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## How to Add Mixed Numbers
A mixed number is a combination of two numbers: a whole number and a proper fraction (A proper fraction is a fraction which has a denominator which is greater than the numerator, i.e., $$\frac{3}{7}$$, $$\frac{9}{11}$$, $$\frac{13}{19}$$ , etc.). Moreover, a mixed number can be converted into a fraction and it always lies between two whole numbers.
For ex: Let’s take the mixed number $$3\frac{1}{6}$$. So, this mixed number comprises of two parts, a whole number which is $$3$$ and a proper fraction $$\frac{1}{6}$$. Now, if we convert this mixed number into an improper fraction which is $$\frac{19}{6}$$ we find that it lies between the two whole numbers $$3$$ and $$4$$.
Some other examples of a mixed number are $$2\frac{1}{2}$$, $$3\frac{1}{3}$$, $$4\frac{1}{5}$$, etc.
#### Parts of a mixed number
A mixed number consists of three distinct parts: a whole number, a numerator and a denominator. Here, the numerator and the denominator are the parts of the proper fraction.
#### How to Convert Improper Fractions to Mixed Fractions
1. First, we need to divide the numerator of the fraction by the denominator.
2. Next, we need to write down the quotient as the whole number of the mixed fraction.
3. Now, the remainder becomes the numerator and the divisor becomes the denominator of the improper part.
Ex: Let’s take the improper fraction $$\frac{7}{2}$$
Now, when we divide $$7$$ by $$2$$, the quotient is $$3$$. Also, the remainder is $$1$$ and the divisor $$2$$. So, the mixed number is $$3\frac{1}{2}$$.
#### Steps to Add Mixed Numbers
1. First, Add the whole parts separately and the fractional parts separately.
2. Next, Now simplify your answer and write it in the lowest terms.
Ex: Let’s add $$4\frac{1}{3} + 2\frac{1}{5}$$
So, now the addition is like $$(4+2) + (\frac{1}{3}+\frac{1}{5}) = 6 + \frac{5 \times 1 + 3 \times 1}{5 \times 3} = 6\frac{8}{15}$$
### Exercises for Add Mixed Numbers
1) $$7 {2 \over 9} \ + \ 3 {3 \over 4} = \$$
2) $$5 {3 \over 7} \ + \ 1 {5 \over 2} = \$$
3) $$4 {6 \over 5} \ + \ 7 {2 \over 4} = \$$
4) $$2 {5 \over 4} \ + \ 5 {4 \over 3} = \$$
5) $$1 {7 \over 8} \ + \ 6 {8 \over 7} = \$$
6) $$7 {3 \over 10} \ + \ 6 {4 \over 3} = \$$
7) $$7 {10 \over 3} \ + \ 7 {5 \over 6} = \$$
8) $$7 {8 \over 6} \ + \ 2 { 3\over 8} = \$$
9) $$4 {9 \over 5} \ + \ 7 {4 \over 3} = \$$
10) $$4 {8 \over 7} \ + \ 6 {4 \over 6} = \$$
1) $$7 {2 \over 9} \ + \ 3 {3 \over 4} = \$$$$\ \color{red}{(7 + 3)+{2 \times 4 \ + \ 3 \times 9 \over 9\times4} = }$$ $$\color{red}{10{35 \over 36}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$7+{2 \over 9}+3+{3 \over 4}$$
Then solve the whole number parts $$7+3=10$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${2 \over 9}+{3 \over 4}$$ = $$2 \times 4 \ + \ 3 \times 9 \over 9 \times4$$ = $$8+27 \over 36$$ = $$35 \over 36$$
At the end Combine the whole and fraction parts $$10 +{35 \over 36}$$ = $$10{35 \over 36}$$
2) $$5 {3 \over 7} \ + \ 1 {5 \over 2} = \$$$$\ \color{red}{(5 + 1)+{3 \times 2 \ + \ 5 \times 7 \over 7\times2} = }$$ $$\color{red}{6{41 \over 14} = 8{13 \over 14}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$5+{3 \over 7}+1+{5 \over 2}$$
Then solve the whole number parts $$5+1=6$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${3 \over 7} +{5 \over 2}$$ = $$3 \times 2 \ + \ 5 \times 7 \over 7 \times 2$$ = $$6 + 35 \over 14$$ = $$41 \over 14$$ = $$2{ 13\over 14}$$
At the end Combine the whole and fraction parts $$6 +2{13\over 14}$$ = $$8{13 \over 14}$$
3) $$4 {6 \over 5} \ + \ 7 {2 \over 4} = \$$$$\ \color{red}{(4 + 7)+{6 \times 4 \ + \ 2 \times 5 \over 5\times4} = }$$ $$\color{red}{11{34 \over 20}}$$$$\color{red}{ = 11 + 1{14 \over 20}}$$$$\color{red}{ = 12{14 \over 20} = 12{ 7 \over 10}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$4+{6 \over 5}+7+{2 \over 4}$$
Then solve the whole number parts $$4+7=11$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${6 \over 5} +{2 \over 4}$$ = $$6 \times 4 \ + \ 2 \times 5 \over 5 \times 4$$ = $$34\over 20$$ = $$1{14 \over 20}$$ = $$1{7 \over 10}$$
At the end Combine the whole and fraction parts $$11 +1{7 \over 10}$$ = $$12{7 \over 10}$$
4) $$2 {5 \over 4} \ + \ 5 {4 \over 3} = \$$$$\ \color{red}{(2 + 5)+{5 \times 3 \ + \ 4 \times 4 \over 4\times3} = }$$ $$\color{red}{7{15 +16 \over 12}}$$$$\color{red}{ = 7{31 \over 12} = 7 + 2{ 7 \over 12}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$2+{5 \over 4}+5+{4 \over 2}$$
Then solve the whole number parts $$2+5=7$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${5 \over 4} +{4 \over 2}$$ = $$5 \times 3 \ + \ 4 \times 4 \over 4 \times 3$$ = $$15 + 16 \over 4$$ = $$31 \over 4$$ =$$2{7 \over 12}$$
At the end Combine the whole and fraction parts $$7 +2{7 \over 12}$$ = $$9{7 \over 12}$$
5) $$1 {7 \over 8} \ + \ 6 {8 \over 7} = \$$$$\ \color{red}{(1 + 6)+{7 \times 7 \ + \ 8 \times 8 \over 8\times7} = }$$ $$\color{red}{7{113 \over 56}}$$$$\color{red}{ = 7 + 2{1 \over 56}}$$$$\color{red}{ = 9{1 \over 56}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$1+{7 \over 8}+6+{8 \over 7}$$
Then solve the whole number parts $$1+6=7$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${7 \over 8} +{8 \over 7}$$ = $$7 \times 7 \ + \ 8 \times 8 \over 8 \times 7$$ = $$113 \over 56$$ = $$2{1 \over 56}$$
At the end Combine the whole and fraction parts $$7 +2{1 \over 56}$$ = $$9{1 \over 56}$$
6) $$7 {3 \over 10} \ + \ 6 {4 \over 3} = \$$$$\ \color{red}{(7 + 6)+{3 \times 3 \ + \ 4 \times 10 \over 10\times3} = 13{ 9+40 \over 30 }}$$ $$\color{red}{= 13{49 \over 30} =13 + 1{19 \over30}= 14{19 \over 30}}$$
Solution:
The first step is to rewrite the equation with parts separated $$7+{3 \over 10}+6+{4 \over 3}$$
Then solve the whole number parts $$7+6=13$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${3 \over 10} +{4 \over 3}$$ = $$3 \times 3 \ + \ 4 \times 10 \over 10 \times 3$$ = $$9+40 \over 30$$ = $$49 \over 30$$ = $$1{ 19 \over 30}$$
At the end Combine the whole and fraction parts $$13 + 1{19 \over30}= 14{19 \over 30}$$
7) $$7 {10 \over 3} \ + \ 7 {5 \over 6} = \$$$$\ \color{red}{(7 + 7)+{10 \times 6 \ + \ 5 \times 3 \over 3\times6} = }$$ $$\color{red}{14{60 + 15 \over 18}}$$ $$\color{red}{=14 +{75 \over 18}}$$ $$\color{red}{=14 + 4{3\over 18}}$$ $$\color{red}{ =18{3 \over 18} = 18{1 \over 6}}$$
Solution:
The first step is to rewrite the equation with parts separated $$7+{10 \over 3}+7+{5 \over 6}$$
Then solve the whole number parts $$7+7=14$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${10 \over 3} +{5 \over 6}$$ = $$10 \times 6 \ + \ 5 \times 3 \over 3 \times 6$$ = $$60+ 15 \over 18$$ = $$75 \over 18$$ = $$4{3 \over 18} = 4{1 \over 6}$$
At the end Combine the whole and fraction parts $$14 + 4{1 \over6}= 18{1 \over 6}$$
8) $$7{8 \over 6} + 2{ 3\over 8} = \$$ $$\ \color{red}{(7 + 2)+{8 \times 8 \ + \ 3 \times 6 \over 6\times8} = }$$ $$\color{red}{9{82 \over 48}}$$$$\color{red}{ = 9 + {41\over 24} = 9+ 1{17\over 24}}$$$$\color{red}{ = 10{17 \over 24}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$7+{8 \over 6}+2+{3 \over 8}$$
Then solve the whole number parts $$7+2=9$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${8 \over 6} +{3 \over 8}$$ = $$8 \times 8 \ + \ 3 \times 6 \over 6 \times 8$$ = $$64 + 18 \over 48$$ = $$82\over 48$$ = $$1{17 \over 24}$$
At the end Combine the whole and fraction parts $$9 +{1{17 \over 24}}$$ = $$10{17 \over 24}$$
9) $$4 {9 \over 5} \ + \ 7 {4 \over 3} = \$$$$\ \color{red}{(4 + 7)+{9 \times 3 \ + \ 4 \times 5 \over 5\times3} = }$$ $$\color{red}{11{47 \over 15}}$$$$\color{red}{ = 11 + 3{2 \over 15}}$$$$\color{red}{ = 14{2 \over 15}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$4+{9 \over 5}+7+{4 \over 3}$$
Then solve the whole number parts $$4+7=11$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${9 \over 5} +{4 \over 3}$$ = $$9 \times 3 \ + \ 4 \times 5 \over 5 \times 3$$ = $$27 + 20\over 15$$ = $$47\over 15$$ = $$3{2 \over 15}$$
At the end Combine the whole and fraction parts $$11 +{3{2 \over 15}}$$ = $$14{2 \over 15}$$
10) $$4 {8 \over 7} \ + \ 6 {4 \over 6} = \$$$$\ \color{red}{(4 + 6)+{8 \times 6 \ + \ 4 \times 7 \over 7\times6} = }$$ $$\color{red}{10{76 \over 42}}$$$$\color{red}{ = 10 + 1{34 \over 42} = 11{76 \over 42}}$$ $$\color{red}{ = 11{17 \over 21}}$$
Solution:
The first step is to rewrite the equation with parts separated, $$4+{8 \over 7}+6+{4 \over 6}$$
Then solve the whole number parts $$4+6=10$$
Now solve the fraction parts, and rewrite to solve with the equivalent fractions. $${8 \over 7} +{4 \over 6}$$ = $$8 \times 6 \ + \ 4 \times 7 \over 7 \times 6$$ = $$48 + 28\over 42$$ = $$76\over 42$$ = $$1{34 \over 42}$$= $$1{17 \over 21}$$
At the end Combine the whole and fraction parts $$10 +1{17 \over 24}$$ = $$11{17 \over 21}$$
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# Plot the line: $3x + 2y = 0$
Last updated date: 19th Sep 2024
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Hint: We have asked to plot a line for $3x + 2y = 0$. This is the linear equation in two variables and every linear equation in two variables is the equation of line. So to plot this firstly we have to find the point which satisfies this equation these are infinitely many points which will satisfy this equation. We will find only two points. These points will be enough to plot the line.
We first put $x = 0$in the equation, this will give the value of $y$. Again we put $y = 0$ in this equation this will give the value of $x$, and we get two points which help to plot the graph.
We have given equation
$3x + 2y = 0$ ------(i)
This is the equation of the line.
Putting $x = 0$in the equation (i)
$3.x.0 + 2y = 0$
$\Rightarrow$ $2y = 0$
$\Rightarrow$ $y = 0$
So for $x = 0$the value of $y = 0$ $(0,0)$ is the point on the line
Putting $x = 2$in the equation (i)
We get
$3.x.2 + 2y = 0$
$\Rightarrow$ $6 + 2y = 0$
$\Rightarrow$ $2y = - 6$
$\Rightarrow$ $y = - 3$
For $x = 2,y = - 3$
$(2,3)$ is the point on the line
Putting $y = - 3$ in the equation (i)
We get
$3x + 2 \times 3 = 0$
$\Rightarrow$ $3x + 6 = 0$
$\Rightarrow$ $3x = - 6$
$x = \dfrac{{ - 3}}{3} = - 2$
For $y = - 3$, $x = 2$
So $( - 2,3)$ is the point on the line
$x$ $0$ $2$ $- 2$ $y$ $0$ $- 3$ $3$
Plotting these points in the graph and joining we get a straight line.
Note: An equation is said to be a linear equation in two variables, if it is written in the form of $ax + by + c = 0$. Where $a, b$ and $c$ are real numbers and the coefficient of $x$ and $y$.
The solution of such equations is a pair of values, one for $x$ and one for $y$ which further moles the two sides of an equation equal. |
# Chapter 1 Linear Equations and Linear Functions.
## Presentation on theme: "Chapter 1 Linear Equations and Linear Functions."— Presentation transcript:
Chapter 1 Linear Equations and Linear Functions
Using Qualitative Graphs to Describe Situations
Section 1.1 Using Qualitative Graphs to Describe Situations
Reading Qualitative Graphs Example Let p be the retail price (in dollars) or Air Jordans and t be the number of years since 1985. What does this graph (or curve) tell us? As t increases, what happens to p? Section 1.1 Slide 3
Identifying Independent and Dependent Variables
Independent and Dependant Variables Definition Assume that an authentic situation can be described by using the variables t and p and that p depends on t: We call t the independent variable. We call p the dependent variable. Section 1.1 Slide 4
Identifying Independent and Dependent Variables
Independent and Dependant Variables Example Identify the independent variable and the dependent variable. You are filling a swimming pool. Let r be the rate (in gallons per hour) at which water is added to a swimming pool, and let t be the number of hours It takes to fill the pool. Section 1.1 Slide 5
Identifying Independent and Dependent Variables
Independent and Dependant Variables Solution Shorter amount of time to fill the pool, the higher the rate at which water is added to the pool Amount of time, t, it takes to fill the pool, depends on the rate at which water is added, r t is the dependent variable and r is the independent variable. The rate at which water is added to the pool does not depend on the amount of time it takes to fill the pool. Section 1.1 Slide 6
Independent and Dependant Variables Example Let A be the average age (in years) when men first marry, and let t be the number of years since The graph describes the relationship between the variables t and A. What does the graph tell us? Section 1.1 Slide 7
Independent and Dependant Variables Solution Graph tells us that the average age when men first marry decreased each year for a while and then increased each year after than Section 1.1 Slide 8
Independent and Dependant Variables
Definitions Independent and Dependant Variables Definition An intercept of a curve is any point where the curve and an axis (or axes) intercept. The shape of the top curve is a parabola. The shape of the bottom curve is a linear line. Section 1.1 Slide 9
Sketching a Qualitative Graphs
Sketching Qualitative Graphs Example Let C be the cost (in dollars) of a 30-second ad during the Super Bowl at t years since For most years the annual increase in cost is more than the previous annual increase in cost. Sketch a qualitative graph that describes the relationship between C and t. Section 1.1 Slide 10
Sketching a Qualitative Graphs
Sketching Qualitative Graphs Solution Cost of an ad varies according to year C is the dependent variable t is the independent variable Ads were not free in 1987 (t=0) C-intercept (or dependent variable intercept) is positive. Section 1.1 Slide 11
The shape of this curve is said to be exponential.
Sketching a Qualitative Graphs Sketching Qualitative Graphs Solution Continued Cost increase, so the curve goes up from left to right Most increases are more than the previous year. Curve should “bend” upward from left to right. The shape of this curve is said to be exponential. Section 1.1 Slide 12
Sketching Qualitative Graphs
Definitions Sketching Qualitative Graphs Definition If a curve goes… … upwards from left to right it is said to be increasing. … downward from left to right it is said to be decreasing. Section 1.1 Slide 13
Which curve is increasing Which is decreasing?
Increasing Decreasing Sketching Qualitative Graphs Example Which curve is increasing Which is decreasing? Section 1.1 Slide 14
Increasing Decreasing
Sketching Qualitative Graphs Solution From left to right the first curve is increasing and the second is decreasing. Increasing curve Decreasing curve Section 1.1 Slide 15
Sketching a Qualitative Graph
Sketching Qualitative Graphs Example The percentage of flight attendants laid-off by top United States airlines increased from 2001 to and decreased thereafter. Let P be the percentage of flight attendants laid-off by top United States airlines, and t be years since 2001. Sketch a quantitative graph that describes the relationship between the variable. Section 1.1 Slide 16
Sketching a Qualitative Graph
Sketching Qualitative Graphs Solution The percentage of flight attendants laid off is determined by the time after 2000 The curve increases then decreases P is the dependant variable and t is the independent variable Both P and t are nonnegative. Section 1.1 Slide 17
Sketching Qualitative Graphs
Quadrants Sketching Qualitative Graphs P and t are both nonnegative, so the curve is in Quadrant I. What quadrant would the curve be in if P and t were both negative? What would the values of P and t be if the curve was only in Quadrant II, or IV> Section 1.1 Slide 18 |
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Math Central Quandaries & Queries
Question from Devon, a student: Having trouble with a math question. I need to put the following question in to y=mx+b form I rent a gym for $150.00 for 30 students. Another time I rent the gym for @270.00 for 70 students. I need to also find a fixed rate. Hi Devon. Let me show you a very similar program and how I solve it. Then you can do the same process with your question. Problem: A bus company took a tour bus on the ferry when there were 30 people aboard. The ferry charged the bus company$180. The following week, the bus had 50 people on board and the ferry charged them $220. How much is the "base rate" for the empty bus? How much does each person cost? Show this using y = mx + b form. Solution: Let x = the number of people on the bus Let y = the total cost for the bus with its passengers to use the ferry. Then, when there are 30 people on the bus (x = 30), the cost was$180 (y = 180). This means the point (x, y) = (30, 180).
Also, when there were 50 people on the bus, the cost was $220. So that is the point (50, 220). You see - we have two points and so we can write the equation of a line that goes through those points. The first thing we have to do is find the slope. The slope of the line joining two points is the rise divided by the run. That means m = (y2 - y1) / (x2 - x1) So we put in the values of (30, 180) and (50, 220): m = (220 - 180) / (50 - 30) m = 40 / 20 m = 2. The slope is 2. Now we can write it in the form y = mx + b, using 2 for m: y = 2x + b. This equation connects both points, so that means it goes through those points. We want to know "b" next, so we can now substitute in either point for the x and the y. I will substitute in x = 50 and y = 220: 220 = 2(50) + b 220 = 100 + b b + 100 = 220 b = 120. Now I know the b and the m, so I can finally write the full y=mx + b form of the equation: y = 2x + 120. For the other questions, I can look at the equation I got. The y-intercept (b value) is$120. That means that we are adding $120 on to the price all the time. That's the price for the empty bus. The 2x means to multiply the number of people by two and add this to the$120. So each person is $2. Here's an example: if you had just 10 people on the bus, you'd pay$120 for the empty bus and $2 for each person. That's$20 for the people and $120 for the bus. That total is$140. Let's see if (10, 140) works in the equation we got:
y = 2x + 120
140 ?=? 2(10) + 120
140 ?=? 20 + 120
yes. 140 = 140. It does work.
Now you try it with your question.
Hope this example helps,
Stephen La Rocque.
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |
# What Are The Factors Of 10: Prime Factors Of 10
Knowing the factors of 10 will help you solve many maths problems. What are the factors? They are just the specific digits that divide the number equally leaving no remainders. Let us save time and know the numbers that are factors of 10.
## What Are The Factors Of 10?
The factors of 10 are 1,2,5 and 10. The factors are the number that divides the number in such a way that there is no remainder left.
So all factors of 10 are 1,2,5 and 10. Now
• 1(10➗1=10)
• 2(10➗2=5)
• 5(10➗5=2)
• 10(10➗10=1)
1,2,5, 10 are factors for 10.
After knowing what all the factors of 10 are let us know more about the 10 factors
### Prime Factors Of 10
To know the prime factors of 10, first, you should know the prime numbers up to 10. After that, we have to divide 10 by all prime numbers. Now, we will calculate the prime factors by factorizing the 10 methods.
• 10➗2 = 5
• 10➗3 = 3.33
• 10➗5 = 2
• 10➗7 = 1.28
So the prime factor for the number 10 is 2, 5. and 7
### Factors Of 10 In Pairs
The product of the number whose result will be 10, that number is nothing but the pair factors of 10.
• 1 x 10 = 10
• 2 x 5 = 10
• 5 x 2 =10
• 10 x 1 = 10
So the pair factors of 10 are 1,2, 5 & 10.
### Factors Of -10
The negative numbers of factors of 10 are called factors of -10.
Below You can see the numbers.
• -1(-10➗-1=-10)
• -2(-10➗-2=-5)
• -5(-10➗-5=-2)
• -10(-10➗-10=-1)
The Factors for -10 are -1, -2, -5, -10
## Factors Of 10 And Other Numbers
Here are the common factors of 10 and other numbers.
### The Factors Of 10 And 6
The factors for 10 and 6 are 1, 2, 5, 10, and 1, 2, 3, 6
• 10 factors – 1, 2, 5, 10
• 6 factors – 1, 2, 3, 6
HCF of 10 and 6 is 3
### The Factors Of 10 And 4
The factors for 10 and 4 are 1, 2, 5, 10, and 1, 2, 4
• 10 factors – 1, 2, 5, 10
• 4 factors – 1, 2, 4
HCF of 10 and 4 is 2
### The Factors Of 10 And 12
The factors for 10 and 12 are 1, 2, 5, 10, and 1, 2, 4, 12
• 10 factors – 1, 2, 5, 10
• 12 factors – 1,2,4,12
HCF of 10 and 12 is 2
### The Factors Of 10 And 14
The factors for 10 and 14 are 1, 2, 5, 10, and 1, 2, 7, 14
• 10 factors – 1, 2, 5, 10
• 14 factors – 1, 2, 7, 14
HCF of 10 and 14 is 2
### The Factors Of 10 And 15
The factors for 10 and 15 are 1, 2, 5, 10, and 1, 3, 5, 15
• 10 factors – 1, 2, 5, 10
• 15 factors – 1, 3, 5, 15
HCF of 10 and 15 is 5
### The Factors Of 10 And 18
The factors for 10 and 18 are 1, 2, 5, 10, and 1, 2, 3, 6, 9, 18
• 10 factors – 1, 2, 5, 10
• 18 factors – 1, 2, 3, 6, 9, 18
HCF of 10 and 18 is 2
### The Factors For 10 And 20
The factors for 10 and 20 are 1, 2, 5, 10, and 1,2,4,5,10, 20
• 10 factors – 1, 2, 5, 10
• 10 factors – 1,2,4,5,10, 20
HCF of 10 and20 is 10
### The Factors Of 10 And 24
The factors for 10 and 24 are 1, 2, 5, 10, and 1, 2, 3, 4, 6, 8, 12
• 10 factors – 1, 2, 5, 10
• 24 factors – 1, 2, 3, 4, 6, 8, 12
HCF of 10 and 24 is 2
### The Factors Of 10 And 25
The factors for 10 and 25 are 1, 2, 5, 10, and 1, 2, 5
• 10 factors – 1, 2, 5, 10
• 25 factors – 1,2,5
HCF of 10 and 25 is 5
### The Factors Of 10 And 28
The factors for 10 and 28 are 1, 2, 5, 10, and 1, 2, 4, 7, 14, 28
• 10 factors – 1, 2, 5, 10
• 28 factors – 1, 2, 4, 7, 14, 28
HCF of 10 and 28 is 2
### The Factors Of 10 20 30
The factors for 10, 20, and 30 are 1, 2, 5, 10, and 1, 2, 4, 5, 10, 20, and 1, 2, 3, 5, 6, 10, 15, 30
• 10 factors – 1, 2, 5, 10
• 20 factors – 1, 2, 4, 5, 10, 20
• 30 factors – 1, 2, 3, 5, 6, 10, 15, 30
HCF of 10, 20, and 30 are 10
### The Factors Of 10 And 40
The factors for 10 and 40 are 1, 2, 5, 10, and 1, 2, 4, 5, 8, 10, 20, 40
• 10 factors – 1, 2, 5, 10
• 40 factors – 1, 2, 4, 5, 8, 10, 20, 40
HCF of 10 and 40 is 10
### The Factors Of 10 And 80
The factors for 10 and 80 are 1, 2, 5, 10, and 1, 2, 4, 5, 8, 10,16, 20, 40, 80
• 10 factors – 1, 2, 5, 10
• 80 factors – 1, 2, 4, 5, 8, 10,16, 20, 40, 80
HCF of 10 and 80 are 10
All factors for the numbers can be seen here on Factorsweb
## FAQ
### Which Factor Of 10 Is A Prime Number?
Prime numbers from 1 to 10 are 2, and 5
### How Many Primes Numbers Are There In 10?
There are 4 prime numbers from 1 to 10 which are 2, 3, 5, 7.
### What Are The Factors Of 10?
The factors of 10 are 1, 2, 5, and 10.
### What Are The Prime Numbers?
The prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
### What Are The Factors And Multiples Of 10?
Thus, multiples of 10 are in the form of 10n, where n is a natural number. Factors are the numbers which when multiplied together, produce the original number. Multiples of 10 are 10, 20, 30, 40, 50, etc. Factors of 10 are 1, 2, 5 and 10.
### What Is The Prime Factor Of 1 And 10?
There are 4 prime numbers from 1 to 10. They are 2, 3, 5 and 7.
## Conclusion
Today we learned to know about factors of 10 by also knowing prime factors for 10, factors for 10 in pairs, factors for -10, and factor numbers of 10 with some other number. As you have read the above article you have the answer to the question for the what are factors of 10.
How many prime numbers are there between 10 and 10
What are the prime factors of 10? |
It"s very common once learning about fractions to want to know how transform a portion like 2/3 into a percentage. In this step-by-step guide, we"ll present you how to turn any portion into a percent really easily. Let"s take a look!
Want to easily learn or show students exactly how to transform 2/3 to a percentage? beat this very quick and fun video now!
Before we acquire started in the portion to percentage conversion, let"s go over some an extremely quick portion basics. Remember that a molecule is the number above the fraction line, and the denominator is the number listed below the portion line. We"ll use this later in the tutorial.
You are watching: 2/3 is what percent
When we space using percentages, what we space really speak is the the percent is a fraction of 100. "Percent" means per hundred, and so 50% is the exact same as saying 50/100 or 5/10 in portion form.
So, since our denominator in 2/3 is 3, we could adjust the fraction to make the denominator 100. To do that, we division 100 by the denominator:
100 ÷ 3 = 33.333333333333
Once we have actually that, we deserve to multiple both the numerator and also denominator through this multiple:
2 x 33.333333333333/3 x 33.333333333333=66.666666666667/100
Now we can see that our portion is 66.666666666667/100, which method that 2/3 together a portion is 66.6667%.
We can additionally work this the end in a simpler way by first converting the fraction 2/3 to a decimal. To execute that, we simply divide the molecule by the denominator:
2/3 = 0.66666666666667
Once we have actually the price to that division, we can multiply the price by 100 to make it a percentage:
0.66666666666667 x 100 = 66.6667%
And there you have actually it! Two various ways to transform 2/3 come a percentage. Both space pretty straightforward and also easy come do, but I personally favor the transform to decimal an approach as that takes less steps.
I"ve watched a the majority of students get perplexed whenever a inquiry comes up around converting a portion to a percentage, but if you monitor the measures laid out right here it must be simple. The said, you might still need a calculator because that more complicated fractions (and you can always use our calculator in the type below).
See more: How Does Walton Describe The Stranger He Takes Onboard, How Does Walton Describe The Stranger
If you want to practice, grab yourself a pen, a pad, and a calculator and try to transform a few fractions come a percent yourself.
Hopefully this tutorial has helped you to understand exactly how to transform a fraction to a percentage. You can now go forth and convert fractions to percentages as lot as your little heart desires!
## Fraction as Percentage
Enter a numerator and also denominator |
# What Is A Scale Factor Of 5?
## What is the scale factor of 4 and 16?
Now we want to figure out the scale factor.
To do this, we simplify the ratio using the greatest common factor.
The greatest common factor of 4 and 16 is 4.
The scale factor is \begin{align*}\frac{1^{\prime\prime}}{4\ ft}\end{align*}..
## What is the scale factor of 5?
3. Scale factor with area and volume. If a figure is enlarged by a factor of five, the length will be five times greater, the width will be five times greater and the height will be five times greater. The area will therefore be five times five times greater.
## What is a scale factor of 2?
The size of an enlargement/reduction is described by its scale factor. For example, a scale factor of 2 means that the new shape is twice the size of the original. A scale factor of 3 means that the new shape is three times the size of the original.
## What is a scale factor in math in 7th grade?
● Scale Factor: The ratio of any two corresponding lengths in two similar. geometric figures.
## What is a scale copy?
A scale copy of a figure is a figure that is geometrically similar to the original figure. This means that the scale copy and the original figure have the same shape but possibly different sizes. … In real life, a scale copy is often smaller than the original figure.
## What is a scale in math?
In math, a scale in graphs can be defined as the system of marks at fixed intervals, which define the relation between the units being used and their representation on the graph. … Each small interval or division measures 100 ml. Fun Facts. A ruler is often called a scale.
## How do u find the scale factor?
To find the scale factor, locate two corresponding sides, one on each figure. Write the ratio of one length to the other to find the scale factor from one figure to the other. In this example, the scale factor from the blue figure to the red figure is 1.6 : 3.2, or 1 : 2.
## What is the scale factor of 12?
Architectural ScalesDrawing ScaleScale FactorViewport Scale3/8″ = 1′-0″321/32xp1/2″ = 1′-0″241/24xp3/4″ = 1′-0″161/16xp1″ = 1′-0″121/12xp7 more rows•Aug 12, 2018
## What does a scale of 5 1 mean?
A 50mm line is to be drawn at a scale of 5:1 (ie 5 times more than its original size). The measurement 50mm is multiplied by 5 to give 250mm. A 250mm line is drawn.
## What is a scale factor math is fun?
more … The ratio of the length in a drawing (or model) to the length on the real thing. Example: in the drawing anything with the size of “1” would have a size of “10” in the real world, so a measurement of 150mm on the drawing would be 1500mm on the real horse.
## What is a scale factor example?
A scale factor is a number which multiplies (“scales”) a quantity. For example,the “C” in y = Cx is the scale factor for x. If the equation were y = 5x, then the factor would be 5.
## What is a scale factor of a triangle?
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. … The ratios of corresponding sides are 6/3, 8/4, 10/5. These all reduce to 2/1. It is then said that the scale factor of these two similar triangles is 2 : 1.
## How do you calculate a scale?
To scale an object to a smaller size, you simply divide each dimension by the required scale factor. For example, if you would like to apply a scale factor of 1:6 and the length of the item is 60 cm, you simply divide 60 / 6 = 10 cm to get the new dimension.
## What does a scale mean?
Definition of scale (Entry 5 of 7) 1 : a graduated series of musical tones ascending or descending in order of pitch according to a specified scheme of their intervals. 2 : something graduated especially when used as a measure or rule: such as.
## How do you calculate scale size?
To convert a measurement to a larger measurement simply multiply the real measurement by the scale factor. For example, if the scale factor is 1:8 and the measured length is 4, multiply 4 × 8 = 32 to convert.
## What’s the difference between a scale and a scale factor?
The scale can be written as a scale factor, which is the ratio of the length or size of the drawing or model to the length of the corresponding side or part on the actual object. Scale Factor needs to be the SAME UNITS! … A scale is the ratio between two sets of measurements. |
# W19 Assignment 3
Updated 2019-09-29
## 1. Orbital Changes 🪐
A small particle of mass $m$ is on a circular orbit of radius $R$ around a much larger mass $M$. Suppose that the speed of mass $m$ is suddenly increased (in the same direction) by a factor $\alpha > 1$, that is, $\vec v _\text{final}=\alpha \vec v_\text{initial}$.
(a) Find expressions for the semi-major axis, the eccentricity, the semi-minor axis and the pericenter and apocentre distances, all in terms of just $R$ and $\alpha$. You may follow the logic used in the examples done in class.
$v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)$
By rearranging the equation, we can obtain the semi-major axis of the new orbit after the velocity change.
$\frac{1}{a}=\frac{2}{R}-\frac{(\alpha v_i)^2}{GM},\quad v_i=\sqrt{\frac{GM}{R}}$
Simplify:
\begin{aligned} \frac{1}{a}&=\frac{2}{R}-\frac{\left(\alpha\sqrt{\frac{GM}{R}}\right)^2}{GM}\\ &=\frac{2}{R}-\frac{\alpha^2}{R}\\ a&=\boxed{\frac{R}{2-\alpha^2}} \end{aligned}
Because $\alpha>1$, then we the small particle mass $m$ is now at the periapsis. We can use the equation to get periapsis orbital altitude to find the new eccentricity of the orbit:
$q=a(1-e)=\boxed{R}\\ e=1-\frac{R}{a}=1-\frac{R}{\frac{R}{2-\alpha^2}}=\boxed{\alpha^2-1}$
The semi-minor axis is derived by Pythagorean theorem, since we already did that for assignment 1, I will jump into using the equation:
\begin{aligned} b^2&=a^2(1-e^2)\\ b&=\left(\frac{R}{2-\alpha^2}\right)\sqrt{1-(\alpha^2-1)^2}\\ &=\frac{R}{2-\alpha^2}(\alpha\sqrt{2-\alpha^2})\\ &=\boxed{\frac{R\alpha}{\sqrt{2-\alpha^2}}} \end{aligned}
Since we increased the speed in the prograde direction, we increased our apoapsis. The new apoapsis is:
\begin{aligned} Q&=a(1+e)\\ &=\frac{R}{2-\alpha^2}(1+(\alpha^2-1))\\ &=\boxed{\frac{R\alpha^2}{2-\alpha^2}} \end{aligned}
(b) There is a maximum value of $\alpha$ for which this is a sensible problem. Why is that, and what is the maximum value of $\alpha$?
Because the geometry of the elliptical orbit is a conic section – a section cut of a cone shape. Given enough $\alpha$. the “tilt of the slice” will be a parabola and then a hyperbola (see figure below – source: voer.edu.vn). In physical sense, if we increase $\alpha$, our particle velocity would be so great that it would reach the escape velocity, and it will escape the orbit.
The escape velocity, the factor we multiply our orbiting speed by $\alpha$ is given as:
\begin{aligned} v_{esc}=v_{f}&=\sqrt{\frac{2GM}{a}}\\ \alpha v_i&=\sqrt{\frac{2GM}{R}},\quad v_i=\sqrt{\frac{GM}{R}}\\ \alpha \sqrt{\frac{GM}{R}}&=\sqrt{\frac{2GM}{R}}\\ \alpha &=\boxed{\sqrt{2}} \end{aligned}
## 2. Kepler’s Equation
Kepler solved his own equation, $E-e\sin E=\phi$ (where $\phi=nt^\star$ is the mean anomaly), by an iterative process; here you will explore that method. Them method relies on rewriting the equation as $E=\phi + e\sin E$. For an initial guess $E_0=\phi$, we can then proceed to a next guess for $E$: $E_1=\phi + e\sin E_0$ and then iterate until $E_n=\phi + e\sin E_{n-1} \simeq E_{n-1}$ to within some tolerance.
Consider an ellipse with eccentricity 0.5. Use mean-anomaly points corresponding to (0, π/10, π/5, 3π/10, 2π/5, π/2, 3π/5, 7π/10, 4π/5, 9π/10, π).
(a) For ONE of these mean-anomaly points (which is not at 0 or π radians), produce a table showing each step in your iteration of Kepler’s equation until you reach a satisfactory value of $E$ to within a difference $\vert E_n-E_{n-1}\vert < 0.01$ rad. Also show the calculation to compute the true anomaly.
I will choose 3π/10 (0.9425 rad) as the mean-anomaly point. I wrote a C program to help me compute the iterations. See the attached Appendix A for the C program code. Here are the results:
Iterations Eccentric anomaly
0 0.9425
1 1.3470
2 1.4300
3 1.4375
The true anomaly ($\theta$) can be found using:
\begin{aligned} \tan\left(\frac{\theta}{2}\right)&=\sqrt{\frac{1+e}{1-e}}\tan\left(\frac{E}{2}\right)\\ \theta&=2\cdot\tan^{-1}\left(\sqrt{\frac{1+e}{1-e}}\tan\left(\frac{E}{2}\right)\right) \end{aligned}
(b) Make a second table with the full set of mean-anomaly points, the corresponding eccentric-anomaly $E$ values and the corresponding true-anomaly $\theta$ values.
Again, I wrote a C program to compute the all the values for the table. See Appendix A for the C program code.
Mean-anomaly $\phi$ Eccentric-anomaly $E$ True-anomaly $\theta$
0.0$\pi$ 0.0000 0.0000
0.1$\pi$ 0.5900 0.9690
0.2$\pi$ 1.0635 1.5896
0.3$\pi$ 1.4375 1.9750
0.4$\pi$ 1.7486 2.2419
0.5$\pi$ 2.0204 2.4462
0.6$\pi$ 2.2692 2.6158
0.7$\pi$ 2.5016 2.7635
0.8$\pi$ 2.7168 2.8938
0.9$\pi$ 2.9345 3.0218
1.0$\pi$ 3.1416 3.1416
(c) Make a plot showing $E$ and $\theta$ as functions of $\phi$ and comment on whether it would be OK to neglect the orbit’s eccentricity when estimating the position of the object.
Using the data from Part (B), I used Excel to create the following plot.
As the graph clearly shows, there is a significant discrepancy between the eccentric anomaly, and the true anomaly of an elliptical orbit. Therefore it is not OK to neglect the eccentricity, in this case $e=0.5$, when estimating the position of the object.
## 3. Temperature Changes in an Eccentric Orbit 🌗
The planet HD 20782b orbits HD 20782, a G3V star with mass 1.05 times that of the Sun and luminosity 1.25 times that of the Sun. The planet’s orbit has a semi-major axis of 1.397 au, an orbital period of 597 days, and an eccentricity of 0.956. The planet’s mass is about 2 times that of Jupiter and you may assume its albedo is similar to that of Jupiter at about 0.5.
(a) (Review) Compute the periapsis and apoapsis distances of the planet from the star.
$q=a(1-e)=(1.397)(1-0.956)=\boxed{0.0615\text{au}}\\ Q=a(1+e)=(1.397)(1+0.956)=\boxed{2.7325\text{au}}$
(b) Assume the planet is spinning rapidly. In class we derived the dependence of planetary blackbody temperature as a function of orbital distance in the Solar system: $T_p=279\text{K}(1-A)^{1/4}\frac{1}{\sqrt{r_\text{au}}}$. Adapt this formula to this new stellar system — that is, work out the new constant that should take the place of 279 K, and take into account the numerical value of the albedo.
First, we need to adapt the luminosity. The sun’s luminosity is the total power outputted by the sun: $L_\odot=4\pi R^2\odot \sigma T\odot^4=3.839\times 10^{26}$W. We multiply this by 1.25 and get:
$L_s=4.799\times 10^{26} \text{W}$
To find the temperature, we need to equate the power intercepted from the star HD 20782 and the power emitted (assuming that temperature on the planet is much lower than the star). Power intercepted is a function of distance from the star ($r$) and is given by flux × surface area × absorption:
$P_{in}(r)=F_s\times A_{p_\text{intercept}}\times (1-A)$
Where $F_s$ is the flux [W/m2] which is the luminosity divided by surface area: $F_s(r)=L_s/4\pi r^2$. The area intercepted by the planet is the “shadow” of the planet which is a disk, given by $A_{p_\text{intercept}}=\pi R_p^2$ where $R_p$ is the radius of the planet.
The power outputted by the planet is by:
$P_{out}=A_p\times\sigma\times T_p^4$
Where $A_p$ is the planet’s surface area, and $T_p$ is the surface temperature of the planet.
We equate power in, and power out and simplify:
\begin{aligned} P_{in}&=P_{out}\\ F_s\cdot A_{p_\text{intercept}}\cdot(1-A)&=A_p\cdot\sigma\cdot T_p^4\\ \frac{L_s}{4\pi r^2}\cdot\pi R_p^2\cdot(1-A)&=4\pi R_p^2 \cdot \sigma \cdot T_p^4\\ \frac{L_s}{4r^2}(1-A)&=4\pi\sigma T_p^4 \end{aligned}
Now we rearrange the equation to match the equation in question by isolating for $T_p$:
\begin{aligned} \frac{L_s}{4r^2}(1-A)&=4\pi\sigma T_p^4\\ T_p^4&=\frac{L_s}{16\pi r^2\sigma}(1-A)\\ T_p&=\left(\frac{L_s}{16\pi\sigma}\right)^{1/4}\frac{1}{\sqrt{r}}(1-A) \end{aligned}
We need to do one more thing. The equation we had at the top has the distance from the star $r$ in units of AU. So we need to account for it here. 149597900000m = 1 AU.
\begin{aligned} T_p&=\left(\frac{L_s}{16\pi\sigma}\right)^{1/4}\left(\frac{1}{\sqrt{r_m}}\times\sqrt{\frac{1\text{au}}{1.496\times10^{11}\text{m}}}\right)(1-A)\\ &=\left[\left(\frac{L_s}{16\pi\sigma}\right)^{1/4}\frac{1\sqrt{\text{au}}}{\sqrt{1.496\times10^{11}\text{m}}}\right](1-A)\frac{1}{\sqrt{r_{au}}} \end{aligned}
Verifying the units: in this section, I’m only showing my work to see that the units for the above derivation and conversion makes sense.
\begin{aligned} \text{K}&=\left[\left(\frac{\text{W}}{\text{Wm}^{-2}\text{K}^{-4}}\right)^{1/4}\frac{\text{au}^{1/2}}{\text{m}^{1/2}}\right]\frac{1}{\text{au}^{1/2}}\\ &=\left[\left(\text{m}^{2}\text{K}^{4}\right)^{1/4}\frac{\text{au}^{1/2}}{\text{m}^{1/2}}\right]\frac{1}{\text{au}^{1/2}}\\ &=\text{m}^{1/2}\text{K}\frac{\text{au}^{1/2}}{\text{m}^{1/2}}\frac{1}{\text{au}^{1/2}}\\ &=\text{K}\text{ au}^{1/2}\frac{1}{\text{au}^{1/2}}\\ &=\text{K}\\ \end{aligned}
👌 Cool. We can proceed.
Plugging in all the values we calculated so far, where $\sigma$ is Stefan-Boltzmann consant. The constant that replaces the 279K is 294.5K:
$T_p=\left[\left(\frac{L_s}{16\pi\sigma}\right)^{1/4}\frac{1\sqrt{\text{au}}}{\sqrt{1.496\times10^{11}\text{m}}}\right](1-A)\frac{1}{\sqrt{r_{au}}}\\ T_p=\boxed{294.5\text{K}}(1-A)\frac{1}{\sqrt{r_{au}}}$
(c) Assuming that there no signi cant internal heating from the planet, determine its temperature and the wavelength of peak emission at both periapsis and apoapsis. In what part of the spectrum do these wavelengths fall?
At Periapsis: the planet’s distance from the star is only 0.0615au. The temperature is 594K. 🔥 $$T_{p_{pe}}=294.5\text{K}(1-(0.5))\frac{1}{\sqrt{0.0615}}=\boxed{594 \text{K}}$$
The wavelength of maximum intensity or peak wavelength $\lambda_p$ is in accordance with Wien’s law. We plug in the numbers here: $$\lambda_{p_{pe}}=\frac{2.9\times 10^3 \mu\text{m}}{T}=\frac{2.9\times 10^3 \mu\text{m}}{594}=\boxed{4.884\mu\text{m}}$$
At Apoapsis: the planet’s distance from the star is 2.7325au. The temperature is 89.1K. ⛄ $$T_{p_{ap}}=294.5\text{K}(1-(0.5))\frac{1}{\sqrt{2.7325}}=\boxed{89.1 \text{K}}$$
Using similar calculations as above, we get peak wavelength at:
$\lambda_{p_{ap}}=\frac{2.9\times 10^3 \mu\text{m}}{89.1}=\boxed{32.6\mu\text{m}}$
So when the planet is at periapsis, the planet is very hot at 594K and emits peak wavelength of 4.9μm in the infrared spectrum. When the planet is at apoapsis, the planet is cold at only 89.1K and emits a peak wavelength of 32.6μm in the far-infrared spectrum.
## Appendix A
### Code Used for Problem 2 Part 1
#include <stdio.h>
#include <math.h>
#define THRES 0.01
#define PI acos(-1)
void part1(void)
{
printf("Part 1 #################\n");
const double starting_mean_anomaly = 3.0 * PI / 10.0;
const double eccentricity = 0.5;
double eccentric_anomaly;
double eccentric_anomaly_prev = starting_mean_anomaly;
int iterations = 1;
printf("| Iterations | Eccentric anomaly |\n");
printf("|--:|--:|\n");
printf("| 0 | %.4f |\n", starting_mean_anomaly);
eccentric_anomaly = starting_mean_anomaly + eccentricity * sin(eccentric_anomaly_prev);
printf("| 1 | %.4f |\n", eccentric_anomaly);
while (fabs(eccentric_anomaly_prev - eccentric_anomaly) > THRES)
{
eccentric_anomaly_prev = eccentric_anomaly;
eccentric_anomaly = starting_mean_anomaly + eccentricity * sin(eccentric_anomaly_prev);
iterations++;
printf("| %d | %.4f |\n", iterations, eccentric_anomaly);
}
printf("E_%d = %.4f rad\n", iterations, eccentric_anomaly);
}
int main(void)
{
part1();
}
### Code Used for Problem 2 Part 2
#include <stdio.h>
#include <math.h>
#define THRES 0.01
#define PI acos(-1)
void part2(void)
{
const double eccentricity = 0.5;
const double mean_anomalies[] = {
0.0, 0.1 * PI, 0.2 * PI, 0.3 * PI, 0.4 * PI,
0.5 * PI, 0.6 * PI, 0.7 * PI, 0.8 * PI, 0.9 * PI, PI
};
const int mean_anomalies_length = 11;
// Ouptut
printf("Part 2 #################\n");
printf("| Mean-anomaly | Eccentric-anomaly $E$ | True-anomaly $\\theta$ |\n");
printf("|--:|--:|--:|\n");
for (int i = 0; i < mean_anomalies_length; i++)
{
// Mean anomaly
const double mult = mean_anomalies[i] / PI;
// Get E
double eccentric_anomaly;
double eccentric_anomaly_prev = mean_anomalies[i];
eccentric_anomaly = mean_anomalies[i] + eccentricity * sin(eccentric_anomaly_prev);
while (fabs(eccentric_anomaly_prev - eccentric_anomaly) > THRES)
{
eccentric_anomaly_prev = eccentric_anomaly;
eccentric_anomaly = mean_anomalies[i] + eccentricity * sin(eccentric_anomaly_prev);
}
// Get theta
double true_anomaly = 2.0 * atan(
sqrt((1 + eccentricity) / (1 - eccentricity)) * tan(eccentric_anomaly / 2)
);
// Print row
printf("| %.1f$\\pi$ | %.4f | %.4f |\n", mult, eccentric_anomaly, true_anomaly);
}
}
int main(void)
{
part2();
} |
Mind Matters Natural and Artificial Intelligence News and Analysis
3. In Infinity, Lines and Squares Have an Equal Number of Points
We can demonstrate this fact with a simple diagram
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In previous posts, we have established that two sets are of the same size if there is a one-to-one correspondence between the elements of both sets. Applying this principle to Cantor’s theory of infinity leads us to the weird but valid conclusion that the number of points on a line segment is the same as the number of points in a square. To show that this is true, here is a picture of a unit length line segment and a unit square.
Let’s choose a point on the line segment. Let’s say 0.6917381276543… . It’s shown with a big blue dot on the line segment on the left. If this point corresponds to an irrational number, it goes on forever without repeating or showing a discernible pattern.
We’re going to break up this number into two numbers. The first number is every other digit. Shown in red below the line segment is this number, namely 0.6131753… . The remaining digits form the second number, shown in green. It is 0.978264… .
A single number between zero and one can thus be broken into two numbers between zero and one. We can take these two numbers and as the x and y coordinates on the unit square shown on the right. They define the blue dot shown in the square on the right. So the blue dot on the line segment maps to the blue dot in the square.
For every point on the line segment, there is one and only one blue dot in the square. The size of the sets of infinity on a line segment and in a square are exactly the same. An extension of this argument shows that the number of points in a cube is the same as the number of points on a line segment. This is counterintuitive and weird.
Backward mapping from the square to the line segment is also possible. Take the two coordinates defining the blue dot in the square and shuffle them together to obtain one number. Two numbers between zero and one can always be combined to make a single number between one and zero. This new number is the point on the line segment.
Since there is a one-to-one mapping of every point on the line segment to every point in the square, the number of points on a line segment is the same as the number of points in a square.
But can’t you draw a line segment and square on a sheet of paper and do the mapping? Isn’t this an illustration of infinity in reality? No. Each point is assumed to be measured with infinite precision. Precision in reality is always finite. So the conclusion that there are no infinities in reality stands strong.
This counterintuitive result, driven by Cantor’s theory of infinities is strange. Nevertheless, it is a valid property of the infinite.
Next: We’ll show that the entire Library of Congress is digitally encoded on almost every randomly chosen number.
We hope you enjoy this series on the unique, reality-defying nature of mathematical infinities. Here are all five parts — and a bonus:
Part 1: Why infinity does not exist in reality. A few examples will show the absurd results that come from assuming that infinity exists in the world around us as it does in math. In a series of five posts, I explain the difference between what infinity means — and doesn’t mean — as a concept.
Part 2. Infinity illustrates that the universe has a beginning. The logical consequences of a literally infinite past are absurd, as a simple illustration will show. The absurdities that an infinite past time would create, while not a definitive mathematical proof, are solid evidence that our universe had a beginning.
Part 3. In infinity, lines and squares have an equal number of points Robert J. Marks: We can demonstrate this fact with simple diagram. This counterintuitive result, driven by Cantor’s theory of infinities is strange. Nevertheless, it is a valid property of the infinite.
Part 4. How almost any numbers can encode the Library of Congress. Robert J. Marks: That’s a weird, counterintuitive — but quite real — consequence of the concept of infinity in math. Math: Almost every number between zero and one, randomly chosen by coin flipping, will at some point contain the binary encoding of the Library of Congress.
and
Part 5: Some infinities are bigger than others but there’s no biggest one Georg Cantor came up with an ingenious proof that infinities can differ in size even though both remain infinite. In this short five-part series, we show that infinity is a beautiful — and provable — theory in math that can’t exist in reality without ludicrous consequences.
You may also wish to read: Yes, you can manipulate infinity in math. The hyperreals are bigger (and smaller) than your average number — and better! (Jonathan Bartlett) |
## Intermediate Algebra (12th Edition)
$x=17$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $3\sqrt{x-1}=2\sqrt{2x+2} ,$ square both sides of the equation and then isolate the variable. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( 3\sqrt{x-1} \right)^2=\left( 2\sqrt{2x+2} \right)^2 \\\\ 9(x-1)=4(2x+2) .\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} 9(x)+9(-1)=4(2x)+4(2) \\\\ 9x-9=8x+8 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 9x-8x=8+9 \\\\ x=17 .\end{array} Upon checking, $x=17$ satisfies the original equation. |
# GMAT Geometry: Triangles on the Coordinate Plane (Basics)
Among all of the concepts explored in GMAT, triangles are a commonly revisited concept. When exploring triangles on the coordinate plane, it is important to understand the following terminology:
• Ordered Pair: identification of a point through its coordinates, and is typically written in the form of (a, b) where a is a point on the x-axis, and b is a point on the y-axis
• Origin: point of intersection of x-axis and y-axis, (0, 0)
• Quadrants: a sector on the graph; there are 4 quadrants with coordinates of (+, +) for quadrant I, (-, +) for quadrant II, (-, -) for quadrant III, and (+, -) for quadrant IV
• Slope: steepness of a line defined by the rise over run which is the units that are present vertically divided by the units present horizontally
Upon understanding the terminology, it is crucial to understand the basic equations regarding the distance between two points, or a line, in order to compose a more complex view on triangles.
The basic equation will be y = mx + b where y is a point on the y-axis, m is the slope, x is a point on the x-axis, and b is the intercept on the y-axis.
Since the GMAT normally provides a coordinate graph, counting the units on the y-axis, and the units on the x-axis will be relatively easy.
For example, if two points on the graph have coordinates (0, 0) and (3, 6), we can find the slope by plugging the coordinates into the theorem of rise over run. The difference between the two points will be a value of 3 for the x-axis, and a value of 6 for the y-axis. The slope will be 6/3, which gives a value of 2. Finding the slope is crucial in determining many factors concerning a triangle, and will be often used to determine the coordinates of the three points.
The most basic equation regarding calculating the length of the sides of the triangle will include a^2 + b ^2 = c ^2 where a and b are the two shorter sides, and c is the longer side. This equation is only viable if the triangle is a right-angled triangle. This means that two of the lines are perpendicular from one another, and form a 90 degree angle.
For example, if we are given three points A at (0, 0), B at (3, 0) and C at (3, 5), we can find the length of each side of the triangle easily.
First of all, the line AB would a horizontal line, as there is no difference on the y-axis; thus, by only taking a look at the coordinates on the x-axis, we can determine its length. 3 – 0 = 3.
The line BC would be a vertical line as there is no difference in the x-axis. Similarly, we can find the length by just looking at the coordinates of the y-axis. 5 – 0 = 5. From there, we can use the equation a^2 + b^2 = c ^2, and plug in the values to get.
(3^2) + (5^2) = c^2
9 + 25 = c^2
34 = c^2
c = sqrt(34)
c is approximately 5.83
It is important to write the units out as well. With this information, it will then be easy to determine the perimeter of the triangle. The perimeter will be the sum of all sides. Multiplying the two sides that are perpendicular to one another, and dividing that number by 2 can find the area of the triangle.
In the scenario above, the perimeter of the triangle would be 3 + 5 + 5.83, which would be approximately 13.83. The area would be 5 x 3 / 2, which would be 15.
It is important to note that most of the questions regarding the triangles on the coordinate plane will generally involve right-angled triangles.
Key Takeaway:
Test-makers generally attempt to complicate the situation by providing two points, and asking candidates to find the third coordinate capable of constructing a triangle that will meet requirements.
Questions regarding isosceles triangles and any other types of triangles will require different equations; however, these are generally not found in coordinate plane questions, as it is difficult for candidates to find the angles involved.
Author :
GoGMAT, founded in 2009, is an adaptive GMAT preparation platform developed by the best instructors in the industry (with 740+ GMAT scores and strong teaching experience). GoGMAT team consists of high profile members, successful entrepreneurs, experts in educational field, top-notch business school alumni with successful track records as investment bankers and management consultants. Our GMAT experts, university professors and successful entrepreneurs, analyzed GMAT test algorithms for almost 12 months. We used their knowledge and skills as the basis to create one of the most efficient solutions in the world for GMAT preparation.
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# Question Video: Using Triangle Congruence Criteria to Establish Congruence Mathematics • 11th Grade
Determine whether the triangles in the given figure are congruent, and, if they are, state which of the congruence criteria proves this.
02:30
### Video Transcript
Determine whether the triangles in the given figure are congruent, and if they are, state which of the congruence criteria proves this.
So we’ve been given a diagram of two triangles, triangle 𝐴𝐵𝐶 and triangle 𝐴 dash 𝐵 dash 𝐶 dash. We’ve been given various pieces of information about these two triangles. And at a glance, we can see that the values are the same. However, we need to look at the congruence criteria to formally prove whether these two triangles are or aren’t congruent.
Let’s begin by looking at the side of length 2.53 that appears in both triangles. This is side 𝐴𝐵 in the first triangle and side 𝐴 dash 𝐶 dash in the second triangle. So we have that 𝐴𝐵 is congruent to 𝐴 dash 𝐶 dash, and the S in brackets indicates that this is a side.
Next, let’s look at the angle of 60.34 degrees that both triangles have. This is angle 𝐵 in the first triangle and angle 𝐶 dash in the second triangle. So we have that angle 𝐵 is congruent to angle 𝐶 dash, and the A in brackets indicates that this is an angle.
Next, let’s look at the side that measures 3.68 units. This is side 𝐵𝐶 in the first triangle and side 𝐶 dash 𝐵 dash in the second triangle. So we have that 𝐵𝐶 is congruent to 𝐶 dash 𝐵 dash. Again, the S in brackets indicates that we’re referring to a side.
So let’s look at what we’ve shown. We’ve shown that these two triangles have two sides and an angle in common. But more specifically, it’s the included angle, the angle that’s between the two known sides. This is sufficient to prove that the two triangles are congruent to each other. By looking at the letters in brackets, we can see which congruence criteria we’ve used, side angle side is the SAS congruency criteria.
So our answer to the problem is that yes, these two triangles are congruent, and the congruency criteria used to prove this is the side angle side criteria. |
# What is commutative property of real numbers?
The word “commutative” comes from “commute” or “move around”, so the Commutative Property is the one that refers to moving stuff around. For addition, the rule is “a + b = b + a”; in numbers, this means 2 + 3 = 3 + 2. For multiplication, the rule is “ab = ba”; in numbers, this means 2×3 = 3×2.
## In this regard, what is the definition of properties of real numbers?
The Properties of Real Numbers. For clarity, “properties” in this context refer to the characteristics or behaviors of real numbers under the operations of addition and/or multiplication that are accepted even without proof.
Additionally, what are the 6 properties of real numbers? Real Numbers are Commutative, Associative and Distributive:
• Commutativeexample.
• a + b = b + a2 + 6 = 6 + 2.
• ab = ba4 × 2 = 2 × 4.
• Associativeexample.
• (a + b) + c = a + ( b + c ) (1 + 6) + 3 = 1 + (6 + 3)
• (ab)c = a(bc)(4 × 2) × 5 = 4 × (2 × 5)
• Distributiveexample.
• a × (b + c) = ab + ac3 × (6+2) = 3 × 6 + 3 × 2.
## People also ask, what is an example of the commutative property?
For example, if you are adding one and two together, the commutative property of addition says that you will get the same answer whether you are adding 1 + 2 or 2 + 1. The commutative property of addition says that you can also add 2 + 1 + 3 or 3 + 2 + 1 and still get the same answer.
## What is associative property and commutative property?
The associative property states that you can re-group numbers and you will get the same answer and the commutative property states that you can move numbers around and still arrive at the same answer.
## What are the 4 properties in math?
There are four basic properties of numbers: commutative, associative, distributive, and identity. You should be familiar with each of these. It is especially important to understand these properties once you reach advanced math such as algebra and calculus.
## What is the distributive property in math?
The distributive property is one of the most frequently used properties in math. In general, this term refers to the distributive property of multiplication which states that the. Definition: The distributive property lets you multiply a sum by multiplying each addend separately and then add the products.
## What are the basic properties?
There are four (4) basic properties of real numbers: namely; commutative, associative, distributive and identity. These properties only apply to the operations of addition and multiplication. That means subtraction and division do not have these properties built in.
## What is a unique real number?
Real numbers are closed under addition, subtraction, and multiplication. That means if a and b are real numbers, then a + b is a unique real number, and a ⋅ b is a unique real number. For example: 3 and 11 are real numbers. 3 + 11 = 14 and 3 ⋅ 11 = 33.
## What are the 5 properties of math?
Commutative Property, Associative Property, Distributive Property, Identity Property of Multiplication, And Identity Property of Addition.
## What are the six properties?
What are the Properties included? Commutative Property of Addition. Commutative Property of Multiplication. Associative Property of Addition. Associative Property of Multiplication. Additive Identity Property. Multiplicative Identity Property. Additive Inverse Property. Multiplicative Inverse Property.
## What are the algebraic properties?
The properties involved in algebra are as follows: Commutative property of Addition: Commutative property of Multiplication: Associativity property of Addition and Multiplication: Distributive property. Additive identity property: Additive inverse property: Multiplicative inverse property:
## What is the identity property in math?
Identity property. The identity property for addition tells us that zero added to any number is the number itself. Zero is called the “additive identity.” The identity property for multiplication tells us that the number 1 multiplied times any number gives the number itself.
## How do you explain commutative property?
The word “commutative” comes from “commute” or “move around”, so the Commutative Property is the one that refers to moving stuff around. For addition, the rule is “a + b = b + a”; in numbers, this means 2 + 3 = 3 + 2. For multiplication, the rule is “ab = ba”; in numbers, this means 2×3 = 3×2.
## What is the formula of commutative property?
Lesson Summary The commutative property of multiplication tells us that it doesn’t matter in what order you multiply numbers. The formula for this property is a * b = b * a. For example, it doesn’t matter if we multiply 5 * 4 or 4 * 5. We will end up with the same answer.
## What are the 4 properties of addition?
Properties of Addition. There are four mathematical properties which involve addition. The properties are the commutative, associative, additive identity and distributive properties. Additive Identity Property: The sum of any number and zero is the original number.
## What are all the properties of addition?
There are four mathematical properties which involve addition. The properties are the commutative, associative, identity and distributive properties. Associative Property: When three or more numbers are added, the sum is the same regardless of the grouping of the addends.
## What is the distributive property in multiplication?
The distributive property of multiplication states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately, then adding the two products together for the same result as multiplying the first number by the sum.
## What is commutative property of division?
The commutative property of an operator is the property that . Multiplication (of integers, rationals, real, and complex numbers) is commutative: Division is not commutative: Subtraction is not commutative: There are some systems, such as quaternions or matrices, where multiplication is not commutative: .
## What is an example of distributive property?
The distributive property of multiplication over addition can be used when you multiply a number by a sum. For example, suppose you want to multiply 3 by the sum of 10 + 2. 3(10 + 2) = ? According to this property, you can add the numbers and then multiply by 3.
## What is the associative property in math?
Definition: The associative property states that you can add or multiply regardless of how the numbers are grouped. By ‘grouped’ we mean ‘how you use parenthesis’. In other words, if you are adding or multiplying it does not matter where you put the parenthesis. Add some parenthesis any where you like!.
## What is a commutative property of multiplication?
The commutative property of multiplication states that two numbers can be multiplied in either order. |
# Perfect Squares and Square Roots
Hello, in this video, we will explore how to simplify square roots and find perfect squares.
The symbol in math that we use to represent square roots is called a radical, and it looks like this: $$\sqrt{}$$. The number that appears under the radical is called the radicand. For example, in this expression, $$\sqrt{20}$$, which we read as, “the square root of twenty,” 20 is the radicand. Just like the fact that 20 is the same as $$\frac{20}{1}$$, or $$20^{1}$$, but we don’t write the 1, because it is a given, in square roots there is something similar. There is a given 2 in the bent arm of the radical that we do not normally write when we are looking for a square root, $$\sqrt[2]{}$$. This number is called the index and it determines what root of the radicand we are trying to find. For example, this: $$\sqrt{20}$$, is asking for the square root, or the second root of 20 and this: $$\sqrt[3]{125}$$, which we read as “the cube root of one hundred twenty-five,” is asking us to find the third root of 125.
I want you to practice this on your own. Pause the video and label the different parts of this expression: $$\sqrt[4]{200}$$.
Think you’ve got it? The root symbol is the radical. The index is 4 because it’s the number in the bent arm of the radical. And the radicand is the number under the radical symbol, so in this case, 200.
$$\sqrt{}$$ – radical
4 – index
Great work!
Now let’s talk about how we find the square root of a number. To do this, we ask ourselves, “what number multiplied by itself will give us the radicand?” For example, what is the square root of 25: $$\sqrt{25}$$? The factors of 25 are $$5\times 5$$, which means that the square root of 25 is 5. This also happens to be what we call a perfect square. A perfect square is when we are simplifying a square root and nothing remains under the radical.
What are the square roots of 36, 81, and 144? Pause the video and try these on your own. When you’re finished, we’ll look over them together.
$$\sqrt{36}=6, \text{ because}\text{ }6\times 6=36$$
$$\sqrt{81}=9, \text{ because}\text{ }9\times 9=81$$
$$\sqrt{144}=12, \text{ because}\text{ }12\times 12=144$$
We get these nice, pretty numbers because we are taking the square roots of perfect squares. But what if we aren’t given a perfect square? Well, then we will have to simplify the square root.
## Simplifying Square Roots
When simplifying a square root, we will get all the perfect squares out from under the radical and whatever is remaining from the factors of the radicand stays under the radical. Let’s look at an example.
Simplify the square root of 40: $$\sqrt{40}$$.
We will start by finding the factors of 40, which are $${2}\times {2}\times {2}\times{5}$$.
Since $$2\times{2}$$ creates a perfect square, we can bring the 2 to the front and what remains under the radical is $$2\times{5}$$.
There are no more perfect squares to take out, so we simply multiply these numbers together to get our new radicand.
Therefore, the square root of 40 is equal to 2 square root of 10: $$\sqrt{40}$$ = $$2\sqrt{10}$$.
Now I want you to try one. Simplify the square root of 96: 96. Pause the video here and simplify. When you’re done, we’ll take a look at it together.
Think you’ve got it? First, we need to find the factors of 96.
$$96 = 2 \times {2}\times {2}\times {2} \times {2}\times{ 3}$$
It is easy to see the perfect squares when the factors are written in this form. We can pull out two sets of 2 from our radicand, which will leave us with $$2\times{2}=4$$ in front of our new radical. We still have a 2 and a 3 from our factor list, so we multiply these numbers together to get 6, and this is our new radicand. Therefore, the most simplified form is:
$$\sqrt{96}= 4\sqrt{6}$$
I hope this video on square roots and perfect squares was helpful. Thanks for watching, and happy studying!
### How do I find a square root?
#### A
The square root is the inverse of squaring a number. When we square a number we multiply the number by itself. 4 squared is $$4\times4$$, which equals 16. 16 is a perfect square, and 4 is the square root of 16. Not all numbers will have a nice whole number square root. For example, a number like 50 is not a perfect square because it does not have an integer square root. There is no way to multiply an integer by itself to create a product of 50. 50 does not have an integer square root, but 49 does. The square root of 49 is 7 because $$7\times7$$ equals 49. 49 is considered the perfect square, and 7 is considered the square root. The radical symbol is used to notate square roots. For example, the square root of 49 is expressed as $$\sqrt{49}$$:$$\sqrt{49}=7$$. This is read as “the square root of 49 equals 7”.
### What are the whole number square roots from 1 to 20?
#### A
The whole number square roots from 1 to 20 are $$\sqrt1=1$$,$$\sqrt4=2$$,$$\sqrt9=3$$, and $$\sqrt{16}=4$$.
Not all numbers will have whole number square roots. For example, 1 has a square root because 1 can be expressed as the product of two equal integers, 1ｘ1. 4 has a whole number square root because it can be expressed as the product of 2ｘ2. However, notice how 2 and 3 will not have whole number square roots. 2 cannot be expressed as an integer multiplied by itself. Similarly, 3 cannot be expressed as an integer multiplied by itself.
### What are numbers with integer square roots?
#### A
Numbers with integer square roots are called perfect squares. For example, 64 is a number with an integer square root. 64 is the product of $$8\times8$$. When a number can be created by multiplying an integer by itself, it is called a perfect square. Examples of numbers with integer square roots are 1, 4, 9, 16, and 25. All of these numbers have integer square roots. 25 is a perfect square which is the product of $$5\times5$$. This means that 5 is the square root of 25.
Not all numbers will have integer square roots. For example, 5 does not have an integer square root. No integer times itself has a product of 5. We can get close: $$1\times=1$$, $$2\times2=4$$, and $$3\times3=9$$, but 5 is not a perfect square, so it does not have an integer square root.
### What are the first 20 perfect squares?
#### A
The first 20 perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, and 400. Perfect squares are found by squaring an integer. $$1\times1=1$$,$$2\times2=4$$,$$3\times3=9$$,$$4\times4=16$$… and so on. Not every number will be a perfect square. Perfect squares can be thought of as literal squares. A square is created using two equal integers as side lengths. This means that the result, or product, will be a perfect square.
### What are the perfect squares from 1 to 100?
#### A
There are only ten perfect squares from 1 to 100. 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. These perfect squares are the result of multiplying a number by itself. For example, 1 is a perfect square because $$1\times1=1$$. 4 is a perfect square because $$2\times2=4$$. 9 is a perfect square because $$3\times3=9$$. Not all numbers are perfect squares. It can be helpful to think of perfect squares as actual squares with a length and a width. For example, 100 is a perfect square built from side lengths of 10 and 10. $$10\times10=100$$, therefore 100 is a perfect square.
### How do you determine perfect squares?
#### A
Numbers that are considered perfect squares are the result of multiplying an integer by itself. For example, 25 is a perfect square because it is the product of $$5\times5$$. A number multiplied by itself creates a perfect square.
In order to determine if a number is a perfect square, you need to identify if it is the product of an integer squared. If the number can be formed by squaring an integer, then it is a perfect square. For example, 33 is not a perfect square because there is no way to square an integer number and have the product be 33. We can be sure this is true since $$5\times5=25$$ and $$6\times6=36$$. Because 33 is between 25 and 36, its square root must be between $$\sqrt{25}=5$$ and $$\sqrt{36}=6$$. Since there is no integer between 5 and 6, 33 cannot have an integer square root. However, 36 is a perfect square because it is the result of squaring 6. $$6^2=36$$, so 36 is considered a perfect square.
### Why are there no perfect squares between 144 and 169?
#### A
Perfect squares are the result of squaring a number. For example, 81 is a perfect square because it is the result of squaring 9: $$9\times9=81$$. 144 is a perfect square because it is the result of squaring 12: $$12\times12=144$$. 169 is a perfect square because it is the result of squaring 13: $$13\times13=169$$. Notice how there are no perfect squares between 144 and 169. If a number is between 144 and 169, its square root must be between $$\sqrt{144}=12$$ and $$\sqrt{169}=13$$. Since there is no integer between 12 and 13, there are no perfect squares between 144 and 169.
We cannot create a perfect square by multiplying something like $$11.5\times11.5$$ because it is not the product of two equal integers. The square roots of perfect squares need to be integers.
## Practice Questions
Question #1:
Which set of numbers contains all perfect squares?
33, 99, 55, 66
33, 99, 55, 66
36, 9, 25, 100
81, 36, 25, 41
The correct answer is 36, 9, 25, 100. Thirty-six is a perfect square composed of $$6×6$$, nine is a perfect square composed of $$3×3$$, twenty-five is a perfect square composed of $$5×5$$, and one hundred is a perfect square composed of $$10×10$$.
Question #2:
Which pair shows a correct match between the perfect square and its whole number square root?
$$\sqrt{144}=12$$
$$\sqrt{36}=2$$
$$\sqrt{25}=4$$
$$\sqrt{100}=11$$
The correct answer is $$\sqrt{144}=12$$. One hundred forty-four is a perfect square composed of $$12×12$$. Thirty-six is a perfect square, but it is not composed of $$2×2$$. It is instead composed of $$6×6$$. Twenty-five is a perfect square, but it is not composed of $$4×4$$. It is instead composed of $$5×5$$. One hundred is a perfect square, but it is not composed of $$11×11$$. It is instead composed of $$10×10$$.
Question #3:
What is the square root of 49?
6
4
5
7
The correct answer is 7. The square root of 49 is 7, because $$7×7=49$$. This also means that 49 is a perfect square.
Question #4:
Which value is NOT a perfect square?
81
100
64
99
The correct answer is 99. Because there is no whole number that can be multiplied by itself to equal 99, it is not a perfect square. Eighty-one is a perfect square composed of $$9×9$$, one hundred is a perfect square composed of $$10×10$$, and sixty-four is a perfect square composed of $$8×8$$.
Question #5:
Which pair shows an incorrect match between the perfect square and its whole number square root?
$$\sqrt{49}$$ and $$7$$
$$\sqrt{4}$$ and $$2$$
$$\sqrt{64}$$ and $$8$$
$$\sqrt{9}$$ and $$6$$
The correct answer is $$\sqrt{9}$$ and $$6$$. The square root of 9 is a perfect square, but it is composed of $$3×3$$, not $$6×6$$. |
## What are 3d (Three- dimensional) Shapes?
What are 3d (Three- dimensional) Shapes?
Geometry basically deals with 2d shapes and 3d shapes. Let us study what are the 3D shapes? Let us draw a picture of a notebook on a piece of paper. What we observe is a plain picture drawn on a paper. It does not occupy any space, but if we keep a real notebook on that piece of paper it occupies some space, and such shapes are called 3D shapes or three-dimensional shapes.
Plane geometry or two-dimensional geometry deal with the flat figures that can be drawn on a piece of paper like line, curves, polygons, quadrilaterals, etc, while solid geometry or three-dimensional geometry deals with solid shapes or three-dimensional shapes. Examples of three-dimensional shapes are sphere, cylinders, cones, etc. Let us study some of the 3D shapes and properties of 3d shapes.
What are 3D Shapes?
The shapes that occupy space are called three-dimensional shapes. Three – dimensional shapes can also be defined as the solid shapes having three dimensions length, width, and height. A football is an example of the sphere which is a three-dimensional figure while a circle drawn on a piece of paper is a two- dimensional figure. Similarly, we have many 3D shapes all around us like a table, chair, notebook, pen, etc. Here are some of the examples of three-dimensional shapes and properties of 3d shapes.
Some of the Attributes of 3D Shapes are:
• Faces: The flat figures of the solid figures are called the faces of the 3D figures.
• Edges: A line segment where the two faces meet are called the edges of the 3D shapes.
• Vertices: A corner point where the edges of the solid figures meet are called vertices.
The below figure represents faces, edges, and vertices of a cube, an example of the 3D shape.
Let us study some of the common examples of three-dimensional shapes used in geometry and properties of 3d shapes.
Cube
A cube is a three-dimensional shape formed by two-dimensional 6 square faces. The cube has the following properties.
• All edges and faces are equal.
• It has 8 vertices, 12 edges, and 6 faces.
• The measures of all the angles is 900.
Cuboid
A cuboid is a three- dimensions shape formed by two- dimensional rectangular faces. It is also called a rectangular prism. The cuboid has the following properties
• Opposite faces and edges are equal.
• It has 8 vertices, 12 edges, and 6 faces.
• All the angles measure 900.
Prism
A solid shape having its base and top as identical polygons and lateral faces as parallelograms are called a prism. Examples: Triangular prism, square prism, Pentagonal prism. The properties of the prism are according to the shapes.
Some of the properties of a triangular prism are
It has 6 vertices, 9 edges and 5 faces – 2 triangles and 3 rectangles
Pyramid
A solid shape having its base as any polygon and side face as triangles with a common vertex is called a pyramid.
The most commonly used pyramid is the square pyramid i.e., it has a square base and four triangular side faces.
Some of the properties of a square pyramid are
It has 5 vertices, 8 edges, and 5 faces
Cylinder
A solid 3D shape which has two circular bases and a curved surface is called a cylinder. Some of the properties of a cylinder are:
• It has no vertex, 2 edges
• 2 flat faces which are circles
• And 1 curved surface
Cone
A solid 3D shape which has one circular base connected with a curved surface and has a single vertex is called a cone.
Which is perfectly round in shape and Some of the important properties of a cone are
• It has 1 vertex and 1 edge
• 1 flat face which is a circle
• And 1 curved face
Sphere
A solid three-dimensional shape that is perfectly round in shape and all its points on the surface is equidistant from the center point is called a sphere.
The distance between the center and any point on the surface is called the radius of the circle. Some of the important properties of the sphere are
• It has no vertex and no edges.
• It has no flat faces.
• It has only 1 curved face.
Surface Area and Volume of 3D shapes
3D shapes are measured in two terms that are:
• Surface Area-The area of all the faces of a solid is called the surface area of the object. Surface area is denoted by SA. Total surface area(TSA) is divided into two parts
• Curved Surface Area is the area of all the curved regions denoted by ‘CSA’.
• Lateral Surface Area is the area of all the curved regions and all the flat surfaces excluding base areas denoted by ‘LSA’.
Volume
The amount of space that encloses a solid shape is called the volume of the shape. And it is denoted by ‘V’
Here is the formula chart for surface area and volume of 3D shapes in a tabular format
## Formula for Surface Area and Volume of 3d shapes(Solid shapes)
S.No Name Abbreviations used Lateral /curved Surface Area Total Surface Area Volume 1. Cuboid h = height, l = length b=breadth 2h(l+b) 6l2 L * b* h 2. Cube a = length of the sides 4a2 6a2 a3 3. Right Prism .. Perimeter of Base × Height Lateral Surface Area + 2(Area of One End) Area of Base × Height 4. Right Circular Cylinder r= radius h=height 2 (π × r × h) 2πr (r + h) πr2h 5. Right pyramid .. ½ (Perimeter of Base × Slant Height) Lateral Surface Area + Area of the Base ⅓ (Area of the Base) × Height 6. Right Circular Cone r = radius l = length πrl πr (l + r) ⅓ (πr2h) 7. Sphere r = radius 4πr2 4πr2 4/3πr3 8. Hemisphere r = radius 2πr2 3πr2 ⅔ (πr3)
Solved examples:
Example 1:
Find the volume and surface area of a cuboid of l= 10cm, b = 8cm and h = 6 cm.
Solution: We have Volume of cuboid = V = l x b x h
=10 x 8 x 6
= 480cm2
Surface area = 2 ( lb + lh + bh)
= 2( 10×8 + 10×6 + 8×6)
=2(80 + 60 + 48)
=376cm2
Quiz Time
Find the area of the right-angled triangle whose base is 12cm and hypotenuse 13cm.
1. 40 cm
2. 85 cm
3. 60 cm
4. 30cm2
The side of a square whose surface area is 600cm is
1. 10cm
2. 20 cm
3. 30 cm
4. 40 cm |
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# Functions and Inverses
## Undo the original function; reflections across y = x.
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Practice Functions and Inverses
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Inverses of Functions
Functions are commonly known as rules that take inputs and produce outputs. An inverse function does exactly the reverse, undoing what the original function does. How can you tell if two functions are inverses?
#### Watch This
http://www.youtube.com/watch?v=qgezKpQYH2w James Sousa: Inverse Functions
#### Guidance
A function is written as f(x)\begin{align*}f(x)\end{align*} and its inverse is written as f1(x)\begin{align*}f^{-1} (x)\end{align*}. A common misconception is to see the -1 and interpret it as an exponent and write 1f(x)\begin{align*}\frac{1}{f(x)}\end{align*}, but this is not correct. Instead, f1(x)\begin{align*}f^{-1} (x)\end{align*} should be viewed as a new function from the range of f(x)\begin{align*}f(x)\end{align*} back to the domain.
It is important to see the cycle that starts with x\begin{align*}x\end{align*}, becomes y\begin{align*}y\end{align*} and then goes back to x\begin{align*}x\end{align*}. In order for two functions to truly be inverses of each other, this cycle must hold algebraically.
f(f1(x))=x\begin{align*}f(f^{-1} (x) )=x\end{align*} and f1(f(x))=x\begin{align*}f^{-1} (f(x) )=x\end{align*}
When given a function there are two steps to take to find its inverse. In the original function, first switch the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}. Next, solve the function for y\begin{align*}y\end{align*}. This will give you the inverse function. After finding the inverse, it is important to check both directions of compositions to make sure that together the function and its inverse produce the value x\begin{align*}x\end{align*}. In other words, verify that f(f1(x))=x\begin{align*}f(f^{-1} (x) )=x\end{align*} and f1(f(x))=x\begin{align*}f^{-1} (f(x) )=x\end{align*}.
Graphically, inverses are reflections across the line y=x\begin{align*}y=x\end{align*}. Below you see inverses y=ex\begin{align*}y =e^x\end{align*} and y=lnx\begin{align*}y= \ln x\end{align*}. Notice how the (x,y)\begin{align*}(x, y)\end{align*} coordinates in one graph become (y,x)\begin{align*}(y, x)\end{align*} coordinates in the other graph.
In order to decide whether an inverse function is also actually a function you can use the vertical line test on the inverse function like usual. You can also use the horizontal line test on the original function. The horizontal line test is exactly like the vertical line test except the lines simply travel horizontally.
Example A
Find the inverse, then verify the inverse algebraically. f(x)=y=(x+1)2+4\begin{align*}f(x)=y=(x+1)^2+4\end{align*}
Solution: To find the inverse, switch x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} then solve for y\begin{align*}y\end{align*}
xx4±x41±x4=(y+1)2+4=(y+1)2=y+1=y=f1(x)
To verify algebraically, you must show x=f(f1(x))=f1(f(x))\begin{align*}x= f(f^{-1} (x) )=f^{-1} (f(x) )\end{align*}:
f(f1(x))=f(1±x4 )=((1±x4 )+1)2+4=(±x4 )2+4=x4+4=x
f1(f(x))=f1((x+1)2+4)=1±((x+1)2+4)4=1±(x+1)2=1+x+1=x
As you can see from the graph, the ±\begin{align*}\pm\end{align*} causes the inverse to be a relation instead of a function. This can be observed in the graph because the original function does not pass the horizontal line test and the inverse does not pass the vertical line test
Example B
Find the inverse of the function and then verify that x=f(f1(x))=f1(f(x))\begin{align*}x= f(f^{-1} (x) )=f^{-1} (f(x) )\end{align*}.
f(x)=y=x+1x1\begin{align*}f(x)=y=\frac{x+1}{x-1}\end{align*}
Solution: Sometimes it is quite challenging to switch x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} and then solve for y\begin{align*}y\end{align*}. You must be careful with your algebra.
xx(y1)xyxxyyy(x1)y=y+1y1=y+1=y+1=x+1=x+1=x+1x1
This function turns out to be its own inverse. Since they are identical, you only need to show that x=f(f1(x))\begin{align*}x= f(f^{-1} (x) )\end{align*}.
f(x+1x1)=(x+1x1)+1(x+1x1)1=x+1+x1x+1(x1)=2x2=x\begin{align*}f\left(\frac{x+1}{x-1}\right)=\frac{\left(\frac{x+1}{x-1}\right)+1}{\left(\frac{x+1}{x-1}\right)-1}=\frac{x+1+x-1}{x+1-(x-1)}=\frac{2x}{2}=x\end{align*}
Example C
What is the inverse of \begin{align*}f(x)=y=\sin x\end{align*}?
Solution: The sine function does not pass the horizontal line test and so its true inverse is not a function.
However, if you restrict the domain to just the part of the \begin{align*}x\end{align*}-axis between \begin{align*}- \frac{\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{2}\end{align*} then it will pass the horizontal line test and the inverse will be a function.
The inverse of the sine function is called the arcsine function, \begin{align*}f(x)=\sin^{-1}(x)\end{align*}, and is shown in black. It is truncated so that it only inverts a part of the whole sine wave. You will study periodic functions and their inverses in more detail later.
Concept Problem Revisited
You can tell that two functions are inverses if each undoes the other, always leaving the original \begin{align*}x\end{align*}
#### Vocabulary
A relation is a general term for functions and non-functions that relate two variables and may or may not pass the vertical line test
Two relations are inverses of each other if they are reflections across the line \begin{align*}y=x\end{align*}.
The horizontal line test is a means of discovering whether the inverse of a function is also a function.
#### Guided Practice
1. Determine the inverse of \begin{align*}f(x)=5+\frac{x}{2}\end{align*}. Verify that the inverse is actually the inverse.
2. Determine if \begin{align*}f(x)=\frac{3}{7}x-21\end{align*} and \begin{align*}g(x)=\frac{7}{3}x+21\end{align*} are inverses of one another.
3. Determine the inverse of \begin{align*}f(x)=\frac{x}{x+4}\end{align*}.
1. To find the inverse,
Verification:
\begin{align*}2\left(5+\frac{x}{2}-5\right)=2\left(\frac{x}{2}\right)=x\end{align*}
\begin{align*}5+\frac{2(x-5)}{2}=5+x-5=x\end{align*}
They are truly inverses of each other.
2. Even though \begin{align*}f(x)=\frac{3}{7}x-21\end{align*} and \begin{align*}g(x)=\frac{7}{3}x+21\end{align*} have some inverted pieces, they are not inverses of each other. In order to show this, you must show that the composition does not simplify to \begin{align*}x\end{align*}. \begin{align*}\frac{3}{7}\left(\frac{7}{3}x+21\right) -21=x+9-21=x-12 \ne x\end{align*}
3. To find the inverse, switch x and y.
#### Practice
Consider \begin{align*}f(x)=x^3\end{align*}.
1. Sketch \begin{align*}f(x)\end{align*} and \begin{align*}f^{-1}(x)\end{align*}.
2. Find \begin{align*}f^{-1}(x)\end{align*} algebraically. It is actually a function?
3. Verify algebraically that \begin{align*}f(x)\end{align*} and \begin{align*}f^{-1}(x)\end{align*} are inverses.
Consider \begin{align*}g(x)=\sqrt{x}\end{align*}.
4. Sketch \begin{align*}g(x)\end{align*} and \begin{align*}g^{-1}(x)\end{align*}.
5. Find \begin{align*}g^{-1}(x)\end{align*} algebraically. It is actually a function?
6. Verify algebraically that \begin{align*}g(x)\end{align*} and \begin{align*}g^{-1}(x)\end{align*} are inverses.
Consider \begin{align*}h(x)=|x|\end{align*}.
7. Sketch \begin{align*}h(x)\end{align*} and \begin{align*}h^{-1}(x)\end{align*}.
8. Find \begin{align*}h^{-1}(x)\end{align*} algebraically. It is actually a function?
9. Verify graphically that \begin{align*}h(x)\end{align*} and \begin{align*}h^{-1}(x)\end{align*} are inverses.
Consider \begin{align*}j(x)=2x-5\end{align*}.
10. Sketch \begin{align*}j(x)\end{align*} and \begin{align*}j^{-1}(x)\end{align*}.
11. Find \begin{align*}j^{-1}(x)\end{align*} algebraically. It is actually a function?
12. Verify algebraically that \begin{align*}j(x)\end{align*} and \begin{align*}j^{-1}(x)\end{align*} are inverses.
13. Use the horizontal line test to determine whether or not the inverse of \begin{align*}f(x)=x^3-2x^2+1\end{align*} is also a function.
14. Are \begin{align*}g(x)=\ln (x+1)\end{align*} and \begin{align*}h(x)=e^{x-1}\end{align*} inverses? Explain.
15. If you were given a table of values for a function, how could you create a table of values for the inverse of the function?
### Vocabulary Language: English
composite function
composite function
A composite function is a function $h(x)$ formed by using the output of one function $g(x)$ as the input of another function $f(x)$. Composite functions are written in the form $h(x)=f(g(x))$ or $h=f \circ g$.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Horizontal Line Test
Horizontal Line Test
The horizontal line test says that if a horizontal line drawn anywhere through the graph of a function intersects the function in more than one location, then the function is not one-to-one and not invertible.
inverse
inverse
Inverse functions are functions that 'undo' each other. Formally: $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$.
Relation
Relation
A relation is any set of ordered pairs $(x, y)$. A relation can have more than one output for a given input.
Vertical Line Test
Vertical Line Test
The vertical line test says that if a vertical line drawn anywhere through the graph of a relation intersects the relation in more than one location, then the relation is not a function. |
# Add Subtract Multiply Complex Numbers Worksheet
This multiplication worksheet concentrates on educating students the way to psychologically increase entire figures. Pupils can use customized grids to suit exactly one concern. The worksheets also protectdecimals and fractions, and exponents. There are also multiplication worksheets having a spread residence. These worksheets can be a should-have for your personal math class. They are often used in course to figure out how to psychologically increase total numbers and line them up. Add Subtract Multiply Complex Numbers Worksheet.
## Multiplication of whole numbers
If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets can help you learn this basic concept. You may go for one particular digit multipliers or two-digit and about three-digit multipliers. Capabilities of 10 are also a great choice. These worksheets will enable you to practice long multiplication and practice looking at the amounts. They are also the best way to support your kids fully grasp the importance of knowing the different kinds of entire numbers.
## Multiplication of fractions
Experiencing multiplication of fractions on a worksheet might help educators plan and put together instruction effectively. Using fractions worksheets allows professors to quickly determine students’ knowledge of fractions. Students may be challenged in order to complete the worksheet within a particular time as well as then tag their strategies to see where they want more instructions. Individuals can usually benefit from phrase things that associate maths to actual-lifestyle situations. Some fractions worksheets involve examples of comparing and contrasting figures.
## Multiplication of decimals
Whenever you increase two decimal numbers, make sure you class them up and down. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. For example, 01 by (11.2) x 2 will be similar to 01 by 2.33 by 11.2 except when the item has decimal areas of lower than two. Then, this product is circular on the closest total variety.
## Multiplication of exponents
A arithmetic worksheet for Multiplication of exponents will help you practice multiplying and dividing numbers with exponents. This worksheet may also provide issues that requires individuals to grow two different exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Apart from, also you can key in special directions about the worksheet alone. When you’re done, it is possible to simply click “Create” along with the worksheet will probably be delivered electronically.
## Section of exponents
The essential rule for section of exponents when multiplying numbers is usually to deduct the exponent from the denominator in the exponent inside the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. As an example, \$23 divided by 4 will the same 27. This method is not always accurate, however. This technique can result in misunderstandings when multiplying amounts which can be too big or not big enough.
## Linear characteristics
If you’ve ever rented a car, you’ve probably noticed that the cost was \$320 x 10 days. So the total rent would be \$470. A linear function of this type provides the develop f(by), where by ‘x’ is the amount of days the auto was hired. Additionally, it has the shape f(by) = ax b, where by ‘b’ and ‘a’ are genuine figures. |
Home > Math Shortcuts > Ration Series Example 1
# Ration Series Example 1
Ration Series Example 1
Ratio number series is a arrangement of numbers in a certain order, This type of series are based on ratio series and you have to find changes and orders that found out in difference between the numbers .
All numbers are arranged in sequent order. we need to observe and find the accurate number to this type series of numbers. Here we learn the ratio series of Example. Find the accurate number to the blank or ? mark series of numbers using calculation.
In ratio Series number is a combination of number in another way it is not a sequential number series number are arranged . In example 1) 11, 22, 33, 44, 88, ? where you need to count them in a one step or two step calculation for obtain the difference common result according with the series of ratio numbers .
you can calculate missing number in ratio series and that you place the actual missing number in the ? or missing place . Be prepared when you calculate differences because it is either one or two step calculation. So when you calculate and get result of two difference numbers you need follow some step wise.
At first calculate the first and second number common difference then follow same steps another two number differences which is carry up to last and after that you get actual missing number by finding the common difference when you put the missing number you have noticed that all series numbers are common difference in between them.
This kind of missing ratio series calculation you go thorough some common calculation shortcut tricks using square or division, cube, addition, multiplication.
In this type series example questions, it is sounds hard, but it really isn’t. Get it? Once you have done this, by practice with more example then you just easily can do in your way as well competitive and as in bank exam also . This type of problem are given in Quantitative Aptitude which is a very essential in banking exam . Under below given some more example for your better practice . So, each of our examples are given below.
Ration Series:
Find the missing term .
Examples: 11, 22, 44, 88, ?
separate each number and
= 11 x 2 = 22,
= 22 x 2 = 44,
= 44 x 2 = 88,
= 88 x 2 = 176.
Examples: 13, 26, 412, ?, 1648
separate each number and
= (1+1 = 2,3+3 = 6) = 26,
= (2+2=4,6+6 = 12) = 412,
= (41+41=82, 2+2=4 = 824,
= (82+82=164,4+4= 8)=1648.
So, the missing term is 824. |
Question
# If a - b = 4 and a + b = 6, then find ab.${\text{A}}{\text{. 5}} \\ {\text{B}}{\text{. 2}} \\ {\text{C}}{\text{. 45}} \\ {\text{D}}{\text{. 3}} \\$
Hint- Here, we will be using elimination method to solve the given two equations in two variables.
Given equations are $a - b = 4{\text{ }} \to {\text{(1)}}$ and $a + b = 6{\text{ }} \to {\text{(2)}}$
The above two given equations have two variables $a$ and $b$.
Adding both the equations (1) and (2), we get
$a - b + a + b = 4 + 6 \Rightarrow 2a = 10 \Rightarrow a = 5$
Put the value of $a$ in equation (1), we get
$5 - b = 4 \Rightarrow b = 5 - 4 \Rightarrow b = 1$
So, the values of the two variables that satisfy the two given equations are $a = 5$ and $b = 1$.
Hence, $ab = 5 \times 1 = 5$
Therefore, the value of $ab$ is 5.
i.e., option A is correct.
Note- These types of problems of two equations in two variables can be solved easily using substitution method or elimination method. In the substitution method, we represent one variable in terms of another variable using one equation and substitute this value in the other equation in order to get the values of both variables. |
# Ex.9.1 Q3 Rational-Numbers Solution - NCERT Maths Class 7
Go back to 'Ex.9.1'
## Question
Give four rational numbers equivalent to:
\begin{align}{{\rm{ (i) }}\;\;\frac{{ - 2}}{7}}\end{align}
\begin{align}{{\rm{(ii) }}\;\;\frac{5}{{ - 3}}}\end{align}
\begin{align}{\rm{(iii)}}\,\,\,\frac{4}{9}\end{align}
Video Solution
Rational Numbers
Ex 9.1 | Question 3
## Text Solution
What is known?
Three rational numbers.
What is unknown?
Four rational numbers equivalent to each of the given rational number.
Reasoning:
To find out the equivalent fraction of any rational number, multiply the numerator and the denominator of the given number by the same numbers. Remember here it is asked for four equivalent rational numbers that means you have to multiply four different numbers, one by one in both numerator and denominator of the given number.
Steps:
(i) \begin{align}\frac{{ - 2}}{7}\end{align}
Multilying both numerator and denominator with the same number, we get
\begin{align} \frac{-2\times 2}{7\times 2}&=\frac{-4}{14}, \\ \frac{-2\times 3}{7\times 3}&=\frac{-6}{21}, \\ \frac{-2\times 4}{7\times 4}&=\frac{-8}{28}, \\ \frac{-2\times 5}{7\times 5}&=\frac{-10}{35} \\ \end{align}
Therefore, the equivalent fractions to the number \begin{align}\frac{{ - 2\;}}{7}\end{align}are,
\begin{align}\frac{{ - 4}}{{14}},\frac{{ - 6}}{{21}},\frac{{ - 8}}{{28}},\frac{{ - 10}}{{35}}\end{align}
(ii) \begin{align}\frac{5}{{ - 3}}\end{align}
Multiplying both numerator and denominator with the same number, we get
\begin{align} & \frac{5\times 2}{-3\times 2}=\frac{10}{-6}, \\ & \frac{5\times 3}{-3\times 3}=\frac{15}{-9}, \\ & \frac{5\times 4}{-3\times 4}=\frac{20}{-12}, \\ & \frac{5\times 5}{-3\times 5}=\frac{25}{-15} \\ \end{align}
Therefore, the equivalent fractions to the number \begin{align}\frac{5}{{ - 3}}\end{align}are,
\begin{align}\frac{{10}}{{ - 6}},\frac{{15}}{{ - 9}},\frac{{20}}{{ - 12}},\frac{{25}}{{ - 15}}\end{align}
(iii) \begin{align} \frac{4}{9}\end{align}
Multiplying both numerator and denominator with the same number, we get
\begin{align} \frac{4\times 2}{9\times 2}&=\frac{8}{18}, \\ \frac{4\times 3}{9\times 3}&=\frac{12}{27}, \\ \frac{4\times 4}{9\times 4}&=\frac{16}{36}, \\ \frac{4\times 5}{9\times 5}&=\frac{20}{45} \\ \end{align}
Therefore, the equivalent fractions to the number \begin{align}\frac{4}{9}\end{align}are,
\begin{align}\frac{8}{{18}},\frac{{12}}{{27}},\frac{{16}}{{36}}\,{\rm{and}}\,\frac{{20}}{{45}}\end{align}
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Area under curve
Find area bound by the curves y=sqrtx, y=-sqrtx, with y=x-6 intersecting it.
• After drawing the curves out we can see that the curves intersect at (0,0). We need to find the other point where they intersect. This can be done by putting y = x and y = x^0.5 equal to each other and solving.
x = x^0.5
If we divide by x^0.5 we get
x^0.5 = 1
x = 1
We will use the shell method for calculating the volume.
Let:
w be the width of our shell piece. The width is the difference between y = x^0.5 and y = x, so w = x^0.5 - x. We subtract y = x from y = x^0.5 since y = x is closest to the rotation point.
r be the radius. Normally in the shell method this would be x, but since we are rotating about x=2, r = 2 - x
The volume for the shell is then calculated by:
V(x) = 2prwdx
V(x) = 2p(2-x)(x^0.5-x)dx
V(x) = 2p(2x^0.5 - x^1.5 - 2x + x^2)dx
V(x) = 2p(2*(2/3)x^1.5 - (2/5)x^2.5 - x^2 + (1/3)x^3)
Evaluate this at our endpoints of x=0 and x=1
V(1) = 2p(2*(2/3)*1^1.5 - (2/5)*1^2.5 - 1^2 + (1/3)*1x^3)
V(1) = 2p(2*(2/3) - (2/5) - 1 + 1/3)
V(1) = 2p(4/3 - 2/5 - 1 + 1/3)
Convert everything to be in 15ths
V(1) = 2p(20/15 - 6/15 - 15/15 + 5/15)
V(1) = 2p(4/15)
V(0) = 2p(2*(2/3)*0^1.5 - (2/5)*0^2.5 - 0^2 + (1/3)*0x^3)
V(0) = 2p(0 - 0 - 0 + 0)
V(0) = 0
The volume is the difference between V(1) and V(0).
V = V(1) - V(0)
V = 2p(4/15) - 0
V = 8p/15
• ? top - minus
y = v(x) & y = x / 2
v(x) = x / 2 ====> square both sides
x = x^2 / 4
4x = x^2
0 = x^2 - 4x
0 = x * (x - 4) =====> x = 0 & 4 ( intersections )
4 . . . . . . . . . . . . . . .9
? [ v(x) - (x / 2) ] dx + ? [ (x / 2) - v(x) ] dx
0 . . . . . . . . . . . . . . 4
4 . . . . . . .. . . . . . . . . . . .9
? [ x^(1/2) - (1/2) * x ] dx + ? [ (1/2) * x - x^(1/2) ] dx
0 . . . . . . . . . . . . . . . . . . 4
. . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . .. . . . . . . . . . . .. . . . . . .9
[ x^(3/2) / (3/2) ] - (1/2) * ( x^2 / 2) ] + (1/2) * ( x^2 / 2) - [ x^(3/2) / (3/2) ] ]
. . . . . . . . . . . . . . . . . . . . . . . . . .0 . . . . . . . . . . . . . . . . . . . . .. . . .4
(2/3) * [ 4^(3/2) - 0^(3/2) ] - (1/4) * [ 4^2 - 0^2 ] + (1/4) * [ 9^2 - 4^2 ] - (2/3) * [ 9^(3/2) - 4^(3/2) ]
(2/3) * [ 8 - 0 ] - (1/4) * [ 16 - 0 ] + (1/4) * [ 81 - 16 ] - (2/3) * [ 27 - 8 ]
(16/3) - 4 + (1/4) * [ 65 ] - (2/3) * [ 19 ]
(16/3) - 4 + (65/4) - 38/3
(16/3) - 4 + (65/4) - 38/3
59/12 unit^2
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# . Prove that the product of three consecutive positive integers is divisible by 6
Explanation:
Let us take the three consecutive positive integers as : a,a + 1,a + 2
A number when divided by 3, will have a remainder = 0 or 1 or 2
We can clearly see the numbers a,a + 1,a + 2 are always divisible by 3.
Similarly, we can show that when a number is divisible by 2, we get the remainder as 0 or 1
Clearly, 2m and 2(m+1) both are divisible by 2.
We can clearly see the numbers a,a + 1,a + 2 are always divisible by 2.
We can say that a,a + 1,a + 2 is always divisible by 2 ,3 ,6
Example:
Let us take: 3,4,5
3 x 4 x 5 = 60
We can see that 60 is divisible by 6 |
# What is 10 percent of 30 + Solution with Free Steps
10 percent of 30 is equal to 3. You can easily calculate it by first dividing 10 by 100 and multiplying the answer with 30 to get 10 percent of 30.
The calculation of 10% of 30 has several applications. Assume your tuition academy is increasing its new student intake by 10%. You are now aware that your academy currently accepts 30 students in each class each year. We have calculated that 10% of 30 is equal to 3. So you can safely assume that 3 students will be admitted this year to your tuition academy.
## What Is 10 percent of 30?
We know that 10% of 30 is equal to 3. It can easily be calculated by dividing 10 by 100 and multiplying the answer by 30 to get 3.
The solution to this question can be found by first dividing 10 by 100 which will result in 0.1 and multiplying this 0.1 with 30 which will result in 3. So these steps can be used to find the 10% of 30 which is 3 in this question.
## How To Calculate 10% of 30?
We can easily calculate the 10 percent of 30 by using the following mathematical steps:
### Step 1
12 percent of 30 can be represented in mathematical form as:
10 percent of 30 = 10% x 12
### Step 2
Substituting the % symbol with the fraction as 1/100:
10 percent of 30 = ( 10 x 1/100 ) x 30
### Step 3
We can rearrange the equation above as:
10 percent of 30 = ( 10 x 30 ) / 100
### Step 4
Now Multiplying 10 with 30:
10 percent of 30 = ( 300 ) / 100
### Step 5
Thus Dividing 300 by 100:
10 percent of 30 = 3
Therefore, we have calculated that the 10 percent of 30 is equal to 3.
Using the pie chart below, we can visualize that 10 percent of 30 equals 3.
Figure 1: Pie Chart of 10 percent of 30
The associated Pie Chart shown above can help us visualize how much 10% of 30 appears to look like. If we divide the total value into 100 equal parts, then 10% of 30 equals 3, which is recognized by the blue area, and 90 percent of 100 is reflected by the yellow area.
A percentage is a number, and possibly a ratio, written as a fraction of 100. “Percentage” is derived from the Latin term “per centum,” which means “by a hundred.” The percentage is represented by the symbol “% “
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# How do u find the perimeter of a kite?
Contents
Therefore, if a kite has two sides of length x and two sides of length y, then the perimeter is found by adding up the lengths of all of the sides: Perimeter of a kite = x + x + y + y = 2x + 2y.
## How do you find a perimeter?
To find the perimeter of a rectangle, add the lengths of the rectangle’s four sides. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Multiply both the height and width by two and add the results.
## How do you calculate the perimeter of a shape?
The perimeter is the distance all the way around the outside of a 2D shape. To work out the perimeter, add up the lengths of all the sides.
## How do I find the perimeter of a shape?
The perimeter of a shape is always calculated by adding up the length of each of the sides. In Year 5 and 6, children might be given shapes like this one and asked to find their perimeter: In this case, they need to work out the lengths of the edges that are unlabelled, by looking at the other labelled edges.
## What is the perimeter of shape B?
Answer: The perimeter of a two-dimensional shape is the distance around the shape. It is found by adding up all the sides (as long as they are all the same unit).
## What is the perimeter of △ LMN?
The perimeter of △LMN is 20.25 units.
## What is the diagonal formula?
The formula to calculate the number of diagonal of an n-sided polygon = n(n-3)/2 where n is the number of sides of the polygon.
## What is the formula of perimeter of cube?
Cube and Cuboid Formulas
Cube Cuboid
Volume of cube = (Side)3 Volume of the cuboid = (length × breadth × height)
Diagonal of a cube = √3l Diagonal of the cuboid =√( l2 + b2 +h2)
Perimeter of cube = 12 x side Perimeter of cuboid = 4 (length + breadth + height)
## How do you find the perimeter of a shape without measurements?
Add the length of all the sides together.
To find the perimeter of non-circular objects, find the sum of all the side lengths to determine the distance around the shape.
## How do you find a missing perimeter?
To find the missing side length, write an addition sentence for the perimeter of the shape. The perimeter of a shape equals the sum of all of its side lengths. Add the lengths of the sides you know. Find the side length that makes the addition sentence true. |
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# 5.E: Trigonometric Functions (Exercises)
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## 5.1: Angles
### Section Exercises
#### Verbal
Draw an angle in standard position. Label the vertex, initial side, and terminal side.
Explain why there are an infinite number of angles that are coterminal to a certain angle.
State what a positive or negative angle signifies, and explain how to draw each.
Whether the angle is positive or negative determines the direction. A positive angle is drawn in the counterclockwise direction, and a negative angle is drawn in the clockwise direction.
How does radian measure of an angle compare to the degree measure? Include an explanation of 1 radian in your paragraph.
Explain the differences between linear speed and angular speed when describing motion along a circular path.
Linear speed is a measurement found by calculating distance of an arc compared to time. Angular speed is a measurement found by calculating the angle of an arc compared to time.
#### Graphical
For the following exercises, draw an angle in standard position with the given measure.
30°
300°
−80°
135°
−150°
$$\frac{2π}{3}$$
$$\frac{7π}{4}$$
$$\frac{5π}{6}$$
$$\frac{π}{2}$$
$$−\frac{π}{10}$$
415°
−120°
240°
−315°
$$\frac{22π}{3}$$
$$\frac{4π}{3}$$
$$−\frac{π}{6}$$
$$−\frac{4π}{3}$$
$$\frac{2π}{3}$$
For the following exercises, refer to Figure. Round to two decimal places.
Find the arc length.
Find the area of the sector.
$$\frac{27π}{2}≈11.00 \text{ in}^2$$
For the following exercises, refer to Figure. Round to two decimal places.
Find the arc length.
Find the area of the sector.
$$\frac{81π}{20}≈12.72\text{ cm}^2$$
#### Algebraic
For the following exercises, convert angles in radians to degrees.
$$\frac{3π}{4}$$ radians
$$\frac{π}{9}$$ radians
20°
$$−\frac{5π}{4}$$ radians
$$\frac{π}{3}$$ radians
60°
$$−\frac{7π}{3}$$ radians
$$−\frac{5π}{12}$$ radians
−75°
$$\frac{11π}{6}$$ radians
For the following exercises, convert angles in degrees to radians.
90°
$$\frac{π}{2}$$ radians
100°
−540°
$$−3π$$ radians
−120°
180°
$$π$$ radians
−315°
150°
$$\frac{5π}{6}$$ radians
For the following exercises, use to given information to find the length of a circular arc. Round to two decimal places.
Find the length of the arc of a circle of radius 12 inches subtended by a central angle of $$\frac{π}{4}$$ radians.
Find the length of the arc of a circle of radius 5.02 miles subtended by the central angle of $$\frac{π}{3}$$.
$$\frac{5.02π}{3}≈5.26$$ miles
Find the length of the arc of a circle of diameter 14 meters subtended by the central angle of 5π6.5π6.
Find the length of the arc of a circle of radius 10 centimeters subtended by the central angle of 50°.
$$\frac{25π}{9}≈8.73$$ centimeters
Find the length of the arc of a circle of radius 5 inches subtended by the central angle of 220°.
Find the length of the arc of a circle of diameter 12 meters subtended by the central angle is 63°.
$$\frac{21π}{10}≈6.60$$ meters
For the following exercises, use the given information to find the area of the sector. Round to four decimal places.
A sector of a circle has a central angle of 45° and a radius 6 cm.
A sector of a circle has a central angle of 30° and a radius of 20 cm.
104.7198 cm2
A sector of a circle with diameter 10 feet and an angle of $$\frac{π}{2}$$ radians.
A sector of a circle with radius of 0.7 inches and an angle of $$π$$ radians.
0.7697 in
or the following exercises, find the angle between 0° and 360° that is coterminal to the given angle.
−40°
−110°
250°
700°
1400°
320°
For the following exercises, find the angle between 0 and 2π 2π in radians that is coterminal to the given angle.
$$−\frac{π}{9}$$
$$\frac{10π}{3}$$
$$\frac{4π}{3}$$
$$\frac{13π}{6}$$
$$\frac{44π}{9}$$
$$\frac{8π}{9}$$
#### Real-World Applications
A truck with 32-inch diameter wheels is traveling at 60 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
A bicycle with 24-inch diameter wheels is traveling at 15 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
A wheel of radius 8 inches is rotating 15°/s. What is the linear speed $$v$$, the angular speed in RPM, and the angular speed in rad/s?
A wheel of radius 14 inches is rotating 0.5 rad/s. What is the linear speed $$v$$, the angular speed in RPM, and the angular speed in deg/s?
7 in./s, 4.77 RPM, 28.65 deg/s
A CD has diameter of 120 millimeters. When playing audio, the angular speed varies to keep the linear speed constant where the disc is being read. When reading along the outer edge of the disc, the angular speed is about 200 RPM (revolutions per minute). Find the linear speed.
When being burned in a writable CD-R drive, the angular speed of a CD is often much faster than when playing audio, but the angular speed still varies to keep the linear speed constant where the disc is being written. When writing along the outer edge of the disc, the angular speed of one drive is about 4800 RPM (revolutions per minute). Find the linear speed if the CD has diameter of 120 millimeters.
$$1,809,557.37 \text{ mm/min}=30.16 \text{ m/s}$$
A person is standing on the equator of Earth (radius 3960 miles). What are his linear and angular speeds?
Find the distance along an arc on the surface of Earth that subtends a central angle of 5 minutes $$(1 \text{ minute}=\frac{1}{60} \text{ degree})$$. The radius of Earth is 3960 miles.
$$5.76$$ miles
Find the distance along an arc on the surface of Earth that subtends a central angle of 7 minutes $$(1 \text{ minute}=\frac{1}{60} \text{ degree})$$. The radius of Earth is $$3960$$ miles.
Consider a clock with an hour hand and minute hand. What is the measure of the angle the minute hand traces in $$20$$ minutes?
$$120°$$
#### Extensions
Two cities have the same longitude. The latitude of city A is 9.00 degrees north and the latitude of city B is 30.00 degree north. Assume the radius of the earth is 3960 miles. Find the distance between the two cities.
A city is located at 40 degrees north latitude. Assume the radius of the earth is 3960 miles and the earth rotates once every 24 hours. Find the linear speed of a person who resides in this city.
794 miles per hour
A city is located at 75 degrees north latitude. Assume the radius of the earth is 3960 miles and the earth rotates once every 24 hours. Find the linear speed of a person who resides in this city.
Find the linear speed of the moon if the average distance between the earth and moon is 239,000 miles, assuming the orbit of the moon is circular and requires about 28 days. Express answer in miles per hour.
2,234 miles per hour
A bicycle has wheels 28 inches in diameter. A tachometer determines that the wheels are rotating at 180 RPM (revolutions per minute). Find the speed the bicycle is travelling down the road.
A car travels 3 miles. Its tires make 2640 revolutions. What is the radius of a tire in inches?
11.5 inches
A wheel on a tractor has a 24-inch diameter. How many revolutions does the wheel make if the tractor travels 4 miles?
### Review Exercises
For the following exercises, convert the angle measures to degrees.
$$\frac{π}{4}$$
$$45°$$
$$−\frac{5π}{3}$$
For the following exercises, convert the angle measures to radians.
-210°
$$−\frac{7π}{6}$$
180°
Find the length of an arc in a circle of radius 7 meters subtended by the central angle of 85°.
10.385 meters
Find the area of the sector of a circle with diameter 32 feet and an angle of $$\frac{3π}{5}$$ radians.
For the following exercises, find the angle between 0° and 360° that is coterminal with the given angle.
$$420°$$
$$60°$$
$$−80°$$
For the following exercises, find the angle between 0 and $$2π$$ in radians that is coterminal with the given angle.
$$− \frac{20π}{11}$$
$$\frac{2π}{11}$$
$$\frac{14π}{5}$$
For the following exercises, draw the angle provided in standard position on the Cartesian plane.
-210°
75°
$$\frac{5π}{4}$$
$$−\frac{π}{3}$$
Find the linear speed of a point on the equator of the earth if the earth has a radius of 3,960 miles and the earth rotates on its axis every 24 hours. Express answer in miles per hour.
1036.73 miles per hour
A car wheel with a diameter of 18 inches spins at the rate of 10 revolutions per second. What is the car's speed in miles per hour?
## 5.2: Unit Circle - Sine and Cosine Functions
### Section Exercises
#### Verbal
Describe the unit circle.
The unit circle is a circle of radius 1 centered at the origin.
What do the x- and y-coordinates of the points on the unit circle represent?
Discuss the difference between a coterminal angle and a reference angle.
Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, $$t$$, formed by the terminal side of the angle $$t$$ and the horizontal axis.
Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.
Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.
The sine values are equal.
#### Algebraic
For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by ttlies.
$$\sin (t)<0$$ and $$\cos (t)<0$$
$$\sin (t)>0$$ and $$\cos (t)>0$$
I
$$\sin (t)>0$$ and $$\cos (t)<0$$
$$\sin (t)<0$$ and $$\cos (t)>0$$
IV
For the following exercises, find the exact value of each trigonometric function.
$$\sin \frac{π}{2}$$
$$\sin \frac{π}{3}$$
$$\frac{\sqrt{3}}{2}$$
$$\cos \frac{π}{2}$$
$$\cos \frac{π}{3}$$
$$\frac{1}{2}$$
$$\sin \frac{π}{4}$$
$$\cos \frac{π}{4}$$
$$\frac{\sqrt{2}}{2}$$
$$\sin \frac{π}{6}$$
$$\sin π$$
0
$$\sin \frac{3π}{2}$$
$$\cos π$$
−1
$$\cos 0$$
$$cos \frac{π}{6}$$
$$\frac{\sqrt{3}}{2}$$
$$\sin 0$$
#### Numeric
For the following exercises, state the reference angle for the given angle.
$$240°$$
$$60°$$
$$−170°$$
$$100°$$
$$80°$$
$$−315°$$
$$135°$$
$$45°$$
$$\frac{5π}{4}$$
$$\frac{2π}{3}$$
$$\frac{π}{3}$$
$$\frac{5π}{6}$$
$$−\frac{11π}{3}$$
$$\frac{π}{3}$$
$$\frac{−7π}{4}$$
$$\frac{−π}{8}$$
$$\frac{π}{8}$$
For the following exercises, find the reference angle, the quadrant of the terminal side, and the sine and cosine of each angle. If the angle is not one of the angles on the unit circle, use a calculator and round to three decimal places.
$$225°$$
$$300°$$
$$60°$$, Quadrant IV, $$\sin (300°)=−\frac{\sqrt{3}}{2}, \cos (300°)=\frac{1}{2}$$
$$320°$$
$$135°$$
$$45°$$, Quadrant II, $$\sin (135°)=\frac{\sqrt{2}}{2}, \cos (135°)=−\frac{\sqrt{2}}{2}$$
$$210°$$
$$120°$$
$$60°$$, Quadrant II, $$\sin (120°)=\frac{\sqrt{3}}{2}$$, $$\cos (120°)=−\frac{1}{2}$$
$$250°$$
$$150°$$
$$30°$$, Quadrant II, $$\sin (150°)=\frac{1}{2}$$, \cos(150°)=−\frac{\sqrt{3}}{2}\)
$$\frac{5π}{4}$$
$$\frac{7π}{6}$$
$$\frac{π}{6}$$, Quadrant III, $$\sin( \frac{7π}{6})=−\frac{1}{2}$$, $$\cos(\frac{7π}{6})=−\frac{\sqrt{3}}{2}$$
$$\frac{5π}{3}$$
$$\frac{3π}{4}$$
$$\frac{π}{4}$$, Quadrant II, $$\sin (\frac{3π}{4})=\frac{\sqrt{2}}{2}$$, $$\cos(\frac{4π}{3})=−\frac{\sqrt{2}}{2}$$
$$\frac{4π}{3}$$
$$\frac{2π}{3}$$
$$\frac{π}{3}$$, Quadrant II, $$\sin (\frac{2π}{3})=\frac{\sqrt{3}}{2}$$, $$\cos (\frac{2π}{3})=−\frac{1}{2}$$
$$\frac{5π}{6}$$
$$\frac{7π}{4}$$
$$\frac{π}{4}$$, Quadrant IV, $$\sin (\frac{7π}{4})=−\frac{\sqrt{2}}{2}$$, $$\cos (\frac{7π}{4})=\frac{\sqrt{2}}{2}$$
For the following exercises, find the requested value.
If $$\cos (t)=\frac{1}{7}$$ and $$t$$ is in the 4th quadrant, find $$\sin (t)$$.
If $$\cos (t)=\frac{2}{9}$$ and $$t$$ is in the 1st quadrant, find $$\sin (t).$$
$$\frac{\sqrt{77}}{9}$$
If $$\sin (t)=\frac{3}{8}$$ and $$t$$ is in the 2nd quadrant, find $$\cos (t)$$.
If $$\sin (t)=−\frac{1}{4}$$ and $$t$$ is in the 3rd quadrant, find $$\cos (t).$$
$$−\frac{\sqrt{15}}{4}$$
Find the coordinates of the point on a circle with radius 15 corresponding to an angle of $$220°$$.
Find the coordinates of the point on a circle with radius 20 corresponding to an angle of $$120°$$.
$$(−10,10\sqrt{3})$$
Find the coordinates of the point on a circle with radius 8 corresponding to an angle of $$\frac{7π}{4}$$.
Find the coordinates of the point on a circle with radius 16 corresponding to an angle of $$\frac{5π}{9}$$.
$$(–2.778,15.757)$$
State the domain of the sine and cosine functions.
State the range of the sine and cosine functions.
$$[–1,1]$$
#### Graphical
For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of $$t$$ .
$$\sin t=\frac{1}{2}, \cos t=−\frac{\sqrt{3}}{2}$$
$$\sin t=− \frac{\sqrt{2}}{2}, \cos t=−\frac{\sqrt{2}}{2}$$
$$\sin t=\frac{\sqrt{3}}{2},\cos t=−\frac{1}{2}$$
$$\sin t=− \frac{\sqrt{2}}{2}, \cos t=\frac{\sqrt{2}}{2}$$
$$\sin t=0, \cos t=−1$$
$$\sin t=−0.596, \cos t=0.803$$
$$\sin t=\frac{1}{2}, \cos t= \frac{\sqrt{3}{2}}$$
$$\sin t=−\frac{1}{2}, \cos t= \frac{\sqrt{3}}{2}$$
$$\sin t=0.761, \cos t=−0.649$$
$$\sin t=1, \cos t=0$$
#### Technology
For the following exercises, use a graphing calculator to evaluate.
$$\sin \frac{5π}{9}$$
$$cos \frac{5π}{9}$$
−0.1736
$$\sin \frac{π}{10}$$
$$\cos \frac{π}{10}$$
0.9511
$$\sin \frac{3π}{4}$$
$$\cos \frac{3π}{4}$$
−0.7071
$$\sin 98°$$
$$\cos 98°$$
−0.1392
$$\cos 310°$$
$$\sin 310°$$
−0.7660
#### Extensions
For the following exercises, evaluate.
$$\sin (\frac{11π}{3}) \cos (\frac{−5π}{6})$$
$$\sin (\frac{3π}{4}) \cos (\frac{5π}{3})$$
$$\frac{\sqrt{2}}{4}$$
$$\sin (− \frac{4π}{3}) \cos (\frac{π}{2})$$
$$\sin (\frac{−9π}{4}) \cos (\frac{−π}{6})$$
$$−\frac{\sqrt{6}}{4}$$
$$\sin (\frac{π}{6}) \cos (\frac{−π}{3})$$
$$\sin (\frac{7π}{4}) \cos (\frac{−2π}{3})$$
$$\frac{\sqrt{2}}{4}$$
$$\cos (\frac{5π}{6}) \cos (\frac{2π}{3})$$
$$\cos (\frac{−π}{3}) \cos (\frac{π}{4})$$
$$\frac{\sqrt{2}}{4}$$
$$\sin (\frac{−5π}{4}) \sin (\frac{11π}{6})$$
$$\sin (π) \sin (\frac{π}{6})$$
0
#### Real-World Applications
For the following exercises, use this scenario: A child enters a carousel that takes one minute to revolve once around. The child enters at the point $$(0,1)$$, that is, on the due north position. Assume the carousel revolves counter clockwise.
What are the coordinates of the child after 45 seconds?
What are the coordinates of the child after 90 seconds?
$$(0,–1)$$
What is the coordinates of the child after 125 seconds?
When will the child have coordinates $$(0.707,–0.707)$$ if the ride lasts 6 minutes? (There are multiple answers.)
37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds
When will the child have coordinates $$(−0.866,−0.5)$$ if the ride last 6 minutes?
### Review Exercises
Find the exact value of $$\sin \frac{π}{3}$$.
$$\frac{\sqrt{3}}{2}$$
Find the exact value of $$\cos \frac{π}{4}$$.
Find the exact value of $$\cos π$$.
–1
State the reference angle for $$300°$$.
State the reference angle for $$\frac{3π}{4}$$.
$$\frac{π}{4}$$
Compute cosine of $$330°$$.
Compute sine of $$\frac{5π}{4}$$.
$$−\frac{\sqrt{2}}{2}$$
State the domain of the sine and cosine functions.
State the range of the sine and cosine functions.
$$[–1,1]$$
## 5.3: The Other Trigonometric Functions
### Section Exercises
#### Verbal
On an interval of $$[ 0,2π )$$, can the sine and cosine values of a radian measure ever be equal? If so, where?
Yes, when the reference angle is $$\frac{π}{4}$$ and the terminal side of the angle is in quadrants I and III. Thus, at $$x=\frac{π}{4},\frac{5π}{4}$$, the sine and cosine values are equal.
What would you estimate the cosine of π π degrees to be? Explain your reasoning.
For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle?
Substitute the sine of the angle in for $$y$$ in the Pythagorean Theorem $$x^2+y^2=1$$. Solve for $$x$$ and take the negative solution.
Describe the secant function.
Tangent and cotangent have a period of $$π$$. What does this tell us about the output of these functions?
The outputs of tangent and cotangent will repeat every $$π$$ units.
#### Algebraic
For the following exercises, find the exact value of each expression.
$$\tan \; \frac{π}{6}$$
$$\sec \; \frac{π}{6}$$
$$\frac{2\sqrt{3}}{3}$$
$$\csc \; \csc \; \frac{π}{6}$$
$$\cot \; \frac{π}{6}$$
$$\sqrt{3}$$
$$\tan \; \frac{π}{4}$$
$$\sec \; \frac{π}{4}$$
$$\sqrt{2}$$
$$\csc \; \csc \; \frac{π}{4}$$
$$\cot \; \frac{π}{4}$$
1
$$\tan \; \frac{π}{3}$$
$$\sec \; \frac{π}{3}$$
2
$$\csc \; \csc \; \frac{π}{3}$$
$$\cot \; \frac{π}{3}$$
$$\frac{\sqrt{3}}{3}$$
For the following exercises, use reference angles to evaluate the expression.
$$\tan \; \frac{5π}{6}$$
$$\sec \; \frac{7π}{6}$$
$$−\frac{2\sqrt{3}}{3}$$
$$\csc \; \csc \; \frac{11π}{6}$$
$$\cot \; \frac{13π}{6}$$
$$\sqrt{3}$$
$$\tan \; \frac{7π}{4}$$
$$\sec \; \frac{3π}{4}$$
$$−\sqrt{2}$$
$$\csc \; \csc \; \frac{5π}{4}$$
$$\cot \; \frac{11π}{4}$$
−1
$$\tan \; \frac{8π}{3}$$
$$\sec \; \frac{4π}{3}$$
−2
$$\csc \; \csc \; \frac{2π}{3}$$
$$\cot \; \frac{5π}{3}$$
$$−\frac{\sqrt{3}}{3}$$
$$\tan \; 225°$$
$$\sec \; 300°$$
2
$$\csc \;150°$$csc \;150°\)
$$\cot \; 240°$$
$$\frac{\sqrt{3}}{3}$$
$$\tan \; 330°$$
$$\sec \; 120°$$
−2
$$\csc \; 210°$$csc \; 210°\)
$$\cot \; 315°$$
−1
If $$\sin t= \frac{3}{4}$$, and $$t$$ is in quadrant II, find $$\cos t, \sec t, \csc t, \tan t, \cot t$$.csc t, \tan t, \cot t \).
If $$\cos t=−\frac{1}{3},$$ and $$t$$ is in quadrant III, find $$\sin t, \sec t, \csc t, \tan t, \cot t.$$csc t, \tan t, \cot t.\)
If $$\sin t=−\frac{2\sqrt{2}}{3}, \sec t=−3, \csc t=−\csc t=−\frac{3\sqrt{2}}{4},\tan t=2\sqrt{2}, \cot t= \frac{\sqrt{2}}{4}$$
If $$\tan t=\frac{12}{5},$$ and $$0≤t< \frac{π}{2}$$, find $$\sin t, \cos t, \sec t, \csc t,$$ and $$\cot t$$.csc t,\) and $$\cot t$$.
If $$\sin t= \frac{\sqrt{3}}{2}$$ and $$\cos t=\frac{1}{2},$$ find $$\sec t, \csc t, \tan t,$$ and $$\cot t.$$csc t, \tan t,\) and $$\cot t.$$
$$\sec t=2, \csc t=\csc t=\frac{2\sqrt{3}}{3}, \tan t= \sqrt{3}, \cot t= \frac{\sqrt{3}}{3}$$
If $$\sin 40°≈0.643 \; \cos 40°≈0.766 \; \sec 40°,\csc 40°,\tan 40°, \text{ and } \cot 40°.$$csc 40°,\tan 40°, \text{ and } \cot 40°.\)
If $$\sin t= \frac{\sqrt{2}}{2},$$ what is the $$\sin (−t)?$$
$$−\frac{\sqrt{2}}{2}$$
If $$\cos t= \frac{1}{2},$$ what is the $$\cos (−t)?$$
If $$\sec t=3.1,$$ what is the $$\sec (−t)?$$
3.1
If $$\csc t=0.34,$$ what is the $$\csc (−t)?$$csc t=0.34,\) what is the $$\csc (−t)?$$
If $$\tan t=−1.4,$$ what is the $$\tan (−t)?$$
1.4
If $$\cot t=9.23,$$ what is the $$\cot (−t)?$$
#### Graphical
For the following exercises, use the angle in the unit circle to find the value of the each of the six trigonometric functions.
$$\sin t= \frac{\sqrt{2}}{2}, \cos t= \frac{\sqrt{2}}{2}, \tan t=1,\cot t=1,\sec t= \sqrt{2}, \csc t= \csc t= \sqrt{2}$$
$$\sin t=−\frac{\sqrt{3}}{2}, \cos t=−\frac{1}{2}, \tan t=\sqrt{3}, \cot t= \frac{\sqrt{3}}{3}, \sec t=−2, \csc t=−\csc t=−\frac{2\sqrt{3}}{3}$$
#### Technology
For the following exercises, use a graphing calculator to evaluate.
$$\csc \csc \frac{5π}{9}$$
$$\cot \frac{4π}{7}$$
–0.228
$$\sec \frac{π}{10}$$
$$\tan \frac{5π}{8}$$
–2.414
$$\sec \frac{3π}{4}$$
$$\csc \csc \frac{π}{4}$$
1.414
$$\tan 98°$$
$$\cot 33°$$
1.540
$$\cot 140°$$
$$\sec 310°$$
1.556
#### Extensions
For the following exercises, use identities to evaluate the expression.
If $$\tan (t)≈2.7,$$ and $$\sin (t)≈0.94,$$ find $$\cos (t)$$.
If $$\tan (t)≈1.3,$$ and $$\cos (t)≈0.61$$, find $$\sin (t)$$.
$$\sin (t)≈0.79$$
If $$\csc (t)≈3.2,$$ and $$\csc (t)≈3.2,$$ and $$\cos (t)≈0.95,$$ find $$\tan (t)$$.
If $$\cot (t)≈0.58,$$ and $$\cos (t)≈0.5,$$ find $$\csc (t)$$.csc (t)\).
$$\csc t≈1.16$$csc t≈1.16\)
Determine whether the function $$f(x)=2 \sin x \cos x$$ is even, odd, or neither.
Determine whether the function $$f(x)=3 \sin ^2 x \cos x + \sec x$$ is even, odd, or neither.
even
Determine whether the function $$f(x)= \sin x −2 \cos ^2 x$$ is even, odd, or neither.
Determine whether the function $$f(x)= \csc ^2 x+ \sec x$$ is even, odd, or neither.csc ^2 x+ \sec x\) is even, odd, or neither.
even
For the following exercises, use identities to simplify the expression.
$$\csc t \tan t$$csc t \tan t\)
$$\frac{\sec t}{ \csc t}$$csc t}\)
$$\frac{ \sin t}{ \cos t}= \tan t$$
#### Real-World Applications
The amount of sunlight in a certain city can be modeled by the function $$h=15 \cos (\frac{1}{600}d),$$ where $$h$$ represents the hours of sunlight, and $$d$$ is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42nd day of the year. State the period of the function.
The amount of sunlight in a certain city can be modeled by the function $$h=16 \cos (\frac{1}{500}d)$$, where $$h$$ represents the hours of sunlight, and $$d$$ is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267th day of the year. State the period of the function.
13.77 hours, period: $$1000π$$
The equation $$P=20 \sin (2πt)+100$$ models the blood pressure, $$P$$, where $$t$$ represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures?
The height of a piston, $$h$$, in inches, can be modeled by the equation $$y=2 \cos x+6,$$ where $$x$$ represents the crank angle. Find the height of the piston when the crank angle is $$55°$$.
7.73 inches
The height of a piston, $$h$$,in inches, can be modeled by the equation $$y=2 \cos x+5,$$ where $$x$$ represents the crank angle. Find the height of the piston when the crank angle is $$55°$$.
### Review Exercises
For the following exercises, find the exact value of the given expression.
$$\cos \frac{π}{6}$$
$$\tan \frac{π}{4}$$
1
$$\csc \frac{π}{3}$$
$$\sec \frac{π}{4}$$
$$\sqrt{2}$$
For the following exercises, use reference angles to evaluate the given expression.
$$\sec \frac{11π}{3}$$
$$\sec 315°$$
$$\sqrt{2}$$
If $$\sec (t)=−2.5$$ , what is the $$\sec (−t)$$?
If $$\tan (t)=−0.6$$, what is the $$\tan (−t)$$?
0.6
If $$\tan (t)=\frac{1}{3}$$, find $$\tan (t−π)$$.
If $$\cos (t)= \frac{\sqrt{2}}{2}$$, find $$\sin (t+2π)$$.
$$\frac{\sqrt{2}}{2}$$ or $$−\frac{\sqrt{2}}{2}$$
Which trigonometric functions are even?
Which trigonometric functions are odd?
sine, cosecant, tangent, cotangent
## 5.4: Right Triangle Trigonometry
### Section Exercises
#### Verbal
For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.
When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?
The tangent of an angle compares which sides of the right triangle?
The tangent of an angle is the ratio of the opposite side to the adjacent side.
What is the relationship between the two acute angles in a right triangle?
Explain the cofunction identity.
For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.
#### Algebraic
For the following exercises, use cofunctions of complementary angles.
$$\cos (34°)= \sin (\_\_°)$$
$$\cos (\frac{π}{3})= \sin (\_\_\_)$$
$$\frac{π}{6}$$
$$\csc (21°) = \sec (\_\_\_°)$$csc (21°) = \sec (\_\_\_°)\)
$$\tan (\frac{π}{4})= \cot (\_\_)$$
$$\frac{π}{4}$$
For the following exercises, find the lengths of the missing sides if side $$a$$ is opposite angle $$A$$, side $$b$$ is opposite angle $$B$$, and side $$c$$ is the hypotenuse.
$$\cos B= \frac{4}{5},a=10$$
$$\sin B= \frac{1}{2}, a=20$$
$$b= \frac{20\sqrt{3}}{3},c= \frac{40\sqrt{3}}{3}$$
$$\tan A= \frac{5}{12},b=6$$
$$\tan A=100,b=100$$
$$a=10,000,c=10,000.5$$
$$\sin B=\frac{1}{\sqrt{3}}, a=2$$
$$a=5, ∡ A=60^∘$$
$$b=\frac{5\sqrt{3}}{3},c=\frac{10\sqrt{3}}{3}$$
$$c=12, ∡ A=45^∘$$
#### Graphical
For the following exercises, use Figure to evaluate each trigonometric function of angle A.
$$\sin A$$
$$\frac{5\sqrt{29}}{29}$$
$$\cos A$$
$$\tan A$$
$$\frac{5}{2}$$
$$\csc A$$csc A \)
$$\sec A$$
$$\frac{\sqrt{29}}{2}$$
$$\cot A$$
For the following exercises, use Figure to evaluate each trigonometric function of angle A.
$$\sin A$$
$$\frac{5\sqrt{41}}{41}$$
$$\cos A$$
$$\tan A$$
$$\frac{5}{4}$$
$$\csc A$$csc A\)
$$\sec A$$
$$\frac{\sqrt{41}}{4}$$
$$\cot A$$
For the following exercises, solve for the unknown sides of the given triangle.
$$c=14, b=7\sqrt{3}$$
$$a=15, b=15$$
#### Technology
For the following exercises, use a calculator to find the length of each side to four decimal places.
$$b=9.9970, c=12.2041$$
$$a=2.0838, b=11.8177$$
$$b=15, ∡B=15^∘$$
$$a=55.9808,c=57.9555$$
$$c=200, ∡B=5^∘$$
$$c=50, ∡B=21^∘$$
$$a=46.6790,b=17.9184$$
$$a=30, ∡A=27^∘$$
$$b=3.5, ∡A=78^∘$$
$$a=16.4662,c=16.8341$$
#### Extensions
Find $$x$$.
Find $$x$$.
188.3159
Find $$x$$.
Find $$x$$.
200.6737
A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is $$36°$$, and that the angle of depression to the bottom of the tower is $$23°$$. How tall is the tower?
A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is $$43°$$, and that the angle of depression to the bottom of the tower is $$31°$$. How tall is the tower?
498.3471 ft
A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is $$15°$$, and that the angle of depression to the bottom of the tower is $$2°$$. How far is the person from the monument?
A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is $$18°$$, and that the angle of depression to the bottom of the monument is $$3°$$. How far is the person from the monument?
1060.09 ft
There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be $$40°$$. From the same location, the angle of elevation to the top of the antenna is measured to be $$43°$$. Find the height of the antenna.
There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be $$36°$$. From the same location, the angle of elevation to the top of the lightning rod is measured to be $$38°$$. Find the height of the lightning rod.
27.372 ft
#### Real-World Applications
A 33-ft ladder leans against a building so that the angle between the ground and the ladder is $$80°$$. How high does the ladder reach up the side of the building?
A 23-ft ladder leans against a building so that the angle between the ground and the ladder is $$80°$$. How high does the ladder reach up the side of the building?
22.6506 ft
The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.
The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.
368.7633 ft
Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be $$60°$$, how far from the base of the tree am I?
### Review Exercises
For the following exercises, use side lengths to evaluate.
$$\cos \frac{π}{4}$$
$$\cot \frac{π}{3}$$
$$\frac{\sqrt{3}}{3}$$
$$\tan \frac{π}{6}$$
$$\cos (\frac{π}{2}) = \sin ( \_\_°)$$
0
$$\csc (18°)= \sec (\_\_°)$$
For the following exercises, use the given information to find the lengths of the other two sides of the right triangle.
$$\cos B= \frac{3}{5}, a=6$$
$$b=8,c=10$$
$$\tan A = \frac{5}{9},b=6$$
For the following exercises, use Figure to evaluate each trigonometric function.
$$\sin A$$
$$\frac{11\sqrt{157}}{157}$$
$$\tan B$$
For the following exercises, solve for the unknown sides of the given triangle.
$$a=4, b=4$$
A 15-ft ladder leans against a building so that the angle between the ground and the ladder is $$70°$$. How high does the ladder reach up the side of the building?
14.0954 ft
The angle of elevation to the top of a building in Baltimore is found to be 4 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.
## Practice Test
Convert $$\frac{5π}{6}$$ radians to degrees.
$$150°$$
Convert $$−620°$$ to radians.
Find the length of a circular arc with a radius 12 centimeters subtended by the central angle of $$30°$$.
6.283 centimeters
Find the area of the sector with radius of 8 feet and an angle of $$\frac{5π}{4}$$ radians.
Find the angle between $$0°$$ and $$360°$$ that is coterminal with $$375°$$.
$$15°$$
Find the angle between 0 and $$2π$$ in radians that is coterminal with $$−\frac{4π}{7}$$.
Draw the angle $$315°$$ in standard position on the Cartesian plane.
Draw the angle $$−\frac{π}{6}$$ in standard position on the Cartesian plane.
A carnival has a Ferris wheel with a diameter of 80 feet. The time for the Ferris wheel to make one revolution is 75 seconds. What is the linear speed in feet per second of a point on the Ferris wheel? What is the angular speed in radians per second?
3.351 feet per second, $$\frac{2π}{75}$$ radians per second
Find the exact value of $$\sin \frac{π}{6}$$.
Compute sine of $$240°$$.
$$−\frac{\sqrt{3}}{2}$$
State the domain of the sine and cosine functions.
State the range of the sine and cosine functions.
$$[ –1,1 ]$$
Find the exact value of $$\cot \frac{π}{4}$$.
Find the exact value of $$\tan \frac{π}{3}$$.
$$\sqrt{3}$$
Use reference angles to evaluate $$\csc \frac{7π}{4}$$.
Use reference angles to evaluate $$\tan 210°$$.
$$\frac{\sqrt{3}}{3}$$
If $$\csc t=0.68$$,what is the $$\csc (−t)$$?
If $$\cos t= \frac{\sqrt{3}}{2}$$,find $$\cos (t−2π)$$.
$$\frac{\sqrt{3}}{2}$$
Which trigonometric functions are even?
Find the missing angle: $$\cos (\frac{π}{6})= \sin (\_\_\_)$$
$$\frac{π}{3}$$
Find the missing sides of the triangle $$ABC: \sin B= \frac{3}{4},c=12$$
Find the missing sides of the triangle.
$$a=\frac{9}{2},b=\frac{9\sqrt{3}}{2}$$
The angle of elevation to the top of a building in Chicago is found to be 9 degrees from the ground at a distance of 2000 feet from the base of the building. Using this information, find the height of the building. |
Surface Area of Prisms and Cylinders Worksheet | Problems & Solutions
# Surface Area of Prisms and Cylinders Worksheet
Surface Area of Prisms and Cylinders Worksheet
• Page 1
1.
Which of the following triangles is similar to the ΔPQR?
a. ΔQRS b. ΔPNS c. ΔSQM d. ΔSRN
#### Solution:
Similar figures have same shape but not necessarily of same size.
ΔPQR is a right triangle.
[From the figure.]
ΔPNS, ΔSRN, and ΔSQM are not similar to ΔPQR as they are not right triangles.
From the given choices the triangle that is similar to ΔPQR is ΔQRS.
2.
What is the surface area of the cylinder shown?
a. 234.68 cm2 b. 192.58 cm2 c. 220 cm d. 212.14 cm2
3.
The diameter of a roller is 110 cm and its length is 150 cm. It takes 800 complete revolutions to move once over a play ground to level it. Find the area of the play ground. [Take $\pi$ = 3.]
a. 1980 m² b. 4686 m² c. 3960 m² d. 1320 m²
#### Solution:
Area of the play ground = No. of revolutions × Curved surface area of the rollers.
[Formula.]
Radius of the roller = d / 2= 55 cm
Length of the roller = 150cm
Curved surface area = 2 × 3 × 55 × 150 = 49500 cm2
[Curved surface area = 2 π r l.]
Area of the play ground = 800×49500 / 10000= 3960 m2
[Substitute in step1 and simplify.]
[1m = 100cm.]
4.
The inner diameter of a circular well is 7 m and it is 6 m deep. Find the cost of plastering its inner curved surface at the rate of $25 per sq m. [Take $\pi$ = 3.] a.$1565 b. $1575 c.$1595 d. $1615 #### Solution: Cost of plastering = (Rate of plastering) × (Inner curved surface area). [Formula.] Inner radius = d / 2 = 3.50 m [Radius = Diamater / 2.] Depth = 6 m [Given.] Inner surface area = 3 × 3.50 × 6 = 63 m2 [Well is in the form of a cylinder.] [Inner surface area of a cylinder = π r h.] Rate of plastering per sq.m =$25
[Given.]
Cost of plastering = 25 × 63 = $1575 [Substitute in step1 and simplify.] Correct answer : (2) 5. A closed circular water tank of outer diameter 4 m and height 2 m is to be painted. Find the total charges for painting if the rate of painting is$10 per sq.m. [Take $\pi$ = 3.]
a. $623 b.$603 c. $437 d.$480
#### Solution:
Cost of painting = (Rate of painting) × (total surface area of the tank).
[Formula.]
Total surface area = Curved surface area + area of the end faces.
[Analysis.]
Radius = Diameter / 2 = 2 m
Height = 2 m.
[Given.]
Curved surface area = 2 × 3 × 2 × 2 = 24 sq.m
[Curved surface area = 2 π r h.]
Area of the end faces = 2 × 3 × 22 = 24 sq.m
[Area of end faces = 2 π r2.]
Total surface area = 24 + 24 = 48 sq.m
[Substitute in step2.]
Rate of painting = $10 per sq.m [Given.] Cost of painting = 10 × 48 =$480
[Substitute in step1 and simplify.]
6.
A cylindrical open vessel is to be made of a thin sheet of uniform thickness. If the diameter of the vessel is 30 cm and the height is 22 cm, then find the area of the sheet required to make it. [Take $\pi$ = 3.]
a. 2778 cm2 b. 2582 cm2 c. 2753 cm2 d. 2655 cm2
#### Solution:
Area of the sheet = Curved surface area of the cylinder + Base area of the vessel.
[Analysis.]
Radius of the vessel = d / 2 = 15 cm
[Given.]
Height of the vessel = 22 cm
[Given.]
Curved surface area = 2πrh = 2 × 3 × 15 × 22 = 1980 sq.cm
[Formula.]
Area of the base = π × r2 = 3 × 152 = 675 sq.cm
[Formula.]
Area of the sheet = 1980 + 675 = 2655 cm2
[From steps 4 and 5.]
7.
The inner and outer diameter of a metallic pipe are 7 cm and 8 cm. Length of the pipe is 4 m. Find the total surface area of the pipe in sq.cm. [Take $\pi$ = 3.]
a. 18022.50 sq.cm b. 17946.50 sq.cm c. 18089.50 sq.cm d. 18198.50 sq.cm
#### Solution:
Total surface area of the pipe = Inner curved surface area + Outer curved surface area + Area of two end faces.
[Formula.]
Inner curved surface area = 2 × 3 × 3.5 × 4 × 100 = 8400 sq.cm
[Inner curved surface area = 2 π a h.]
Outer curved surface area = 2 × 3 × 4 × 4 × 100 = 9600 sq.cm
[Outer curved surface area = 2 π b h.]
Area of the end faces = 2(3 × 42 - 3 × 3.52) = 22.50 sq.cm
[Area of the circular ring = 2(π b2 - π a2).]
Total surface area = 8400 + 9600 + 22.50 = 18022.50 sq.cm
[Substitute in step1 and simplify.]
So, the total surface area of the pipe is 18022.50 sq.cm
8.
Find the surface area of the cylinder.
a. 16$\pi$ b. 128$\pi$ c. 52$\pi$ d. 96$\pi$
9.
Find the area of the aluminium sheet required to make a cylindrical can measuring a base diameter of 8 cm. and a height of 12 cm. Round the answer to the nearest ten.
a. 401 cm2 b. 401.92 cm2 c. 402.92 cm2 d. 402 cm2
#### Solution:
The surfaces of a cylinder include two circles and one rectangle.
Therefore, the total surface area = 2 × area of the circle + area of the rectangle.
= 2 π r2 + b h
= 2 π (4 × 4) + 2 π (4 × 12)
[Diameter = radius / 2and substituting.]
= π (32 + 96) = π (128) = 401.92 cm.2
[Simplify.]
The area of the aluminium sheet needed to make the cylindrical can is 401.92 cm.2 and to the nearest ten is 402 cm.2.
10.
The figure shows a rectangular prism with a cylinder which has been removed from it. What is the total surface area of the given solid? Round the answer to the nearest ten.
a. 263 m2 b. 232 m2 c. 231 m2 d. 230 m2
#### Solution:
Total surface area of the rectangular prism = surface area of the rectangular prism + lateral area of the cylinder - 2 × base area of the cylinder.
Dimensions of the rectangular prism are 4m. × 6m. × 8m.
Therefore, the surface area of the rectangular prism = 2 (lw + wh + lh)
= 2 (8 × 4 + 4 × 6 + 6 × 8) = 208
[Substituting the values.]
Lateral area of the cylinder = 2 π r h
[Formulae.]
2 (π × 2.5 × 4) = 20 π = 62.8
[Substituting the values and simplify.]
Base area of the cylinder = π r2
[Formulae.]
= π (2.5 × 2.5) = 6.25 π = 19.625
[Substituting the values.]
The total surface area of the rectangular prism = 208 + 62.8 - 2 × 19.625
= 208 + 62.8 - 39.25 = 231.55 m.2
[Simplify.]
231.55 to the nearest ten is 232 m.2.
Therefore, the surface area of the solid to the nearest ten is 232 m.2. |
# Three Common Algebra Mistakes You MUST Avoid
Whether it is a calculus class or a first algebra class, I have seen students at every level and every natural ability make these mistakes. Why would so many fall into these traps? Well, it really comes down to working too quickly and applying rules without stopping to think about whether or not they apply. If you can avoid these mistakes, you will find yourself well ahead of the curve. In fact, these are the kinds of things people writing multiple choice tests think about.
1. Distributing a Power
It seems as though everyone wants $(a+b)^2=a^2+b^2$, and who wouldn’t? Every math exercise in the world becomes much easier if you use this! Unfortunately, in general, $(a+b)^2\neq a^2+b^2$. Its true in a trivial case – when a or b equals zero (plug zero in for either one to check). Otherwise, this is a no go and if you aren’t sure, try it out with some nonzero numbers: What if a=1 and b=1?
$(1+1)^2=2^2=4$ while $1^2+1^2=1+1=2$. Clearly we get two different things. Now this isn’t a mathematical proof since I used two specific numbers but the fact that it does not work here should give you pause. In fact, you would have to FOIL the left hand side to find its equivalent expression. In other words, $(a+b)^2=(a+b)(a+b)^2=a^2+2ab+b^2$.
2. Forgetting to Completely Distribute
What if you needed to simplify $-(x+2y-5)$. Many students would write $-x+2y-5$. Can you see what is wrong?
The negative in front of the parentheses is basically a negative 1 multiplying all of the terms within those parentheses. That means it must be distributed to EVERYTHING, not just the first term! $-(x+2y-5)=-x-2y+5$. This works regardless of what is in front of the parentheses. For instance, $2(5x+4x^2-1)=10x+8x^2-2$.
Most often, it seems that people understand this but make this mistake when they are going too fast. This is why I advocate writing down every single step when you do any math problem. (or just double check yourself anytime you have had to distribute)
3. Applying the Zero-Product Rule to Everything
The zero product rule is what allows you to solve an equation like $(x+2)(x-3)=0$. By realizing that the only way two things can multiply to give us zero is if one or both are zero, we can say $x+2=0$ and $x-3=0$. This is all based on the fact that there is a zero on the right hand side!
If instead, we had $(x+2)(x-3)=2$ we couldn’t immediately take this approach. This does not imply that $x+2=2$ and $x-3=2$. Again, you can only apply this if you manage to get a product of terms equal to zero. (to solve the new equation, you could foil the left hand side and then bring the two over – going from there)
There are plenty of other mistakes that I have seen but these are by far the most common. In the end, it comes down to applying properties in situations where they do not apply. Learn to ask yourself if the rule you are applying makes sense in a given situation and you will be well on your way to avoiding these! |
# We Try To Predict Benedict Cumberbatch’s Future (10/20/2019)
How will Benedict Cumberbatch fare on 10/20/2019 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is of questionable accuracy – take it with a grain of salt. I will first calculate the destiny number for Benedict Cumberbatch, and then something similar to the life path number, which we will calculate for today (10/20/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 10/20/2019: We will consider the month (10), the day (20) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 10 and add the digits together: 1 + 0 = 1 (super simple). Then do the day: from 20 we do 2 + 0 = 2. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 1 + 2 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 10/20/2019.
DESTINY NUMBER FOR Benedict Cumberbatch: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Benedict Cumberbatch we have the letters B (2), e (5), n (5), e (5), d (4), i (9), c (3), t (2), C (3), u (3), m (4), b (2), e (5), r (9), b (2), a (1), t (2), c (3) and h (8). Adding all of that up (yes, this can get tedious) gives 77. This still isn’t a single-digit number, so we will add its digits together again: 7 + 7 = 14. This still isn’t a single-digit number, so we will add its digits together again: 1 + 4 = 5. Now we have a single-digit number: 5 is the destiny number for Benedict Cumberbatch.
CONCLUSION: The difference between the path number for today (6) and destiny number for Benedict Cumberbatch (5) is 1. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is of questionable accuracy. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
Courses
# NCERT Solutions(Part- 1)- Practical Geometry Class 8 Notes | EduRev
## Class 8 : NCERT Solutions(Part- 1)- Practical Geometry Class 8 Notes | EduRev
The document NCERT Solutions(Part- 1)- Practical Geometry Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8
Question: Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.
Solution: No, the quadrilateral ABCD cannot be constructed with the given combination of measurements. Here, knowledge of BC or DC is must.
Question: (i) We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?
(ii) Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5cm?
Why?
(iii) Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?
(iv) A student attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it. What is the reason?
Hint: Discuss it using a rough sketch.
Solution:
(i) No, any 5 measurements (elements) cannot determine a quadrilateral. Actually to construct a quadrilateral, we need a specific combination of measurements such as:
(a) Four sides and one diagonal
or
(b) Three sides and two diagonals
or
(c) Two adjacent sides and three angles
or
(d) Three sides and two included angles.
or
(e) Some special properties are given.
(ii) Let us draw a rough sketch of BATS as given.
Here, we cannot locate the points T and B without knowing the measurements ST and SB respectively.
Thus, we cannot draw a parallelogram with the given measurements.
(iii) Let us draw a rough sketch of the (rhombus) quadrilateral ZEAL and mark the given measurements on it.
Yes, we can draw the required rhombus, because it's all sides are equal to 3.5 cm and a diagonal EL is known.
(iv) Let us draw the rough sketch of the quadrilateral PLAY and mark its measurements, No, this quadrilateral cannot be drawn
∵ Point P cannot be located. In D LPY, sum of the lengths of PL and PY is less than LY, i.e. (2 cm + 3 cm) < 6 cm.
EXERCISE 4.1
Question 1. Construct the following quadrilaterals.
AB = 45 cm
BC = 5.5 cm
CD = 4 cm
AC = 7 cm
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm
(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm
(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
Solution:
Note: Before making a fair construction for each of the above quadrilaterals, a rough sketch must be drawn and the given measurements be marked.
(i) Steps of construction:
I. Draw a line segments AB = 4.5 cm.
II. With centre at A and radius = 7 cm, draw an arc.
III. With centre at B and radius = 5.5 cm, draw another arc to intersect the previous arc at C.
IV. With centre at A and radius = 6 cm, draw an arc on the side opposite to that of B.
V. With centre at C and radius = 4 cm, draw another arc to intersect the previous arc at D.
VI. Join AC, BC, DA and DC.
Thus, ABCD is the required quadrilateral.
(ii) Steps of construction:
I. Draw a line segments JU = 3.5 cm.
II. With centre at U and radius UP = 6.5 cm, draw an arc.
III. With centre at J and radius = 4.5 cm, draw another arc to intersect the previous arc at P.
IV. With centre at U and radius = 4 cm, draw an arc on the side opposite to that of J.
V. With centre at P and radius = 4 cm, draw another arc intersecting the previous arc at M.
VI. Join PJ, PU, PM and UM.
Thus, JUMP is the required quadrilateral.
(iii) Steps of construction:
I. Draw a line segments ER = 4.5 cm.
II. With centre at E and radius = 7.5 cm, draw an arc.
III. With centre at R and radius = 6 cm, draw another arc to intersect the previous are at O.
IV. With centre at R and radius = 6 cm, draw an are on the side opposite to R.
V. With centre at O and radius = 4.5 cm, draw another arc to intersect the previous arc at M.
VI. Join RO, EM, EO and MO.
Thus, EROM or MORE is the required parallelogram.
(iv) Steps of construction:
I. Draw a line segment BE = 4.5 cm.
II. With centre at E and radius = 6 cm, draw an arc.
III. With centre at B and radius = 4.5 cm, draw another arc to intersect the previous arc at T.
Note: All the 4 sides of a rhombus are equal.
IV. Join TB and TE.
V. With centre at T and radius = 4.5 cm, draw an arc on the side opposite to that of B.
VI. With centre at E and radius = 4.5 cm, draw another arc to intersect the previous arc at S.
VII. Join ST and SE Thus, BEST is the required quadrilateral.
CONSTRUCTION OF A QUADRILATERAL WHEN ITS TWO DIAGONALS AND THREE SIDES ARE KNOWN
In this case, we take one of the diagonal as the base of a triangle, then taking any two the given sides, we locate the vertex (third point of the quadrilateral) of the triangle. Again, using the second diagonal and third given side, we locate the fourth point of the quadrilateral.
Question 1. “Construct a quadrilateral ABCD, given that BC = 4.5 cm, AD = 5.5 cm, CD = 5 cm, the diagonal AC = 5.5 cm and diagonal BD = 7 cm.” Can we draw the quadrilateral by drawing D ABD first and then find the fourth point C?
Solution: Since, the measurement of AB is not given.
∴ We cannot locate the point B. Thus, the quadrilateral cannot be drawn.
Question 2. Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm? Justify your answer
Solution: Yes a quadrilateral PQRS can be drawn using PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm.
EXERCISE 4.2
Question 1. Construct the following quadrilaterals.
LI = 4 cm
I F = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm
(iii) Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm
Solution: (i) Steps of constructions:
I. Draw a line segment LI = 4 cm.
II. With centre at I and radius = 4 cm, draw an arc.
III. With centre at L and radius = 2.5 cm, draw another arc to intersect the previous arc at T.
IV. Join TI and TL.
V. With centre I and radius = 3 cm, draw an arc on the side opposite to that of L.
VI. With centre at L and radius = 4.5 cm, draw another arc to intersect the previous arc at F.
VII. Join FL and FI.
Thus, LIFT is the required quadrilateral.
(ii) Steps of construction:
I. Draw a line segment LD = 5 cm.
II. With centre at L and radius = 6 cm, draw an arc.
III. With centre at D and radius = 6 cm, draw another arc to intersect the previous arc at G.
IV. Join GL and GD.
V. With centre at D and radius = 10 cm, draw an arc on the same side as that of G.
VI. With centre at L and radius = 7.5 cm, draw another arc to intersect the previous arc at O.
VII. Join OL and OD.
Thus, GOLD is the required quadrilateral.
(iii) Steps of construction:
I. Draw a line segment DE = 6.5 cm.
II. Draw perpendicular bisector PQ of DE such that M being the mid-point of DE.
III. With centre M and radius = 2.8 cm, draw arcs to intersect PQ at B and N.
IV. Join ND, NE, BE and BD.
Thus, BEND is the required rhombus.
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## Mathematics (Maths) Class 8
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# algebra
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Two of the real roots of x^3 - Mx + 15 = 0 sum to 5. What is M?
Oct 3, 2020
#1
+345
+1
Have you learned Vieta's formula yet? You could really use it here.
$$\text {root}_1 + \text {root}_2 = 5\\\text {in your situation. Vieta's formula says that this will be equal to} \\-\frac ba \\\text {where }b=-M \text { and } a=x \text { in your case}$$
So this gives us $$-\frac {-M}x = 5 \\\implies \frac Mx = 5 \\\implies \mathbf {M = 5x}$$
So M=5x.
Try to solve the equation now that you know this.
Oct 4, 2020
#2
0
$$\displaystyle \text{If the equation }\: x^{3}+bx^{2}+cx+d=0,\: \text{has roots}\: \alpha, \beta, \gamma,\\ \text{then} \\ \alpha + \beta + \gamma = -b, \\ \alpha \beta + \beta \gamma + \gamma \alpha = c, \text{ and}\\ \alpha \beta \gamma = -d.$$
$$\displaystyle \text{So, for the equation} \: x^{3}-Mx +15=0,\\ \alpha + \beta + \gamma = 0 \dots \dots(1)\\ \alpha \beta + \beta \gamma +\gamma \alpha = -M \dots \dots(2)\\ \alpha\beta \gamma=-15 \dots \dots(3).$$
$$\displaystyle \text{If now } \: \alpha + \beta = 5, \text{then from (1)}, \gamma = -5.$$
$$\displaystyle \text{Substituting into (2)},\: \alpha \beta -5\beta-5\alpha=\alpha \beta -5(\alpha + \beta)=\alpha \beta -25=-M, \\ \text{from which} \: M =25-\alpha\beta \dots \dots (4).$$
$$\displaystyle \text{From (3)} \: \alpha \beta(-5)=-15\quad\text{so } \: \alpha \beta = 3 \: \: \text{and so from (4)}, \\ M=25-3 = 22.$$
.
Oct 4, 2020 |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Published by Pearson
# Chapter 1-9 - Cumulative Review: 2
#### Answer
$c-6$
#### Work Step by Step
$\bf{\text{Solution Outline:}}$ Use the Distributive Property to simplify the given expression, $3c-[8-2(1-c)] .$ Then combine like terms. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3c-[8-2(1)-2(-c)] \\\\= 3c-[8-2+2c] \\\\= 3c-1[8]-1[-2]-1[2c] \\\\= 3c-8+2-2c .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 3c-8+2-2c \\\\= (3c-2c)+(-8+2) \\\\= c-6 .\end{array}
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. |
# Greatest Common Factor (G.C.F)
H. C. F that is highest common factor is also called Greatest Common Factor (G.C.F). There is actually no difference between the two. Greatest Common Factor is actually the greatest factor of two or more numbers.
Here are few examples on Greatest Common Factor
1. Find the greatest common factor of 56 and 63
Solution:
The prime factors of 56 are: 2 × 2 × 2 × 7 = 23×7
The prime factors of 63 are: 3 × 3 × 7 = 32× 7
The common prime factor of 56 and 63 is 7
The lowest power of 7 is 7
Therefore, G .C.F = 7
2. Find the greatest common factor of 42 and 44
Solution:
The prime factors of 42 are: 2 × 3 × 7
The prime factors of 44 are: 2 × 2 × 11 = 2× 11
The common prime factor of 42 and 44 is 2
The lowest power of 2 is 2
Hence, G. C. F = 2
3. Find the greatest common factor of 88 and 110
Solution:
The prime factors of 88 are: 2 × 2 × 2 × 11 = 2× 11
The prime factors of 110 are: 2 × 5 × 11
The common prime factors of 8 and 110 are 2 and 11
The lowest power of 2 is 2
The lowest power of 11 is 11
Hence, G. C. F = 2 × 11 = 22
4. Find the greatest common factor of 450 and 500
Solution:
The prime factors of 450 are: 2 × 5 × 5 × 3 × 3 = 2 × 5× 32
The prime factors of 500 are: 2 × 2 × 5 × 5 × 5 = 22 × 53
The common prime factors of 450 and 500 are 2 and 5
The lowest power of 2 is 2
The lowest power of 5 is 52
Therefore, G. C. F = 2 × 5=2 ×5 ×5=50
5. Find the greatest common factor of 18 and 20
Solution:
The prime factors of 18 are: 2 × 3 × 3 = 2 × 32
The prime factors of 20 are: 2 × 2 × 5 = 2× 5
The common prime factor of 18 and 20 is 2
The lowest power of 2 is 2
Therefore, G. C. F = 2
6. Find the greatest common factor of 35, 28 and 70
Solution:
The prime factors of 35 are: 5 ×7
The prime factors of 28 are: 2 × 2 × 7 = 2×7
The prime factors of 70 = 2 × 5 × 7
The common prime factor of 35, 28 and 70 is
The lowest power of 7 is 7
Therefore, G. C. F = 7
7. Find the greatest common factor of 56, 28 and 128
Solution:
The prime factors of 56 are: 2 × 2 × 2 × 7 = 2× 7
The prime factors of 28 are: 2 × 2 × 7 = 2× 7
The prime factors of 128 are: 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27
The common prime factor of 56, 28 and 128 is 2
The lowest power of 2 is 22
Therefore, G. C. F = 2 × 2 = 4
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# Number Patterns In Whole Numbers
Numbers have fascinated humans since ages, be it the mathematicians or statisticians. There is so much that one can do with them, and there’s so much that is yet to be discovered. For example, we know that the whole numbers represent the set of all positive numbers, including zero, without any decimal or fractional parts. But did we know that we can derive relationships between the whole numbers by finding some kind of patterns between them? This is why numbers are so interesting or fascinating. In this article, we are going to discuss what are number patterns in Mathematics, charts, examples in detail.
## What are Number Patterns?
In Mathematics, number patterns are the patterns in which a list number that follows a certain sequence. Generally, the patterns establish the relationship between two numbers. It is also known as the sequences of series in numbers. In order to solve the problems on the number pattern, first, we have to understand the rule being followed in the pattern. Let us take a simple example to understand the principle behind a number pattern.
### Number Pattern Example
Consider an example given here. The given sequence of numbers is 11, 17, 23, 29, 35, 41, 47, and 53. The following figure helps to understand the relationship between the numbers.
In the given pattern, the sequence is increased by 6. It means the addition of the number 6 to the previous number gives the succeeding number. Also, the difference between the two consecutive number is 6.
## Number Patterns Using Dots
Let’s start representing each whole number with a set of dots and arranging these dots in some elementary shape to find number patterns. For arranging these dots, we take strictly four shapes into account. Numbers can be arranged into:
1. A line
2. A rectangle
3. A square
4. A triangle
## Line
Every number can be arranged in a line. Examples:
The number 2 can be represented byÂ
The number 3 can be represented by Â
All other numbers can be represented in a similar pattern.
## Rectangle
Some numbers can be arranged as a rectangle. Examples:
The number 6 can be arranged as a rectangle with 2 rows and 3 columns as Â
Similarly, 12 can be arranged as a rectangle with 3 rows and 4 columns as   Â
Or as a rectangle with 2 rows and 6 columns as Â
Similar it can be formed by 8, 10, 14, 15, etc.
## Square
Some numbers can be arranged as squares. Examples:
The number 4 can be represented as  and 9 as Â
Similar it can be formed by 16, 25, 36, 49 and so on.
## Triangle
Some numbers can be arranged as triangles. Examples:
The number 3 can be represented as   and 6 as Â
Similar it can be formed by 10, 15, 21, 28, etc. It is to be noted that the triangle should have its 2 sides equal. Hence, the number of dots in the rows starting from the bottom row should be like 4,3,2,1. The top row should always have one dot.
## Number Pattern Types
There are different types of number patterns in Mathematics. They are:
• Arithmetic Sequence
• Geometric Sequence
• Square Numbers
• Cube Numbers
• Triangular Numbers
• Fibonacci Numbers
### Number Patterns Observation
Observation of number patterns can guide to simple processes and make the calculations easier.
Consider the following examples which help the addition and subtraction with numbers like 9, 99, 999, etc. simpler.
• 145 + 9 = 145 + 10 – 1 = 155 – 1 = 154
• 145 – 9 = 145 – 10 + 1 = 135 + 1 =Â 136
• 145 + 99 = 145 + 100 – 1 = 245 – 1 = 244
• 145 – 99 = 145 – 100 + 1 = 45 + 1 = 46
Consider another pattern which simplifies multiplication with 9, 99, 999, and so on:
• 62 x 9 = 62 x (10 – 1) = 558
• 62 x 99 = 62 x (100 – 1) = 6138
• 62 x 999 = 62 x (1000-1)Â = 61938
Consider the following pattern which simplifies multiplication with numbers like 5, 25, 15, etc.:
• 48 x 5 = 48 x 10/2 = 480/2 = 240
• 48 x 25 = 48 x 100/4 = 4800/4 = 1200
• 48 x 125 = 48 x 1000/8 = 48000/8 = 6000
A number of patterns of similar fashion can be observed in the whole numbers which simplifies calculations to quite an extent. |
# Statistics – Mean, Distribution and Variance
## Foreword
I wrote the following many years ago when I learning basic statistics. It still serves as a reference occasionally and I thought I’d put it online. There is also a page on Statistics – Confidence Intervals.
## Arithmetic Mean
The arithmetic mean, or average, is calculated simply by summing individual values in a list, and dividing by the number of values.
If the list contains all possible values, then the list is a statistical population and the mean is called a population mean or true mean and is often designated by the symbol μ, which can be calculated precisely by:
```μ = 1/N * (a1 + ... + aN) = 1/N * ∑ai where i = [1, N]
Equation 1a
Population Mean```
where N is the number of values in the population.
In practice, however, it is usually impossible to know all possible N values because populations are typically continuous in nature, and therefore, the number of possible values infinite. However, it is possible to estimate the population mean by drawing n random samples from a larger population of N possible values. The estimation is known as the sample mean, because it is the mean of the sample, rather than the population.
In this case, we can write:
```ā = 1/n * (a1 + ... + an) = 1/n * ∑ai where i = [1, n]
Equation 1b
Sample Mean```
The difference between equations 1a and 1b is subtle, but it is important to recognise that the sample mean ā represents an estimate of population mean μ, and that the larger the number of samples n taken, the closer the sample mean approximates the population mean μ.
With regard to performance testing, we can say that the number of measurements (N) which we could potentially observe is infinite*, but in practice we can only sample a finite number (n). Therefore, a mean value of such measurements will provides us with an estimate (ā) of the system’s true performance (μ), and the more measurements made–the more accurate the estimate will be.
* Strictly speaking, due to numerical precision restrictions, the number of measurements made with computers are possibly finite, but very large nevertheless.
The distribution of a population (i.e. the possible values) describes the probability of observing (or measuring) any value within a specific range. Many natural phenomenon exhibit, what is referred to as, a normal or Gaussian distribution.
In a normal distribution, it is most expected to observes values close to the population mean, while values far from the mean are expected less often. For example, if a population mean is estimated to be 10, values such as 9.8, 9.9, 10.1 etc. may be observed frequently, while values such as 3 or 17.5, while possible, may be seen infrequently.
The shape of a normal distribution is shown below.
Note that a normal distribution peaks at the mean μ and is symmetrical.
## Central Limit Theorem
The Central Limit Theorem states that if the sum of the variables has a finite variance, then it will be approximately normally distributed. Since many real processes yield distributions with finite variance, this explains the ubiquity of the normal probability distribution.
The Central Limit Theorem implies that if the sample size n is “large,” then the distribution of the sample mean is approximately normal. Of course, the term “large” is relative. Roughly, the more “abnormal” the basic distribution, the larger n must be for normal approximations to work well. The rule of thumb is that a sample size n of at least 30 will suffice.
It is important to understand that the Central Limit Theorem refers to the distribution of the sample mean, not the distribution of data itself. However, the normality of the sample mean is of fundamental importance, because it means that we can approximate the distribution of certain statistics (i.e. standard error), even if we know very little about the underlying sampling distribution.
## Variance
The variance of a random variable (or somewhat more precisely, of a probability distribution) is a measure of its statistical dispersion, indicating how its possible values are spread around the expected value. Where the population mean shows the location of the distribution, the variance indicates the scale of the values.
The variance of a finite population of size N is given by:
```σ2 = 1/N * ∑(ai - μ)2 where i = [1, N]
Equation 1a
Population Variance```
This is known as population variance and represents the case where all possible values of the population are known, which is rarely the case.
More likely, the statistical population will be large, or infinite, and the calculation of the population variance impossible. In this case, it is possible to estimate the population variance by calculating a sample variance, as determined by drawing n samples at random from the larger (or infinite) population.
A source of confusion is that there two commonly used formulas for calculating sample variance:
```sn2 = 1/n * ∑(ai - ā)2 where i = [1, n]
Equation 1b
Biased Sample Variance```
or:
```sn-12 = 1/(n - 1) * ∑(ai - ā)2 where i = [1, n]
Equation 1c
Unbiased Sample Variance```
Here we are using “s2“, rather than “σ2“, to denote that the result is an estimate calculated only from a sample n, rather than the population N.
In addition, note the different “n” and “n-1” in the above equations. Equation 1b is equivalent to equation 1a, but the sample is treated as if it where the population, and the sample mean is used in place of the population mean. Whereas, equation 1c is known as the unbiased estimator of population variance, and is the method used by convention in applied statistics where the population size N is known to be large or infinite.
The results of both equations can be referred to as a sample variance, but will give slightly different results due to the different “n” and “n-1” terms. In practice, however, the difference becomes negligible as n becomes larger.
## Standard Deviation
A more understandable measure of variance is called standard deviation–simply the square root of variance, as denoted by the symbol σ. As its name implies it gives, in a standard form, an indication of the possible deviations from the mean, as illustrated below for normally distributed data.
### Normally Distributed Data
One often assumes that the data is from an approximately normally distributed population. If this assumption is justified, then one standard deviation (σ) equates to a range around the mean where the probability of observing a value is approximately 68.27%.
In addition, for normally distributed data, it is known that about 95% of the values are within two standard deviations of the mean and about 99.7% lie within 3 standard deviations. This is known as the “68-95-99.7 rule”, or “the empirical rule”.
The population standard deviation is defined by:
```σ = √( 1/N * ∑(ai - μ)2 ) where i = [1, N]
Equation 2a
Population Standard Deviation```
In practice the standard deviation must be estimated by calculating a value based up on a sample of the population, rather than the population itself. The result is known as the sample standard deviation, and as with variance, there are two commonly used formulas for calculating it:
```sn = √( 1/n * ∑(ai - ā)2 ) where i = [1, n]
Equation 2b
Population Standard Deviation ```
and:
```sn-1 = √( 1/(n - 1) * ∑(ai - ā)2 ) where i = [1, n]
Equation 2c
Sample Standard Deviation ```
Here we are using “s”, rather than “σ”, to denote that the result is an estimate calculated from a sample. Equation 2c is known as the sample standard deviation, and as with the equivalent unbiased estimator of variance, it is the method used by convention in applied statistics where the population size N is known to be large or infinite. Note, that it is not correct to refer to equation 2c as the “unbiased standard deviation”, as the ‘unbiasedness’ of the variance does not survive the square root term.
### Pooled Standard Deviation
Pooled standard deviation is a way to find a better estimate of the true standard deviation given several different samples taken in different circumstances where the mean may vary between samples but the true or population standard deviation is assumed to remain the same. It is estimated by:
```s = √( ∑( si*(ni - 1) ) / ∑(ni - 1) ) where i = [1, L]
Equation 3
Pooled Standard Deviation ```
where ni is the sample size of the i’th sample, si is the standard deviation of the i’th sample, and L is the number of samples being combined.
### Arbitrarily Distributed Data
If it is known that the data is not normally distributed, or it is not known whether the distribution is normal or not, one can always make use of the Chebyshev Theorem to determine how much data lies close to the mean.
This states that, for any positive k, where k >= 1, the proportion r of data lying within k standard deviations of the mean is at least:
```r = 1 - 1/k2
Equation 4
Chebyshev's Theorem ```
This holds true for any population no matter what the shape of distribution, and maybe taken to be a worst case for any scenario.
For example, for a population which is not normally distributed, we can be sure that at least the proportion 0.56 (or 56%) of the data lies within 1.5 standard deviations. Whereas, if we knew that we were dealing with a normally distributed population, it can be shown that 56% of the data would lie within only 0.77 standard distributions.
## Standard Error
The standard error of a method of measurement or estimation is the estimated standard deviation of the error in that method. Namely, it is the standard deviation of the difference between the measured or estimated values and the true values. Notice that the true value of the measurement is often unknown and, this implies that, the standard error of an estimate is itself an estimated value.
### Standard Error of the Mean
The standard error of the mean describes the probable error associated with the sample mean. It is a measure of how close the sample mean is likely be to the true or population mean.
The standard error of the mean is an estimate of the standard deviation of the sample mean based on the population mean. Given that a standard deviation equates to a 68.27% probability, we could say that the sample mean (an estimated value) is within +/-1 standard error of the population mean (true value) with a 68.27% level of confidence.
One method of standard error calculation would be to determine the sample mean multiple times using different randomly drawn samples, and then estimate the standard deviation around the mean of the means.
In practice, we do not need to estimate the mean multiple times as a statistical method already exists:
```SE(ā) = s / √n
Equation 5
Standard Error of the Mean```
where s is an estimate of the standard deviation as determined, for example, by equation 2c. Note that because s is an estimate, then it follows that SE(ā) is an estimate also.
The Central Limit Theorem implies that if the sample size n is “large,” then the distribution of the sample mean is approximately normal, irrespective of the distribution of the underlying data. Therefore, the standard error is not dependent upon an assumption that the population is normally distributed.
### Standard Error of the Standard Deviation
As discussed, where the calculation of standard deviation is based upon a sample of a larger population, the result is an estimation. The standard error of the standard deviation gives the probable error associated with the sample standard deviation, and can be calculated by:
```SE(s) = s / √(2n)
Equation 6
Standard Error of the Standard Deviation```
By Andrew Thomas
Andrew Thomas is a software author and writer in the north of England. He holds a degree in Physics and Space Physics and began a career in spacecraft engineering but later moved into programming and telecommunications.
View all posts by Andrew Thomas |
University Physics Volume 1
# Summary
## 2.1Scalars and Vectors
• A vector quantity is any quantity that has magnitude and direction, such as displacement or velocity. Vector quantities are represented by mathematical objects called vectors.
• Geometrically, vectors are represented by arrows, with the end marked by an arrowhead. The length of the vector is its magnitude, which is a positive scalar. On a plane, the direction of a vector is given by the angle the vector makes with a reference direction, often an angle with the horizontal. The direction angle of a vector is a scalar.
• Two vectors are equal if and only if they have the same magnitudes and directions. Parallel vectors have the same direction angles but may have different magnitudes. Antiparallel vectors have direction angles that differ by $180°180°$. Orthogonal vectors have direction angles that differ by $90°90°$.
• When a vector is multiplied by a scalar, the result is another vector of a different length than the length of the original vector. Multiplication by a positive scalar does not change the original direction; only the magnitude is affected. Multiplication by a negative scalar reverses the original direction. The resulting vector is antiparallel to the original vector. Multiplication by a scalar is distributive. Vectors can be divided by nonzero scalars but cannot be divided by vectors.
• Two or more vectors can be added to form another vector. The vector sum is called the resultant vector. We can add vectors to vectors or scalars to scalars, but we cannot add scalars to vectors. Vector addition is commutative and associative.
• To construct a resultant vector of two vectors in a plane geometrically, we use the parallelogram rule. To construct a resultant vector of many vectors in a plane geometrically, we use the tail-to-head method.
## 2.2Coordinate Systems and Components of a Vector
• Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectors have two components. In three dimensions (in space), vectors have three components.
• A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vector of an axis with its scalar component along this axis. A vector is the resultant of its vector components.
• Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted from end point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum of the squares of its components.
• In a plane, the direction of a vector is given by an angle the vector has with the positive x-axis. This direction angle is measured counterclockwise. The scalar x-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle, and the scalar y-component can be expressed as the product of its magnitude with the sine of its direction angle.
• In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors $i^i^$ and $j^j^$ along the x-axis and the y-axis, respectively. The polar coordinate system is defined by the radial unit vector $r^r^$, which gives the direction from the origin, and a unit vector $t^t^$, which is perpendicular (orthogonal) to the radial direction.
## 2.3Algebra of Vectors
• Analytical methods of vector algebra allow us to find resultants of sums or differences of vectors without having to draw them. Analytical methods of vector addition are exact, contrary to graphical methods, which are approximate.
• Analytical methods of vector algebra are used routinely in mechanics, electricity, and magnetism. They are important mathematical tools of physics.
## 2.4Products of Vectors
• There are two kinds of multiplication for vectors. One kind of multiplication is the scalar product, also known as the dot product. The other kind of multiplication is the vector product, also known as the cross product. The scalar product of vectors is a number (scalar). The vector product of vectors is a vector.
• Both kinds of multiplication have the distributive property, but only the scalar product has the commutative property. The vector product has the anticommutative property, which means that when we change the order in which two vectors are multiplied, the result acquires a minus sign.
• The scalar product of two vectors is obtained by multiplying their magnitudes with the cosine of the angle between them. The scalar product of orthogonal vectors vanishes; the scalar product of antiparallel vectors is negative.
• The vector product of two vectors is a vector perpendicular to both of them. Its magnitude is obtained by multiplying their magnitudes by the sine of the angle between them. The direction of the vector product can be determined by the corkscrew right-hand rule. The vector product of two either parallel or antiparallel vectors vanishes. The magnitude of the vector product is largest for orthogonal vectors.
• The scalar product of vectors is used to find angles between vectors and in the definitions of derived scalar physical quantities such as work or energy.
• The cross product of vectors is used in definitions of derived vector physical quantities such as torque or magnetic force, and in describing rotations.
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# Matrix Multiplication: Product of Two Matrices
Matrix multiplication is the “messy type” because you will need to follow a certain set of procedures in order to get it right. This is the “messy type” because the process is more involved. However, you will realize later after going through the procedure and some examples that the steps required are manageable. Don’t worry, I will help you with this!
But first, we need to ensure that the two matrices are “allowed” to be multiplied together. Otherwise, the given two matrices are “incompatible” to be multiplied. If this is the case, we say that the solution is undefined.
## Matrix to Matrix Multiplication a.k.a “Messy Type”
Always remember this!
In order for matrix multiplication to work, the number of columns of the left matrix MUST EQUAL to the number of rows of the right matrix.
Suppose we are given the matrices $A$ and $B$, find $AB$ (do matrix multiplication, if applicable). Determine which one is the left and right matrices based on their location. It is a very important step.
To determine if I can multiply the two given matrices, I need to pay attention to the number of columns of matrix $A$ and the number of rows of matrix $B$. If they are equal, then I can proceed with Matrix Multiplication. Otherwise, I will conclude that the answer is undefined!
Because Matrix A has the number of columns of 2, and Matrix B has the number of rows of 3, and they are not equal (2 ≠3), I conclude that $AB$ = undefined. That means their product can’t be found.
### Examples of Matrix Multiplication a.k.a. “Messy Type”
Directions: Given the following matrices, perform the indicated operation.
Example 1: Calculate, if possible, the product of $B$ and $E$.
In order for matrices $B$ and $E$ to have a product, the number of columns of left matrix B must equal the number of rows of right matrix E.
• Matrix B (left)
number of columns = 3
• Matrix E (right)
number of rows = 3
Since this is the case, then it is okay to multiply them together. Now, these are the steps:
Step 1: Place them side by side.
Step 2: Multiply the rows of $B$ into the columns of $E$ by multiplying the corresponding elements of each row to each element of the column, and then add them together.
Please watch the animated solution carefully.
If you have no patience watching the animated solution above on how to perform matrix multiplication, you can view the regular solution I have included below.
Example 2: Calculate, if possible, the product of $E$ and $F$.
Check first if the product of the two matrices exists by making sure that the number of columns of left matrix E equals the number of rows of right matrix F.
• Matrix E (left)
number of columns = 2
• Matrix F (right)
number of rows = 2
This is wonderful since the number of columns of matrix $E$ equals the number of rows of matrix $F$. This means the product of $EF$ is defined so we can go ahead and perform matrix multiplication. See below for the animated step-by-step solution of matrix multiplication.
Example 3: Calculate, if possible, the product of $F$ and $E$.
In our previous example, we have successfully obtained the product of $EF$. This time around, we want to find if we can find the product of $E[latex] and [latex]F$, in that order.
Just to remind you, real numbers are commutative under multiplication operation which means that the order of multiplication does not affect the final product. For instance…
So the big question becomes, does it work also in matrix multiplication?
Let’s check if the number of columns of matrix F equals the number of rows of matrix $E$.
• Matrix F (left)
number of columns = 2
• Matrix E (right)
number of rows = 3
Obviously, the number of columns of Matrix $F$ does not equal the number of rows of Matrix $E$. The implication is that the product of $FE$ cannot be calculated, and therefore undefined!
In general, matrix multiplication is not commutative.
Example 4: Calculate, if possible, the product of $AE$.
The standard way to describe the size or dimension of a matrix is to…
(state number of rows) x (state number of columns)
…read as “the number of rows by the number of columns”.
3 x (three by three matrix)
3 x 2 (three by two matrix)
Since the number of columns of matrix A equals the number of rows of matrix E then we conclude that the product of $AE$ is defined.
Let’s work it out. See the animated solution below.
Example 5: Calculate, if possible, the product of $E$ and $A$.
3 x 2 (three by two matrix)
3 x 3 (three by three matrix)
Obviously, the number of columns of matrix E does not equal the number of columns of matrix A. Therefore, the product of $EA$ cannot be calculated, or undefined.
Example 6: Calculate, if possible, the product of $D$ and $F$.
Since the number of columns of matrix $D$ equals the number of rows of matrix $F$, the product of $DF$ is defined.
Example 7: What is the product of matrix C when multiplied by itself?
This is rather simple. We will simply multiply matrix $C$ by matrix $C$ which can be written as $CC$ or ${C^2}$. In other words, we are squaring matrix $C$.
We need to be cautious here. Notice that only a square matrix can be squared. Just to remind you, a square matrix is a matrix where the number of its row is equal to the number of its column.
I will leave it to you to verify that the solution below is correct. For math problems such as this, although tedious, I always recommend doing it by hand using a pencil and paper.
You may also be interested in these related math lessons or tutorials: |
# How To Find The Height When The Length And Width Are Known
## Video: How To Find The Height When The Length And Width Are Known
Many geometric shapes are based on rectangles and squares. The most common among them is a parallelepiped. They also include the cube, pyramid, and truncated pyramid. All four of these shapes have a parameter called height.
## Instructions
### Step 1
Draw a simple isometric shape called a rectangular parallelepiped. It got its name from the fact that its faces are rectangles. The base of this parallelepiped is also a rectangle of width a and length b.
### Step 2
The volume of a rectangular parallelepiped is equal to the product of the base area by the height: V = S * h. Since there is a rectangle at the base of the parallelepiped, the area of this base is S = a * b, where a is the length and b is the width. Hence, the volume is V = a * b * h, where h is the height (moreover, h = c, where c is the edge of the parallelepiped). If in the problem you need to find the height of the box, transform the last formula as follows: h = V / a * b.
### Step 3
There are rectangular parallelepipeds with squares at their bases. All of its faces are rectangles, of which two are squares. This means that its volume is V = h * a ^ 2, where h is the height of the parallelepiped, a is the length of the square, equal to the width. Accordingly, find the height of this figure as follows: h = V / a ^ 2.
### Step 4
For a cube, all six faces are squares with the same parameters. The formula for calculating its volume looks like this: V = a ^ 3. It is not required to calculate any of its sides, if the other is known, since they are all equal to each other.
### Step 5
All of the above methods assume the calculation of the height through the volume of the parallelepiped. However, there is another way to calculate the height for a given width and length. It is used if the area is given in the problem statement instead of the volume. The area of the parallelepiped is S = 2 * a ^ 2 * b ^ 2 * c ^ 2. Hence, c (the height of the parallelepiped) is equal to c = sqrt (s / (2 * a ^ 2 * b ^ 2)).
### Step 6
There are other problems in calculating the height for a given length and width. Some of them feature pyramids. If the problem gives the angle at the plane of the base of the pyramid, as well as its length and width, find the height using the Pythagorean theorem and the properties of angles.
### Step 7
To find the height of the pyramid, first determine the diagonal of the base. From the drawing, we can conclude that the diagonal is equal to d = √a ^ 2 + b ^ 2. Since the height falls to the center of the base, find half the diagonal as follows: d / 2 = √a ^ 2 + b ^ 2/2. Find the height using the properties of the tangent: tgα = h / √a ^ 2 + b ^ 2/2. It follows that the height is h = √a ^ 2 + b ^ 2/2 * tgα. |
# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 5 : (1/2) = 10/1 = 10
Spelled result in words is ten.
### How do we solve fractions step by step?
1. Divide: 5 : 1/2 = 5/1 · 2/1 = 5 · 2/1 · 1 = 10/1 = 10
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/2 is 2/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators.
In other words - five divided by one half is ten.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
# Some Applications Solving Equations Which Have Linear Expressions on One Side and Numbers on the Other Side
#### Example
Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers?
The sum of these two numbers x and x + 10 is 74.
This means that x + (x + 10) = 74.
or 2x + 10 = 74
Transposing 10 to RHS,
2x = 74 - 10
or 2x = 64
Dividing both sides by 2,
x = 32
This is one number (x) = 32
The other number is x + 10 = 32 + 10 = 42.
The desired numbers are 32 and 42.
#### Example
What should be added to twice the rational number (-7)/3 "to get" 3/7?
Twice the rational number ((-7)/3) "is" 2 xx ((-7)/3) = (-14)/3.
Suppose x added to this number gives 3/7; i.e.,
x + ((-14)/3) = 3/7
or x - 14/3 = 3/7
or x = 3/7 + 14/3 .......(transposing 14/3 to RHS)
or x= ((3 xx 3) + (14 xx 7))/21
or x = (9 + 98)/21
or x = 107/21.
Thus, 107/21 "should be added to" 2 xx ((-7)/3) "to give" 3/7.
#### Example
The perimeter of a rectangle is 13 cm and its width is 2 3/4 cm. Find its length.
Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 × (length + width)
= 2 xx (x + 2 3/4)
= 2(x + 11/4)
The perimeter is given to be 13 cm. Therefore,
2(x + 11/4) = 13
or x + 11/4 = 13/2 ......(Divinding both sides by 2)
or x = 13/2 - 11/4
or x = 26/4 - 11/4
or x = 15/4
or x = 3 3/4
The length of the rectangle is 3 3/4 cm.
#### Example
The present age of Sahil's mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.
Let Sahil’s present age be x years.
Sahil Mother Sum Present age x 3x Age 5 years later x + 5 3x + 5 4x + 10
It is given that this sum is 66 years.
Therefore, 4x + 10 = 66.
This equation determines Sahil's present age which is x years.
We transpose 10 to RHS,
4x = 66 - 10
or 4x = 56
or x = 56/4 = 14.
Thus, Sahil's present age is 14 years, and his mother's age is 42 years.
#### Example
Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs. 77, how many coins of each denomination does he have?
Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x.
The amount Bansi has:
(i) from 5 rupee coins, Rs. 5 × x = Rs. 5x.
(ii) from 2 rupee coins, Rs. 2 × 3x = Rs. 6x
Hence the total money he has = Rs. 11x
But this is given to be Rs. 77; therefore,
11x = 77
x = 77/11 = 7
Thus, number of five-rupee coins = x = 7. and
number of two-rupee coins = 3x = 21.
#### Example
The sum of three consecutive multiples of 11 is 363. Find these multiples.
If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22.
So we can take three consecutive multiples of 11 as x, x+ 11, and x+ 22.
It is given that the sum of these consecutive multiples of 11 is 363.
x + (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363.
or 3x + 33 = 363.
or 3x = 363 - 33
or 3x = 330
or x = 330/3
or x = 110
hence, the three consecutive multiples are 110, 121, 132.
#### Example
The difference between two whole numbers is 66. The ratio of the two numbers is 2: 5. What are the two numbers?
Since the ratio of the two numbers is 2: 5, we may take one number to be 2x and the other to be 5x.
The difference between the two numbers is (5x – 2x). It is given that the difference
is 66.
Therefore,
5x - 2x = 66
or 3x = 66
or x = 22
Since the numbers are 2x and 5x, they are 2 × 22 or 44 and 5 × 22 or 110, respectively.
The difference between the two numbers is 110 – 44 = 66 as desired.
#### Example
Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20, and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3: 5. If she has a total of 25 notes, how many notes of each denomination she has?
Let the number of ₹ 50 notes and ₹ 20 notes be 3x and 5x, respectively. But she has 25 notes in total.
Therefore, the number of ₹ 10 notes = 25 - (3x + 5x) = 25 - 8x
The amount she has
from ₹ 50 notes: 3x × 50 = ₹ 150x
from ₹ 20 notes: 5x × 20 = ₹ 100x
from ₹ 10 notes: (25 - 8x) × 10 = ₹ (250 - 80x)
Hence the total money she has = 150x + 100x + (250 - 80x) = ₹ (170x + 250)
But she has ₹ 590. Therefore,
170x + 250 = 590
or 170x = 590 - 250 = 340
or x = 340/170 = 2
The number of ₹ 50 notes she has = 3x = 3 × 2 = 6.
The number of ₹ 20 notes she has = 5x = 5 × 2 = 10.
The number of ₹ 10 notes she has = 25 - 8x = 25 - (8 × 2) = 25 - 16 = 9.
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# Linear Systems with Addition or Subtraction
## Solve systems using elimination of one variable
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Linear Systems with Addition or Subtraction
What if you were given a system of linear equations like and ? How could you solve for one of the variables by eliminating the other? After completing this Concept, you'll be able to solve a system of linear equations by elimination.
### Guidance
In this lesson, we’ll see how to use simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns ( and ) to a single unknown (either or ), this method is often referred to by solving by elimination. We eliminate one variable in order to make our equations solvable! To illustrate this idea, let’s look at the simple example of buying apples and bananas.
#### Example A
If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs$2.00, how much does one banana cost? One apple?
It shouldn’t take too long to discover that each banana costs $0.75. After all, the second purchase just contains 1 more banana than the first, and costs$0.75 more, so that one banana must cost $0.75. Here’s what we get when we describe this situation with algebra: Now we can subtract the number of apples and bananas in the first equation from the number in the second equation, and also subtract the cost in the first equation from the cost in the second equation, to get the difference in cost that corresponds to the difference in items purchased. That gives us the cost of one banana. To find out how much one apple costs, we subtract$0.75 from the total cost of one apple and one banana.
So an apple costs 50 cents.
To solve systems using addition and subtraction, we’ll be using exactly this idea – by looking at the sum or difference of the two equations we can determine a value for one of the unknowns.
Solving Linear Systems Using Addition of Equations
Often considered the easiest and most powerful method of solving systems of equations, the addition (or elimination) method lets us combine two equations in such a way that the resulting equation has only one variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute the value we get for that variable back into either one of the original equations to solve for the other variable.
#### Example B
Solve this system by addition:
Solution
We will add everything on the left of the equals sign from both equations, and this will be equal to the sum of everything on the right:
A simpler way to visualize this is to keep the equations as they appear above, and to add them together vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST be sure to keep the s and s in their own columns. You may also wish to use terms like as a placeholder!
Again we get , or . To find a value for , we simply substitute our value for back in.
Substitute into the second equation:
The reason this method worked is that the coefficients of the two equations were opposites of each other: 2 and -2. Because they were opposites, they canceled each other out when we added the two equations together, so our final equation had no term in it and we could just solve it for .
Solving Linear Systems Using Subtraction of Equations
Another, very similar method for solving systems is subtraction. When the or coefficients in both equations are the same (including the sign) instead of being opposites, you can subtract one equation from the other.
If you look again at Example 3, you can see that the coefficient for in both equations is +1. Instead of adding the two equations together to get rid of the s, you could have subtracted to get rid of the s:
So again we get , and we can plug that back in to determine .
The method of subtraction is just as straightforward as addition, so long as you remember the following:
• Always put the equation you are subtracting in parentheses, and distribute the negative.
• Don’t forget to subtract the numbers on the right-hand side.
• Always remember that subtracting a negative is the same as adding a positive.
### Vocabulary Language: English
elimination
elimination
The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. |
# How can base ten blocks be used in the classroom?
How can base ten blocks be used in the classroom?
## How can base ten blocks be used in the classroom?
Base Ten Blocks can be used to develop an understanding of the meaning of addition, subtraction, multiplication, and division. Modeling addition on a place-value mat provides students with a visual basis for the concept of regrouping.
### How do you explain base-10 to kids?
In base-10, each digit of a number can have an integer value ranging from 0 to 9 (10 possibilities) depending on its position. The places or positions of the numbers are based on powers of 10. Each number position is 10 times the value to the right of it, hence the term base-10.
How do you explain Base Ten Blocks?
Base 10 blocks are a set of four different types of blocks that, when used together, can help you to see what a number looks like and understand its value. Additionally, base 10 blocks can be used to help understand addition, subtraction, multiplication, division, volume, perimeter, and area.
What is a base ten numeral form?
What are Base Ten Numerals? In math, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are base-ten numerals. We can only count to nine without the need for two numerals or digits. All numbers in the number system are made by combining these 10 numerals or digits.
## What is a challenge of using base ten blocks?
The need to manipulate the materials for exchanges between place values is an important process. And when using base-ten blocks, the trades that must take place for grouping and ungrouping (aka, carrying and borrowing) obstruct student understanding rather than support it.
### What is the meaning of base ten?
In base 10, each digit in a position of a number can have an integer value ranging from 0 to 9 (10 possibilities). This system uses 10 as its base number, so that is why it is called the base 10 system.
What is a base 10 diagram?
Base-ten diagrams represent collections of base-ten units—tens, ones, tenths, hundredths, etc. We can use them to help us understand sums of decimals. Suppose we are finding . Here is a diagram where a square represents 0.01 and a rectangle (made up of ten squares) represents 0.1.
What is 10 block model?
Base Ten Blocks provide a spatial model of our base ten number system. The smallest blocks—cubes that measure 1 cm on a side—are called units. The long, narrow blocks that measure 1 cm by 1 cm by 10 cm are called rods. The flat, square blocks that measure 1 cm by 10 cm by 10 cm are called flats. |
# A linear combination is a sum of scalars times quantities. Such expressions arise quite frequently and have the form
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## Transcription
1 Section 1.3 Matrix Products A linear combination is a sum of scalars times quantities. Such expressions arise quite frequently and have the form (scalar #1)(quantity #1) + (scalar #2)(quantity #2) (scalar #n)(quantity #n) This pattern of a sum of products arises so frequently we give it a special name. Definition The dot product or inner product of two n-vectors of scalars a= a a a and b= b b b is 1 2 n 1 2 n n a b= ab 11+ a2b2+ + anbn = ab j j. j= 1
2 Example An instructor gives four exams worth 20%, 25%, 25% and 30% respectively. A student gets the following scores out of 100; 78, 65, 85, 82. Determine the course average using dot products. Example Find the dot product of each of the following pairs of vectors. a) v= 7 2 1, w = b) v= , w = 1 2 \$ #
3 Every linear combination of vectors can be written as a product in a special way. Many times we will be using a linear combination of columns like 2 5 c1 3 + c2 7 2# 0 # \$ \$ # Lets show how to write this as a special kind of product. Definition Let A be an m n matrix and c a column vector with n entries, then the matrix-vector product Ac is the linear combination c 1col1( A) + c 2 col2( A) + + c n coln( A) and the entries of Ac can be computed directly as dot products using row A c 1( ) row A c Ac = 2( ). rown( A) c# \$
4 Example Let A = 3 6 1\$ # and c = Write Ac as a linear combination of columns of A. Write Ac in terms of dot products. \$ #
5 Every system of equations can be written as a matrix equation involvng a matrix-vector product. 5x1 3x2+ x3 = 1 4x1+ 2x2+ 6x3 = 8
6 We can form the product of a pair of matrices under certain circumstances. Definition Let A = [a ij ] be an m n matrix and B = [b ij ] an n p matrix. Then the product of matrices A and B, denoted AB, is the m p matrix C = [c ij ] where c ij = row i (A) col j (B) = a a a i1 i2 in b b b 1j 2j nj \$ # n k= 1 a b 1 i m and 1 j p. (Product AB is defined only when the number of columns of A equals the number of rows of B.) Figure 1 illustrates a matrix product AB, where A is m n and B is n p. = ik kj Figure 1.
7 Examples: Form the following products, if possible A = B= C= D= E= F= 0 5 \$ # \$ # # 2 0\$ # 0 1 # \$ # \$ # AB BF AD EA
8 Products of Matrices vs. Products of Real Numbers Example 6 in the text illustrates several major distinctions between multiplication of matrices and multiplication of scalars. 1. Multiplication of matrices is not defined for all pairs of matrices. The number of columns of the first matrix must equal the number of rows of the second matrix. 2. Matrix multiplication is not commutative; that is, AB need not equal BA. 3. The product of two matrices can be zero when neither of the matrices consists of all zeros.
9 Special Terminology The definition of a matrix product includes row-by-column products as a 1 n matrix times an n 1 matrix and it also includes matrix-vector products as an m n matrix times an n 1 matrix. Thus we will only refer to matrix products from here on, except when we want to emphasize the nature of a product as one of these special cases. In addition note that C= AB= A col1( B) col2( B) colp( B) = Acol1( B) Acol2( B) Acolp( B) which says that col j (AB) is a linear combination of the columns of A with scalars from col j (B) since Acol (B) = b col ( A) + b col ( A) + + b col ( A) j 1j 1 2j 2 nj n Hence linear combinations are the fundamental building blocks of matrix products.
10 Matrix Powers If A is a square matrix, then products of A with itself like AA, AAA,... are defined and we denote them as powers of the matrix A in the form AA = A 2, AAA = A 3,... If p is a positive integer, then we define p = p factors A AA A For an n n matrix A we define A 0 = I n, the n n identity matrix. For nonnegative integers p and q, matrix powers obey familiar rules for exponents, namely A p A q = A p+q and (A p ) q = A pq. However, since matrix multiplication is not commutative, for general square matrices (AB) p A p B p..
11 Example A government employee at an obscure warehouse keeps busy by shifting portions of stock between two rooms. Each week two-thirds of the stock from room R1 is shifted to room R2 and two-fifths of the stock from R2 to R1. The following transition matrix T provides a model for this activity: Entry t ij is the fraction of stock that is shipped to Ri from Rj each week. If R1 contains 3000 items and R2 contains 9000 items, at the end of the week the stock in a room is the amount that remains in the room plus the amount that is shipped in from the other room. We have 1/3(stock in R1) + 2/5(stock in R2) = 1/3(3000) + 2/5(9000) = /3(stock in R1) + 3/5(stock in R2) = 2/3(3000) + 3/5(9000) = 7400 So the new stock in each room is a linear combination of the stocks at the start of the week and this can be computed as the product of the transition matrix and S 0 = \$# first week is TS, the original stock vector. Thus the stock vector at the end of the 4600 = S = computed as TS T S S 1 \$# = =. The stock vector at the end of the second week is \$# = 3 and it follows that S3 TS2 T S 0 = = = \$# (We have rounded the entries of the stock vectors to whole numbers.) The table is a portion of the records that the employee keeps. Original stock Stock at end of week #1 Stock at end of week #2 Stock at end of week #3 Stock at end of week #4 Stock at end of week #
12 We see that in just a few weeks the employee can claim to do the shifting of the stock without actually performing the task since the number of items in each room does not change after the third week. We say the process has reached a steady state. This type of iterative process is called a Markov chain or Markov process. The basic question that we pose for such a process is to determine the limiting behavior of the sequence of vectors S0, S1, S2, S3, We show later that the ultimate behavior depends upon the information contained in the transition matrix and we show how to determine the steady state, if it exists.
13 Properties of Matrix Products Property of Scalar Multiplication and Matrix Products. Let r be a scalar and A and B be matrices. Then r(ab) = (ra)b = A(rB). Properties of Matrix Products and Linear Combinations. Let A, B, and C be matrices, x and y be vectors, and r and s be scalars. Then A(B + C) = AB + AC and A(rx + sy) = rax + say. Each of these properties is called a distributive rule. Property of Matrix Products and the Transpose. If A and B are matrices, then (AB) T = B T A T.
14 The Angle Between 2-vectors. The dot product of a pair of n-vectors provided us with an algebraic tool for computing matrix-vector products and matrix products. However, the dot product is more than just an algebraic construction, it arises naturally when we determine the angle between geometric n-vectors; see Section 1.1. Here we show how a b is related to the angle between the 2-vectors a = [a 1 a 2 ] and b = [b 1 b 2 ]. Using the geometric representation for 2-vectors given in Section 1.1 we can view vectors a and b as shown in Figure 2 where α is the angle between vector a and the horizontal axis and β is the angle between vector b and the horizontal axis.hence the angle between vectors a and b shown in Figure 2 is α-β. Figure 2. From basic trigonometric relationships we have cos( α) = sin( α) = a 1 a1 2 + a2 2 a 2 a1 2 + a2 2 cos( β) = sin( β) = b 1 b1 2 + b2 2 b 2 b1 2 + b2 2 From the trigonometric identity cos(α-β) = cos(α)cos(β) + sin(α)sin(β) it follows that
15 cos( α - β ) = = = = ab ab 1 1+ a2b a + a b + b a + a b + b a b a + a b + b a b a a b b + a b 2 2 a + a b + b Thus the cosine of the angle between 2-vectors a and b has been expressed in terms of dot products. Later we generalize this result to a pair of n-vectors.
16 Application: Path Lengths In Section 1.1 we introduced an incidence matrix as a mathematical representation of a graph. The graph in Figure 3 is represented by the incidence matrix A = \$ #. We say there is a path between nodes P i and P j if we can move along edges of the graph to go from P i to P j. For example, there is a path between P 1 and P 4 in Figure 3; traverse the edge from P 1 to P 3 and then from P 3 to P 4. However, there is no edge from P 1 to P 4. For the graph in Figure 3, there is a path connecting every pair of nodes. This is not always the case; see Figure 4. There is no path between P 1 and P 4. The graph in Figure 4 has two separate, or disjoint, pieces. In some applications that can be modeled by using a graph there can be more than one edge between a pair of nodes. For the graph in Figure 5 there are two edges between P 2 and P 3. In this case the incidence matrix is B = \$ #. We use the term path of length 1 to mean that there are no intervening nodes of the graph between the ends of the path. (The term length here does not mean a measure of the actual distance between nodes, rather it is a count of the number of edges traversed between the starting node and the ending node
17 of a path.) The entries of the incidence matrix of a graph count the number of paths of length 1 between a pair of nodes. In certain applications it is important to determine the number of paths of a certain length k between each pair of nodes of a graph. Also, one frequently must calculate the longest path length that will ever be needed to travel between any pair of nodes. Both of these problems can be solved by manipulating the incidence matrix associated with a graph. Our manipulation tools will be matrix addition and multiplication. Since the incidence matrix A of a graph is square, its powers A 2, A 3, etc. are defined. We have We know that ij i j i1 1j i2 2j in nj ent A = row A col A = a a + a a + + a a. a i1 = number of paths of length 1 from P i to P 1 a 1j = number of paths of length 1 from P 1 to P j so the only way there could be a path from P i to P j through node P 1 is if both a i1 and a 1j are not zero. Thus product a i1 a 1j is the number of paths of length two between P i and P j through P 1. Similarly, a is a sj is the number of paths of length 2 between P i and P j through P s. Hence the dot product row i (A) col j (A) gives the total number of paths of length 2 between P i and P j. It follows that matrix A 2 provides a mathematical model whose entries count the number of paths of length 2 between nodes of the graph. By corresponding arguments, the number of paths of length k from P i to P j will be given by the (i,j)-entry of matrix A k. In order to calculate the longest path length that will ever be required to travel between any pair of nodes in a graph, we proceed as follows. The (i,j)-entry of matrix A + A 2 is the number of paths of length less than or equal to 2 from P i to P j. Similarly, the entries of A + A 2 + A 3 are the number of paths of length less than or equal to 3 between nodes of the graph. We continue this process until we find a value of k such that all the nondiagonal entries of A + A 2 + A A k are nonzero. This implies that there is a path of length k or less between distinct nodes of the graph. (Why can the diagonal entries be ignored?) The value of k
18 determined in this way is the length of the longest path we would ever need to traverse to go from one node of the graph to any other node of the graph.
19 True/False Review Questions 1. If x and y are both 3x1 matrices then x y is the same as x T y. 2. If A is 3x2 and x is 2x1 then Ax is a linear combination of the columns of A. 3. If A is 4x2 and B is 2x4, then AB = BA. 4. If A is 10x7 and B is 7x8, then ent 5,2 (AB) = row 5 (A)col 2 (B). 5. Matrix multiplication is commutative. 6. If A is 3x3 and b is 3x1 so that Ab = a zero column \$ # then either A is a zero matrix or b is 7. If A and B are the same size, then AB = BA. 8. Powers of a matrix are really successive matrix products. 9. The dot product of a pair of real n-vectors x and y is commutative; that is, x y = y x. 10. Let x =[1 1 1], then (x T x) 2 = 3x T x.
20 Terminology Dot product; inner product. Matrix-vector product. Matrix product. Markov chain; steady state. Row-by-column-product. Matrix equation. Matrix powers. Angle between 2-vectors. There is a connection between some of the terms listed above. Understanding and being able to explicitly state such connections is important in later sections. For practice with the concepts of this section formulate responses to the following questions and statements in your own words. How are dot products and row-by-column products related? Describe how matrix-vector products are computed using dot products. Explain how we to compute a matrix-vector product as a linear combination of columns. Explain how we formulate every linear system of equations as a matrix equation. (Explicitly describe all matrices used in the equation.) Explain how to express the matrix product AB in terms of matrix-vector products. Explain how to express the matrix product AB in terms of dot products. For what type of matrix A can we compute powers, A 2, A 3,, etc. Explain why there is such a restriction. What do we mean when we say that matrix multiplication distributes over matrix addition? Complete the following: The transpose of a product of two matrices is. Explain what we mean by the steady state of a Markov chain. (Express this relationship in words and using a matrix equation.)
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## Search This Blog
### Algebra Notes 3
3.1) Permutations
Every ordering of the first n natural numbers $1,2,\dots,n$ is called a permutation.
There are $n!$ permutations of the first n natural numbers.
3.2) Inversions
If we have the permutation $i_1, i_2, \dots i_n$ (of the first n natural numbers) then the numbers $i_k$ and $i_s$ are said to form an inversion if $k \lt s$ but $i_k > i_s$
In other words, an inversion in a permutation is a pair of numbers such that the larger number appears to the left of the smaller one in the permutation. The inversion number of a permutation is the total number of inversions. The inversion number of the permutation $i_1, i_2, \dots i_n$ is denoted by $[i_1, i_2, \dots ,i_n]$.
3.3) Parity of a permutation
Odd permutations - the inversion number is odd
Even permutations - the inversion number is even
3.4) Sign of a permutation
The sign of a permutation $i_1, i_2, \dots i_n$ is defined as $(-1)^{[i_1, i_2, \dots ,i_n]}$, where $[i_1, i_2, \dots ,i_n]$ is the inversion number of the permutation. It is obvious that the sign is $+1$ for even permutations and $-1$ for odd permutations.
3.5) Transposition
We say that we have applied a transposition if in a permutation P we swap the order of any two elements and the other elements we leave in place. When we do this, we get (from P) a new permutation R.
Lemma 1: The permutations P and R are of different parity.
Lemma 2: When $n \ge 2$, the number of the even permutations is equal to the number of the odd permutations.
3.6) Determinant
The concepts introduced here are important building blocks for introducing the concept of determinant (of a square matrix). The concept of determinant is usually defined using the Leibniz formula for determinants
### Algebra Notes 2
2.1) Systems of linear equations
$a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1$
$a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2$
$\dots$
$a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m$
This is a system of m linear equations with n unknowns.
a) The matrix of this system is defined as follows
$$A = \begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{bmatrix}$$
На български: матрица на системата
b) The augmented matrix of this system is defined as follows
$$B = \begin{bmatrix}a_{11}&a_{12}&...&a_{1n}&b_1\\a_{21}&a_{22}&...&a_{2n}&b_2\\...&...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}&b_m\end{bmatrix}$$
На български: разширена матрица на системата
2.2) Types of systems of linear equations
a) Independent system: has exactly one solution
b) Inconsistent system: has no solutions
c) Consistent system: has at least one solution
d) Dependent system: has infinitely many solutions
a) Определена система: има точно едно решение
b) Несъвместима система: няма решения
c) Съвместима систeмa: има поне едно решение
d) Неопределена система: има безбройно много решения
2.3) Elementary transformations applied to a system of linear equations
a) swapping two rows
b) multiplying a row of the system with a non-zero number
$R := \lambda R$, where $\lambda \neq 0$
c) adding to a row another row (multiplied by a number)
$R_2 := R_2 + \lambda R_1$
Note: If we apply (to a given system of linear equations) a finite number of elementary transformations, the resulting system is equivalent to the original system.
2.4) Gaussian elimination - a general method for solving systems of linear equations |
# Question: What Is The Ratio Of 60% To 40 %?
## What is 25% as a ratio?
Following the same logic 25% is 1:3 and not 1:4, else 50% would end up being 1:2 (but 1 in 2 is the proportion and not ratio)..
## What number is 40% of 300?
120What is 40 percent (calculated percentage %) of number 300? Answer: 120.
## What is the ratio of 3 to 4?
The simplified or reduced ratio “3 to 4” tells us only that, for every three men, there are four women. The simplified ratio also tells us that, in any representative set of seven people (3 + 4 = 7) from this group, three will be men. In other words, the men comprise 73 of the people in the group.
## What is the ratio of 15 to 100?
15%Convert fraction (ratio) 15 / 100 Answer: 15%
## What is the ratio of 15 to 5?
5 is the third part of 15, just as 8 is the third part of 24. We will now introduce this symbol 5 : 15 to signify the ratio of 5 to 15. A proportion will then appear as follows: 5 : 15 = 8 : 24.
## What is a 4 to 1 ratio?
If your mix ratio is 4:1 or 4 parts water to 1 part solution, there are (4 + 1) or 5 parts.
## What is the ratio of 70% to 30%?
Latest decimal numbers, fractions, rations or proportions converted to percentages80 / 94 = 85.106382978723%Nov 30 08:24 UTC (GMT)70 / 30 = 233.333333333333%Nov 30 08:24 UTC (GMT)498 / 1,200 = 41.5%Nov 30 08:24 UTC (GMT)34 / 53 = 64.150943396226%Nov 30 08:24 UTC (GMT)0.125 = 12.5%Nov 30 08:24 UTC (GMT)9 more rows
## What is a 50% ratio?
Example. Example: Convert the ratio 2:4 into a percentage: 2 : 4 can be written as 2 / 4 = 0.5; Multiplied 0.5 by 100, 0.5 × 100 = 50, so the percentage of ratio 2 : 4 is 50%.
## What percentage is a 2 to 1 ratio?
Simple percentage and ratio conversions.RatioPercentage – %3 to 125%2.5 to 129%2 to 133%1.5 to 140%8 more rows
## What is 35% as a ratio?
Convert fraction (ratio) 35 / 100 Answer: 35%
## What is the ratio of 7 3?
This ratio is written as 7/3. Would a mix of 14 parts copper and 6 parts zinc make the same brass? Well, this ratio would be written as 14/6. Using your knowledge of equivalent fractions, you can see that 7/3 = 14/6 (multiply the numerator and denominator of 7/3 by 2 and you get 14/6).
## What does the ratio 75/25 simplify to?
What is 75/25 Simplified? – 3/1 is the simplified fraction for 75/25. Simplify 75/25 to the simplest form.
## How do you find the ratio of a percentage?
Ratios are often expressed in the form m:n or m/n. To convert a ratio into the form of a percentage, simply divide m by n and then multiply the result by 100.
## What is the ratio of 60 percent?
Convert fraction (ratio) 60 / 100 Answer: 60%
## What is the ratio of 5 25?
20%Convert fraction (ratio) 5 / 25 Answer: 20%
## What percentage is a 3 to 2 ratio?
150%Convert fraction (ratio) 3 / 2 Answer: 150%
## What is 100% as a ratio?
Percentspercent notationratio notationnumber notation30%30 : 10030 / 100 = 0.38%8 : 1008 / 100 = 0.0863.7%63.7 : 10063.7 / 100 = 0.637100%100 : 100100 / 100 = 12 more rows
## What is the ratio of 25 to 100?
25%Convert fraction (ratio) 25 / 100 Answer: 25%
## What is 10% as a ratio?
Finally, just like fractions, ratios are ideally given in their simplest terms. But they don’t always start out that way. So just as a fraction of 3/30 can be simplified to 1/10, a ratio of 3:30 (or 4:40, 5:50, 6:60 and so on) can be simplified to 1:10.
## What is 40% as a ratio?
Percent – a special type of fraction100%=100/100=150%= 50/100= 5/1025%= 25/100= 5/2040%= 40/100= 4/105%= 5/100= 1/201 more row
## What is 70% as a ratio?
Convert fraction (ratio) 70 / 100 Answer: 70% |
# The vector $\overrightarrow{a}+\overrightarrow{b}$ bisects the angle between the non-collinear vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ if__________.
$\begin{array}{1 1} |\overrightarrow a|=|\overrightarrow b| \\ |\overrightarrow a|=2|\overrightarrow b| \\ 2 |\overrightarrow a|=|\overrightarrow b| \\ If \;\overrightarrow a\; and \;\overrightarrow b\; are\; \perp\end{array}$
Toolbox:
• $\cos \theta=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let us consider two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$
Let $\overrightarrow a+\overrightarrow b$ be the vector which bisects the angle between the two vectors.
Hence $\theta_1=\theta_2$
Therefore $\cos \theta_1=\large\frac{\overrightarrow a.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow a||\overrightarrow a+\overrightarrow b|}$
$\cos \theta_2=\large\frac{\overrightarrow b.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow b||\overrightarrow a+\overrightarrow b|}$
Since $\theta_1=\theta_2$=> $\cos \theta_1=\cos \theta_2$
$\large\frac{\overrightarrow a.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow a||\overrightarrow a+\overrightarrow b|}=\large\frac{\overrightarrow b.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow b||\overrightarrow a+\overrightarrow b|}$
$=>\overrightarrow a=\overrightarrow b$
The vector $\overrightarrow{a}+\overrightarrow{b}$ bisects the angle between the non-collinear vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are equal vectors |
# Solving Math Problems with Number Lines on the SAT
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Lesson Transcript
Instructor: Elizabeth Foster
Elizabeth has been involved with tutoring since high school and has a B.A. in Classics.
In this lesson, you'll get an overview of what number lines are, some tips for tackling them on the SAT and a preview of the kinds of number line problems on the test.
## Number Lines
A number line is a straight line on which every real number is represented by one point. Number lines show the relative sizes of all the numbers on the line: numbers to the left are progressively smaller, and numbers to the right are progressively greater. In elementary school, number lines were there to help you solve problems - but on the SAT, they work a little differently. The number line isn't part of the solution, it's part of the problem, and you're the one in charge of figuring out what on Earth it all means.
The heart of a number line problem is the relationship among all the quantities on the line - this is what you'll have to manipulate in various ways to solve the problems correctly. In this section, we'll take a look at how that works, what kind of number line problems you'll see on the test and how to interpret them.
## SAT Number Lines
You're probably used to number lines that look something like this.
One line, a bunch of numbers, usually with zero in the middle. Sometimes the scale will be different, but the pattern is generally the same. That's the classic number line. But on the SAT, the number lines look more like this.
One line, a bunch of numbers and some variables inserted as the basis of the question. Some of them even look like this. No numbers at all, just variables floating around in space. These tick marks could be units of one or units of a thousand; you have no way to know.
And just to make it even more complicated, the tick marks on the number line aren't always the same distance apart; you'll have to pay very close attention to the individual question to avoid making unjustified assumptions about the picture.
## Example Question
So, what kind of questions are you likely to see on a number line? How does the SAT take such a simple tool and come up with problems that can stump high school students? Here's an example:
On the number line shown, the tick marks are evenly spaced. Which of the following must be true of x?
(A) X is between 0 and 1
(B) X is between -1 and 0
(C) X is between 1 and 2
(D) X equals -1
(E) X equals 1
You can see how this one is a souped-up version of the 'variables in place of numbers' theme. Now let's take a look at how we'll solve it.
## Example Solution
Remember that the core of a number line question is the relationship among the numbers. In this case, we have no absolute standard: there are no real numbers on the line. So we'll use inequalities to get a handle on the relationships shown. Let's start on the left. We can see that 2x is less than x because it is to the left of x on the number line.
This tells us that whatever x is, it will be greater than 2x. This is great: it lets us ignore the number line for a minute and just work with this equation. Any answer choice that doesn't satisfy this inequality is automatically out. So, we'll plug them in and see.
We'll start with answer choice (A). Pick a number between 0 and 1 as a test. Let's take 0.5. If you multiply it by 2, do you get a smaller number?
Nope. 0.5 times 2 is 1, which is bigger. So, cross off (A). You don't need to test every number in the group because the question asks you which of the statements in the answers must be true. If even one number in the range is false, the whole answer is bunk.
Now we'll move on to (B). For a number between -1 and 0, we'll try -0.5. Multiply -0.5 by 2, and we get -1. That checks out, so we'll keep (B) for now.
What about (C)? If we multiply 1.5 by 2, we get 3, which is bigger than 1.5. Cross it off and move on.
In choice (D), we see that -1 times 2 is -2: smaller than -1, so that definitely works. For (E), multiplying 1 times 2 gives us 2, which is bigger than 1, so this is an easy elimination.
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# Pre-algebra
Community Questions
Fractions
All content in “Fractions”
### Understanding fractions
If you don't understand fractions, you won't be even 1/3 educated. Glasses will seem half empty rather than half full. You'll be lucky to not be duped into some type of shady real-estate scheme or putting far too many eggs in your cake batter. Good thing this tutorial is here. You'll see that fractions allow us to view the world in entirely new ways. You'll see that everything doesn't have to be a whole. You'll be able to slice and dice and then put it all back together (and if you order now, we'll throw in a spatula warmer for no extra charge).
### Visualizing equivalent fractions
Do you want 2/3 or 4/6 of this pizza? Doesn't matter because they are both the same fraction. This tutorial will help us explore this idea by really visualizing what equivalent fractions represent.
### Equivalent fractions and simplified form
There are literally infinite ways to represent any fraction (or number for that matter). Don't believe us? Let's take 1/3. 2/6, 3/9, 4/12 ... 10001/30003 are all equivalent fractions (and we could keep going)! If you know the basics of what a fraction is, this is a great tutorial for recognizing when fractions are equivalent and then simplifying them as much as possible!
### Comparing fractions
In this tutorial, we'll practice understanding what quantities fractions actually represent and comparing those to each other.
### Decomposing fractions
In this tutorial, we'll see that a fraction can be broken up (or decomposed) into a bunch of other fractions. You might see the world in a completely different way after this.
### Adding and subtracting fractions with unlike denominators
We've already had some good practice adding fractions with like denominators. We'll now add fractions with unlike denominators. This is a very big deal. After this tutorial, you'll be able to add, pretty much, any two (or three or four or... ) fractions!
You've already got 2 cups of sugar in the cupboard. Your grandmother's recipe for disgustingly-sweet-fudge-cake calls for 3 and 1/3 cups of sugar. How much sugar do you need to borrow from you robot neighbor? Adding and subtracting fractions is key. It might be a good idea to look at the equivalent fractions tutorial before tackling this one.
### Adding and subtracting with unlike denominator word problems
You know what a fraction is and are now eager to apply this knowledge to real-world situations (especially ones where the denominators aren't equal)? Well, you're about to see that adding and subtracting fractions is far more powerful (and fun) then you've ever dreamed possible!
### Multiplying fractions
What is 2/3 of 2/3? If 4/7 of the class are boys, how many boys are there? Multiplying fractions is not only super-useful, but super-fun as well.
### Multiplying fractions word problems
Multiplying fractions is useful. Period. That's all we really have to say. Believe us don't believe us. You'll learn eventually. This tutorial will have you multiplying in real-world scenarios (which is almost as fun as completely artificial, fake scenarios).
### Mixed numbers and improper fractions
We can often have fractions whose numerators are not less than the denominators (like 23/4 or 3/2 or even 6/6). These top-heavy friends are called improper fractions. Since they represent a whole or more (in absolute terms), they can also be expressed as a combination of a whole number and a "proper fraction" (one where the numerator is less than the denominator) which is called a "mixed number." They are both awesome ways of representing a number and getting acquainted with both (as this tutorial does) is super useful in life!
### Mixed number addition and subtraction
You know the basics of what mixed numbers are. You're now ready to add and subtract them. This tutorial gives you plenty of examples and practice in this core skill!
### Mixed number multiplication and division
My recipe calls for a cup and a half of blueberries and serves 10 people. But I have 23 people coming over. How many cups of blueberries do I need? You know that mixed numbers and improper fractions are two sides of the same coin (and you can convert between the two). In this tutorial we'll learn to multiply and divide mixed numbers (mainly by converting them into improper fractions first).
### Decimals and fractions
If you already know a bit about both decimals and fractions, this tutorial will help build a bridge between the two. Through a bunch of examples and practice, you'll be able operate in both worlds. Have fun!
### Dividing fractions
This is one exciting tutorial. In it, we will understand that fractions can represent division (and the other way around). Then we will create fractions by dividing whole numbers and then start dividing the fractions themselves. We'll see that dividing by something is the exact same thing as multiplying by that thing's reciprocal!
### Dividing fractions by fractions
In this tutorial, we'll become fraction dividing experts! In particular, we'll understand what it means to divide a fraction by another fraction. Too much fun!!!
### Number sets
The world of numbers can be split up into multiple "sets", many of which overlap with each other (integers, rational numbers, irrational numbers, etc.). This tutorial works through examples that expose you to the terminology of the various sets and how you can differentiate them. |
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 122 regularly, students can get better problem-solving skills.
Directions: Solve each problem.
Question 1.
Subtract 57 from 81.
24.
Explanation:
The subtraction of 57 from 81 is 24.
Question 2.
7 • 36 = __________
252
Explanation:
The multiplication of 7 • 36 is 252.
Question 3.
117.86
Explanation:
The root multiplication is 117.86.
Question 4.
What is the value of the digit 6 in 246,307?
6,000
Explanation:
The value of the digit 6 in 246,307 is 6,000.
Question 5.
Double $3.65. _________ Answer:$7.30.
Explanation:
The addition of $3.65 and$3.65 is $7.30. Question 6. 30 – 50 ÷ 2 = ___________ Answer: 5. Explanation: 30-50÷2 The division of 50 and 2 is 25 Now this 25 should be subtracted from 30 30-25 =5. Question 7. Answer: 9 Explanation: The addition of 16 + 9 is 25. Question 8. Is 250 mL the same as $$\frac{12}{4}$$ L? Answer: No Explanation: No , 250 mL the same as $$\frac{12}{4}$$ L . Question 9. Name the angle as right, obtuse. or acute. Answer: obtuse angle. Explanation: The angle which is given is obtuse angle The angle of obtuse angle is 117 , 121 , 166 ….. Question 10. The chart below shows how many cups of lemonade Marcia sold each hour she had her lemonade stand set up. 1st Hour 2nd Hour 3rd Hour 4th Hour 6 5 11 15 If Marcia charged 35 cents for each cup of lemonade, how much money did she make in the four hours? Answer:$12.95
Explanation:
Marcia charged 35 cents for each cup.
Total cups of lemonade are 37
The money she made in the four hours is 37 × 35 = 1295 cents.
1 dollar = 100 cents
Therefore $$\frac{1295}{100}$$
\$12.95
Question 11.
A scout leader is going to pair a new member with one of the existing 15 troop members. Five of the boys love to go camping, ten like to fish, three enjoy archery, twelve like to go hiking, and one boy enjoys carving. What is the probability the new boy will be paired with a boy who likes camping or carving?
The total number of troop memebers = 15
The number of boys love to go camping =5
The number of boys like to fishing =10
The number of boys enjoy archery =3
The number of boys like hiking =12
The number of boys enjoys carving=1
camping =5 ;carving =1
5+1=6
p(E)=number of favourable outcomes /total number of outcomes
p(E)= $$\frac{6}{15}$$
p(E)= $$\frac{2}{5}$$
Question 12.
Complete the input/output table. Look for a pattern and write the rule.
Input 93 83 73 63 53 43 Output 74 64 54
Input 93 83 73 63 53 43 Output 74 64 54 44 34 24
Rule :
Subract 19 from the input to get the output .
93-19=74
83-19=64
73-19=54
63-19=44
53-19=34
43-19=24 .
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# Section A: Practice Problems Size and Location of Fractions
## Section Summary
Details
In this section, we used fraction strips to represent fractions with denominators of 2, 3, 4, 5, 6, 8, 10, and 12. We also used the strips to reason about relationships between fifths and tenths, and between sixths and twelfths.
We learned that 2 tenths are equivalent to 1 fifth, or that splitting 5 fifths into two will produce 10 equal parts or tenths. When the denominator is larger, there are more parts in a whole.
We used what we learned about fraction strips to partition number lines and represent different fractions.
## Problem 1 (Pre-Unit)
What fraction of each figure is shaded?
## Problem 2 (Pre-Unit)
Explain why the shaded portion represents of the full rectangle.
## Problem 3 (Pre-Unit)
Label each tick mark with the number it represents. Explain your reasoning.
## Problem 4 (Pre-Unit)
Explain or show why and are equivalent fractions.
## Problem 5 (Lesson 1)
1. The entire diagram represents 1 whole. Shade the diagram to represent .
2. To represent on the tape diagram, would we shade more or less than what we did for ? Explain your reasoning.
## Problem 6 (Lesson 2)
1. The entire diagram represents 1 whole. What fraction does the shaded portion represent? Explain your reasoning.
2. Shade this diagram to represent .
## Problem 7 (Lesson 3)
For each pair of fractions, decide which is greater. Explain or show your reasoning.
1. or
2. or
3. or
## Problem 8 (Lesson 4)
Use the fraction strips to name three pairs of equivalent fractions. Explain how you know the fractions are equivalent.
## Problem 9 (Lesson 5)
1. Explain or show why the point on the number line describes both and .
2. Explain why and are equivalent fractions.
## Problem 10 (Lesson 6)
For each question, explain your reasoning. Use a number line if you find it helpful.
1. Is more or less than ?
2. Is more or less than 1?
## Problem 11 (Exploration)
Make fraction strips for each of these fractions. How did you fold the paper to make sure you have the right-size parts?
## Problem 12 (Exploration)
1. Andre looks at these fraction strips and says “Each is and another half of . Do you agree with Andre? Explain your reasoning.
2. What relationship do you see between and ? Explain your reasoning.
3. Can you find a relationship between and using fraction strips? |
# Domain of One Variable Function – A function with log – Exercise 5749
Exercise
Determine the domain of the function:
$$y=\sqrt{\log_3 [(3-2x)(1-x)]}$$
$$x\geq 2\text{ or } x\leq\frac{1}{2}$$
Solution
Let’s find the domain of the function:
$$y=\sqrt{\log_3 [(3-2x)(1-x)]}$$
Because there is a log, we need the expression inside the log to be greater than zero:
$$(3-2x)(1-x)>0$$
It also has a square root, so the expression inside the root must be non-negative:
$$\log_3 [(3-2x)(1-x)]\geq 0$$
Solve the first inequality:
$$(3-2x)(1-x)>0$$
It is a square inequality. Let’s look at the quadratic equation:
$$(3-2x)(1-x)=0$$
Because it is broken down into factors, its roots are easy to find. The first root is
$$3-2x=0$$
$$3=2x$$
$$x=\frac{3}{2}$$
The second root is
$$1-x=0$$
$$x=1$$
Because we are looking for the section above the x-axis or on it and the parabola “smiles”, we get that the solution of the inequality is
$$x>\frac{3}{2}\text{ or } x<1$$
Solve the second inequality:
$$\log_3 [(3-2x)(1-x)]\geq 0$$
We got a log in the inequality. By log definition, we get that our inequality is equivalent to:
$$(3-2x)(1-x)\geq 3^0=1$$
Note: If the log base was less than one, we would turn over the inequality sign.
Open brackets:
$$3-3x-2x+2x^2\geq 1$$
$$2x^2-5x+3\geq 1$$
$$2x^2-5x+2\geq 0$$
It is a square inequality. Its coefficients are
$$a=2, b=-5, c=2$$
The coefficient of the squared expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us
$$x_{1,2}=\frac{5\pm \sqrt{{(-5)}^2-4\cdot 2\cdot 2}}{2\cdot 2}=$$
$$=\frac{5\pm \sqrt{9}}{4}=$$
$$=\frac{5\pm 3}{4}$$
Hence, we get the solutions:
$$x_1=\frac{5+ 3}{4}=2$$
$$x_2=\frac{5- 3}{4}=\frac{1}{2}$$
Since the parabola “smiles” and we are interested in the sections above the x-axis or on it, we get
$$x\geq 2\text{ or } x\leq\frac{1}{2}$$
Finally, we intersect both results ( “and”) and get:
$$x>\frac{3}{2}\text{ or } x<1$$
and
$$x\geq 2\text{ or } x\leq\frac{1}{2}$$
The intersection is the final answer:
$$x\geq 2\text{ or } x\leq\frac{1}{2}$$
Have a question? Found a mistake? – Write a comment below!
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# In a Certain A.P. the 24th Term is Twice the 10th Term. Prove that the 72nd Term is Twice the 34th Term. - Mathematics
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
#### Solution
Given:
$a_{24} = 2 a_{10}$
$\Rightarrow a + \left( 24 - 1 \right)d = 2\left[ a + \left( 10 - 1 \right)d \right]$
$\Rightarrow a + 23d = 2(a + 9d)$
$\Rightarrow a + 23d = 2a + 18d$
$\Rightarrow 5d = a . . . (i)$
$\text { To prove }:$
$a_{72} = 2 a_{34}$
$\text { LHS: } a_{72} = a + \left( 72 - 1 \right)d$
$\Rightarrow a_{72} = a + 71d$
$\Rightarrow a_{72} = 5d + 71d \left( \text { From }(i) \right)$
$\Rightarrow a_{72} = 76d$
$\text { RHS }: 2 a_{34} = 2\left[ a + \left( 34 - 1 \right)d \right]$
$\Rightarrow 2 a_{34} = 2\left( a + 33d \right)$
$\Rightarrow 2 a_{34} = 2(5d + 33d) \left( \text { Form }(i) \right)$
$\Rightarrow 2 a_{34} = 2\left( 38d \right)$
$\Rightarrow 2 a_{34} = 76d$
∴ RHS = LHS
Hence, proved.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 19 Arithmetic Progression
Exercise 19.2 | Q 12 | Page 12 |
# 6.4 Graphs of logarithmic functions (Page 7/8)
Page 7 / 8
Translations of the Parent Function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)$
Translation Form
Shift
• Horizontally $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units to the left
• Vertically $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units up
$y={\mathrm{log}}_{b}\left(x+c\right)+d$
Stretch and Compress
• Stretch if $\text{\hspace{0.17em}}|a|>1$
• Compression if $\text{\hspace{0.17em}}|a|<1$
$y=a{\mathrm{log}}_{b}\left(x\right)$
Reflect about the x -axis $y=-{\mathrm{log}}_{b}\left(x\right)$
Reflect about the y -axis $y={\mathrm{log}}_{b}\left(-x\right)$
General equation for all translations $y=a{\mathrm{log}}_{b}\left(x+c\right)+d$
## Translations of logarithmic functions
All translations of the parent logarithmic function, $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right),$ have the form
where the parent function, $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right),b>1,$ is
• shifted vertically up $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units.
• shifted horizontally to the left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units.
• stretched vertically by a factor of $\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|>0.$
• compressed vertically by a factor of $\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}0<|a|<1.$
• reflected about the x- axis when $\text{\hspace{0.17em}}a<0.$
For $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{log}\left(-x\right),$ the graph of the parent function is reflected about the y -axis.
## Finding the vertical asymptote of a logarithm graph
What is the vertical asymptote of $\text{\hspace{0.17em}}f\left(x\right)=-2{\mathrm{log}}_{3}\left(x+4\right)+5?$
The vertical asymptote is at $\text{\hspace{0.17em}}x=-4.$
What is the vertical asymptote of $\text{\hspace{0.17em}}f\left(x\right)=3+\mathrm{ln}\left(x-1\right)?$
$x=1$
## Finding the equation from a graph
Find a possible equation for the common logarithmic function graphed in [link] .
This graph has a vertical asymptote at $\text{\hspace{0.17em}}x=–2\text{\hspace{0.17em}}$ and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:
$f\left(x\right)=-a\mathrm{log}\left(x+2\right)+k$
It appears the graph passes through the points $\text{\hspace{0.17em}}\left(–1,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,–1\right).\text{\hspace{0.17em}}$ Substituting $\text{\hspace{0.17em}}\left(–1,1\right),$
Next, substituting in $\text{\hspace{0.17em}}\left(2,–1\right)$ ,
This gives us the equation $\text{\hspace{0.17em}}f\left(x\right)=–\frac{2}{\mathrm{log}\left(4\right)}\mathrm{log}\left(x+2\right)+1.$
Give the equation of the natural logarithm graphed in [link] .
$f\left(x\right)=2\mathrm{ln}\left(x+3\right)-1$
Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?
Yes, if we know the function is a general logarithmic function. For example, look at the graph in [link] . The graph approaches $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ (or thereabouts) more and more closely, so $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, $\text{\hspace{0.17em}}\left\{x\text{\hspace{0.17em}}|\text{\hspace{0.17em}}x>-3\right\}.\text{\hspace{0.17em}}$ The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as $\text{\hspace{0.17em}}x\to -{3}^{+},f\left(x\right)\to -\infty \text{\hspace{0.17em}}$ and as $\text{\hspace{0.17em}}x\to \infty ,f\left(x\right)\to \infty .$
Access these online resources for additional instruction and practice with graphing logarithms.
## Key equations
General Form for the Translation of the Parent Logarithmic Function
## Key concepts
• To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ See [link] and [link]
• The graph of the parent function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ has an x- intercept at $\text{\hspace{0.17em}}\left(1,0\right),$ domain $\text{\hspace{0.17em}}\left(0,\infty \right),$ range $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ vertical asymptote $\text{\hspace{0.17em}}x=0,$ and
• if $\text{\hspace{0.17em}}b>1,$ the function is increasing.
• if $\text{\hspace{0.17em}}0 the function is decreasing.
• The equation $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x+c\right)\text{\hspace{0.17em}}$ shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ horizontally
• left $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c>0.$
• right $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}c<0.$
• The equation $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)+d\text{\hspace{0.17em}}$ shifts the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ vertically
• up $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}d>0.$
• down $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if $\text{\hspace{0.17em}}d<0.$
• For any constant $\text{\hspace{0.17em}}a>0,$ the equation $\text{\hspace{0.17em}}f\left(x\right)=a{\mathrm{log}}_{b}\left(x\right)$
• stretches the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|>1.$
• compresses the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|<1.$
• When the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the x -axis. When the input is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the y -axis.
• The equation $\text{\hspace{0.17em}}f\left(x\right)=-{\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ represents a reflection of the parent function about the x- axis.
• The equation $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(-x\right)\text{\hspace{0.17em}}$ represents a reflection of the parent function about the y- axis.
• A graphing calculator may be used to approximate solutions to some logarithmic equations See [link] .
• All translations of the logarithmic function can be summarized by the general equation See [link] .
• Given an equation with the general form we can identify the vertical asymptote $\text{\hspace{0.17em}}x=-c\text{\hspace{0.17em}}$ for the transformation. See [link] .
• Using the general equation $\text{\hspace{0.17em}}f\left(x\right)=a{\mathrm{log}}_{b}\left(x+c\right)+d,$ we can write the equation of a logarithmic function given its graph. See [link] .
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In these worksheets, students will work with arithmetic and geometric sequences.
#### An arithmetic sequence is a sequence of numbers in which the interval between the consecutive terms is constant. A geometric sequence is a sequence of numbers in which after the first term, consecutive ones are derived from multiplying the term before by a fixed, non-zero number called the common ratio. We come across the terms 'sequence' and 'series' very often in our lives. By sequence, we mean a list of things that obey a specific order. Series, on the other hand, is the arrangement of similar things one after the other, without following a fixed order. When talking about sequence and series in mathematics, a sequence is a collection of numbers that are placed, following a specific order with repetitions allowed. The series, on the other hand, is a process of adding infinitely many numbers without a fixed order. There are a variety of different types of these sequences and series. The most basic ones are arithmetic and geometric. So, what is the difference between these two basic types of sequences and series? The first difference is that the arithmetic sequence follows a constant difference between consecutive terms. In geometric sequence or series, there is a constant ratio being followed between consecutive terms. To find the next term in an arithmetic sequence, we use the following formula; tn = t1 - (n-1)d Here, t_1 is the first term of the sequence, n is the term number that we need to find, and d is the common difference between two consecutive terms. The common difference can be calculated by subtracting any two consecutive terms. To find the next term in a geometric sequence, we use the following formula; tn =t1 x r(n-1) Here, r is the common ratio between the consecutive terms. The ratio, r, can be calculated by dividing any two consecutive terms in the sequence.
These worksheets introduce the concepts of arithmetic and geometric series. In these worksheets, students will determine if a series is arithmetic or geometric. They will find the common difference in arithmetic sequences. They will find the common ratio in geometric sequences. This set of worksheets contains step-by-step solutions to sample problems, both simple and more complex problems, a review, and a quiz. It also includes ample worksheets for students to practice independently. When finished with this set of worksheets, students will be able to recognize arithmetic and geometric sequences and calculate the common difference and common ratio. These worksheets explain how to use arithmetic and geometric sequences and series to solve problems. Sample problems are solved and practice problems are provided.
# Arithmetic and Geometric Sequences and Series Worksheets
## Arithmetic and Geometric Sequences and Series Lesson
This worksheet explains the differences and use of arithmetic and geometric sequence and series to solve for terms. A sample problem is solved, and two practice problems are provided.
## Lesson and Practice
This sheet will walk students through this problem: Find the sum of the first 5 term of the sequence 2, 4, 6, 8, ...
## Worksheet
Students will use arithmetic and geometric sequences and series to solve problems. Ten problems are provided.
## Practice
We will work on these problems in a variety of ways.
## Review
The concept of how to use arithmetic and geometric sequences and series to solve problems is reviewed. A sample problem is solved and six practice problems are provided.
## Quiz
Students will demonstrate their proficiency with these concepts and skills. Ten problems are provided.
## Check
For each series you will find the d or r, an and Sn. Three problems are provided, and space is included for students to copy the correct answer when given. |
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```Chapter 7: Factorization Exercise – 7.7
Question: 1
Factories
x2 + 12x – 45
Solution:
To factories x2 + 12x – 45, we will find two numbers p and q such that p + q = 12 and pq = - 45.
Now,
15 + (-3) = 12 And 15 x (-3) = - 45
Splitting the middle term 12x in the given quadratic as -3x + 15x, we get:
x2 + 12x – 45
= x2 – 3x + 15x – 45
= (x2 – 3x) + (15x – 45)
= x(x – 3) + 15(x – 3)
= (x – 3) (x + 15)
Question: 2
Factories
40 + 3x – x2
Solution:
We have:
40 + 3x – x2
= -(x2 – 3x – 40)
To factories (x2 – 3x – 40), we fill find two number p and q such p + q = -3 and pq = - 40
Now,
5 + (-8) = -3 And 5 x (- 8) = - 40
Splitting the middle term -3x in the given quadratic as 5x – 8x, we get:
40 + 3x – x2 = -(x2 – 3x – 40)
= -(x2 + 5x – 8x – 40)
= - [(x2 + 5x) – (8x + 40)]
= - [x(x + 5) – 8 (x + 5)]
= -(x – 8)(x + 5)
= (x + 5)(- x + 8)
Question: 3
Factories
a2 + 3a – 88
Solution:
To factories a2 + 3a – 88, we will find two numbers p and q such that p + q = 3 and pq = - 88.
Now, 11 + (-8) = 3
And 11 x (-8) = - 88
Splitting the middle term 3a in the given quadratic as 11a – 8a, we get:
a2 + 3a – 88 = a2 + 11a – 8a – 88
= (a2 + 11a) – (8a + 88)
= a(a + 11 ) – 8(a + 11)
= (a – 8)(a + 11)
Question: 4
Factories
a2 – 14a – 51
Solution:
To factories a2 – 14a – 51, we will find two numbers p and q such that p + q = -14 and pq = - 51
Now,
3 + (-17) = -14 And 3 x (-17) = - 51
Splitting the middle term -14a in the given quadratic as 3a – 17a, we get:
a2 – 14a – 51 = a2 + 3a -17a – 51
= (a2 + 3a) – (17a + 51)
= a(a + 3) – 17(a + 3)
= (a – 17) (a + 3)
Question: 5
Factories
x2 + 14x + 45
Solution:
To factories x2 + 14x + 45, we will find two numbers p and q such that p + q = 14 and pq = 45
Now,
9 + 5 = 14 And 9 x 5 = 45
Splitting the middle term 14x in the given quadratic as 9x + 5x, we get:
x2 + 14x + 45 = x2 + 9x + 5x + 45
= (x2 + 9x) + (5x + 45)
= x(x + 9) + 5(x + 9)
= (x + 5)(x + 9)
Question: 6
Factories
x2 – 22x + 120
Solution:
To factories x2 – 22x + 120, we will find two numbers p and q such that p + q = -22 and pq = 120
Now, (-12) + (-10) = – 22 And (-12) x (-10) = 120
Splitting the middle term -22x in the given quadratic as -12x – 10x, we get:
x2 - 22x + 12 = x2 – 12x – 10x + 120
= (x2 – 12x) + (-10x + 120)
= x(x – 12) – 10(x – 12)
= (x – 10)(x – 12)
Question: 7
Factories
x2 – 11x – 42
Solution:
To factories x2 – 11x – 42, we will find two numbers p and q such that p + q = -11 and pq = - 42
Now,
3 + (-14) = -22 And 3 x (-14) = 42
Splitting the middle term -11x in the given quadratic as -14x + 3x, we get:
x2 - 11x – 42 = x2 - 14x + 3x – 42
= (x2 – 14x) + (3x – 42)
= x(x – 14) + 3(x – 14)
= (x – 14)(x + 3)
Question: 8
Factories
a2 – 2a – 3
Solution:
To factories a2 – 2a – 3, we will find two numbers p and q such that p + q = 2 and pq = – 3
Now,
3 + (-1) = 2 And 3 x (-1) = -3
Splitting the middle terms 2a in the given quadratic as – a + 3a, we get:
a2 + 2a – 3 = a2 – a + 3a – 3
= (a2 – a) + (3a – 3)
= a(a – 1) + 3(a – 1)
Question: 9
Factories
a2 + 14a + 48
Solution:
To factories a2 + 14a + 48, we will find two numbers p and q such that p + q = 14 and pq = 48
Now,
8 + 6 = 14 And 8 x 6 = 48
Splitting the middle terms 14a in the given quadratic as 8a + 6a, we get:
a2 + 14a + 48 = a2 + 8a + 6a +48
= (a2 + 8a) + (6a + 48)
= a(a + 8) + 6(a + 8)
= (a + 6)(a + 8)
Question: 10
Factories
x2 – 4x – 21
Solution:
To factories x2 – 4x – 21, we will find two numbers p and q such that p + q = -4 and pq = -21
Now,
3 + (-7) = - 4 And 3 x (-7) = -21
Splitting the middle terms -4x in the given quadratic as -7x + 3x, we get:
x2 – 4x – 21 = x2 – 7x + 3x – 21
= (x2 – 7x) + (3x – 21)
= x(x – 7) + 3(x – 7)
= (x – 7) (x + 3)
Question: 11
Factories
y2 + 5y – 36
Solution:
To factories y2 + 5y – 36, we will find two numbers p and q such that p + q = 5 and pq = -36
Now,
9 + (-4) = 5 And 9 x (-4) = -36
Splitting the middle terms 5y in the given quadratic as -7y + 9y, we get:
y2 + 5y – 36 = y2 – 4y + 9y – 36
= (y2 – 4y) + (9y – 36)
= y(y – 4) + 9(y – 4)
= (y – 4)(y – 4)
Question: 12
Factories
(a2 - 54)2 - 36
Solution:
(a2 - 54)2 - 36
= (a2 - 5a)2 - 62
= [(a2 - 5a) - 6] [(a2 - 5a) + 6]
= (a2 - 5a - 6) (a2 - 5a + 6)
In order to factories a2 - 5a - 6, we will find two numbers p and q such that p + q = - 5 and pq = -6
Now,
(-6) + 1 = - 5 And (- 6) x 1 = - 6
Splitting the middle term – 5 in the given quadratic as – 6a + a, we get:
a2 - 5a - 6 = a2 - 6a + a - 6
= (a2 - 6a) + (a - 6)
= a(a - 6) + (a -6)
= (a + 1) (a - 6)
Now, In order to factories a2 - 5a + 6, we will find two numbers p and q such that p + q = - 5 and pq = 6
Clearly,
(-2) + (-3) = - 5 And (-2) x (-3) = 6
Splitting the middle term – 5 in the given quadratic as – 2a – 3a, we get:
a2 – 5a + 6 = a2 – 2a – 3a + 6
= (a2 – 2a) – (3a – 6)
= a (a – 2) – 3(a – 2)
= (a – 3) (a – 2)
∴ (a2 – 5a – 6) (a2 – 5a + 6)
= (a – 6) (a + 1) (a – 3) (a – 2)
= (a + 1) (a – 2) (a – 3) (a – 6)
Question: 13
Factories
(a + 7)(a – 10) + 16
Solution:
(a + 7)(a – 10) + 16
= a2 – 10a + 7a – 70 + 16
= a2 - 3a – 54
To factories a2 – 3a – 54, we will find two numbers p and q such that p + q = -3 and pq = - 54
Now,
6 + (-9) = -3 And 6 x (-9) = -54
Splitting the middle term -3a in the given quadratic as – 9a + 6a, we get:
a2 – 3a – 54 = a2 – 9a + 6a – 54
= (a2 - 9a) + (6a – 54)
= a(a – 9) + 6(a – 9)
= (a + 6)(a – 9)
```
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## LESSON 8. Newmark’s Influence Chart
8.1 Use of Newmark’s Influence Chart
Newmark (1942) developed influence chart to determine the vertical stress due to loaded of area of any shape, irregular geometry at any point below the loaded area. A uniformly loaded circular area of radius r1 is divided into 20 divisions (say the first circle in Figure 8.1). Now if q is the intensity of loading then each small unit of the first circle will produce a vertical stress equal to ${{{\sigma _z}}\over {20}}$ at any depth of z below the center of the loaded area (as first circle is divided into 20 small divisions). Now from Eq. 7.5 of lesson 7 one can write that at any depth z vertical stress due to each small unit of the circle as:
$\frac{{{\sigma _Z}}}{{20}}=\frac{q}{{20}}\left[{1-{{\left( {\frac{1}{{1+{{\left( {\frac{{{r_1}}}{z}}\right)}^2}}}}\right)}^{\frac{3}{2}}}}\right]$ (8.1)
Let the right hand side of the Eq. 8.1 is equal to an arbitrary fixed value which is called influence value (say 0.005q). Thus,
$\frac{q}{{20}}\left[{1-{{\left({\frac{1}{{1+{{\left({\frac{{{r_1}}}{z}}\right)}^2}}}}\right)}^{\frac{3}{2}}}}\right]=0.005q$ (8.2)
The influence value of the chart is equal to 0.005 and each small unit is producing a vertical stress equal to 0.005q. After solving Eq. 8.2, one can get ${{{r_1}} \over z}=0.27$ . Thus, the radius of the first circle is 0.27z. In the influence chart (Figure 8.1), AB line representing the value of z (in the figure it is 2.5 cm). Thus, according to the chart shown in the Figure 8.1, any depth is represented by 2.5cm and based on that the scale has to be decided. According to that scale the loaded area has to be drawn for stress calculation. In the Figure 1, the radius of the first circle is 0.27 x 2.5 = 0.675 cm. Similarly, the second concentric circle of radius r2 is also divided in to 20 divisions. Including first and second circles, there are 40 divisions and again each unit is producing a vertical stress equal to 0.005q at a depth of z below the centre of the loaded area. Therefore, the each small unit in the second circle has two units and each unit will produce a vertical stress equal to 0.005q a depth of z below the centre of the loaded area. Thus, each small unit in the second circle will produce 2 x 0.005q amount of vertical stress a depth of z below the centre of the loaded area. Vertical stress due to each unit of the second circle is:
$\frac{q}{{20}}\left[{1-{{\left({\frac{1}{{1+{{\left({\frac{{{r_2}}}{z}}\right)}^2}}}}\right)}^{\frac{3}{2}}}}\right]=2\times 0.005q$ (8.3)
Again After solving Eq. 8.3, one can get ${{{r_2}} \over z}=0.40$ . Thus, the radius of the second circle is 0.40z. The general expression of the vertical stress produce by each unit of the each circle at a depth of z below the centre of the loaded area can be written as:
$\frac{q}{{20}}\left[{1-{{\left({\frac{1}{{1+{{\left({\frac{{{r_i}}}{z}}\right)}^2}}}}\right)}^{\frac{3}{2}}}}\right] = i\times 0.005q$ (8.4)
where i = 1, 2, ……..9. After solving Eq. 8.4, the radius of the third to ninth circle can be determined as: 0.52z, 0.64z, 0.77z, 0.92z, 1.11z, 1.39z, 1.91z, respectively. The radius of the tenth circle is infinity and cannot be drawn.
Problem 1
A raft foundation of dimension 11m x 6.2m is placed at a depth of 2m below the ground level. Determine the net stress due to the raft foundation (stress due to applied load only) at a depth of 7m below the ground level. Also determine the total stress due to raft (stress due to applied load) and stress due to soil (overburden pressure) at a depth of 7m below the ground level. The raft is subjected to total load of 10000 kN. The unit weight of the soil is 18 kN/m3. Neglect the pore water pressure (assumed soil is completely dry).
Fig. 8.1. Newmark’s influence chart to calculate vertical stress
Solution:
The total stress acting at the base of the raft = ${{100000} \over {11 \times 6.2}}=146.6$ kN/m2. The net stress at the base of raft = 146.6 – 18 x 2 = 110.6 kN/m2 (net stress means total stress minus the stress due to the soil above the base of the foundation as before the application of load soil was existing there. Thus, stress due to soil has to be deducted to calculate the net stress).
Now, depth of the point below the base of the raft is (z) = 7 – 2 = 5m.
Thus, according to the Newmark’s chart (Figure 1), 2.5 cm = 5m. Scale is 1: 200. Now, the raft (CDEF) is drawn with a scale of 1: 200 and placed on the Newmark’s chart (as shown in Figure 8.1) such that the centre of the raft is coincided with the centre of the Newmark’s chart. This is to be noted that, here the stress below the centre of the raft is determined. Thus, centre of the raft is coincided with the centre of the Newmark’s chart. If the vertical stress below the corner or any other point within the raft is to be determined than the corner or the point on interest has to be coincided with the centre of the Newmark’s chart. The total number of influence area covered by the raft = 116 (as shown in Figure 8.1). The net stress at a depth of 7 m below the ground level or 5 m below the base of the raft = 110.6 x 0.005 x 116 = 64.2 kN/m2. Thus, vertical stress due to applied load only is 64.2 kN/m2.
The vertical stress due to the overburden pressure at a depth of 7 m below the ground level = 18 x 7 = 126 kN/m2. Thus total vertical stress due to the applied load and overburden pressure at a depth of 7 m below the ground level = (64.2 + 126) = 190.2 kN/m2.
References
Ranjan, G. and Rao, A.S.R. (2000). Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India |
Mathematics Algebra of Matrices For CBSE-NCERT
Click for Only Video
### Topic Covered
star Addition/Subtraction of matrices
star Multiplication of a matrix by a scalar
star Multiplication of matrices
### Operations on Matrices
color{green} ✍️ we shall study certain operations on matrices, namely, addition of matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices.
color{green} ✍️ The sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. Furthermore, the two matrices have to be of the same order.
color{green}{A = [a_(ij)]_(mxxn) \ \ \ \ B = [b_(ij)]_(mxxn)}
color{green}{A+B = [a_(ij) + b_(ij)]_(mxxn)}
color{green} ✍️ Thus, if A = [(a_(11),a_(12),a_(13)),(a_(21),a_(2,2),a_(2,3))] and B = [(b_(11),b_(12),b_(13)),(b_(21),b_(2,2),b_(2,3))] A+B =[(a_(11) + b_(11) ,a_(12)+b_12,a_(13)+b_13),(a_(21)+b_21,a_(22)+b_22,a_(23)+b_23)]
color{red}{"Key Concept :"} Subtraction(Difference) peform same way as addition in matrix.
color{green}{A = [a_(ij)]_(mxxn) ;\ \ \ \ B = [b_(ij)]_(mxxn)}
color{green}{A-B = [a_(ij) - b_(ij)]_(mxxn)}
Point to remember : The addition And subtraction is not defined for different order matrices.
Q 3184578457
Given A = [ ( sqrt (3) ,1, -1 ), ( 2,3, 0) ] and B = [ ( 2 , sqrt (5) ,1 ), ( -2,3, 1/2) ] , find A+B
Class 12 Chapter 3 Example 6
Solution:
Since A, B are of the same order 2 × 3. Therefore, addition of A and B is defined
and is given by
A+B = [ ( 2+sqrt (3) , 1+ sqrt (5) , 1-1 ), ( 2-2 , 3+3 , 0+ 1/2 ) ] = [ ( 2+ sqrt (3) , 1+ sqrt (5) , 0 ), ( 0, 6 , 1/2) ]
Q 3164678555
Find X and Y , if X+ Y = [ ( 5,2),( 0,9 ) ] and X - Y = [ (3,6), ( 0,-1) ]
Class 12 Chapter 3 Example 9
Solution:
We have (X+ Y) + (X-Y) = [( 5,2), ( 0, 9 ) ] + [ ( 3,6), ( 0,-1) ]
or (X+X ) + (Y-Y) = [ (8,8 ), (0,8) ] => 2X = [ (8,8), (0,8) ]
or X= 1/2 [ (8,8), ( 0,8) ] = [ (4,4),(0,4) ]
Also (X+ Y ) - (X- Y ) = [ (5,2), (0,9) ] - [ (3,6),( 0,-1) ]
or (X-X) + (Y+Y) = [ (5-3 , 2-6 ) , (0, 9+1) ] = 2Y = [ ( 2,-4), (0,10) ]
OR y=1/2 [ (2,-4), ( 0,10) ] = [ (1,-2), (0,5) ]
(i) color{red}{"Commutative Law :"} If A = [a_(ij)], B = [b_(ij)],C = [c_(ij)] are matrices of the same order, say m × n,
Then color{orange}{A + B = B + A}
Now A + B = [a_(ij)] + [b_(ij)] = [a_(ij) + b_(ij)] = [b_(ij) + a_(ij)]
(addition of numbers is commutative) = ([b_(ij)] + [a_(ij)]) = B + A
(ii) color{red}{"Associative Law :"}
color{orange}{A + B) + C = A + (B + C}
Now (A + B) + C = ([a_(ij)] + [b_(ij)]) + [c_(ij)]
= [a_(ij) + b_(ij)] + [c_(ij)]
= [a_(ij) + (b_(ij) + c_(ij))]
= [a_(ij)] + ([b_(ij)] + [c_(ij)]) = A + (B + C)
(iii) color{red}{"Existence of additive identity :"}
Let A = [a_(ij)] be an m xx n matrix and O be an m × n zero matrix,
Then color{orange}{A + O = O + A = A.}
(iv) color{red}{"The existence of additive inverse :"}
A = [a_(ij)]_(m xx n) be any matrix, then we have another matrix as – A = [– a_(ij)]_(m xx n) such
That color{orange}{A + (– A) = (– A) + A= O.} So – A is the additive inverse of A or negative of A
### Multiplication of a matrix by a scalar
color{green} ✍️if A = [a_(ij)]_(mxxn) is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar k.
color{green} ✍️ In other words, kA = k[a_(ij)]_(mxxn) = [k(a_(ij))_(mxxn))] that is, (i, j)^(th) element of kA is ka_(ij) for all possible values of i and j.
color{green}{"Negative of a matrix : The negative of a matrix is denoted by" – A."} We define –A = (– 1) A.
Q 3114678550
If A = [ (1,2,3), ( 2,3,1) ] and B= [ ( 3,-1,3 ), ( -1,0,2 ) ] , then find 2A – B .
Class 12 Chapter 3 Example 7
Solution:
We have
2A- B = 2 [( 1,2,3), ( 2,3,1) ] - [ ( 3,-1,3),( -1,0,2) ]
= [ ( 2,4,6), ( 4,6,2) ] + [ (-3,1,-3), ( 1,0,-2) ]
= [ (2-3 ,4+1, 6-3), ( 4+1 , 6+0 , 2-2 ) ] = [ ( -1,5,3 ), ( 5,6,0) ]
Q 3124778651
Find the values of x and y from the following equation:
2 [ ( x,5), ( 7,y-3) ] + [ (3,-4), ( 1,2 ) ] = [ ( 7,6), ( 15,14) ]
Class 12 Chapter 3 Example 10
Solution:
We have
2 [ (x,5), ( 7, y-3) ] + [ ( 3,-4),( 1,2) ] = [ (7,6), ( 15,14) ] => [ (2x ,10 ), ( 14, 2y-6) ] + [ ( 3,-4), ( 1,2) ] = [ (7,6), ( 15,14) ]
or [ (2x+3, 10-4), ( 14+1 , 2y -6 +2 ) ] = [ (7,6), ( 15,14) ] => [ (2x+3, 6), ( 15, 2y -4) ] = [ (7,6), ( 15,14) ]
or 2x +3 = 7 and 2y -4 =14 (Why?)
or 2x = 7 – 3 and 2y = 18
or x= 4/2 and y= 18/2
i.e., x= 2 and y = 9
### Properties of scalar multiplication of a matrix
color{green} ✍️ If A = [a_(ij)] and B = [b_(ij)] be two matrices of the same order, say m × n, and k and I are scalars, then
(i)color{orange}{ k(A +B) = k A + kB},
(ii) color{orange}{(k + I)A = k A + I A = kA +A}
Q 3144678553
If A = [ (8,0), ( 4,-2), ( 3,6) ] and B= [ ( 2,-2),(4,2),(-5,1) ] , then find the matrix X, such that
2A + 3X = 5B.
Class 12 Chapter 3 Example 8
Solution:
We have 2A + 3X = 5B
or 2A + 3X – 2A = 5B – 2A
or 2A – 2A + 3X = 5B – 2A (Matrix addition is commutative)
or O + 3X = 5B – 2A (– 2A is the additive inverse of 2A)
or 3X = 5B – 2A (O is the additive identity)
or X = 1/3 (5B-2A)
or X = 1/3 (5 [ (2,-2), (4,2), (-5,1) ] -2 [ (8,0), ( 4,-2), (3,6) ] ) = 1/3 ( [ (10 , -10), (20,10), (-25,5) ] + [ (-16,0), (-8,4), (-6,-12) ] )
= 1/3 [ (10-16 , -10+0 ), ( 20-8 , 10+4), ( -25-6, 5-12) ] = 1/3 [ (-6,-10),( 12,14),( -31,-7) ] = [ ( -2, (-10)/3), ( 4, 14/3) , ( (-31)/3 , (-7)/3) ]
Q 3164778655
Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these
varieties of rice by both the farmers in the month of September and October are given
by the following matrices A and B.
(i) Find the combined sales in September and October for each farmer in each
variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October.
Class 12 Chapter 3 Example 11
Solution:
(i) Combined sales in September and October for each farmer in each variety is
given by
(ii) Change in sales from September to October is given by
(iii) 2% of B = 2/100 xx B = 0.02 xx B
Thus, in October Ramkishan receives Rs 100, Rs 200 and Rs 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of Rs
400, Rs 200 and Rs 200 in the sale of each variety of rice, respectively.
### Multiplication of matrices
color{green} ✍️ The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B.
color{green} ✍️ Let A = [a_(ij)] be an m xx n matrix and B = [b_(jk)] be an n × p matrix. Then the product of the matrices A and B is the matrix C of order color{green}{m × p.}
color{green} ✍️ To get product, we take the i^(th) row of A and k^(th) column of B, multiply them element wise and take the sum of all these products.
color{green} ✍️ In other words, if A = [a_(ij)]_(m xx n), B = [b_(jk)]_(n xx p), then the i^(th) row of A is [a_(i1) a_(i2) a_(i3).....a_(i n)] and the k^(t h) column of B is
color{green} ✍️ The matrix C = [c_(ik)]_(m × p) is the product of A and B.
Q 3144878753
Find AB, if A = [ ( 6,9), ( 2,3) ] and B = [ (2,6,0 ), ( 7,9,8) ]
Class 12 Chapter 3 Example 12
Solution:
The matrix A has 2 columns which is equal to the number of rows of B.
Hence AB is defined. Now
AB = [ ( 6(2) +9 (7 ) , 6 (6) + 9 (9) , 6 (0) + 9 ( 8) ), ( 2(2) +3 (7) , 2 ( 6) + 3 (9) ,2 (0) +3(8 ) ) ]
= [ ( 12+63 , 36 +81, 0+72 ), ( 4+21 ,12+27 , 0+24) ] = [ ( 75, 117 ,72), ( 25 ,39 ,24) ]
Q 3114878759
If A= [ ( 1,-2,3), (-4,2,5) ] and B = [ (2,3), ( 4,5), (2,1) ] , then find AB, BA. Show that
AB ≠ BA.
Class 12 Chapter 3 Example 13
Solution:
Since A is a 2 × 3 matrix and B is 3 × 2 matrix. Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively. Note that
AB = [ (1,-2,3), (-4,2,5) ] [ (2,3),( 4,5),( 2,1) ] = [ (2-8+6, 3-10+3), (-8+8 +10 , -12+10 + 5 ) ] = [ (0,-4), (10,3) ]
and BA = [ (2,3), (4,5), (2,1) ] [ (1,-2,3), (-4,2,5) ] = [ ( 2-12, -4+6, 6+15 ), ( 4-20 , -8 +10 ,12+25), ( 2-4, -4+2,6+5) ] = [ ( -10,2,21), (-16,2,37 ), ( -2, -2, 11 ) ]
Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA. But
one may think that perhaps AB and BA could be the same if they were of the same
order. But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same.
Q 3144078853
If A = [ (1,0 ), ( 0,-1) ] and B = [ ( 0,1), (1,0 ) ] , then AB = [ (0,1), ( -1, 0 ) ]
and BA = [ ( 0 ,-1), (1,0) ] Clearly AB ≠ BA.
Thus matrix multiplication is not commutative.
Class 12 Chapter 3 Example 14
Solution:
Zero matrix as the product of two non zero matrices
We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. This need
not be true for matrices, we will observe this through an example.
Q 3164078855
Find AB, if A = [ (0,-1), (0,2) ] and B = [ ( 3,5), ( 0, 0 ) ]
Class 12 Chapter 3 Example 15
Solution:
We have AB = [ (0,-1), (0,2) ] [ (3,5), (0,0) ] = [ (0,0), (0,0 ) ]
Thus, if the product of two matrices is a zero matrix, it is not necessary that one of
the matrices is a zero matrix.
### Properties of multiplication of matrices
1. The associative law For any three matrices A, B and C. We have color{orange}{(AB) C = A (BC),}
whenever both sides of the equality are defined.
2. The distributive law For three matrices A, B and C.
(i)color{orange}{ A (B+C) = AB + AC}
(ii)color{orange}{ (A+B) C = AC + BC,} whenever both sides of equality are defined.
3. The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that color{orange}{IA = AI = A} Now, we shall verify these properties by examples.
Q 3134178952
If A = [ ( 1,1,-1), (2,0,3), ( 3,-1,2) ] , B = [ (1,3), (0,2), (-1,4) ] and C= [ ( 1,2,3,-4), ( 2,0,-2,1) ] , find
A(BC), (AB)C and show that (AB)C = A(BC).
Class 12 Chapter 3 Example 16
Solution:
We have AB = [ (1,1,-1), (2,0,3 ), ( 3,-1,2) ] [ (1,3), ( 0,2), (-1,4) ] = [ ( 1+0 +1 , 3+2-4), ( 2+0 -3, 6+0+12), ( 3+0-2, 9-2+8) ] = [ (2,1), (-1,18), (1,15) ]
(AB) (C) = = [ (2,1), (-1,18), (1,15) ] [ (1,2,3,-4), ( 2,0,-2,1),] = [ (2+2, 4+0 , 6-2 , -8 +1 ), ( -1+36 , -2+0 , -3-36 , 4 +18 ), ( 1+30, 2+0 , 3-30 , -4+15) ]
= [ ( 4,4,4,-7), ( 35, -2,-39 ,22), (31,2,-27 , 11) ]
Now BC = [ (1,3), (0,2), (-1,4) ] [ (1,2,3,-4), ( 2,0,-2,1) ] = [ (1+6 , 2+0 , 3-6 , -4+3), ( 0+4, 0 +0 , 0-4, 0+2), ( -1+8 , -2+0 , -3-8, 4+4) ]
= [ ( 7,2,-3,-1), (4,0,-4,2) , ( 7,-2,-11,8) ]
Therefore A(BC) = [ (1,1,-1), (2,0,3), ( 3,-1,2) ] [ (7,2,-3,-1), ( 4,0,-4,2), ( 7,-2,-11, 8) ]
= [ ( 7+4-7 , 2+0 +2, -3-4+11, -1+2-8 ), ( 14 + 0 + 21, 4 + 0 -6 , -6+0 -33 , -2 + 0 + 24), ( 21 -4 +14 , 6+0 -4 , -9 +4-22 , -3-2+16) ]
= [ ( 4,4,4,-7), ( 35,-2,-39,22), (31,2,-27 ,11) ] Clearly, (AB) C = A (BC)
Q 3124180051
If A = [ (0,6,7 ), (-6,0 ,8 ), (7, -8 , 0 ) ] , B = [ (0,1,1 ), (1,0,2), (1,2, 0 ) ] ,C = [ (2), (-2), (3) ]
Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC
Class 12 Chapter 3 Example 17
Solution:
Now , A+ B = [ (0,7,8 ), (-5,0,10 ), ( 8,-6,0) ]
so, (A+B ) C = [ ( 0,7,8) , (-5,0,10 ), ( 7, -8 , 0 ) ] [ (2), (-2), (3) ] = [ ( 0 -12 +21),(-12+0+24) ,(14+16 +0) ] = [ ( 9), (12), (30) ]
and BC = [ (0,1,1), ( 1,0,2), ( 1,2, 0) ] [ (2,), (-2), (3) ] = [ (0-2+3), ( 2+0+6), ( 2-4 + 0) ] = [ (1), (8), (-2) ]
So AC + BC = [ (9), (12), ( 30) ] + [ (1), (8), (-2) ] = [ (10), (20), (28) ]
Clearly, (A + B) C = AC + BC
Q 3154380254
If A= [ (1,2,3), (3,-2,1), (4,2,1) ] , then show that A^3 – 23A – 40 I = O
Class 12 Chapter 3 Example 18
Solution:
We have A^2 = A * A= [ (1,2,3), (3,-2,1), (4,2,1) ] [ (1,2,3), (3,-2,1), (4,2,1) ] = [ (19,4,8), (1,12,8), ( 14,6,15 ) ]
So, A^3 = A A^2 = [ (1,2,3), (3,-2,1), (4,2,1) ] [ (19,4,8 ), ( 1,12,8 ), ( 14,6, 15) ] = [ (63,46,69 ), ( 69,-6, 23), ( 92,46,63) ]
Now,
A^3 -23 A -40 I = [ (63,46 ,69), ( 69 , -6,23), (92,46,63) ] -23 [ (1,2,3), ( 3,-2,1), (4,2,1) ] -40 [ (1,0,0 ), (0,1,0 ), ( 0 ,0 ,1) ]
= [ (63,46 ,69 ), (69,-6 ,23 ), ( 92,46,63) ] + [ (-23 , -46, -69), ( -69 ,46, -23), ( -92,-46, -23) ] + [ (-40,0,0 ), ( 0, -40 , 0 ), ( 0,0, -40 ) ]
= [ ( 63-23-40 ,46 -46 +0 , 69-69 +0 ), (69 -69 +0 , -6 +46-40 , 23-23 +0 ), ( 92-92+0 , 46 -46 +0 , 63 -23 -40) ]
= [ (0,0,0), (0,0,0), (0,0,0) ] =0
Q 3114480350
In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters. The
cost per contact (in paise) is given in matrix A as
The number of contacts of each type made in two cities X and Y is given by
Class 12 Chapter 3 Example 19
Solution:
We have
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i.e., Rs 3400 and Rs 7200, respectively. |
blog
# In a Triangle ABC: Exploring the Properties and Relationships
A triangle is one of the fundamental shapes in geometry, consisting of three sides and three angles. It is a versatile figure that appears in various fields, from architecture to mathematics. In this article, we will delve into the properties and relationships within a triangle ABC, uncovering fascinating insights and practical applications.
## The Basics: Understanding Triangle ABC
Before we dive into the intricacies of triangle ABC, let’s establish a common understanding of its basic components:
• Triangle: A polygon with three sides and three angles.
• Vertices: The points where the sides of a triangle intersect.
• Sides: The line segments that connect the vertices of a triangle.
• Angles: The measures of rotation between the sides of a triangle.
Now that we have a foundation, let’s explore the properties and relationships within triangle ABC.
## 1. Triangle Classification
Triangles can be classified based on their side lengths and angle measures. Let’s examine the different types:
### 1.1 Scalene Triangle
A scalene triangle has three unequal side lengths and three different angle measures. It is the most general type of triangle, as no sides or angles are congruent.
### 1.2 Isosceles Triangle
An isosceles triangle has two congruent side lengths and two equal angles. The third side and angle are different. The base angles, formed by the congruent sides, are always equal.
### 1.3 Equilateral Triangle
An equilateral triangle has three congruent side lengths and three equal angles. All angles in an equilateral triangle measure 60 degrees.
## 2. Angle Relationships
The angles within a triangle have fascinating relationships that can help us solve various problems. Let’s explore these relationships:
### 2.1 Triangle Sum Theorem
The sum of the interior angles in any triangle is always 180 degrees. This theorem allows us to find missing angles when we know the measures of the other angles.
### 2.2 Exterior Angle Theorem
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. In other words, if we extend one side of a triangle, the exterior angle formed is equal to the sum of the two opposite interior angles.
### 2.3 Congruent Angles
If two angles in one triangle are congruent to two angles in another triangle, the third pair of angles must also be congruent. This property is known as the Angle-Angle (AA) similarity criterion.
## 3. Side Relationships
The sides of a triangle also have interesting relationships that can aid in problem-solving. Let’s explore these relationships:
### 3.1 Pythagorean Theorem
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is widely used in various fields, such as construction and engineering.
### 3.2 Triangle Inequality Theorem
The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This theorem helps us determine if a given set of side lengths can form a valid triangle.
### 3.3 Similar Triangles
Two triangles are similar if their corresponding angles are congruent and their corresponding sides are proportional. Similar triangles have the same shape but may differ in size. This property is useful in various applications, such as map scaling and shadow calculations.
## 4. Practical Applications
The properties and relationships within triangle ABC have practical applications in various fields. Let’s explore a few examples:
### 4.1 Architecture and Construction
Architects and construction professionals use triangle properties to ensure stability and balance in structures. They rely on the Pythagorean theorem to calculate diagonal lengths, determine angles for roof trusses, and create aesthetically pleasing designs using the principles of symmetry and proportion.
In navigation and surveying, understanding triangle relationships is crucial for accurate measurements and calculations. Triangulation, a technique that uses the properties of similar triangles, is used to determine distances, heights, and positions of objects or landmarks.
### 4.3 Engineering and Mechanics
Engineers and mechanics utilize triangle properties to design and analyze structures, machines, and mechanisms. The Triangle Inequality Theorem helps ensure the integrity of load-bearing structures, while the concept of similar triangles aids in scaling down models for testing and analysis.
## Summary
Triangle ABC is a fascinating geometric figure with numerous properties and relationships. By understanding these properties, we can solve problems, make accurate measurements, and create stable structures. From the classification of triangles to the relationships between angles and sides, the knowledge of triangle properties finds applications in various fields, including architecture, navigation, and engineering. So, the next time you encounter a triangle, remember the wealth of information it holds and the practical insights it offers.
## Q&A
1. What is the sum of the interior angles in a triangle?
The sum of the interior angles in any triangle is always 180 degrees.
2. What is the Pythagorean theorem?
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
3. How can triangle properties be applied in architecture?
Architects use triangle properties to ensure stability, calculate diagonal lengths, determine angles for roof trusses, and create aesthetically pleasing designs using principles of symmetry and proportion.
4. What is the Triangle Inequality Theorem?
The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
5. How are similar triangles useful in real-world applications?
Similar triangles are used in map scaling, shadow calculations, and various measurement techniques in navigation, surveying, and engineering.
Kabir Sharma is a tеch еnthusiast and cybеrsеcurity analyst focusing on thrеat intеlligеncе and nеtwork sеcurity. With еxpеrtisе in nеtwork protocols and cybеr thrеat analysis, Kabir has contributеd to fortifying nеtwork dеfеnsеs. |
# Variable Separable
## Some differential equations can be solved by the method of separation of variables. (dy/dx) = f(x, y) is a first order and first degree differential equation. If f(x, y) can be written as h(x).g(y), where h(x) is a function of x and g(y) is a function of y, then we can separate h(x) and g(y). Solutions of differential equations of the type (dy/dx) = f(ax + by + c): Step1: Substitute ax + by + c = v in the given differential equation and change the differential coefficients accordingly.Step2: Obtain the relation f(v)dv = g(x)dxStep3: Integrate the obtained equation to get a relation between x, v and arbitrary constants.Step4: Substitute the value of v in the relation obtained in step-3 to get a relation between x, y and arbitrary constants.Keywords: Variable Separable method, Solutions of differential equations
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• Q1
Marks:1
• Q2
Marks:1
##### Explanation:
• Q3
The general solution of the differential equation
(1 + x2) dy = (1 + y2) dx is
Marks:1
tan-1y = tan-1x + C.
##### Explanation:
• Q4
The general solution of the differential equation y(x2y + ex)dx – exdy = 0 is
Marks:1
x3y + 3ex = Cy
##### Explanation:
Hence, the solution is
$v·{e}^{x}=-\int {x}^{2}{e}^{-x}·{e}^{x}dx+\frac{C}{3}$
• Q5
The differential equation, (ex + 1)ydy = (y + 1)exdx has the solution
Marks:1 |
## Fraction Models, and a Card Game
Models give us a way to form and manipulate a mental image of an abstract concept, such as a fraction. There are three basic ways we can imagine a fraction: as partially-filled area or volume, as linear measurement, or as some part of a given set. Teach all three to give your students a well-rounded understanding.
When teaching young students, we use physical models — actual food or cut-up pieces of construction paper. Older students and adults can firm up the foundation of their understanding by drawing many, many pictures. As we move into abstract, numbers-only work, these pictures remain in our minds, an always-ready tool to help us think our way through fraction problems.
## How to Read a Fraction
Fraction notation and operations may be the most abstract math monsters our students meet until they get to algebra. Before we can explain those frustrating fractions, we teachers need to go back to the basics for ourselves. First, let’s get rid of two common misconceptions:
• A fraction is not two numbers.
Every fraction is a single number. A fraction can be added to other numbers (or subtracted, multiplied, etc.), and it has to obey the Distributive Law and all the other standard rules for numbers. It takes two digits (plus a bar) to write a fraction, just as it takes two digits to write the number 18 — but, like 18, the fraction is a single number that names a certain amount of whatever we are counting or measuring.
• A fraction is not something to do.
A fraction is a number, not a recipe for action. The fraction 3/4 does not mean, “Cut your pizza into 4 pieces, and then keep 3 of them.” The fraction 3/4 simply names a certain amount of stuff, more than a half but not as much as a whole thing. When our students are learning fractions, we do cut up models to help them understand, but the fractions themselves are simply numbers.
## How Shall We Teach Fractions?
How did you fare on the Frustrating Fractions Quiz? With so many apparent inconsistencies, we can all see why children (and their teachers) get confused. And yet, fractions are vital to our children’s test scores — and scores are important to college admissions officers. What is a teacher to do? Must we tell our children, “Do it this way, and don’t ask questions”?
Parents and teachers are tempted to wonder if the struggle is worth it. After all, how often do you divide by a fraction in your adult life? If only we could skip the hard stuff…
## Quiz: Those Frustrating Fractions
[Photo by jimmiehomeschoolmom.]
Fractions confuse almost everybody. In fact, fractions probably cause more math phobia among children (and their parents) than any other topic before algebra. Middle school textbooks devote a tremendous number of pages to teaching fractions, and still many students find fractions impossible to understand. Standardized tests are stacked with fraction questions.
Fractions are a filter, separating the math haves from the luckless have nots. One major source of difficulty with fractions is that the rules do not seem to make sense. Can you explain these to your children?
## Question #1
If you need a common denominator to add or subtract fractions…
• Why don’t you need a common denominator when you multiply?
## How to Solve Math Problems
### That’s a Tough One!
What can you do when you are stumped? Too many students sit and stare at the page, waiting for inspiration to strike — and when the solution doesn’t crack their heads open and step out, fully formed, they complain: “Math is too hard!”
So this year I have given my Math Club students a couple of mini-posters to put up on the wall above their desk, or wherever they do their math homework. The first gives four questions to ask yourself as you think through a math problem, and the second is a list of problem-solving strategies.
## Trouble with Percents
Can your students solve this problem?
There are 20% more girls than boys in the senior class.
What percent of the seniors are girls?
This is from a discussion of the semantics of percent problems and why students have trouble with them, going on over at MathNotations. (Follow-up post here.)
Our pre-algebra class just finished a chapter on percents, so I thought Chickenfoot might have a chance at this one. Nope! He leapt without thought to the conclusion that 60% of the class must be girls.
After I explained the significance of the word “than”, he solved the follow-up problem just fine.
## How Old Are You, in Nanoseconds?
Conversion factors are special fractions that contain problem-solving information. Why are they called conversion factors? “Conversion” means change, and conversion factors help you change the numbers and units in your problem. “Factors” are things you multiply with. So to use a conversion factor, you will multiply it by something.
For instance, if I am driving an average of 60 mph on the highway, I can use that rate as a conversion factor. I may use the fraction $\frac{60 \: miles}{1 \: hour}$, or I may flip it over to make $\frac{1 \: hour}{60 \: miles}$. It all depends on what problem I want to solve.
After driving two hours, I have traveled:
$\left(2 \: hours \right) \times \frac{60 \: miles}{1 \: hour} = 120$miles so far.
But if I am planning to go 240 more miles, and I need to know when I will arrive:
$\left(240 \: miles \right) \times \frac{1 \: hour}{60 \: miles} = 4$hours to go.
Any rate can be used as a conversion factor. You can recognize them by their form: this per that. Miles per hour, dollars per gallon, cm per meter, and many, many more.
Of course, you will need to use the rate that is relevant to the problem you are trying to solve. If I were trying to figure out how far a tank of gas would take me, it wouldn’t be any help to know that an M1A1 Abrams tank gets 1/3 mile per gallon. I won’t be driving one of those.
## Using Conversion Factors Is Like Multiplying by One
If I am driving 65 mph on the interstate highway, then driving for one hour is exactly the same as driving 65 miles, and:
$\frac{65 \: miles}{1 \: hour} = the \: same \: thing \: divided \: by \: itself = 1$
This may be easier to see if you think of kitchen measurements. Two cups of sour cream are exactly the same as one pint of sour cream, so:
$\frac{2 \: cups}{1 \: pint} = \left(2 \: cups \right) \div \left(1 \:pint \right) = 1$
If I want to find out how many cups are in 3 pints of sour cream, I can multiply by the conversion factor:
$\left(3 \: pints \right) \times \frac{2 \: cups}{1 \: pint} = 6 \: cups$
Multiplying by one does not change the original number. In the same way, multiplying by a conversion factor does not change the original amount of stuff. It only changes the units that you measure the stuff in. When I multiplied 3 pints times the conversion factor, I did not change how much sour cream I had, only the way I was measuring it.
## Conversion Factors Can Always Be Flipped Over
If there are $\frac{60 \: minutes}{1 \: hour}$, then there must also be $\frac{1 \: hour}{60 \: minutes}$.
If I draw house plans at a scale of $\frac{4 \: feet}{1 \: inch}$, that is the same as saying $\frac{1 \: inch}{4 \: feet}$.
If there are $\frac{2\: cups}{1 \: pint}$, then there is $\frac{1\: pint}{2 \: cups} = 0.5 \: \frac{pints}{cup}$.
Or if an airplane is burning fuel at $\frac{8\: gallons}{1 \: hour}$, then the pilot has only 1/8 hour left to fly for every gallon left in his tank.
This is true for all conversion factors, and it is an important part of what makes them so useful in solving problems. You can choose whichever form of the conversion factor seems most helpful in the problem at hand.
How can you know which form will help you solve the problem? Look at the units you have, and think about the units you need to end up with. In the sour cream measurement above, I started with pints and I wanted to end up with cups. That meant I needed a conversion factor with cups on top (so I would end up with that unit) and pints on bottom (to cancel out).
## You Can String Conversion Factors Together
String several conversion factors together to solve more complicated problems. Just as numbers cancel out when the same number is on the top and bottom of a fraction (2/2 = 2 ÷ 2 = 1), so do units cancel out if you have the same unit in the numerator and denominator. In the following example, quarts/quarts = 1.
How many cups of milk are there in a gallon jug?
$\left(1\: gallon \right) \times \frac{4\: quarts}{1\: gallon} \times \frac{2\: pints}{1\: quart} \times \frac{2\: cups}{1\: pint} = 16\: cups$
As you write out your string of factors, you will want to draw a line through each unit as it cancels out, and then whatever is left will be the units of your answer. Notice that only the units cancel — not the numbers. Even after I canceled out the quarts, the 4 was still part of my calculation.
## Let’s Try One More
The true power of conversion factors is their ability to change one piece of information into something that at first glance seems to be unrelated to the number with which you started.
Suppose I drove for 45 minutes at 55 mph in a pickup truck that gets 18 miles to the gallon, and I wanted to know how much gas I used. To find out, I start with a plain number that I know (in this case, the 45 miles) and use conversion factors to cancel out units until I get the units I want for my answer (gallons of gas). How can I change minutes into gallons? I need a string of conversion factors:
$\left(45\: min. \right) \times \frac{1\: hour}{60\: min.} \times \frac{55\: miles}{1\: hour} \times \frac{1\: gallon}{18\: miles} = 2.3\: gallons$
## How Old Are You, Anyway?
If you want to find your exact age in nanoseconds, you need to know the exact moment at which you were born. But for a rough estimate, just knowing your birthday will do. First, find out how many days you have lived:
$Days\: I\:have\: lived = \left(my\: age \right) \times \frac{365\: days}{year}$
$+ \left(number\: of\: leap\: years \right) \times \frac{1\: extra\: day}{leap\: year}$
$+ \left(days\: since\: my\: last\: birthday,\: inclusive \right)$
Once you know how many days you have lived, you can use conversion factors to find out how many nanoseconds that would be. You know how many hours are in a day, minutes in an hour, and seconds in a minute. And just in case you weren’t quite sure:
$One\: nanosecond = \frac{1}{1,000,000,000} \: of\: a\: second$
Have fun playing around with conversion factors. You will be surprised how many problems these mathematical wonders can solve.
## Bill Gates Proportions II
[Feature photo above by Remy Steinegger via Wikimedia Commons (CC BY 2.0).]
Another look at the Bill Gates proportion… Even though I couldn’t find any data on his real income, I did discover that the median American family’s net worth was $93,100 in 2004 (most of that is home equity) and that the figure has gone up a bit since then. This gives me another chance to play around with proportions. So I wrote a sample problem for my Advanced Math Monsters workshop at the APACHE homeschool conference: The median American family has a net worth of about$100 thousand. Bill Gates has a net worth of $56 billion. If Average Jane Homeschooler spends$100 in the vendor hall, what would be the equivalent expense for Gates?
## Putting Bill Gates in Proportion
[Feature photo above by Baluart.net.]
A friend gave me permission to turn our email discussion into an article…
Can you help us figure out how to figure out this problem? I think we have all the information we need, but I’m not sure:
The average household income in the United States is $60,000/year. And a man’s annual income is$56 billion. Is there a way to figure out what this man’s value of $1mil is, compared to the person who earns$60,000/year? In other words, I would like to say — \$1,000,000 to us is like 10 cents to Bill Gates.
## Percents: The Search for 100%
[Rescued from my old blog.]
Percents are one of the math monsters, the toughest topics of elementary and junior high school arithmetic. The most important step in solving any percent problem is to figure out what quantity is being treated as the basis, the whole thing that is 100%. The whole is whatever quantity to which the other things in the problem are being compared. |
DISCOVER
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# How do I calculate minutes into a fraction of an hour?
Updated July 20, 2017
There are 60 minutes in an hour. This is the simple, man-made concept to keep in mind when calculating minutes as a fraction of an hour. The best way to understand the calculation of minutes as fractions of an hour is to first visualise some common hour-fractions which we see every day.
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## Visualising the Fraction
Visualise the face of a clock as a pie, seeing the quarters that mark 15 minutes, 30 minutes, 45 minutes, and 60 minutes in your mind's eye. When you imagine the 15 minute mark, what do you see? It is a quarter (or 1/4) of a pie! Thus, 15 minutes equate to 1/4 hour. Now, if you further halved the quarter pie, what would you get? Yes, one-eighth of a pie. Thus, seven-and-a-half minutes equate to one-eighth (1/8) of an hour.
## Understanding the Unit
To calculate fractions of hours for other numbers of minutes, we need to understand the unit of calculation. If 60 minutes make an hour, then each minute is one-sixtieth (1/60) of an hour. Therefore, this is the unit of calculation, and forms the basis of converting other numbers of minutes into fractions of an hour.
## Applying the Unit to Calculations
Now that we have established the unit of calculation as one minute, which equals one-sixtieth of an hour, we can apply this to other quantities. If one minute is one-sixtieth of an hour, then five minutes must be five times as much. Therefore, five minutes equals five times one-sixtieth (i.e., 5 x 1/60), which is one-twelfth (5/60 = 1/12) of an hour. Similarly, six minutes equals six times one-sixtieth (i.e., 6 x 1/60), which is one-tenth (6/60 = 1/10) of an hour. And twenty minutes equals twenty times one-sixtieth (i.e., 20 x 1/60), which is one-third (20/60 = 1/3) of an hour.
We have now mastered the art of calculating minutes as fractions of the hour.
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# Can parallelograms similar?
## Can parallelograms similar?
No , we can say that all parallelograms are similar , because if we consider a rectangle and a square side by side , they are not similar to each other , we can take rhombus and square as another example .
## How do you prove that a parallelogram is similar?
Well, we must show one of the six basic properties of parallelograms to be true!
1. Both pairs of opposite sides are parallel.
2. Both pairs of opposite sides are congruent.
3. Both pairs of opposite angles are congruent.
4. Diagonals bisect each other.
5. One angle is supplementary to both consecutive angles (same-side interior)
Are two Rhombi always similar?
Are all rhombuses similar? In a Rhombus, all the sides are equal. So, it can very much happen that two rhombuses have different angles. Hence, all rhombuses are not similar.
Are two parallelograms always congruent?
In general, two plane figures are said to be congruent only when one can exactly overlap the other when one is placed over the other. Two parallelograms will be congruent only when all four corresponding sides are equal in length & one corresponding internal angle is equal.
### Are 2 parallelograms similar?
A parallelogram has adjacent sides with the lengths of and . Find a pair of possible adjacent side lengths for a similar parallelogram. Explanation: Since the two parallelogram are similar, each of the corresponding sides must have the same ratio.
### Do similar parallelograms have the same angles?
The opposite angles of a parallelogram are equal. The opposite sides of a parallelogram are equal. The diagonals of a parallelogram bisect each other.
Are the opposite sides of a parallelogram always equal?
The opposite sides of a parallelogram are equal. The diagonals of a parallelogram bisect each other.
Which geometric figures are always similar?
Answer: The two geometrical figures which are always similar are circles, squares or line segment.
## Are any two squares always similar?
All squares are similar. Two figures can be said to be similar when they are having the same shape but it is not always necessary to have the same size. The size of every square may not be the same or equal but the ratios of their corresponding sides or the corresponding parts are always equal.
## Do all parallelograms equal 360?
Parallelograms have angles totalling 360 degrees, but also have matching pairs of angles at the ends of diagonals.
Do all angles of a parallelogram have the same measure?
A parallelogram must have equivalent opposite interior angles. Additionally, the sum of all four interior angles must equal degrees. And, the adjacent interior angles must be supplementary angles (sum of degrees). Since, angles and are opposite interior angles, thus they must be equivalent.
How many sides does a parallelogram have?
A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other. It is a type of polygon having four sides (also called quadrilateral), where the pair of parallel sides are equal in length. Also, the interior opposite angles of a parallelogram are equal in measure.
### Which of the following shapes have similar properties of a parallelogram?
Square and Rectangle: A square and a rectangle are two shapes which have similar properties of a parallelogram. Both have their opposite sides equal and parallel to each other. Diagonals of both the shapes bisect each other. Rhombus: If all the sides of a parallelogram are congruent or equal to each other, then it is a rhombus.
### How do you know if two parallelograms are congruent?
Two parallelograms will be congruent only when all four corresponding sides are equal in length & one corresponding internal angle is equal. All the sides and all the angles must be the same to be congruent. For a triangle if the sides are the same then the angles must also be the same (congruent).
Are parallelograms on the same base equal?
Theorem 1: Parallelograms on the same base and between the same parallel sides are equal in area. Proof: Two parallelograms ABCD and ABEF, on the same base DC and between the same parallel line AB and FC. To prove that area (ABCD) = area (ABEF). |
# Reciprocal Function – Properties, Graph, and Examples
Finding the reciprocal function will return a new function – the reciprocal function. These functions exhibit interesting properties and unique graphs. When two expressions are inversely proportional, we also model these behaviors using reciprocal functions. Reciprocal functions consist of two components: a constant on the numerator and an algebraic expression in the denominator. From its definition to its graph, we’ll learn extensively about reciprocal functions in this article. We’ll learn the following:
• Understanding the properties of reciprocal functions.
• Graphing reciprocal functions using different methods.
• Determining the function’s expression based on its graph.
We use reciprocal functions when describing relationships inversely proportional to each other such as the stress and elasticity of an object, time and speed’s relationship, and more. This is why learning about reciprocal functions can help us in our current and future advanced math classes. So, why don’t we go ahead and begin with its form and definition?
## What is a reciprocal function?
The common form of reciprocal functions that we may encounter is $y = \dfrac{k}{x}$, where $k$ is a real number. This means that reciprocal functions are functions that contain constant on the numerator and algebraic expression in the denominator. Here are some examples of reciprocal functions:
• $f(x) = \dfrac{2}{x^2}$
• $g(x) = \dfrac{1}{x + 1} – 4$
• $h(x) = -\dfrac{2}{x + 4} + 3$
As we can see from the three examples, all functions have numerator constants and denominators containing polynomials. The general form of reciprocal functions is $\mathbf{y}$ $\mathbf{= \dfrac{x}{(x – h)} + k}$, where $a$, $h$, and $k$ are real number constants.
## How to find the reciprocal of a function?
As we have learned in the past, we can determine a number’s reciprocal by dividing 1 by the given number. The same concept applies when we find a function’s reciprocal function – we divide 1 by the function’s expression. Here’s a table to compare the reciprocal that we learned in the past and reciprocal functions:
Reciprocal Reciprocal Function Given a number, $k$, its reciprocal is $\dfrac{1}{k}$. Given a function, $f(x)$, its reciprocal function is $\dfrac{1}{f(x)}$. The product of $k$ and its reciprocal is equal to $k$ · $\dfrac{1}{k} = 1$. The product of $f(x)$ and its reciprocal is equal to $f(x)$ · $\dfrac{1}{f(x)} = 1$. Given $\dfrac{1}{k}$, its value is undefined when $k = 0$. Given $\dfrac{1}{f(x)}$, its value is undefined when $f(x) = 0$.
The table shows that reciprocals and reciprocal functions share similar characteristics and properties. Let’s go ahead and observe the function $y = 2x – 1$.
• This means that if we want to find the reciprocal of $y = 2x – 1$, its reciprocal can be expressed as $y = \dfrac{1}{2x – 1}$.
• We can also confirm the product of $2x – 1$ and its reciprocal:
$(2x – 1) \cdot \dfrac{1}{2x – 1} = 1$
• This also means that $2x – 1$ must never be zero, so $x$ must never be $\frac{1}{2}$.
## How to graph reciprocal functions?
There are different ways for us to graph reciprocal functions. In this article, we’ll focus on two methods:
1. Graphing reciprocal functions by finding the function’s table of values first.
2. Graphing reciprocal functions using different transformation techniques.
### How to graph functions using their tables of values
Let’s go ahead and start with graphing the parent function, $y = \frac{1}{x}$ by first finding its table of values.
• Find the value of the function at different values of $x$.
• Plot these points on the $xy$-coordinate system.
• Graph the curves using these points.
$\mathbf{x}$ $-3$ $-2$ $-1$ $-\frac{1}{2}$ $-\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{2}$ $1$ $2$ $3$ $\mathbf{y}$ $-\frac{1}{3}$ $-\frac{1}{2}$ $-1$ $-2$ $-3$ $3$ $2$ $1$ $\frac{1}{2}$ $\frac{1}{3}$
As we have expected, $\dfrac{1}{x}$ can never be equal to $0$, and $x$ can never be zero. From the graph, we can see other than the restrictions, and the function $y = x$ can be any real number. Why don’t we summarize its domain and range?
Domain $(-\infty, 0) \cup (0, \infty)$ Range $(-\infty, 0) \cup (0, \infty)$
Since we have restrictions at $x = 0$ and $y = 0$, these equations also represent the vertical and horizontal asymptotes of $f(x) = \frac{1}{x}$. Let’s include these asymptotes on our final graph: We can also use the graph of $y = \dfrac{1}{x}$ to graph other reciprocal functions. We can do this by applying different transformation techniques.
### How to graph reciprocal functions by transformation
Here are some common graph transformations that we’ve already learned in the past:
These techniques can be applied as well on the graph of $y = \dfrac{1}{x}$ so that we can graph different reciprocal functions. Let’s try graphing $y = \dfrac{2}{x – 1} + 4$ by first identifying the transformations done on $y = \dfrac{1}{x}$ to attain our given function. $y = \mathbf{2} \left(\frac{1}{x-\mathbf{1}}\right) \mathbf{+4}$ From this, we can see that we’ll need to perform the following transformations on the graph of $y = \dfrac{1}{x}$:
• Translate the graph one unit to the right.
• Stretch the graph vertically by two units.
• Translate the graph four units upward.
Let’s start by moving the graph of $y = \frac{1}{x}$ one unit to the right. Use reference points to guide you. The $x$-coordinates must be one unit greater. Now, stretch the graph vertically by two units. Check our resulting graph by making sure that the $y$-coordinate’s values are doubled. Move the resulting graph four units upward. Double-check your resulting graphs by making sure our reference points’ $y$-coordinates have increased by $4$. Hence, we’ve shown that we can graph reciprocal functions using transformations. Notice that the asymptotes have shifted too? This makes sense since our new asymptotes can be calculated as shown below:
Vertical Asymptote Horizontal Asymptote The expression $\frac{2}{x-1}$ can’t be $0$, so $y \neq 4$. Asymptote:$y = 4$ The denominator $x – 1$ can’t be $0$, so $x \neq 1$. Asymptote: $x = 1$
Using the general form of reciprocal functions, the vertical asymptote can be expressed as y = k, and the horizontal asymptote can be expressed as x = h. From the graph and the asymptotes, we can also find the function’s domain and range:
Domain $(-\infty, 1) \cup (1, \infty)$ Range $(-\infty, 4) \cup (4, \infty)$
Here’s a quick exercise for you:
• Try graphing $y = -\dfrac{1}{x}$ on your own and compare this with the graph of $y = \dfrac{1}{x}$.
• Sketch $y = x$ and $y = -x$ on the graphs of $y = \dfrac{1}{x}$ and $-\dfrac{1}{x}$.
• What can you say about each pair of graphs?
Observe that when the function is positive, it is symmetric with respect to the equation $\mathbf{y = x}$. Meanwhile, when the function is negative (i.e., has a negative constant), it is symmetric with respect to the equation $\mathbf{y = -x}$.
### Summary of reciprocal function definition and properties
Before we try out some more problems that involve reciprocal functions, let’s summarize everything that we have learned so far about these unique functions.
• Reciprocal functions are functions that have a constant on their denominator and a polynomial on their denominator.
• The reciprocal of a function, $f(x)$, can be determined by finding the expression for $\dfrac{1}{f(x)}$.
• We can graph a reciprocal function using the function’s table of values and transforming the graph of $y = \dfrac{1}{x}$.
• Make sure to find the vertical and horizontal asymptotes of the function.
• The domain and range of a reciprocal function will depend on the asymptotes’ values.
• The symmetry of the reciprocal function’s graph will depend on the constant’s sign.
Example 1 If $g(x)$ is the reciprocal of $f(x)$, what is the value of $g(x) \cdot f(x)$? Solution Find the expression for $g(x)$ in terms of $f(x)$. Since it is the reciprocal of $f(x)$, we have $g(x) = \frac{1}{f(x)}$. Multiply the two expressions to find their product. \begin{aligned} g(x) \cdot f(x)&= \frac{1}{f(x)}\cdot f(x)\\&=\mathbf{1}\end{aligned} We can also use the fact that a function’s product and its reciprocal will always be equal to 1. Example 2 What are the reciprocals of the following functions? a. $f(x) = x^2 – 4$ b. $g(x) = \dfrac{x}{2}$ c. $h(x) = \dfrac{3(x – 1)}{5}$ Solution To find the reciprocal of a function, we can divide 1 by the expression. a. For $f(x) = x^2 – 4$, we simply divide $1$ by the $f(x)$’s expression, hence its reciprocal is equivalent to $\mathbf{\dfrac{1}{x^2-4}}$. b. When finding the reciprocal of a rational expression, we switch the numerator and denominator positions. This means that its reciprocal is $\mathbf{\dfrac{2}{x}}$. We apply the same process when finding the reciprocal of $h(x)$. Hence, $\frac{1}{h(x)}\mathbf{=\dfrac{5}{3(x-1)}}$. Example 3 Complete the table below by finding the reciprocal functions’ symmetry, domain, range, vertical asymptote, and horizontal asymptote.
Function Symmetry Vertical Asymptote Horizontal Asymptote Domain Range $y=-\dfrac{3}{x}$ $y=\dfrac{2}{2x+1}$ $y=\dfrac{4}{x-4}+2$ $y=-\dfrac{3}{x-2}+5$
Solution When finding the symmetry of the reciprocal function, we base it on the constant’s sign.
• If the constant is negative, its graph is symmetric with respect to the line $y = -x$.
• If the constant is positive, the graph is symmetric with respect to $y = x$.
Inspect the equation’s expression and compare it with the general form, $\dfrac{a}{x-h} + k$.
• The vertical asymptote will be $x = h$.
• The horizontal asymptote will be $y = k$.
The domain for each expression will be $(-\infty,h)\cup(h, \infty)$ and the range will b e$(-\infty,k)\cup(k, \infty)$. Let’s use all the information to fill in the table:
Function Symmetry Vertical Asymptote Horizontal Asymptote Domain Range $y=-\dfrac{3}{x}$ $y = -x$ $x = 0$ $y = 0$ $(-\infty,0)\cup(0, \infty)$ $(-\infty,0)\cup(0, \infty)$ $y=\dfrac{2}{2x+1}$ $y = x$ $x = 0$ $y = 1$ $(-\infty,0)\cup(0, \infty)$ $(-\infty,0)\cup(0, \infty)$ $y=\dfrac{4}{x-4}+2$ $y = x$ $x = 4$ $y = 2$ $(-\infty,4)\cup(4, \infty)$ $(-\infty,2)\cup(2, \infty)$ $y=-\dfrac{3}{x-2}+5$ $y = -x$ $x = 2$ $y = 5$ $(-\infty,2)\cup(2, \infty)$ $(-\infty,5)\cup(5, \infty)$
Example 4 Choose from the four functions given in Example 3 and graph the two of them:
• One function is to be graphed by finding the table of values.
• The second function is to be graphed by transforming $y=\dfrac{1}{x}$.
Let’s graph the first function, $y=-\dfrac{3}{x}$ by finding its table of values to graph the first half of its curve.
$\mathbf{x}$ $\mathbf{y = -\dfrac{3}{x}}$ $\dfrac{1}{3}$ $-9$ $\dfrac{1}{2}$ $-6$ $1$ $-3$ $3$ $-1$
Reflect the curve over its line of symmetry to graph the remaining half of the curve. Include the vertical and horizontal asymptotes in your final graph. Let’s try to graph the third function from Example 3: $y = \dfrac{4}{x-4} + 2$. We can start by identifying the transformations we need to perform on $y = \dfrac{1}{x}$. $\frac{4}{x-4}+2=\mathbf{4}\left(\dfrac{1}{x\mathbf{-4}}\right)\mathbf{+2}$ Hence, we have the following transformations:
• Translate $y = \dfrac{1}{x}$ to the right by $4$ units.
• Vertically stretch the function’s graph by $4$.
• Translate the resulting function by $2$ units upwards.
Let’s perform the following transformations in order to graph $y = \dfrac{4}{x-4} + 2$. Include the vertical and horizontal asymptotes of the function as well in its final graph. The graph confirms that our domain and range from Example 3 is consistent for this function:
Domain $(-\infty,4)\cup(4, \infty)$ Range $(-\infty,2)\cup(2, \infty)$
This problem shows that there are advantages in using both methods when graphing reciprocal functions.
### Practice Questions
1. If $h(x)$ is the reciprocal of $g(x)$, which of the following shows the value of $h(x) \cdot g(x)$?
2. If $h(x)$ is the reciprocal of $g(x)$, which of the following shows the value of $2h(x)\cdot g(x)$?
3. If $h(x)$ is the reciprocal of $g(x)$, which of the following shows the value of $h(x) \cdot \dfrac{g(x)}{3}$?
4. Suppose that $g(x)$ is the reciprocal of $f(x)$, which of the following shows the expression of $g(x)$ when $f(x) = 3x – 1$?
5. Suppose that $g(x)$ is the reciprocal of $f(x)$, which of the following shows the expression of $g(x)$ when $f(x) =\dfrac{3x}{5}$?
6. Suppose that $g(x)$ is the reciprocal of $f(x)$, which of the following shows the expression of $g(x)$ when $f(x) =\dfrac{4(2x + 1)}{9}$?
7. Suppose that $g(x)$ is the reciprocal of $f(x)$, which of the following shows the expression of $g(x)$ when $f(x) =x + 1$?
8. Suppose that $g(x)$ is the reciprocal of $f(x)$, which of the following shows the graph of $g(x)$ when $f(x) =x + 1$?
9. Using the expression for $g(x)$ from the previous problem, which of the following shows the expression when we stretch $g(x)$ two units vertically and translate the resulting function three units downward?
10. What is the vertical asymptote of $y=-\dfrac{5}{3x + 1}$?
11. What is the horizontal asymptote of $y=-\dfrac{5}{3x + 1}$?
12. Which of the following shows the graph of $y=-\dfrac{5}{3x + 1}$?
Images/mathematical drawings are created with GeoGebra.
5/5 - (14 votes) |
# Heather Locklear – Future Predictions (03/27/2020)
How will Heather Locklear perform on 03/27/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not at all guaranteed – don’t get too worked up about the result. I will first work out the destiny number for Heather Locklear, and then something similar to the life path number, which we will calculate for today (03/27/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology people.
PATH NUMBER FOR 03/27/2020: We will analyze the month (03), the day (27) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 27 we do 2 + 7 = 9. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 9 + 4 = 16. This still isn’t a single-digit number, so we will add its digits together again: 1 + 6 = 7. Now we have a single-digit number: 7 is the path number for 03/27/2020.
DESTINY NUMBER FOR Heather Locklear: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Heather Locklear we have the letters H (8), e (5), a (1), t (2), h (8), e (5), r (9), L (3), o (6), c (3), k (2), l (3), e (5), a (1) and r (9). Adding all of that up (yes, this can get tedious) gives 70. This still isn’t a single-digit number, so we will add its digits together again: 7 + 0 = 7. Now we have a single-digit number: 7 is the destiny number for Heather Locklear.
CONCLUSION: The difference between the path number for today (7) and destiny number for Heather Locklear (7) is 0. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is not scientifically verified. If you want a reading that people really swear by, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
CLAT > Decimals
# Decimals - Notes | Study Quantitative Techniques for CLAT - CLAT
Table of contents Addition of Decimals. Subtraction of Decimals. Converting a Decimal into Fraction. Converting a Fraction into a Decimal Multiplication of Decimals. Division of Decimal
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Decimal Fractions and Decimals: The fractions in which the denominators are 10, 100, 1000 etc. are known as decimal fractions. Numbers written in decimal form are called decimals.
• A decimal has two parts, namely, the whole number part and the decimal part.
• The number of digits contained in the decimal part of a decimal is called the number of its decimal places.
We have 0.1 = 0.10 = 0.100 etc
Also, 0.2 = 0.20 = 0.200 etc.
• Unlike decimals may be converted into like decimals by annexing the requisite number of zeros at the end of the decimal part.
Step 1: Write down the decimal numbers, one number under the other number and line up the decimal points.
Step 2: Convert the given decimals to like decimals.
Step 3: Arrange the addends in such a way that the digits of the same place are in the same column.
Step 4: Add the numbers from right as we carry addition usually.
Step 5: Remember to place the decimal point down in the answer in the same place as the numbers above it.
(i) 1.83, 21.105, 236.8 and 0.9
Solution:
We have 1.83 + 21.105 + 236.8 + 0.9
Converting all the addends to like decimals
1.830 + 21.105 + 236.800 + 0.900
Subtraction of Decimals.
Step 1: Write down the two decimal numbers, one number under the other number and line up the decimal points.
Step 2: Convert the given decimals to like decimals.
Step 3: Write the smaller decimal number under the larger decimal number in the column.
Step 4: Arrange the decimal numbers in the column in such a way that the digits of the same place are in the same column.
Step 5: Subtract the numbers in the column from the right.
Remember to place the decimal point down in the answer in the same place as the numbers above it.
Subtract the decimals:
(i) 27.59 from 31.4
Solution:
We have 27.59 from 31.4
Converting both the decimal numbers into like decimals,
31.40 - 27.59
Now arrange them in columns and subtract we get
Converting a Decimal into Fraction.
Step 1: Write the given decimal without the decimal point as the numerator of the fraction.
Step 2: In the denominator write 1 followed by as many zeros as there are decimal places in the given decimal.
Step 3: Convert the above fraction into its simplest form.
Convert the following into fractions.
(i) 3.91
Solution:
Write the given decimal number without the decimal point as numerator.
In the denominator, write 1 followed by two zeros as there are 2 digits in the decimal part of the decimal number.
= 391/100
Converting a Fraction into a Decimal
a) If the denominator of the fraction is 10 or a power of 10.
• To convert a fraction having 10 in the denominator, we put the decimal point one place left of the first digit in the numerator.
• To convert a fraction having 100 in the denominator, we put the decimal point two places left of the first digit in the numerator.
• To convert a fraction having 1000 in the denominator, we put the decimal point three places left of the first digit in the numerator.
b) If the denominator is other than a power of 10
Step I Divide the num. by the den. till a non-zero remainder is obtained
Step II Put a decimal point in the dividend as well as in the quotient
Step III Put a zero on the right of the decimal point as well as on the right of the remainder
Step IV Divide again as we do in whole numbers
Step V Repeat step IV till the remainder is zero
Example 1 Convert 29/4 into a decimal fraction
Solution: On dividing we get
Example 2 Convert into a decimal fraction.
We find that in the above two examples, the process of division has terminated after a few decimal places i.e there are finite places after the decimal point. Such fractions are said to be terminating decimals.
In the case of the fractions the process of division is unending and there are infinite number of places after the decimal point. Such fractions are said to be non-terminating decimals.
Decimal representation of is 0.888 ------ or is represented as 0.
Also
So every positive rational number is either a terminating or a non-terminating repeating decimal.
To search whether a rational number is a terminating or non-terminating repeating decimal.
If the denominator of a rational number (which is in its lowest terms) as no prime factor other than 2 or 5 or both, the rational number is a terminating decimal. Otherwise the rational number is a non-terminating repeating decimal.
Example: etc are terminating
decimals where as are non-terminating repeating decimals.
Multiplication of Decimals.
a) Multiplication of a decimal by a whole number.
To multiply a decimal by a whole number, we multiply the two numbers as if they are whole numbers. We put the decimal point in the product so that there are as many places of decimal in it as there are in the given decimal.
b) Multiplication of a decimal by 10, 100, 1000 ----- etc.
To multiply a decimal by 10, 100, 1000 etc, we have to give a jump to the decimal by one. Two or three etc places to the right
For example 2.96 x 10 = 29.6
2.96 x 100 = 296
2.96 x 1000 = 2960 etc
c) Multiplication of a decimal by a decimal
To multiply two decimals, we first multiply them as if we are multiplying two whole numbers
(ignoring the decimal points). Then the decimal is placed in the product such that it has as many decimal places as there are in multiplicand and the multiplier taken together.
For example: To find the product of 34.17 x 3.2 we first multiply 3417 by 32 and get 109344
The multiplicand and the multiplier taken together have 3 decimal places, the product should also have 3 decimal places. Hence 34.17 x 3.2 = 109.344
Division of Decimal
a) Division of decimal by a whole number is the same as for whole numbers. The decimal point in the quotient is placed just above the decimal point in the dividend.
Example:
Therefore 442.4/ 7 = 63.2
b) Division of a decimal by 10,100,1000 etc.
To divide a decimal by 10,100,1000 etc we have to give a jump to the decimal by one, two, three etc places to the left.
For example :
c) Division of a decimal by a decimal
If we want to divide 6.25 by 2.5 we write
Since the divisor has one decimal place we multiply both numerator and denominator by 10 and get
Now the problem reduces to the division of a decimal by a whole number.
Similarly
d) Division of a whole number by a decimal Convert the decimal of divisor to a whole number and divide.
The document Decimals - Notes | Study Quantitative Techniques for CLAT - CLAT is a part of the CLAT Course Quantitative Techniques for CLAT.
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## Quantitative Techniques for CLAT
56 videos|35 docs|91 tests
## Quantitative Techniques for CLAT
56 videos|35 docs|91 tests
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More Lessons for Basic Algebra
Math Worksheets
A series of free Basic Algebra Lessons.
How to find the vertex and axis of symmetry of a quadratic equation or quadratic function?
How to solve word problems using quadratic equations?
The following diagrams show how to find the vertex of a quadratic function and use that to convert from the general form to the vertex form. Scroll down the page for more examples and solutions for quadratic equations.
The Vertex and Axis of Symmetry
In a parabola, the vertex is the highest or lowest point on a parabola.
To calculate the vertex of a parabola, we can first calculate the x-value by dividing the opposite of the " b" value by 2 times the "a" value. The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.
How to find find the vertex, axis of symmetry, domain and range, and x and y intercepts for a quadratic function algebraically?
Example:
Given f(x) = -4x2 + 10x + 9
Find the following
a) vertex
b) axis of symmetry
c) domain and range
d) x-intercepts
e) y-intercepts
Finding the Vertex and Axis of Symmetry
Example:
Find algebraically the equation of the axis of symmatry and the coordinates of the vertex of the parabola whose equation is y = -2x2 - 8x + 3
In Algebra I and Algebra II, we sometimes need to solve word problems using quadratic equations. When solving word problems, some common quadratic equation applications include projectile motion problems and Geometry area problems. The most important thing when solving these types of problems is to make sure that they are set up correctly so we can use the quadratic equation to easily solve them.
How to solve a quadratic word problem that involves the area of a rectangle?
Example:
Suppose the area of a rectangle is 114.4 m2 and the length is 14 m longer than the width. Find the length and width of the rectangle. Solving Word Problems involving Distance, Rate, and Time Using Quadratics
Example:
Suppose Jessica drove 100 miles and then increased her speed by 30 mph for the following 200 miles. If the second part took 1 hour less than the first part, what was her average speed?
Solving projectile problems with quadratic equations
Example:
Suppose a projectile is launched from a tower into the air with an initial velocity of 48 feet per second. Its height, h, in feet, above the ground is modeled by the function
h = -16t2 + v0t + 64
where t is the time, in second, since the projectile was launched and v0 is the initial velocity
How long was the projectile in the air?
When did it reach its maximum height?
What was its maximum height? Application Problem with Quadratic Formula (Projectile Problem)
A ball is shot into the air from the edge of a building 50 feet above the ground. Its initial velocity is 20 feet per second.
The equation is h = -16t2 + 20t + 50 can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
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Home > College of Sciences > Institute of Fundamental Sciences > Maths First > Algebra > Coordinate Systems and Graphs of Functions > Cartesian Coordinate System in 2D SEARCH MASSEY
# Graphs
## Ordered Pairs
An ordered pair is simply a pair of elements written in a certain order (order is important). For example, the numbers 5 and 2 can form two ordered pairs:
(5,2) and (2,5).
In general, given two numbers x and y, the two ordered pairs (x,y) and (y,x) are different, unless x = y.
The number x is called the first component and y the second component of the ordered pair (x,y).
Ordered pairs have many uses. It can be used to represent an object with two distinguishable parts. For example, a complex number x+iy can be represented by the ordered pair (x,y). In an Excel spreadsheet we use a letter and a number to denote a cell on the sheet.
On a street map we use the same method to locate a street.
## Cartesian Coordinate System in 2-D
Ordered pairs are useful in representing relationships between two sets of data, in the form of a graph. These can then be represented by points on a grid known as the Cartesian Coordinate System in 2-D.
to see how to draw a grid:
The following example illustrates how to represent an ordered pair as a point on this grid.
### Example 1.
To plot the point represented by (2,4):
First draw a vertical line through x =2.
Next draw a horizontal line through y =4.
The point (2,4) is at the intersection of these two lines, P.
More Exercises (opens in another window)
If a point P is represented by (a,b) we call a and b the coordinates of P. In 2-D Cartesian Coordinates, the first coordinate in the ordered pair is called the x-coordinate and the second coordinate in the ordered pair is called the y-coordinate.
### Exercise 1.
Enter the correct coordinates for the points plotted below, A-L.
Remember to separate your x and y coordinates by a comma.
A: ( ) B: ( ) C: ( ) D: ( ) E: ( ) F: ( ) G: ( ) H: ( ) I: ( ) J: ( ) K: ( ) L: ( )
The two axes partition the plane into 4 regions, called quadrants.
They are referred to as the 1st, 2nd, 3rd, and 4th quadrant as shown.
• If the x and y coordinates are both positive, the point is located in the 1st quadrant.
• If the x coordinate is negative and the y coordinate is positive, the point is located in the 2nd quadrant.
• If the x and y coordinates are both negative, the point is located in the 3rd quadrant.
• If the x coordinate is positive and the y coordinate is negative, the point is located in the 4th quadrant.
### Example 2.
Look at the following example to see which quadrants various points are located in. Check the grid below to confirm the solution. |
Teachmint
## Area of Triangle
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## Solution:
Let us find the area using the area of triangle formula:
Area of triangle = (1/2) x b x h
A = 1/2 x 10 x5
A = 1/2 x 50
Therefore, the area of triangle (A) = 25 in2
## Solution:
Let us find the area using the area of triangle formula:
Area of triangle = (1/2) x b x h
A = 1/2 x 10 x5
A = 1/2 x 50
Therefore, the area of triangle (A) = 25 in2
### Practice Questions
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### What is the Area of Triangle
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### Therefore, the area of the triangle (A) = 25 in2
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### Therefore, the area of the triangle (A) = 25 in2
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### Therefore, the area of the triangle (A) = 25 in2
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### Therefore, the area of the triangle (A) = 25 in2
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# How do you simplify \frac { b ^ { 95} c ^ { - 4} } { b ^ { 58} c \cdot b ^ { - 8} c ^ { - 1} }?
Apr 17, 2017
${b}^{45} / {c}^{4}$
#### Explanation:
First of all simplify the numerator and the denominator by multiplying numbers with same base (i.e. $c$ with $c$ and $b$ with $b$). (Remember, when you are multiplying you are adding the exponents, and when you are dividing you are subtracting them).
$\frac{{b}^{95} {c}^{-} 4}{{b}^{58} \textcolor{m a \ge n t a}{c} {b}^{-} 8 {c}^{-} 1}$
This $\textcolor{m a \ge n t a}{c}$ is the same as ${c}^{1}$
$\frac{{b}^{95} {c}^{-} 4}{{b}^{\textcolor{red}{58}} {b}^{\textcolor{g r e e n}{- 8}} {c}^{-} 1 {c}^{\textcolor{b l u e}{1}}}$
$\frac{{b}^{95} {c}^{-} 4}{\left({b}^{\textcolor{red}{58} \textcolor{g r e e n}{- 8}}\right) \left({c}^{- 1 + \textcolor{b l u e}{1}}\right)}$
$= \frac{{b}^{95} {c}^{-} 4}{{b}^{50} {c}^{0}}$
Anything to the power of $0$ is one, so it can be removed
$= \frac{{b}^{\textcolor{red}{95}} {c}^{-} 4}{b} ^ \textcolor{red}{50}$
Subtract the smaller exponent from the same base
$= \frac{{b}^{\textcolor{red}{95} - 50} {c}^{-} 4}{b} ^ \left(\textcolor{red}{50} - 50\right)$
$= \frac{{b}^{45} {c}^{-} 4}{b} ^ 0$
$= {b}^{45} {c}^{-} 4$
The definition of negative exponents is ${a}^{-} m = \frac{1}{a} ^ m$ so,
${b}^{45} {c}^{-} 4 = {b}^{45} \cdot \frac{1}{c} ^ 4 = \textcolor{p u r p \le}{{b}^{45} / {c}^{4}}$
This is the final answer, note that when you simplify expressions with negative exponents you need to remove them and convert them to a positive exponents. |
# COMMON LOGARITHMS -- DEFINITION, SOLVED EXAMPLES, EXERCISES, LINKS TO FURTHER STUDY
Logarithm Formulas before Common Logarithms
if you have not already done so.
Having the knowledge of the Formulae
is a prerequisite here.
### Definition of Common logarithm :
The base 10 of a logarithm has a significance.
In Experimental Science and Engineering,
the results of the investigation are presented in graphs.
How one variable varies with the other is studied by
plotting the variables on X Axis and Y Axis.
Some times it is useful to take one parameter
on one Axis (usually X-Axis) with logarithmic scale,
the other being ordinary scale.
Such a plot is called Semi-Log plot.
Some times, we also take the parameters on both Axes
to logarithmic scale which is called Log-Log plot.
In all these cases the base of the logarithm is
invariably 10.
The Logarithms with base 10 are called
Common Logarithms.
Here, In Common Logarithms, we give some
Solved Examples and Exercise in Logarithms.
### Solved Example 1 of Common Logarithms :
Example 1 of Common Logarithms :
(i) If log12 27 = a, compute log6 32 in terms of a
(ii) If log30 3 = a and If log30 5 = b, find log30 8
Solution to Example 1 of Common Logarithms :
Solution to 1(i) of Common Logarithms :
If log12 27 = a, compute log6 32
We have a = log12 27 = log12 33
We know, in log of a power (See Formula 7), the exponent multiplies the log.
a = 3 log12 3
We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
a = 3 {(log 3)⁄(log 12)}
We know 12 = 3 x 4 = 3 x 22 ⇒ log 12 = log (3 x 22)
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log 12 = log 3 + log (22)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log (22) = 2 log 2
Thus log 12 = log 3 + 2 log 2
a = 3 {(log 3)⁄(log 3 + 2 log 2 )}
Dividing numerator and denominator by log 3, we get
a = 3 [(1)⁄{1 + 2 (log 2⁄log 3)}]
= 3⁄(1 + 2p) where p = (log 2⁄log 3)
⇒ 1 + 2p = 3⁄a ⇒ 2p = 3⁄a - 1 = (3 - a)⁄a
p = (3 - a)⁄2a ...........(i)
Let x = log6 32
We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
x = (log 32)⁄(log 6)
We know 32 = 2 x 2 x 2 x 2 x 2 = 25; 6 = 2 x 3;
x = {log (25}⁄{log (2 x 3)}
Numerator is logarithm of a power and Denominator is logarithm of a product.
We know, in log of a power (See Formula 7), the exponent multiplies the log.
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
x = {5 log 2}⁄{log 2 + log 3}
Dividing numerator and denominator by log 3, we get
x = 5 (log 2⁄log 3)⁄{(log 2⁄log 3) + 1}
= 5p⁄(p + 1) [Since p = (log 2⁄log 3)]
We have p = (3 - a)⁄2a ...........(i)
p + 1 = (3 - a)⁄2a + 1 = (3 - a + 2a)⁄2a = (3 + a)⁄2a........(ii)
Substituting the value of p from (i) and (p + 1) from (ii) in x, we get
x = 5 {(3 - a)⁄2a}⁄{(3 + a)⁄2a} = 5 {(3 - a)⁄(3 + a)}. Ans.
Solution to 1(ii) of Common Logarithms :
we know 30 = 3 x 5 x 2; log30 30 = 1
∴ log30 30 = 1 ⇒ log30 (3 x 5 x 2) = 1
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log30 3 + log30 5 + log30 2 = 1
a + b + log30 2 = 1 ⇒ log30 2 = 1 - a - b
log30 8 = log30 (23)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log30 8 = 3(log30 2) = 3(1 - a - b). Ans.
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### Solved Example 2 of Common Logarithms :
Example 2 of Common Logarithms :
If b = √(ac), prove that
(loga n)⁄(logc n) = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}
Solution to Example 2 of Common Logarithms :
By data b = √(ac) ⇒ b2 = ac ⇒ ba = cb.........(i)
Let x = logn a then 1⁄x = 1⁄logn a;
Let y = logn b then 1⁄y = 1⁄logn b;
Let z = logn c then 1⁄z = 1⁄logn c;
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ 1⁄x = loga n; 1⁄y = logb n; 1⁄z = logc n
To prove (loga n)⁄(logc n) = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}
L.H.S. = (loga n)⁄(logc n) = (1⁄x)⁄(1⁄z) = zx ......(ii)
R.H.S. = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}
= {(1⁄x) - (1⁄y)}⁄{(1⁄y) - (1⁄z)} = {(y - x)⁄(xy)}⁄{(z - y)⁄(yz)}
= {(y - x)⁄(xy)} x {(yz)⁄(z - y)} = (zx){(y - x)⁄(z - y)}.........(iii)
(y - x)⁄(z - y) = (logn b - logn a)⁄(logn c - logn b)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ (y - x)⁄(z - y) = {logn (ba)}⁄{logn (cb)} = 1 [Since ba = cb from (i)]
Using this value in (iii), we get
R.H.S. = (zx) (1) = zx
From (ii), L.H.S. = zx
∴ L.H.S. = R.H.S. (Proved.)
Thus Example 2 of Common Logarithms is solved.
### More Solved Examples
to more Solved Examples.
More Solved Examples
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### Exercise : Common Logarithms
Exercise Problems : Common Logarithms
1. If 1⁄x = 1 + loga bc, 1⁄y = 1 + logb ca, 1⁄z = 1 + logc ab,
then prove that x + y + z = 1.
2. If a = log24 12, b = log36 24, c = log48 36,
then prove that 1 + abc = 2bc.
## Progressive Learning of Math : Common Logarithms
Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.
This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.
These build a student’s new knowledge of concepts
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These provide many pages of practice that gradually
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These also provide teachers and parents with lessons
on how to work with the child on the concepts.
The series is low to reasonably priced and include
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# Geometry: Using and Proving Angle Supplements
## Using and Proving Angle Supplements
When acute angles need a supplement, they turn to an obtuse angle. Even though I am an authority on the subject, and I make that statement with a certain level of confidence and conviction, skeptics should demand that I put my money where my mouth is and prove it. And to set a good example, I will.
• Example 6: Prove that the supplement of an acute angle is an obtuse angle.
• Solution: You know what to do, so just do it.
• 1. State the theorem.
• Theorem 9.5: The supplement of an acute angle is an obtuse angle.
• 2. Draw a picture. Figure 9.6 shows an acute angle ?ABC and its supplement ?CBD. Together ?ABC and ?CBD form the straight angle ?ABD.
• 3. Interpret what you are given in terms of your drawing. You are given ?ABC and its supplement ?CBD, with ?ABC acute.
• 4. Interpret what you are trying to prove in terms of your drawing. Prove that ?CBD is obtuse.
• 5. Prove the theorem. What's the game plan? This proof involves acute, obtuse, and supplementary angles, so you'll probably use their definitions somewhere. Because you'll be dealing with inequalities (acute angles have measure less than 90 and obtuse angles have measure greater than 90), you might need your definitions of < or >, and you might need our Protractor Postulate. And there's always algebra.
StatementsReasons
1. ?ABC is acute, and ?ABC and ?CBD are supplementary. Given
2. m?ABC + m?CBD = 180 Definition of supplementary angles
3. m?ABC < 90 Definition of acute angle
4. m?ABC + 90 < 180 Addition property of inequality
5. m?ABC + 90 < m?ABC + m?CBD Substitution (steps 2 and 4)
6. 90 < m?CBD Subtraction property of inequality
7. ?CBD is obtuse Definition of obtuse angle
Now that you're starting to crank out those formal proofs, it's time to open things up and see how you perform on the open road. Take a look at Figure 9.7. ?ABC and ?CBD are adjacent supplementary angles. If you construct the bisectors of each of these two angles, then together the bisectors will form a new angle. But not just any angle. This new angle will be right. And you'll prove it.
Figure 9.7?ABC and ?CBD are adjacent supplementary angles; ?BE bisects ?ABC, and ?BF bisects ?CBD.
• Example 7: Prove that the bisectors of two adjacent supplementary angles form a right angle.
• 1. State the theorem.
• Theorem 9.6: The bisectors of two adjacent supplementary angles form a right angle.
• 2. Draw a picture (see Figure 9.7).
• 3. Interpret the given information in terms of the picture. ?ABC and ?CBD are adjacent supplementary angles; ?BE bisects ?ABC, and ?BF bisects ?CBD.
• 4. Interpret what to prove in terms of the picture. Prove that ?EBF is a right angle.
• 5. Prove the theorem. Your game plan: You'll need some definitions in this proof: supplementary angles, right angles, and angle bisectors. Because you will be breaking up angles, the Angle Addition Postulate might be useful. Let's see how it all unfolds.
StatementsReasons
1.?ABC and ?CBD are adjacent supplementary angles; ?BE bisects ?ABC, and ?BF bisects ?CBD Given
2. ?ABE ~= ?EBC, ?CBF ~= ?FBD Definition of angle bisector
3. m?ABE = m?EBC, m?CBF = m?FBD Definition of ~=
4. m?ABC + m?CBD = 180 Definition of supplementary angles
5. m?ABE + m?EBC = m?ABC, m?CBF + m?FBD = m?CBD, and m?EBC + m?CBF = m?EBF Angle Addition Postulate
6. m?ABE + m?EBC + m?CBF + m?FBD = 180 Substitution (steps 4 and 5)
7. 2m?EBC + 2m?CBF = 180 Substitution (steps 3 and 6)
8. m?EBC + m?CBF = 90 Algebra
9. m?EBF = 90 Substitution (steps 5 and 8)
10. ?EBF is right Definition of right angle
Keep in mind that there is more than one way to construct a proof. If you put three mathematicians in a room and have them prove the same theorem, you will probably get three different proofs. They would all be valid (assuming they did it right), though they might have taken different steps along the way. Variety is the spice of life. Just be sure to avoid using cheap reasons in your proofs. Trust me: It will show.
Put Me in, Coach!
Here's your chance to shine. Remember that I am with you in spirit and have provided the answers to these questions in Answer Key.
• 1. If E is between D and F, write a formal proof that DE = DF ? EF.
• 2. Given ?ABC and ?BD as in Figure 9.8, write a formal proof that m?ABD = m?ABC ? m?DBC.
• 3. Prove that the angle bisector of an angle is unique.
• 4. Prove that the supplement of a right angle is a right angle.
Excerpted from The Complete Idiot's Guide to Geometry 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.
To order this book direct from the publisher, visit the Penguin USA website or call 1-800-253-6476. You can also purchase this book at Amazon.com and Barnes & Noble. |
# Multiplying Even and Odd Numbers
Today I’m going to be showing you what would happen if you were to multiply:
a) An even number by an even number;
b) An odd number by an odd number;
c) An even number by an odd number.
Firstly, let us define what an even number is:
An even number can be described using the expression $2n$, whereby (n) would be a whole number ranging from 0 upwards.
Next, let us define what an odd number is:
An odd number can be described using the expression $2n+1$, and similarly (as is the case with even numbers), (n) would be a whole number ranging from 0 upwards.
Now, since we’ve defined how both even numbers and odd numbers can be described in terms of mathematical expressions, let’s focus our attention on multiplying even numbers by even numbers, odd numbers by odd numbers and even numbers by odd numbers…
Multiplying even numbers by even numbers:
Let’s produce two whole numbers which could be equal to one another or not equal to one another… Let’s call these numbers ${ n }_{ 1 }$ and ${ n }_{ 2 }$.
Using these two whole numbers we can multiply two unknown even numbers by each other in such a manner:
$2{ n }_{ 1 }\cdot 2{ n }_{ 2 }$
This would invariably give us the result $4{ n }_{ 1 }{ n }_{ 2 }=2\cdot 2{ n }_{ 1 }{ n }_{ 2 }$. Now, the product of ${ n }_{ 1 }{ \cdot n }_{ 2 }$ would be a whole number and since this is the case, you would have to say that an even number multiplied by an even number would produce an even number. Let’s not forget that even numbers are multiples of 2.
Multiplying odd numbers by odd numbers:
Once again, let’s come up with two whole numbers which could be equal to one another or not equal to one another… These whole numbers will be ${ n }_{ 3 }$ and ${ n }_{ 4 }$.
This would mean that two odd numbers being multiplied by one another would produce an expression as such:
$\left( 2{ n }_{ 3 }+1 \right) \left( 2{ n }_{ 4 }+1 \right)$
And if we expand the expression above, we’ll get:
$4{ n }_{ 3 }{ n }_{ 4 }+2{ n }_{ 3 }+2{ n }_{ 4 }+1$
Now if we re-arrange the expression above, we can get:
$2\left( 2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 } \right) +1$
Since the expression $2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 }$ must be a whole number, you would be forced to conclude that an odd number multiplied by an odd number would produce an odd number.
Multiplying even numbers by odd numbers:
We will for the last time come up with two whole numbers ${ n }_{ 5 }$ and ${ n }_{ 6 }$.
An even number and an odd number being multiplied by one another could be shown using the mathematical expression below:
$2{ n }_{ 5 }\cdot \left( 2{ n }_{ 6 }+1 \right)$
Lazily, we could conclude that an even number multiplied by an odd number would produce an even number. This is because even numbers are all multiples of 2.
Ok… So, let’s summarise what we’ve discovered:
i) An even number multiplied by an even number would produce an even number;
ii) An odd number multiplied by an odd number would produce an odd number;
iii) An even number multiplied by an odd number would produce an even number.
Knowing this we can further strengthen our mathematical reasoning. 🙂
# Binomial Expansions on Python
The other day I was asked to solve a complex probability problem. This probability problem was related to a vocabulary quiz which was being circulated around the web. It is a vocabulary quiz which contains 15 questions with 2 two possible answers to each question. My goal was to simply state the chances of someone (with a terrible vocabulary) answering at least 7/15 questions in this quiz correctly without referring to Google or any other search engine… Their answers to these questions were going to be completely random.
Now, in order solve this probability problem I first had to acknowledge seven essential facts:
1) The person taking this quiz has a terrible vocabulary and just wouldn’t be able to answer any of the questions in this quiz with confidence;
2) The person answering these questions would be providing random answers to them;
3) The person taking this quiz isn’t allowed to use any tools/resources which would help him/her answer these questions correctly;
4) There were only two possible answers to each question (right (R) or wrong (W) answers);
5) Each answer to these questions were mutually exclusive, in other words, selecting the right answer to one question wouldn’t increase your chances of answering any other question correctly ;
6) There were 15 questions in total;
7) A binomial distribution would be required to solve the problem as there were two mutually exclusive outcomes to each question being answered.
With these facts in mind I knew exactly what I had to do to solve this problem, however, expanding the expression (R+W)^15 was going to be a very tedious task indeed and would require plenty of manual labour.
Was there another way to solve this problem?? Could a computer program help me solve the problem more easily??
Well it turned out that I would be able to solve the problem more easily, but that I’d have to use Python (a high level general purpose programming language) including an “add on” called Sympy to achieve such a feat. So what I essentially did was install both Python(x, y) and Sympy on my Windows laptop.
*You can find out how to download both Python and Sympy by watching the video below:
I then searched “Python IDLE” on my laptop and opened up Python’s shell application. This is what I typed into the application:
If you look at the image above carefully you will see that this Python shell application (thanks to the help of the “add on” Sympy, including some handy code) was able to expand (R+W)^15 for me… In fact, it did it in a split second.
From the moment (R+W)^15 had been expanded, to solve the problem I had been asked to solve, all I had to do was:
i) Sum up the coefficients of the variables (R, W) included in the expansion;
ii) Place this sum (a) under the value of the sum of coefficients sitting beside R variables with exponentials greater or equal to 7 (b) – in a fraction b/a.
This fraction (b/a) would give me the probability / chances of someone randomly being able to answer at least 7/15 questions correctly in the vocabulary test under the conditions which had been set.
a = 2(1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 6435) = 32,768
b = 1 + 15 + 105 + 455 + 1365 + 3003 + 5005 + 2(6435) = 22,819
b/a = 22,819/32,768 = 0.70 (to 2 decimal places)
Solution to problem: 70% chance
Now the irony of the story is this… I gave myself this problem to solve because my “posh” vocabulary is horrendous. It turned out that I managed to get 9/15 questions in the vocabulary test correct and all my answers were guesses. 🙂 |
# 2.1 Linear functions
Page 1 / 17
In this section, you will:
• Represent a linear function.
• Determine whether a linear function is increasing, decreasing, or constant.
• Calculate and interpret slope.
• Write the point-slope form of an equation.
• Write and interpret a linear function.
Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( [link] ). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes. http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm
Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.
## Representing linear functions
The function describing the train’s motion is a linear function , which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
## Representing a linear function in word form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.
• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.
The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.
## Representing a linear function in function notation
Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where $x$ is the input value, $m$ is the rate of change, and $b$ is the initial value of the dependent variable.
In the example of the train, we might use the notation $D\left(t\right)$ in which the total distance $D$ is a function of the time $t.$ The rate, $m,$ is 83 meters per second. The initial value of the dependent variable $b$ is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani |
New
New
Year 3
# Subtracting small numbers
I can subtract numbers below 10 using efficient strategies.
New
New
Year 3
# Subtracting small numbers
I can subtract numbers below 10 using efficient strategies.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. When the subtrahend is bigger than the ones value of the minuend, we will have to bridge through 10
2. Composition of numbers to 10 allows us to partition numbers when we are subtracting.
3. Composition of numbers to 10 allows us to bridge through ten when subtracting.
### Common misconception
Difference can only be found by counting back.
Difference can be calculated by subtracting the subtrahend from the minuend or by counting between the subtrahend and the minuend.
### Keywords
• Minuend - The minuend is the number being subtracted from.
• Subtrahend - A subtrahend is a number subtracted from another.
• Difference - The difference is the result after subtracting one number from another.
Pupils need to understand that subtraction is the inverse of addition. Addition facts can help with solving subtraction calculations. Tens frames, bar models and number lines can all be used to represent subtraction problems.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
What is one less than 7?
Q2.
What is the missing number in the equation?
Q3.
Match the one less questions to the correct answer.
Correct Answer:What is one less than 6?,5
5
Correct Answer:What is one less than 8? ,7
7
Correct Answer:What is one less than 10? ,9
9
Correct Answer:What is one less than 9? ,8
8
Q4.
Which equation could Alex use to find the missing number in this bar model?
Correct answer: 8 − 3 =
8 + 3 =
3 − 8 =
Q5.
What is the answer to this equation?
Q6.
Izzy has nine cherries. She eats three. How many cherries are left?
12
8
7
## Exit quiz
### 6 Questions
Q1.
Complete: 5 + 4 = 9 so 9 − 5 =
Q2.
Alex has six cherries. He eats three of them. How many are left?
4
5
6
Q3.
What is 9 in this equation? 9 − 2 = 7
the subtrahend
the difference
Q4.
Which equation is represented by the number line?
9 − 3 = 6
9 + 3 = 6
9 + 6 = 3
Correct answer: 9 − 6 = 3
Q5.
What is the difference between 9 and 11?
Q6.
Match each equation to the answer.
4 |
# 9. 1 Modeling with Differential Equations Spring 2010 Math 2644 Ayona Chatterjee.
## Presentation on theme: "9. 1 Modeling with Differential Equations Spring 2010 Math 2644 Ayona Chatterjee."— Presentation transcript:
9. 1 Modeling with Differential Equations Spring 2010 Math 2644 Ayona Chatterjee
Motivation Mathematical models often take the form of a differential equation (DE). – That is, an equation that contains an unknown function and some of its derivatives. – In real world problems we are often interested in the changes that occur and if we can predict the behavior of the future based on the current changes.
Models of Population Growth One model for the growth of a population is based on the assumption that the population grows at a rate proportional to the size of the population. Let’s identify and name the variables in this model: – t time the independent variable – P the number of individuals in the population the dependent variable The rate of growth of the population is the derivative. So our assumption that the rate of growth of the population is proportional to the population size is written as 1
Model continued where k is the proportionality constant. Equation 1 is our first model for population growth; it is a differential equation because it contains an unknown function P and its derivative dP/dt. If we rule out a population of 0, then P(t)>0 for all t. So, if k >0, then Equation 1 shows that P’(t) > 0 for all t. This means that the population is always increasing. In fact, as P(t) increases, Equation 1 shows that dP/dt becomes larger. In other words, the growth rate increases as the population increases.
Family of Solutions Thus any exponential function of the form P(t)= Ce kt is a solution of Equation 1. Allowing C to vary through all the real numbers, we get the family of solutions P(t)= Ce kt whose graphs are shown in Figure 1.
Logistic Differential Equation Many populations start by increasing in an exponential manner, but the population levels off when it approaches its carrying capacity K (or decreases toward K if it ever exceeds K). For a model to take into account both trends, we make two assumptions:
Equation 2 is called the logistic differential equation and was proposed by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s as a model for world population growth. We first observe that the constant functions P(t)=0 and P(t)= K are solutions because, in either case, one of the factors on the right side of Equation 2 is zero. (This certainly makes physical sense: If the population is ever either 0 or at the carrying capacity, it stays that way.) These two constant solutions are called equilibrium solutions.
General Differential Equations In general, a differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the order of the highest derivative that occurs in the equation. Thus, Equations 1 and 2 are first-order equations In all two of those equations the independent variable is called t and represents time, but in general the independent variable doesn’t have to represent time.
Solution of general DE A function f is called a solution of a differential equation if the equation is satisfied when y=f(x) and its derivatives are substituted into the equation. Thus f is a solution of Equation 4 if
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Introduction to Algebra
Document Sample
``` ONE STEP
EQUATIONS
ONE STEP EQUATIONS
An equation is like a balance scale
because it shows that two
quantities are equal.
What you do to one side of the
equation must also be done to the
other side to keep it balanced.
ONE STEP EQUATIONS
To solve one step equations, you need
equation:
• What is the variable?
• What operation is performed on the variable?
• What is the inverse operation?
(The one that will undo what is being
done to the variable)
ONE STEP EQUATIONS
Example 1 Solve x + 4 = 12
What is the variable? The variable is x.
What operation is being performed on the variable? Addition.
What is the inverse operation (the one that will undo what is
being done to the variable)? Subtraction.
Using the subtraction property of equality, subtract 4 from both
sides of the equation.
The subtraction property of equality tells us to subtract
the same thing on both sides to keep the equation equal.
x + 4 = 12
-4 -4
x = 8
ONE STEP EQUATIONS
Example 2 Solve y - 7 = -13
What is the variable? The variable is y.
What operation is being performed on the variable? Subtraction.
What is the inverse operation (the one that will undo what is being
Using the addition property of equality, add 7 to both sides of
the equation.
thing on both sides to keep the equation equal.
y - 7 = -13
+7 +7
y = -6
ONE STEP EQUATIONS
Example 3 Solve –6a = 12
What is the variable? The variable is a.
What operation is being performed on the variable? Multiplication.
What is the inverse operation (the one that will undo what is being done
to the variable)? Division
Using the division property of equality, divide both sides of the
equation by –6.
The division property of equality tells us to divide the same
thing on both sides to keep the equation equal.
–6a = 12
-6 -6
a = -2
ONE STEP EQUATIONS
Example 4 Solve b = -10
What is the variable?
2 The variable is b.
What operation is being performed on the variable? Division.
What is the inverse operation (the one that will undo what is
being done to the variable)? Multiplication
Using the multiplication property of equality, multiply both sides
of the equation by 2.
The multiplication property of equality tells us to multiply
the same thing on both sides to keep the equation equal.
b
= -10
2
b
2 • = -10 • 2
2
b = -20
Solve the following
equations with one step
1. r + 4 = 17
2. g - 15 = -11
3. 13 + p = - 9
4. -5b = 85
5. - 7v = - 91
6. w = -2
3
1. r = 13
2. g = 4
3. p = - 22
4. b = - 17
5. v = 13
6. w=-6
Word Problems
with One Step Equations
You will need to know how to solve word problems involving
these equations, by using the four-step problem solving
plan:
2. Plan the solution
3. Solve the problem
4. Examine the solution
1. Explore the problem
problem
carefully, then:
• Identify what
information is
given
• Identify what you
2. Plan the Solution
• Choose a variable to represent the unknown that
you are trying to solve for
• Create an algebraic equation for solving
3. Solve the Problem 4. Examine the
Solution
• use algebraic
methods
“reasonable” ?
• Did you check your
work to be sure?
Try it !
Write an Equation and Solve the Problem:
some athletic shoes for \$112, he had \$96 left.
How much was his paycheck?
• Let p = paycheck
• p - 112 = 96
• + 112 +112
• p = 208
• Noah’s paycheck was \$208.
```
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In trigonometry, the angle addition and subtraction formulas provide a way to express the sine and cosine of the sum or difference of two angles in terms of the sines and cosines of the individual angles. These formulas are incredibly useful for simplifying expressions and solving trigonometric equations.
## Angle Addition and Subtraction Formulas for Sine and Cosine
The angle addition and subtraction formulas for sine and cosine are as follows:
$\sin(A + B) = \sin A \cos B + \cos A \sin B$
$\sin(A – B) = \sin A \cos B – \cos A \sin B$
$\cos(A + B) = \cos A \cos B – \sin A \sin B$
$\cos(A – B) = \cos A \cos B + \sin A \sin B$
## Using the Formulas to Solve for Specific Angles
### Solving for $$\tan\left(\frac{\pi}{12}\right)$$
Step 1: Find two angles on the unit circle that you can add or subtract to get $$\frac{\pi}{12}$$
$\tan(\frac{\pi}{12}) = \tan(\frac{4\pi}{12} – \frac{3\pi}{12}) = \tan(\frac{\pi}{3} – \frac{\pi}{4})$
Step 2: Apply the fact $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$.
$\tan(\frac{\pi}{3} – \frac{\pi}{4}) = \frac{\sin(\frac{\pi}{3} – \frac{\pi}{4})}{\cos(\frac{\pi}{3} – \frac{\pi}{4})}$
Step 3: Apply the angle subtraction formulas for sine and cosine:
$\sin(\frac{\pi}{3} – \frac{\pi}{4}) = \sin\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{4}\right) – \cos\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{4}\right)$
$\cos(\frac{\pi}{3} – \frac{\pi}{4}) = \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{4}\right)$
Step 4: Use the unit circle to replace the sines and cosines with their numerical values:
$\sin(\frac{\pi}{3} – \frac{\pi}{4}) = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} – \frac{1}{2} \times \frac{\sqrt{2}}{2}$
$\cos(\frac{\pi}{3} – \frac{\pi}{4}) = \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}$
Step 5: Simplify to find $$\tan\left(\frac{\pi}{12}\right)$$:
$\tan\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} – \sqrt{2}}{4}$
## Conclusion
The angle addition and subtraction formulas are powerful tools in trigonometry. They not only simplify complex trigonometric expressions but also pave the way for solving intricate problems, much like the ones we’ve tackled today. As St. Augustine once said, “The world is a book, and those who do not travel read only one page.” Similarly, the realm of trigonometry is vast, and these formulas are but stepping stones in understanding this fascinating subject. |
# Common Core: High School - Geometry : Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures
## Example Questions
### Example Question #521 : High School: Geometry
Choose the answer that describes a congruent triangle
Triangles with corresponding sides of the same lengths and corresponding angles of the same measures
Triangles with 2 of the 3 corresponding sides of the same lengths but differing corresponding angles
Triangles with corresponding sides of the same lengths but differing corresponding angles
Triangles with corresponding angles of the same measure but differing corresponding sides
Triangles with corresponding sides of the same lengths and corresponding angles of the same measures
Explanation:
To be congruent, triangles must have all corresponding angles and sides be of the same measure. Corresponding sides of triangles are sides of the same measure in the same positions on different triangles. This is shown in red on the two triangles below. Corresponding angles of triangles are angles of the same measure in the same position on different triangles. This is shown in blue on the two triangles below.
### Example Question #1 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
Choose the answer that describes similar triangles
Triangles with corresponding sides of the same lengths and corresponding angles of the same measures
Triangles with 2 of the 3 corresponding sides of the same lengths but differing corresponding angles
Triangles with corresponding sides of the same lengths but differing corresponding angles
Triangles with corresponding angles of the same measure but differing corresponding sides
Triangles with corresponding angles of the same measure but differing corresponding sides
Explanation:
Shapes, in general, are similar when they are the same shape but have different sizes. If they were the exact same size, they would be congruent. To have the same shape, even when the sizes differ, the shape needs to have the same angles. Take the two triangles below for example, they have equal angles but are different sizes. Therefore these triangles are similar.
### Example Question #1 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
True or False: When considering right triangles, if two right triangles have a congruent hypotenuse and a congruent leg then these triangles are congruent.
False
True
True
Explanation:
If two right triangles have a congruent leg and hypotenuse we can say they have two congruent sides. Since both of these triangles are right triangles, we also know that they have a congruent angle (their 90-degree angle). So these two triangles have two pairs of corresponding congruent sides and one pair of corresponding congruent angles. By SAS Theorem, these right triangles are congruent. When using this fact that two right triangles are congruent when they have a congruent hypotenuse and a congruent leg, this is called the HL Theorem.
### Example Question #10 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
True or False: Triangles that have all three corresponding sides equal in length can still have differing corresponding angles, therefore triangles with all three corresponding sides equal in length are not guaranteed to be either similar or congruent.
False
True
False
Explanation:
The Side-Side-Side (SSS) Theorem states that two triangles are congruent if the three corresponding sides of each triangle are congruent. If you think about it, in order for all three corresponding sides to be equal they must fit together at the same angles as well making these triangles congruent. A detailed proof based on Philoh’s approach is below for a better explanation. Note that there are stronger proofs than this for this theorem, but this proof is best visually for this level of understanding of the theorem.
Proof:
(We want to show that if all three sides of two triangles are equal, then these two triangles are congruent)
Consider two triangles and . Assume the following:
Since we are able to rotate triangle to coincide and . We will just call this common line segment for simplification.
Now we will draw a line from vertex to vertex . This gives us two isosceles triangles. Recall from the definition of an isosceles triangle that the two adjacent sides and angles are equivalent. So and .
We can then deduce that . We now know that at least two sides of the two given triangles are equal and the angle enclosed by these sides is congruent between the two triangles. By SAS Theorem, these two triangles are congruent.
### Example Question #1 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
Consider the two triangles below (ABE and CBE). Given that sides and are equal and bisects , prove triangles ABE and CBD are congruent. and are equal and bisects , prove triangles ABE and CBD are congruent.
Proof:
Proof:
Proof:
Proof:
Explanation:
Explanation: For the explanation follow the detailed proof below:
### Example Question #5 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
Triangle is similar to triangle . Solve for and .
Explanation:
In order to solve for and below, we need to use the fact that similar triangles are proportional. This allows us to set up ratios of the lengths of the sides to solve for the unknown variables:
To solve for :
So we can set up the following ratios
by cross-multiplying to get rid of the fractions
So
To solve for :
So we can set up the following ratios
by cross-multiplying to get rid of the fractions
So
### Example Question #531 : High School: Geometry
Consider the group of line segments below. is parallel to . What is the relationship between triangles and ?
Similar
No relationship
Congruent
Similar
Explanation:
Since is parallel to , the point where intersects forms two pairs of opposite vertical angles. We can now say that and are equal by the definition of opposite vertical angles. Notice that and are alternate interior angles. By definition, .
We have two equal corresponding angles. The Angle-Angle (AA) Theorem for similar triangles says that if two triangles have two pairs of congruent corresponding angles, the triangles are similar. So by AA Theorem, triangles and are similar.
### Example Question #2 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
and are parallel. Are triangles and triangle are similar? If so, solve for .
There is not enough information to determine if these triangles are similar
No, these triangles are not similar
Explanation:
Since is parallel to , the point where intersects forms two pairs of opposite vertical angles. We can now say that and are equal by the definition of opposite vertical angles. Notice that and are alternate interior angles. By definition, . We have two equal corresponding angles. The Angle-Angle (AA) Theorem for similar triangles says that if two triangles have two pairs of congruent corresponding angles, the triangles are similar. So by AA Theorem, triangles and are similar. In order to solve for , we need to use the fact that similar triangles are proportional. This allows us to set up ratios of the lengths of the sides to solve for :
We are able to set up the following ratios:
cross multiplying to get rid of the fractions
So the length of
### Example Question #532 : High School: Geometry
Consider the parallelogram below. From what you know about parallelograms and our theorems for congruence in triangles, prove that triangles and are congruent.
Proof:
Proof:
Proof:
Proof:
Explanation:
### Example Question #1 : Use Congruence And Similarity Criteria For Triangles To Solve Problems And To Prove Relationships In Geometric Figures
True or False: The triangles below are similar, NOT congruent.
True
False |
# Parabolas
Let us now consider the graphs of polynomial functions of degree two -- that is to say, quadratic functions of the form $$f(x)=ax^2 + bx + c$$ for some constants $a$, $b$, and $c$.
Plotting a bunch of points for some randomly chosen quadratic functions reveals some interesting properties, as seen below. All of the quadratic functions, when plotted, appear to take on a "U"-like shape -- although sometimes this "U" is upside down. As a result, the quadratic functions appear to have either a maximum or minimum value, depending on whether the "U" is upside-down or right-side-up, respectively. The graph of a quadratic function (which we call a parabola) seems to sometimes be "wide", sometimes "narrow", and still other times something between those two extremes. They sometimes cross the $x$-axis producing two $x$-intercepts. Other times, the only touch the $x$-axis once, or not at all. They always have a $y$-intercept (granted, our graph has to be big enough to show it).
Let's see if we can't consider each of these ideas a bit more carefully...
First, with regards to deducing the shape of the graph of this function from it's equation -- note, things would probably be easier to analyze if we could rewrite the equation for the function using only a single occurrence of $x$. Recall the trick employed to the same end when solving general quadratic equations -- namely, "completing the square"...
$$\begin{array}{rcl} f(x) &=& ax^2 + bx + c\\\\ &=& a(x^2 + \frac{b}{a}x\, + \,?\,) + c\\\\ &=& a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) + c - \frac{b^2}{4a}\\\\ &=& a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a} \end{array}$$
Now, let us consider the factor of the first term seen in the last line above: $(x + \frac{b}{2a})^2$. This term is certainly positive, by virtue of being a square -- and thus equivalent to
$$\left| x + \frac{b}{2a} \right|^2 = \left| x - (-\frac{b}{2a}) \right|^2$$
This might initially seem to be a strange way to rewrite $f(x)$, but remember that $|p-q|$ gives the distance between $p$ and $q$ on the real number line. So the expression above is the square of the distance between $x$ and $-\frac{b}{a}$. Armed with this interpretation, we can see that as $x$ gets farther and farther from $-\frac{b}{2a}$, the expression above gets larger and larger in magnitude.
Also note that in finding the height of the graph of $f(x)$ at some particular $x$, $$\begin{array}{rcl} f(x) &=& a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}\\\\ &=& a \left| x - (-\frac{b}{2a}) \right|^2 + \frac{4ac - b^2}{4a} \end{array}$$ we see that the amount added to $\frac{4ac - b^2}{4a}$ depends only on the distance between $x$ and $-\frac{b}{a}$, and not whether $x$ is to the right or left of this value.
This confirms our suspicion that parabolas are symmetric. In fact, we can say even more -- the axis of symmetry of a parabola described by $f(x)=ax^2+bx+c$ is the vertical line $x=-\frac{b}{a}$.
Now consider the impact of the sign of $a$. When $a$ is positive, for any $x \ne -\frac{b}{2a}$, we are adding a positive value to $\frac{4ac - b^2}{4a}$ -- and one that grows as $x$ gets farther from $-\frac{b}{2a}$. Consequently, if $a > 0$, the function $f(x)$ has a minimum value of $\frac{4ac - b^2}{4a}$ at $x = -\frac{b}{2a}$, and the graph is basically "U"-shaped -- where the direction of opening is "up".
Alternatively, when $a$ is negative, then for any $x \ne -\frac{b}{2a}$, we are always subtracting some positive value from $\frac{4ac - b^2}{4a}$ -- again, one that grows in magnitude as $x$ gets farther from $-\frac{b}{2a}$. So, if $a < 0$, the function $f(x)$ has a maximum value of $\frac{4ac - b^2}{4a}$ at $x = -\frac{b}{2a}$, and the graph is also shaped like a "U", only upside-down (i.e., the direction of opening is "down").
The reader should reflect on the impact of the magnitude of $a$ as well. Upon doing so, one can quickly conclude that the greater the magnitude of $a$, the more steeply the curve ascends (or descends, as appropriate). Consequently, large magnitudes of $a$ correspond to "narrow" parabolas, while small magnitudes of $a$ correspond to "wide" parabolas.
As a matter of verbiage, the special point $(-\frac{b}{2a}, \frac{4ac - b^2}{4a})$ where either the maximum or minimum occurs we refer to as the vertex of the parabola.
It's worth noting that if we are given the quadratic function $f(x) = ax^2 + bx + c$ and we complete the square to rewrite the function in the form $f(x)=a(x-h)^2 + k$ as we did above, then $h = -\frac{b}{2a}$ and $k = \frac{4ac - b^2}{4a}$.
Consequently, if someone has already done some of the work for us and given us a quadratic function in the form $f(x)=a(x-h)^2 + k$, then the vertex must be given by $(h,k)$.
Finally, with regards to the $x$-intercepts, recall that points on the $x$-axis have a $y$-coordinate of zero. As such, we can find all $x$ intercepts by solving
$$ax^2 + bx + c = 0$$
Being a quadratic equation, we know the solution to the above takes the form
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
This results in $0$, $1$, or $2$ solutions for $x$, depending on the value of the discriminant $b^2-4ac$.
As an immediate consequence -- there are two $x$-intercepts when the discriminant is positive; one $x$-intercept when the discriminant is zero; and no $x$-intercepts when the discriminant is negative. |
# 180 Days of Math for Fifth Grade Day 89 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 89 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 89
Directions: Solve each problem.
Question 1.
Calculate the sum of 46 and 8.
54
Explanation:
Question 2.
3 • 28 = ___________
84
Explanation:
Question 3.
7.857
Explanation:
Question 4.
Write the even number after 5,367.
5368
Explanation:
An even number is any number which can be divided by 2.
The division of 5,367 and 2 is 2,683.5
The next number is 5368, when divided by 2 gives 2684.
Question 5.
$$\frac{1}{3}$$ of 60 is _________.
20
Explanation:
(1/3) of 60
= (1/3) × 60
= 60 ÷ 3
= 20
Question 6.
25 + 35 ÷ 7 = ___________
30
Explanation:
First, Divide 35 and 7
35 ÷ 7 = 5
Now add 25 and 5; 25 + 5 = 30
Question 7.
37 + = 74
37
Explanation:
Let us consider the blank space be ‘x’.
37 + x = 74
Subtract 37 on both sides.
37 + x – 37 = 74 – 37
x = 37
Question 8.
Write the time in words.
Three nineteen (3 : 19)
Explanation:
As the small hand is near three and the long hand is near nineteen.
Therefore, the time represents 3 : 19.
Question 9.
Name the polygon.
Explanation:
It is a quadrilateral as it has four sides and all are unequal sides.
Question 10.
True or false?
The mode is the number that occurs most often in a set of data.
True
Explanation:
Yes, it is true the number that occurs most often in a set of data is mode.
Question 11.
Imagine that you write each letter of the word CALIFORNIA on individual cards. You shuffle them, turn them facedown on a table, and turn over the top card. What is the probability of turning over a R? |
# Inverse of a Rational Function – Understanding the Basics
An inverse of a rational function is a new function that effectively reverses the original function’s inputs and outputs.
Considering a rational function, which is typically given as a fraction of two polynomials, finding its inverse involves exchanging the roles of the independent variable (usually x) and the dependent variable (usually y), and then solving for y.
The process requires ensuring that the function is one-to-one, meaning that each output is generated by exactly one input. This is crucial because a function must be one-to-one to have an inverse function that is also a function.
Working through the inverse process, I focus on the original function’s domain and range since these sets switch places in the inverse.
For the inverse function to exist, I make certain that the original function doesn’t map multiple inputs to the same output.
Having this property ensures that every output in the function’s range corresponds to one unique input within its domain, forming a mirror image across the line (y = x) when graphed. Stay tuned as we explore this intriguing world of inverses and how they unfold within the realm of rational functions.
## Understanding Inverse of Rational Functions
In my exploration of rational functions, I’ve found that these expressions are fractions where both the numerator and the denominator are polynomials.
Identifying the inverse of a rational function involves a few interesting steps that are quite intriguing. It’s like solving a puzzle where I swap the input and output to find a new function that undoes the original one.
The first thing I look at is the domain and range of the function. The domain consists of all possible input values, while the range is all possible output values.
For the inverse to exist, I need the original function to be one-to-one; each element of the domain corresponds to a unique element in the range. One aspect that’s crucial to remember is that for any rational function, the values that make the denominator equal to zero are excluded from the domain.
When I graph a rational function, I pay close attention to the asymptotes. An asymptote is a line that the graph approaches but never actually touches. I typically find two types: vertical and horizontal asymptotes.
The vertical ones are located where the denominator is zero. As for the horizontal asymptote, it gives me a good indication of the function’s end behavior, essentially telling me where the graph levels off as the inputs get very large or very small.
Here’s a simple breakdown of what to consider when finding an inverse:
• Ensure the function is one-to-one.
• Swap the variables ( x ) and ( y ) in the equation.
• Solve for ( y ) to get the inverse.
• Find the new domain and range for the inverse function.
• Look out for any new asymptotes.
By keeping these steps and features in mind, I’ve enjoyed a clearer understanding of the inverses of rational functions and how they behave.
## Finding Inverses of Rational Functions
In my journey with mathematics, I’ve found that understanding the concept of an inverse of a rational function is quite fascinating. To clarify, a function is one-to-one if each output is determined by exactly one input. Why is this important? Well, only one-to-one functions pass the horizontal line test, which is necessary for a function to be invertible.
Imagine you have a function, and you want to find its mirror image along the line ( y = x ). This reflection represents the inverse.
To ensure functions are invertible, you can perform a simple horizontal line test: if any horizontal line cuts the function’s graph more than once, then the function isn’t one-to-one and doesn’t have an inverse.
When finding the inverse, I swap the inputs and outputs, effectively interchanging the ( x ) and ( y ) in the function’s equation. This swap is followed by solving the ensuing equation for ( y ), which gives me the algebraic form of the inverse.
Here’s a step-by-step method to find the inverse algebraically:
1. Write the original function as ( y = f(x) ).
2. Interchange ( x ) and ( y ), forming the equation ( x = f(y) ).
3. Solve for ( y ), which gives the inverse function, denoted $f^{-1}(x)$.
Let’s take an example. Suppose I have the function $h(x) = \frac{x – 1}{x + 4}$. To find its inverse, I follow these steps:
1. Write it as $y = \frac{x – 1}{x + 4}$.
2. Interchange ( x ) and ( y ): $x = \frac{y – 1}{y + 4}$.
3. Solve this equation for ( y ) to obtain the inverse function.
After solving, my inverse function would be $h^{-1}(x) = \frac{4x + 1}{1 – x}$, which I can verify by the composition of the functions to see if $h(h^{-1}(x)) = x$.
By using a table, I can list down pairs of corresponding inputs and outputs for both functions to further confirm their relationship.
Here’s a tip: the inverse of a rational function might not always exist or be easily found but exploring these intricacies is all part of the fun in mathematics.
## Conclusion
In my exploration of inverse functions, I’ve established that rational functions are a pivotal topic in algebra. I’ve seen that finding the inverse of a rational function involves a series of steps that, while they might seem complex at first, become more intuitive with practice.
I understand that to find an inverse, I usually start by replacing the function notation ( f(x) ) with ( y ), and then I swap the roles of ( x ) and ( y ).
The next hurdle is to solve for ( y ) after this switch, which will give me the inverse function. This process is crucial because it reveals a relationship where each output of the original function corresponds to an input of its inverse.
The methods described in this article for finding the inverse of a rational function are systematic and applicable in various mathematical and real-life situations.
However, it’s important to remember that not all functions have an inverse, particularly when they are not one-to-one.
Summing up, mastering the calculation of an inverse function can aid in understanding how functions behave and interact. It’s a vital skill that applies to multiple areas, including geometry, physics, engineering, and economics.
Through a combination of algebraic manipulation and understanding the foundational concepts, anyone can grasp the method to find the inverse of a given rational function. |
## Engage NY Eureka Math Algebra 1 Module 5 End of Module Assessment Answer Key
### Eureka Math Algebra 1 Module 5 End of Module Assessment Task Answer Key
Question 1.
In their entrepreneurship class, students are given two options for ways to earn a commission selling cookies. For both options, students will be paid according to the number of boxes they are able to sell, with commissions being paid only after all sales have ended. Students must commit to one commission option before they begin selling.
Option 1: The commission for each box of cookies sold is 2 dollars.
Option 2: The commission will be based on the total number of boxes of cookies sold as follows: 2 cents is the total commission if one box is sold, 4 cents is the commission if two boxes are sold,
8 cents if three boxes are sold, and so on, doubling the amount for each additional box sold. (This option is based upon the total number of boxes sold and is paid on the total, not each individual box.)
a. Define the variables and write function equations to model each option. Describe the domain for each function.
Let C represent the commission for each option in dollars for Option 1 and in cents for Option 2. (Note: Students may try to use 0.02 for the exponential base but will find that the decimals present problems. They might also use 200 for Option 1 so that both can use cents as the unit. However, as long as they are careful, they can use different units for each function.)
Let x represent the number of boxes sold.
C1 = 2x (in dollars)
C2 = 2x (in cents)
Domain: Positive integers
b. If Barbara thinks she can sell five boxes of cookies, should she choose Option 1 or 2?
5 boxes
Option 1: C1 = 2(5) = 10
Option 2: C2 = 25 = 32
With Option 1, she will make $10 ; with option 2, she will make 32¢ or$0.32. She should choose Option 1 because she will make more money.
c. Which option should she choose if she thinks she can sell ten boxes? Explain.
10 boxes
Option 1: C1 = 2(10) = 20
Option 2: C2 = 210= 1024
With Option 1, she will make $20; with option 2, she will make 1024¢ or$10.24. She should still choose Option 1 because she will make more money.
d. How many boxes of cookies would a student have to sell before Option 2 pays more than Option 1? Show your work and verify your answer graphically.
Using tables: We see that at 11 boxes Option 1 is still more than Option 2, but after that it reverses.
When graphing both functions on the same coordinate plane, it is important to remember to use the same units for both equations and that the graph will be discrete.
Question 2.
The table shows the average sale price, p, of a house in New York City, for various years, t, since 1960.
a. What type of function most appropriately represents this set of data? Explain your reasoning.
Quadratic The first differences are not the same, but the second differences are the same.
b. In what year is the price at the lowest? Explain how you know.
The lowest price was when t = 5 or 1965. The lowest price in the data set is $20, 000; this is the vertex or minimum. c. Write a function to represent the data. Show your work. Answer: We use the general vertex form: f(t) = a(t – h)2+ k … f(t) =a(t – 5)2+ 20 Substituting an ordered pair that we know, (0, 45), we get 45 = a(–5)2+ 20 45 = 25a + 20 25 = 25a a = 1 So, f(t) =(t – 5)2+ 20 . d. Can this function ever be equal to zero? Explain why or why not. Answer: No, the lowest price is at the vertex:$20, 000.
e. Mr. Samuels bought his house in New York City in 1970. If the trend continued, how much was he likely to have paid? Explain and provide mathematical evidence to support your answer.
1970 would be when t = 10. If we substitute 10 into the function equation in part (c), we get f(10) = (5)2 + 20 = 45. So, he would have paid \$45, 000 for his house.
Question 3.
Veronica’s physics class is analyzing the speed of a dropped object just before it hits the ground when it is dropped from different heights. They are comparing the final velocity, in feet/second, versus the height, in feet, from which the object was dropped. The class comes up with the following graph.
a. Use transformations of the parent function, f(x)=$$\sqrt{x}_{\iota}$$, to write an algebraic equation that represents this graph. Describe the domain in terms of the context.
The graph represents a square root function. The parent function of the square root function is f(x) = $$\sqrt{x}$$. From the image of the graph, I can tell the graph has not shifted left or right but, from the points given, it has been stretched. I will use the point (4, 16) to find the symbolic representation of the graph and then use the point (49, 56) to check that my function is correct. The domain is the set of all real numbers greater than or equal to 0. However, realistically there is a limit to how big the numbers can go since there are limits to the heights from which an object can be dropped.
f(x)=$$\sqrt{ax}$$, → 16 =$$\sqrt{a 4}$$ → 16 = 2$$\sqrt{a}$$ → $$\frac{16}{2}$$ =$$\sqrt{a}$$ → 8 =$$\sqrt{a}$$ → 64 = a
(Note: Starting with f(x) = a$$\sqrt{x}$$, will reach the same end result.)
Check:
f(x)=$$\sqrt{64x}$$ → f(x) = 8$$\sqrt{x}$$ → 56 = 8$$\sqrt{49}$$ → 56 = 8(7) → 56 = 56 → f(x)= 8$$\sqrt{x}$$, (x ≥ 0)
b. Veronica and her friends are planning to go cliff diving at the end of the school year. If she dives from a position that is 165 ft. above the water, at what velocity will her body be moving right before she enters the water? Show your work and explain the level of precision you chose for your answer.
f(165) = 8$$\sqrt{165}$$ ≈ 102.8 . Her velocity will be approximately 103 ft/sec just before she enters the water. Since the information in the problem is given to the nearest whole number of feet and seconds, I decided to do the same for my answer.
c. Veronica’s friend, Patrick, thinks that if she were able to dive from a 330 ft. position, she would experience a velocity that is twice as fast. Is he correct? Explain why or why not.
He is not correct. Patrick is describing the relationship between the velocity and the height as if it were a linear function, but it is not. The graph that represents the relationship between the two is a square root, which has average rates of change on different intervals that are different from a linear function. I rounded to the nearest whole number because this is a model and only approximates a real-world phenomenon. f(330 ) = 8$$\sqrt{330}$$ = 145.3 . Her velocity would be approximately 145 ft/sec which is not double the 103 ft/sec speed calculated before.
Question 4.
Suppose that Peculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply through a mechanism in which each single bacterial cell splits into four. However, they split at different rates: Peculiar Purples split every 12 minutes, while Outrageous Oranges split every 10 minutes.
a. If the multiplication rate remains constant throughout the hour and we start with three bacterial cells of each, after one hour, how many bacterial cells will there be of each type? Show your work and explain your answer.
Let n = the number of 10 -minute or 12 -minute time intervals. Then P(n) represents the number of Purples and O(n) represents the number of Oranges at the end of any time period. The tables below show the number of bacterial cells after 1 hour for each:
b. If the multiplication rate remains constant for two hours, which type of bacteria is more abundant? What is the difference between the numbers of the two bacterial types after two hours?
Continuing the table from part (a) we find that the Oranges will have one more split in the first hour, so two more after two hours. The Oranges will have 47, 185, 920 more bacterial cells than the Purples. (At 2 hours, Oranges = 50, 331, 648 and Purples = 3, 145, 728.)
c. Write a function to model the growth of Peculiar Purples and explain what the variable and parameters represent in the context.
P(n) = 3(4n) , where n represents the number of 12 -minute splits, 3 is the initial value, and 4 is the number of Purples created for each split.
d. Use your model from part (c) to determine how many Peculiar Purples there will be after three splits (i.e., at time 36 minutes). Do you believe your model has made an accurate prediction? Why or why not?
P(3) = 3(43) = 3(64) = 192. Yes, this matches the values I found in the table for part (a).
e. Write an expression to represent a different type of bacterial growth with an unknown initial quantity but in which each cell splits into two at each interval of time.
F(n) = a(2n), where n represents the number of time interval and a represents the initial number of bacterial cells.
Question 5.
In a study of the activities of dolphins, a marine biologist made a slow-motion video of a dolphin swimming and jumping in the ocean with a specially equipped camera that recorded the dolphin’s position with respect to the slow-motion time in seconds. Below is a piecewise quadratic graph, made from the slow-motion dolphin video, which represents a dolphin’s vertical height (in feet, from the surface of the water) while swimming and jumping in the ocean, with respect to the slow-motion time (in seconds). Use the graph to answer the questions. (Note: The numbers in this graph are not necessarily real numbers from an actual dolphin in the ocean.)
a. Given the vertex (11, -50), write a function to represent the piece of the graph where the dolphin is underwater. Identify your variables and define the domain and range for your function.
Using the vertex form for a quadratic function equation: h(t) = a(t – h)2 + k , we know the vertex (h, K) to be (11, -50) . Now, to find the leading coefficient, we can substitute a point we know, say (6, 0) , and solve for a :
0 = a (6 – 11 )2 – 50
a (- 5) 2 = 50
25a = 50
a = 2
So, h(t) = 2(t – 11) 2 – 50
Domain (interval of time in seconds): [6, 16]
Range (distance from the surface): [-50, 0]
b. Calculate the average rate of change for the interval from 6 to 8 seconds. Show your work and explain what your answer means in the context of this problem.
Average rate of change:
$$\frac{h(8)-h(6)}{8-6}$$ = $$\frac{-32-0}{2}$$ = -16
The dolphin is moving downward at an average rate of 16 feet per second.
c. Calculate the average rate of change for the interval from 14 to 16 seconds. Show your work and explain what your answer means in the context of this problem.
Average rate of change:
$$\frac{h(16)-h(14)}{16-14}$$ = $$\frac{0- -32}{2}$$ = + 16
The dolphin is moving upward at a rate of 16 feet per second.
d. Compare your answers for parts (b) and (c). Explain why the rates of change are different in the context of the problem.
The two average rates show that the dolphin’s rate is the same for each interval except that in the first it is moving downward and in the second upward. They are different because of the symmetric nature of the quadratic graph. The intervals chosen are symmetric, so they will have the same y-values.
Question 6.
The tables below represent values for two functions, f and g, one absolute value and one quadratic.
a. Label each function as either absolute value or quadratic. Then explain mathematically how you identified each type of function.
In the first table (f(x)) the rates on any interval on the same side of the vertex (0, 0) is – $$\frac{1}{2}$$ , and on the other side of the vertex the rates of change are all $$\frac{1}{2}$$ .
In the second table (g(x)) the rates vary on each side of the vertex: The intervals closest to the vertex have average rates of – $$\frac{1}{2}$$ and $$\frac{1}{2}$$, but the next intervals have –1.5 and +1.5, then -2.5 and +2.5, etc.
For this reason, the first is absolute value (linear piecewise) and the second is quadratic.
b. Represent each function graphically. Identify and label the key features of each in your graph (e.g., vertex, intercepts, axis of symmetry).
a = $$\frac{1}{2}$$
So, g (x ) = $$\frac{1}{2}$$ x 2 . |
# Formula of Kinetic Energy
Kinetic energy refers to a type of energy that an object or body possesses by virtue of its motion. It is possessed by every particle performing any kind of motion. Kinetic energy is a property of a moving body or object that does not depend on motion only, but it also depends upon the body's mass. Mechanical energy can be divided into two different types, the first one is Kinetic energy, and the second one is potential energy. Potential energy refers to a type of energy that an object or body possessed by virtue of its position. For example, when an object is a free fall, the object possesses kinetic energy. Kinetic energy can be due to rotation, translation, and vibration. We can calculate KE using the mass and velocity of that body.
In this article, we will discuss the formula to calculate the kinetic energy with certain examples.
## Method 1:
### Understanding Kinetic Energy
By knowing the Formula
The formula using for calculating the Kinetic energy is given below
KE = ½ mv2
Where
m = mass of an object or body.
v = velocity of an object or body.
Let's understand what is the mass and what is the velocity of an object.
Mass: Mass of an object or body refers to how much matter is in the object. It is denoted by m. Unit of mass is the joule.
Velocity: The velocity of an object refers to the rate at which the object changes its position. It is denoted by v. Unit of velocity is m/s.
Calculate the mass of an object:
When we solve a problem where the mass is not given, then first, you need to calculate the mass to proceed. This can be done by obtaining the mass in kg.
Find the velocity of an object:
You often face the problem while solving any problem, it does not give you the velocity of an object, but distance and time are given. Here, you need to calculate the velocity by applying the given formula.
Velocity = Distance/time = d/t
Example 1: If an object travels a distance of 8 meters in 2 sec, find the velocity of an object?
Solution:
Given,
Distance = 8 meter
Time = 2 sec
We know that,
V = distance/ time = 8/2 = 4 m/sec
Example 2: If an object travels a distance of 10 meters in 5 sec, find the velocity of an object?
Solution:
Given,
Distance = 10 meter
Time = 5 sec
We know that,
V = distance/ time = 10/5 = 2 m/sec
## Method 2
### Finding the Kinetic Energy
Write the equation:
The formula used to calculate the Kinetic energy is given below.
KE = ½ mv2
Where
m = mass of an object or body.
v = velocity of an object or body.
The value of KE should always be in joules J, which is the standard unit of measurement of KE.
Put the value of mass and velocity.
To calculate the KE, you need to put the value of mass and velocity in the formula. If the value of mass and velocity are not given in the questions, you need to calculate it.
Example 1:
Find the KE of an 80 kg men running with a velocity of 8 m/s.
Solution:
Given;
Mass = 80 kg
Velocity = 8 m/s
We know that,
KE = ½ mv2
KE = ½ × 80 × 82
= ½ × 80 × 64
= 40 × 64
= 256 J
Example 2:
Find the KE of a 120 kg man running with a velocity of 10 m/s.
Solution:
Given;
Mass = 120 kg
Velocity = 10 m/s
We know that,
KE = ½ mv2
KE = ½ × 120 × 102
= ½ × 120 × 100
= 60× 100
= 6000 J
Using KE calculate velocity or Mass
The formula using for calculating the Kinetic energy is given below
KE = ½ mv2
Where,
m = mass of an object or body
v = velocity of an object or body.
Put the Unknown Variables:
In some questions, you have the value of KE and mass. In that case, you need to calculate the value of the velocity.
Let's understand this concept with the help of an example
Example 1: What is the velocity of a body with a mass of 50 kg, and the value of kinetic energy is 2500 J?
Solution:
Given;
mass = 50 kg
Kinetic Energy = 2500 J
We know that
KE = ½ mv2
2500 = ½ × 50 × v2
v2 = 2500/25
v2 = 100
v = 10
Example 2: What is the velocity of a body with a mass of 60 kg and the value of kinetic energy is 480 J?
Solution:
Given;
mass = 60 kg
Kinetic Energy = 480 J
We know that
KE = ½ mv2
480 = ½ × 60 × v2
v2 = 480/30
v2 = 16
v = √16 = 4
Let's understand this concept with the help of an example
Suppose you have the value of KE and velocity. In that case, you need to find the value of the mass.
Example 1: Find the mass of an object with a kinetic energy of 500 J and a velocity of 10 m/s?
Solution:
Given;
Kinetic energy = 500 J
Velocity = 10 m/s
We know that
KE = ½ mv2
500 = ½ × m ×10 2
500 = ½ × m ×100
500 = 50 × m
m = 10 kg
Example 2: Find the mass of an object with a kinetic energy of 600 J and a velocity of 20 m/s?
Solution:
Given;
Kinetic energy = 600 J
Velocity = 20 m/s
We know that
KE = ½ mv2
600 = ½ × m ×20 2
500 = ½ × m ×400
500 = 200 × m
m = 2.5 kg
### Questions based on kinetic energy
Question 1: If a football has a mass of 3 kg, travels with a speed of 12 m/s, calculate the kinetic energy possessed by it.
Solution:
Given;
Mass = 3 kg
Velocity = 12 m/s
We know that,
Kinetic Energy = ½ mv2
KE = ½ ×3× (12)2
= ½ ×3× 144
= 216 J
Question 2: If a basketball has a mass of 4 kg, travels with a speed of 16 m/s, calculate the kinetic energy possessed by it.
Solution:
Given;
Mass = 4 kg
Velocity = 16 m/s
We know that,
Kinetic Energy = ½ mv2
KE = ½ ×4× (16)2
= ½ ×4× 256
= 512 J
Question 3: If a 12 kg object was traveling at a speed of 120 m/s. Now, this mass transfers all its energy to a mass of 24 k. What will be the velocity of the 24 kg mass after being hit by it?
Solution:
We know that,
KE = ½ mv2
First, we will calculate the KE of the heavier object
Given,
M = 12 kg and v = 120 m/s
We know that,
KE = ½ mv2
= ½ × 12 × (120)2
= ½ × 12 × 14400
= 86400 J
Now, this energy is transferred to another ball
Given;
Mass = 24 Kg
Velocity =?
We have calculated the value of kinetic energy
KE = 86400
So,
We know that,
KE = ½ mv2
86400 = ½ (24) v2
v2 = 86400/12 = 7200
v = 60 m/s
Question 4: If an 8 kg object was traveling at a speed of 80 m/s. Now, this mass transfers all its energy to a mass of 16 k. What will be the velocity of the 16 kg mass after being hit by it?
Solution:
We know that,
KE = ½ mv2
First, we will calculate the KE of the heavier object
Given,
M = 8 kg and v = 80 m/s
We know that,
KE = ½ mv2
= ½ × 8 × (80)2
= ½ × 8 × 6400
= 25600 J
Now, this energy is transferred to another ball
Given;
Mass = 16 Kg
Velocity =?
We have calculated the value of kinetic energy
KE = 25600
So,
We know that,
KE = ½ mv2
25600 = ½ (16) v2
v2 = 25600/16 = 16
v = 4 m/s
Next TopicAnime.js |
Newton's method for finding roots
This is an iterative method invented by Isaac Newton around 1664. However, sometimes this method is called the Raphson method, since Raphson invented the same algorithm a few years after Newton, but his article was published much earlier.
The task is as follows. Given the following equation:
$$f(x) = 0$$
We want to solve the equation, more precisely, to find one of its roots (it is assumed that the root exists). It is assumed that $f(x)$ is continuous and differentiable on an interval $[a; b]$.
Algorithm
The input parameters of the algorithm consist of not only the function $f(x)$ but also the initial approximation - some $x_0$, with which the algorithm starts.
Suppose we have already calculated $x_i$, calculate $x_{i+1}$ as follows. Draw the tangent to the graph of the function $f(x)$ at the point $x = x_i$, and find the point of intersection of this tangent with the $x$-axis. $x_{i+1}$ is set equal to the $x$-coordinateof the point found, and we repeat the whole process from the beginning.
It is not difficult to obtain the following formula:
$$x_{i+1} = x_i - \frac{f(x_i)}{f^\prime(x_i)}$$
It is intuitively clear that if the function $f(x)$ is "good" (smooth), and $x_i$ is close enough to the root, then $x_{i+1}$ will be even closer to the desired root.
The rate of convergence is quadratic, which, conditionally speaking, means that the number of exact digits in the approximate value $x_i$ doubles with each iteration.
Application for calculating the square root
Let's use the calculation of square root as an example of Newton's method.
If we substitute $f(x) = \sqrt{x}$, then after simplifying the expression, we get:
$$x_{i+1} = \frac{x_i + \frac{n}{x_i}}{2}$$
The first typical variant of the problem is when a rational number $n$ is given, and its root must be calculated with some accuracy eps:
double sqrt_newton(double n) {
const double eps = 1E-15;
double x = 1;
for (;;) {
double nx = (x + n / x) / 2;
if (abs(x - nx) < eps)
break;
x = nx;
}
return x;
}
Another common variant of the problem is when we need to calculate the integer root (for the given $n$ find the largest $x$ such that $x^2 \le n$). Here it is necessary to slightly change the termination condition of the algorithm, since it may happen that $x$ will start to "jump" near the answer. Therefore, we add a condition that if the value $x$ has decreased in the previous step, and it tries to increase at the current step, then the algorithm must be stopped.
int isqrt_newton(int n) {
int x = 1;
bool decreased = false;
for (;;) {
int nx = (x + n / x) >> 1;
if (x == nx || nx > x && decreased)
break;
decreased = nx < x;
x = nx;
}
return x;
}
Finally, we are given the third variant - for the case of bignum arithmetic. Since the number $n$ can be large enough, it makes sense to pay attention to the initial approximation. Obviously, the closer it is to the root, the faster the result will be achieved. It is simple enough and effective to take the initial approximation as the number $2^{\textrm{bits}/2}$, where $\textrm{bits}$ is the number of bits in the number $n$. Here is the Java code that demonstrates this variant:
public static BigInteger isqrtNewton(BigInteger n) {
BigInteger a = BigInteger.ONE.shiftLeft(n.bitLength() / 2);
boolean p_dec = false;
for (;;) {
For example, this code is executed in $60$ milliseconds for $n = 10^{1000}$, and if we remove the improved selection of the initial approximation (just starting with $1$), then it will be executed in about $120$ milliseconds. |
# How to multiply
Multiplication is one of the four basic arithmetic operations, with the other three being subtraction, addition, and division. Learning how to multiply is a necessary aspect of studying mathematics. For whole numbers, it can be thought of as repeated addition. Learning how to multiply largely involves memorizing a multiplication chart, also referred to as a times table or multiplication table. For larger values, a multiplication algorithm, sometimes referred to as "long multiplication," can be used.
## Multiplication symbols
Multiplication is an operation that, unlike addition and subtraction, can be indicated in a number of ways. The following are all multiplication symbols: ×, *, ·. Furthermore, multiplication can also be indicated using parenthesis; if there is no operator separating two numerals in parenthesis, the assumed operation is multiplication.
2 × 2 = 4
2 * 2 = 4
2 · 2 = 4
(2)(2) = 4
Using parenthesis to indicate multiplication can be particularly helpful for negative values:
(-2)(-2) = 4
The above example makes it clear that the -2's are being multiplied. Otherwise, something like -2 · -2 could be mistaken for a subtraction problem.
Multiplying whole numbers can be thought of as performing repeated addition, which is simply adding equal groups of objects together a given number of times. For example, the multiplication problem
2 × 4
can be read as "two groups of four," meaning that we would add a set of 4 objects twice to find the result. It is also equivalent to four groups of two.
Referencing the above figure,
4 × 2 = 4 + 4 = 8
2 × 4 = 2 + 2 + 2 + 2 = 8
In both cases, we can perform the multiplication problem by repeatedly adding. For larger values, or when multiplying decimals, we can use long multiplication.
## Long multiplication
Long multiplication is a multiplication algorithm we can use to multiply larger numbers or decimal numbers once we've memorized the multiplication chart. Use the following steps along with the example below to understand the process.
1. Write the numerals being multiplied, aligning their ones places, with the largest numeral on top. Draw a line below the numerals being multiplied.
2. Multiply the digit in the ones place in the bottom numeral by the digit in the ones place of the top numeral. Write the result below the line. If the product of the column is greater than 9, write the ones-place digit of the result below the line in the same column, and the tens-place digit above the top of the following column.
3. Continue the process, multiplying the digit in the ones place in the bottom numeral with the digit in each column of the top numeral, moving from right to left; if there is a digit above the column, add it after multiplying the appropriate digits. The product of the last 2 digits, even if greater than 9, is simply written below the line in the appropriate result row.
4. If the bottom numeral in the multiplication problem has more than one digit, continue the same process above for each digit, aligning the ones place of the result with the tens place of the previous result (shift the result one place value to the left).
5. Once each digit in the top and bottom numerals have been multiplied, add the results of each row vertically. Refer to the addition algorithm if necessary.
Example
Multiply 1437 × 28:
The problem above is separated just to make multiplication of the rows, as well as the carried digits, clearer. On top, shown in blue, is the first row of the result, obtained by multiplying 8 by each digit in the top numeral, and following the steps detailed above.
The bottom half shows the same process to multiply the digits shown in green.
The last line in black is the result of adding the blue and green rows, and is the solution. Therefore:
1437 × 28 = 40,236
## How to multiply fractions
Multiplying fractions is similar to multiplying whole numbers, except that we need to pay attention to which digits we multiply. When multiplying fractions, multiply the numerators of all the fractions involved. Then multiply the denominators of all the fractions involved. The solution is the product of the numerators over the product of the denominators.
Example
Multiply :
In the above example, we simplified the problem as we multiplied. Otherwise, we would've had to simplify the fraction after finding the solution. |
### Home > GB8I > Chapter 3 Unit 4 > Lesson INT1: 3.2.3 > Problem3-101
3-101.
Write an equation for each situation below showing that the area as a product is equal to the area as a sum.
1. $12x$ $+1$ $x$ $-5$
• Review the Math Notes box in Lesson 3.2.2.
$(12x+1)(x−5)=12x^2−59x−5$
1. $2m^2$ $-4m$ $-1$ $3m$ $+5$
• See part (a).
1. $(2x+5)(x+6)$
• Draw a generic rectangle and label the length and width.
$(2x + 5)(x + 6) = 2x^2 + 17x + 30$
$\color{red}{2x}$ $\color{red}{+5}$ $\color{red}{x}$ $\color{red}{+6}$
$\color{red}{2x}$ $\color{red}{+5}$ $\color{red}{x}$ $\color{red}{2x^2}$ $\color{red}{5x}$ $\color{red}{+6}$ $\color{red}{12x}$ $\color{red}{30}$
Fill in the grid by multiplying the length and width of each box.
Write an equation showing the product is the sum of the parts.
Area as a sum $= 2x^2 + 12x + 5x + 30 = 2x^2 + 17x + 30$
Area as a product = $(2x + 5)(x + 6)$
1. $( 3 - 5 y ) ( 2 + y )$
• See the help for part (c). |
Mathematics » Functions III » Exponential Functions
# Logarithms
## Logarithms
### Definition: Logarithm
If $$x = {b}^{y}$$, then $$y = {\log}_{b}(x)$$, where $$b>0$$, $$b \ne 1$$ and $$x>0$$.
Note that the brackets around the number $$(x)$$ are not compulsory, we use them to avoid confusion.
The logarithm of a number $$(x)$$ with a certain base $$(b)$$ is equal to the exponent $$(y)$$, the value to which that certain base must be raised to equal the number $$(x)$$.
For example, $$\log_{2}(8)$$ means the power of $$\text{2}$$ that will give $$\text{8}$$. Since $${2}^{3}=8$$, we see that $${\log}_{2}(8)=3$$. Therefore the exponential form is $${2}^{3}=8$$ and the logarithmic form is $${\log}_{2}{8}=3$$.
### Restrictions on the definition of logarithms
$\begin{array}{rll} \text{Restriction:}& & \text{Reason: } \\ & & \\ b > 0 & & \text{If } b \text{ is a negative number. then } b^{y} \text{ will oscillate between:} \\ & & \text{positive values if } y \text{ is even } \\ & & \text{negative values if } y \text{ is odd } \\ & & \\ b \ne 1 & & \text{Since } 1^{\text{(any value)}} = 1 \\ & & \\ x > 0 & & \text{Since } \text{(positive number)}^{\text{(any value)}} > 0\end{array}$
## Optional Investigation: Exponential and logarithmic form
Discuss the following statements and determine whether they are true or false:
1. $$p = a^{n}$$ is the inverse of $$p = \log_{a}{n}$$.
2. $$y = 2^{x}$$ is a one-to-one function, therefore $$y = \log_{2}{x}$$ is also a one-to-one function.
3. $$x = \log_{5}{y}$$ is the inverse of $$5^{x} = y$$.
4. $$k = b^{t}$$ is the same as $$t = \log_{b}{k}$$.
### To determine the inverse function of $$y=b^{x}$$:
$\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = b^{y} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & y = \log_{b}{x} \end{array}$
Therefore, if we have the exponential function $$f(x) = b^{x}$$, then the inverse is the logarithmic function $$f^{-1}(x) = \log_{b}{x}$$.
The “common logarithm” has a base $$\text{10}$$ and can be written as $$\log_{10}{x} = \log{x}$$. In other words, the $$\log$$ symbol written without a base is interpreted as the logarithm to base $$\text{10}$$. For example, $$\log{\text{25}} = \log_{10}{\text{25}}$$.
## Example
### Question
Write the following exponential expressions in logarithmic form and express each in words:
1. $$5^{2} = 25$$
2. $$10^{-3} = \text{0.001}$$
3. $$p^{x} = q$$
### Determine the inverse of the given exponential expressions
Remember: $$m = a^{n}$$ is the same as $$n = \log_{a}{m}$$.
1. $$2 = \log_{5}{25}$$
2. $$-3 = \log_{10}{(\text{0.001})}$$
3. $$x = \log_{p}{q}$$
### Express in words
1. $$\text{2}$$ is the power to which $$\text{5}$$ must be raised to give the number $$\text{25}$$.
2. $$-\text{3}$$ is the power to which $$\text{10}$$ must be raised to give the decimal number $$\text{0.001}$$.
3. $$x$$ is the power to which $$p$$ must be raised to give $$q$$.
## Example
### Question
Write the following logarithmic expressions in exponential form:
1. $$\log_{2}{128} = 7$$
2. $$-2 = \log_{3}{( \cfrac{1}{9} )}$$
3. $$z = \log_{w}{k}$$
### Determine the inverse of the given logarithmic expressions
For $$n = \log_{a}{m}$$, we can write $$m = a^{n}$$.
1. $$2^{7} = \text{128}$$
2. $$3^{-2} = \cfrac{1}{9}$$
3. $$w^{z} = k$$
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# AREA OF RECTANGLE WORKSHEET
1. Find the area of a rectangle, if its length is 15 cm and width is 20 cm.
2. The length of a rectangle is 3 times its width. If its perimeter is 32 ft., then find the area of the rectangle.
3. If the length of each diagonal of a rectangle is 13 cm and its width is 12 cm, then find the area of the rectangle.
4. The difference between the length and width of a rectangle is 1 cm. If the length of one of the diagonals is 5 cm, then find the area of the rectangle.
5. The length and width of a rectangle are in the ratio 3 : 4 and its area is 588 square inches. Find its length and width.
6. Find the cost of carpeting the floor of a room whose length is 13 m and width is 9 m. If the cost of the carpet is \$12.40 per square meter, find the total cost of carpeting the floor of the room.
7. The length of a rectangle is 70 cm and width is 30 cm. If the length is increased by 10% and width is by 20%, then find the percentage increase in area.
8. The length of a rectangle is 80 cm and width is 40 cm. If the length is increased by 20% and width is decreased by 10%, By what percentage will the area be increased or decreased ?
Formula for area of a rectangle :
l ⋅ w
Substitute 15 for l and 20 for w.
= 15 ⋅ 20
= 300 cm2
Let x be the width of the rectangle.
Then, its length is 3x.
Perimeter of the rectangle is 32 ft
2(l + w) = 32
Divide each side by 2.
l + w = 16
Substitute 3x for l and x for w.
3x + x = 16
4x = 16
Divide each side by 4.
x = 4
Therefore, width of the rectangle is 4 ft.
And length of the rectangle is
= 3(4)
= 12 ft
Formula for area of a rectangle :
l ⋅ w
Substitute 12 for l and 4 for w.
= 12 ⋅ 4
= 48 ft2
To find the area of a rectangle, we have to know its length and width. Width is given in the question, that is 12 cm. So, find its length.
Draw a sketch.
In the figure shown above, consider the right triangle ABC.
By Pythagorean Theorem, we have
AB2 + BC2 = AC2
Substitute.
122 + l2 = 132
Simplify and solve for l.
144 + l2 = 169
Subtract 144 from each side.
l2 = 25
Find positive square root on both sides.
l2 = 25
l = 5
Therefore, the length of the rectangle is 5 cm.
Formula for area of a rectangle :
l ⋅ w
Substitute 5 for l and 12 for w.
= 5 ⋅ 12
= 60 cm2
Because the difference between the length and width of a rectangle is 1 cm, we can assume the length of the rectangle as x cm and width as (x + 1) cm.
Draw a sketch.
In the figure shown above, consider the right triangle ABC.
By Pythagorean Theorem, we have
AB2 + BC2 = AC2
Substitute.
(x + 1)2 + x2 = 52
Simplify and solve for x.
x2 + 2 ⋅ x ⋅ 1 + 12 + x2 = 25
x2 + 2x + 1 + x2 = 25
2x2 + 2x + 1 = 25
Subtract 25 from each side.
2x2 + 2x - 24 = 0
Divide each side by 2.
x2 + x - 12 = 0
Factor.
(x + 4)(x -3) = 0
x + 4 = 0 or x - 3 = 0
x = - 4 or x = 3
Because the length of a rectangle can not be negative, we can ignore x = - 4.
So, length is 3 cm.
Then the width is
= x + 1
= 3 + 1
= 4 cm
Formula for area of a rectangle :
l ⋅ w
Substitute 3 for l and 4 for w.
= 3 ⋅ 4
= 12 cm2
From the ratio 3 : 4, let the length and width of the rectangle be 3x and 4x respectively.
Area of the rectangle = 588 in2
l ⋅ w = 588
Substitute 3x for l and 4x for w.
3x ⋅ 4x = 588
12x2 = 588
Divide each side by 12.
x2 = 49
Find positive square root on both sides.
√x2 = √49
x = 7
Length = 3x = 3(7) = 21 in
Width = 4x = 4(7) = 28 in
To find the total cost of carpeting the floor of the room, we have to know its area. Because the floor of the room is in rectangle shape, we can use the formula for area of a rectangle to find the area of the floor.
Formula for area of a rectangle :
l ⋅ w
Substitute 13 for l and 9 for w.
= 13 ⋅ 9
= 117
So, the area of the floor is 117 square meters.
The cost of carpet is \$12.40 per square meter.
Then, the total cost of carpet for 117 square meters :
= 117 ⋅ 12.40
= \$1450.80
Before increase in length and width :
Formula for area of a rectangle :
= l ⋅ w
Substitute 70 for l and 30 for w.
= 70 ⋅ 30
= 2100
Therefore, the area of the rectangle is 2100 square cm.
After increase in length and width :
Length = (100 + 10)% of 70 = 1.1 ⋅ 70 = 77 cm
Width = (100 + 20)% of 30 = 1.2 ⋅ 30 = 36 cm
Formula for area of a rectangle :
= l ⋅ w
Substitute 77 for l and 36 for w.
= 77 ⋅ 36
= 2772 cm2
Percentage increase in area :
Increase in area = 2772 - 2100
= 672 cm2
Percentage increase in area = (672 / 2100) ⋅ 100 %
= 32%
Before changes in length and width :
Formula for area of a rectangle :
= l ⋅ w
Substitute 80 for l and 40 for w.
= 80 ⋅ 40
= 3200
Therefore, the area of the rectangle is 3200 square cm.
After changes in length and width :
Length = (100 + 20)% of 80 = 1.2 ⋅ 80 = 96 cm
Width = (100 - 10)% of 40 = 0.9 ⋅ 40 = 36 cm
Formula for area of a rectangle :
= l ⋅ w
Substitute 96 for l and 36 for w.
= 96 ⋅ 36
= 3456 cm2
Difference in area :
= 3456 - 3200
= 256 cm2
After 20% increase in length and 10% decrease in width, the area of the rectangle is increase by 256 square cm.
Percentage increase in area :
= (256/3200) ⋅ 100 %
= 8%
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# A square plate $0.1m$ side moves parallel to a second plate with a velocity of $0.1m{s^{ - 1}}$. Both plates are immersed in water. If the viscous force is $0.002N$ and the coefficient of viscosity $0.001poise$, distance between the plates is:(A) $0.1m$(B) $0.05m$(C ) $0.005m$(D) $0.0005m$
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Hint:We have the values of all required quantities in the question. We would calculate the area of the square plate at first as the dimension of one of the sides is given.
Using that value of area, and the formula for viscous force we can obtain the required value of distance between the plates.
As per the question, the following values of given to us:
Side of the plate $= 0.1m = a$
Velocity of the plate $= dv = 0.1m{s^{ - 1}}$
Viscous force $= F = 0.002N$
Coefficient of viscosity $= 0.001poise$
From here, we can calculate the area of the plate is $= {a^2} = (0.1 \times 0.1){m^2}$
Thus, area $= 0.01{m^2}$
Coefficient of Viscosity $= \eta = 0.001poise$
But, we need to take the value in decaPoise, therefore we divide it by $10$
Therefore, $\eta = \dfrac{{0.001}}{{10}}decapoise = 0.0001decapoise$
We have the velocity of the plate given,
Now, using the expression for viscous force, we get:
$F = \eta A\dfrac{{dv}}{{dx}}$
Where:
$F =$ Viscous force
$\eta =$ Coefficient of viscosity
$A =$ Area of the plate
$dv =$ Velocity with which the plate moves
$dx =$ Distance between the plates.
For the above expression, we need to find the value of $dx$.
Putting the values, in the expression we get:
$0.002 = \dfrac{{0.0001 \times 0.01 \times 0.1}}{{dx}}$
Rearranging the equation we get:
$dx = \dfrac{{0.0001 \times 0.01 \times 0.1}}{{0.002}}$
Thus, we obtain:
$dx = 0.00005m$
This is the required solution.
Note:The unit poise is in the MKS unit, therefore, we need to convert it to deca poise, and otherwise we may get erroneous results. Viscosity is defined as a resistance experienced by a fluid while it flows. This is a property of the fluid. Viscous force is defined as the force between a body and the fluid, while it flows. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 10.9: Vectors
Skills to Develop
• View vectors geometrically.
• Find magnitude and direction.
• Perform vector addition and scalar multiplication.
• Find the component form of a vector.
• Find the unit vector in the direction of v.
• Perform operations with vectors in terms of i and j.
• Find the dot product of two vectors.
An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 10.9.1. What are the ground speed and actual bearing of the plane?
Figure 10.9.1
Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s ground speed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors.
### A Geometric View of Vectors
A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:
• Lower case, boldfaced type, with or without an arrow on top such as $$u$$, $$w$$, $$\overrightarrow{v}$$, $$\overrightarrow{u}$$, $$\overrightarrow{w}$$.
• Given initial point $$P$$ and terminal point $$Q$$, a vector can be represented as $$\overrightarrow{PQ}$$ . The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
• Given an initial point of $$(0,0)$$ and terminal point $$(a,b)$$, a vector may be represented as $$⟨a,b⟩$$.
This last symbol $$⟨a,b⟩$$ has special significance. It is called the standard position. The position vector has an initial point $$(0,0)$$ and a terminal point $$⟨a,b⟩$$. To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector $$\overrightarrow{CD}$$ is $$C(x_1,y_1)$$ and the terminal point is $$D(x_2,y_2)$$, then the position vector is found by calculating
\begin{align} \overrightarrow{AB} &= ⟨x_2−x_1,y_2−y_1⟩ \\ &= ⟨a,b⟩ \end{align}
In Figure 10.9.2, we see the original vector $$\overrightarrow{CD}$$ and the position vector $$\overrightarrow{AB}$$.
Figure 10.9.2
A General Note: PROPERTIES OF VECTORS
A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at $$(0,0)$$ and is identified by its terminal point $$⟨a,b⟩$$.
Example
Find the Position Vector
Consider the vector whose initial point is $$P(2,3)$$ and terminal point is $$Q(6,4)$$. Find the position vector.
Solution:
The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus
\begin{align} v &= ⟨6−2,4−3⟩ \\ &=⟨4,1⟩ \end{align}
The position vector begins at $$(0,0)$$ and terminates at $$(4,1)$$. The graphs of both vectors are shown in Figure 10.9.3.
Figure 10.9.3
We see that the position vector is $$⟨4,1⟩$$.
Example
Drawing a Vector with the Given Criteria and Its Equivalent Position Vector
Find the position vector given that vector $$v$$ has an initial point at $$(−3,2)$$ and a terminal point at $$(4,5)$$, then graph both vectors in the same plane.
Solution:
The position vector is found using the following calculation:
\begin{align} v &= ⟨4−(−3),5−2⟩ \\ &= ⟨7,3⟩ \end{align}
Thus, the position vector begins at $$(0,0)$$ and terminates at $$(7,3)$$. See Figure 10.9.4.
Figure 10.9.4
Exercise
Draw a vector $$v$$ that connects from the origin to the point $$(3,5)$$.
Solution:
### Finding Magnitude and Direction
To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.
A General Note: MAGNITUDE AND DIRECTION OF A VECTOR
Given a position vector $$v=⟨a,b⟩$$,the magnitude is found by $$| v |=\sqrt{a^2+b^2}$$.The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by $$\tan \theta=(\dfrac{b}{a})⇒\theta={\tan}^{−1}(\dfrac{b}{a})$$, as illustrated in Figure 10.9.5.
Figure 10.9.5
Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.
Example
Finding the Magnitude and Direction of a Vector
Find the magnitude and direction of the vector with initial point $$P(−8,1)$$ and terminal point $$Q(−2,−5)$$.Draw the vector.
Solution:
First, find the position vector.
\begin{align} u &= ⟨−2,−(−8),−5−1⟩ \\ &= ⟨6,−6⟩ \end{align}
We use the Pythagorean Theorem to find the magnitude.
\begin{align} |u| &= \sqrt{{(6)}^2+{(−6)}^2} \\ &= \sqrt{72} \\ &=\sqrt{62} \end{align}
The direction is given as
\begin{align} \tan \theta & =\dfrac{−6}{6}=−1\rightarrow \theta={\tan}^{−1}(−1) \\ &= −45° \end{align}
However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, $$−45°+360°=315°$$. See Figure 10.9.6.
Figure 10.9.6
Example
Showing That Two Vectors Are Equal
Show that vector v with initial point at $$(5,−3)$$ and terminal point at $$(−1,2)$$ is equal to vector u with initial point at $$(−1,−3)$$ and terminal point at $$(−7,2)$$. Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector.
Solution:
As shown in Figure 10.9.7, draw the vector $$v$$ starting at initial $$(5,−3)$$ and terminal point $$(−1,2)$$. Draw the vector $$u$$ with initial point $$(−1,−3)$$ and terminal point $$(−7,2)$$. Find the standard position for each.
Next, find and sketch the position vector for v and u. We have
\begin{align} v &= ⟨−1−5,2−(−3)⟩ \\ &= ⟨−6,5⟩u \\ &= ⟨−7−(−1),2−(−3)⟩ \\ & =⟨−6,5⟩ \end{align}
Since the position vectors are the same, v and u are the same.
An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.
\begin{align} |v| &= \sqrt{{(−1−5)}^2+{(2−(−3))}^2} \\ &= \sqrt{{(−6)}^2+{(5)}^2} \\ &= \sqrt{36+25} \\ &= \sqrt{61} \\ |u| &= \sqrt{{(−7−(−1))}^2+{(2−(−3))}^2} \\ &=\sqrt{{(−6)}^2+{(5)}^2} \\ &= \sqrt{36+25} \\ &= \sqrt{61} \end{align}
As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives
\begin{align} \tan \theta &= −\dfrac{5}{6}⇒\theta={\tan}^{−1}(−\dfrac{5}{6}) \\ & = −39.8° \end{align}
However, we can see that the position vector terminates in the second quadrant, so we add $$180°$$. Thus, the direction is $$−39.8°+180°=140.2°$$.
Figure 10.9.7
### Performing Vector Addition and Scalar Multiplication
Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector $$u=⟨x,y⟩$$ as an arrow or directed line segment from the origin to the point $$(x,y)$$, vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector uv, the resultant vector.
To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 10.9.8.
Figure 10.9.8
Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 10.9.9 for a visual that compares vector addition and vector subtraction using parallelograms.
Figure 10.9.9
Example
Adding and Subtracting Vectors
Given $$u=⟨3,−2⟩$$ and $$v=⟨−1,4⟩$$, find two new vectors u + v, and u − v.
Solution:
To find the sum of two vectors, we add the components. Thus,
\begin{align} u+v &= ⟨3,−2⟩+⟨−1,4⟩ \\ &= ⟨3+(−1),−2+4⟩ \\ &=⟨2,2⟩ \end{align}
See Figure 10.9.10 (a).
To find the difference of two vectors, add the negative components of v to u. Thus,
\begin{align}u+(−v) &=⟨3,−2⟩+⟨1,−4⟩ \\ &= ⟨3+1,−2+(−4)⟩ \\ &= ⟨4,−6⟩ \end{align}
See Figure 10.9.10 (b).
Figure 10.9.10 (a) Sum of two vectors (b) Difference of two vectors
# Multiplying By a Scalar
While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.
A General Note: SCALAR MULTIPLICATION
Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply $$v=⟨a,b⟩$$ by $$k$$, we have
$$kv=⟨ka,kb⟩$$
Only the magnitude changes, unless $$k$$ is negative, and then the vector reverses direction.
Example
Performing Scalar Multiplication
Given vector $$v=⟨3,1⟩$$, find 3v, $$\dfrac{1}{2}$$, and −v.
Solution:
See Figure 10.9.11 for a geometric interpretation. If $$v=⟨3,1⟩$$, then
\begin{align} 3v &= ⟨3⋅3,3⋅1⟩ \\ &= ⟨9,3⟩ \\ \dfrac{1}{2}v &= ⟨\dfrac{1}{2}⋅3,\dfrac{1}{2}⋅1⟩ \\ &=⟨\dfrac{3}{2},\dfrac{1}{2}⟩ \\ −v &=⟨−3,−1⟩ \end{align}
Figure 10.9.11
Analysis
Notice that the vector 3v is three times the length of v, $$\dfrac{1}{2}v$$ is half the length of v, and –v is the same length of v, but in the opposite direction.
Exercise
Find the scalar multiple $$3u$$ given $$u=⟨5,4⟩$$.
Solution:
$$3u=⟨15,12⟩$$
Example
Find a linear equation to solve for the following unknown quantities: One number exceeds another number by $$17$$ and their sum is $$31$$. Find the two numbers.
Solution:
First, we must multiply each vector by the scalar.
\begin{align} 3u &= 3⟨3,−2⟩ \\ &= ⟨9,−6⟩ \\ 2v &= 2⟨−1,4⟩ \\ &= ⟨−2,8⟩ \end{align}
Then, add the two together.
\begin{align} w &= 3u+2v \\ &=⟨9,−6⟩+⟨−2,8⟩ \\ &= ⟨9−2,−6+8⟩ \\ &= ⟨7,2⟩ \end{align}
So, $$w=⟨7,2⟩$$.
# Finding Component Form
In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the $$x$$ direction, and the vertical component is the $$y$$ direction. For example, we can see in the graph in Figure 10.9.12 that the position vector $$⟨2,3⟩$$ comes from adding the vectors $$v_1$$ and $$v_2$$. We have $$v_2$$ with initial point $$(0,0)$$ and terminal point $$(2,0)$$.
\begin{align} v_1 &= ⟨2−0,0−0⟩ \\ &= ⟨2,0⟩ \end{align}
We also have $$v_2$$ with initial point $$(0,0)$$ and terminal point $$(0, 3)$$.
\begin{align} v_2 &= ⟨0−0,3−0⟩ \\ &= ⟨0,3⟩ \end{align}
Therefore, the position vector is
\begin{align} v &= ⟨2+0,3+0⟩ \\ &= ⟨2,3⟩ \end{align}
Using the Pythagorean Theorem, the magnitude of $$v_1$$ is $$2$$, and the magnitude of $$v_2$$ is $$3$$. To find the magnitude of $$v$$, use the formula with the position vector.
\begin{align} |v| &= \sqrt{{|v_1|}^2+{|v_2|}^2} \\ &= \sqrt{2^2+3^2} \\ &= \sqrt{13} \end{align}
The magnitude of $$v$$ is $$\sqrt{13}$$. To find the direction, we use the tangent function $$\tan \theta=\dfrac{y}{x}$$.
\begin{align} \tan \theta &= \dfrac{v_2}{v_1} \\ \tan \theta &= \dfrac{3}{2} \\ \theta &={\tan}^{−1}(\dfrac{3}{2})=56.3° \end{align}
Figure 10.9.12
Thus, the magnitude of $$v$$ is $$\sqrt{13}$$ and the direction is $$56.3∘$$ off the horizontal.
Example
Finding the Components of the Vector
Find the components of the vector $$v$$ with initial point $$(3,2)$$ and terminal point $$(7,4)$$.
Solution:
First find the standard position.
\begin{align} v &= ⟨7−3,4−2⟩ \\ &= ⟨4,2⟩ \end{align}
See the illustration in Figure 10.9.13.
Figure 10.9.13
The horizontal component is $$v_1=⟨4,0⟩$$ and the vertical component is $$v_2=⟨0,2⟩$$.
# Finding the Unit Vector in the Direction of v
In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations.
Unit vectors are defined in terms of components. The horizontal unit vector is written as $$i=⟨1,0⟩$$ and is directed along the positive horizontal axis. The vertical unit vector is written as $$j=⟨0,1⟩$$ and is directed along the positive vertical axis. See Figure 10.9.14.
Figure 10.9.14
A General Note: THE UNIT VECTORS
If $$v$$ is a nonzero vector, then $$\dfrac{v}{| v |}$$ is a unit vector in the direction of $$v$$. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.
Example
Finding the Unit Vector in the Direction of v
Find a unit vector in the same direction as $$v=⟨−5,12⟩$$.
Solution:
First, we will find the magnitude.
\begin{align} |v| &= \sqrt{{(−5)}^2+{(12)}^2} \\ &= \sqrt{25+144} \\ &=\sqrt{169} \\ &= 13 \end{align}
Then we divide each component by $$| v |$$, which gives a unit vector in the same direction as v:
$\dfrac{v}{| v |} = −\dfrac{5}{13}i+\dfrac{12}{13}j$
or, in component form
$\dfrac{v}{| v |}=⟨ −\dfrac{5}{13},\dfrac{12}{13} ⟩$
See Figure 10.9.15.
Figure 10.9.15
Verify that the magnitude of the unit vector equals 1. The magnitude of $$−\dfrac{5}{13}i+\dfrac{12}{13}j$$ is given as
\begin{align} \sqrt{ {(−\dfrac{5}{13})}^2+{ (\dfrac{12}{13}) }^2 } &= \sqrt{\dfrac{25}{169}+\dfrac{144}{169}} \\ &= \sqrt{\dfrac{169}{169}}=1 \end{align}
The vector $$u=\dfrac{5}{13}i+\dfrac{12}{13}j$$ is the unit vector in the same direction as $$v=⟨−5,12⟩$$.
# Performing Operations with Vectors in Terms of i and j
So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j.
A General Note: VECTORS IN THE RECTANGULAR PLANE
Given a vector$$v$$with initial point $$P=(x_1,y_1)$$ and terminal point $$Q=(x_2,y_2)$$, v is written as
$v=(x_2−x_1)i+(y_1−y_2)j$
The position vector from $$(0,0)$$ to $$(a,b)$$, where $$(x_2−x_1)=a$$ and $$(y_2−y_1)=b$$, is written as v = aibj. This vector sum is called a linear combination of the vectors i and j.
The magnitude of v = ai + bj is given as $$| v |=\sqrt{a^2+b^2}$$. See Figure 10.9.16.
Figure 10.9.16
Example
Writing a Vector in Terms of i and j
Given a vector v with initial point $$P=(2,−6)$$ and terminal point $$Q=(−6,6)$$, write the vector in terms of i and j.
Solution:
Begin by writing the general form of the vector. Then replace the coordinates with the given values.
\begin{align} v &= (x_2−x_1)i+(y_2−y_1)j \\ &=(−6−2)i+(6−(−6))j \\ &= −8i+12j \end{align}
Example
Writing a Vector in Terms of i and j Using Initial and Terminal Points
Given initial point $$P_1=(−1,3)$$ and terminal point $$P_2=(2,7)$$, write the vector v in terms of i and j.
Solution:
Begin by writing the general form of the vector. Then replace the coordinates with the given values.
\begin{align} v &= (x_2−x_1)i+(y_2−y_1)j \\ v &= (2−(−1))i+(7−3)j \\ &= 3i+4j \end{align}
Exercise
Write the vector u with initial point $$P=(−1,6)$$ and terminal point $$Q=(7,−5)$$ in terms of i and j.
Solution:
$$u=8i−11j$$
### Performing Operations on Vectors in Terms of i and j
When vectors are written in terms of $$i$$ and $$j$$, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.
A General Note: ADDING AND SUBTRACTING VECTORS IN RECTANGULAR COORDINATES
Given $$v = ai + bj$$ and $$u = ci + dj$$, then
\begin{align} v+u &= (a+c)i+(b+d)j \\ v−u &= (a−c)i+(b−d)j \end{align}
Example
Finding the Sum of the Vectors
Find the sum of $$v_1=2i−3j$$ and $$v_2=4i+5j$$.
Solution:
\begin{align} v_1+v_2 &= (2+4)i+(−3+5)j \\ &= 6i+2j \end{align}
### Calculating the Component Form of a Vector: Direction
We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using $$i$$ and $$j$$. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.
Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with $$| v |$$ replacing $$r$$.
A General Note: VECTOR COMPONENTS IN TERMS OF MAGNITUDE AND DIRECTION
Given a position vector $$v=⟨x,y⟩$$ and a direction angle $$\theta$$,
\begin{align} \cos \theta &= \dfrac{x}{|v|} \text{ and } \sin \theta=y|v| \\ x &= |v| \cos \theta \\ y &= |v| \sin \theta \end{align}
Thus, $$v=xi+yj=| v | \cos \theta i+| v | \sin \theta j$$, and magnitude is expressed as $$| v |=\sqrt{x^2+y^2}$$.
Example
Writing a Vector in Terms of Magnitude and Direction
Write a vector with length 7 at an angle of 135° to the positive x-axis in terms of magnitude and direction.
Solution:
Using the conversion formulas $$x=| v | \cos \theta i$$ and $$y=| v | \sin \theta j$$, we find that
\begin{align} x &= 7\cos(135°)i \\ &= −\dfrac{7\sqrt{2}}{2} \\ y &=7 \sin(135°)j \\ &= \dfrac{7\sqrt{2}}{2} \end{align}
This vector can be written as $$v=7\cos(135°)i+7\sin(135°)j$$ or simplified as
$v=−\dfrac{7\sqrt{2}}{2}i+\dfrac{7\sqrt{2}}{2}j$
Exercise
A vector travels from the origin to the point $$(3,5)$$. Write the vector in terms of magnitude and direction.
Solution:
$$v=\sqrt{34}\cos(59°)i+\sqrt{34}\sin(59°)j$$
Magnitude = 34
$$\theta={\tan}^{−1}(\dfrac{5}{3})=59.04°$$
### Finding the Dot Product of Two Vectors
As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.
The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.
A General Note: DOT PRODUCT
The dot product of two vectors $$v=⟨a,b⟩$$ and $$u=⟨c,d⟩$$ is the sum of the product of the horizontal components and the product of the vertical components.
$v⋅u=ac+bd$
To find the angle between the two vectors, use the formula below.
$\cos \theta=\dfrac{v}{| v |}⋅\dfrac{u}{| u |}$
Example
Finding the Dot Product of Two Vectors
Find the dot product of $$v=⟨5,12⟩$$ and $$u=⟨−3,4⟩$$.
Solution:
Using the formula, we have
\begin{align} v⋅u &= ⟨5,12⟩⋅⟨−3,4⟩ \\ &= 5⋅(−3)+12⋅4 \\ &= −15+48 \\ &= 33 \end{align}
Example
Finding the Dot Product of Two Vectors and the Angle between Them
Find the dot product of $$v_1 = 5i + 2j$$ and $$v_2 = 3i + 7j$$. Then, find the angle between the two vectors.
Solution:
Finding the dot product, we multiply corresponding components.
\begin{align} v_1⋅v_2 &= ⟨5,2⟩⋅⟨3,7⟩ \\ &= 5⋅3+2⋅7 \\ &= 15+14 \\ &= 29 \end{align}
To find the angle between them, we use the formula $$\cos \theta=\dfrac{v}{|v|}⋅\dfrac{u}{|u|}$$.
\begin{align} \dfrac{v}{|v|}⋅\dfrac{u}{|u|} &= ⟨ \dfrac{5}{\sqrt{29}}+\dfrac{2}{\sqrt{29}} ⟩⋅⟨ \dfrac{3}{\sqrt{58}}+\dfrac{7}{\sqrt{58}} ⟩ \\ &=\dfrac{5}{\sqrt{29}}⋅\dfrac{3}{\sqrt{58}}+\dfrac{2}{\sqrt{29}}⋅\dfrac{7}{\sqrt{58}} \\ &= \dfrac{15}{\sqrt{1682}}+\dfrac{14}{\sqrt{1682}}=\dfrac{29}{\sqrt{1682}} \\ =0.707107 \\ {\cos}^{−1}(0.707107) &= 45° \end{align}
See Figure 10.9.17.
Figure 10.9.17
Example
Finding the Angle between Two Vectors
Find the angle between $$u=⟨−3,4⟩$$ and $$v=⟨5,12⟩$$.
Solution:
Using the formula, we have
\begin{align} \theta &= {\cos}^{−1}(\dfrac{u}{|u|}⋅\dfrac{v}{|v|}) \\ (\dfrac{u}{|u|}⋅\dfrac{v}{|v|}) &= \dfrac{−3i+4j}{5}⋅\dfrac{5i+12j}{13} \\ &= (− \dfrac{3}{5}⋅ \dfrac{5}{13})+(\dfrac{4}{5}⋅ \dfrac{12}{13}) \\ &= −\dfrac{15}{65}+\dfrac{48}{65} \\ &= \dfrac{33}{65} \\ \theta &= {\cos}^{−1}(\dfrac{33}{65}) \\ &= 59.5∘ \end{align}
See Figure 10.9.18.
Figure 10.9.18
Example
Finding Ground Speed and Bearing Using Vectors
We now have the tools to solve the problem we introduced in the opening of the section.
An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 10.9.19.
Figure 10.9.19
Solution:
The ground speed is represented by $$x$$ in the diagram, and we need to find the angle $$\alpha$$ in order to calculate the adjusted bearing, which will be $$140°+\alpha$$ .
Notice in Figure 10.9.19, that angle $$BCO$$ must be equal to angle $$AOC$$ by the rule of alternating interior angles, so angle $$BCO$$ is 140°. We can find $$x$$ by the Law of Cosines:
\begin{align} x^2 &= {(16.2)}^2+{(200)}^2−2(16.2)(200) \cos(140°) \\ x^2 &= 45,226.41 \\ x &= \sqrt{45,226.41} \\ x &= 212.7 \end{align}
The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.
\begin{align} \dfrac{\sin \alpha}{16.2} &= \dfrac{\sin(140°)}{212.7} \\ \sin \alpha &= \dfrac{16.2 \sin(140°)}{212.7} \\ &=0.04896 \\ {\sin}^{−1}(0.04896) &= 2.8° \end{align}
Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.
Media: Access these online resources for additional instruction and practice with vectors.
### Key Concepts
• The position vector has its initial point at the origin. See Example.
• If the position vector is the same for two vectors, they are equal. See Example.
• Vectors are defined by their magnitude and direction. See Example.
• If two vectors have the same magnitude and direction, they are equal. See Example.
• Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example.
• Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. SeeExample and Example.
• Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See Example.
• The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
• The magnitude of a vector in the rectangular coordinate system is $$| v |=\sqrt{a^2+b^2}$$. See Example.
• In the rectangular coordinate system, unit vectors may be represented in terms of ii and jj wherei i represents the horizontal component andj j represents the vertical component. Then, v = ai + bj is a scalar multiple of v by real numbers a and b. SeeExample and Example.
• Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See Example.
• A vector v = ai + bj is written in terms of magnitude and direction as $$v=| v |\cos \theta i+| v |\sin \theta j$$. See Example.
• The dot product of two vectors is the product of thei i terms plus the product of the j terms. See Example.
• We can use the dot product to find the angle between two vectors. Example and Example.
• Dot products are useful for many types of physics applications. See Example. |
# Kinematics – Graphing Motion
## Presentation on theme: "Kinematics – Graphing Motion"— Presentation transcript:
Kinematics – Graphing Motion
Unit #2 Kinematics
Objectives and Learning Targets
Construct and interpret graphs of position, velocity, and acceleration versus time. Determine and interpret slopes and areas of motion graphs. Unit #2 Kinematics
Position-Time (x-t) Graphs
Graphing Motion Position-Time (x-t) Graphs The position-time graph shows the displacement (or, in the case of scalar quantities, distance) of an object as a function of time. Positive displacements indicate the object's position is in the positive direction from its starting point, while negative displacements indicate the object's position is opposite the positive direction. Unit #2 Kinematics
Example Suppose Cricket the WonderDog wanders away from her house at a constant velocity of 1 m/s, stopping only when she's 5m away (which, of course, takes 5 seconds). She then decides to take a short five-second rest in the grass. After her five second rest, she hears the dinner bell ring, so she runs back to the house at a speed of 2 m/s. The displacement-time graph for her motion would look something like this: Unit #2 Kinematics
Example Unit #2 Kinematics
Example Cricket's displacement begins at zero meters at time zero. Then, as time progresses, Cricket's displacement increases at a rate of 1 m/s, so that after one second, Cricket is one meter away from her starting point. After two seconds, she's two meters away, and so forth, until she reaches her maximum displacement of five meters from her starting point at a time of five seconds. Cricket then remains at that position for 5 seconds while she takes a rest. Following her rest, at time t=10 seconds, Cricket hears the dinner bell and races back to the house at a speed of 2 m/s, so the graph ends when Cricket returns to her starting point at the house, a total distance traveled of 10m, and a total displacement of zero meters. Unit #2 Kinematics
Example As you look at the position-time graph, notice that at the beginning, when Cricket is moving in a positive direction, the graph has a positive slope. When the graph is flat (has a zero slope) Cricket is not moving. And when the graph has a negative slope, Cricket is moving in the negative direction. It's also easy to see that the steeper the slope of the graph, the faster Cricket is moving. Unit #2 Kinematics
Sample Problem #1 The graph below represents the displacement of an object moving in a straight line as a function of time. What was the total distance traveled by the object during the 10-second time interval? Unit #2 Kinematics
Sample Problem #1 Answer: Total distance traveled is 8 meters forward from 0 to 4 seconds, then 8 meters forward from 6 to 8 seconds, then 8 meters backward from 8 to 10 seconds, for a total of 24 meters. Unit #2 Kinematics
Velocity-Time (v-t) Graphs
Another type motion graph is the velocity-time graph, which shows the velocity of an object on the y-axis, and time on the x-axis. Positive values indicate velocities in the positive direction, while negative values indicate velocities in the opposite direction. In reading these graphs, it’s important to realize that a straight horizontal line indicates the object maintaining a constant velocity – it can still be moving, its velocity just isn’t changing. A value of 0 on the v-t graph indicates the object has come to a stop. If the graph crosses the x-axis, the object was moving in one direction, came to a stop, and switched the direction of its motion. Unit #2 Kinematics
Cricket’s v-t Graph Describe what is happening with Cricket during this Velocity- Time Graph. Unit #2 Kinematics
Cricket’s v-t Graph For the first five seconds of Cricket’s journey, you can see she maintains a constant velocity of 1 m/s. Then, when she stops to rest, her velocity changes to zero for the duration of her rest. Finally, when she races back to the house for dinner, she maintains a negative velocity of 2 m/s. Because velocity is a vector, the negative sign indicates that Cricket’s velocity is in the opposite direction (initially the direction away from the house was positive, so back toward the house must be negative!) Unit #2 Kinematics
Graph Tranformations Looking at a position-time graph, the faster an object’s position/displacement changes, the steeper the slope of the line. Since velocity is the rate at which an object’s position changes, the slope of the position-time graph at any given point in time gives you the velocity at that point in time. You can obtain the slope of the position-time graph using the following formula: Unit #2 Kinematics
Graph Tranformations Realizing that the rise in the graph is actually ∆x, and the run is ∆t, you can substitute these variables into the slope equation to find: With a little bit of interpretation, it’s easy to show that the slope is really just change in position over time, which is the definition of velocity. Put directly, the slope of the position- time graph gives you the velocity. Unit #2 Kinematics
Graph Tranformations It makes sense that if you can determine velocity from the position-time graph, you should be able to work backward to determine change in position (displacement) from the v-t graph. If you have a v-t graph, and you want to know how much an object’s position changed in a time interval, take the area under the curve within that time interval. So, if taking the slope of the position-time graph gives you the rate of change of position, which is called velocity, what do you get when you take the slope of the v-t graph? You get the rate of change of velocity, which is called acceleration! The slope of the v-t graph, therefore, tells you the acceleration of an object. Unit #2 Kinematics
Sample Problem #2 a (t=5.0s) = slope (t=5.0s) = 0 m/s2
Total distance = area under the curve = AreaTriangle + AreaRectangle = 20m + 20m = 40m Unit #2 Kinematics
Sample Problem #3 Unit #2 Kinematics
Sample Problem #3 Unit #2 Kinematics
Sample Problem #4 Unit #2 Kinematics
Sample Problem #4 Unit #2 Kinematics
Sample Problem #5 Unit #2 Kinematics
Sample Problem #5 Unit #2 Kinematics
Acceleration-Time (a-t) Graphs
Similar to velocity, you can make a graph of acceleration vs. time by plotting the rate of change of an object’s velocity (its acceleration) on the y-axis, and placing time on the x-axis. When you took the slope of the position-time graph, you obtained the object’s velocity. In the same way, taking the slope of the v-t graph gives you the object’s acceleration. Going the other direction, when you analyzed the v- t graph, you found that taking the area under the v-t graph provided you with information about the object’s change in position. In similar fashion, taking the area under the a-t graph tells you how much an object’s velocity changes. Putting it all together, you can go from position-time to velocity-time by taking the slope, and you can go from velocity-time to acceleration-time by taking the slope. Or, going the other direction, the area under the acceleration-time curve gives you an object’s change in velocity, and the area under the velocity-time curve gives you an object’s change in position. Unit #2 Kinematics
Moving from One Graph to Another
Unit #2 Kinematics
Sample Problem #6 Unit #2 Kinematics
Sample Problem #6 Unit #2 Kinematics |
Arithmetic Algorithms in Different Bases
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# Arithmetic Algorithms in Different Bases - PowerPoint PPT Presentation
Arithmetic Algorithms in Different Bases. Addition, Subtraction, Multiplication and Division. To add in any base – Step 1 Add the digits in the “ones ”column to find the number of “1s”. If the number is less than the base place the number under the right hand column.
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### Arithmetic Algorithms in Different Bases
To add in any base – Step 1
Add the digits in the “ones ”column to find the number of “1s”.
If the number is less than the base place the number under the right hand column.
If the number is greater than or equal to the base, express the number as a base numeral. The first digit indicates the number of the base in the sum so carry that digit to the “base” column and write the second digit under the right hand column.
14
+ 32 5
2 + 4 = 6
6 > 5
6 = 11 5
Place 1 under 4 + 2 and carry the 1 five to the “fives” column
Step 2
Add the digits in the “base” column.
If the sum is less than the base enter under the “base” column.
If the sum is greater than or equal to the base express the sum as a base numeral. Carry the first digit to the “base squared” column and place the second digit under the “base” column.
Step 3
Repeat this process to the end.
1
14
+ 32 5
1
1+1+3 = 5
5 = 10 5
Carry the 1 five to the 52 column.
Place 0 under the “five” column
The sum is 101 5
Find: 1
123 123
342 6 342 6
5 0 5
1 1
123
342 6
5056
3 + 2 = 5 < 6 →no “6”s to carry.
2 + 4 = 6 6 = 10 6
One 6 2 to carry.
0 to go in the 6 column.
1 + 1 + 3 = 5 < 6 → no 62 to carry.
The Subtraction Algorithm
Subtraction in any base, b
• Start as “1”s column.
• If the subtraction is possible enter the number under the “1”s column.
• If the subtraction is not possible borrow 1 of the groups of b in the minuend.
• The one b being borrowed is worth b in the “1”s column so add b to the digit in the minuend in the “1”s column .
• Continue in this manner through the problem
2−4 can not do.
Borrow 1 of the 5 “7”s in the minuend and add 7 to the 2 then subtract.
2−4 = 2
3−1 = 2
A Subtraction Problem
The Multiplication Algorithm
To Multiply in any base
• The multiplication algorithm is the same as in base 10. Things you need to remember.
• Carry groups of the base
• Indent for each multiplication row
Multiplying in Base 3
2 x 2 = 4 = 11 3
Remember to
indent
Carrying in Multiplication
2 x 5 =10 =146
2 x 2 =4,
4 + 1 =5,
5 = 5 6
2 x 4 =8,
8 +1 =9
9 = 1 3 6
The Division Algorithm
• To divide in any base
• The division algorithm in any base is the same as in base 10. Things to remember –
• Digits that are entered in the quotient are at first estimates.
• Do all the multiplications on the side so that the digits used are not estimates.
• Multiply and subtract in the base used in the problem.
A Division Problem
1 x 45 = 45
2 x 45 = 135
3 x 45 = 225
4 x 45 = 315
0 x 45 =0
Subtract inbase 5
3 x 45 =225 |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$t^{24}-4t^{18}+6t^{12}-4t^6+1$
RECALL: $x^my^m=(xy)^m$ Use the rule above to obtain: $=[(t^3-1)(t^3+1)]^4 \\=(t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1)$ Use the formula $(a-b)(a+b) = a^2-b^2$ with $a=t^3$ and $b=1$ to obtain: $=[(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \\=(t^6-1) \cdot (t^6-1) \cdot (t^6-1) \cdot (t^6-1) \\=[(t^6-1) \cdot (t^6-1)] \cdot [(t^6-1) \cdot (t^6-1)]$ Use the formula $(a-b)(a-b) = a^2-2ab+b^2$ with $a=t^6$ and $b=1$ to obtain: $=[(t^6)^2-2(t^6)(1)+1^2] \cdot [(t^6)^2-2(t^6)(1)+1^2] \\=(t^{12}-2t^6+1) \cdot (t^{12}-2t^6+1)$ Distribute each term of the first trinomial to the second trinomial to obtain: $=t^{12}(t^{12}-2t^6+1)-2t^6(t^{12}-2t^6+1)+1(t^{12}-2t^6+1) \\=t^{12}(t^{12}) -t^{12}2t^6+t^{12}(1)-2t^6(t^{12})-2t^6(-2t^6)-2t^6(1)+(t^{12}-2t^6+1) \\=t^{24}-2t^{18}+t^{12}-2t^{18}+4t^{12}-2t^6+t^{12}-2t^6+1$ Combine like terms to obtain: $\\=t^{24}+(-2t^{18}-2t^{18})+(t^{12}+4t^{12}+t^{12})+(-2t^6-2t^6)+1 \\=t^{24}+(-4t^{18})+6t^{12}+(-4t^6)+1 \\=t^{24}-4t^{18}+6t^{12}-4t^6+1$ |
# EQUATION OF A TRANSLATED CIRCLE FROM STANDARD FORM
Problem 1 :
(x - 16)2 + (y - 6)2 = 1
Translated 4 left, 2 up
Solution:
The standard form of a circle is
(x - h)2 + (y - k)2 = r2
Where r is the radius of the circle and (h, k) is the center of the circle.
Center (h, k) = (16, 6)
When the circle is translated left 4 units and up 2 units.
The center should be changed as,
((x - 16) - (-4))2 + ((y - 6) - (2))2 = 1
(x - 16 + 4)2 + (y - 6 - 2)2 = 1
(x - 12)2 + (y - 8)2 = 1
Problem 2 :
(x + 5)2 + (y + 7)2 = 36
Translated 5 left, 4 down
Solution:
The standard form of a circle is
(x - h)2 + (y - k)2 = r2
Where r is the radius of the circle and (h, k) is the center of the circle.
(x - (-5))2 + (y - (-7))2 = 36
Center (h, k) = (-5, -7)
Translating 5 units left and 4 units down
((x + 5) - (-5))2 + ((y + 7) - (-4))2 = 36
(x + 5 + 5)2 + (y + 7 + 4)2 = 36
(x + 10)2 + (y + 11)2 = 36
Problem 3 :
x2 + y2 + 14x + 12y + 76 = 0
Translated 2 right, 4 down
Solution:
x2 + y2 + 14x + 12y + 76 = 0
x2 + 14x + y2 + 12y + 76 = 0
x2 + 2 ⋅ x ⋅ 7 + 72 - 72 + y2 + 2 ⋅ y ⋅ 6 + 62 - 62 + 76 = 0
(x + 7)2 - 49 + (y + 6)2 - 36 + 76 = 0
(x + 7)2 + (y + 6)2 - 85 + 76 = 0
(x + 7)2 + (y + 6)2 = 9
Before translation :
Center of the circle is (7, -6).
(x + 7)2 + (y + 6)2 = 9
After translation :
(x + 7 - 2)2 + (y + 6 - (-4))2 = 9
Translated 2 right, 4 down
So, equation of new circle is,
(x + 5)2 + (y + 10)2 = 9
Problem 4 :
x2 + y2 - 10x + 20y + 61 = 0
Translated 1 left, 2 down
Solution:
x2 + y2 - 10x + 20y + 61 = 0
x2 - 2 ⋅ x ⋅ 5 + 52 - 52+ y2+ 2 ⋅ y ⋅ 10 + 102 - 102 + 61 = 0
(x - 5)2 - 25 + (y + 10)2 - 100 + 61 = 0
(x - 5)2 + (y + 10)2 - 25 - 100 + 61 = 0
(x - 5)2 + (y + 10)2 - 64 = 0
(x - 5)2 + (y + 10)2 = 64
Before translation, the center will be at :
(5, -10)
(x - 5)2 + (y + 10)2 = 64
After translation, the center will be at :
Translated 1 left, 2 down
(x - 5 - (-1))2 + ((y + 10) - (-2))2 = 64
(x - 5 + 1)2 + (y + 10 + 2)2 = 64
(x - 4)2 + ((y + 12)2 = 64
Problem 5 :
x2 + y2 + 14x - 8y + 29 = 0
Translated 3 right, 4 down
Solution:
x2 + y2 + 14x - 8y + 29 = 0
x2 + 14x + y2 - 8y + 29 = 0
x2 + 2⋅x⋅7 + 7- 72 + y2 - 2⋅y⋅4 + 42- 42 + 29 = 0
(x + 7)2 - 49 + (y - 4)2 - 16 + 29 = 0
(x + 7)2 + (y - 4)2 - 65 + 29 = 0
(x + 7)2 + (y - 4)2 - 36 = 0
(x + 7)2 + (y - 4)2 = 36
Before the translation, center will be at :
(-7, 4)
After the translation, center will be at :
(x + 7 - 3)2 + (y - 4 - (-4))2 = 36
(x + 4)2 + (y - 4 + 4)2 = 36
(x + 4)2 + y2 = 36
Problem 6 :
4y + y2 = -28x - x2 - 191
Translated 4 right
Solution:
4y + y2 = -28x - x2 - 191
x2 + 28x + y24y + 191 = 0
(x + 14)2 + (y - 2)2 - 142 - 22+ 191 = 0
(x + 14)2 + (y - 2)2 - 196 - 4 + 191 = 0
(x + 14)2 + (y - 2)2 - 200 + 191 = 0
(x + 14)2 + (y - 2)2 - 9 = 0
(x + 14)2 + (y - 2)2 = 9
Thus the coordinates of the center at (-14, -2). When the circle is translated right 4 units,
h = -14 + 4 = -10
So, equation of new circle is,
(x + 10)2 + (y + 2)2 = 9
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# 1.1 Functions and function notation (Page 10/21)
Page 10 / 21
Why does the horizontal line test tell us whether the graph of a function is one-to-one?
When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.
## Algebraic
For the following exercises, determine whether the relation represents a function.
$\left\{\left(a,b\right),\left(b,c\right),\left(c,c\right)\right\}$
function
For the following exercises, determine whether the relation represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
$5x+2y=10$
$y={x}^{2}$
function
$x={y}^{2}$
$3{x}^{2}+y=14$
function
$2x+{y}^{2}=6$
$y=-2{x}^{2}+40x$
function
$y=\frac{1}{x}$
$x=\frac{3y+5}{7y-1}$
function
$x=\sqrt{1-{y}^{2}}$
$y=\frac{3x+5}{7x-1}$
function
${x}^{2}+{y}^{2}=9$
$2xy=1$
function
$x={y}^{3}$
$y={x}^{3}$
function
$y=\sqrt{1-{x}^{2}}$
$x=±\sqrt{1-y}$
function
$y=±\sqrt{1-x}$
${y}^{2}={x}^{2}$
not a function
${y}^{3}={x}^{2}$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the indicated values
$f\left(x\right)=2x-5$
$\begin{array}{cccc}f\left(-3\right)=-11;& f\left(2\right)=-1;& f\left(-a\right)=-2a-5;& -f\left(a\right)=-2a+5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=2a+2h-5\end{array}$
$f\left(x\right)=-5{x}^{2}+2x-1$
$f\left(x\right)=\sqrt{2-x}+5$
$\begin{array}{cccc}f\left(-3\right)=\sqrt{5}+5;& f\left(2\right)=5;& f\left(-a\right)=\sqrt{2+a}+5;& -f\left(a\right)=-\sqrt{2-a}-5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=\end{array}$ $\sqrt{2-a-h}+5$
$f\left(x\right)=\frac{6x-1}{5x+2}$
$f\left(x\right)=|x-1|-|x+1|$
Given the function $\text{\hspace{0.17em}}g\left(x\right)=5-{x}^{2},\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}\frac{g\left(x+h\right)-g\left(x\right)}{h},\text{\hspace{0.17em}}h\ne 0.$
Given the function $\text{\hspace{0.17em}}g\left(x\right)={x}^{2}+2x,\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}\frac{g\left(x\right)-g\left(a\right)}{x-a},\text{\hspace{0.17em}}x\ne a.$
$\frac{g\left(x\right)-g\left(a\right)}{x-a}=x+a+2,\text{\hspace{0.17em}}x\ne a$
Given the function $\text{\hspace{0.17em}}k\left(t\right)=2t-1:$
1. Evaluate $\text{\hspace{0.17em}}k\left(2\right).$
2. Solve $\text{\hspace{0.17em}}k\left(t\right)=7.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)=8-3x:$
1. Evaluate $\text{\hspace{0.17em}}f\left(-2\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=-1.$
a. $\text{\hspace{0.17em}}f\left(-2\right)=14;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}x=3$
Given the function $\text{\hspace{0.17em}}p\left(c\right)={c}^{2}+c:$
1. Evaluate $\text{\hspace{0.17em}}p\left(-3\right).$
2. Solve $\text{\hspace{0.17em}}p\left(c\right)=2.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3x:$
1. Evaluate $\text{\hspace{0.17em}}f\left(5\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
a. $\text{\hspace{0.17em}}f\left(5\right)=10;\text{\hspace{0.17em}}$ b. or
Given the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x+2}:$
1. Evaluate $\text{\hspace{0.17em}}f\left(7\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
Consider the relationship $\text{\hspace{0.17em}}3r+2t=18.$
1. Write the relationship as a function $\text{\hspace{0.17em}}r=f\left(t\right).$
2. Evaluate $\text{\hspace{0.17em}}f\left(-3\right).$
3. Solve $\text{\hspace{0.17em}}f\left(t\right)=2.$
a. $\text{\hspace{0.17em}}f\left(t\right)=6-\frac{2}{3}t;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}f\left(-3\right)=8;\text{\hspace{0.17em}}$ c. $\text{\hspace{0.17em}}t=6\text{\hspace{0.17em}}$
## Graphical
For the following exercises, use the vertical line test to determine which graphs show relations that are functions.
not a function
function
function
function
function
function
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(-1\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=3.$
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(0\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=-3.$
a. $\text{\hspace{0.17em}}f\left(0\right)=1;\text{\hspace{0.17em}}$ b. or
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(4\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=1.$
For the following exercises, determine if the given graph is a one-to-one function.
not a function so it is also not a one-to-one function
one-to- one function
function, but not one-to-one
## Numeric
For the following exercises, determine whether the relation represents a function.
$\left\{\left(-1,-1\right),\left(-2,-2\right),\left(-3,-3\right)\right\}$
$\left\{\left(3,4\right),\left(4,5\right),\left(5,6\right)\right\}$
function
$\left\{\left(2,5\right),\left(7,11\right),\left(15,8\right),\left(7,9\right)\right\}$
For the following exercises, determine if the relation represented in table form represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.$
$x$ 5 10 15 $y$ 3 8 14
function
$x$ 5 10 15 $y$ 3 8 8
$x$ 5 10 10 $y$ 3 8 14
not a function
For the following exercises, use the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ represented in [link] .
$x$ $f\left(x\right)$ 0 74 1 28 2 1 3 53 4 56 5 3 6 36 7 45 8 14 9 47
Evaluate $\text{\hspace{0.17em}}f\left(3\right).$
Solve $\text{\hspace{0.17em}}f\left(x\right)=1.$
$f\left(x\right)=1,\text{\hspace{0.17em}}x=2$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the values $f\left(-2\right),\text{\hspace{0.17em}}f\left(-1\right),\text{\hspace{0.17em}}f\left(0\right),\text{\hspace{0.17em}}f\left(1\right),$ and $\text{\hspace{0.17em}}f\left(2\right).$
$f\left(x\right)=4-2x$
$f\left(x\right)=8-3x$
$\begin{array}{ccccc}f\left(-2\right)=14;& f\left(-1\right)=11;& f\left(0\right)=8;& f\left(1\right)=5;& f\left(2\right)=2\end{array}$
$f\left(x\right)=8{x}^{2}-7x+3$
$f\left(x\right)=3+\sqrt{x+3}$
$\begin{array}{ccccc}f\left(-2\right)=4;\text{ }& f\left(-1\right)=4.414;& f\left(0\right)=4.732;& f\left(1\right)=4.5;& f\left(2\right)=5.236\end{array}$
$f\left(x\right)=\frac{x-2}{x+3}$
$f\left(x\right)={3}^{x}$
$\begin{array}{ccccc}f\left(-2\right)=\frac{1}{9};& f\left(-1\right)=\frac{1}{3};& f\left(0\right)=1;& f\left(1\right)=3;& f\left(2\right)=9\end{array}$
For the following exercises, evaluate the expressions, given functions $f,\text{\hspace{0.17em}}\text{\hspace{0.17em}}g,$ and $\text{\hspace{0.17em}}h:$
• $f\left(x\right)=3x-2$
• $g\left(x\right)=5-{x}^{2}$
• $h\left(x\right)=-2{x}^{2}+3x-1$
$3f\left(1\right)-4g\left(-2\right)$
$f\left(\frac{7}{3}\right)-h\left(-2\right)$
20
## Technology
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-100,100\right]$
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{3}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $\text{\hspace{0.17em}}y=\sqrt{x}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $y=\sqrt[3]{x}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-0.001,\text{0.001}\right]$
$\left[-0.1,\text{0.1}\right]$
$\left[-1000,\text{1000}\right]$
$\left[-1,000,000,\text{1,000,000}\right]$
## Real-world applications
The amount of garbage, $\text{\hspace{0.17em}}G,\text{\hspace{0.17em}}$ produced by a city with population $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is given by $\text{\hspace{0.17em}}G=f\left(p\right).\text{\hspace{0.17em}}$ $G\text{\hspace{0.17em}}$ is measured in tons per week, and $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is measured in thousands of people.
1. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}f\left(5\right)=2.$
The number of cubic yards of dirt, $\text{\hspace{0.17em}}D,\text{\hspace{0.17em}}$ needed to cover a garden with area $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ square feet is given by $\text{\hspace{0.17em}}D=g\left(a\right).$
1. A garden with area 5000 ft 2 requires 50 yd 3 of dirt. Express this information in terms of the function $\text{\hspace{0.17em}}g.$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}g\left(100\right)=1.$
a. $\text{\hspace{0.17em}}g\left(5000\right)=50;$ b. The number of cubic yards of dirt required for a garden of 100 square feet is 1.
Let $\text{\hspace{0.17em}}f\left(t\right)\text{\hspace{0.17em}}$ be the number of ducks in a lake $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ years after 1990. Explain the meaning of each statement:
1. $f\left(5\right)=30$
2. $f\left(10\right)=40$
Let $\text{\hspace{0.17em}}h\left(t\right)\text{\hspace{0.17em}}$ be the height above ground, in feet, of a rocket $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds after launching. Explain the meaning of each statement:
1. $h\left(1\right)=200$
2. $h\left(2\right)=350$
a. The height of a rocket above ground after 1 second is 200 ft. b. the height of a rocket above ground after 2 seconds is 350 ft.
Show that the function $\text{\hspace{0.17em}}f\left(x\right)=3{\left(x-5\right)}^{2}+7\text{\hspace{0.17em}}$ is not one-to-one.
difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott |
Do girlfriend have any kind of idea about the factorization the polynomials? since you now have some basic information about polynomials, we will learn exactly how to deal with quadratic polynomials by factorization.
You are watching: Factor x2-2x-1
First the all, let’s take it a quick review of the quadratic equation. A quadratic equation is a polynomial of a second degree, normally in the kind of f(x) = ax2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The hatchet ‘a’ is described as the leading coefficient, if ‘c’ is the absolute term the f (x).
Every quadratic equation has two values of the unknown variable, usually well-known as the root of the equation (α, β). We can achieve the roots of a quadratic equation by factoring the equation.
For this reason, factorization is a fundamental step towards solving any kind of equation in mathematics. Let’s discover out.
## How to aspect a Quadratic Equation?
Factoring a quadratic equation deserve to be characterized as the process of break the equation into the product the its factors. In various other words, us can additionally say that factorization is the turning back of multiplying out.
To resolve the quadratic equation ax 2 + bx + c = 0 by factorization, the following steps are used:
Expand the expression and also clear every fractions if necessary.Move every terms to the left-hand side of the equal to sign.Factorize the equation through breaking down the middle term.Equate each factor to zero and solve the direct equations
Example 1
Solve: 2(x 2 + 1) = 5x
Solution
Expand the equation and move every the terms to the left that the same sign.
⟹ 2x 2 – 5x + 2 = 0
⟹ 2x 2 – 4x – x + 2 = 0
⟹ 2x (x – 2) – 1(x – 2) = 0
⟹ (x – 2) (2x – 1) = 0
Equate each factor equal to zero and solve
⟹ x – 2 = 0 or 2x – 1 = 0
⟹ x = 2 or x = 1212
Therefore, the remedies are x = 2, 1/2.
Example 2
Solve 3x 2 – 8x – 3 = 0
Solution
3x 2 – 9x + x – 3 = 0
⟹ 3x (x – 3) + 1(x – 3) = 0
⟹ (x – 3) (3x + 1) = 0
⟹ x = 3 or x = -13
Example 3
Solution
Expand the equation (2x – 3)2 = 25 to get;
⟹ 4x 2 – 12x + 9 – 25 = 0
⟹ 4x 2 – 12x – 16 = 0
Divide every term by 4 to get;
⟹ x 2 – 3x – 4 = 0
⟹ (x – 4) (x + 1) = 0
⟹ x = 4 or x = -1
The are countless methods of factorizing quadratic equations. In this article, our focus will be based on how to aspect quadratic equations, in which the coefficient of x2 is either 1 or greater than 1.
Therefore, we will use the trial and also error technique to acquire the right factors for the given quadratic equation.
## Factoring once the Coefficient of x 2 is 1
To factorize a quadratic equation that the type x 2 + bx + c, the leading coefficient is 1. You need to determine two numbers whose product and also sum room c and also b, respectively.
CASE 1: once b and also c space both positive
Example 4
Solve the quadratic equation: x2 + 7x + 10 = 0
List under the components of 10:
1 × 10, 2 × 5
Identify two determinants with a product that 10 and a sum of 7:
1 + 10 ≠ 72 + 5 = 7.
Verify the factors using the distributive property of multiplication.
(x + 2) (x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10
The components of the quadratic equation are:(x + 2) (x + 5)
Equating each aspect to zero gives;
x + 2 = 0 ⟹x= -2
x + 5 = 0 ⟹ x = -5
Therefore, the systems is x = – 2, x = – 5
Example 5
x 2 + 10x + 25.
Solution
Identify two determinants with the product of 25 and sum the 10.
5 × 5 = 25, and also 5 + 5 = 10
Verify the factors.
x 2 + 10x + 25 = x 2 + 5x + 5x + 25
= x (x + 5) + 5x + 25
= x (x + 5) + 5(x + 5)
= (x + 5) (x + 5)
Therefore, x = -5 is the answer.
CASE 2: once b is positive and c is negative
Example 6
Solve x2 + 4x – 5 = 0
Solution
Write the determinants of -5.
1 × –5, –1 × 5
Identify the factors whose product is – 5 and sum is 4.
1 – 5 ≠ 4–1 + 5 = 4
Verify the components using the distributive property.
(x – 1) (x + 5) = x2 + 5x – x – 5 = x2 + 4x – 5(x – 1) (x + 5) = 0
x – 1 = 0 ⇒ x = 1, orx + 5 = 0 ⇒ x = -5
Therefore, x = 1, x = -5 are the solutions.
CASE 3: as soon as b and also c space both negative
Example 7
x2 – 5x – 6
Solution
Write down the factors of – 6:
1 × –6, –1 × 6, 2 × –3, –2 × 3
Now identify components whose product is -6 and sum is –5:
1 + (–6) = –5
Check the determinants using the distributive property.
(x + 1) (x – 6) = x2 – 6 x + x – 6 = x2 – 5x – 6
Equate each aspect to zero and solve to get;(x + 1) (x – 6) = 0
x + 1 = 0 ⇒ x = -1, orx – 6 = 0 ⇒ x = 6
Therefore, the solution is x=6, x = -1
CASE 4: once b is negative and c is positive
Example 8
x2 – 6x + 8 = 0
Solution
Write under all components of 8.
–1 × – 8, –2 × –4
Identify components whose product is 8 and sum is -6–1 + (–8) ≠ –6–2 + (–4) = –6
Check the components using the distributive property.
(x – 2) (x – 4) = x2 – 4 x – 2x + 8 = x2 – 6x + 8
Now equate each aspect to zero and also solve the expression come get;
(x – 2) (x – 4) = 0
x – 2 = 0 ⇒ x = 2, orx – 4 = 0 ⇒ x = 4
Example 9
Factorize x2 +8x+12.
Solution
Write under the factors of 12;
12 = 2 × 6 or = 4 × 3Find factors whose sum is 8:
2 + 6 = 82 × 6 ≠ 8
Use distributive property to check the factors;
= x2+ 6x +2x + 12 = (x2+ 6x) +(2x + 12) = x(x+6) +2(x+6)
= x (x + 6) +2 (x + 6) = (x + 6) (x + 2)
Equate each factor to zero to get;
(x + 6) (x + 2)
x = -6, -2
### Factoring as soon as the coefficient that x 2 is higher than 1
Sometimes, the top coefficient the a quadratic equation might be greater than 1. In this case, we can not resolve the quadratic equation through the usage of usual factors.
Therefore, we require to take into consideration the coefficient the x2 and also the factors of c to discover numbers whose amount is b.
Example 10
Solve 2x2 – 14x + 20 = 0
Solution
Determine the common factors of the equation.
2x2 – 14x + 20 ⇒ 2(x2 – 7x + 10)
Now us can find the factors of (x2 – 7x + 10). Therefore, write down factors of 10:
–1 × –10, –2 × –5
Identify factors whose amount is – 7:
1 + (–10) ≠ –7–2 + (–5) = –7
Check the factors by applying distributive property.
See more: How To Make Glass In Doodle God ? How Do You Make A Void In Doodle God
2(x – 2) (x – 5) = 2(x2 – 5 x – 2x + 10)= 2(x2 – 7x + 10) = 2x2 – 14x + 20
Equate each aspect to zero and solve;2(x – 2) (x – 5) = 0
x – 2 = 0 ⇒ x = 2, orx – 5 = 0 ⇒ x = 5
Example 11
Solve 7x2 + 18x + 11 = 0
Solution
Write under the factors of both 7 and 11.
7 = 1 × 7
11 = 1 × 11
Apply distributive property to check the factors as displayed below:
(7x + 1) (x + 11) ≠ 7x2 + 18x + 11
(7x + 11) (x + 1) = 7x2 + 7x + 11x + 11 = 7x2 + 18x + 11
Now equate each variable to zero and solve to get;
7x2 + 18x + 11= 0(7x + 11) (x + 1) = 0
x = -1, -11/7
Example 12
Solve 2x2 − 7x + 6 = 3
Solution
2x2 − 7x + 3 = 0
(2x − 1) (x − 3) = 0
x=1/2 or x=3
Example 13
Solve 9x 2 +6x+1=0
Solution
Factorize come give:
(3x + 1) (3x + 1) = 0
(3x + 1) = 0,
Therefore, x = −1/3
Example 14
Factorize 6x2– 7x + 2 = 0
Solution
6x2 – 4x – 3x + 2 = 0
Factorize the expression;
⟹ 2x (3x – 2) – 1(3x – 2) = 0
⟹ (3x – 2) (2x – 1) = 0
⟹ 3x – 2 = 0 or 2x – 1 = 0
⟹ 3x = 2 or 2x = 1
⟹ x = 2/3 or x = ½
Example 15
Factorize x2 + (4 – 3y) x – 12y = 0
Solution
Expand the equation;
x2 + 4x – 3xy – 12y = 0
Factorize;
⟹ x (x + 4) – 3y (x + 4) = 0
x + 4) (x – 3y) = 0
⟹ x + 4 = 0 or x – 3y = 0
⟹ x = -4 or x = 3y
Thus, x = -4 or x = 3y
### Practice Questions
Solve the complying with quadratic equations by factorization:
3x 2– 20 = 160 – 2x 2(2x – 3) 2 = 4916x 2 = 25(2x + 1) 2 + (x + 1) 2 = 6x + 472x 2+ x – 6 = 03x 2 = x + 4(x – 7) (x – 9) = 195x 2– (a + b) x + abdominal muscle = 0x2+ 5x + 6 = 0x2− 2x − 15 = 0 |
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# How do you find exponential decay rate?
Last updated date: 13th Jun 2024
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373.2k+ views
Hint:Here in this question we have to find the exponential decay rate. That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time t. So we can introduce a proportionality constant. Then further applying an integration on both sides and on simplification we get the required result.
Exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula $y = a{\left( {1 - b} \right)^x}$ where y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed. Exponential decays typically start with a differential equation of the form:
$\Rightarrow \,\,\dfrac{{dN}}{{dt}}\alpha \, - N(t)$
That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time t. So we can introduce a proportionality constant $\alpha$:
$\Rightarrow \,\,\dfrac{{dN}}{{dt}} = - \alpha \,N(t)$
We will now solve the equation to find a function of $N(t)$:
$\Rightarrow \,\,\dfrac{{dN}}{{N(t)}} = - \alpha \,dt$
$\Rightarrow \,\,\dfrac{{dN}}{N} = - \alpha \,dt$
Apply integration both sides, then
$\Rightarrow \,\,\int {\dfrac{{dN}}{N}} = \int { - \alpha } \,dt$
$\Rightarrow \,\,\int {\dfrac{{dN}}{N}} = - \alpha \int {dt} \,$
Using the integration formula $\int {\dfrac{1}{x}dx = \ln x + c}$ and $\int {dx = x + c}$, where c is an integrating constant.
$\Rightarrow \,\,\ln N = - \alpha t + c\,$
As we know the logarithm function is the inverse form of exponential function, then
$\Rightarrow \,\,N = {e^{ - \alpha \,\,t + c}}\,$
Or it can be written as:
$\therefore \,\,N = A{e^{ - \alpha \,t}}$
Where A is a constant.
Hence, the general form of exponential decay rate is $N = A{e^{ - \alpha \,t}}$.
Note:Exponential decay describes the process of reducing an amount by a constant percentage rate over a period of time. The integration is inverse of the differentiation so to cancel differentiation the integration is applied. Likewise the logarithm is inverse of exponential. Hence by using these concepts we obtain answers. |
# How to Calculate 1/1 Minus 18/20
Are you looking to work out and calculate how to subtract 1/1 from 18/20? In this really simple guide, we'll teach you exactly what 1/1 - 18/20 is and walk you through the step-by-process of how to subtract one fraction from another.
Want to quickly learn or show students how to calculate 1/1 minus 18/20? Play this very quick and fun video now!
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because to subtract a fractions from another we need to first make sure both fractions have the same denominator.
Let's set up 1/1 and 18/20 side by side so they are easier to see:
1 / 1 - 18 / 20
Our denominators are 1 and 20. What we need to do is find the lowest common denominator of the two numbers, which is 20 in this case.
If we multiply the first denominator (1) by 20 we will get 20. If we we multiply the second denominator (20) by 1 we will also get 20. We also need to multiply the numerators above the line by the same amounts so that the fraction values are correct:
1 x 20 / 1 x 20 - 18 x 1 / 20 x 1
This is what 1/1 minus 18/20 looks like with the same denominator:
20 / 20 - 18 / 20
Now that these fractions have been converted to have the same denominator, we can subtract one numerator from the other to make one fraction:
20 - 18 / 20 = 2 / 20
You're done! You now know exactly how to calculate 1/1 - 18/20. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
1/10
## Convert 1/1 minus 18/20 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
2 / 20 = 0.1
### Cite, Link, or Reference This Page
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• "How to Calculate 1/1 minus 18/20". VisualFractions.com. Accessed on January 31, 2023. http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-18-20/.
• "How to Calculate 1/1 minus 18/20". VisualFractions.com, http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-18-20/. Accessed 31 January, 2023.
• How to Calculate 1/1 minus 18/20. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-18-20/.
### Preset List of Fraction Subtraction Examples
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# What Are the Very Basic Concepts of Number Lines Which Everybody Should Know?
General » What Are the Very Basic Concepts of Number Lines Which Everybody Should Know?
by | Sep 2, 2021
In the world of mathematics, the number lines are considered to be the horizontal straight lines in which the integers will be placed at regular intervals. Every number into a particular sequence can be perfectly represented in the number line and this particular line will extend its self indefinitely on both sides.
The number line is the pictorial representation of the numbers on a straight line and it can be considered as the best possible reference for comparing and ordering the numbers. It can even be utilized in terms of representing any kind of real number that will include every whole number and every natural number. Just to recollect the whole number is considered to be the set of numbers that will include all the counting numbers as well as zero and the natural numbers will be including all the counting numbers without zero. Writing the numbers on the number line is a very easy process so that comparison of the numbers can be carried out very easily and there is no problem at any point in time.
The operations of the numbers can also be explained on the number line very easily. It is very much important for the kids to be clear about locating the numbers on the number nine and zero will always be the middle point of the number line. Every positive natural number will occupy the right side of the zero and every negative number will be occupying the left side of the number line. As the people will be moving on the left side value of the number will decrease for example one will be greater than -2. In the cases of number nine the integers, fractions, and decimals can be easily represented without any kind of problem.
## Important Facts Which the Kids Should Know About the Number Line:
On the number line, the number on the left will always be less than the number on the right
Similarly, the number on the right will always be greater than the number on the left.
It is very much important for the kids to have a good command over the number line of one- hundred so that they never get confused and can deal with things very easily.
Following are some of the very basic operations which the people can perform on the number line:
1. Adding the positive numbers: Whenever people will add two positive numbers the result will always be a positive number for example 1+5 will always come out to be six.
2. Adding negative numbers: Whenever the people will go with the option of adding to negative numbers the resultant will be the negative number for example -2 and -3 will come out to be -5.
3. Subtraction on the number line of positive numbers: Whenever people will subtract two positive numbers for example two from five the result will be positive which will be three.
4. Subtracting the negative numbers: Whenever the people will subtract to negative numbers and will move to the right as far as the value of the second number for example -4 from -2 the result will come out to be +2.
It is very much important for the kids to be clear about all these kinds of things and they should also have a good idea about the whole number and natural numbers as well as basic concepts associated with both of them. Apart from this kids also need to be clear about the concept of parallel lines as well as angle formation of all these kinds of lines so that there is no issue and everything is perfectly undertaken throughout the process.
Hence, depending upon platforms like Cuemath is the best way of ensuring that people have a good command over the entire thing very easily and there is no problem at any point in time. This is considered to be the best way of ensuring that people will be able to score well in the exams.
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# The moon subtends an angle of 57 minute at the base-line equal to the radius of the earth. What is the distance of the moon from the earth? [Radius of earth=$6.4\times {{10}^{6}}m$]A. $11.22\times {{10}^{8}}m$B. $3.86\times {{10}^{8}}m$C. $3.68\times {{10}^{-3}}cm$D. $3.68\times {{10}^{8}}cm$
Last updated date: 25th Jun 2024
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Hint: To solve this problem first we will convert the given angle in degree than in radians to calculate the required distance, because for small angles , angle can be written as arc length divided by radius. In the above problem arc is radius of earth and radius is distance of the moon from earth.
Formula used:
$\theta =\dfrac{arc}{radius}$
Complete step by step solution:
Given, moon subtends an angle of 57 minute at the base-line equal to radius of earth so, by converting it into degrees we have,
$\theta =57'$
$\Rightarrow \theta =\dfrac{57}{60}=\dfrac{19}{20}={{0.95}^{\circ }}$
$\Rightarrow \theta ={{0.95}^{\circ }}$
Now, to change angle theta into radian form we need to multiply the above value of theta with $\dfrac{\pi }{180}$so,
$\Rightarrow \theta ={{0.95}^{\circ }}\times \dfrac{\pi }{180}$
Now, we know that the above angle is very small so we can write angle theta by,
$\Rightarrow \theta =\dfrac{arc}{radius}$
Here, arc (Radius of earth R) and radius (Distance of the moon from earth D).
$\Rightarrow \theta =\dfrac{R}{D}$
$\Rightarrow D=\dfrac{R}{\theta }$
By putting the value of R from the problem and of theta calculated above we have,
$\Rightarrow D=\dfrac{6.4\times {{10}^{6}}}{{{0.95}^{\circ }}\times \dfrac{\pi }{180}}m$
$\Rightarrow D=\dfrac{11.52\times {{10}^{8}}}{{{0.95}^{\circ }}\times \pi }m$
$\Rightarrow D=3.86\times {{10}^{8}}m$
$\therefore$ The distance of the moon from the earth is $3.86\times {{10}^{8}}m$ so option (B) is correct.
In mathematics both radians and degrees represent angles in different notations, in which radian value can be defined using arc of circle and we know that $\pi$ radians is equal to ${{180}^{\circ }}$ so we can easily calculate the value of 1 degree which is given by,
$\Rightarrow {{1}^{\circ }}=\dfrac{\pi }{180}$ |
Benita and Chloe, from Drumbowie, suggested the following method:
We found the differences between the numbers and looked at that times table, and then we found out if the table we had was shifted up or down from the times table it was meant to be, and how much by.
Nikita commented how much easier Levels 1 and 2 were because you simply need to find the nth term rule.
Jamie gave an example:
14, 24, 34, 44, 54
If the unit digits are identical, the table will be a multiple of ten and the shifted up number will be the same as the unit digit - here, 4.
He also gave some rules for helping to determine the times table:
• If the numbers are all odd, the original table will be even and the shift will be odd.
• If the numbers are all even, both the table and the shift will be even.
• If the numbers are a mixture of odd and even, the table will be odd and the shifted up number will be even.
• If there are only two different unit digits then the table is probably a multiple of five.
• If the difference between two numbers is prime, then the table will be that prime number.
Lucia and Matt gave similar rules. Matt went on to share a harder example:
348, 92, 252, 284, 124
The differences are even, so we can see that the times table is even. We can also see that this sequence has more than 2 different unit digits, so cannot be a multiple of either 5 or 10.
To work out exactly what the times table is we need to start by subtracting the lowest number from the second lowest number: 124 - 92 = 32. From this we can tell that the times table must be a factor of 32. We then need to subtract the second lowest number from the third lowest number (252 - 124 = 128), and then the next pair (284 - 252 = 32), and then the final pair (348 -284 = 64).
This gives us four numbers: 32,128,32 and 64. The highest common factor of these numbers is 32. From this we can see that the times table is 32 as all the numbers are a multiple of 32 apart.
To find the shift up or down we can divide one of the numbers in our times table and look at the remainder, e.g. 92/32 = 2 remainder 28. From this, we can see that if we were to move our number down 28 it would be a multiple of 32. This gives us our solution - the times table is 32 and you need to move it down 28.
Well done to Sharanya who used the same method as Matt, and to everyone else who found a method. |
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