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# Linear Equation Problem Solving
Solving linear equations in algebra is done with multiplication, division, or reciprocals.
Solving linear equations in algebra is done with multiplication, division, or reciprocals.
Tags: Research Proposal Time ScheduleShort Narrative Stories EssaysBible And Mythology Creation EssaysTextile Business PlanDiscipline Life EssayEconomic Causes Civil War EssayDissertations HelpShort Form Business PlanProblem Solving In Linear Equation
In these equations, we will need to undo two operations in order to isolate the variable.
In each of the examples above, there was a single step to perform before we had our answer.
Solve: $$2x-7=13$$ Notice the two operations happening to $$x$$: it is being multiplied by 2 and then having 7 subtracted. But, only the $$x$$ is being multiplied by 2, so the first step will be to add 7 to both sides. Adding 7 to both sides: $$\begin 2x-7 &= 13\\ 2x-7 \color & =13 \color\\ 2x&=20\end$$ Now divide both sides by 2: $$\begin 2x &=20 \\ \dfrac&=\dfrac\\ x&= \boxed\end$$ Just like with simpler problems, you can check your answer by substituting your value of $$x$$ back into the original equation.
$$\begin2x-7&=13\\ 2(10) – 7 &= 13\\ 13 &= 13\end$$ This is true, so we have the correct answer.
Solve: $$3x 2=4x-1$$ Since both sides are simplified (there are no parentheses we need to figure out and no like terms to combine), the next step is to get all of the x’s on one side of the equation and all the numbers on the other side.
The same rule applies – whatever you do to one side of the equation, you must do to the other side as well!
It is possible to either move the $$3x$$ or the $$4x$$. Since it is positive, you would do this by subtracting it from both sides: $$\begin3x 2 &=4x-1\ 3x 2\color &=4x-1\color\ -x 2 & =-1\end$$ Now the equation looks like those that were worked before.
The next step is to subtract 2 from both sides: $$\begin-x 2\color &= -1\color\-x=-3\end$$ Finally, since $$-x= -1x$$ (this is always true), divide both sides by $$-1$$: $$\begin\dfrac &=\dfrac\ x&=3\end$$ You should take a moment and verify that the following is a true statement: $$3(3) 2 = 4(3) – 1$$ In the next example, we will need to use the distributive property before solving.
This isn’t 100% necessary for every problem, but it is a good habit so we will do it for our equations.
In this example, our original equation was $$4x = 8$$.
## Comments Linear Equation Problem Solving
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Introductory Statistics
# 4.8Discrete Distribution (Lucky Dice Experiment)
Introductory Statistics4.8 Discrete Distribution (Lucky Dice Experiment)
## Stats Lab
### Discrete Distribution (Lucky Dice Experiment)
Class Time:
Names:
Student Learning Outcomes
• The student will compare empirical data and a theoretical distribution to determine if a Tet gambling game fits a discrete distribution.
• The student will demonstrate an understanding of long-term probabilities.
Supplies
• one “Lucky Dice” game or three regular dice
Procedure
Round answers to relative frequency and probability problems to four decimal places.
1. The experimental procedure is to bet on one object. Then, roll three Lucky Dice and count the number of matches. The number of matches will decide your profit.
2. What is the theoretical probability of one die matching the object?
3. Choose one object to place a bet on. Roll the three Lucky Dice. Count the number of matches.
4. Let X = number of matches. Theoretically, X ~ B(______,______)
5. Let Y = profit per game.
Organize the DataIn Table 4.17, fill in the y value that corresponds to each x value. Next, record the number of matches picked for your class. Then, calculate the relative frequency.
1. Complete the table.
x y Frequency Relative Frequency
0
1
2
3
Table 4.17
2. Calculate the following:
1. $x ¯ x ¯$ = _______
2. sx = ________
3. $y ¯ y ¯$ = _______
4. sy = _______
3. Explain what $x ¯ x ¯$ represents.
4. Explain what $y ¯ y ¯$ represents.
5. Based upon the experiment:
1. What was the average profit per game?
2. Did this represent an average win or loss per game?
3. How do you know? Answer in complete sentences.
6. Construct a histogram of the empirical data.
Figure 4.8
Theoretical DistributionBuild the theoretical PDF chart for x and y based on the distribution from the Procedure section.
1. x y P(x) = P(y)
0
1
2
3
Table 4.18
2. Calculate the following:
1. μx = _______
2. σx = _______
3. μx = _______
3. Explain what μx represents.
4. Explain what μy represents.
5. Based upon theory:
1. What was the expected profit per game?
2. Did the expected profit represent an average win or loss per game?
3. How do you know? Answer in complete sentences.
6. Construct a histogram of the theoretical distribution.
Figure 4.9
Use the Data
## Note
RF = relative frequency
Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
1. P(x = 3) = _________________
2. P(0 < x < 3) = _________________
3. P(x ≥ 2) = _________________
Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.
1. RF(x = 3) = _________________
2. RF(0 < x < 3) = _________________
3. RF(x ≥ 2) = _________________
Discussion QuestionFor questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences.
2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions.
3. Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not.
4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.17 or Table 4.18 to change, and how would it change? Why? Why wouldn’t the other table change?
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## Elementary Geometry for College Students (6th Edition)
$\angle$5 is congruent to $\angle$6.
Let's call the unmarked angle, $\angle$7 (in the diagram, between angles 5 and 6). Because all right angles equal 90$^{\circ}$, we can add angles 5 and 7, creating m$\angle$5 + m$\angle$7 = 90$^{\circ}$. For the same reason (all right angles equal 90$^{\circ}$) we can add angles 6 and 7, creating m$\angle$6 + m$\angle$7 = 90$^{\circ}$. Using the Property of Substitution, m$\angle$5 + m$\angle$7 = m$\angle$6 + m$\angle$7 (since both sets of angles equal ninety degrees we can put them equal to each other). Using the Subtraction Property, we can subtract angle 7 from both sides of our equation, like this m$\angle$5 + m$\angle$7 - m$\angle$7 = m$\angle$6 + m$\angle$7 - m$\angle$7. This proves that m$\angle$5 = m$\angle$6. And lastly, using the Definition of Congruence $\angle$5 is congruent to $\angle$6. |
# Thread: Solving Applied Problems (word problem)
1. ## Solving Applied Problems (word problem)
Hi, If you could show how you solve this, that would be great. Thanks.
Mr. Silvester is 5yr older than his wife. Five years
ago his age was 4/3 her age. What are their ages now?
2. ## Re: Solving Applied Problems (word problem)
Let s be the age of Mr. Silvester and w be the age of his wife
s-5= w
s-5 = (4/3)(w-5)
Hence
w = (4/3)w - 20/3
-(1/3)w = -20/3
w = 20, s=25
3. ## Re: Solving Applied Problems (word problem)
I thought I was over this, but apparently not. I have completed all the rest of the hw without a bother, but when it comes to these word problems, I don't know how to form the problem. Here's another one: Chris is 10yr older than Josh. Next year, Chris will be twice as old as Josh. What are their ages?
So then...
Chris = (J+10) -1
Josh = (C/2) -1
How do I write the equation?
4. ## Re: Solving Applied Problems (word problem)
Hello, Sprinkledozer!
Chris is 10 years older than Josh.
Next year, Chris will be twice as old as Josh.
What are their present ages?
We can solve this with one variable, one equation.
Let $\displaystyle J$ = Josh's age (now).
Then Chris' age is $\displaystyle J + 10$
Next year, they both will be one year older.
Chris will be $\displaystyle (J +10) + 1 \,=\,J+11$
Josh will be: $\displaystyle J + 1$
$\displaystyle \text{At that time: }\:\underbrace{\text{Chris}}_{J+11}\;\underbrace{ \text{will be}}_{=}\;\underbrace{\text{twice as old as Josh}}_{2(J+1)}$
Solve the equation: .$\displaystyle J + 11 \:=\:2(J+1)$
. . . . . and we get: .$\displaystyle J = 9$
Then Chris $\displaystyle = J+10 = 19$ |
## Finding a basis for a kernel or image
A := M C C ( T) = ( 1 0 0 0 − 1 2) Studying the rank of A you can find the linearly independent columns of the matrix which span I m ( T) = T ( R 2 [ x]) and solving the system ( A
## 9.8: The Kernel and Image of a Linear Map
So suppose a polynomial a + b x + c x 2 is in the kernel, so we have T ( a + b x + c x 2) = 0. Then by the way the linear transformation is defined, − ( a + b + c) + ( a − c) x + ( b − c) x 2 + c x 3 = 0.
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## 2.2 Kernel and Range of a Linear Transformation
A basis for the kernel of A is Question : (1 point) Let -6 -2 4-6 A= -9-3 6 -9 -] Find a basis for the kernel of A (or equivalently, for the linear transformation T(x) = Ax). This problem has been
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## The Kernel and the Range of a Linear Transformation
Finding a basis for the kernel of the following linear transformation: $T(p(x))=p'(x)$,$T:P_3[\mathbb{R}]\rightarrow P_3[\mathbb{R}]$
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## MATH 304 Linear Algebra Lecture 15: Kernel and range
GIS. [Math] Finding Kernel, Range and Dimensions of a linear transformation. linear algebra. Find the kernel and range, and state their dimensions, of the following linear transformation \$L :
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Word Lesson: Exponential Growth
In order to solve problems involving exponential growth, it is necessary to
Exponential growth is generally applied to word problems such as compound interest problems and population growth problems. To grow exponentially means that the topic being studied is increasing in proportion to what was previously there. For example, money deposited in the bank earns interest that is added to the money previously in the bank.
Suppose you want to know how long will it take \$1200 to double if it is invested at compounded continuously.
First, we will need to use the exponential growth formula for compounding interest:
In the formula, A represents the amount of money that will be in the account when \$1200 is doubled. P represents principal - the amount of money currently being invested. The letter r stands for rate of interest, and t stands time in years. In this formula e represents the irrational number 2.71828…..
Now, we need to substitute known values for the variables in the formula. The problem asks how long it will take \$1200 to double. Therefore, A is 2400 (the value of 1200 doubled) in this problem. P is the money to be invested, so P is 1200. The rate, r, is which is or 0.105 as a decimal. Time t is what we are trying to find. So we have the following:
Finally we must solve the equation for time t. To do so, first divide both sides by 1200 to simplify the equation.
Now, we take the natural log of each side of the equation. For a reminder on taking the log of both sides as well as the properties of logs, please examine the material in this companion lesson.
Using the following property of logs, , we have:
Since the is equal to 1, we will substitute 1 for to give us the following:
Use a calculator to find the value for the ln 2 and then divide each side by 0.105 to obtain the final answer:
Therefore, we have determined that if \$1200 is invested at compounded continuously, it will take 6.6 years for the money to double.
Examples
Growth of bacteria in food products causes a need to “time-date” some products (like milk) so that shoppers will buy the product and consume it before the number of bacteria grows too large and the product goes bad. Suppose that the formula represents the growth of bacteria in a food product. The variable t represents time in days and represents the number of bacteria in millions. If the product cannot be eaten after the bacteria count reaches 4,000,000, how long will it take? What is your answer?
Victor wants to buy a new car that costs \$90,000. He has saved \$20,000. Determine how many years it will take his \$20,000 to grow to \$90,000 at 6% interest compounded continuously. What is your answer?
Examples
In a given year, the minimum wage was only \$1.60 per hour. Use the exponential growth formula to predict when that minimum wage in the United States will reach 8.50 per hour if the rate of growth in the minimum wage is 3.9%. In the formula 1.6 represents a current minimum wage; A represents the amount of minimum wage you wish to obtain, and t represents time in years. 4.3 years 42.8 years 136.2 years -42.8 years What is your answer?
Scientific research has shown that the risk of having a car accident increases exponentially as the concentration of alcohol in the blood increases. A formula that models the risk of an accident is the following: . In the formula, R represents the % of risk. [R will be given as a percent and should be used as a percent rather than a decimal in working the problem.] Find the blood alcohol concentration ( ) that corresponds to a 25% risk of a car accident. -0.25 11.15 -0.11 0.11 What is your answer?
As you can see, this type of problem requires that you write an exponential growth function based on given information. You must then correctly substitute given values for variables and solve the equation you obtain. In solving the equation you must convert the exponential equation to a log equation and correctly use the log property . At the conclusion of the problem, you should always check for the reasonableness of your solution.
D Saye
Show Related AlgebraLab Documents |
# Video: Integrating Functions Numerically β Left Endpoints
Calculate the left endpoint estimate of β«_(0)^(4) π₯Β² + 2 dπ₯ with π = 2 subintervals. Is the result an overestimate or underestimate of the actual value?
03:13
### Video Transcript
Calculate the left endpoint estimate of the definite integral between zero and four of π₯ squared plus two with respect to π₯ with π equals two subintervals. Is the result an overestimate or underestimate of the actual value?
Remember, we can estimate the solution to definite integrals by considering the area between the curve and the π₯-axis in terms of rectangles. These are called Riemann sums. And here weβre looking to find the left Riemann sum. When working with left Riemann sums, we take values of π from zero to π minus one. And an approximation to our definite integral is the sum of Ξπ₯ times π of π₯π for values of π between zero and π minus one, where Ξπ₯ is π minus π over π. Remember, π is the number of subintervals and π₯π is equal to π plus π lots of Ξπ₯.
If we looked carefully at our integral, we see that the lower limit is zero and the upper limit is four. So weβre going to let π be equal to zero and π be equal to four. And also, weβre told that π is equal to two. The first step with any Riemann sum is usually to work out the value of Ξπ₯. π is zero, π is four. So Ξπ₯ is four minus zero divided by π, which is two. And that tells us that Ξπ₯ is equal to two. Once weβve worked out Ξπ₯, we can then usually work out an expression for π₯π. Itβs π which we know to be zero plus Ξπ₯, which is two lots of π. Well, that simplifies to two π.
And, of course, for the summation formula, we actually need to work out π of π₯π. Well, it follows that π of π₯π must be equal to π of two π. So letβs substitute two π into our function. Thatβs two π squared plus two. Thatβs four π squared plus two. And we now have everything we need to approximate our integral. We take values of π in our sum from zero to π minus one. Well, two minus one is one. Then, itβs Ξπ₯ times π of π₯π, which we just found to be four π squared plus two.
Itβs worth knowing that we can take out any constant factors independent of π. Well, four is independent of π. So we can write this as four times the sum of two π squared plus one for values of π between zero and one. Letβs now substitute π equals zero and π equals one into this expression and find its sum. Thatβs four lots of two times zero squared plus one plus two times one squared plus one, which is 16. And so, we found an estimation of this definite integral.
Our next step is to work out whether this is an overestimate or underestimate of the exact value. Well, we need to decide whether π₯ squared plus two is an increasing or decreasing function over our interval. To do this, weβre going to differentiate π₯ squared plus two. And we get two π₯. Now, of course, this is greater than zero for values of π₯ greater than zero. So this tells us the function π₯ squared plus two must be increasing for values of π₯ greater than zero. What this says for our curve is that itβs sloping upwards for values of π₯ greater than zero.
Well, when working with increasing functions, the left Riemann sum gives an underestimate, whereas the right Riemann sum gives an overestimate for the approximation of the integral. The reverse, of course, is true for decreasing functions. And so, this means the left endpoint estimate of our definite integral is an underestimate. |
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# 11.4: Simplification of Radical Expressions
Difficulty Level: At Grade Created by: CK-12
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Progress
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What if you wanted to perform an operation on two radical expressions, like $\sqrt{32x} - \sqrt{8x}$ , in which the numbers under the radical signs were different? How could you find the difference? After completing this Concept, you'll be able to add, subtract, multiply, and divide radical expressions.
### Guidance
When we add and subtract radical expressions, we can combine radical terms only when they have the same expression under the radical sign. This is a lot like combining like terms in variable expressions.
#### Example A
Simplify the following expressions as much as possible.
a.) $4 \sqrt{2} + 5 \sqrt{2}$
b.) $2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2}$
$4 \sqrt{2} + 5 \sqrt{2} & = 9 \sqrt{2}\\& \text{or} \\ 2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2} & = 7 \sqrt{3} + 9 \sqrt{2}$
It’s important to reduce all radicals to their simplest form in order to make sure that we’re combining all possible like terms in the expression. For example, the expression $\sqrt{8} - 2\sqrt{50}$ looks like it can’t be simplified any more because it has no like terms. However, when we write each radical in its simplest form we get $2\sqrt{2} - 10 \sqrt{2}$ , and we can combine those terms to get $-8 \sqrt{2}$ .
#### Example B
Simplify the following expressions as much as possible.
a) $4 \sqrt[3]{128} - \sqrt[3]{250}$
b) $3 \sqrt{x^3} - 4x \sqrt{9x}$
Solution
a) $\text{Re-write radicals in simplest terms:} && & = 4 \sqrt[3]{2 \cdot 64} - \sqrt[3]{2 \cdot 125} = 16 \sqrt[3]{2} - 5 \sqrt[3]{2}\\\text{Combine like terms:} && & = 11 \sqrt[3]{2}$
b) $\text{Re-write radicals in simplest terms:} && 3 \sqrt{x^2 \cdot x} - 12x \sqrt{x} & = 3x \sqrt{x} - 12x \sqrt{x}\\\text{Combine like terms:} && & =-9x \sqrt{x}$
When we multiply radical expressions, we use the “raising a product to a power” rule: $\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}$ . In this case we apply this rule in reverse.
#### Example C
Simplify the expression $\sqrt{6} \cdot \sqrt{8}$ .
Solution:
$\sqrt{6} \cdot \sqrt{8} = \sqrt{6 \cdot 8} = \sqrt{48}$
Or, in simplest radical form: $\sqrt{48} = \sqrt{16 \cdot 3} = 4 \sqrt{3}.$
We’ll also make use of the fact that: $\sqrt{a} \cdot \sqrt{a} = \sqrt{a^2} = a$ .
When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately. For example, $a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}$ .
#### Example D
Multiply the following expressions.
a) $\sqrt{2}\left(\sqrt{3} + \sqrt{5}\right )$
b) $2 \sqrt{x}\left (3 \sqrt{y} - \sqrt{x}\right )$
c) $\left (2 + \sqrt{5}\right )\left (2 - \sqrt{6}\right )$
d) $\left (2 \sqrt{x} + 1\right )\left (5 - \sqrt{x}\right )$
Solution
In each case we use distribution to eliminate the parentheses.
a) $\text{Distribute} \ \sqrt{2} \ \text{inside the parentheses:} && \sqrt{2}\left (\sqrt{3} + \sqrt{5}\right ) & = \sqrt{2} \cdot \sqrt{3} + \sqrt{2} \cdot \sqrt{5}\\\text{Use the raising a product to a power'' rule:} && & = \sqrt{2 \cdot 3} + \sqrt{2 \cdot 5}\\\text{Simplify:} && & =\sqrt{6} + \sqrt{10}$ b) $\text{Distribute} \ 2 \sqrt{x} \ \text{inside the parentheses:} && & =(2 \cdot 3)\left (\sqrt{x} \cdot \sqrt{y}\right ) - 2 \cdot \left ( \sqrt{x} \cdot \sqrt{x}\right )\\\text{Multiply:} && & =6 \sqrt{xy} - 2 \sqrt{x^2}\\\text{Simplify:} && & =6 \sqrt{xy} - 2x$ c) $\text{Distribute:} && (2 + \sqrt{5})(2 - \sqrt{6}) & = (2 \cdot 2) - \left (2 \cdot \sqrt{6}\right ) + \left( 2 \cdot \sqrt{5} \right ) - \left ( \sqrt{5} \cdot \sqrt{6} \right )\\\text{Simplify:}&& & =4 - 2 \sqrt{6} + 2 \sqrt{5} - \sqrt{30}$ d) $\text{Distribute:} && \left (2 \sqrt{x} - 1\right )\left (5 - \sqrt{x}\right ) &=10 \sqrt{x} - 2x - 5 + \sqrt{x}\\\text{Simplify:} && & =11 \sqrt{x} - 2x - 5$
Rationalize the Denominator
Often when we work with radicals, we end up with a radical expression in the denominator of a fraction. It’s traditional to write our fractions in a form that doesn’t have radicals in the denominator, so we use a process called rationalizing the denominator to eliminate them.
Rationalizing is easiest when there’s just a radical and nothing else in the denominator, as in the fraction $\frac{2}{\sqrt{3}}$ . All we have to do then is multiply the numerator and denominator by a radical expression that makes the expression inside the radical into a perfect square, cube, or whatever power is appropriate. In the example above, we multiply by $\sqrt{3}$ :
$\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$
Cube roots and higher are a little trickier than square roots.
#### Example E
How would we rationalize $\frac{7}{\sqrt[3]{5}}$ ?
Solution:
We can’t just multiply by $\sqrt[3]{5}$ , because then the denominator would be $\sqrt[3]{5^2}$ . To make the denominator a whole number, we need to multiply the numerator and the denominator by $\sqrt[3]{5^2}$ :
$\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}} = \frac{7 \sqrt[3]{25}}{\sqrt[3]{5^3}} = \frac{7 \sqrt[3]{25}}{5}$
Trickier still is when the expression in the denominator contains more than one term.
#### Example F
Consider the expression $\frac{2}{2 + \sqrt{3}}$ . We can’t just multiply by $\sqrt{3}$ , because we’d have to distribute that term and then the denominator would be $2 \sqrt{3} + 3$ .
Instead, we multiply by $2 - \sqrt{3}$ . This is a good choice because the product $\left (2 + \sqrt{3}\right )\left (2 - \sqrt{3}\right )$ is a product of a sum and a difference, which means it’s a difference of squares. The radicals cancel each other out when we multiply out, and the denominator works out to $\left (2 + \sqrt{3} \right )\left (2 - \sqrt{3}\right ) = 2^2 - \left ( \sqrt{3}\right )^2 = 4 - 3 = 1$ .
When we multiply both the numerator and denominator by $2 - \sqrt{3}$ , we get:
$\frac{2}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2\left (2 - \sqrt{3}\right )}{4 - 3} = \frac{4 - 2 \sqrt{3}}{1} = 4 - 2 \sqrt{3}$
Now consider the expression $\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}}$ .
In order to eliminate the radical expressions in the denominator we must multiply by $\sqrt{x} + 2 \sqrt{y}$ .
We get: $\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}} \cdot \frac{\sqrt{x} + 2 \sqrt{y}}{\sqrt{x} + 2 \sqrt{y}} = \frac{\left (\sqrt{x} - 1\right )\left (\sqrt{x} + 2 \sqrt{y}\right )} {\left (\sqrt{x} - 2 \sqrt{y} \right ) \left ( \sqrt{x} + 2 \sqrt{y} \right )} = \frac{x - 2 \sqrt{y} - \sqrt{x} + 2 \sqrt{xy}}{x - 4y}$
Watch this video for help with the Examples above.
### Vocabulary
• When we multiply radical expressions , we use the “raising a product to a power” rule:
$\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}$ .
• When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately:
$a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}$
### Guided Practice
Simplify the following expressions as much as possible.
a) $4 \sqrt{3} + 2 \sqrt{12}$
b) $10 \sqrt{24} - \sqrt{28}$
Solutions:
a) $\text{Simplify} \ \sqrt{12} \ \text{to its simplest form:} && & =4 \sqrt{3} + 2 \sqrt{4 \cdot 3} = 4 \sqrt{3} + 6 \sqrt{3}\\\text{Combine like terms:} && & =10 \sqrt{3}$
b) $\text{Simplify} \ \sqrt{24} \ \text{and} \ \sqrt{28} \ \text{to their simplest form:} && & =10 \sqrt{6 \cdot 4} - \sqrt{7 \cdot 4} = 20 \sqrt{6} - 2 \sqrt{7}\\\text{There are no like terms.}$
### Practice
Simplify the following expressions as much as possible.
1. $3\sqrt{8} - 6 \sqrt{32}$
2. $\sqrt{180} + \sqrt{405}$
3. $\sqrt{6} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}$
4. $\sqrt{8x^3} - 4x \sqrt{98x}$
5. $\sqrt{48a} + \sqrt{27a}$
6. $\sqrt[3]{4x^3} + x \cdot \sqrt[3]{256}$
Multiply the following expressions.
1. $\sqrt{6}\left (\sqrt{10} + \sqrt{8}\right )$
2. $\left (\sqrt{a} - \sqrt{b}\right )\left (\sqrt{a} + \sqrt{b}\right )$
3. $\left (2 \sqrt{x} + 5\right )\left (2 \sqrt{x} + 5\right )$
Rationalize the denominator.
1. $\frac{7}{\sqrt{5}}$
2. $\frac{9}{\sqrt{10}}$
3. $\frac{2x}{\sqrt{5x}}$
4. $\frac{\sqrt{5}}{\sqrt{3y}}$
5. $\frac{12}{2 - \sqrt{5}}$
6. $\frac{6 + \sqrt{3}}{4 - \sqrt{3}}$
7. $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{x}}$
8. $\frac{5y}{2 \sqrt{y} - 5}$
### Vocabulary Language: English
A radical expression is an expression with numbers, operations and radicals in it.
Rationalize the denominator
Rationalize the denominator
To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
Aug 13, 2012
Oct 28, 2014 |
Question
Use the divisibility tests to determine whether each number is divisible by 2,3,4,5,6 and 10.
Factors and multiples
In the following exercises, use the divisibility tests to determine whether each number is divisible by 2,3,4,5,6 and 10.
800
2021-08-11
To determine
Whether the given number is divisible by 2,3,4,5,6 and 10 or not.
Explanation of Solution
Given:
The given number is 800.
Concept Used: We use the concept of divisibility test.
A number is divisible by 2 if it has 0,2,4,6 or 8 as its unit place.
A number is divisible by 3 if the sum of its digits is a multiple of 3.
A number is divisible by 4 if its last two digits are a multiple of 4.
A number is divisible by 5 if it has 0 or 5 as its unit place.
A number is divisible by 6 if it is divisible by 2 and 3 both.
A number is divisible by 10 if it has 0 as its unit place.
Calculation:
As the given number is 800, 0 being its unit place, it is divisible by 2, 5 and 10. Now to check for 3, the sum of its digits is $$\displaystyle{8}+{0}+{0}={8}$$, which is not a multiple of 3, hence 800 is not divisible by 3. As 00 is a multiple of 4, it means 800 is divisible by 4. Also, as 800 is not divisible by 3, it is not divisible by 6
Conclusion:
The given number is divisible by 2,4,5 and 10 but not 3 and 6. |
1. ## Modular Arithmetic examples
Just picked out a few examples from a practice exam.
1)Solving the equation az=1 mod m
(a) a=3 m = 10 (ANS = 7)
(b) a=5 m = 18
2)Solve the following congruence, equations for x.(Giving the solution in the form a mod m)
(a) 3x - 1 = 7 mod 10 (ANS = x = 6 mod 10)
Any chance of a step by step explanation on a few of them .
2. Hello,
Originally Posted by Ciachyou
Just picked out a few examples from a practice exam.
1)Solving the equation az=1 mod m
(a) a=3 m = 10 (ANS = 7)
(b) a=5 m = 18
In fact, you're asked to find the inverse of numbers modulo m.
Let's do it for these two, I guess they'll give you quite the general idea. The key is to use the extended Euclidean algorithm (you can look for it in google)
Find z in $3z \equiv 1 \bmod 10$
You know that since 3 and 10 are coprime, you'll have a remainder 1 while continuing doing the algorithm. And this is why it is useful.
Do the algorithm with 10 and 3 as starting points :
$10=3 \times 3+1$
Hence $1=10-3 \times 3$
You can note that $10-3 \times 3 \equiv -3 \times 3 \bmod 10$ because $10 \equiv 0 \bmod 10$
So we have $-3 \times 3 \equiv 1 \bmod 10$
$z=-3 \bmod 10$
In general, we want z to be the least positive integer satisfying the condition.
So you can add 10 : $z=7$
For the second one, you're looking for z in $5z \equiv 1 \bmod 18$
Algorithm for 5 and 18 :
$18=5 \times 3+3$
My method is to write successively the remainders.
$3=18-5 \times 3$
Now, do the division for 5 and the new remainder, 3.
$5=3 \times 1+2$
$\implies 2=5-3$
Division for 3 and the new remainder, 2.
$3=2+1$
$\implies 1=3-2$
But we saw that $2=5-3$ :
$\implies 1=3-(5-3)=2 \times 3-5$
$\implies 1=2 \times (18-5 \times 3) -5=18 \times 2-5 \times 6-5=\boxed{18 \times 2-5 \times 7}$
So we have $-5 \times 7 \equiv 1 \bmod 18$
$\implies z \equiv -7 \bmod 18 \equiv 11 \bmod 18$
3. Thanks, that helped me out alot.
Got a b+ |
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## Sunday, September 12, 2021
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## Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers
Plus Two Maths Chapter 5 Continuity and Differentiability question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus Two Maths Chapter 5 Continuity and Differentiability chapter wise questions and answers to help them secure good marks in class tests and exams.
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Question 1.
Consider the function g(x) = $$\frac{|x-2|}{x-2}$$
1. Find the domain and range of the function g(x). (2)
2. Check whether the g(x) is continuous at x = 2. (1)
1. The function is not defined at points where the denominator is zero.
i.e., x – 2 = 0 ⇒ x = 2.
∴ domain is R – {2}.
g(x) = $$\frac{|x-2|}{x-2}$$ = $$\left\{\begin{array}{c}{\frac{x-2}{x-2}, \quad x-2>0} \\{\frac{-(x-2)}{x-2}, \quad x-2<0}\end{array}=\left\{\begin{array}{ll}{1,} & {x>2} \\{-1,} & {x<2}\end{array}\right.\right.$$
∴ range is {-1, 1}
2. The function g(x) is not defined at x = 2. Therefore discontinuous.
Question 2.
(i) If f(x) = x+|x| + 1, then which of the follow represents f (x) (1)
(ii) Test whether f (x) is continuous at x=0. Explain. (2)
(i) (b) Since, f(x) = $$\left\{\begin{array}{c}{x+x+1, x \geq 0} \\{x-x+1, \quad x<0} \end{array}=\left\{\begin{array}{c}{2 x+1, x \geq 0} \\{1, x<0}\end{array}\right.\right.$$
(ii) We have ,f (0) = 1,
Continuous at x = 0.
Question 3.
Consider the function f(x) = $$\left\{\begin{array}{ll}{\frac{\sin x}{x}} & {, x<0} \\{x+1} & {, x \geq 0}\end{array}\right.$$
1. Find $$\lim _{x \rightarrow 0} f(x)$$ (2)
2. Is f (x) continuous at x= 0? (1)
1. To find $$\lim _{x \rightarrow 0} f(x)$$ we have to find f(0 )and f(0+)
f(0) = $$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$$,
f(0+) = $$\lim _{x \rightarrow 0}$$ + 1 = 0 + 1 = 1
f(0) = f(0+) = 1 .Therefore $$\lim _{x \rightarrow 0} f(x)$$ = 1
2. Here, f (0) = 0 + 1 = 1 = f(0) = f(0+) = 1
Therefore continuous at x = 0.
Question 4.
Consider the figure and answer the following questions.
(i) Identify the graphed function. (1)
(ii) Discuss the continuity of the above function at x = 2. (2)
(i) (b) f(x) = $$\left\{\begin{array}{ll}{\frac{|x-2|}{x-2},} & {x \neq 2} \\{0,} & {x=2}\end{array}\right.$$
(ii) From the figure we can see that
f(2) = 1, f(2+) = -1 and f(2) = 0
Therefore, f(2) = 1 ≠ f(2+) = -1 ≠ f(2) = 0. Discontinuous.
Question 5.
Consider f(x) = $$\left\{\begin{array}{ll}{2 x} & {\text { if } x<2} \\{2} & {\text { if } x=2} \\ {x^{2}} & {\text { if } x>2}\end{array}\right.$$
(a) Find $$\lim _{x \rightarrow 2^{-}} f(x)$$ and $$\lim _{x \rightarrow 2^{+}} f(x)$$ (2)
(b) f(x) is continuous. If not so, how can you make it continuous. (1)
Therefore f(x) is not continuous at x = 2.
If f(2) = 4, then f(x) becomes continuous.
Question 6.
If y = log10x + logx10 + logxx + log1010. Find $$. Answer: y = logx log10e + loge10 logxe + 1 + 1 Question 7. Examine the continuity of the function Answer: In both the intervals x = 1 and x < 1 the function f(x) is a polynomial so continuous. So we have to check the continuity at x = 1. f(1) = 2 Since f(x) is continuous at x = 1. Question 8. Find [latex]\frac{d y}{d x}$$ of the following (3 score each)
1. 2x + 3y = sinx
2. xy + y2 = tanx + y
3. x3 + x2y + xy2 + y3 = 81
4. sin2x + cos2y = 1
5. $$\sqrt{x}$$ + $$\sqrt{y}$$ = 1
6. x2 + xy + y2 = 7
7. x2(x – y) = y2(x + y)
8. xy2 + x2y = 2
9. sin y = xcos (a + y)
1. Given; 2x + 3y = sinx
Differentiating with respect to x;
2. Given; xy + y2 = tanx + y
Differentiating with respect to x;
3. Given; x3 + x2y + xy2 + y3 = 81
Differentiating with respect to x;
4. Given; sin2x + cos2y = 1
Differentiating with respect to x;
5. $$\sqrt{x}$$ + $$\sqrt{y}$$ = 1
Differentiating with respect to x;
6. x2 + xy + y2 = 7
Differentiating with respect to x;
7. x2(x – y) = y2(x + y)
Differentiating with respect to x;
8. xy2 + x2y = 2
Differentiating with respect to x;
9. sin y = xcos (a + y)
⇒ x = $$\frac{\sin y}{\cos (a+y)}$$
Diff. with respect to y.
Question 9.
Find $$\frac{d y}{d x}$$ of the following (3 score each)
(viii) x = a(t – sint), y = a(1 – cost)
(ix) y = et cost, x = et sint.
(i) We know; y = sin-1$$\left(\frac{2 x}{1+x^{2}}\right)$$
⇒ y = = 2 tan-1(x)
Differentiating with respect to x;
(ii) We know; y = tan-1$$\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$$
Put x = tanθ ⇒ θ = tan-1x
⇒ y = tan-1(tan3θ)
⇒ y = 3 tan-1(x)
Differentiating with respect to x;
(iii) We know; y = sin-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
Put x = tanθ ⇒ θ = tan-1x
(iv) We know; y = sec-1 = $$\left(\frac{1}{2 x^{2}-1}\right)$$
Put x = cosθ ⇒ θ = cos-1x
⇒ y = sec-1 sec2θ = 2θ
⇒ y = 2cos-1(x)
Differentiating with respect to x;
(v) ∴ y = tan-1$$\sqrt{\tan ^{2} x / 2}$$ = tan-1 tanx/2 = x/2
$$\frac{d y}{d x}=\frac{1}{2}$$.
= π – 2tan-1x
$$\frac{d y}{d x}=\frac{-2}{1+x^{2}}$$.
(vii) Let x = sinθ and $$\sqrt{x}$$ = sinφ
= sin-1 (sinθcosφ + cosφsinφ)
= sin-1 (sin(θ + φ)) = θ + φ
= sin-1x + sin-1$$\sqrt{x}$$
$$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}}$$.
(ix) y = etcost ⇒ $$\frac{d y}{d t}$$ = – et sin t + et
x = et sint ⇒ $$\frac{d x}{d t}$$ = et cos t + et sin t
Question 10.
If y = log$$\left(\frac{1}{x}\right)$$, Show that $$\frac{d y}{d x}$$ + ey = 0.
Given, y = log$$\left(\frac{1}{x}\right)$$ ? $$\left(\frac{1}{x}\right)$$ = ey …….(1)
Question 11.
If ey (x+1) = 1. Show that
1. $$\frac{d y}{d x}$$ = -ey (2)
2. $$\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$$ (1)
1. ey (x+1) = 1
Differentiating ey +ey(x +1) $$\frac{d y}{d x}$$ = 0
2.
Question 12.
(i) Evaluate $$\lim _{x \rightarrow 0} \frac{k \cos x}{\pi-2 x}$$ (2)
(Hint: Put π – 2x = y , where Iris a constant)
(ii) Find the value of k if f (x) is a continuous function given by (1)
(i) $$\lim _{x \rightarrow \pi / 2} \frac{k \cos x}{\pi-2 x}=k \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\pi-2 x}$$,
Put π – 2x = y when Put x → π/2, y → 0
(ii) Since f (x) is continuous
Question 13.
If f(x) $$f(x)=\left\{\begin{array}{cc}{x-[x]} & {, x<.2} \\{0} & {; x=2} \\{3 x-5} & {, x>2} \end{array}\right.$$
1. Find $$\lim _{x \rightarrow 2} f(x)$$ (2)
2. Is f(x) continuous at x = 2? (1)
1. To find $$\lim _{x \rightarrow 2} f(x)$$
we have to find f(2) and f(2+)
f(2) = $$\lim _{x \rightarrow 2} x-[x]$$ = 2 -1 = 1,
f(2+) = $$\lim _{x \rightarrow 2}$$ 3x – 5 = 6 -5 = 1
f(2) = f(2+) = 1.
Therefore $$\lim _{x \rightarrow 2}$$ f(x) = 1
2. Here, f(2) = 0 ≠ f(2) = f(2+) = 1.
Therefore discontinuity at x = 2.
### Plus Two Maths Continuity and Differentiability Four Mark Questions and Answers
Question 1.
If x = 2cosθ; y = 3sinθ
1. Find $$\frac{d y}{d x}$$.
2. Find $$\frac{d^{2} y}{d x^{2}}$$
1. x = 2cosθ ⇒ $$\frac{d x}{d θ}$$ = -2sinθ
y = 3sinθ ⇒ $$\frac{d x}{d θ}$$ = 3cosθ
2.
Question 2.
If y = (tan-1 x)2, show that (x2 +1)2 y2 + 2x(x2 +1) y1 = 2.
y = (tan-1 x)2
⇒ y1 = 2(tan-1 x) $$\frac{1}{1+x^{2}}$$
⇒ (1 + x2)y1 = 2(tan-1 x)
⇒ (1 + x2)y2 + y12x = 2 $$\frac{1}{1+x^{2}}$$
⇒ (1 + x2)2 y2 + x(1 + x2)y1 = 2.
Question 3.
Find $$\frac{d y}{d x}$$ if y = sin-1 $$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$, 0 < x < 1.
Put x = tanθ
Question 4.
Let f(x) = $$\left\{\begin{array}{ll}{\cos x,} & {0 \leq x \leq c} \\{\sin x,} & {c<x \leq \pi}\end{array}\right.$$
1. Find the value of c if / is continuous on [0, π].
2. Show that is f not differentiable at the point c.
1. Since f is continuous on [0, π], we have;
$$\lim _{x \rightarrow c^{+}}$$ f(x) = $$\lim _{x \rightarrow c^{-}}$$ f(x) = f(c)
⇒ $$\lim _{x \rightarrow c^{+}}$$ sinx = $$\lim _{x \rightarrow c^{-}}$$ cosx = cosc
⇒ sinc = cosc ⇒ c = $$\frac{π}{4}$$.
2. $$f^{\prime}(x)=\left\{\begin{array}{ll}{-\sin x,} & {0 \leq x \leq c} \\{\cos x,} & {c<x \leq \pi}\end{array}\right.$$
Left derivative at $$\frac{\pi}{4}$$ = – sin $$\frac{\pi}{4}$$ = –$$\frac{1}{\sqrt{2}}$$
Right derivative at $$\frac{\pi}{4}$$ = cos $$\frac{\pi}{4}$$ = $$\frac{1}{\sqrt{2}}$$
Left derivative at $$\frac{\pi}{4}$$ ≠ Right derivative at $$\frac{\pi}{4}$$
Therefore is not differentiable at the point c.
Question 5.
1. Find $$\frac{d y}{d x}$$ if x = 2sinθ; y = 3cosθ
2. Which among the following functions is differentiable on R?
(a) |sinx|
(b) |cosx|
(c) cos|x|
(d) sin|x|
1. $$\frac{d x}{d θ}$$ = 2cosθ; $$\frac{d y}{d θ}$$ = -3 sinθ ⇒ $$\frac{d y}{d x}$$ = $$-\frac{3}{2}$$tanθ
2. (c) cos|x|
(Since cos x is an even function, it treats x and -x in the same way).
Question 6.
(i) Examine whether the function defined by $$f(x)=\left\{\begin{array}{ll}{x+5,} & {x \leq 1} \\{x-5,} & {x>1}\end{array}\right.$$ is continuous or not. (2)
(ii) If x = sin-1t, y = cos-1t, a > 0, show that $$\frac{d y}{d x}=-\frac{y}{x}$$
f(x) is not continuous.
Question 7.
(i) If $$f(x)=\left\{\begin{array}{ll}{1-x,} & {0 \leq x \leq 1} \\{1+x,} & {1<x \leq 2}\end{array}\right.$$ then which of the following is not true (1)
(a) f is continuous in ( 0, 1 )
(b) f is continuous in (1, 2 )
(c) f is continuous in [ 0, 2 ]
(d) f is continuous in [ 0,1 ]
if $$\left\{\begin{array}{cc}{1,} & {x \leq 3} \\{a x+b} & {, \quad 3<x<5} \\{7,} & {5 \leq x}\end{array}\right.$$
(ii) Find f(3+) and f(5) (1)
(iii) Hence find the value of ‘a’ and ‘b’ so that f(x) is continuous. (2)
(i) (c) Since f is not continuous at x = 1.
(iii) Since f (x) is continuous, it is continuous at x = 3 and x = 5
∴ f(3+) = f(3) ⇒ 3a + b = 1 ____(1)
and f(5) = f(5) ⇒ 5a + b = 7 ____(2)
(2) – (1) ⇒ 2a = 6, a = 3
(1) ⇒ b = 1 – 3 a ⇒ b = -8
∴ a = 3, b = – 8.
Question 8.
Consider f(x) = $$\left\{\begin{array}{ll}{2 x+3,} & {x \leq 2} \\{x+2 k,} & {x>2}\end{array}\right.$$
1. Find f(2) (1)
2. Evaluate $$\lim _{x \rightarrow 2^{+}}$$f(x) (1)
3. Find the value of k, if is continuous at x = 2. (2)
1. f(2) = 2(2) + 3 = 7
2. Here, f(x) = x + 2k for x > 2.
$$\lim _{x \rightarrow 2^{+}}$$f(x) = $$\lim _{x \rightarrow 2}$$(x + 2k) = 2 + 2k.
3. Since f (x) is continuous at x = 2
We have, f(2) = $$\lim _{x \rightarrow 2^{+}}$$f(x)
⇒ 7 = 2 + 2k ⇒ k = $$\frac{5}{2}$$
Question 9.
Find $$\frac{d y}{d x}$$ of the following (4 score each)
1. y = (logx)cosx
2. x = 2at2, y = at4
3. x = a(cosθ + θsinθ), y = a(sinθ – θcosθ)
4. y=xx
5. y =(x log x)log(logx)
6. y = $$\sqrt{\sin x \sqrt{\sin x+\sqrt{\sin x+\ldots .}}}$$
7. yx = xsiny
8. y =(log x)x + xlogx
9. y = (sinx)x + sin-1 $$\sqrt{x}$$
1. Given; y = (logx)cosx, taking log on both sides;
log y = cosxlog(logx),
Differentiating with respect to x;
2. Given; x = 2at2 ⇒ $$\frac{d x}{d t}$$ = 4at
3. Given; x = a(cosθ + θsinθ)
$$\frac{d x}{d \theta}$$ = a(-sinθ + θcosθ + sinθ) = aθcosθ
y = a(sinθ – θcosθ)
$$\frac{d x}{d \theta}$$ = a(cosθ – θ(-sinθ) – cosθ) = aθ sinθ
$$\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$$
4. y= xx; Taking log on both sides;
log y = x log x
5. y = (x log x)log logx
Taking log on both sides;
log y = (log log x) [log (xlogx)]
6. y = $$\sqrt{\sin x+y}$$ ⇒ y2 = sinx + y
⇒ y2 – y = sinx
7. yx = xsin y;
Taking log on both sides;
xlogy = siny log x
8. y = (log x)x + xlogx = u + v
9. y = (sinx)x + sin-1$$\sqrt{x}$$
Let u = (sinx)x ⇒ log u = x log sinx
Question 10.
Find $$\frac{d^{2} y}{d x^{2}}$$ of the following
1. y = x2 + 3x + 2 (2)
2. y = tan-1x (2)
1. Given; y = x2 + 3x + 2
Differentiating with respect to x;
$$\frac{d y}{d x}$$ = 2x + 3
Differentiating again with respect to x;
$$\frac{d^{2} y}{d x^{2}}$$ = 2.
2. Given; y = tan-1x
Differentiating with respect to x; $$\frac{d y}{d x}$$ = $$\frac{1}{1+x^{2}}$$
Differentiating again with respect to x;
$$\frac{d^{2} y}{d x^{2}}$$ = $$-\frac{1}{\left(1+x^{2}\right)^{2}} \cdot 2 x$$.
Question 11.
Match the following. (4)
Question 12.
If x – sint and y = sinmt show that
(i) y = sin(m sin-1x) (1)
(ii) $$\frac{d y}{d x}$$ (1)
(iii) (1 – x2) y2 – xy1 + m2y = 0 (2)
(i) x = sint, y = sinmt
t = sin-1x ⇒ y = sin(msin-1x).
multiplying with $$\sqrt{1-x^{2}}$$
(1 – x2) y2 – xy1 = -m2y
(1 – x2) y2 – xy1 + m2y = 0.
Question 13.
Consider the function f(x) = x(x – 2), x ∈ [1, 3]. Verify mean value theorem for the function in[1, 3].
f(x) = x(x – 2) = x2 – 2x ⇒ f'(x) = 2x – 2.
As f is a polynomial, it is continuous in the interval [1, 3] and differentiable in the interval (1, 3).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (1,3)such that.
Hence MVT is verified.
Question 14.
Verify Lagranges’ Mean value theorem for the function f(x) = 2x2 – 10x + 29 in [2, 9]
f(x) = 2x2 – 10x + 29; f'(x) = 4x – 10.
As f is a polynomial, it is continuous in the interval [2, 9] and differentiable in the interval (2, 9).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (2, 9) such that.
Hence MVT is verified.
Question 15.
Let f(x) = x(x – 1)(x – 2), x ∈ [0, 2]
1. Find f(0) and f(2) (1)
2. Find f'(x) (1)
3. Find the values of x when f'(x) = 0 verify Rolle’s theorem. (2)
1. f(0) = 0, f(2) = 2(2 – 1)(2 – 2) = 0
2. We have, f(x) = x3 – 3x2 + 2x
⇒ f'(x) = 3x2 – 6x + 2.
3. f'(x) = 3x2 – 6x + 2 = 0
Clearly all the three conditions of Rolle’s theorem are satisfied and 1 ± $$\frac{1}{\sqrt{3}}$$ ∈ (0, 2).
Question 16.
Verify Rolle’s Theorem for the function
f(x) = x2 + 2x – 8, x ∈ [-4, 2]
f(x) = x2 + 2x – 8, f'(x) = 2x + 2.
As f is a polynomial, it is continuous in the interval [-4, 2] and differentiable in the interval (-4, 2).
f(-4) = 16 – 8 – 8 = 0
f(2) = 4 + 4 – 8 = 0
f'(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1 ∈ (-4, 2) Hence Rolle’s Theorem is verified.
Question 17.
Examine that Rolle’s Theorem is applicable to the following function in the given intervals, justify your answer.
1. f(x) = [x], x ∈ [5, 9]
2. f(x) = x2 – 1, x ∈ [1, 2]
1. The function f(x) = [x] is not differentiable and continuous at integral values. So in the given interval [5, 9] the function is neither differentiable nor continuous at x = 6, 7 ,8. Therefore Rolle’s Theorem is not applicable.
2. The function f(x) = x2 – 1 is a polynomial function so differentiable and continuous.
f(1) = 1 – 1 = 0, f(2) = 4 – 1 = 3
f(1) ≠ f(2) . Therefore Rolle’s Theorem is not applicable.
Question 18.
Examine the continuity of the function
$$f(x)=\left\{\begin{array}{cc}{|x|+3,} & {x \leq-3} \\{-2 x,} & {-3<x<3} \\{6 x+2,} & {x \geq 3}\end{array}\right.$$
In the intervals x ≤ -3, f(x) is the sum of a constant function and modulus function so continuous. In the intervals -3 < x < 3 and x ≥ 3the function f(x) is a polynomial so continuous. Hence we have to check the continuity at x = -3, x = 3.
At x = -3
f(-3) = 6
f(x) is continuous at x = -3.
At x = 3
f(3) = 6(3) + 2 = 20
Since $$\lim _{x \rightarrow 3^{-}}$$f(x) = f(3), f(x) is not continuous at x = 3.
Question 19.
Test continuity for the following functions.
= 0 × a finite quantity between -1 and 1 = 0 Also f(0) = 0
Therefore f(x) is continuous at x = 0.
Therefore f(x) is discontinuous at x = 0
But f(1) = 2
∴ f(x) is discontinuous at x = 1.
Question 20.
If y = $$\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$$ prove that
(i) (1 – x2) y2 = (sin-1x)2 (1)
(ii) (1 – x2)y1 – xy = 1 (1)
(iii) (1 – x2) y2 – 3xy1 – y = 0 (2)
(ii) Differentiating
(1 – x2) 2y.y1 + y2x – 2x = $$\frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}}$$
(1 – x2)2yy1 – 2xy2 = 2y
(1 – x2) y1 – xy = 1.
(iii) Again differentiating
(1 – x2) y2 + y1 x – 2x – xy1 – y = 0
(1 – x2) y2 – 3xy1 – y = 0.
Question 21.
At what point on the curve y = x2, x ∈ [-2, 2] at which the tangent is parallel to x-axis?
Y = x2, a continuous function on [-2, 2] and differentiable on [-2, 2] f(2) = 4 = f(-2). All conditions of Rolles theorem is satisfied. Given the tangent is parallel to x-axis.
f1 (x) = 2x
f1(c) = 2c
f1(c) = 0 ⇒ 2c = 0 ⇒ c = 0 ∈ [-2, 2]
where c = 0, y = 0
Therefore (0, 0) is the required point.
Question 22.
is continuous in the interval [-1 1].
(a) Find $$\lim _{x \rightarrow 0}$$f(x) (2)
(b) Find f(0). (1)
(c) Find P. (1)
(c) Since f is continuous in [-1 1] it is continuous at 0.
Therefore P = $$-\frac{1}{2}$$.
Question 23.
If ax2 + 2hxy + by2 = 1
1. Find $$\frac{d y}{d x}$$ (1)
2. Find $$\frac{d^{2} y}{d x^{2}}$$ (3)
1. We have, ax2 + 2hxy + by2 = 1 ___(1)
Differentiating w.r.t.x, we get,
2. Differentiating (2) w.r.tx, we get,
Question 24.
Consider the function y = x$$\sqrt{x}$$
1. Express the above function as logy = $$\left(x+\frac{1}{2}\right)$$ logx (2)
2. Find $$\frac{d y}{d x}$$ (2)
1. Given, y = x$$\sqrt{x}$$. Take log on both sides,
2. We have, logy = $$\left(x+\frac{1}{2}\right)$$ logx
Differentiating w.r.t x, we get,
### Plus Two Maths Continuity and Differentiability Six Mark Questions and Answers
Question 1.
1. Verify mean value theorem for the function f(x) = (x – 2)2 in [1, 4].
2. Find a point on the curve y = (x – 2)2 at which the tangent is parallel to the chord joining the points (1, 1) and (4, 2)
3. Find a point on the above curve at which the tangent is parallel to the x-axis.
1. f(x) = (x – 1)2, x ∈ [1, 4]
f(x) is continuous in [1, 4]
f'(x) = 2(x – 2) is differentiable in [1, 4]
Then there exists c ∈ [1, 4] so that
Hence Mean Value Theorem is verified.
2. c = $$\frac{5}{2}$$ will be the x-coordinate to the point of contact of tangent and the curve, then y = (x – 2)2 ⇒ y = ($$\frac{5}{2}$$ – 2)2 = $$\frac{1}{4}$$
Therefore the point is ($$\frac{5}{2}$$, $$\frac{1}{4}$$).
3. The tangent parallel to x- axis will have
f'(c) = 0 ⇒ 2(c – 2) = 0 ⇒ c = 2
Then; x = 2 ⇒ y = (2 – 2)2 = 0
Therefore the point is (2, 0).
Question 2.
1. Differentiate xsinx w.r.t.x (2)
2. If x = at2, y = 2at, then find $$\frac{d y}{d x}$$ (2)
3. If y = sin-1(cosx) + cos-1(sinx), then find $$\frac{d y}{d x}$$. (2)
1. Let y = xsinx, take log on both sides,
log y = sinx logx differentiate w. r.t.x, we get
2.
3. Given, y = sin-1(cosx) + cos-1(sinx)
Differentiate w.r.t. x, we get $$\frac{d y}{d x}$$ = -2.
Question 3.
1. Differentiate $$\frac{x-1}{x-3}$$ with respect to x.(2)
2. Differentiate $$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$ with respect to x. (4)
1. Let y = $$\frac{x-1}{x-3}$$ Differentiate w.r.t. x, we get;
2. Given, y = $$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Take log on both sides;
Differentiate w.r.t. x, we get;
Question 4.
(i) Define |x|
(a) |x| = $$\sqrt{x^{2}}$$
(b) |x| = x
(c) |x| = -x
(d) |x| = x2
(ii) At which point $$\frac{d}{d x}$$|x| does not exist?
Find $$\frac{d}{d x}$$ |x|. (2)
(iii) Find $$\frac{d}{d x}$$|x3 – 7x| . Also, find the point at which the derivative exists. (3)
(i) (a) |x| = $$\sqrt{x^{2}}$$.
(ii) At x = 0, $$\frac{d}{d x}$$ |x| does not exist.
Does not exist at
x3 – 7x = 0 ⇒ x(x2 – 7) = 0
⇒ x = 0, x2 – 7 = 0 ⇒ x = ±$$\sqrt{7}$$.
Question 5.
(i) Match the following (4)
(ii) If log (x2 + y2) = 2 tan-1$$\left(\frac{y}{x}\right)$$, then, show that $$\frac{d y}{d x}=\frac{x+y}{x-y}$$ (2)
(i)
(ii) Given, log (x2 + y2) = 2 tan-1$$\left(\frac{y}{x}\right)$$.
Differentiate w.r.to x, we get;
Question 6.
If x = a sec3θ and y = a tan3θ
(i) Given, x = a sec3θ
Differentiate w.r.to θ, we get;
$$\frac{d x}{d \theta}$$ = 3a sec2θ. secθ. tanθ = 3a sec3 θ. tan θ
Given, y = a tan3θ .
Differentiating w.r.to θ, we get
$$\frac{d x}{d \theta}$$ = 3a tan2 θ. sec2θ.
(iii) We have, $$\frac{d y}{d x}$$ = sinθ
Differentiating w.r.to x, we get
(iv) We have,
Question 7.
Consider the function $$f(x)=\left\{\begin{array}{cc}{1-x} & {, \quad x<0} \\{1} & {x=0} \\ {1+x} & {, \quad x>0}\end{array}\right.$$
(i) Compete the following table. (2)
(ii) Draw a rough sketch of f (x). (2)
Since, f (- 2) = 1 – (- 2) = 3, f (-1) = 1 – (-1) = 2,
f(1) = 1 + (1) = 2, f (2) = 1 + (2) = 3.
(ii)
(iii) From the graph we can see that there is no break or jump at x = 0. Therefore continuous.
From the figure we can see that
f(0) = 1 f(0+) = 1 and f(0) = 1
Hence, f(0) = f(0+) = f(0) = 1.
Therefore continuous.
Question 8.
Consider the equation $$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$$
(i) Simplify the above equation to sin-1x – sin-1y = 2cot-1 a by giving suitable substitution.
(ii) Prove that $$\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$$
(ii) We have; sin-1 x – cos-1y = 2cot-1a.
Differentiating w.r.t x, we get,
Question 9.
(i) Match the following. (4)
(ii) If y = ea cos-1x, then show that (1 – x2)y2 – xy1 – a2y = 0 (2)
(i)
(ii) Given, y = ea cos-1x ____(1)
Differentiating w.r.to x,
Again differentiating w. r.to x
⇒ (1 – x2)y2 – xy1 = a2. ea cos-1x
⇒ (1 – x2)y2 – xy1 = a2y
⇒ (1 – x2)y2 – xy1 – a2y = 0.
Question 10.
(i) Match the following (3)
Differentiate the following
(ii) y =$$\frac{1}{5 x^{2}+3 x+7}$$ (1)
(iii) y = 3cosec4(7x) (1)
(iv) y = e2log tan 5x (1)
(i)
(iii) Given,
(iv) Given,
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# How do you find the slope of a line with x-intercept 3 and y-intercept -4?
Jul 19, 2018
The slope is $\frac{4}{3}$.
#### Explanation:
An $x$-intercept is the value of $x$ when $y = 0$, so an $x$-intercept of $3$ can be written as a coordinate on the graph as $\left(3 , 0\right)$.
Likewise, an $y$-intercept is the value of $y$ when $x = 0$, so an $y$-intercept of $- 4$ can be written as a coordinate on the graph as $\left(0 , - 4\right)$
Now we have two points, $\left(3 , 0\right)$ and $\left(0 , - 4\right)$.
To find the slope given two points, we use the formula $\text{rise"/"run}$, or $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$.
Plug in the given points into the formula:
$\frac{- 4 - 0}{0 - 3} = - \frac{4}{-} 3 = \frac{4}{3}$
Therefore, the slope is $\frac{4}{3}$.
Hope this helps! |
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## properties of subtraction dez 30, 2020 Sem categoria
Properties Of Subtraction. The method of complements is a technique used to subtract one number from another using only addition of positive numbers. For instance, Example 1- Let us consider two numbers 3 and 5. Example 1: 3 … For example, you can do: $$12-5$$ (since $$12$$ is greater than $$5$$), ⦠Properties of subtraction of rational numbers. VIEW THIS YOUTUBE LINK There are some facts related to subtraction. 5 - 0 = 5 11 - 0 = 11 2. Note: Associative does not hold for subtraction. VIEW THIS YOUTUBE LINK There are some facts related to subtraction. Multiplication of Rational Numbers. Then, Intuitively, this is obvious. r Properties of subtraction of matrices. For example, in the adjacent picture, there are 5 â 2 applesâmeaning 5 apples with 2 taken away, resulting in a total of 3 apples. This movement to the left is modeled by subtraction: Now, a line segment labeled with the numbers 1, 2, and 3. The European method corrects by increasing the subtrahend digit si+1 by one. Another method that is useful for mental arithmetic is to split up the subtraction into small steps.[19]. One simply adds the amount needed to get zeros in the subtrahend.[20]. A column of two numbers, with the lower number in red, usually indicates that the lower number in the column is to be subtracted, with the difference written below, under a line. Subtractive property states that if we subtract zero (0) from any number, the answer or difference will be the non-zero number. If a < b, then subtraction a â b is not possible in whole numbers. To subtract a binary number y (the subtrahend) from another number x (the minuend), the ones' complement of y is added to x and one is added to the sum. The "Properties of Addition and Rules of Subtraction Poster and Sort" set is perfect when introducing this concept to your students. From 3, it takes 3 steps to the left to get to 0, so 3 â 3 = 0. r There are some facts related to subtraction. 119 - 59 = 60. Example: 0 - 7 = -7 7 - 0 = 7 Numbers Page 6th Grade Page From Properties of Subtraction to HOME PAGE. Subtraction : Observe the following examples: (i) 12 - 5 = 7 (ii) 5 - 12 = -7 (vi) 18 - (-13) = 18 + 13 = 31 From the above examples, it is clear that subtraction of any two integers is again an integer. The additive inverse is the opposite (negative) of the number. Properties More challenging questions presented on the second page. Properties of Properties of Addition and Subtraction of Integers Closure property. 23 + 12 = 35 (Result is an integer) 5 + (-6) = -1 (Result is an integer) Therefore, the difference of 5 and 2 is 3, that is, 5 â 2 = 3. Addends are the numbers that are being added. c If a < b, then subtraction a – b is not possible in whole numbers. All of these rules can be proven, starting with the subtraction of integers and generalizing up through the real numbers and beyond. The subtraction of whole numbers is not commutative, that is, if a and b are two whole numbers, then in general a – b is not equal to (b – a). Use this Google Search to find what you need. What's the Addition Property of Equality? Click The Property You Want 119 - 59 = 60. If we reverse the order of the matrices and subtract both of them with the same order/dimensions, the result will differ. Then we can check subtraction by adding. [7] Using the gerundive suffix -nd results in "subtrahend", "thing to be subtracted". With a little bit of help from a handful of Star Wars characters these posters informative and fun to look at. 1. The subtraction then proceeds in the hundreds place, where 6 is not less than 5, so the difference is written down in the result's hundred's place. Closure property under addition states that the sum of any two integers will always be an integer. There are also situations where subtraction is "understood", even though no symbol appears: Formally, the number being subtracted is known as the subtrahend,[4][5] while the number it is subtracted from is the minuend. i We cannot interchange the order of ⦠When three or more numbers are added, the sum is the same regardless of the order of addition. a The following properties of the expected value are also very important. It states that addition of two Integers always results in an Integer. Closure Property: The product of two rational numbers is always a rational number. Properties of Subtraction. Property 3:If a is any whole number other than zero, then a – 0 = a but 0 – a is not defined.Verification: We know that 15 – 0 = 15, but 0 – 15 is not possible. ln(x) = log e (x) = y . The minuend is 704, the subtrahend is 512. If a/b and c/d are any two rational numbers, then a×b/b×db d is also a rational number. d : Subtraction doesn´t have the same properties as addition. Associative & Commutative Properties Practice recognizing and working with the properties of addition problems. Subtraction is anti-commutative, meaning that if one reverses the terms in a difference left-to-right, the result is the negative of the original result. This is an addition equation: Addition equations have addends and a sum. When subtracting two numbers with units of measurement such as kilograms or pounds, they must have the same unit. These facts are called the properties of subtractions. Use this Google Search to find what you need. This allows an easier use of associativity and commutativity. {\displaystyle {\begin{array}{rrrr}&\color {Red}-1\\&C&D&U\\&7&0&4\\&5&1&2\\\hline &1&9&2\\\end{array}}{\begin{array}{l}{\color {Red}\longleftarrow {\rm {carry}}}\\\\\longleftarrow \;{\rm {Minuend}}\\\longleftarrow \;{\rm {Subtrahend}}\\\longleftarrow {\rm {Rest\;or\;Difference}}\\\end{array}}}. Place the posters on the board during your lesson or o To subtract arbitrary natural numbers, one begins with a line containing every natural number (0, 1, 2, 3, 4, 5, 6, ...). Conclude that 26 cannot be subtracted from 11; subtraction becomes a. Brownell, W.A. The Distributive Property is easy to remember, if you recall that "multiplication distributes over addition". This method was commonly used in mechanical calculators, and is still used in modern computers. 10 - 1 = 9 455 - 1 = 454 3. The figure above illustrates the addition property of zero and it can be written as 2 + 0 = 2. This helps to keep the ring of real numbers "simple", by avoiding the introduction of "new" operators such as subtraction. Closure under Subtraction Closure property under subtraction states that the difference of any two integers will always be an integer. Addition property of zero: The addition property of zero says that a number does not change when adding or subtracting zero from that number. Such a case uses one of two approaches: Subtraction of real numbers is defined as addition of signed numbers. Percentage change represents the relative change between the two quantities as a percentage, while percentage point change is simply the number obtained by subtracting the two percentages.[8][9][10]. o Properties of subtraction of rational numbers. The expected value of is a weighted average of the values that can take on. To resolve this issue, one must establish an order of operations, with different orders yielding different results. And adding 1 to get the two's complement can be done by simulating a carry into the least significant bit. All of this terminology derives from Latin. Instead of finding the difference digit by digit, one can count up the numbers between the subtrahend and the minuend.[18]. Example: Also, 8 + 17 = 25Therefore, 25 – 8 = 17 or, 8 + 17 = 25Similarly 89 – 74 = 15 because 74 + 15 = 89. 1. There are a few properties that are applicable while dealing with the Subtraction of Rational Numbers. y ⟵ Adding exponents and subtracting exponents really doesnât involve a rule. Beginning at the one's place, 4 is not less than 2 so the difference 2 is written down in the result's one's place. D A-B B-A; The negative of matrix A is written as (-A) such that if the addition of matrix with the negative matrix will always produce a null matrix. This page was last edited on 17 December 2020, at 23:15. e Subtraction is not commutative. In general, the expression. Therefore, the set of integers is closed under subtraction. A number - 1 = its predecessor Eg. Thus, to subtract is to draw from below, or to take away. In your example of the slope formula, you're just multiplying the numerator and denominator by -1. 1 â 3 = not possible.We add a 10 to the 1. • there are two subtraction properties. n Multiplication of Rational Numbers. s For this reason, we call $0$ the additive identity. n We are now done, the result is 192. [11][12] Although a method of borrowing had been known and published in textbooks previously, the use of crutches in American schools spread after William A. Brownell published a studyâclaiming that crutches were beneficial to students using this method. There are four mathematical properties which involve addition. That is, if a, b, c are three whole numbers, then in general a – (b – c) is not equal to (a – b) – c.Verification:We have,20 – (15 – 3) = 20 – 12 = 8,and, (20 – 15) – 3 = 5 – 3 = 2Therefore, 20 – (15 – 3) ≠ (20 – 15) – 3.Similarly, 18 – (7 – 5) = 18 – 2 = 16,and, (18 – 7) – 5 = 11 – 5 = 6.Therefore, 18 – (7 – 5) ≠ (18 – 7) – 5. Expectation of a positive random variable. Because the 10 is "borrowed" from the nearby 5, the 5 is lowered by 1. When a child is adding or subtracting large groups of numbers, remind her that the number zero … Let be an integrable random variable defined on a sample space.Let for all (i.e., is a positive random variable). 87 - 36 = 51. formula, you're using the property the square of any non-zero real number is positive. COMMUTATIVE PROPERTY. 28 - 0 = 28 u 87 - 36 = 51. Closure Property: The product of two rational numbers is always a rational number. Any number added to 0 gives the original number. subtraction properties ⢠there are two subtraction properties. If a number is raised to a power, add it to another number raised to a power (with either a different base or different exponent) by calculating the result of the exponent term and then directly adding this to the other. This is a good introductory worksheet that contains only simple 1 and 2-digit numbers. Closure under Addition. Because 0 is the additive identity, subtraction of it does not change a number. Changes in percentages can be reported in at least two forms, percentage change and percentage point change. 9 Other properties. u D Subtraction also obeys predictable rules concerning related operations, such as addition and multiplication. Basically, the rational numbers are the fractions which can be represented in the number line. It takes 2 steps to the left to get to position 1, so 3 â 2 = 1. That is, if a, b, c are three whole numbers, then in general a – (b – c) is not equal to (a – b) – c. If a, b and c are whole numbers such that a – b = c, then b + c = a. Then, Intuitively, this is obvious. Otherwise, mi is increased by 10 and some other digit is modified to correct for this increase. The answer is 1, and is written down in the result's hundred's place. We used these properties again each time we introduced a new system of numbers. The difference is written under the line. Knowing all the addition number facts will help with subtraction. U Ex: (â 21) â (â 9) = (â 12); 8 â 3 = 5. Letâs consider the following pairs of integers. A number - 0 = The same number Eg. Properties - Table of Contents. 1 130 - 60 = 70. So, we add 10 to it. Subtraction is signified by the minus sign, â. Then base e logarithm of x is. The method of complements is especially useful in binary (radix 2) since the ones' complement is very easily obtained by inverting each bit (changing "0" to "1" and vice versa). Rather it increases the subtrahend hundred's digit by one. ⢠the subtraction property of equality states that when the same quantity is subtracted from both sides of an equation, the two sides remain equal. Closure property under multiplication: Integers are closed under multiplication, i.e. A number - 0 = The same number Eg. To help you understand each and every property we have taken enough examples and explained all of them step by step. These facts are called the properties of subtractions. This means subtracting zero from any number does not change the answer or the number sign (+, -). The identity property also applies to subtraction since 3 - 0 = 3. The subtrahend digits are s3 = 5, s2 = 1 and s1 = 2. Subtraction of natural numbers and its properties Unlike the sum, when subtracting two natural numbers, the first one has to be greater than the second (otherwise you do not get a natural number). Then the subtraction proceeds by asking what number when increased by 1, and 5 is added to it, makes 7. The System of Integers under Subtraction A variant of the American method where all borrowing is done before all subtraction.[16]. Properties of Multiplication Commutative property of multiplication. remaining un-declined as in, Paul E. Peterson, Michael Henderson, Martin R. West (2014), Susan Ross and Mary Pratt-Cotter. In what is known in the United States as traditional mathematics, a specific process is taught to students at the end of the 1st year (or during the 2nd year) for use with multi-digit whole numbers, and is extended in either the fourth or fifth grade to include decimal representations of fractional numbers. e n Subtraction is usually written using the minus sign "â" between the terms;[3] that is, in infix notation. 7 r Commutative Property: When two numbers are added, the sum is the same regardless of the order of the addends. e Recognize the Identity Properties of Addition and Multiplication. Then we move on to subtracting the next digit and borrowing as needed, until every digit has been subtracted. The commutative property and associative property are not applicable to subtraction, but subtraction has a property called subtractive property of zero. 2 The solution is to consider the integer number line (..., â3, â2, â1, 0, 1, 2, 3, ...). Among the various properties of integers, closure property under addition and subtraction states that the sum or difference of any two integers will always be an integer i.e. For example, 26 cannot be subtracted from 11 to give a natural number. This property is a subset of Properties Of Subtraction. Commutative property: When two numbers are added, the sum is the same regardless of the order of the addends. Subtraction Facts. For example: 9 - 5 = 4. From position 3, it takes no steps to the left to stay at 3, so 3 â 0 = 3. 15 â 9 = ... Now the subtraction works, and we write the difference under the line. Or want to know more information Closure Property The System of Integers in Addition. "Subtraction" is an English word derived from the Latin verb subtrahere, which in turn is a compound of sub "from under" and trahere "to pull". 1234 â 567 = can be solved in the following way: The same change method uses the fact that adding or subtracting the same number from the minuend and subtrahend does not change the answer. "Subtraction in the United States: An Historical Perspective,", Learn how and when to remove this template message, "List of Arithmetic and Common Math Symbols", The Many Ways of Arithmetic in UCSMP Everyday Mathematics. Integers can be added and subtracted to each other. Check out the Closure, Commutative, Distributive and Associative Properties of Rational Numbers under Subtraction Operation. e y = x. EXAMPLE: (2 + … for any two integers a and b, a â b is an integer. Properties of natural number subtraction. In most cases, the difference will have the same unit as the original numbers. For Addition The sum of two or more real numbers is always the same regardless of the order in which they are added. Symbolically, if a and b are any two numbers, then The explanation of each of the integer properties is given below. Addition Properties. Notice how it mirrors the Subtraction Property of Equality. Examples: a) a+b=b+aa + b = b + aa+b=b+a b) 5+7=7+55 + 7 = 7 + 55+7=7+5 c) −4+3=3+−4{}^ - 4 + 3 = 3 + {}^ - 4−4+3=3+−4 d) 1+2+3=3+2+11 + 2 + 3 = 3 + 2 + 11+2+3=3+2+1 For Multiplication The product of two or more real numbers is not affected by the order in which they are being multiplied. The result is expressed with an equals sign. Some properties of subtraction of whole numbers are: Property 1: If a and b are two whole numbers such that a > b or a = b, then a â b is a whole number. 7 â 4 = 3This result is only penciled in. Properties - Topics. 1 Math Only Math is based on the premise that children do not make a distinction between play and work and learn best when learning becomes play and play becomes learning.However, suggestions for further improvement, from all quarters would be greatly appreciated. When. Natural logarithm (ln) rules & properties. ⟵ Changing the order of multiplication doesnât change the product. Some properties of subtraction of whole numbers are: If a and b are two whole numbers such that a > b or a = b, then a – b is a whole number. 4 t Subtraction is not commutative. Example: There is an additional subtlety in that the student always employs a mental subtraction table in the American method. But 3 â 4 is still invalid, since it again leaves the line. Property 4:The subtraction of whole numbers is not associative. Associative & Commutative Properties Practice recognizing and working with the properties of addition problems. We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Solving equations can be tough, especially if you've forgotten or have … In their place, one places plus or minus signs depending on whether the minuend is greater or smaller than the subtrahend. 1 Hence, in general (a – b) is not equal to (b – a). The "Properties of Addition and Rules of Subtraction Poster and Sort" set is perfect when introducing this concept to your students. Example: Some of the worksheets for this concept are 76274 r 03 l1 l8 2603 814 pm 1 addition, Properties of addition and subtraction, Addition properties, Properties of real numbers, Addition properties, The addition and subtraction properties of equality, Addition properties 1, Commutative property of addition 1. 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Than two numbers, the difference of 5 and 2 is 3, it takes 2 to... Thus, to subtract is to draw from below, or to take away leaves the line 20 ] property... Know more information about Math only Math widgets are defective +, - ) = 25, Did n't what. -7 7 - 0 = 7 properties of rational numbers is always rational... Carry into the original set properties is given below Wars characters these posters informative and to. 'Ve forgotten or have … 2 … subtraction is an additional subtlety in that section, we keep this.. Of position 3, it takes 3 steps to the left is anticommutative, meaning changing! Six months later, 20 % of widgets made in a factory are.. -Nd results in an integer subtract zero ( 0 ) from any number, the we! '' of the partial differences is the difference will be the non-zero number ''. Variable defined on a sample space.Let for all ( i.e., is a subset properties... Number facts will help with subtraction. 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West ( 2014 ), which goes down 1. Caught on rapidly, displacing the other methods of subtraction. [ 20 ] properties of subtraction.. Number minus or plus zero is the subtractive identity '' good introductory worksheet that contains only simple 1 s1! Leading digit 1 '' of the addends as kilograms or pounds, they must have same. - properties of subtraction = -7 7 - 3 = 4 asking what number when increased by 10 some. Addition of two approaches: subtraction doesn´t have the same regardless of the integers, these are and! Sign, â this Google Search to find what you were looking?... Applied them to solving equations can be proven, starting with the property. Look at. [ 20 ] â b contains two terms, namely a and b, ab is integer... 9 ) = log e ( x ) = y while dealing the. Of complements is a good introductory worksheet that contains only simple 1 and properties of subtraction numbers pounds, must. Youtube LINK There are also very important symmetric, addition, subtraction,.... 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The addition number facts will help with subtraction. [ 14 ] [ 6 ] predictable concerning! Results in an addition equation: addition equations have addends and a....
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# 3.3 Rates of change and behavior of graphs (Page 6/15)
Page 6 / 15
Estimate the intervals where the function is increasing or decreasing.
Estimate the point(s) at which the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ has a local maximum or a local minimum.
local maximum: local minimum:
For the following exercises, consider the graph in [link] .
If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing.
If the complete graph of the function is shown, estimate the absolute maximum and absolute minimum.
absolute maximum at approximately absolute minimum at approximately
## Numeric
[link] gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004?
Year Sales
(millions of dollars)
1998 201
1999 219
2000 233
2001 243
2002 249
2003 251
2004 249
2005 243
2006 233
[link] gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006?
Year Population
(thousands)
2000 87
2001 84
2002 83
2003 80
2004 77
2005 76
2006 78
2007 81
2008 85
a. –3000; b. –1250
For the following exercises, find the average rate of change of each function on the interval specified.
$f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ on
$h\left(x\right)=5-2{x}^{2}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[-2,\text{4}\right]$
-4
$q\left(x\right)={x}^{3}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[-4,\text{2}\right]$
$g\left(x\right)=3{x}^{3}-1\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[-3,\text{3}\right]$
27
$y=\frac{1}{x}\text{\hspace{0.17em}}$ on
$p\left(t\right)=\frac{\left({t}^{2}-4\right)\left(t+1\right)}{{t}^{2}+3}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[-3,\text{1}\right]$
–0.167
$k\left(t\right)=6{t}^{2}+\frac{4}{{t}^{3}}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[-1,3\right]$
## Technology
For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing.
$f\left(x\right)={x}^{4}-4{x}^{3}+5$
Local minimum at $\text{\hspace{0.17em}}\left(3,-22\right),\text{\hspace{0.17em}}$ decreasing on increasing on
$h\left(x\right)={x}^{5}+5{x}^{4}+10{x}^{3}+10{x}^{2}-1$
$g\left(t\right)=t\sqrt{t+3}$
Local minimum at $\text{\hspace{0.17em}}\left(-2,-2\right),\text{\hspace{0.17em}}$ decreasing on $\text{\hspace{0.17em}}\left(-3,-2\right),\text{\hspace{0.17em}}$ increasing on
$k\left(t\right)=3{t}^{\frac{2}{3}}-t$
$m\left(x\right)={x}^{4}+2{x}^{3}-12{x}^{2}-10x+4$
Local maximum at local minima at $\text{\hspace{0.17em}}\left(-3.25,-47\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2.1,-32\right),\text{\hspace{0.17em}}$ decreasing on $\text{\hspace{0.17em}}\left(-\infty ,-3.25\right)\text{\hspace{0.17em}}$ and increasing on and
$n\left(x\right)={x}^{4}-8{x}^{3}+18{x}^{2}-6x+2$
## Extension
The graph of the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is shown in [link] .
Based on the calculator screen shot, the point is which of the following?
1. a relative (local) maximum of the function
2. the vertex of the function
3. the absolute maximum of the function
4. a zero of the function
A
Let $f\left(x\right)=\frac{1}{x}.$ Find a number $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ such that the average rate of change of the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left(1,c\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}-\frac{1}{4}.$
Let $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}$ . Find the number $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ such that the average rate of change of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left(2,b\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}-\frac{1}{10}.$
$b=5$
## Real-world applications
At the start of a trip, the odometer on a car read 21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip?
A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank?
2.7 gallons per minute
Near the surface of the moon, the distance that an object falls is a function of time. It is given by $\text{\hspace{0.17em}}d\left(t\right)=2.6667{t}^{2},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in seconds and $\text{\hspace{0.17em}}d\left(t\right)\text{\hspace{0.17em}}$ is in feet. If an object is dropped from a certain height, find the average velocity of the object from $\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}t=2.$
The graph in [link] illustrates the decay of a radioactive substance over $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ days.
Use the graph to estimate the average decay rate from $\text{\hspace{0.17em}}t=5\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}t=15.$
approximately –0.6 milligrams per day
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has |
Notes on Trigonometric Ratios | Grade 7 > Optional Maths > Trigonometry | KULLABS.COM
• Note
• Things to remember
• Videos
• Exercise
• Quiz
#### Trigonometric Ratios:
As we know, in the earlier chapter that right angled triangle consists of three sides i.e ; Perpendicular , Base and Hypotenuse then as we talk about ratios 6 ratios each of them obtained from these 3 sides they are :
1. $$\frac{p} {h}$$
2. $$\frac{b}{h}$$
3. $$\frac{p}{b}$$
4. $$\frac{b}{p}$$
5. $$\frac{h}{b}$$
6. $$\frac{h}{p}$$
Above ratios are called trigonometric ratios and they are given certain names .Detail information is given below ;
Here,
Let Δ MNO be right angled triangle where ∠MNO = 90 ,
∠MNO = θ be the reference angle .
Then,
Side MN = Perpendicular ( p )
Side NO = Base ( b )
Side MO= Hypotenuse ( h )
Now , we can get six ratios here ;
1. $$\frac{p}{h}$$ = $$\frac{MN}{MO}$$
2. $$\frac{b}{h}$$ = $$\frac{NO}{MO}$$
3. $$\frac{p}{b}$$ = $$\frac{MN}{NO}$$
4. $$\frac{b}{p}$$ = $$\frac{NO}{MN}$$
5. $$\frac{h}{b}$$ = $$\frac{MO}{NO}$$
6. $$\frac{h}{p}$$ = $$\frac{MO}{MN}$$
Now , lets introduce the names for these ratios.
S.No. Ratio Nomenclature Abbreviation 1. $$\frac{p}{h}$$ sine sin 2. $$\frac{b}{h}$$ cosine cos 3. $$\frac{p}{b}$$ tangent tan 4. $$\frac{b}{p}$$ cotangent cot 5. $$\frac{h}{b}$$ secant sec 6. $$\frac{h}{p}$$ cosecant cosec/csc
Again , In the earlier statement if we try to link these ratio we get,
Sinθ = $$\frac{p}{h}$$ =$$\frac{MN}{MO}$$
cosθ = $$\frac{b}{p}$$ =$$\frac{NO}{MO}$$
tanθ = $$\frac{p}{b}$$ =$$\frac{MN}{NO}$$
Cotθ =$$\frac{b}{p}$$ =$$\frac{NO}{MN}$$
Secθ = $$\frac{h}{b}$$ =$$\frac{MO}{NO}$$
Cosecθ=$$\frac{h}{p}$$ = $$\frac{MO}{MN}$$
Operation Of Trignometric Ratios
Here, in trignometry ratio also, addition , subtraction , multiplication and division take place in sameway as in algebra .It is important to have the prior knowledge on algebra before operating trignometric ratios.
for eg;
1. a + a = 2a
sin θ+ sinθ = 2sinθ
2. a + b = a + b
sinθ + cosθ = sinθ + cosθ
3. a2+ a = a2 + a
sec2 θ+ secθ = sec2θ + secθ
Subtraction
We can subtract the trigonometric ratios as we subtract in algebra.
1. 2a - a = a
2tanθ - tanθ = tanθ
2. a - b = a - b
cosecθ - cotθ = cosecθ - cotθ
3. a2- a = a2- a
sec2θ - secθ = sec2θ - secθ
Multiplication and Division
For, multiplication trigonometric ratios will also follow the laws of indices. Let's review laws of indicesonce .
a) a2× a3 = a ×a ×a ×a ×a = 5
Generalizing this statement : ap×aq= ap+q
b) a6÷ a2 = $$\frac{a×a ×a×a×a×a}{a×a}$$
= a4
Generaliziting ,
ap÷ aq = apq
c) ( a2)3= a2×a2×a2
= a2+2+2
=a6
Similary,
sin θ× sinθ = sin2θ
2sin2θ ×3sin3θ = 6sin5θ [ ap × aq = ap+q ]
6cos4θ ÷ 2cos2θ =$$\frac{6cos^4θ}{2cos^2θ}$$ [ ap÷ aq = apq ]
( cot2θ)3 = cotθ6θ [ (ap)q÷ aq= ap×q]
3cosecθ× 4secθ = 12cosecθ.secθ
Some basic algebraical formula
S.No. Formula Expanded form Factorized form 1. (a + b)2 a2+2ab+b2 (a+b) (a+b) 2. (a - b)2 a2-2ab+b2 (a-b) (a-b) 3. a2 -b2 (a+b) (a-b) 4. a2+b2 (a+b)2 - 2ab (a-b)2 +2ab 5. (a + b )3 a3+3a2b+3ab2+b3 a3+b3+3ab(a+b) (a+b)(a+b)(a+b) 6. ( a- b)3 a3-3a2b+3ab2-b3 a3-b3-3ab(a-b) (a-b)(a-b)(a-b) 7. a3+ b3 (a+b)3-3ab(a+b) (a+b) (a2-ab+b2) 8. a3- b3 (a-b)3+3ab(a-b) (a-b)(a2+ab+b2)
• Multiplication trigonometric ratios follow the laws of indices.
• We can subtract the trigonometric ratios as we subtract in algebra.
• 6 ratios each of them obtained from these 3 sides they are :
1. $$\frac{p} {h}$$
2. $$\frac{b}{h}$$
3. $$\frac{p}{b}$$
4. $$\frac{b}{p}$$
5. $$\frac{h}{b}$$
6. $$\frac{h}{p}$$
.
#### Click on the questions below to reveal the answers
Solution ;
Here, In ΔABC, ∠ABC = 90 ∠BAC = θ be the referance angle .
Then ,
Side BC = perpendicular ( p )
Side AC = hypotenuse ( h )
Side AB = base (b )
Now, using trignometry ratios formula we get;
sinθ = $$\frac{p}{h}$$ coscθ = $$\frac{b}{h}$$
tanθ = $$\frac{p}{b}$$ cotθ = $$\frac{b}{p}$$
secθ = $$\frac{h}{b}$$ cosecθ = $$\frac{h}{p}$$
a )
Solution;
sinθ + 2sinθ
=sinθ ( 1 + 2 )
= sinθ ( 3 )
=3 sinθ
b )
Solution;
5tanθ - 2tanθ
=tanθ ( 5 - 2 )
=tanθ ( 3 )
=3 tanθ
a )sin θ × sin 2θ
= sin θ × sin 2θ ( a× a n= am + n )
=sin3 θ
b ) ( cosec2θ - sec2θ ) by ( cosecθ + secθ )
= $$\frac{cosec^2θ-sec^2θ}{cosecθ+secθ}$$
= $$\frac{(cosecθ)-(secθ)^2}{cosecθ+secθ}$$
=$$\frac{(cosecθ+secθ)(cosecθ-secθ)}{cosecθ+secθ}$$
=cosecθ-secθ
The methods of providing a trigonometric identity are as follows:
a. Take the identity on the left-hand side(L.H.S) and show it equal to right-hand side(R.H.S).
b. Take the identity on the R.H.S and show it equal to L.H.S.
Solution
= cos θ ( cosθ + sinθ) - sinθ ( cosθ + sinθ)
= cos 2 θ + cosθ. sinθ - sinθ . cosθ - sin2θ
= cos 2 θ + cosθ. sinθ - cosθ . sinθ - sin2θ
= cos2θ - sin2θ
0%
two
six
three
five
sine
tan
cos
cos
sin
tan
cot
cos
tan
cot
sin
(frac{h}{b})
(frac{p}{h})
(frac{b}{h})
(frac{p}{b})
(frac{b}{h})
(frac{b}{p})
(frac{h}{p})
(frac{p}{h})
(frac{p}{h})
(frac{h}{p})
(frac{h}{b})
(frac{p}{b})
sec
sin
cot
tan
sec
sin
cosec
cos
cosine
sine
secant
cot
## ASK ANY QUESTION ON Trigonometric Ratios
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Equilateral and Isoceles Triangle
Chapter 6 Class 7 Triangle and its Properties
Concept wise
### Find angle x
In ∆ABC,
AB = AC (Given)
Therefore,
∠C = ∠B (Angles opposite to equal sides are equal)
40° = x
x = 40°
### Find angle x
In ∆PQR,
PQ = QR (Given)
Therefore,
∠R = ∠P (Angles opposite to equal sides are equal)
45° = ∠P
∠P = 45°
Now, by Angle sum property,
∠P + ∠Q + ∠R = 180°
45° + x + 45° = 180°
x + 90° = 180°
x = 180° − 90°
x = 90 °
### Find angle x
In ∆XYZ,
XZ = ZY (Given)
Therefore,
∠Y = ∠X (Angles opposite to equal sides are equal)
x = 50°
### Find angle x
In ∆ABC,
AB = CA (Given)
Therefore,
∠C = ∠B (Angles opposite to equal sides are equal)
∠C = x
Now, By angle sum property,
∠A + ∠B +∠C = 180°
100° + x + x = 180°
100° + 2x = 180°
2x = 180° − 100°
2x = 80°
x = 80° /2
x = 40°
### Find angle x
In ∆PQR,
QR = PQ (Given)
Therefore,
∠P = ∠R (Angles opposite to equal sides are equal)
x = ∠R
∠R = x
Now, by angle sum property,
∠P + ∠Q + ∠R = 180°
x + 90° + x = 180°
90° + 2x = 180°
2x = 180° − 90°
2x = 90°
x = (90°)/2
x = 45°
### Find angle x
In ∆XYZ,
XY = YZ (Given)
Therefore,
∠Z = ∠X (Angles opposite to equal sides are equal)
x = ∠X
∠X = x
Now, by angle sum property,
∠X + ∠Y + ∠Z = 180°
x + 40° + x = 180°
40° + 2x = 180°
2x = 180° − 40°
2x = 140°
x = (140°)/2
x = 70°
### Find angle x
In ∆PQR,
PQ = PR (Given)
Therefore,
∠PRQ = ∠Q (Angles opposite to equal sides are equal)
∠PRQ = x
Now,
∠PRQ + ∠PRS = 180° (Linear pair)
x + 120° = 180°
x = 180° − 120°
x = 60 °
### Find angle x
In ∆PQR,
PR = PQ (Given)
Therefore,
∠Q = ∠R (Angles opposite to equal sides are equal)
∠Q = x
We know that,
Exterior angle is equal to sum of interior opposite angles
∠SPR = ∠Q + ∠R
110° = x + x
110° = 2x
(110°)/2 = x
55° = x
x = 55°
### Find angle x
Here
∠MYN = ∠ZYX (Vertically opposite angles)
30° = ∠ZYX
∠ZYX = 30°
In ∆XYZ,
YZ = ZX (Given)
Therefore,
30° = ∠ZYX (Angles opposite to equal sides are equal)
∠X = ∠ZYX
x = 30°
### Find angle x and y
In ∆ABC,
AC = AB (Given)
Therefore,
∠B = ∠ACB (Angles opposite to equal sides are equal)
y = ∠ACB
∠ACB = y
Now,
∠ACB + ∠ACD = 180° (Linear pair)
y + 120° = 180°
y = 180° − 120°
y = 60°
We know that,
Exterior angle is equal to sum of interior opposite angles.
∠ACD = ∠A + ∠B
120 ° = x + y
120 ° = x + 60°
120 ° − 60° = x
60° = x
x = 60°
### Find angle x and y
In ∆PQR,
QR = PR (Given)
Therefore,
∠QPR = ∠Q (Angles opposite to equal sides are equal)
∠QPR = x
Now,
In Δ PQR
By Angle sum property
∠ QPR + ∠ Q + ∠ R = 180°
x + x + 90° = 180°
2x + 90° = 180°
2x = 180° – 90°
2x = 90°
x = (90°)/2
x = 45°
Also,
Exterior angle is sum of interior opposite angles
y = x + ∠ R
y = 45° + 90°
y = 135°
### Find angle x and y
Here,
∠JIK = ∠MIN (Vertically opposite angle)
∠JIK = 92°
In ∆IJK,
IJ = IK (Given)
Therefore,
∠IKJ = ∠IJK (Angles opposite to equal sides are equal)
∠IKJ = x
Now, By angle sum property,
∠IJK + ∠JKI + ∠KIJ = 180°
x + x + 92° = 180°
2x + 92° = 180°
2x = 180° − 92°
2x = 88°
x = (88°)/2
x = 44°
Hence,
IKJ = x
∠IKJ = 44°
Now,
∠IJK + ∠IKL = 180° (Linear pair)
44° + y = 180°
y = 180° − 44°
y = 136°
x = 44° and y = 136°
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# What is the equation of the line that passes through the point (-2,3) and that is perpendicular to the line represented by 3x-2y= -2?
Jan 13, 2017
$\left(y - 3\right) = - \frac{3}{2} \left(x + 2\right)$
Or
$y = - \frac{3}{2} x$
#### Explanation:
First, we need to convert the line into slope-intercept form to find the slope.
The slope-intercept form of a linear equation is:
$y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and color(blue)(b is the y-intercept value.
We can solve the equation in the problem for $y$:
$3 x - 2 y = - 2$
$3 x - \textcolor{red}{3 x} - 2 y = - 2 - \textcolor{red}{3 x}$
$0 - 2 y = - 3 x - 2$
$- 2 y = - 3 x - 2$
$\frac{- 2 y}{\textcolor{red}{- 2}} = \frac{- 3 x - 2}{\textcolor{red}{- 2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} y}{\cancel{\textcolor{red}{- 2}}} = \frac{- 3 x}{\textcolor{red}{- 2}} - \frac{2}{\textcolor{red}{- 2}}$
$y = \frac{3}{2} x + 1$
So for this equation the slope is $\frac{3}{2}$
A line perpendicular to this line will have a slope which is the negative inverse of our line or $- \frac{3}{2}$
We can now use the point-slope formula to write the equation for the perpendicular line:
The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the point from problem and the slope we calculated gives:
$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{3}{2}} \left(x - \textcolor{red}{- 2}\right)$
$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{3}{2}} \left(x + \textcolor{red}{2}\right)$
Or, we can put the equation in the more familiar slope-intercept form by solving for $y$:
$y - \textcolor{red}{3} = \textcolor{b l u e}{- \frac{3}{2}} x + \left(\textcolor{b l u e}{- \frac{3}{2}} \times \textcolor{red}{2}\right)$
$y - \textcolor{red}{3} = - \frac{3}{2} x - 3$
$y - \textcolor{red}{3} + 3 = - \frac{3}{2} x - 3 + 3$
$y = - \frac{3}{2} x + 0$
$y = - \frac{3}{2} x$ |
# Significant Figures in Calculation
Significant figures in calculations are used to keep the precision of the calculation in check. The more significant figures the more precise the measurement is. It is understood that only the last digit is uncertain. However, let us say that we have two numbers that have different significant figures, or even containing different decimal places. When calculating the answers for any problem with significant figures, how does one know how many significant figures to use for the answer? One golden rule to follow is that the answer cannot be more precise, containing more significant figures, than any of the measurements used to find that answer. When carrying out computations involving significant figures, there are two sets of rules to abide by that cover the four basic math computations.
In addition and subtraction the result can have no more decimal places than the measurement with the fewest number of decimal places. In other words, find the number with the lowest decimals and that will show you the number of decimals to use in your answer.
The sum in the addition problem was 49.
335 in normal calculations (Containing 3 decimal places). However, out of the set of numbers, 12 had the lowest numbers of decimal places (Containing 0 decimal places), the answer must also contain zero decimal places therefore, making the answer 49.
The difference of the subtraction problem was 7.
116 in normal calculations(Containing 3 decimal places). However with .10 having only two decimal places, the answer can only contain two decimal places. Due to the fact that the digit after the cut off point was a 6, which is more than the middle digit 5, one can round up the cut off number up, thus making the answer 7.12 (Containing 2 decimal places).
### Multiplication and Division Rule
In multiplication and division the result must be reported with the same number of significant figures as the measurement with the fewest significant figures. This is different compared to addition and subtraction because it is not decimal places that are used, but rather significant figures.
Example:
Calculate the density of a sample with mass 25.624 g and a volume of 25 mL.
D=M/V, so 25.624g/25mL=1.02496 g/mL, but we record the answer simply as 1.0 g/mL to meet the significant figure requirement.
The answer through simple calculations contained 6 significant figures, but the lowest significant figure in number set, 25 mL, contained 2 significant figures, thus the answer must also contain 2 significant figures.
### Using Scientific Notation:
If your calculation gives you a long number that needs to be condensed to show the correct number of significant figures, one can use scientific notation to display the correct number of significant figures.
Ex.1) 100,000 needs to be shown with 3 significant figures in a calculation. One can show 100,000 = 1.00 x 10^5. The 1.00 portion shows the 3 significant figures. The 10^5 does not count in the significant figures.
(See above diagram as well.)
### Rounding
When you are at the cut off number, you round it based on the next digit to the right of the cut off number. If it's 4 and below keep the cut off digit the same. If the number after the cut off digit is 5 or above, round it up.
Ex. 1.456 = 1.46
The cut off number is 5, and the number after the 5 is a 6, which is greater than 5, so the cut off digit is rounded up to a 6.
### Rounding to Even
Normally when the number after the cut off number is a 5, there is usually always another number there that will make it "weigh" more, which is why the number 5 usually makes things round up.
If by some chance the number is 4.655(000000000 etc.) and you needed 3 significant figures, the number after the cut off digit is 5, but all the digits after that are assumed as 0000 infinitely. The 5 in this case, doesn't have any digits behind it to make it "weigh" enough to round up. In this case we use the round to even rule, where we round the cut off number to the closest even number.
Ex. 4.655 = 4.66
The 5 is rounded to 6 because it is the closest even number. 4 is also just as close, but its a rule to round up.
Ex. 4.645 = 4.64
The 4 is rounded to 4 because it is the closest even number, since it is already an even number.
### Resources
Brown, LeMay, Bursten. Chemistry: The Central Science. Prentice Hall, 2003. |
# Finite Sets & Infinite Sets
In these lessons, we will learn about finite sets and infinite sets.
Finite sets are sets that have a finite number of members. If the elements of a finite set are listed one after another, the process will eventually “run out” of elements to list.
Example:
A = {0, 2, 4, 6, 8, …, 100}
C = {x : x is an integer, 1 < x < 10}
An infinite set is a set which is not finite. It is not possible to explicitly list out all the elements of an infinite set.
Example:
T = {x : x is a triangle}
N is the set of natural numbers
A is the set of fractions
The number of elements in a finite set A is denoted by n(A).
Example:
If A is the set of positive integers less than 12 then
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} and n(A) = 11
If C is the set of numbers which are also multiples of 3 then
C = {3, 6, 9, …} and C is an infinite set
If D is the set of integers x defined by –3 < x < 6 then
D = {–2, –1, 0, 1, 2, 3, 4, 5} and n(D) = 8 If Q is the set of letters in the word ‘HELLO’ then
Q = {H, E, L, O } , n(Q) = 4 ← ‘L’ is not repeated.
Infinite vs. Finite Number Sets
The word finite means
• having bounds
• a set that contains a countable number of elements.
• the set starts and stops
The word infinite means
• the set in which the number of elements cannot be counted or determined (never ending)
• the set can continue forever in the beginning, or the end, or both Examples of Infinite and Finite Number Sets
Errata (submitted by Tim): At 3:20, the set {x ∈ whole numbers: 0 < x < 10} should be {1, 2, 3, 4, 5, 6, 7, 8, 9} Examples of finite and infinite sets
Set of objects, Natural numbers, Real numbers and how to find distance between two points. How to classify Finite and Infinite Number Sets?
What is a set? Finite Sets, Infinite Sets, and the Null Set
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.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Trigonometric Identities
Trigonometric identities are the correspondences including geometrical capacities and remain constant for each worth of the factors in question, with the end goal that the two sides of the fairness are characterised. In this small example, we will investigate mathematical characters. There are three essential geometrical proportions: sin, cos, and tan. The three other geometrical proportions sec, cosec, and cot in geometry are the reciprocals of sin, cos, and tan separately. How are these geometrical proportions (sin, cos, tan, sec, cosec, and bed) associated with one another? They are associated through Trigonometric identities(or in short trig identities).
## Let’s understand some basics of Trigonometric Identities
Trigonometric identities are conditions that connect with various mathematical capacities and are valid for any value of the variable that is there in the space. Essentially, identity is a condition that remains constant for every one of the upsides of the variable(s) present in it.
## Now let’s focus on some of the algebraic identities
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab+ b2
(a + b)(a-b)= a2 – b2
The algebraic identity is an interrelation between the variables whereas the trig identities relate the 6 trigonometric functions sine, cosine, tangent, cosecant, secant, and cotangent.
## Basics of Reciprocal Trigonometric Identities
We already observed that the reciprocals of sin, cosine, and tangent are cosecant, secant, and cotangent respectively.
Due to that, reciprocal trigonometric identities are:
• sin θ = 1/cosecθ (OR) cosec θ = 1/sinθ
• cos θ = 1/secθ (OR) sec θ = 1/cosθ
• tan θ = 1/cotθ (OR) cot θ = 1/tanθ
Now let’s learn about Pythagorean Trigonometric Identities:
Pythagoras Theorem is used in Trigonometry to derive Pythagorean trigonometric identities. If we apply Pythagoras theorem to a Right Angled Triangle then the given below formulae are used:
Dividing both sides by Hypotenuse2
• sin2θ + cos2θ = 1
This is one of the Pythagorean identities. Similarly, we can derive 2 other Pythagorean trigonometric identities as follows:
• 1 + tan2θ = sec2θ
• 1 + cot2θ = cosec2θ
Now let’s understand the other part of trigonometric identities – Complementary and Supplementary Trigonometric Ideas:A pair of two angles such that their sum is equal to 90° is called a Complementary Angle. θ is the complement angle of (90 – θ). Now let’s learn the ratios of Complementary Angles:
• sin (90°- θ) = cos θ
• cos (90°- θ) = sin θ
• cosec (90°- θ) = sec θ
• sec (90°- θ) = cosec θ
• tan (90°- θ) = cot θ
• cot (90°- θ) = tan θ
A pair of two angles such that their sum is equal to 180° is called a Supplementary Angle. θ is the supplement angle of (180 – θ). Now let’s learn the ratios of Supplementary Angles:
• sin (180°- θ) = sinθ
• cos (180°- θ) = -cos θ
• cosec (180°- θ) = cosec θ
• sec (180°- θ)= -sec θ
• tan (180°- θ) = -tan θ
• cot (180°- θ) = -cot θ
Now let’s understand The Sum And Difference of various Trigonometric Ideas:
The sum and difference identities consist of the formulas of sin(A+B), cos(A-B), cot(A+B), etc.
• 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
• tan (A+B) = (tan A + tan B)/(1 – tan A tan B)
• tan (A-B) = (tan A – tan B)/(1 + tan A tan B)
Now let’s move on to the another interesting concept-
Double and Half Angle Trigonometric Ideas:
The sum and difference formulae can be applied to obtain the double angle trigonometric identities.
For example, from the above formulas:
sin (A+B) = sin A cos B + cos A sin B
Substitute A = B = θ on both sides here, we get:
sin (θ + θ) = sinθ cosθ + cosθ sinθ
sin 2θ = 2 sinθ cosθ
Similarly, we can apply the other double-angle identities.
• sin 2θ = 2 sinθ cosθ
• cos 2θ = cos2θ – sin 2θ
= 2 cos2θ – 1
= 1 – sin 2 θ
• tan 2θ = (2tanθ)/(1 – tan2θ)
## Half Angle Formulas
Using one of the stated above double angle formulas,
cos 2θ = 1 – 2 sin2θ
2 sin2θ = 1- cos 2θ
sin2θ = (1 – cos2θ)/(2)
sin θ = ±√[(1 – cos 2θ)/2]
Replacing θ by θ/2 on both sides,
sin (θ/2) = ±√[(1 – cos θ)/2]
This is the half-angle formula of sin.
Similarly, we can apply the other half-angle formulas.
sin (θ/2) = ±√[(1 – cosθ)/2]
cos (θ/2) = ±√(1 + cosθ)/2
tan (θ/2) = ±√[(1 – cosθ)(1 + cosθ)]
Now let’s learn the Cosine and Sine Rule Trigonometric Identities:
The relation between the angles and the corresponding sides of a triangle is called the Sine Rule.
We use the sine rule and the cosine rule for non-right angled triangles. Sine rule can be given as,
• a/sinA = b/sinB = c/sinC
• sinA/a = sinB/b = sinC/c
• a/b = sinA/sinB; a/c = sinA/sinC; b/c = sinB/sinC
The relation between the angles and the sides of a triangle which is usually used when two sides and the included angle of a triangle are given.In this case we can say that this is Cosine Rule.
• a2 = b2 + c2 – 2bc·cosA
• b2 = c2 + a2 – 2ca·cosB
• c2 = a2 + b2 – 2ab·cosC
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# Difference between revisions of "Topology Glossary"
This is a Helper Page
## Abstract (Manifold)
In geometry, we have many shapes with specific names, dimensions and properties. For instance, we are all familiar with a square, it has 2 dimensions, a width, a base, and other cool math properties. In topology, we have manifolds. A manifold is a broad definition of a shape. Manifolds are thought of as surfaces without any boundaries or edges.
Manifolds can be categorized by their dimensions. A one-dimensional manifold is just a one-dimensional shape or surface. This means each section of a one-dimensional manifold looks like a line. A two-dimensional manifold is just a two dimensional shape or surface. This means each section of a two dimensional manifold looks like a plane. In fact, a surface is a two dimensional manifold. Manifolds are the first step in understanding what type of surface Boy's Surface actually is.
An example of a manifold could be a tossed blanket. The tossed blanket is a shape. It has dimensions just like the typical square, even if we do not really think about it in that way. We don't have a particular name for this shape, but it is a shape nonetheless with many properties.
## Boundary
In Topology, the term boundary refers to the edge or edges of a figure. The sphere has no boundary, while the circle has one, and the has two.
The edge of a 1 dimensional object is a corner, while that of a 2 dimensional object is a side, like the sides of a square. The edges of a 3 dimensional object is what is commonly called a surface. The boundary of a table is the table-top, sides, and bottom.
Boundary is not the same as bounded.
## Bounded
A bounded entity is one in which there is a limit to how far away two points can be from each other. In a certain sense, it means that the shape does not extend infinity in any direction.
Bounded is not the same as boundary.
## Closed
Closed sets contain all of their boundaries. If a set contains no boundaries, then this is satisfied by default. Thus, a topological set, or entity, is closed if it contains all points which are part of the figure. Hence, if there are edges on a shape, then the figure must include these edges in order to be closed.
## Compact
A compact figure is, in essence, one that is both closed and bounded.
## Manifold
Simply put, a manifold is a shape without edges. Manifolds are classified by their dimensionality; for instance, a line segment is 1 dimensional, so is called a 1 manifold. A cube is a 3 dimensional manifold, or 3 manifold. A manifold with n dimensions is an n manifold. There are manifolds with any and every number of dimensions.
The main characteristic of a manifold is that a small, n - 1 dimensional ball can be drawn around any point on the manifold without hitting a boundary. In effect, this means that manifolds have no boundaries, or edges.
### Surface
A surface is a 2 dimensional manifold.
## Orientability
```this page] from Plus Magazine. When viewing this, remember that the face is within the surface, not merely pasted on one of the sides.
```
Orientability and non-orientability are most often discussed in relation to surfaces. Nevertheless, the properties are descriptive of higher dimensional manifolds as well. One can consider a 3 dimensional space where certain paths would lead back to their start points, but flip the right and left sides of things that traveled along them.
Here is a more technical explanation: In a non-orientable manifold, there exists at least one path around the manifold such that, if we take a set of basis vectors for the manifold and move them along this path, they will arrive at their starting point with the following result. The determinant of the matrix composed of these basis vectors will have the opposite sign (positive/ negative) when it arrives back at the starting point as it did when it left.
Not all image pages will discuss in depth how non-orientability arises for a particular surface; the phenomenon is discussed, however, for the Mobius Strip and Real Projective Plane, with images to explain the phenomenon. |
# 8.4 Confidence intervals: confidence interval, single population (Page 2/9)
Page 2 / 9
Calculators and computers can easily calculate any Student's-t probabilities. The TI-83,83+,84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
A probability table for the Student's-t distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's-t distribution.) When using t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.
A Student's-t table (See the Table of Contents Tables ) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's-t probabilities.
## The notation for the student's-t distribution is (using t as the random variable) is
• $T$ ~ ${t}_{\text{df}}$ where $\text{df}=n-1$ .
• For example, if we have a sample of size n=20 items, then we calculate the degrees of freedom as df=n−1=20−1=19 and we write the distribution as $T$ ~ ${t}_{\text{19}}$
If the population standard deviation is not known , the margin of error for a population mean is:
• $\mathrm{ME}={t}_{\frac{\alpha }{2}}\cdot \left(\frac{s}{\sqrt{n}}\right)\phantom{\rule{20pt}{0ex}}$
• ${t}_{\frac{\alpha }{2}}$ is the t-score with area to the right equal to $\frac{\alpha }{2}$
• use $\text{df}=n-1$ degrees of freedom
• $s$ = sample standard deviation
The format for the confidence interval is:
$\left(\overline{x}-\mathrm{ME},\overline{x}+\mathrm{ME}\right)$ .
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects withthe results given below. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) fromwhich you took the data.
The solution is shown step-by-step and by using the TI-83, 83+ and 84+ calculators.
• 8.6
• 9.4
• 7.9
• 6.8
• 8.3
• 7.3
• 9.2
• 9.6
• 8.7
• 11.4
• 10.3
• 5.4
• 8.1
• 5.5
• 6.9
• Check the assumptions and conditions.
• The first solution is step-by-step (Solution A).
• The second solution uses an two column model(Solution B).
## Solution a
Start by discussing the assumptions and conditions that support you model.
• Randomization-The problem does not state that the sample was randomly selected. I will assume that it is in order to meet the conditions necessary to calculate a confidence interval.
• Independence- The problem does not state that the 15 subjects are independent. I will assume that they are in order to meet the conditions necessary to calculate a confidence interval.
• 10% condition-A sample size of 15 must be less than 10% of the total population receiving acupuncture to relieving pain. (A reasonable assumption).
• Nearly Normal-The problem states that the population should be assumed to be nearly normal (skewed to the left in histogram but somewhat evenly distributed about the median). The histogram and boxplot below support that statement.
To find the confidence interval, you need the sample mean, $\overline{x}$ , and the ME.
$\overline{x}=8.2267\phantom{\rule{20pt}{0ex}}s=1.6722\phantom{\rule{20pt}{0ex}}n=15$
$\text{df}=15-1=14$
$\text{CL}=0.95\phantom{\rule{5pt}{0ex}}$ so $\phantom{\rule{5pt}{0ex}}\alpha =1-\text{CL}=1-0.95=0.05$
$\frac{\alpha }{2}=0.025\phantom{\rule{20pt}{0ex}}{t}_{\frac{\alpha }{2}}={t}_{.025}$
The area to the right of ${t}_{.025}$ is 0.025 and the area to the left of ${t}_{.025}$ is 1−0.025=0.975
${t}_{\frac{\alpha }{2}}={t}_{.025}=2.14$ using invT(.975,14) on the TI-84+ calculator.
$\mathrm{ME}={t}_{\frac{\alpha }{2}}\cdot \left(\frac{s}{\sqrt{n}}\right)$
$\mathrm{ME}=2.14\cdot \left(\frac{\mathrm{1.6722}}{\sqrt{\mathrm{15}}}\right)=0.924$
$\overline{x}-\text{ME}=8.2267-0.9240=7.3$
$\overline{x}+\text{ME}=8.2267+0.9240=9.15$
The 95% confidence interval is (7.30, 9.15) .
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
When calculating the margin of error, a probability table for the Student's-t distribution can be used to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table.
**With contributions from Roberta Bloom
#### Questions & Answers
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Geometric Sequences*
Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be$26,520 after one year; $27,050.40 after two years;$27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.
## Terms of Geometric Sequences
### Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
### A General Note: Definition of a Geometric Sequence
A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If ${a}_{1}$ is the initial term of a geometric sequence and $r$ is the common ratio, the sequence will be
$\left\{{a}_{1}, {a}_{1}r,{a}_{1}{r}^{2},{a}_{1}{r}^{3},...\right\}$.
### How To: Given a set of numbers, determine if they represent a geometric sequence.
1. Divide each term by the previous term.
2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.
### Example: Finding Common Ratios
Is the sequence geometric? If so, find the common ratio.
1. $1\text{,}2\text{,}4\text{,}8\text{,}16\text{,}..$.
2. $48\text{,}12\text{,}4\text{, }2\text{,}..$.
Answer: Divide each term by the previous term to determine whether a common ratio exists.
1. $\begin{array}{llllllllll}\frac{2}{1}=2\hfill & \hfill & \hfill & \frac{4}{2}=2\hfill & \hfill & \hfill & \frac{8}{4}=2\hfill & \hfill & \hfill & \frac{16}{8}=2\hfill \end{array}$The sequence is geometric because there is a common ratio. The common ratio is 2.
2. $\begin{array}{lllllll}\frac{12}{48}=\frac{1}{4}\hfill & \hfill & \hfill & \frac{4}{12}=\frac{1}{3}\hfill & \hfill & \hfill & \frac{2}{4}=\frac{1}{2}\hfill \end{array}$The sequence is not geometric because there is not a common ratio.
#### Analysis of the Solution
The graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.
### Q & A
#### If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?
No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.
### Try It
Is the sequence geometric? If so, find the common ratio. $5,10,15,20,..$.
Answer: The sequence is not geometric because $\frac{10}{5}\ne \frac{15}{10}$ .
### Try It
Is the sequence geometric? If so, find the common ratio. $100,20,4,\frac{4}{5},..$.
Answer: The sequence is geometric. The common ratio is $\frac{1}{5}$ .
### Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is ${a}_{1}=-2$ and the common ratio is $r=4$, we can find subsequent terms by multiplying $-2\cdot 4$ to get $-8$ then multiplying the result $-8\cdot 4$ to get $-32$ and so on.
$\begin{array}{l}{a}_{1}=-2\hfill \\ {a}_{2}=\left(-2\cdot 4\right)=-8\hfill \\ {a}_{3}=\left(-8\cdot 4\right)=-32\hfill \\ {a}_{4}=\left(-32\cdot 4\right)-128\hfill \end{array}$
The first four terms are $\left\{-2\text{, }-8\text{, }-32\text{, }-128\right\}$.
### How To: Given the first term and the common factor, find the first four terms of a geometric sequence.
1. Multiply the initial term, ${a}_{1}$, by the common ratio to find the next term, ${a}_{2}$.
2. Repeat the process, using ${a}_{n}={a}_{2}$ to find ${a}_{3}$ and then ${a}_{3}$ to find ${a}_{4,}$ until all four terms have been identified.
3. Write the terms separated by commons within brackets.
### Example: Writing the Terms of a Geometric Sequence
List the first four terms of the geometric sequence with ${a}_{1}=5$ and $r=-2$.
Answer: Multiply ${a}_{1}$ by $-2$ to find ${a}_{2}$. Repeat the process, using ${a}_{2}$ to find ${a}_{3}$, and so on.
$\begin{array}{l}{a}_{1}=5\hfill \\ {a}_{2}=-2{a}_{1}=-10\hfill \\ {a}_{3}=-2{a}_{2}=20\hfill \\ {a}_{4}=-2{a}_{3}=-40\hfill \end{array}$
The first four terms are $\left\{5,-10,20,-40\right\}$.
### Try It
List the first five terms of the geometric sequence with ${a}_{1}=18$ and $r=\frac{1}{3}$.
Answer: $\left\{18,6,2,\frac{2}{3},\frac{2}{9}\right\}$
## Explicit Formulas for Geometric Sequences
### Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
### A General Note: Recursive Formula for a Geometric Sequence
The recursive formula for a geometric sequence with common ratio $r$ and first term ${a}_{1}$ is [latex-display]{a}_{n}=r{a}_{n - 1},n\ge 2[/latex-display]
### How To: Given the first several terms of a geometric sequence, write its recursive formula.
1. State the initial term.
2. Find the common ratio by dividing any term by the preceding term.
3. Substitute the common ratio into the recursive formula for a geometric sequence.
### Example: Using Recursive Formulas for Geometric Sequences
Write a recursive formula for the following geometric sequence.
$\left\{6\text{, }9\text{, }13.5\text{, }20.25\text{, }...\right\}$
Answer: The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
$r=\frac{9}{6}=1.5$
Substitute the common ratio into the recursive formula for geometric sequences and define ${a}_{1}$.
$\begin{array}{l}{a}_{n}=r{a}_{n - 1}\\ {a}_{n}=1.5{a}_{n - 1}\text{ for }n\ge 2\\ {a}_{1}=6\end{array}$
#### Analysis of the Solution
The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function.
### Q & A
#### Do we have to divide the second term by the first term to find the common ratio?
No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.
### Try It
Write a recursive formula for the following geometric sequence. [latex-display]\left\{2\text{, }\frac{4}{3}\text{, }\frac{8}{9}\text{, }\frac{16}{27}\text{, }...\right\}[/latex-display]
Answer: $\begin{array}{l}{a}_{1}=2\\ {a}_{n}=\frac{2}{3}{a}_{n - 1}\text{ for }n\ge 2\end{array}$
### Using Explicit Formulas for Geometric Sequences
Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.
${a}_{n}={a}_{1}{r}^{n - 1}$
Let’s take a look at the sequence $\left\{18\text{, }36\text{, }72\text{, }144\text{, }288\text{, }...\right\}$. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
${a}_{n}=18\cdot {2}^{n - 1}$
### A General Note: Explicit Formula for a Geometric Sequence
The nth term of a geometric sequence is given by the explicit formula:
${a}_{n}={a}_{1}{r}^{n - 1}$
### Example: Writing Terms of Geometric Sequences Using the Explicit Formula
Given a geometric sequence with ${a}_{1}=3$ and ${a}_{4}=24$, find ${a}_{2}$.
Answer: The sequence can be written in terms of the initial term and the common ratio $r$.
$3,3r,3{r}^{2},3{r}^{3},..$.
Find the common ratio using the given fourth term.
$\begin{array}{ll}{a}_{n}={a}_{1}{r}^{n - 1}\hfill & \hfill \\ {a}_{4}=3{r}^{3}\hfill & \text{Write the fourth term of sequence in terms of }{\alpha }_{1}\text{and }r\hfill \\ 24=3{r}^{3}\hfill & \text{Substitute }24\text{ for}{a}_{4}\hfill \\ 8={r}^{3}\hfill & \text{Divide}\hfill \\ r=2\hfill & \text{Solve for the common ratio}\hfill \end{array}$
Find the second term by multiplying the first term by the common ratio.
$\begin{array}{ll}{a}_{2}\hfill & =2{a}_{1}\hfill \\ \hfill & =2\left(3\right)\hfill \\ \hfill & =6\hfill \end{array}$
#### Analysis of the Solution
The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.
### Try It
Given a geometric sequence with ${a}_{2}=4$ and ${a}_{3}=32$ , find ${a}_{6}$.
Answer: ${a}_{6}=16,384$
### Example: Writing an Explicit Formula for the nth Term of a Geometric Sequence
Write an explicit formula for the $n\text{th}$ term of the following geometric sequence.
$\left\{2\text{, }10\text{, }50\text{, }250\text{, }...\right\}$
Answer: The first term is 2. The common ratio can be found by dividing the second term by the first term.
$\frac{10}{2}=5$
The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
$\begin{array}{l}{a}_{n}={a}_{1}{r}^{\left(n - 1\right)}\hfill \\ {a}_{n}=2\cdot {5}^{n - 1}\hfill \end{array}$
The graph of this sequence shows an exponential pattern.
### Try It
Write an explicit formula for the following geometric sequence.
$\left\{-1\text{, }3\text{, }-9\text{, }27\text{, }...\right\}$
Answer: ${a}_{n}=-{\left(-3\right)}^{n - 1}$
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of ${a}_{0}$ instead of ${a}_{1}$. In these problems, we can alter the explicit formula slightly by using the following formula:
${a}_{n}={a}_{0}{r}^{n}$
### Example: Solving Application Problems with Geometric Sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
1. Write a formula for the student population.
2. Estimate the student population in 2020.
1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.Let $P$ be the student population and $n$ be the number of years after 2013. Using the explicit formula for a geometric sequence we get ${P}_{n} =284\cdot {1.04}^{n}$
2. We can find the number of years since 2013 by subtracting. [latex-display]2020 - 2013=7[/latex-display] We are looking for the population after 7 years. We can substitute 7 for $n$ to estimate the population in 2020. [latex-display]{P}_{7}=284\cdot {1.04}^{7}\approx 374[/latex-display] The student population will be about 374 in 2020.
### Try It
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
1. Write a formula for the number of hits.
2. Estimate the number of hits in 5 weeks.
1. ${P}_{n} = 293\cdot 1.026{a}^{n}$
2. The number of hits will be about 333.
The following video provides a short lesson on some of the topics covered in this lesson. https://youtu.be/XHyeLKZYb2w
## Key Equations
recursive formula for $nth$ term of a geometric sequence ${a}_{n}=r{a}_{n - 1},n\ge 2$ explicit formula for $nth$ term of a geometric sequence ${a}_{n}={a}_{1}{r}^{n - 1}$
## Key Concepts
• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term.
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly.
• A recursive formula for a geometric sequence with common ratio $r$ is given by ${a}_{n}=r{a}_{n - 1}$ for $n\ge 2$ .
• As with any recursive formula, the initial term of the sequence must be given.
• An explicit formula for a geometric sequence with common ratio $r$ is given by ${a}_{n}={a}_{1}{r}^{n - 1}$.
• In application problems, we sometimes alter the explicit formula slightly to ${a}_{n}={a}_{0}{r}^{n}$.
## Glossary
common ratio the ratio between any two consecutive terms in a geometric sequence geometric sequence a sequence in which the ratio of a term to a previous term is a constant |
# Eureka Math Algebra 1 Module 3 Lesson 19 Answer Key
## Engage NY Eureka Math Algebra 1 Module 3 Lesson 19 Answer Key
### Eureka Math Algebra 1 Module 3 Lesson 19 Exploratory Challenge Answer Key
Exploratory Challenge 1
Let f(x) = x2 and g(x) = f(2x), where x can be any real number.
a. Write the formula for g in terms of x2 (i.e., without using f(x) notation).
g(x) = (2x)2
b. Complete the table of values for these functions.
c. Graph both equations: y = f(x) and y = f(2x).
See the discussion below for an explanation of the steps and arrows.
d. How does the graph of y = g(x) relate to the graph of y = f(x)?
The corresponding x-value of y = g(x) is half of the corresponding x-value of y = f(x) when g(x) = f(x); the points of the graph of g are $$\frac{1}{2}$$ the distance to the y-axis as the corresponding points of the graph of f, which makes the graph of g appear to “shrink horizontally.”
e. How are the values of f related to the values of g?
For equal outputs of f and g, the input of g only has to be half as big as the input of f.
Exploratory Challenge 2
Let f(x) = x2 and h(x) = f($$\frac{1}{2}$$ x), where x can be any real number.
a. Rewrite the formula for h in terms of x2 (i.e., without using f(x) notation).
h(x) = ($$\frac{1}{2}$$ x)2
b. Complete the table of values for these functions.
c. Graph both equations: y = f(x) and y = f($$\frac{1}{2}$$ x).
d. How does the graph of y = f(x) relate to the graph of y = h(x)?
Since the corresponding x-value of y = h(x) is twice the corresponding x-value of y = f(x) when h(x) = f(x), the points of the graph of h are 2 times the distance to the y-axis as the corresponding points of the graph of f, which makes the graph of h appear to “stretch horizontally.”
e. How are the values of f related to the values of h?
To get equal outputs of each function, the input of h has to be twice the input of f.
Exploratory Challenge 3
a. Look at the graph of y = f(x) for the function f(x) = x2 in Exploratory Challenge 1 again. Would we see a difference in the graph of y = g(x) if -2 were used as the scale factor instead of 2? If so, describe the difference. If not, explain why not.
There would be no difference. The function involves squaring the value within the parentheses; so, both the graph of y = f(2x) and the graph of y = f(-2x) are the same set as the graph of y = g(x), but both correspond to different transformations: The first is a horizontal scaling with scale factor $$\frac{1}{2}$$, and the second is a horizontal scaling with scale factor $$\frac{1}{2}$$ and a reflection across the y-axis.
b. A reflection across the y-axis takes the graph of y = f(x) for the function f(x) = x2 back to itself. Such a transformation is called a reflection symmetry. What is the equation for the graph of the reflection symmetry of the graph of y = f(x)?
y = f(-x).
c. Deriving the answer to the following question is fairly sophisticated; do this only if you have time. In Lessons 17 and 18, we used the function f(x) = |x| to examine the graphical effects of transformations of a function. In this lesson, we use the function f(x) = x2 to examine the graphical effects of transformations of a function. Based on the observations you made while graphing, why would using f(x) = x2 be a better option than using the function f(x) = |x|?
Not all of the effects of multiplying the input of a function are as visible with an absolute function as they are with a quadratic function. For example, the graph of y = 2|x| is the same as y = |2x|. Therefore, it is easier to see the effect of multiplying a value to the input of a function by using a quadratic function than it is by using the absolute value function.
### Eureka Math Algebra 1 Module 3 Lesson 19 Exercise Answer Key
Exercise
Complete the table of values for the given functions.
a.
b. Label each of the graphs with the appropriate functions from the table.
c. Describe the transformation that takes the graph of y = f(x) to the graph of y = g(x).
The graph of y = g(x) is a horizontal scale with scale factor $$\frac{1}{2}$$ of the graph of y = f(x).
d. Consider y = f(x) and y = h(x). What does negating the input do to the graph of f?
The graph of h is a reflection over the y-axis of the graph of f.
e. Write the formula of an exponential function whose graph would be a horizontal stretch relative to the graph of g.
Answers vary. Example: y = 2(0.5x).
### Eureka Math Algebra 1 Module 3 Lesson 19 Problem Set Answer Key
Let f(x) = x2, g(x) = 2x2, and h(x) = (2x)2, where x can be any real number. The graphs above are of the functions y = f(x), y = g(x), and y = h(x).
Question 1.
Label each graph with the appropriate equation.
Question 2.
Describe the transformation that takes the graph of y = f(x) to the graph of y = g(x). Use coordinates to illustrate an example of the correspondence.
The graph of y = g(x) is a vertical stretch of the graph of y = f(x) by scale factor 2; for a given x-value, the value of g(x) is twice as much as the value of f(x).
OR
The graph of y = g(x) is a horizontal shrink of the graph of y = f(x) by scale factor $$\frac{1}{\sqrt{2}}$$. It takes $$\frac{1}{\sqrt{2}}$$ times the input for y = g(x) as compared to y = f(x) to yield the same output.
Question 3.
Describe the transformation that takes the graph of y = f(x) to the graph of y = h(x). Use coordinates to illustrate an example of the correspondence.
The graph of y = h(x) is a horizontal shrink of the graph of y = f(x) by a scale factor of $$\frac{1}{2}$$. It takes $$\frac{1}{2}$$ the input for y = h(x), as compared to y = f(x) to yield the same output.
OR
The graph of y = h(x) is a vertical stretch of the graph of y = f(x) by scale factor 4; for a given x-value, the value of h(x) is four times as much as the value of f(x).
### Eureka Math Algebra 1 Module 3 Lesson 19 Exit Ticket Answer Key
Let f(x) = x2, g(x) = (3x)2, and h(x) = ($$\frac{1}{3}$$ x)2, where x can be any real number. The graphs above are of y = f(x), y = g(x), and y = h(x).
Question 1.
Label each graph with the appropriate equation.
The graph of y = g(x) is a horizontal shrink of the graph of y = f(x) with scale factor $$\frac{1}{3}$$. The corresponding x-value of y = g(x) is one-third of the corresponding x-value of y = f(x) when g(x) = f(x). This can be illustrated with the coordinate (1, 9) on g(x) and the coordinate (3, 9) on f(x). |
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# ${{C}_{1}}$and ${{C}_{2}}$ are two circles with center ${{O}_{1}}$ and ${{O}_{2}}$ and intersect each other at points $A$ and $B$. If ${{O}_{1}}{{O}_{2}}$ intersect $AB$ at $M$ then show that $M$ is the midpoint of $AB$.
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Hint: We will try to show $\Delta {{O}_{1}}AB$$\cong$$\Delta {{O}_{2}}AB$ using ‘SSS’ type of triangle congruency and then we will show $\Delta {{O}_{1}}AM\cong \Delta {{O}_{1}}BM$ using ‘SAS’ type of triangle congruency and finally we will show $M$ is the midpoint of $AB$.
Given that two circles ${{C}_{1}}$and ${{C}_{2}}$ with the center ${{O}_{1}}$ and ${{O}_{2}}$ intersect each other at points $A$ and $B$.
Also, ${{O}_{1}}{{O}_{2}}$ intersects $AB$ at $M$.
Then, we have to show that $M$ is the midpoint of $AB$.
Let us assume that the radius of the circle ${{C}_{1}}$ be $r$ and the radius of the circle ${{C}_{2}}$ be $s$.
Proof:
In $\Delta {{O}_{1}}A{{O}_{2}}$ and $\Delta {{O}_{1}}B{{O}_{2}}$, we have
${{O}_{1}}A={{O}_{1}}B.....\left( i \right)$
Both are radii of the same circle ${{C}_{1}}$.
$\Rightarrow {{O}_{2}}A={{O}_{2}}B.....\left( ii \right)$
Both are radii of the same circle ${{C}_{2}}$.
Also, ${{O}_{1}}{{O}_{2}}={{O}_{2}}{{O}_{1}}....\left( iii \right)$
Common sides of both the triangles $\Delta {{O}_{1}}A{{O}_{2}}$and $\Delta {{O}_{1}}B{{O}_{2}}$
So, from the equation $\left( i \right),\left( ii \right)$and$\left( iii \right)$, we get both triangles $\Delta {{O}_{1}}A{{O}_{2}}$ and $\Delta {{O}_{1}}B{{O}_{2}}$ are congruent with each other by ‘SSS’ type of triangle congruency.
Or, $\Delta {{O}_{1}}A{{O}_{2}}\cong \Delta {{O}_{1}}B{{O}_{2}}$ by SSS type of triangle congruency.
(Here, ‘SSS’ type means side – side – side type of triangle congruency)
SSS – Theorem
Side-side - side postulate (SSS) states that two triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle.
Here, from $\Delta ABC$ and $\Delta DEF$, we can say that
$AB=DF....\left( a \right)$
$AC=DE....\left( b \right)$
$BC=EF....\left( c \right)$
So, from the equation $\left( a \right),\left( b \right)$ and $\left( c \right)$, we have $\Delta ABC\cong \Delta DEF$ by SSS – type triangle congruency which clearly shows that
$\Rightarrow M\angle {{O}_{2}}{{O}_{1}}A=M\angle {{O}_{2}}{{O}_{1}}B....\left( iv \right)$ by CPCT
(Here, CPCT is corresponding parts of the congruent triangle)
Also, we have,
$\Rightarrow M\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B....\left( v \right)$
From the equation $\left( iv \right)$, both are the same angle.
$\Rightarrow {{O}_{1}}A={{O}_{1}}B.....\left( vi \right)$ by CPCT
Now, in $\Delta AM{{O}_{1}}$and $\Delta BM{{O}_{1}}$, we have
$\Rightarrow {{O}_{1}}A={{O}_{1}}B....\left( vii \right)$
(Both are radii of the same circle)
$\Rightarrow m\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B.....\left( viii \right)$
From corresponding parts of the congruent triangle.
$\Rightarrow {{O}_{1}}M=M{{O}_{1}}.....\left( ix \right)$
The common side of both the triangles $\Delta AM{{O}_{1}}$and $\Delta BM{{O}_{1}}$.
So, from the equation $\left( vii \right),\left( viii \right)$and $\left( ix \right)$, we get both triangles $\Delta AM{{O}_{1}}$and $\Delta BM{{O}_{1}}$ are congruent with each other by the SAS test of triangle congruency.
Or, $\Delta AM{{O}_{1}}\cong BM{{O}_{1}}$ by ’SAS’ test of triangle congruency.
(Here, ‘SAS’ means Side – Angle – Side type of triangle congruency)
SAS – theorem
If any two sides and the angle included between one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.
Here in the diagram,
Side $AB=DE$
Also side $BC=EF$
And $\angle ABC=\angle DEF$
Thus, $\Delta ABC\cong \Delta DEF$ by SAS type of triangle congruency.
Also, from this, we have$AM=BM$ by the corresponding part of congruent triangles $\Delta AM{{O}_{1}}\cong \Delta BM{{O}_{{}}}$ which means that $M$ is the midpoint of $AB$.
Hence proved.
Note: Visualize the geometry first before attempting the question. Make a clear diagram of the required question which reduces the probability of error in your solution using the SSS and SAS theorems to prove triangles are congruent, we prove the required statement. |
# Quilt Blocks: Simple Math Part 1
## Just the word “math” scares some people away from quilting, but it’s simple when you break it down into parts.
The March/April ’13 issue of Quiltmaker is #150, a milestone! To celebrate, we featured a Block Bonanza of 150 quilt blocks on page 75.
The Block Bonanza is diagrams of quilt blocks you can make. There aren’t cutting dimensions or sewing instructions. So…how can you make those blocks? It’s easy if you know a few simple recipes (that’s a familiar word used in place of the more scary word “formulas”). See? We’re making it simpler already.
Today let’s talk about one simple block, the very first one shown. And let’s talk about just the first step in figuring out the block math.
Look at the diagram and notice how the block is divided. How many equal sections across are there? Another way of looking at it is in “columns”: How many equal columns of patches do you see? For our sample block, there are just two.
Being able to see how many divisions there are across the block is the first (easy!) step in knowing how to make these blocks. Here are a few more blocks for you to practice on.
How many equal columns or divisions do you see in this block?
You can see by the lines I’ve drawn that there are five.
If you drew the lines like this, you could say there are 10 divisions or columns. That would be another way of looking at it. But for simplicity’s sake, let’s stick with the idea of five divisions. (If your lines are cutting some patches in half, you’re probably using too many lines.)
Here’s another practice block.
How many columns or divisions? Again, without cutting patches into awkward parts, you’d have to say two.
Here’s another practice block.
This one’s a bit more tricky. I would draw the lines like so:
And I would call it four columns or divisions. I did cut the center patch in half, but that’s okay, you’ll see how that works in Part 2 next week.
This is the Big Block Baby Quilt from our Jan/Feb ’13 issue. It’s really just one big block. How would you divide it?
I’d do this and call it five divisions.
Practice on the blocks below. In Part 2 next week, I’ll tell you what to do with this information—how it can be used to figure the size of your patches. This is exciting stuff!!
The block:
The block’s divisions:
The block:
The block’s divisions:
The block:
And the block’s divisions:
I can hardly wait to show you more!
I'm an editor for Quiltmaker magazine in Golden, Colorado, USA. For six years, I wrote pattern instructions, product reviews and how-to articles. Then I spent four years as QM's Interactive Editor, working to generate much of our online content. Now I'm back to patterns and how-tos, which is a great fit for me. I still love writing about quilt-related topics for Quilty Pleasures, and I always have my finger on the pulse of the quilting world. I teach a variety of quilt classes and give guild programs, too. Reach me by email: editor@quiltmaker.com.
This entry was posted in Quilting 101, Scrapbag and tagged , , , . Bookmark the permalink.
### 9 Responses to Quilt Blocks: Simple Math Part 1
1. Jan Duggan says:
I’ve always loved the look of a border made up of continuous little blocks but I have always shy away from doing them as I cannot figure out how to get the correct amount of little blocks to make it look proper as I would always seem to have to have a quarter block or half block and that really shows up… hence to say I gave up trying.
Yes I am no good at math either lets say I had the world’s first for a test result receiving a minas 13 on a test result in grade 8.
Also when marking the quilting designs around the border on a quilt I cant get the corners to have that continuous flow any suggestions for this of is this not a quilt math question. Thank you
2. Soozie says:
How about one on figuring out how much material you will need for your borders of the quilt.
3. Ella says:
Very helpful – this is easy math – great explanation
4. Karen D. says:
I have never been good at math but somehow figuring out how quilt blocks go together has always been relatively easy for me. This simplifies it even more. We are so fortunate to keep using our minds, as well as our hands, as we age. And we are rewarded with producing beautiful quilts that will last for generations, hopefully.
5. Kaye M. says:
I took geometry in high school and wondered what I would ever do with this information in my life time. Many years later when I started quilting, I had an “aha” moment! Finally, now I could put that geometry knowledge to use! |
# Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
## Class 10: Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
The document Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
All you need of Class 10 at this link: Class 10
Class-X
Time: 120 Minutes
M.M: 40
General Instructions:
Read the following instructions very carefully and strictly follow them:
1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. All questions are compulsory.
3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
5. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.
Section - A
Q.1. How many two digits numbers are divisible by 3?
OR
Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Lowest two digit number divisible by 3 is 12. and highest two digit number divisible by 3 is 99.
Hence, the sequence start with 12 and ends with 99 and common difference is 3.
So, the A.P. will be 12, 15, 18, ----, 96, 99.
Here, a = 12, d = 3, l = 99
∴ l = a + (n – 1)d
∴ 99 = 12 + (n – 1)3
⇒ 99 – 12 = 3(n – 1)
⇒ n – 1 = (87/3)
⇒ n – 1 = 29
⇒ n = 30
Therefore, there are 30, two digit numbers which are divisible by 3.
OR
The number which ends with 0 is divisible by 2 and 5 both.
∴ Such numbers between 102 and 998 are:
110, 120, 130, .........., 990.
Last term, an = 990
a + (n + 1)d = 990
110 + (n – 1) × 10 = 990 (∵ a = first term = 110)
110 + 10n – 10 = 990
10n + 100 = 990
10n = 990 – 100
10n = 890
n = 890/10 = 89.
Q.2. Solve the following quadratic equation for x:
9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0
Given,
9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0
First, we solve,
2a2 + 5ab + 2b2 = 2a2 + 4ab + ab + 2b2
Here, = 2a[a + 2b] + b[a + 2b]
= (a + 2b) (2a + b)
Hence, the equation becomes
9x2 – 9(a + b)x + (a + 2b)(2a + b) = 0
⇒ 9x2 – 3[3a + 3b]x + (a + 2b)(2a + b) = 0
⇒ 9x2 – 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a +b) = 0
⇒ 9x2 – 3(a + 2b)x – 3(2a + b)x +(a + 2b) (2a + b)= 0
⇒ 3x[3x – (a + 2b)] – (2a + b) [3x – (a + 2b)] = 0
⇒ [3x – (a + 2b)][3x – (2a + b)] = 0
⇒ 3x – (a + 2b) = 0 or 3x – (2a + b) = 0
⇒ x = ((a + 2b)/3) or x = ((2a + b)/3)
Hence, the roots =
Q.3. Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Given, AP and BP are tangents of circle having centre O.
To Prove : AP = BP
Construction : Join OP, AO and BO
Proof : In DOAP and DOBP
OA = OB (Radius of circle)
OP = OP (Common side)
∠OAP = ∠OBP = 90° (Radius ⊥ tangent)
ΔOAP ≅ ΔOBP (RHS congruency rule)
AP = BP (By cpct) 1.
Hence Proved.
Q.4. The volume of a right circular cylinder with its height equal to the radius, is 25(1/7) cm3
Find the height of the cylinder. (Use π = 22/7)
Given,
Volume of a right circular cylinder = 25(1/7) cm3
i.e., πr2h = 176/7
Since, r = h ...given
(22/7) x h2 x h = 176/7
⇒ h3 = 176/22 = 8 = 23.
Hence, height of the cylinder = 2 cm.
Q.5. Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.
Here, Modal Class = 80 – 100
l = 80, f1 = 18, f2 = 5, f0 = 12 and h = 20
∴ Mode = l +
= 80 + (6/19) x 20
= 80 + 6.31 = 86.31 (approx.)
Hence, mode = 86.31.
Q.6. For what value(s) of 'a' quadratic equation 3ax2 – 6x + 1 = 0 has no real roots ?
OR
If one root of the equation (k – 1)x2 – 10x + 3 = 0 is the reciprocal of the other, then find the value of k.
Given that,
3ax2 – 6x + 1 = 0
For no real roots b2 – 4ac < 0
Discriminant D < 0 so,
(– 6)2 – 4(3a) (1) < 0
12a > 36
a > 3
Detailed Solution:
Given, 3ax2 – 6x + 1 = 0
On Comparing with AX2 + BX + C = 0,
we get A = 3a, B = – 6 and C = 1
Discriminant, D = B2 – 4AC
= (– 6)2 – 4 × 3a × 1
= 36 – 12a
For condition of 'no real roots',
B2 – 4AC < 0
⇒ 36 – 12a < 0
⇒ 12a > 36
⇒ a > 3.
OR
Let one root = α
and the other root = 1/α
Product of roots = α x (1/α) = 1
Given equation, (k – 1)x2 – 10x + 3 = 0
Comparing it with standard quadratic equation ax2 + bx + c = 0
we get, a = (k – 1), b = – 10 & c = 3
Product of roots = c/a = 3/(k - 1)
1 = 3/(k - 1)
or, 3 = k – 1
or, k = 4.
Section - B
Q.7. Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median
n/2 = 250/2
= 125 ⇒ median class is 2 – 3,
l = 2, h = 1, cf = 120, f = 62
Median =
= 2 + (5/62)
= 129/62 = 2(5/62) m or 2.08 m
50% of students jumped below 2(5/62) m and 50% above it.
Detailed Solution:
Take N/2 = 250/2 = 125
∴ Median class = 2 – 3
Lower limit of median class (l) = 2
Size of median class (h) = 1
Frequency corresponding to median class (f) = 62
Total number of observations
Frequency (N) = 250
Cumulative frequency preceding
Median class (c.f.) = 120
Median =
= 2 + (5/62)
= 129/62
= 2(5/62) m or 2.08 m
∴ 50% of students jumped below 2.08 m and 50% of students jumped above 2.08 m.
Q.8. Construct a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Draw circle of radius 4cm
Draw OA and construct ∠AOB = 120°
Draw ∠OBP = ∠OBP = 90°
PA and PB are required tangents
Detailed Solution:
Steps of construction:
1. Draw a circle of radius 4 cm.
2. Draw two radii having an angle of 120°.
3. Let the radii intersect circle at A and B.
4. Draw angle of 90° on both A and B.
5. The point where both rays of 90° intersect is P.
6. PA and PB are the required tangents.
Q.9. The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Mean
Q.10. Two vertical poles of different heights are standing 20 m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole from the foot of the first pole is 30°. Find the difference between the heights of two poles. (Take √3 = 1.73)
OR
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take √3 = 1.73)
In ∆PQS, tan60° = y/20
⇒ y = 20√3 m
In ∆RSQ, tan30° = x/20
⇒ x = (20/√3) m
y – x =
OR
Let PR be the building and AB be the boy
In ∆PQR, tan 60° = PQ/50 ⇒ PQ = 50√3m
Height of the building = ( 50√3 +1.7 )m = = 88.2 m
Detailed Solution:
Let the heights of two pole be y and x.
Distance between the poles is QS = 20 m.
In ∆PQS, ∠Q = 90°
tan 60° = PQ/QS
√3 =(y/20)
y = 20√3 m
In ∆RSQ, ∠S = 90°
tan 30° = RS/QS
1/√3 = x/20
20/√3 = x
x = (20/3)√3 m
Difference between their heights
= (20/3)√3 x 2
= (40/3)√3.
= 23.07m
OR
Height of Boy = AB = 1.7 m
= QR = 1.7 m
Distance between Boy and building
BR = 50 m
⇒ AQ = 50 m
In ∆PQA, ∠Q = 90°
tan 60° = PQ/AQ
√3 = PQ/50
50√3 = PQ
Total height of the building = PQ + QR
= 86.5 + 1.7 = 88.2 m.
Section - C
Q.11. A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, then find the ratio of the radius and the height of the conical part.
Let ABC be a cone, which is mounted on a hemisphere.
Given, OC = OD = r cm
Curved surface area of the hemispherical part = (1/2)4πr2
= 2pr2
Slant height of a cone,
and curved surface area of a cone = πrl
According to the problem,
(Given)
on squaring both of the sides, we get
4r2 = h2 + r2
⇒ 4r2 – r2 = h2
⇒ 3r2 = h2
Hence, the ratio of the radius and the height of conical part = 1 : √3.
Q.12. PQ is a tangent to a circle with centre O at point P. If ΔOPQ is an isosceles triangle, then find ∠OQP.
OR
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.
In ΔOPQ,
∠P+∠Q+∠O =180°
(∠O = ∠Q, isosceles triangle)
2∠Q+∠P = 180°
2∠Q+90° = 180°
2∠Q = 90°
∠Q = 45°
Detailed Solution:
Since, ∠OPQ = 90° (Angle between tangent and radius)
Let ∠PQO be x°, then ∠QOP = x°
(Since OPQ is an isosceles triangle) (OP = OQ) (given)
In ΔOPQ,
∠OPQ + ∠PQO + ∠QOP = 180° (Sum of the angles of a triangle)
∴ 90° + x° + x° = 180°
⇒ 2x° = 180° – 90° = 90°
⇒ x = 90°/2 = 45°.
Hence, ∠OQP is 45°.
OR
∠APB = 90° (angle in semi-circle) and ∠ODB = 90° (radius is perpendicular to tangent)
ΔABP ~ ΔOBD
⇒ 26/13 = AP/8
Hence, AP = 16 cm.
Case Study-1
Q.13. ‘Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively.
Based on the following figure related to sky sailing answer the questions:
(i) In the given figure, if tan θ = cot (30° + θ), where Q and 30° + θ are acute angles, then the value of θ.
(ii) Find the value of tan 30°. cot 60°.
(i) Given, tan θ = cot(30° + θ)
= tan[90° – (30° + θ)]
= tan(90° – 30° – θ)
⇒ tan θ = tan(60° – θ)
⇒ θ = 60° – θ
⇒ 2θ = 60°
⇒ θ = 30°.
(ii) tan 30° = 1/√3
cot 60° = 1/√3
Now, tan 30° × cot 60° =
= 1/3.
Case Study-2
Q.14. A ladder has rungs 25 cm apart. (see the below).
The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and the bottom rungs are 2(1/2)m apart.
(i) Find the number of the rungs:
(ii) What is the length of the wood required for the rungs?
(i) The distance between the two rungs is 25 cm.
Hence, the total number of rungs
(ii) Here, length of first rung be a and last rung be l, then
S= (n/2)(a+l)
Here, a = 25, l = 45
and n = 11
Then, the required length of the wood,
S11 = (11/2)[25 + 45]
= (11/2) x 70
= 385 cm.
The document Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
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# Question e8b51
May 29, 2017
One has to do with instantaneous rates of change, while the other mainly deals with the sums of very small components. Or when you have to find the opposite of a derivative.
#### Explanation:
Covering the obvious ones...
Let's suppose you have a distance-time graph:
graph{0.25x^2 [-6.06, 16.44, -2.43, 8.815]}
And you had to find the velocity at $t = 5$.
Since velocity is a rate of change, or the slope, and we have to find the slope of a point $\left(t , s \left(t\right)\right)$ we use a derivative. In this case the rate of change is $\frac{\mathrm{ds}}{\mathrm{dt}}$
or the derivative of the function $s$ in respect to $t$.
Now, let's suppose you have a question like:
What is the change in velocity of an object in free fall from $0$ to $10$ seconds since release?
Well, we know that all objects accelerate at $9.8 \setminus \quad \frac{m}{s} ^ 2$ towards the earth. We also know that the derivative of acceleration is velocity.
So, the derivative of what will give us $9.8$?
d/dt[?]=9.8
In that case, the what is the integral of $9.8$
?=int \quad 9.8 \quad dt
?=9.8x + C#
But we wanted to find the change in velocity.
So, we take the difference in the velocity from 0 to 10.
$\Delta V = 9.8 \cdot 10 - 9.8 \cdot 0 = 98$
Writing it in shorthand would be:
$\Delta V = {\int}_{0}^{10} 9.8 \setminus \quad \mathrm{dt}$
What's fascinating is that the area under that line 9.8 represents the change in velocity.
Integrals can also be used to find the sum of things divided into very small parts. I'm pretty sure you're familiar with the ol' finding the area under the curve thing.
The sum of all the blue rectangles would be the integral of an interval of the function.
Those are the basics. |
## 2.2. Equation technique in CVP analysis
We just saw how to calculate the volume of minimum sales in units and dollars to break-even (i.e., have zero profits). What if we want to know how many valves need to be sold to earn a \$30,000 profit? What about a \$40,000 profit?
The answer can be found by using the same equation we used before:
Sales = Variable Costs + Fixed Costs + Profits
Note that:
Sales = Sales Price per Unit x Unit Sales
Variable Costs = Variable Cost per Unit x Unit Sales
If we replace Sales and Variable Costs in the equation, we will get the following result:
Selling Price per Unit x Unit Sales = Variable Costs per Unit x Unit Sales + Fixed Costs + Profits
When we further re-arrange the equation, we will obtain the following:
Unit Sales = Fixed Costs + Profits Selling Price per Unit - Variable Costs per Unit
Using this formula, we can calculate how many units should be sold to generate \$30,000 in profits for Friends Company:
Unit Sales = \$10,000 + \$30,000 = 20,000 units \$5 - \$3
The same formula can be used when the target profit is \$40,000:
Unit Sales = \$10,000 + \$40,000 = 25,000 units \$5 - \$3
The 20,000 and 25,000 units can be converted to dollars as follows: \$100,000 (\$5 x 20,000 units) and \$125,000 (\$5 x 25,000 units), respectively.
We can check our answers for the desired \$30,000 and \$40,000 profits by using the same formula we applied before:
Profits = Selling Price per Unit x Unit Sales - Variable Costs per Unit x Unit Sales - Fixed Costs
For the 20,000 units sold, the profits will be calculated as follows:
Profits = \$5 x 20,000 - \$3 x 20,000 - \$10,000 = \$30,000
And for the 25,000 units sold, the profits can be determined as shown below:
Profits = 25,000 x \$5 - 25,000 x \$3 - 10,000 = \$40,000
In both cases, we arrived at the desired level of profits and this proves that our calculations were correct.
Now when we know that we've got to sell 20,000 units (or generate \$100,000 in sales) to earn the \$30,000 profits and 25,000 units (or \$125,000) to make \$40,000 of profits, Friends Company can assess (a) if it is possible considering the market situation; (b) what level of sales is more realistic in the company's situation; (c) what amount of resources the company needs; and (d) if Friends Company needs to hire more employees to generate a desired level of profits.
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# How can you find cubes of numbers without using multiplication?
Oct 10, 2016
Use logarithms and exponents.
#### Explanation:
Hmmm. Well I suppose you could use addition, logarithms and exponents...
${x}^{3} = {e}^{3 \ln x} = {e}^{{e}^{\ln 3 + \ln \ln x}}$
Or:
${x}^{3} = {10}^{3 \log x} = {10}^{{10}^{\log 3 + \log \log x}}$
Is that what you were looking for?
$\textcolor{w h i t e}{}$
Some more explanation
The cube of a number $x$ is formed by multiplying:
${x}^{3} = x \cdot x \cdot x$
You can convert multiplication into addition using logs and exponents.
We can use any base of logarithm, but the most commonly used ones are natural logarithm $\ln$, which is the inverse of exponentiation base $e$ or common logarithm $\log$, which is the inverse of exponentiation base $10$.
Note that exponentiation relates multiplication and addition like this:
${a}^{m + n} = {a}^{m} \cdot {a}^{n}$
So we can use it in combination with logarithm to do multiplication using addition:
$x y = {a}^{{\log}_{a} x + {\log}_{a} y}$
As an extension of this we find that:
${\log}_{a} {x}^{n} = n {\log}_{a} x$
and hence:
${x}^{n} = {a}^{n {\log}_{a} x}$
We can express the multiplication $n {\log}_{a} x$ in terms of exponentiation and logarithms too:
$n {\log}_{a} x = {a}^{{\log}_{a} n + {\log}_{a} {\log}_{a} x}$
Hence putting $a = e$ and $n = 3$ we find:
${x}^{3} = {e}^{{e}^{\ln 3 + \ln \ln x}}$
Putting $a = 10$ and $n = 3$ we find:
${x}^{3} = {10}^{{10}^{\log 3 + \log \log x}}$
Alternatively, we could use addition instead of multiplication by $3$ to get the forms:
${x}^{3} = {e}^{\ln x + \ln x + \ln x}$
${x}^{3} = {10}^{\log x + \log x + \log x}$
Oct 11, 2016
#### Explanation:
Think of multiplication as addition. So explaining by example.
$2 \times 3$ is 2 lots of 3 added together $\to 3 + 3 = 6$
so lets see if we can change ${4}^{3}$ into addition
We know that
4^3=4xx4^2" "=" "4xx(4" lots of "4)=4xx(4+4+4+4)
But $4 \times \left(4 + 4 + 4 + 4\right)$ is the same as:
$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\underline{\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4} \leftarrow \text{ "Add}$
$16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$16$
$16$
$16$
$\underline{16} \leftarrow \text{ Add}$
$64$
So by using addition we have demonstrated the ${4}^{3} = 64$
Oct 11, 2016
If you are happy to square numbers, add, subtract and divide by $2$ then:
${x}^{3} = {\left(\frac{{x}^{2} + x}{2}\right)}^{2} - {\left(\frac{{x}^{2} - x}{2}\right)}^{2}$
#### Explanation:
Suppose you are not happy with multiplying two arbitrary numbers but:
• You know the squares of numbers.
• You are happy to add, subtract and divide by $2$.
Note first that:
${\left(a + b\right)}^{2} - {\left(a - b\right)}^{2} = \left({a}^{2} + 2 a b + {b}^{2}\right) - \left({a}^{2} - 2 a b + {b}^{2}\right) = 4 a b$
So:
${\left(\frac{a + b}{2}\right)}^{2} - {\left(\frac{a - b}{2}\right)}^{2} = a b$
Put $a = {x}^{2}$ and $b = x$ to find:
${x}^{3} = {\left(\frac{{x}^{2} + x}{2}\right)}^{2} - {\left(\frac{{x}^{2} - x}{2}\right)}^{2}$
For example:
${4}^{3} = {\left(\frac{{4}^{2} + 4}{2}\right)}^{2} - {\left(\frac{{4}^{2} - 4}{2}\right)}^{2}$
$\textcolor{w h i t e}{{4}^{3}} = {\left(\frac{16 + 4}{2}\right)}^{2} - {\left(\frac{16 - 4}{2}\right)}^{2}$
$\textcolor{w h i t e}{{4}^{3}} = {\left(\frac{20}{2}\right)}^{2} - {\left(\frac{12}{2}\right)}^{2}$
$\textcolor{w h i t e}{{4}^{3}} = {10}^{2} - {6}^{2}$
$\textcolor{w h i t e}{{4}^{3}} = 100 - 36$
$\textcolor{w h i t e}{{4}^{3}} = 64$
Oct 12, 2016
Here's a way to construct cubes using just addition and subtraction...
#### Explanation:
Here's a way to construct the sequence of cubes of positive integers without using multiplication...
Write down the first four cubes of positive integers in a line with spaces between them (leaving space on the right too)...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
In the gaps under each pair of numbers, write the difference between them to make a second line...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$
Add a third line consisting of the differences between each pair of numbers in the second line...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$
Add a fourth line containing the difference between the pair of numbers in the third line...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$
$\textcolor{w h i t e}{00000000} 6$
Extend the fourth line by repeating the number $6$ we arrived at as many times as you want more cubes (I will just do two to save typing)...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$
$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$
$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$
Repeat to get additional terms for the second line...
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37 \textcolor{w h i t e}{000} \textcolor{red}{61} \textcolor{w h i t e}{000} \textcolor{red}{91}$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$
$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$
Finally repeat for the first row to get:
$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64 \textcolor{w h i t e}{00} \textcolor{red}{125} \textcolor{w h i t e}{00} \textcolor{red}{216}$
$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37 \textcolor{w h i t e}{000} \textcolor{red}{61} \textcolor{w h i t e}{000} \textcolor{red}{91}$
$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$
$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$ |
# Radius, Diameter and Circumference
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The radius of a circle is a line segment that goes from the center point to a point on the circle. Thus, it goes halfway through the circle.
## Diameter
The diameter of circle is a line segment that goes all the way across a circle through the center point.
## Circumference
The circumference of a circle is the distance around the outside of the circle.
The radius, diameter, and circumference of a circle always have the same relationships. Thus, if you know one of these, you can calculate the other two.
The diameter is always twice as long as the radius. This makes sense because the diameter is all the way across the circle while the radius is only halfway across.
When circumference is involved, the relationship is a little bit more complex because of the number (Pi).
### Did you know that if you divide the circumference of any circle by its diameter, you will always get the same exact number every time?
The number you get is called (Pi) and is approximately equal to 3.14, but the decimal places actually keep going and going.
Computers have calculated millions and even billions of digits to the number , but they haven't been able to find the end of it yet.
When you use in a problem, typically you can just remember is approximately 3.14.
There are three formulas you can use when dealing with the circumference:
In these formulas, C = circumference, r = radius, and d = diameter
1) Use the formula given below to find the circumference when you know the radius.
C = 2x xr
2) Use the formula given below to find the circumference when you know the diameter.
C = xd
3) Use the formula given below to find the diameter when you know the circumference.
C = c ÷
Let's take a look at some examples:
### Example 1
Find the radius, diameter, and circumference of each circle.
Radius = 4 inches (Remember the radius goes halfway across the circle.)
The diameter is always twice as long as the radius.
So, the diameter will be:
4 x 2 = 8
The diameter is 8 inches.
For the circumference, we need to choose a formula.
It's usually better to use the information we are given (instead of the information we figured out ourselves - just in case we made a mistake).
We started with r = 4. The first formula C = 2x xr has r in it, so we will start with that formula and substitute in the information we know.
C = 2x xr
This formula says we must multiply 2 times Pi (3.14) times the radius
C = 2x 3.14 xr
Then we just need to multiply:
C = 25.12 in.
Answer: Radius = 4 in., diameter = 8 in., circumference =25.12 in.
Diameter = 6 feet (as shown in the above picture)
The diameter is always twice as long as the radius.
So, if our diameter is 6 feet, the radius must be half as long as that.
6 ÷ 2 = 3
So the radius is 3 feet long.
For the circumference, we need to choose a formula. Remember it's usually better to use the information we are originally given. We started with d = 6.
The first formula C = xd has d in it, so we will start with that formula and substitute in the information we know.
C = xd
This formula says we must multiply Pi (3.14) by the diameter.
C = 3.14x6
Then we just need to multiply:
C = 18.84 ft.
Diameter = 6 ft.,
Circumference = 18.84 ft.
We know that the circumference = 6.28 meters because we are told this, but we need to find the diameter or radius.
The formula given below should help us do that:
d = c ÷
This formula says that to find the diameter, we must take the circumference and divide it by Pi (3.14)
d = 6.28 ÷ 3.14
On simplifying we get :
d = 2 meters
Now, as we know the diameter, it is easy to find the radius. The diameter is twice as long as the radius, so the radius is half as long as the diameter.
2 ÷ 2 = 1
So, the radius is 1 meter long.
(And this answer makes sense. If halfway across the circle is 1 meter, all the way across is 2 meters).
Answer: C = 6.28 m., d ? 2 m., r ? 1 m.
## Radius, Diameter and Circumference
Parts of circle
A line segment that goes from the center point to a point on the circle. Thus, it goes halfway through the circle.
2. Diameter
A line segment that goes all the way across a circle through the center point.
3. Circumference
The distance around the outside of the circle.
Finding radius, diameter, and circumference:
The diameter is always twice as long as the radius ; d = 2 x r.
Important formulas: C = 2x xr , C = xd , d = C ÷
(Pi) is a special number that occurs in circles. Its decimal places keep going and going, but we can use the approximation 3.14. So, in calculations, think ? 3.14.
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#### Explain Solution R.D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 21 maths Textbook Solution.
Answer:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
Hint: $\text { First we will make two elements of } \mathrm{C}_{3} \text { zero. then we will expand it }$
Given:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
Solution:
$\text { L.H.S }\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|$
\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & (a+2)-(a+3) & 0 \\ (a+2)(a+3)-(a+3)(a+4) & (a+3)-(a+4) & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(a+1-a-3) & a+2-a-3 & 0 \\ (a+3)(a+2-a-4) & a+3-a-4 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(-2) & -1 & 0 \\ (a+3)(-2) & -1 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \end{aligned}
\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{3}\\ &=0\left|\begin{array}{cc} (a+3)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|-0\left|\begin{array}{cc} (a+2)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|+1 \mid\left(\begin{array}{ll} a+2)(-2) & -1 \\ (a+3)(-2) & -1 \end{array} \mid\right.\\ &=0-0+1\{(a+2)(-2)(-1)-(-1)(-2)(a+3)\}\\ &=1\{2(a+2)-2(a+3)\}\\ &=2 a+4-2 a-6\\ &=-2\\ &=R \cdot H \cdot S \end{aligned}
Hence it is proved that
$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$ |
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The pentagonal prism is a three-dimensional figure formed by two pentagonal bases. The pentagonal bases are congruent and parallel to each other. These bases are connected by five rectangular side faces. These prisms are characterized by having a total of 7 faces, 10 vertices, and 15 edges. Pentagonal prisms are also considered as heptahedra.
Here, we will learn some of the most important characteristics of this type of prism. In addition, we will learn about its most used formulas and we will apply them to solve some problems.
##### GEOMETRY
Relevant for
Learning about the characteristics of pentagonal prisms.
See characteristics
##### GEOMETRY
Relevant for
Learning about the characteristics of pentagonal prisms.
See characteristics
## Definition of a pentagonal prism
A pentagonal prism is a prism that has two pentagonal bases and five lateral rectangular faces. These prisms are a type of heptahedron with 7 faces, 10 vertices, and 15 edges.
Each vertex of the pentagonal prism is a point where three faces, two rectangular faces, and one pentagonal face meet. The following is a diagram of a pentagonal prism.
There are two main types of pentagonal prisms:
• Regular Pentagonal Prisms
• Right Pentagonal Prisms
### Regular Pentagonal Prisms
Regular pentagonal prisms are formed when all sides of the pentagonal base are the same length. In these prisms, all the rectangular side faces are congruent, that is, they have the same dimensions. The lateral faces are often called the sides of the prism.
### Right Pentagonal Prisms
A pentagonal prism is right when it has two pentagonal bases that are congruent and parallel to each other and when the lateral faces form a right angle (90 degrees) with the bases.
## Fundamental characteristics of a pentagonal prism
The following are the most important characteristics of a pentagonal prism:
• The bases of the prism are pentagons.
• The bases are congruent and parallel with each other
• The lateral faces of the prism are rectangles.
• If the prism is regular, the lateral faces are congruent.
• These prisms have 7 faces, 10 vertices, and 15 edges.
• Two rectangular faces and a pentagonal face meet at each vertex.
• A pentagonal prism is right when the bases form 90 degree angles with the lateral faces.
## Important pentagonal prism formulas
Pentagonal prisms are 3D figures, so their most important formulas are the volume formula and the surface area formula.
### Formula for the volume of a pentagonal prism
The volume of a prism is equal to the area of the base times the height of the prism. In this case, we have the formula:
where a represents the length of the apothem of the hexagon, l is the length of one of the sides of the hexagon and h is the height of the prism.
### Formula for the surface area of a pentagonal prism
The surface area of a prism is equal to the sum of the areas of all its faces. We have two pentagonal faces and five rectangular faces, so its formula for the surface area is:
## Examples of pentagonal prism problems
The following exercises are solved by applying the pentagonal prism formulas seen above.
### EXAMPLE 1
If a pentagonal prism has an apothem of 5.5 m, a height of 6 m, and sides of length 8 m, what is its volume?
Solution: We have the lengths , and . Using these values in the volume formula, we have:
The volume of the prism is 660 m³.
### EXAMPLE 2
What is the volume of a prism that has a height of 10 m and a pentagonal base with sides of a length of 9 m and an apothem of 6.2 m?
Solution: We have the values , and . We use the volume formula with these values:
The volume of the prism is 1395 m³.
### EXAMPLE 3
What is the surface area of a prism that has a height of 10 m, an apothem of 5.5 m, and sides of length 6 m?
Solution: We have the lengths , , and , so we use these values in the formula for surface area:
The surface area is 465 m².
### EXAMPLE 4
What is the surface area of a prism that has a height of 6 m, an apothem of 4.8 m, and sides of length 8 m?
Solution: We can use the values given in the formula for surface area:
The surface area is 432 m². |
Let’s recall the definition of an equivalence relation:
A relation R on a set A is termed an equivalence relation if it is simultaneously reflexive, symmetric and transitive.
Let’s look at more examples:
Example One: Let $A = \{2, 11, 17, 20\}$ be a set with the following relation: $R = \{ (2,2) (11,11) (17,17) (20,20) (2,20) (20,2) (11,17) (17,11) \}.$
The relation described by R is termed “the same parity.” Elements x and y are said to have the same parity if they are both odd or both even. In our case, the elements 11 and 17 are both odd – hence have the same parity. Similarly, 20 and 2 have the same parity because they are both even. An element will always have the same parity as itself.
The elements that share the same parity as 11 can be grouped together to form a set: $O = \{ 11, 17 \}.$ This is the set of all odd elements from A.
Similarly, the even elements can be grouped together to form the set: $E = \{ 2, 20\}.$
The new sets, O and E, form the equivalence classes of the relation R on set A.
Equivalence classes are defined in the following way: Suppose R is an equivalence relation on a set A. Given any element a from set A, the equivalence class containing a is the subset $\{ x \in A : xRa \}$ of A consisting of all the elements of A that relate to a. This set is denoted as [a].
In other words, the equivalence class of some element a in set A is a subset of set A. The subset contains all the elements related to a under the given relation R. Let’s look at an example.
Example Two: $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ is a set. Let x $\sim$ y if and only if $x - y$ is divisible by $3$, where x and y are elements of S.
First, we need to show that $\sim$ is an equivalence relation. Then we can find the equivalence classes.
Note: To prove this if and only if statement, it is enough to assume the left hand side and prove the right.
Show that $\sim$ is reflexive.
Let x be an element from S. Since $x-x = 0$ and 0 is divisible by 3, then we know $\sim$ is reflexive as required.
Show that $\sim$ is symmetric.
Let x and y be elements from S such that x $\sim$ y.
We want to show that y $\sim$ x. This is equivalent to saying that $y - x$ is divisible by $3.$ In other words, we can find some integer (let’s call it p) such that $3p = y - x.$
Since x $\sim$ y, then the definition of $\sim$ (in our question) tells us that we have $x - y$ is divisible by $3.$
This is the same as saying we can find at least one integer $t,$ such that $3t = x -y$
We can multiply both sides of the equation by -1 and get: $-3t = y - x.$ We do this because we want the right hand side to have the form: $y - x$
We can finally write that y $\sim$ x because $-3t = y - x$ is the same as writing $3(-t) = y - x.$
So the integer we wanted is $p = -t.$ In conclusion, $\sim$ is, indeed, symmetric.
Show that $\sim$ is transitive.
Let x, y and z be elements from S such that x$\sim$ y and y $\sim$ z hold true.
We want to show that x $\sim$ z holds, too. In other words, $x - z$ is divisible by $3$ so there exists some integer p such that $3p = x -z$
Since x $\sim$ y, then there exists some integer $t$ such that $3t = x - y.$
Similarly, y $\sim$ z so there exists some integer $r$ such that $3r = y -z.$
We can add the above equations to get
$3t + 3r = (x - y) + (y - z)$
$3(t + r) = x + (-y + y) - z$
$3(t + r) = x + 0 - z$
$3(r + t) = x - z$
So since $r, t \in \mathbb{Z},$ then their sum $r + t$ is also some integer. In fact, this is the integer we were looking for: $p = r + t$
In other words, $x - z$ is divisible by $3$ as required! This tells us x $\sim$ z, hence $\sim$ is transitive.
So $\sim$ is an equivalence relation! Now we can go on to find its equivalence classes.
Find the equivalence classes of $\sim$
Recall our set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9 \}$. An equivalence class for $a \in S$ is defined as the subset $\{ x \in S : xRa \}.$ This subset has all the elements that relate to a under the relation R. In our case, R is denated $\sim.$
Lets find the equivalence class for 1. We need to find a subset of S such that all the elements in the subset are related to 1 under $\sim$. In other words, we want to find elements $x$ in S such that 1 $\sim$ x:
Since 1 $\sim$ x means $1 - x$ is divisible by $3,$
Then we know there exists some $p \in \mathbb{Z}$ such that $3p = 1 - x$
We can rewrite this as $3p -1 = -x$
$1 - 3p = x$
We now substitute different values of $p$ (e.g. …, -2, -1, 0 , 1, 2,…) to find values of $x$ that are in set S.
For instance, we get the set $\{.. , -5, -2, 1, 4, 7, 10, ..\}$ for different values of $p.$ However, since an equivalence class is defined as a subsets of S and S only consists of positive integers 1 to 9, then we must remove the negative values from the set obtained. In conclusion, the equivalence class of 1 is the set $\{ 1, 4, 7\}$
So the equivalence class for 1 is the same the equivalence class for 4 and 7.
Let’s look at the equivalence class for element 2:
We want to find elements $x \in S$ such that 2 $\sim$ x. This means $2 - x$ is divisible by $3.$
Let $t \in \mathbb{Z}.$ We can write $3t = 2 - x$ which is equivalent to $2 - 3t = x.$
To find the equivalence class for 2, we need to substitute for values of $t = -2,-1, 0, 1.$
The set $\{.. , -1, 2,5,8, ..\}$ is condensed to $\{2, 5, 8 \}$ to fit the requirements for an equivalence class.
Again, note that the equivalence classes for 2, 5 and 8 are the same.
Repeat the process to obtain the last equivalence class: $\{3, 6, 9\}$
How clear is this post? |
## What is a system of equations?
A system of equations is a set of equations with a common solution.
Example 1:
$$\begin{eqnarray*} x^2 - 3 x + 2 &=& 0\\ x^2 - x - 2 &=& 0 \end{eqnarray*}$$
The first equation has solution $$x = 2$$ or $$x = 1$$, while the second has solution $$x = 2$$ or $$x = -1$$. The common solution is $$x = 2$$. Thus, that is the solution of the system.
## Solving a system of equations by substitution
Example 2:
$$\mathbb{S}_1$$:
$$\begin{eqnarray*} x + xy &=& 1\\ y - xy &=& - 4 \end{eqnarray*}$$
This system of equations has two unknown quantities, $$x$$ and $$y$$. The solution will consist of ordered pairs of numbers $$( x, y )$$ satisfying both equations.
A useful method for solving systems of equations is the method of substitution. In this method, one solves one of the equations in the system for a selected variable in terms of the other variables. Then all occurrences of the selected variable in the other equations of the system are replaced by the solution obtained.
In the example at hand, for the first equation in $$\mathbb{S}_1$$ (denoted $$E_1$$ ) the left side may be factored to yield $$x(1+y)=1$$. So neither $$x$$ nor $$1+y$$ may equal $$0$$. Thus we may divide both sides by $$1+y$$ to get the equivalent equation $$x = \dfrac{1}{ 1 + y }$$.
The $$x$$ in $$E_2$$ is replaced with $$\dfrac{1}{ 1 + y }$$ to produce an equation in only one variable, $$y$$. This yields a second system of equations which is equivalent to the first system of equations.
Substitution yields a second system $$\mathbb{S}_2$$ of equations $$E^\prime_1$$ and $$E^\prime_2$$ equivalent to the first.
$$\mathbb{S}_2$$:
$$\begin{eqnarray*} x &=& \dfrac{1}{ 1 + y }\\ y - \dfrac{y}{ 1 + y } &=& - 4 \end{eqnarray*}$$
Now, $$E^\prime_2$$ can be replaced by the equivalent equation $$( y + 2 )^2 = 0$$, which, in turn is equivalent to the equation $$y = - 2$$. Thus, the system
$$\mathbb{S}_3$$:
$$\begin{eqnarray*} x &=& \dfrac{1}{ 1 + y }\\ y &=& -2 \end{eqnarray*}$$
is equivalent to the original system of equations. Substituting the value $$y=-2$$ into the first equation yields $$x = -1$$ giving us the system of equations
$$\mathbb{S}_4$$:
$$\begin{eqnarray*} x &=& -1\\ y &=& -2 \end{eqnarray*}$$
which is equivalent to the original system. That means it has the same solution as the original system. But the solution of this last system is transparent. Thus, we have solved the original system.
In this example we see a general approach to solving all systems of equations. It consists of replacing one system by another equivalent system in a systematic way in order to produce (eventually) an equivalent system whose solution is obvious.
We have seen a couple of principles involved in replacing one system with an equivalent system:
1. Any equation in the system may be replaced by an equivalent equation. (The Equivalent Equation Principle)
2. Any variable or expression in an equation of the system may be replaced by another expression found to be equivalent to that variable or expression in one of the other equations of the system. (The Substitution Principle)
There are other principles which allow us to find equivalent systems, some of which will be discussed in the >.
## Exercise 7.1.1
Use the equivalent equation principle and the substitution principle to solve the system of equations:
$$\mathbb{S}_1$$:
$$\begin{eqnarray*} x + y &=& 0\\ x + y^2 &=& 0 \end{eqnarray*}$$
See Solution
## Exercise 7.1.2
Solve the system of equations:
$$\mathbb{S}_1$$:
$$\begin{eqnarray*} x + y &=& 1\\ 2x + y &=& -1 \end{eqnarray*}$$
See Solution |
We've updated our
TEXT
# Ordered Pairs as Solutions to Systems
### Learning Outcomes
• Evaluate ordered pairs as solutions to systems
A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time.
## Determine whether an ordered pair is a solution for a system of linear equations
In this section, we will look at systems of linear equations in two variables which consist of two equations that each contain two different variables. For example, consider the following system of linear equations in two variables.
$\begin{array}{l}2x+y=\text{ }15\\3x-y=\text{ }5\end{array}$
The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair $(4, 7)$ is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly, we will investigate methods of finding such a solution if it exists.
$\begin{array}{l}2\left(4\right)+\left(7\right)=15\text{ }\text{True}\hfill \\ 3\left(4\right)-\left(7\right)=5\text{ }\text{True}\hfill \end{array}$
### How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution
1. Substitute the ordered pair into each equation in the system.
2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.
The lines in the graph above are defined as
$\begin{array}{r}2x-y=-4\\ x-y=-1\end{array}$.
They cross at what appears to be $\left(-3,-2\right)$. Using algebra, we can verify that this shared point is actually $\left(-3,-2\right)$ and not $\left(-2.999,-1.999\right)$. By substituting the $x$- and $y$-values of the ordered pair into the equation of each line, you can test whether the point is on both lines. If the substitution results in a true statement, then you have found a solution to the system of equations! Since the solution of the system must be a solution to all the equations in the system, you will need to check the point in each equation. In the following example, we will substitute $-3$ for $x$ and $-2$ for $y$ in each equation to test whether it is actually the solution.
### Example
Is $\left(-3,-2\right)$ a solution of the system [latex-display]\begin{array}{r}2x-y=-4\\ x-y=-1\end{array}[/latex-display]
Answer: Test $2x-y=-4$ first: [latex-display]\begin{array}{r}2(-3)-(-2) = -4\\-6+2=-4\\-4 = -4\\\text{TRUE}\end{array}[/latex-display] Now test $x-y=-1$. [latex-display]\begin{array}{r}(-3)-(-2) = -1\\-3+2=-1\\-1 = -1\\\text{TRUE}\end{array}[/latex-display] [latex-display]\left(-3,-2\right)[/latex] is a solution of $x-y=-1[/latex-display] Since[latex]\left(-3,-2\right)$ is a solution of each of the equations in the system,$\left(-3,-2\right)$ is a solution of the system.
$\left(-3,-2\right)$ is a solution to the system.
### Example
Is $(3, 9)$ a solution of the system [latex-display]\begin{array}{r}y=3x\\2x–y=6\end{array}[/latex-display]
Answer: Since the solution of the system must be a solution to all the equations in the system, check the point in each equation. Substitute $3$ for $x$ and $9$ for $y$ in each equation. [latex-display]\begin{array}{l}y=3x\\9=3\left(3\right)\\\text{TRUE}\end{array}[/latex-display] $(3, 9)$ is a solution of $y=3x$. [latex-display]\begin{array}{r}2x–y=6\\2\left(3\right)–9=6\\6–9=6\\-3=6\\\text{FALSE}\end{array}[/latex-display] $(3, 9)$ is not a solution of $2x–y=6$. Since $(3, 9)$ is not a solution of one of the equations in the system, it cannot be a solution of the system.
$(3, 9)$ is not a solution to the system.
### Example
Determine whether the ordered pair $\left(5,1\right)$ is a solution to the given system of equations.
$\begin{array}{l}x+3y=8\hfill \\ 2x - 9=y\hfill \end{array}$
Answer: Substitute the ordered pair $\left(5,1\right)$ into both equations.
$\begin{array}{ll}\left(5\right)+3\left(1\right)=8\hfill & \hfill \\ \text{ }8=8\hfill & \text{True}\hfill \\2\left(5\right)-9=\left(1\right)\hfill & \hfill \\ \text{ }\text{1=1}\hfill & \text{True}\hfill \end{array}$
The ordered pair $\left(5,1\right)$ satisfies both equations, so it is the solution to the system. We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.
Is $(−2,4)$ a solution for the system [latex-display]\begin{array}{r}y=2x\\3x+2y=1\end{array}[/latex-display] Before you do any calculations, look at the point given and the first equation in the system. Can you predict the answer to the question without doing any algebra?
Answer: Substitute $-2$ for $x$, and $4$ for $y$ into the first equation: [latex-display]\begin{array}{l}y=2x\\4=2\left(-2\right)\\4=-4\\\text{FALSE}\end{array}[/latex-display] You can stop testing because a point that is a solution to the system will be a solution to both equations in the system. $(−2,4)$ is NOT a solution for the system [latex-display]\begin{array}{r}y=2x\\3x+2y=1\end{array}[/latex-display]
In the following video, we will show another example of how to verify whether an ordered pair is a solution to a system of equations. https://youtu.be/2IxgKgjX00k Remember that in order to be a solution to the system of equations, the values of the point must be a solution for both equations. Once you find one equation for which the point is false, you have determined that it is not a solution for the system.
### Try It
[ohm_question]48988[/ohm_question]
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# Dice
In probability and statistics, dice are commonly used to construct simple experiments. When a fair, six-sided die is rolled, the sample space can be denoted as:
{1, 2, 3, 4, 5, 6}
Each of the six possible outcomes has an equal likelihood of occurring. Given that A represents any one of the six outcomes, its probability can be denoted as:
P(A) = 1/6 = 0.167
There are many different ways that a die (or multiple dice) can be used to construct probability experiments, and it is common to define these experiments and events using set notation. For example, given the possible outcomes of rolling a fair, six-sided die,
• A = rolling an odd number
• B = rolling a number greater than 4
• C = rolling a prime number
determine the outcomes that satisfy the following conditions, then calculate their probabilities for a single roll of a die.
1. The union of A and B (A ∪ B)
2. The intersection of B and C (B ∩ C)
3. The union of A, B, and C (A ∪ B ∪ C)
i. There are 3 values that satisfy event A: {1, 3, 5}; there are 2 values that satisfy event B: {5, 6}. Thus:
A ∪ B = {1, 3, 5, 6}
ii. Two values satisfy event B, {5, 6}, and three values satisfy event C, {2, 3, 5}, so:
B ∩ C = {5}
iii. The outcomes that satisfy events A, B, and C are {1, 3, 5}, {5, 6}, and {2, 3, 5} respectively, so:
A ∪ B ∪ C = {1, 2, 3, 5, 6}
The probability of an event (E) occurring can be calculated using the formula:
Thus, the probabilities of the events above occurring can be computed as follows.
i. P(A ∪ B):
ii. P(B ∩ C):
iii. P(A ∪ B ∪ C):
There are many other ways that dice can be used to demonstrate simple probability experiments. Refer to the roll a die page for more examples. |
# Logarithms - Laws of Operations
## (Simplifying Logarithmic Expressions)
In this section we learn the rules for operations with logarithms, which are commonly called the laws of logarithms.
These rules will allow us to simplify logarithmic expressions, those are expressions involving logarithms.
For instance, by the end of this section, we'll know how to show that the expression: $3.log_2(3)-log_2(9)+log_2(5)$ can be simplified and written: $log_2(15)$
To do this we learn three rules:
• the addition rule for logarithms
• the subtraction rule for logarithms
• the power rule for logarithms
we'll state the rule and see a detailed tutorial for each of these rules as well as learn a few "must-know tricks" along the way.
Let's get started.
When adding two logarithms, in the same base $$b$$, the following simplification can always be made: $log_b(a)+log_b(c) = log_b(a\times c)$
## Example
The expression: $log_3(5)+log_3(8)$ can be simplied and written: \begin{aligned} log_3(5)+log_3(8) & = log_3(5\times 8) \\ & = log_3(40) \end{aligned}
## Exercise 1
Simplify each of the following as much as possible:
1. $$log_3(5)+log_3(2)$$
2. $$log_7(8)+log_7(10)$$
3. $$log_5(6)+log_5(2)+log_5(3)$$
4. $$log_b \begin{pmatrix} p \end{pmatrix} +log_b\begin{pmatrix} q \end{pmatrix}$$
5. $$log_b \begin{pmatrix} p \end{pmatrix} +log_b\begin{pmatrix} p^2 \end{pmatrix}$$
We find the following results:
1. $$log_3(5)+log_3(2) = log_3(10)$$
2. $$log_7(8)+log_7(10) = log_7(80)$$
3. $$log_5(6)+log_5(2)+log_5(3) = log_5(36)$$
4. $$log_b \begin{pmatrix} p \end{pmatrix} +log_b\begin{pmatrix} q \end{pmatrix} = log_b \begin{pmatrix} pq \end{pmatrix}$$
5. $$log_b \begin{pmatrix} p \end{pmatrix} +log_b\begin{pmatrix} p^2 \end{pmatrix} = log_b \begin{pmatrix} p^3 \end{pmatrix}$$
## Subtraction Law
When subtracting with logarithms, in the same base $$b$$, the following simplification can always be made: $log_b(a)-log_b(c) = log_b \begin{pmatrix} \frac{a}{c}\end{pmatrix}$
## Example
The expression: $log_2(50)-log_2(10)$ can be simplied and written: \begin{aligned} log_2(50)-log_2(10) & = log_2\begin{pmatrix}\frac{50}{10}\end{pmatrix} \\ & = log_2(5) \end{aligned}
## Exercise 2
Simplify each of the following as much as possible:
1. $$log_3(8)-log_3(4)$$
2. $$log_{10}(24)-log_{10}(6)$$
3. $$log_b\begin{pmatrix} p^3 \end{pmatrix} - log_b\begin{pmatrix} p^2 \end{pmatrix}$$
4. $$log_5(64)-log_5(4)-log_5(2)$$
5. $$log_b\begin{pmatrix} p \end{pmatrix} + log_b\begin{pmatrix} p^3 \end{pmatrix} - log_b\begin{pmatrix} p^2 \end{pmatrix}$$
We find the following results:
1. $$log_3(8)-log_3(4) = log_3(2)$$
2. $$log_{10}(24)-log_{10}(6) = log_{10}(4)$$
3. $$log_b\begin{pmatrix} p^3 \end{pmatrix} - log_b\begin{pmatrix} p^2 \end{pmatrix} = log_b\begin{pmatrix} p \end{pmatrix}$$
4. $$log_5(64)-log_5(4)-log_5(2) = log_5(8)$$
5. $$log_b\begin{pmatrix} p \end{pmatrix} + log_b\begin{pmatrix} p^3 \end{pmatrix} - log_b\begin{pmatrix} p^2 \end{pmatrix} = log_b\begin{pmatrix} p^2 \end{pmatrix}$$
## Power Rule for Logarithms (multiplication by a scalar)
When a logarithm, base $$b$$ is multiplied by a scalar, $$x$$, the following simplification can always be made: $x.log_b(a)= log_b \begin{pmatrix} a^x\end{pmatrix}$
## Example
The following expression: $4.log_6(2)$ can be simplified as: \begin{aligned} 4.log_6(2) & = log_6\begin{pmatrix}2^4\end{pmatrix}\\ & = log_6(16) \end{aligned}
## Exercise 3
Simplify each of the following as much as possible:
1. $$3. log_2(5)$$
2. $$2. log_5(3) + 3. log_5(2)$$
3. $$6. log_b\begin{pmatrix}p \end{pmatrix} - log_b\begin{pmatrix}p^2 \end{pmatrix}$$
4. $$2.log_4(a)-5.log_4(b)$$
5. $$5.log_2(m)+2.log_2 \begin{pmatrix}n^3\end{pmatrix}$$
Simplify each of the following as much as possible:
1. $$3. log_2(5) = log_5\begin{pmatrix} 125\end{pmatrix}$$
2. $$2. log_5(3) + 3. log_5(2) = log_5 \begin{pmatrix} 72 \end{pmatrix}$$
3. $$6. log_b\begin{pmatrix}p \end{pmatrix} - log_b\begin{pmatrix}p^2 \end{pmatrix} = log_b\begin{pmatrix}p^4 \end{pmatrix}$$
4. $$2.log_4(a)-5.log_4(b) = log_4\begin{pmatrix} \frac{a^2}{b^5}\end{pmatrix}$$
5. $$5.log_2(m)+2.log_2 \begin{pmatrix}n^3\end{pmatrix} = log_2\begin{pmatrix} m^5.n^6\end{pmatrix}$$
## Some "must-know" results & tricks
We now learn how to deal with numbers being added or subtracted to a logarithm. In particular, we learn how to write any number as a logarithm.
For instance we may be required to simplify the expression: $3 + log_3\begin{pmatrix}5\end{pmatrix}$
## Writing any number as a Logarithm
Any number $$k$$ can be written as a logarithm in any base $$b$$ using the following result: $k = log_b\begin{pmatrix}b^k\end{pmatrix}$
## Example
Say we wish to simplify the expression: $3+log_2(5)$ Then the trick is to write $$3$$ as a logarithm in base $$2$$ and then use the addition rule to simplify.
Using the result, written above, we can state: $3 = log_2\begin{pmatrix}2^3\end{pmatrix}$ And so we can rewrite and simplify the expression as follows: \begin{aligned} 3+log_2(5) & =log_2\begin{pmatrix}2^3 \end{pmatrix} + log_2\begin{pmatrix} 5 \end{pmatrix} \\ & = log_2 \begin{pmatrix}8 \end{pmatrix} + log_2 \begin{pmatrix} 5 \end{pmatrix} \\ & = log_2 \begin{pmatrix}8\times 5 \end{pmatrix} \\ 3+log_2(5) & = log_2 \begin{pmatrix}40 \end{pmatrix} \end{aligned} This technique is further illustrated in the tutorial below.
## Exercise 4
Simplify each of the following as much as possible:
1. $$2+log_3(4)$$
2. $$4 - log_2(3)$$
3. $$1+2.log_3(5)$$
4. $$1+3.log_5(2)$$
5. $$3 - log_{10}(5)$$
We find the following:
1. $$2+log_3(4) = log_3(36)$$
2. $$4 - log_2(3) = log_2\begin{pmatrix} \frac{16}{3}\end{pmatrix}$$
3. $$1+2.log_3(5) = log_3(75)$$
4. $$1+3.log_5(2) = log_5(40)$$
5. $$3 - log_{10}(5) = log_{10}(200)$$ |
# Solving Right Triangle in Trigonometry
Solving right triangle in trigonometry is really easy. We just need to remember some mnemonics and a little background in Pythagorean Theorem.
This topic is very powerful in many ways. Right triangle has a very huge application in both basic and advanced mathematics. The foundation in solving this easily is to use SOH-CAH-TOA formula.
Given the right triangle,
Right Triangle in Trigonometry
Parts:
$\Theta$ – this is the angle itself
Adjacent Side (A) – this is the side adjacent to the angle, so if $\theta$ is in $\angle C$, the adjacent side will be the side $\overline{BC}$
Opposite Side (O) – this is the side always opposite to the given angle
Hypotenuse (H) – this is constant part, the hypotenuse is always the longest side of the triangle opposite the right angle.
When solving this problem, we need to make sure the location of the angle. If it is on the 1st, 2nd, 3rd or 4th quadrant because the sign of cosine and sine varies depending on the location of the angle. Take a look of the photo below.
Signs of Sine and cosine in each quadrant
SOH-CAH-TOA mnemonics:
$sin\theta=\displaystyle\frac{O}{H}$
$cos\theta=\displaystyle\frac{A}{H}$
$tan\theta=\displaystyle\frac{O}{A}$
Worked Problem 1:
Given that $sin\theta =\displaystyle\frac{3}{5}$ and $\theta$ is in quadrant 2. Find the value of the following:
a. $cos\theta$
b. $tan\theta$
c. $sin2\theta$
Solution:
Draw a quick draft of the figure so that you can identify where is the adjacent and opposite sides.
Since we’re dealing with right triangle, we can always use Pythagorean Theorem to solve for the unknown side.
$H^2=A^2+O^2$
$5^2=A^2+3^2$
$A^2=25-9=16$
$A=\sqrt{16}=\pm4$
Solving for $cos\theta$: We use CAH, consider that cosine in quadrant 2 is negative
$cos\theta=\displaystyle\frac{A}{H}$
$cos\theta=\displaystyle\frac{-4}{5}$
Solving for $tan\theta$: In quadrant 2, sine is positive and cosine is negative
$tan\theta=\displaystyle\frac{O}{A}$
$tan\theta=\displaystyle\frac{3}{-4}=-\displaystyle\frac{3}{4}$
Solving for $sin2\theta$:
Now, we need to break up the a little $sin2\theta$
Recall that $sin2\theta=2sin\theta cos\theta$, from double angle identity.
$sin2\theta=2sin\theta cos\theta$
$sin2\theta=2\cdot \displaystyle\frac{3}{5}\cdot -\displaystyle\frac{4}{5}$
$sin2\theta=-\displaystyle\frac{24}{25}$
Worked Problem 2:
Find $sin(x-y)$ if $sin(x)=\displaystyle\frac{3}{5}$ and $sin(y)=\displaystyle\frac{5}{13}$. If $x,y$ are on the 1st quadrant.
Solution:
I know what you’re thinking, $sin(x-y)\ne sin(x)-sin(y)$. We cannot use the distributive property because we’re dealing with angles. Not algebraic quantities.
Since we have two angles x and y, we also need to create 2 triangles.
In figure 1, we need to complete the side of the triangle.
$H^2=A^2+O^2$
$5^2=3^2+O^2$
$A^2=5^2-3^2=16$
$A^2=\pm 4$
Same goes in figure 2,
$13^2=A^2+5^2$
$A^2=13^2-5^2$
$A^2=\pm 12$
Now, recall the sum and difference formula of $sin(x\pm y)=sin(x)cos(y)\pm cos(x)sin(y)$ (sincoscosin mnemonic)
Thus, $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$
Now, we don’t have $cos(x)$ and $cos(y)$, so let’s look for it.
Solving for $cos(x)$: Refer to figure 1, remember that location of the angle, since it is on quadrant 1, all values of A and O are positive.
$cos(x) =\displaystyle\frac{A}{H}$
$cos(x) =\displaystyle\frac{4}{5}$
Solving for $cos(y)$: Refer to figure 2, again the angle is in quadrant 1 so the values of opposite and adjacent are all positive.
$cos(y) =\displaystyle\frac{A}{H}$
$cos(y) =\displaystyle\frac{12}{13}$
Going back to $sin(x-y)$
$sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$
$sin(x-y)=\displaystyle\frac{3}{5}\cdot\displaystyle\frac{12}{13}-\displaystyle\frac{4}{5}\cdot\displaystyle\frac{5}{13}$
$sin(x-y)=\displaystyle\frac{16}{65}$
To outline the steps and guide you to solve right triangle in trigonometry, here are some few tips and steps that you might want to consider so that you would where to start.
Step 1: DRAW the figure. Unless you have the power of wild imagination of the images, you can skip this, but if you’re in the learning process, always draw!
Step 2: Complete the parts of right triangle using Pythagorean Theorem. Easy right?
Step 3: Location of angle. The sign of your answer is dependent on the location of the angle. Mathematics is so strict; always take into account the sign after solving the problem. If your sign is wrong, you are as good as your 3-year old brother. Your answer is wrong.
Step 4: Give the answer to the question if what is being asked. Remember that, because you might be pre occupied with the series of steps and you have forgotten that the problem is just asking for the hypotenuse. Do not overthink!
Step 5: Be cool. The key to making your life easier in trigonometry is to memorize the basic trigonometric identities and learn how to use them properly. Trigonometric Identities is like knowing how to add, subtract, multiply, and divide in arithmetic.
So, you think you can do these simple steps? |
# Print
## A. Prior to the lesson:
1. Arrange students into groups. Each group needs at least ONE person who has a mobile device.
2. If their phone camera doesn't automatically detect and decode QR codes, ask students to
• Bring these devices into the lesson.
4. Cut them out and place them around your class / school.
## B. The lesson:
1. Give each group a clipboard and a piece of paper so they can write down the decoded questions and their answers to them.
2. Explain to the students that the codes are hidden around the school. Each team will get ONE point for each question they correctly decode and copy down onto their sheet, and a further TWO points if they can then provide the correct answer and write this down underneath the question.
3. Away they go! The winner is the first team to return with the most correct answers in the time available. This could be within a lesson, or during a lunchbreak, or even over several days!
## C. TIPS / OTHER IDEAS
4. A detailed case study in how to set up a successful QR Scavenger Hunt using this tool can be found here.
### Question
1. Fraction - 3/4+2/4=5/4
2. Fraction - 7/8-1/2=3/8
3. Fraction - True or False? 1/2 is between 1/3 & 3/4True
4. Fraction - 3/4-1/4=2/4
5. Fraction - 1/2÷2/3=3/4
6. Fraction - Write 2 1/4 as an improper fraction9/4
7. Fraction - Write 19/2 as a proper fraction9 1/2
8. Fraction - What is the lowest common denominator for 3/4 and 2/6?12
9. Angles - What is the total internal angles of a square360
10. Angles - What is the total internal angles of a triangle180
11. Angles - How many degrees can a reflex angle be?180-360
12. Angles - How many degrees can an acute angle be?<90
13. Angles - How many degrees can an obtuse angle be?>90but<180
14. Fractions - 3/5 x 1/2=3/10
15. Fractions - 10/15 ÷ 2/3=1
16. Fractions - 10/16 - 1/8=8/16or1/2
17. Fractions - 12/15 + 1/3=1and2/15
18. BODMAS - 4^2+4-(3x4)=8
19. BODMAS - ((3+4)-1)x2-1=11
20. BODMAS - 3x4-4+1=7
21. Area - What is the area of a square with lengths of 7cm?49cm^2
22. Volume- What is the volume of a rectangular prism with W=7cm L=2cm & H=3cm42cm3
23. Volume - What is the volume of a cube with sides of 5cm?125cm3
24. Weight - How much would a cube with sides of 4cm in length hold of water in grams?64g
25. Weight - If I had 100ml of water how much would it weigh?100g
26. Weight - If I had a rectangular prism with H=8cm L=4cm W=2cm what would the height of the water be if I wanted to put 40g of water into the container?5cm
27. Weight - Convert: 27cm3 = ???ml = ???L27ml0.027L
28. Weight - Convert: 138cm3 = ???g = ???kg138g0.138kg
29. Weight - Convert: 4200cm3 = ???ml = ???L4200ml4.2L
30. Weight - Convert: 420cm3 = ???g = ???ml420g420ml
31. Measurement - Convert: 322mm into cm3.22cm
32. Measurement - Convert: 321cm into m0.321m
33. Weight - Convert: 42cm3 = ???kg = ???L0.042kg0.042L
34. BODMAS - 3^3-7x3+7=13
35. BODMAS - 5+6x7+6x5=77
36. BODMAS - 4+(45-3x11)+(7-3x2)=15
37. BODMAS - ((7+6)x2)+10=36
38. BODMAS - 90-((102)/5)/2=
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Difference Between Similar Terms and Objects
# Difference Between Area and Surface Area
Area vs Surface Area
Mathematics has ways to make us think, and rethink, and do it all over again. As if mathematics is not confusing enough, brought about by its formulas, operations, and derivations – people can also get confused with definitions, especially with similar terms.
Most of us know geometry is the mathematics of measuring earth, spaces, shape, and figures, and when one thinks of geometry, it is most likely that the term ‘area’ comes to mind.
Area is, commonly, an expression of the size of a 2-dimensional plane. It is expressed in many different units. These units include: square meter, hectare, square kilometer, square foot, square yard, square perch, acre, and square mile, just to name a few.
One of the most basic known formulas of area, is that of a rectangle, which is length multiplied by width (l x w), and in the case of the square, it is length of a side squared (s²).
Other formulas include:
Triangle ‘“ ½ bh; where b is base and h is height.
Rhombus ‘“ ½ ab; where a and b are lengths of the two diagonals.
Parallelogram ‘“ bh; where b is the base length, and h is the perpendicular height.
Trapezoid ‘“ ½ (a + b)h; where a and b are the length of parallel sides, and h is the height.
Circle ‘“ pr²; where r is the length of the radius (the square of radius time pi).
Area is often confused with ‘surface area’, which is technically the same if it is in terms of 2-dimensional surfaces. However, it is more appropriately used to express the size of a surface exposed, by a particular solid, that is 3-dimensional. For example, a cube will have a surface area equal to the sum of the areas of all six sides (6s²).
Like area, surface area is also expressed in square units.
Formulas of surface areas of some solids:
Cylinder – 2pr²(r + h); where r is radius, and h is the height of the cylinder.
Cone – pr(r + l); where r is the radius, and l is the slant height of the cone.
Sphere ‘“ 4pr²; where r is the radius.
Summary:
1. The term area is a general term that expresses the size measurement of a surface, while surface area is more appropriately used to express the measurement of the exposed surface of a particular solid object.
2. Area is for 2-dimensional flat surfaces, while surface area is for 3-dimensional solids.
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# Ch 02
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### Ch 02
1. 1. Pythagoras’theorem 2 George is a builder and needs to be sure that the walls in his building are square with the floor. This means that the wall and floor meet at right angles. The walls are 240 centimetres high. The builder runs a string from the top of the wall to a point on the floor 1 metre from the foot of the wall. The string is measured as being 260 centimetres long. Are the walls square with the floor? This chapter looks at working with right-angled triangles and, in particular, using a relationship between the sides of a right-angled triangle.
2. 2. 44 Maths Quest 9 for Victoria Right-angled triangles This chapter investigates one of the most important ideas in geometry related to right- angled triangles. Triangles can be classified according to the length of their sides. They can be classified as either equilateral, isosceles or scalene. Triangles can be classified by their angles as well. A triangle with one right angle is said to be a right-angled triangle. Types of triangles Isosceles Equilateral Scalene Right-angled In any triangle, the longest side is opposite the largest angle; so Hy in a right-angled triangle, the longest side is opposite the right po ten us angle. This side has a special name. It is called the hypotenuse. e The theorem, or rule, that you will learn more about was named after an ancient Greek mathematician, called Pythagoras (580–501 BC). It will enable you to solve all kinds of practical problems related to right-angled triangles. The following exercise will help you to understand what Pythagoras has been credited with discovering 2500 years ago. remember remember 1. The longest side of a right-angled triangle is called the hypotenuse. 2. The hypotenuse is always situated opposite the right angle. 2A Right-angled triangles 2.1 1 For each of the following triangles, carefully measure the length of each side, in milli- HEET metres, and record your results in the table which follows. Note that the hypotenuse isSkillS always marked as c. a a b c c a Pythagoras’ c b b theorem b c a
3. 3. Chapter 2 Pythagoras’ theorem 45 d a e f b a b a c c c b g a h a b c b c a b c d e f g h a b c a2 b2 a 2 + b2 c22 What do you notice about the results in the table in question 1?3 On a sheet of paper, carefully draw 6 right-angled triangles of different sizes. You will need to use a protractor, set square or template to make sure the triangles are right- angled. Carefully measure the sides of each triangle, and complete a table like the one in question 1. What do you notice about these results?4 Now draw some triangles that are not right-angled, measure their sides and complete the same kind of table. Remember to label the longest side as c. What do you notice this time?5 For this activity you will need graph paper, coloured pencils, glue and scissors. a On a sheet of graph paper, draw a right-angled triangle with a base of 4 cm and a height of 3 cm. b Carefully draw a square on each of the three sides of the triangle and mark a grid on each so that the square on the base is divided into 16 small squares, while the square on the height is divided into 9 small squares.
5. 5. Chapter 2 Pythagoras’ theorem 47Using Pythagoras’ theorem Pythagoras was perhaps the first mathematician to recognise and investigate a very important property of right-angled triangles. The theorem, or rule, named after him states that: In any right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The rule is written as c2 = a2 + b2 where a and b are the two shorter sides and c is the hypotenuse. The hypotenuse is the longest side of a right-angled triangle c a and is always the side that is opposite the right angle. b Pythagoras’ theorem gives us a way of finding the length of the third side in a triangle, if we know the lengths of the two x other sides. 4 Finding the hypotenuse 7 We are able to find the length of the hypotenuse when we are given the length of the two shorter sides by substituting into the formula c2 = a2 + b2. WORKED Example 1 For the triangle at right, find the length of the hypotenuse, x, correct to 1 decimal place. x 4 THINK WRITE 7 1 Copy the diagram and label the sides a, b and c. Remember to label the c=x hypotenuse as c. a=4 b=7 2 Write Pythagoras’ theorem. c2 = a 2 + b 2 3 Substitute the values of a, b and c into x2 = 4 2 + 7 2 this rule and simplify. = 16 + 49 = 65 4 Calculate x by taking the square root x = 65 of 65. Round your answer correct to x = 8.1 1 decimal place. In many cases we are able to use Pythagoras’ theorem to solve practical problems. We can model the problem by drawing a diagram, and use Pythagoras’ theorem to solve the right-angled triangle. We then use the result to give a worded answer.
6. 6. 48 Maths Quest 9 for Victoria WORKED Example 2A fire is on the twelfth floor of a building. A child needs to be rescued from a window thatis 20 metres above ground level. If the rescue ladder can be placed no closer than 6 m fromthe foot of the building, what is the minimum length ladder needed to make the rescue?Give your answer correct to the nearest metre.THINK WRITE 1 Draw a diagram and label the sides a, b and c. a c Remember to 20 m x label the hypotenuse as c. 6m b 2 Write Pythagoras’ c2 = a2 + b2 theorem. 3 Substitute the x2 = 202 + 62 values of a, b and c = 400 + 36 into this rule and = 436 simplify. 4 Calculate x by taking x = 436 the square root of 436. = 20.8806 Round your answer ≈ 21 correct to the nearest metre. 5 Give a worded The ladder answer. needs to be 21 metres long. remember remember 1. The hypotenuse is the longest side of the triangle and is opposite the right angle. 2. The length of the hypotenuse can be found if we are given the length of the two shorter sides by using the formula: c2 = a2 + b2. 3. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. 4. Worded problems should be given a worded answer.
7. 7. Chapter 2 Pythagoras’ theorem 49 2B Using Pythagoras’ theoremWORKED 1 For the following triangles, find the length of the hypotenuse, x, correct to 1 decimal 2.2Example place. HEET 1 SkillS a b c x x 7 x 3 5 L Spread XCE sheet E 24 4 12 Finding the d e 896 f 1.3 length of the x hypotenuse 6.2 Math 17.5 cad 742 x x 12.2 Pythagoras’ theorem 2 For each of the following triangles, find the length of the hypotenuse, giving answers GC pro correct to 2 decimal places. gram a 4.7 b 19.3 c Pythagoras’ theorem 804 6.3 27.1 562 d e 0.9 f 152 7.4 87 10.3 2.7 3 A right-angled triangle has a base of 4 cm and a height of 12 cm. Find the length of the hypotenuse to 2 decimal places. 4 Find the lengths of the diagonals of squares that have side lengths of: a 10 cm b 17 cm c 3.2 cm 5 What is the length of the diagonal of a rectangle whose sides are: a 10 cm and 8 cm? b 620 cm and 400 cm? c 17 cm and 3 cm? 6 An isosceles triangle has a base of 30 cm and a height of 10 cm. Find the length of the two equal sides. 7 A right-angled triangle has a height of 17.2 cm, and a base that is half the height. FindWORKED the length of the hypotenuse, correct to 2 decimal places.Example 8 A ladder leans against a vertical wall. The foot of the ladder is 1.2 m from the wall, 2 and the top of the ladder reaches 4.5 m up the wall. How long is the ladder?
8. 8. 50 Maths Quest 9 for Victoria 9 A flagpole, 12 m high, is supported by three wires, attached from the top of the pole to the ground. Each wire is pegged into the ground 5 m from the pole. How much wire is needed to support the pole? 3.8 km me 10 Sarah goes canoeing in a large lake. She paddles 2.1 km to E ti the north, then 3.8 km to the west. Use the triangle at rightGAM 2.1 km Pythagoras’ to find out how far she must then paddle to get back to her theorem — starting point in the shortest possible way. 001 Starting point 11 A baseball diamond is a square of side length 27 m. When a runner on first base tries to steal second base, the catcher has to throw the ball from home base to second base. How far is that throw? Second base 27 m Catcher Finding a shorter side Sometimes a question will give you the length of the hypotenuse and ask you to find one of the shorter sides. In such examples, we need to rearrange Pythagoras’ formula. Given that c2 = a2 + b2, we can rewrite this as: a2 = c2 − b2 or b2 = c2 − a2. It is easier to know what to do in this case if you remember that: finding a short side means subtract.
9. 9. Chapter 2 Pythagoras’ theorem 51 WORKED Example 3Find the length, correct to 1 decimal place, of the unmarked side ofthe triangle at right. 14 cmTHINK WRITE 8 cm1 Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse a as c. c = 14 b=82 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b23 Substitute the values of a, b and c into this a2 = 142 − 82 rule and simplify. = 196 − 64 = 1324 Find a by taking the square root of 132. a = 132 Round to 1 decimal place. = 11.5 cm Practical problems may also involve you being required to find the shorter side of a right-angled triangle. WORKED Example 4A ladder that is 4.5 m long leans up against a vertical wall. The foot of the ladder is 1.2 mfrom the wall. How far up the wall does the ladder reach? Give your answer correct to1 decimal place.THINK WRITE1 Draw a diagram and label the sides a, b and c. Remember to label the hypotenuse as c. c = 4.5 m a b = 1.2 m2 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b23 Substitute the values of a, b and c into this a2 = 4.52 − 1.22 rule and simplify. = 20.25 − 1.44 = 18.814 Find a by taking the square root of 18.81. a = 18.81 Round to 1 decimal place. = 4.3 m5 Give a written answer. The ladder will reach a height of 4.3 m up the wall. Some questions will require you to decide which method is needed to solve the problem. A diagram will help you decide whether you are finding the hypotenuse or one of the shorter sides.
10. 10. 52 Maths Quest 9 for Victoria remember remember 1. A shorter side of the triangle can be found if we are given the length of the hypotenuse and the other shorter side. 2. When finding a shorter side, the formula used becomes a2 = c2 − b2 or b2 = c2 − a2. 3. On your diagram check whether you are finding the length of the hypotenuse or one of the shorter sides. 2C Finding a shorter side reads L Sp he WORKED 1 Find the length, correct to 1 decimal place, of the unmarked side in each of the Example etEXCE 3 following triangles. Finding a b c the length of 14 3.2 the shorter side 10 d hca 17 8.4Mat 8 Pythagoras’ theorem d e 1.2 f 51.8 ogram 382GC pr Pythagoras’ 457 theorem 75.2 3.8 2 Find the value of the pronumeral, correct to 2 decimal places. a a b c c 1.98 8.4 30.1 47.2 2.56 17.52 b d 0.28 e f 2870 d 468 1920 f e 0.67 114 For questions 3 to 13, give your answers correct to 2 decimal places.
11. 11. Chapter 2 Pythagoras’ theorem 53 3 The diagonal of the rectangular sign at right is 34 cm. If the height of this sign is 25 cm, find the width. 4 The diagonal of a rectangle is 120 cm. One side has a length of 70 cm. Find: a the length of the other side b the perimeter of the rectangle c the area of the rectangle. 5 An equilateral triangle has sides of length 20 cm. Find the height of the triangle. 6 The road sign shown at left is in the form of an equilateral triangle. Find the height of the sign and, hence, find its area. 76 cmWORKED 7 A ladder that is 7 metres long leans up against a verticalExample 4 wall. The top of the ladder reaches 6.5 m up the wall. How far from the wall is the foot of the ladder? 8 A tent pole that is 1.5 m high is to be supported by ropes attached to the top. Each rope is 2 m long. How far from the base of the pole can each rope be pegged? 9 A kite is attached to a string 150 m long. Sam holds the 150 m end of the string 1 m above the ground, and the horizontal distance of the kite from Sam is 80 m as 80 m shown at left. How far above the ground is the kite? 1m 10 Ben’s dog ‘Macca’ has wandered onto a frozen pond, and is too frightened to walk back. Ben estimates that the dog is 3.5 m from the edge of the pond. He finds a plank, 4 m long, and thinks he can use it to rescue Macca. The pond is surrounded by a bank that is 1 m high. Ben uses the plank to make a ramp for Macca to walk up. Will he be able to rescue his dog? 11 Penny, the carpenter, is building a roof for a new house. The roof has a gable end in the form of an isosceles triangle, with a base 7.5 m 7.5 m of 6 m and sloping sides of 7.5 m. She decides to put 5 evenly spaced vertical strips of wood as decoration on the gable as shown at right. How many metres of this decorative wood does she need? 6m 12 Wally is installing a watering system in his garden. The pipe is to go all around the edge of the rectangular 56 cm garden, and have a branch diagonally across the garden. The garden measures 5 m by 7.2 m. If the pipe costs 40 cm \$2.40 per metre (or part thereof), what will be the total cost of the pipe? 13 The size of a rectangular television screen is given by the length of its diagonal. What is the size of the screen at left?
12. 12. 54 Maths Quest 9 for Victoria Shortest path A surf lifesaving contest involves swimming A from marker A to the shoreline BC, then on to D 100 m 80 m marker D as shown in the diagram at right. Suppose a lifesaver swims the course A–E–D B E C where E is 60 m from B. 60 m 150 m 1 What total distance will the lifesaver swim? 2 Suppose point E was 100 m from B. Determine how far the lifesaver would swim in total. 3 Suppose point E is halfway between B and C. How far would the lifesaver swim in total? 4 Complete the table below. Length BE Length AE Length ED Total length (m) (m) (m) (AE + ED) (m) 50 60 70 80 90 100 110 5 Use the data in the table to plot a graph of (AE + ED) (m) length BE (horizontal axis) versus total Total length 240 length (AE + ED) (vertical axis). 238 6 What is the shortest distance the lifesaver 236 could swim from marker A to the shoreline 234 then to marker D? 7 Draw a scale diagram showing the path taken 50 70 90 110 Length BE (m) by the lifesaver to swim this shortest distance. Measure the angles AEB and DEC. What do you notice?
13. 13. Chapter 2 Pythagoras’ theorem 55 The shortest distance covered by the lifesaver obeys the Law of Reflection. The angle of A D incidence (AEB) equals the angle of reflection 100 m 80 m (DEC). We can use this law and a scale diagram to find the point, E, for situations B E C where the markers are at different distances D from the shoreline. Reflect the line segment, DC, about the line, BC (shown dotted), then connect A to D′. The intersection point with the shoreline gives the position of E. 8 Use the Law of Reflection to solve A the following problem. Remember G to use a scale diagram. D This time the race involves 100 m 80 m touching the shoreline in two places E and F. What is the shortest 60 m distance the lifesaver should swim? B E F C 100 m 200 m 1 1 True or false? a The hypotenuse is the longest side of a right-angled triangle. x b The hypotenuse forms part of the right angle. 2 cm c Pythagoras’ theorem involves the formula a2 − b2 = c2. 2 Find the length of the hypotenuse of the triangle above right, 5 cm correct to 2 decimal places. 11 m 3 Find the value of the pronumeral in the triangle at right, correct to 1 decimal place. x 16 m 4 multiple choice The length of a diagonal of a square which has sides of 8 mm is: A 10.5 mm B 11.31 mm C 17.3 mm D 19.6 mm E 22.23 mm 5 A roof is 3 m above the ground. A ladder is placed 1 m from the foot of the wall. How long does the ladder need to be in order to reach the roof? (Give your answer correct to the nearest cm.) 6 A flagpole is supported by a wire that is 5 m long. The wire is pegged 2.5 m from the foot of the pole. Find the height of the pole, correct to 2 decimal places. 7 The diagonal of a rectangle is 9.2 cm. The length is 6.3 cm. Find the width, correct to 1 decimal place. 8 Find the area of the rectangle in question 7. 9 Keith drives 5.3 km east, then 9.2 km north. How far in, metres, is he from his starting point? (Give your answer correct to 1 decimal place.)10 An isosceles triangle has sloping edges equal to 10 cm and a base equal to 5 cm. Calculate the height of the triangle, correct to 1 decimal place.
14. 14. 56 Maths Quest 9 for VictoriaWorking with different units When we use Pythagoras’ theorem, we are usually working with a practical situation where measurements have been given. In any calculation, it is essential that all of the measurements are in the same units (for example, cm). Do you remember the relationship between the units of length? 10 mm = 1 cm 100 cm = 1 m To change to a larger unit, divide by the conversion factor. To change to a smaller unit, multiply by the conversion factor. 1000 m = 1 km For example, to change kilometres to metres, multiply by 1000. To change centimetres to metres, divide by 100. The following chart shows all the conversions of length that you are likely to need. ÷ 10 ÷ 100 ÷ 1000 millimetres centimetres metres kilometres (mm) (cm) (m) (km) × 10 × 100 × 1000 When using Pythagoras’ theorem, always check the units given for each measurement. If necessary, convert all measurements to the same units before using the rule. WORKED Example 5 Find the length, in mm, of the hypotenuse of a right-angled triangle if the 2 shorter sides are 7 cm and 12 cm. THINK WRITE 1 Draw a diagram and label the sides a, b and c. Remember to label the c hypotenuse as c. a = 12 cm b = 7 cm 2 Check that all measurements are in the same units. They are. 3 Write Pythagoras’ theorem for the c2 = a2 + b2 hypotenuse. 4 Substitute the values of a and b into this c2 = 122 + 72 rule and simplify. = 144 + 49 = 193 5 Find c by taking the square root. Give c = 193 the units in the answer. = 13.89 cm 6 Check the units required in the answer = 13.89 × 10 mm and convert if necessary. = 138.9 mm
15. 15. Chapter 2 Pythagoras’ theorem 57 WORKED Example 6 The hypotenuse and one other side of a right-angled triangle are 450 cm and 3.4 m respectively. Find the length of the third side, in cm, correct to the nearest whole number. THINK WRITE 1 Draw a diagram and label the sides a, b and c. Remember to label the c = 450 cm hypotenuse as c. a b = 3.4 m 2 Check that all measurements are in 3.4 m = 3.4 × 100 cm the same units. They are different, = 340 cm so convert 3.4 m into cm. 3 Write Pythagoras’ theorem for a a2 = c2 − b2 shorter side. 4 Substitute the values of b and c a2 = 4502 − 3402 into this rule and simplify. = 202 500 − 115 600 = 86 900 5 Find a by taking the square root of a = 86 900 86 900. Round to the nearest whole = 295 number. 6 Give a worded answer. The third side will be approximately 295 cm long. remember remember 1. When using Pythagoras’ theorem, always check the units given for each measurement. 2. If necessary, convert all measurements to the same units before using the rule. Working with different 2D units Where appropriate, give answers correct to 2 decimal places.WORKED 1 Find the length, in mm, of the hypotenuse of a right-angled triangle, if the two shorter 2.3Example HEET sides are 5 cm and 12 cm. SkillS 5 2 Find the length of the hypotenuse of the following right-angled triangles, giving the answer in the units specified. a Sides 456 mm and 320 mm, hypotenuse in cm. XCE L Spread b Sides 12.4 mm and 2.7 cm, hypotenuse in mm. sheet E c Sides 32 m and 4750 cm, hypotenuse in m. Pythagoras’ d Sides 2590 mm and 1.7 m, hypotenuse in mm. theorem e Sides 604 cm and 249 cm, hypotenuse in m. f Sides 4.06 km and 4060 m, hypotenuse in km. g Sides 8364 mm and 577 cm, hypotenuse in m. h Sides 1.5 km and 2780 m, hypotenuse in km.
16. 16. 58 Maths Quest 9 for VictoriaWORKED 3 The hypotenuse and one other side of a right-angled triangle are given for each caseExample below. Find the length of the third side in the units specified. 6 a Sides 46 cm and 25 cm, third side in mm. b Sides 843 mm and 1047 mm, third side in cm. c Sides 4500 m and 3850 m, third side in km. d Sides 20.3 cm and 123 mm, third side in cm. e Sides 6420 mm and 8.4 m, third side in cm. f Sides 0.358 km and 2640 m, third side in m. g Sides 491 mm and 10.8 cm, third side in mm. h Sides 379 000 m and 82 700 m, third side in km. 4 Two sides of a right-angled triangle are given. Find the third side in the units specified. The diagram shows how each triangle c is to be labelled. Remember: c is always the hypotenuse. a a a = 37 cm, c = 180 cm, find b in cm. b a = 856 mm, b = 1200 mm, find c in cm. b c b = 4950 m, c = 5.6 km, find a in km. d a = 125 600 mm, c = 450 m, find b in m. e a = 0.0641 km, b = 0.153 km, find c in m. f a = 639 700 cm, b = 2.34 km, find c in m. 4 cm 5 multiple choice a What is the length of the hypotenuse in this triangle? 3 cm A 25 cm B 50 cm C 50 mm D 500 mm E 2500 mm 82 cm b What is the length of the third side in this triangle? A 48.75 cm B 0.698 m C 0.926 m 43 cm D 92.6 cm E 69.8 mm c The most accurate measure for the length of the third side in the triangle at right is: A 4.83 m B 23.3 cm C 3.94 m 5.6 m D 2330 mm E 4826 mm 2840 mm d What is the length of the third side in this triangle? A 34.71 m B 2.97 m C 5.89 m D 1722 cm E 4.4 m 394 cm 4380 mm 6 A rectangle measures 35 mm by 4.2 cm. Find the length of the diagonal in mm. 7 A sheet of A4 paper measures 210 mm by 297 mm. Find the length of the diagonal in centimetres. 8 A rectangular envelope has a length of 21 cm and a diagonal measuring 35 cm. Find: a the width of the envelope b the area of the envelope. 9 A right-angled triangle has a hypotenuse of 47.3 cm and one other side of 30.8 cm. Find the area of the triangle.
17. 17. Chapter 2 Pythagoras’ theorem 5910 A horse race is 1200 m. The track is straight, and 35 m wide. How much further than 1200 m will a horse run if it starts on the outside and finishes on the inside as shown? Finishing Starting gate post 1200 m 35 m11 A ramp is 9 metres long, and rises to a height of 250 cm. What is the horizontal distance, in metres, between the bottom and the top of the ramp?12 Sarah is making a gate, which has to be 1200 mm wide. It must be braced with a diagonal strut made of a different type of timber. She has only 2 m of this kind of timber available. What is the maximum height of the gate that she can make?13 A rectangular park is 260 m by 480 m. Danny usually trains by running around the edge of the park. After heavy rain, two adjacent sides are too muddy to run along, so he runs a triangular path along the other two sides and the diagonal. Danny does 5 circuits of this path for training. How far does he run? Give your answer in km.14 A swimming pool is 50 m by 25 m. Peter is bored by his usual training routine, and decides to swim the diagonal of the pool. How many diagonals must he swim to com- plete his normal distance of 1200 m?15 A hiker walks 4.5 km west, then 3.8 km south. How far in metres is she from her SHE ET 2.1 Work starting point?16 A square has a diagonal of 10 cm. What is the length of each side?
18. 18. 60 Maths Quest 9 for VictoriaWhat is Bagheera in ‘The Jungle Book’? Bagheera Jungle Book’ Calculate the lengths of the lettered sides in the triangles. Place the lengths with their letters above them, in ascending order in the boxes below. The letters will spell out the puzzle answer. Give lengths accurate to 2 decimal places. 130 cm 6m 1.2 m R 1m B 9m L 1.5 m 6m 7m A T 6.31 m E 520 cm 155 cm 180 cm 7.2 m N 170 cm A 2m 6.2 m 4m 13 cm C 12.98 cm 5.8 m A 4m K 3m H 3m 578 cm P 621 cm 567 cm
19. 19. Chapter 2 Pythagoras’ theorem 61Composite shapes In all of the exercises so far, we have been working with only one right-angled triangle. Many situations involve more complex diagrams, where the right-angled triangle is not as obvious. A neat diagram is essential for these questions. WORKED Example 7 8 cmFind the length of the side, x. Give your x 4 cmanswer correct to 2 decimal places.THINK 10 cm WRITE 1 Copy the diagram. On the diagram, create a right-angled triangle 8 and use the given measurements to work out the lengths of 2 sides. 2 Label the sides of your right-angled triangle as a, b and c. c a 4 Remember to label the hypotenuse as c. b 2 8 3 Check that all measurements are in the same units. They are the same. 4 Write Pythagoras’ theorem for the hypotenuse. c 2 = a2 + b2 5 Substitute the values of a, b and c into this rule and simplify. x2 = 4 2 + 2 2 = 16 + 4 = 20 6 Find x by taking the square root of 20. Round your answer correct x = 20 to 2 decimal places. = 4.47 cm Some situations will involve shapes that contain more than one triangle. In this case it is a good idea to split the diagram into separate right-angled triangles first. WORKED Example 8For the diagram at right, find 15the length of the sides marked 6 xx and y to 2 decimal places.THINK 2 WRITE y 1 Copy the diagram. 15 a 2 Find and draw any right-angled triangles c a x 6 x contained in the diagram. Label their sides. c 3 To find an unknown side in a right-angled b b y 2 triangle, we need to know 2 sides, so find x first. 4 For the triangle containing x, write down a2 = c2 − b2 Pythagoras’ theorem for a shorter side. x2 = 62 − 22 Use it to find x. = 36 − 4 = 32 x = 32 = 5.66 5 We now know 2 sides for the other triangle because we can substitute x = 5.66. 6 For the triangle containing y, write down b2 = c2 − a2 Pythagoras’ theorem for a shorter side and y2 = 152 − 5.662 use it to find y. = 225 − 32 = 193 y = 193 = 13.89
20. 20. 62 Maths Quest 9 for Victoria remember remember 1. To examine more complex situations involving Pythagoras’ theorem, a diagram is essential. 2. When a diagram is given, try to create or separate any right-angled triangles using further diagrams. 2E Composite shapes 8 cm Where appropriate, give answers correct to 2 decimal places. WORKED Example 1 Find the length of the side x in the figure at right. x reads 7 cm L Sp he 7 etEXCE 2 For the following diagrams, find the length of the sides Pythagoras’ 12 cm theorem marked x. a 12 b x c x d x hca 7 37 20 120Mat 200 Pythagoras’ 15 52 theorem 100 WORKED 3 For each of the following diagrams, find the length of the sides marked x and y. d Example hca a b 8Mat Pythagoras’ 10 x 5 theorem DIY 12 8 x 3 y y 4 c 5 d y 18 x x 5 12 20 10 2 7 4 multiple choice a The length of the diagonal of the rectangle at right is: 12 A 9.7 B 19 C 13.9 D 12.2 E 5 b The area of the rectangle at right is: A 12.1 B 84.9 C 15.7 D 109.6 E 38 14 7
21. 21. Chapter 2 Pythagoras’ theorem 63 20 c The value of x in this shape is: x A 24 B 24.7 C 26 D 38.4 E 10 24 30 d What is the value of x in this figure? A 5.4 B 7.5 C 10.1 D 10.3 E 4 x 5 2 75 A hobby knife has a blade in the shape of a right-angled trapezium with the sloping edge 20 mm, and parallel sides of 32 mm and 48 mm. Find the width of the blade and, hence, the area.6 Two buildings, 10 and 18 m high, are directly opposite each other on either side of a 6 m wide European street. What is the distance between the top of the two buildings? 18 m 10 m 6m7 Jess paddles a canoe 1700 m to the west, then 450 m south, and then 900 m to the east. She then stops for a rest. How far is she from her starting point?8 A yacht race starts and finishes at A and consists of 6 legs; AB, BC, CA, AE, EC, CA, in that order as shown in the B 4 km A figure at right. If AB = 4 km, BC = 3 km and CE = 3 km, find: 3 km a AE b AC C c the total length of the race. 3 km E
22. 22. 64 Maths Quest 9 for Victoria 9 A painter uses a trestle to stand on in order to paint a ceiling. It consists of 2 stepladders connected by a 4 m long plank. The inner feet of the 2 stepladders are 3 m apart, and each ladder has sloping sides of 2.5 m. How high off the ground is the plank? 10 A feature wall in a garden is in the shape of a 4.7 m trapezium, with parallel sides of 6.5 m and 4.7 m. The wall is 3.2 m high. It is to have fairy lights around the 3.2 m perimeter (except for the base). How many metres of 6.5 m lighting are required? 10 m 11 A garden bed is in the shape of a rectangle, with a triangular pond at one end as shown in the figure at right. 5m Garden Tony needs to cover the garden to a depth of 20 cm with Pond topsoil. How much soil (in cubic metres) does he need? 3m 12 Katie goes on a hike, and walks 2.5 km north, then 3.1 km east. She then walks 1 km north and 2 km west. How far is she, in a straight line, from her starting point? 13 A rectangular gate is 3.2 m long and 1.6 m high, and 3.2 m consists of three horizontal beams, and five vertical beams as shown in the diagram at right. Each section is 1.6 m braced with diagonals. How much timber is needed for the gate? 14 The diagram at right shows the cross-section through a roof. 5200 mm a Find the height of the roof, h, to the nearest millimetre. h b The longer supports are 5200 mm long. Find the length of the shorter supports, to the nearest millimetre. 9000 mm A 10 15 Find the distance, AB, in the plan of the paddock at right. 10 B 17 1 16 Calculate the value of a, b, c, d. Leave your answers in surd 1 (square root) form. Can you see a pattern? 1 a b c 1 d 1 QUEST SM AT H GE 1 If your watch is running 2 minutes late and EN loses 5 seconds each hour, how many CH L hours will it take for your watch to AL be running 5 minutes late? 2 Sixteen matches are used to create the pattern shown at right. Remove 4 matches to leave exactly 4 triangles. Draw your answer. |
# Rhombus – Definition, Properties, Area, Perimeter, Examples
Home » Math Vocabulary » Rhombus – Definition, Properties, Area, Perimeter, Examples
## What Is Rhombus?
A rhombus is a quadrilateral with four equal sides. Its opposite sides are parallel, and opposite angles are equal.
We know that opposite sides of a parallelogram are equal, a rhombus is a special type of parallelogram with all sides equal. All rhombuses are parallelograms, but not all parallelograms are rhombuses.
## How Is a Rhombus Different from a Square?
The difference between a square and a rhombus is that all angles of a square are right angles, but the angles of a rhombus need not be right angles.
So, a rhombus with four right angles becomes a square.
We can say, “Every square is a rhombus, but all rhombus are not squares.
## Real-life Examples of a Rhombus Shape
You can find rhombus shapes in various everyday objects around us, such as kites, finger rings, rhombus-shaped earrings, the structure of a window glass pane, etc.
## Properties of a Rhombus
Some of the important properties of a rhombus are stated below.
• All sides of a rhombus are equal. Here, AB = BC = CD = DA.
• Diagonals bisect each other at 90°. Here, diagonals AC and BD bisect each other at 90°.
• Opposite sides are parallel in a rhombus. Here, AB ∥ CD and AD ∥ BC.
• Opposite angles are equal in a rhombus. ∠A = ∠C and ∠B = ∠D.
∠A + ∠B = 180°
∠B + ∠C = 180°
∠C + ∠D = 180°
∠A + ∠D = 180°
• All the interior angles of a rhombus add up to 360°.
• The diagonals of a rhombus are perpendicular to each other. Here, AC ⟂ BD.
• The diagonals of a rhombus bisect each other. Here, DI = BI and AI = CI.
• A rhombus has a rotational symmetry of 180 degrees (order 2). A rhombus retains its original orientation when rotated by an angle of 180 degrees.
• The diagonals of a rhombus coincide with the 2 lines of symmetry. They divide the rhombus into 2 identical halves.
## Rhombus Formulas
Let’s discuss important formulas associated with a rhombus.
### Area of a Rhombus
The area of a rhombus is the region enclosed by the 4 sides of a rhombus.
There are two ways to find the area of a rhombus.
• Area of a rhombus when its base and altitude are known
The area of a rhombus is calculated by finding the product of its base and corresponding altitude (height).
Note that you can choose any of its sides as the base, and the perpendicular distance from this side to the opposite side becomes the height.
Area of rhombus = base × height = (b × h) square units.
• Area of a rhombus when the lengths of its diagonals are known
When the lengths of the diagonals of a rhombus are known, then its area is given by half of their product.
Area of rhombus $= \frac{d_{1} \times d_{2}}{2}$ square units
where
d1 and d2 are the lengths of diagonals of a rhombus.
### Perimeter of Rhombus
The perimeter of a rhombus is the total length of its boundaries. As all four sides of a rhombus are equal, its perimeter is calculated by multiplying the length of its side by 4.
Perimeter of a rhombus = 4 × a units
where ‘a’ is the length of the side of the rhombus.
## Solved Examples on Rhombus
Example 1: The lengths of the two diagonals of a rhombus are 18 cm and 12 cm. Find the area of the rhombus.
Solution:
Diagonal (d1 ) = 18 cm
Diagonal (d2 ) = 12 cm
Area of rhombus $= \frac{d_{1}\times d_{2}}{2} = \frac{18 \times 12}{2}$ sq. cm $= 108$ sq. cm
Example 2: Find the perimeter of the rhombus if one of its sides measures 15 cm.
Solution:
Length of a side of the rhombus (a) = 15 cm
Perimeter of rhombus = 4 × a = 4 × 15 cm = 60 cm
Example 3: The area of a rhombus is 56 sq. cm. If the length of one of its diagonals is 14 cm, find the length of the other diagonal.
Solution:
Area of rhombus = 56 sq. cm
d1 = 14 cm
Area of rhombus $= \frac{d_{1} \times d_{2}}{2}$
⇒ 56 = $\frac{14 \times d2}{2}$
⇒ 56 = 7 × d2
⇒ d2 = 56 ÷ 7
⇒ d2 = 8 cm
So, the second diagonal of the given rhombus measures 8 cm.
Example 4: In a rhombus ABCD, if ∠A = 60°, find the measure of remaining angles.
Solution:
60° + ∠B = 180° (Given, A = 60°)
∠B = 180° – 60°
∠B = 120°
∠C = ∠A = 60° (Opposite angles are equal in a rhombus)
∠D = ∠B = 120° (Opposite angles are equal in a rhombus)
## Practice Problems on Rhombus
1
### Which of the following quadrilaterals is definitely a rhombus?
Trapezium
Rectangle
Square
Parallelogram
CorrectIncorrect
All sides of a square are equal, so all squares are rhombus.
2
### If the length of one of the sides of the rhombus is 10 cm. What will be the length of the opposite side of the given rhombus?
5 cm
10 cm
20 cm
40 cm
CorrectIncorrect
All sides of the rhombus are equal in length.
3
### What will be the altitude of the rhombus, whose area is 320 sq. cm and side is 40 cm?
4 cm
6 cm
8 cm
10 cm
CorrectIncorrect
Area = base × altitude
⇒ 320 = 40 × altitude
⇒ altitude = 320 ÷ 40 = 8 cm
4
### The floor area of a room is 500,000 sq. cm. If the floor is to be covered with rhombus-shaped tiles, with each tile having diagonals 40 cm and 25 cm, find the number of tiles required.
50
500
1000
5000
CorrectIncorrect
Area of floor = 500,000 sq. cm
Area of each rhombus-shaped tile = $\frac{d_{1} \times d_{2}}{2}$ = $\frac{40 \times 25}{2}$ = 500 sq. cm
Number of tiles = Area of floor ÷ Area of 1 tile
= 500,000 ÷ 500
= 1,000 tiles
So, 1,000 tiles are required to cover the floor. |
# Percent And Fraction: Math Test Quiz!
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Amy Phillips
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Amy Phillips
Community Contributor
Quizzes Created: 5 | Total Attempts: 1,406
Questions: 10 | Attempts: 358
Settings
.
• 1.
### If I eat 7/10 M&M's, what percent of the M&M's did I eat?
• A.
80%
• B.
54%
• C.
70%
• D.
30%
• E.
Option5
C. 70%
Explanation
If you eat 7/10 of the M&M's, it means you ate 70% of the total amount. This can be calculated by dividing 7 by 10 and then multiplying by 100 to convert it into a percentage.
Rate this question:
• 2.
### What is 3/4 equal to?
• A.
75%
• B.
12%
• C.
90%
• D.
31%
• E.
235%
A. 75%
Explanation
The fraction 3/4 is equal to 75% because to convert a fraction to a percentage, you divide the numerator by the denominator and multiply by 100. In this case, 3 divided by 4 is 0.75, and when multiplied by 100, it becomes 75%.
Rate this question:
• 3.
### Is 30% equal to?
• A.
3 4/10
• B.
4/10
• C.
3/10
• D.
2/10
• E.
23 2/10
C. 3/10
Explanation
The correct answer is 3/10 because 30% can be written as a fraction by dividing 30 by 100, which simplifies to 3/10.
Rate this question:
• 4.
### 50% is equal to what fraction?
• A.
1/2
• B.
3/4
• C.
1/4
A. 1/2
Explanation
50% is equal to 1/2 because percent means "per hundred" and 50% represents 50 per hundred. To convert a percentage to a fraction, we divide the percentage by 100. Therefore, 50/100 simplifies to 1/2.
Rate this question:
• 5.
### If apples are 2 for \$1.00. How much are 5 apples?
• A.
\$0.50
• B.
\$2.00
• C.
\$10.00
• D.
\$2.50
D. \$2.50
Explanation
If apples are 2 for \$1.00, it means that each apple costs \$0.50. Therefore, if we want to know the cost of 5 apples, we multiply the cost of one apple (\$0.50) by the number of apples (5). This gives us a total of \$2.50.
Rate this question:
• 6.
### If I get 67 points on a test that is scored out of 100 points, what percent did I answer correctly? ________
67%, 67 percent
Explanation
To find the percentage of correct answers, divide the number of points obtained (67) by the total possible points (100) and multiply by 100. This will give the percentage of correct answers, which is 67%.
Rate this question:
• 7.
### 53% of people have iPhones. What percent of people DO NOT have iPhones?
• A.
53%
• B.
47%
• C.
100%
B. 47%
Explanation
The correct answer is 47%. Since the question states that 53% of people have iPhones, it implies that the remaining percentage of people do not have iPhones. To determine the percentage of people who do not have iPhones, we subtract 53% from 100%, which equals 47%. Therefore, 47% of people do not have iPhones.
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• 8.
### If 13 out of 100 people are bankers, what percent of people are bankers?
• A.
100%
• B.
87%
• C.
113%
• D.
13%
D. 13%
Explanation
The given question states that 13 out of 100 people are bankers. To find the percentage of people who are bankers, we divide the number of bankers (13) by the total number of people (100) and multiply by 100. This calculation gives us 13%. Therefore, 13% of people are bankers.
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• 9.
### 88% of people go shopping on Black Friday. Does that mean that 12 people out of 100 do not go Black Friday shopping, true or false?
• A.
True
• B.
False
A. True
Explanation
The statement "88% of people go shopping on Black Friday" implies that out of a group of 100 people, 88 of them go shopping on Black Friday. Therefore, it can be deduced that 12 people out of the 100 do not go Black Friday shopping, making the answer true.
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Related Topics |
How do you determine the oblique asymptote?
A slant (oblique) asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator. To find the slant asymptote you must divide the numerator by the denominator using either long division or synthetic division.
How do you find the oblique asymptote using synthetic division?
Following are answers to the practice questions:
1. The answer is y = x– 2. Use synthetic division or long division to divide the denominator into the numerator:
2. The answer is y = x+ 1. Use synthetic division or long division to divide the denominator into the numerator:
3. The answer is y = x –1.
4. The answer is y = –3x+ 13.
How do you find the horizontal oblique asymptote?
1. 2) If the degree of the denominator is equal to the degree of the numerator, there will be a horizontal asymptote at the ratio between the coefficients of the highest degree of the function.
2. Oblique asymptotes occur when the degree of denominator is lower than that of the numerator.
Can functions cross oblique asymptotes?
Note that your graph can cross over a horizontal or oblique asymptote, but it can NEVER cross over a vertical asymptote.
Is oblique asymptote a hole?
The oblique asymptote is y=x−2. The vertical asymptotes are at x=3 and x=−4 which are easier to observe in last form of the function because they clearly don’t cancel to become holes.
Can you have an oblique and horizontal asymptote?
There may be no vertical, horizontal or oblique asymptotes. A function cannot have both horizontal & oblique asymptotes.
Why do oblique asymptotes occur?
Oblique asymptotes only occur when the numerator of f(x) has a degree that is one higher than the degree of the denominator. When you have this situation, simply divide the numerator by the denominator, using polynomial long division or synthetic division. The quotient (set equal to y) will be the oblique asymptote.
What is an oblique asymptote?
An oblique or slant asymptote is an asymptote along a line , where . Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator. For example, the function has an oblique asymptote about the line and a vertical asymptote at the line .
What does oblique asymptote mean?
Oblique Asymptote. An oblique or slant asymptote is an asymptote along a line , where . Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator. For example, the function has an oblique asymptote about the line and a vertical asymptote at the line …
What is the difference between horizontal and oblique asymptotes?
Horizontal asymptotes occur when the numerator of a rational function has degree less than or equal to the degree of the denominator. Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator.
Is it possible to have no vertical horizontal and oblique asymptotes?
Explanation: A rational function y=P(x)Q(x) , where P(x) and Q(x) are non-zero polynomials, may have 0 or more vertical asymptotes, but the number of asymptotes must be finite. The function has no horizontal or oblique asymptotes. |
# 4.3: Classic Fourier Series
Learning Objectives
• Signals can be composed by a superposition of an infinite number of sine and cosine functions.
• The coefficients of the superposition depend on the signal being represented and are equivalent to knowing the function itself.
The classic Fourier series as derived originally expressed a periodic signal (period T) in terms of harmonically related sines and cosines.
$s(t)=a_{0}+\sum_{k=1}^{\infty }a_{k}\cos \left ( \frac{2\pi kt}{T} \right )+\sum_{k=1}^{\infty }b_{k}\sin \left ( \frac{2\pi kt}{T}\right )$
The complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum. The Fourier coefficients, ak and bk, express the real and imaginary parts respectively of the spectrum while the coefficients ck of the complex Fourier series express the spectrum as a magnitude and phase. Equating the classic Fourier series to the complex Fourier series, an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each.
$c_{k}=\frac{1}{2}\left ( a_{k}-ib_{k} \right )$
Exercise $$\PageIndex{1}$$
Derive this relationship between the coefficients of the two Fourier series.
Solution
Write the coefficients of the complex Fourier series in Cartesian form as:
$c_{k}=A_{k}+iB_{k}$
Now substitute into the expression for the complex Fourier series.
$\sum_{k=-\infty }^{\infty }c_{k}e^{i\frac{2\pi kt}{T}}=\sum_{k=-\infty }^{\infty }\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}$
Simplifying each term in the sum using Euler's formula,
$\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}=\left ( A_{k}+iB_{k}\right )\left ( \cos \left ( \frac{2\pi kt}{T} \right )+i\sin \left ( \frac{2\pi kt}{T}\right )\right )$
$\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}=A_{k}\cos \left ( \frac{2\pi kt}{T}\right )-B_{k} \sin \left ( \frac{2\pi kt}{T}\right )+i\left ( A_{k}\sin \left ( \frac{2\pi kt}{T}\right )+B_{k} \cos \left ( \frac{2\pi kt}{T}\right )\right )$
We now combine terms that have the same frequency index in magnitude. Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry:
$c_{-k}=\overline{c_{k}}\; or \; A_{-k}=\overline{A_{k}}\; and\; B_{-k}=\overline{B_{k}}$
After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes:
$2A_{k}\cos \left ( \frac{2\pi kt}{T}\right )-2B_{k} \sin \left ( \frac{2\pi kt}{T}\right )$
To obtain the classic Fourier series, we must have:
$2A_{k}=a_{k}\; and\; 2B_{k}=-b_{k}$
Just as with the complex Fourier series, we can find the Fourier coefficients using the orthogonality properties of sinusoids. Note that the cosine and sine of harmonically related frequencies, even the same frequency, are orthogonal.
$\forall k,l,k\in \mathbb{Z}l\in \mathbb{Z}:\left ( \int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right ) \cos \left ( \frac{2\pi lt}{T} \right )dt=0\right )$
$\int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right ) \sin \left ( \frac{2\pi lt}{T} \right )dt=\begin{cases} \frac{T}{2} & \text{ if } (k=l)\wedge (k\neq 0)\wedge (l\neq 0) \\ 0 & \text{ if } (k\neq l)\vee (k=0=l) \end{cases}$
$\int_{0}^{T}\cos \left ( \frac{2\pi kt}{T} \right ) \cos \left ( \frac{2\pi lt}{T} \right )dt=\begin{cases} \frac{T}{2} & \text{ if } (k=l)\wedge (k\neq 0)\wedge (l\neq 0) \\ T & \text{ if } k=0=l \\ 0 & \text{ if } k\neq l \end{cases}$
These orthogonality relations follow from the following important trigonometric identities.
$\sin (\alpha )\sin (\beta )=\frac{1}{2}\left ( \cos (\alpha -\beta )-\cos (\alpha +\beta )\right )$
$\cos (\alpha )\cos (\beta )=\frac{1}{2}\left ( \cos (\alpha +\beta )+\cos (\alpha -\beta )\right )$
$\sin (\alpha )\cos (\beta )=\frac{1}{2}\left ( \sin (\alpha +\beta )+\sin (\alpha -\beta )\right )$
These identities allow you to substitute a sum of sines and/or cosines for a product of them. Each term in the sum can be integrated by noticing one of two important properties of sinusoids.
• The integral of a sinusoid over an integer number of periods equals zero.
• The integral of the square of a unit-amplitude sinusoid over a period T equals T/2.
To use these, let's, for example, multiply the Fourier series for a signal by the cosine of the lth harmonic:
$\cos \left ( \frac{2\pi lt}{T} \right )$
and integrate.
The idea is that, because integration is linear, the integration will sift out all but the term involving al.
$\int_{0}^{T}s(t)\cos \left ( \frac{2\pi lt}{T} \right )dt=\int_{0}^{T}a_{0}\cos \left ( \frac{2\pi lt}{T} \right )dt+\sum_{k=1}^{\infty }a_{k}\int_{0}^{T}\cos \left ( \frac{2\pi kt}{T} \right )\cos \left ( \frac{2\pi lt}{T} \right )dt+\sum_{k=1}^{\infty }b_{k}\int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right )\cos \left ( \frac{2\pi lt}{T} \right )dt$
The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices k and l are equal (but not zero), in which case we obtain:
$\frac{a_{1}T}{2}$
$If\; \; k=0=l\; \; a_{0}T\; is\; obtained$
Consequently,
$\forall l,l\neq 0:\left ( a_{l}=\frac{2}{T} \int_{0}^{T}s(t)\cos \left ( \frac{2\pi lt}{T} \right )dt\right )$
All of the Fourier coefficients can be found similarly.
$a_{0}=\frac{2}{T} \int_{0}^{T}s(t)dt$
$\forall k,k\neq 0:\left ( a_{k}=\frac{2}{T} \int_{0}^{T}s(t)\cos \left ( \frac{2\pi kt}{T} \right )dt\right )$
$b_{k}=\frac{2}{T} \int_{0}^{T}s(t)\sin \left ( \frac{2\pi kt}{T} \right )dt$
Exercise $$\PageIndex{1}$$
The expression for a0 is referred to as the average value of s(t). Why?
Solution
The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit of a Riemann sum, the integral corresponds to the average.
Exercise $$\PageIndex{1}$$
What is the Fourier series for a unit-amplitude square wave?
Solution
We found that the complex Fourier series coefficients are given by
$c_{k}=\frac{2}{i\pi k}$
The coefficients are pure imaginary, which means ak=0. The coefficients of the sine terms are given by:
$b_{k}=-\left ( 2\Im (c_{k}) \right )$
so that:
$b_{k}=\begin{cases} \frac{4}{\pi k} & \text{ if } k\; odd \\ 0 & \text{ if } k\; even \end{cases}$
Thus, the Fourier series for the square wave is
$sq(t)=\sum_{k\in \left \{ 1,3,... \right \}}\frac{4}{\pi k}\sin \left ( \frac{2\pi kt}{T} \right )$
Example $$\PageIndex{1}$$:
Let's find the Fourier series representation for the half-wave rectified sinusoid.
$s(t)=\begin{cases} \sin \frac{2\pi t}{T} & \text{ if } 0\leq t\leq \frac{T}{2} \\ 0 & \text{ if } \frac{T}{2}\leq t< T \end{cases}$
Begin with the sine terms in the series; to find bk we must calculate the integral
$b_{k}=\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt$
Using our trigonometric identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids, which are much easier to evaluate.
$\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt=\frac{1}{2}\int_{0}^{\frac{T}{2}}\cos \left ( \frac{2\pi (k-1)t}{T} \right ) -\cos \left ( \frac{2\pi (k+1)t}{T} \right )dt$
$\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt=\begin{cases} \frac{1}{2} & \text{ if } k=1 \\ 0 & \text{ if } otherwise \end{cases}$
Thus,
$b_{1}=\frac{1}{2}$
$b_{2}=b_{3}=...=0$
On to the cosine terms. The average value, which corresponds to a0, equals 1/π. The remainder of the cosine coefficients are easy to find, but yield the complicated result:
$a_{k}=\begin{cases} -\left ( \frac{2}{\pi } \frac{1}{k^{2}-1}\right ) & \text{ if } k\in \left \{ 2,4,... \right \} \\ 0 & \text{ if } k\; odd \end{cases}$
Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and the even harmonics.
## Contributor
• ContribEEOpenStax |
# How to Graph Transformation on the Coordinate Plane: Dilation?
This article teaches you how to graph Dilations on the coordinate plane in a few simple steps.
## Step by step guide to graph Transformation: Dilation
A dilation is a type of transformation that creates an image that is the same shape as the original but in a different size.
In a Dilation, each point of an object is moved along a straight line. The straight line is drawn from a fixed point called the center of dilation. The distance the points move depends on the scale factor. The center of dilation is the only invariant point. The scale factor tells us how much this figure is stretched or reduced.
$$Scale\space factor:$$ $$\frac{image\space length}{original\space length}=\frac{distance\space of \space image \space from\space center\space of\space dilation}{distance\space of\space object\space from\space center\space of\space dilation}$$
### Transformation: Dilation – Example 1:
Dilate the image with a scale factor of $$2.5$$.
Solution:
: First, find the original coordinates:
$$A=(-2, -2)$$ $$B=(0, 2)$$ $$C=(2, -2)$$
Next, take all of the coordinates, and multiply them by $$2.5$$:
$$A^\prime=(-5, -5)$$ $$B^\prime=(0, -5)$$ $$C^\prime=(5, -5)$$
Now, graph the new image. Since the new figure is larger and our scale factor is greater than $$1$$, the new image is an enlargement.
### Transformation: Dilation – Example 2:
Dilate the image with a scale factor of $$0.5$$.
Solution:
: First, find the original coordinates:
$$A=(-4, 0)$$ $$B=(0, 2)$$ $$C=(2, 0)$$ $$D=(-2, -4)$$
Next, take all of the coordinates, and multiply them by $$0.5$$:
$$A^\prime=(-2, 0)$$ $$B^\prime=(0, 1)$$ $$C^\prime=(1, 0)$$ $$D^\prime=(-1, -2)$$
Now, graph the new image.
## Exercises for Transformation: Dilation
### Graph the image of the figure using the transformation given.
1) Dilation of $$2$$
2) Dilation of $$0.5$$
1)
2)
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Literal Equations
Home > Lessons > Literal Equations Search | Updated May 22nd, 2021
Introduction
This lesson page will inform you how to solve literal equations, which is a STEM (Science, Technology, Engineering, and Mathematics) topic. Here are the sections within this lesson page:
Before continuing with this lesson, it is important that you understand how to linear solve equations. Use this before proceeding if you are uncertain how to solve linear equations.
esson: Solving Linear Equations
Literal Equations If you want to know what a literal equation is, consider this well known equation. This equation is the relationship between the circumference of a circle with its diameter. The equation indicates how to determine the circumference of a circle -- by multiplying its diameter and the value pi. However, what if we knew the circumference of a circle and wanted to know its diameter? It would be helpful to know how to calculate the diameter by having a better equation. If we were able to solve the equation for the d-value, the equation would be more handy. The process of rearranging terms and performing mathematical operations on an equation to solve for a different variable is called solving a literal equation. Likewise, the area of a square is written like so. This means the area of a square is equal to the length of its side times itself. What is the length of a side of a square if you know the area of that square? The answer to that question demands that we are able to solve a literal equation to get the s-value alone. Being able to solve literal equations like the ones above is required to fully understand and utilize all mathematical relations, like those found within several science disciplines. The next section will continue down that road and demonstrate how to solve literal equations. Solving Literal Equations: Basic Level Let us examine a few equations to get a feel for how to solve literal equations. Example 1: Solve for the diameter. To get the letter-d alone, we need to find a way to remove the pi from the right side. Since pi is being multiplied to the letter-d, the inverse operation would be to divide by pi. So, dividing both sides by pi gives us this. The diameter of a circle is equal to its circumference divided by pi. Example 2: Solve for the side. The equation is s-squared is equal to area. The inverse of squaring is square-rooting. So, we need to take the square root of both sides to simplify the right side of the equation, like so. This tells us that the square root of a square's area is equal to the length of its side. Try this quizmaster to determine if you understand basic literal equations. Solving Literal Equations: Moderate Level Here are more difficult problems to consider. Example 1: Solve for the length. This formula indicates exactly what to do with the length and width of a rectangle to gain its perimeter. On the other hand, if we want to get the length -- the letter-l alone -- we have to first get the 2w term to the opposite side of the equation. Since we are adding 2w, the inverse is to subtract 2w from both sides, like so. Now, to cancel the two that is being multiplied to the l-variable, we need to divide both sides of the equation by two. This equation now tells us how to use the perimeter of a rectangle given its perimeter and width. Example 2: Solve for the height. This equation tells us how to find the surface area of a cylinder using its radius and height. Let's use the equation to solve for h, the height of a cylinder. First, we need to do the distributive property. Now, we need to get the h-term alone, which means canceling the term with the r-squared. So, we need to subtract that term from both sides. For our last step, we need to divide both sides of the equation by 2(pi)r to get the h-variable alone, like so. This tells us how to determine the height of a cylinder given its surface area (S) and radius (r). The following quizmaster contains six moderate level problems. Give it a look to see if you understand how to manipulate literal equations. Interactive Quizzes Try these lessons, which are related to the sections above. Related Lessons Try these lessons, which are related to the sections above. esson: Average Velocity esson: Graphing Lines esson: Systems of Equations esson: Solving Inequalities esson: Linear Programming esson: Conic Sections: General Form to Standard Form |
Topics
# Find at what time between 7 and 8 o'clock will the hands of a clock will be the same straight line but not together
A) 30 minutes past 7 B) 5 minutes past 7 C) $$\Large 5\frac{5}{11}$$ minutes past 7 D) 6 minutes past 7
C) $$\Large 5\frac{5}{11}$$ minutes past 7
Fig. (1) shows the position of the hands of the clock. It is clear tht they are 335 minutes apart. To be in straight line They have to be 40 minutes apart. So the minute hand will have to move 5 minutes space.
55 minutes space gained in 60 minutes.
Therefore, 5 minutes space will be gained in
$$\Large \frac{60}{55} \times 5=\frac{60}{11}=5\frac{5}{11}$$ minutes
The hands will be in a straight line at $$\Large 5\frac{5}{11}$$ minutes part 7.
Part of solved Clocks questions and answers : >> Elementary Mathematics >> Clocks
Similar Questions
1). How many times do the hands of a clock point towards each other in a day.
A). 6 B). 11 C). 18 D). 22
2). A clock is right at 9 a. m. The clock gains 15 minutes in 24 hours. What will be the right time when the clock indicates 12 a.m. on the following day.
A). 43 minutes past 11 B). 45 minutes past 11 C). 44 minutes past 11 D). $$\Large 44\frac{1}{2}$$ minutes past 11
3). The time of a clock as seen in a mirror is 3.15 The correct time is.
A). 8.45 B). 9.45 C). 8.15 D). None of these
4). What is the angle between the hour hand and minute hand at 3.40.
A). $$\Large 120 ^{\circ}$$ B). $$\Large 130 ^{\circ}$$ C). $$\Large 140 ^{\circ}$$ D). $$\Large 150 ^{\circ}$$
5). 6, 12, 18, 24, 30, 36, ?
A). 42 B). 48 C). 52 D). 56
6). 10, 15, 25, 40, 65, ?
A). 66 B). 105 C). 72 D). 84 |
# NCERT Exemplar Solutions Class 9 Science Solutions for Motion - Exercise in Chapter 8 - Motion
Question 27 Motion - Exercise
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig.8.4. Plot a velocity-time graph for the same.
Video transcript
"hi kids welcome to lido india's best live online learning platform so let's see what's the today's question is a girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position her displacement time graph is shown in the figure eight point four plot a velocity time graph for the same so this is the question so what's the question is telling that a girl is there and she is traveling in a straight path to drop a letter and again she comes back to her initial position that means from where she started she came back to the same position so you can see first the graph is increasing and then it is decreasing this is your displacement and here is time so let's try to calculate the velocity for the increasing line right increasing graph increasing means that the velocity is getting greater but what's the formula for the velocity so if you are going to write the formula so velocity is equals to your displacement upon time so for the very first increasing line what's the displacement you can see displacement is 100 and the time taken as 50 so what you will write 50 sorry the displacement is basically 100 not 50. so you will write the displacement value as 100 and time as you can see from here to here it is 50 so you will write 50 and you get the velocity as 2 meter per second for the very first part so here you can see if we want to plot a graph we have taken the velocity and we have taken the time so as the velocity is 2 so we have drawn the velocity 2 and the time is 50. so this graph for the very first half this graph will come like this for the velocity time and now what happens the velocity starts decreasing in this graph what happens your displacement time graph is showing a decrement over here the graph is go going down and then it is becoming complete 0 so now let's try to solve this part again your displacement will be 100 and from 50 to 100 the time is going to be 50 again you will get 2 but this is getting decreased so ultimately it will come in the negative cartesian plane so what will happen again you will mark minus 2 and again you will take the time as what 50 from 50 to 100 so if you are going to plot you will get this particular velocity time graph and here you can see it is minus 2 so you just took the minus 2 part and for 50th it is the 50th second and here you have drawn the 100 second and you simply just joined whole and ultimately you got this graph as an answer so kids this is it for today and if you have any doubts please drop it in the comment section below and do subscribe to leader happy learning "
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Transcript Area PowerPoint
```Warm-Up
1/29
1. Create a vocabulary and formulas flipbook!
2. Use the papers on your desk and the example on the
board to make a flipbook with 6 pages.
3. The tabs should be labeled as follows
1.
2.
3.
4.
5.
6.
Area
What is Area?
Square/Rectangle
All Triangles
Parallelogram/Trapezoid
Irregular Figures
Warm-Up
1/30
How do we find the area of different shapes?
Standard and Essential Question
MCC6.G.1: Find the area of right triangles, other
triangles, special quadrilaterals, and polygons by
composing into rectangles and decomposing into
triangles and other shapes; apply these techniques in
the context of solving real-world and mathematical
problems.
How do I find the area of a square and rectangle
without a formula and decompose those to find the
area of a triangle?
area? (Brainstorm)
What are the important terms?
Area: the number of square units it takes to
completely fill a shape or surface.
Polygon: a two dimensional (2-D) figure made up of
line segments that are connected to form a closed
shape.
Quadrilateral: a four sided polygon
Vertex: the end point of two or more line segments
Unit Squares
You can count unit squares to find the area of a figure.
Height (h)
(width)
Base (b)
(length)
Area= 9 square units
Base= 3 units
Height= 3 units
Area= 4.5 square units
Base= 3 units
Height= 3 units
Geoboard Squares
Square: a quadrilateral that has 4 congruent sides and 4
right angles (90˚).
Create a square that is 9 units long by 9 units wide.
Copy that onto your dot paper!
How many unit squares make up this quadrilateral?
What is the area of the square?
What is a “shortcut”/formula to find the area?
Now put a diagonal to divide the square in half.
When you divide the square in half what two
shapes do you get?
What is the area of one of those shapes?
Area= 12 square units
Base= 4 units
Height= 3 units
Area= 6 square units
Base= 4 units
Height= 3 units
Geoboard Rectangles
Rectangle: a 4-sided polygon with 4 angles that measure
90˚.
Make a rectangle that is 8 units long and 7 units wide.
Copy onto your dot paper!
How many unit squares make up this shape?
What is the area of the figure?
What is a “shortcut”/formula to find the area?
Now put a diagonal to divide the rectangle in half.
When you divide the rectangle in half what two shapes
do you get?
What is the area of one of those shapes?
So what’s the formula?
Warm-Up
1/31
Stacy is planting a square garden in his backyard and
needs to know how much soil to buy. One bag of soil
covers 9 ft2.
If the height of the garden is 14 ft., what is the area of the
garden?
How many bags of soil does he need?
Standard and Essential Question
MCC6.G.1: Find the area of right triangles, other
triangles, special quadrilaterals, and polygons by
composing into rectangles and decomposing into
triangles and other shapes; apply these techniques in
the context of solving real-world and mathematical
problems.
How are areas of geometric figures related to each
other?
Area of Square/Rectangle
Area of Triangle
9 units
4.5 units
12 units
6 units
14 units
16 units
What pattern do you notice?
Types of Triangles
By Sides:
Isosceles: a triangle that has 2 equal sides
Scalene: a triangle that has no equal sides
Equilateral: a triangle that has 3 equal sides
By Angles:
Right: a triangle with one right angle
Acute: a triangle with only acute angles
Obtuse: a triangle with one obtuse angle
Other Important Terms
Height of a triangle: The perpendicular distance
from the base to the highest vertex.
It can be
measured outside
the triangle!
It can be
measured inside
the triangle!
Does the area change depending
on the type of triangle?
Use the link below to determine whether the area
formula of a triangle changes depending on the
type of triangle.
toria/TriangleArea.html
How can I
determine the
area by only
counting unit
squares?
Let’s do some examples…
What’s the area?
9.4 ft.
7 ft.
Area= 32.9 ft2
-----------------------------
What’s the area?
6 cm.
4 cm.
Area= 12 cm2
So what are the 2 ways to find the
area of a triangle?
Warm-Up
2/1
1.
-----------------------------
Find the area of the following triangles.
2.
11.7 cm.
5m
9m
6 cm.
Standard and Essential Question
MCC6.G.1: Find the area of right triangles, other
triangles, special quadrilaterals, and polygons by
composing into rectangles and decomposing into
triangles and other shapes; apply these techniques in
the context of solving real-world and mathematical
problems.
How can we use our knowledge of the area of one
figure to determine the area of another?
---------------------------
---------------------------
Parallelogram
of opposite sides parallel
Do you think you can determine how to
find the area of the parallelogram using
How do you get the area?
http://learnzillion.com/l
A=bh
essons/1058-find-thearea-of-a-parallelogramby-decomposing
Height (h)
Base (b)
Let’s do some examples…
What is the area?
-------------------
12 m
15 m
4.8 m
-------------------
What is the area?
24 m
Area= 360 m2
Area= 57.6 m2
---------------------------
Trapezoid
---------------------------
pair of parallel sides
Do you think you can determine how to find
the area of the trapezoid using the shapes
How do you get the area?
---------------------------
---------------------------
Base 2 (b2)
Base 1 (b1)
Height (h)
Let’s do some examples…
What’s the area?
What’s the area?
9 cm
10 cm
15 cm
Area= 60 m2
Area= 120 cm2
1 ft.
---------------------------
Daniel has a room in his house shaped like a trapezoid.
Use the picture of Daniel’s room to answer the following
questions.
7 ft.
4 ft.
8 ft.
1. What is the area of the room? 32 ft.2
2. If one tile can cover 1.5 square feet, how many tiles
does Daniel need to cover the entire room?
22 tiles
Warm-Up
2/4
---------------------------
---------------------------
A square poster board has sides that are 40 inches long.
When the triangular flaps at the sides are opened, the
poster board takes the shape of a trapezoid. The base of
each of the triangles is 24 inches. What is the area of the
trapezoid poster board?
Standard and Essential Question
MCC6.G.1: Find the area of right triangles, other
triangles, special quadrilaterals, and polygons by
composing into rectangles and decomposing into
triangles and other shapes; apply these techniques in
the context of solving real-world and mathematical
problems.
How can you find the area of irregular polygons
when you don’t have a specific formula?
Irregular Shapes
So how am I supposed to
find the area for that?
Method 1: Counting Squares
Area=
18 units2
How can you count squares?
Hint!
Create two
squares
using the
triangles to
help find
the area!
Area=
16 units2
Method 2: Find the area of the whole figure
and then subtract the shapes that aren’t
included.
12 ft.
2 ft.
7 ft.
3 ft.
3 ft.
10 ft.
120
ft2
–9
ft2
12 ft.
10 ft.
Area= 111 ft2
Method 3: Find the area of the shapes
separately and add their areas together.
25 mm.
12 mm.
625 mm2
(square area)
25 mm.
+ 150 mm2
(triangle area)
Area= 775 mm2
Counting Squares Challenge
Area=
38.5 units2
Method 2 Challenge
15 ft.
6 ft.
4 ft.
4 ft.
11 ft.
2 ft.
4 ft.
4 ft.
4 ft.
Area= 235 ft2
Online Area Games
Triangle Area Game
http://www.shodor.org/interactivate/activities/Triangl
eExplorer/
Area Explorer Game
http://www.shodor.org/interactivate/activities/AreaEx
plorer/
``` |
# Search by Topic
#### Resources tagged with Making and proving conjectures similar to An Introduction to Magic Squares:
Filter by: Content type:
Stage:
Challenge level:
### There are 30 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### An Introduction to Magic Squares
##### Stage: 2, 3 and 4
Find out about Magic Squares in this article written for students. Why are they magic?!
### Roll over the Dice
##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Spirals, Spirals
##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Three Neighbours
##### Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### What Was in the Box?
##### Stage: 1 Challenge Level:
This big box adds something to any number that goes into it. If you know the numbers that come out, what addition might be going on in the box?
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Primary Proof?
##### Stage: 1
Proof does have a place in Primary mathematics classrooms, we just need to be clear about what we mean by proof at this level.
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Move Those Halves
##### Stage: 2 Challenge Level:
For this task, you'll need an A4 sheet and two A5 transparent sheets. Decide on a way of arranging the A5 sheets on top of the A4 sheet and explore ...
### Open Squares
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Always, Sometimes or Never? KS1
##### Stage: 1 Challenge Level:
Are these statements relating to calculation and properties of shapes always true, sometimes true or never true?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Walking Round a Triangle
##### Stage: 1 Challenge Level:
This ladybird is taking a walk round a triangle. Can you see how much he has turned when he gets back to where he started?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Six Numbered Cubes
##### Stage: 2 Challenge Level:
This task combines spatial awareness with addition and multiplication.
### Six Ten Total
##### Stage: 2 Challenge Level:
This challenge combines addition, multiplication, perseverance and even proof.
### Sheep Talk
##### Stage: 2 Challenge Level:
In sheep talk the only letters used are B and A. A sequence of words is formed by following certain rules. What do you notice when you count the letters in each word?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Always, Sometimes or Never? Shape
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Tables Without Tens
##### Stage: 2 Challenge Level:
Investigate and explain the patterns that you see from recording just the units digits of numbers in the times tables.
### Dining Ducks
##### Stage: 2 Challenge Level:
Use the information about the ducks on a particular farm to find out which of the statements about them must be true.
### Become Maths Detectives
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? Don't forget to keep visiting NRICH projects site for the latest developments and questions. |
Related Articles
Distance Formula & Section Formula – Three-dimensional Geometry
• Last Updated : 19 Jan, 2021
One of the most powerful tools in any engineer or scientist’s toolkit is 3D geometry. 3D geometry comes to picture into model real-world quantities such as velocity, fluid flows, electrical signals, and many other physical quantities. In this article, we are going to discuss two important formulas Distance and Section along with some important examples. One can understand any 3D space in terms of 3 coordinates- x, y, and z. Below is a simple representation of 3D space.
## Distance Formula and Its Use in 3D Geometry
The distance formula states the distance between any two points in the XYZ space.
Formula:
For two points P1(x1, y1, z1) and P2(x2, y2, z2)
Distance (d) =
This distance formula is used to calculate the distance between two points in any 3D space. When we know the coordinates of the two points in the plane (in form of ordered pair (x, y, z)) we can easily get the distance between two points by substituting in the distance formula.
### The Distance Between Two Points: Formulas, and Solved Examples
Suppose there are two points P(x1, y1, z1) and Q(x2, y2, z2) in the 3D space. To find the distance between them, the formula is,
Distance (d) =
Example 1: Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2)?
Solution:
Using the formula to calculate the distance between point P and Q,
Distance (d) =
d =
Example 2: Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear?
Solution:
We know that points are said to be collinear if they lie on a line.
As PQ + QR = PR
Hence P, Q and R are collinear.
### The Distance of Point From a Line: Meaning, Formulas, and Examples
The distance of a point from a line is the perpendicular distance from the point to the line. Suppose we have to find the distance of a point P(x0, y0, z0) from line l, then the formula is,
Distance (d) =
Where ‘s’ is the directing vector of line l.
Example: Find the distance from point P(-6, 1, 21) to a line ?
Solution:
### The Distance of a Point from a Plane: Formulas, Equations, and Examples
The distance from a point to a plane is the perpendicular distance from a point on a plane. If Qx + Ry + Sz + T = 0 is a plane equation, then the distance from point P(Px, Py, Pz) to plane can be found using the following formula:
Distance (d) =
Example: To find a distance between plane 2x + 4y – 4z – 6 = 0 and point P(0, 3, 6)?
Solution:
Using the formula:
Distance (d) =
### The Distance Between Parallel Lines: Formulas and Examples
The distance between any two parallel lines is the perpendicular distance from any point on one line to the other line. Suppose there are two parallel lines y = mx + c1 and y = mx + c2, then the formula is,
Distance (d) =
If the equation of two parallel lines is given as:
ax + by + d1 = 0 and ax + by + d2 = 0, then the formula is,
Distance (d) =
Where a and b are the coefficients of variables x and y in the line.
Example 1: Find the distance between lines y = 2x + 10 and y = 2x + 12? (Note: Both lines are parallel to each other)
Solution:
The lines y = 2x + 10 and y = 2x + 12 are in form y = mx + c.
Where c1 = 10, c2 = 12, m = 2
Using formula, the distance (d) =
Example 2: Find the distance between two parallel lines 4x + 3y + 6 = 0 and 4x + 3y – 3 = 0?
Solution:
The lines given are 4x + 3y + 6 = 0 and 4x + 3y – 3 = 0. Both lines are in form ax + by + d = 0.
Hence, d1 = 6, d2 = −3, a = 4, b = 3
Using formula for this case, distance (d) will be calculated as:
## Section Formula: Definition, Vector Formula, Case, and Examples
Section formula is the concept that can be implemented in 2D and 3D space as well. In three dimension system, we have to choose a coordinate system. Suppose two points P(x1, y1, z1) and Q (x2, y2, z2) are given and let the point R (x, y, z) divide PQ in the given ratio m: n internally. Hence, the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n can be calculated using the formula,
If the point R divides PQ externally in the ratio m: n, then its coordinates are obtained by replacing n by – n, and the formula for the same is,
Case 1: Coordinates of the mid-point: In case R is the mid-point of PQ, then m : n = 1 : 1 so that
x = , y = and z =
These are the coordinates of the midpoint of the segment joining P (x1, y1, z1)and Q (x2, y2, z2).
Case 2: The coordinates of the point R which divides PQ in the ratio k: 1 are obtained by taking k= which are as given below,
Example 1: Find the coordinates of the point which divides the line segment joining the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 (i) internally, and (ii) externally?
Solution:
(i) Let P (x, y, z) be the point that divides the line segment joining A(1, – 2, 3)
and B (3, 4, –5) internally in the ratio 2 : 3. Therefore
Thus, the required point is
(ii) Let P (x, y, z) be the point which divides segment joining A (1, –2, 3) and
B (3, 4, –5) externally in the ratio 2 : 3. Then,
Therefore, the required point is (–3, –14, 19).
Example 2: Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6), and (14, 0, –2) are collinear?
Solution:
Let A (– 4, 6, 10), B (2, 4, 6), and C(14, 0, – 2) be the given points. Let the
point P divides AB in the ratio k: 1. Then coordinates of the point P are
Let us examine whether, for some value of k, the point P coincides with point C.
On putting , When k=then,
Therefore, C (14, 0, –2) is a point that divides AB externally in the ratio 3: 2 and is
same as P. Hence A, B, C are collinear.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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# Circle Problems With Solutions
Clocks, wheels, dining plates, and coins. What do these things have in common?
They are all circular.
We are all familiar with what a circle looks like – it is round, has no sides, and has no start or end. Aside from identifying what it looks like, we also know how important it is to us. As early as the Mesopotamian civilization, people already used circular wheels to do their work and complete their journeys faster.
In this chapter, let us view circles from a “geometric lens” and explore how they’re helpful in our everyday lives.
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## Definition of a Circle
In geometry, we formally define a circle as the set of all points that are equidistant (same distance) from a fixed point. That fixed point is called the center of the circle.
In the image above, point C is the center of the circle. If you take any two points in the circle, you are assured that these points have the same distance from point C.
Note that when we mention the phrase “on the circle,” it means “in the frame of the circle.”
Suppose that points A and B are both in the circle. The distance from point C to point A is similar to that from point C to point B, or AC = BC.
Now, suppose that you take point D instead; the distance from point C to point D is certainly equal to the distance from point C to A and from point C to point B. Hence, AC = BC = CD.
Again, remember that all points on the circle have the same distance from the center.
## Radius and Diameter of a Circle
The line segment formed by connecting the center to one of the points on the circle is called the radius.
If we draw a line segment from point C to point A in the figure above, we have formed a circle radius, which is segment AC since we have created a segment that connects the center (point C) and a point on the circle (point A).
Since all points on the circle are equidistant (same distance) from the center, all radii (plural of radius) have equal lengths. In other words, all radii are congruent segments. If someone tells you that one of the radii of a circle is longer than the other radius of the same circle, you know that it is false since all radii are equal in length.
Meanwhile, if you draw a line segment from one point on a circle, pass through the center, and end on another point on the circle, then you have formed a diameter. Segment AQ below is an example of a diameter. The diameter of a circle is a line segment that divides the circle into two equal parts.
As you have noticed, a diameter consists of two radii (AC and CQ). The length of a circle’s diameter is equal to twice the length of a radius.
d = 2r (where d is the diameter and r is the radius)
This also implies that the radius of a circle is ½ of the length of the diameter.
Also, like radii, all circle diameters are congruent or have the same length.
## Circumference of a Circle
Suppose that you are watching a race in a circular field. How will you know the distance covered by the runners in that field?
The distance covered by the runners is equal to the total distance around the circular field. The total distance around a circle is called the circumference.
The circumference is the counterpart of the perimeter for circles. If squares, triangles, or rectangles have a perimeter as the total distance around them, the circumference is the total distance around a circle.
How do we determine the circumference of a circle?
We use the formula for the circumference of a circle which was derived as early as the ancient Greek civilization:
C = πd
where C represents the circle’s circumference, d is the circle’s diameter, and π is a constant irrational number approximately equal to 3.1416.
The number π (the Greek letter for “pi,” which is read as “pie”) is one of the most important numbers in mathematics since it’s used extensively not only in geometry but also in trigonometry and calculus. Many mathematicians provided an estimate for this irrational number, but one of the earliest estimations is from the Greek mathematician Archimedes.
π is the result when you try to divide the circumference of any circle on a flat surface by its diameter. It is an irrational number, meaning we cannot express it as a fraction with integers. One of the most common estimations of π used for calculation is 3.14 or 3.1416. However, these estimations are not the actual values of the number since the digits of π are never-ending. One of the most recent estimates of π has 62.8 trillion digits!
Going back to the circumference of a circle, the formula C = πd means that the distance around a circle is equal to the product of the irrational number and the circle’s diameter. Since the diameter of a circle is equivalent to twice the radius of a circle or d = 2r, we can rewrite the formula for the circumference of a circle as:
C = 2πr
where r is the radius of the circle.
This means that to calculate the circumference of a circle, you can use the following:
• Formula 1: C = πd (if the diameter is given)
• Formula 2: C = 2πr (if the radius is given)
Use formula 1 if the given in the problem is the circle’s diameter. However, if the radius is given, use formula 2.
Sample Problem 1: Determine the circumference of the circle below.
Solution: The circle above has a radius PQ with a length of 3 cm. Since the radius of the circle is given, let us use the formula C = 2πr.
C = 2πr
C = 2π(3) since r = 3
C = 6π
Thus, the circumference of the circle is 6π cm.
Sometimes, the problem provides us with an estimated value of π that we should use in the problem. For instance, look at the given problems below.
Sample Problem 2: Compute the circumference of the circle below. Use π = 3.14.
Solution: The circle above has a radius of XY with a length of 5 cm. Since the radius of the circle is given, let’s use the formula C = 2πr.
C = 2πr
C = 2(3.14)(5) Take note that the problem provides the value of π
C = 31.4
Hence, the circumference of the circle is 31.4 cm.
Sample Problem 3: Calculate the circumference of a wheel with a diameter of 20 centimeters. Use π = 3.14.
Solution: We are given the diameter of the wheel. Thus, it is more convenient to use the formula C = πd.
C = πd
C = (3.14)(20)
C = 62.8
Hence, the circumference of the circle is 62.8 cm.
Sample Problem 4: The rim of a basketball ring has a circumference of 56 inches. Determine the diameter of the basketball ring.
Solution: The problem states that the rim of the basketball ring has a computed circumference of 56 inches.
Recall that the circumference of a circle can be calculated using the formula:
C = πd
We will use this formula instead of the other one since we are looking for the diameter of the ring.
The circumference is 56 inches. So, we have C = 56:
56 = πd
The problem doesn’t provide a precise estimate of π, so we leave it as it is.
Let us now solve for the diameter (d) by dividing both sides by π:
56 = πd
πd = 56 Symmetric Property of Equality
πd⁄π = 56⁄π Dividing both sides of the equation by π
d = 56⁄π
Thus, the diameter of the basketball ring is 56⁄π inches.
Sample Problem 5: The motorcycle’s wheel has a radius of 30 cm. How many revolutions must the wheel make to cover a total distance of 18.84 meters (1,884 centimeters)? Use π = 3.14.
Solution: When a wheel makes one revolution, it revolves at a total distance equivalent to its circumference. In other words, one revolution of the wheel = circumference of the circle.
Therefore, to find the number of revolutions it takes for the wheel to cover a total distance of 1,884 centimeters, we must divide 1,884 by the wheel’s circumference.
Let us first compute the circumference of the wheel.
The problem states that the wheel has a radius of 30 cm. Thus, we need to use the formula C = 2πr. Note that the problem requires us to use π = 3.14.
C = 2πr
C = 2(3.14)(30) Substitute r = 30 and π = 3.14
C = 188.4
Hence, the circumference of the wheel is 188.4 centimeters.
Now, to determine the number of revolutions the wheel will make to cover a total distance of 1,884 centimeters, let us divide 1,884 by the wheel’s circumference (188.4).
Number of revolutions = 1,884 ÷ 188.4 = 10
Thus, the answer to this problem is 10 revolutions.
## Area of a Circle
The area of a circle refers to the amount of space a circle occupies on a flat surface.
To calculate the area of a circle, we use the formula below.
Acircle = πr2
where r is the circle’s radius, and π is the constant irrational number approximately equal to 3.1416.
The formula above states that the area of a circle is the product of the square of a circle’s radius and π.
How does the circumference differ from the area of a circle? Take a look at the image below.
The circumference is the total distance around a circle. It measures how long the boundary of that circle is. Meanwhile, the area measures how much space is occupied inside the circle.
Recall that when providing the area of a plane figure, we write the units in square unit form. For instance, if the circle’s radius is given in centimeters (cm), its area must be in square centimeters (cm2). Similarly, if the unit of measurement is given in inches (in), then the area must be written in square inches (in2). Take note of this reminder because even if your calculation is correct, but your unit is not, your final answer will be considered incorrect.
Sample Problem 1: A circle has a radius of 2 meters. Determine its area.
Solution: We only have to use the formula for the area of a circle to solve the problem. Furthermore, take note that the problem doesn’t provide an estimate of π. Thus, we use π as it is.
Acircle = πr2
Acircle = π(2)2 Substitute r = 2
Acircle = 4π
Thus, the area of the circle is 4π m2.
Sample Problem 2: The radius of a circle is ½ inches long. Determine its area (Use π = 3.14).
Solution: Let’s use the formula for the area of a circle for this problem.
Please take note that we have r = ½ for this problem. However, converting ½ into a decimal form is much easier, so all of our figures in the formula will be decimal numbers. If you convert ½ into decimal form, you will get 0.5.
Acircle = πr2
Acircle = (3.14)(0.5)2 Substitute r = 0.5
Acircle = (3.14)(0.25)
Acircle = 0.785
Therefore, the area of the circle is 0.785 square inches.
Sample Problem 3: Determine the area of the circle below (Use π = 3.14):
Solution: The image above provides us with the circle’s diameter of 10 cm. We cannot use the diameter to compute the area of the circle since the formula requires us to have the circle’s radius and not the diameter.
Recall that the diameter is equivalent to twice the radius of the circle. Hence, if the circle above has a diameter of 10 cm, then its radius is 10⁄2 = 5 cm.
Now that we have r = 5 cm. Let us compute the area of the circle using the formula:
Acircle = πr2
Acircle = (3.14)(5)2 Substitute r = 5
Acircle = (3.14)(25)
Acircle = 78.5
Thus, the area of the circle above is 78.5 cm2.
Sample Problem 4: The inscribed square (square inside the circle) in the image below has a perimeter of 24 cm. Meanwhile, the circle has a radius of 10 cm. Determine the area of the shaded region in the image below.
Solution: The problem above takes a little bit of analysis. How will you find the shaded region’s area in the image above?
If you try to remove the inscribed square in the image, you might notice that the remaining figure will be the shaded region. Hence, the area of the shaded region can be calculated if we are going to subtract the area of the square from the area of the circle:
Ashaded region = Acircle – Asquare
So, to find the area of the shaded region, we must first find the areas of the square and the circle.
Let us compute the area of the square first.
The problem states that the perimeter of the square is 24 cm. Recall that a square’s perimeter equals four times its side or P = 4s. Thus, if 24 is the perimeter of the square, then its side is:
24 = 4s
24⁄4 = 4s⁄4 Dividing both sides of the equation by 4
6 = s
s = 6
Thus, the side of the square in the figure is 6 cm. We can now calculate the area of the square.
Recall the formula for the area of the square: Asquare = s2
Asquare = s2
Asquare = (6)2 Since s = 6
Asquare = 36
Therefore the area of the square is 36 cm2.
Let us now compute the area of the circle.
The radius of the circle is 10 cm. Thus, we have r = 10. Let us use the formula for the area of a circle. Note that the problem does not provide us with an estimation of π that we should use, so we use π as it is.
Acircle = πr2
Acircle = π(10)2
Acircle = 100π
Hence, the area of the circle is 100π cm2.
We are not done yet. Remember that we are looking for the area of the shaded region. We have stated earlier that the area of the shaded region is equivalent to the difference between the area of the circle and the area of the square:
Ashaded region = Acircle – Asquare
Ashaded region = 100π – 36
That’s it! The area of the shaded region is 100π – 36 cm2.
Let us now move on to other concepts related to circles, such as the arcs and angles of a circle.
## Arcs of a Circle
An arc is a portion of the circumference of a circle. It is formed by two points that are on the circle.
In the image above, the red-colored part of the circumference of a circle is an example of an arc. Points X and Y, which are on the circle, are the endpoints of this arc. We call this arc XY. In symbols:
### Measurement of an Arc
The measurement of an arc refers to the measure of the central angle of the circle that intercepts that arc.
Let us go back to the previous example. You can create an angle that intercepts (or touches) the arc with the circle’s center as the vertex. That angle is called a central angle
In the figure above, we draw ∠XCY with C as the vertex. This is a central angle since its vertex is the circle’s center. Now, notice that this central angle intercepts or touches the arc XY.
The degree measurement of the central angle that touches the arc is the degree measurement also of that arc. Hence, in the figure above, if m∠XCY = 30°, then the measurement of the arc that it intercepts is also 30°.
Just like angles, we use degrees (°) as a unit of measurement for arcs.
The measurement of an arc can be any real number from 0° to 360°. The whole circle (the entire circumference) measures 360°.
We will discuss more central angles in the latter part of this review module, but in the meantime, this intuitive meaning of the measurement of an arc should be enough for you to continue to the next section.
### Types of Arcs
We can classify arcs according to the degree measurement: minor arcs, semicircles, or major arcs.
#### 1. Minor Arc
A minor arc is an arc with a measure that is less than 180°.
Arc XY is an example of a minor arc since its degree measurement is only 30°, less than 180°.
#### 2. Semicircle
A semicircle is an arc with a measure that is exactly 180°.
In the image above, arc PQ is a semicircle since it has a degree measurement of 180°.
You might also realize that the semicircle is half of the circle. Thus, we can state that the distance of the arc created by a semicircle is equal to ½ of the circumference of the circle. Moreover, all circles have two semicircles.
#### 3. Major Arc
A major arc has a degree measurement greater than 180° but less than 360°.
To name a major arc, we use three letters: two letters for the arc’s endpoints, and the third one is the point between these endpoints.
Arc ABC is a major arc since it has a measurement of 200°.
The measurement of a major arc is equivalent also to 360 minus the measurement of the minor arc with the same endpoints as the major arc.
For instance, if arc AC is 160°, we can compute the measurement of arc ABC by subtracting the 160° from 360°. Thus, the measurement of arc ABC is 200°.
If an arc is formed by two adjacent arcs, then the measurement of the arc that is formed is equivalent to the sum of measurements of the adjacent arcs.
The arc addition postulate is analogous to the angle and segment addition postulates.
In the figure above, arc AC is formed by adjacent arcs AB and BC. The arc addition postulate tells us that
m= m+ m
Sample Problem: Determine the measure of if m = 42° and m= 72°.
Solution: By the arc addition postulate,
m= m + m
m= 42° + 72°
m= 114°
## Chords, Secants, and Tangents of a Circle
This section discusses other circle parts, such as chords, secants, and tangents.
A chord is a segment with endpoints that are points on a circle. A secant is a line that intersects a circle in two points. Meanwhile, a tangent is a line intersecting a circle at exactly one point.
In the figure above, segment AB is a chord since its endpoints are points on the circle (A and B). Meanwhile, Line l1 is a secant since it is a line that intersects the circle at two points (which are C and D). Lastly, Line l is a tangent since it is a line that intersects the circle at exactly one point (which is at point Q). The point where the tangent and the circle intersect is called the point of tangency (Q is a point of tangency).
Also, note that the diameter is also a chord of a circle.
### Some Theorems on Chords, Secants, and Tangents
Here are some important theorems that you must know about these three features of a circle:
#### 1. Theorem: In a circle, two minor arcs are congruent if and only if their corresponding chords are congruent.
To better explain the theorem above, look at the given figure above.
PR and QS are congruent chords. This means that these chords have equal measurements or lengths. Since these chords are congruent, the theorem above tells us that their corresponding arcs, arc PQ and arc RS, are also congruent.
So, in the figure above, if m = 55°, then m = 55° also.
#### 2. Theorem: Tangent segments from a common external point are congruent.
In the figure above, line segments JK and JL have a common external point of J. As per the theorem above, since both of these segments are tangent to the circle (this means that each intersects the circle at exactly one point only), JK and JL are congruent or have equal lengths.
So suppose JK = 15 cm, then JL must be 15 cm since JK and JL are congruent according to the above theorem.
#### 3. Theorem (Segment of Chords Theorem): If two chords of a circle intersect, then the product of the segments of the first chord is equal to the product of the segments of the second chord.
To understand this theorem better, please refer to the image above.
Chords AB and CD intersect at point R. The theorem tells us that the product of AB (AR and RB) segments is equal to the product of CD (CR and RD). In symbols: AR x RB = CR x RD.
Sample Problem 1: Find the value of x in the figure below.
Solution: In the figure above, x indicates the RB (or BR) measurement, a chord AB segment.
The Segment of Chords Theorem tells us that AR x RB = CR x RD (we have discussed this above).
Let us use this theorem to find x:
AR ⋅ RB = CR ⋅ RD Replace the “x” sign with”⋅” since we are using x as a variable
12 ⋅ x = 10 ⋅ 6 Substituting the given values in the problem
12x = 60 Simplifying the equation
12x⁄12 = 60⁄12 Dividing both sides of the equation by 12
x = 5
Hence, the value of x is 5.
#### 4. Theorem (Segment of Secants Theorem): If two secant segments have a common external endpoint, then the product of the lengths of the entire secant segment and its external segment is equal to the product of the lengths of the other entire secant segment and its external segment.
Let us take a look at the image above. Segments AC and AE are secant segments since both intersect two points of the circle. Furthermore, these segments share a common external endpoint which is A. Hence, the theorem above applies to these secant segments.
The theorem states that if we multiply the length of the entire secant segment AC by its external segment, which is AB (it is external since AB is outside the circle), then the result will be equal to the result when we multiply the entire segment AE by its external segment AD.
In symbols:
AC (entire secant segment) ⋅ AB(external segment of AC) = AE(entire secant segment) ⋅ AD (external segment of AE)
Sample Problem: Determine how long the segment PU is in the image below.
If you look carefully at the image above, PR and PU are secant segments that share a common external endpoint which is P. Thus, we can apply the Segment of Secants Theorem to find the length of segment PU.
The Segment of the Secants Theorem allows us to conclude that the product of the entire secant segment PR and its external segment (which is PQ) is equal to the product of the entire secant segment PU and its external segment (which is PT).
Thus, we can write this equation:
PR ⋅ PQ = PU ⋅ PT
Let x be the length of segment PU:
PR ⋅ PQ = x ⋅ PT
Note that the segment PR is ten units long (since PR = PQ + QR by the segment addition postulate).
Using the given values in the image above:
10 ⋅ 3 = x ⋅ 2
Simplifying the equation above:
30 = 2x
By the symmetric property of equality
2x = 30
If we divide both sides of the equation by 2, we will obtain the following:
2x⁄2 = 30⁄2
x = 15
Since x represents the length of the secant segment PU, then PU = 15 units.
## Central Angles and Inscribed Angles
Angles can also be observed in a circle and have interesting properties you must know.
### 1. Central Angle
You have already encountered central angles in our earlier discussion about measuring an arc of a circle. Again, a central angle is an angle such that its vertex is the center of a circle, and its sides are the circle’s radii.
In the figure below, ABC is a central angle since its vertex is the circle’s center, B, and its sides are radii of the circle (AB and AC).
To reiterate what you have also learned earlier, the central angle and its intercepted arc have the same measurement. In other words, the central angle and its intercepted arc have the same measurement. Thus, in the figure above, ABC is congruent with its intercepted arc AC.
Sample Problem: Solve for the value of x below.
In the figure above, angle XYZ is a central angle that intercepts the arc XZ. We know that the measurement of the central angle is equal to its intercepted arc, thus:
m∠XYZ = m
Using the given values in the image above:
x + 5 = 30
We can now solve for x by simply transposing 5 to the right-hand side of the equation:
x + 5 = 30
x = -5 + 30
x = 25
Thus, the value of x is 25.
### 2. Inscribed Angle
If the vertex of an angle is a point on the circle and its sides contain chords of the circle, then that angle is inscribed.
In the image above, FOR is an inscribed angle since its vertex is F, a point on the circle, and its sides, ray FO and FR, contain chords of the circles.
#### Inscribed Angle Theorem: “The measurement of an inscribed angle is equal to ½ of the measurement of its intercepted arc.”
The inscribed angle theorem provides us with a way to determine the measurement of an inscribed angle given its intercepted arc (or the arc it touches). According to the theorem, the measurement of an inscribed angle is just half of the measurement of its intercepted arc.
In the figure above, is 40°. is the intercepted arc of FOR. Thus, the measurement of ∠FOR is half the measurement of which is 40⁄2 = 20°.
Thus, m∠FOR = 20°.
Using the inscribed angle theorem, can you prove that the angle will be right when you inscribe an angle in a semicircle?
Recall that a semicircle has a degree measurement of 180°. Thus, if you inscribe an angle in it, its measure will be half of the measurement of the semicircle. Half of 180° is 90°. Therefore, the inscribed angle is a right angle.
Hence, if an angle is inscribed in a semicircle, that angle is a right angle.
## Inscribed Polygon
An inscribed polygon is one in which all vertices lie on a circle. Meanwhile, the circle that contains that polygon and touches its vertices is called a circumscribed circle.
In the image below, the trapezoid is an inscribed polygon, while the circle that contains it is a circumscribed circle.
If you inscribe a right triangle in a circle, the longest side of the right triangle (called the “hypotenuse”) will be the diameter of that circle.
Next topic: Right Triangles
Previous topic: Volume of Solid Figures |
## Engage NY Eureka Math Grade 6 Module 5 Lesson 17 Answer Key
### Eureka Math Grade 6 Module 5 Lesson 17 Opening Exercise Answer Key
Opening Exercise:
a. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.
i.
A = $$\frac{1}{2}$$ (14 cm)(12 cm) = 84cm2
14 cm represents the bose of the figure because 5 cm + 9 cm = 14 cm, and 12 cm represents the altitude of the figure because it forms a right angle with the base.
ii. How would you write an equation that shows the area of a triangle with base b and height h?
A = $$\frac{1}{2}$$ bh
b. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.
A = (28 ft.)(18 ft) = 504 ft2
28 ft. represents the base of the rectangle, 18 ft. and 18 ft. represents the height of the rectangle.
ii. How would you write an equation that shows the area of a rectangle with base b and height h?
A = bh.
### Eureka Math Grade 6 Module 5 Lesson 17 Example Answer Key
Example 1:
Use the net to calculate the surface area of the figure. (Note: all measurements are in centimeters.)
→ When you are calculating the area of a figure, what are you finding?
The area of a figure is the amount of space inside a two-dimensional figure.
→ The surface area is similar to the area, but the surface area is used to describe three-dimensional figures. What do you think is meant by the surface area of a solid?
The surface area of a three-dimensional figure is the area of each face added together.
→ What type of figure does the net create? How do you know?
It creates a rectangular prism because there are six rectangular faces.
If the boxes on the grid paper represent a 1 cm × 1 cm box, label the dimensions of the net.
→ The surface area of a figure is the sum of the areas of all faces. Calculate the area of each face, and record this value inside the corresponding rectangle.
→ In order to calculate the surface area, we have to find the sum of the areas we calculated since they represent the area of each face. There are two faces that have an area of 4 cm2 and four faces that have an area of 2 cm2. How can we use these areas to write a numerical expression to show how to calculate the surface area of the net?
The numerical expression to calculate the surface area of the net would be
(1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (2 cm × 2 cm)+ (2 cm × 2 cm).
→ Write the expression more compactly, and explain what each part represents on the net.
4(1 cm × 2 cm) + 2(2 cm × 2 cm)
The expression means there are 4 rectangles that have dimensions 1 cm × 2 cm on the net and 2 rectangles that have dimensions 2 cm × 2 cm on the net.
→ What is the surface area of the net?
The surface area of the net is 16 cm2.
Example 2:
Use the net to write an expression for surface area. (Note: all measurements are in square feet.)
→ What type of figure does the net create? How do you know?
It creates a square pyramid because one face is a square and the other four faces are triangles.
→ If the boxes on the grid paper represent a 1 ft. × 1 ft. square, label the dimensions of the net.
→ How many faces does the rectangular pyramid have?
5
→ Knowing the figure has 5 faces, use the knowledge you gained in Example ito calculate the surface area of the rectangular pyramid.
Area of Base: 3 ft. × 3 ft. = 9 ft2
Area of Triangles: $$\frac{1}{2}$$ × 3 ft. × 2 ft. = 3 ft2
Surface Area: 9 ft2 + 3 ft2 + 3 ft2 + 3 ft2 + 3 ft2 = 21 ft2
### Eureka Math Grade 6 Module 5 Lesson 17 Exercise Answer Key
Exercises:
Name the solid the net would create, and then write an expression for the surface area. Use the expression to determine the surface area. Assume that each box on the grid paper represents a 1 cm × 1 cm square. Explain how the expression represents the figure.
Exercise 1.
Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Surface Area: 4 cm × 4 cm + 4($$\frac{1}{2}$$ × 4 cm × 3 cm)
Work: 16 cm2 + 4(6 cm2) = 40 cm2
The surface area is 40 cm2. The figure is made up of a square base that measures 4 cm × 4 cm and four triangles, each with a base of 4 cm and a height of 3 cm.
Exercise 2.
Name of Shape: Rectangular Prism
Surface Area: 2(5 cm × 5 cm) + 4(5 cm × 2 cm)
Work: 2(25 cm2 ) + 4(10 cm2) = 90 cm2
The surface area is 90 cm2. The figure has 2 square faces, each of which measures 5 cm × 5 cm, and 4 rectangular faces, each of which measures 5 cm × 2 cm.
Exercise 3.
Name of Shape: Rectangular Pyramid
SurfaceArea: 3 cm × 4 cm + 2($$\frac{1}{2}$$ × 4 cm × 4 cm)+ 2($$\frac{1}{2}$$ × 4 cm × 3 cm)
Work: 12 cm2 + 2(8 cm2) + 2(6 cm2) = 40 cm2
The surface area is 40 cm2. The figure has 1 rectangular base that measures 3 cm × 4 cm, 2 triangular faces, each with a bose of 4 cm and a height of 4 cm, and 2 other triangular faces, each with a base of 3 cm and a height of 4 cm.
Exercise 4.
Name of Shape: Rectangular Prism
Surface Area: 2(6 cm × 5 cm) + 2(5 cm × 1 cm) + 2(6 cm × 1 cm)
Work: 2(30 cm2) + 2(5 cm2) + 2(6 cm2) = 82 cm2
The surface area is 82 cm2. The figure has two 6 cm × 5 cm rectangular faces, two 5 cm × 1 cm rectangular faces, and two 6 cm × 1 cm rectangular faces.
### Eureka Math Grade 6 Module 5 Lesson 17 Problem Set Answer Key
Name the shape, and write an expression for surface area. Calculate the surface area of the figure. Assume each box on the grid paper represents a 1 ft. × 1 ft. square.
Question 1.
Name of Shape: Rectangular Prism
SurfaceArea: (2 ft. × 4 ft.) + (2 ft. × 4 ft.)+ (4 ft. × 7 ft.) + (4 ft. × 7 ft.) + (7 ft. × 2 ft.)+ (7 ft. × 2 ft.)
Work: 2(2 ft. × 4 ft.) + 2(4 ft. × 7 ft.) + 2(7 ft. × 2 ft.)
= 16 ft2 + 56 ft2 + 28 ft2
= 100 ft2
Question 2.
Name of Shape: Rectangular Pyramid
SurfaceArea: (2 ft. × 5 ft.) + ($$\frac{1}{2}$$ × 2ft. × 4ft.)+ ($$\frac{1}{2}$$ × 2 ft. × 4ft.) + ($$\frac{1}{2}$$ × 5 ft. × 4ft.) + $$\frac{1}{2}$$ × 5ft. × 4 ft.)
Work: 2 ft. × 5 ft. + 2($$\frac{1}{2}$$ × 2 ft. × 4 ft.) + 2($$\frac{1}{2}$$ × 5 ft. × 4ft.)
= 10 ft2 + 8 ft2 + 20 ft2 = 38 ft2
Explain the error in each problem below. Assume each box on the grid paper represents a 1 m × 1 m square.
Question 3.
Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Area of Base: 3m × 3m = 9m2
Area of Triangles: 3 m × 4m = 12 m2
SurfaceArea: 9m2 + 12m2 + 12m2 + 12m2 + 12m2 = 57m2
The error in the solution is the area of the triangles. In order to calculate the correct area of the triangles, you must use the correct formula A = $$\frac{1}{2}$$bh. Therefore, the area of each triangle would be 6 m2 and not 12 m2.
Question 4.
Name of Shape: Rectangular Prism or, more specifically. a Cube
Area of Faces: 3m × 3m = 9m2
Surface Area: 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9 m2 = 45m2
The surface area is incorrect because the student did not find the sum of all 6 faces. The solution is shown above only calculates the sum of 5 faces. Therefore, the correct surface area should be 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9m2 = 54m2 and not 45m2.
Question 5.
Sofia and Ella are both writing expressions to calculate the surface area of a rectangular prism. However, they wrote different expressions.
a. Examine the expressions below, and determine if they represent the same value. Explain why or why not. Sofia’s Expression:
(3 cm × 4 cm) + (3 cm × 4 cm) + (3 cm × 5 cm) + (3 cm × 5 cm) + (4 cm × 5 cm) + (4 cm × 5 cm)
Ella’s Expression:
2(3 cm × 4 cm) + 2(3 cm × 5 cm) + 2(4 cm × 5 cm)
Sofia’s and Ella’s expressions are the same, but Ella used the distributive property to make her expression more compact than Sofia’s.
b. What fact about the surface area of a rectangular prism does Ella’s expression show more clearly than Sofia’s?
A rectangular prism is composed of three pairs of sides with identical areas.
### Eureka Math Grade 6 Module 5 Lesson 17 Exit Ticket Answer Key
Question 1.
Name the shape, and then calculate the surface area of the figure. Assume each box on the grid paper represents a 1 in. × 1 in. square.
Name of Shape: Rectangular Pyramid
Area of Base: 5 in. × 4 in. = 20 in2
Area of Triangles: $$\frac{1}{2}$$ × 4 in. × 4 in. = 8 in2, $$\frac{1}{2}$$ × 5 in. × 4 in. = 10 in2
SurfaceArea: 20 in2 +8 in2 + 8in2 + 10 in2 + 10 in2 = 56 in2
Addition and Subtraction Equations – Round 1:
Directions: Find the value of m in each equation.
Question 1.
m + 4 = 11
m = 7
Question 2.
m + 2 = 5
m = 3
Question 3.
m + 5 = 8
m = 3
Question 4.
m – 7 = 10
m = 17
Question 5.
m – 8 = 1
m = 9
Question 6.
m – 4 = 2
m = 6
Question 7.
m + 12 = 34
m = 22
Question 8.
m + 25 = 45
m = 20
Question 9.
m + 43 = 89
m = 46
Question 10.
m – 20 = 31
m = 51
Question 11.
m – 13 = 34
m = 47
Question 12.
m – 45 = 68
m = 113
Question 13.
m + 34 = 41
m = 7
Question 14.
m + 29 = 52
m = 23
Question 15.
m + 37 = 61
m = 24
Question 16.
m – 43 = 63
m = 106
Question 17.
m – 21 = 40
valuem = 61
Question 18.
m – 54 = 37
m = 91
Question 19.
4 + m = 9
m = 5
Question 20.
6 + m = 13
m = 7
Question 21.
2 + m = 31
m = 29
Question 22.
15 = m + 11
m = 4
Question 23.
24 = m + 13
m = 11
Question 24.
32 = m + 28
m = 4
Question 25.
4 = m – 7
m = 11
Question 26.
3 = m – 5
m = 8
Question 27.
12 = m – 14
m = 26
Question 28.
23.6 = m – 7.1
m = 30.7
Question 29.
14.2 = m – 33.8
m = 48
Question 30.
2.5 = m -41.8
m = 44.3
Question 31.
64.9 = m + 23.4
m = 41.5
Question 32.
72.2 = m + 38.7
m = 33.5
Question 33.
1.81 = m – 15.13
m = 16.94
Question 34.
24.68 = m – 56.82
m = 81.5
Addition and Subtraction Equations – Round 2:
Directions: Find the value of m in each equation.
Question 1.
m + 2 = 7
m = 5
Question 2.
m + 4 = 10
m = 6
Question 3.
m + 8 = 15
m = 7
Question 4.
m + 7 = 23
m = 16
Question 5.
m + 12 = 16
m = 4
Question 6.
m – 5 = 2
m = 7
Question 7.
m – 3 = 8
m = 11
Question 8.
m – 4 = 12
m = 16
Question 9.
m – 14 = 45
m = 59
Question 10.
m + 23 = 40
m = 17
Question 11.
m + 13 = 31
m = 18
Question 12.
m – 23 = 48
m = 25
Question 13.
m + 38 = 52
m = 14
Question 14.
m – 14 = 27
m = 41
Question 15.
m – 23 = 35
m = 58
Question 16.
m – 17 = 18
m = 35
Question 17.
m – 64 = 1
m = 65
Question 18.
6 = m + 3
m = 3
Question 19.
12 = m + 7
m = 5
Question 20.
24 = m + 16
m = 8
Question 21.
13 = m + 9
m = 4
Question 22.
32 = m – 3
m = 35
Question 23.
22 = m – 12
m = 34
Question 24.
34 = m – 10
m = 44
Question 25.
48 = m + 29
m = 19
Question 26.
21 = m + 17
m = 4
Question 27.
52 = m + 37
m = 15
Question 28.
$$\frac{6}{7}$$ + m = $$\frac{4}{7}$$
m = $$\frac{2}{7}$$
Question 29.
$$\frac{2}{3}$$ = m – $$\frac{5}{3}$$
m = $$\frac{7}{3}$$
Question 30.
$$\frac{1}{4}$$ – m = $$\frac{8}{3}$$
m = $$\frac{35}{12}$$
Question 31.
$$\frac{5}{6}$$ = m – $$\frac{7}{12}$$
m = $$\frac{17}{12}$$
Question 32.
$$\frac{7}{8}$$ = m – $$\frac{5}{12}$$
m = $$\frac{31}{24}$$
$$\frac{7}{6}$$ + m = $$\frac{16}{3}$$
m = $$\frac{25}{6}$$
$$\frac{1}{3}$$ + m = $$\frac{13}{15}$$
m = $$\frac{8}{15}$$ |
SOME SPECIAL LIMITS - Functions, Rates, and Limits - The Calculus Primer
## The Calculus Primer (2011)
### Chapter 4. SOME SPECIAL LIMITS
1—18. The Limit of (1 + x)1/x. One of the most useful limits in the calculus is the
Consider the function
y = (1 + x)1/x.
x y 10 1.001 2 1.732 1 2.000 .5 2.250 .1 2.594 .01 2.705 .001 2.717 −.5 4.000 −.1 2.868 −.01 2.732 −.001 2.720
From graphic considerations, therefore, it appears that as x → 0, the limit of y, or of (1 + x)1/x, lies between 2.717 and 2.720. By using more advanced methods (Chapter Eleven), we shall learn how to compute this limit e to any number of decimal places. Note that y approaches e as a limit not only as x approaches 0 from the right, but also as x approaches 0 from the left as well.
1—19. The Limit of . Another useful limit is
Reference to a five-place (or more) table of values of trigonometric functions will reveal that for all angles less than 2°, the sine of the angle and the angle itself (expressed in radians) are very nearly equal. In fact, for all angles less than 10°, the angle in radians and its sine are equal to three decimal places. Thus for small angles, the sine and the angle are nearly equal; the smaller the angle, the less the difference between them becomes. Hence we see that as x→ 0, appears to be equal to 1.
That this actually is the limit can be proved from the following geometrical proof. Let O be the center of a circle whose radius is unity.
Let = = x; PT1 and PT2 are tangents at T1 and T2. From trigonometry, we have:
T1ST2 < T1MT2 < T1PT2;
hence2 sin x < 2x < 2 tan x.(1)
Dividing (1) through by 2 sin x, we get:
Now as x → 0, the above inequality (2) holds true for all values of x, however small. At the same time, as x → 0,
The is therefore seen to lie between the two values 1, and the inequality (2) can hold only if so, by the properties of inequalities and reciprocals.
1—20. Continuity of a Function in an Interval. We are now in a position to formulate what is meant by the continuity of a function in an interval more precisely than was stated in §1—6.
A single-valued function f(x) is said to be continuous at x = a if :
(1)f(a) is defined;
(2) exists;
(3) = f(a).
If any one of these three conditions is not satisfied, the function is said to be discontinuous at the point where x = a. If the reader should find this formal definition somewhat complicated at first, he will find that it will become more significant as he becomes more familiar with the subject.
In general, a function is said to be continuous throughout an interval if it is continuous at every point in the interval. More specifically, polynomial functions, such as
f(x) = a0xn + a1xn−1 + a2xn−2 + ··· + an,
are continuous for every value of x.
EXERCISE 1—3
Evaluate each of the following limits:
|
When I first left undergraduate school I took a job teaching 8th and 9th grade math. My 8th graders were introduced to integer multiplication for the first time at that level. They certainly knew natural number multiplication and even knew integer addition which includes subtraction by the time we took up integer multiplication.
When teaching integer addition I presented it as addition of vectors directed along the number line. So +2 + -4 would be an arrow starting at zero of length 2 pointing to the right then append an arrow of length 4 pointing to the left. The tip of the second arrow wound up at the point -2 on the number line.
To extend that idea to integer multiplication I introduced items I called ‘Number Line Frogs’. Each frog had a number associated with it and a direction. So a +2 frog could jump two places at a time and it pointed to the right. The number -3 was represented by a frog that could jump three places and pointed to the left.
I then asked the students to think of integer multiplication problems in two parts. The problem a x b was to be visualized as an ‘a-frog’ being asked to do something and that something was to perform ‘b’ jumps. The problem always started at the origin or ‘0’ on the number line. The sign of the number ‘b’ indicated which direction the frog was to jump. The absolute value of the number ‘b’ indicated how many jumps the frog was to take. The sign of the number ‘a’ indicated which direction the frog was pointing and the absolute value of the number ‘a’ indicated how far the frog could jump.
In this way of thinking about it the problem +2 x -3 was the same as saying ‘Allow a +2 frog to jump three jumps backward (-3)’. I stressed that the problem -3 x +2 was the same problem as could be verified by saying ‘Allow a -3 frog to take two jumps forward (+2)’ In both cases the end result was the number -6.
The rule for ‘positive-a times positive-b’ was to say ‘let a postive-a-frog take b-jumps-forward’ which came out to be the same problem as saying ‘let a positive-b-frog take a-jumps-forward’.
The rule for ‘positive-a times negative-b’ was to say ‘let a positive-a-frog take b-jumps-backward, while the rule for ‘negative-b times positive-a’ was to say ‘let a negative-b frog take a-jumps-forward’.
Finally the rule for ‘negative-a times negative-b’ was to say ‘let a ‘negative-a frog take ‘b-jumps-backward’.
When viewed this way and taking into account the ‘frog’ always started at zero the sign table for integer multiplication was justified. Additionally the commutative property of multiplication was illustrated by noting any number that could be decomposed into ‘b-groups of length a’ was equivalent to the observation that the same number could be decomposed into ‘a-groups of length b’.
I will say that the 8th grade group I had were sorted in an inhomogeneous manner with all the top students put into one class and that class was assigned to me. I did worry about this sorting process and felt that it was wrong on multiple levels but given there was nothing a first year math teacher could do to change the situation I made the most of it and presented many mathematical concepts using non standard methods (i.e. presented not-by-the-book). I think that math class did enjoy my presentations. |
# 3.4: Solve Systems of Equations with Three Variables
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##### Learning Objectives
By the end of this section, you will be able to:
• Determine whether an ordered triple is a solution of a system of three linear equations with three variables
• Solve a system of linear equations with three variables
• Solve applications using systems of linear equations with three variables
Before you get started, take this readiness quiz.
1. Evaluate $$5x−2y+3z$$ when $$x=−2, y=−4,$$ and $$z=3.$$
If you missed this problem, review [link].
2. Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. $$\left\{ \begin{array} {l} −2x+y=−11 \\ x+3y=9 \end{array} \right.$$
If you missed this problem, review [link].
3. Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. $$\left\{ \begin{array} {l} 7x+8y=4 \\ 3x−5y=27 \end{array} \right.$$
If you missed this problem, review [link].
## Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables
In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.
We learned earlier that the graph of a linear equation, $$ax+by=c$$, is a line. Each point on the line, an ordered pair $$(x,y)$$, is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions
We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Similarly, for a linear equation with three variables ax+by+cz=d,ax+by+cz=d, every solution to the equation is an ordered triple, (x,y,z)(x,y,z)that makes the equation true.
##### LINEAR EQUATION IN THREE VARIABLES
A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form
$ax+by+cz=d\nonumber$
Every solution to the equation is an ordered triple, $$(x,y,z)$$ that makes the equation true.
All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.
When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.
To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple $$(x,y,z)$$ that makes all three equations true. These are called the solutions of the system of three linear equations with three variables.
##### SolutionS OF A SYSTEM OF LINEAR EQUATIONS WITH THREE VARIABLES
Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple $$(x,y,z)$$.
To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.
##### Example $$\PageIndex{1}$$
Determine whether the ordered triple is a solution to the system: $$\left\{ \begin{array} {l} x−y+z=2 \\ 2x−y−z=−6 \\ 2x+2y+z=−3 \end{array} \right.$$
ⓐ $$(−2,−1,3)$$ ⓑ $$(−4,−3,4)$$
##### Example $$\PageIndex{2}$$
Determine whether the ordered triple is a solution to the system: $$\left\{ \begin{array} {l} 3x+y+z=2 \\ x+2y+z=−3 \\ 3x+y+2z=4 \end{array} \right.$$
ⓐ $$(1,−3,2)$$ ⓑ $$(4,−1,−5)$$
ⓐ yes ⓑ no
##### Example $$\PageIndex{3}$$
Determine whether the ordered triple is a solution to the system: $$\left\{ \begin{array} {l} x−3y+z=−5 \\ −3x−y−z=1 \\ 2x−2y+3z=1 \end{array} \right.$$
ⓐ $$(2,−2,3)$$ ⓑ $$(−2,2,3)$$
ⓐ no ⓑ yes
## Solve a System of Linear Equations with Three Variables
To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!
Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.
##### Example $$\PageIndex{4}$$: How to Solve a System of Equations With Three Variables by Elimination
Solve the system by elimination: $$\left\{ \begin{array} {l} x−2y+z=3 \\ 2x+y+z=4 \\ 3x+4y+3z=−1 \end{array} \right.$$
##### Example $$\PageIndex{5}$$
Solve the system by elimination: $$\left\{ \begin{array} {l} 3x+y−z=2 \\ 2x−3y−2z=1 \\ 4x−y−3z=0 \end{array} \right.$$
$$(2,−1,3)$$
##### Example $$\PageIndex{6}$$
Solve the system by elimination: $$\left\{ \begin{array} {l} 4x+y+z=−1 \\ −2x−2y+z=2 \\ 2x+3y−z=1 \end{array} \right.$$
$$(−2,3,4)$$
The steps are summarized here.
##### SOLVE A SYSTEM OF LINEAR EQUATIONS WITH THREE VARIABLES.
1. Write the equations in standard form
• If any coefficients are fractions, clear them.
2. Eliminate the same variable from two equations.
• Decide which variable you will eliminate.
• Work with a pair of equations to eliminate the chosen variable.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable
3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
4. The two new equations form a system of two equations with two variables. Solve this system.
5. Use the values of the two variables found in Step 4 to find the third variable.
6. Write the solution as an ordered triple.
7. Check that the ordered triple is a solution to all three original equations.
##### Example $$\PageIndex{7}$$
Solve: $$\left\{ \begin{array} {l} 3x−4z=0 \\ 3y+2z=−3 \\ 2x+3y=−5 \end{array} \right.$$
$\left\{ \begin{array} {ll} 3x−4z=0 &(1) \\ 3y+2z=−3 &(2) \\ 2x+3y=−5 &(3) \end{array} \right. \nonumber$
We can eliminate $$z$$ from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.
Notice that equations (3) and (4) both have the variables $$x$$ and $$y$$. We will solve this new system for $$x$$ and $$y$$.
To solve for y, we substitute $$x=−4$$ into equation (3).
We now have $$x=−4$$ and $$y=1$$. We need to solve for z. We can substitute $$x=−4$$ into equation (1) to find z.
We write the solution as an ordered triple. $$(−4,1,−3)$$
We check that the solution makes all three equations true.
$$\begin{array} {lll} {3x-4z=0 \space (1)} &{3y+2z=−3 \space (2)} &{2x+3y=−5 \space (3)} \\ {3(−4)−4(−3)\overset{?}{=} 0} &{3(1)+2(−3)\overset{?}{=} −3} &{2(−4)+3(1)\overset{?}{=} −5} \\ {0=0 \checkmark} &{−3=−3 \checkmark} &{−5=−5 \checkmark} \\ {} &{} &{\text{The solution is }(−4,1,−3)} \end{array}$$
##### Example $$\PageIndex{8}$$
Solve: $$\left\{ \begin{array} {l} 3x−4z=−1 \\ 2y+3z=2 \\ 2x+3y=6 \end{array} \right.$$
$$(−3,4,−2)$$
##### Example $$\PageIndex{9}$$
Solve: $$\left\{ \begin{array} {l} 4x−3z=−5 \\ 3y+2z=7 \\ 3x+4y=6 \end{array} \right.$$
$$(−2,3,−1)$$
When we solve a system and end up with no variables and a false statement, we know there are no solutions and that the system is inconsistent. The next example shows a system of equations that is inconsistent.
##### Example $$\PageIndex{10}$$
Solve the system of equations: $$\left\{ \begin{array} {l} x+2y−3z=−1 \\ x−3y+z=1 \\ 2x−y−2z=2 \end{array} \right.$$
$\left\{ \begin{array} {ll} x+2y−3z=−1 &(1) \\ x−3y+z=1 &(2) \\ 2x−y−2z=2 &(3) \end{array} \right.\nonumber$
Use equation (1) and (2) to eliminate z.
Use (2) and (3) to eliminate $$z$$ again.
Use (4) and (5) to eliminate a variable.
There is no solution.
We are left with a false statement and this tells us the system is inconsistent and has no solution.
##### Example $$\PageIndex{11}$$
Solve the system of equations: $$\left\{ \begin{array} {l} x+2y+6z=5 \\ −x+y−2z=3 \\ x−4y−2z=1 \end{array} \right.$$
no solution
##### Example $$\PageIndex{12}$$
Solve the system of equations: $$\left\{ \begin{array} {l} 2x−2y+3z=6 \\ 4x−3y+2z=0 \\ −2x+3y−7z=1 \end{array} \right.$$
no solution
When we solve a system and end up with no variables but a true statement, we know there are infinitely many solutions. The system is consistent with dependent equations. Our solution will show how two of the variables depend on the third.
##### Example $$\PageIndex{13}$$
Solve the system of equations: $$\left\{ \begin{array} {l} x+2y−z=1 \\ 2x+7y+4z=11 \\ x+3y+z=4 \end{array} \right.$$
$\left\{ \begin{array} {ll} x+2y−z=1 &(1) \\ 2x+7y+4z=11 &(2) \\ x+3y+z=4 &(3) \end{array} \right.\nonumber$
Use equation (1) and (3) to eliminate x.
Use equation (1) and (2) to eliminate x again.
Use equation (4) and (5) to eliminate $$y$$.
There are infinitely many solutions. Solve equation (4) for y. Represent the solution showing how x and y are dependent on z. \begin{aligned} y+2z &= 3 \\ y &= −2z+3 \end{aligned} Use equation (1) to solve for x. $$x+2y−z=1$$ Substitute $$y=−2z+3$$. \begin{aligned} x+2(−2z+3)−z &= 1 \\ x−4z+6−z &= 1 \\ x−5z+6 &= 1 \\ x &= 5z−5 \end{aligned}
The true statement $$0=0$$ tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form (x,y,z)(x,y,z) where $$x=5z−5;\space y=−2z+3$$ and z is any real number.
##### Example $$\PageIndex{14}$$
Solve the system by equations: $$\left\{ \begin{array} {l} x+y−z=0 \\ 2x+4y−2z=6 \\ 3x+6y−3z=9 \end{array} \right.$$
infinitely many solutions $$(x,3,z)$$ where $$x=z−3;\space y=3;\space z$$ is any real number
##### Example $$\PageIndex{15}$$
Solve the system by equations: $$\left\{ \begin{array} {l} x−y−z=1 \\ −x+2y−3z=−4 \\ 3x−2y−7z=0 \end{array} \right.$$
infinitely many solutions $$(x,y,z)$$ where $$x=5z−2;\space y=4z−3;\space z$$ is any real number
## Solve Applications using Systems of Linear Equations with Three Variables
Applications that are modeled by a systems of equations can be solved using the same techniques we used to solve the systems. Many of the application are just extensions to three variables of the types we have solved earlier.
##### Example $$\PageIndex{16}$$
The community college theater department sold three kinds of tickets to its latest play production. The adult tickets sold for $15, the student tickets for$10 and the child tickets for $8. The theater department was thrilled to have sold 250 tickets and brought in$2,825 in one night. The number of student tickets sold is twice the number of adult tickets sold. How many of each type did the department sell?
We will use a chart to organize the information. Number of students is twice number of adults. Rewrite the equation in standard form. \begin{aligned} y &= 2x \\ 2x−y &= 0 \end{aligned} Use equations (1) and (2) to eliminate z. Use (3) and (4) to eliminate $$y$$. Solve for x. $$x=75$$ adult tickets Use equation (3) to find y. $$−2x+y=0$$ Substitute $$x=75$$. \begin{aligned} −2(75)+y &= 0 \\ −150+y &= 0 \\ y &= 150\text{ student tickets}\end{aligned} Use equation (1) to find z. $$x+y+z=250$$ Substitute in the values $$x=75, \space y=150.$$ \begin{aligned} 75+150+z &= 250 \\ 225+z &= 250 \\ z &= 25\text{ child tickets} \end{aligned} Write the solution. The theater department sold 75 adult tickets, 150 student tickets, and 25 child tickets.
##### Example $$\PageIndex{17}$$
The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for $20, the student tickets for$12 and the child tickets for $10.The fine arts department was thrilled to have sold 350 tickets and brought in$4,650 in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?
The fine arts department sold 75 adult tickets, 200 student tickets, and 75 child tickets.
##### Example $$\PageIndex{18}$$
The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for $10, the student tickets for$8 and the child tickets for $5. The soccer team was thrilled to have sold 600 tickets and brought in$4,900 for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?
The soccer team sold 200 adult tickets, 300 student tickets, and 100 child tickets.
Access this online resource for additional instruction and practice with solving a linear system in three variables with no or infinite solutions.
• Solving a Linear System in Three Variables with No or Infinite Solutions
• 3 Variable Application
## Key Concepts
• Linear Equation in Three Variables: A linear equation with three variables, where a, b, c, and d are real numbers and a, b,and c are not all 0, is of the form
$ax+by+cz=d\nonumber$
Every solution to the equation is an ordered triple, $$(x,y,z)$$ that makes the equation true.
• How to solve a system of linear equations with three variables.
1. Write the equations in standard form
If any coefficients are fractions, clear them.
2. Eliminate the same variable from two equations.
Decide which variable you will eliminate.
Work with a pair of equations to eliminate the chosen variable.
Multiply one or both equations so that the coefficients of that variable are opposites.
Add the equations resulting from Step 2 to eliminate one variable
3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
4. The two new equations form a system of two equations with two variables. Solve this system.
5. Use the values of the two variables found in Step 4 to find the third variable.
6. Write the solution as an ordered triple.
7. Check that the ordered triple is a solution to all three original equations.
## Glossary
solutions of a system of linear equations with three variables
The solutions of a system of equations are the values of the variables that make all the equations true; a solution is represented by an ordered triple (x,y,z).
This page titled 3.4: Solve Systems of Equations with Three Variables is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax. |
# How do you prove a surjective injective?
## How do you prove a surjective injective?
To show that g ◦ f is injective, we need to pick two elements x and y in its domain, assume that their output values are equal, and then show that x and y must themselves be equal.
## How do you prove an injective function?
Proving a function is injective
1. To prove that a function is injective, we start by: “fix any with ” Then (using algebraic manipulation etc) we show that .
2. To prove that a function is not injective, we demonstrate two explicit elements and show that .
Can something be Injective and Surjective?
A function is bijective if it is both injective and surjective. A bijective function is also called a bijection or a one-to-one correspondence. A function is bijective if and only if every possible image is mapped to by exactly one argument.
Is 2x 1 injective or surjective?
So range of f(x) is same as domain of x. So it is surjective. Hence, the function f(x) = 2x + 1 is injective as well as surjective.
### What is Surjective function example?
The function f : R → R defined by f(x) = x3 − 3x is surjective, because the pre-image of any real number y is the solution set of the cubic polynomial equation x3 − 3x − y = 0, and every cubic polynomial with real coefficients has at least one real root.
### How do you prove Bijective?
According to the definition of the bijection, the given function should be both injective and surjective. In order to prove that, we must prove that f(a)=c and f(b)=c then a=b. Since this is a real number, and it is in the domain, the function is surjective.
How do you prove not surjective?
To show a function is not surjective we must show f(A) = B. Since a well-defined function must have f(A) ⊆ B, we should show B ⊆ f(A). Thus to show a function is not surjective it is enough to find an element in the codomain that is not the image of any element of the domain.
Is 2x a Bijection?
Example: The function f(x) = 2x from the set of natural numbers N to the set of non-negative even numbers E is one-to-one and onto. Thus it is a bijection.
## What is the inverse of 2x 1?
Answer: The Inverse of the Function f(x) = 2x + 1 is f-1(x) = x/2 – 1/2.
## Which functions are surjective?
Surjective functions, also called onto functions, is when every element in the codomain is mapped to by at least one element in the domain. In other words, nothing in the codomain is left out. This means that for all “bs” in the codomain there exists some “a” in the domain such that a maps to that b (i.e., f(a) = b).
Which is an example of proof of a surjection?
Proof. We prove part (b), leaving part (a) as an exercise. Suppose c ∈ C. We wish to show h 1 ( c) = h 2 ( c). By hypothesis g is surjective, so there is a b ∈ B such that g ( b) = c. So as desired. Ex 4.4.1 Show by example that if g ∘ f is injective, then g need not be injective.
Who was the author of proofs of a conspiracy?
In this work, Proofs of a Conspiracy, Robison laid the groundwork for modern conspiracy theorists by implicating the Bavarian Illuminati as responsible for the excesses of the French Revolution. The Bavarian Illuminati, a rationalist secret society, was founded by Adam Weishaupt in 1776 in what is today Germany.
### What are the properties of injection and surjection?
Theorem 4.4.2 Suppose f 1, f 2: A → B, g: B → C, h 1, h 2: C → D are functions. a) If g is injective and g ∘ f 1 = g ∘ f 2 then f 1 = f 2 . b) If g is surjective and h 1 ∘ g = h 2 ∘ g then h 1 = h 2 . Proof. We prove part (b), leaving part (a) as an exercise. Suppose c ∈ C. We wish to show h 1 ( c) = h 2 ( c).
### How to prove that a function is a surjective?
On topic: Surjective means that every element in the codomain is “hit” by the function, i.e. given a function f: X → Y the image im(X) of f equals the codomain set Y. To prove that a function is surjective, take an arbitrary element y ∈ Y and show that there is an element x ∈ X so that f(x) = y. |
# How to write an expression for the nth term of a geometric sequence
This also works for any pair of consecutive numbers. Given the sequence 2, 6, 18, 54.
## Arithmetic sequence nth term
To find the 10th term of any sequence, we would need to have an explicit formula for the sequence. Because these sequences behave according to this simple rule of multiplying a constant number to one term to get to another, they are called geometric sequences. This geometric sequence has a common ratio of 3, meaning that we multiply each term by 3 in order to get the next term in the sequence. Find the explicit formula for 0. The formula for calculating r is Site Navigation Geometric Sequences This lesson will work with arithmetic sequences, their recursive and explicit formulas and finding terms in a sequence. If neither of those are given in the problem, you must take the given information and find them. The formula says that we need to know the first term and the common ratio. So the explicit or closed formula for the geometric sequence is. Find a6, a9, and a12 for problem 8. To find the explicit formula, you will need to be given or use computations to find out the first term and use that value in the formula. Find the recursive formula for 0. In this lesson, it is assumed that you know what an arithmetic sequence is and can find a common difference. Your formulas should be simplified if possible, but be very careful when working with exponential expressions. This also works for any pair of consecutive numbers.
This will give us Notice how much easier it is to work with the explicit formula than with the recursive formula to find a particular term in a sequence. For example, when writing the general explicit formula, n is the variable and does not take on a value.
Luckily, there is a way to arrive at the th term without the need for calculating terms 1 through If you need to review these topics, click here.
Notice that the an and n terms did not take on numeric values. But if you want to find the 12th term, then n does take on a value and it would be Find the explicit formula for a geometric sequence where and. If we match each term with it's corresponding term number, we get: n.
Since we already found that in our first example, we can use it here. This sounds like a lot of work. So the explicit or closed formula for the geometric sequence is.
## Find the 8th term of the geometric sequence
The formula says that we need to know the first term and the common ratio. This means that if we refer to the tenth term of a certain sequence, we will label it a Sequence C is a little different because it seems that we are dividing; yet to stay consistent with the theme of geometric sequences, we must think in terms of multiplication. Notice this example required making use of the general formula twice to get what we need. Using the recursive formula, we would have to know the first 49 terms in order to find the 50th. Your formulas should be simplified if possible, but be very careful when working with exponential expressions. Site Navigation Geometric Sequences This lesson will work with arithmetic sequences, their recursive and explicit formulas and finding terms in a sequence. And there is! The third number times 6 is the fourth number: 0. If we had to find the th term of sequence A above, we would undertake a tedious task had we decided to multiply by two each step of the way all the way to the th term. Find a6, a9, and a12 for problem 4. Rather than write a recursive formula, we can write an explicit formula.
The recursive formula for a geometric sequence is written in the form For our particular sequence, since the common ratio r is 3, we would write So once you know the common ratio in a geometric sequence you can write the recursive form for that sequence.
Rated 7/10 based on 22 review |
# Factorize Each of the Following Algebraic Expression: 40 + 3x − X2 - Mathematics
Sum
Factorize each of the following algebraic expression:
40 + 3x − x2
#### Solution
We have:
$40 + 3x - x^2$
$\Rightarrow - ( x^2 - 3x - 40)$
$\text{ To factorise }( x^2 - 3x - 40),\text{ we will find two numbers p and q such that }p + q = - 3\text{ and pq }= - 40 .$
Now,
$5 + ( - 8) = - 3$
and
$5 \times ( - 8) = - 40$
$\text{ Splitting the middle term }- 3x \text{ in the given quadratic as }5x - 8x,\text{ we get: }$
$40 + 3x - x^2 = - ( x^2 - 3x - 40)$
$= - ( x^2 + 5x - 8x - 40)$
$= - [( x^2 + 5x) - (8x + 40)]$
$= - [x(x + 5) - 8(x + 5)]$
$= - (x - 8)(x + 5)$
$= (x + 5)( - x + 8)$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 7 Factorization
Exercise 7.7 | Q 2 | Page 27 |
# How Many Different Combinations Are Possible Using Four Numbers?
When a number is used more than once in a combination, there are 10,000 possible combinations of four numbers. When a number is used just once, there are 5,040 possible combinations of four numbers.
How so? For each combination of numbers, there are ten options, numbered from 0 to 9. There are a total of 10 options for each of the four integers because there are four numbers in the combination. In other words, there are 10,000 different combinations out of 101010*10 or 104.
The number of combinations can be determined generally using the binomial coefficient formula. The number of ways to combine k items from a set of n elements is shown here as n!/(k!*(n-k)!), where the exclamation point denotes a factorial. Do we need to elaborate more? We have your back.
## Formula for Number of Combinations
There is a straightforward equation that may be used to determine how many possible possibilities there are for four numbers. Consider each digit in the combination as a person, and each position as a seat. There can only be one person in each seat, and each seat can only hold ten people. Because single-digit numbers range from 0 to 9, there are ten numbers.
Any one of the 10 numbers may occupy one of the four seats in any particular combination. There are ten possible configurations for the first seat. Additionally, there are 10 possible combinations for the second seat.
The third and fourth chairs also fit this description. Multiply the number of options for the first seat by the number of options for the second seat by the number of options for the third seat by the number of options for the fourth seat to get the total options for all possible combinations.
Therefore, you must multiply 10 by 10 by 10 by 10. There are 10,000 possible combinations of four numbers, you’ll discover in the end.
Formulas for Combinations of Numbers When They Don’t Repeat
You would be correct and incorrect if you said that there were 10,000 potential combinations using just four integers. In other words, the 10,000 solution allows any one of the 10 numbers to occupy any one of the four seats.
According to this hypothesis, one of the 10,000 combinations may be one of the following: 1111, 0000, 2222, or 3333. Let’s introduce a complication to the situation.
Four-digit combinations frequently lack repeated digits in the real world. In fact, a lot of businesses forbid users from creating four-digit passwords that contain the same number repeatedly. So how many non-repeating four-digit number combinations are there in total?
For a moment, ignore the chairs and focus on the binomial coefficient formula, a handy-dandy mathematical formula. The equation reads as follows:
n!/(k! x (n-k)!)
The factorial is represented by each exclamation point, in case you didn’t know. Although the term and the formula both appear difficult, they are actually considerably simpler in use. It turns out that the idea of people sitting in seats will be useful for this one as well. The number of individuals who can fit in any one of the seats is denoted by “K,” and the number of seats that any one of those persons can take is denoted by “n.”
When trying to determine how many possible combinations there are of four numbers, k=10 and n=4. This is how the equation looks:
4!/(10! x (4-10)!)
That translates to: without factorization
10 x 9 x 8 x 7 = 5,040
Do you see a pattern here? Any one of the 10 numbers may take the first seat. There are now just nine numbers available for the second seat. The third seat can only hold eight additional people after the final number is down, while the fourth seat can only hold seven numbers in total.
See? Contrary to appearances, the binomial coefficient is much easier. The binomial coefficient ensures that any number selected for one seat is automatically discarded from consideration for the other seats. This roughly reduces the number of possible combinations.
Let’s be real here. You probably didn’t look up the amount of possible combinations of four digits unless you have a serious interest in numbers. Actually, the reason you’ve probably arrived at this section of the internet is because you’re attempting to create a four-digit password. And it’s admirable that you’re considering your passcode.
Given that they are among the smallest passwords you’re likely to use, four-digit passwords can appear to be quite straightforward. They also frequently rank among the most significant, though.
Four-digit number combinations can be used to open your phone or log in to some apps more quickly, but how else do you use them? To authorise transactions and utilise ATMs, the majority of banks require their customers to choose a PIN with four digits.
Four-digit number combinations are frequently used as passwords for items you probably care less about securing than your bank card PIN, which gives hackers an opening. When it comes to passwords, people are not nearly as creative as they should be.
Since there is a very high likelihood that the numbers will match, if someone can guess the code on your lock screen, it is possible that they can also authorise a transaction on your debit card.
Banks also contribute little to the issue. People frequently have 10,000 options for PINs because many banks permit repeating numbers. You’ll only have 5,040 choices to pick from if your bank is a little more security-conscious.
Many people employ four-digit sequences that are repetitive or sequential. For instance, many people choose 1234, whereas some people repeatedly combine the same number, such as 1111 or 2222.
Don’t waste your understanding of the binomial coefficient. There are literally tens of thousands of possible four-number combinations. Do not simply choose your birth year or day. Please, for the love of everything good, don’t choose 1234 either.
You’ll need to work much more than that to prevent a certain someone from prying eyes your smartphone. Be careful when selecting passwords to protect your identity and personal data. |
# Third Equation of Motion by Graphical Method
Here in this article, we will derive the third equation of motion by graphical method. This equation relates initial and final velocity with acceleration and displacement. While solving problems you must remember that these equations can only be used when acceleration is constant. Here we will use v-t graph to derive the third equation of motion.
## Formula for the third equation of motion
The third equation of motion is given by the relation
$$v^2=u^2+2as$$
Where,
$$v=$$ final velocity,
$$u=$$ initial velocity,
$$a=$$ acceleration and
$$s=$$ distance travelled.
Note: – This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.
## Derivation of 3rd equation of motion by graphical method
To derive the 3rd equation of motion we will make the following assumptions
• Object under consideration is moving with acceleration $$a\,\, m/s^2$$
• At time $$t=0$$ object have some initial velocity. Let’s denote it by $$u\,\, m/s$$
• At time $$ts$$ object have some final velocity. Let’s denote it by $$v\,\, m/s$$
• Total distance covered by object in time $$t$$ seconds is $$s$$ meters.
Object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time $$t = 0$$ and $$t = t\,\, s$$ are $$u$$ and $$v$$ respectively. During time $$t$$, let $$s$$ be the total distance travelled by the object.Figure given below show the velocity-time graph for the object whose initial velocity is $$u$$ at time $$t=0$$ and velocity $$v$$ at time $$t$$.
We know that the distance covered by the object moving with uniform acceleration is given by the area under the velocity-time graph. Here area under the velocity-time graph is equal to area of trapezium OPQS
∴ Area of trapezium OPQS $= \frac{1}{2}\text{(Sum of Parallel Slides + Distance between Parallel Slides)}$Or, $s=\frac{OP+SQ}{2}\times PR$
We have knowledge about acceleration of the moving object. So, Acceleration $a=\frac{\text{Change in velocity}}{time}=\frac{QR}{PR}$Or,$PR=\frac{QR}{a}$From graph we can see that $QR=v-u$$t=PR=\frac{v-u}{a}$From above figure we can see that$OP=u$$SQ=v$$OP+SQ=u+v$Substituting these values, we get$s=\left(\frac{u+v}{2}\right)\times\left(\frac{v-u}{a}\right)=\frac{v^2-u^2}{2a}$Rearranging it we get$v^2=u^2+2as$Which is third equation of motion. |
Kettering University Mathematics Olympiad For High School Students 2004, Sample Solutions
Transcription
Kettering University Mathematics Olympiad For High School Students 2004, Sample Solutions
```Kettering University Mathematics Olympiad For High
School Students 2004, Sample Solutions
1. (Solution by Mr. Dan Pan, a 4th-6th finisher)
By inspection, we see that (x, y) = (0, 1) and (x, y) = (1, 0) are solutions.
The solutions of x6 + y 6 = 1 are bounded by −1 ≤ 1 and −1 ≤ y ≤ 1
because both x6 and y 6 are nonnegative numbers. Thus, its graph is
bounded similarly as well.
1
x6 + y 6 = 1 → y = ±(1 − x6 ) 6 . The corresponding graph is shown in
1
Figure 1. x5 + y 5 = 1 → y = (1 − x5 ) 5 . The corresponding graph is
1
0.5
0
-0.5
-1
-1
-0.5
0
0.5
1
Figure 1: Graph of x6 + y 6 = 1
shown in Figure 2. From above we see that when x < 0, y > 1 and when
y < 0, x > 1. These conditions mean that the solutions of the system are
1
bounded by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (because the graph of y = (1 − x5 ) 5
1
will not intersect with y = (1 − x6 ) 6 outside of these bounds)
Now we find all solutions within the bounds:
• (0, 1), (1, 0) are known solutions
• Note 1: for x ∈ (0, 1), x6 < x5
• Note 2: for x ∈ (0, 1), 1 − x6 > 1 − x5
1
1
• Note 3: for a ∈ (0, 1), a 6 > a 5
1
1.2
1
0.8
0.6
0.4
0.2
0
-1
-0.5
0
0.5
1
Figure 2: Graph of x5 + y 5 = 1
1
1
• Note 2 and 3 implies that for x ∈ (0, 1), (1 − x6 ) 6 > (1 − x5 ) 5
1
1
• Thus, the curve y = (1−x6 ) 6 > y = (1−x5 ) 5 for all x, y ∈ (0, 1) and
thus there are no intersections/solutions. Therefore, (x, y) = (0, 1)
and (x, y) = (1, 0) are the only solutions of the system.
2
2. (Solution by Mr. Rahul Ramesh, a 4th-6th finisher)
Using the diagram drawn in Figure 3 we can conclude the following:
E
A
D
O
C
G
B
F
B
G a/2
D
C
H
O
R
F
F
Figure 3: Diagram For Question 2
r
a2
a2
GF = R −
⇒ GF = R2 −
4
4
√
2
a
a 3
2
GA = a2 −
⇒ GA =
4
√ 2
1
a 3
GO = GA ⇒ GO =
3
r 6
√
a 3
a2
2
+
F O = F G + GO = R −
4
6
r
√
2
a
3
3
3
a
+
FH = FO =
R2 −
2
2
4
4
r
r
2
a √
a
2
3
a
a2
2
+ = 3 R2 −
+
FE = √ FH = √
R −
4
2
4
2
3
3
2
2
3
E
3. (Solution by Ms. Amy Palmgren, a 4th-6th finisher)
Since any integer power of 2 is even, the number we are looking for must
ends in 2,4,6 or 8. Suppose there exists a positive integer power of 2 that
ends with four equal digits. Let’s call this number p = 2n . We note that
n ≥ 10. We have 4 cases to consider:
. . . 1111
Case 1: p is of the form . . . 2222. Then pn−1 = p2 =
or
. . . 6111.
Neither case is possible.
. . . 2222
or
Case 2: p is of the form . . . 4444. Then pn−1 = p2 =
. . . 7222.
The first situation reduces to Case 1 which we already know is impossible.
If p2 is of the form . . . 7222 then pn−2 = p4 is of the form . . . 611 which is
not possible.
Case 3: p is of the form . . . 6666. Then pn−1 =
impossible.
p
2
= . . . 333 which is
. . . 4444
or
Case 4: p is of the form . . . 8888. Then pn−1 = p2 =
. . . 9444.
The first situation reduces to Case 2 which wealready know is impossible.
. . . 4722
If p2 is of the form . . . 9444 then pn−2 = p4 =
or
. . . 9722.
. . . 2361
or
If p4 is of the form . . . 4722 then pn−3 = p8 =
. . . 7361.
Neither situation is possible.
. . . 4861
or
If p4 is of the form . . . 9722 then pn−3 = p8 =
. . . 9861.
Again neither situation is possible.
All outcomes with the same last digit cannot come from 2n . Repeatedly
dividing by 2 leaves a non 2,4,6 or 8 last digit eventually.
4
4. (Solution by Mr. Brandon Long, Mr. Long placed second in the 2004 Kettering Math Olympiad.)
It is not possible to move all coins into a sector with 2004 moves. Moves
can only be wasted in pairs. (for example, move a coin out of a region and
then return the same coin to its original region) This means the process
to get all coins into a region must have an even number of moves to be
accomplished in 2004 moves.
Figure 4 divides the sectors. We note the final sector by X. Sectors denoted
by E are even, meaning an even number of moves must be made to get
the coin to X. Sectors denoted by O are odd, meaning an odd numbers
of moves is required to get the coin to X. Since there are 5 odd sectors,
a total of 5 × 9 = 45 coins requires an odd number of moves to reach X.
This results in an odd number of moves being required. Making 2004, or
any other even number of moves, an impossible solution.
E
O
O
E
E
O
O
x
E
O
Figure 4: Diagram For Question 4
5
5. (Solution by Mr. Dan Schultz, Mr. Schultz placed third in the 2004 Kettering Math Olympiad.)
Yes, it is possible to divide a convex polygon with an arbitrary number of
points inside into smaller convex polygons that each contain one point.
Lemma: Any rectangle is convex.
Proof: By definition all of its angles = 900 < 1800 .
We can draw an arbitrary convex polygon as in Figure 5. Let the arbitrary
points be P : {p1 , p2 , . . . , pn } have coordinates (x1 , y1 ), (x2 , y2 ), . . . (xn , yn ).
Figure 5: Diagram For Question 5
Now for A, B ∈ < and A 6= B, there exists D such that A < D < B.
Group P into j sets with x coordinates the same and in increasing order.
That is
P ={
with
j
X
(x1 , y1,1 ), (x1 , y1,2 ), . . . , (x1 , y1,k1 )
(x2 , y2,1 ), (x2 , y2,2 ), . . . , (x2 , y2,k2 )
..
.
(xj , yj,1 ), (xj , yj,2 ), . . . , (xj , yj,kj )}
= k1 = n and xi < xi+1 .
l=1
Next find c0i s such that xi < ci < xi+1 . Now divide the polygon into
columns with edges with the equation x = ci .
Now we are left with columns with points inside that have different x
coordinates. We can then divide these columns with horizontal lines in
between the y coordinates of the points. If the column contains only one
point then no further division is necessary.
6
6. (Solution by Mr. Colin Clarke, Mr. Clark is the winner of the 2004 Kettering Math Olympiad.)
It is possible for the grasshopper to visit every square. Let the grasshopper
start from a point in the upper right square and then do a snake pattern
between each of the squares as shown in Figure 6. Each time the grasshop-
no
pq
rs
tu
vw
xy
z{
|}
~
!
"#
\$%
&'
()
*+
,-
./
01
23
45
67
89
:;
<=
>?
@A
BC
DE
FG
HI
JK
LM
NO
PQ
RS
TU
VW
XY
Z[
\]
^_
`a
bc
de
fg
hi
jk
lm
Figure 6: Diagram For Question 6
per lands it is in the equivalent spot as in every other square. This path
is valid from any point in the first square. The total amount of paint that
the troll uses is 0.015 × 64 = 0.96 of a square. Since the grasshopper can
start from any part of the first square, even if the troll paints the board
to stop as many of these paths as possible, he will not be able to.
7
``` |
How do you find dy/dx by implicit differentiation of (x+y)^3=x^3+y^3 and evaluate at point (-1,1)?
Apr 6, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x y + {y}^{2}}{{x}^{2} + 2 x y}$ and at $\left(- 1 , 1\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$
Explanation:
As ${\left(x + y\right)}^{3} = {x}^{3} + {y}^{3}$, taking differential on both sides implicitly, we get
$3 {\left(x + y\right)}^{2} \times \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$
or $3 \frac{\mathrm{dy}}{\mathrm{dx}} = 3 \left\{{x}^{2} - {\left(x + y\right)}^{2}\right\}$
or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - {\left(x + y\right)}^{2}}{{\left(x + y\right)}^{2} - {y}^{2}}$
or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x y + {y}^{2}}{{x}^{2} + 2 x y}$
and at $\left(- 1 , 1\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left(- 1\right) \times 1 + {1}^{2}}{{1}^{2} + 2 \left(- 1\right) \left(1\right)} = - \frac{- 1}{- 1} = - 1$ |
# Math Tips for Exams
I created and developed this article with the intention of giving others access to these fundamental tricks necessary for improving mental math speed and arithmetic skills.
Of course, if you pay for math homework and do not understand anything in math, my advice will help you prepare for the exam faster.
This Math Tips and Tricks can be helpful for all kinds of people. If you are a cashier and would like to become much faster at your job, our tricks can help you. If you are studying for a standardized exam and are struggling to get through the math section on time, these ideas and tips can help you improve your time. These tricks have been helping me since I discovered several of them in my 8th grade math class. This helped me do an outstanding job in all my math courses, the math sections on SAT and ACT, and for everyday use since I began discovering these tricks.
## Exam Tip # 1
I have seen this type of problem come up dozens of times before from math classes to the SAT’s. The problem gives you a range of numbers. It will provide you with at least 2 relatively small numbers and will ask you to figure out how many integers in that range are divisible by the provided numbers. The majority of people pull out their calculators to count or take a sheet and list each number. I will show you a much faster way to solve this problem.
### Example 1:
How many numbers greater than 1 and less than or equal to 1,000 are divisible by 4 and 9.
Step 1: Find out how many numbers between 1 and 1,000 are divisible by 4:
Considering 1,000 is included, we can just divide 1000 by 4 to get the solution.
1000 / 4 = 250
Step 2: Find out how many numbers between 1 and 1,000 are divisible by 9:
1000 / 9 = 111
Remember to always round down. We can’t round up because the next number 112 is greater then 1,000 when multiplied by 9 so it doesn’t fit.
Step 3: Multiply 9 and 4: 9 * 4 = 36
Step 4: Find out how many numbers between 1 and 1,000 are divisible by 36:
1000 / 36 = 27
Step 5: Add the solutions to steps 1 and 2, then subtract the solution from step 4.
250 + 111 – 27 = 334
Review: The correct answer is 334. For these types of problems, it may be necessary to use calculators when needed to deal with large numbers especially for exams. To explain the steps, we found out how many different numbers between 1 and 1000 are divisible by 4 and 9, but what we need to take into account for, which is step 4, is that every number divisible by 36 between 1 and 1000 is counted twice because it is both divisible by 4 and 9. So therefore we subtract 361, the original total, with 27 and we get 334 as our final answer.
### Example 2:
How many numbers greater than 1 and less than 8,000 are divisible by 3, 4, and 5.
This is as difficult as this problem can get basically. Once you understand how to do this, all these types of problems will be very easy.
Step 1: We must consider that we are really looking from 2 to 7,999 because we are not including the 1 or the 8,000. Our first step is to find out how many numbers are divisible by 3:
7,999 / 3 = 2,666
Step 2: Find out how many numbers are divisible by 4:
7,999 / 4 = 1,999
Step 3: Find out how many numbers are divisible by 5:
7,999 / 5 = 1,599
Step 4: For this problem, we are dealing with 3 numbers so there are more duplicates to consider. Every number divisible by ( 3 and 4 ) are counted twice, as well as numbers divisible by ( 3 and 5 ), ( 4 and 5 ), and ( 3, 4 and 5). So we must subtract all these duplicates.
3 * 4 = 12 ——————— 7,999 / 12 = 666
3 * 5 = 15 ——————— 7,999 / 15 = 533
4 * 5 = 20 ——————— 7,999 / 20 = 399
3 * 4 * 5 = 60 —————– 7,999 / 60 = 133
Step 5: Combine steps 1, 2 and 3, then subtract the sums from step 4:
2,666 + 1,999 + 1,599 = 6,264
6,264 – 666 – 533 – 399 – 133 = 4,533
Review: Hopefully this makes sense and is useful to everyone. I used to do this the long way by writing down each term and then adding up the number of terms until I discovered this method. Know this for the SAT, ACT, AMC, AIME or other exams in case you are asked.
## Math Tip # 2
The least common multiple is the smallest number that any two or more numbers are divisible by. For example, if you are given 8 and 6, the least common multiple is 24. To find the least common multiple of any two numbers, we simply factor the two numbers into prime numbers, then we combine the factors together and multiply them to come up with the solution. I’ll provide detailed explanations for each example.
### Example 1
Find the least common multiple of 48 and 36
Step 1: Factor the two numbers until you are left with only prime numbers.
48 = 24 * 2 = 12 * 2 * 2 = 6 * 2 * 2 * 2 = 3 * 2 * 2 * 2 * 2
36 = 18 * 2 = 9 * 2 * 2 = 3 * 3 * 2 * 2
Step 2: Rewrite using powers
48 = 3 * 2 ^ 4
36 = 3 ^2 * 2 ^2
Step 3: For both numbers provided, look at the factors and see if the two numbers have the same factor. If they do, take the one with the higher power. For this example, 48 and 36 both have 2′s as factors, but 48 has 2 ^ 4 whereas 36 has 2 ^ 2, so we take the 2 ^ 4 and multiply it by the other factors to find the least common multiple. Every unique factor must be used to find the least common multiple.
2 ^ 4 * 3 ^ 2 = 144
Therefore 144 is the least common multiple of 48 and 36
### Example 2
Find the least common multiple of 7, 210, and 25
Step 1: Factor the three numbers until you are left with only prime numbers.
7 = 7 * 1
25 = 5 * 5
210 = 70 * 3 = 7 * 10 * 3 = 7 * 2 * 5 * 3
Step 2: Rewrite using powers:
25 = 5 ^ 2
7 = 7 * 1
210 = 2 * 3 * 5 * 7
Step 3: We must take each unique factor from every number as well as the larger power for similar factors. For this problem, the unique factors are 2 and 3 since they only exist as factors for 210. The repeat factors are 7 and 5. We will take the 5 ^ 2 factor from 25 for the least common multiple because it is larger than the 5 factor for 210.
Therefore 7 * 3 * 2 * 5 ^ 2 = 1050
### Example 3
Find the least common multiple of 14, 49, 32 and 80
Step 1: Break down each term:
14 = 2 * 7
49 = 7 * 7 = 7 ^ 2
32 = 2 * 16 = 2 * 2 * 8 = 2 * 2 * 2 * 4 = 2 * 2 * 2 * 2 * 2 = 2 ^ 5
80 = 2 * 40 = 2 * 2 * 20 = 2 * 2 * 2 * 10 = 2 * 2 * 2 * 2 * 5 = 5 * 2 ^ 4
Step 2: Find each unique factor, then take the largest power for each factor:
Our factors are 2, 5 and 7. The largest factor of 2 is 2 ^5. The largest factor of 5 is just 5. The largest factor of 7 is 7 ^ 2
Step 3: Multiply the largest factors:
2 ^ 5 * 5 * 7 ^ 2 = 32 * 5 * 49 = 7,840
This means the least common multiple of 14, 49, 32 and 80 is 7,840
Hopefully this made sense. If not let me know by leaving a comment and i’ll try to explain it some more the best, I can. |
# How do you solve the separable differential equation dy/dx=(cosx)e^(y+sinx)?
Mar 13, 2017
$y = \ln \left(\frac{1}{A - {e}^{\sin} x}\right)$ is the General Solution
#### Explanation:
We have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos x\right) {e}^{y + \sin x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos x\right) {e}^{y} {e}^{\sin} x$
Which is a First Order Separable Differential Equation, which we can rewrite as:
$\frac{1}{e} ^ y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos x\right) {e}^{\sin} x$
We can then "separate the variables" to get:
$\int \setminus {e}^{-} y \setminus \mathrm{dy} = \int \setminus \left(\cos x\right) {e}^{\sin} x \setminus \mathrm{dx}$
Which we can directly (and easily) integrate to get:
$- {e}^{-} y = {e}^{\sin} x + B$
$\therefore {e}^{-} y = A - {e}^{\sin} x$
$\therefore - y = \ln \left(A - {e}^{\sin} x\right)$
$\therefore y = - \ln \left(A - {e}^{\sin} x\right)$
$\therefore y = \ln \left(\frac{1}{A - {e}^{\sin} x}\right)$
Mar 13, 2017
$y = - \ln \left\mid C - {e}^{\sin} x \right\mid$
#### Explanation:
To "separate" a differential equation means to move all the terms with $y$ to one side of the equation, and all the terms with $x$ to the other side of the equation.
We treat $\frac{\mathrm{dy}}{\mathrm{dx}}$ as a division problem, so we can move the $\mathrm{dx}$ to the other side of the equation, leaving just $\mathrm{dy}$.
To separate this, we also need to split up ${e}^{y + \sin x}$ as ${e}^{y} \left({e}^{\sin} x\right)$.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\cos x\right) {e}^{y + \sin x}$
$\mathrm{dy} = \left(\cos x\right) {e}^{y} \left({e}^{\sin} x\right) \mathrm{dx}$
$\frac{\mathrm{dy}}{e} ^ y = \left(\cos x\right) {e}^{\sin} x \mathrm{dx}$
Now we can integrate both sides:
$\int {e}^{-} y \mathrm{dy} = \int {e}^{\sin} x \left(\cos x\right) \mathrm{dx}$
For the left-hand side, use the substitution $u = - y$, implying that $\mathrm{du} = - \mathrm{dy}$.
$- \int {e}^{-} y \left(- \mathrm{dy}\right) = \int {e}^{\sin} x \cos x \mathrm{dx}$
$- \int {e}^{u} = \int {e}^{\sin} x \cos x \mathrm{dx}$
$- {e}^{u} = \int {e}^{\sin} x \cos x \mathrm{dx}$
$- {e}^{-} y = \int {e}^{\sin} x \cos x \mathrm{dx}$
Following a similar process on the right, let $t = \sin x$ so $\mathrm{dx} = \cos x \mathrm{dx}$.
$- {e}^{-} y = \int {e}^{t} \mathrm{dt}$
$- {e}^{-} y = {e}^{\sin} x + C$
Solving for $y$:
${e}^{-} y = - {e}^{\sin} x + C$
$- y = \ln \left\mid C - {e}^{\sin} x \right\mid$
$y = - \ln \left\mid C - {e}^{\sin} x \right\mid$ |
1. ## Linear Approximation Application
I'm having some trouble applying linear approximation and changes, so I'm stuck on this question.
The diameter of a tree = 10 in.
Circumference increases by 2 in
About how much did the tree's diameter grow? How much did the tree's cross section area grow?
I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time.
2. Originally Posted by Wyau
I'm having some trouble applying linear approximation and changes, so I'm stuck on this question.
The diameter of a tree = 10 in.
Circumference increases by 2 in
About how much did the tree's diameter grow? How much did the tree's cross section area grow?
I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time.
The linear approximation to a change in the argument of a function f(x) is
$f(x + \Delta x) \approx f(x) + f^{\prime}(x) \cdot \Delta x$
So we know that the initial diameter of the tree is d = 10 in. And we know that the circumference of the tree grows by 2 in.
The formula for the circumference is
$C = \pi d$
so
$d(C) = \frac{C}{\pi}$
I'm assuming you have to use the linear approximation equation here (though the answer turns out to be exact, not an approximation) so
$C = \pi d = \pi \cdot 10~in = 31.415~in$
and
$d^{\prime}(C) = \frac{1}{\pi}$
Thus
$d(C + \Delta C) = \frac{C}{\pi} + \frac{1}{\pi} \cdot \Delta C$
$d(31.415~in + 2~in) = \frac{31.415~in}{\pi} + \frac{1}{\pi} \cdot 2~in = 10.637~in$
For the second problem, we know that the cross-section is a circle and the area of a circle is:
$A = \pi r^2 = \pi \left ( \frac{d}{2} \right )^2 = \frac{\pi}{4}d^2$
So
$A^{\prime}(d) = \frac{\pi}{2}d$
The change in d is:
$\Delta d = 10.637~in - 10~in = 0.637~in$
So
$A(d + \Delta d) \approx A(d) + A^{\prime}(d) \cdot \Delta d$
$\approx \frac{\pi}{4}(10~in)^2 + \frac{\pi}{2}(10~in) \cdot (0.637~in)$
$\approx 88.540 ~ in^2$
(The actual area is $A = 88.8581~in^2$, so this is a good approximation.)
-Dan |
Tuesday, July 16, 2024
# What Is Another Word For Average In Math
## How To Find The Mode
Pre Algebra Lesson: Mean, Median, and Mode
Of all the measures, finding the mode requires the least amount of mathematical calculation. Instead, since the mode is simply the most frequently occurring score in a distribution, all you do is look at all your scores and select the most common one.
• Step 1: Look at all the data scores
• Step 2: Identify the data score that appears most often
As an example, consider the following number distribution: 2, 3, 6, 3, 7, 5, 1, 2, 3, 9. The mode of these numbers would be 3 since this is the most frequently occurring number .
If no number in a set occurs more than once, there is no mode for that set of data. It’s also possible for a data set to have two modes. This is known as bi-modal distribution.
Bi-modal distribution occurs when there are two numbers that are tied in frequency. For example, consider the following set of numbers: 13, 17, 20, 20, 21, 23, 23, 26, 29, 30. In this set, both 20 and 23 occur twice . Therefore, they are both modes.
## How To Find The Mean
Take these two steps to calculate the mean:
• Step 1: Add all the scores together
• Step 2: Divide the sum by the number of scores used
As an example, imagine that your psychology experiment returned the following number set: 3, 11, 4, 6, 8, 9, 6. To calculate the mean, you first add all the numbers together . Then you divide the total sum by the number of scores used . In this example, the mean or average of the number set is 6.7.
## What Is The Difference Between Average And Mean
Average and mean are usually confused with one another as both the mathematical terms are used to explain the set of numbers. Mean can simply be calculated by adding the set of values and divided by the number of quantities. Thus, this is the core definition of what mean is. You can find the tabular column below to learn the difference between the average and mean.
Difference between Average and Mean
Average
Mean
Average can simply be defined as the sum of all the numbers divided by the total number of values. A mean is defined as the mathematical average of the set of two or more data values.
Average is usually defined as mean or arithmetic mean. Mean is simply a method of describing the average of the sample.
Average can be calculated for any discrete numbers where it assumes uniform distribution. It is mainly used in Statistics, and it is applied for any distribution such as geometric, binomial, Poisson distribution, and so on.
The arithmetic mean is considered as a form of average. There are various types of mean.
Average is usually used in conversations in general day to day English. Mean is used in a more technical and mathematical sense.
The average is capable of giving us the median and the mode. Mean, on the other hand, cannot give us the median or mode.
Recommended Reading: What Are The 4 Major Goals Of Psychology
## Recap Of How To Find The Median
The median is calculated by arranging the scores in numerical order, dividing the total number of scores by two, then rounding that number up if using an odd number of scores to get the position of the median or, if using an even number of scores, by averaging the number in that position and the next position.
## Examples Of When Means Are Important In Investing
Within business and investing, mean is used extensively to analyze performance. Examples of situations in which you may encounter mean include:
• Determining whether an equity is trading above or below its average over a specified time period.
• Looking back to see how comparative trading activity may determine future outcomes. For example, seeing the average rate of return for broad markets during prior recessions may guide decision-making in future economic downturns.
• Seeing whether trading volume or the quantity of market orders is in line with recent market activity.
• Analyzing the operational performance of a company. For instance, some financial ratios like the days sales outstanding require determining the average accounts receivable balance for the numerator.
• Quantifying macroeconomic data like average unemployment over a period of time to determine general health of an economy.
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## Definition Of Mean Median And Mode
To understand the differences between the mean, median, and mode, let’s start by defining these three terms.
• The mean is the arithmetic average of a set of given numbers. Therefore, the mean in math is often referred to as simply the “average.”
• The median is the middle score in a set of given numbers. As the median, half of the scores are above this number and half are below.
• The mode is the most frequently occurring score in a set of given numbers. Put another way, it is the score that appears the greatest number of times.
## Average Formula In Maths
The formula to find the average of given numbers or values is very easy. We just have to add all the numbers and then divide the result by the number of values given. Hence, the average formula in Maths is given as follows:
Average = Sum of Values/ Number of values
Suppose, we have given with n number of values such as x1, x2, x3 ,.., xn. The average or the mean of the given data will be equal to:
Average = /n
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## What Is A Mean
Mean is the simple mathematical average of a set of two or more numbers. The mean for a given set of numbers can be computed in more than one way, including the arithmetic mean method, which uses the sum of the numbers in the series, and the geometric mean method, which is the average of a set of products. However, all the primary methods of computing a simple average produce the same approximate result most of the time.
## What Does Average Mean In Math
Math Antics – What Percent Is It?
Average is a term that is used, mis-used and often overused. Typically, many individuals refer to average when they really mean the arithmetic average . Average can mean the mean, the median, and the mode, it can refer to a geometric mean and weighted averages.
Although most people use the term average for this type of calculation:
Four tests results: 15, 18, 22, 20The sum is: 75Divide 75 by 4: 18.75The ‘Mean’ is 18.75
The truth of the matter is that the above calculation is considered the arithmetic mean, or often referred to as the mean average.
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## Exploring Some Measures Of Central Tendency
Knowing how to find the mean, median, and mode can help you interpret data collected through psychological research. These values provide more insight into what may be considered “normal” or “abnormal” for a specific group of people in terms of cognitive processes or behaviors, for instance.
Because they are all measures of central tendency, psychology students often find it easy to confuse the three. Yet, there are differences in what each one is and how it is found. Here are some useful tips to help you distinguish between these measures, as well as how to calculate mean, median, and mode.
## Average Of Negative Numbers
If there are negative numbers present in the list, then also the process or formula to find out the average is the same. Lets understand this with an example.
Example:
Find the average of 3, 7, 6, 12, 2.
Solution:- The sum of these numbers
= 3 + + 6 + 12 +
= 3 7 + 6 + 12 2
= 12
Total Units = 5
Hence, average = 12/5 = 2.4
How does this whole idea of average or mean works? Average helps you to calculate how to make all the units present in a list equal.
Find the average of 6, 13, 17, 21, 23.
Solution:
= 6 + 13 + 17 + 21 + 23 = 80
Total units = 5
Hence, average = 80/5 = 16
Example 3:
If the age of 9 students in a team is 12, 13, 11, 12, 13, 12, 11, 12, 12. Then find the average age of students in the team.
Solution:
Given, the age of students are 12, 13, 11, 12, 13, 12, 11, 12, 12.
Average = Sum of ages of all the students/Total number of students
A = /9
A = 108/9
Hence, the average age of students in a team is 12 years.
Example 4:
If the heights of males in a group are 5.5, 5.3, 5.7, 5.9, 6, 5.10, 5.8, 5.6, 5.4, 6. Then find the average height.
Solution:
Given the height of males: 5.5, 5.3, 5.7, 5.9, 6, 5.10, 5.8, 5.6, 5.4 and 6
Average = Sum of heights of males/total number of males
A = /10
A = 56.3/10
A = 5.63
Recommended Reading: Chapter 9 Review Algebra 2
## Definition Of Average And Mean
Average: The term Average describes a value that should represent the sample. An average is defined as the sum of all the values divided by the total number of values in a given set. It is also known as the arithmetic mean. Let us consider a simple data to find the average.
Given, the set of values are 1, 2, 3, 4, 5.
Average = Sum of all the values/ Total number of values
Average = /5 = 15/5 = 3
Mean: A mean is a mathematical term, that describes the average of a sample. In Statistics, the definition of the mean is similar to the average.
## Is It A Good Idea To Go Above And Beyond
An above-average student is driven and wants to go somewhere. If you want to be above average, above average is a good idea.
Its not better or worse than average its just that its not average, and that cant be good or bad on its own. Good and bad are subjective opinions that are based on a comparison with something.
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## What Is A Mean Definition In Math And Formula For Calculation
Adam Hayes, Ph.D., CFA, is a financial writer with 15+ years Wall Street experience as a derivatives trader. Besides his extensive derivative trading expertise, Adam is an expert in economics and behavioral finance. Adam received his master’s in economics from The New School for Social Research and his Ph.D. from the University of Wisconsin-Madison in sociology. He is a CFA charterholder as well as holding FINRA Series 7, 55 & 63 licenses. He currently researches and teaches economic sociology and the social studies of finance at the Hebrew University in Jerusalem.
## Difference Between Average And Mean
To understand the difference between average and mean, one must be aware of what separates one from the other. Average and mean are used interchangeably. In Statistics, instead of the term average, the term mean is used. Average can simply be defined as a quantity or a rate which usually fall under the centre of the data. The average is quite similar to mean but also has its key differences from mean as well. If one can understand the arithmetic mean and range, it can be incredibly helpful in understanding and solving math topics.
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## An Example Of Mean Median And Mode In Psychology
Imagine a research study in which psychologists are interested in learning the typical age at which someone might be diagnosed with schizophrenia. To collect this data, they send a questionnaire to mental health providers, asking that they share their patients’ ages upon formal diagnosis.
The responses received indicate that the practitioners’ patients were the following ages:
Using the calculations above, you would find that the mean, median, and mode for this data set are all around 27 years . In this case, any of these measures could be used to help you arrive at the typical age of onset.
But what if you had an additional score of 13? In this case, the calculation of the mean would be 25.6, while the median and mode would both be 27. Since the mean includes an outlier, median and mode would be more accurate as they aren’t skewed by this number.
In case you are curious, the National Alliance on Mental Health reports that the average age of schizophrenia onset for men is late teens to early 20s, while women tend to be diagnosed with this condition in their late 20s to early 30s.
## What Is The Mean Median And Mode
Algebra Basics: What Are Polynomials? – Math Antics
The mean is the number you get by dividing the sum of a set of values by the number of values in the set.
In contrast, the median is the middle number in a set of values when those values are arranged from smallest to largest.
The mode of a set of values is the most frequently repeated value in the set.
To illustrate the difference, lets look at a very simple example.
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## Averages: Mean Median And Mode
The term average occurs frequently in all sorts of everyday contexts. For example, you might say Im having an average day today, meaning your day is neither particularly good nor bad, it is about normal. We may also refer to people, objects and other things as average.
The term ‘average’ refers to the middle or central point. When used in mathematics, the term refers to a number that is a typical representation of a group of numbers . Averages can be calculated in different ways – this page covers the mean, median and mode. We include an averages calculator, and an explanation and examples of each type of average.
The most widely used method of calculating an average is the mean. When the term average is used in a mathematical sense, it usually refers to the mean, especially when no other information is given.
Quick Guide:
Count how many times each value occurs the value that occurs most often is the mode.
## When To Use Mean Median And Mode
How do you determine whether to use the mean, median, or mode when analyzing psychology research? The one you select can depend on the data scores themselves.
If there are no outliers in your data set, the mean may be the best choice in terms of accuracy since it takes into account each individual score and finds the average. Conversely, if outliers exist, the median or mode may be more accurate since the results won’t be skewed.
Also consider what you are trying to measure. Are you looking for the average , do you want to identify the middle score , or are you looking for the score that appears most often ? While they are all measures of central tendency, each one looks at this tendency from a slightly different point of view.
Also Check: Geometry 1.1 1.3 Answers
## How To Calculate Average
We can easily calculate the average for a given set of values. We just have to add all the values and divide the outcome by the number of given values.
Average can be calculated using three simple steps. They are:
Step 1: Sum of Numbers:
The first step in finding the average of numbers is to find the sum of all the given numbers.
Step 2: Number of Observations:
Next, we have to count how many numbers are in the given dataset.
Step 3: Average Calculation:
The final step in calculating the average is to divide the sum by the number of observations.
Now, let us consider an example to calculate the average.
If there are a group of numbers say, 20, 21, 23, 22, 21, 20, 23. Then find the average of these values.
Average = /No.of values
= /7
= 150/7
## Pros And Cons Of Mean Median And Mode
Each measure of central tendency has its own strengths and weaknesses. Here are a few to consider.
• The mean utilizes all numbers in a set to express the measure of central tendency. However, outliersor data that lies well outside of the data setcan distort the overall measure. For example, a couple of extremely high scores can skew the mean, so that the average score appears much higher than most of the scores actually are.
• The median gets rid of outliers or disproportionately high or low scores. At the same time, this could be an issue because it may not adequately represent the full set of numbers.
• The mode may be less influenced by outliers as well and is good at representing what is “typical” for a given group of numbers. But it also may be less useful in cases where no number occurs more than once.
While the mean in math is theoretically neutral, some contend that the use of the mean in psychology can lead to inappropriate conclusions if care is not taken with its application. This is due, in part, to behavior and cognition being both complex and variable in nature.
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## What Is Another Word For On Average
Synonyms for on average include generally, normally, ordinarily, usually, generally speaking, in general, overall, typically, as a rule and in most cases. Find more similar words at wordhippo.com!
How is the word average distinct from other similar nouns?
How is the word average distinct from other similar nouns? Some common synonyms of average are mean, median, and norm. While all these words mean something that represents a middle point, average is the quotient obtained by dividing the sum total of a set of figures by the number of figures.
## Averages As A Rhetorical Tool
Due to the aforementioned colloquial nature of the term “average”, the term can be used to obfuscate the true meaning of data and suggest varying answers to questions based on the averaging method used. In his article “Framed for Lying: Statistics as In/Artistic Proof”, University of Pittsburgh faculty member Daniel Libertz comments that statistical information is frequently dismissed from rhetorical arguments for this reason. However, due to their persuasive power, averages and other statistical values should not be discarded completely, but instead used and interpreted with caution. Libertz invites us to engage critically not only with statistical information such as averages, but also with the language used to describe the data and its uses, saying: “If statistics rely on interpretation, rhetors should invite their audience to interpret rather than insist on an interpretation.” In many cases, data and specific calculations are provided to help facilitate this audience-based interpretation.
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0 like 0 dislike
simplify sqrt72 -3sqrt12 + sqrt 192
0 like 0 dislike
6√2+2√3
Step-by-step explanation:
We want to simplify the following radical expression
$$\displaystyle \sqrt{72} - 3 \sqrt{12} + \sqrt{192}$$
Recall that
$$\sqrt{ab} = \sqrt{a} \sqrt{b} , \forall \text{a and b such that a\geq0,b\geq0}$$
Utilizing the formula yields,
$$\sqrt{72} \implies \sqrt{36 \cdot 2} \implies 6 \sqrt{2}$$
$$\sqrt{12} \implies \sqrt{4\cdot 3} \implies 2 \sqrt{3}$$
$$\sqrt{192} \implies \sqrt{64\cdot 3} \implies 8 \sqrt{3}$$
So,
$$6 \sqrt{2}-3\cdot2 \sqrt{3}+8 \sqrt{3}$$
Carry out multiplication:
$$\implies 6 \sqrt{2}-6 \sqrt{3}+8 \sqrt{3}$$
$$\boxed{6 \sqrt{2}+2 \sqrt{3}}$$
and we're done!
by
0 like 0 dislike
$$6\sqrt{2}+2{\sqrt3}$$
Step-by-step explanation:
Given expression:
$$\sqrt{72}-3\sqrt{12}+\sqrt{192}$$
Rewrite 72 as (36 · 2), 12 as (4 · 3), and 192 as (64 · 3):
$$\implies \sqrt{36 \cdot 2}-3\sqrt{4 \cdot 3}+\sqrt{64 \cdot 3}$$
Apply the radical rule $$\sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$$ :
$$\implies \sqrt{36}\sqrt{2}-3\sqrt{4}\sqrt{3}+\sqrt{64}{\sqrt3}$$
Rewrite 36 as 6², 4 as 2², and 64 as 8²:
$$\implies \sqrt{6^2}\sqrt{2}-3\sqrt{2^2}\sqrt{3}+\sqrt{8^2}{\sqrt3}$$
Apply the radical rule $$\sqrt{a^2}=a$$ :
$$\implies 6\sqrt{2}-3\cdot 2\sqrt{3}+8{\sqrt3}$$
Simplify:
$$\implies 6\sqrt{2}-6\sqrt{3}+8{\sqrt3}$$
$$\implies 6\sqrt{2}+2{\sqrt3}$$
by |
# Lesson 8
The $n^{\text{th}}$ Term
### Problem 1
A sequence is defined by $$f(0) = \text-20, f(n) = f(n-1) -5$$ for $$n \ge1$$.
1. Explain why $$f(1) = \text- 20 - 5$$.
2. Explain why $$f(3) = \text- 20 - 5 - 5 - 5$$.
3. Complete the expression: $$f(10)=\text-20-\underline{\hspace{.5in}}$$. Explain your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
A sequence is defined by $$f(0) = \text- 4, f(n) = f(n-1) - 2$$ for $$n\ge1$$. Write a definition for the $$n^{\text{th}}$$ term of the sequence.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
Here is the recursive definition of a sequence: $$f(1) = 3,f(n) = 2 \boldcdot f(n-1)$$ for $$n\ge2$$.
1. Find the first 5 terms of the sequence.
2. Graph the value of the term as a function of the term number.
3. Is the sequence arithmetic, geometric, or neither? Explain how you know.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 1, Lesson 7.)
### Problem 4
Here is a graph of sequence $$M$$. Define $$M$$ recursively using function notation.
### Solution
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(From Unit 1, Lesson 6.)
### Problem 5
Write the first five terms of each sequence. Determine whether each sequence is arithmetic, geometric, or neither.
1. $$a(1) = 5, a(n) = a(n-1) + 3$$ for $$n\ge2$$.
2. $$b(1) = 1, b(n) = 3 \boldcdot b(n-1)$$ for $$n\ge2$$.
3. $$c(1) = 3, c(n) = \text-c(n-1) + 1$$ for $$n\ge2$$.
4. $$d(1) = 5, d(n) = d(n-1) + n$$ for $$n\ge2$$.
### Solution
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(From Unit 1, Lesson 5.)
### Problem 6
Here is the graph of a sequence:
1. Is this sequence arithmetic or geometric? Explain how you know.
2. List at least the first five terms of the sequence.
3. Write a recursive definition of the sequence.
### Solution
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(From Unit 1, Lesson 7.) |
Lesson 2
# HOW TO READ AND WRITEWHOLE NUMBERS
## THE POWERS OF 10
1. What are the ten digits? The ten symbols: 0 1 2 3 4 5 6 7 8 9
Examples.
105 is a three-digit number. The digits are 1, 0, and 5.
28 ends in the digit 8.
\$364 has the same digits as \$3.64.
Those ten marks are also known as the Arabic numerals, because it was the Arab mathematicians who introduced them into Europe from India, where their forms evolved.
We saw in Lesson 1 that '364' is a numeral, which is a symbol for a number. A number is the actual collection of units.
The powers of 10
Number Ten is a collection of ten Ones.
One Hundred is a collection of ten Tens.
The number we call One Thousand is a collection of ten One Hundreds.
Ten One Thousands are called Ten Thousand.
The numbers in that sequence are called the powers of 10.
2. Which numbers are the powers of 10? They are the numbers produced when, starting with One, we repeatedly collect them into groups of 10. 10 Ones. 10 Tens. 10 Hundreds. 10 Thousands. And so on.
Here are their names and numerals.
The Powers of 10
Class of One 1 Ones Ten 10 Hundred 100 Class of One thousand 1,000 Thousands Ten thousand 10,000 Hundred thousand 100,000 Class of One million 1,000,000 Millions Ten million 10,000,000 Hundred million 100,000,000 Class of One billion 1,000,000,000 Billions Ten billion 10,000,000,000 Hundred billion 100,000,000,000
Each power is composed of ten of the one above.
(The metric system is the system of measurement based on the powers of 10; see Lesson 4.)
Strictly, 1 is not a power of 10. The first power of 10 is 10 itself. Its numeral is a 1 followed by one 0. The second power of 10 is 100; it has two 0's. The third power has three 0's. And so on.
Notice how the names fall into groups of three:
One thousand, Ten thousand, Hundred thousand.
One million, Ten million, Hundred million.
Each group of three -- Ones, Tens, Hundreds -- is called a class.
Starting with Billions (bi for two), each class has a Latin prefix. To read a number more easily, we separate each class -- each group of three digits -- by commas.
Note that each class is 1000 times the previous class; the Thousands are 1000 times the Ones; the Millions are 1000 times the Thousands; and so on.
In Lesson 1 we showed how to read and write any number from 1 to 999, which are the numbers in the class of Ones. Together with knowing the sequence of class names, that is all that is necessary to be able to name or read any whole number.
4. How do we read a whole number, however large? 256,312,785,649,408,163 Starting from the left, read each three-digit group; then say the name of its class.
256,312,785,649,408,163
Answer. Starting from the left, 256, read each three-digit group. Then say the name of the class.
Say:
"256 Quadrillion, 312 Trillion, 785 Billion, 649 Million, 408 Thousand, 163."
Do not say the class name "Ones."
Example 2. To distinguish the classes, place commas in this number:
8792456
Answer. Starting from the right, place commas every three digits:
8,792,456
"8 million, 792 thousand, 456."
Example 3. Read this number: 7,000,020,002
Answer. "Seven billion , twenty thousand, two."
When a class is absent, we do not say its name; we do not say, "Seven billion, no million, ..."
Also, every class has three digits and so we must distinguish the following:
002 "Two" 020 "Twenty" 200 "Two hundred"
As for "and," in speech it is common to say "Six hundred and nine," but in writing we should reserve "and" for the decimal point, as we will see in the next Lesson. (For example, we should write \$609.50 as "Six hundred nine dollars and fifty cents." Not "Six hundred and nine dollars.")
Example 4. Write in numerals:
Four hundred eight million, twenty-nine thousand, three hundred fifty-six.
Answer. Pick out the classes: "million""thousand". Each class (except perhaps the first class on the left) has exactly three digits:
Example 5. Write in numerals:
Five billion, sixteen thousand, nine.
Answer. After the billions, we expect the millions, but it is absent. Therefore write
5,000,016,009
Again, we must write "sixteen thousand" as 016; and "nine" as 009; because each class must have three digits. The exception is the class on the extreme left. We may write "Five" as 5 rather than 005.
When writing a four-digit number, such as Four thousand five hundred, it is permissible to omit the comma and write 4500. In fact, we often read that as "Forty-five hundred." But when a number has more than four digits, then for the sake of clarity we should always place the commas.
Example 6. Distinguish the following:
a) Two hundred seventeen million b) Two hundred million seventeen Answers. a) 217,000,000 b) 200,000,017
At this point, please "turn" the page and do some Problems.
or
Continue on to Section 2: Place value
Please make a donation to keep TheMathPage online.
Even \$1 will help. |
Mensuration Class 8 Extra Questions Maths Chapter 11
# Mensuration Class 8 Extra Questions Maths Chapter 11
## Mensuration Class 8 Extra Questions Maths Chapter 11
Extra Questions for Class 8 Maths Chapter 11 Mensuration
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### Mensuration Class 8 Extra Questions Very Short Answer Type
Question 1.
Find the perimeter of the following figures:
Solution:
(i) Perimeter of the rectangle = 2(l + b) = 2(8 + 6) = 2 × 14 = 28 cm
(ii) Perimeter of the square = 4 × side = 4 × 6 = 24 cm
(iii) Perimeter of the circle = 2πr = 2 × $$\frac { 22 }{ 7 }$$ × 7 = 44 cm.
Question 2.
The length and breadth of a rectangle are 10 cm and 8 cm respectively. Find its perimeter if the length and breadth are (i) doubled (ii) halved.
Solution:
Length of the rectangle = 10 cm
Breadth of the rectangle = 8 cm
(i) When they are doubled,
l = 10 × 2 = 20 cm
and b = 8 × 2 = 16 cm
Perimeter = 2(l + b) = 2(20 + 16) = 2 × 36 = 72 cm
(ii) When they are halved,
l = $$\frac { 10 }{ 2 }$$ = 5 cm
b = $$\frac { 8 }{ 2 }$$ = 4 cm
Perimeter = 2(l + b) = 2(5 + 4) = 2 × 9 = 18 cm
Question 3.
A copper wire of length 44 cm is to be bent into a square and a circle. Which will have a larger area?
Solution:
(i) When the wire is bent into a square.
Side = $$\frac { 44 }{ 4 }$$ = 11 cm
Area of the square = (side)2 = (11)2 = 121 cm2
(ii) When the wire is bent into a circle.
Circumference = 2πr
44 = 2πr
So, the circle will have a larger area.
Question 4.
The length and breadth of a rectangle are in the ratio 4 : 3. If its perimeter is 154 cm, find its length and breadth.
Solution:
Let the length of the rectangle be 4x cm and that of breadth = 3x cm
Perimeter = 2(l + b) = 2(4x + 3x) = 2 × 7x = 14x cm
14x = 154
x = 11
Length = 4 × 11 = 44 cm
and breadth = 3 × 11 = 33 cm
Question 5.
The area of a rectangle is 544 cm2. If its length is 32 cm, find its breadth.
Solution:
Area = 544 cm2
Length = 32 cm
Breadth of the rectangle = $$\frac { Area }{ Length }$$
= $$\frac { 544 }{ 32 }$$
= 17 cm
Hence, the required breadth = 17 cm
Question 6.
If the side of a square is doubled then how much time its area becomes?
Solution:
Let the side of the square be x cm.
Area = (side)2 = x2 sq. cm
If its side becomes 2x cm then area = (2x)2 = 4x2 sq. cm
Ratio is x2 : 4x2 = 1 : 4
Hence, the area would become four times.
Question 7.
The areas of a rectangle and a square are equal. If the length of the rectangle is 16 cm and breadth is 9 cm, find the side of the square.
Solution:
Area of the square = Area of the rectangle = 16 × 9 = 144 cm2
Side of the square = √Area of the square = √144 = 12 cm
Hence, the side of square = 12 cm.
Question 8.
If the lengths of the diagonals of a rhombus are 16 cm and 12 cm, find its area.
Solution:
Given:
First diagonal d1 = 16 cm
Second diagonal d2= 12 cm
Hence, the required area = 96 cm2.
Question 9.
The area of a rhombus is 16 cm2. If the length of one diagonal is 4 cm, find the length of the other diagonal.
Solution:
Given: Area of the rhombus = 16 cm2
Length of one diagonal = 4 cm
Hence, the required length = 8 cm.
Question 10.
If the diagonals of a rhombus are 12 cm and 5 cm, find the perimeter of the rhombus.
Solution:
Given: d1 = 12 cm, d2 = 5 cm
The perimeter = 4 × side = 4 × 6.5 = 26 cm
Hence, the perimeter = 26 cm.
### Mensuration Class 8 Extra Questions Short Answer Type
Question 11.
The volume of a box is 13400 cm3. The area of its base is 670 cm2. Find the height of the box.
Solution:
Volume of the box = 13400 cm3
Area of the box = 670 cm2
Hence, the required height = 20 cm.
Question 12.
Complete the following table; measurement in centimetres.
Solution:
Question 13.
Two cubes are joined end to end. Find the volume of the resulting cuboid, if each side of the cubes is 6 cm.
Solution:
Length of the resulting cuboid = 6 + 6 = 12 cm
Height = 6 cm
Volume of the cuboid = l × b × h = 12 × 6 × 6 = 432 cm3
Question 14.
How many bricks each 25 cm by 15 cm by 8 cm, are required for a wall 32 m long, 3 m high and 40 cm thick?
Solution:
Converting into same units, we have,
Length of the wall = 32 m = 32 × 100 = 3200 cm
Breadth of the wall = 3 m = 3 × 100 = 300 cm
and the height = 40 cm
v, length of the brick = 25 cm
and height = 8 cm
Number of bricks required
Hence, the required number of bricks = 12800.
Question 15.
MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.
Solution:
Method I: Divide the given hexagon into two similar trapezia by joining QN.
Area of the hexagon MNOPQR = 2 × area of trapezium MNQR
= 2 × $$\frac { 1 }{ 2 }$$ (6 + 11) × 4
= 17 × 4
= 68 cm2
Method II: The hexagon MNOPQR is divided into three parts, 2 similar triangles and 1 rectangle by joining MO, RP.
Question 16.
The area of a trapezium is 400 cm2, the distance between the parallel sides is 16 cm. If one of the parallel sides is 20 cm, find the length of the other side.
Solution:
Given: Area of trapezium = 400 cm2
Height = 16 cm
Hence, the required length = 30 cm.
Question 17.
Find the area of the hexagon ABCDEF given below. Given that: AD = 8 cm, AJ = 6 cm, AI – 5 cm, AH = 3 cm, AG = 2.5 cm and FG, BH, EI and CJ are perpendiculars on diagonal AD from the vertices F, B, E and C respectively.
Solution:
Given:
FG = 3 cm
AJ = 6 cm
EI = 4 cm
AI = 5 cm
BH = 3 cm
AH = 3 cm
CJ = 2 cm
AG = 2.5 cm
Area of hexagon ABCDEF = Area of ΔAGF + Area of trapezium FGIE + Area of ΔEID + Area of ΔCJD + Area of trapezium HBCJ + Area of ΔAHB
= 3.75 cm2 + 8.75 cm2 + 6 cm2 + 2 cm2 + 7.5 cm2 + 4.5 cm2
= 32.50 cm2.
Question 18.
Three metal cubes of sides 6 cm, 8 cm and 10 cm are melted and recast into a big cube. Find its total surface area.
Solution:
Volume of the cube with side 6 cm = (side)3 = (6)3 = 216 cm3
Volume of the cube with side 8 cm = (side)3 = (8)3 = 512 cm3
Volume of the cube with side 10 cm = (side)3 = (10)3 = 1000 cm3
Volume of the big cube = 216 cm3 + 512 cm3 + 1000 cm3 = 1728 cm3
Side of the resulting cube = $$\sqrt [ 3 ]{ 1728 }$$ = 12 cm
Total surface area = 6 (side)2 = 6(12)2 = 6 × 144 cm2 = 864 cm2.
Question 19.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Solution:
Given: Diameter of the roller = 84 cm
Radius = $$\frac { 84 }{ 2 }$$ = 42 cm
Height = 120 cm
Curved surface area of the roller = 2πrh
Area covered by the roller in one complete revolution = 3.168 m2
Area covered in 500 complete revolutions = 500 × 3.168 = 1584 m2
Hence, the required area = 1584 m2.
Question 20.
A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder. Find its volume.
Solution:
Circumference of the base = 2πr
= 423.5 cm3
Hence, the required volume = 423.5 cm3.
Question 21.
160 m3 of water is to be used to irrigate a rectangular field whose area is 800 m2. What will be the height of the water level in the field? (NCERT Exemplar)
Solution:
Volume of water = 160 m3
Area of rectangular field = 800 m2
Let h be the height of water level in the field.
Now, the volume of water = volume of cuboid formed on the field by water.
160 = Area of base × height = 800 × h
⇒ h = 0.2
So, required height = 0.2 m
Question 22.
Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm. (NCERT Exemplar)
Solution:
Let ABCD be the rhombus as shown below.
DO = OB = 4 cm, since diagonals of a rhombus are perpendicular bisectors of each other.
Therefore, using Pythagoras theorem in ΔAOB, AO2 + OB2 = AB2
Question 23.
The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Solution:
Let ABCD be the trapezium such that
AB = 40 cm and CD 20 cm and AD = BC = 26 cm.
Then ALCD is a parallelogram.
So AL = CD = 20 cm
and CL = AD = 26 cm.
In ΔCLB, we have CL = CB = 26 cm
Therefore, ΔCLB is an isosceles triangle.
Draw altitude CM of ΔCLB.
Since ΔCLB is an isosceles triangle. So, CM is also the median.
Then LM = MB = $$\frac { 1 }{ 2 }$$ BL = $$\frac { 1 }{ 2 }$$ × 20 cm = 10 cm
[as BL = AB – AL = (40 – 20) cm = 20 cm].
Applying Pythagoras theorem in ΔCLM, we have
CL2 = CM2 + LM2
262 = CM2 + 102
CM2 = 262 – 102 = (26 – 10) (26 + 10) = 16 × 36 = 576
CM = √576 = 24 cm
Hence, the area of the trapezium = $$\frac { 1 }{ 2 }$$ (sum of parallel sides) × height
= $$\frac { 1 }{ 2 }$$ (20 + 40) × 24
= 30 × 24
= 720 cm2.
Question 24.
Find the area of polygon ABCDEF, if AD = 18 cm, AQ = 14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP, QC and BN are perpendiculars to diagonal AD. (NCERT Exemplar)
Solution:
In the figure
MP = AP – AM = (12 – 4) cm = 8 cm
PD = AD – AP = (18 – 12) cm = 6 cm
NQ = AQ – AN = (14 – 8) cm = 6 cm
QD = AD – AQ = (18 – 14) cm = 4 cm
Area of the polygon ABCDEF = area of ∆AFM + area of trapezium FMPE + area of ∆EPD + area of ∆ANB + area of trapezium NBCQ + area of ∆QCD.
## Related content
NCERT Exemplar for Class 6 Maths Solutions CBSE Notes for Class 8 Maths CBSE Notes for Class 7 Science CBSE Notes for Class 8 Science Lines and Angles Class 9 Extra Questions Maths Chapter 6 CBSE Notes for Class 7 Maths Class 7 CBSE Notes AMU Class 11 Entrance Exam Sample Papers CBSE Notes Class 4 Maths Harappan Civilization
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# 2007 AMC 12B Problems/Problem 24
Also refer to the 2007 AMC 10B #25 (same problem)
## Problem
How many pairs of positive integers $(a,b)$ are there such that $\text{gcd}(a,b)=1$ and $\frac{a}{b} + \frac{14b}{9a}$ is an integer?
$\mathrm {(A)}\ 4\quad\mathrm {(B)}\ 6\quad\mathrm {(C)}\ 9\quad\mathrm {(D)}\ 12\quad\mathrm {(E)}\ \text{infinitely many}$
## Solution 1
Combining the fraction, $\frac{9a^2 + 14b^2}{9ab}$ must be an integer.
Since the denominator contains a factor of $9$, $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$
Since $b = 3n$ for some positive integer $n$, we can rewrite the fraction(divide by $9$ on both top and bottom) as $\frac{a^2 + 14n^2}{3an}$
Since the denominator now contains a factor of $n$, we get $n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2$.
But since $1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)$, we must have $n=1$, and thus $b=3$.
For $b=3$ the original fraction simplifies to $\frac{a^2 + 14}{3a}$.
For that to be an integer, $a$ must be a factor of $14$, and therefore we must have $a\in\{1,2,7,14\}$. Each of these values does indeed yield an integer.
Thus there are four solutions: $(1,3)$, $(2,3)$, $(7,3)$, $(14,3)$ and the answer is $\mathrm{(A)}$
## Solution 2
Let's assume that $\frac{a}{b} + \frac{14b}{9a} = m$ We get
$9a^2 - 9mab + 14b^2 = 0$
Factoring this, we get $4$ equations-
$(3a-2b)(3a-7b) = 0$
$(3a-b)(3a-14b) = 0$
$(a-2b)(9a-7b) = 0$
$(a-b)(9a-14b) = 0$
(It's all negative, because if we had positive signs, $a$ would be the opposite sign of $b$)
Now we look at these, and see that-
$3a=2b$
$3a=b$
$3a=7b$
$3a=14b$
$a=2b$
$9a=7b$
$a=b$
$9a=14b$
This gives us $8$ solutions, but we note that the middle term needs to give you back $9m$.
For example, in the case
$(a-2b)(9a-7b)$, the middle term is $-25ab$, which is not equal by $-9m$ for any integer $m$.
Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total $\mathrm {(A)}$
## Solution 3
Let $u = \frac{a}{b}$. Then the given equation becomes $u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}$.
Let's set this equal to some value, $k \Rightarrow \frac{9u^2 + 14}{9u} = k$.
Clearing the denominator and simplifying, we get a quadratic in terms of $u$:
$9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}$
Since $a$ and $b$ are integers, $u$ is a rational number. This means that $\sqrt{(9k)^2 - 504}$ is an integer.
Let $\sqrt{(9k)^2 - 504} = x$. Squaring and rearranging yields:
$(9k)^2 - x^2 = 504$
$(9k+x)(9k-x) = 504$.
In order for both $x$ and $a$ to be an integer, $9k + x$ and $9k - x$ must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let $9k + x = 2m$ and $9k - x = 2n$.
Then:
$2m \cdot 2n = 504$
$mn = 126$.
Factoring 126, we get $6$ pairs of numbers: $(1,126), (2,63), (3,42), (6,21), (7,18),$ and $(9,14)$.
Looking back at our equations for $m$ and $n$, we can solve for $k = \frac{2m + 2n}{18} = \frac{m+n}{9}$. Since $k$ is an integer, there are only $2$ pairs of $(m,n)$ that work: $(3,42)$ and $(6,21)$. This means that there are $2$ values of $k$ such that $u$ is an integer. But looking back at $u$ in terms of $k$, we have $\pm$, meaning that there are $2$ values of $u$ for every $k$. Thus, the answer is $2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}$.
## Solution 4
Rewriting the expression over a common denominator yields $\frac{9a^2 + 14b^2}{9ab}$. This expression must be equal to some integer $m$.
Thus, $\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm$. Taking this $\pmod{a}$ yields $14b^2 \equiv 0\pmod{a}$. Since $\gcd(a,b)=1$, $14 \equiv 0\pmod{a}$. This implies that $a|14$ so $a = 1, 2, 7, 14$.
We can then take $9a^2 + 14b^2 = 9abm \pmod{b}$ to get that $9 \equiv 0 \pmod{b}$. Thus $b = 1, 3, 9$.
However, taking $9a^2 + 14b^2 = 9abm \pmod{3}$, $b^2 \equiv 0\pmod{3}$ so $b$ cannot equal 1.
Also, note that if $b = 9$, $\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}$. Since $a|14$, $\frac{14}{a}$ will be an integer, but $\frac{a}{9}$ will not be an integer since none of the possible values of $a$ are multiples of 9. Thus, $b$ cannot equal 9.
Thus, the only possible values of $b$ is 3, and $a$ can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is $\mathrm{(A)}$.
## Solution 5 (Similar to Solution 1)
Rewriting $\frac{a}{b} + \frac{14b}{9a}$ over a common denominator gives $\frac{9a^2 + 14b^2}{9ab}.$
Thus, we have $9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.$
Next, we have $ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.$
Thus, $a \in (1,2,7,4).$
Next, we have $b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.$
Thus, $b \in (1,3,9).$
Now, we simply do casework on $b.$
Plugging in $b = 1,3$ and $9$ gives that there are $4$ total solutions for $(a,b).$
~coolmath2017
### Solution 6 (Similar to solution 3)
Let $\frac{a}{b} = r.$ So $r + \frac{14}{9r}=I$, where I is an integer. Algebraic manipulations yield: $r^2-Ir+\frac{14}{9}=0$. The discriminant of this must be the square of a rational number, call this R. So $I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}$. I is $\frac{1}{2}$ the sum of $I-R$ and $I+R$. To have an integer sum, $I-R$ and $I+R$ must have the same denominator, namely 3. We proceed with casework.
Case 1. $I+R=56/3$, $I-R=1/3$. This yields $I=19/2$, which is not an integer. This case produces 0 solutions.
Case 2. $I+R=28/3$, $I-R=2/3$. This yields $I=5$. Substituting into our original equation yields: $r^2-5r+\frac{14}{9}=0$. Factoring gives: $r=\frac{1}{3}$, $r=\frac{14}{3}$. This case produces 2 solutions, namely (1,3) and (14,3).
Case 3. $I+R=14/3$, $I-R=4/3$. This yields $I=3$. Substituting into our original equation yields: $r^2-3r+\frac{14}{9}=0$. Factoring gives: $r=\frac{2}{3}$, $r=\frac{7}{3}$. This case produces 2 solutions, namely (2,3) and (7,3).
Case 4. $I+R=8/3$, $I-R=7/3$. This yields $I=5/2$, which is not an integer. This case produces 0 solutions.
Altogether, we have 4 solutions, so our answer is $\boxed{(A)}$.
~Math4Life2020
## Solution 7
Rewrite the equation$$\frac{a}{b}+\frac{14b}{9a}=k$$in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain$$\frac{14b^2}{9a}=bk-a.$$Because $a$, $b$, and $k$ are integers, $14b^2$ must be a multiple of $a$, and because $a$ and $b$ have no common factors greater than 1, it follows that 14 is divisible by $a$. Next, multiply both sides of the original equation by $9a$ and subtract $14b$ to obtain$$\frac{9a^2}{b}=9ak-14b.$$This shows that $9a^2$ is a multiple of $b$, so 9 must be divisible by $b$. Thus if $(a,b)$ is a solution, then $b=1$, $3$, or $9$, and $a=1$, 2, 7, or 14. This gives a total of twelve possible solutions $(a,b)$, each of which can be checked quickly.$$\frac{a}{b}+\frac{14b}{9a}$$ (Error compiling LaTeX. Unknown error_msg) will only be an integer when $(a,b) \in \{(1,3),(2,3), (7,3), (14,3)\},$ for a total of $\boxed{4}$ pairs. |
# MTH 252 Final Lab Bonus 3
Justin Drawbert August 5, 2010
In our Final Lab Bonus question 3, we are asked to consider the triangle formed by the x1 axis, y-axis, and the tangent line to the curve y = 2x at x = a. (image) We are then asked to find an expression for the area, A, of the triangle in terms of a and to determine what happens as a → ∞. We start by finding the derivative to our original function. y= 1 , 2x y′ = − 1 2x2
We then use the following form to find the equation for the line tangent to our curve at point a. y − f (a) = f ′ (a)(x − a) By this we get our equation for our line y− 1 1 = − 2 (x − a) 2a 2a 1 1 1 y− = − 2x + 2a 2a 2a 1 1 y=− 2 + 2a a
So we now have our equation for our line. We can see that the triangle formed by the tangent line is a right triangle so we know the equation for the area is simply 1 b·h 2 Where b is the base, or in this case our x-intercept minus 0 and h is the height, or our y-intercept minus 0. We solve for our x-intercept by setting the equation of the line equal to 0 and then solving for x. 0=− 1 1 x+ 2 2a a
1 1 x= 2 2a a 1 x = · 2a2 a x = 2a We then find our y-intercept by letting x = 0 y=− 1 1 1 (0) + = 2a2 a a 1
1 1 · 2a · = 1 2 a This is interesting, or perhaps completely the opposite. What this says is that no matter what, for 1 any number a, the area of the triangle formed by the line tangent to y = 2x , at the point x = a, will always be 1. There’s no need to even evaluate a limit here. Because before you could, the a’s would cancel, and you’d be left evaluating the limit of a constant, which in our case happens to be 1. And the limit as any variable approaches anything of the constant 1 is 1. What is interesting, is I did play around with some other functions, to make sure this wasn’t a weird case, and found that the equation for the area of the triangle formed by the slope of a line, (in the cases I looked at) turns out to be an equation who’s degree is 1 degree higher than the original function, which sorta makes sense.
So the formula for the the area A, is
2 |
# How do you integrate ((x^2-1)/sqrt(2x-1) )dx?
May 28, 2018
$I = \frac{1}{60} \left[3 {\left(\sqrt{2 x - 1}\right)}^{5} + 10 {\left(\sqrt{2 x - 1}\right)}^{3} - 45 \sqrt{2 x - 1}\right] + c$
#### Explanation:
Here,
$I = \int \left(\frac{{x}^{2} - 1}{\sqrt{2 x - 1}}\right) \mathrm{dx}$
Subst, $\sqrt{2 x - 1} = u \implies 2 x - 1 = {u}^{2} \implies 2 x = {u}^{2} + 1$
$\implies x = \frac{1}{2} \left({u}^{2} + 1\right) \implies \mathrm{dx} = \frac{1}{2} \left(2 u\right) \mathrm{du} = u \mathrm{du}$
So,
$I = \int \frac{\frac{1}{4} {\left({u}^{2} + 1\right)}^{2} - 1}{u} \times u \mathrm{du}$
$I = \int \left[\frac{1}{4} \left({u}^{4} + 2 {u}^{2} + 1\right) - 1\right] \mathrm{du}$
$= \frac{1}{4} \int \left[{u}^{4} + 2 {u}^{2} + 1 - 4\right] \mathrm{du}$
$= \frac{1}{4} \int \left[{u}^{4} + 2 {u}^{2} - 3\right] \mathrm{du}$
$= \frac{1}{4} \left[{u}^{5} / 5 + 2 {u}^{3} / 3 - 3 u\right] + c$
$= \frac{1}{4} \times \frac{1}{15} \left[3 {u}^{5} + 10 {u}^{3} - 45 u\right] + c$
$= \frac{1}{60} \left[3 {\left(u\right)}^{5} + 10 {\left(u\right)}^{3} - 45 u\right] + c$
Subst, back , $u = \sqrt{2 x - 1}$
$I = \frac{1}{60} \left[3 {\left(\sqrt{2 x - 1}\right)}^{5} + 10 {\left(\sqrt{2 x - 1}\right)}^{3} - 45 \sqrt{2 x - 1}\right] + c$ |
Top
Fractions as we know are used to denote some part out of the whole one. And it’s very necessary, to know how to apply the 4 mathematical operations (addition, subtraction, multiplication and division) on fractions.
We will first consider the addition of fractions, which will form the base for applying the other operations on fractions.
• The procedure to add two fractions are, firstly, to check whether the fractions are in mixed form or not.
• If they are mixed fraction, we will convert them into simple fractions by multiplying the denominator of the mixed fraction by its initial number and then adding the numerator to that multiplied number.
• Secondly, if all the fractions are in simple form, the next thing to check is their denominators. Either denominators can be alike or they could be different.
• When denominators are same, to add fractions, the resulting sum of the fractions will have the same denominator and the numerator will consist of the sum of the individual numerator of each fraction.
• If denominators of fractions are different, the first step would be to find the least common denominator (LCD) of the fractions, secondly, changing each fraction by multiplying its numerator and denominator by the same concerned multiple to get that LCD.
• Once, we will get the same denominators of each fraction, we will follow the same procedure mentioned above, where the sum of fractions, will be the same denominator (LCD) and the resulting numerator will be the sum of individual numerators of each fraction.
Solved examples: -
1.Add $\frac{2}{3}+\frac{8}{3}$
To add fractions with common denominators, we simply add their numerators and put it over the common denominator
We get $\frac{2+8}{3}= \frac{10}{3}$.
2.Add $\frac{5}{4}+\frac{25}{20}$
$\frac{25}{20}$ can be simplified to $\frac{5}{4}$ by dividing both the numerator and the denominator by 5.
Therefore we get the equation as $\frac{5}{4}$ + $\frac{5}{4}$ and this is equal to $\frac{5+5}{4}$ = $\frac{10}{4}$.
Now we can divide both the numerator and the denominator by 2, which makes the fraction as $\frac{5}{2}$.
3.Add $\frac{7}{5}+\frac{1}{2}$
This is the case of fractions with different denominators
First find the least common denominator. Here we have 2 and 5, so the least common denominator will be 10.
Now we multiply the numerator and denominator of $\frac{7}{5}$ by 2 to get 10 in the denominator, we get
$\frac{7}{5} \times \frac{2}{2}$ = $\frac{14}{10}$.
Similarly with $\frac{1}{2}$, we multiply both 1 and 2 by 5 to get 10 in the bottom of the fraction.
$\frac{1}{2} \times \frac{5}{5}$ = $\frac{5}{10}$
Thus it will be $\frac{14}{10}+\frac{5}{10}$ which can be added to give $\frac{19}{10}$. |
# A triangle has corners A, B, and C located at (4 ,5 ), (3 ,6 ), and (8 ,4 ), respectively. What are the endpoints and length of the altitude going through corner C?
Aug 14, 2018
$N \left(\frac{13}{2} , \frac{5}{2}\right) \mathmr{and} C \left(8 , 4\right)$ are the endpoints of altitudes
going through corner $C$
The length of the altitude $= \therefore C N = \frac{3}{\sqrt{2}} \approx 2.12$
#### Explanation:
Let $\triangle A B C \text{ be the triangle with corners at}$
$A \left(4 , 5\right) , B \left(3 , 6\right) \mathmr{and} C \left(8 , 4\right)$
Let $\overline{A L} , \overline{B M} \mathmr{and} \overline{C N}$ be the altitudes of sides $\overline{B C} , \overline{A C} \mathmr{and} \overline{A B}$ respectively.
Let $\left(x , y\right)$ be the intersection of three altitudes
Slope of $\overline{A B} = \frac{5 - 6}{4 - 3} = - \frac{1}{1} = - 1$
$\overline{A B} \bot \overline{C N} \implies$slope of $\overline{C N} = 1$ ,
$\overline{C N}$ passes through $C \left(8 , 4\right)$
$\therefore$The equn. of $\overline{C N}$ is $: y - 4 = 1 \left(x - 8\right)$
$\implies y - 4 = x - 8$
$\implies y = x - 8 + 4$
i.e. color(red)(y=x-4.....to (1)
Now, Slope of $\overline{A B} = - 1$ and $\overline{A B}$ passes through
$A \left(4 , 5\right)$
So, eqn. of $\overline{A B}$ is: $y - 5 = - 1 \left(x - 4\right)$
$\implies y - 5 = - x + 4$
=>color(red)(y=9-x...to(2)
from $\left(1\right) \mathmr{and} \left(2\right)$
$x - 4 = 9 - x$
=>x+x=9+4=>2x=13=>color(blue)(x=13/2=6.5
From $\left(1\right) ,$
y=13/2-4=>color(blue)(x=5/2=2.5
$\implies N \left(\frac{13}{2} , \frac{5}{2}\right) \mathmr{and} C \left(8 , 4\right)$ are the endpoints of altitudes
going through corner $C$
Using Distance formula,
$C N = \sqrt{{\left(\frac{13}{2} - 8\right)}^{2} + {\left(\frac{5}{2} - 4\right)}^{2}} = \sqrt{\frac{9}{4} + \frac{9}{4}}$
$\therefore C N = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}}$
$\therefore C N = \frac{3}{\sqrt{2}} \approx 2.12$ |
# Using Abstract Algebra To Understand Basic Combinatorics
Pascal’s triangle has many fascinating properties. One of them is for any given prime number p, the number of k-element subsets (0< k < p) of a p-element set is divisible by p:
$$p | \binom{p}{k}\qquad \qquad \qquad (*)$$
You can think for a while how would you prove it. There might well be an elementary combinatorial proof of $$(*)$$, although I suspect that it would be rather long and tedious. Fortunately there exists a simple proof using a bit of algebra!:
If $$k=1$$ or $$k=p-1$$, then $$\binom{p}{k}$$ equals $$p$$, so the task is trivial. Suppose now that 1<k<p-1 we will use the following simple facts about a group G acting on a set X:
Fact 1. X is the disjoint union of the orbits of the action.
Fact 2. The size of an orbit divides the size of the group.
Once you know what is an action, the above facts are almost immediate. To prove $$(*)$$, consider the multiplicative cyclic group of size $$p-1$$,
$$G=\mathbb{F}_p^*=\langle\{1,\dots,p-1\},\cdot\rangle$$
where $$m\cdot n$$ is calculated mod p.
Let X be the set of all k-element subsets of $$\{0,\dots,p-1\}$$ and let G act on X by $$gA=\{ga\mid a\in A\}.$$ Let us show that the orbit of any $$A\in X$$ contains more than one element. Suppose $$A\in X$$. Since 1< k < p-1, there are non-zero elements $$m\in A$$ and $$n\in \{0,\dots, p-1\}\setminus A$$. Moreover since G is a group, there exists $$g\in G$$ such that $$g\cdot m=n$$ if you think of m and n as elements of G. But this is the same as saying $$n\in gA$$, so the orbit of A contains some set that is not A, i.e. we proved what we wanted.
By Fact 2, the orbit has exactly size p and so Fact 1 implies that |X| is divisible by p.
For details see this
This entry was posted in Algebra, Combinatorics, Mathematics and tagged . Bookmark the permalink.
### 5 Responses to Using Abstract Algebra To Understand Basic Combinatorics
1. janka says:
Why p divides all the numbers in one line in Pascal triangle?
I would look at the expression of (p over k) as the ratio of three factorial expressions. Obviously, p divides the numerator p!. In the denominator there is a product of k! and (p-k)! and none of the factors nor their product is divisible by p because they are produced from small numbers (< p) and p is prime.
• of course.. you are right. But let me like the algebraic proof still :P
• In the algebraic way you do not need to prove that (p over k)=p!/((p-k)!(k!))
2. janka says:
In the “divisibiity proof” you do not need to know what a group action is :-)
As for me, to prove that (p over k) = p!/k!(p-k)! is not harder
than to explain what a group action is.
3. janka says:
but of course, you may like the algebraic proof.
There’s no accounting for taste. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
## Categories of quadrilaterals based on sides and angles.
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What if you were given a quadrilateral in the coordinate plane? How could you determine if that quadrilateral qualifies as one of the special quadrilaterals: parallelograms, squares, rectangles, rhombuses, kites, or trapezoids? After completing this Concept, you'll be able to make such a determination.
### Guidance
When working in the coordinate plane, you will sometimes want to know what type of shape a given shape is. You should easily be able to tell that it is a quadrilateral if it has four sides. But how can you classify it beyond that?
First you should graph the shape if it has not already been graphed. Look at it and see if it looks like any special quadrilateral. Do the sides appear to be congruent? Do they meet at right angles? This will give you a place to start.
Once you have a guess for what type of quadrilateral it is, your job is to prove your guess. To prove that a quadrilateral is a parallelogram, rectangle, rhombus, square, kite or trapezoid, you must show that it meets the definition of that shape OR that it has properties that only that shape has.
If it turns out that your guess was wrong because the shape does not fulfill the necessary properties, you can guess again. If it appears to be no type of special quadrilateral then it is simply a quadrilateral.
The examples below will help you to see what this process might look like.
#### Example A
Determine what type of parallelogram TUNE\begin{align*}TUNE\end{align*} is: T(0,10),U(4,2),N(2,1)\begin{align*}T(0, 10), U(4, 2), N(-2, -1)\end{align*}, and E(6,7)\begin{align*}E(-6, 7)\end{align*}.
This looks like a rectangle. Let’s see if the diagonals are equal. If they are, then TUNE\begin{align*}TUNE\end{align*} is a rectangle.
EU=(64)2+(72)2=(10)2+52=100+25=125TN=(0+2)2+(10+1)2 =22+112 =4+121 =125
If the diagonals are also perpendicular, then \begin{align*}TUNE\end{align*} is a square.
\begin{align*}\text{Slope of}\ EU = \frac{7 - 2}{-6 - 4} = -\frac{5}{10} = -\frac{1}{2} \quad \text{Slope of}\ TN = \frac{10 - (-1)}{0-(-2)} = \frac{11}{2}\end{align*}
The slope of \begin{align*}EU \neq\end{align*} slope of \begin{align*}TN\end{align*}, so \begin{align*}TUNE\end{align*} is a rectangle.
#### Example B
A quadrilateral is defined by the four lines \begin{align*}y=2x+1\end{align*}, \begin{align*}y=-x+5\end{align*}, \begin{align*}y=2x-4\end{align*}, and \begin{align*}y=-x-5\end{align*}. Is this quadrilateral a parallelogram?
To check if its a parallelogram we have to check that it has two pairs of parallel sides. From the equations we can see that the slopes of the lines are \begin{align*}2\end{align*}, \begin{align*}-1\end{align*}, \begin{align*}2\end{align*} and \begin{align*}-1\end{align*}. Because two pairs of slopes match, this shape has two pairs of parallel sides and is a parallelogram.
#### Example C
Determine what type of quadrilateral \begin{align*}RSTV\end{align*} is. Simplify all radicals.
There are two directions you could take here. First, you could determine if the diagonals bisect each other. If they do, then it is a parallelogram. Or, you could find the lengths of all the sides. Let’s do this option.
From this we see that the adjacent sides are congruent. Therefore, \begin{align*}RSTV\end{align*} is a kite.
Algebra Review: When asked to “simplify the radical,” pull all square numbers (1, 4, 9, 16, 25, ...) out of the radical. Above \begin{align*}\sqrt{50}=\sqrt{25 \cdot 2}\end{align*}. We know \begin{align*}\sqrt{25}=5\end{align*}, so \begin{align*}\sqrt{50}=\sqrt{25 \cdot 2}=5\sqrt{2}\end{align*}.
Watch this video for help with the Examples above.
#### Example D
Is the quadrilateral \begin{align*}ABCD\end{align*} a parallelogram?
We have determined there are four different ways to show a quadrilateral is a parallelogram in the \begin{align*}x-y\end{align*} plane. Let's check if a pair of opposite sides are congruent and parallel. First, find the length of \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.
\begin{align*}AB = CD\end{align*}, so if the two lines have the same slope, \begin{align*}ABCD\end{align*} is a parallelogram.
Slope \begin{align*}AB = \frac{5-3}{-1-3}=\frac{2}{-4}=-\frac{1}{2}\end{align*} Slope \begin{align*}CD = \frac{-2+4}{2-6}=\frac{2}{-4}=-\frac{1}{2}\end{align*}
Therefore, \begin{align*}ABCD\end{align*} is a parallelogram.
### Guided Practice
1. A quadrilateral is defined by the four lines \begin{align*}y=2x+1\end{align*}, \begin{align*}y=-2x+5\end{align*}, \begin{align*}y=2x-4\end{align*}, and \begin{align*}y=-2x-5\end{align*}. Is this quadrilateral a rectangle?
2. Determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. \begin{align*}A(-3, 3), \ B(1, 5), \ C(4, -1), \ D(1, -5)\end{align*}. Simplify all radicals.
3. Determine what type of quadrilateral \begin{align*}EFGH\end{align*} is. \begin{align*}E(5, -1), F(11, -3), G(5, -5), H(-1, -3)\end{align*}
1. To be a rectangle a shape must have four right angles. This means that the sides must be perpendicular to each other. From the given equations we see that the slopes are \begin{align*}2\end{align*}, \begin{align*}-2\end{align*}, \begin{align*}2\end{align*} and \begin{align*}-2\end{align*}. Because the slopes are not opposite reciprocals of each other, the sides are not perpendicular, and the shape is not a rectangle.
2. First, graph \begin{align*}ABCD\end{align*}. This will make it easier to figure out what type of quadrilateral it is. From the graph, we can tell this is not a parallelogram. Find the slopes of \begin{align*}\overline{BC}\end{align*} and \begin{align*}\overline{AD}\end{align*} to see if they are parallel.
Slope of \begin{align*}\overline{BC}=\frac{5-(-1)}{1-4}=\frac{6}{-3}=-2\end{align*}
Slope of \begin{align*}\overline{AD}=\frac{3-(-5)}{-3-1}=\frac{8}{-4}=-2\end{align*}
We now know \begin{align*}\overline{BC} \ || \ \overline{AD}\end{align*}. This is a trapezoid. To determine if it is an isosceles trapezoid, find \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.
\begin{align*}AB \neq CD\end{align*}, therefore this is only a trapezoid.
3. We will not graph this example. Let’s find the length of all four sides.
All four sides are equal. That means, this quadrilateral is either a rhombus or a square. The difference between the two is that a square has four \begin{align*}90^\circ\end{align*} angles and congruent diagonals. Let’s find the length of the diagonals.
The diagonals are not congruent, so \begin{align*}EFGH\end{align*} is a rhombus.
### Explore More
1. If a quadrilateral has exactly one pair of parallel sides, what type of quadrilateral is it?
2. If a quadrilateral has two pairs of parallel sides and one right angle, what type of quadrilateral is it?
3. If a quadrilateral has perpendicular diagonals, what type of quadrilateral is it?
4. If a quadrilateral has diagonals that are perpendicular and congruent, what type of quadrilateral is it?
5. If a quadrilateral has four congruent sides and one right angle, what type of quadrilateral is it?
Determine what type of quadrilateral \begin{align*}ABCD\end{align*} is.
1. \begin{align*}A(-2, 4), B(-1, 2), C(-3, 1), D(-4, 3)\end{align*}
2. \begin{align*}A(-2, 3), B(3, 4), C(2, -1), D(-3, -2)\end{align*}
3. \begin{align*}A(1, -1), B(7, 1), C(8, -2), D(2, -4)\end{align*}
4. \begin{align*}A(10, 4), B(8, -2), C(2, 2), D(4, 8)\end{align*}
5. \begin{align*}A(0, 0), B(5, 0), C(0, 4), D(5, 4)\end{align*}
6. \begin{align*}A(-1, 0), B(0, 1), C(1, 0), D(0, -1)\end{align*}
7. \begin{align*}A(2, 0), B(3, 5), C(5, 0), D(6, 5)\end{align*}
\begin{align*}SRUE\end{align*} is a rectangle and \begin{align*}PRUC\end{align*} is a square.
1. What type of quadrilateral is \begin{align*}SPCE\end{align*}?
2. If \begin{align*}SR = 20\end{align*} and \begin{align*}RU = 12\end{align*}, find \begin{align*}CE\end{align*}.
3. Find \begin{align*}SC\end{align*} and \begin{align*}RC\end{align*} based on the information from part b. Round your answers to the nearest hundredth.
### Vocabulary Language: English
Kite
Kite
Parallelogram
Parallelogram
A parallelogram is a quadrilateral with two pairs of parallel sides.
A quadrilateral is a closed figure with four sides and four vertices.
Rectangle
Rectangle
A rectangle is a quadrilateral with four right angles.
Rhombus
Rhombus
A rhombus is a quadrilateral with four congruent sides.
Trapezoid
Trapezoid
A trapezoid is a quadrilateral with exactly one pair of parallel opposite sides. |
# Algebra II
41 %
59 %
Education
Published on June 15, 2007
Author: Woodwork
Source: authorstream.com
Algebra II: Algebra II By Monica Yuskaitis Definitions: Definitions Equation – A mathematical sentence stating that 2 expressions are equal. 12 – 3 = 9 8 + 4 = 12 Definitions: Definitions Equation – A mathematical sentence with an equals sign. 16 – 5 = 11 14 + 3 = 17 Definitions: Definitions Equals Sign (=) Means that the amount is the same on both sides. 4 + 2 = 6 5 – 2 = 3 An Equation is like a balance scale. Everything must be equal on both sides.: An Equation is like a balance scale. Everything must be equal on both sides. 10 5 + 5 = When the amounts are equal on both sides it is a true equation.: When the amounts are equal on both sides it is a true equation. 12 6 + 6 = When the amounts are unequal on both sides it is a false equation.: When the amounts are unequal on both sides it is a false equation. 8 2 + 2 = When an amount is unknown on one side of the equation it is an open equation.: When an amount is unknown on one side of the equation it is an open equation. 7 n + 2 = When you find a number for n you change the open equation to a true equation. You solve the equation.: When you find a number for n you change the open equation to a true equation. You solve the equation. 7 n + 2 = 5 Are these equations true, false or open?: Are these equations true, false or open? 11 - 3 = 5 13 + 4 = 17 N + 4 = 7 12 – 3 = 8 3 + v = 13 15 – 6 = 9 false true open false open true Definitions: Definitions Inverse operation – the opposite operation used to undo the first. 4 + 3 = 7 7 – 4 = 3 6 x 6 = 36 36 / 6 = 6 How to solve an addition equation: How to solve an addition equation Use the inverse operation for addition which is subtraction m + 8 = 12 12 - 8 = 4 m = 4 4 + 8 = 12 How to solve a subtraction equation: How to solve a subtraction equation Use the inverse operation for subtraction which is addition m - 3 = 5 5 + 3 = 8 m = 8 8 - 3 = 5 Solve these equations using the inverse operations: Solve these equations using the inverse operations n + 4 = 7 n – 5 = 4 n + 4 = 17 n – 6 = 13 n + 7 = 15 n – 8 = 17 3 9 13 19 8 9 Commutative Property: Commutative Property 5 + 4 = 9 4 + 5 = 9 a + b = c b + a = c 6 + 3 = 9 3 + 6 = 9 x+ y = z y + x = z 3 + 4 + 1 = 8 1 + 3 + 4 = 8 Solve these equations using the commutative property: Solve these equations using the commutative property n + 7 = 7 + 4 m + 2 = 2 + 5 z + 3 = 3 + 9 g + 6 = 6 + 11 s + 4 = 4 + 20 c + 8 = 8 + 32 n = 4 m = 5 z = 9 g = 11 s = 20 c = 32 The Identity Property of Addition: The Identity Property of Addition 7 + 0 = 7 a + 0 = a 8 + 0 = 8 c + 0 = c 2 + 0 = 2 Use the Identity Property of addition to solve these problems: Use the Identity Property of addition to solve these problems n + 0 = 8 b + 0 = 7 m + 0 = 3 v + 0 = 5 w + 0 = 4 r + 0 = 2 n = 8 b = 7 m = 3 v = 5 w = 4 r = 2 Subtraction Rules of zero: Subtraction Rules of zero 7 – 7 = 0 n – n = 0 4 – 0 = 4 n – 0 = n Find the value of n using the rules of subtraction: Find the value of n using the rules of subtraction n - 8 = 0 n – 9 = 0 n – 0 = 7 n – 0 = 9 n – 7 = 0 n – 0 = 5 n = 8 n = 9 n = 7 n = 9 n = 7 n = 5 Write an equation for these problems using a variable: Write an equation for these problems using a variable Timothy got 72 right on his timed test in July. He got 99 right on this same test in November. Jasmin runs 15 minutes before school and 30 minutes after school. One zinger costs 25 cents. Issak bought 4. 72 + n = 99 or 99 – 72 = n 15 + 30 = n 4 x 25 = n
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# Learning Division on YouTube and with Spirit Lake 1
Last time we got started on the basic of division and practiced division of one two-digit numbers, like 28 ÷ 7. Now we’re going to move on to long division. All of these videos can be found on our youtube play list on, you guessed it, division. Watching all of the videos together takes less than five minutes. After your
## What’s the first step in long division?
This video explains the first step in long division, which is to guess, and walks through step by step solving a problem of dividing a three-digit number by a two-digit number. A hundred years ago when I was in elementary school, they called this method “guess and check”.
## Division with remainders
The next video explains and gives an example of division with remainders. As you might have guessed by now, we’re a big believer that students need examples and also that having information presented both verbally and visually helps.
## Another long division example
Here is another video which uses guessing, again dividing a three-digit number by a two-digit number. Why do we have two videos? Because it’s helpful to go over a concept or skill more than once.
The last video uses division in word problems. The first problem is a simple one, dividing a two-digit number by a one-digit number. The others are dividing three-digit numbers by two-digit numbers. You may want to use a calculator for this.
## Spirit Lake: The Game teaches multiplication and division.
After watching these videos, you may want to have your child play that game. While U.S. schools are out, we are making Spirit Lake available free. You can click the link below to download it for Mac and Windows computers. As an added bonus, the game story line is based on the history of the Spirit Lake Dakota, so your child can learn social studies while learning math.
If you don’t have a computer you can spare for school work or you are just think worksheets aren’t a bad idea , you can find a link to several sources for free worksheets here. Be aware that many of these do require the worksheet to be printed, so you would need access to a printer.
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# Free Decimals Subjective Test 01 Practice Test - 6th grade
Explain the expression "a tenth". How do we represent a tenth using decimals? [1 MARK]
#### SOLUTION
Solution :
If a block of one unit is divided into 10 equal parts, then each part is 110 (one-tenth) of the unit. It is written as 0.1 in decimal representation. The dot denotes the decimal point.
For example, a tenth of 5 = 5×(110) = 0.5.
Convert the given fractions into decimals and mention the place values of the digits of each number. [2 MARKS]
1. 7100
2. 91000
#### SOLUTION
Solution :
Fraction to decimal: 1 Mark
Place values: 1 Mark
1. 1100 = 0.07
Place values:
OnesTenthsHundredths7
2. 91000 = 0.009
Place values:
OnesTenthsHundredthsThousandths9
Represent the following numbers on a number line: [2 MARKS]
1. 1.5
2. 2.35
#### SOLUTION
Solution :
Number line: 1 Mark each
1.
2.
Change the units of the following measures into the measures given below. [3 MARKS]
MeasureChange fromChange to7KilogramsGrams274MillimetersCentimeters145PaiseRupees
#### SOLUTION
Solution :
Measures: 1 Mark each
1. 7 Kilograms = 7000 Grams [ as 1 kilogram=1000 grams]
2. 274 Millimeters = 27.4 Centimeters [ 1 cm = 10 mm]
3. 145 Paise = 1.45 Rupees [ 1 rupee = 100 Paise ]
Arrange the following sets of decimals according to their ascending order. [3 MARKS]
1) 0.6, 0.4, 5.4, 3.2, 1.5
2) 0.7, 1.8, 1.09, 2.1, 2.02
3) 1.5, 2.1, 0.21, 2.03, 1.35
#### SOLUTION
Solution :
Ascending order: 1 Mark each
1) 0.4 < 0.6 < 1.5 < 3.2 < 5.4
2) 0.7 < 1.09 < 1.8 < 2.02 < 2.1
3) 0.21 < 1.35 < 1.5 < 2.03 < 2.1
Add up the following decimals: [3 MARKS]
1) 2.1 + 3.7 + 1.2
2) 2.02 + 1.80
3) 1.03+ 7.002
#### SOLUTION
Solution :
Sum: 1 Mark each
1) 2.1 + 3.7 + 1.2 = 7
2) 2.02 + 1.80 = 3.82
3) 1.03+ 7.002 = 8.032
Perform the following arithmetic operations. [4 MARKS]
1) 26.5 - 3.14 + 9.47 - 2.1 = ?
2) (5.3 × 2.1) - (6.2 × 1.4) = ?
3) (0.001×100) + (89.3 + 10.6) = ?
4) 4.05 + 5.13 - 9 = ?
#### SOLUTION
Solution : Operations: 1 Mark each
1) 26.5 - 3.14 + 9.47 - 2.1 = 30.73
2) (5.3 × 2.1) - (6.2 × 1.4) = 2.45
3) (0.001×100) + (89.3 + 10.6) = 100
4) 4.07 + 5.13 - 9.2 = 0
Find the product of the following pairs of numbers: [4 MARKS]
1) 3.6 × 2.1 = ?
2) 4.2 × 1.4 = ?
3) 2 × 7.1 = ?
4) 2.3 × 3.22 = ?
#### SOLUTION
Solution :
Product: 1 Mark each
1) 3.6 × 2.1 = 7.56
2) 4.2 × 1.4 = 5.88
3) 2 × 7.1 = 14.2
4) 2.3 × 3.22 = 7.406
Find the difference between the following numbers and represent it in decimals: [4 MARKS]
1) 4.5 - 2.3 = ?
2) 5 - 2.5 = ?
3) 6.2 - 3.7 = ?
4) 5.6 - 4.7 = ?
#### SOLUTION
Solution :
Differences: 1 Mark each
1) 4.5 - 2.3 = 2.2
2) 5 - 2.5 = 2.5
3) 6.2 - 3.7 = 2.5
4) 5.6 - 4.7 = 0.9
Perform the following arithmetic operations and express the answer in decimals: [4 MARKS]
i) 4102100 = ?
ii) 7100 + 310 = ?
iii) 210
x 52100 = ?
iv) 412100
x 7241000 = ?
#### SOLUTION
Solution :
Problems: 1 Mark each
i) 4102100 = 401002100 = 23100=0.38
ii) 7100 + 310 =7100 + 30100=37100=0.37
iii) 210 x52100 =1041000= 0.104
iv) 412100
x 7241000 = 298288100000=2.98288 |
# How to Calculate the Percentage of Something
Percentage calculations come into play in many real-world situations, such as determining how much to tip in a restaurant and figuring out the discounted prices of clothing. Here’s how to do these calculations.
Know the Basics
Before going gung-ho on performing calculations, know the basics about percentages. The word “percent” means “out of 100” or “for every 100.” The sentence “It rained 14 percent of the time” is much less cumbersome than the statement “It rained 14 days out of every 100,” but it means essentially the same thing. In math equations, the % symbol is used to represent percentages. To write a percentage as a decimal, move the decimal point two spaces to the left. To write a decimal as percentage, move the decimal point two spaces to the right. Further, to convert a fraction to a percentage, simply multiply by 100; to convert a fraction to a decimal, divide by 100 and then reduce the fraction, if necessary, according to Helping With Math. Read on for some examples of how percentage calculations can come in handy in everyday life.
Calculating Percentages Based on Ratios
Ratios with denominators other than 100 are often used in calculating percentages. Imagine that 14 out of 50 students in a class received an “A” on a test. What percentage got an “A”? To find the percentage represented by the ratio, divide the numerator by the denominator, then multiply by 100. Because the ratio is 14 out of 50, the percentage is calculated by dividing 14 by 50 to get 0.28, and then multiplying 0.28 by 100 to get 28 percent. In this example, 28 percent of the 50 students got an “A.”
Working Backwards from a Percentage
Working backwards can be a useful method for calculating loan interest, tips and taxes, or in any other situation wherein the initial total and the percentage are known but the numerical value of the percentage is not. First, convert the percentage into a decimal by multiplying the percentage by .01. Then, multiply the initial total by the decimal. For instance, to calculate daily interest of 3 percent on a \$15 personal loan, multiply 3 by .01 to get .03. Then multiply 15 by .03 to get .045. This means \$0.45 is the amount of interest earned each day. To add a 20 percent tip to a \$50 restaurant bill, multiply 20 by .01 to get .20, or .2. Then multiply \$50 by .2 to get \$10. \$10 is the amount of the tip.
Calculating Discounts from the Opposite of the Percent
Retail stores frequently put out racks of clothing with signs saying “30 percent off” or “40 percent off,” and the original price tags are still attached to the clothing. To find the discounted price of such a garment, first determine the opposite of the discount percent. If a shirt is on sale for 30 percent off, for example, the opposite is 70 percent. Convert the opposite percent into a decimal, so that 70 percent then becomes .7, in this example. Finally, multiply the original price by the decimal. If the shirt originally cost \$20, multiply \$20 by .7. to get \$14. The discounted price is \$14. |
# Lesson 1
Lesson 1
1 / 43
Slide 1: Tekstslide
MathematicsLower Secondary (Key Stage 3)
In deze les zitten 43 slides, met interactieve quizzen en tekstslides.
Lesduur is: 50 min
## Onderdelen in deze les
Lesson 1
#### Slide 1 -Tekstslide
Calculating y Coordinate Values
If y = 12x, calculate the value value of y for each value of x
timer
5:00
x = 0
x = 1
x = 2
x = -1
x = -2
x = 0.5
x = 0.25
x = 0.1
#### Slide 2 -Tekstslide
Calculating y Coordinate Values
If y = 12x, calculate the value value of y for each value of x
x = 0
x = 1
x = 2
x = -1
y = 0
y = 12
y = 24
y = -12
x = -2
x = 0.5
x = 0.25
x = 0.1
#### Slide 3 -Tekstslide
Calculating y Coordinate Values
If y = 12x, calculate the value value of y for each value of x
x = 0
x = 1
x = 2
x = -1
y = 0
y = 12
y = 24
y = -12
x = -2
y = -24
x = 0.5
x = 0.25
y = 3
x = 0.1
y = 1.2
y = 6
#### Slide 6 -Tekstslide
Drag the colour disc that is part of a line on the circle beneath the correct equation.
x = 2
y = 1
y = 3
x = 9
#### Slide 7 -Sleepvraag
Drag the correct coordinate pair onto the box that matches the missing values in the table.
x
x
x
x
0
1
3
0
1
3
4
4
4
-1
0
3
y
0
1
3
y
0
1
3
y
1
2
4
y
1
-0
-3
(2, 3)
(2, -2)
(2, 2)
(4, 2)
#### Slide 8 -Sleepvraag
Given y = x, explain what will happen to y for each value of x?
#### Slide 9 -Woordweb
Explain what the graph of y = x will look like?
#### Slide 11 -Tekstslide
Exercise 1
Complete the table for the equation y = x and write down the coordinate pairs.
x
-2
-1
0
1
2
3
y
-2
(x, y)
(-2 ,-2)
#### Slide 12 -Tekstslide
Exercise 1
Complete the table for the equation y = x and write down the coordinate pairs
x
-2
-1
0
1
2
3
y
-2
-1
0
1
2
3
(x, y)
(-2 ,-2)
#### Slide 13 -Tekstslide
Exercise 1
Complete the table for the equation y = x and write down the coordinate pairs
x
-2
-1
0
1
2
3
y
-2
-1
0
1
2
3
(x, y)
(-2 ,-2)
(-1 ,-1)
(0 , 0)
(1 , 1)
(2 , 2)
(3 , 3)
#### Slide 14 -Tekstslide
Exercise 2.
Draw the graph using the coordinates in the table below.
Use suitable axes and label the line.
x
-2
-1
0
1
2
3
y
-2
-1
0
1
2
3
(x, y)
(-2 ,-2)
(-1 , -1)
(0 , 0)
(1 , 1)
(2 , 2)
(3 , 3)
#### Slide 15 -Tekstslide
NEW LEARNING
TITLE - Unit 10 Week 2 - Lesson 1 - NEW LEARNING
INSTRUCTIONS - Write down the following notes in your exercise book including the graph.
#### Slide 17 -Tekstslide
Explain in no more than 20 words, what does y = -x mean?
#### Slide 18 -Woordweb
Exercise 3
Complete the table for the equation y = -x and write down the coordinate pairs
x
-2
-1
0
1
2
3
y
(x, y)
#### Slide 19 -Tekstslide
Exercise 3
Complete the table for the equation y = -x and write down the coordinate pairs
x
-2
-1
0
1
2
3
y
2
1
0
-1
-2
-3
(x, y)
#### Slide 20 -Tekstslide
Exercise 3
Complete the table for the equation y = -x and write down the coordinate pairs
x
-2
-1
0
1
2
3
y
2
1
0
-1
-2
-3
(x, y)
(-2 , 2)
(-1 , 1)
(0 , 0)
(1 , -1)
(2 , -2)
(3 , -3)
#### Slide 21 -Tekstslide
Exercise 4
Use the table below to draw the equation y = -x.
Use suitable axes and label the line.
x
-2
-1
0
1
2
3
y
2
1
0
-1
-2
-3
(x, y)
(-2 , 2)
(-1 , 1)
(0 , 0)
(1 , -1)
(2 , -2)
(3 , -3)
A
(2, 2)
B
(-3.2, -3.2)
C
(1800, 1800)
D
(-4, 4)
#### Slide 26 -Tekstslide
Which line is NOT the line of y = x?
A
B
C
D
A
(7, 4)
B
(5, 4)
C
(-2, 2)
D
(7, 6)
#### Slide 30 -Tekstslide
Which graph is the line y = -x?
A
B
C
D
#### Slide 33 -Tekstslide
Reflection on the work covered
I CAN recognise each of the graphs: x = 3, y = 3 and y = x
I can recognize the difference between x = 3 and y = 3
I CANNOT recognise any of the graphs: x = 3, y = 3 and y = x
#### Slide 34 -Poll
INDEPENDENT STUDY
TITLE - Unit 10 Week 1 - Lesson 4 - NEW LEARNING
INSTRUCTIONS - Write down the questions in your book and then answer. Show all of your working out.
Minimum target - Concept corner and questions 1 to 3 |
# 2TH §5.1 Removing single brackets
1. Calculate the perimeter of the figure. Write your answer as short as possible.
2. Calculate the area of the figure. Write your answer as a single square root.
Entry Ticket
timer
6:00
1 / 19
Slide 1: Slide
WiskundeMiddelbare schoolhavo, vwoLeerjaar 2
This lesson contains 19 slides, with text slides.
Lesson duration is: 50 min
## Items in this lesson
1. Calculate the perimeter of the figure. Write your answer as short as possible.
2. Calculate the area of the figure. Write your answer as a single square root.
Entry Ticket
timer
6:00
#### Slide 1 -Slide
Starting with something new
Removing single brackets
Something like this ..... 5(2+4) =
#### Slide 2 -Slide
Removing single brackets
5(2+4)=
11 -1
30 something else
#### Slide 3 -Slide
Rectangles
You can think of the calculation 5(2+4) as working out the area of this rectangle.
#### Slide 4 -Slide
Rectangles
The area of the rectangle is
length x width =
5 x (2+4) =
5(2+4)
#### Slide 5 -Slide
Rectangles
area rectangle = 5(2+4) = 5 x 6 = 30
You could also calculate the area of the
blue rectangle and red rectangle first
area rectangle = 5 x 2 + 5 x 4 =
10 + 20 = 30
#### Slide 6 -Slide
Rectangles with variables
If at least one of the sides of the rectangle is given as a variable, you can make a formula for the area of the rectangle.
#### Slide 7 -Slide
Rectangles with variables
You can make a formula with brackets and one without.
Area rectangle = 5 x (t+4) = 5(t+4)
OR
Area rectangle = 5 x t + 5 x 4 =
5t + 20
#### Slide 8 -Slide
Rectangles with variables
Because both formulas give the same area, we now now that
5(t+4) = 5t + 20
We can use this to rewrite formulas
with brackets as formulas without
brackets.
#### Slide 9 -Slide
Removing brackets
Write the formula y = 2(3t + 4) without brackets.
#### Slide 10 -Slide
Removing brackets
Write the formula y = 2(3t + 4) without brackets.
Think of a rectangle with a side of 2 and a side of 3t + 4
#### Slide 11 -Slide
Removing brackets
Write the formula y = 2(3t + 4) without brackets.
Think of a rectangle with a side of 2 and a side of 3t + 4
Area blue rectangle = 2 x 3t = 6t
Area red rectangle = 2 x 4 = 8
y = 6t + 8
#### Slide 12 -Slide
Removing brackets
Instead of a rectangle, you can also use a multiplication table.
y = 2(3t + 4) = 6t + 8
#### Slide 13 -Slide
More variables, negative numbers
If you want to write 2t(5t - 3) without brackets, it is hard to imagine a rectangle, because a lenght of -3 is not possible.
We can still use a multiplication table!
#### Slide 14 -Slide
More variables, negative numbers
Write y = 2t(5t - 3) without brackets.
What should be written over here?
7t
10t
7t2
10t2
#### Slide 15 -Slide
More variables, negative numbers
Write y = 2t(5t - 3) without brackets.
What should be written over here?
t
6t
6t
2t3
#### Slide 16 -Slide
More variables, negative numbers
Write y = 2t(5t - 3) without brackets.
#### Slide 17 -Slide
Notation
If a 1 is written in front of a variable or bracket, you do not write down the 1.
Instead of 1p, you write p
Instead of -1t, you write -t
Instead of -1(5t+2), you write -(5t+2)
#### Slide 18 -Slide
Homework Tuesday
§5.1 Do exercises S4, 4, 5, S6 and 6
Don't forget to check and correct your answers with the solutions book! |
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# Write the numbers whose multiplicative inverses are the numbers themselves
A Number System is a way of representing numbers. Representation of numbers is done by using digits or symbols. The numbers that are represented by digits or symbols have the value and the value depends on the place, base, and value of the digits used. There are many types of numbers in the decimal number system, for instance, real numbers, complex numbers, natural numbers, whole numbers, and so on. Let’s take a look at their definitions,
### Types of numbers
In the number system, the decimal number system is majorly used. In the decimal number system, there are many types of numbers based on their different characteristics. Lets take a brief look at their definitions,
The Real Numbers
The Real Numbers are the Numbers which consist of All the numbers ie., all of the Rational Numbers and Irrational Numbers. Some example of real numbers are 3, 1.444…, 55, 9.73409…, etc.
The Complex Numbers
The Complex Numbers are the numbers that are represented in (a+ib) form where b≠0 , where a and b are real numbers and i is an imaginary unit with value √-1, when the value is b. It is a real number since in (a+ib) b=0 then a+i*0 = a which is a real Number.
The Natural Numbers
The Natural Numbers are the numbers that start from 1 and counts to Infinity. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11…..up to Infinity are Natural numbers. It is a Subset of Whole numbers as whole number contains all natural numbers and 0 as well.
The Whole Numbers
The Whole numbers are the Natural numbers with Including an extra number zero ie., the numbers that start from 0 and counts to infinity are called Whole Numbers. It is a superset of natural numbers. Some examples are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,….upto Infinity.
The Integers
Integers set are the number sets that consist of negative and positive numbers including zero. All the basic operations of number system i.e., Addition, Multiplication, Subtraction except Division will result in an Integer. The division may or may not result in Integer because when a smaller numerator and bigger denominator are used in division. It results in a fraction, that may be rational Or irrational
Irrational Number
A Number is called irrational if it cannot be expressed in the form of p/q, where p and q are Integers. In other words, if it cannot be expressed as a ratio of two numbers. When a number is expressed in decimal form, if it never terminates it is an irrational number. For example,
• The value π(PI) is 3.1415926…. the value never terminates. It is rounded off to 3.14 and used in calculations like this.
• √2 is 1.41421356237309… which is an irrational number. So, square root of those numbers which are not perfect squares come under this category.
• While, square Roots of Perfect Squares are always Rational
Rational Number
A Number is called Rational if it can be expressed in the form of p/q, where p and q are Integers. In other words, if it can be expressed as a ratio of two Numbers. When a number is expressed in decimal form, if it terminates it is a Rational Number. Some numbers that are never terminating But are recurring are Rational Numbers. For example, 0.0833333….. it is the decimal representation of 1/12 is a rational Number. √4 is rational it can be expressed in p/q form i.e., 2/1.
There are many operations done on these numbers and one of them is the multiplicative inverse, lets see what are multiplicative Inverses and if it is possible to obtain the same term after performing multiplicative inverse on a number,
### What are the numbers whose multiplicative inverses are the numbers themselves?
An Multiplicative Inverse of a number is defined as a value we multiply with the number to get the product as 1. In other words, Multiplicative Inverse is defined as the reciprocal of a number. For example, Multiplicative Inverse of number x : As per definition the multiplicative Inverse is Reciprocal of the Number i.e, multiplicative Inverse of x is 1/x or x-1. Similarly,
Multiplicative Inverse of numbers,
• 11 is 1/11 or 11-1
• -20 is -(1/20) or (-20)-1
• 3 is 1/3 or 3-1
The Numbers that are equal to its Multiplicative Inverse are -1 and 1.
Because Reciprocal of 1 is 1/1 which is 1.
i.e, Multiplicative Inverse of 1 = 1/1 = 1.
and,
Reciprocal of -1 is -1/1 is -1
i.e, Multiplicative Inverse of -1 = -1/1 = -1
-1 and 1 are the Only Numbers which are equal to its Multiplicative Inverse.
Note There is No Multiplicative Inverse of 0 because Reciprocal of 0 is not defined. hence, Multiplicative Inverse of 0 does not exist.
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# Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1
Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.
## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1
Question 1.
Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x2 – x + 1 at (1, 3)
Solution:
y = 3x2 – x + 1
∴ $$\frac{d y}{d x}=\frac{d}{d x}$$ (3x2 – x + 1)
= 3 × 2x – 1 + 0
= 6x – 1
∴ $$\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}$$ = 6(1) – 1
= 5
= slope of the tangent at (1, 3).
∴ the equation of the tangent at (1, 3) is
y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0.
The slope of the normal at (1, 3) = $$\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}$$
∴ the equation of the normal at (1, 3) is
y – 3 = $$-\frac{1}{5}$$(x – 1)
∴ 5y – 15 = -x + 1
∴ x + 5y – 16 = 0
Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.
(ii) 2x2 + 3y2 = 5 at (1, 1)
Solution:
2x2 + 3y2 = 5
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = $$\frac{-2}{3}$$(x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) = $$\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}$$
∴ the equation of the normal at (1, 1) is
y – 1 = $$\frac{3}{2}$$(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.
(iii) x2 + y2 + xy = 3 at (1, 1)
Solution:
x2 + y2 + xy = 3
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) = $$\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}$$
= $$\frac{-1}{-1}$$
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.
Question 2.
Find the equations of the tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Differentiating y = x2 + 5 w.r.t. x, we get
$$\frac{d y}{d x}=\frac{d}{d x}$$(x2 + 5) = 2x + 0 = 2x
$$\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}$$
= slope of the tangent at (x1, y1)
Let m1 = 2x1
The slope of the line 4x – y + 1 = 0 is
m2 = $$\frac{-4}{-1}$$ = 4
Since, the tangent at P(x1, y1) is parallel to the line 4x – y + 1 = 0,
m1 = m2
∴ 2x1 = 4
∴ x1 = 2
Since, (x1, y1) lies on the curve y = x2 + 5, y1 = $$x_{1}^{2}$$ + 5
∴ y1 = (2)2 + 5 = 9 ……[x1 = 2]
∴ the coordinates of the point are (2, 9) and the slope of the tangent = m1 = m2 = 4.
∴ the equation of the tangent at (2, 9) is
y – 9 = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x – y + 1 = 0
Slope of the normal = $$\frac{-1}{m_{1}}=-\frac{1}{4}$$
∴ the equation of the normal at (2, 9) is
y – 9 = $$-\frac{1}{4}$$(x – 2)
∴ 4y – 36 = -x + 2
∴ x + 4y – 38 = 0
Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.
Question 3.
Find the equations of the tangent and normal to the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Differentiating y = 3x2 – 3x – 5 w.r.t. x, we get
$$\frac{d y}{d x}=\frac{d}{d x}$$(3x2 – 3x – 5)
= 3 × 2x – 3 × 1 – 0
= 6x – 3
∴ $$\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3$$
= slope of the tangent at (x1, y1)
Let m1 = 6x1 – 3
The slope of the line 3x – y + 1 = 0
m2 = $$\frac{-3}{-1}$$ = 3
Since, the tangent at P(x1, y1) is parallel to the line 3x – y + 1 = 0,
m1 = m2
∴ 6x1 – 3 = 3
∴ 6x1 = 6
∴ x1 = 1
Since, (x1, y1) lies on the curve y = 3x2 – 3x – 5,
$$y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5$$, where x1 = 1
= 3(1)2 – 3(1) – 5
= 3 – 3 – 5
= -5
∴ the coordinates of the point are (1, -5) and the slope of the tangent = m1 = m2 = 3.
∴ the equation of the tangent at (1, -5) is
y – (-5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x – y – 8 = 0
Slope of the normal = $$-\frac{1}{m_{1}}=-\frac{1}{3}$$
∴ the equation of the normal at (1, -5) is
y – (-5) = $$-\frac{1}{3}$$(x – 1)
∴ 3y + 15 = -x + 1
∴ x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively. |
# What are the Techniques of Integration?
## Introduction
Integration is a fundamental concept in calculus that deals with finding the antiderivative of a function. It plays a crucial role in various branches of mathematics and is extensively used in solving problems in physics, engineering, economics, and other fields. The process of finding antiderivatives involves applying different techniques of integration, each suited to specific types of functions.
## Integration Methods
The different methods of integration include:
1. Integration by Substitution
2. Integration by Parts
3. Integration Using Trigonometric Identities
4. Integration of Some particular function
5. Integration by Partial Fraction
## 1- Integration by Substitution
Integration by substitution, also known as u-substitution, is a powerful technique used to simplify the process of integration. When faced with complex integrals, this method allows us to transform the integral into a more manageable form by introducing a new variable, typically represented as ‘u’.
The steps for integration by substitution are straightforward:
1. Identify a part of the integrand that can be simplified with a new variable ‘u’.
2. Calculate the derivative of ‘u’ with respect to the independent variable, usually denoted as ‘du/dx’.
3. Rearrange the integral in terms of ‘u’ and ‘du’, using the chain rule to express the integrand solely in terms of ‘u’.
4. Evaluate the new integral with respect to ‘u’.
5. Finally, reintroduce ‘x’ by substituting ‘u’ back into the expression, completing the integration process.
Integration by substitution is particularly useful when dealing with complex functions, trigonometric expressions, or nested functions. It is an essential technique in calculus, facilitating the evaluation of a wide range of integrals that may be challenging or impossible to solve using basic integration rules alone. Mastering this method empowers mathematicians and students to tackle more advanced mathematical problems and deepen their understanding of calculus.
## 2- Integration by Parts
Integration by parts is a part of any A Level maths revision course as well as a powerful technique used to evaluate integrals of products of functions. When faced with an integral in the form ∫u dv, integration by parts allows us to transform it into a more manageable form by choosing two parts of the integrand and applying a specific formula.
The integration by parts formula is given as:
∫u dv = uv – ∫v du
where u and v are functions of the independent variable.
The steps for integration by parts are as follows:
1. Identify two parts of the integrand, usually denoted as ‘u’ and ‘dv’. It is crucial to choose ‘u’ in a way that simplifies the integral when differentiated and ‘dv’ such that its integral is easy to compute.
2. Differentiate ‘u’ to find ‘du’ and integrate ‘dv’ to find ‘v’.
3. Substitute the values of ‘u’, ‘dv’, ‘du’, and ‘v’ into the integration by parts formula.
4. Evaluate the new integral on the right-hand side of the formula.
5. If necessary, repeat the process by applying integration by parts again to the new integral obtained.
The versatility of integration by parts makes it an essential tool in various scientific disciplines, including mathematics, physics, and engineering. Many real-world problems require the evaluation of complex integrals, and mastering this technique empowers mathematicians and students to handle such challenges effectively. A solid understanding of this technique enhances problem-solving skills and enables researchers and students to obtain accurate solutions efficiently.
## 3- Integration by Using Trigonometric Identities
Integration by using trigonometric identities is a specialised technique employed to evaluate integrals involving trigonometric functions. Trigonometric identities, such as Pythagorean identities and double-angle formulas, play a crucial role in simplifying these integrals and transforming them into solvable forms.
When faced with trigonometric integrals, the key is to recognize the appropriate trigonometric identities that can be applied to rewrite the integrand. By using these identities, we can manipulate the expression and often reduce it to a more manageable form for integration.
Some common trigonometric identities used in integration include:
• Pythagorean identities: sin^2(x) + cos^2(x) = 1 and sec^2(x) – tan^2(x) = 1.
• Double-angle formulas: sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) – sin^2(x).
By skillfully applying these trigonometric identities, mathematicians and students can simplify complex trigonometric integrals and evaluate them more easily. Mastery of these identities and their applications is crucial for tackling advanced calculus problems that involve trigonometric functions. This method plays an essential role in calculus, particularly in trigonometry-intensive problems encountered in physics, engineering, and other scientific disciplines. A solid grasp of trigonometric identities empowers individuals to explore and solve a wide range of mathematical challenges.
## 4- Integration of Some Particular Function
The integration of particular functions is a significant aspect of calculus that involves finding the antiderivative or indefinite integral of specific types of functions. While the integration techniques discussed earlier apply to a wide range of functions, certain functions have special properties that lead to more straightforward integration methods.
1. Polynomials: The integration of polynomials is straightforward, following the power rule of integration. For example, the integral of x^n with respect to x is (x^(n+1))/(n+1) + C, where n is a constant not equal to -1. For instance, the integral of 3x^2 + 2x + 5 is (3x^3)/3 + (2x^2)/2 + 5x + C, which simplifies to x^3 + x^2 + 5x + C.
2. Exponential Functions: The integral of exponential functions like e^x follows a simple integration rule. The integral of e^x with respect to x is e^x + C. For example, the integral of 2e^x is 2e^x + C.
3. Trigonometric Functions: The integration of trigonometric functions depends on their type. For instance, the integral of sin(x) with respect to x is -cos(x) + C, and the integral of cos(x) with respect to x is sin(x) + C. The integral of sec^2(x) is tan(x) + C, and the integral of csc^2(x) is -cot(x) + C.
4. Inverse Trigonometric Functions: The integral of inverse trigonometric functions also follows specific rules. For example, the integral of 1/(1 + x^2) with respect to x is arctan(x) + C.
These are just a few examples of particular functions and their corresponding integration rules. It is essential for mathematicians and students to become familiar with these specific integration techniques, as they can significantly simplify the evaluation of integrals involving these functions. A solid understanding of these integration methods empowers individuals to tackle a wide array of mathematical problems effectively.
## 5- Integration by Partial Fraction
Integration by partial fractions is a powerful technique used to evaluate integrals of rational functions, which are the quotient of two polynomials. When faced with a rational function that cannot be easily integrated using other methods, partial fractions provide a systematic approach to break it down into simpler fractions, making the integration process more manageable.
The steps for integration by partial fractions are as follows:
1. Factor the denominator of the rational function into linear and irreducible quadratic factors.
2. Express the rational function as the sum of simpler fractions, each with a specific factor in the denominator.
3. Determine the unknown constants in the partial fractions using a system of equations or comparing coefficients.
4. Integrate each partial fraction separately using basic integration rules.
By using partial fractions, mathematicians can transform complex integrals into a series of simpler integrals that can be easily solved. This technique is particularly useful in various scientific and engineering applications, where the evaluation of integrals involving rational functions is common.
## Final Thoughts
By understanding these diverse integration techniques, individuals can navigate through complex mathematical challenges efficiently and accurately. Each method serves a specific purpose, and mastering them empowers mathematicians and students to tackle real-world problems across various scientific disciplines, including physics, engineering, economics, and more.
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# Composite Functions
Related Pages
Math Functions
Composite Functions
Lessons with videos, examples and solutions looking at the composition of functions or composite functions.
### What Is A Composite Function?
A composite function is a function that depends on another function. A composite function is created when one function is substituted into another function.
For example, f(g(x)) is the composite function that is formed when g(x) is substituted for x in f(x).
f(g(x)) is read as “f of g of x”.
f(g(x)) can also be written as (f ∘ g)(x) or fg(x),
In the composition (f ∘ g)(x), the domain of f becomes g(x).
The following diagram shows some examples of composite functions. Scroll down the page for more examples and solutions.
Example:
Given f(x) = x2 + 6 and g(x) = 2x – 1, find
a) (f ∘ g)(x)
b) (g ∘ f)(x)
Solution:
a) (f ∘ g)(x)
= f(2x – 1)
= (2x – 1)2 + 6
= 4x2 – 4x + 1 + 6
= 4x2 – 4x + 7
b) (g ∘ f)(x)
= g(x2 + 6)
= 2(x2 + 6) – 1
= 2x2 + 12 – 1
= 2x2 + 11
#### Composite Functions
This lesson explains the concept of composite functions. An example is given demonstrating how to work algebraically with composite functions and another example involves an application that uses the composition of functions.
Examples:
1. If f(x) = x + 5 and g(x) = 3x2 find
(a) (f ∘ g)(x)
(b) (f ∘ g)(2)
(c) g(f(x))
2. A newspaper company creates routes with 50 subscribers(n) for each delivery person(d). There is a supervisor (s) for every 10 delivery persons.
(a) Write d as a function of n.
(b) Write s as a function of d.
(c) Substitute to write s as a function of n.
#### How To Determine The Value Of A Composite Function And How To Determine A Composite Function Given Two Functions?
Examples:
1. Given the functions, determine the value of each composite function.
f(x) = 2x - 1, g(x) = x3 - 5, h(x) = 5 - x2
(a) (f ∘ g)(3)
(b) (g ∘ f)(3)
(c) (h ∘ g)(-1)
2. Given the functions, determine the value of each composite function.
f(x) = 4x + 1, g(x) = x2 - x + 5
(a) (f ∘ g)(x)
(b) (g ∘ f)(x)
#### How To Find The Composition Of Functions?
Example:
f(x) = x2 + x and g(x) = 4 - x
Find
(a) (f ∘ g)(x)
(b) (g ∘ f)(x)
#### What Is The Composition Of Two Functions?
Example:
f(x) = 2x4 + x4 + 1, g(x) = √x
Find f(g(x))
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# CN#5 Objectives Prove certain triangles are similar by using AA, SSS, and SAS. Use triangle similarity to solve problems.
## Presentation on theme: "CN#5 Objectives Prove certain triangles are similar by using AA, SSS, and SAS. Use triangle similarity to solve problems."— Presentation transcript:
CN#5 Objectives Prove certain triangles are similar by using AA, SSS, and SAS. Use triangle similarity to solve problems.
There are several ways to prove certain triangles are similar
There are several ways to prove certain triangles are similar. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent.
Example 1: Using the AA Similarity Postulate
Explain why the triangles are similar and write a similarity statement. BCA ECD by the Vertical Angles Theorem. Also, A D by the Right Angle Congruence Theorem. Therefore ∆ABC ~ ∆DEC by AA~.
Example 2A: Verifying Triangle Similarity
Verify that the triangles are similar. ∆PQR and ∆STU Therefore ∆PQR ~ ∆STU by SSS ~.
Example 2B: Verifying Triangle Similarity
Verify that the triangles are similar. ∆DEF and ∆HJK D H by the Definition of Congruent Angles. Therefore ∆DEF ~ ∆HJK by SAS ~.
Example 3: Finding Lengths in Similar Triangles
Explain why ∆ABE ~ ∆ACD, and then find CD. Step 1 Prove triangles are similar. A A by Reflexive Property of , and B C since they are both right angles. Therefore ∆ABE ~ ∆ACD by AA ~.
Example 3 Continued Step 2 Find CD. Corr. sides are proportional. Seg. Add. Postulate. Substitute x for CD, 5 for BE, 3 for CB, and 9 for BA. x(9) = 5(3 + 9) Cross Products Prop. 9x = 60 Simplify. Divide both sides by 9.
Example 4: Writing Proofs with Similar Triangles
Given: 3UT = 5RT and 3VT = 5ST Prove: ∆UVT ~ ∆RST
Statements Reasons 1. 3UT = 5RT 1. Given 2.
2. Divide both sides by 3RT. 3. 3VT = 5ST 3. Given. 4. 4. Divide both sides by3ST. 5. RTS VTU 5. Vert. s Thm. 6. ∆UVT ~ ∆RST 6. SAS ~ Steps 2, 4, 5
You learned in Chapter 2 that the Reflexive, Symmetric, and Transitive Properties of Equality have corresponding properties of congruence. These properties also hold true for similarity of triangles.
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A quadratic equation is an equation where the highest power of x is x². There are various methods of solving quadratic equations, as shown below.
NOTE: If x² = 36, then x = +6 or -6 (since squaring either of these numbers will give 36). However, Ö36 = + 6 only.
Completing the Square
9 and 25 can be written as 3² and 5² whereas 7 and 11 cannot be written as the square of another exact number. 9 and 25 are called perfect squares. Another example is (9/4) = (3/2)². In a similar way, x² + 2x + 1 = (x + 1)².
To make x² + 6x into a perfect square, we add (6²/4) = 9. The resulting expression, x² + 6x + 9 = (x + 3)² and so is a perfect square. This is known ascompleting the square. To complete the square in this way, we take the number before the x, square it, and divide it by 4. This technique can be used to solve quadratic equations, as demonstrated in the following example.
Example:
Solve x² - 6x +…
## Similar Mathematics resources:
See all Mathematics resources »See all Quadratic equations resources »
A quadratic equation is an equation where the highest power of x is x². There are various methods of solving quadratic equations, as shown below.
NOTE: If x² = 36, then x = +6 or -6 (since squaring either of these numbers will give 36). However, Ö36 = + 6 only.
Completing the Square
9 and 25 can be written as 3² and 5² whereas 7 and 11 cannot be written as the square of another exact number. 9 and 25 are called perfect squares. Another example is (9/4) = (3/2)². In a similar way, x² + 2x + 1 = (x + 1)².
To make x² + 6x into a perfect square, we add (6²/4) = 9. The resulting expression, x² + 6x + 9 = (x + 3)² and so is a perfect square. This is known ascompleting the square. To complete the square in this way, we take the number before the x, square it, and divide it by 4. This technique can be used to solve quadratic equations, as demonstrated in the following example.
Example:
Solve x² - 6x +… |
# How to Write Numbers in Standard Form? Explained with Examples
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In the realm of mathematics, expressions, and numbers can be represented in various forms. One such essential representation is the standard form of numbers. The standard form of numbers plays a key role in the calculations of very large or small numbers.
The standard form of numbers helps simplify complex numerical values, making them easier to comprehend and work with across various disciplines. The standard form of numbers is employed in various scientific sectors, and its applications span scientific research, engineering innovation, financial analysis, etc.
Embracing standard form equips us with a universal language to comprehend and manipulate numbers efficiently. In this article, we will address the core concept of the standard form of numbers. We will elaborate on its definition, converting numbers to the standard form, and applications, and we will solve some examples.
Inside Story
## Defining Standard Form of Numbers:
Standard form is a compact and organized way to represent large or small numbers. In this format, a number is expressed as a product of a coefficient and a power of ten. This power of ten indicates how many places the decimal point should be moved to achieve the original number. Mathematically
$P\times10^q$ where $1\leq P < 10 \text{ and } q \in \mathbb{Z}$ (a positive or negative integer) that is the exponent of 10.
The standard form of numbers simplifies the representation, calculations, and comparisons of both large and small numbers with different magnitudes using powers of 10, making them easier to work with and compare.
## How to convert Numbers to Standard Form?
A standard form calculator is an easy and effective way to convert numbers in standard form. You can also convert numbers in standard form manually by following the below-mentioned steps.
• First of all, identify the coefficient “P” by shifting the decimal point in the original number so that it lies to the left of the first non-zero digit i.e. between 1 and 10.
• Next, determine the exponent “q” by counting the number of places in the original number that the decimal point shifted. This count will be the exponent “q” of the 10.
• If the original number is higher than or equal to 1, then the exponent (power) of 10 will be positive, and if the original number is less than that of 1, then the exponent of 10 will be negative.
• Now write the coefficient as a product with 10 raised to the power of the exponent.
• Moreover, when adding or subtracting numbers in standard form, the exponents must align before performing the operation.
• Multiplying numbers in standard form is accomplished by multiplying the coefficients and adding the exponents of the powers of ten.
• In the division, coefficients are divided, and subtracting the exponents of the powers of 10.
## Examples:
Let’s explore some examples to better grasp the concept of writing numbers in a standard form better:
Example 1:
Express the number $920 000 000 000 000 000$ in standard form.
Solution:
Step 1: The non-zero digits (92) make up the coefficient, which we are able to determine.
Step 2: The decimal point will be located after the first non-zero number, like 9.2.
Step 3: Determine the number of digits after 9. There are 17 digits to which the decimal point has crossed to come in the standard position. This will be the exponent of 10 i.e. 1017.
Step 4: So, the given number in standard form will be expressed as $9.2 \times 10^{17}$.
Example 2:
Express the number $0.000 000 000 000 000 000 063$ in standard form.
Solution:
Step 1: We identify the coefficient i.e. the non-zero digits (63) will form the coefficient.
Step 2: Place the decimal point after the first non-zero digit i.e. 6.3
Step 3: We count the number of digits before 6. There are 20 digits to which the decimal point has crossed to come in standard position from left to right. This will be the exponent of 10 i.e. 10-20
Step 4: So, the given number standard form will be expressed as $6.3 \times 10^{-20}$.
## Applications of Standard Form in Real Life:
The standard form of numbers finds its application in various real-life scenarios.
### Use in Scientific Notation:
In scientific fields, standard form aids in expressing measurements and quantities that span vast ranges. Scientific notation is a common application of the standard form in scientific research. It allows scientists to express extremely large or small values encountered in various scientific disciplines, from astronomy to physics.
### Use in Financial Statements:
In finance, the standard form is used to represent large monetary figures in a compact and easily understandable manner.
It’s particularly useful in annual reports, where revenues, expenses, and other financial metrics are often presented in millions or billions. In economics, it helps express values like GDP and national debt. So, the standard form is used to represent large financial figures like national debts and economic indicators.
### Use in Engineering:
Engineers frequently utilize the standard form to express measurements and quantities in fields such as electronics, mechanics, and materials science.
This form allows them to maintain precision while working with numbers that can vary across wide ranges. Engineers use it to quantify measurements and calculations in a concise manner, aiding in accurate design and analysis.
## Advantages of Using Standard Form of Numbers:
The standard form offers several advantages that contribute to its widespread usage.
Easy Comparison of Numbers: Numbers in standard form are much easier to compare since the focus shifts to the order of magnitude rather than the actual values. This makes it straightforward to determine which value is larger or smaller.
Easy comparison helps in quickly understanding the relative magnitudes of numbers, which is crucial for making informed decisions and drawing conclusions.
Simplifies Calculations: Performing calculations with numbers in standard form is far more convenient. Multiplication and division become simpler as you work with the coefficients and simply add or subtract the exponents of 10.
Limitations of Standard Form of Numbers:
While the standard form is highly useful, it does have some limitations.
• Limited Use for Small and Simple Numbers: For small numbers that are close to 1, the standard form may not provide significant advantages. It’s generally more practical for values that vary across several orders of magnitude.
# Conclusion:
We can summarize that in the world of numbers, the standard form of numbers emerges as a powerful tool for simplifying, representing, calculating, and comparing complicated numbers. in this article, we have explored the idea of the standard form of numbers. We have discussed its definition, useful steps to convert ordinary numbers into the standard form of numbers, important implications, and examples.
Hopefully, apprehending this article will enhance your ability to compare, compute, and communicate, as well as simplify complex numerical numbers across different fields. Whether you’re dealing with colossal astronomical distances or minuscule atomic scales, standard form bridges the gap and simplifies complex numbers.
Stay tuned with Laws Of Nature for more useful and interesting content.
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Home / General Studies / Mathematics / Mastering Arithmetic Operations: 4 Fundamental Rules for Success
Arithmetic operations
# Mastering Arithmetic Operations: 4 Fundamental Rules for Success
In the realm of mathematics, mastering the four fundamental rules of arithmetic operations is essential for solving a wide range of problems. These rules—addition, subtraction, multiplication, and division—are the building blocks of mathematical calculations. In this comprehensive guide, we’ll explore these rules in detail and discuss various types of numbers to deepen your understanding of mathematics.
Mathematics forms the cornerstone of countless fields and disciplines, and at its heart lie the four fundamental rules of arithmetic operations: addition, subtraction, multiplication, and division. These fundamental operations are the bedrock upon which mathematical problem-solving is built. In this comprehensive guide, we will delve into these rules, gaining a thorough understanding of each. Additionally, we will explore different types of numbers, each with its own unique characteristics, providing you with a holistic view of the mathematical landscape.
### Arithmetic Operations and Their Significance:
Before we embark on our journey to understand the fundamental rules of arithmetic, let’s take a moment to appreciate the significance of these operations in our daily lives. Whether you’re calculating your monthly expenses, determining the cooking time for a recipe, or solving complex engineering problems, arithmetic operations are at play. They are the language of mathematics that allows us to quantify, measure, and analyze various aspects of the world around us.
Now, let’s dive into the four fundamental rules that underpin arithmetic:
1. Addition (+): Addition is the process of combining two or more numbers to find their total. It’s represented by the plus sign (+). For example, when you add 3 and 5, you get 8. Addition is commutative, meaning that changing the order of the numbers doesn’t affect the result. In other words, 3 + 5 and 5 + 3 yield the same sum, 8.
2. Subtraction (-): Subtraction involves taking away one number from another to find the difference. It’s represented by the minus sign (-). For instance, subtracting 4 from 7 gives you 3. Unlike addition, subtraction is not commutative. In this case, changing the order of the numbers will yield different results. 7 – 4 is not the same as 4 – 7.
3. Multiplication (x or ): Multiplication is the process of repeated addition. It allows you to find the result of adding the same number multiple times. Multiplication can be represented by the “x” symbol or an asterisk (). For example, 3 multiplied by 4 is 12, which can be expressed as 3 x 4 or 3 * 4. Multiplication is also commutative, so 3 x 4 is equivalent to 4 x 3.
4. Division (/): Division is the operation of splitting a number into equal parts or groups. It’s represented by the division sign (/). When you divide 8 by 2, you get 4. Division is not commutative, similar to subtraction. In this case, reversing the order of the numbers will yield different results. 8 / 2 is not the same as 2 / 8.
Understanding the properties and applications of these fundamental arithmetic operations is essential for solving mathematical problems accurately and efficiently. Whether you’re working with whole numbers, fractions, decimals, or even complex mathematical expressions, these rules remain constant and serve as the foundation of all mathematical calculations.
Types of Numbers and Their Characteristics:
To fully grasp the world of mathematics, it’s crucial to acquaint ourselves with various types of numbers, each with its own set of properties and characteristics. Here are some key categories of numbers:
1. Even Numbers: Even numbers are those that are divisible by 2. Examples include 2, 4, 6, and 8. They are characterized by the absence of a remainder when divided by 2.
2. Odd Numbers: Odd numbers are not divisible by 2 and always leave a remainder of 1 when divided by 2. Examples include 3, 5, 7, and 9.
3. Prime Numbers: Prime numbers are integers greater than 1 that are only divisible by 1 and themselves. Examples of prime numbers include 2, 3, 5, 7, and 11.
4. Composite Numbers: Composite numbers are positive integers greater than 1 that have divisors other than 1 and themselves. For instance, 4, 6, and 9 are composite numbers because they can be divided evenly by numbers other than just 1 and 4, 6, and 9, respectively.
5. Whole Numbers: Whole numbers encompass all positive integers, including zero (0). They form a complete set of numbers without fractions or decimals. Examples include 0, 1, 2, 3, and so on, including their negative counterparts: -1, -2, -3, and so forth.
6. Natural Numbers: Natural numbers, often referred to as counting numbers, are the set of positive integers used for counting and ordering. These numbers begin at 1 and continue infinitely: 1, 2, 3, 4, 5, and so forth.
7. Rational Numbers: Rational numbers are those that can be expressed as a fraction, where the numerator (p) and denominator (q) are integers, and q is not equal to zero (q ≠ 0). Examples of rational numbers include 3/5, -3/5, 0/4, 2/-5, and -3/7.
8. Irrational Numbers: Irrational numbers cannot be expressed as fractions. They are decimal numbers that continue infinitely without repeating. Examples of irrational numbers include √5, √5 + √3, and π (pi).
9. Real Numbers: Real numbers encompass both rational and irrational numbers. They are represented on the real number line, which extends infinitely in both directions. Real numbers are used to describe quantities and measurements in the physical world.
In conclusion, arithmetic operations and an understanding of different types of numbers are fundamental to mathematics. These concepts serve as the building blocks upon which more complex mathematical ideas and calculations are constructed. Whether you’re performing basic calculations or exploring advanced mathematical theories, a solid grasp of these fundamental principles is invaluable.
### Arithmetic operations – FAQ
Q1: What are the four fundamental rules of arithmetic operations? A1: The four fundamental rules of arithmetic operations are addition, subtraction, multiplication, and division. These operations form the foundation of mathematical calculations and are essential for solving a wide range of mathematical problems.
Q2: Are addition and multiplication commutative operations? A2: Yes, addition and multiplication are commutative operations. This means that changing the order of the numbers being added or multiplied does not affect the result. For addition, a + b is the same as b + a. Similarly, for multiplication, a x b is the same as b x a.
Q3: Are subtraction and division commutative operations? A3: No, subtraction and division are not commutative operations. Changing the order of the numbers in subtraction or division will yield different results. For subtraction, a – b is not the same as b – a. For division, a ÷ b is not the same as b ÷ a.
Q4: What are even and odd numbers? A4: Even numbers are those that are divisible by 2 and leave no remainder when divided by 2. Examples include 2, 4, 6, and 8. Odd numbers, on the other hand, are not divisible by 2 and always leave a remainder of 1 when divided by 2. Examples include 3, 5, 7, and 9.
Q5: What is the difference between prime and composite numbers? A5: Prime numbers are positive integers greater than 1 that are only divisible by 1 and themselves. Examples include 2, 3, 5, and 7. Composite numbers are positive integers greater than 1 that have divisors other than 1 and themselves. For instance, 4, 6, and 9 are composite numbers because they can be divided evenly by numbers other than just 1 and 4, 6, and 9, respectively.
Q6: What are rational and irrational numbers? A6: Rational numbers can be expressed as a fraction, where the numerator (p) and denominator (q) are integers, and q is not equal to zero (q ≠ 0). Examples of rational numbers include 3/5, -3/5, 0/4, 2/-5, and -3/7. Irrational numbers, on the other hand, cannot be expressed as fractions. They are decimal numbers that continue infinitely without repeating. Examples of irrational numbers include √5, √5 + √3, and π (pi).
Q7: What is the significance of real numbers? A7: Real numbers encompass both rational and irrational numbers. They are used to describe quantities and measurements in the physical world and are represented on the real number line, which extends infinitely in both directions. Real numbers play a crucial role in various scientific and mathematical applications.
Q8: Why are these fundamental rules and types of numbers important? A8: These fundamental rules of arithmetic operations and the understanding of various types of numbers are essential for solving mathematical problems accurately and efficiently. Whether you’re performing basic calculations in everyday life or tackling complex mathematical theories, a solid foundation in these principles is crucial. They provide the tools and language to quantify, measure, and analyze various aspects of the world around us.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 21/4 + 57/8 = 65/8 = 8 1/8 = 8.125
Spelled result in words is sixty-five eighths (or eight and one eighth).
### How do you solve fractions step by step?
1. Conversion a mixed number 2 1/4 to a improper fraction: 2 1/4 = 2 1/4 = 2 · 4 + 1/4 = 8 + 1/4 = 9/4
To find new numerator:
a) Multiply the whole number 2 by the denominator 4. Whole number 2 equally 2 * 4/4 = 8/4
b) Add the answer from previous step 8 to the numerator 1. New numerator is 8 + 1 = 9
c) Write a previous answer (new numerator 9) over the denominator 4.
Two and one quarter is nine quarters
2. Conversion a mixed number 5 7/8 to a improper fraction: 5 7/8 = 5 7/8 = 5 · 8 + 7/8 = 40 + 7/8 = 47/8
To find new numerator:
a) Multiply the whole number 5 by the denominator 8. Whole number 5 equally 5 * 8/8 = 40/8
b) Add the answer from previous step 40 to the numerator 7. New numerator is 40 + 7 = 47
c) Write a previous answer (new numerator 47) over the denominator 8.
Five and seven eighths is forty-seven eighths
3. Add: 9/4 + 47/8 = 9 · 2/4 · 2 + 47/8 = 18/8 + 47/8 = 18 + 47/8 = 65/8
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 8) = 8. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 8 = 32. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - nine quarters plus forty-seven eighths = sixty-five eighths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for 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.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Samuel
Samuel has 1/3 of a bag of rice and Isabella has a 1/2 bag of rice. What fraction of are bag of rice do they have altogether?
• Berry Smoothie
Rory has 5/8 cup of milk. How much milk does she have left after she doubles the recipe of the smoothie? Berry Smoothie: 2 cups strawberries 1 cup blueberries 1/4 cup milk 1 tbsp (tablespoon) sugar 1/2 tsp (teaspoon) lemon juice 1/8 tsp (teaspoon) vanilla
Peter bought 1/2 a pound of chocolate at rocky mountain chocolate factory. Later he went to the sweet shoppie and he bought 6/9 of a pound more chocolate. How much chocolate did he buy that day?
• Two pizzas
Jacobs mom bought two whole pizzas. He ate 2/10 of the pizza and his dad ate 1 1/5. How much is left.
• Carrie
Carrie picked 2/5 of the raspberries from the garden, and Robin picked some too. When they were finished, 1/3 of the raspberries still needed to be picked. What fraction of the raspberries did Robin pick? Use pictures, numbers or words and write your fi
• Expressions
Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n
• Evaluate expression
Evaluate expression using BODMAS rule: 1 1/4+1 1/5÷3/5-5/8
• Complicated sum minus product
What must be subtracted from the sum of 3/8 and 5/16 to get difference equal to the product of 5/8 and 3/16?
• Simplify 3
Simplify mixed numerals expression: 8 1/4- 3 2/5 - (2 1/3 - 1/4) Show your solution.
• Honey
Ila collected the honey from 3 of her beehives. From the first hive, she collected 2/3 gallons of honey. The last two hives yielded 1/4 gallon each. After using some of the honey she collected for baking, Lila found that she only had 3/4 gallon of honey l
• Stock market 2
For the week of July 22, the following day to day changes in the stock market was recorded a certain stock: -2 on Monday; +4 Tuesday; -8 Wednesday; + 2 1/2 Thursday; - 3 1/4 Friday. The stock began the week at 78 points. How many points did it finish with
• Chestnuts
Three divisions of nature protectors participated in the collection of chestnut trees.1. the division harvested 1250 kg, the 2nd division by a fifth more than the 1st division and the 3rd division by a sixth more than the second division. How many tons of
• Circular garden
Alice creates a circular vegetable garden. Tomatoes are planted in 1/3 of the circular garden, carrots are planted in 2/5 of the circular garden, and green peppers are planted in 1/10 of the circular garden. What fraction represents the remaining unplante |
# 6.7 SPM Practice (Long Questions)
Question 9 (12 marks):
The data in the Diagram shows the mass, in g, of 30 strawberries plucked by a tourist from a farm.
Diagram
(a)
Based on the Diagram, complete Table 3 in the answer space.
(b)
Based on Table, calculate the estimated mean mass of a strawberry.
(c)
For this part of the question, use graph paper.
By using the scale of 2 cm to 10 g on the horizontal axis and 2 cm to 1 strawberry on the vertical axis, draw a histogram for the data.
(d)
Based on the histogram drawn in 14(c), state the number of strawberries with the mass of more than 50 g.
Solution:
(a)
(b)
(c)
(d)
Number of strawberries with the mass more than 50 g
= 3 + 2
= 5
# 6.7 SPM Practice (Long Questions)
Question 8 (12 marks):
Diagram shows the marks obtained by a group of 36 students in a Mathematics test.
Diagram
(a)
Based on the data in Diagram, complete Table in the answer space.
(b)
Based on the Table, calculate the estimated mean mark of a student.
(c)
For this part of the question, use graph paper.
By using the scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.
(d)
Based on the frequency polygon in 8(c), state the number of students who obtained more than 40 marks.
Table
Solution:
(a)
(b)
(c)
(d)
Number of students who obtained more than 40 marks
= 10 + 8 + 4
= 22 students
# SPM Practice Question 2
Question 2:
The third term and the sixth term of a geometric progression are 24 and $7\frac{1}{9}$ respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rn ≈ 0.
Solution:
(a)
(b)
(c)
Therefore, sum of the first n terms with n is very big approaching rn ≈ 0 is 162.
# Short Questions (Question 7 & 8)
Question 7:
The masses of mangoes in a stall have a normal distribution with a mean of 200 g and a standard deviation of 30 g.
(a) Find the mass, in g, of a mango whose z-score is 0.5.
(b) If a mango is chosen at random, find the probability that the mango has a mass of at least 194 g.
Solution:
µ = 200 g
σ = 30 g
Let X be the mass of a mango.
(a)
$\begin{array}{l}\frac{X-200}{30}=0.5\\ X=0.5\left(30\right)+200\\ X=215g\end{array}$
(b)
$\begin{array}{l}P\left(X\ge 194\right)\\ =P\left(Z\ge \frac{194-200}{30}\right)\\ =P\left(Z\ge -0.2\right)\\ =1-P\left(Z>0.2\right)\\ =1-0.4207\\ =0.5793\end{array}$
Question 8:
Diagram below shows a standard normal distribution graph.
The probability represented by the area of the shaded region is 0.3238.
(a) Find the value of k.
(b) X is a continuous random variable which is normally distributed with a mean of 80 and variance of 9.
Find the value of X when the z-score is k.
Solution:
(a)
P(Z > k) = 0.5 – 0.3238
= 0.1762
k = 0.93
(b)
µ = 80,
σ2 = 9, σ = 3
$\begin{array}{l}\frac{X-80}{3}=0.93\\ X=3\left(0.93\right)+80\\ X=82.79\end{array}$
# Short Questions (Question 5 & 6)
Question 5:
Diagram below shows the graph of a binomial distribution of X.
(a) the value of h,
(b) P (X ≥ 3)
Solution:
(a)
P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1
$\begin{array}{l}\frac{1}{16}+\frac{1}{4}+h+\frac{1}{4}+\frac{1}{16}=1\\ h=1-\frac{5}{8}\\ h=\frac{3}{8}\end{array}$
(b)
P (X ≥ 3) = P (X = 3) + P (X = 4)
$P\left(X\ge 3\right)=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$
Question 6:
The random variable X represents a binomial distribution with 10 trails and the probability of success is ¼.
(a) the standard deviation of the distribution,
(b) the probability that at least one trial is success.
Solution:
(a)
n = 10, p = ¼
$\begin{array}{l}\text{Standard deviation}=\sqrt{npq}\\ \text{}=\sqrt{10×\frac{1}{4}×\frac{3}{4}}\\ \text{}=1.875\end{array}$
(b)
$\begin{array}{l}P\left(X=r\right)=C{}_{r}{\left(\frac{1}{4}\right)}^{r}{\left(\frac{3}{4}\right)}^{10-r}\\ P\left(X\ge 1\right)\\ =1-P\left(X<1\right)\\ =1-P\left(X=0\right)\\ =1-C{}_{0}{\left(\frac{1}{4}\right)}^{0}{\left(\frac{3}{4}\right)}^{10}\\ =0.9437\end{array}$
# SPM Practice 2 (Question 11 & 12)
Question 11 (3 marks):
Diagram 6 shows the graph of a straight line
Diagram 11
Based on Diagram 6, express y in terms of x.
Solution:
Question 12 (3 marks):
The variables x and y are related by the equation $y=x+\frac{r}{{x}^{2}}$ , where r is a constant. Diagram 8 shows a straight line graph obtained by plotting
Diagram 12
Express h in terms of p and r.
Solution:
# SPM Practice 2 (Linear Law) – Question 1
Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of is plotted.
(a) Based on Table 1, construct a table for the values of
(c) Using the graph in 1(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.
Solution:
(a)
(b)
(c)(i)
(c)(ii)
# SPM Practice 3 (Linear Law) – Question 6
Question 6
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation $\frac{a}{y}=\frac{b}{x}+1$ , where k and p are constants.
(a) Based on the table above, construct a table for the values of $\frac{1}{x}$ and $\frac{1}{y}$ . Plot $\frac{1}{y}$ against $\frac{1}{x}$ , using a scale of 2 cm to 0.1 unit on the $\frac{1}{x}$ - axis and 2 cm to 0.2 unit on the $\frac{1}{y}$ - axis. Hence, draw the line of best fit.
(b) Use the graph from (b) to find the value of
(i) a,
(ii) b.
Solution
Step 1 : Construct a table consisting X and Y.
Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit
Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph
Step 4 : Rewrite the original equation given and reduce it to linear form
Step 5 : Compare with the values of m and c obtained, find the values of the unknown required
# SPM Practice 3 (Linear Law) – Question 5
Question 5
The following table shows the corresponding values of two variables, x and y, that are related by the equation $y=p{k}^{\sqrt{x}}$ , where p and k are constants.
(a) Plot $log 10 y$ against $x$ . Hence, draw the line of best fit
(b) Use your graph in (a) to find the values of p and k.
Solution
Step 1 : Construct a table consisting X and Y.
Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit
Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph
Step 4 : Rewrite the original equation given and reduce it to linear form
Step 5
:
Compare with the values of m and c obtained, find the values of the unknown required
# SPM Practice 2 (Linear Law) – Question 4
Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation $y=qx+\frac{p}{qx}$ , where p and q are constants.
One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against ${x}^{2}$ . Hence, draw the line of best fit
(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.
Solution
Step 1 : Construct a table consisting X and Y.
Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit |
## What table comes in 70 and 100?
Tables 2 to 10
Table of 2 Table of 3 Table of 10
2×7=14 3×7=21 10×7=70
2×8=16 3×8=24 10×8=80
2×9=18 3×9=27 10×9=90
2×10=20 3×10=30 10×10=100
## What is the calculation of 1 to 100?
How to Find the Sum of Natural Numbers 1 to 100? The sum of all natural numbers from 1 to 100 is 5050.
#### What is the 200 times table?
What is the 200 Times Table?
200×1 = 200 200
200×4 = 800 200 + 200 + 200 + 200 = 800
200×5 = 1000 200 + 200 + 200 + 200 + 200 = 1000
200×6 = 1200 200 + 200 + 200 + 200 + 200 + 200 = 1200
200×7 = 1400 200 + 200 + 200 + 200 + 200 + 200 + 200 = 1400
### Does 4 tables have 100?
Multiplication Table of 4 | Read and Write the Table of 4 | 4 Times Table
• Repeated addition by 4’s means the multiplication table of 4.
• (i) When 3 candle-stands having four candles each.
• By repeated addition we can show 4 + 4 + 4 = 12
• Then, four 3 times or 3 fours
• 3 × 4 = 12
• Therefore, there are 12 candles.
• (ii) When there are 7 packets of cookies having four cookies in each.
• By repeated addition we can show 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28
• Then, four 7 times or 7 fours
• 7 × 4 = 28
• Therefore, there are 28 cookies.
We will learn how to use the number line for counting the multiplication table of 4. (i) Start at 0. Hop 4, five times. Stop at 20,5 fours are 20 5 × 4 = 20 (ii) Start at 0. Hop 4, eight times. Stop at _. Thus, it will be 32 8 fours are 32 8 × 4 = 32 (iii) Start at 0. Hop 4, eleven times.
1. Stop at _. Thus, it will be 44
2. 11 fours are 44 11 × 4 = 44
3. How to read and write the table of 4?
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The above diagram will help us to read and write the 4 times table.
Read 1 four is 4 2 fours are 8 3 fours are 12 4 fours are 16 5 fours are 20 6 fours are 24 7 fours are 28 8 fours are 32 9 fours are 36 10 fours are 40 11 fours are 44 12 fours are 48 Write 1 × 4 = 4 2 × 4 = 8 3 × 4 = 12 4 × 4 = 16 5 × 4 = 20 6 × 4 = 24 7 × 4 = 28 8 × 4 = 32 9 × 4 = 36 10 × 4 = 40 11 × 4 = 44 12 × 4 = 48
ul>
• Now we will learn how to do forward counting and backward counting by 4’s.
• Forward counting by 4’s: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100,
• Backward counting by 4’s:, 100, 96, 92, 88, 84, 80, 76, 72, 68, 64, 60, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20, 16, 12, 8, 4, 0.
• Didn’t find what you were looking for? Or want to know more information about, Use this Google Search to find what you need.
Multiplication Table of 4 | Read and Write the Table of 4 | 4 Times Table
### Is there 100 in 5 table?
Multiplication Table of 5 | Read and Write the Table of 5 | 5 Times Table
• Repeated addition by 5’s means the multiplication table of 5.
• (i) When 6 bowls of five fruits each.
• By repeated addition we can show 5 + 5 + 5 + 5 + 5 + 5 = 30
• Then, five 6 times or 6 fives
• 6 × 5 = 30
• Therefore, there are 30 fruits.
• (ii) When 9 baskets each having 5 shirts.
• By repeated addition we can show 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45
• Then, five 9 times or 9 fives
• 9 × 5 = 45
• Therefore, there are 45 shirts.
We will learn how to use the number line for counting the multiplication table of 5. (i) Start at 0. Hop 5, four times. Stop at 20,4 fives are 20 4 × 5 = 20 (ii) Start at 0. Hop 5, seven times. Stop at _. Thus, it will be 35 7 fives are 35 7 × 5 = 35 (iii) Start at 0. Hop 5, twelve times.
1. Stop at _. Thus, it will be 60
2. 12 fives are 60 12 × 5 = 60
3. How to read and write the table of 5?
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The above chart will help us to read and write the 5 times table.
Read 1 five is 5 2 fives are 10 3 fives are 15 4 fives are 20 5 fives are 25 6 fives are 30 7 fives are 35 8 fives are 40 9 fives are 45 10 fives are 50 11 fives are 55 12 fives are 60 Write 1 × 5 = 5 2 × 5 = 10 3 × 5 = 15 4 × 5 = 20 5 × 5 = 25 6 × 5 = 30 7 × 5 = 35 8 × 5 = 40 9 × 5 = 45 10 × 5 = 50 11 × 5 = 55 12 × 5 = 60
ul>
• Now we will learn how to do forward counting and backward counting by 5’s.
• Forward counting by 5’s: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120,
• Backward counting by 5’s:, 120, 115, 110, 105, 100, 95, 90, 85, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 10, 5, 0.
• Didn’t find what you were looking for? Or want to know more information about, Use this Google Search to find what you need.
Multiplication Table of 5 | Read and Write the Table of 5 | 5 Times Table
## What is 1 * 2 * 3 * 4 * 5 all the way to 100?
Here a = 1. ➡ 5050. Sum of 1+2+3+4 100 is 5050.
### What is the sum of 1 to 99?
Answer: The number series 1, 2, 3, 4,., 98, 99. Therefore, 4950 is the sum of positive integers upto 99
### How do you solve 1.5 100?
The 1.5 percent of 100 is equal to the number 1.5. To compute this answer quickly, just multiply the fraction 0.015 by the number 100. Another way to find the exact answer is by taking the fraction 1.5/100 and multiplying it by 100. The final answer will come out to be 1.5 when you solve the equation.
#### How many times 3 comes in 1 to 100?
It appears in the numbers 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, and 38. And each number we will get 20 times (1,2,3,4,5,6,7,8,9). Hence number of 3 occur between 1 – 100 is 20 times.
You might be interested: Ccl Points Table 2023
## What number goes into 1 and 100?
What is the GCF of 1 and 100? | Thinkster Math To get started using this method of finding the Greatest Common Factor of and, we should first find and list all the factors of each number. You can see how this is done by looking at the “Factors of” articles that are linked to below.
Let’s take a look at the factors for each of these numbers, 1 and 100:: 1: 1, 2, 4, 5, 10, 20, 25, 50, 100
When you compare the two lists of factors, you can see that the only common factor is 1. So, in this case, the GCF of 1 and 100 is 1. : What is the GCF of 1 and 100? | Thinkster Math
## What does it mean for a table to be one to one?
What is a one-to-one relationship? – One-to-one relationships are frequently used to indicate critical relationships so you can get the data you need to run your business. A one-to-one relationship is a link between the information in two tables, where each record in each table only appears once.
1. For example, there might be a one-to-one relationship between employees and the cars they drive.
2. Each employee appears only once in the Employees table, and each car appears only once in the Company Cars table.
3. You might use one-to-one relationships if you have a table containing a list of items, but the specific information you want to capture about them varies by type.
For example, you might have a contacts table in which some people are employees and other people are subcontractors. For the employees, you want to know their employee number, their extension, and other key information. For subcontractors, you want to know their company name, phone number, and bill rate, among other things.
## How many between 1 and 100 includes 1 and 100?
Since we’re dealing with numbers between 1 and 100, we cannot include 1 and 100 in our calculations. So, there are 98 numbers between 1 and 100. |
# Irrational numbers definition explained with example
Do you know what are the irrational numbers and how to give the irrational numbers definition? Irrational numbers are one of the types of numbers that we have in mathematics.
By the name irrational, we get some idea that it is not a rational number. Actually, after learning rational numbers we are going to study about irrational numbers. So if you know rational numbers definition then writing irrational numbers definition is not that much difficult. Now let’s start learning about irrational numbers.
## Irrational numbers definition and example:
Irrational numbers definition can be stated as “the numbers which we cannot write in the $\frac { p }{ q }$ form is called as irrational numbers”. Another definition we can give as “non terminating non recurring decimal numbers are irrational numbers”. The meaning of non terminating Is division process does not end i.e. remainder won’t become zero and recurring means some set of digits will be repeating continuously.
Examples:
$\sqrt { 23 }$ , $\sqrt { 91 }$ ,$\sqrt { 147 }$ ,$\sqrt { 253 }$ , ………………
1.232621247536510245237890123698017845211…………..
0.1014578129658702384308705957198124587………………
Some Example problems on irrational numbers:
Is $\sqrt { 27 }$ is an irrational number? If yes give the reason.
Solution:
Yes, it is an irrational number because it satisfies irrational numbers definition.
Check Whether $\sqrt { 49 }$ is irrational or not?
Solution:
$\sqrt { 49 }$ Is not an irrational number because its value is equals to 7 which is a rational number.
Note:
Under root, if there is a perfect square number then it is not an irrational number.
Example:
$\sqrt { 81 }$,$\sqrt { 64 }$ ,$\sqrt { 4 }$, …………are not irrational numbers because under root all are perfect square numbers.
Suppose under root if there is a non perfect square number then it is an irrational number.
Example:
$\sqrt { 7 }$ ,$\sqrt { 55 }$ ,$\sqrt { 61 }$ , ………………are irrational numbers because under root all are non perfect square numbers.
Hope that irrational numbers definition is clear and I hope you understood it. |
# MATHEMATICS FORM FIVE TOPIC 9: DIFFERENTIATION
DERIVATIVES
Slope of a curve
A curve has different slopes at each point. Let A, B, and C be different points of a curve f (x)
Where ðx is the small increase in x
ðy is the small increase in y
The slope of chord AC =
If C moves right up to A the chord AC becomes the tangent to the curve at A and the slope at A is the limiting value of
This is known as differentiating by first principle
From the first principle
Example
Differentiate the following with respect to x
i)y = x2+3x
Solution
Differentiation of products functions [ product rule]
Let y =uv
Where u and v are functions of x
If x → x+ðx
u → u +ðu
v → v+ðv
y → ðy +y
y= uv ……i)
Therefore
y+ðy = [ u+ðu][v+ðv]
y+ðy = uv +uðv+vðu +ðuðv….ii
Subtract (i) from (ii)
δy =uðv +vðu +ðuðv
Therefore
Therefore
Therefore
If y= uv
It is the product rule
Examples
Differentiate the following with respect to x
i) y = [ x2+3x] [4x+3]
ii) y = [ +2] [x2+2]
Solution
Y = [x2+3x] [4x+3]
DIFFERENTIATION OF A QUOTIENT [QUOTIENT RULE]
Exercise
Differentiate the following with respect to x
I.
II.
DIFFERENTIATION OF A FUNCTION [CHAIN RULE]
If y = f(u), where u = f(x)
PARAMETRIC EQUATIONS
IMPLICIT FUNCTION
Implicit function is the one which is neither x nor y a subject e.g.
1) x2+y2 = 25
2) x2+y2+2xy=5
One thing to remember is that y is the function of x
Exercise
Find when x+ y3 – 3xy2 = 8
Differentiation of trigonometric functions
3) Let y = tan -1 x
x = tan y
Let y =
X =
∴
Exercises
Differentiate the following with respect to x.
i) Sin 6x
ii) Cos (4x2+5)
iii) Sec x tan 2 x
Differentiate sin2 (2x+4) with respect to x
Differentiate the following from first principle
i) Tan x
Differentiation of logarithmic and exponential functions
1- Let y = ln x
Differentiation of Exponents
1) Let y = ax
If a function is in exponential form apply natural logarithms on both sides
i.e. ln y = ln ax
ln y =x ln a
Since â„®x does not depend on h,then
Therefore
Example
Find the derivative of y = 105x
Solution
Y = 105x
Iny = In105x
Therefore
Exercise
Find the derivatives of the following functions
a) a) Y =
b) b) Y =
c) c)Y=
APPLICATION OF DIFFERENTIATION
Differentiation is applied when finding the rates of change, tangent of a curve, maximum and minimum etc
i) The rate of change
Example
The side of a cube is increasing at the rate of 6cm/s. find the rate of increase of the volume when the length of a side is 9cm
Solution
A hollow right circular cone is held vertex down wards beneath a tap leaking at the rate of 2cm3/s. find the rate of rise of the water level when the depth is 6cm given that the height of the cone is 18cm and its radius is 12cm.
Solution
V
A horse trough has triangular cross section of height 25cm and base 30cm and is 2m long. A horse is drinking steadily and when the water level is 5cm below the top is being lowered at the rate of 1cm/min find the rate of consumption in litres per minute
Solution
Volume of horse trough
From the ratio of the corresponding sides
A rectangle is twice as long as it is broad find the rate change of the perimeter when the width of the rectangle is 1m and its area is changing at the rate of 18cm2/s assuming the expansion is uniform
Solution
TANGENTS AND NORMALS
From a curve we can find the equations of the tangent and the normal
Example
i. Find the equations of the tangents to the curve y =2x2 +x-6 when x=3
Solution
(x. y)= (3, 5) is the point of contact of the curve with the tangent
But
Gradient of the tangent at the curve is
Example
ii. Find the equation of the tangent and normal to the curve y = x2 – 3x + 2 at the point where it cuts y axis
Solution
The curve cuts y – axis when x = 0
Slope of the tangent [m] = -3
Equation of the tangent at (0, 2) is
Slope of the normal
From; m1m2 = -1,Given m1=-3
Equation of the normal is
Exercise
Find the equation of the tangent to 2x2 – 3x which has a gradient of 1
Find the equations of the normal to the curve y = x2-5x +6 at the points where the curve cuts the x axis
Stationary points [turning points]
A stationary point is the one where by = 0 it involves:
Minimum turning point
Maximum turning point
Point of inflection
#### Nature of the curve of the function
At point A, a maximum value of a function occurs
At point B, a minimum value of a function occurs
At point C, a point of inflection occurs
At the point of inflection is a form of S bend
Note that
Points A, B and C are called turning points on the graph or stationary values of the function
Investigating the nature of the turning point
Minimum points
At turning points the gradient changes from being negative to positive i.e.
Increasing as x- increases
Is positive at the minimum point
Is positive for minimum value of the function of (y)
Maximum points
At maximum period the gradient changes from positive to negative
i.e.
Decreases as x- increases
Is negative at the maximum value of the function (y)
Point of inflection
This is the changes of the gradient from positive to positive
Is positive just to the left and just to the left
This is changes of the gradient from negative to negative.
Is negative just to the left and just to the right
Is zero for a point of inflection i.e
Is zero for point of inflection
Examples
Find the stationary points of the and state the nature of these points of the following functions
Y = x4 +4x3-6
Solution
At stationary points
Therefore,
Then the value of a function
At x = 0, y = -6
X= -3, y =-33
Stationary point at (0,-6) and (-3,-33)
At (0,-6)
Point (0,-6) is a point of inflection
At (-3,-33)
At (-3, -33) is a minimum point
Alternatively,
You test by taking values of x just to the right and left of the turning point
Exercise
1) 1. Find and classify the stationary points of the following curves
a) (i) y = 2x-x2
b) (ii) y = +x
c) (iii) y= x2(x2– 8x)
2) 2. Determine the smallest positive value of x at which a point of inflection occurs on the graph of y = 3â„®2x cos (2x-3)
3) 3. If 4x2 + 8xy +9y2 8x – 24y +4 =0 show that when = 0,
x + y = 1. Hence find the maximum and minimum values of y
Example
1. A farmer has 100m of metal railing with which to form two adjacent sides of a rectangular enclosure, the other two sides being two existing walls of the yard meeting at right angles, what dimensions will give the maximum possible area?
Solution
Where, W is the width of the new wall
L is the length of the new wall
The length of the metal railing is 100m
2. An open card board box width a square base is required to hold 108cm3 what should be the dimensions if the area of cardboard used is as small as possible
Solution
Exercise
The gradient function of y = ax2 +bx +c is 4x+2. The function has a maximum value of 1, find the values of a, b, and c
MACLAURIN’S SERIES [from power series ]
Let f(x) = a1 +a2x+a3x2 +a4x3 +a5x4+ a6x5…….i
In order to establish the series we have to find the values of the constant co efficient a1, a2, a3, a4, a5, a6 etc
Put x = 0 in …i
Putting the expressions a1,a2,a3,a4,a5,………back to the original series and get
which is the maclaurin series.
Examples
Expand the following
i) â„®x
ii) f(x) = cos x
Solution
Exercise
Write down the expansion of
If x is so small that x3 and higher powers of x may be neglected, show that
TAYLOR’S SERIES
Taylor’s series is an expansion useful for finding an approximation for f(x) when x is close to a
By expanding f(x) as a series of ascending powers of (x-a)
f(x) = a +a1(x-a) +a2(x-a)2+a3(x-a)3 +……..
This becomes
Example
Expand in ascending powers of h up to the h3 term, taking as
1.7321 And 5.5 as 0.09599c find the value of cos 54.5 to three decimal places
Solution
Obtain the expansion of in ascending powers of x as far as the x3term
Introduction to partial derivative
Let f (x, y) be a differentiable function of two variables. If y kept constant and differentiates f (assuming f is differentiable with respect to x
Keeping x constant and differentiate f with respect to y
Example
find the partial derivatives of fx and fy
If f(x, y) = x2y +2x+y
Solution
Find fx and fy if f (x,y) is given by
f(x, y) = sin(xy) +cos x
Solution
Exercise
1. find fx and fy if f(x,y) is given by
Suppose compute
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Most mathematical activity involves the use of pure reason to discover or prove the properties of abstract objects, which consist of either abstractions from nature or—in modern mathematics—entities that are stipulated with certain properties, called axioms. A mathematical proof consists of a succession of applications of some deductive rules to already known results, including previously proved theorems, axioms and (in case of abstraction from nature) some basic properties that are considered as true starting points of the theory under consideration.
Mathematics is used in science for modeling phenomena, which then allows predictions to be made from experimental laws. The independence of mathematical truth from any experimentation implies that the accuracy of such predictions depends only on the adequacy of the model. Inaccurate predictions, rather than being caused by incorrect mathematics, imply the need to change the mathematical model used. For example, the perihelion precession of Mercury could only be explained after the emergence of Einstein’s general relativity, which replaced Newton’s law of gravitation as a better mathematical model.
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Integral Calculus
### Line, Surface, and Volume Integrals
(a) Line Integrals
A line integral is an expression of the form
whereis a vector function, is the infinitesimal displacement vector and the integral is to be carried out along a prescribed path P from point a to point b. If the path in question forms a closed loop (that is, if b = a), put a circle on the integral sign:
At each point on the path we take the dot product of (evaluated at that point) with the displacement to the next point on the path. The most familiar example of a line integral is the work done by a force
Ordinarily, the value of a line integral depends critically on the particular path taken from a to b, but there is an important special class of vector functions for which the line integral is independent of the path, and is determined entirely by the end points(A force that has this property is called conservative.)
Example 14: Calculate the line integral of the function from the point a = (1, 1, 0) to the point b = (2, 2, 0), along the paths (1) and (2) as shown in figure. What is for the loop that goes from a to b along (1) and returns to a along (2)?
Since Path (1) consists of two parts. Along the “horizontal” segment dy = dz = 0, so
(i)
On the “vertical” stretch dx = dz = 0, so
(ii)
By path (1), then,
Meanwhile, on path (2) x = y, dx = dy, and dz = 0, so
so
For the loop that goes out (1) and back (2), then,
Example 15: Find the line integral of the vector around a square of side ‘b’ which has a corner at the origin, one side on the x axis and the other side on the y axis.
In a Cartesian coordinate syste
Along OP, y = 0, dy = 0 ⇒
Along PQ , x = b, dx = 0 ⇒
Along QR, y = b, dy = 0 ⇒
Along RO, x = 0, dx = 0 ⇒
Example 16: Compute the line integralalong the triangular path shown in the figure.
Line Integral
On path C1, x = 0, y = 0,
On path C2, x = 0, z = 0,
On path C3 the slope of line is -2 and intercept on z axis is 2 ⇒ z = -2y + 2 = 2 1 (1 - y) and the connecting points are (0, 1, 0) and (0, 0, 2)
On C3, x=0, dx = 0
Example 17: Given in cylindrical coordinates. Find where c1 and c2 are contours shown in the figure.
In cylindrical coordinate system
In figure on curve c1,Ф varies from 0 to 2π, r = b and dr = 0
On curve c2 , r = a,Ф varies from 0 to - 2π , and dr = 0 ⇒
So,
(b) Surface Integrals
A surface integral is an expression of the form
where is again some vector function, and is an infinitesimal patch of area, with direction perpendicular to the surface(as shown in figure). There are, of course, two directions perpendicular to any surface, so the sign of a surface integral is intrinsically ambiguous. If the surface is closed then “outward” is positive, but for open surfaces it’s arbitrary. If describes the flow of a fluid (mass per unit area per unit time), then arepresents the total mass per unit time passing through the surface-hence the alternative name, “flux.”
Ordinarily, the value of a surface integral depends on the particular surface chosen, but there is a special class of vector functions for which it is independent of the surface, and is determined entirely by the boundary line.
Example 18: Calculate the surface integral of A over five sides (excluding the bottom) of the cubical box (side 2) as shown in figure. Let “upward and outward” be the positive direction, as indicated by the arrows.
Taking the sides one at a time:
(i)
(ii)
(iii)
(iv)
(v)
Evidently the total flux is
Example 19: Given a vector Evaluate over the surface of the cube with the centre at the origin and length of side ‘a’.
The surface integral is performed on all faces. The differential surface on the different faces are Face abcd,
Face efgh,
Face cdfe,
Face aghb,
Similarly for the other two faces adfg and bceh we can find the surface integral withrespectively. The addition of these two surface integrals will be zero.
In the present case sum of all the surface integral
Example 20: Use the cylindrical coordinate system to find the area of a curved surface on the right circular cylinder having radius = 3 m and height = 6 m and 30º << 120º.
From figure, surface area is required for a cylinder when r = 3m, z = 0 to 6m,
In cylindrical coordinate system, the elemental surface area as scalar is
Taking the magnitude only
Example 21: Use spherical coordinate system to find the area of the strip on the spherical shell of radius ‘a’. Calculate the area when α = 0 and β = π.
Sphere has radius ‘a’ and θ varies between α and β.
For fixed radius the elemental surface is da = (rsinθ d∅)(rdθ) = r2 sinθdθd∅
For α = 0, β = π, Area = 2πa2 (1 + 1) = 4πa2 , is surface area of the sphere.
(c) Volume Integrals
A volume integral is an expression of the form
where T is a scalar function and dτ is an infinitesimal volume element. In Cartesian coordinates,
dτ = dx dy dz.
For example, if T is the density of a substance (which might vary from point to point) then the volume integral would give the total mass. Occasionally we shall encounter volume integrals of vector functions:
because the unit vectors are constants, they come outside the integral.
The document Integral Calculus | Mathematical Models - Physics is a part of the Physics Course Mathematical Models.
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## FAQs on Integral Calculus - Mathematical Models - Physics
1. What is integral calculus?
Ans. Integral calculus is a branch of mathematics that deals with the concept of integration and finding the area under curves. It involves finding antiderivatives and evaluating definite integrals to solve problems related to accumulation of quantities, area, volume, and other physical and mathematical applications.
2. What is the importance of integral calculus in IIT JAM exam?
Ans. Integral calculus plays a crucial role in the IIT JAM exam, particularly in subjects like Mathematics and Physics. Many questions in these subjects require the application of integral calculus to solve problems related to rate of change, optimization, finding areas and volumes, and analyzing the behavior of functions. A good understanding of integral calculus is essential to score well in the exam.
3. What are the basic integration techniques required for IIT JAM?
Ans. Some of the basic integration techniques required for IIT JAM include: - Power rule: Integration of functions of the form xn. - Substitution method: Replacing variables to simplify the integrand. - Integration by parts: Breaking down the integrand using the product rule. - Trigonometric identities: Utilizing trigonometric identities to simplify trigonometric integrals. - Partial fractions: Breaking down rational functions into simpler fractions. Knowing these techniques and practicing them extensively will help in solving integration problems effectively in the IIT JAM exam.
4. How can I improve my skills in integral calculus for the IIT JAM exam?
Ans. To improve your skills in integral calculus for the IIT JAM exam, you can follow these steps: 1. Understand the fundamental concepts: Make sure you have a clear understanding of the basic concepts, definitions, and properties of integration. 2. Practice regularly: Solve a variety of integration problems from textbooks, previous year question papers, and online resources to enhance your problem-solving skills. 3. Learn integration techniques: Familiarize yourself with various integration techniques such as substitution, integration by parts, trigonometric identities, and partial fractions. 4. Seek guidance: Consult your teachers, professors, or subject experts to clarify any doubts or difficulties you may have while studying integral calculus. 5. Mock tests and sample papers: Solve mock tests and sample papers to get acquainted with the exam pattern and time management. By following these steps consistently, you can significantly improve your skills in integral calculus for the IIT JAM exam.
5. Can I score well in the IIT JAM exam without a strong grasp of integral calculus?
Ans. It is highly unlikely to score well in the IIT JAM exam without a strong grasp of integral calculus. Integral calculus is a fundamental concept in various subjects covered in the exam, and many questions require its application. Scoring well in subjects like Mathematics and Physics, which heavily rely on integral calculus, would be challenging without a thorough understanding of the topic. Therefore, it is crucial to dedicate sufficient time and effort to learn and practice integral calculus to increase your chances of scoring well in the IIT JAM exam.
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