 months and since it's for four months we are calculating for four months your time period would be for three quarter okay so time is three quarter now what we are going to do is we are quickly going dotting down the formula for amount so let's write down the formula for amount so my amount formula would be p principle 1 plus r upon 100 raised to n so my principle as i know that it is 1 lakh twenty five thousand one plus now i know that my new rate of interest would be two percent so i'm writing here as two upon hundred since i want it for four months and my time is three quarters so this is nothing but t or i can say n so this is three quarter so it is three so one lakh twenty five thousand sorry one lakh twenty five thousand now what i am going to do is i am going to calculate the lcm for this particular part so lcm would be 100 so 100 plus 2 would be 102 divided by 100 so this bracket becomes 102 divided by 100 raised to 3 now since this is for thrice i am going to multiply this rise so i am going to get 102 upon 100 into 102 upon 100 into 102 upon 100 so on calculation of this whole thing that is numerators multiply with the numerators and denominators multiply with the denominators and dividing by this or i can just directly take the decimal points i am going to get my amount as 1 lakh 32 600 651 okay so kindly note that we have received our amount as rupees 1 lakh 32 651 so now i am going to write down in the statement form that is amount is rupees 1 lakh 32 651 this is what my answer is okay so that's all for today see you all next time take care till then please subscribe to lilo and do comments don't forget to comments we are waiting for your comments bye
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Mathematics » Introducing Polynomials » Using Multiplication Properties of Exponents
# Simplifying Expressions Using the Product to a Power Property
## Simplifying Expressions Using the Product to a Power Property
We will now look at an expression containing a product that is raised to a power. Look for a pattern.
$${\left(2x\right)}^{3}$$ What does this mean? $$2x·2x·2x$$ We group the like factors together. $$2·2·2·x·x·x$$ How many factors of 2 and of $$x?$$ $${2}^{3}·{x}^{3}$$ Notice that each factor was raised to the power. $${\left(2x\right)}^{3}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}{2}^{3}·{x}^{3}$$ We write: $${\left(2x\right)}^{3}$$ $${2}^{3}·{x}^{3}$$
The exponent applies to each of the factors. This leads to the Product to a Power Property for Exponents.
### Definition: Product to a Power Property of Exponents
If $$a$$ and $$b$$ are real numbers and $$m$$ is a whole number, then
$${\left(ab\right)}^{m}={a}^{m}{b}^{m}$$
To raise a product to a power, raise each factor to that power.
An example with numbers helps to verify this property:
$$\begin{array}{ccc}\hfill {\left(2·3\right)}^{2}& \stackrel{?}{=}& {2}^{2}·{3}^{2}\hfill \\ \hfill {6}^{2}& \stackrel{?}{=}& 4·9\hfill \\ \hfill 36& =& 36✓\hfill \end{array}$$
## Example
Simplify: $${\left(-11x\right)}^{2}.$$
### Solution
$${\left(-11x\right)}^{2}$$ Use the Power of a Product Property, $${\left(ab\right)}^{m}={a}^{m}{b}^{m}.$$ Simplify. $$121{x}^{2}$$
## Example
Simplify: $${\left(3xy\right)}^{3}.$$
### Solution
$${\left(3xy\right)}^{3}$$ Raise each factor to the third power. Simplify. $$27{x}^{3}{y}^{3}$$ |
# How do you solve 2(3 - a) - (-2)(a - 3) = - (3a - 2)?
May 18, 2016
$a = \frac{2}{3}$
#### Explanation:
The first step is to multiply out to get rid of the brackets, remembering that multiplying a negative by a negative gives a positive.
$2 \left(3 - a\right) - \left(- 2\right) \left(a - 3\right) = - \left(3 a - 2\right)$
$= 6 - 2 a - \left(- 2 a + 6\right) = - 3 a + 2$
$= 6 - 2 a + 2 a - 6 = - 3 a + 2$
Move all the items to the left hand side, remembering to change the sign.
$= 6 - 2 a + 2 a - 6 + 3 a - 2 = 0$
Group the items.
$- 2 a + 2 a + 3 a + 6 - 6 - 2 = 0$
$\therefore 3 a - 2 = 0$
$3 a = 2$
$a = \frac{2}{3}$ |
# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q: 21 $x^2^n-y^2^n$ is divisible by $x+y.$
Let the given statement be p(n) i.e.
$p(n):x^2^n-y^2^n$
For n = 1 we have
$p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y)$ , which is divisible by $(x+y)$ , hence true $(using \ a^2-b^2=(a+b)(a-b))$
For n = k we have
$p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by $(x+y)$ $=(x+y)m$
Now,
For n = k + 1 we have
$p(k+1):x^{2(k+1)}-y^{2(k+1)}$ $=x^{2k}.x^2-y^{2k}.y^2$
$=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2$
$=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2$
$=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))$
$=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )$
$=(x+y)l$ where $l = (x^2.m+(x-y).y^{2k})$ some natural number
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by $(x+y)$ for all natural numbers n
Exams
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Completing the Square Examples MathBitsNotebook.com
See Completing the Square for a discussion of the process.
Creating a perfect square trinomial on the left side of a quadratic equation, with a constant (number) on the right, is the basis of a method called completing the square.
Find the solutions for: x2 = 3x + 18
(The leading coefficient is one.) Get the x-related terms on the left side. Keep the constant term on the right side. Prepare the equation to receive the added value (boxes). Take half of the x-term's coefficient and square it. Add this value to both sides (fill the boxes). Combine terms on the right. Factor the perfect square trinomial on the left side. Take the square root of both sides. Be sure to consider "plus and minus", as we need two answers. Solve for x. Clearly indicate your answers. Prepare a check of the answers. 62 - 3(6) = 18 check (-3)2 - 3(-3) = 18 check
Find the x-intercepts for: 4x2 - 8x - 32 = y
Finding the x-intercepts requires setting y = 0 and solving for the x-values. This question is really asking you to solve 4x2 - 8x - 32 = 0. The leading coefficient is NOT 1. Divide all terms by 4 (the leading coefficient). [ Note: In some problems, this division process may create fractions, which is OK. Just be careful when working with the fractions.] Move the constant to the right hand side. Prepare the equation to receive the added value (boxes). Take half of the x-term's coefficient and square it. Add this value to both sides (fill the boxes). Combine like terms. Factor the perfect square trinomial on the left side. Take the square root of both sides. Be sure to consider "plus and minus". Solve for x. Clearly indicate your answers. Prepare a check of the answers. 4(4)2 - 8(4) - 32 = 0 check 4(-2)2 - 8(-2) - 32 = 0 check
Find the solutions for: x2 = 4x -1
Get the x-related terms on the same side (move 4x). The leading coefficient is 1. Prepare the equation to receive the added value (boxes). Take half of the x-term's coefficient and square it. Add this value to both sides (fill the boxes). Combine like terms. Factor the perfect square trinomial on the left side. Take the square root of both sides. Be sure to consider "plus and minus". Solve for x. Clearly indicate your answers. Prepare a check of the answers.
Find the solutions for: x2 - 5x + 7 = 0
Notice that this example involves the imaginary "i", and has complex roots of the form a + bi. These answers are not "real number" solutions. They do not have a place on the x-axis. (The leading coefficient is one.) Move the constant to the right hand side. Prepare the equation to receive the added value (boxes). Take half of the x-term's coefficient and square it. Add this value to both sides (fill the boxes). Get a common denominator on the right. Factor the perfect square trinomial on the left side. Combine terms on the right. At this point, you have a squared value on the left, equal to a negative number. We know that it is not possible for a "real" number to be squared and equal a negative number. ____________________________________________ This problem involves "imaginary" numbers. Take the square root of both sides. Be sure to consider "plus and minus". Notice the negative under the radical. Solve for x. Prepare a check of the answers.
Completing the Square with Algebra Tiles
Starting with x2 + 6x - 16 = 0, we rearrange x2 + 6x = 16 and attempt to complete the square on the left-hand side.
KEY: See more about Algebra Tiles.
x2 + 6x = 16 Arrange the x2-tile and 6x-tiles to start forming a square. Notice how many 1-tiles are needed to complete the square. Read the sides of the completed square. (x + 3)2 = 16 + 9 |
# 1996 AHSME Problems/Problem 25
## Problem
Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
## Solution 1
Complete the square to get $$(x-7)^2 + (y-3)^2 = 64.$$ Applying Cauchy-Schwarz directly, $$64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.$$ $$40 \ge 3x+4y-33$$ $$3x+4y \le 73.$$ Thus our answer is $\boxed{(B)}$.
## Solution 2 (Geometric)
The first equation is a circle, so we find its center and radius by completing the square: $x^2 - 14x + y^2 - 6y = 6$, so $$(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.$$
So we have a circle centered at $(7,3)$ with radius $8$, and we want to find the max of $3x + 4y$.
The set of lines $3x + 4y = A$ are all parallel, with slope $-\frac{3}{4}$. Increasing $A$ shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether. This means $3x + 4y = A$ will be tangent to the circle.
Imagine that this line hits the circle at point $(a,b)$. The slope of the radius connecting the center of the circle, $(7,3)$, to tangent point $(a,b)$ will be $\frac{4}{3}$, since the radius is perpendicular to the tangent line.
So we have a point, $(7,3)$, and a slope of $\frac{4}{3}$ that represents the slope of the radius to the tangent point. Let's start at the point $(7,3)$. If we go $4k$ units up and $3k$ units right from $(7,3)$, we would arrive at a point that's $5k$ units away. But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$.
Thus, $k = \frac{8}{5}$, and we want to travel $4\cdot \frac{8}{5}$ up and $3\cdot \frac{8}{5}$ over from the point $(7,3)$ to reach our maximum. This means the maximum value of $3x + 4y$ occurs at $\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)$, which is $\left(\frac{59}{5}, \frac{47}{5}\right).$
Plug in those values for $x$ and $y$, and you get the maximum value of $3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}$, which is option $\boxed{(B)}$.
## Solution 2B
Let the tangent point be $P$, and the tangent line's x-intercept be $Q$. Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle $OPK$ is 3-4-5, $OP=8$, so $OK = \frac{5}{3}*8 = \frac{40}{3}$. Note that the horizontal distance from $O$ to the origin is $7$, and the horizontal distance from K to Q is 4, ($\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$. The value of $3x+4y$ is 73 at point $Q$. Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$. $\boxed{(B)}$.
## Solution 3
Let $z = 3x + 4y$. Solving for $y$, we get $y = (z - 3x)/4$. Substituting into the given equation, we get $$x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,$$ which simplifies to $$25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.$$
This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so $$(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,$$ which simplifies to $$-64z^2 + 4224z + 32704 \ge 0,$$ which can be factored as $$-64(z + 7)(z - 73) \ge 0.$$ The largest value of $z$ that satisfies this inequality is $\boxed{73}$, which is $\boxed{(B)}$.
## Solution 4 (Using Answer Choice + Calculus)
Implicitly differentiating the given equation with respect to $x$ yields:
$2x + 2y\frac{dy}{dx} = 14 + 6\frac{dy}{dx}$
Now solve for $\frac{dy}{dx}$ to obtain:
$\frac{dy}{dx} = -\frac{x - 7}{y - 3}$
Set the equation equal to zero to find the maximum occurs at $x = 7$
Plug this back into the equation that we are trying to maximize and see that we are left with: $21 + 4y$.
The only answer choice that can be obtained from this equation is $\bf{73}$
## Solution 5 (Lagrange Multipliers)
First, we move all the non-constant terms of the constraint to one side and assign it to the function $g(x,y)$: $$g(x,y)=x^2+y^2-14x-6y.$$ Since we are trying to maximize $f(x,y)=3x+4y$, we need to solve for $x$ and $y$ in the system $$\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}$$ We have that \begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*} and \begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}. To solve the original system, we can solve for $x$ and $y$ in terms of $\lambda$ using our equations from the gradients, then substitute them into the first equation. We have that $x=\frac{3}{2}\lambda+7$ and $y=2\lambda+3$. Substituting into the first equation, we have that \begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*} Using the solutions of $x$ and $y$ in terms of $\lambda$ that we found earlier, we have that $$3x+4y=\frac{25}{2}\lambda+33.$$ Because we are trying to maximize this function, we will use the positive solution for $\lambda$. Therefore, after substituting, we have that the largest value of $g(x,y)$ that satisfies $f(x,y)=6$ is $\boxed{\textbf{(B) }73}$.
~qianqian07
## Solution 6 (polar coordinates)
Completing the square gives the equation of a circle, $(x-7)^2+(y-3)^2=64.$ Seeing that we would like to maximize $3x+4y,$ we parameterize the circle using polar coordinates: \begin{align*} x&=7+8\cos\theta&y&=3+8\sin\theta. \end{align*} Then, we have $3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.$ Since $a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},$ the desired answer is $33+\sqrt{24^2+32^2}=33+40=\boxed{73}.$ - Ultroid999OCPN
## Solution 6 (Alcumus)
The equation $x^2 + y^2 = 14x + 6y + 6$ can be written$(x-7)^2 + (y-3)^2 = 8^2,$which defines a circle of radius 8 centered at $(7,3)$. If $k$ is a possible value of $3x + 4y$ for $(x,y)$ on the circle, then the line $3x + 4y = k$ must intersect the circle in at least one point. The largest value of $k$ occurs when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. Because the slope of the tangent line is $-3/4$ the slope of the radius is $\ 4/3$. It follows that the point on the circle that yields the maximum value of $3x + 4y$ is one of the two points of tangency,$x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \hspace{.3in} y = 3 + \frac{4 \cdot 8}{5} = \frac{47}{5},$or$x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \hspace{.3in} y = 3 - \frac{4 \cdot 8}{5} = - \frac{17}{5}.$[asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); draw((0,15)--origin--(15,0)); dot("$(7,3)$",(7,3),S); draw(Circle((7,3),8)); line a = line((0,9),(12,0)); line b = line((0,12),(16,0)); line c = line((0,18.3),(18.3*4/3,0)); draw(a^^b^^c); [/asy]
The first point of tangency gives$3x + 4y = 3 \cdot \frac{59}{5} + 4 \cdot \frac{47}{5} = \frac{177}{5}+\frac{188}{5}= 73,$and the second one gives$$\, 3x + 4y = \frac{33}{5}-\frac{68}{5} = -7.$$Thus, $\boxed{73}$ is the desired maximum, while $-7$ is the minimum. |
# Cardinal Numbers
Home » Math Vocabulary » Cardinal Numbers
## What Are Cardinal Numbers?
Cardinal numbers are numbers that are used for counting real objects or counting things. They are also known as “counting numbers” or “cardinals.” We commonly use cardinal numbers or cardinals to answer the question starting with “How many?”
So, how to find the cardinal numbers around us? Let’s see an example.
Example:
How many dogs are in the given picture?
To know the total number of dogs, we need to count the dogs given in the image.
There are 8 dogs in all. Here, 8 is the cardinal number.
## Cardinal Numbers: Definition
The numbers that we use for counting are called cardinal numbers. They tell us the quantity of objects.
Cardinal Numbers Examples: 2 bananas, 5 suitcases, 100 points, a million dollars, etc.
Cardinal numbers do not include fractions or decimals. Cardinal numbers are natural numbers or positive integers. The smallest cardinal number is 1.
Examples of cardinal numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, and so on.
The smallest cardinal number is 1.
If we count a bag filled with a million dollars, we get a cardinal number: “one million!”
Example:
## Cardinal Numbers in Words
We can also write cardinal numbers in words. For the first 10 numbers, we can write the cardinal numbers as:
• 1 – One
• 2 – Two
• 3 – Three
• 4 – Four
• 5 – Five
• 6 – Six
• 7 – Seven
• 8 – Eight
• 9 – Nine
• 10 – Ten
## List of Cardinal Numbers
Here’s a list of cardinal numbers up to 100, in figures and in words:
Cardinal Numbers as Multiples of 100:
We can count further and list a few larger cardinal numbers as well:
## What Is Cardinality?
The cardinality of a group or a set represents how many elements or numbers are present in a group or in a set.
For example, if a pencil set has 10 pencils in it, then the cardinality of pencils is 10.
## Cardinal Number of a Set
The number of elements in a set is known as the cardinal number of that set. If A is a finite set and it has N elements, then the cardinal number of set A is given by n(A) $=$ N.
Note: The cardinal number of an empty set is always zero.
For example, what is the cardinal number of a set if the set is defined as A $=$ {2, 5, 7, 9, 11, 15, 19, 22, 24}?
The cardinal number of set A is 9. Hence, n(A) $=$ 9
## Cardinal Numbers vs. Ordinal Numbers
As we discussed, the cardinal numbers are used for counting. They include all the natural numbers. However, an ordinal number is a number that is used to represent the position or order of an object. Examples:
1st rank, 2nd in a queue, 3rd row, tenth floor, etc.
Ordinal numbers are used for ranking or ordering.
Here’s an example that explain the difference between cardinal and ordinal numbers:
In the image given above, we can see a group of 8 children on the picnic spot. Here, 8 represents the cardinal number.
In the above image, we can see the position of the kids in a running event. Here, first, second, third, fourth, fifth are ordinal numbers.
## Nominal Numbers
Nominal numbers, as the name suggests, are used to name an object or a thing in a set of groups. They help us in the identification of objects. It is not used to represent the quantity or the position of an object.
For example:
• Social security number
• Zip code number
• Cell phone number
In the image given below, the numbers 22, 18 are nominal numbers since they are used on the jerseys to represent players or to identify players.
## Conclusion
In this article, we learnt about the Cardinal Numbers. Cardinal Numbers are the numbers used for counting numbers. To read more such informative articles on other concepts, do visit SplashLearn. We, at SplashLearn, are on a mission to make learning fun and interactive for all students.
## Solved Examples
1. Kathy has a list of numbers as shown: 10, 9, 7th, 22, Third, Five, 21st. Identify the cardinal numbers.
Solution: The cardinal numbers are used for counting. The cardinal numbers are 10, 9, 22 and Five.
2. Help Mark calculate the number of consonants in “INDEPENDENCE.” Also identify the number of alphabets used to form this word.
Solution: The consonants are alphabets other than vowels. The vowels are a, e, i, o and u. The number of consonants INDEPENDENCE are 7. The number of alphabets in the word INDEPENDENCE are 12.
3. Write the cardinality of the flowers in the vase.
Solution: There are 5 flowers in the vase. So, the cardinality is 5.
4. What is the position of the strawberry from the left?
Solution: The strawberry is at the second position from the left. Here, second is the ordinal number.
5. How many kites are there in all? What is a cardinal number here?
Solution: On counting, we get:
There are 8 kites. 8 is the cardinal number since we used it for counting.
## Practice Problems
1
### How many two-digit numbers are there?
89
90
91
92
CorrectIncorrect
Correct answer is: 90
The smallest two-digit number is 10 and the greatest two-digit number is 99.
From 1 to 99, there are 99 numbers. Out of them 9 numbers (1 to 9) are one-digit numbers.
So, we will subtract 9 from 99 to find the two-digit numbers.
Number of two-digit numbers $= 99 – 9 = 90$
2
### The cardinality of the set A $= {12, 18, 23, 65}$ is
4
10
18
Cannot be defined.
CorrectIncorrect
Correct answer is: 4
Since the number of elements in the set is 4, the cardinality of the set is 4.
3
### Which of the following is not a cardinal number?
23
Sixty-one
4th
77
CorrectIncorrect
Correct answer is: 4th
4th is not a cardinal number as it tells us about the position of something.
4
### For which of the following are cardinal numbers used?
To find the number of objects
To identify the rank or order of something
To identify objects
All of these
CorrectIncorrect
Correct answer is: To find the number of objects
We use cardinal or natural numbers for finding the number of objects.
5
### 76 is a $\underline{}$ number.
Ordinal
Cardinal
Nominal
None of these
CorrectIncorrect
Correct answer is: Cardinal
76 is a natural number and hence, it is cardinal.
## Frequently Asked Questions
The smallest cardinal number is 1.
No, cardinal numbers cannot be negative. They are positive integers or natural numbers, as we count the number of items starting from number 1, and then it goes up to infinity.
No, we can not count 0 things.
The biggest cardinal number is infinity.
Yes, we can do operations with the cardinal numbers. We can add, subtract, multiply, and divide the cardinal numbers.
All whole numbers are not cardinal numbers. 0 is a whole number but not a cardinal number as we can never say that the number of objects is 0.
When you count a number of objects in a group, the total number of items is the last word said while counting. |
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• Graphing Quadratic Functions y = ax 2 + bx + c
• All the slides in this presentation are timed.
• You do not need to click the mouse or press any keys on the keyboard for the presentation on each slide to continue.
• However, in order to make sure the presentation does not go too quickly, you will need to click the mouse or press a key on the keyboard to advance to the next slide.
• You will know when the slide is finished when you see a small icon in the bottom left corner of the slide.
Click the mouse button to advance the slide when you see this icon.
• Quadratic Functions The graph of a quadratic function is a parabola . A parabola can open up or down. If the parabola opens up, the lowest point is called the vertex. If the parabola opens down, the vertex is the highest point. NOTE: if the parabola opened left or right it would not be a function! y x Vertex Vertex
• Standard Form y = ax 2 + bx + c The parabola will open down when the a value is negative. The parabola will open up when the a value is positive. The standard form of a quadratic function is y x a > 0 a < 0
• Line of Symmetry Parabolas have a symmetric property to them. If we drew a line down the middle of the parabola, we could fold the parabola in half. We call this line the line of symmetry . The line of symmetry ALWAYS passes through the vertex. Or, if we graphed one side of the parabola, we could “fold” (or REFLECT ) it over, the line of symmetry to graph the other side. y x Line of Symmetry
• Finding the Line of Symmetry Find the line of symmetry of y = 3 x 2 – 18 x + 7 When a quadratic function is in standard form The equation of the line of symmetry is y = ax 2 + bx + c , For example… Using the formula… This is best read as … the opposite of b divided by the quantity of 2 times a . Thus, the line of symmetry is x = 3.
• Finding the Vertex We know the line of symmetry always goes through the vertex. Thus, the line of symmetry gives us the x – coordinate of the vertex. To find the y – coordinate of the vertex, we need to plug the x – value into the original equation. STEP 1: Find the line of symmetry STEP 2: Plug the x – value into the original equation to find the y value. y = –2 x 2 + 8 x –3 y = –2(2) 2 + 8(2) –3 y = –2(4)+ 8(2) –3 y = –8+ 16 –3 y = 5 Therefore, the vertex is (2 , 5)
• A Quadratic Function in Standard Form The standard form of a quadratic function is given by y = ax 2 + bx + c There are 3 steps to graphing a parabola in standard form. STEP 1 : Find the line of symmetry STEP 2 : Find the vertex STEP 3 : Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve. Plug in the line of symmetry ( x – value) to obtain the y – value of the vertex. MAKE A TABLE using x – values close to the line of symmetry. USE the equation
• STEP 1 : Find the line of symmetry Let's Graph ONE! Try … y = 2 x 2 – 4 x – 1 A Quadratic Function in Standard Form Thus the line of symmetry is x = 1 y x
• Let's Graph ONE! Try … y = 2 x 2 – 4 x – 1 STEP 2 : Find the vertex A Quadratic Function in Standard Form Thus the vertex is (1 ,–3). Since the x – value of the vertex is given by the line of symmetry, we need to plug in x = 1 to find the y – value of the vertex. y x
• 5 – 1 Let's Graph ONE! Try … y = 2 x 2 – 4 x – 1 STEP 3 : Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve. A Quadratic Function in Standard Form y x 3 2 y x |
# Multiply Whole Number by Fraction
Video solutions to help Grade 5 students learn how to multiply any whole number by a fraction using tape diagrams.
Common Core Standards: 5.NF.4a
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 5
Common Core For Grade 5
## New York State Common Core Math Module 4, Grade 5, Lesson 7
Lesson 7 Application Problem
Mr. Peterson bought a case (24 boxes) of fruit juice. One-third of the drinks were grape and two-thirds were cranberry. How many boxes of each flavor did Mr. Peterson buy? Show your work using a tape diagram or an array.
Lesson 7 Concept Development
Problem 1: What is 3/5 of 35?
Problem 2
Aurelia buys 2 dozen roses. Of these roses, 3/4 are red and the rest are white. How many white roses did she buy?
Problem 3
Rosie had 17 yards of fabric. She used one-third of it to make a quilt. How many yards of fabric did Rosie use for the quilt?
Problem 4
2.3 of a number is 8. What is the number?
Problem 5
Tiffany spent 4/7 of her money on a teddy bear. If the teddy bear cost \$24, how much money did she have at first?
Lesson 7 Problem Set
1. Solve using a tape diagram.
a. 1/3 of 18
c. 3/4 of 24
e. 4/5 x 25
f. 1/4 x 9
i. 2/3 of a number is 10. What’s the number?
j. 3/4 of a number is 24. What’s the number?
Lesson 7 Problem Set
2. Solve using tape diagrams.
a. There are 48 students going on a field trip. One-fourth are girls. How many boys are going on the trip?
b. Three angles are labeled below with arcs. The smallest angle is 3/8 as large as the 160° angle. Find the value of angle a.
c. Abbie spent 5/8 of her money and saved the rest. If she spent \$45, how much money did she have at first?
d. Mrs. Harrison used 16 ounces of dark chocolate while baking. She used 2/5 of the chocolate to make some frosting and used the rest to make brownies. How much more chocolate did Mrs. Harrison use in the brownies than in the frosting?
Lesson 7 Homework
2. Solve using tape diagrams.
b. A straight angle is split into two smaller angles as shown. The smaller angle’s measure is 1/6 of straight angle. What is the value of angle a?
c. Annabel and Eric made 17 ounces of pizza dough. They used 5/8 of the dough to make a pizza and used the rest to make calzones. What is the difference between the amount of dough they used to make pizza and the amount of dough they used to make calzones?
Lesson 7 Homework
1. Solve using a tape diagram.
1/4 of 24
2/3 of 18
This video shows how to multiply whole numbers and fractions using tape diagrams.
1. Solve using a tape diagram.
a) 1/4 of 24
c) 2/3 of 18
k) 3/4 of a number is 27. What’s the number?
2. Solve using tape diagrams.
Homework Question 1
Homework Question 2
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# Properties of addition of rational number
In this chapter we will discuss important properties of addition of rational numbers with examples.
The following properties are discussed in this chapter;
(a) Closure Property
(b) Commutative Property
(c) Associative Property
(d) Distributive Property
We will discuss all these property one by one.
## Addition Property of rational numbers
Important properties of addition of rational number has been discussed below with examples.
### Closure Property
The property states that ” addition of two rational numbers result in another rational number“.
For example;
Let 2/3 and 6/3 are the rational numbers.
The addition of these numbers is expressed as;
\mathtt{\Longrightarrow \ \frac{2}{3} \ +\ \frac{6}{3}}\\\ \\ \mathtt{\Longrightarrow \frac{2+6}{3} \ }\\\ \\ \mathtt{\Longrightarrow \frac{8}{3}}
The final result is 8/3 which is also a rational number.
Example 02
Solution
\mathtt{\Longrightarrow \ \frac{1}{5} \ +\ \frac{2}{7}}\\\ \\ \mathtt{\Longrightarrow \frac{7+10}{35} \ }\\\ \\ \mathtt{\Longrightarrow \frac{17}{35}}
The final result is 17 / 35 which is also a rational number.
Conclusion
Hence, the sum of two rational numbers is always a rational number.
### Commutative Property
The Commutative property of rational number states that changing the position of rational number in addition will not change the final result.
So if (a/b) & (c/d) are two rational numbers then;
(a/b) + (c/d) = (c/d) + (a/b)
For example;
Consider the rational numbers \mathtt{\frac{13}{11} \ \&\ \ \frac{5}{11}}
\mathtt{\frac{13}{11} \ +\ \frac{5}{11} \ \Longrightarrow \ \frac{18}{11}}
Now interchanging the position of numbers.
\mathtt{\frac{5}{11} \ +\ \frac{13}{11} \ \Longrightarrow \ \frac{18}{11}}
Hence, interchanging the position of rational number in addition will not change the final result.
### Associative property
It states that during addition, changing the group of rational number will not affect the end result.
If a/b, c/d & d/e are the three rational numbers. Then according to the additive property.
( a/b + c/d ) + d/e = a/b + ( c/d + d/e )
For Example;
Let 2/3, 5/3 and 7/3 are the given rational numbers.
Then according to associative property.
(2/3 + 5/3) + 7/3 = 2/3 + (5/3 + 7/3)
7/3 + 7/3 = 2/3 + 12/3
14/3 = 14/3
Hence Proved.
### Distributive Property
The multiplication of rational number with sum of two rational number is equal to sum of product of rational numbers.
If a/b, c/d & d/e are three rational numbers.
Then according to distributive property.
a/b ( c/d + d/e ) = a/b . c/d + a/b . d/e
Let’s prove the distributive property with the help of example.
Given are three rational numbers 1/5, 4/5 and 3/5.
According to distributive property;
\mathtt{\frac{1}{5} \ \left(\frac{4}{5} +\frac{3}{5}\right) =\ \frac{1}{5} .\frac{4}{5} \ +\ \frac{1}{5} .\frac{3}{5} \ }\\\ \\ \mathtt{\frac{1}{5} \ .\frac{7}{5} \ \ =\frac{4}{25} +\frac{3}{25}}\\\ \\ \mathtt{\frac{7}{25} \ =\ \frac{7}{25}}
Since, LHS = RHS.
We have proved the distributive property of addition of rational number.
The property states that ” when we add any rational number with 0 we get the same rational number
Hence by adding with number 0, the identity of rational number will be preserved.
a/b + 0 = a/b
The number which addition with rational number produce 0 result is called additive inverse.
If a/b is the rational number then -a/b is the additive inverse.
a/b + (-a/b) = 0
Hence, for any rational number if we insert negative sign in the front it becomes its additive inverse.
Example 01
Find the additive inverse of 2/3
-2/3 is the additive inverse as it adds up with 2/3 to produce 0.
⟹ 2/3 + (-2/3)
⟹ 0
Example 02 |
Practice
MEDIUM
Math
# PROBLEM:
Sachin and Sakshi are Newly Married couple in the town. And since our couple are bit romantic the air in the town is filled up with the love. I hope you can feel it like me. Because they belive that “Love is Life” .
Today our couple is going to buy some Food for the childs living in orfanage. The food items are being sold at price Z rupees per food item. Let’s consider Sachin has A rupees and Sakshi has R rupees with them. Both i.e Sachin and Sakshi will buy as many food items they can from their own money not by sharing each other because both belive in independent idologies. This way each of them will buy an integer non-negative number of food items. But after some time they both came up with good idea and found that the total number of food items they buy can increase (or decrease) if one of them gives several rupees to the other.
(NOTE : The rupee will always strictly be a non-negative integer only).
So Sachin and Sakshi want to exchange with rupees in such a way that they will buy the maximum possible number of food items. so among all possible ways to buy the maximum possible number of food items find such a way that minimizes the number of rupees one can share with other.
Your job is to print the Minimum number of rupees one will share with other to buy the maximum number of food items.
# QUICK EXPLANATION:
Suppose Sachin has 50 rupees and Sakshi has 40 rupees, and the price for one food item is 30 rupees. If they don’t exchange with rupees, they will buy 1+1=2 food items. BUT if Sakshi gives Sachin ten rupee, then Sachin will have 60 rupees, Sakshi will left with 30 rupees, and with this exchange they will buy 2+1=3 food items. (which is more than previous one i.e 2).
# EXPLANATION:
Let’s consider Sachin has A rupees and Sakshi has R rupees with them. Both i.e Sachin and Sakshi will buy as many food items they can from their own money not by sharing each other because both belive in independent idologies. This way each of them will buy an integer non-negative number of food items. But after some time they both came up with good idea and found that the total number of food items they buy can increase (or decrease) if one of them gives several rupees to the other.
Suppose Sachin has 50 rupees and Sakshi has 40 rupees, and the price for one food item is 30 rupees. If they don’t exchange with rupees, they will buy 1+1=2 food items. BUT if Sakshi gives Sachin ten rupee, then Sachin will have 60 rupees, Sakshi will left with 30 rupees, and with this exchange they will buy 2+1=3 food items. (which is more than previous one i.e 2).
(NOTE : The rupee will always strictly be a non-negative integer only).
# SOLUTIONS:
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
ll t;
cin>>t;
while(t–)
{
ll a,r,z;
cin>>a>>r>>z;
ll transfer = 0;
if(a%z + r%z >=z)
{
transfer = z - max(a%z , r%z);
}
cout<<transfer<<endl;
}
``````return 0;
``````
}
Tester's Solution
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
ll t;
cin>>t;
while(t–)
{
ll a,r,z;
cin>>a>>r>>z;
ll transfer = 0;
if(a%z + r%z >=z)
{
transfer = z - max(a%z , r%z);
}
cout<<transfer<<endl;
}
``````return 0;
``````
}
Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
ll t;
cin>>t;
while(t–)
{
ll a,r,z;
cin>>a>>r>>z;
ll transfer = 0;
if(a%z + r%z >=z)
{
transfer = z - max(a%z , r%z);
}
cout<<transfer<<endl;
}
``````return 0;
``````
} |
P. 1
08 Comparing Quantities
# 08 Comparing Quantities
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# COMPARING QUANTITIES 117
8.1 Recalling Ratios and Percentages
We know, ratio means comparing two quantities.
A basket has two types of fruits, say, 20 apples and 5 oranges.
Then, the ratio of the number of oranges to the number of apples = 5 : 20.
The comparison can be done by using fractions as,
5
20
=
1
4
The number of oranges are
1
4
th the number of apples. In terms of ratio, this is
1 : 4, read as, “1 is to 4”
Number of apples to number of oranges =
20 4
5 1
=
which means, the number of apples
are 4 times the number of oranges. This comparison can also be done using percentages.
There are 5 oranges out of 25 fruits.
So percentage of oranges is
5 4 20
20%
25 4 100
× = = OR
Since contains only apples and oranges,
So, percentage of apples + percentage of oranges = 100
or percentage of apples + 20 = 100
or percentage of apples = 100 – 20 = 80
Thus the basket has 20% oranges and 80% apples.
Example 1: A picnic is being planned in a school for Class VII. Girls are 60% of the
total number of students and are 18 in number.
The picnic site is 55 km from the school and the transport company is charging at the rate
of Rs 12 per km. The total cost of refreshments will be Rs 4280.
Comparing Quantities
CHAPTER
8
By unitary method:
Out of 25 fruits, number of oranges are 5.
So out of 100 fruits, number of oranges
=
5
100
25
×
= 20.
OR
118 MATHEMATICS
Can you tell.
1. The ratio of the number of girls to the number of boys in the class?
2. The cost per head if two teachers are also going with the class?
3. If their first stop is at a place 22 km from the school, what per cent of the total
distance of 55 km is this? What per cent of the distance is left to be covered?
Solution:
1. To find the ratio of girls to boys.
Ashima and John came up with the following answers.
They needed to know the number of boys and also the total number of students.
Ashima did this John used the unitary method
Let the total number of students There are 60 girls out of 100 students.
be x. 60% of x is girls. There is one girl out of
100
60
students.
Therefore, 60% of x = 18 So, 18 girls are out of how many students?
60
100
x ×
= 18 OR Number of students =
100
18
60
×
or, x =
18 100
60
×
= 30 = 30
Number of students = 30.
So, the number of boys = 30 – 18 = 12.
Hence, ratio of the number of girls to the number of boys is 18 : 12 or
18
12
=
3
2
.
3
2
is written as 3 : 2 and read as 3 is to 2.
2. To find the cost per person.
Transportation charge = Distance both ways × Rate
= Rs (55 × 2) × 12
= Rs 110 × 12 = Rs 1320
Total expenses = Refreshment charge
+ Transportation charge
= Rs 4280 + Rs 1320
= Rs 5600
Total number of persons =18 girls + 12 boys + 2 teachers
= 32 persons
Ashima and John then used unitary method to find the cost per head.
For 32 persons, amount spent would be Rs 5600.
The amount spent for 1 person = Rs
5600
32
= Rs 175.
3. The distance of the place where first stop was made = 22 km.
COMPARING QUANTITIES 119
To find the percentage of distance:
Ashima used this method: John used the unitary method:
22 22 100
40%
55 55 100
= × =
Out of 55 km, 22 km are travelled.
OR Out of 1 km,
22
55
km are travelled.
Out of 100 km,
22
55
× 100 km are travelled.
That is 40% of the total distance is travelled.
She is multiplying
100
the ratio by =1
100
and converting to
percentage.
TRY THESE
Both came out with the same answer that the distance from their school of the place where
they stopped at was 40% of the total distance they had to travel.
Therefore, the percent distance left to be travelled = 100% – 40% = 60%.
In a primary school, the parents were asked about the number of hours they spend per day
in helping their children to do homework. There were 90 parents who helped for
1
2
hour
to
1
1
2
hours. The distribution of parents according to the time for which,
they said they helped is given in the adjoining figure ; 20% helped for
more than
1
1
2
hours per day;
30% helped for
1
2
hour to
1
1
2
hours; 50% did not help at all.
(i) How many parents were surveyed?
(ii) How many said that they did not help?
(iii) How many said that they helped for more than
1
1
2
hours?
EXERCISE 8.1
1. Find the ratio of the following.
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km (c) 50 paise to Rs 5
2. Convert the following ratios to percentages.
(a) 3 : 4 (b) 2 : 3
3. 72% of 25 students are good in mathematics. How many are not good in mathematics?
4. A football team won 10 matches out of the total number of matches they played. If
their win percentage was 40, then how many matches did they play in all?
5. If Chameli had Rs 600 left after spending 75% of her money, how much did she have
in the beginning?
120 MATHEMATICS
6. If 60% people in a city like cricket, 30% like football and the remaining like other
games, then what per cent of the people like other games? If the total number of
people are 50 lakh, find the exact number who like each type of game.
8.2 Finding the Increase or Decrease Per cent
We often come across such information in our daily life as.
(i) 25% off on marked prices (ii) 10% hike in the price of petrol
Let us consider a few such examples.
Example 2: The price of a scooter was Rs 34,000 last year. It has increased by 20%
this year. What is the price now?
Solution:
OR
Amita said that she would first find
the increase in the price, which is 20% of
Rs 34,000, and then find the new price.
20% of Rs 34000 = Rs
20
34000
100
×
= Rs 6800
New price = Old price + Increase
= Rs 34,000 + Rs 6,800
= Rs 40,800
Similarly, a percentage decrease in price would imply finding the actual decrease
followed by its subtraction the from original price.
Suppose in order to increase its sale, the price of scooter was decreased by 5%.
Then let us find the price of scooter.
Price of scooter = Rs 34000
Reduction = 5% of Rs 34000
= Rs
5
34000
100
×
= Rs 1700
New price = Old price – Reduction
= Rs 34000 – Rs 1700 = Rs 32300
We will also use this in the next section of the chapter.
8.3 Finding Discounts
Discount is a reduction given on the Marked Price
(MP) of the article.
This is generally given to attract customers to buy
goods or to promote sales of the goods. You can find
the discount by subtracting its sale price from its
marked price.
So, Discount = Marked price – Sale price
Sunita used the unitary method.
20% increase means,
Rs 100 increased to Rs 120.
So, Rs 34,000 will increase to?
Increased price = Rs
120
34000
100
×
= Rs 40,800
COMPARING QUANTITIES 121
TRY THESE
Example 3: An item marked at Rs 840 is sold for Rs 714. What is the discount and
discount %?
Solution: Discount = Marked Price – Sale Price
= Rs 840 – Rs 714
= Rs 126
Since discount is on marked price, we will have to use marked price as the base.
On marked price of Rs 840, the discount is Rs 126.
On MP of Rs 100, how much will the discount be?
Discount =
126
100
840
× = 15%
You can also find discount when discount % is given.
Example 4: The list price of a frock is Rs 220.
A discount of 20% is announced on sales. What is the amount
of discount on it and its sale price.
Solution: Marked price is same as the list price.
20% discount means that on Rs 100 (MP), the discount is Rs 20.
By unitary method, on Re 1 the discount will be Rs
20
100
.
On Rs 220, discount = Rs
20
220
100
× = Rs 44
The sale price = (Rs 220 – Rs 44) or Rs 176
Rehana found the sale price like this —
A discount of 20% means for a MP of Rs 100, discount is Rs 20. Hence the sale price
is Rs 80. Using unitary method, when MP is Rs 100, sale price is Rs 80;
When MP is Re 1, sale price is Rs
80
100
.
Hence when MP is Rs 220, sale price = Rs
80
220
100
× = Rs 176.
1. A shop gives 20% discount. What would the sale price of each of these be?
(a) A dress marked at Rs 120 (b) A pair of shoes marked at Rs 750
(c) A bag marked at Rs 250
2. A table marked at Rs 15,000 is available for Rs 14,400. Find the discount given
and the discount per cent.
3. An almirah is sold at Rs 5,225 after allowing a discount of 5%. Find its marked price.
Even though the
discount was not
found, I could find
the sale price
directly.
122 MATHEMATICS
8.3.1 Estimation in percentages
Your bill in a shop is Rs 577.80 and the shopkeeper gives a discount of 15%. How would
you estimate the amount to be paid?
(i) Round off the bill to the nearest tens of Rs 577.80, i.e., to Rs 580.
(ii) Find 10% of this, i.e., Rs
10
580 Rs 58
100
× = .
(iii) Take half of this, i.e.,
1
58 Rs 29
2
× = .
(iv) Add the amounts in (ii) and (iii) to get Rs 87.
You could therefore reduce your bill amount by Rs 87 or by about Rs 85, which will
be Rs 495 approximately.
1. Try estimating 20% of the same bill amount. 2. Try finding 15% of Rs 375.
8.4 Prices Related to Buying and Selling (Profit and Loss)
For the school fair (mela) I am going to put a stall of lucky dips. I will charge Rs 10 for
one lucky dip but I will buy items which are worth Rs 5.
So you are making a profit of 100%.
No, I will spend Rs 3 on paper to wrap the gift and tape. So my expenditure is Rs 8.
This gives me a profit of Rs 2, which is,
2
100 25%
8
× = only.
before selling it. These expenses have to be included in the cost price.
These expenses are sometimes referred to as overhead charges. These may include
expenses like amount spent on repairs, labour charges, transportation etc.
8.4.1 Finding cost price/selling price, profit %/loss%
Example 5: Sohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500
on its repairs and sold it for Rs 3,300. Find his loss or gain per cent.
Solution: Cost Price (CP) = Rs 2500 + Rs 500 (overhead expenses are added to give CP)
= Rs 3000
Sale Price (SP) = Rs 3300
As SP > CP, he made a profit = Rs 3300 – Rs 3000 = Rs 300
His profit on Rs 3,000, is Rs 300. How much would be his profit on Rs 100?
Profit
300 30
100% % 10%
3000 3
= × = = P% =
P
100
CP
×
COMPARING QUANTITIES 123
TRY THESE
TRY THESE
1. Find selling price (SP) if a profit of 5% is made on
(a) a cycle of Rs 700 with Rs 50 as overhead charges.
(b) a lawn mower bought at Rs 1150 with Rs 50 as transportation charges.
(c) a fan bought for Rs 560 and expenses of Rs 40 made on its repairs.
Example 6: A shopkeeper purchased 200 bulbs for Rs 10 each. However 5 bulbs
were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the
gain or loss %.
Solution: Cost price of 200 bulbs = Rs 200 × 10 = Rs 2000
5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195
These were sold at Rs 12 each.
The SP of 195 bulbs = Rs 195 × 12 = Rs 2340
He obviously made a profit (as SP > CP).
Profit = Rs 2340 – Rs 2000 = Rs 340
On Rs 2000, the profit is Rs 340. How much profit is made on Rs 100?
Profit =
340
100
2000
×
= 17%.
Example 7: Meenu bought two fans for Rs 1200 each. She sold one
at a loss of 5% and the other at a profit of 10%. Find the selling price of
each. Also find out the total profit or loss.
Solution: Overall CP of each fan = Rs 1200. One is sold at a loss of 5%.
This means if CP is Rs 100, SP is Rs 95.
Therefore, when CP is Rs 1200, then SP = Rs
95
1200
100
× = Rs 1140
Also second fan is sold at a profit of 10%.
It means, if CP is Rs 100, SP is Rs 110.
Therefore, when CP is Rs 1200, then SP = Rs
110
1200
100
× = Rs 1320
Was there an overall loss or gain?
We need to find the combined CP and SP to say
whether there was an overall profit or loss.
Total CP = Rs 1200 + Rs 1200 = Rs 2400
Total SP = Rs 1140 + Rs 1320 = Rs 2460
Since total SP > total CP, a profit of Rs (2460 – 2400) or Rs 60 has been made.
1. A shopkeeper bought two TV sets at Rs 10,000 each. He sold one at a profit 10%
and the other at a loss of 10%. Find whether he made an overall profit or loss.
124 MATHEMATICS
TRY THESE
The teacher showed the class a bill in which the following heads were written.
Bill No. Date
S.No. Item Quantity Rate Amount
Bill amount
+ ST (5%)
Total
ST means Sales Tax, which we pay when we buy items.
This sales tax is charged by the government on the sale of an item.
It is collected by the shopkeeper from the customer and given to the government.
This is, therefore, always on the selling price of an item and is added to the value of the bill.
These days however, the prices include the tax known as Value Added Tax (VAT).
Example 8: (Finding Sales Tax) The cost of a pair of
roller skates at a shop was Rs 450. The sales tax charged was
5%. Find the bill amount.
Solution: On Rs 100, the tax paid was Rs 5.
On Rs 450, the tax paid would be = Rs
5
450
100
×
= Rs 22.50
Bill amount = Cost of item + Sales tax = Rs 450 + Rs 22.50 = Rs 472.50.
Example 9: (Value Added Tax (VAT)) Waheeda bought an air cooler for Rs 3300
including a tax of 10%. Find the price of the air cooler before VAT was added.
Solution: The price includes the VAT, i.e., the value added tax. Thus, a 10% VAT
means if the price without VAT is Rs 100 then price including VAT is Rs 110.
Now, when price including VAT is Rs 110, original price is Rs 100.
Hence when price including tax is Rs 3300, the original price = Rs.
1. Find the buying price of each of the following when 5% ST is added on the
purchase of
(a) A towel at Rs 50 (b) Two bars of soap at Rs 35 each
(c) 5 kg of flour at Rs 15 per kg
COMPARING QUANTITIES 125
THINK, DISCUSS AND WRITE
2. If 8% VAT is included in the prices, find the original price of
(a) A TV bought for Rs 13,500 (b) A shampoo bottle bought for Rs 180
1. Two times a number is a 100% increase in the number. If we take half the number
what would be the decrease in per cent?
2. By what per cent is Rs 2,000 less than Rs 2,400? Is it the same as the per cent by
which Rs 2,400 is more than Rs 2,000?
EXERCISE 8.2
1. A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his
original salary.
2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What
is the per cent decrease in the people visiting the Zoo on Monday?
3. A shopkeeper buys 80 articles for Rs 2,400 and sells them for a profit of
16%. Find the selling price of one article.
4. The cost of an article was Rs 15,500. Rs 450 were spent on its repairs. If
it is sold for a profit of 15%, find the selling price of the article.
5. A VCR and TV were bought for Rs 8,000 each. The shopkeeper made a
loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss
percent on the whole transaction.
6. During a sale, a shop offered a discount of 10%
on the marked prices of all the items. What would a
customer have to pay for a pair of jeans marked at
Rs 1450 and two shirts marked at Rs 850 each?
7. A milkman sold two of his buffaloes for Rs 20,000 each.
On one he made a gain of 5% and on the other a loss of
10%. Find his overall gain or loss. (Hint: Find CP of each)
8. The price of a TV is Rs 13,000. The sales tax charged on
it is at the rate of 12%. Find the amount that Vinod will
have to pay if he buys it.
9. Arun bought a pair of skates at a sale where the discount given was 20%. If the
amount he pays is Rs 1,600, find the marked price.
10. I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT
8.6 Compound Interest
You might have come across statements like “one year interest for FD (fixed deposit) in
the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’.
126 MATHEMATICS
TRY THESE
Interest is the extra money paid by institutions like banks or post offices on money
deposited (kept) with them. Interest is also paid by people when they borrow money.
We already know how to calculate Simple Interest.
Example 10: A sum of Rs 10,000 is borrowed at a rate of interest 15% per annum for
2 years. Find the simple interest on this sum and the amount to be paid at the end of 2 years.
Solution: On Rs 100, interest charged for 1 year is Rs 15.
So, on Rs 10,000, interest charged =
15
10000
100
×
= Rs 1500
Interest for 2 years = Rs 1500 × 2 = Rs 3000
Amount to be paid at the end of 2 years = Principal + Interest
= Rs 10000 + Rs 3000 = Rs 13000
Find interest and amount to be paid on Rs 15000 at 5% per annum after 2 years.
My father has kept some money in the post office for 3 years. Every year the money
increases as more than the previous year.
We have some money in the bank. Every year some interest is added to it, which is
shown in the passbook. This interest is not the same, each year it increases.
Normally, the interest paid or charged is never simple. The interest is calculated on the
amount of the previous year. This is known as interest compounded or Compound
Interest (C.I.).
Let us take an example and find the interest year by year. Each year our sum or
principal changes.
Calculating Compound Interest
A sum of Rs 20,000 is borrowed by Heena for 2 years at an interest of 8% compounded
annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of
2 years.
Aslam asked the teacher whether this means that they should find the interest year by
year. The teacher said ‘yes’, and asked him to use the following steps :
1. Find the Simple Interest (S.I.) for one year.
Let the principal for the first year be P
1
. Here, P
1
= Rs 20,000
SI
1
= SI at 8% p.a. for 1st year = Rs
20000 8
100
×
= Rs 1600
2. Then find the amount which will be paid or received. This becomes principal for the
next year.
Amount at the end of 1st year = P
1
+ SI
1
= Rs 20000 + Rs 1600
= Rs 21600 = P
2
(Principal for 2nd year)
COMPARING QUANTITIES 127
3. Again find the interest on this sum for another year.
SI
2
= SI at 8% p.a.for 2nd year = Rs
21600 8
100
×
= Rs 1728
4. Find the amount which has to be paid or received at the end of second year.
Amount at the end of 2nd year = P
2
+ SI
2
= Rs 21600 + Rs 1728
= Rs 23328
Total interest given = Rs 1600 + Rs 1728
= Rs 3328
Reeta asked whether the amount would be different for simple interest. The teacher
told her to find the interest for two years and see for herself.
SI for 2 years = Rs
20000 8 2
100
× ×
= Rs 3200
Reeta said that when compound interest was used Heena would pay Rs 128 more.
Let us look at the difference between simple interest and compound interest. We start
with Rs 100. Try completing the chart.
Under Under
Simple Interest Compound Interest
First year Principal Rs 100.00 Rs 100.00
Interest at 10% Rs 10.00 Rs 10.00
Year-end amount Rs 110.00 Rs 110.00
Second year Principal Rs 100.00 Rs 110.00
Interest at 10% Rs 10.00 Rs 11.00
Year-end amount Rs(110 + 10) = Rs 120 Rs 121.00
Third year Principal Rs 100.00 Rs 121.00
Interest at 10% Rs 10.00 Rs 12.10
Year-end amount Rs(120 + 10) = Rs 130 Rs 133.10
Note that in 3 years,
Interest earned by Simple Interest = Rs (130 – 100) = Rs 30, whereas,
Interest earned by Compound Interest = Rs (133.10 – 100) = Rs 33.10
Note also that the Principal remains the same under Simple interest, while it changes
year after year under compound interest.
Which
means you
pay interest
on the
interest
accumulated
till then!
128 MATHEMATICS
8.7 Deducing a Formula for Compound Interest
Zubeda asked her teacher, ‘Is there an easier way to find compound interest?’ The teacher
said ‘There is a shorter way of finding compound interest. Let us try to find it.’
Suppose P
1
is the sum on which interest is compounded annually at a rate of R%
per annum.
Let P
1
= Rs 5000 and R = 5% per annum. Then by the steps mentioned above
1. SI
1
= Rs
5000 5 1
100
× ×
or SI
1
= Rs
1
P R 1
100
× ×
so, A
1
= Rs 5000 +
5000 5 1
100
× ×
or A
1
= P
1
+ SI
1
=
1
1
P R
P
100
+
= Rs 5000
5
1
100
| |
+
|
\ ¹
= P
2
=
1 2
R
P 1 P
100
| |
+ =
|
\ ¹
2. SI
2
= Rs 5000
5 5 1
1
100 100
× | |
+ ×
|
\ ¹
or SI
2
=
2
P R 1
100
× ×
= Rs
5000 5 5
1
100 100
× | |
+
|
\ ¹
=
1
R R
P 1
100 100
| |
+ ×
|
\ ¹
=
1
P R R
1
100 100
| |
+
|
\ ¹
A
2
= Rs
5 5000 5 5
5000 1 Rs 1
100 100 100
× | | | |
+ + +
| |
\ ¹ \ ¹
A
2
= P
2
+ SI
2
= Rs
5 5
5000 1 1
100 100
| | | |
+ +
| |
\ ¹ \ ¹
=
1 1
R R R
P 1 P 1
100 100 100
| | | |
+ + +
| |
\ ¹ \ ¹
= Rs
2
5
5000 1
100
| |
+
|
\ ¹
= P
3
=
1
R R
P 1 1
100 100
| | | |
+ +
| |
\ ¹ \ ¹
=
2
1 3
R
P 1 P
100
| |
+ =
|
\ ¹
Proceeding in this way the amount at the end of n years will be
A
n
=
1
R
P 1
100
n
| |
+
|
\ ¹
Or, we can say A =
R
P 1
100
n
| |
+
|
\ ¹
COMPARING QUANTITIES 129
So, Zubeda said, but using this we get only the formula for the amount to be paid at
the end of n years, and not the formula for compound interest.
Aruna at once said that we know CI = A – P, so we can easily find the compound
interest too.
Example 11: Find CI on Rs 12600 for 2 years at 10% per annum compounded
annually.
Solution: We have, A = P
R
1
100
n
| |
+
|
\ ¹
, where Principal (P) = Rs 12600, Rate (R) = 10,
Number of years (n) = 2
= Rs
2
10
12600 1
100
| |
+
|
\ ¹
= Rs
2
11
12600
10
| |
|
\ ¹
= Rs
11 11
12600
10 10
× ×
= Rs 15246
CI = A – P = Rs 15246 – Rs 12600 = Rs 2646
8.8 Rate Compounded Annually or Half Yearly
(Semi Annually)
You may want to know why ‘compounded
annually’ was mentioned after ‘rate’. Does it
mean anything?
It does, because we can also have interest
rates compounded half yearly or quarterly. Let
us see what happens to Rs 100 over a period
of one year if an interest is compounded
annually or half yearly.
TRY THESE
1. Find CI on a sum of Rs 8000
for 2 years at 5% per annum
compounded annually.
P = Rs 100 at 10% per P = Rs 100 at 10% per annum
annum compounded annually compounded half yearly
The time period taken is 1 year The time period is 6 months or
1
2
year
I = Rs
100 10 1
Rs 10
100
× ×
=
I = Rs
1
100 10
2
Rs 5
100
× ×
=
A = Rs 100 + Rs 10 A = Rs 100 + Rs 5 = Rs 105
= Rs 110 Now for next 6 months the P = Rs 105
So, I = Rs
1
105 10
2
100
× ×
= Rs 5.25
and A = Rs 105 + Rs 5.25 = Rs 110.25
Rate
becomes
half
Time period and rate when interest not compounded
annually
The time period after which the interest is added each
time to form a new principal is called the conversion
period. When the interest is compounded half yearly,
there are two conversion periods in a year each after 6
months. In such situations, the half yearly rate will be
half of the annual rate. What will happen if interest is
compounded quarterly? In this case, there are 4
conversion periods in a year and the quarterly rate will
be one-fourth of the annual rate.
130 MATHEMATICS
THINK, DISCUSS AND WRITE
TRY THESE
Do you see that, if interest is compounded half yearly, we compute the interest two
times. So time period becomes twice and rate is taken half.
Find the time period and rate for each .
1. A sum taken for
1
1
2
years at 8% per annum is compounded half yearly.
2. A sum taken for 2 years at 4% per annum compounded half yearly.
A sum is taken for one year at 16% p.a. If interest is compounded after every three
months, how many times will interest be charged in one year?
Example 12: What amount is to be repaid on a loan of Rs 12000 for 1
1
2
years at
10% per annum compounded half yearly.
Solution:
Principal for first 6 months = Rs 12,000 Principal for first 6 months = Rs 12,000
There are 3 half years in
1
1
2
years. Time = 6 months =
6 1
year year
12 2
=
Therefore, compounding has to be done 3 times. Rate = 10%
Rate of interest = half of 10% I = Rs
1
12000 10
2
100
× ×
= Rs 600
= 5% half yearly A = P + I = Rs 12000 + Rs 600
A =
R
P 1
100
n
| |
+
|
\ ¹
= Rs 12600. It is principal for next 6 months.
= Rs 12000
3
5
1
100
| |
+
|
\ ¹
I = Rs
1
12600 10
2
100
× ×
= Rs 630
= Rs
21 21 21
12000
20 20 20
× × × Principal for third period = Rs 12600 + Rs 630
= Rs 13,891.50 = Rs 13,230.
I = Rs
1
13230 10
2
100
× ×
= Rs 661.50
A = P + I = Rs 13230 + Rs 661.50
= Rs 13,891.50
COMPARING QUANTITIES 131
TRY THESE
Find the amount to be paid
1. At the end of 2 years on Rs 2,400 at 5% per annum compounded annually.
2. At the end of 1 year on Rs 1,800 at 8% per annum compounded quarterly.
Example 13: Find CI paid when a sum of Rs 10,000 is invested for 1 year and
3 months at 8
1
2
% per annum compounded annually.
Solution: Mayuri first converted the time in years.
1 year 3 months =
3
1
12
year =
1
1
4
years
Mayuri tried putting the values in the known formula and came up with:
A = Rs 10000
1
1
4 17
1
200
| |
+
|
\ ¹
Now she was stuck. She asked her teacher how would she find a power which is fractional?
The teacher then gave her a hint:
Find the amount for the whole part, i.e., 1 year in this case. Then use this as principal
to get simple interest for
1
4
year more. Thus,
A = Rs 10000
17
1
200
| |
+
|
\ ¹
= Rs 10000 ×
217
200
= Rs 10,850
Now this would act as principal for the next
1
4
year. We find the SI on Rs 10,850
for
1
4
year.
SI = Rs
1
10850 17
4
100 2
× ×
×
= Rs
10850 1 17
800
× ×
= Rs 230.56
132 MATHEMATICS
Interest for first year = Rs 10850 – Rs 10000 = Rs 850
And, interest for the next
1
4
year = Rs 230.56
Therefore, total compound Interest = 850 + 230.56 = Rs 1080.56.
8.9 Applications of Compound Interest Formula
There are some situations where we could use the formula for calculation of amount in CI.
Here are a few.
(i) Increase (or decrease) in population.
(ii) The growth of a bacteria if the rate of growth is known.
(iii) The value of an item, if its price increases or decreases in the intermediate years.
Example 14: The population of a city was 20,000 in the year 1997. It increased at
the rate of 5% p.a. Find the population at the end of the year 2000.
Solution: There is 5% increase in population every year, so every new year has new
population. Thus, we can say it is increasing in compounded form.
Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st
year)
Increase at 5% =
5
20000 1000
100
× =
Population in 1999 = 20000 + 1000 = 21000
Increase at 5% =
5
21000 1050
100
× =
Population in 2000 = 21000 + 1050
= 22050
Increase at 5% =
5
22050
100
×
= 1102.5
At the end of 2000 the population = 22050 + 1102.5 = 23152.5
or, Population at the end of 2000 = 20000
3
5
1
100
| |
+
|
\ ¹
=
21 21 21
20000
20 20 20
× × ×
= 23152.5
So, the estimated population = 23153.
Treat as
the Principal
for the
2nd year.
Treat as
the Principal
for the
3rd
year.
COMPARING QUANTITIES 133
TRY THESE
Aruna asked what is to be done if there is a decrease. The teacher then considered
the following example.
Example 15: A TV was bought at a price of Rs 21,000. After one year the value of
the TV was depreciated by 5% (Depreciation means reduction of value due to use and
age of the item). Find the value of the TV after one year.
Solution:
Principal = Rs 21,000
Reduction = 5% of Rs 21000 per year
= Rs
21000 5 1
100
× ×
= Rs 1050
value at the end of 1 year = Rs 21000 – Rs 1050 = Rs 19,950
Alternately, We may directly get this as follows:
value at the end of 1 year = Rs 21000
5
1
100
| |
|
\ ¹
= Rs 21000 ×
19
20
= Rs 19,950
1. A machinery worth Rs 10,500 depreciated by 5%. Find its value after one year.
2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate
of increase is 4%.
EXERCISE 8.3
1. Calculate the amount and compound interest on
(a) Rs 10,800 for 3 years at 12
1
2
% per annum compounded annually.
(b) Rs 18,000 for 2
1
2
years at 10% per annum compounded annually.
(c) Rs 62,500 for 1
1
2
years at 8% per annum compounded half yearly.
(d) Rs 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify).
(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.
2. Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a.
compounded yearly. What amount will she pay at the end of 2 years and 4 months to
clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the
2nd year amount for
4
12
years).
134 MATHEMATICS
3. Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and
Radha borrows the same amount for the same time period at 10% per annum,
compounded annually. Who pays more interest and by how much?
4. I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years.
Had I borrowed this sum at 6% per annum compound interest, what extra amount
would I have to pay?
5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded
half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
6. Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum,
find the difference in amounts he would be paying after
1
1
2
years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum
compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
8. Find the amount and the compound interest on Rs 10,000 for
1
1
2
years at 10% per
annum, compounded half yearly. Would this interest be more than the interest he
would get if it was compounded annually?
9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at
1
12 %
2
per annum, interest being compounded half yearly.
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the
rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially
5, 06,000.
12. A scooter was bought at Rs 42,000. Its value
depreciated at the rate of 8% per annum.
Find its value after one year.
COMPARING QUANTITIES 135
WHAT HAVE WE DISCUSSED?
1. Discount is a reduction given on marked price.
Discount = Marked Price – Sale Price.
2. Discount can be calculated when discount percentage is given.
Discount = Discount % of Marked Price
3. Additional expenses made after buying an article are included in the cost price and are known
4. Sales tax is charged on the sale of an item by the government and is added to the Bill Amount.
Sales tax = Tax% of Bill Amount
5. Compound interest is the interest calculated on the previous year’s amount (A = P + I)
6. (i) Amount when interest is compounded annually
=
R
P 1
100
n
| |
+
|
\ ¹
; P is principal, R is rate of interest, n is time period
(ii) Amount when interest is compounded half yearly
=
2
R
P 1
200
n
| |
+
|
\ ¹
R
is half yearly rate and
2
2 = number of 'half-years' n
¦
¦
´
¦
¹
136 MATHEMATICS
NOTES
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# Timestables in 21 Days
## Multiplication Facts
By
Multiplication Tables in 21 days!
Let's face in, when you don't know your times tables, it slows down your progress in math. Some things you just have to know and committing the times tables to memory is one of them. Today, we're in an information age, information is doubling faster than it ever used to and our math teachers no longer have the luxury of assisting us to learn the times tables. In case you haven't noticed, the math curriculum is much larger than it ever was. Students and parents are now left with the task of helping to commit the times tables to memory. So let's get started:
Step 1
First of all, you will need to be able to skip count or count by a certain number. For instance 2,4,6,8,10 or 5, 10, 15, 20, 25. Now you will need to use your fingers when skip counting. Remember back in grade 1 when you used to use your fingers to count to 10? Now you'll need them to skip-count. For example, use your fingers to count by 10. First finger or thumb is 10, second is 20, third is 30. Therefore 1 x 10 = 10, 2 x 10 = 20 and so on and so forth. Why use your fingers? Because it's an effective strategy. Any strategy that improves speed with your tables is worth using!
Step 2
How many skip counting patterns do you know? Probably the 2's, 5's and 10's. Practice tapping these out on your fingers.
Step 3
Now you're ready for the 'doubles'. Once you learn the doubles, you have the 'counting up' strategy. For instance, if you know that 7 x 7 = 49, then you'll count up 7 more to quickly determine that 7 x 8 = 56. Once again, effective strategies are almost as good as memorizing your facts. Remember, you already know the 2's, 5's and 10's. Now you need to concentrate on 3x3, 4x4, 6x6, 7x7, 8x8 and 9x9. That's only committing 6 facts to memory! You're three-quarters of the way there. If you memorize those doubles, you'll have an effective strategy to quickly obtain most of the remaining facts!
Step 4
Not counting the doubles, you have the 3's, 4's, 6's, 7's and 8's. Once you know what 6x7 is, you'll also know what 7x6 is. For the remaining facts (and there aren't many) you will want to learn by skip-counting, in fact use a familiar tune while skip counting! Remember to tap your fingers (just as you did when counting) each time you skip count, this enables you to know which fact you're on. When skip counting by 4's and when you've tapped on the fourth finger, you'll know that it's the 4x4=16 fact. Think of Mary Had A Little Lamb in your mind. Now apply 4,8, 12, 16, (Mary had a....)and continue on! Once you've learned to skip-count by 4's as easily as you can by 2's, you're ready for the next fact family. Don't worry if you forget the odd one, you will be able to fall back on your doubling strategy and counting up.
Remember, being able to do math well means having great strategies. The above strategies will help you learn the times tables. However, you will need to commit daily time to these strategies to learn your tables in 21 days.
Try some of the following:
• Each day when you wake up, skip count the fact family you're working on.
• Each time you walk through a doorway, skip count again (silently)
• Each time you use the washroom, skip count!
• Each time the phone rings, skip count!
• During every commercial when you're watching TV, skip count! When you go to bed each night, skip count for 5 minutes.
If you stick it out, you'll have your tables memorized in 21 days!
Here's a few multiplication tricks to help you. Try these worksheets which are developed to correspond to the 'correct' way of learning your multiplication tables. |
# Perimeter Area Volume.
## Presentation on theme: "Perimeter Area Volume."— Presentation transcript:
Perimeter Area Volume
The measure of the distance around a closed figure.
Perimeter The measure of the distance around a closed figure.
Think of perimeter like a fence around a yard.
On crime shows you often hear law enforcement say, “We need to set up a perimeter.”
How do you find the perimeter of a figure?
To find the perimeter of any closed figure simply add the length of each side of the figure together. 2 cm 3 cm Perimeter = s1 + s2 + s3 P = 2 cm + 2 cm + 3 cm P = 7 cm
write the FORMULA you are using
F.S.S.L F. write the FORMULA you are using S. SUBSTITUTE in the values S. SOLVE the equation L. LABEL your answer
The amount of surface inside a closed figure.
Area The amount of surface inside a closed figure.
Think of area like the amount of carpet you would need in a room
Parallelograms What polygons are parallelograms? 1) Square
2) Rectangle 3) Rhombus 4) Parallelogram
Parallelograms Do all parallelograms have the same formula for area?
We know that squares and rectangles use the formula Area = base • height Let’s use our GeoBoards to see if the area of a rhombus or parallelogram is also base • height
How Do You Find the Area of a Parallelogram?
To find the area of any parallelogram use the formula Area = base ● height and solve using the dimensions given. REMEMBER F.S.S.L!!! 5 in 3 in Area = bh A = 5 in • 3 in A = 15 in2
How Do You Find the Area of a Parallelogram?
To find the area of any parallelogram use the formula Area = base ● height and solve using the dimensions given. 6 in REMEMBER F.S.S.L!!! Area = bh A = 6 in • 4 in A = 24 in2 5 in 4 in
Triangles We can use the formula for a parallelogram to help us find the area of any triangle. Let’s use our GeoBoards to help us determine this formula.
How do you find the area of a triangle?
To find the area of a triangle use the formula Area = ½ base ● height and solve using the dimensions given. REMEMBER F.S.S.L!!! A = ½bh A= ½ • 8ft • 10ft A = 4ft • 10ft A = 40 ft2 6 ft 10 ft 8 ft
8 cm 8 cm 16 cm 8 cm 10 cm
How do you find the area of a trapezoid?
base2 base1 base2
The amount of space in a 3-D object.
Volume The amount of space in a 3-D object.
Think of volume like the amount of liquid you could pour into a cup
How do you find the volume of a rectangular prism?
To find the volume of a rectangular prism use the formula Volume = length ● width ● height and solve using the dimensions given. REMEMBER F.S.S.L!!! Volume = lwh V = 6 in • 2 in • 4 in V = 48 in3 4 in 2 in 6 in
Labels Depending on the measurement you are solving for (area, volume, or perimeter) your label will change. Perimeter = unit Area = unit2 Volume = unit3
Find the perimeter & area of…
15 cm 5 cm
Find the perimeter & area of…
15 cm 5 cm 4 cm
Find the perimeter & area of…
12 cm 12 cm 9 cm 12 cm
Find the volume of… 3 cm 6 cm
Find the perimeter & area of…
13 cm 9 cm 5 cm 12 cm
Construct the following shape…
I have a rectangle that has a perimeter of 24 mm and an area of 32 mm2. What are the dimensions of my rectangle?
Construct the following shape…
I have a triangle that has a perimeter of 23 ft and an area of 15 ft2. What are the dimensions of my triangle? Draw this triangle.
Find the area of the gold region…
3m 8m 6m 6m
Find the area of the green region…
Find the area of the red region…
12mm 10mm 6mm 3mm
Find the area of the blueish region… |
# How to Multiply Percentages in Excel
• Home
• / How to Multiply Percentages in Excel
In today’s digital age, proficiency in Microsoft Excel is crucial for effective data management and analysis. Excel offers numerous powerful features, including the ability to calculate percentages and perform mathematical operations with ease.
One common task is multiplying percentages, which can be particularly useful in various scenarios, such as calculating discounts, taxes, or growth rates. In this article, we will provide you with a step-by-step guide on how to multiply percentages in Excel, enabling you to streamline your calculations and improve your productivity.
## Understanding Percentages in Excel
Before diving into the specifics of multiplying percentages, it’s essential to understand how Excel handles percentages. In Excel, percentages are represented as decimal numbers between 0 and 1. For instance, 50% is represented as 0.5, and 75% is represented as 0.75. By converting percentages to decimal form, Excel can perform mathematical operations accurately.
## Different Methods for Multiplying Percentages in Excel
### Method 1: Using the Multiplication Operator
A multiplication operator is a straightforward approach for multiplying values by a percentage in Excel.
#### Increasing Values:
To increase a value by a certain percentage, use the formula:
Amount * (1 + Percentage%)
For example, if you have a price of \$1,500 (in cell C5) and want to increase it by 10%, you can use the formula:
=C5 * (1 + D5)
#### Decreasing Values:
To decrease a value by a certain percentage, use the formula:
Amount * (1 – Percentage%)
For instance, if you have a price of \$1,500 (in cell C5) and want to decrease it by 10%, you can use the formula:
=C5 * (1 – D5)
### Method 2: Utilizing the Addition Operator
The addition operator provides another method for multiplying values by a percentage in Excel.
#### Increasing Values:
To increase a value by a certain percentage, use the formula:
Amount + (Amount * Percentage%)
For example, if you have a price of \$1,500 (in cell C5) and want to increase it by 10%, you can use the formula:
=C5 + (C5 * D5)
#### Decreasing Values:
To decrease a value by a certain percentage, use the formula:
Amount – (Amount * Percentage%)
For instance, if you have a price of \$1,500 (in cell C5) and want to decrease it by 10%, you can use the formula:
=C5 – (C5 * D5)
### Method 3: Calculating Percentage Differences
Calculating the percentage difference between two values can be useful for data analysis.
#### Step 1: Select the Target Cells
Start by selecting the cell or cells where you want to display the output.
#### Step 2: Calculate the Percentage Difference
To calculate the percentage difference, use the formula:
=(New Value – Old Value) / Old Value
For example, if you want to calculate the percentage difference between a new value (cell D5) and an old value (cell C5), you can use the formula:
=(D5 – C5) / C5
#### Step 3: Format the Result
• After performing the calculation, select the output cell (E5) and navigate to the Home tab.
• Choose the “Percent Style” option in the Number section or press Ctrl+Shift+% to format the result as a percentage.
• This will generate your desired result like this:
## Troubleshooting Common Percentage Calculation Issues
When working with percentages in Excel, you may encounter certain issues that can affect the accuracy of your calculations. Here are some common problems and their solutions to help you troubleshoot percentage calculation issues effectively:
• Incorrect Formulas: Double-check your formulas to ensure they are written correctly and reference the correct cells. Incorrect formulas can lead to inaccurate results.
• Formatting Inconsistencies: Check the formatting of your cells to ensure they are set to the correct number format. Formatting inconsistencies, such as using text instead of numbers, can cause unexpected results.
• Decimal Point Errors: Be mindful of decimal points when entering percentages. A misplaced decimal point can significantly affect the accuracy of your calculations.
• Missing Percentage Symbols: Ensure that you include the percentage symbol (%) when entering or referencing percentages in formulas. Excel treats numbers without the percentage symbol as decimals, which can lead to incorrect calculations.
• Inconsistent Data Types: Confirm that all the data involved in your calculations is of the same type (either percentages or decimals). Mixing different data types can produce incorrect results.
## Conclusion
Multiplying percentages in Excel is a fundamental skill that can greatly enhance your data analysis and calculation capabilities. By following the techniques and tips outlined in this article, you can confidently multiply percentages and overcome any common issues that may arise during the process.
Excel’s versatility and robust functionality make it an invaluable tool for handling percentage calculations in various fields and scenarios. Mastering the art of multiplying percentages in Excel empowers you to make informed decisions, conduct comprehensive data analysis, and improve your overall productivity. |
## Intermediate Algebra (12th Edition)
$2+2i$
$\bf{\text{Solution Outline:}}$ To divide the given expression, $\dfrac{8i}{2+2i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. $\bf{\text{Solution Details:}}$ Multiplying both the numerator and the denominator by the complex conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{8i}{2+2i}\cdot\dfrac{2-2i}{2-2i} \\\\= \dfrac{8i(2-2i)}{(2+2i)(2-2i)} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{8i(2-2i)}{(2)^2-(2i)^2} \\\\= \dfrac{8i(2-2i)}{4-4i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{8i(2-2i)}{4-4(-1)} \\\\= \dfrac{8i(2-2i)}{4+4} \\\\= \dfrac{8i(2-2i)}{8} \\\\= \dfrac{\cancel8i(2-2i)}{\cancel8} \\\\= i(2-2i) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} i(2)+i(-2i) \\\\= 2i-2i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2i-2(-1) \\\\= 2i+2 \\\\= 2+2i .\end{array} |
When you’re looking for help with a math problem, it can feel like every possible solution has been exhausted. But rest assured that there are people out there who are willing to lend a hand. While getting homework help from a tutor or certified teacher might seem like the obvious answer, it’s not always the best one.
If you’re ready to solve that particular mathematical problem on your own, take a look at these helpful hints. Whether you’re working through an algebra book or studying the basics of geometry, there is a step you can take each time to get better at solving problems. These seven steps will help you solve any math problem with ease.
## Understand the Problem
Some math problems are easier to solve if you first understand how they work. This can involve a general understanding of what you’re trying to solve and why, or it can mean that you break down the formula and figure into smaller parts.
For example, let’s say you’re looking at a math problem about a certain kind of plant that can grow in different soil conditions. You want to know the number of plants that will be able to grow in a certain area. In order to solve the problem, you need to first understand the soil conditions.
Let’s say you’re given a certain area that you want to see how many plants will be able to grow in. If you didn’t break down the formula and figure, you might solve the problem blindly. You might come to the final result without really knowing why.
## Break down the formula or figure
Breaking down the formula or figure of a math problem into smaller parts will help you understand it better. This will also lead you to a new understanding of the problem and may even get you to a different solution. For example, let’s look at a math problem about the years that are left before a certain person’s retirement. In order to solve this problem, you first need to break down the formula or figure.
## Identify the variables
Before you can make any meaningful progress in solving a problem, you need to understand what the variables are. These are the particular values you’re trying to find.
For example, let’s say you need to find the number of years that a certain person will live if they retire at the age of 50. The problem asks for the person’s age, so that’s a variable. The person’s retirement age is another variable. If you want to find out the person’s remaining years, that’s another variable.
Maths is very difficult subject so many students face problems in their math assignment , so i will suggest them they can take help from many websites like math homework help
## Find the value of each variable
Before you can find the value of any variable, though, you need to find out what the value of every other variable is. As you work through the formulas and figures used in a math problem, you may find that some of the variables are unknown.
For example, let’s say that you need to find the number of years that a certain person will live if they retire at the age of 50. One of the variables is the person’s age. But the other variable is their retirement age, which you don’t know.
Once you have the value of each of the variables, you can start working on your problem. In order to plug in your values, you’ll first need to write them down. This will help you get all the calculations and numbers down in one place, so you don’t forget them.
Every step you take in a mathematical problem is not only important, but it should be verified. To do this, you should always walk through your work and check everything.
If you’re working through a math problem, you should go through each step to make sure that you’re doing it right. This should be done for each part of the problem.
After you have all the results of your mathematical calculations, you should be able to get a better sense of what the problem is asking. If you’re working through a math problem, you should try to make sense of the results you got. You should consider what the problem is asking, and how your results relate to that.
## Try another solution or ask for help
As you work through the steps of solving a math problem, you should keep in mind that there is no one correct answer. You should always consider other possible solutions, and you should always check to make sure that you’ve made no mistakes.
As you work through the seven steps to solve a math problem, you should keep in mind that there is no one correct answer. You should always consider other possible solutions, and you should always check to make sure that you’ve made no mistakes.
If you find that your problem is too difficult for you to solve, you might want to consider seeing a tutor or certified teacher. This way, you can get help without affecting your grades.
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Home | | User Interface Design | N-queens Problem
# N-queens Problem
The problem is to place it queens on an n-by-n chessboard so that no two queens attack each other by being in the same row or in the same column or on the same diagonal.
N-QUEENS PROBLEM
The problem is to place it queens on an n-by-n chessboard so that no two queens attack each other by being in the same row or in the same column or on the same diagonal. For n = 1, the problem has a trivial solution, and it is easy to see that there is no solution for n = 2 and n =3. So let us consider the four-queens problem and solve it by the backtracking technique. Since each of the four queens has to be placed in its own row, all we need to do is to assign a column for each queen on the board presented in the following figure.
Steps to be followed
We start with the empty board and then place queen 1 in the first possible position of its row, which is in column 1 of row 1.
Then we place queen 2, after trying unsuccessfully columns 1 and 2, in the first acceptable position for it, which is square (2,3), the square in row 2 and column 3. This proves to be a dead end because there i no acceptable position for queen 3. So, the algorithm backtracks and puts queen 2 in the next possible position at (2,4).
Then queen 3 is placed at (3,2), which proves to be another dead end.
The algorithm then backtracks all the way to queen 1 and moves it to (1,2). Queen 2 then goes to (2,4), queen 3 to (3,1), and queen 4 to (4,3), which is a solution to the problem.
(x denotes an unsuccessful attempt to place a queen in the indicated column. The numbers above the nodes indicate the order in which the nodes are generated)
If other solutions need to be found, the algorithm can simply resume its operations at the leaf at which it stopped. Alternatively, we can use the board‘s symmetry for this purpose.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Analysis and Design of Algorithm : Backtracking : N-queens Problem | |
# Area of a Regular Octagon - PowerPoint PPT Presentation
1 / 7
Area of a Regular Octagon. By Carson Dial. Now, for the basics. T his is a regular octagon, an 8-sided figure whose sides are congruent. The small lines on the edges indicate the congruency of said edges . Now, we can begin….
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Area of a Regular Octagon
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## Area of a Regular Octagon
By Carson Dial
Now, for the basics
This is a regular octagon, an 8-sided figure whose sides are congruent.
The small lines on the edges indicate the
congruency of said edges.
Now, we can begin…
First, there is no way to find the area of an octagon by itself, you must divide the octagon into equilateral triangles, eight of them.
Continuing on now…
We need to find the area of a single equilateral triangle.
So we’ll say the length of
this triangle’s base is 5
units.
5
Next up…
We’ll need the height of this triangle,
so for convenience's sake, we’ll call it 6.5 units.
6.5
5
Now, to work this through…
The equation looks something like this
B*h/2 or,
Base times height divided by 2
So, we’ll substitute our numbers in…
5*6.5/2
6.5
5
6.5
* 5
= 32.5 / 2= 16.25
So, the area of this triangle is 16.25 units.
Up next, the full area…
Now, we will multiply the area of the triangle by eight to find the area of the octagon.
• 16.25
• 8
• --------
• 132 |
# Section 1. Inequalities
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1 Worksheet 2.4 Introduction to Inequalities Section 1 Inequalities The sign < stands for less than. It was introduced so that we could write in shorthand things like 3 is less than 5. This becomes 3 < 5. The sign > stands for greater than. In a similar way we can write 5 > 3 in this shorthand form. These statements are called inequalities. Recall the number line (introduced in an earlier worksheet): It is drawn so that the numbers increase from left to right. Alternatively the numbers decrease from right to left. Any number which lies to the right of another on the number line is greater than it and any number which lies to the left of another on the number line is less than it. Example 1 : 6 < 0 5 > 2 5 < < 3 We can also use inequalities in algebraic expressions. So 21a is less than 30a is written 21a < 30a (only true if a > 0) The expression a > 1 means that a is one of all the numbers to the right of 1 on the number line. We draw this using an open circle and an arrow heading to the right. The open circle sits over 1 and indicates that the actual number 1 is not included. That is a is bigger than 1 but not equal to 1: Example 2 : x > 2
2 Example 3 : x < 4 This indicates all numbers to the left of 4 but not including 4. If we wish to include 4 in the above example we would write x 4. This is shorthand for less than or equal to 4. x a x < a or x = a and x a x > a or x = a For, read implies. A closed circle or dot is used on the number line when the actual number is included in the inequality. Example 4 : x This indicates all the numbers to the right of 7 and 7 itself. We can combine signs to make further algebraic expressions. For example 3 x < 2 means all numbers greater than or equal to 3 but smaller than or equal to 2. That 3 and 2 and all the numbers in between. You may also see the notation [ 3, 2] used to represent this interval. Example 5 : 0 < x 1 means all numbers greater than zero and less than or equal to one. In interval notation this is represented by (0, 1]. To draw this type of inequality on the number line we use a line drawn between the open or closed circles over the numbers. So 0 < x 1 is drawn as Example 6 : 1 x 1 Page 2
3 Example 7 : The number of matches in a box could be as little as 47 or as many as 58. If x stands for the number of matches in a box then we can write that 47 x 58 Exercises: 1. Write inequalities for the following: (a) Numbers less than or equal to 6 (b) Numbers between 1 and 4 inclusive (c) Numbers between 2 and 5 exclusive 2. Which of the numbers indicated satisfy the accompanying inequality? (a) x 3 1, 2, 3, 3 1, (b) x < 4 2, 7, 8 1, 0 2 (c) 5 x < 3 6, 4, 0, 1 2, 3 (d) 2 < x , 2 1 2, 4, 5 1 4, 5.5, 7 3. Graph the following inequalities on number lines: (a) x 1 (b) x > 4 (c) 1 < x < 5 Section 2 Solving Inequalities Solving inequalities is similar to solving equations as in worksheet 2.2. We may add or subtract numbers or algebraic terms from both or all sides of the inequality to isolate the variable from the rest of the expression. Similarly we can multiply and divide each side with one very important qualification. When multiplying or dividing both sides of an inequality by a negative number the sign must be reversed. Page 3
4 Example 1 : Given 2x < 4 we want to solve for x. Divide both sides by 2 and reverse the inequality. We get 2x 2 > 4 2 and after cancelling we get x > 2 Only do this when multiplying or dividing both sides of the inequality by a negative number. Let s see what happens when we don t do this. Example 2 : Observe that 1 < 2 is a true statement. 1 < < < 2 We have ended up with 1 < 2, which is a false statement. Some examples of solving inequalities follow. Example 3 : x x x 1 Example 4 : x 2 5 x x 3 Example 5 : 5x x 10 5 x 2 Page 4
5 Example 6 : 3 2x x x 2 2x x 1 Example 7 : 5x + 3 2x + 2 5x + 3 2x 2x + 2 2x 3x x 2 3 3x 1 x 1 3 Example 8 : 3(x + 2 ) < 5(2x + 5) 3 3x + 2 < 10x x 10x + 2 < 25 7x < x < 23 x > 23 7 x > 23 7 Exercises: 1. Solve the following inequalities: (a) x 7 < 5 (b) 2x + 3 > 8 (c) 4(m + 1) m 3 (d) m (e) 6 m < 8 4m (f) 3y + 2 5y + 10 Page 5
6 (g) y > y (h) 4(x + 1 ) 2(x + 3 ) x+1 (i) x+3 < 4x (j) 4(m + 3) + 5(2m 1) > 7m + 6 Section 3 Absolute Values and Inequalities The absolute value of a number was discussed in Worksheet 1.7. Recall that the absolute value of a, written a, is the distance in units that a is away from the origin. For example 7 = 7. Alternatively you could define it as a = + a 2. So when we combine absolute values and inequalities we are looking for all numbers that are either less than or greater than a certain distance away from the origin. Example 1 : The equation x < 3 represents all the numbers whose distance away from the origin is less than 3 units. We could rewrite this inequality as 3 < x < 3. If we draw this on the number line we get To solve inequalities involving absolute values we often need to rewrite the inequality without the absolute value signs as we have just done in the above example. Example 2 : x > 1 can be rewritten as x > 1 or x < 1 In other words all the numbers whose distance away from the origin is greater than 1 unit. Notice that the solution is just the converse of 1 x 1. Since x 1 is easier to solve than x > 1 we can begin by solving 1 x 1 and the solution will be its converse. Page 6
7 Example 3 : x x x 2 this is the same as 2 x 2 Example 4 : x + 3 < 5 5 < x + 3 < 5 Here we note that the expression inside the absolute value sign is treated as a single entity, as if it were in brackets, and must be rewritten before you try to solve it. So 5 < x + 3 < < x < < x < 2 When dealing with a three-sided inequality like the above we follow the same rules as if it were two-sided, i.e. everything that we do to one side of the inequality we must do to all the other sides. Example 5 : Solve: x 2 4 Consider the converse: x 2 < 4. Then 4 < x 2 < 4 2 < x < 6 Hence the solution is the converse x 2 or x 6 Page 7
8 This can also be viewed graphically: 4 y = Notice that the function y = x 2 is at or above the line y = 4 for x 2 and x 6. To graph y = x 2, we set up a table of values as follows: x y Exercises: 1. Solve the following inequalities, and graph the solutions on a number line: (a) x 3 (b) x > 4 (c) x (d) x + 1 < 6 (e) x 4 2 Page 8
9 Exercises 2.4 Introduction to Inequalities 1. (a) Show the following inequalities on a number line: i. x > 3 ii. x 2.5 iii. 1 < x 4 iv. x > 6 or x < 8 v. [0, 5) vi. ( 2, 4] (b) Write down the appropriate inequalities from the following information. i. ii iii. Sophie s blood alcohol level was at least 0.07 iv. Tony had no more than 2 traffic offences. v. The number of drug offences was no more than 200. vi. To enter the theatre, you must be at least 18 years old or under 12 months old. vii. To join the veteran s team, you must be 35 or more. (c) i. Is x = 1.5 a solution to 7x + 5 > 19? ii. Is x = 2 a solution to 4 x < 6? 2. Solve: (a) x 3 2 (b) y > 6 (c) 2x < 24 (d) 1 < x + 2 < 5 (e) 2a + 4 > 3a 11 (f) 2x + 7 > 10 (g) 3(t + 5) 2(t + 1) (h) 1(x + 3) 3 (x 2) 2 4 (i) x = 7 (j) 6 x = 7 (k) y > 2 (l) 2x + 1 < 3 (m) x 5 5 (n) x + 13 < 2 Page 9
10 Answers 2.4 Section 1 1. (a) x 6 (b) 1 x 4 (c) 2 < x < 5 2. (a) 3, 3 1, 4 1 (b) 7, 8 1 (c) 4, 0, 1 (d) 2 1, 4, 5 1, (a) (b) (c) Section 2 1. (a) x < 12 (b) x > (c) m 7 3 (d) m 7 (e) m < 2 3 (f) y 4 (g) y < 19 (h) x 3 (i) x > (j) m > 1 7 Section 3 1. (a) (b) (c) (d) (e) 3 x x < 4 or x > 4 3 x < x < x 2 or x 6 Page 10
11 Exercises (a) i. ii. iii. iv. v. vi (b) i. x > 1 or x 0.5 ii. 1 < x 3 iii. S 0.07 iv. T 2 v. D 200 vi. x 18 or x < 1 vii. A 35 (c) (i) No (ii) No 2. (a) x 5 (b) y < 6 (c) x < 12 (d) 3 < x < 3 (e) a < 15 (f) x < 3 2 (g) t 13 (h) x 12 (i) x = 7 or x = 7 (j) x = 1 or x = 13 (k) y > 2 or y < 2 (l) 2 < x < 1 (m) x 10 or x 0 (n) No solutions Page 11
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# Weekly Dose 16 - Maths Olympiad Practice
Weekly Dose 16 - Maths Olympiad Practice
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### Weekly Dose 16 - Maths Olympiad Practice
1. 1. Train A and Train B travel towards each other from Town a and Town b respectively, at a constant speed. The two towns are 1320km apart. After the two trains meet, Train A takes 5 hours to reach Town b while Train B takes 7.2 hours to reach Town a. How many km does Train A run per hour? Solution: 𝑥 ∶ 5 = 7.2 ∶ 𝑥 𝑥2= 7.2 × 5 = _____ 𝑥 = _____ Speed for A = 1320 ÷ 𝑥 + 5 = _____ km/h Answer: 120𝑘𝑚
2. 2. In the figure below, in a right-angled triangle ACD, the area of shaded region is 10 cm2. AD = 5 cm, AB = BC, DE = EC. Find the length of AB, in cm. Solution: Because DE = EC, ∆𝐵𝐸𝐷 = ∆𝐵𝐸𝐶 = 10 𝑐𝑚2 Because AB = BC and DE = EC, AD = 2 × BE Since AD = 5cm, BE = 2.5 cm ∆𝐵𝐸𝐶 = 1 2 × BE× BC = 10 cm2 1 2 × 2.5 × BC = 10 cm2 BC = 8 cm AD = _____ cm Answer: 8 cm
3. 3. Eve said to her mother, “If I reverse the two-digits of my age, I will get your age.” Her mother said, “Tomorrow is my birthday, and my age will then be twice your age.” It is known that their birthdays are not on the same day. How old is Eve? Solution: If Eve’s age is 𝑎𝑏, her mother’s age is 𝑏𝑎. 𝑎 and 𝑏 is whole number between 1 and 9 And 𝑏𝑎 + 1 = 2 × 𝑎𝑏. 10𝑏 + 𝑎 + 1 = 2 × (10𝑎 + 𝑏) 8𝑏 = 19𝑎 − 1 ---- ① Because all the multiples for 8 are even numbers, 19𝑎 must be odd number. If 𝑏 is the biggest number 9, 8𝑏 = 72, which mean 19𝑎 − 1 cannot be greater than 72, 19𝑎 cannot be greater than 73, 𝑎 cannot be greater than 73 19 which is 3 3 19 . When 𝑎 = 1, cannot fulfill ①, so 𝑎 cannot be 1 When 𝑎 = 3, 𝑏 = ____ Answer: Eve is 37 years old
4. 4. Balls of the same size and weight are placed in a container. There are 8 different colors and 90 balls in each color. What is the minimum number of balls that must be drawn from the container in order to get balls of 4 different colors with at least 9 balls for each color? ** Note: Always treat this kind of question as finding worst case scenario, the bad luck case ** Let’s find the largest number of balls we can drawn without achieving the desired result. We may draw all 90 balls of each of 3 colors ⇒ 3 × 90 = 270 Then we may drawn 8 balls of each of the remaining colors ⇒ 5 × 8 = 40 If we draw one more ball, unavoidable the desired result will be met. Therefore by drawing 270 + 40 + 1 = ___ balls, we are guaranteed to get at least 9 balls of each of 4 colors. Answer: 311 Solution: |
# Rhombus Lesson for Kids: Definition & Facts
Instructor: Jeremy Cook
Jeremy has been teaching in elementary education for 13 years and holds a master's degree in Education
This lesson will teach you all the attributes you need to have a rhombus. Examine how to find the perimeter and area, as well as discover some interesting things about rhombuses.
## What is a Rhombus?
What kind of shape do you think you'd get if you kicked a square hard enough for it to lean a bit to one side? Don't try that at home, but that leaning square actually has a name. It's called a rhombus.
A rhombus is a four-sided polygon. A polygon is a many sided shape where the lines connect. A rhombus is in the broad category of polygons with four sides and four angles we call quadrilaterals. This group consists of common named figures such as rectangles, squares, trapezoids and parallelograms.
## What Makes a Rhombus
To have a rhombus, there are certain things that the shape must have.
• All four sides must be equal, which is why I talked about kicking a square. So a rhombus is like a square in that the sides are all the same length.
• The opposite inside angles are equal.
• The opposite sides are parallel. This makes the rhombus a special kind of parallelogram.
Finding the perimeter of a rhombus is done the same way you would any other quadrilateral; by adding up the four sides. Finding the area is different though.
## Finding the Area
Because the sides are tilted, you can't multiply the length by the width like you do with a square to find the area of a rhombus. You must find the height of the rhombus and then multiply that by the length. To do that, you must draw a straight, vertical line from the top outside corner down and draw a horizontal line that extends out from the base until it meets the vertical line you drew. Then measure the vertical line from the point it meets the horizontal line you drew to the corner of the rhombus. That will give you the height. Then multiply the height times the length of either the top or bottom side.
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help-with-math-problems-geometry-similar-triangles-criteria-for-similarity-of-triangles
# Interactive video lesson plan for: Help with math problems | Geometry similar triangles | Criteria for Similarity of Triangles
#### Activity overview:
Help with math problems Geometry similar triangles | Criteria for Similarity of Triangles
http://www.learncbse.in/ncert-solutions-class-10th-maths-chapter-6-triangles-exercise-6-1-question-1/
http://www.learncbse.in/rd-sharma-class-10-solutions/
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Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion)
AAA similarity criterion
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
SSS (Side–Side–Side) similarity criterion for two triangles
If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
SAS (Side–Angle–Side) similarity criterion for two triangles
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar
00:03 CBSE class 10 maths NCERT Solutions chapter 6 Triangles Q1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
00:47 Tests for Similar Triangles AA similarity criterion
01:18 (ii) | 02:58 (iii) | 03:59 (iv) | 04:54 (v) | 06:10 (vi) |
02:17 Solving similar triangles SSS (Side–Side–Side) similarity criterion
05:43 SAS (Side–Angle–Side) similarity criterion for two triangles
06:36 Angle sum prpperty of triangle.
07:22 Solving similar triangles AA similarity criterion
07:47 CBSE class 10 maths NCERT Solutions chapter 6 Triangles Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.
08:16 constructions procedure to solve geometry problems
09:20 Vertically opposite angles are equal
09:33 Alrenate interior angles
10:18 AAA similarity criterion
11:05 CBSE class 10 maths NCERT Solutions chapter 6 Triangles Q4. In Fig. 6.36, QR/QS = QT/PR and angle 1 = angle 2. Show that triangle PQS ~ triangle TQR.
12:03 It tow angles are equal then their sides opposite to equal angles are equal.
13:55 SAS (Side–Angle–Side) similarity criterion for two triangles
14:14 Q5. S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Show that triangle RPQ ~ triangle RTS.
16:08 AAA similarity criterion
16:38 Q6. In Fig. 6.37, if triangle ABE ~ triangle ACD, show that triangle ADE ~ triangle ABC.
18:33 SAS (Side–Angle–Side) similarity criterion
19:30 Q7. In Fig. 6.38, altitudes AD and CE of triangle ABC intersect each other at the point P. Show that:
20:32 (i) triangle AEP ~ triangle CDP
21:13 Vertically opposite angles are equal
22:08 (ii) triangle ABD ~ triangle CBE
22:52 use AAA similarity criterion
23:37 (iii) triangle AEP ~ triangle ADB
25:03 AAA similarity criterion
25:35 (iv) triangle PDC ~ triangle BEC
26:18 AA similarity criterion
26:38 CBSE class 10 maths NCERT Solutions chapter 6 Triangles Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that triangle ABE ~ triangle CFB.
28:52 Apply AA similarity criterion.
29:41 CBSE class 10 maths NCERT Solutions chapter 6 Triangles Q9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
30:08 (i) triangle ABC ~ triangle AMP 31:22 (ii) CA/PA = BC/MP.
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CBSE solutions for class 10 maths Chapter 6 Triangles Exercise 6.3
CBSE class 10 maths NCERT Solutions chapter 6 Triangles Exercise 6.2 | SIMILAR TRIANGLES
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## Rational Functions
### Learning Objectives
In this section, you will:
• Use arrow notation.
• Solve applied problems involving rational functions.
• Find the domains of rational functions.
• Identify vertical asymptotes.
• Identify horizontal asymptotes.
• Graph rational functions.
Suppose we know that the cost of making a product is dependent on the number of items,$\,x,\,$produced. This is given by the equation$\,C\left(x\right)=15,000x-0.1{x}^{2}+1000.\,$If we want to know the average cost for producing$\,x\,$items, we would divide the cost function by the number of items,$\,x.$
The average cost function, which yields the average cost per item for$\,x\,$items produced, is
$f\left(x\right)=\frac{15,000x-0.1{x}^{2}+1000}{x}$
Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.
In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.
### Using Arrow Notation
We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in (Figure), and notice some of their features.
Figure 1.
Several things are apparent if we examine the graph of$\,f\left(x\right)=\frac{1}{x}.$
1. On the left branch of the graph, the curve approaches the x-axis$\,\left(y=0\right) \text{as} x\to –\infty .$
2. As the graph approaches$\,x=0\,$from the left, the curve drops, but as we approach zero from the right, the curve rises.
3. Finally, on the right branch of the graph, the curves approaches the x-axis$\,\left(y=0\right) \text{as} x\to \infty .$
To summarize, we use arrow notation to show that$\,x\,$or$\,f\left(x\right)\,$is approaching a particular value. See (Figure).
Symbol Meaning
$x\to {a}^{-}$ $x\,$approaches$\,a\,$from the left ($x<a\,$but close to$\,a$)
$x\to {a}^{+}$ $x\,$approaches$\,a\,$from the right ($x>a\,$but close to$\,a$)
$x\to \infty$ $x\,$approaches infinity ($x\,$increases without bound)
$x\to -\infty$ $x\,$approaches negative infinity ($x\,$decreases without bound)
$f\left(x\right)\to \infty$ the output approaches infinity (the output increases without bound)
$f\left(x\right)\to -\infty$ the output approaches negative infinity (the output decreases without bound)
$f\left(x\right)\to a$ the output approaches$\,a$
#### Local Behavior of$\,f\left(x\right)=\frac{1}{x}$
Let’s begin by looking at the reciprocal function,$\,f\left(x\right)=\frac{1}{x}.\,$We cannot divide by zero, which means the function is undefined at$\,x=0;\,$so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in (Figure).
$x$ –0.1 –0.01 –0.001 –0.0001 $f\left(x\right)=\frac{1}{x}$ –10 –100 –1000 –10,000
We write in arrow notation
$\text{as }x\to {0}^{-},f\left(x\right)\to -\infty$
As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in (Figure).
$x$ 0.1 0.01 0.001 0.0001 $f\left(x\right)=\frac{1}{x}$ 10 100 1000 10,000
We write in arrow notation
$\text{As }x\to {0}^{+}, f\left(x\right)\to \infty .$
See (Figure).
Figure 2.
This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line$\,x=0\,$as the input becomes close to zero. See (Figure).
Figure 3.
### Vertical Asymptote
A vertical asymptote of a graph is a vertical line$\,x=a\,$where the graph tends toward positive or negative infinity as the inputs approach$\,a.\,$We write
$\text{As }x\to a,f\left(x\right)\to \infty , \text{or as }x\to a,f\left(x\right)\to -\infty .$
#### End Behavior of$\,f\left(x\right)=\frac{1}{x}$
As the values of$\,x\,$approach infinity, the function values approach 0. As the values of$\,x\,$approach negative infinity, the function values approach 0. See (Figure). Symbolically, using arrow notation
$\text{As }x\to \infty ,f\left(x\right)\to 0,\text{and as }x\to -\infty ,f\left(x\right)\to 0.$
Figure 4.
Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line$\,y=0.\,$See (Figure).
Figure 5.
### Horizontal Asymptote
A horizontal asymptote of a graph is a horizontal line$\,y=b\,$where the graph approaches the line as the inputs increase or decrease without bound. We write
$\,\text{As }x\to \infty \text{ or }x\to -\infty ,\text{ }f\left(x\right)\to b.$
### Using Arrow Notation
Use arrow notation to describe the end behavior and local behavior of the function graphed in (Figure).
Figure 6.
### Try It
Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.
### Using Transformations to Graph a Rational Function
Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.
#### Analysis
Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.
### Try It
Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.
### Solving Applied Problems Involving Rational Functions
In (Figure), we shifted a toolkit function in a way that resulted in the function$\,f\left(x\right)=\frac{3x+7}{x+2}.\,$This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.
### Rational Function
A rational function is a function that can be written as the quotient of two polynomial functions$\,P\left(x\right) \text{and} Q\left(x\right).$
$f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{a}_{p}{x}^{p}+{a}_{p-1}{x}^{p-1}+…+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q-1}{x}^{q-1}+…+{b}_{1}x+{b}_{0}},Q\left(x\right)\ne 0$
### Solving an Applied Problem Involving a Rational Function
A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the ratio of sugar to water, in pounds per gallon in the tank after 12 minutes. Is that a greater ratio of sugar to water, in pounds per gallon than at the beginning?
### Try It
There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.
### Finding the Domains of Rational Functions
A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.
### Domain of a Rational Function
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.
### How To
Given a rational function, find the domain.
1. Set the denominator equal to zero.
2. Solve to find the x-values that cause the denominator to equal zero.
3. The domain is all real numbers except those found in Step 2.
### Finding the Domain of a Rational Function
Find the domain of$\,f\left(x\right)=\frac{x+3}{{x}^{2}-9}.$
#### Analysis
A graph of this function, as shown in (Figure), confirms that the function is not defined when$\,x=±3.$
Figure 8.
There is a vertical asymptote at$\,x=3\,$and a hole in the graph at$\,x=-3.\,$We will discuss these types of holes in greater detail later in this section.
### Try It
Find the domain of$\,f\left(x\right)=\frac{4x}{5\left(x-1\right)\left(x-5\right)}.$
### Identifying Vertical Asymptotes of Rational Functions
By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.
#### Vertical Asymptotes
The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.
Given a rational function, identify any vertical asymptotes of its graph.
1. Factor the numerator and denominator.
2. Note any restrictions in the domain of the function.
3. Reduce the expression by canceling common factors in the numerator and the denominator.
4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”
### Identifying Vertical Asymptotes
Find the vertical asymptotes of the graph of$\,k\left(x\right)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}.$
#### Removable Discontinuities
Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity.
For example, the function$\,f\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}-2x-3}\,$may be re-written by factoring the numerator and the denominator.
$f\left(x\right)=\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-3\right)}$
Notice that$\,x+1\,$is a common factor to the numerator and the denominator. The zero of this factor,$\,x=-1,\,$is the location of the removable discontinuity. Notice also that$\,x–3\,$is not a factor in both the numerator and denominator. The zero of this factor,$\,x=3,\,$is the vertical asymptote. See (Figure). [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]
Figure 10.
### Removable Discontinuities of Rational Functions
A removable discontinuity occurs in the graph of a rational function at$\,x=a\,$if$\,a\,$is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.
### Identifying Vertical Asymptotes and Removable Discontinuities for a Graph
Find the vertical asymptotes and removable discontinuities of the graph of$\,k\left(x\right)=\frac{x-2}{{x}^{2}-4}.$
### Try It
Find the vertical asymptotes and removable discontinuities of the graph of$\,f\left(x\right)=\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}.$
### Identifying Horizontal Asymptotes of Rational Functions
While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.
There are three distinct outcomes when checking for horizontal asymptotes:
Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at$\,y=0.$
$\text{Example: }f\left(x\right)=\frac{4x+2}{{x}^{2}+4x-5}$
In this case, the end behavior is$\,f\left(x\right)\approx \frac{4x}{{x}^{2}}=\frac{4}{x}.\,$This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function$\,g\left(x\right)=\frac{4}{x},\,$and the outputs will approach zero, resulting in a horizontal asymptote at$\,y=0.\,$See (Figure). Note that this graph crosses the horizontal asymptote.
Figure 12. Horizontal asymptote$\,y=0\,$when$\,f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},\,q\left(x\right)\ne 0\,\text{where degree of}\,p<\text{degree of }q.$
Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.
$\text{Example: }f\left(x\right)=\frac{3{x}^{2}-2x+1}{x-1}$
In this case, the end behavior is$\,f\left(x\right)\approx \frac{3{x}^{2}}{x}=3x.\,$This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function$\,g\left(x\right)=3x.\,$As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of$\,g\left(x\right)=3x\,$looks like a diagonal line, and since$\,f\,$will behave similarly to$\,g,\,$it will approach a line close to$\,y=3x.\,$This line is a slant asymptote.
To find the equation of the slant asymptote, divide$\,\frac{3{x}^{2}-2x+1}{x-1}.\,$The quotient is$\,3x+1,\,$and the remainder is 2. The slant asymptote is the graph of the line$\,g\left(x\right)=3x+1.\,$See (Figure).
Figure 13. Slant asymptote when$\,f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},\,q\left(x\right)\ne 0\,$where degree of$\,p>\text{degree of }q\,\text{by}\,\text{1}\text{.}\,$
Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at$\,y=\frac{{a}_{n}}{{b}_{n}},\,$where$\,{a}_{n}\,$and$\,{b}_{n}\,$are the leading coefficients of$\,p\left(x\right)\,$and$\,q\left(x\right)\,$for$\,f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0.$
$\text{Example: }f\left(x\right)=\frac{3{x}^{2}+2}{{x}^{2}+4x-5}$
In this case, the end behavior is$\,f\left(x\right)\approx \frac{3{x}^{2}}{{x}^{2}}=3.\,$This tells us that as the inputs grow large, this function will behave like the function$\,g\left(x\right)=3,\,$which is a horizontal line. As$\,x\to ±\infty ,f\left(x\right)\to 3,\,$resulting in a horizontal asymptote at$\,y=3.\,$See (Figure). Note that this graph crosses the horizontal asymptote.
Figure 14. Horizontal asymptote when$\,f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},\,q\left(x\right)\ne 0\,\text{where degree of }p=\text{degree of }q.$
Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.
It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function
$f\left(x\right)=\frac{3{x}^{5}-{x}^{2}}{x+3}$
with end behavior
$f\left(x\right)\approx \frac{3{x}^{5}}{x}=3{x}^{4},$
the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.
$x\to ±\infty , f\left(x\right)\to \infty$
### Horizontal Asymptotes of Rational Functions
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.
• Degree of numerator is less than degree of denominator: horizontal asymptote at$\,y=0.$
• Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
• Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.
### Identifying Horizontal and Slant Asymptotes
For the functions listed, identify the horizontal or slant asymptote.
1. $g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}$
2. $h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}$
3. $k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}$
### Identifying Horizontal Asymptotes
In the sugar concentration problem earlier, we created the equation$\,C\left(t\right)=\frac{5+t}{100+10t}.$
Find the horizontal asymptote and interpret it in context of the problem.
### Identifying Horizontal and Vertical Asymptotes
Find the horizontal and vertical asymptotes of the function
$f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x-5\right)}$
### Try It
Find the vertical and horizontal asymptotes of the function:
$f\left(x\right)=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-2\right)\left(x+3\right)}$
### Intercepts of Rational Functions
A rational function will have a y-intercept at$\,f\left(0\right)$, if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero.
Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero.
### Finding the Intercepts of a Rational Function
Find the intercepts of$\,f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x-5\right)}.$
### Try It
Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x– and y-intercepts and the horizontal and vertical asymptotes.
### Graphing Rational Functions
In (Figure), we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.
The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See (Figure).
Figure 17.
When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See (Figure).
Figure 18.
For example, the graph of$\,f\left(x\right)=\frac{{\left(x+1\right)}^{2}\left(x-3\right)}{{\left(x+3\right)}^{2}\left(x-2\right)}\,$is shown in (Figure).
Figure 19.
• At the x-intercept$\,x=-1\,$corresponding to the$\,{\left(x+1\right)}^{2}\,$factor of the numerator, the graph “bounces”, consistent with the quadratic nature of the factor.
• At the x-intercept$\,x=3\,$corresponding to the$\,\left(x-3\right)\,$factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
• At the vertical asymptote$\,x=-3\,$corresponding to the$\,{\left(x+3\right)}^{2}\,$factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function$\,f\left(x\right)=\frac{1}{{x}^{2}}.$
• At the vertical asymptote$\,x=2,\,$corresponding to the$\,\left(x-2\right)\,$factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function$\,f\left(x\right)=\frac{1}{x}.$
### How To
Given a rational function, sketch a graph.
1. Evaluate the function at 0 to find the y-intercept.
2. Factor the numerator and denominator.
3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts.
4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points.
5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
8. Sketch the graph.
### Graphing a Rational Function
Sketch a graph of$\,f\left(x\right)=\frac{\left(x+2\right)\left(x-3\right)}{{\left(x+1\right)}^{2}\left(x-2\right)}.$
### Try It
Given the function$\,f\left(x\right)=\frac{{\left(x+2\right)}^{2}\left(x-2\right)}{2{\left(x-1\right)}^{2}\left(x-3\right)},\,$use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.
### Writing Rational Functions
Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.
### Writing Rational Functions from Intercepts and Asymptotes
If a rational function has x-intercepts at$\,x={x}_{1},{x}_{2},…,{x}_{n},\,$vertical asymptotes at$\,x={v}_{1},{v}_{2},\dots ,{v}_{m},\,$and no$\,{x}_{i}=\text{any }{v}_{j},\,$then the function can be written in the form:
$f\left(x\right)=a\frac{{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}}{{\left(x-{v}_{1}\right)}^{{q}_{1}}{\left(x-{v}_{2}\right)}^{{q}_{2}}\cdots {\left(x-{v}_{m}\right)}^{{q}_{n}}}$
where the powers$\,{p}_{i}\,$or$\,{q}_{i}\,$on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor$\,a\,$can be determined given a value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero.
### How To
Given a graph of a rational function, write the function.
1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
3. Use any clear point on the graph to find the stretch factor.
### Writing a Rational Function from Intercepts and Asymptotes
Write an equation for the rational function shown in (Figure).
Figure 22.
Access these online resources for additional instruction and practice with rational functions.
### Key Equations
Rational Function $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{a}_{p}{x}^{p}+{a}_{p-1}{x}^{p-1}+…+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q-1}{x}^{q-1}+…+{b}_{1}x+{b}_{0}}, Q\left(x\right)\ne 0$
### Key Concepts
• We can use arrow notation to describe local behavior and end behavior of the toolkit functions$\,f\left(x\right)=\frac{1}{x}\,$and$\,f\left(x\right)=\frac{1}{{x}^{2}}.\,$See (Figure).
• A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See (Figure).
• Application problems involving rates and concentrations often involve rational functions. See (Figure).
• The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See (Figure).
• The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See (Figure).
• A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See (Figure).
• A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See (Figure), (Figure), (Figure), and (Figure).
• Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See (Figure).
• If a rational function has x-intercepts at$\,x={x}_{1},{x}_{2},\dots ,{x}_{n},\,$vertical asymptotes at$\,x={v}_{1},{v}_{2},\dots ,{v}_{m},\,$and no$\,{x}_{i}=\text{any }{v}_{j},\,$then the function can be written in the form
$\begin{array}{l}\begin{array}{l}\hfill \\ f\left(x\right)=a\frac{{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}}{{\left(x-{v}_{1}\right)}^{{q}_{1}}{\left(x-{v}_{2}\right)}^{{q}_{2}}\cdots {\left(x-{v}_{m}\right)}^{{q}_{n}}}\hfill \end{array}\hfill \end{array}$
See (Figure).
### Section Exercises
#### Verbal
What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?
What is the fundamental difference in the graphs of polynomial functions and rational functions?
If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?
Can a graph of a rational function have no vertical asymptote? If so, how?
Can a graph of a rational function have no x-intercepts? If so, how?
#### Algebraic
For the following exercises, find the domain of the rational functions.
$f\left(x\right)=\frac{x-1}{x+2}$
$f\left(x\right)=\frac{x+1}{{x}^{2}-1}$
$f\left(x\right)=\frac{{x}^{2}+4}{{x}^{2}-2x-8}$
$f\left(x\right)=\frac{{x}^{2}+4x-3}{{x}^{4}-5{x}^{2}+4}$
For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.
$f\left(x\right)=\frac{4}{x-1}$
$f\left(x\right)=\frac{2}{5x+2}$
$f\left(x\right)=\frac{x}{{x}^{2}-9}$
$f\left(x\right)=\frac{x}{{x}^{2}+5x-36}$
$f\left(x\right)=\frac{3+x}{{x}^{3}-27}$
$f\left(x\right)=\frac{3x-4}{{x}^{3}-16x}$
$f\left(x\right)=\frac{{x}^{2}-1}{{x}^{3}+9{x}^{2}+14x}$
$f\left(x\right)=\frac{x+5}{{x}^{2}-25}$
$f\left(x\right)=\frac{x-4}{x-6}$
$f\left(x\right)=\frac{4-2x}{3x-1}$
For the following exercises, find the x– and y-intercepts for the functions.
$f\left(x\right)=\frac{x+5}{{x}^{2}+4}$
$f\left(x\right)=\frac{x}{{x}^{2}-x}$
$f\left(x\right)=\frac{{x}^{2}+8x+7}{{x}^{2}+11x+30}$
$f\left(x\right)=\frac{{x}^{2}+x+6}{{x}^{2}-10x+24}$
$f\left(x\right)=\frac{94-2{x}^{2}}{3{x}^{2}-12}$
For the following exercises, describe the local and end behavior of the functions.
$f\left(x\right)=\frac{x}{2x+1}$
$f\left(x\right)=\frac{2x}{x-6}$
$f\left(x\right)=\frac{-2x}{x-6}$
$f\left(x\right)=\frac{{x}^{2}-4x+3}{{x}^{2}-4x-5}$
$f\left(x\right)=\frac{2{x}^{2}-32}{6{x}^{2}+13x-5}$
For the following exercises, find the slant asymptote of the functions.
$f\left(x\right)=\frac{24{x}^{2}+6x}{2x+1}$
$f\left(x\right)=\frac{4{x}^{2}-10}{2x-4}$
$f\left(x\right)=\frac{81{x}^{2}-18}{3x-2}$
$f\left(x\right)=\frac{6{x}^{3}-5x}{3{x}^{2}+4}$
$f\left(x\right)=\frac{{x}^{2}+5x+4}{x-1}$
#### Graphical
For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.
The reciprocal function shifted up two units.
The reciprocal function shifted down one unit and left three units.
The reciprocal squared function shifted to the right 2 units.
The reciprocal squared function shifted down 2 units and right 1 unit.
For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.
$p\left(x\right)=\frac{2x-3}{x+4}$
$q\left(x\right)=\frac{x-5}{3x-1}$
$s\left(x\right)=\frac{4}{{\left(x-2\right)}^{2}}$
$r\left(x\right)=\frac{5}{{\left(x+1\right)}^{2}}$
$f\left(x\right)=\frac{3{x}^{2}-14x-5}{3{x}^{2}+8x-16}$
$g\left(x\right)=\frac{2{x}^{2}+7x-15}{3{x}^{2}-14x+15}$
$a\left(x\right)=\frac{{x}^{2}+2x-3}{{x}^{2}-1}$
$b\left(x\right)=\frac{{x}^{2}-x-6}{{x}^{2}-4}$
$h\left(x\right)=\frac{2{x}^{2}+ x-1}{x-4}$
$k\left(x\right)=\frac{2{x}^{2}-3x-20}{x-5}$
$w\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)\left(x-5\right)}{{\left(x+2\right)}^{2}\left(x-4\right)}$
$z\left(x\right)=\frac{{\left(x+2\right)}^{2}\left(x-5\right)}{\left(x-3\right)\left(x+1\right)\left(x+4\right)}$
For the following exercises, write an equation for a rational function with the given characteristics.
Vertical asymptotes at$\,x=5\,$and$\,x=-5,\,$x-intercepts at$\,\left(2,0\right)\,$and$\,\left(-1,0\right),\,$y-intercept at$\,\left(0,4\right)$
Vertical asymptotes at$\,x=-4\,$and$\,x=-1,\,$x-intercepts at$\,\left(1,0\right)\,$and$\,\left(5,0\right),\,$y-intercept at$\,\left(0,7\right)$
Vertical asymptotes at$\,x=-4\,$and$\,x=-5,\,$x-intercepts at$\,\left(4,0\right)\,$and$\,\left(-6,0\right),\,$Horizontal asymptote at$\,y=7$
Vertical asymptotes at$\,x=-3\,$and$\,x=6,\,$x-intercepts at$\,\left(-2,0\right)\,$and$\,\left(1,0\right),\,$Horizontal asymptote at$\,y=-2$
Vertical asymptote at$\,x=-1,\,$Double zero at$\,x=2,\,$y-intercept at$\,\left(0,2\right)$
Vertical asymptote at$\,x=3,\,$Double zero at$\,x=1,\,$y-intercept at$\,\left(0,4\right)$
For the following exercises, use the graphs to write an equation for the function.
#### Numeric
For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote
$f\left(x\right)=\frac{1}{x-2}$
$f\left(x\right)=\frac{x}{x-3}$
$f\left(x\right)=\frac{2x}{x+4}$
$f\left(x\right)=\frac{2x}{{\left(x-3\right)}^{2}}$
$f\left(x\right)=\frac{{x}^{2}}{{x}^{2}+2x+1}$
#### Technology
For the following exercises, use a calculator to graph$\,f\left(x\right).\,$Use the graph to solve$\,f\left(x\right)>0.$
$f\left(x\right)=\frac{2}{x+1}$
$f\left(x\right)=\frac{4}{2x-3}$
$f\left(x\right)=\frac{2}{\left(x-1\right)\left(x+2\right)}$
$f\left(x\right)=\frac{x+2}{\left(x-1\right)\left(x-4\right)}$
$f\left(x\right)=\frac{{\left(x+3\right)}^{2}}{{\left(x-1\right)}^{2}\left(x+1\right)}$
#### Extensions
For the following exercises, identify the removable discontinuity.
$f\left(x\right)=\frac{{x}^{2}-4}{x-2}$
$f\left(x\right)=\frac{{x}^{3}+1}{x+1}$
$f\left(x\right)=\frac{{x}^{2}+x-6}{x-2}$
$f\left(x\right)=\frac{2{x}^{2}+5x-3}{x+3}$
$f\left(x\right)=\frac{{x}^{3}+{x}^{2}}{x+1}$
#### Real-World Applications
For the following exercises, express a rational function that describes the situation.
A large mixing tank currently contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after$\,t\,$minutes.
A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after$\,t\,$minutes.
For the following exercises, use the given rational function to answer the question.
The concentration$\,C\,$of a drug in a patient’s bloodstream$\,t\,$hours after injection in given by$\,C\left(t\right)=\frac{2t}{3+{t}^{2}}.\,$What happens to the concentration of the drug as$\,t\,$increases?
The concentration$\,C\,$of a drug in a patient’s bloodstream$\,t\,$hours after injection is given by$\,C\left(t\right)=\frac{100t}{2{t}^{2}+75}.\,$Use a calculator to approximate the time when the concentration is highest.
For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question.
An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let$\,x\,$= length of the side of the base.
A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/ square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let$\,x\,$= length of the side of the base.
A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let$\,x\,$= radius.
A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let$\,x\,$= radius.
A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let$\,x\,$= radius.
### Glossary
arrow notation
a way to represent symbolically the local and end behavior of a function by using arrows to indicate that an input or output approaches a value
horizontal asymptote
a horizontal line$\,y=b\,$where the graph approaches the line as the inputs increase or decrease without bound.
rational function
a function that can be written as the ratio of two polynomials
removable discontinuity
a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function
vertical asymptote
a vertical line$\,x=a\,$where the graph tends toward positive or negative infinity as the inputs approach$\,a$ |
# Section 4.5. Matrix Inverses
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## Transcription
1 Section 4.5 Matrix Inverses
2 The Definition of Inverse Recall: The multiplicative inverse (or reciprocal) of a nonzero number a is the number b such that ab = 1. We define the inverse of a matrix in almost the same way. Definition Let A be an n n square matrix. We say A is invertible (or nonsingular) if there is a matrix B of the same size, such that identity matrix AB = I n and BA = I n In this case, B is the inverse of A, and is written A 1... Example A = ( ) B = ( ) I claim B = A 1. Check: ( ) ( ) ( ) AB = = ( ) ( ) ( ) " BA = =
3 Poll Poll Do there exist two matrices A and B such that AB is the identity, but BA is not? If so, find an example. (Both products have to make sense.) Yes, for instance: A = ( 1 0 ) B = However If A and B are square matrices, then AB = I n if and only if BA = I n. So in this case you only have to check one.
4 Solving Linear Systems via Inverses Solving Ax = b by dividing by A Theorem If A is invertible, then Ax = b has exactly one solution for every b, namely: x = A 1 b. Why? Divide by A! Ax = b A 1 (Ax) = A 1 b (A 1 A)x = A 1 b I nx = A 1 b x = A 1 b. I nx = x for every x Important If A is invertible and you know its inverse, then the easiest way to solve Ax = b is by dividing by A : x = A 1 b. This is very convenient when you have to vary b!
5 Solving Linear Systems via Inverses Example Example Solve the system 2x 1 + 3x 2 + 2x 3 = 1 x 1 + 3x 3 = 1 2x 1 + 2x 2 + 3x 3 = 1 Answer: using x x 2 = x = = = The advantage of using inverses is it doesn t matter what s on the right-hand side of the = : 2x 1 + 3x 2 + 2x 3 = b 1 1 x b 1 x 1 + 3x 3 = b 2 = x 2 = b 2 2x 1 + 2x 2 + 3x 3 = b x b 3 3 6b 1 5b 2 + 9b 3 = 3b 1 + 2b 2 4b 3. 2b 1 + 2b 2 3b 3
6 Some Facts Say A and B are invertible n n matrices. 1. A 1 is invertible and its inverse is (A 1 ) 1 = A. 2. AB is invertible and its inverse is (AB) 1 = A 1 B 1 B 1 A 1. Why? (B 1 A 1 )AB = B 1 (A 1 A)B = B 1 I nb = B 1 B = I n. 3. A T is invertible and (A T ) 1 = (A 1 ) T. Why? A T (A 1 ) T = (A 1 A) T = In T = I n. Question: If A, B, C are invertible n n matrices, what is the inverse of ABC? i. A 1 B 1 C 1 ii. B 1 A 1 C 1 iii. C 1 B 1 A 1 iv. C 1 A 1 B 1 Check: (ABC)(C 1 B 1 A 1 ) = AB(CC 1 )B 1 A 1 = A(BB 1 )A 1 = AA 1 = I n. In general, a product of invertible matrices is invertible, and the inverse is the product of the inverses, in the reverse order.
7 Computing A 1 The 2 2 case ( ) a b Let A =. The determinant of A is the number c d ( ) a b det(a) = det = ad bc. c d Facts: 1. If det(a) 0, then A is invertible and A 1 = 1 det(a) 2. If det(a) = 0, then A is not invertible. Why 1? ( ) ( ) a b d b = c d c a ( ) ad bc 0 = 0 ad bc So we get the identity by dividing by ad bc. ( d b c a ). ( ) ( ) d b a b c a c d Example ( ) 1 2 det = = ( ) = ( )
8 Computing A 1 In general Let A be an n n matrix. Here s how to compute A Row reduce the augmented matrix ( A I n ). 2. If the result has the form ( I n B ), then A is invertible and B = A Otherwise, A is not invertible. Example A = [interactive]
9 Computing A 1 Example R 3 = R 3 + 3R R 1 = R 1 2R 3 R 2 = R 2 R R 3 = R /2 1/ So = /2 1/2 Check: = " /2 1/
10 Why Does This Work? We can think of the algorithm as simultaneously solving the equations Ax 1 = e 1 : Ax 2 = e 2 : Ax 3 = e 3 : Now note A 1 e i = A 1 (Ax i ) = x i, and x i is the ith column in the augmented part. Also A 1 e i is the ith column of A 1.
11 Invertible Transformations Definition A transformation T : R n R n is invertible if there exists another transformation U : R n R n such that T U(x) = x and U T (x) = x for all x in R n. In this case we say U is the inverse of T, and we write U = T 1. In other words, T (U(x)) = x, so T undoes U, and likewise U undoes T. Fact A transformation T is invertible if and only if it is both one-to-one and onto. If T is one-to-one and onto, this means for every y in R n, there is a unique x in R n such that T (x) = y. Then T 1 (y) = x.
12 Invertible Transformations Examples Let T = counterclockwise rotation in the plane by 45. What is T 1? T T 1 T 1 is clockwise rotation by 45. [interactive: T 1 T ] [interactive: T T 1 ] Let T = shrinking by a factor of 2/3 in the plane. What is T 1? T T 1 T 1 is stretching by 3/2. [interactive: T 1 T ] [interactive: T T 1 ] Let T = projection onto the x-axis. What is T 1? It is not invertible: you can t undo it.
13 Invertible Linear Transformations If T : R n R n is an invertible linear transformation with matrix A, then what is the matrix for T 1? Let B be the matrix for T 1. We know T T 1 has matrix AB, so for all x, Hence AB = I n, so B = A 1. ABx = T T 1 (x) = x. Fact If T is an invertible linear transformation with matrix A, then T 1 is an invertible linear transformation with matrix A 1.
14 Invertible Linear Transformations Examples Let T = counterclockwise rotation in the plane by 45. Its matrix is ( ) cos(45 ) sin(45 ) A = sin(45 ) cos(45 = 1 ( ) 1 1. ) Then T 1 = counterclockwise rotation by 45. Its matrix is ( ) cos( 45 ) sin( 45 ) B = sin( 45 ) cos( 45 = 1 ( ) 1 1. ) Check: AB = 1 ( ) ( ) ( ) = Let T = shrinking by a factor of 2/3 in the plane. Its matrix is ( ) 2/3 0 A = 0 2/3 Then T 1 = stretching by 3/2. Its matrix is ( ) 3/2 0 B = 0 3/2 ( ) ( ) 2/3 0 3/2 0 Check: AB = = 0 2/3 0 3/2 ( ) " "
15 The Invertible Matrix Theorem A.K.A. The Really Big Theorem of Math 1553 The Invertible Matrix Theorem Let A be an n n matrix, and let T : R n R n be the linear transformation T (x) = Ax. The following statements are equivalent. 1. A is invertible. 2. T is invertible. 3. The reduced row echelon form of A is the identity matrix I n. 4. A has n pivots. 5. Ax = 0 has no solutions other than the trivial solution. 6. Nul(A) = {0}. 7. nullity (A) = The columns of A are linearly independent. 9. The columns of A form a basis for R n. 10. T is one-to-one. 11. Ax = b is consistent for all b in R n. 12. Ax = b has a unique solution for each b in R n. 13. The columns of A span R n. 14. Col A = R n. 15. dim Col A = n. 16. rank A = n. 17. T is onto. 18. There exists a matrix B such that AB = I n. 19. There exists a matrix B such that BA = I n. you really have to know these
16 The Invertible Matrix Theorem Summary There are two kinds of square matrices: 1. invertible (non-singular), and 2. non-invertible (singular). For invertible matrices, all statements of the Invertible Matrix Theorem are true. For non-invertible matrices, all statements of the Invertible Matrix Theorem are false. Strong recommendation: If you want to understand invertible matrices, go through all of the conditions of the IMT and try to figure out on your own (or at least with help from the book) why they re all equivalent. You know enough at this point to be able to reduce all of the statements to assertions about the pivots of a square matrix.
17 The Invertible Matrix Theorem Example Question: Is this matrix invertible? A = The second column is a multiple of the first, so the columns are linearly dependent. A does not satisfy condition (8) of the IMT, so it is not invertible.
18 The Invertible Matrix Theorem Another Example Problem: Let A be a 3 3 matrix such that 1 2 A 7 = A Show that the rank of A is at most 2. If we set 1 2 b = A 7 = A 0, 0 1 then Ax = b has multiple solutions, so it does not satisfy condition (12) of the IMT. Hence it also does not satisfy condition (16), so the rank is not 3. In any case the rank is at most 3, so it must be less than 3.
19 Summary The inverse of a square matrix A is a matrix A 1 such that AA 1 = I n (equivalently, A 1 A = I n). If A is invertible, then you can solve Ax = b by dividing by A : b = A 1 x. There is a unique solution x = A 1 b for every x. You compute A 1 (and whether A is invertible) by row reducing ( A ) I n. There s a trick for computing the inverse of a 2 2 matrix in terms of determinants. A linear transformation T is invertible if and only if its matrix A is invertible, in which case A 1 is the matrix for T 1. The Invertible Matrix theorem is a list of a zillion equivalent conditions for invertibility that you have to learn (and should understand, since it s well within what we ve covered in class so far).
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# Triangle A has an area of 6 and two sides of lengths 5 and 8 . Triangle B is similar to triangle A and has a side with a length of 19 . What are the maximum and minimum possible areas of triangle B?
Feb 2, 2018
Case 1. Maximum Area of $\Delta B$ = $\textcolor{g r e e n}{225.3902}$ sq. units
Case 2. Minimum Area of $\Delta B$ = $\textcolor{red}{13.016}$ sq. units
#### Explanation:
$\Delta A = 6 , p = 5 , q = 8 , x = 19$
Side r can have values between q-p) , (q + p)
$r > \left(8 - 5\right) , < \left(8 + 5\right)$
$r > 3 < 13$
${r}_{\min} = 3.1 , {r}_{\max} = 12.9$, rounded to one decemal.
Case 1. Maximum Area of $\Delta B$
x should correspond to least side of A, viz. r = 3.1 to get minimum area of B.
$\Delta \frac{B}{\Delta} A = {\left(\frac{x}{r}\right)}^{2}$
$\Delta B = 6 \cdot {\left(\frac{19}{3.1}\right)}^{2} = \textcolor{g r e e n}{225.3902}$ sq. units
Case 2. Minimum Area of $\Delta B$
x should correspond to longest side of A, viz. r = 12.9 to get maximum area of B.
$\Delta \frac{B}{\Delta} A = {\left(\frac{x}{r}\right)}^{2}$
$\Delta B = 6 \cdot {\left(\frac{19}{12.9}\right)}^{2} = \textcolor{red}{13.016}$ sq. units |
# In figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region. - Mathematics
Sum
In figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.
#### Solution
Let r be the radius of each sector = 21 cm
Area of the shaded region = Area of the four sectors
Let angles subtended at A, B, C and D be x°, y°, z° and w° respectively.
Angle subtended at A, B, C, D (in radians, (θ)) be (xpi)/180, (ypi)/180, (zpi)/180, (wpi)/180 respectively.
∴ Area of a sector with central angle at A = 1/2 r^2 theta
= 1/2 xx (21)^2 xx (xpi)1/0 = (441xpi)/360 cm^2
∴ Area of a sector with central angle at B = 1/2 r^2 theta
= 1/2 xx (21)^2 xx (ypi)/180 = (441ypi)/360 cm^2
Area of a sector with central angle at C = 1/2 r^2 theta
= 1/2 xx (21)^2 xx (zoi)/180 = (441zpi)/360 cm^2
∴ Area of a sector with central angle at D = 1/2 r^2 theta
= 1/2 xx (21)^2 xx (wpi)/180 = (441wpi)/360 cm^2
∴ Area of four sectors = ((441xpi)/360 + (441ypi)/360 + (441zpi)/360 + (441wpi)/360) cm^2
Since, sum of all interior angles in any quadrilateral is 360°
∴ x + y + z +w = 360°
Thus, Area of four sectors = ((441pi)/360) (x + y + z + w)
= (441pi)/360 xx 360
= 441π cm2
= 1386 cm2
Hence, required area of the shaded region is 1386 cm2.
Concept: Areas of Sector and Segment of a Circle
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 10
Chapter 11 Area Related To Circles
Exercise 11.3 | Q 15 | Page 128
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Home » Mathematics » Why Do We Learn Percentage?
# Why Do We Learn Percentage?
### Start here
This is the second post in the “Why Do We Learn …?” series. The main purpose of this series is to show the real life application of different Mathematics concepts. In this post, we continue with the topic of buying a commodity.
Below is table showing the price of the commodity with different weights:
Weight Sell Out Price (S\$) Buy Back Price (S\$) Difference between SOP and BBP (S\$) Difference in Percentage (%) 1 g 56.50 56.10 0.40 0.71 1 kg 56300 56100 200 0.36 1 ounce 1820 1750 70 3.85 10 ounce 17900 17600 300 1.68
Concept 1: Money
The sell out price is the money that you pay to the bank when you buy the commodity.
The buy back price is the money that the bank pays you when you sell the commodity back to the bank.
From the table above, we can see that there is difference between the sell out price and buy back price and the buy back price is always lower than the sell out price.
Concept 2: Percentage
If we invest in a commodity, we would like to earn money, that’s common sense.
With the assumption that the price is the same as the above table when you buy and sell the commodity, which weight would you choose to make maximum profit?
The difference in price does not make a good indicator because the weight difference is big and thus the difference in price is big too. Thus, we should choose the difference in percentage as a better indicator.
We choose the smallest difference in percentage to make maximum profit. To calculate the difference in percentage for 1 g:
The price difference = 56.50 – 56.10
= 0.40
The percentage = (0.40/56.50) x 100%
= 0.71%
From the table, we know that the smallest difference in percentage is for 1 kg, 0.36% and the biggest difference in percentage is for 1 ounce, 3.85%.
If the price of the commodity has increased by 2%, investment in 1 kg has positive return while investment in 1 ounce has negative return.
In a nutshell, we can decide which weight will give you a good return by using the difference in percentage, i.e. 1 kg gives the best return among the four weights while 1 ounce gives the least return.
Disclaimer: This post is not encouraging children and / or adults to invest blindly. This is only a simple example where you can use the knowledge learnt in real life. Investment in real life is more complicated and involves more risks. |
## Engage NY Eureka Math 3rd Grade Module 1 End of Module Assessment Answer Key
Question 1.
Mr. Lewis arranges all the desks in his classroom into 6 equal groups of 4. How many desks are in his classroom? Show a picture and multiplication sentence in your work.
a. What does the product in your multiplication sentence represent?
24 desks are there in Mr. Lewis classroom,
Multiplication Sentence: 4 X 6 = 24, the product in your multiplication sentence represent there are 24 desks.
Explanation:
Given Mr. Lewis arranges all the desks in his classroom into 6 equal groups of 4.
So total desks his classroom are 24,
Shown a picture and multiplication sentence is 4 × 6 = 24.
b. Fill in the blanks below to complete a related division sentence.
_24_ ÷ 4 = ___6___,
Explanation:
Related division sentence is 24 ÷ 4 = 6,
when 24 is divided into 4 times we get 6 parts each.
c. What does the quotient in Part (b) represent?
Quotient represents 6 parts,
Explanation:
The quotient in the division sentence 24 ÷ 4 represents 24 is divided by 4 then we get quotient as 6 equal parts.
Question 2.
a. Draw an array that shows 9 rows of 2. Write a multiplication sentence to represent the array, and circle the factor that represents the number of rows.
b. Draw another array that shows 2 rows of 9. Write a different multiplication sentence, and circle the factor that represents the size of the row.
c. Explain the relationship between the two arrays using number sentences and words.
a.
Explanation:
Drawn an array that shows 9 rows of 2,
Wrote a multiplication sentence to represent the array as 9 × 2 = 18 and circled the factor 9 that represents the number of rows.
b.
Explanation:
Drawn another array that shows 2 rows of 9.
Wrote a different multiplication sentence as 2 × 9 = 18 and circled the factor that 2 represents the size of the row.
c. The relationship between the two arrays is the results is same,
Explanation:
The relationship between the two arrays is same as number sentences is 9 × 2 = 18, 2 × 9 = 18, means if 9 rows are multiplied by 2 times we get 18 and if 2 is multiplied by 9 times also we get result as 18 only. therefore if x rows and y columns are there then x × y = y × x the product result will be the same.
Question 3.
Ms. Park buys a tray of apples for a class party. There are 5 rows of 4 red apples. There is 1 row of 4 green apples.
a. The picture below shows Ms. Park’s apples. Fill in the blanks to complete the expressions.
Explanation:
Given Ms. Park buys a tray of apples for a class party.
There are 5 rows of 4 red apples.
There is 1 row of 4 green apples,
The expression is Total apples: 6 × 4,
Red apples: 5 × 4, Green apples: 1 × 4.
b. Fill in the unknowns in the equation below to match the picture of the apples in Part (a). Use the break apart and distribute strategy to find the total number of apples Ms. Park bought.
_____ × 4 = _____ × 4 + _____ × 4
Ms. Park bought ________ apples.
Equation is 6 × 4 = ((5 × 4) + (1 × 4)),
A total number of apples Ms. Park bought 24 apples.
Explanation:
The equation below to match the picture of the apples in Part (a). Using the break apart and distribute strategy is Equation is 6 × 4 = ((5 × 4) + (1 × 4)),
The total number of apples Ms. Park bought are 24 apples.
c. Lilly brings 8 green apples for the class party. Show Lilly’s green apples on the picture in Part (a). Then, fill in the unknowns in the equation below to match the new picture. Solve to find the total number of apples.
_____ × 4 = _____ × 4 + _____ × 4
There are ________ apples in all.
Equation is 8 × 4 = ((5 × 4) + (3 × 4)),
Total number of apples are 32.
Explanation:
Given Lilly brings 8 green apples for the class party.
Shown Lilly’s green apples in the picture in Part (a) as above, Then, filled the unknowns in the equation below to match the new picture as 8 × 4 = ((5 × 4) + (3 × 4)), on solving to find the total number of apples are 20 + 12 = 32 apples.
Question 4.
Mr. Myer’s class plays a game. The class earns 5 points each time they answer a question correctly. The class earns 50 points playing the game on Monday.
a. How many questions did the class answer correctly? Show a picture and division sentence in your work.
The class answered 10 questions correctly,
The division sentence is 50 ÷ 5= 10,
Explanation:
Given Mr. Myer’s class plays a game, The class earns 5 points each time they answer a question correctly. The class earns 50 points playing the game on Monday. So the number of questions the class answered correctly is (50 ÷ 5= 10) 10 questions.
Shown a picture and division sentence in the picture above,
Division Sentence : 50 ÷ 5= 10.
b. Mr. Myer uses the equation 5 × _____ = 50 to find how many questions the class answered correctly. Is his method correct? Why or why not?
His method is correct only but it is time consuming,
Explanation:
Given Mr. Myer uses the equation 5 × _____ = 50 to find how many questions the class answered correctly. His method is correct, Why because he goes on
substituting numbers till he gets the answer, Mr. Myer gets the answer as 5 X 10 = 50, But his method is time consuming and little long.
c. The class answered 7 questions correctly on Tuesday. What is the total number of points the class earned on both days?
The total number of points the class earned on both days is 50 + 35 = 85 points,
Explanation:
Given the class answered 7 questions correctly on Tuesday. So the total number of points the class earned on both days is Monday – 50 points, Tuesday – 5 × 7 = 35 points, In total 50 + 35 = 85 points.
Question 5.
Complete as many problems as you can in 100 seconds. Your teacher will time you and tell you when to stop. |
# LCM and HCF Questions
Practice Problem on HCF and LCM by solving the important Aptitude Test Questions and Answers given here. Here in this article we have provided LCM and HCF Questions and Answers with their explanations. It will help you in understanding the basic concepts. These questions are useful in all competitive examinations such as placement or entrance test.
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For the easiness of contenders who are preparing for the examination, we have mentioned LCM and HCF Questions with Solutions, so that in case they are not able to solve the questions at their own, they can take help from there. So, let’s start solving the questions one by one:
## Problem on HCF and LCM
Formulas and Quick Tricks for LCM and HCF
• Least Common Multiple is the full form of LCM.
• The smallest number which is exactly divisible by each one of the given numbers is known as their Least Common Multiple (LCM)
• HCF is the short form of Highest Common Factor.
• HCF are also known as Greatest Common Divisor (GCD) and Greatest Common Measure (GCM)
• The Highest Common Factor of two or more numbers is the greatest number that divides each one of them exactly.
• Two numbers are said to be co-prime if their HCF is 1
• HCF of fractions = HCF of numerators/LCM of denominators
• LCM of fractions = GCD of numerators/HCF of denominators
LCM and HCF Questions and Answers
Question 1) Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
(a) 40
(b) 80
(c) 120
(d) 200
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Question 2) The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
(a) 74
(b) 94
(c) 184
(d) 364
Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Question 3) The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
(a) 276
(b) 299
(c) 322
(d) 345
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Question 4) The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
(a) 9000
(b) 9400
(c) 9600
(d) 9800
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 – 399) = 9600.
Question 5) The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
(a) 3
(b) 13
(c) 23
(d) 33
Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
Question 6) What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
(a) 196
(b) 630
(c) 1260
(d) 2520
Explanation:
L.C.M. of 12, 18, 21
= 2 x 3 x 2 x 3 x 7 x 5 = 1260.
Required number = (1260 / 2) = 630.
Question 7) 252 can be expressed as a product of primes as:
(a) 2 x 2 x 3 x 3 x 7
(b) 2 x 2 x 2 x 3 x 7
(c) 3 x 3 x 3 x 3 x 7
(d) 2 x 3 x 3 x 3 x 7
Answer 7) (a) 2 x 2 x 3 x 3 x 7
Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7
Question 8) Find the highest common factor of 36 and 84.
(a) 4
(b) 6
(c) 12
(d) 18
Explanation:
36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
Question 9) The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
(a) 504
(b) 536
(c) 544
(d) 548
Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
Question 10) Which of the following has the most number of divisors?
(a) 99
(b) 101
(c) 176
(d) 182
Explanation:
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Question 11) The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
(a) 28
(b) 32
(c) 40
(d) 64
Explanation:
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
Question 12) Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
(a) 4
(b) 5
(c) 6
(d) 8
Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Question 13) The G.C.D. of 1.08, 0.36 and 0.9 is:
(a) 0.03
(b) 0.9
(c) 0.18
(d) 0.108
Explanation:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
Question 14) The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
(a) 101
(b) 107
(c) 111
(d) 185
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111)
Greater number = 111.
Question 15) The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
(a) 1677
(b) 1683
(c) 2523
(d) 3363
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
Question 16) The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
(a) 1008
(b) 1015
(c) 1022
(d) 1032
Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Must Check: Aptitude Questions & Answers
Question 17) The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
(a) 279
(b) 283
(c) 308
(d) 318
Explanation: (c) 308
Other number = (11 x 7700)/ 275 = 308.
Question 18) Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
(a) 75
(b) 81
(c) 85
(d) 89
Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = (551/29) = 19;
Third number = (1073/29) = 37.
Required sum = (19 + 29 + 37) = 85.
Question 19) The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
(a) 1
(b) 2
(c) 3
(d) 4
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
Start Now: Quantitative Aptitude Quiz
Question 20) A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
(a) 26 minutes and 18 seconds
(b) 42 minutes and 36 seconds
(c) 45 minutes
(d) 46 minutes and 12 seconds
Answer 20) (d) 46 minutes and 12 seconds
Explanation:
L.C.M. of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Question 21: LCM of 1/3, 5/6, 5/4, 10/7 is:
1. 10/11
2. 11/10
3. 10/7
4. 10
Explanation: LCM of numerators = 10
HCF of denominators = 1
=> 10/1 = 10
Question 22: HCF of 3/16, 5/12, 7/8 is:
1. 1/48
2. 2/47
3. 3/47
4. 5/48
Explanation: HCF of numerators = 1
LCM of denominators = 48
=> 1/48
Question 23: The least square number which divides 8, 12 and 18 is?
1. 64
2. 100
3. 144
4. 196
Explanation: LCM = 72
72 * 2 = 144
Question 24: The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?
1. 51
2. 61
3. 69
4. 71
Explanation: LCM = 48 + 3 = 51
Question 25: The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?
1. 1261
2. 1265
3. 1267
4. 1269 |
# 10 Times Table (Times Tables) – Learning Drives
10 Times Table (Times Tables) – Learning Drives .The 10 times table is the easiest table to remember. The easiest way to learn the table 10 is to add one zero to every number you multiply with. Let’s take a look at the 10-multiplication table and learn more in this lesson plan.
## Multiplication Table for 10
It is helpful to learn the multiplication table 10 when solving math problems or understanding number patterns. To help you solve math problems faster, take a look at the below 10 times table.
## Tips for 10 Times Table
The easiest to remember is Table of 10. The digit in the place of multiples of 10 is always 0.
To get the 10 times table, simply write the natural number followed with a 0
Ten times we read the table as:
• One time ten is 10
• Two times ten is 20
• Three times ten is 30
• Four times ten is 40
• Five times ten is 50
• Six times ten equals 60
• Seven times ten is 70
• Eight times ten is 80
• Nine times ten is 90
• Ten times ten is 100
• Eleven times ten is 110
Twelve times ten is 12
We write 10 times as many tables as:
• 1 x 10 = 10.
• 2 x 10 = 20
• 3 x 10 = 30,
• 4 x 10 =40
• 5 x 10 = 50
• 6 x 10 = 60
• 7 x 10 =70
• 8 x 10 =80
• 9 x 10 =90
• 10 x 10 = 100
• 11 x 10 =110
• 12 x 10 = 120
### What is the table with 10?
Multiplication Table of 10: 10×1 = 10, 10×2 = 20, 10×3 = 40,10×4 = 40,10×5 = 50,10×6 = 60,10×7 = 70,10×7 = 70,10×8 = 80,10×9 = 90,10×10 = 100
### Is there a trick that makes it easy to master the 10 times table
It is easiest to learn the multiplication table of 10, by remembering that the digit in column one of multiples of 10 is always zero. To get multiples of 10, you can also write the multiplication table for 1 and then add 0 to the end of each multiple of 1. |
In this chapter, we will develop certain techniques that help solve problems stated in words. These techniques involve sầu rewriting problems in the khung of symbols. For example, the stated problem
"Find a number which, when added to 3, yields 7"
may be written as:
3 + ? = 7, 3 + n = 7, 3 + x = 1
và so on, where the symbols ?, n, & x represent the number we want lớn find. We Gọi such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = 7 are first-degree equations, since the variable has an exponent of 1. The terms khổng lồ the left of an equals sign trang điểm the left-hvà thành viên of the equation; those to the right ảo diệu the right-hvà member. Thus, in the equation x + 3 = 7, the left-h& thành viên is x + 3 and the right-hvà member is 7.
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## SOLVING EQUATIONS
Equations may be true or false, just as word sentences may be true or false. The equation:
3 + x = 7
will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable và determining the truth or falsity of the result.
Example 1 Determine if the value 3 is a solution of the equation
4x - 2 = 3x + 1
Solution We substitute the value 3 for x in the equation and see if the left-hvà thành viên equals the right-hand member.
4(3) - 2 = 3(3) + 1
12 - 2 = 9 + 1
10 = 10
Ans. 3 is a solution.
The first-degree equations that we consider in this chapter have sầu at most one solution. The solutions lớn many such equations can be determined by inspection.
Example 2 Find the solution of each equation by inspection.
a.x + 5 = 12b. 4 · x = -20
Solutions a. 7 is the solution since 7 + 5 = 12.b.-5 is the solution since 4(-5) = -trăng tròn.
## SOLVING EQUATIONS USING ADDITION AND SUBTRACTION PROPERTIES
In Section 3.1 we solved some simple first-degree equations by inspection. However, the solutions of most equations are not immediately evident by inspection. Hence, we need some mathematical "tools" for solving equations.
EQUIVALENT EQUATIONS
Equivalent equations are equations that have sầu identical solutions. Thus,
3x + 3 = x + 13, 3x = x + 10, 2x = 10, and x = 5
are equivalent equations, because 5 is the only solution of each of them. Notice in the equation 3x + 3 = x + 13, the solution 5 is not evident by inspection but in the equation x = 5, the solution 5 is evident by inspection. In solving any equation, we transsize a given equation whose solution may not be obvious khổng lồ an equivalent equation whose solution is easily noted.
The following property, sometimes called the addition-subtraction property, is one way that we can generate equivalent equations.
If the same quantity is added to or subtracted from both membersof an equation, the resulting equation is equivalent lớn the originalequation.
In symbols,
a - b, a + c = b + c, và a - c = b - c
are equivalent equations.
Example 1 Write an equation equivalent to
x + 3 = 7
by subtracting 3 from each member.
Solution Subtracting 3 from each member yields
x + 3 - 3 = 7 - 3
or
x = 4
Notice that x + 3 = 7 & x = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation.
Example 2 Write an equation equivalent to
4x- 2-3x = 4 + 6
by combining lượt thích terms và then by adding 2 khổng lồ each thành viên.
Combining lượt thích terms yields
x - 2 = 10
Adding 2 lớn each thành viên yields
x-2+2 =10+2
x = 12
To solve an equation, we use the addition-subtraction property khổng lồ transform a given equation lớn an equivalent equation of the khung x = a, from which we can find the solution by inspection.
Example 3 Solve sầu 2x + 1 = x - 2.
We want khổng lồ obtain an equivalent equation in which all terms containing x are in one member and all terms not containing x are in the other. If we first add -1 lớn (or subtract 1 from) each member, we get
2x + 1- 1 = x - 2- 1
2x = x - 3
If we now add -x to (or subtract x from) each member, we get
2x-x = x - 3 - x
x = -3
where the solution -3 is obvious.
The solution of the original equation is the number -3; however, the answer is often displayed in the form of the equation x = -3.
Since each equation obtained in the process is equivalent khổng lồ the original equation, -3 is also a solution of 2x + 1 = x - 2. In the above sầu example, we can kiểm tra the solution by substituting - 3 for x in the original equation
2(-3) + 1 = (-3) - 2
-5 = -5
The symmetric property of echất lượng is also helpful in the solution of equations. This property states
If a = b then b = a
This enables us to interchange the members of an equation whenever we please without having to lớn be concerned with any changes of sign. Thus,
If 4 = x + 2thenx + 2 = 4
If x + 3 = 2x - 5then2x - 5 = x + 3
If d = rtthenrt = d
There may be several different ways khổng lồ apply the addition property above. Sometimes one method is better than another, & in some cases, the symmetric property of equality is also helpful.
Example 4 Solve 2x = 3x - 9.(1)
Solution If we first add -3x khổng lồ each thành viên, we get
2x - 3x = 3x - 9 - 3x
-x = -9
where the variable has a negative sầu coefficient. Although we can see by inspection that the solution is 9, because -(9) = -9, we can avoid the negative sầu coefficient by adding -2x và +9 khổng lồ each member of Equation (1). In this case, we get
2x-2x + 9 = 3x- 9-2x+ 9
9 = x
from which the solution 9 is obvious. If we wish, we can write the last equation as x = 9 by the symmetric property of equality.
## SOLVING EQUATIONS USING THE DIVISION PROPERTY
Consider the equation
3x = 12
The solution khổng lồ this equation is 4. Also, note that if we divide each member of the equation by 3, we obtain the equations
whose solution is also 4. In general, we have the following property, which is sometimes called the division property.
If both members of an equation are divided by the same (nonzero)quantity, the resulting equation is equivalent lớn the original equation.
In symbols,
are equivalent equations.
Example 1 Write an equation equivalent to
-4x = 12
by dividing each thành viên by -4.
Solution Dividing both members by -4 yields
In solving equations, we use the above sầu property lớn produce equivalent equations in which the variable has a coefficient of 1.
Example 2 Solve 3y + 2y = trăng tròn.
We first combine like terms khổng lồ get
5y = 20
Then, dividing each member by 5, we obtain
In the next example, we use the addition-subtraction property & the division property lớn solve an equation.
Example 3 Solve sầu 4x + 7 = x - 2.
Solution First, we add -x và -7 lớn each member lớn get
4x + 7 - x - 7 = x - 2 - x - 1
Next, combining like terms yields
3x = -9
Last, we divide each member by 3 to lớn obtain
## SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY
Consider the equation
The solution khổng lồ this equation is 12. Also, note that if we multiply each thành viên of the equation by 4, we obtain the equations
whose solution is also 12. In general, we have the following property, which is sometimes called the multiplication property.
If both members of an equation are multiplied by the same nonzero quantity, the resulting equation Is equivalent to lớn the original equation.
In symbols,
a = b & a·c = b·c (c ≠ 0)
are equivalent equations.
Example 1 Write an equivalent equation to
by multiplying each thành viên by 6.
Solution Multiplying each member by 6 yields
In solving equations, we use the above property lớn produce equivalent equations that are free of fractions.
Example 2 Solve
Solution First, multiply each thành viên by 5 khổng lồ get
Now, divide each member by 3,
Example 3 Solve
.
Solution First, simplify above sầu the fraction bar khổng lồ get
Next, multiply each member by 3 khổng lồ obtain
Last, dividing each member by 5 yields
## FURTHER SOLUTIONS OF EQUATIONS
Now we know all the techniques needed lớn solve most first-degree equations. There is no specific order in which the properties should be applied. Any one or more of the following steps listed on page 102 may be appropriate.
Steps lớn solve sầu first-degree equations:Combine lượt thích terms in each thành viên of an equation.Using the addition or subtraction property, write the equation with all terms containing the unknown in one member & all terms not containing the unknown in the other.Combine like terms in each member.Use the multiplication property khổng lồ remove sầu fractions.Use the division property to obtain a coefficient of 1 for the variable.
Example 1 Solve 5x - 7 = 2x - 4x + 14.
Solution First, we combine lượt thích terms, 2x - 4x, lớn yield
5x - 7 = -2x + 14
Next, we add +2x and +7 khổng lồ each thành viên & combine like terms to lớn get
5x - 7 + 2x + 7 = -2x + 14 + 2x + 1
7x = 21
Finally, we divide each member by 7 khổng lồ obtain
In the next example, we simplify above the fraction bar before applying the properties that we have been studying.
Example 2 Solve sầu
Solution First, we combine like terms, 4x - 2x, to lớn get
Then we add -3 to each member và simplify
Next, we multiply each thành viên by 3 khổng lồ obtain
Finally, we divide each member by 2 khổng lồ get
## SOLVING FORMULAS
Equations that involve sầu variables for the measures of two or more physical quantities are called formulas. We can solve sầu for any one of the variables in a formula if the values of the other variables are known. We substitute the known values in the formula & solve for the unknown variable by the methods we used in the preceding sections.
Example 1 In the formula d = rt, find t if d = 24 và r = 3.
Solution We can solve sầu for t by substituting 24 for d và 3 for r. That is,
d = rt
(24) = (3)t
8 = t
It is often necessary lớn solve sầu formulas or equations in which there is more than one variable for one of the variables in terms of the others. We use the same methods demonstrated in the preceding sections.
Example 2 In the formula d = rt, solve sầu for t in terms of r và d.
Solution We may solve for t in terms of r and d by dividing both members by r to lớn yield
from which, by the symmetric law,
In the above sầu example, we solved for t by applying the division property to lớn generate an equivalent equation. Sometimes, it is necessary khổng lồ apply more than one such property.
Bài viết liên quan |
# How do you solve 4x=sqrt(6-4x) and find any extraneous solutions?
Apr 8, 2017
You can detect any extraneous solutions, if you restrict the argument of the square root to be greater than or equal to zero, then you can square both sides of the equation.
#### Explanation:
Given: $4 x = \sqrt{6 - 4 x}$
Restrict the argument of the square root to be greater than or equal to zero:
4x=sqrt(6-4x); 6-4x>=0
Simplify the restriction:
4x=sqrt(6-4x); 6>=4x
4x=sqrt(6-4x); 3/2>=x
4x=sqrt(6-4x); x<=3/2
Square both sides:
16x^2= 6-4x; x<=3/2
8x^2+2x-3=0; x<=3/2
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$
x = (-2+-sqrt(2^2-4(8)(-3)))/(2(8)); x<=3/2
x = (-2+-sqrt(100))/16; x<=3/2
$x = \frac{- 12}{16} \mathmr{and} x = \frac{8}{16}$
$x = - \frac{3}{4} \mathmr{and} x = \frac{1}{2}$ |
Saturday, July 20
# Easy Way to Learn Geometry Formulas
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Mathematics is an integral part of learning for all children. Introducing Maths to children from an early age helps to develop their problem-solving skills, reasoning skills and logical thinking in a broad range of contexts. Numeracy and shapes improve spatial awareness. It helps every child to recognise, create and describe patterns, which is necessary for the early problem-solving skills.
Geometry is a part of a student’s curriculum from kindergarten to higher secondary education. It also continues through college and throughout the entire lifetime. Even without opening the geometry book, it is used daily by almost everyone. It explores geometric reasoning and spatial sense. It is involved in all the disciplines such as Engineering, Space, Art and Architecture, Astronomy, Robotics etc.
Geometry is the central concept that links some topics in Maths, especially in measurement. In schooling, the main focus of geometry is learning about the mensuration formulas. But children find difficulties in learning those formulas. Here, the article provides some easy ways to learn geometry formulas.
## Understand the Concept
To learn the geometry formulas, first, understand the concepts and parameters of different shapes. Clear with concepts such as two-dimensional and three-dimensional geometry. The parameters such as length, breadth, height, altitude, slant height, area, volume, etc. are used in geometry.
## Learn the Relationship Between Different Shapes
Each shape has different characteristics. But some properties of geometrical shape relates to the other form. If you know the relationship between the shapes, it is easy to learn the formulas on your own. If you are learning 3D shape formulas, try to relate with the respective 2D shape. Because 3D shapes are obtained from the rotation of 2D shapes.
## Practice
Once you understand the formulas, write it down and practice it. Practising the formulas, again and again, makes you so familiar with that. For many shapes, the pi value defines the property of that shape. For example, the area of a circle is πr2. If you remove π, it does not represent the circle area.
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### Learning Objectives
Graph a circle in traditional form. Recognize the equation of a circle provided its graph. Rewrite the equation of a circle in standard form.
## The one in conventional Form
A circleA circle is the collection of points in a plane that lie a fixed distance native a given point, referred to as the center. Is the collection of point out in a aircraft that lied a fixed distance, referred to as the radiusThe fixed distance from the facility of a circle to any allude on the circle., from any kind of point, called the center. The diameterThe length of a heat segment passing through the facility of a circle whose endpoints room on the circle. Is the size of a line segment passing with the facility whose endpoints are on the circle. In addition, a circle can be developed by the intersection of a cone and also a plane that is perpendicular to the axis the the cone:
In a rectangular coordinate plane, wherein the facility of a circle through radius r is (h,k), we have
Calculate the distance in between (h,k) and also (x,y) using the street formula,
(x−h)2+(y−k)2=r
Squaring both political parties leads us to the equation of a circle in traditional formThe equation the a circle composed in the kind (x−h)2+(y−k)2=r2 whereby (h,k) is the center and r is the radius.,
(x−h)2+(y−k)2=r2
In this form, the center and radius space apparent. Because that example, provided the equation (x−2)2+ (y + 5)2=16 we have,
(x−h)2+ (x−k)2=r2↓↓↓(x−2)2+2=42
In this case, the center is (2,−5) and r=4. More examples follow:
Equation
Center
(x−3)2+(y−4)2=25
(3,4)
r=5
(x−1)2+(y+2)2=7
(1,−2)
r=7
(x+4)2+(y−3)2=1
(−4,3)
r=1
x2+(y+6)2=8
(0,−6)
r=22
The graph that a circle is totally determined by its center and radius.
You are watching: What is a solution point of a circle
### Example 1
Graph: (x−2)2+(y+5)2=16.
Solution:
Written in this form we have the right to see that the center is (2,−5) and also that the radius r=4 units. Native the center mark clues 4 systems up and also down as well as 4 systems left and right.
Then attract in the circle v these 4 points.
As with any graph, we space interested in recognize the x- and also y-intercepts.
### Example 2
Find the intercepts: (x−2)2+(y+5)2=16.
Solution:
To find the y-intercepts collection x=0:
(x−2)2+(y+5)2=16(0−2)2+(y+5)2=164+(y+5)2=16
For this equation, we deserve to solve by extract square roots.
(y+5)2=12y+5=±12y+5=±23y=−5±23
Therefore, the y-intercepts space (0,−5−23) and (0,−5+23). To discover the x-intercepts set y=0:
(x−2)2+(y+5)2=16(x−2)2+(0+5)2=16(x−2)2+25=16(x−2)2=−9x−2=±−9x=2±3i
And due to the fact that the services are facility we conclude the there space no real x-intercepts. Keep in mind that this does do sense provided the graph.
Answer: x-intercepts: none; y-intercepts: (0,−5−23) and (0,−5+23)
Given the center and radius that a circle, us can find its equation.
### Example 3
Graph the circle v radius r=3 units centered at (−1,0). Give its equation in standard form and determine the intercepts.
Solution:
Given the the center is (−1,0) and also the radius is r=3 we map out the graph as follows:
Substitute h, k, and r to find the equation in standard form. Because (h,k)=(−1,0) and also r=3 we have,
(x−h)2+(y−k)2=r22+(y−0)2=32(x+1)2+y2=9
The equation that the one is (x+1)2+y2=9, usage this to recognize the y-intercepts.
(x+1)2+y2=9 Set x=0 to and solve for y.(0+1)2+y2=91+y2=9y2=8y=±8y=±22
Therefore, the y-intercepts room (0,−22) and also (0,22). To discover the x-intercepts algebraically, set y=0 and also solve because that x; this is left because that the reader as an exercise.
Answer: Equation: (x+1)2+y2=9; y-intercepts: (0,−22) and also (0,22); x-intercepts: (−4,0) and (2,0)
Of certain importance is the unit circleThe circle focused at the beginning with radius 1; that equation is x2+y2=1.,
x2+y2=1
Or,
(x−0)2+(y−0)2=12
In this form, it have to be clear the the center is (0,0) and that the radius is 1 unit. Furthermore, if we deal with for y we acquire two functions:
x2+y2=1y2=1−x2y=±1−x2
The function defined by y=1−x2 is the top fifty percent of the circle and the duty defined by y=−1−x2 is the bottom fifty percent of the unit circle:
Try this! Graph and also label the intercepts: x2+(y+2)2=25.
(click to see video)
## The circle in general Form
We have actually seen that the graph of a circle is fully determined by the center and also radius which can be review from its equation in typical form. However, the equation is not always given in conventional form. The equation of a one in general formThe equation of a circle composed in the kind x2+y2+cx+dy+e=0. Follows:
x2+y2+cx+dy+e=0
Here c, d, and also e are real numbers. The steps for graphing a circle given its equation in general type follow.
### Example 4
Graph: x2+y2+6x−8y+13=0.
Solution:
Begin by rewriting the equation in traditional form.
Step 1: group the terms v the exact same variables and also move the consistent to the best side. In this case, subtract 13 on both sides and also group the terms entailing x and also the terms including y together follows.
x2+y2+6x−8y+13=0(x2+6x+___)+(y2−8y+___)=−13
Step 2: finish the square for each grouping. The idea is to include the value that completes the square, (b2)2, come both sides for both groupings, and then factor. Because that the terms including x usage (62)2=32=9 and for the terms involving y use (−82)2=(−4)2=16.
(x2+6x +9)+(y2−8y+16)=−13 +9+16(x+3)2+(y−4)2=12
Step 3: identify the center and also radius native the equation in traditional form. In this case, the center is (−3,4) and also the radius r=12=23. Step 4: native the center, note the radius vertically and horizontally and also then lay out the circle through these points.
### Example 5
Determine the center and also radius: 4x2+4y2−8x+12y−3=0.
Solution:
We can achieve the general kind by first dividing both political parties by 4.
4x2+4y2−8x+12y−34=04x2+y2−2x+3y−34=0
Now that we have the general form for a circle, wherein both state of level two have actually a top coefficient of 1, we deserve to use the measures for rewriting it in conventional form. Begin by adding 34 to both sides and group variables that room the same.
(x2−2x+___)+(y2+3y+___)=34
Next complete the square for both groupings. Use (−22)2=(−1)2=1 because that the first grouping and also (32)2=94 for the second grouping.
(x2−2x +1)+(y2+3y+94)=34 +1+94(x−1)2+(y+32)2=164(x−1)2+(y+32)2=4
In summary, to transform from standard form to general type we multiply, and also to transform from general form to standard form we finish the square.
Try this! Graph: x2+y2−10x+2y+21=0.
(click to watch video)
### Key Takeaways
The graph the a one is totally determined through its center and also radius. Standard type for the equation the a one is (x−h)2+(y−k)2=r2. The facility is (h,k) and also the radius measures r units. Come graph a circle note points r systems up, down, left, and also right indigenous the center. Draw a circle with these four points. If the equation that a one is given in general form x2+y2+cx+dy+e=0, team the terms v the exact same variables, and also complete the square for both groupings. This will result in conventional form, native which we deserve to read the circle’s center and also radius. We acknowledge the equation that a one if the is quadratic in both x and y whereby the coefficient that the squared terms space the same.
### Part A: The circle in traditional Form
Determine the center and radius provided the equation that a circle in traditional form.
(x−5)2+(y+4)2=64
(x+9)2+(y−7)2=121
x2+(y+6)2=4
(x−1)2+y2=1
(x+1)2+(y+1)2=7
(x+2)2+(y−7)2=8
Determine the standard form for the equation of the circle provided its center and radius.
Graph.
(x−1)2+(y−2)2=9
(x+3)2+(y−3)2=25
(x−2)2+(y+6)2=4
(x+6)2+(y+4)2=36
x2+(y−4)2=1
(x−3)2+y2=4
x2+y2=12
x2+y2=8
(x−7)2+(y−6)2=2
(x+2)2+(y−5)2=5
(x+3)2+(y−1)2=18
(x−3)2+(y−2)2=15
Find the x- and also y-intercepts.
(x−1)2+(y−2)2=9
(x+5)2+(y−3)2=25
x2+(y−4)2=1
(x−3)2+y2=18
x2+y2=50
x2+(y+9)2=20
(x−4)2+(y+5)2=10
(x+10)2+(y−20)2=400
Find the equation the the circle.
Circle with facility (1,−2) passing with (3,−4).
Circle with facility (−4,−1) passing v (0,−3).
Circle whose diameter is characterized by (5,1) and (−1,7).
Circle who diameter is identified by (−5,7) and (−1,−5).
Circle with center (5,−2) and also area 9π square units.
Circle with center (−8,−3) and also circumference 12π square units.
Find the area that the circle with equation (x+12)2+(x−5)2=7.
Find the one of the circle through equation (x+1)2+(y+5)2=8.
### Part B: The circle in general Form
Rewrite in standard form and graph.
x2+y2+4x−2y−4=0
x2+y2−10x+2y+10=0
x2+y2+2x+12y+36=0
x2+y2−14x−8y+40=0
x2+y2+6y+5=0
x2+y2−12x+20=0
x2+y2+8x+12y+16=0
x2+y2−20x−18y+172=0
4x2+4y2−4x+8y+1=0
9x2+9y2+18x+6y+1=0
x2+y2+4x+8y+14=0
x2+y2−2x−4y−15=0
x2+y2−x−2y+1=0
x2+y2−x+y−12=0
4x2+4y2+8x−12y+5=0
9x2+9y2+12x−36y+4=0
2x2+2y2+6x+10y+9=0
9x2+9y2−6x+12y+4=0
Given a one in general form, determine the intercepts.
x2+y2−5x+3y+6=0
x2+y2+x−2y−7=0
x2+y2−6y+2=2
x2+y2−6x−8y+5=0
2x2+2y2−3x−9=0
3x2+3y2+8y−16=0
Determine the area the the circle who equation is x2+y2−2x−6y−35=0.
Determine the area that the circle whose equation is 4x2+4y2−12x−8y−59=0.
Determine the one of a circle whose equation is x2+y2−5x+1=0.
Determine the one of a circle who equation is x2+y2+5x−2y+3=0.
Find general kind of the equation of a circle centered at (−3,5) passing v (1,−2).
Find general form of the equation that a circle centered at (−2,−3) passing v (−1,3).
Given the graph that a circle, determine its equation in general form.
### Part C: conversation Board
Is the center of a circle component of the graph? Explain.
Make up your own circle, compose it in general form, and graph it.
Explain exactly how we deserve to tell the difference between the equation the a parabola in general type and the equation that a one in general form. Provide an example.
See more: What Decimal Is 1/10 Of 0.08, What Decimal Is \$\$ Frac { 1 } { 10 } \$\$ Of 0
Do every circles have actually intercepts? What space the possible numbers the intercepts? highlight your explanation v graphs. |
August 9, 2024
# Solved examples on rational numbers
Here we are giving some practice questions based on rational numbers.
1. Are the following statements are true or false? Give reason for your answers.
(i) every whole number is a natural numbers.
(ii) every integer is a rational number .
(iii)There are infinitely many rational numbers between any two given rational numbers.
(iv) 0 is a rational number.
(V)Every rational number is a whole number.
ANS. (i) False , beacuse ‘0’ is a whole number not a natural number.
(ii)True, because every integer $m$ can be expressed in the form $\frac{m}{1}$
, so it is a rational number.
(iii) True
(iv) True , since we can write $0=\frac{0}{1}$
(v)false
2. Find six rational numbers between 3 and 4.
Sol. To find a rational number between $r$ and $s$ we can add $r$ and $s$ and divide the sum by 2 that is $\frac{r+s}{2}$ lies between $r$ and $s$ .
Now let $r$ =3 and $s$ =4 , then $\frac{r+s}{2}$ = $\frac{7}{2}$ is a number lying between 3 and 4 .
Next let $r$ =3 and $s$ =$\frac{7}{2}$, then $\frac{3+\frac{7}{2}}{2}=&space;\frac{13}{4}$ is also a number lying between 3 and 4.
Rational no. between $\frac{7}{2}&space;\;&space;and&space;\;&space;4$ is $\frac{\frac{7}{2}+4}{2}=&space;\frac{15}{4}$ .
Proceeding in the same manner the six rational numbers between 3 and 4 are $\frac{7}{2},$ $\frac{13}{4}$ , $\frac{15}{4},\frac{27}{8},&space;\frac{28}{8},$ $\frac{29}{8}$.
3. Express the decimal $32.12\bar{35}$ in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q\neq&space;0$.
Sol. let $x=32.12353535......$. Since two digits are repeating, we multiply by $x$ by 100 to get
$100x=&space;3212.353535..........$
So, $100x=3180.23+32.123535...........&space;\:&space;=3180.23+x$
Therefore $100x-x&space;=3180.23$
i.e. $99x=&space;\frac{318023}{100}$ , which gives $x=\frac{318023}{9900}$
Hence , $32.12\bar{35}$ =$\frac{318023}{9900}$.
#### Bina singh
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# Integers Class 7 Maths Formulas
For those looking for help on Integers Class 7 Math Concepts can find all of them here provided in a comprehensive manner. To make it easy for you we have jotted the Class 7 Integers Maths Formulae List all at one place. You can find Formulas for all the topics lying within the Integers Class 7 Integers in detail and get a good grip on them. Revise the entire concepts in a smart way taking help of the Maths Formulas for Class 7 Integers.
## Maths Formulas for Class 7 Integers
The List of Important Formulas for Class 7 Integers is provided on this page. We have everything covered right from basic to advanced concepts in Integers. Make the most out of the Maths Formulas for Class 7 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Integers Class 7 covering numerous concepts and use them to solve your Problems effortlessly.
Representation of integers on the number line.
Integers are closed under addition. In general, for any two integers a and b, a + b is an integer.
Integers are closed under subtraction. Thus, if a and b are two integers then a – b is also an integer.
Addition is commutative for integers. In general, for any two integers a and b, we can say a + b = b + a
Subtraction is not commutative for integers.
Addition is associative for integers.
In general for any integers a, b and c, we can say a + (b + c) = (a + b) + c
Zero is an additive identity for integers. In general, for any integer a
a + 0 = a = 0 + a
While multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (-) before the product. We thus get a negative integer. In general, for any two positive integers a and b we can say a × (-b) = (-a) × b = -(a × b)
Product of two negative integers is a positive integer. We multiply the two negative integers as whole numbers and put positive sign before the product. In general, for any two positive integers a and b, (-a) × (-b) = a × b
Integers are closed under multiplication. a × b is an integer, for all integers a and b,
Multiplication is commutative for integers. In general, for any two integers a and b, a × b = b × a
The product of a negative integer and zero is zero a × 0 = 0 × a=0
1 is the multiplicative identity for integers.
a × 1 = 1 × a = a
Multiplication is associative for integers, (a × b) × c = a × (b × c)
The distributivity of multiplication over addition is true for integers.
a × (b + c) = a × b + a × c
The distributivity of multiplication over subtraction is true for integers.
a × (b – c) = a × b – a × c
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (-) before the quotient.
a ÷ (-b) = (-a) ÷ b where b ≠ 0
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).
(-a) ÷ (-b) = a ÷ b where b ≠ 0
Any integer divided by 1 gives the same number.
a ÷ 1 = a
For any integer a, we have a ÷ 0 is not defined.
Natural numbers, whole numbers and integers: The numbers 1, 2, 3,……… which we use for counting are known as natural numbers. The natural numbers along with zero forms the collection of whole numbers.
The numbers……., -3, -2, -1, 0, 1, 2, 3, form the collection of integers.
Integers Whole numbers 1. The integers form a bigger group which contains whole numbers and negative numbers. 1. The whole numbers do not form a group as big as integers because they do not contain negative numbers. 2. The group of integers includes all the whole numbers. 2. The group of whole numbers does not include all the integers. 3. There is no smallest integer. 3. 0 is the smallest whole number. 4. Integers are closed under subtraction. 4. Whole numbers are not closed under subtraction.
In this chapter, we shall learn more about integers, their properties and operations.
Properties of Addition and Subtraction of Integers
Closure Under Addition
We know that the addition of two whole numbers is again a whole number. For example, 17 + 24 = 41 which is a whole number. This property is known as the closure property for the addition of whole numbers.
This property is true for integers also, i.e., the sum of two integers is always an integer. We cannot find a pair of integers whose addition is not an integer. Since additions of integers give integers, we can say integers are closed under’addition just like whole numbers. In general, for any two integers a and b, a + b is also an integer.
Closure Under Subtraction
If we subtract two integers, then their difference is also an integer. We cannot find any pair of integers whose difference is not an integer. Since subtraction of integers gives integers, we can say integers are closed under subtraction. In general, for any two integers a and b, a – b is also an integer.
Note: The whole numbers do not satisfy this property.
For example: 5 – 7 = -2 which is not a whole number.
Commutative Property
Commutativity of Addition: We know that 3+ 5 = 5 + 3 = 8, that is, the whole numbers can be added in any order. In other words, addition is commutative for whole numbers. Similarly, the addition is commutative for integers.
We cannot find any pair of integers for which the sum is different when the order is changed. So, we conclude that addition is commutative for integers also. In general, for any two integers a and b, we can say that a + b = b + a.
Commutativity of Subtraction: We know that the subtraction is not commutative for whole numbers.
For example, 10 – 20 = -10 and 20 – 10 = 10
So, 10 – 20 ≠ 20 – 10
Similarly, the subtraction is not commutative for integers.
Associative Property
We cannot find any example for which sum is different when the order of addition is changed. This shows that addition is associative for integers.
In general, for any integers a, b and c, we can say that a + (b + c) = (a + b) + c
Additive Identity
When we add zero to any whole number {i.e., zero and positive integer), we get the same whole number. So, zero is an additive identity for whole numbers. In particular, we can say that zero is an additive identity for positive integers.
Consider the following examples:
(-8) + 0 = -8
(-23) + 0 = -23
0 + (-37) = -37
0 + (-59) = -59
0 + (-43) = -43
-61 + 0 = -61
-50 + 0 = -50
These examples show that zero is an additive identity for negative integers also. Thus, we can say that zero is an additive identity for integers. In general, for any integer a, a + 0 = a = 0 + a
Product of Three or More Negative Integers
We find that if the number of negative integers in a product is even, the product is a positive integer; if the number of negative integers in a product is odd, the product is a negative integer.
Properties of Multiplication of Integers
Closure Under Multiplication
Closure: Let us observe the following table:
We observe that the product of two integers is an integer. We cannot find a pair of integers whose product is not an integer. This gives an idea that the product of two integers is again an integer. So, we say that integers are closed under multiplication. In general, a × b is an integer, for all integers a and b.
Commutativity of Multiplication
We know that multiplication is commutative for whole numbers (i.e., zero and positive
integers). Now, let us observe the following table:
We observe that two integers can be multiplied in any order. The above examples suggest commutativity of multiplication of integers. So, in general, we can say that for any two integers a and b, a × b = b × a.
Multiplication by Zero
We know that any whole number [i.e., zero and positive integers] multiplied by zero gives zero. Let us observe the following table showing the product of a negative integer and zero.
(-3) × 0 = 0
0 × (-4) = 0
(-5) × 0 = 0
0 × (-6) = 0
This table shows that the product of a negative integer and zero is again zero.
In general, for any integer a, a × 0 = 0 × a = 0
Multiplicative Identity
We know that 1 is the multiplicative identity for whole numbers (i.e., zero and positive integers). Let us observe the following table showing the product of a negative integer and 1.
(-3) × 1 = -3
(-4) × 1 = -4
1 × (-5) = -5
1 × (-6) = -6
This table shows that 1 is the multiplicative identity for negative integers also. In general, for any integer a, we have,
a × 1 = 1 × a = a
Multiplication with (-1): Let us observe the following table showing the product of an integer and (-1).
(-3) × (-1) = 3
3 × (-1) = – 3
(-6) × (-1) = 6
(-1) × 13 = -13
(-1) × (-25) = 25
18 × (-1) = -18.
This table shows that (-1) is not the multiplicative identity for integers because when we multiply an integer with (-1) or (-1) with an integer, the result is the integer with the sign changed, i.e., we do not get the same integer. Therefore, for any integer a, we have, a × (-1) = (-1) × a = -a ≠ a
Note: 0 is the additive identity whereas 1 is the multiplicative identity for integers. We get additive inverse of an integer a when we multiply (-1) to a,
i.e., a × (-1) = (-1) × a = -a.
Associativity for Multiplication
Take the integer (- 3). Multiply it with (- 2) to get 6, i.e., (-3) × (-2) = 6.
Then, multiply the product 6 with 5 to get 30, i.e., [(-3) × (-2)] × 5 = 6 × 5 = 30.
Also, (-2) × 5 = (-10).
Multiply integer (-3) with (-10) to get 30.
i.e., (-3) × [(-2) × 5] = (-3) × (-10) = 30.
So, we get the same answer in both the processes, i.e., we get [(-3) × (-2)] × 5 = (-3) × [(-2) × 5]
We observe that the arrangement of integers does not affect the product of integers.
In general, for any three integers a, b and c, (a × b) × c = a × (b × c)
Thus, like whole numbers, the product of three integers does not depend upon the arrangement of integers and this is called associative property for multiplication of integers.
Distributive Property
(i) Distributivity of Multiplication Over Addition: We know that the property of distributivity of multiplication over addition is true for whole numbers.
For example: 16 × (10 + 2) = (16 × 10) + (16 × 2).
(ii) Distributivity of Multiplication Over Subtraction: We know that the property of distributivity of multiplication over subtraction is true for whole numbers (i.e. zero and positive integers).
For example: 4 × (3 – 8) = 4 × 3 – 4 × 8
This property is also true for integers.
For example:
(-9) × [10-(-3)] = (-9) × 13 = -117
and, -9 × 10 – (-9) × (-3) = -90 – 27 = -117
So, (-9) × [10-(-3)]=(-9) × 10 – (-9) × (-3).
We find that these are also equal.
In general, for any three integers a, b and c, a × (b – c) = a × b – a × c.
Division of Integers
1. The division is the inverse operation of multiplication.
Observing the entries in the above table, we find that
• When we divide a negative integer by a positive integer, we get a negative integer.
• When we divide a positive integer by a negative integer, we get a negative integer.
• When we divide a negative integer by a negative integer, we get a positive integer.
2. Division of a negative integer by a positive integer
We observe that
(-12) ÷ 6 = -2 = -(12 ÷ 6)
(-32) ÷ 4 = -8 = -(32 ÷ 4)
(-45) ÷ 5 = -9 = -(45 ÷ 5)
(-12) ÷ 2 = -6 = -(12 ÷ 2)
(- 20) ÷ 5 = -4 = -(20 ÷ 5)
So, we find that while dividing a negative integer by a positive integer, we divide them as whole numbers and put a minus sign (-) before the quotient (i.e. we get a negative integer).
3. Division of a positive integer by a negative integer
We also observe that
72 ÷ (- 8) = -9 = – (72 ÷ 8)
21 ÷ 7 = -3 = -(27 ÷ 7)
This shows that while dividing a positive integer by a negative integer, we divide them as whole numbers and put a minus sign (-) before the quotient (i.e., we get a negative integer).
4. If the dividend and divisor are of opposite sign, then the quotient is negative integer.
Wehave, (—48) ÷ 8= -(48 ÷ 8) = -6
(48) ÷ (-8) = -(48÷8) = -6
So, (-48) ÷ 8 = -6 = 48 ÷ (-8)
5. Division of a negative integer by a negative integer
Lastly we observe that
(-20) ÷ (-4) = 5 = 20 ÷ 4
(-12) ÷ (-6) = 2 = 12 ÷ 6
(-32) ÷ (-8) = 4 = 32 ÷ 8
(-45) ÷ (-9) = 5 = 45 ÷ 9
Here, we notice that while dividing a negative integer by a negative integer, we divide them as whole numbers and put a positive sign i.e. we get a positive integer. We can say that if dividend and divisor are of same signs, then the quotient is a positive integer.
Properties of Division of Integers
(i) Closure: We know that integers are closed under addition, subtraction and multiplication. However, the integers are not closed under division. It can be observed from the following table:
(ii) Commutativity: We know that division is not commutative for whole numbers. For example 16 ÷ 4 ≠ 4 ÷ 16.
Similarly, the division is not commutative for integers.
Note: The division is commutative for integers when the dividend and divisor are equal.
(iii) Like whole numbers, any integer divided by zero is meaningless and zero divided by any integer (other than zero) is equal to zero, i.e., for any integer a, a + 0 is not defined but 0 ÷ a (≠0) = 0.
(iv) When we divide a whole number (i.e., zero and positive integers) by 1, it gives the same whole number.
It is true for negative integers also. For example:
(-8) ÷ 1 = -8
(-11) ÷ 1 = -11
These examples show that negative integer divided by one gives the same negative integer. So, any integer divided by 1 gives the same integer. In general, we can say that for any integer a, a ÷ 1 = a. |
## Section10.4Vertex Form
The graphing form of the equation of a parabola is
\begin{equation*} y=a(x-h)^2+k \end{equation*}
where $a\text{,}$ $h\text{,}$ and $k$ are all real numbers, $a \ne 0\text{.}$
When written in graphing form, we can infer the vertex of the parabola directly from the equation, it is the point $(h,k)\text{.}$ This fact is actually somewhat intuitive. No matter what values you choose for $x\text{,}$ the smallest value you'll ever get for the expression $(x-h)^2$ is zero, and that will occur when $x=h\text{.}$ So if $a$ is positive, the expression $a(x-h)^2+k$ will have its least value when $x=h$ and if the value of $a$ is negative, the expression $a(x-h)^2+k$ will have its greatest value when $x=h\text{.}$ So for any non-zero value of $a\text{,}$ the parabola will have either its lowest point or its highest point when $x=h\text{.}$ That low or high point is precisely the vertex. The fact that the $y$-coordinate has a value of $k$ when $x=h$ comes directly from substitution.
\begin{align*} y\amp=a(\highlight{h}-h)^2+k\\ \amp=a(0)^2+k\\ \amp=k \end{align*}
For example, given the equation
\begin{equation*} y=\frac{1}{2}(x-3)^2-7 \end{equation*}
we know immediately that the vertex of the parabola is the point $(3,-7)\text{.}$ Sometimes folks get a little uncertain about the sign of the value $h\text{.}$ If you start second guessing yourself, try to remember that $h$ is the value that causes $(x-h)^2$ to equal zero. In the last equation that gives us the following.
Direct substitution gives us the $y$-coordinate when $x=3\text{.}$
\begin{align*} y\amp=\frac{1}{2}(\highlight{3}-3)^2-7\\ \amp=\frac{1}{2}(0)^2-7\\ \amp=-7 \end{align*}
The other information about the parabola is determined in the same way that it was determined when we worked with the standard form of the equation. To summarize:
When presented with an equation of form $y=a(x-h)^2+k\text{,}$ $a \ne 0\text{,}$ the equation graphs to a parabola with the following properties.
• The vertex of the parabola is the point $(h,k)\text{.}$
• The axis of symmetry for the parabola is the vertical line with equation $x=h\text{.}$
• The parabola is concave up (opens upward) when $a \gt 0$ and the parabola is concave down (opens downward) when $a \lt 0\text{.}$
• If there are $x$-intercept(s) on the parabola, the $x$-coordinates of those (or that) point(s) are the real number solutions to the equation $a(x-h)^2+k=0\text{.}$ When that equation has no real number solutions, the parabola has no $x$-intercepts.
• The $y$-coordinate of the $y$-intercept is determined by replacing $x$ with zero in the equation $y=a(x-h)^2+k\text{.}$
###### Example10.4.2.
Determine the vertex, axis of symmetry, concavity, and all intercepts for the parabola with equation
\begin{equation*} y=-2(x+6)^2+8\text{.} \end{equation*}
Solution
he vertex is the point $(-6,8)$ and the axis of symmetry is the vertical line with equation $x=-6\text{.}$ Note that $(x+6)^2$ is equivalent to $(x-(\highlight{-6}))\text{,}$ which is another way of seeing that the $x$-coordinate of the vertex is $-6\text{.}$
Because the value of $a\text{,}$ $-2\text{,}$ is negative, we know that the parabola is concave down.
To determine the $x$-coordinates of the $x$-intercepts, we need to solve the equation $-2(x+6)^2+8=0\text{.}$ Because of the form of this equation, by far the fastest route to solution is the square root method.
\begin{align*} -2(x+6)^2+8\amp=0\\ -2(x+6)^2+8\subtractright{8}\amp=0\subtractright{8}\\ -2(x+6)^2\amp=-8\\ \divideunder{-2(x+6)^2}{-2}\amp=\divideunder{-8}{-2}\\ (x+6)^2\amp=4\\ x+6\amp=\pm\sqrt{4}\\ x+6\amp=\pm 2 \end{align*}
\begin{align*} x+6\amp=-2 \amp\amp\text{ or }\amp x+6\amp=2\\ x+6\subtractright{6}\amp=-2\subtractright{6} \amp\amp\text{ or }\amp x+6\subtractright{6}\amp=2\subtractright{6}\\ x\amp=-8 \amp\amp\text{ or }\amp x\amp=-4 \end{align*}
The $x$-intercepts on the parabola are $(-8,0)$ and $(-4,0)\text{.}$ Note that the parabola having its an axis of symmetry with equation $x=-6$ is consistent with these intercepts — their $x$-coordinates are equidistant from $x=-6\text{.}$
The value of the $y$-coordinate of the $y$-intercept is determined as follows.
\begin{align*} y\amp=-2(\highlight{0}+6)^2+8\\ y\amp=-2 \cdot 36 +8\\ y\amp=-72+8\\ y\amp=-64 \end{align*}
The $y$-intercept of the parabola is $(0,-64)\text{.}$
###### Example10.4.3.
Determine the vertex, axis of symmetry, concavity, and all intercepts for the parabola with equation
\begin{equation*} y=\frac{2}{3}(x-4)^2+1\text{.} \end{equation*}
Solution
The vertex is the point $(4,1)$ and the axis of symmetry is the vertical line with equation $x=4\text{.}$
Because the value of $a\text{,}$ $\frac{2}{3}\text{,}$ is positive, the parabola is concave up.
To determine the $x$-coordinates of the $x$-intercepts, we need to solve the equation $\frac{2}{3}(x-4)^2+1=0\text{.}$ Because of the form of this equation, by far the fastest route to solution is the square root method.
\begin{align*} \frac{2}{3}(x-4)^2+1\amp=0\\ \frac{2}{3}(x-4)^2+1\subtractright{1}\amp=0\subtractright{1}\\ \frac{2}{3}(x-4)^2\amp=-1\\ \multiplyleft{\frac{3}{2}}\frac{2}{3}(x-4)^2\amp=\multiplyleft{\frac{3}{2}}-1\\ (x-4)^2\amp=-\frac{3}{2}\\ x-4\amp=\pm\sqrt{-\frac{3}{2}} \end{align*}
Because $\sqrt{-\frac{3}{2}}$ is not a real number, the parabola has no $x$-intercepts. Note that this is consistent with the fact that the parabola opens upward and the vertex of the parabola is above the $x$-axis.
The value of the $y$-coordinate of the $y$-intercept is determined as follows.
\begin{align*} y\amp=\frac{2}{3}(\highlight{0}-4)^2+1\\ y\amp=\frac{2}{3}(16)-4\\ y\amp=\frac{32}{3}-\frac{12}{3}\\ y\amp=\frac{20}{3} \end{align*}
The $y$-intercept of the parabola is $\left(0,\frac{20}{3}\right)\text{.}$
###### Example10.4.4.
Determine an equation that would produce the parabola shown in Figure 10.4.5. State the equation in both graphing form and standard form.
Solution
From the vertex of the parabola, $(-2,6)\text{,}$ we know that the graphing form of the equation of the parabola is
\begin{equation*} y=a(x+2)^2+6 \end{equation*}
for some unknown value of $a\text{.}$ We can use the ordered pair $(2,-6)$ to determine the value of $a\text{.}$ Note that the presence of the point $(2,-6)$ on the parabola tells us that the value of $y$ is $-6$ when the value of $x$ is $2\text{.}$
\begin{align*} \highlightr{y}\amp=a(\highlight{x}+2)^2+6\\ \highlightr{-6}\amp=a(\highlight{2}+2)^2+6\\ -6\amp=16a+6\\ -6\subtractright{6}\amp=16a+6\subtractright{6}\\ -12\amp=16a\\ \divideunder{-12}{16}\amp=\divideunder{16a}{16}\\ -\frac{3}{4}\amp=a \end{align*}
So we now know that the graphing form of the the equation for the parabola is
\begin{equation*} y=\frac{3}{4}(x+2)^2+6\text{.} \end{equation*}
We need to expand the right side of the equation to determine the standard form of the equation.
\begin{align*} y\amp=\frac{3}{4}(x+2)^2+6\\ y\amp=\frac{3}{4}(x+2)(x+2)+6\\ y\amp=\frac{3}{4}(x^2+4x+4)+6\\ y\amp=\frac{3}{4}x^2+3x+3+6\\ y\amp=\frac{3}{4}x^2+3x+9 \end{align*}
The standard form of the equation for the parabola is
\begin{equation*} y=\frac{3}{4}x^2+3x+9\text{.} \end{equation*}
### ExercisesExercises
For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.
###### 1.
$y=2(x-3)^2-12$
Solution
The equation $y=2(x-3)^2-12$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=2\text{,}$ $h=3\text{,}$ and $k=-12\text{.}$ This tells us that the vertex of the parabola is $(3,-12)$ and that the axis of symmetry is $x=3\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.
\begin{align*} y\amp=2(\highlight{0}-3)^2-12\\ \amp=2(-3)^2-12\\ \amp=6 \end{align*}
So the $y$-intercept is $(0,6)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.
So, the $x$-intercepts are $(3-\sqrt{6},0)$ and $(3-\sqrt{6},0)\text{.}$
###### 2.
$y=-(x+5)^2$
Solution
The equation $y=-(x+5)^2$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=-1\text{,}$ $h=-5\text{,}$ and $k=0\text{.}$ This tells us that the vertex of the parabola is $(-5,0)$ and that the axis of symmetry is $x=-5\text{.}$ Because $a$ is negative, we also know that the parabola is concave down. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.
\begin{align*} y\amp=-(\highlight{0}+5)^2\\ \amp=-(5)^2\\ \amp=-25 \end{align*}
So the $y$-intercept is $(0,-25)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.
\begin{align*} -(x+5)^2\amp=0\\ x+5\amp=0\\ x+5\subtractright{5}\amp=0\subtractright{5}\\ x\amp=-5 \end{align*}
So, the only $x$-intercept is $(-5,0)\text{.}$
###### 3.
$y=-4x^2+3$
Solution
The equation $y=-4x^2+3$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=-4\text{,}$ $h=0\text{,}$ and $k=3\text{.}$ This tells us that the vertex of the parabola is $(0,3)$ and that the axis of symmetry is $x=0$ (i.e. the $y$-axis). Because $a$ is negative, we also know that the parabola is concave down. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.
\begin{align*} y\amp=-4 \cdot (\highlight{0})^2+3\\ \amp=3 \end{align*}
So the $y$-intercept is $(0,3)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.
\begin{align*} -4x^2+3\amp=0\\ -4x^2+3\subtractright{3}\amp=0\subtractright{3}\\ -4x^2\amp=-3\\ \divideunder{-4x^2}{-4}\amp=\divideunder{-3}{-4}\\ x^2\amp=\frac{3}{4}\\ x\amp=\pm\sqrt{\frac{3}{4}}\\ x\amp=\pm\frac{\sqrt{3}}{2} \end{align*}
So, the $x$-intercepts are $\left(-\frac{\sqrt{3}}{2},0\right)$ and $\left(\frac{\sqrt{3}}{2},0\right)\text{.}$
###### 4.
$y=3(x-1)^2+4$
Solution
The equation $y=3(x-1)^2+4$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=3\text{,}$ $h=1\text{,}$ and $k=4\text{.}$ This tells us that the vertex of the parabola is $(1,4)$ and that the axis of symmetry is $x=1\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.
\begin{align*} y\amp=3(\highlight{0}-1)^2+4\\ \amp=3(-1)^2+4\\ \amp=7 \end{align*}
So the $y$-intercept is $(0,7)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.
\begin{align*} 3(x-1)^2+4\amp=0\\ 3(x-1)^2+4\subtractright{4}\amp=0\subtractright{4}\\ 3(x-1)^2\amp=-4\\ \divideunder{3(x-1)^2}{3}\amp=\divideunder{-4}{3}\\ (x-1)^2\amp=-\frac{4}{3}\\ x-1=\pm\sqrt{-\frac{4}{3}} \end{align*}
Because the square roots of negative numbers are not real numbers, this parabola has no $x$-intercepts. This is concistent with the fact that the parabola opens upward and that its vertex is already above the $x$-axis.
###### 5.
$y=5(x+6)^2-44$
Solution
The equation $y=5(x+6)^2-44$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=5\text{,}$ $h=-6\text{,}$ and $k=-44\text{.}$ This tells us that the vertex of the parabola is $(-6,-44)$ and that the axis of symmetry is $x=-6\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.
\begin{align*} y\amp=5(\highlight{0}+6)^2-44\\ \amp=5(6)^2-44\\ \amp=136 \end{align*}
So the $y$-intercept is $(0,136)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.
\begin{align*} 5(x+6)^2-44\amp=0\\ 5(x+6)^2-44\addright{44}\amp=0\addright{44}\\ 5(x+6)^2\amp=44\\ \divideunder{5(x+6)^2}{5}\amp=\divideunder{44}{5}\\ (x+6)^2\amp=\frac{44}{5}\\ x+6\amp=\pm\sqrt{\frac{44}{5}}\\ x+6\amp=\pm\frac{\sqrt{44}}{\sqrt{5}}\\ x+6\amp=\pm\frac{\sqrt{44}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}\\ x+6\amp=\pm\frac{\sqrt{44} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{\sqrt{4 \cdot 11} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{\sqrt{4} \cdot \sqrt{11} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{2 \cdot \sqrt{11} \sqrt{5}}{5}\\ x+6\amp=\pm\frac{2\sqrt{55}}{5}\\ x+6\subtractright{6}\amp=\pm\frac{2\sqrt{55}}{5}\subtractright{6}\\ x\amp=-6\pm\frac{2\sqrt{55}}{5}\\ x\amp=\frac{-30\pm 2\sqrt{55}}{5} \end{align*}
So, the $x$-intercepts are $\left(\frac{-30-2\sqrt{55}}{5},0\right)$ and $\left(\frac{-30+2\sqrt{55}}{5},0\right)$
For each graph, determine an equation that would produce the parabola. State the equation in both graphing form and standard form.
###### 6.
Solution
The vertex of the parabola shown in Figure 10.4.6 is $(3,-6)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x-3)^2-6$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{0},\highlight{-3})$ in that equation to determine the value of $a\text{.}$
So the graphing form of the equation of the parabola is
\begin{equation*} y=\frac{1}{3}(x-3)^2-6\text{.} \end{equation*}
To get the standard for of the equation, we simply need to expand and simplify the graphing form of the equation.
\begin{align*} y\amp=\frac{1}{3}(x-3)^2-6\\ y\amp=\frac{1}{3}(x-3)(x-3)-6\\ y\amp=\frac{1}{3}(x^2-6x+9)-6\\ y\amp=\frac{1}{3}x^2-2x+3-6\\ y\amp=\frac{1}{3}x^2-2x-3 \end{align*}
So the standard for of the equation is
\begin{equation*} y=\frac{1}{3}x^2-2x-3\text{.} \end{equation*}
###### 7.
Solution
The vertex of the parabola shown in Figure 10.4.7 is $(0,5)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x-0)^2+5$ or merely $y=ax^2+5$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{2},\highlight{1})$ in that equation to determine the value of $a\text{.}$
\begin{align*} \highlight{1}\amp=(\highlightr{2})^2+5\\ 1\amp=4a+5\\ 1\subtractright{5}\amp=4a+5\subtractright{5}\\ -4\amp=4a\\ \divideunder{-4}{4}\amp=\divideunder{4a}{4}\\ -1\amp=a \end{align*}
So the graphing form of the equation of the parabola is $y=-1 \cdot x^2+5$ or simply
\begin{equation*} y=-x^2+5\text{.} \end{equation*}
We'll note that the last stated equation is also the standard form of the equation of the parabola.
###### 8.
Solution
The vertex of the parabola shown in Figure 10.4.8 is $(-2,-1)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x+2)^2-1$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{0},\highlight{3})$ in that equation to determine the value of $a\text{.}$
So the graphing form of the equation of the parabola is $y=1 \cdot (x+2)^2-1$ or simply |
# 1.6 Add and subtract fractions (Page 3/4)
Page 3 / 4
Simplify: $\frac{3a}{4}-\phantom{\rule{0.2em}{0ex}}\frac{8}{9}$ $\frac{3a}{4}·\frac{8}{9}.$
$\frac{27a-32}{36}$ $\frac{2a}{3}$
Simplify: $\frac{4k}{5}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$ $\frac{4k}{5}·\frac{1}{6}.$
$\frac{24k-5}{30}$ $\frac{2k}{15}$
## Use the order of operations to simplify complex fractions
We have seen that a complex fraction is a fraction in which the numerator or denominator contains a fraction. The fraction bar indicates division . We simplified the complex fraction $\frac{\frac{3}{4}}{\frac{5}{8}}$ by dividing $\frac{3}{4}$ by $\frac{5}{8}.$
Now we’ll look at complex fractions where the numerator or denominator contains an expression that can be simplified. So we first must completely simplify the numerator and denominator separately using the order of operations. Then we divide the numerator by the denominator.
## How to simplify complex fractions
Simplify: $\frac{{\left(\frac{1}{2}\right)}^{2}}{4+{3}^{2}}.$
## Solution
Simplify: $\frac{{\left(\frac{1}{3}\right)}^{2}}{{2}^{3}+2}.$
$\frac{1}{90}$
Simplify: $\frac{1+{4}^{2}}{{\left(\frac{1}{4}\right)}^{2}}.$
$272$
## Simplify complex fractions.
1. Simplify the numerator.
2. Simplify the denominator.
3. Divide the numerator by the denominator. Simplify if possible.
Simplify: $\frac{\frac{1}{2}+\frac{2}{3}}{\frac{3}{4}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}}.$
## Solution
It may help to put parentheses around the numerator and the denominator.
$\begin{array}{cccccc}& & & & & \frac{\left(\frac{1}{2}+\frac{2}{3}\right)}{\left(\frac{3}{4}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}\right)}\hfill \\ \\ \\ \begin{array}{c}\text{Simplify the numerator (LCD = 6)}\hfill \\ \text{and simplify the denominator (LCD = 12).}\hfill \end{array}\hfill & & & & & \hfill \frac{\left(\frac{3}{6}+\frac{4}{6}\right)}{\left(\frac{9}{12}-\phantom{\rule{0.2em}{0ex}}\frac{2}{12}\right)}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & & & \hfill \frac{\left(\frac{7}{6}\right)}{\left(\frac{7}{12}\right)}\hfill \\ \\ \\ \text{Divide the numerator by the denominator.}\hfill & & & & & \hfill \frac{7}{6}÷\frac{7}{12}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & & & \hfill \frac{7}{6}·\frac{12}{7}\hfill \\ \\ \\ \text{Divide out common factors.}\hfill & & & & & \hfill \frac{7·6·2}{6·7}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & & & \hfill 2\hfill \end{array}$
Simplify: $\frac{\frac{1}{3}+\frac{1}{2}}{\frac{3}{4}-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}}.$
2
Simplify: $\frac{\frac{2}{3}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}}{\frac{1}{4}+\frac{1}{3}}.$
$\frac{2}{7}$
## Evaluate variable expressions with fractions
We have evaluated expressions before, but now we can evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.
Evaluate $x+\frac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}$ $x=-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}.$
1. To evaluate $x+\frac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{1}{3},$ substitute $-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}$ for $x$ in the expression.
Simplify. $\phantom{\rule{18em}{0ex}}$ 0
2. To evaluate $x+\frac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{3}{4},$ we substitute $-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$ for x in the expression.
Rewrite as equivalent fractions with the LCD, 12. Simplify. Add. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
Evaluate $x+\frac{3}{4}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{7}{4}$ $x=-\phantom{\rule{0.2em}{0ex}}\frac{5}{4}.$
$-1$ $-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
Evaluate $y+\frac{1}{2}$ when $y=\frac{2}{3}$ $y=-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}.$
$\frac{7}{6}$ $-\phantom{\rule{0.2em}{0ex}}\frac{1}{12}$
Evaluate $-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}-y$ when $y=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}.$
## Solution
Rewrite as equivalent fractions with the LCD, 6. Subtract. Simplify. $-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$
Evaluate $-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}-y$ when $y=-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}.$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$
Evaluate $-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}-y$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{5}{2}.$
$-\phantom{\rule{0.2em}{0ex}}\frac{17}{8}$
Evaluate $2{x}^{2}y$ when $x=\frac{1}{4}$ and $y=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}.$
## Solution
Substitute the values into the expression.
$2{x}^{2}y$ Simplify exponents first. $2\left(\frac{1}{16}\right)\left(-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}\right)$ Multiply. Divide out the common factors. Notice we write 16 as $2\cdot 2\cdot 4$ to make it easy to remove common factors. $-\phantom{\rule{0.2em}{0ex}}\frac{\overline{)2}\cdot 1\cdot \overline{)2}}{\overline{)2}\cdot \overline{)2}\cdot 4\cdot 3}$ Simplify. $-\phantom{\rule{0.2em}{0ex}}\frac{1}{12}$
Evaluate $3a{b}^{2}$ when $a=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$ and $b=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}.$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
Evaluate $4{c}^{3}d$ when $c=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$ and $d=-\phantom{\rule{0.2em}{0ex}}\frac{4}{3}.$
$\frac{2}{3}$
The next example will have only variables, no constants.
Evaluate $\frac{p+q}{r}$ when $p=-4,q=-2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=8.$
## Solution
To evaluate $\frac{p+q}{r}$ when $p=-4,q=-2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=8,$ we substitute the values into the expression.
$\frac{p+q}{r}$ Add in the numerator first. $\frac{-6}{8}$ Simplify. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$
Evaluate $\frac{a+b}{c}$ when $a=-8,b=-7,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c=6.$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{2}$
Evaluate $\frac{x+y}{z}$ when $x=9,y=-18,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=-6.$
$\frac{3}{2}$
## Key concepts
• Fraction Addition and Subtraction: If $a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$ are numbers where $c\ne 0,$ then
$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ and $\frac{a}{c}-\phantom{\rule{0.2em}{0ex}}\frac{b}{c}=\frac{a-b}{c}.$
To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.
• Strategy for Adding or Subtracting Fractions
1. Do they have a common denominator?
Yes—go to step 2.
No—Rewrite each fraction with the LCD (Least Common Denominator). Find the LCD. Change each fraction into an equivalent fraction with the LCD as its denominator.
2. Add or subtract the fractions.
3. Simplify, if possible. To multiply or divide fractions, an LCD IS NOT needed. To add or subtract fractions, an LCD IS needed.
• Simplify Complex Fractions
1. Simplify the numerator.
2. Simplify the denominator.
3. Divide the numerator by the denominator. Simplify if possible.
## Practice makes perfect
Add and Subtract Fractions with a Common Denominator
$\frac{6}{13}+\frac{5}{13}$
$\frac{11}{13}$
$\frac{4}{15}+\frac{7}{15}$
$\frac{x}{4}+\frac{3}{4}$
$\frac{x+3}{4}$
$\frac{8}{q}+\frac{6}{q}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{16}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{7}{16}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{8}$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{16}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{9}{16}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{8}{17}+\frac{15}{17}$
$\frac{7}{17}$
$-\phantom{\rule{0.2em}{0ex}}\frac{9}{19}+\frac{17}{19}$
$\frac{6}{13}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{10}{13}\right)+\left(-\phantom{\rule{0.2em}{0ex}}\frac{12}{13}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{16}{13}$
$\frac{5}{12}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{7}{12}\right)+\left(-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\right)$
In the following exercises, subtract.
$\frac{11}{15}-\phantom{\rule{0.2em}{0ex}}\frac{7}{15}$
$\frac{4}{15}$
$\frac{9}{13}-\phantom{\rule{0.2em}{0ex}}\frac{4}{13}$
$\frac{11}{12}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
$\frac{1}{2}$
$\frac{7}{12}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
$\frac{19}{21}-\phantom{\rule{0.2em}{0ex}}\frac{4}{21}$
$\frac{5}{7}$
$\frac{17}{21}-\phantom{\rule{0.2em}{0ex}}\frac{8}{21}$
$\frac{5y}{8}-\phantom{\rule{0.2em}{0ex}}\frac{7}{8}$
$\frac{5y-7}{8}$
$\frac{11z}{13}-\phantom{\rule{0.2em}{0ex}}\frac{8}{13}$
$-\phantom{\rule{0.2em}{0ex}}\frac{23}{u}-\phantom{\rule{0.2em}{0ex}}\frac{15}{u}$
$-\phantom{\rule{0.2em}{0ex}}\frac{38}{u}$
$-\phantom{\rule{0.2em}{0ex}}\frac{29}{v}-\phantom{\rule{0.2em}{0ex}}\frac{26}{v}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}\right)$
$\frac{1}{5}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{7}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{7}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{7}{9}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{2}{9}$
$-\phantom{\rule{0.2em}{0ex}}\frac{8}{11}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{11}\right)$
Mixed Practice
In the following exercises, simplify.
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{18}·\frac{9}{10}$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{14}·\frac{7}{12}$
$\frac{n}{5}-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}$
$\frac{n-4}{5}$
$\frac{6}{11}-\phantom{\rule{0.2em}{0ex}}\frac{s}{11}$
$-\phantom{\rule{0.2em}{0ex}}\frac{7}{24}+\frac{2}{24}$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{24}$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{18}+\frac{1}{18}$
$\frac{8}{15}÷\frac{12}{5}$
$\frac{2}{9}$
$\frac{7}{12}÷\frac{9}{28}$
Add or Subtract Fractions with Different Denominators
In the following exercises, add or subtract.
$\frac{1}{2}+\frac{1}{7}$
$\frac{9}{14}$
$\frac{1}{3}+\frac{1}{8}$
$\frac{1}{3}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}\right)$
$\frac{4}{9}$
$\frac{1}{4}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}\right)$
$\frac{7}{12}+\frac{5}{8}$
$\frac{29}{24}$
$\frac{5}{12}+\frac{3}{8}$
$\frac{7}{12}-\phantom{\rule{0.2em}{0ex}}\frac{9}{16}$
$\frac{1}{48}$
$\frac{7}{16}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
$\frac{2}{3}-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}$
$\frac{7}{24}$
$\frac{5}{6}-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$
$-\phantom{\rule{0.2em}{0ex}}\frac{11}{30}+\frac{27}{40}$
$\frac{37}{120}$
$-\phantom{\rule{0.2em}{0ex}}\frac{9}{20}+\frac{17}{30}$
$-\phantom{\rule{0.2em}{0ex}}\frac{13}{30}+\frac{25}{42}$
$\frac{17}{105}$
$-\phantom{\rule{0.2em}{0ex}}\frac{23}{30}+\frac{5}{48}$
$-\phantom{\rule{0.2em}{0ex}}\frac{39}{56}-\phantom{\rule{0.2em}{0ex}}\frac{22}{35}$
$-\phantom{\rule{0.2em}{0ex}}\frac{53}{40}$
$-\phantom{\rule{0.2em}{0ex}}\frac{33}{49}-\phantom{\rule{0.2em}{0ex}}\frac{18}{35}$
$-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}\right)$
$\frac{1}{12}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}\right)$
$1+\frac{7}{8}$
$\frac{15}{8}$
$1-\phantom{\rule{0.2em}{0ex}}\frac{3}{10}$
$\frac{x}{3}+\frac{1}{4}$
$\frac{4x+3}{12}$
$\frac{y}{2}+\frac{2}{3}$
$\frac{y}{4}-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}$
$\frac{4y-12}{20}$
$\frac{x}{5}-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$
Mixed Practice
In the following exercises, simplify.
$\frac{2}{3}+\frac{1}{6}$ $\frac{2}{3}÷\frac{1}{6}$
$\frac{5}{6}$ 4
$-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}$ $-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}·\frac{1}{8}$
$\frac{5n}{6}÷\frac{8}{15}$ $\frac{5n}{6}-\phantom{\rule{0.2em}{0ex}}\frac{8}{15}$
$\frac{25n}{16}$ $\frac{25n-16}{30}$
$\frac{3a}{8}÷\frac{7}{12}$ $\frac{3a}{8}-\phantom{\rule{0.2em}{0ex}}\frac{7}{12}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}÷\left(-\phantom{\rule{0.2em}{0ex}}\frac{3}{10}\right)$
$\frac{5}{4}$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}÷\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\right)$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}+\frac{5}{12}$
$\frac{1}{24}$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}+\frac{7}{12}$
$\frac{5}{6}-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}$
$\frac{13}{18}$
$\frac{5}{9}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$
$-\phantom{\rule{0.2em}{0ex}}\frac{7}{15}-\phantom{\rule{0.2em}{0ex}}\frac{y}{4}$
$\frac{-28-15y}{60}$
$-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}-\phantom{\rule{0.2em}{0ex}}\frac{x}{11}$
$\frac{11}{12a}·\frac{9a}{16}$
$\frac{33}{64}$
$\frac{10y}{13}·\frac{8}{15y}$
Use the Order of Operations to Simplify Complex Fractions
In the following exercises, simplify.
$\frac{{2}^{3}+{4}^{2}}{{\left(\frac{2}{3}\right)}^{2}}$
54
$\frac{{3}^{3}-{3}^{2}}{{\left(\frac{3}{4}\right)}^{2}}$
$\frac{{\left(\frac{3}{5}\right)}^{2}}{{\left(\frac{3}{7}\right)}^{2}}$
$\frac{49}{25}$
$\frac{{\left(\frac{3}{4}\right)}^{2}}{{\left(\frac{5}{8}\right)}^{2}}$
$\frac{2}{\frac{1}{3}+\frac{1}{5}}$
$\frac{15}{4}$
$\frac{5}{\frac{1}{4}+\frac{1}{3}}$
$\frac{\frac{7}{8}-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}}{\frac{1}{2}+\frac{3}{8}}$
$\frac{5}{21}$
$\frac{\frac{3}{4}-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}}{\frac{1}{4}+\frac{2}{5}}$
$\frac{1}{2}+\frac{2}{3}·\frac{5}{12}$
$\frac{7}{9}$
$\frac{1}{3}+\frac{2}{5}·\frac{3}{4}$
$1-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}÷\frac{1}{10}$
$-5$
$1-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}÷\frac{1}{12}$
$\frac{2}{3}+\frac{1}{6}+\frac{3}{4}$
$\frac{19}{12}$
$\frac{2}{3}+\frac{1}{4}+\frac{3}{5}$
$\frac{3}{8}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}+\frac{3}{4}$
$\frac{23}{24}$
$\frac{2}{5}+\frac{5}{8}-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$
$12\left(\frac{9}{20}-\phantom{\rule{0.2em}{0ex}}\frac{4}{15}\right)$
$\frac{11}{5}$
$8\left(\frac{15}{16}-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}\right)$
$\frac{\frac{5}{8}+\frac{1}{6}}{\frac{19}{24}}$
1
$\frac{\frac{1}{6}+\frac{3}{10}}{\frac{14}{30}}$
$\left(\frac{5}{9}+\frac{1}{6}\right)÷\left(\frac{2}{3}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\right)$
$\frac{13}{3}$
$\left(\frac{3}{4}+\frac{1}{6}\right)÷\left(\frac{5}{8}-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}\right)$
Evaluate Variable Expressions with Fractions
In the following exercises, evaluate.
$x+\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}\right)$ when
$x=\frac{1}{3}$
$x=-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$ $-1$
$x+\left(-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\right)$ when
$x=\frac{11}{12}$
$x=\frac{3}{4}$
$x-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}$ when
$x=\frac{3}{5}$
$x=-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}$
$\frac{1}{5}$ $-1$
$x-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}$ when
$x=\frac{2}{3}$
$x=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$
$\frac{7}{10}-w$ when
$w=\frac{1}{2}$
$w=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
$\frac{1}{5}$ $\frac{6}{5}$
$\frac{5}{12}-w$ when
$w=\frac{1}{4}$
$w=-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$
$2{x}^{2}{y}^{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$ and $y=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
$-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}$
$8{u}^{2}{v}^{3}$ when $u=-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$ and $v=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
$\frac{a+b}{a-b}$ when $a=-3,b=8$
$-\phantom{\rule{0.2em}{0ex}}\frac{5}{11}$
$\frac{r-s}{r+s}$ when $r=10,s=-5$
## Everyday math
Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs $\frac{1}{2}$ yard of print fabric and $\frac{3}{8}$ yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?
$\frac{7}{8}$ yard
Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs $\frac{1}{2}$ cup of sugar for the chocolate chip cookies and $\frac{1}{4}$ of sugar for the oatmeal cookies. How much sugar does she need altogether?
## Writing exercises
Why do you need a common denominator to add or subtract fractions? Explain.
How do you find the LCD of 2 fractions?
## Self check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After looking at the checklist, do you think you are well-prepared for the next chapter? Why or why not?
4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
At 1:30 Marlon left his house to go to the beach, a distance of 5.625 miles. He rose his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 1.5 times his walking speed. Find his speed when skateboarding and when walking.
divide 3x⁴-4x³-3x-1 by x-3
how to multiply the monomial
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55\$. The number of quarters is five less than three times the number of dimes
app is wrong how can 350 be divisible by 3.
June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
Georgie
@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
Ashley
@Geogie my bad that was meant for u
Ashley
Hi everyone, I'm glad to be connected with you all. from France.
I'm getting "math processing error" on math problems. Anyone know why?
Can you all help me I don't get any of this
4^×=9
Did anyone else have trouble getting in quiz link for linear inequalities?
operation of trinomial |
# High School Math : How to find the perimeter of a right triangle
## Example Questions
### Example Question #1 : How To Find The Perimeter Of A Right Triangle
What is the perimeter of a triangle with side lengths of 5, 12, and 13?
Explanation:
To find the perimeter of a triangle you must add all of the side lengths together.
In this case our equation would look like
### Example Question #11 : Right Triangles
Three points in the xy-coordinate system form a triangle.
The points are .
What is the perimeter of the triangle?
Explanation:
Drawing points gives sides of a right triangle of 4, 5, and an unknown hypotenuse.
Using the pythagorean theorem we find that the hypotenuse is .
### Example Question #2 : How To Find The Perimeter Of A Right Triangle
Find the perimeter of the following triangle:
Explanation:
The formula for the perimeter of a right triangle is:
where is the length of a side.
Use the formulas for a a triangle to find the length of the base. The formula for a triangle is .
Our triangle is:
Plugging in our values, we get:
### Example Question #3 : How To Find The Perimeter Of A Right Triangle
Find the perimeter of the following right triangle:
Explanation:
The formula for the perimeter of a right triangle is:
where is the length of a side.
Use the formulas for a triangle to find the length of the base and height. The formula for a triangle is
Our triangle is:
Plugging in our values, we get:
### Example Question #1 : How To Find The Perimeter Of A Right Triangle
Based on the information given above, what is the perimeter of triangle ABC?
Explanation:
Consult the diagram above while reading the solution. Because of what we know about supplementary angles, we can fill in the inner values of the triangle. Angles A and B can be found by the following reductions:
A + 120 = 180; A = 60
B + 150 = 180; B = 30
Since we know A + B + C = 180 and have the values of A and B, we know:
60 + 30 + C = 180; C = 90
This gives us a 30:60:90 triangle. Now, since 17.5 is across from the 30° angle, we know that the other two sides will have to be √3 and 2 times 17.5; therefore, our perimeter will be as follows: |
# Chapters 1 Through 9 Cumulative Review Quiz
Approved & Edited by ProProfs Editorial Team
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| By Huber
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Community Contributor
Quizzes Created: 2 | Total Attempts: 171
Questions: 21 | Attempts: 70
Settings
This is a math quiz to help you review what we have learned so far this year.
• 1.
### What is 4/12 + 6/12 in simplest form?
• A.
1/6
• B.
10/24
• C.
11/12
• D.
5/6
• E.
None of these
D. 5/6
Explanation
To find the sum of 4/12 and 6/12 in simplest form, we add the numerators (4 + 6 = 10) and keep the denominator the same (12). Therefore, the sum is 10/12. To simplify this fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 2. Dividing 10 by 2 gives us 5, and dividing 12 by 2 gives us 6. Therefore, the simplified form of 10/12 is 5/6.
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• 2.
### Which is the closest estimate of 31/3 - 13/8
• A.
1
• B.
1 1/2
• C.
2 3/4
• D.
4
• E.
None of these
B. 1 1/2
Explanation
To find the closest estimate of 31/3 - 13/8, we can convert both fractions to have a common denominator. The common denominator for 3 and 8 is 24. So, we can rewrite 31/3 as 248/24 and 13/8 as 39/24. Subtracting these values, we get 248/24 - 39/24 = 209/24. Simplifying this fraction, we get 8 17/24. Since 8 is closer to 9 than to 7, the closest estimate is 8 17/24, which is approximately equal to 1 1/2.
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• 3.
### Adina has a 4-foot piece of rope and she wants to cut it into 3 1/2 -foot lengths. How many 3 1/2 -foot pieces can she make from the original rope?
• A.
3 pieces
• B.
4 pieces
• C.
5 pieces
• D.
49 pieces
• E.
None of these
B. 4 pieces
Explanation
Adina has a 4-foot piece of rope and wants to cut it into 3 1/2-foot lengths. To determine how many 3 1/2-foot pieces she can make, we need to divide the total length of the rope (4 feet) by the desired length of each piece (3 1/2 feet). When we divide 4 by 3 1/2, we get a quotient of 1 with a remainder of 1/2. This means Adina can make 1 full 3 1/2-foot piece and have 1/2 foot of rope left over. Therefore, she can make 1 full piece and an additional piece from the remaining 1/2 foot, resulting in a total of 2 pieces.
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• 4.
### Use mental math to find the value of y.7/18 = y/36
• A.
7
• B.
14
• C.
21
• D.
28
• E.
None of these
B. 14
Explanation
To find the value of y, we can set up a proportion using the given equation. Cross-multiplying, we get 7 * 36 = 18 * y. Simplifying further, we have 252 = 18y. Dividing both sides by 18, we find that y = 14. Therefore, the correct answer is 14.
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• 5.
### What is 37 out of every 100 throws expressed as a percent?
• A.
100%
• B.
74%
• C.
37%
• D.
3.7%
• E.
None of these
C. 37%
Explanation
The correct answer is 37% because out of every 100 throws, 37 of them are being referred to. To express this as a percent, we divide 37 by 100 and then multiply by 100 to get the percentage. This calculation gives us 37%.
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• 6.
### How many lines of symetry does a rectangle have?
• A.
0
• B.
1
• C.
2
• D.
4
• E.
None of these
C. 2
Explanation
A rectangle has two lines of symmetry because it can be divided into two equal halves by a vertical line passing through its center and a horizontal line passing through its center.
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• 7.
### What is the perimeter of an equilateral triangle whose sides are 16 centimeters?
• A.
19 cm
• B.
32 cm
• C.
48 cm
• D.
64 cm
• E.
None of these
C. 48 cm
Explanation
An equilateral triangle has all three sides equal in length. In this case, the sides of the triangle are given as 16 centimeters. To find the perimeter, we need to add up the lengths of all three sides. Since all three sides are equal, we can simply multiply the length of one side by 3. Therefore, the perimeter of the equilateral triangle is 16 cm x 3 = 48 cm.
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• 8.
### What is the perimeter of a square with an area of 16 square centimeters?
• A.
4 cm
• B.
8 cm
• C.
12 cm
• D.
16 cm
• E.
None of these
D. 16 cm
Explanation
The perimeter of a square is the sum of all its sides. In this case, since the area of the square is given as 16 square centimeters, we can find the length of one side by taking the square root of the area. The square root of 16 is 4, so each side of the square is 4 centimeters. Since a square has four equal sides, the perimeter of this square would be 4 + 4 + 4 + 4 = 16 cm.
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• 9.
### What is the perimeter of this figure?
• A.
38 ft
• B.
40 ft
• C.
60 ft
• D.
75 ft
• E.
None of these
C. 60 ft
• 10.
### What is the area of the figure below?
• A.
18 square inches
• B.
43 square inches
• C.
49 square inches
• D.
63 square inches
• E.
None of these
C. 49 square inches
• 11.
### What is the compliment of an angle whose measure is 32 degrees?
• A.
58 degrees
• B.
90 degrees
• C.
148 degrees
• D.
180 degrees
• E.
None of these
A. 58 degrees
Explanation
The complement of an angle is the angle that, when added to the given angle, equals 90 degrees. Since the given angle measures 32 degrees, the complement would be 90 - 32 = 58 degrees.
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• 12.
### The length of a bridge on a map is 3 inches. The actual length of the bridge is 12 miles. What is the scale of the map?
• A.
1 in : 4 mi
• B.
1 in : 8 mi
• C.
1 in : 12 mi
• D.
1 in : 24 mi
• E.
None of these
A. 1 in : 4 mi
Explanation
The scale of the map is 1 inch represents 4 miles. This means that for every inch on the map, the actual distance is 4 miles. Since the length of the bridge on the map is 3 inches, the actual length of the bridge is 3 inches x 4 miles/inch = 12 miles.
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• 13.
### What is the area of a circle whose radius is 3 meters? Use 3.14 for pi
• A.
9.42 square meters
• B.
18.84 square meters
• C.
28.26 square meters
• D.
56.52 square meters
• E.
None of these
C. 28.26 square meters
Explanation
The area of a circle can be calculated using the formula A = πr^2, where A is the area and r is the radius. In this case, the radius is given as 3 meters. Plugging this value into the formula, we get A = 3.14 * 3^2 = 3.14 * 9 = 28.26 square meters.
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• 14.
### What is the name of a polygon with six sides?
• A.
Sixagon
• B.
Decagon
• C.
Hexagon
• D.
Octagon
• E.
None of these
C. Hexagon
Explanation
A polygon with six sides is called a hexagon.
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• 15.
### Write 2/5 as a percent and as a decimal.
• A.
20%, 0.2
• B.
25%, 0.25
• C.
40%, 0.4
• D.
52%, 0.52
• E.
None of these
C. 40%, 0.4
Explanation
The correct answer is 40%, 0.4. To convert a fraction to a percent, we divide the numerator by the denominator and multiply by 100. In this case, 2 divided by 5 is 0.4, and when multiplied by 100, it becomes 40%. To convert a fraction to a decimal, we divide the numerator by the denominator. Again, 2 divided by 5 is 0.4. Therefore, the fraction 2/5 can be expressed as 40% or 0.4.
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• 16.
### What is 35% of 46?
• A.
161
• B.
16.1
• C.
156.4
• D.
16
• E.
None of these
B. 16.1
Explanation
To find 35% of 46, we can multiply 46 by 0.35. This gives us 16.1, which is the correct answer.
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• 17.
### Find the surface area of a rectangular prism with a length of 5 cm, a width of 4 cm, and a height of 3 cm.
• A.
3 cm
• B.
47 square cm
• C.
94 square cm
• D.
104 square cm
• E.
None of these
C. 94 square cm
Explanation
To find the surface area of a rectangular prism, we need to calculate the area of each face and then add them together. The formula for the surface area of a rectangular prism is 2lw + 2lh + 2wh, where l is the length, w is the width, and h is the height. Plugging in the given values, we get 2(5)(4) + 2(5)(3) + 2(4)(3) = 40 + 30 + 24 = 94 square cm. Therefore, the correct answer is 94 square cm.
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• 18.
### What is the area of a parallelogram with a height of 3.7 m and a base of 4.2 m?
• A.
12.14 square meters
• B.
15.54 square meters
• C.
16 square meters
• D.
16.25 square meters
• E.
None of these
B. 15.54 square meters
Explanation
The area of a parallelogram is calculated by multiplying the base length by the height. In this case, the base length is 4.2 m and the height is 3.7 m. Therefore, the area is 4.2 m * 3.7 m = 15.54 square meters.
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• 19.
### How many sides does a pentagon have?
five
5
Explanation
A pentagon is a polygon with five sides. Therefore, the correct answer is five or 5.
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• 20.
### Kelly baby-sits weekly and earns \$3.75 an hour. Click here to see a spreadsheet that shows a typical weekly schedule for her. What formula would you type in cell D4?
• A.
= B4 - C4
• B.
= C4 - B4
• C.
= B4 + C4
• D.
= D2 + D3
• E.
None of these
B. = C4 - B4
Explanation
The formula = C4 - B4 would be typed in cell D4. This formula calculates the difference between the value in cell C4 (the total earnings) and the value in cell B4 (the hours worked). This would give Kelly her total earnings for the week.
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• 21.
### Kelly baby-sits weekly and earns \$3.75 an hour. Click here to see a spreadsheet that shows a typical weekly schedule for her. What formula would you type in cell E6?
• A.
= E2 + E3 + E4 + E5
• B.
= A6 + B6 + C6 + D6
• C.
= (C6 - B6) x 3.75
• D.
= 4 x D6
• E.
None of these
A. = E2 + E3 + E4 + E5
Explanation
The correct formula to type in cell E6 is = E2 + E3 + E4 + E5. This formula will add up the values in cells E2, E3, E4, and E5, which likely represent the number of hours Kelly baby-sits each day of the week. By summing these values, the formula calculates the total number of hours Kelly baby-sits in a week.
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• Current Version
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# Optimization
## Optimization
Lecture notes coming soon...
• Q** Your room has a window whose height is 1.5 meters.. The bottom edge of the window is 10 cm above your eye level. How far away from the window should you stand to get the best view? ("Best view" means the largest visual angle, ie. angle between the lines of sight to the bottom and to the top of the window.)
• A** First we formulate a couple relationships: tan(alpha) = 0.1/x, and tan(alpha + theta) = 1.6/x. Then we use the multiple angle formula for tangent to make it tan(alpha + theta) = tan(alpha) + tan(theta)/ (1 - tan(alpha)tan(theta) ) = 1.6/x. Simplifying a bit, and sub in tan(alpha) = 0.1/x to get tan(theta) = 1.5x/(x^2 + 0.16). Because tan(theta) is a function of x, we define f(x) = tan(theta) = 1.5x/(x^2 + 0.16). As we know that tan(theta) is an increasing function of theta, so maximizing theta is the same as maximizing tan(theta) (THIS IS THE KEY INSIGHT TO SOLVING THIS QUESTION). Therefore all we need to do is to maximize tan(theta), or equivalently, maximize f(x). To maximize, we differentiate f'(x) = (-1.5x^2 + 0.24)/(x^2 + 0.16)^2 = 0, and solving yields x = 0.4.
Note: another method is to write theta = arctan(1.6/x) - arctan(0.1/x), and then use derivative formulas for arctan to differentiate.
• Q** Jack and Jill have an on-again off-again love affair. The sum of their love for one another is given by the function
y( t) = sin(2t) + cos(2t).
( a) Find the times when their total love is at a maximum.
(b ) Find the times when they dislike each other the most.
• A** We need to find the max and the min, so we differentiate. y' = 2cos(2t) - 2sin(2t) = 0. To solve this, we have cos(2t) = sin(2t). We need to find the angles (2t) where sine and cosine are the same. From high school (if you don't remember this, you have to review), sine and cosine are equal at angles pi/4 and 5pi/4. However, because the function oscillates, we have to find the general representation of the solution, which is pi/4 + 2n*pi, and 5pi/4 + 2n*pi, for any integer n. The reason we add this 2n*pi term is because the period of trig functions are 2pi. So we have 2t = pi/4 + 2n*pi and 5pi/4 + 2n*pi. Then, t = pi/8 + n*pi and 5pi/8 + n*pi. Checking second derivative to see that t = pi/8 + n*pi is the max and t = 5pi/8 + n*pi is the min.
• Q** A farmer wants to construct a fenced enclosure for his Alpacas. one side of the enclosure will be his existing farm house. Assuming that a rectangular geometry, what is the largest possible area for the enclosure if he only has 100 meter of fencing?
• A**we may first draw what the fencing will be like to help us understand the question.
We assume two sides of the fencing is x meter and y meter respectively. so we can write the Area as
```A= x*y
```
and from the question, we know that
2y+x=100
so now we can rewrite Area so that there's only one variable y
A=x*y=(100-2y)*y=100y-2y^2 (0<=y<=50)
Let's find the global maximum of A on this interval
First we find the critical points, so when the first derivative of A is equal to 0
A'=100-4y
So A'=0, y=25
Let's compute the area A, at the possible maxima.
y=0, A=0
y=25,A=1250
y=50,A=0
So we can see that when y=25, the area A, is maximized
At y=25, we find x=100-2y=50
Thus, the alpacas will be most comfortable in an enclosure that is 25 meter by 50 meter.
• Q** An open box is formed from a square sheet of cardboard by cutting equal squares from each corner and folding up the edges. If the dimensions of the cardboard are 18cm by 18cm, What should be the dimensions of the box so as to maximize the volume, and what is the maximum volume?
• A**
Area of the box = Area of the cardboard sheet => Area of the box = 18*18 = 324cm^2 Volume of the box = height * side ^2 (since sides are equal) = h*s^2 also 2*h + s = 18 => s = 18 - 2*h
V = h*(18-2h)^2
dV/dh = (18-2h)^2 + 2*h*(-2)*(18-2h) = 0
18-2h-4h=0 => h=3cm => s=12cm => V=432cm^3 |
How to calculate limit of function with no use of derivatives? f(x)=(x^2+12x-13)/(x-1), x approaches 1
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to calculate the value of lim x-->1 [(x^2+12x-13)/(x-1)].
We cannot substitute x = 1 directly as that yields an indeterminate form. Instead we do the following.
lim x-->1 [(x^2+12x-13)/(x-1)]
=> lim x-->1 [(x^2 +13x - x -13)/(x-1)]
=> lim x-->1 [(x(x +13) - 1(x + 13))/(x-1)]
=> lim x-->1 [(x - 1)(x +13)/(x-1)]
=> lim x-->1 [(x +13)]
substitute x = 1
=> 1 + 13
=> 14
The required value of lim x-->1 [(x^2+12x-13)/(x-1)] = 14
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
First, we'll substitute x by 1 and we'll verify if it is an indetermination:
lim (x^2+12x-13)/(x-1) = (1+12-13)/(1-1) = (13-13)/(0) = 0/0
Since we've get an indetermination, that means that x = 1 represents a root for both numerator and denominator.
We'll determine the 2nd root of the numerator, using Viete's relations:
1 + x = -12
x = -12-1
x = -13
We'll rewrite the numerator as a product of linear factors:
x^2+12x-13 = (x-1)(x+13)
We'll re-write the limit
lim (x^2+12x-13)/(x-1) = lim (x-1)(x+13)/(x-1)
We'll simplify inside limit:
lim (x-1)(x+13)/(x-1)= lim (x+13)
We'll substitute again x by 1:
lim (x+13) = 1 + 13 = 14
The limit of the function, if x approaches to 1, is:lim (x^2+12x-13)/(x-1) = 14. |
• Accueil
• Math
• Without using a calculator, find the value of the...
Without using a calculator, find the value of the following logarithmic expression.
• Réponse publiée par: maledabacuetes
(a) log2 32 = answer (b) log9 729 = answer (c) log5 5
step-by-step explanation:
• Réponse publiée par: kuanjunjunkuan
let b = log₂ 32
then 2ᵇ =32
2ᵇ = 2⁵
b = 5
• Réponse publiée par: maledabacuetes
Step-by-step explanation:
That would be
x=1
• Réponse publiée par: 20201947
the value of log7 1 is
7=1
• Réponse publiée par: shannel99
Without using a calculator, the value of the logarithmic expression is equal to 1.
Step-by-step explanation:
To answer the given question, we will be using the concept of logarithm. Here is the definition of logarithm.
What is a logarithm? In simple explanation, logarithm tells us what the exponent of a certain number (base) is, to be able to arrive at a certain value. In general, = c is the just the same as a^{c} = b.
Based on the definition of logarithm indicated above, here is the step-by-step explanation on how to find the value of the logarithmic expression :
Based on the definition of logarithm above, the given logarithmic expression in the question, , is therefore equivalent to the expression = 5, where x is the value we are looking for. To get the value of 5, we must raise the base 5 to the first power. Therefore, the answer is 1.
That is the explanation and answer in finding out the value of the logarithmic expression . If you want to read more information about the topic of logarithms, here are other links that are related to this topic.
What are exponential and logarithmic functions? Definition and some examples of logarithmic functions: What are natural logarithms?
• Réponse publiée par: tayis
Definition of Logarithms Let a, b and c be positive real numbers such that b≠ 1. The Logarithm of a with base b is denoted by log_ba and is defined as c = logьa if and only if a= b^c
The following are the reminders for logarithm
In both logarithmic and exponential forms, b is the base. In the exponential form, c is an exponent; this implies that the logarithm is actually an exponent. Hence, logarithmic and exponential functions are inverse.
In the logarithmic form log_bx, x cannot be negative.
The value of log_bx can be negative
STEPS IN SOLVING LOGARITHMIC EXPRESSION
1. Rewriting to exponential form
2. Using logarithmic properties
3. Applying the one- one property of logarithmic functions
The zero factor Property: if ab=0 then a =0 or b=0.
Read the details on how to solve logarithmic equation in
Given: log2 32 = x
Solution : 2^ x = 32
2^ x = 2^ 5
X = 5 |
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# The Class Size of a Distribution is 25 and the First Class-interval is 200-224. There Are Seven Class-intervals. (I) Write the Class-intervals. (Ii) Write the Class-marks of Each Interval. - CBSE Class 9 - Mathematics
ConceptCollection of Data
#### Question
The class size of a distribution is 25 and the first class-interval is 200-224. There are seven
class-intervals.
(i) Write the class-intervals.
(ii) Write the class-marks of each interval.
#### Solution
Given,
Class size = 25
First class interval =200-224
(i) Seven class intervals are:
200-240, 225-249, 250-274, 275-299, 300-324, 325-349, 350-374.
(ii) Class mark of 200-224=(200+224)/2=424/2=212
Class mark of 225-249=(225+249)/2=474/2=237
Class mark of 250-274=(250+274)/2=524/2=262
Class mark of 300-324=(300+324)/2=624/2=312
Class mark of 325-349=(325+349)/2=674/2=337
Class mark of 350-374=(350+374)/2=724/2=362
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 22: Tabular Representation of Statistical Data
Ex.22.10 | Q: 12 | Page no. 17
Solution The Class Size of a Distribution is 25 and the First Class-interval is 200-224. There Are Seven Class-intervals. (I) Write the Class-intervals. (Ii) Write the Class-marks of Each Interval. Concept: Collection of Data.
S |
# How do I evaluate int (arcsin x)/sqrt (1+x) dx?
##### 1 Answer
Feb 17, 2015
We will integrate this using integration by parts.
Recall that $u v - \int v \mathrm{du}$
Let $u = \arcsin \left(x\right)$ and $\mathrm{dv} = \frac{\mathrm{dx}}{\sqrt{1 + x}}$
Differentiating $u = \arcsin x$ we have
$\mathrm{du} = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$
Integrate $\mathrm{dv} = \frac{\mathrm{dx}}{\sqrt{1 + x}}$
$\int \mathrm{dv} = \int \frac{\mathrm{dx}}{\sqrt{1 + x}}$
$v = 2 \sqrt{1 + x}$
Therefore,
$\arcsin \left(x\right) \left(2 \sqrt{1 + x}\right) - \int 2 \sqrt{1 + x} \left(\frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$
Some rewriting and factor $1 - {x}^{2}$
$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{\sqrt{1 + x}}{\sqrt{\left(1 + x\right) \left(1 - x\right)}} \mathrm{dx}$
$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{\sqrt{1 + x}}{\sqrt{1 + x} \sqrt{1 - x}} \mathrm{dx}$
$2 \arcsin \left(x\right) \sqrt{1 + x} - 2 \int \frac{1}{\sqrt{1 - x}} \mathrm{dx}$
Finally, we integrate
$2 \arcsin \left(x\right) \sqrt{1 + x} + 4 \sqrt{1 - x} + C$ |
# Percentage Confusion
## PERCENTAGE COMPARISONS
With percentages, the wording is very important.
Let me illustrate this with a couple of examples:
• Fred has 10 candy bars. Sally has 8 candy bars. Fred has 25% more candy bars than Sally, but Sally has 20% fewer candy bars than Fred. In one case it’s 25%; in the other case it’s 20%. The distinction is important.
• Jenny has 65 dollars saved. Mike has 50 dollars saved. Jenny has saved 30% more money than Mike, yet we could also say that Jenny’s savings is 130% of Mike’s savings. In one case it’s 30%; in the other case it’s 130%. The wording makes all the difference.
## REVIEW OF PERCENTS
A percentage is an alternate way to express a decimal or a fraction.
100% corresponds to one unit.
So, for example, if Linda ate 50% of the donuts, this means that Linda ate half of the donuts, since 50% is half of 100%.
Here are a few more examples:
• 200% means double, since 200% is twice 100%.
• 25% is one-fourth, as 25% is a quarter of 100%.
• 150% is one and one-half, since it’s 1.5 times 100%
The purpose of this article isn’t to teach percents, but to explain the importance of how it is worded. This quick review was intended just as a brief refresher to illustrate the basic concept.
## INTERPRETING PERCENTS
Let’s look at the two original examples more closely.
(1) Fred has 10 candy bars. Sally has 8 candy bars.
When Fred compares his candy bars to Sally’s, Fred divides 10 by 8 to get 1.25, which equates to 125%.
(Recall that any decimal value can be converted into a percentage by multiplying by 100%, since 100% means one.)
Since 125% is 25% more than 100%, this means that Fred has 25% more candy bars than Sally.
However, when Sally compares her candy bars to Fred’s, Sally divides 8 by 10 to get 0.8, which equates to 80%.
Since 80% is 20% less than 100%, this means that Sally has 20% fewer candy bars than Fred.
The distinction here is that in the first case Fred used Sally’s candy bars for the comparison, so Fred divided by Sally’s number (8) to see how his compared to hers.
In the second case, Sally used Fred’s candy bars for the comparison, so Sally divided by Fred’s number (10) to see how hers compared to his.
In either case, divide by the number that you’re comparing with.
(2) Jenny has 65 dollars saved. Mike has 50 dollars saved.
This time, we’ll only compare Jenny’s savings to Mike’s savings, so we’ll definitely divide by Mike’s 50 dollars.
Therefore, we divide 65 by 50 to get 1.3, which equates to 130%.
This means that Jenny’s savings is 130% of Mike’s savings. That’s one way to put it.
There is another way to say the same thing. Since 130% is 30% more than 100%, we could instead say that Jenny has 30% more savings than Mike has.
In the second case, we used the word ‘more.’ If you follow the percentage by the word ‘more’ (or by the word ‘less’βor their synonyms, like ‘fewer’) you’re comparing the overall percentage (which we obtained by dividing the two values) to 100%.
Saying that Jenny has 130% of Mike’s savings means to multiply Mike’s savings by 1.3 (\$50 times 1.3 equals \$65).
Saying that Jenny has 30% more than Mike means to find 30% of Mike’s savings and then add that to Mike’s savings (\$15 plus \$50 equals \$65).
## COMPARISON EXAMPLES
Here are a few more examples:
• 3 bananas is 75% of 4 bananas.
• 4 bananas is 133% of 3 bananas. (Technically, it’s 133 and 1/3 percent, but I rounded.)
• 3 bananas is 25% less than 4 bananas.
• 4 bananas is 33% more than 3 bananas. (Really, 33 and 1/3 percent.)
• \$800 is 400% of \$200.
• \$800 is 300% more than \$200.
• \$200 is 25% of \$800.
• \$200 is 75% less than \$800. (Subtract 25% from 100%.)
• 90 cents is 150% of 60 cents.
• 90 cents shows a 50% improvement over 60 cents.
Copyright Β© 2014 Chris McMullen, author of the Improve Your Math Fluency Series
## 7 comments on “Percentage Confusion”
1. I just might learn something here. π
• I hope to convince several science students to admit as much tomorrow. π
2. I cannot even begin to think about this, Chris. You used to be a safe haven for me. Now, I get nervous just coming into your class. I’ll try this again tomorrow. π
• Don’t worry. There won’t be any tests. π
• I’m taking the worst pop quiz at the moment, so thanks for going easy on me. There are far too many numbers up there, Sir. They started dancing…
• Gee, I could replace them with letters, like x and y. π
• :} No, they dance even faster… |
## Insight Things
### A scientific blog revealing the hidden links which shape our world
I’ve already explained in a demonstrative way how the formula for sum squared numbers arises. Not only will I only show in this article how to calculate simple series like 1+2+3+4+…., but you will also see how we enhance our findings to tackle more complicated series like 1³+2³+3³+…. These formulas are applied in many different contexts.
### Simple Sum: 1+2+3+4+…
The most simple series (or in mathematical terms: partial sum) we could calculate is the sum of the first $n$ integers. Therefore we define three series:
$\displaystyle a_{n}=1+1+1+1+\dots=\sum\limits_{k=1}^{n}1=n$
$\displaystyle b_{n}=1+2+3+4+\dots=\sum\limits_{k=1}^{n}k$
$\displaystyle c_{n}=1^{2}+2^{2}+3^{2}+4^{2}+\dots=\sum\limits_{k=1}^{n}k^{2}$
You may wonder why $c_{n}$ is needed, but often in mathematics you need to generate information from almost nothing and this is what we do here 😉 See where this approach takes us.
$\displaystyle c_{n}-c_{n-1}=\sum\limits_{k=1}^{n}k^{2}-\sum\limits_{k=1}^{n-1}k^{2}=n^{2}$
Now we come to the point where the magic happens 😀 The right-hand sum with upper limit $n-1$ can be modified. Therefore let’s add 1 to the upper limit. To make the value of the sum stay constant, we change the term in the sum from $k$ to $k-1$ . The lower limit stays at it is, because the value for $k=1$ is zero and, in that, doesn’t affect the sum’s value.
$\displaystyle n^{2}=\sum\limits_{k=1}^{n}k^{2}-\sum\limits_{k=1}^{n}(k-1)^{2}=\sum\limits_{k=1}^{n}\big( k^{2}-(k-1)^{2}\big)$
$\displaystyle n^{2}=\sum\limits_{k=1}^{n}(2k-1)=2\sum\limits_{k=1}^{n}k-\sum\limits_{k=1}^{n}1$
It is almost obvious how to proceed. The equation can be rearranged to make the demanded series the subject. Mission complete!
$\displaystyle 2\sum\limits_{k=1}^{n}k=n^{2}+\sum\limits_{k=1}^{n}1$
$\displaystyle \sum\limits_{k=1}^{n}k=1+2+3+4+\dots=\frac{n^{2}+n}{2}=\frac{n(n+1)}{2}$
### Sum of Squares: 1²+2²+3²+4²+…
It is very important that you followed the ideas of the previous section, because – believe it or not – the shown pattern can be extended to tackle the summation of squared numbers! In addition to the series, which I already covered in the previous section, let’s define $d_{n}$.
$\displaystyle d_{n}=1^{3}+2^{3}+3^{3}+4^{3}+\dots=\sum\limits_{k=1}^{n}k^{3}$
Again we apply the magic trick involving subtraction of the highest series.
$\displaystyle d_{n}-d_{n-1}=\sum\limits_{k=1}^{n}k^{3}-\sum\limits_{k=1}^{n-1}k^{3}=n^{3}$
After shifting the limits like shown in the previous section, we can simplify.
$\displaystyle n^{3}=\sum\limits_{k=1}^{n}\Big(k^{3}-(k-1)^{3}\Big)=\sum\limits_{k=1}^{n} 3k^{2}- \sum\limits_{k=1}^{n}3k+\sum\limits_{k=1}^{n}1=3c_{n}- 3b_{n}+a_{n}$
But what’s this? We know $n^{3}$ on the left-hand side as well as $a_{n}$ and $b_{n}$ on the right-hand side! We just have to make $c_{n}$ the subject of the formula.
$\displaystyle n^{3}=3c_{n}- \frac{3n(n+1)}{2}+n$
$\displaystyle 3c_{n}=\frac{2n^{3} + 3n^{2}+n}{2}$
$\displaystyle c_{n}=1^{2}+2^{2}+3^{2}+4^{2}+\dots=\frac{n(n+1)(2n+1)}{6}$
### Sum of Cubes: 1³+2³+3³+4³+…
As expected, we will introduce a new series. We are looking for the sum of cube numbers, so the new series comprise numbers raised to the fourth power. I will skip the definition of $e_{n}$ and directly show the approach:
$\displaystyle e_{n}-e_{n-1}=\sum\limits_{k=1}^{n}\Big(k^{4}-(k-1)^{4}\Big)=4d_{n}-6c_{n}+4b_{n}-a_{n}=n^{4}$
Making $d_{n}$ the subject leads us to the solution.
$\displaystyle d_{n}=\frac{n^{4}+6c_{n}-4b_{n}+a_{n}}{4}=\frac{n^{4}+n(n+1)(2n+1)-2n(n+1)+n}{4}$
Just simplify this and receive a handy formula which allows you to add cube numbers together 😉
$\displaystyle d_{n}=1^{3}+2^{3}+3^{3}+4^{3}+.\dots=\frac{n^{4}+2n^{3}+n^{2}}{4}$
### Sum of Higher Powers
You can extend the pattern to find formulas for sums of even higher powers. Just bear in mind that you have to introduce a series (partial sum) whose summands are raised to the power you are searching for + 1. Next, set up the difference between the elements with number $n$ and $n-1$, then simplify. 😉 |
Online Data Interpretation Test - Data Interpretation Test 2
Instruction:
• Total number of questions : 20.
• Time alloted : 30 minutes.
• Each question carry 1 mark, no negative marks.
• DO NOT refresh the page.
• All the best :-).
Direction (for Q.Nos. 1 - 5):
The following pie-charts show the distribution of students of graduate and post-graduate levels in seven different institutes in a town.
1.
What is the total number of graduate and post-graduate level students is institute R?
A.
8320
B.
7916
C.
9116
D.
8099
Explanation:
Required number = (17% of 27300) + (14% of 24700) = 4641 + 3458 = 8099.
2.
What is the ratio between the number of students studying at post-graduate and graduate levels respectively from institute S?
A.
14 : 19
B.
19 : 21
C.
17 : 21
D.
19 : 14
Explanation:
Required ratio = (21% of 24700) = (21 x 24700) = 19 . (14% of 27300) 14 x 27300 14
3.
How many students of institutes of M and S are studying at graduate level?
A.
7516
B.
8463
C.
9127
D.
9404
Explanation:
Students of institute M at graduate level= 17% of 27300 = 4641.
Students of institute S at graduate level = 14% of 27300 = 3822.
Total number of students at graduate in institutes M and S = (4641 + 3822) = 8463.
4.
What is the ratio between the number of students studying at post-graduate level from institutes S and the number of students studying at graduate level from institute Q?
A.
13 : 19
B.
21 : 13
C.
13 : 8
D.
19 : 13
Explanation:
Required ratio = (21% of 24700) = (21 x 24700) = 19 . (13% of 27300) 13 x 27300 13
5.
Total number of students studying at post-graduate level from institutes N and P is
A.
5601
B.
5944
C.
6669
D.
8372
Explanation:
Required number = (15% of 24700) + (12% of 24700) = 3705 + 2964 = 6669.
Direction (for Q.Nos. 6 - 10):
The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.
Percentage of Candidates Qualified to Appeared in an Examination Over the Years
6.
The difference between the percentage of candidates qualified to appeared was maximum in which of the following pairs of years?
A.
1994 and 1995
B.
1997 and 1998
C.
1998 and 1999
D.
1999 and 2000
Explanation:
The differences between the percentages of candidates qualified to appeared for the give pairs of years are:
For 1994 and 1995 = 50 - 30 = 20.
For 1998 and 1999 = 80 - 80 = 0.
For 1994 and 1997 = 50 - 30 = 20.
For 1997 and 1998 = 80 - 50 = 30.
For 1999 and 2000 = 80 - 60 = 20.
Thus, the maximum difference is between the years 1997 and 1998.
7.
In which pair of years was the number of candidates qualified, the same?
A.
1995 and 1997
B.
1995 and 2000
C.
1998 and 1999
D.
Explanation:
The graph gives the data for the percentage of candidates qualified to appeared and unless the absolute values of number of candidates qualified or candidates appeared is know we cannot compare the absolute values for any two years.
Hence, the data is inadequate to solve this question.
8.
If the number of candidates qualified in 1998 was 21200, what was the number of candidates appeared in 1998?
A.
32000
B.
28500
C.
26500
D.
25000
Explanation:
The number of candidates appeared in 1998 be x.
Then, 80% of x = 21200 x = 21200 x 100 = 26500 (required number). 80
9.
If the total number of candidates appeared in 1996 and 1997 together was 47400, then the total number of candidates qualified in these two years together was?
A.
34700
B.
32100
C.
31500
D.
Explanation:
The total number of candidates qualified in 1996 and 1997 together, cannot be determined until we know at least, the number of candidates appeared in any one of the two years 1996 or 1997 or the percentage of candidates qualified to appeared in 1996 and 1997 together.
10.
The total number of candidates qualified in 1999 and 2000 together was 33500 and the number of candidates appeared in 1999 was 26500. What was the number of candidates in 2000?
A.
24500
B.
22000
C.
20500
D.
19000
Explanation:
The number of candidates qualified in 1999 = (80% of 26500) = 21200.
Number of candidates qualified in 2000 = (33500 - 21200) = 12300.
Let the number of candidates appeared in 2000 be x.
Then, 60% of x = 12300 x = 12300 x 100 = 20500. 60
Direction (for Q.Nos. 11 - 15):
A school has four sections A, B, C, D of Class IX students.
The results of half yearly and annual examinations are shown in the table given below.
Result No. of Students Section A Section B Section C Section D Students failed in both Exams 28 23 17 27 Students failed in half-yearlybut passed in Annual Exams 14 12 8 13 Students passed in half-yearlybut failed in Annual Exams 6 17 9 15 Students passed in both Exams 64 55 46 76
11.
If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context?
A.
Half yearly examinations were more difficult.
B.
Annual examinations were more difficult.
C.
Both the examinations had almost the same difficulty level.
D.
The two examinations cannot be compared for difficulty level.
Explanation:
Number of students who passed half-yearly exams in the school
= (Number of students passed in half-yearly but failed in annual exams)
+ (Number of students passed in both exams)
= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)
= 288.
Also, Number of students who passed annual exams in the school
= (Number of students failed in half-yearly but passed in annual exams)
+ (Number of students passed in both exams)
= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)
= 288.
Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.
Thus Statements (a), (b) and (d) are false and Statement (c) is true.
12.
How many students are there in Class IX in the school?
A.
336
B.
189
C.
335
D.
430
Explanation:
Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:
= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)
= 430.
13.
Which section has the maximum pass percentage in at least one of the two examinations?
A.
A Section
B.
B Section
C.
C Section
D.
D Section
Explanation:
Pass percentages in at least one of the two examinations for different sections are:
For Section A (14 + 6 + 64) x 100 % = 84 x 100 % = 75%. (28 + 14 + 6 + 64) 112
For Section B (12 + 17 + 55) x 100 % = 84 x 100 % = 78.5%. (23 + 12 + 17 + 55) 107
For Section C (8 + 9 + 46) x 100 % = 63 x 100 % = 78.75%. (17 + 8 + 9 + 46) 80
For Section D (13 + 15 + 76) x 100 % = 104 x 100 % = 79.39%. (27 + 13 + 15 + 76) 131
Clearly, the pass percentage is maximum for Section D.
14.
Which section has the maximum success rate in annual examination?
A.
A Section
B.
B Section
C.
C Section
D.
D Section
Explanation:
Total number of students passed in annual exams in a section
= [ (No. of students failed in half-yearly but passed in annual exams)
+ (No. of students passed in both exams)
] in that section
Success rate in annual exams in Section A
= No. of students of Section A passed in annual exams x 100 % Total number of students in Section A
= (14 + 64) x 100 % (28 + 14 + 6 + 64)
= 78 x 100 % 112
= 69.64%.
Similarly, success rate in annual exams in:
Section B (12 + 55) x 100 % = 67 x 100 % = 62.62%. (23 + 12 + 17 + 55) 107
Section C (8 + 46) x 100 % = 54 x 100 % = 67.5%. (17 + 8 + 9 + 46) 80
Section D (13 + 76) x 100 % = 89 x 100 % = 67.94%. (27 + 13 + 15 + 76) 131
Clearly, the success rate in annual examination is maximum for Section A.
15.
Which section has the minimum failure rate in half yearly examination?
A.
A section
B.
B section
C.
C section
D.
D section
Explanation:
Total number of failures in half-yearly exams in a section
= [ (Number of students failed in both exams)
+ (Number of students failed in half-yearly but passed in Annual exams)
] in that section
Failure rate in half-yearly exams in Section A
= Number of students of Section A failed in half-yearly x 100 % Total number of students in Section A
= (28 + 14) x 100 % (28 + 14 + 6 + 64)
= 42 x 100 % 112
= 37.5%.
Similarly, failure rate in half-yearly exams in:
Section B (23 + 12) x 100 % = 35 x 100 % = 32.71%. (23 + 12 + 17 + 55) 107
Section C (17 + 8) x 100 % = 25 x 100 % = 31.25%. (17 + 8 + 9 + 46) 80
Section D (27 + 13) x 100 % = 40 x 100 % = 30.53%. (27 + 13 + 15 + 76) 131
Clearly, the failure rate is minimum for Section D.
Direction (for Q.Nos. 16 - 20):
The following bar chart shows the trends of foreign direct investments(FDI) into India from all over the world.
Trends of FDI in India
16.
What was the ratio of investment in 1997 over the investment in 1992 ?
A.
5.5
B.
5.36
C.
5.64
D.
5.75
Explanation:
The 1997 figure of investment as a factor of 1992 investment = (31.36/5.70) = 5.50
17.
What was absolute difference in the FDI to India in between 1996 and 1997 ?
A.
7.29
B.
7.13
C.
8.13
D.
None of these
Explanation:
The difference in investments over 1996-1997 was
31.36 - 24.23 = € 7.13 millions.
18.
If India FDI from OPEC countries was proportionately the same in 1992 and 1997 as the total FDI from all over the world and if the FDI in 1992 from the OPEC countries was Euro 2 million. What was the amount of FDI from the OPEC countries in 1997 ?
A.
11
B.
10.72
C.
11.28
D.
11.5
Explanation:
Let x be the FDI in 1997.
Then: (2/5.7) = (x/31.36)
x = (2/5.7) x 31.36
x = 11
19.
Which year exhibited the highest growth in FDI in India over the period shown ?
A.
1993
B.
1994
C.
1995
D.
1996
Explanation:
It can be seen that the FDI in 1996 more than doubles over that of 1995. No other year is close to that rate of growth.
20.
What was India's total FDI for the period shown in the figure ?
A.
93.82
B.
93.22
C.
93.19
D.
None of these
Explanation:
Total FDI investment in the figure shown is = 5.7 + 10.15 + 12.16 + 10.22 + 24.23 + 31.36 = 93.82 billion. |
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# 4.10: Add and Subtract Mixed Numbers (Part 1)
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Skills to Develop
• Model addition of mixed numbers with a common denominator
• Add mixed numbers with a common denominator
• Model subtraction of mixed numbers
• Subtract mixed numbers with a common denominator
• Add and subtract mixed numbers with different denominators
be prepared!
Before you get started, take this readiness quiz.
1. Draw figure to model $$\dfrac{7}{3}$$. If you missed this problem, review Example 4.1.6.
2. Change $$\dfrac{11}{4}$$ to a mixed number. If you missed this problem, review Example 4.1.9.
3. Change $$3 \dfrac{1}{2}$$ to an improper fraction. If you missed this problem, review Example 4.1.11.
## Model Addition of Mixed Numbers with a Common Denominator
So far, we’ve added and subtracted proper and improper fractions, but not mixed numbers. Let’s begin by thinking about addition of mixed numbers using money.
If Ron has $$1$$ dollar and $$1$$ quarter, he has $$1 \dfrac{1}{4}$$ dollars. If Don has $$2$$ dollars and $$1$$ quarter, he has $$2 \dfrac{1}{4}$$ dollars. What if Ron and Don put their money together? They would have $$3$$ dollars and $$2$$ quarters. They add the dollars and add the quarters. This makes $$3 \dfrac{2}{4}$$ dollars. Because two quarters is half a dollar, they would have $$3$$ and a half dollars, or $$3 \dfrac{1}{2}$$ dollars.
$\begin{split} & 1 \dfrac{1}{4} \\ + & 2 \dfrac{1}{4} \\ \hline \\ & 3 \dfrac{2}{4} = 3 \dfrac{1}{2} \end{split} \nonumber$
When you added the dollars and then added the quarters, you were adding the whole numbers and then adding the fractions.
$1 \dfrac{1}{4} + 2 \dfrac{1}{4} \nonumber$
We can use fraction circles to model this same example:
Start with $$1 \dfrac{1}{4}$$. one whole and one $$\dfrac{1}{4}$$ pieces $$1 \dfrac{1}{4}$$ Add $$2 \dfrac{1}{4}$$ more. two wholes and one $$\dfrac{1}{4}$$ pieces $$\begin{split}+ & 2 \dfrac{1}{4} \\ & \hline \end{split}$$ The sum is: three wholes and two $$\dfrac{1}{4}$$'s $$3 \dfrac{2}{4} = 3 \dfrac{1}{2}$$
Example $$\PageIndex{1}$$: model
Model $$2 \dfrac{1}{3} + 1 \dfrac{2}{3}$$ and give the sum.
Solution
We will use fraction circles, whole circles for the whole numbers and $$\dfrac{1}{3}$$ pieces for the fractions.
two wholes and one $$\dfrac{1}{3}$$ $$2 \dfrac{1}{3}$$ plus one whole and two $$\dfrac{1}{3}$$s $$\begin{split}+ & 1 \dfrac{2}{3} \\ & \hline \end{split}$$ sum is three wholes and three $$\dfrac{1}{3}$$s $$3 \dfrac{3}{3} = 4$$
This is the same as $$4$$ wholes. So, $$2 \dfrac{1}{3} + 1 \dfrac{2}{3} = 4$$.
Exercise $$\PageIndex{1}$$
Use a model to add the following. Draw a picture to illustrate your model. $$1 \dfrac{2}{5} + 3 \dfrac{3}{5}$$
$$5$$
Exercise $$\PageIndex{2}$$
Use a model to add the following. Draw a picture to illustrate your model. $$2 \dfrac{1}{6} + 2 \dfrac{5}{6}$$
$$5$$
Example $$\PageIndex{2}$$: model
Model $$1 \dfrac{3}{5} + 2 \dfrac{3}{5}$$ and give the sum as a mixed number.
Solution
We will use fraction circles, whole circles for the whole numbers and $$\dfrac{1}{5}$$ pieces for the fractions.
one whole and three $$\dfrac{1}{5}$$s $$1 \dfrac{3}{5}$$ plus two wholes and three $$\dfrac{1}{5}$$s $$\begin{split}+ & 2 \dfrac{3}{5} \\ & \hline \end{split}$$ sum is three wholes and six $$\dfrac{1}{5}$$s $$3 \dfrac{6}{5} = 4 \dfrac{1}{5}$$
Adding the whole circles and fifth pieces, we got a sum of $$3 \dfrac{6}{5}$$. We can see that $$\dfrac{6}{5}$$ is equivalent to $$1 \dfrac{1}{5}$$, so we add that to the $$3$$ to get $$4 \dfrac{1}{5}$$.
Exercise $$\PageIndex{3}$$
Model, and give the sum as a mixed number. Draw a picture to illustrate your model. $$2 \dfrac{5}{6} + 1 \dfrac{5}{6}$$
$$4\dfrac{2}{3}$$
Exercise $$\PageIndex{4}$$
Model, and give the sum as a mixed number. Draw a picture to illustrate your model. $$1 \dfrac{5}{8} + 1 \dfrac{7}{8}$$
$$3\dfrac{1}{2}$$
## Add Mixed Numbers
Modeling with fraction circles helps illustrate the process for adding mixed numbers: We add the whole numbers and add the fractions, and then we simplify the result, if possible.
HOW TO: ADD MIXED NUMBERS WITH A COMMON DENOMINATOR
Step 1. Add the whole numbers.
Step 2. Add the fractions.
Step 3. Simplify, if possible.
Example $$\PageIndex{3}$$: add
Add: $$3 \dfrac{4}{9} + 2 \dfrac{2}{9}$$.
Solution
Add the whole numbers. $$\begin{split} & \textcolor{red}{3} \dfrac{4}{9} \\ + & \textcolor{red}{2} \dfrac{2}{9} \\ \hline \\ & \textcolor{red}{5} \end{split}$$ Add the fractions. $$\begin{split} & 3 \textcolor{red}{\dfrac{4}{9}} \\ + & 2 \textcolor{red}{\dfrac{2}{9}} \\ \hline \\ & 5 \textcolor{red}{\dfrac{6}{9}} \end{split}$$ Simplify the fraction. $$\begin{split} & 3 \dfrac{4}{9} \\ + & 2 \dfrac{2}{9} \\ \hline \\ & \textcolor{red}{5 \dfrac{6}{9}} = 5 \dfrac{2}{3} \end{split}$$
Exercise $$\PageIndex{5}$$
Find the sum: $$4 \dfrac{4}{7} + 1 \dfrac{2}{7}$$.
$$5\dfrac{6}{7}$$
Exercise $$\PageIndex{6}$$
Find the sum: $$2 \dfrac{3}{11} + 5 \dfrac{6}{11}$$.
$$7\dfrac{9}{11}$$
In Example $$\PageIndex{3}$$, the sum of the fractions was a proper fraction. Now we will work through an example where the sum is an improper fraction.
Example $$\PageIndex{4}$$: add
Find the sum: $$9 \dfrac{5}{9} + 5 \dfrac{7}{9}$$.
Solution
Add the whole numbers and then add the fractions. $$\begin{split} & 9 \dfrac{5}{9} \\ + & 5 \dfrac{7}{9} \\ \hline \\ & 14 \dfrac{12}{9} \end{split}$$ Rewrite $$\dfrac{12}{9}$$ as an improper fraction. $$14 + 1 \dfrac{3}{9}$$ Add. $$15 \dfrac{3}{9}$$ Simplify. $$15 \dfrac{1}{3}$$
Exercise $$\PageIndex{7}$$
Find the sum: $$8 \dfrac{7}{8} + 7 \dfrac{5}{8}$$.
$$16\dfrac{1}{2}$$
Exercise $$\PageIndex{8}$$
Find the sum: $$6 \dfrac{7}{9} + 8 \dfrac{5}{9}$$.
$$15\dfrac{1}{3}$$
An alternate method for adding mixed numbers is to convert the mixed numbers to improper fractions and then add the improper fractions. This method is usually written horizontally.
Example $$\PageIndex{5}$$: add
Add by converting the mixed numbers to improper fractions: $$3 \dfrac{7}{8} + 4 \dfrac{3}{8}$$.
Solution
Convert to improper fractions. $$\dfrac{31}{8} + \dfrac{35}{8}$$ Add the fractions. $$\dfrac{31 + 35}{8}$$ Simplify the numerator. $$\dfrac{66}{8}$$ Rewrite as a mixed number. $$8 \dfrac{2}{8}$$ Simplify the fraction. $$8 \dfrac{1}{4}$$
Since the problem was given in mixed number form, we will write the sum as a mixed number.
Exercise $$\PageIndex{9}$$
Find the sum by converting the mixed numbers to improper fractions: $$5 \dfrac{5}{9} + 3 \dfrac{7}{9}$$
$$9\dfrac{1}{3}$$
Exercise $$\PageIndex{10}$$
Find the sum by converting the mixed numbers to improper fractions: $$3 \dfrac{7}{10} + 2 \dfrac{9}{10}$$
$$6\dfrac{3}{5}$$
Table $$\PageIndex{1}$$ compares the two methods of addition, using the expression $$3 \dfrac{2}{5} + 6 \dfrac{4}{5}$$ as an example. Which way do you prefer?
Table $$\PageIndex{1}$$
Mixed Numbers Improper Fractions
$$\begin{split} & 3 \dfrac{2}{5} \\ + & 6 \dfrac{4}{5} \\ \hline \\ & 9 \dfrac{6}{5} \end{split}$$ $$3 \dfrac{2}{5} + 6 \dfrac{4}{5}$$
$$9 + \dfrac{6}{5}$$ $$\dfrac{17}{5} + \dfrac{34}{5}$$
$$9 + 1 \dfrac{1}{5}$$ $$\dfrac{51}{5}$$
$$10 \dfrac{1}{5}$$ $$10 \dfrac{1}{5}$$
## Model Subtraction of Mixed Numbers
Let’s think of pizzas again to model subtraction of mixed numbers with a common denominator. Suppose you just baked a whole pizza and want to give your brother half of the pizza. What do you have to do to the pizza to give him half? You have to cut it into at least two pieces. Then you can give him half.
We will use fraction circles (pizzas!) to help us visualize the process. Start with one whole.
Figure $$\PageIndex{1}$$
Algebraically, you would write:
Example $$\PageIndex{6}$$: subtract
Use a model to subtract: $$1 − \dfrac{1}{3}$$.
Solution
Exercise $$\PageIndex{11}$$
Use a model to subtract: $$1 − \dfrac{1}{4}$$.
$$\dfrac{3}{4}$$
Exercise $$\PageIndex{12}$$
Use a model to subtract: $$1 − \dfrac{1}{5}$$.
$$\dfrac{4}{5}$$
What if we start with more than one whole? Let’s find out.
Example$$\PageIndex{7}$$: subtract
Use a model to subtract: $$2 − \dfrac{3}{4}$$.
Solution
Exercise $$\PageIndex{13}$$
Use a model to subtract: $$2 − \dfrac{1}{5}$$.
$$\dfrac{9}{5}$$
Exercise $$\PageIndex{14}$$
Use a model to subtract: $$2 − \dfrac{1}{3}$$.
$$\dfrac{5}{3}$$
In the next example, we’ll subtract more than one whole.
Example $$\PageIndex{8}$$: subtract
Use a model to subtract: $$2 − 1 \dfrac{2}{5}$$.
Solution
Exercise $$\PageIndex{15}$$
Use a model to subtract: $$2 − 1 \dfrac{1}{3}$$.
$$\dfrac{2}{3}$$
Exercise $$\PageIndex{16}$$
Use a model to subtract: $$2 − 1 \dfrac{1}{4}$$.
$$\dfrac{3}{4}$$
What if you start with a mixed number and need to subtract a fraction? Think about this situation: You need to put three quarters in a parking meter, but you have only a $$\1$$ bill and one quarter. What could you do? You could change the dollar bill into $$4$$ quarters. The value of $$4$$ quarters is the same as one dollar bill, but the $$4$$ quarters are more useful for the parking meter. Now, instead of having a $$\1$$ bill and one quarter, you have $$5$$ quarters and can put $$3$$ quarters in the meter.
This models what happens when we subtract a fraction from a mixed number. We subtracted three quarters from one dollar and one quarter.
We can also model this using fraction circles, much like we did for addition of mixed numbers.
Example $$\PageIndex{9}$$: subtract
Use a model to subtract: $$1 \dfrac{1}{4} − \dfrac{3}{4}$$
Solution
Rewrite vertically. Start with one whole and one fourth. $$\begin{split} & \textcolor{red}{1 \dfrac{1}{4}} \\ - & \dfrac{3}{4} \\ \hline \end{split}$$ Since the fractions have denominator 4, cut the whole into 4 pieces. You now have $$\dfrac{4}{4}$$ and $$\dfrac{1}{4}$$ which is $$\dfrac{5}{4}$$. $$\begin{split} & \textcolor{red}{\dfrac{5}{4}} \\ - & \dfrac{3}{4} \\ \hline \end{split}$$ Take away $$\dfrac{3}{4}$$. There is $$\dfrac{1}{2}$$ left. $$\begin{split} & \dfrac{5}{4}\\\textcolor{red}{- & \dfrac}3}{4}}\\ \hline \\ & \dfrac{2}{4} = \dfrac{1}{2} \end{split}$$
Exercise $$\PageIndex{17}$$
Use a model to subtract. Draw a picture to illustrate your model. $$1 \dfrac{1}{3} − \dfrac{2}{3}$$
Exercise $$\PageIndex{18}$$
Use a model to subtract. Draw a picture to illustrate your model. $$1 \dfrac{1}{5} − \dfrac{4}{5}$$ |
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# What is the Prime Factorization Of 35?
• Equcation for number 35 factorization is: 5 * 7
• It is determined that the prime factors of number 35 are: 5, 7
## Is 35 A Prime Number?
• No the number 35 is not a prime number.
• Thirty-five is a composite number. Because 35 has more divisors than 1 and itself.
## How To Calculate Prime Number Factorization
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 3 5
• About Number 3. Three is the first odd prime number and the second smallest right after number two. At the same time it is the first Mersenne prime (2 ^ 2-1), the first Fermat prime (2 ^ {2 ^ 0} +1), the second Sophie Germain prime and the second Mersenne prime exponent. It is the fourth number of the Fibonacci sequence and the second one that is unique. The triangle is the simplest geometric figure in the plane. With the calculation of its sizes deals trigonometry. Rule of three: If the sum of the digits of a number is a multiple of three, the underlying number is divisible by three.
• About Number 5. Integers with a last digit as a zero or a five in the decimal system are divisible by five. Five is a prime number. All odd multiples of five border again with the five (all even with zero). The fifth number of the Fibonacci sequence is a five. Five is also the smallest prime number that is the sum of all other primes which are smaller than themselves. The Five is a Fermat prime: 5 = 2 ^ {2 ^ 1} +1 and the smallest Wilson prime. Number five is a bell number (sequence A000110 in OEIS). There are exactly five platonic bodies. There are exactly five tetrominoes.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What is Prime Number Factorization?
• In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
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# Similar triangles - Similarity
## Triangle similarity
### define similarity
When you hear that two triangles are similar what does that actually mean? It means that their only difference is their size. Otherwise, their angles are all identical when you match them up! You may see triangles that are flipped, or rotated, but they can still be similar if there’s only a difference in their size.
Another thing to note is that with two similar triangles, their corresponding sides have the same ratio. So for example, one triangle may be 1:2 to another triangle, so all their respective sides will be 1:2 to the other triangle.
Let’s delve into different ways to prove that two triangles are similar.
## SAS similarity theorem
The SAS similarity theorem stands for side angle side. When you’ve got two triangles and the ratio of two of their sides are the same, plus one of their angles are equal, you can prove that the two triangles are similar. You’ve just learned the SAS definition! But there’s more...
## SSS similarity theorem
In the SSS similarity theorem, you’re looking at proving for the side side side. When you’ve got two triangles with three sides that have the same ratio, you once again can prove that you’ve got two similar SSS triangles.
## AA similarity theorem
The AA similarity theorem is named after angle angle. In this case, you can prove that two triangles are similar if two of their corresponding angles are equal. This isn’t hard to understand since you know that every triangle’s interior angles must equal to 180 degrees. If you’ve got 2 of the angles figured out, then you know that the last one’s value as well. This means essentially that if you know 2 of the angles in two respective triangle are the same, the last angle from the two triangles will be the same as well.
## Example problems
Question 1:
Determine if the triangles are similar:
Solution:
To help you more easily deal with this question, try reorienting the triangles so that they’re in a similar orientation. This will help you compare the sides and angles more easily. When you’ve got this done, let’s look at the ratio of the sides of these two triangles.
Ratio of the corresponding sides:
$\frac{DE}{AB}=\frac{1.8}{0.9}=2$
$\frac{EF}{BC}=\frac{2}{1}=2$
Two sides of the triangles have the same ratio. You can also see that there is a 90 angle in both triangles. Therefore, you’ve proven that they are similar based on the SAS triangle rule.
Question 2:
Which triangles are similar?
A similar triangle must have all the angles equal/congruent. So, the triangle GHI is out of the question. Now, calculate the ratios of the sides. For triangle JKL and DEF, the ratio of their sides is 2. For triangle DEF and ABC, the ratio is 1.25. We can conclude that triangle ABC, DEF , and JKL are similar triangles.
Not convinced about the proofs for similar triangles? See this online interactive diagram that shows you how the angles and the ratio of the sides of two similar triangles change as one triangle gets smaller or bigger.
To help you understand more about triangles, feel free to review the pythagorean theorem and how to use the pythagorean relationship. It’s a good stepping stone to help you understand the sides of triangles.
Later on, you’ll learn about congruent triangles, how to prove congruence by SSS, as well as SAS and HL, and last but not least, ASA and AAS.
### Similar triangles
Learn how to determine similar triangles by using side-side-side (SSS) similarity, side-angle-side (SAS) similarity and the properties of similar triangles to solve questions. It's always a good idea to refresh your memory on scale factors and proportions because we will be using them a lot in this lesson. |
# Rational Numbers Worksheet Pdf Grade 7 Pdf
A Realistic Numbers Worksheet might help your son or daughter become more knowledgeable about the concepts behind this rate of integers. In this particular worksheet, students should be able to fix 12 different issues linked to reasonable expression. They may discover ways to increase a couple of phone numbers, group of people them in couples, and figure out their items. They are going to also training simplifying reasonable expression. Once they have perfected these principles, this worksheet might be a important instrument for continuing their studies. Rational Numbers Worksheet Pdf Grade 7 Pdf.
## Rational Figures are a proportion of integers
The two main kinds of numbers: rational and irrational. Rational figures are described as whole numbers, in contrast to irrational figures usually do not perform repeatedly, and possess an endless amount of numbers. Irrational phone numbers are low-absolutely no, non-terminating decimals, and sq . roots that are not perfect squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To determine a reasonable quantity, you must know what a realistic number is. An integer can be a whole number, as well as a logical amount can be a ratio of two integers. The ratio of two integers will be the amount at the top divided up with the amount at the base. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They are often produced into a small percentage
A rational amount features a denominator and numerator which are not absolutely nothing. Consequently they may be indicated being a small fraction. In addition to their integer numerators and denominators, realistic phone numbers can furthermore have a bad worth. The negative importance needs to be put left of and its particular total importance is its length from absolutely nothing. To streamline this illustration, we will state that .0333333 is really a small percentage that may be created being a 1/3.
Together with negative integers, a realistic number can even be manufactured right into a small fraction. For instance, /18,572 is a logical amount, whilst -1/ is not really. Any portion composed of integers is realistic, as long as the denominator does not include a and may be published for an integer. Similarly, a decimal that ends in a position is also a logical quantity.
## They are sense
Regardless of their brand, realistic phone numbers don’t make very much sensation. In mathematics, they may be individual organizations using a exclusive duration on the amount line. Which means that when we matter one thing, we can purchase the size by its proportion to the unique number. This holds real even when you can find endless reasonable numbers involving two certain amounts. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the duration of a pearl, by way of example, we could matter its size. A single pearl weighs in at 10 kgs, which is a realistic number. Furthermore, a pound’s bodyweight is equal to ten kilograms. Therefore, we must be able to break down a lb by 15, with out worry about the duration of one particular pearl.
## They may be expressed as a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity could be published as being a several of two integers, so four times 5 various is the same as 8-10. A similar problem involves the repetitive portion 2/1, and either side must be split by 99 to get the right solution. But how do you make your transformation? Below are a few cases.
A reasonable number can be developed in many forms, which includes fractions along with a decimal. One way to stand for a rational variety in a decimal is to split it into its fractional counterpart. There are actually three ways to split a reasonable number, and every one of these methods yields its decimal equal. One of these techniques is to break down it into its fractional counterpart, and that’s what’s referred to as a terminating decimal. |
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Times tables
What is a times table?
A times table is a list of multiples of a number.
Learning times tables
Times tables can be learned by practicing counting in multiples, learning by rote and applying your knowledge of number facts.
The English National Curriculum sets out the expectation that children will know their times tables up to 12 x 12 by the end of Year 4.
Scroll down the page to learn about the zero times table all the way up to the thirteen times table!
0 times table
The zero times table is:
1 x 0 = 0
2 x 0 = 0
3 x 0 = 0
4 x 0 = 0
5 x 0 = 0
6 x 0 = 0
7 x 0 = 0
8 x 0 = 0
9 x 0 = 0
10 x 0 = 0
11 x 0 = 0
12 x 0 = 0
13 x 0 = 0
The product of the zero times table is always zero. Think about these number sentences as a word problem.
You go to six shops looking for a toy. None of the shops sell toys. How many toys do you find? Six lots of zero is zero!
You play a game ten times. You win zero games. How many times did you win? Ten lots of zero equals… zero!
How many apples do you have in fifteen empty boxes of apples? 15 x 0 = ?
1 times table
The one times table is:
1 x 1 = 1
2 x 1 = 2
3 x 1 = 3
4 x 1 = 4
5 x 1 = 5
6 x 1 = 6
7 x 1 = 7
8 x 1 = 8
9 x 1 = 9
10 x 1 = 10
11 x 1 = 11
12 x 1 = 12
13 x 1 = 13
What do you notice about the one times table? Can you explain what is happening in each number sentence?
You might have noticed the products in the one times table are the same as the other factor or multiplier in the number sentence. The products alternate between odd and even numbers because they go up in ones. Each product is one more than the product before.
What are 20 lots of 1? 20 x 1 = ?
2 times table
The two times table is:
1 x 2 = 2
2 x 2 = 4
3 x 2 = 6
4 x 2 = 8
5 x 2 = 10
6 x 2 = 12
7 x 2 = 14
8 x 2 = 16
9 x 2 = 18
10 x 2 = 20
11 x 2 = 22
12 x 2 = 24
13 x 2 = 26
The products in the two times table are all even numbers.
You can practice learning your two times tables by counting in twos or knowing your doubles. Use your skills to make your way through this free two-times table maze. Then try colouring in all the even numbers on a hundred square or jumping in twos along a number line. Do you notice anything about the pattern?
Learning the two times table is a part of the year 2 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 2, 5 and 10 times tables.
3 times table
The three times table is:
1 x 3 = 3
2 x 3 = 6
3 x 3 = 9
4 x 3 = 12
5 x 3 = 15
6 x 3 = 18
7 x 3 = 21
8 x 3 = 24
9 x 3 = 27
10 x 3 = 30
11 x 3 = 33
12 x 3 = 36
13 x 3 = 39
The products in the three times table alternate between odd and even numbers.
Putting the three times table in a three-by-three grid is a handy way to see the number pattern. What do you notice about the green ones numbers?
What do you notice about the red tens numbers?
Practice learning your three times tables by making your way through this free three-times table maze.
Learning the three times table is a part of the year 3 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 3, 4 and 8 times tables.
4 times tables
The four times table is:
1 x 4 = 4
2 x 4 = 8
3 x 4 = 12
4 x 4 = 16
5 x 4 = 20
6 x 4 = 24
7 x 4 = 28
8 x 4 = 32
9 x 4 = 36
10 x 4 = 40
11 x 4 = 44
12 x 4 = 48
13 x 4 = 52
The products in the four times table are all even numbers. Can you notice a link between the two times table and the four times table?
Knowing your doubles is really helpful when solving problems involving the four times table. To multiply a number by 4, simply double it and then double the answer.
Six times four is the same as double six times two, or double twelve. Six times four is twenty-four.
Practice learning your four times tables by making your way through this free four-times table maze.
Learning the four times table is a part of the year 3 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 3, 4 and 8 times tables.
5 times table
The five times table is:
1 x 5 = 5
2 x 5 = 10
3 x 5 = 15
4 x 5 = 20
5 x 5 = 25
6 x 5 = 30
7 x 5 = 35
8 x 5 = 40
9 x 5 = 45
10 x 5 = 50
11 x 5 = 55
12 x 5 = 60
13 x 5 = 65
What do you notice about the pattern of the five times table?
Did you notice the products of the number sentences alternate between odd and even numbers? All the odd numbers have five ones, all the even numbers have zero ones.
Practice learning your five times tables by making your way through this free five-times table maze.
Learning the five times table is a part of the year 2 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 2, 5 and 10 times tables.
6 times table
The six times table is:
1 x 6 = 6
2 x 6 = 12
3 x 6 = 18
4 x 6 = 24
5 x 6 = 30
6 x 6 = 36
7 x 6 = 42
8 x 6 = 48
9 x 6 = 54
10 x 6 = 60
11 x 6 = 66
12 x 6 = 72
13 x 6 = 78
‘If you know your three times table you also know your six times table’. Do you think this statement is true? Explain your answer.
Did you notice that the ones digits in the six times table repeat themselves after five times? Spotting patterns like this can help you remember and check your six times tables.
Practice learning your six times tables by making your way through this free six-times table maze.
Learning the six times table is a part of the year 4 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know all the multiplication and division facts up to 12 x 12.
7 times table
The seven times table is:
1 x 7 = 7
2 x 7 = 14
3 x 7 = 21
4 x 7 = 28
5 x 7 = 35
6 x 7 = 42
7 x 7 = 49
8 x 7 = 56
9 x 7 = 63
10 x 7 = 70
11 x 7 = 77
12 x 7 = 84
13 x 7 = 91
The seven times table is the hardest to learn. This is because seven is a prime number, which means the pattern takes longer to repeat. In fact you have to reach 10 x 7 before the ones digits start to repeat.
Practice learning your seven times tables by making your way through this free seven-times table maze.
Learning the seven times table is a part of the year 4 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know all the multiplication and division facts up to 12 x 12.
8 times table
The eight times table is:
1 x 8 = 8
2 x 8 = 16
3 x 8 = 24
4 x 8 = 32
5 x 8 = 40
6 x 8 = 48
7 x 8 = 56
8 x 8 = 64
9 x 8 = 72
10 x 8 = 80
11 x 8 = 88
12 x 8 = 96
13 x 8 = 104
How do you think the two, the four and the eight times table are linked?
How could you use the four times table to solve 3 x 8? You could solve 3 x 4 and then double it! 3 x 4 = 12, then double 12 is 24.
Can you explain why 3 x 8 is the same as double 3 x 4?
Practice learning your eight times tables by making your way through this free eight-times table maze.
Learning the eight times table is a part of the year 3 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 3, 4 and 8 times tables.
9 times table
The nine times table is:
1 x 9 = 9
2 x 9 = 18
3 x 9 = 27
4 x 9 = 36
5 x 9 = 45
6 x 9 = 54
7 x 9 = 63
8 x 9 = 72
9 x 9 = 81
10 x 9 = 90
11 x 9 = 99
12 x 9 = 108
13 x 9 = 117
A handy way to check your nine times table is to use your hands. Hold both your hands up with palms facing you. Number the fingers from left to right as one to ten. Hold down your first finger to work out 1 x 9. The fingers to the left of the folded down finger are the tens. The fingers to the right of the folded down finger are the ones. When you fold down your first finger, there are no fingers to the left so zero tens and nine to the right, so nine ones: 1 x 9 = 9.
Now let’s try 4 x 9.
Hold down your fourth finger to work out 4 x 9. The fingers to the left of the folded down finger are the tens. The fingers to the right of the folded down finger are the ones. When you fold down your fourth finger, there are three fingers to the left so three tens and six to the right, so six ones: 4 x 9 = 36.
Another way to work out the nine times table is to multiply by 10 and then take away the number you multiplied by. Let’s use this method to solve 5 x 9.
5 x 10 = 50
Now take away 5
50 - 5 = 45
5 x 9 = 45
Five multiplied by nine is the same as five multiplied by ten minus one lot of five.
Now let’s try 8 x 9.
8 x 10 = 80
80 - 8 = 72
8 x 9 = 72
Practice learning your nine times tables by making your way through this free nine-times table maze.
Learning the nine times table is a part of the year 4 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know all the multiplication and division facts up to 12 x 12.
10 times tables
The ten times table is:
1 x 10 = 10
2 x 10 = 20
3 x 10 = 30
4 x 10 = 40
5 x 10 = 50
6 x 10 = 60
7 x 10 = 70
8 x 10 = 80
9 x 10 = 90
10 x 10 = 100
11 x 10 = 110
12 x 10 = 120
13 x 10 = 130
What do you notice about the pattern of the ten times table?
Did you notice the products of the number sentences are all even numbers? All the answers have zero ones. The tens number increases by ten each time.
Practice learning your ten times tables by making your way through this free ten-times table maze.
Learning the ten times table is a part of the year 2 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know their 2, 5 and 10 times tables.
11 times table
The eleven times table is:
1 x 11 = 11
2 x 11 = 22
3 x 11 = 33
4 x 11 = 44
5 x 11 = 55
6 x 11 = 66
7 x 11 = 77
8 x 11 = 88
9 x 11 = 99
10 x 11 = 110
11 x 11 = 121
12 x 11 = 132
13 x 11 = 143
What do you notice about the products of the eleven times table? Do you spot any patterns?
Eleven is one more than ten. We can use this information to help us solve a problem involving multiplying by eleven. Three multiplied by eleven is the same as three multiplied by ten plus one more lot of three.
3 x 11 = 3 x 10 + 3 = 30 + 3 = 33
3 x 11 = 33
Did you know you can use a simple trick to work out the answer to eleven multiplied by a two-digit number?
Let’s work out 12 x 11 using this trick.
First split the two numbers that make up 12 into the first and third digit of a three digit number:
1_2
Then work out the missing tens digit of the three digit number by adding the two numbers together: 1+2 = 3
Now place this tens digit in the correct column: 132
12 x 11 = 132
This method becomes more complicated when the product of the two numbers you add is more than 9. When this happens you need to do an extra step.
19 x 11 =
First split the two numbers that make up 19 into the first and third digit of a three digit number:
1_9
Then work out the missing tens digit of the three digit number by adding the two numbers: together. 1+9 = 10
If we followed the previous method and inserted the product of 1+9 into the number we would get 1109. This isn’t correct. So rather than placing the two digits into the middle of the number, we add the 1 from the 10 and the 1 from the 100 and then place the 0 in as our missing digit: 209
19 x 11 = 209
Practice learning your eleven times tables by making your way through this free eleven-times table maze.
Learning the eleven times table is a part of the year 4 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know all the multiplication and division facts up to 12 x 12.
12 times table
The twelve times table is:
1 x 12 = 12
2 x 12 = 24
3 x 12 = 36
4 x 12 = 48
5 x 12 = 60
6 x 12 = 72
7 x 12 = 84
8 x 12 = 96
9 x 12 = 108
10 x 12 = 120
11 x 12 = 132
12 x 12 = 144
13 x 12 = 156
What do you notice about the products of the twelve times table?
What are the tens numbers increasing by?
What are the ones numbers increasing by?
Can you explain what you noticed, and why this pattern happens?
Twelve is two more than ten. We can use this information to help us solve a problem involving multiplying by twelve. Six multiplied by twelve is the same as six multiplied by ten plus two more lots of six.
6 x 12 = 6 x 10 + 6 + 6 = 60 + 6 + 6 = 60 + 12 = 72
6 x 12 = 72
We can also use partitioning to solve this problem. We know that twelve is one ten and two ones. Six multiplied by twelve is the same as six multiplied by ten plus six multiplied by two.
6 x 12 = 6 x 10 + 6 x 2 = 60 + 12 = 72
6 x 12 = 72
Practice learning your twelve times tables by making your way through this free twelve-times table maze.
Learning the twelve times table is a part of the year 4 national curriculum for primary schools in England. The expectation is that by the end of the year, all students will know all the multiplication and division facts up to 12 x 12.
13 times table
The thirteen times table is:
1 x 13 = 13
2 x 13 = 26
3 x 13 = 39
4 x 13 = 52
5 x 13 = 65
6 x 13 = 78
7 x 13 = 91
8 x 13 = 104
9 x 13 = 117
10 x 13 = 130
11 x 13 = 143
12 x 13 = 156
13 x 13 = 169
We can use partitioning to solve thirteen times table problems. Think back to how we worked out the twelve times table and apply the same methods.
Solve 5 x 13 using partitioning:
13 is 10 + 3
5 x 13 = 5 x 10 + 5 x 3 = 50 + 15 = 65
5 x 13 = 65
Solve 12 x 13 using partitioning:
13 is 10 + 3
12 x 13 = 12 x 10 + 12 x 3 = 120 + 36 = 156
12 x 13 = 156
If solving 12 x 3 is too tricky, we can partition twelve further:
12 x 13
= 12 x 10 + 12 x 3
= 12 x 10 + 10 x 3 + 2 x 3
= 120 + 30 + 6
= 156
12 x 13 = 156
If you can solve these thirteen times table number sentences you can solve just about any multiplication problem, simply apply your knowledge of multiplication and partitioning.
When are times tables taught in the English National Curriculum?
Children are formally taught multiplication tables from Year 2 onwards. In the Early Years and Year 1, children develop number knowledge, which gives them strong foundations to build on as they move into a more abstract understanding of numbers.
The Maths National Curriculum document outlines the following statutory requirements regarding times tables:
Year 2 times tables
In Year 2, children are taught to:
‘recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers’ (page 13).
Year 3 times tables
In Year 3, children are taught to:
‘recall and use multiplication and division facts for the 3, 4 and 8 multiplication tables’ (page 19).
Year 4 times tables
In Year 4, children are taught to:
‘recall multiplication and division facts for multiplication tables up to 12 × 12’ (page 25).
PlanBee have a wide selection of maths schemes dedicated to teaching multiplication and times tables. We have a whole collection of multiplication and division schemes of work you can view, or alternatively check out our curriculum objective checker to focus on a specific national curriculum objective. |
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