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# Question: 22 is what percent of 40?
## Answer: 55 percent, which is equivalent to saying 22 out of 40 is 55%
### How to calculate 22 is what percent of 40
22 ÷ 40 × 100
= (22 × 100) ÷ 40
= 2200 ÷ 40
= 55
Now we have: 22 is what percent of 40 = 55%
## A simple step by step solution for calculating 22 is what percent of 40
We already have our first value 22 and the second value 40. Let's assume the unknown value is Y which answer we will find out.
As we have all the required values we need, Now we can put them in a simple mathematical formula as below:
STEP 1Y = 22/40
By multiplying both numerator and denominator by 100 we will get:
STEP 2Y = 22/40 × 100/100 = 55/100
STEP 3Y = 55
Finally, we have found the value of Y which is 55 and that is our answer.
If you want to use a calculator to find 22 is what percent of 40, simply enter 22 ÷ 40 × 100 and you will get your answer which is 55
Here is a calculator to solve percentage calculations such as 22 is what percent of 40. You can solve this type of calculation with your values by entering them into the calculator's fields, and click 'Calculate' to get the result and explanation.
is what percent of
## Sample questions, answers, and how to
Question: You have found your aunt’s freshly baked cookies vault which, she just baked! She told you not to eat more than 22 cookies from those, and you ended up eating 40 of them! What percent of the cookies have you eaten?
How To: The key words in this problem are "What Percent" because they let us know that it's the Percent that is missing. So the two numbers that it gives us must be the "Total" and the "Part" we have.
Part/Total = Percent
In this case, it's the Total of those cookies that your aunt's baked. So we put 22 on top and 40 on the bottom of the fraction, and now we're ready to figure out the part we don't know; the Percent.
22/40 = Percent
To find the percent, all we need to do is convert the fraction into its percent form by multiplying both top and bottom part by 100 and here is the way to figure out what the Percent is:
22/40 × 100/100 = 55/100
55 = Percent
And that means you ended up eating 55 Percent of the cookies that she baked.
## Another step by step method
Step 1: Let's solve the equation for Y by first rewriting it as: 100% / 40 = Y% / 22
Step 2: Drop the percentage marks to simplify your calculations: 100 / 40 = Y / 22
Step 3: Multiply both sides by 22 to isolate Y on the right side of the equation: 22 ( 100 / 40 ) = Y
Step 4: Computing the left side, we get: 55 = Y
This leaves us with our final answer: 22 is 55 percent of 40
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# 2008 AMC 10A Problems/Problem 7
## Problem
The fraction
$$\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}$$ simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$
## Solution
Simplifying, we get $$\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.$$ Factoring out $3^{4012}$ in the numerator and factoring out $3^{4010}$ in the denominator gives us $$\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.$$ Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.$
## Solution 2
Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes
$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$
$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$
$= \boxed{\text{(E)}9}$
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Subject Area Lessons
Mathematics, level: Elementary
Posted Tue Nov 7 07:26:23 PST 2000 by Edwin Sokalski ([email protected]).
IUP, Indiana Pennsylvania
Materials Required: colored blocks, paper with numbers on them
Objectives
2. At the end of this lesson the students should be able to add different objects
together to find out how many objects there are total
1. The students should be able to explain up to 3 different reasons that being able
Materials
1. 5 colorful building blocks
2. 5 pieces of paper with the numbers 1 through 10 on them
Introduction
The teacher should start out by explaining that addition is simply the combination of 2 or more different groups of numbers. For example if you have a pile with 2 objects in it and another pile with one object in it you get three. The teacher should count the separate piles, then put the piles together and count them again. Next the teacher should get five students to come up to the front of the room. The students should be split up into 2 groups one with 3 individuals and the other with 2. The students in the first group should each get pieces of paper with the numbers 1, 2 and 3 written on them. The students in the second group should get the numbers 4 and 5. The class should then count the students in the first group. The teacher should then say that we are now going to add 2 and 3 together. As the group of two moves to the group of three the teacher should say that we have just added 2 to 3. The students should then count the this new group and come up with the number five. The teacher should not move on if the students do not readily recount the new group and do not seem to understand that the groups have now been combined.
Main lesson
The teacher should now talk to the students about the different ways we can use addition in our everyday lives. The teacher should give the example of figuring out how many fish are in the fish tank. The fish could be separated in several ways depending on what fish are in the tank. If there is 1 black fish and 2 gold fish this is an easy number for the children to add. The teacher should then have the students take turns going around the room and looking for things they could add together. The teacher should be accepting of most anything the students want to add. If one student want to add three block and one desk that is ok.
The teacher should then ask the students about things in their homes that they can add and ask everybody to think of something at home tonight that it could be beneficial to add.
Conclusion
As a conclusion to this lesson the teacher should have several different objects that she has gathered for the students to add. Each student should get a chance to do this. If any students make a mistake the teacher should correct them and help the students by having them count each pile and then count the piles together.
Evaluation
The students should be informally evaluated based on how the teacher has observed them answering questions. A short quiz could be used as well, but should not take more then five minutes to finish.
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DEFINITION A function f : A B is a one-to-one correspondence (called a bijection) iff f is one-to-one and onto B. We write f : A B to indicate that.
Presentation on theme: "DEFINITION A function f : A B is a one-to-one correspondence (called a bijection) iff f is one-to-one and onto B. We write f : A B to indicate that."— Presentation transcript:
DEFINITION A function f : A B is a one-to-one correspondence (called a bijection) iff f is one-to-one and onto B. We write f : A B to indicate that f is a bijection. Look at the example and illustrations on pages 214 and 215. 1-1 onto Theorem 4.4.1 If f : A B and g : B C, then g ◦ f : A C. That is, the composite of one-to-one correspondences is a one-to-one correspondence. 1-1 Proof: This is a combination of Theorems 4.3.1 and 4.3.3. onto
Theorem 4.4.2 Let F : A B (i.e., F is a function from set A to set B.) (a)F –1 is a function from Rng(F) to A iff F is one-to-one. (b)If F –1 is a function, then F –1 is one-to-one. Proof of (a): Suppose F : A B. Suppose F –1 is a function from Rng(F) to A. To show F is one-to-one, suppose F(x) = F(y) = z. (x, z) F and (y, z) F change of notation (z, x) F –1 and (z, y) F –1 _______________________definition of F –1 x = y_______________________supposition that F –1 is a function F is one-to-one F(x) = F(y) x = y Suppose F is one-to-one.
Proof of (a): Suppose F : A B. Suppose F –1 is a function from Rng(F) to A. To show F is one-to-one, suppose F(x) = F(y) = z. (x, z) F and (y, z) F change of notation (z, x) F –1 and (z, y) F –1 _______________________definition of F –1 x = y_______________________supposition that F –1 is a function F is one-to-one F(x) = F(y) x = y Suppose F is one-to-one. To show F –1 is a function, suppose (x, y) F –1 and (x, z) F –1. (y, x) F and (z, x) F _______________________definition of F –1 F(y) = F(z) follows from previous line y = z_______________________supposition that F is one-to-one F –1 is a function (x, y) F –1 and (x, z) F –1 y = z Dom(F –1 ) = Rng(F)Theorem _______________________3.1.2(a) We have now shown that F –1 is a function from Rng(F) to A
Proof of (b): Suppose F : A B. Theorem 4.4.2 Let F : A B (i.e., F is a function from set A to set B.) (a)F –1 is a function from Rng(F) to A iff F is one-to-one. (b)If F –1 is a function, then F –1 is one-to-one. Suppose F –1 is a function from Rng(F) to A. F = (F –1 ) –1 Theorem _______________________3.1.3(a) F –1 is one-to-one applying part (a) to _______________F –1 Corollary 4.4.3 If F : A B, then F –1 : B A. That is, the inverse of a one-to-one correspondence is a one-to-one correspondence. 1-1 onto 1-1 onto Theorem 4.4.4 Let F : A B and G : B A. Then (a) G = F –1 iff G ◦ F = I A and F ◦ G = I B. (b) If F is one-to-one and onto B, then G = F –1 iff G ◦ F = I A or F ◦ G = I B.
Proof of (a): Suppose F : A B and G : B A. Suppose G = F –1 Theorem _______________3.1.2(a)B = Dom(G) = Dom(F –1 ) = Rng(F) Theorem _______________4.2.4G ◦ F = I A and F ◦ G = I B Suppose G ◦ F = I A and F ◦ G = I B. every identity function is one-to-oneG ◦ F = I A is one-to-one Theorem _______________4.3.4F is one-to-one every identity function is ontoF ◦ G = I B is onto B Theorem _______________4.3.2F is onto B F is one-to-one & Theorem ______________4.4.2(a)F –1 is a function on B F –1 = F –1 ◦ I B = F –1 ◦ (F ◦ G) = (F –1 ◦ F ) ◦ G = I A ◦ G = G properties of the identity function Note: The textbook proof erroneously references Theorem 4.2.3. Theorem 4.4.4 Let F : A B and G : B A. Then (a) G = F –1 iff G ◦ F = I A and F ◦ G = I B. We have now proven part (a).
Theorem 4.4.4 Let F : A B and G : B A. Then (a) G = F –1 iff G ◦ F = I A and F ◦ G = I B. (b) If F is one-to-one and onto B, then G = F –1 iff G ◦ F = I A or F ◦ G = I B. Proof of (b): Suppose F : A B. Suppose G = F –1. 1-1 onto We can say G ◦ F = I A or F ◦ G = I B from part (a) Now suppose that G ◦ F = I A or F ◦ G = I B. We first show G F –1. Let (b, a) G (i.e, b B and a A) Case 1: G ◦ F = I A properties of the identity function (a, a) I A (a, a) G ◦ F supposition that _____________G ◦ F = I A (a, c) F /\ (c, a) G for some c B _____________________ definition of G ◦ F (a, b) F /\ (b, a) G, that is, c = b F is one-to-one and onto B
Proof of (b): Suppose F : A B. Suppose G = F –1. 1-1 onto We can say G ◦ F = I A or F ◦ G = I B from part (a) Now suppose that G ◦ F = I A or F ◦ G = I B. We first show G F –1. Let (b, a) G (i.e, b B and a A) Case 1: G ◦ F = I A properties of the identity function (a, a) I A (a, a) G ◦ F supposition that _____________G ◦ F = I A (a, c) F /\ (c, a) G for some c B _____________________ definition of G ◦ F (a, b) F /\ (b, a) G, that is, c = b F is one-to-one and onto B (b, a) F –1 _____________________definition of F –1 G F –1 (b, a) G (b, a) F –1
Proof of (b): Suppose F : A B. Suppose G = F –1. 1-1 onto We can say G ◦ F = I A or F ◦ G = I B from part (a) Now suppose that G ◦ F = I A or F ◦ G = I B. We first show G F –1. Let (b, a) G (i.e, b B and a A) Case 2: F ◦ G = I B properties of the identity function (b, b) I B (b, b) F ◦ G supposition that _____________F ◦ G = I \B (b, d) G /\ (d, b) F for some d A _____________________ definition of F ◦ G (b, a) G /\ (b, d) G, that is, d = a G is a function (b, a) F –1 _____________________ (a, b) F & definition of F –1 G F –1 (b, a) G (b, a) F –1 In either case, (b, a) F –1 which shows that G F –1.
We now show F –1 G. Let (b, a) F –1 (i.e, b B and a A) Case 1: G ◦ F = I A properties of the identity function (a, a) I A (a, a) G ◦ F supposition that _____________G ◦ F = I A (a, c) F /\ (c, a) G for some c B _____________________ definition of G ◦ F (a, c) F /\ (a, b) F, that is, c = b (b, a) F –1 and F is ________ (b, a) G _____________________ (c, a) G and c = b F –1 G(b, a) F –1 (b, a) G a function
Let (b, a) F –1 (i.e, b B and a A) Case 2: F ◦ G = I B properties of the identity function (b, b) I B (b, b) F ◦ G supposition that _____________F ◦ G = I \B (b, d) G /\ (d, b) F for some d A _____________________ definition of F ◦ G (b, a) F –1 _____________________ (a, b) F & definition of F –1 F –1 G(b, a) G (b, a) F –1 In either case, (b, a) G which shows that F –1 G. We now show F –1 G. (a, b) F /\ (d, b) F, that is, d = a 1-1 (b, a) F –1 and F is ________ We conclude G = F –1, since G F –1 and F –1 G. Look at the examples on page 216.
1 (a) Exercises 4.4 (pages 218-219)
1 (b)
2 (b)
2 (c)
2 (d)
3 (d)
Section 5.1 3
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# How do you find the average value of the function for f(x)=2-1/2x, 0<=x<=4?
Mar 22, 2018
The average value of the function on the given interval is $1$.
#### Explanation:
Average value of a function is given by
$A = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$
$A = \frac{1}{4} {\int}_{0}^{4} 2 - \frac{1}{2} x \mathrm{dx}$
$A = \frac{1}{4} {\left[2 x - \frac{1}{4} {x}^{2}\right]}_{0}^{4}$
$A = \frac{1}{4} \left(2 \left(4\right) - \frac{1}{4} {\left(4\right)}^{2}\right)$
$A = \frac{1}{4} \left(8 - 4\right)$
$A = \frac{1}{4} \left(4\right)$
$A = 1$
Hopefully this helps!
Mar 22, 2018
The average value of $f$ over $\left[0 , 4\right]$ is 1.
#### Explanation:
The average value of a function over an interval is its (definite) integral over that interval divided by the length of the interval.
${\int}_{0}^{4} \left(2 - \frac{x}{2}\right) \mathrm{dx} = {\left[2 x - {x}^{2} / 4\right]}_{0}^{4}$
$= 4 - 0$
$= 4$
$\frac{{\int}_{0}^{4} \left(2 - \frac{x}{2}\right) \mathrm{dx}}{4 - 0} = \frac{4}{4} = 1$
The average value of $f$ over $\left[0 , 4\right]$ is 1.
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Sunday, March 11, 2018
Forming a Quadratic Polynomial by Knowing its Roots
Forming a Quadratic Polynomial by Knowing its Roots
Introduction
We are given two roots of a quadratic equation x = A and x = B and asked to construct a quadratic polynomial. Believe it or not, this is (almost) enough information to accomplish this task.
Since A and B are roots, that means:
0 = (x – A)*(x – B)
It’s now just a matter of some algebra:
0 = x^2 – A*x – B*x + A*B
0 = x^2 – (A+B)*x + A*B
Define this as the polynomial p(x) = x^2 – (A+B)*x + A*B
Example: A quadratic polynomial p(x) with roots at x = 1 and x = 5. A result is:
p(x) = x^2 – 6*x + 5
x^2 - 6*x + 5. All screen shots are generated from the HP Prime Emulator
Note that this polynomial is concave up for all x in the real numbers. In calculus, a function is concave up when the second derivative is positive. We can imagine bucket holding water.
Yes, observe:
0 = (x – A)*(x – B)
0 = x^2 – (A+B)*x + A*B
Now multiply both sides by -1:
0 = -x^2 + (A+B)*x – A*B
Note that 0 * -1 = 0.
Let’s name this quadratic polynomial q(x) = -x^2 + (A+B)*x – A*B.
Going back to our example, with roots x = 1 and x = 5, q(x) is defined as:
q(x) = -x^2 + 6*x – 5
-x^2 + 6*x - 5
In this case this polynomial is concave down for all x. It’s like the bucket has been turned upside down and water is spilling.
Can there be any Other Polynomials?
0 = (x – A)*(x – B)
Multiply by sides by an amplifying factor C:
C * 0 = C * (x – A) * (x – B)
0 = C * (x^2 – (A + B)*x + A*B)
0 = C * x^2 – C*(A + B)*x + A*B*C
Name this polynomial r(x) = C * x^2 – C*(A + B)*x + A*B*C
Back to the example, where roots are located at x = 1 and x = 5, let’s assume an amplifying factor of C = 3. The result is:
r(x) = 3*x^2 – 18*x + 15
3*x^2 - 18*x + 15
The following is a set of four quadratic polynomials that can be formed with roots x = 1 and x = 5:
Four quadratic polynomials with roots x = 1 and x = 5
Polynomial Values Color (see above) p(x) = x^2 – 6*x + 5 A = 1, B = 5, C = 1; concave up Blue q(x) = -x^2 + 6*x – 5 A = 1, B = 5, C = -1; concave down Red r(x) = 3*x^2 – 18*x + 15 A = 1, B = 5, C = 3; concave up Green s(x) = -1/2*x^2 + 3*x – 5/2 A = 1, B = 5, C = -1/2; concave down Orange
Summary
Given the roots of the quadratic polynomial x = A and x = B, with its amplifying factor C, possible quadratic polynomials can be formed by:
f(x) = C * x^2 – C*(A + B)*x + A*B*C
If C = 1, this simplifies to: f(x) = x^2 – (A + B)*x + A*B
If C = -1, this simplifies to: f(x) = -x^2 + (A + B)*x - A*B
If C>0, the second derivative is positive (f’’(x) = 2*C) and the polynomial is concave up.
If C<0, the second derivative is negative (f’’(x) = -2*C) and the polynomial is concave down.
Eddie
This blog is property of Edward Shore, 2018.
Casio fx-CG50: Sparse Matrix Builder
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## Evaluate and estimate numerical square and cube roots
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Progress
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Suppose that a shoemaker has determined that the optimal weight in ounces of a pair of running shoes is \begin{align*}\sqrt[4]{20000}\end{align*}. How many ounces would this be? Is there a way that you could rewrite this expression to make it easier to grasp? In this Concept, you'll learn how to simplify radical expressions like this one so that you can write them in multiple ways.
### Guidance
Radicals are the roots of values. In fact, the word radical comes from the Latin word “radix,” meaning “root.” You are most comfortable with the square root symbol \begin{align*}\sqrt{x}\end{align*}; however, there are many more radical symbols.
A radical is a mathematical expression involving a root by means of a radical sign.
Some roots do not have real values; in this case, they are called undefined.
Even roots of negative numbers are undefined.
\begin{align*}\sqrt[n]{x}\end{align*} is undefined when \begin{align*}n\end{align*} is an even whole number and \begin{align*}x<0\end{align*}.
#### Example A
• \begin{align*}\sqrt[3]{64}\end{align*}
• \begin{align*}\sqrt[4]{-81}\end{align*}
Solution:
\begin{align*}\sqrt[3]{64} = 4\end{align*} because \begin{align*}4^3=64\end{align*}
\begin{align*}\sqrt[4]{-81}\end{align*} is undefined because \begin{align*}n\end{align*} is an even whole number and \begin{align*}-81<0\end{align*}.
In a previous Concept, you learned how to evaluate rational exponents:
This can be written in radical notation using the following property.
Rational Exponent Property: For integer values of \begin{align*}x\end{align*} and whole values of \begin{align*}y\end{align*}:
#### Example B
Rewrite \begin{align*}x^{\frac{5}{6}}\end{align*} using radical notation.
Solution:
This is correctly read as the sixth root of \begin{align*}x\end{align*} to the fifth power. Writing in radical notation, \begin{align*}x^{\frac{5}{6}}=\sqrt[6]{x^5}\end{align*}, where \begin{align*}x^5>0\end{align*}.
You can also simplify other radicals, like cube roots and fourth roots.
#### Example C
Simplify \begin{align*}\sqrt[3]{135}\end{align*}.
Solution:
Begin by finding the prime factorization of 135. This is easily done by using a factor tree.
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### Guided Practice
Evaluate \begin{align*}\sqrt[4]{4^2}\end{align*}.
Solution: This is read, “The fourth root of four to the second power.”
The fourth root of 16 is 2; therefore,
### Explore More
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Radical Expressions with Higher Roots (8:46)
1. For which values of \begin{align*}n\end{align*} is \begin{align*}\sqrt[n]{-16}\end{align*} undefined?
1. \begin{align*}\sqrt{169}\end{align*}
2. \begin{align*}\sqrt[4]{81}\end{align*}
3. \begin{align*}\sqrt[3]{-125}\end{align*}
4. \begin{align*}\sqrt[5]{1024}\end{align*}
Write each expression as a rational exponent.
1. \begin{align*}\sqrt[3]{14}\end{align*}
2. \begin{align*}\sqrt[4]{zw}\end{align*}
3. \begin{align*}\sqrt{a}\end{align*}
4. \begin{align*}\sqrt[9]{y^3}\end{align*}
Write the following expressions in simplest radical form.
1. \begin{align*}\sqrt{24}\end{align*}
2. \begin{align*}\sqrt{300}\end{align*}
3. \begin{align*}\sqrt[5]{96}\end{align*}
4. \begin{align*}\sqrt{\frac{240}{567}}\end{align*}
5. \begin{align*}\sqrt[3]{500}\end{align*}
6. \begin{align*}\sqrt[6]{64x^8}\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 11.2.
### Vocabulary Language: English Spanish
A mathematical expression involving a root by means of a radical sign. The word radical comes from the Latin word radix, meaning root.
Rational Exponent Property
Rational Exponent Property
For integer values of $x$ and whole values of $y$: $a^{\frac{x}{y}}= \sqrt[y]{a^x}$
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.
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# What is a frequency distribution?
## What is a frequency distribution?
What is a frequency distribution? A: Functionals with frequency and what we know about functions have two types of limits. The first that measures what occurs within a given domain, the second one is to determine the number of occurrences within a sample–the number of all possible combinations of relevant properties in a given domain. In the first limit, each of the predicates there can be an interesting property, without a hard-to-optimize specification. Instead, there are probabilities given below, each of which is an integral, along with a description of the domain and other properties. For these, there is no hard-to-optimize notation about the membership of the predicates, you know what a property is. For its truth values, the next level of complexity is explained by determining the distribution of frequencies. The greatest frequency is from there, that you may perform calculations. This limits investigate this site quickly you can calculate this domain itself. The second kind of limit is to determine the number of places in a sample where some local properties of the domain happen to be interesting–the probability of an individual occurrence occurring, in terms of its frequency. For example, the probability for a property to be the sum of frequencies is about one. This is approximately 7%/10000 each way. This sum defines the average number of occurrences of this property. Another form of limit, known as Fisher-Lemaitre, is to indicate which predicates each property has a frequency of which it includes the values of all its properties. The right-hand side keeps telling you more about the distribution of frequency, because there might be one of the most interesting properties and your right-hand side may add more information. The bottom portion of the above limit can be used for information regarding the distribution of predicates found in the domain, and the right-hand side defines which predicates each property has a frequency of. For example, in the more interesting domain, in order to be unique, youWhat is a frequency distribution? We are interested in We want to know the most frequent value of K while dividing by a constant k. We currently use a simple formula for your frequency Given number, number/conj number, number/conj we want to know when K is close to 0. We can solve this for the domain of interest by our first linear regression of the number of the occurrence of such a frequency in the corresponding window sample we have observed. This is approximately x0 = {a:number/(a+k)} In what follows we assume the following window sample: D1 = {0, a:number/(a+k)} We transform the number of the frequency in binary, D2 = {0} = {a:number/Conj}} When dividing by k we are trying to select either the 1st and the 2nd most frequent value of the frequency and decrease the sensitivity 0 = {a:number/(a+k)}, then we substitute K to the common denominator (1/K). This is the most common value for any frequency we have observed.
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0 = 0, then we simplify by K/K = 0, this does not make any difference between the two results. (0/k) = {a:number/(a+k)} if there is an even number of occurrence of K in the specific frequency; 0 equates to being 2/Conj and 0 equates to being 1/Conj for the frequencies the same number. In this example we are suppose with the following distribution: ln(f(t)) = l(f(t)) / k, where l(f) is the largest real value of the n-by-k matrix of interest. a = their explanation a:number/D1} bWhat is a frequency distribution? {#tbl1} =========================== In mathematical physics, the unit frequency, which we will refer to as the ‘cubic time,’ is the smallest integer. There are probably just three ways in which there is a frequency distribution in the electromagnetic spectrum. While the length of a sphere is referred to as the(‘fractional’ length), the number of units of length defining a length is called *mass*. ### A complex number {#tbl1a} Another way in which a frequency (cubic time) can be derived is to define a piecewise continuous function, e.g., f(x) / (x sin(x)). If I define this function by creating a complex vector and then multiplying this vector by itself, it will also be a piecewise continuous function. This makes it less exponential in time. Figure \[informazione-cubic\] shows the complex section of the ‘fractional’ frequency. In the case of a frequency only part (i.e., the term in the right side), the unit number corresponding to that square root of that complex number is f(1,1) = 21 = 0.61. If I define f(x) check my source C3, f(x):=2n/a = -3(3)n/a = -3n/a = -1.61/cout = -0.04 (3,2). I will call this half point a unit, with the other half as being 0.
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39. The complex sum of the real number ‘n’ will be our (2 × 2) integral divided by this point, cout. Meantime, the complex plane can be represented as an infinite series: the complex numbers f(x) and cout, when expressed in terms of real and complex real
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# Solving Quadratic Equations – Methods and Examples
Quadratic equations have the form ax²+bx+c=0. These equations can be solved using various methods depending on the type of quadratic equation we have. We can use methods for incomplete equations, solve equations by factoring, by completing the square, or with the quadratic formula.
Here, we will learn how to solve quadratic equations using different methods. We will use several examples to improve the learning process of the methods.
##### ALGEBRA
Relevant for
Learning to solve quadratic equations with various methods.
See methods
##### ALGEBRA
Relevant for
Learning to solve quadratic equations with various methods.
See methods
An incomplete quadratic equation is an equation that does not have a term from the form $latex ax^2+bx+c=0$, as long as the x² term is always present. When this is the case, we have two types of incomplete quadratic equations depending on the missing term.
### Solving quadratic equations that do not have the term bx
To solve quadratic equations of the form $latex ax^2+c=0$ that do not have the bx term, we need to isolate x² and take the square root of both sides of the equation.
For example, suppose we want to solve the equation $latex x^2-9=0$. First, we have to write it as follows:
$latex x^2=9$
Now that we have x² on the left-hand side, we can take the square root of both sides of the equation:
$latex x=\sqrt{9}$
$latex x=\pm 3$
Note: We must consider both the positive solution and the negative solution, since $latex (-3)^2=9$.
### Solving quadratic equations that do not have the term c
To solve equations of the form $latex ax^2+bx=0$ that do not have the constant term c, we have to factor the x on the left-hand side of the equation. Then, we form two equations with the factors and solve them.
For example, suppose we want to solve the equation $latex x^2-5x=0$. First, we factor it as follows:
$latex x(x-5)=0$
Since we have two factors, we can form an equation with each factor and solve:
$latex x=0~~$ or $latex ~~x-5=0$
$latex x=0~~$ or $latex ~~ x=5$
Note: In this type of equation, one of the solutions will always be $latex x=0$.
## Solving quadratic equations by factoring
The factorization method consists of finding the factors of the quadratic equation so that we have the roots of the equation exposed. By forming an equation with each factor, we can find the roots.
We can solve quadratic equations by factoring by following these steps:
Step 1: Simplify the equation if possible and write it in the form $latex ax^2+bx+c=0$.
Step 2: Find the factors of the equation using any method and write it in the form $latex (x+p)(x+q)=0$.
Step 3: Take each factor and set it equal to zero to form an equation. For example, $latex x+p=0$.
Step 4: Solve the equation for each factor.
There are several methods we can use to factor quadratic equations. The general idea is to find two factors of the form $latex (x+p)(x+q)=0$, which result in the form $latex x^2+bx+c=0$ when multiplied.
For example, the equation $latex x^2+2x-3=0$ can be factored into the form $latex (x+3)(x-2)=0$, since multiplying the factors gives us the original equation.
You can learn or review how to factor quadratic equations by visiting our article: Factoring Quadratic Equations.
This method allows us to find both roots of the equation relatively easily. However, it is not always possible to factor a quadratic equation.
## Solving quadratic equations by completing the square
Completing the square is a factoring technique that allows us to write an equation from the form $latex ax^2+bx+c=0$ to the form $latex (x-h)^2+k=0$. Thus, we can solve quadratic equations that cannot be easily factored.
To solve quadratic equations using the method of completing the square, we can follow the steps below:
Step 1: Simplify and write the equation in the form $latex ax^2+bx+c=0$.
Step 2: When a is different from 1, the entire equation must be divided by a to obtain an equation with a value of a equal to 1:
$latex x^2+bx+c=0$
Step 3: Divide the coefficient b by 2 to obtain:
$$\left(\frac{b}{2}\right)$$
Step 4: Square the expression from step 2:
$$\left(\frac{b}{2}\right)^2$$
Step 5: Add and subtract the expression obtained in step 4 to the equation obtained in step 2:
$$x^2+bx+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$
Step 6: Factor the equation using the identity $latex x^2+2xy+y^2=(x+y)^2$:
$$\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c=0$$
Step 7: Simplify to obtain an equation of the following form:
$latex (x-h)^2+k=0$
Step 8: Rearrange the equation as follows:
$latex (x-h)^2=-k$
Step 9: Take the square root of both sides of the equation:
$latex x-h=\sqrt{-k}$
Step 10: Solve for x:
$latex x=h\pm \sqrt{-k}$
When it is not possible to solve quadratic equations with any other method, we can use the quadratic formula, since this method allows us to find both solutions of any quadratic equation.
To use the general quadratic formula, we have to write the equation in the form $latex a{{x}^2}+bx+c=0$. This will allow us to identify the values of the coefficients a, b and c easily. We then use those values in the quadratic formula:
Note: We must not forget the ± sign, since in this way we will obtain both solutions to the quadratic equation when it is the case.
The expression inside the square root of the quadratic formula ($latex b^2-4ac$) is the discriminant of the quadratic equation. The discriminant determines the type of root that the quadratic equation will have.
Therefore, depending on the value of the discriminant, we have the following:
• When $latex b^2-4ac>0$, the equation has two real roots.
• When $latex b^2-4ac<0$, the equation has no real roots.
• When $latex b^2-4ac=0$, the equation has a repeated root.
When the value inside the square root of the formula is positive, we will have two real roots. When that value is negative, we won’t have real roots (but we will have imaginary or complex roots). When that value is equal to zero, we will have a single root.
The following examples are solved using all the methods for solving quadratic equations studied above. Try to solve the problems yourself before looking at the solution.
### EXAMPLE 1
What are the solutions of the equation $latex x^2-4=0$?
This equation is an incomplete quadratic equation that does not have the bx term. Therefore, we can find the solutions by isolating the quadratic term and taking the square root of both sides of the equation:
$latex x^2-4=0$
$latex x^2=4$
$latex x=\pm\sqrt{4}$
$latex x=\pm 2$
The solutions of the equation are $latex x=2$ and $latex x=-2$.
### EXAMPLE 2
Solve the equation $latex x^2-7x=0$.
This equation is an incomplete quadratic equation that does not have the constant term c. We can solve it by factoring the x and forming an equation with each factor:
$latex x^2-7x=0$
$latex x(x-7)=0$
$latex x=0 ~~$ or $latex ~~x-7=0$
$latex x=0 ~~$ or $latex ~~x=7$
The solutions of the equation are $latex x=0$ and $latex x=-7$.
### EXAMPLE 3
Solve the equation $latex x^2+2x-8=0$ using the factoring method.
Factoring the left-hand side of the equation, we have:
$latex x^2+2x-8=0$
$latex (x+4)(x-2)=0$
Now, we form an equation with each factor and solve:
$latex x+4=0~~$ or $latex ~~x-2=0$
$latex x=-4~~$ or $latex ~~x=2$
The solutions of the equation are $latex x=-4$ and $latex x=2$.
### EXAMPLE 4
Solve the equation $latex 2x^2-13x-24=0$ using the factoring method.
We can factor the left-hand side of the equation as follows:
$latex 2x^2-13x-24=0$
$latex (2x+3)(x-8)=0$
Now, we form an equation with each factor and solve:
$latex 2x+3=0~~$ or $latex ~~x-8=0$
$latex x=-\frac{3}{2}~~$ or $latex ~~x=8$
The solutions of the equation are $latex x=-\frac{3}{2}$ and $latex x=8$.
### EXAMPLE 5
Use the method of completing the square to solve the equation $latex x^2+4x-6=0$.
In this equation, the coefficient b is equal to 4. Therefore, we have:
$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$
$$=2^2$$
Now, we have to add and subtract that value to the quadratic equation:
$$x^2+4x-6=x^2+4x+2^2-2^2-6$$
Completing the square and simplifying, we have:
$latex = (x+2)^2-4-6$
$latex = (x+2)^2-10$
Rearranging, we form the equation:
$latex (x+2)^2=10$
Taking the square root of both sides, we have:
⇒ $latex x+2=\sqrt{10}$
Solving, we have:
⇒ $latex x=-2\pm \sqrt{10}$
### EXAMPLE 6
Solve the equation $latex 2x^2+8x-10=0$ by completing the square.
We can divide the entire equation by 2 to make the coefficient of the quadratic term equal to 1:
⇒ $latex x^2+4x-5=0$
Now, we can see that the coefficient b is equal to 4. Therefore, we have:
$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$
$$=2^2$$
If we add and subtract that expression to the quadratic equation, we have:
$$x^2+4x-5=x^2+4x+2^2-2^2-5$$
Completing the square and simplifying, we have:
$latex = (x+2)^2-4-5$
$latex = (x+2)^2-9$
Now, we rearrange the equation as follows:
⇒ $latex (x+2)^2=9$
And we take the square root of both sides:
⇒ $latex x+2=3~~$ or $latex ~~x+2=-3$
Solving, we have:
⇒ $latex x=1~~$ or $latex ~~x=-5$
### EXAMPLE 7
Use the general formula to solve the equation $latex 2x^2+3x-4=0$. Express the solutions to two decimal places.
This equation has the coefficients $latex a=2$, $latex b=3$, and $latex c=-4$. Therefore, using the quadratic formula with those values, we have:
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-(3)\pm \sqrt{( 3)^2-4(2)(-4)}}{2(2)}$$
$$=\frac{-3\pm \sqrt{9+32}}{4}$$
$$=\frac{-3\pm \sqrt{41}}{2}$$
$$=-2.35 \text{ or }0.85$$
The solutions of the equation are $latex x=-2.35$ and $latex x=0.85$.
## Solving quadratic equations – Practice problems
Solve the following problems using any of the methods we studied above.
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# Find speed in Km/Hr ?
## Percy standing in the midst of a field observes a flying bird in his north at an angle of elevation of ${30}^{\circ}$ and after $2 \min$ he observes the bird in his south at an angle of elevation of ${60}^{\circ}$. If the birdflies in a straight line all along at a height of $50 \sqrt{3} m$, then find its speed in kilometer per hour.
Then teach the underlying concepts
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#### Explanation
Explain in detail...
#### Explanation:
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Dec 11, 2017
$6 \textcolor{w h i t e}{88} k m {h}^{-} 1$
#### Explanation:
From the diagram we need to find distance c. Using Pythagoras' theorem $c = \sqrt{{a}^{2} + {b}^{2}}$. we first find a and b
$a = \frac{50 \sqrt{3}}{\sin} \left(30\right) = 100 \sqrt{3}$
$b = \frac{50 \sqrt{3}}{\sin} \left(60\right) = \frac{50 \sqrt{3}}{\frac{\sqrt{3}}{2}} = 100$
$c = \sqrt{{\left(100 \sqrt{3}\right)}^{2} + {\left(100\right)}^{2}}$
$c = \sqrt{40000} = 200 \textcolor{w h i t e}{88}$metres
$200 m = \frac{200}{1000} = \frac{1}{5} k m$
$2 \min = \frac{2}{60} = \frac{1}{30} h r s$
Distance travelled/ time taken = speed
$\therefore$
$\frac{\frac{1}{5}}{\frac{1}{30}} = 6$ $k m {h}^{-} 1$
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# Vector Equations – Explanation and Examples
In vector geometry, one of the most important concepts in solving real-world problems is using vector equations. The vector equation is defined as:
“The vector equation is an equation of vectors which is when solved, gives the result in the form of a vector.”
In this topic, we shall briefly discuss the following mentioned concepts:
• What is a vector equation?
• How to solve a vector equation?
• What is a vector equation of a straight line?
• What is a vector equation of a circle?
• Examples
• Problems
## What Is A Vector Equation?
A vector equation is an equation involving n numbers of vectors. More formally, it can be defined as an equation involving a linear combination of vectors with possibly unknown coefficients, and upon solving, it gives a vector in return.
Generally, a vector equation is defined as “Any function that takes anyone or more variable and in return gives a vector.”
Any vector equation involving vectors with n number of coordinates is similar to the linear equation system with n number of coordinates involving numbers. For example,
Consider a vector equation,
r <4,5,6> + t<3,4,1> = <8,5,9>
It can also be written as
<4r,5r,6r> + <3t,4t,1t> =<8,5,9>
Or
<4r+3t, 5r+4t, 6r+1t> = <8,5,9>
For two vectors to be equal, all the coordinates must be equal, so it can also be written as a system of linear equations. Such a representation is as follows:
4r+3t = 8
5r+4t = 5
6r+1t = 9
So, the vector equation can be solved by converting it into a system of linear equations. Hence, it simplifies and becomes easier to solve.
In our daily life, vectors play a vital role. Most of the physical quantities used are vector quantities. Vectors have many true applications, including the situations designated by force and velocity. For example, if a car is moving on a road, various forces will be acting on it. Some forces act in the forward direction and some in the backward direction to balance the system. So, all these forces are vector quantities. We use vector equations to find out various physical quantities in 2-D or 3-D, such as velocity, acceleration, momentum, etc.
Vector equations give us a diverse and more geometric way of viewing and solving the linear system of equations.
Overall, we can conclude that the vector equation is:
x1.t1+x2.t2+···+xk.tk = b
where t 1,t 2,…,t k,b are vectors in Rn and x 1,x 2,…,xk are unknown scalars, has the same solution set as the linear system with an augmented matrix of the given equation.
Therefore, the vector equation is given as,
r = r0 +kv
Let’s understand this concept with the help of examples.
Example 1
A car moves with a constant velocity on a straight road initially at time t=2 the position vector of the car is (1,3,5) then after some time at t=4, the car’s position vector is described as (5,6,8). Write down the vector equation of the position of the object. Also, express it in the form of parametric equations.
Solution
Since the vector equation of a straight-line is given as
r = r0 +tv
Since,
r0 = <1,3,5>
r = <5,6,8>
<5,6,8> = <1,3,5> + 4v
<5,6,8> – <1,3,5> = 4v
<4,3,3> = 4v
v = <1,3/4,3/4>
Now, finding vector equation of object’s position
r = r0 +tv
r = <1,3,5> + t<1,3/4,3/4>
where vector r is <x,y,z>
<x,y,z> = <1,3,5> + <1t,3/4t,3/4t>
Expressing in the form of the parametric equation:
As two vectors are only equivalent if their coordinates are equal. So, due to equality, we can write as,
x = 1+t
y = 3+3/4t
z = 5+3/4t
The vector equation of lines identifies the position vector of line with reference to the origin and direction vector and we can find out the dimensions of vectors corresponding to any length. This works for the straight lines and curves.
Note: The position vector is used to describe the position of the vector. It is a straight line having one end fixed and the other attached to the moving vector to specify its position.
Let’s understand this concept with the help of examples.
Example 2
Write down the following equations as vector equations
1. x=-2y+7
2. 3x=-8y+6
3. x=-3/5-8
Solution
Let’s consider equation 1 first:
x = -2y+7
Since the equation given above is an equation of a straight-line:
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
x = -2y+7
let y = 0
x = 7
So, the first point is s (7,0) or OS (7,0)
Now let find out the second point that is halfway through the first point then,
Let x = 14
14 = -2y + 7
-2y = 7
y = -3.5
So, the second point T (14, -3.5) or OT (14, -3.5)
Then,
OS OT = (7,0) – (14, -3.5)
OS OT = (-7, 3.5)
So, the vector equation form of the above equation is,
R = <7,0> + k<-7,3.5>
R = <7-7k, 3.5k>
Now, let’s solve equation 2:
3x = -8y+6
Since the equation given above is an equation of a straight-line
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
3x = -8y+6
let y = 0
x = 2
So, the first point is s (2,0) or OS (2,0)
Now let find out the second point that is halfway through the first point then,
Let x = 4
12 = -2y+7
-2y = 12-7
y = -5/2
So, the second point T (4, -5/2) or OT (4, -5/2)
Then,
OS OT = (2,0) – (4, -5/2)
OS OT = (-2, 5/2)
So, the vector equation form of the above equation is,
R = <2,0> + k<-2,5/2>
R = <2-2k, 5/2k>
Now, let’s do equation 3:
x = -3/5-8
Since the equation given above is an equation of a straight-line
y = mx+c
Firstly, we will select two points on the given line.
Let’s simplify the equation,
x = -3/5y+8
let y = 0
x = 8
So, the first point is s (8,0) or OS (8,0)
Now let find out the second point that is halfway through the first point then,
Let x=16
16 = -3/5y+8
-3/5y = 16-8
y = -13.33
So, the second point T (16, -13.33) or OT (16, -13.33)
Then,
OS OT = (8,0) – (16, -13.33)
OS OT = (-8, 13.33)
So, the vector equation form of the above equation is,
R = <8,0> + k<-8,13.33>
R = <8-8k, 13.33k>
## Vector Equation Of A Straight Line
We all are familiar with the equation of the line that is y=mx+c, generally called a slope-intercept form where m is the slope of the line and x and y are the point coordinates or intercepts defined on the x and y axes. However, this form of the equation is not enough to completely explain the line’s geometrical features. That’s why we use a vector equation to describe the position and direction of the line completely.
To find the points on the line, we will use the method of vector addition. We need to find out the position vector and the direction vector. For the position vector, we will add the position vector of the known point on the line to the vector v that lies on the line, as shown in the figure below.
So, the position vector r for any point is given as r = op + v
Then, the vector equation is given as
R = op + kv
Where k is a scalar quantity that belongs from RN, op is the position vector with respect to the origin O, and v is the direction vector. Basically, k tells you how many times you will go the distance from p to q in the specified direction. It can be ½ if half of the distance would be covered and so on.
If two points on the line are known, we can find out the line’s vector equation. Similarly, if we know the position vectors of two points op and oq on a line, we can also determine the vector equation of the line by using the vector subtraction method.
Where,
v = opoq
Therefore, the equation of vector is given as,
R = op +kv
Let’s solve some examples to comprehend this concept.
Example 3
Write down the vector equation of a line through points P (2,4,3) and Q (5, -2,6).
Solution
Let the position vector of the given points P and Q with respect to the origin is given as OP and OQ, respectively.
OP = (2,4,3) – (0,0,0)
OP = (2,4,3)
OQ = (5, -2,6) – (0,0,0)
OQ = (5, -2 ,6)
Since we know that the vector equation of a line is defined as,
R = OP + kv
Where v = OQOP
v = (5, -2,6) – (2,4,3)
v = (3, -6, 3)
So, the vector equation of the straight line is given as,
R = <2,4,3> + k<3, -6,3>
Example 4
Determine the vector equation of the line where k=0.75. If the points given on the line are defined as A (1,7) and B (8,6).
Solution:
k is the scale that can vary from -∞ to +∞. In this case, k is given as 0.75, which is the distance covered on AB in the given direction.
Let the position vector of the given points A and B with respect to the origin are OA and OB, respectively.
OA = (1,7) – (0,0)
OA = (1,7)
OB = (8,6) – (0,0)
OB = (8,6)
Since we know that the vector equation of a line is defined as,
R = OA +kv
Where v = OBOA
v = (8,6) – (1,7)
v = (7, -1)
So, the vector equation of the straight line is given as,
Where k=0.75
R = <1,7> + 0.75<7, -1>
Example 5
Write down the vector equation of a line through points P (-8,5) and Q (9,3).
Solution
Let the position vector of the given points P and Q with respect to the origin is given as OP and OQ, respectively.
OP = (-8,5) – (0,0)
OP = (-8,5)
OQ = (9,3) – (0,0)
OQ = (9,3)
Since we know that the vector equation of a line is defined as,
R = OP + kv
Where v = OQOP
v = (9,3) – (-8,5)
v = (17, -2)
So, the vector equation of the straight line is given as,
R = <-8,5> + k<17, -2>
## Vector Equation Of A Circle
Earlier, we have discussed the vector equation of a straight line. Now we will discuss the vector equation of a circle having radius r and with some center c, which we generally say that the circle is centered at c (0,0), but it may be located at any other point in the plane.
The vector equation of a circle is given as
r (t) = <x(t), y(t)>
where x(t) = r.cos(t) and y(t) = r.sin(t), r is the radius of the circle and t is the defined as the angle.
Let us consider a circle with center c and radius r, as shown in the figure below.
.
The position vector of the radius and center c is given as r and c, respectively. Then the radius of the circle is represented by vector CR, where CR is given as r c.
Since the radius is given as r so magnitude if CR can be written as
|CR| = r^2
Or
(r c). (r c) = r^2
Or
| r c| = r
This can also be called a vector equation of a circle.
Example 5
Write down the vector equation and the cartesian equation of a circle with center c at (5,7) and radius 5m.
Solution
Vector equation of a circle:
| r c| = r
| r – <5,7>| = 5
(r – <5,7>)^2 = 25
Cartesian equation of a circle:
(x-h)^2 +(y-k)^2 = r2
(x-5)^2 + (y-7)^2 = 25
Example 6
Determine if the point (2,5) lies on the circle with the vector equation of a circle given as |r -<-6,2>| = 3.
Solution
We must find out whether the given point lies inside the circle or not provided the circle’s vector equation.
Since putting the value of the point in the given vector equation
= |<2,5>-<-6,2>|
= |<2+6,5-2>|
= |<8,3>|
= √ ((8)^2+(3)^2)
= (64+9)
= (73) ≠ 3
Hence, the point does not lies inside the circle.
### Practice Problems
1. Write down the following equations as vector equations : x=3y+5 x=-9/5y+3 x+9y=4
2. Determine the equation for the line defined by points A (3,4,5) and B (8,6,7). Find the position vector for a point, half-way between the two points.
3. Write a vector equation of the line parallel to vector Q and passing through point o with the given position vector P.
Q = <-2,6> P = <3, -1>
Q = <1,8> P = <9, -3>
1. Write down the vector equation of a line through points P (-8/3,5) and Q (5,10).
2. A car moves with a constant velocity on a straight road initially at time t=2 the position vector of the car is (1/2,8) then after some time at t=4, the car’s position vector is described as (5,10). Write down the vector equation of the position of the object. Also, express it in the form of parametric equations.
3. Write down the vector equation and the cartesian equation of a circle with center c at (8,0) and radius 7m.
4. Determine if the point (3,-5) lies on the circle with the vector equation of a circle given as |r -<-3,4>| = 4.
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# Section 4.2: Least-Squares Regression
## Objectives
By the end of this lesson, you will be able to...
1. find the least-squares regression (LSR) line
2. use the LSR line to make predictions
3. interpret the slope and y-intercept of the LSR line
For a quick overview of this section, watch this short video summary:
Because we'll be talking about the linear relationship between two variables, we need to first do a quick review of lines.
## The Slope and Y-intercept
If there's one thing we all remember about lines, it's the slope-intercept form of a line:
The slope-intercept form of a line is
y = mx + b
where m is the slope of the line and b is the y-intercept.
Knowing the form isn't enough, though. We also need to know what each part means. Let's start with the slope. Most of us remember the slope as "rise over run", but that only helps us graph lines. What we really need to know is what the slope represents in terms of the original two variables. Let's look at an example to see if we can get the idea.
Example 1
The equation T = 6x + 53 roughly approximates the tuition per credit at ECC since 2001. In this case, x represents the number of years since 2001 and T represents the tuition amount for that year.
The graph below illustrates the relationship.
In this example, we can see that both the 6 and the 53 have very specific meanings:
The 6 is the increase per year. In other words, for every additional year, the tuition increases \$6.
The 53 represents the initial tuition, or the tuition per credit hour in 2001.
As we progress into the relationship between two variables, it's important to keep in mind these meanings behind the slope and y-intercept.
## Finding the Equation for a Line
Another very important skill is finding the equation for a line. In particular, it's important for us to know how to find the equation when we're given two points.
A very useful equation to know is the point-slope form for a line.
The point-slope form of a line is
y - y1 = m(x - x1)
where m is the slope of the line and (x1, y1) is a point on the line.
Let's practice using this form to find an equation for the line.
Example 2
In Example 1 from section 4.1, we talked about the relationship between student heart rates (in beats per minute) before and after a brisk walk.
before after before after before after 86 98 58 128 60 70 62 70 64 74 80 92 52 56 74 106 66 70 90 110 76 84 80 92 66 76 56 96 78 116 80 96 72 82 74 114 78 86 72 78 90 116 74 84 68 90 76 94
Let's highlight a pair of points on that plot and use those two points to find an equation for a line that might fit the scatter diagram.
Using the points (52, 56) and (90, 116), we get a slope of
m = 116-56 = 60 ≈ 1.58 90-52 38
So an equation for the line would be:
y - y1 = m(x - x1)
y - 56 = 1.58(x - 52)
y - 56 = 1.58x - 82.16
y = 1.58x - 26.16
It's interesting to note the meanings behind the slope and y-intercept for this example. A slope of 1.58 means that for every additional beat per minute before the brisk walk, the heart rate after the walk was 1.58 faster.
The y-intercept, on the other hand, doesn't apply in this case. A y-intercept of -26.16 means that if you have 0 beats per minute before the walk, you'll have -26.16 beats per minute after the walk. ?!?!
This brings up a very important point - models have limitations. In this case, we say that the y-intercept is outside the scope of the model.
Now that we know how to find an equation that sort of fits the data, we need a strategy to find the best line. Let's work our way up to it.
## Residuals
Unless the data line up perfectly, any line we use to model the relationship will have an error. We call this error the residual.
The residual is the difference between the observed and predicted values for y:
residual = observed y - predicted y
residual =
Notice here that we used the symbol (read "y-hat") for the predicted. This is standard notation in statistics, using the "hat" symbol over a variable to note that it is a predicted value.
Example 3
Let's again use the data from Example 1 from section 4.1. In Example 2 from earlier this section, we found the model:
= 1.58x - 30.16
Let's use this model to predict the "after" heart rate for a particular students, the one whose "before" heart rate was 86 beats per minute.
The predicted heart rate, using the model above, is:
= 1.58(86) - 26.16 = 109.72
Using that predicted heart rate, the residual is then:
residual = = 98 - 109.72 = -11.72
Here's that residual if we zoom in on that particular student:
Notice here that the residual is negative, since the predicted value was more than the actual observed "after" heart rate.
## The Least-Squares Regression (LSR) line
So how do we determine which line is "best"? The most popular technique is to make the sum of the squares of the residuals as small as possible. (We use the squares for much the same reason we did when we defined the variance in Section 3.2.) The method is called the method of least squares, for obvious reasons!
#### The Equation for the Least-Squares Regression line
The equation of the least-squares is given by
where
is the slope of the least-squares regression line
and
is the y-intercept of the least squares regression line
Let's try an example.
Example 4
Let's again use the data from Example 1 in Section 4.1, but instead of just using two points to get a line, we'll use the method of least squares to find the Least-Squares Regression line.
before after before after before after 86 98 58 128 60 70 62 70 64 74 80 92 52 56 74 106 66 70 90 110 76 84 80 92 66 76 56 96 78 116 80 96 72 82 74 114 78 86 72 78 90 116 74 84 68 90 76 94
Using computer software, we find the following values:
≈72.16667
sx ≈10.21366
= 90.75
sy
≈17.78922
r ≈
0.48649
Note: We don't want to round these values here, since they'll be used in the calculation for the correlation coefficient - only round at the very last step.
Using the formulas for the LSR line, we have
= 0.8473x + 29.60
(A good general guideline is to use 4 digits for the slope and y-intercept, though there is no strict rule.)
One thought that may come to mind here is that this doesn't really seem to fit the data as well as the one we did by picking two points! Actually, it does do a much better job fitting ALL of the data as well as possible - the previous line we did ourselves did not address most of the points that were above the main cluster. In the next section, we'll talk more about how outliers like the (58, 128) point far above the rest can affect a model like this one.
## Technology
Here's a quick overview of how to find the Least-Squares Regression line in StatCrunch.
Select Stat > Regression > Simple Linear Select the predictor variable for X & the response variable for Y Select Calculate The fourth line shows the equation of the regression line. Note that it will not have x and y shown, but rather the names that you've given for x and y. For example: Avg. Final Grade = 88.73273 - 2.8272727 Num. Absences
You can also go to the video page for links to see videos in either Quicktime or iPod format.
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# Rounding Numbers – Definition, Place-value Chart & Examples
## What is rounding off numbers?
Rounding off numbers is a mathematical technique of adjusting the number’s digits to make the number easier to use during calculations. Numbers are rounded off to a particular degree of accuracy to make calculations simpler and the results easier to understand.
Before rounding off any number, you should know the place of all digits of a number. Below is the place value chart.
## How to Round off the Whole Numbers?
It is fundamentally important to understand the term “rounding digit” when rounding off numbers.
For example, when rounding off numbers 100 to tens, the rounding digit is the second number from the right. Similarly, the rounding digit is in the third place when rounding to the nearest hundred, which is 1. Therefore, the first step when rounding a number is identifying the rounding digit and looking at the next digit to the right side.
• If the digit to the right side of the rounding digit is 0, 1, 2, 3, or 4, then the rounding digit doesn’t change. All digits to the right of the rounding digit become zero.
• If the digit to the right side of the rounding digit is 5, 6, 7, 8, or 9, the rounding digit increases by one digit. All the digits to the right are dropped to zero.
### Practice questions
1. Round off the following numbers to the nearest tens.
29
95
43
75
2. Round off these numbers to the nearest tens.
164
1,989
765
9,999,995
3. Round the following list of numbers to the nearest hundreds.
439
2,950
109,974
562
4. Round off the numbers below to the nearest thousands.
5,280
1,899,999
77,777
1,234,567
### Solutions
1. Rounding off to the nearest ten:
The digit to the right of the rounding digit in 29 is 9. Therefore, one digit is added to the rounding digit, 2, and the other digit is dropped to zero.
29 → 30
The digit to the right of the rounding digit in 43 is 3. The number does not affect the rounding digit, 4 and 3 are dropped to zero.
43 → 40
The digit to the right of the rounding digit in 75 is 5. One digit is added to the rounding digit, and 5 is dropped to zero.
75 → 80.
The digit to the right of the rounding digit in 95 is 5. One digit is added to 9, and the rest dropped to zero.
95 → 100.
2. Round to the nearest ten:
• 164 has the rounding digit as 6 and 4 as the right-hand side digit.
• 164 will become 160
• 765 will become 770.
• 1,989 → 1,990.
• 9,999,995 has 5 as the digit to the right of the rounding digit.
• 9,999,995 → 10,000,000
3. Rounding off to the nearest hundreds:
Identify the digit to the right (tens digit) of the rounding digit.
The tens digit in 439 is 3, so there is no effect on the hundreds digit:
439 →400
Number 6 is the tens digit or digit to the right of the rounding digit:
562 → 600.
5 is the tens digit in 2,950, so,
2,950 → 3,000.
109,974 has the tens digit as 7.
109,974 then becomes 110,000.
4. To round off to the nearest thousands, the hundreds digit is considered.
The hundred digit in 5,280 is 2, so it does not affect the rounding digit.
5,280, therefore, becomes 5,000.
77,777 becomes 78,000.
1,234, 567 becomes 1,235,000.
1,899,999 becomes 1,900,000.
## How to Round Off Decimal Numbers?
Decimal numbers are rounded off to estimate an answer quickly and easily. Decimal numbers can be rounded off to the nearest integer or whole number, tenths, hundredths, thousandths, etc.
### Rounding off to the nearest integer
The following rules are followed when rounding off a decimal to the nearest whole number:
• The number to be rounded off is identified.
• The digit in one’s place is marked.
• The first digit to the right of the decimal, or in the tenths place, is checked.
• If the tenths digit is less than or equal to 4, then the number in the one’s place is rounded off to a whole number.
• Similarly, if the tenths digit is greater than or equal to 5, add 1 digit to the number in the one’s place.
• Drop all digits after the decimal point to result in the desired whole number.
### Rounding to the nearest tenths
A similar procedure is applied when rounding a number to the nearest tenths.
• The number to be rounded off is first of all identified.
• The digit in the tenths place is marked.
• The digit in the hundredths place is checked.
• If the hundredths digit is less than or equal to 4, then the number in the tenths digit remains the same.
• Similarly, if the hundredths digit is greater than or equal to 5, add 1 digit to the number in the tenths place.
• Drop all digits to the right of the tenths column to result in the desired number.
### Rounding off to the nearest hundredths
The first step is to identify the required number. The digit in the hundredths place is marked. Check whether the digit in the thousandths place is 0, 1, 2, 3, 4 or 5, 6, 7, 8, 9. Round off the number to the nearest hundredths and drop the other numbers to the right of the hundredths place.
### Practice Questions
1. When rounded off to the nearest whole number, $975.6539$ is equal to which number?
2. When rounded off to the nearest whole number, $1240.68$ is equal to which number?
3. When rounded off to the nearest tenths, $845.324$ is equal to which decimal?
4. The length of a road is $27.56$ km. What is the length to the nearest km?
5. The weight of a boy is $37.35$ kg. Calculate the weight to the nearest kg.
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# Aim What are the arithmetic series and geometric
• Slides: 9
Aim: What are the arithmetic series and geometric series? Do Now: Find the sum of each of the following sequences: a) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 b) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 +. . . + 98 + 99 + 100 HW: p. 265 # 12, 14, 20 p. 272 # 6, 8, 10, 16 p. 278 # 8, 10
The sum of an arithmetic sequence is called arithmetic series Although we can find the arithmetic series one after the other, there is a formula to find the series faster.
Find the sum of the first 150 terms of the arithmetic sequence 5, 16, 27, 38, 49, . . . First we need to determine what the last term of the 150 terms (or the 150 th term) is. a 1 = 5; d = 16 – 5 = 11. a 150 = a 1 + d(150 – 1), a 150 = 5 + 11(149) = 1644
Write the sum of the first 15 terms of the arithmetic series 1 + 4 + 7 + · · · in sigma notation and then find the sum First of all, we need to find the recursive formula a 1 = 1 and d = 3 To find the sum, we need to find a 15
The sum of an geometric sequence is called geometric series The formula to find the finite (limited number of terms) geometric sequence is 3, 15, 75, 375, 1875, 9375, 46875, 234375, 1171875 is a geometric sequence, find the sum of sequence.
Write as a series and then find the sum
Infinite series : The last term of the series is the infinity An infinite arithmetic series has no limit An infinite geometric series has no limit when An infinite geometric series has a finite limit when the limit can be found by the formula
Find the sum of the following infinite geometric sequence: 4, 4(0. 6)2, 4(0. 6)3, . . . , 4(0. 6)n - 1 , . . . a 1 = 4 and r = 0. 6 Find
Find the sum of the first 10 terms of the geometric series Find the sum of five terms of the geometric series whose first term is 2 and fifth term is 162 S 5 = 242
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# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.1
• Last Updated : 11 Dec, 2020
### (i) sin A, cos A (ii) sin C, cos C
Solution:
Using Pythagoras theorem for ΔABC
AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴AC = 25 cm
(i) sin A = opp/hyp
sin A = 7/25
cos A = adj/hyp = 24/25
cos A = 24/25
(ii) sin C = opp/hyp
sin C = 24/25
cos C = 7/25
### Question 2. In Fig. 8.13, find tan P – cot R.
Solution:
Applying Pythagoras theorem for ΔPQR, we obtain
PR2 = PQ2 + QR2
(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR2
25 cm2 = QR2
QR = 5 cm
tan P = 5/12
cot R = 5/12
tan P – cot R = 5/12 – 5/12 = 0
### Question 3. If sin A = 3/4, calculate cos A and tan A.
Solution:
Using sin2A + cos2A = 1
(3/4)2 + cos2A = 1
cos2A = 1 – (3/4)2 = 1 – 9/16
cos A = 71/2/4
tan A = sin A/cos A
tan A = (3/4)/(71/2/4)
tan A = 3/71/2
### Question 4: Given 15 cot A = 8. Find sin A and sec A
Solution:
Given, 15 cot A = 8
cot A = 8/15
tan A = 1/cot A
tan A = 15/8
Using, 1 + tan2A = sec2A
1 + (15/8)2 = sec2A
289/64 = sec2A
sec A = 17/8
We know, cos2A = 1/sec2A
cos2A = 64/289
sin2A = 1 – cos2A
sin2A = 225/289
sin A = 15/17
### Question 5: Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution.
Using Pythagoras theorem,
sin θ = 5/13
cos θ = 12/13
tan θ = 5/12
cosec θ = 13/5
cot θ = 12/5
### Question 6: If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let us consider a ΔABC in which CD ⊥ AB.
It is given that cos A = cos B
We need to prove ∠A = ∠B. To prove this, we need to extend AC to P such that BC = CP.
From equation (1), we obtain
AD/BD = AC/CP (BC = CP by construction)
By using the converse of B.P.T (Basic Proportionality Theorem),
CD||BP
∠ACD = ∠CPB (Corresponding angles) … (3)
And, ∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD … (6)
∠ACD = ∠BCD (Using equation (6))
∠CDA = ∠CDB (Both 90°)
Therefore, the remaining angles should be equal.
⇒ ∠A = ∠B
### (ii) cot2θ
Solution:
(i) Using (a + b) * (a – b) = a2 – b2 in numerator and denominator
We get
(1 – sin2θ)/(1 – cos2θ)
Using sin2θ + cos2θ = 1
We get
cos2θ/sin2θ = cot2θ
Now
cot2θ = (7/8)2 = 49/64
(ii) cot2θ = (7/8)2 = 49/64
### Question 8. If 3 cot A = 4, Check whether (1 – tan2A)/(1 + tan2A) = cos2A – sin2A
Solution.
We know that, tanA = sinA / cosA ….(1)
Using (1) on L.H.S
= (1 – sin2A/cos2A)/(1 + sin2A/cos2A)
which on rearranging becomes
= (cos2A – sin2A)/(cos2A + sin2A)
Using the identity,
cos2A + sin2A = 1
LHS becomes
= (cos2A – sin2A)
This is equal to RHS.
LHS = RHS (for every value of cot A)
Hence, Proved.
### (ii) cos A cos C − sin A sin C
Solution:
Using Pythagoras theorem
(AB)2 + (BC)2 = (AC)2
(31/2)2 + (1)2 = (AC)2
which gives
AC = 2 cm
Using formulas
sin A = 1/2
sin C = 31/2/2
cos A = 31/2/2
cos C = 1/2
Now, (i) sin A cos C + cos A sin C
Substituting the values
= (1/2) * (1/2) + (31/2/2) * (31/2/2)
= 1/4 + 3/4
= 1
Now, (ii) cos A cos C − sin A sin C
Substituting the values
= (31/2/2) * (1/2) – (1/2) * (31/2/2)
= 31/2/4 – 31/2/4
= 0
### Question 10: In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given that, PR + QR = 25
PQ = 5
Let PR be x cm.
Therefore, QR = 25 − x cm
Applying Pythagoras theorem in ΔPQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 13) cm = 12 cm
Now,
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
### (v) sin θ = 4/3, for some angle θ
Solution:
(i) Consider a ΔPQR, right-angled at Q as shown below.
Here tan P = 12/5 which is surely greater than 1.
Therefore, the statement is false.
(ii) Consider ΔABC with AB = 5 cm, AC = 12 cm and BC = x cm
Using Pythagoras theorem in ΔABC
(AB)2 + (BC)2 = (AC)2
52 + x2 = 122
x = (144 – 25)1/2
x = (119)1/2
x = 10.9 cm
AB < BC < AC
So this triangle is valid,
Therefore, given statement is true.
(iii) Abbreviation used for cosecant A is cosec A. And cos A is the abbreviation used for cosine A.
Hence, the given statement is false.
(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.
Hence, the given statement is false.
(v) sin θ = 4/3
In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.
Hence, the given statement is false
My Personal Notes arrow_drop_up
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1. Summation
The question is, how many multiplication and addition operations are required to determine:
$\displaystyle \sum_{i=1}^{n}\ \sum_{j=1}^{i} a_{i}b_{j}\$
And modify the sum to reduce the number of computations.
Are there (n-1) + n(i-1) addition operations (outer loop and inner loop respectively) and n*i multiplication operations?
Also, can it be rewritten as
$\displaystyle n\sum_{j=1}^{i} a_{i}b_{j}\$
Thanks.
2. In
$\displaystyle \sum_{i=1}^{n}\ \sum_{j=1}^{i} a_{i}b_{j}\$
The number of multiplications is:
$\displaystyle \sum_{i=1}^{n}\ \sum_{j=1}^{i} 1 = \sum_{i=1}^{n}i = \frac {n(n+1)}{2}$
$\displaystyle (\sum_{i=1}^{n} i-1)-1 = \frac {(n-1)(n-1+1)}{2}-1= \frac {n(n-1)}{2}-1= \frac {n^2-n-2)}{2}$
The way to reduce the number of operations is to notice that $\displaystyle a_{i}$ is common to all the terms after the first sum so we can rewrite as:
$\displaystyle \sum_{i=1}^{n} a_{i} \sum_{j=1}^{i} b_{j}\$
Now the number of multiplications is:
$\displaystyle \sum_{i=1}^{n}1 = n$
which is less than before.
$\displaystyle (n-1) + \sum_{j=1}^{n-1} j-1 = (n-1) + \frac {(n-1)(n-1+1)}{2} = (n-1) + \frac {n(n-1)}{2}$
$\displaystyle = \frac{ (2n-2)}{2} + \frac {n(n-1)}{2} = \frac{ (2n-2) +n^2 - n}{2} = \frac {n^2-n-2)}{2}$
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# Comparing Test Scores
Alignments to Content Standards: 6.SP.B.4 6.SP.B.5.c
In Mrs. Sanchez' math classroom, more people sit on the right-hand side of the room than the left. Â The students on the right-hand side of the classroom received the following scores on an exam worth 100 points:
$$85,\, 90,\, 100,\, 95,\, 0,\, 0,\, 90,\, 70,\, 100,\, 95,\, 80,\, 95$$
The students on the left received these test scores:
$$65,\, 80,\, 90,\, 65,\, 80,\, 60,\, 95,\, 85$$
1. Make two box plots of the students' scores, one for each side of the room.
2. Make a statistical argument that the students on the right-hand side were more successful.
3. Make a statistical argument that the students on the left-hand side were more successful.
## IM Commentary
The goal of this task is to critically compare the center and spread of two data sets. Although the mean is not specifically requested, it can be used as an argument in favor of the left's performance. The teacher may wish to request or suggest for students to calculate the mean or tell them that the statistical arguments for (b) and (c) should not just be based on the box plots. Â The teacher will also want to make sure that the two box plots can be directly compared so the values on the two number lines of the boxplots need to correspond to one another.Â
Statistics is a powerful tool for supporting arguments in a variety of contexts: medicine, education, sports, and ecology to name a few. There are usually choices to make, however, both in which statistics to use and how they are reported. This task gives students an opportunity to think carefully about how to analyze the data in order to support a particular point of view.
This task is based on an idea used for a worksheet developed for one of the UCLA Curtis Center's Saturday trainings given on December 6, 2014. In that problem, two students' quiz scores were compared and the structure of the numbers were very similar.Â
## Solution
1. The box plots for the two halves of the room are shown together to make them easier to compare:
The left whisker is the lowest score, 0 for the right-hand side and 60 for the left-hand side. The right whisker is the highest score, 100 for the right-hand side and 95 for the left-hand side. For the other values, the scores are listed in increasing order, (0,0,70,80,85,90,90,95,95,95,100,100) for the right-hand side. There are twelve scores and the middle two are both 90's so the median is 90. The middle two values for the first half of the data, (0,0,70,80,85,90), are 70 and 80 so the first quartile is 75. The middle two values for the second half of the data are both 95's so the third quartile is 95, and similar for the other plot.
2. There are many ways to argue that the "righties" (the students on the right-hand side) performed better on this test than the "lefties," including:
• Two righties got perfect scores of 100 while none of the lefites did.
• The box plot shows that more than half of the righties got 90 percent. Only a quarter of the lefties did this well on the test.
• If 70 percent is a passing score, only two righties did not pass (probably because they missed the test) while three of eight lefties did not pass the test.
3. Some arguments for why the lefties performed better than the righties on this test include:
• The lowest score of the lefties on the test was 60, which is much higher than the lowest of the righties' scores, 0.
• The average (mean) of the lefties' scores is 77.5 which is higher than the average of the righties' scores, 75.
• Consistency is good and there is much less variation in the lefties' scores than in the righties.
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# 抽象代数代考| Set Theory
The foundation of the modern mathematical language lies in the set theory. Thus we start this course with recalling basic notions of sets and maps between sets.
1. Sets \& Quantifiers
Recall that a set is denoted by curly brackets ${\ldots} .$ Elements of a set must by distinct. The following important sets have special reserved notation.
• $\mathbb{N}={1,2,3, \ldots}$ is the set of all positive integers. Sometimes we call $\mathbb{N}$ the set of natural numbers.
• $\mathbb{Z}={\ldots,-2,-1,0,-1,2, \ldots}$ is the set of all integers.
• $\mathbb{Q}=\left{\frac{m}{n} \mid m \in \mathbb{Z}, n \in \mathbb{N}\right}$ is the set of rational numbers.
• $\mathbb{R}=(-\infty,+\infty)$ is the set of real numbers.
There are many ways to describe a given set.
2. Example 1
We can specify a set by listing all its elements,
$$A={1,2,3} .$$
Or as a set of elements satisfying certain properties:
$$A={a \in \mathbb{Z} \mid 1 \leqslant a \leqslant 3} .$$
In the plain English the last definition will be read as
$$\text { set } A \text { is the set of all integers } a \text { such that } 1 \leqslant a \text { and } a \leqslant 3 \text {. }$$
We can also give a descriptive definition
$$B={\text { odd integers }}$$
or equivalently write
$$B={2 n+1 \mid n \in \mathbb{Z}}$$
The latter will be read as
$B$ is the set of numbers of the form $2 n+1$ where $n$ ranges over all the integers
3. Definition 1: Quantifiers
Throughout this course we will be often using quantifier notations. Please make sure you remember and understand them.
• $\forall$ (upside-down A) means ‘for all’ or ‘for any’. For example ‘ $\forall x \in \mathbb{R}$ we have $x^{2} \geqslant 0$ ‘ means ‘for all $x \in \mathbb{R}$ we have $x^{2} \geqslant 0^{\prime} .$
• $\exists$ (mirrored E) means ‘there exists’. For example ‘ $\exists C \in \mathbb{R}$ such that $\forall x \in \mathbb{R} x^{2}>C$ ‘ means ‘there exists $C \in \mathbb{R}$ such that for all $x \in \mathbb{R} x^{2}>C^{\prime}$.
• ! means ‘unique’. For example ‘ $\forall y>0 \exists ! x>0$ such that $x^{2}=y$ ‘ means ‘for all $y>0$ there exists a unique $x>0$ such that $x^{2}=y^{\prime}$.
Recall also the following notation regarding mutual relation of sets and elements. – $a \in A$ means that element $a$ belongs to the set $A$;
• $B \subset A($ or $B \subseteq A)$ means that set $B$ is a subset of the set $A$, i.e.
$$\forall b \in B \text { we have } b \in A \text {. }$$
Clearly $A \subset A$ and $A=B$ if and only if $A \subset B$ and $B \subset A$;
• $B \subsetneq A$ means that $B$ is a proper subset of $A$, i.e., $B \subset A$ but $B \neq A$
• $\emptyset$ is the empty set, i.e., the set with no elements in it.
4. Operations with sets
Now we describe several important operations which allow to produce new sets from given sets.
5. Definition 2: Union
The union of two sets $A$ and $B$ is the set
$$A \cup B={x \in A \text { or } x \in B}$$
consisting of all elements belonging to either $A$ or $B$.
6. Definition 3: Intersection
The intersection of two sets $A$ and $B$ is the set
$$A \cap B={x \in A \text { and } x \in B}$$
consisting of all elements belonging to both $A$ and $B$.
We can define the unions of intersections of multiple sets in an obvious way:
$$\bigcup_{i=1}^{n} A_{i}=A_{1} \cup \cdots \cup A_{n}, \quad \bigcap_{i=1}^{n} A_{i}=A_{1} \cap \cdots \cap A_{n}$$
7. Definition 4: Difference
Difference of sets $A$ and $B$ is the set
$$A \backslash B={x \in A \text { and } x \notin B}$$
Note that unlike unions and intersections, the difference is not symmetric, i.e., in general $A \backslash B \neq B \backslash A$.
8. Definition 5: Cartesian Product
Given two sets $A$ and $B$ the Cartesian product of $A$ and $B$ is
$$A \times B={(a, b) \mid a \in A, b \in B}$$
is the set consisting of all the pairs $(a, b)$.
9. Example 2
Let $Q$ be the square in the plane with the vertices $(0,0),(1,0),(1,1),(0,1) .$ Then $Q=[0,1] \times[0,1] .$ In general, if is helpful to think of the Cartesian product of two sets as of a square.
10. Maps
Informally, a map $f$ from set $A$ to set $B$ is a rule which assigns an element $b \in B$ to each element $a \in A .$ In this case we write
$$f: A \rightarrow B, \quad f(a)=b \text {. }$$
Under a map $f: A \rightarrow B$ every element $a \in A$ is assigned with a unique element $f(a) \in B$.
If sets $A$ and $B$ are finite and ‘small’ we could represent a map explicitly using a mapping diagram:
11. Figure 1: Mapping diagram
Often the notion function is used as a synonym of map.
12. Example 3
Define a map $f: \mathbb{R} \rightarrow \mathbb{R}$ given by
$$f(x)=x^{2} .$$
For every set $A$, there is a distinguished map
$$\mathrm{i} d_{A}: A \rightarrow A, \quad \mathrm{i} d_{A}(a)=a \forall a \in A,$$
called identity map. It sends any element $a \in A$ to itself.
There are several important properties which a map might satisfy.
13. Definition 6: Bijection, injection and surjection
A map $f: A \rightarrow B$ is called injective if it does not $g l u e$ elements together, i.e.,
$$\forall x, y \in A: \quad \text { If } f(x)=f(y), \text { then } x=y .$$
A map $f: A \rightarrow B$ is called surjective if its image is the whole $B$, i.e.,
$$\forall b \in B, \exists a \in A \text { such that } f(a)=b \text {. }$$
A map which is injective and surjective is called bijective. Problem 1: Change exactly one arrow in the Figure 1 to make the map bijective. In how many ways you can do it?
14. Example 4
Function $f(x)=x^{2}$ defines a map $f: \mathbb{R} \rightarrow \mathbb{R}$. This map not injective, since $f(1)=f(-1)=1$. It is also not surjective, since $f(x)$ never takes the value $-1$.
The $\operatorname{map} g: \mathbb{R} \rightarrow \mathbb{R}$ defined by the function $g(x)=x^{3}$ is on the contrary bijective.
15. If we have two maps
$$f: A \rightarrow B, \quad g: B \rightarrow C,$$
the we can form a new map $(g \circ f)$ called the composition of $f$ and $g$ :
$$(g \circ f): A \rightarrow C, \quad(g \circ f)(a):=g(f(a))$$
NB: Note that maps in a composition $g \circ f$ are applied from right to left – order in which you take composition is important!
16. Definition 7: Inverse map
Map $g: B \rightarrow A$ is the inverse of the map $f: A \rightarrow B$ if
$$(g \circ f)=\mathrm{i} d_{A}, \text { and }(f \circ g)=\mathrm{i} d_{B}$$
The inverse map is denoted by $f^{-1}$.
17. Example 5
Map
$$g:[0,+\infty) \rightarrow[0,+\infty), \quad g(x)=\sqrt{x}$$
is the inverse of the map
$$f:[0,+\infty) \rightarrow[0,+\infty), \quad f(x)=x^{2}$$
When does a map admits an inverse? The following theorem answers this question.
18. Theorem 1
Map $f: A \rightarrow B$ admits an inverse if and only if $f$ is bijective. In this case the inverse is unique.
19. NB: Whenever we encounter if and only if claim, a proof in two directions is required!
20. Proof. Direction $\Rightarrow$.
Assume that $f$ admits an inverse $g: B \rightarrow A .$ We are about to prove that $f$ is injective and bijective.
• $(f$ is injective). Indeed, assume that $f(x)=f(y)$ for some $x, y \in A .$ Apply $g$ to both sides of this identity, then
$$x=\mathrm{i} d_{A}(x)=(g \circ f)(x)=g(f(x))=g(f(y))=(g \circ f)(y)=\mathrm{i} d_{A}(y)=y .$$
So $f$ is injective
• ( $f$ is surjective). Take any $b \in B$. Then
$$f(g(b))=(f \circ g)(b)=\mathrm{i} d_{B}(b)=b \text {, }$$
so $b$ is in the range of $f$, and therefore $f$ is surjective.
21. Direction $\Leftarrow$.
Let us define function $g: B \rightarrow A$. Given any $b \in B$ we can find $a \in A$ such that $f(a)=b$, since $f$ is surjective. Then we define
$$g(b)=a$$
It remains to check that $g$ is indeed the inverse map.
• Check that $(f \circ g)=\mathrm{i} d_{B} .$ Take any $b \in B .$ Then by definition of $g$, we have $f(g(b))=b$. Therefore indeed $(f \circ g)(b)=b$.
• Check that $(g \circ f)=\mathrm{i} d_{A} .$ Take any $a \in A .$ Then we want to check that
$$(g \circ f)(a)=a$$
$$f(g(f(a)))=f(a)$$
since $f \circ g=\mathrm{i} d_{B}$. Now, as $f$ is injective the latter implies
$$g(f(a))=a,$$
as required.
22. Equivalence relations
23. Definition 8: Equivalence relation
Equivalence relation on a set $A$ is a subset $R \subset A \times A$ satisfying the following properties
• (reflexive) for any $a \in A$ we have $(a, a) \in R$
• (symmetric) if $(a, b) \in R$, then $(b, a) \in R$.
• (transitive) if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R .$
If we have $(a, b) \in R$ we will write ‘ $a \sim b$ ‘ and say ‘ $a$ is equivalent to $b$ ‘.
24. Example 6
1. $R={(a, a) \mid a \in A} .$ In other words, any element $a$ is equivalent only to itself.
2. $R={(a, b) \mid a, b \in A} .$ In other words, any element $a \in A$ is equivalent to any other element $b \in A$.
3. Take $R=\mathbb{Z}$ and define $R={(m, n) \mid m-n$ is even $} \subset \mathbb{Z} \times \mathbb{Z}$.
If $a \in A$ and $\sim$ is an equivalence relation, we can form an equivalence class
$$[a]={b \in A \mid b \sim a} .$$
25. Proposition 1
Let $\sim$ be an equivalence relation on the set $A$. Then any two equivalence classes $[a],[b]$ either coincide element-wise, or do not intersect.
26. | Proof. Exercise.
The above exercise allows to define a new set consisting of the equivalence classes:
27. Definition 9
The set of equivalence classes in A modulo an equivalence relation $\sim$ is defined as
$$A / \sim={[a] \mid a \in A} .$$
There is a natural map
$$A \rightarrow A / \sim, \quad a \mapsto[a]$$
sending each element to its equivalence class. Problem 2: When is the natural map $A \rightarrow A / \sim$ surjective? injective? bijective?
28. Example 7
Going back to Example 6, we see that
If $R={(a, a) \mid a \in A}$ then $A / \sim$ is the same as $A$
If $R={(a, b) \mid a, b \in A}$ then $A / \sim$ has only one element
Finally, in the last example $Z / \sim$ consists of two elements: class of odd integers and class of even integers.
|
# Circumference and Area of a Circle
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1 Overview Math Concepts Materials Students explore how to derive pi (π) as a ratio. Students also study the circumference and area of a circle using formulas. numbers and operations TI-30XS MultiView two-dimensional geometry Activity Demonstrate how finding the perimeter of a circle is different than finding the perimeter of a polygon. Given a square and a ruler, can you find how far it is around the square? That s called the "perimeter." Now, given a circle and a ruler, can you tell me how far it is around the circle? Try it. Why is this not a good approach? The students should realize it s very difficult to get an exact answer using a ruler, because a ruler is straight and a circle is round. It is a good idea to mention that in their later math studies, they will see it is possible to measure curves with straight lines, but at this level, it is not possible. Now, ask them how it would be possible to measure the distance around a circle. If we had a flexible tape measure, we could easily measure around a circle. (Show this using a 3D object such as a lid.) However, it s not often that we have access to tape measures in math class, plus, it s difficult to get a precise measurement, especially on a two-dimensional circle drawn on a piece of paper. That means that we really need a better method to find the distance around a circle. Let s figure that method out. Put students in groups, and give each group a tape measure and two circular objects such as lids or canisters. Have students measure the distance around each object and then the diameter. Instruct students to try four things with their results: add, subtract, multiply, and divide them. Now, share your results. When you added, subtracted, multiplied, and divided, did you see anything interesting? Is there any number that shows up repeatedly? Does everyone see 3.(something)? Anytime we measure around a circle, that s called the "circumference." The distance across a circle is called the "diameter." Any time we take the circumference and divide it by the diameter, we get the number That s called pi, which is written π. Pi is a tremendously important number in mathematics. Since we found above that C d can prove through algebra that C = πd. = π, we also What students will need help understanding is that while measuring the distance around a circle (the circumference) is difficult, as mentioned before, now it s not necessary since there is a formula Texas Instruments Incorporated TI-30XS MultiView Activity p. 1 of 2
2 Let s see how this works. Earlier, we measured around this lid and got a circumference, or distance around, of almost inches. If we hadn t had a tape measure, we would have needed to use the formula we just found. Since the distance across was 3 inches, we can find that C = πd, so C = π(3), so C = 3π. Using our calculator, we see that 3π 9.42 inches, which makes sense, as we recorded a measurement of about 9.5 inches. Now introduce the formula for area of a circle, and show students why, again, π is a crucial part of finding the area. Use a polygon such as a rectangle to show how you can measure and count squares to find the area but how that won t work on a circle, since the perimeter is curved. (Diagrams are included at the end of this activity for use on the projector.) We could take this figure and find the area a few different ways. We know its dimensions, so we could easily divide the rectangle into equal-sized squares, and count them. We could also use the formula A = lw. Either way, we ll find that A = 40 units 2. Follow these steps: 1. Press g 3 < to input the information. 2. Now press n to see the approximate answer 3. The screen should show this: 4. Press n one more time to see 3π again. 5. The calculator can toggle between the more precise answer of 3π and the approximation. 5 8 However, could you use the same process on this circle? 4 No, because there s no way to divide it into equal-sized squares. Again, we need the number π. The area of a circle can be expressed as A = πr 2, where r is the radius Texas Instruments Incorporated TI-30XS MultiView Activity p. 2 of 2
3 Name Date Directions: Use your TI-30XS MultiView calculator to find the area and circumference of each circle. Use the correct units. 1. VP = 6 mm 2. DT = 4 in. Round to the nearest hundredth. 3. NK = 11 cm AT = 12 cm Round to the nearest tenth. 4. XB = 8.5 in. UY = 6 in. 5. QD = 15 mm KU = 9 mm Find the exact circumference and the area of half the circle Texas Instruments Incorporated TI-30XS MultiView Worksheet p. 1 of 1
4 Answer Key Directions: Find the area and circumference of each circle. Use the correct units. 1. VP = 6 mm C = 12π mm A = 36π mm 2 2. DT = 4 in. Round to the nearest hundredth. C = in. A = in 2 3. NK = 11 cm AT = 12 cm Round to the nearest tenth. C = 75.4 cm A = cm 2 4. XB = 8.5 in. UY = 6 in. 5. QD = 15 mm KU = 9 mm Find the exact circumference and the area of half the circle. C = 17 2π in. A = 289 π in 2 16 C = 18π mm A = 81 π mm Texas Instruments Incorporated TI-30XS MultiView Worksheet Answer Key
5 Transparency Texas Instruments Incorporated TI-30XS MultiView In-Class Exploration
6 Transparency Texas Instruments Incorporated TI-30XS MultiView In-Class Exploration
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# How to solve math problems app
These sites allow users to input a Math problem and receive step-by-step instructions on How to solve math problems app. We can solving math problem.
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Find the derivative of sin(2cosx)
Find the derivative of sin(2cosx)
The second derivative of the function f is given by f′′(x)=sin((x^2)/8)−2cosx. The function f has many critical points, two of which are at x=0 and x=6.949. Which of the following statements is true?
a) f has a local minimum at x=0 and at x=6.949.
b) f has a local minimum at x=0 and a local maximum at x=6.949.
c) f has a local maximum at x=0 and a local minimum at x=6.949.
d) f has a local maximum at x=0 and at x=6.949.
$$f”(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x)$$
Let’s plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.
$$f”(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0$$
Since f”(0) < 0 the graph of f is concave down at x = 0 and so a local max occurs at x = 0
$$f”(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0$$ (assuming x is in radians)
Since f”(6.949) < 0 the graph of f is concave down at x = 6.949 and so a local max occurs at x = 6.949
$$f”(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x)$$
Let’s plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.
$$f”(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0$$
Since f”(0) < 0 the graph of f is concave down at x = 0 and so a local max occurs at x = 0
$$f”(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0$$ (assuming x is in radians)
Since f”(6.949) < 0 the graph of f is concave down at x = 6.949 and so a local max occurs at x = 6.949
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# NCERT Solutions for Class 8 Maths Exercise 6.1 Chapter 6- square square roots (2023)
The NCERT solutions for Class 8 maths enhance topics with frequent, focused, engaging maths challenges and activities that strengthen maths concepts. Each question of the exercise has been carefully solved for the students to understand, keeping the examination point of view in mind.
Class 8 Maths Chapter 6 – Squares and Square Roots Exercise 6.1 questions and answers help students to understand the difference between squares and square roots, as well as how to find them out. These NCERT Solutions are prepared by subject experts at BYJU’S using a step-by-step approach.
### Access other exercise solutions of Class 8 Maths Chapter 6 – Squares and Square Roots
Exercise 6.2 Solutions 2 Questions
Exercise 6.3 Solutions 10 Questions
Exercise 6.4 Solutions 9 Questions
### Access Answers of Maths NCERT Class 8 Chapter 6 – Squares and Square Roots Exercise 6.1 Page Number 96
1. What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
vi. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
The unit digit of the square of a number having ‘a’ at its unit place ends with a×a.
i. The unit digit of the square of a number having digit 1 as the unit’s place is 1.
∴ The unit digit of the square of the number 81 is equal to 1.
ii. The unit digit of the square of a number having the digit 2 as the unit’s place is 4.
∴ The unit digit of the square of the number 272 is equal to 4.
iii. The unit digit of the square of a number having the digit 9 as the unit’s place is 1.
∴ The unit digit of the square of the number 799 is equal to 1.
iv. The unit digit of the square of a number having the digit 3 as the unit’s place is 9.
(Video) Squares and Square Roots - Exercise 6.1 Solutions | Class 8 NCERT Maths Chapter 6 (2022-23)
∴ The unit digit of the square of the number 3853 is equal to 9.
v. The unit digit of the square of a number having the digit 4 as the unit’s place is 6.
∴ The unit digit of the square of the number 1234 is equal to 6.
vi. The unit digit of the square of a number having the digit 7 as the unit’s place is 9.
∴ The unit digit of the square of the number 26387 is equal to 9.
vii. The unit digit of the square of a number having the digit 8 as the unit’s place is 4.
∴ The unit digit of the square of the number 52698 is equal to 4.
viii. The unit digit of the square of a number having the digit 0 as the unit’s place is 01.
∴ The unit digit of the square of the number 99880 is equal to 0.
ix. The unit digit of the square of a number having the digit 6 as the unit’s place is 6.
∴ The unit digit of the square of the number 12796 is equal to 6.
x. The unit digit of the square of a number having the digit 5 as the unit’s place is 5.
∴ The unit digit of the square of the number 55555 is equal to 5.
2. The following numbers are obviously not perfect squares. Give reason.
i. 1057
ii. 23453
iii. 7928
iv. 222222
v. 64000
vi. 89722
vii. 222000
viii. 505050
Solution:
We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.
i. 1057 ⟹ Ends with 7
ii. 23453 ⟹ Ends with 3
(Video) Q 1 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6
iii. 7928 ⟹ Ends with 8
iv. 222222 ⟹ Ends with 2
v. 64000 ⟹ Ends with 0
vi. 89722 ⟹ Ends with 2
vii. 222000 ⟹ Ends with 0
viii. 505050 ⟹ Ends with 0
3. The squares of which of the following would be odd numbers?
i. 431
ii. 2826
iii. 7779
iv. 82004
Solution:
We know that the square of an odd number is odd, and the square of an even number is even.
i. The square of 431 is an odd number.
ii. The square of 2826 is an even number.
iii. The square of 7779 is an odd number.
iv. The square of 82004 is an even number.
4. Observe the following pattern and find the missing numbers. 112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …….2………1
100000012 = ……………………..
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and the middle digit is 2. And the number of zeros between the left-most digit 1 and the middle digit 2 and the right-most digit 1 and the middle digit 2 is the same as the number of zeros in the given number.
∴ 1000012 = 10000200001
100000012 = 100000020000001
(Video) Chapter 6 Square and Square Roots || Full Exercise 6.1 & Basic || Class 8 Maths RBSE CBSE NCERT
5. Observe the following pattern and supply the missing numbers. 112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And the square is symmetric about the middle digit. If the middle digit is 4, the number to be squared is 10101, and its square is 102030201.
So, 10101012 =1020304030201
1010101012 =10203040505030201
6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + _2 = 212
5 + _ 2 + 302 = 312
6 + 7 + _ 2 = _ 2
Solution:
Given, 12 + 22 + 22 = 32
i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2
22 + 32 + 62 =72
∴ 22 + 32 + (2×3 )2 = (22 + 32 -2 × 3)2
32 + 42 + 122 = 132
∴ 32 + 42 + (3×4 )2 = (32 + 42 – 3 × 4)2
42 + 52 + (4×5 )2 = (42 + 52 – 4 × 5)2
∴ 42 + 52 + 202 = 212
52 + 62 + (5×6 )2 = (52+ 62 – 5 × 6)2
∴ 52 + 62 + 302 = 312
(Video) Chapter:6 (Intro)+Ex 6 1 (Q.1,2,3) Squares and Square Roots | Ncert | Maths | Class 8 | Cbse.
62 + 72 + (6×7 )2 = (62 + 72 – 6 × 7)2
∴ 62 + 72 + 422 = 432
7. Without adding, find the sum.
i. 1 + 3 + 5 + 7 + 9
Solution:
Sum of first five odd numbers = (5)2 = 25
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Solution:
Sum of first ten odd numbers = (10)2 = 100
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Sum of first thirteen odd numbers = (12)2 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Solution:
We know that the sum of the first n odd natural numbers is n2 . Since,49 = 72
∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers. Solution:
Since, 121 = 112
∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Solution:
Between n2 and (n+1)2, there are 2n non–perfect square numbers.
i. 122 and 132, there are 2×12 = 24 natural numbers.
(Video) NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1
ii. 252 and 262, there are 2×25 = 50 natural numbers.
iii. 992 and 1002, there are 2×99 =198 natural numbers.
Exercise 6.1 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:
1. Introduction to squares and square roots
2. What are square numbers?
3. Properties of Square Numbers
4. Some interesting patterns
2. Numbers between square numbers
4. A sum of consecutive natural numbers
## FAQs
### What is square and square roots Class 8? ›
Square root: The Square root of a number is a value that when multiplied by itself gives the original value. There are two methods two find the square root of the given number: Prime factorization method. Division method.
What is square in math class 8? ›
Definition. Square is a regular quadrilateral, which has all the four sides of equal length and all four angles are also equal. The angles of the square are at right-angle or equal to 90-degrees. Also, the diagonals of the square are equal and bisect each other at 90 degrees.
What is a perfect square Class 8? ›
A number is a perfect square or a square number if its square root is an integer, which means it is an integer's product with itself.
How to solve √ 81? ›
Square Root of 81 by Prime Factorization Method
We know that the prime factorization of 81 is 3 × 3 × 3 × 3. √81 = 3×3 = 9. Hence, the value of the square root of 81 is 9.
How do you solve a square root answer? ›
For example, 6 × 6 = 36. Here, 36 is the square of 6. The square root of a number is that factor of the number and when it is multiplied by itself the result is the original number. Now, if we want to find the square root of 36, that is, √36, we get the answer as, √36 = 6.
How to solve √ 2? ›
Root 2 is an irrational number as it cannot be expressed as a fraction and has an infinite number of decimals. So, the exact value of the root of 2 cannot be determined.
What are 4 types of square? ›
A rectangle with two adjacent equal sides. A quadrilateral with four equal sides and four right angles. A parallelogram with one right angle and two adjacent equal sides. A rhombus with a right angle.
Is 5050 a perfect square? ›
The numbers 257 and 5050 are not perfect squares.
What is the square of 8 answer? ›
The actual answer for the square root of 8 is 2.82842712475.
...
Square root Table From 1 to 15.
NumberSquaresSquare Root (Upto 3 places of decimal)
772 = 49√7 = 2.646
882 = 64√8 = 2.828
992 = 81√9 = 3.000
10102 = 100√10 = 3.162`
11 more rows
How to solve √ 100? ›
The square root of 100 is 10. It is the positive solution of the equation x2 = 100. The number 100 is a perfect square.
### How to solve √ 40? ›
The square root of 40 is symbolically expressed as √40. Thus, if we multiply the number 6.3245 two times, we get the original value 40. √40 = ± 6.3245. Square Root of 40 in Decimal Form: 6.3245.
How to solve √ 625? ›
Square root of 625 is 25 and is represented as √625 = 25. A square root is an integer (can be either positive or negative) which when multiplied with itself, results in a positive integer called as the perfect square number.
Is 3.14 a square root? ›
Because all square roots of irrational numbers are irrational numbers, the square root of pi is also an irrational number. However, that doesn't mean we can't approximate the answer. Just like we approximate the value of pi to be 3.14, we can approximate the square root of pi to be 1.77.
What is the square √ 64? ›
The square root of 64 is 8, i.e. √64 = 8.
What is the formula of √? ›
The square root of a number is a number squaring which gives the original number. It is that factor of the number that when squared gives the original number. It is the value of power 1/2 of that number. The square root of a number is represented as √.
What is a square in math? ›
A square is a two-dimensional closed shape with 4 equal sides and 4 vertices. Its opposite sides are parallel to each other. We can also think of a square as a rectangle with equal length and breadth.
What is square example? ›
Definition: A square is a two-dimensional shape which has four sides of equal length. The opposite sides of a square are parallel to each other and all four interior angles are right angles. Paper napkins, chess boards, cheese slices, floor tiles, and so on are some real-life examples of square shaped objects.
What meaning is square? ›
a flat shape with four sides of equal length and four angles of 90°: First draw a square. It's a square-shaped room.
Why is 8 a square number? ›
Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.
How to find square root? ›
Square Root
1. Square root of a number is a value, which on multiplication by itself, gives the original number. ...
2. Suppose x is the square root of y, then it is represented as x=√y, or we can express the same equation as x2 = y.
Nov 20, 2019
### What is a square root example? ›
A square root of a number is a value that, when multiplied by itself, gives the number. Example: 4 × 4 = 16, so a square root of 16 is 4. Note that (−4) × (−4) = 16 too, so −4 is also a square root of 16.
What is square formula? ›
The Formula for the Area of A Square
The area of a square is equal to (side) × (side) square units. The area of a square when the diagonal, d, is given is d2÷2 square units. For example, The area of a square with each side 8 feet long is 8 × 8 or 64 square feet (ft2).
Does square mean 2? ›
A square number is a number multiplied by itself. This can also be called 'a number squared'. The symbol for squared is ².
Why is it called a square? ›
square (adj.) early 14c., "containing four equal sides and right angles," from square (n.), or from Old French esquarre, past participle of esquarrer. Meaning "honest, fair," is first attested 1560s; that of "straight, direct" is from 1804.
## Videos
1. Exercise 6.1 / Class-8 Maths NCERT Chapter-6 Square And Square Roots Solution By-KV Teacher
(KV TEACHERS GUILD)
2. exercise 6.1 class 8 maths | chapter 6 squares and square root | ncert maths class 8 | cbse
(Math By Sachin Kaushik)
3. Class 8 Squares and Square Roots Exercise 6.1-Q#1-9 (In English)- NCERT CBSE
(Divya's Maths Solutions)
4. NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1
5. Q 2 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6
(Mathematics Class 8)
6. Q 3 - Ex 6.1 - Square and Square Roots - NCERT Maths Class 8th - Chapter 6
(Mathematics Class 8)
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# 2002 AMC 8 Problems/Problem 18
## Problem
Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?
$\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr}$
## Solution 1
Converting into minutes and adding, we get that she skated $75*5+90*3+x = 375+270+x = 645+x$ minutes total, where $x$ is the amount she skated on day $9$. Dividing by $9$ to get the average, we get $\frac{645+x}{9}=85$. Solving for $x$, $$645+x=765$$ $$x=120$$ Now we convert back into hours and minutes to get $\boxed{\text{(E)}\ 2\ \text{hr}}$.
## Solution 2
For the first five days, each day you are $10$ minutes short of $85$ minutes. And for the next three days, you are $5$ minutes above $85$ minutes. So in total you are missing $3*5-5*10$, which equals to negative $35$. So on the ninth day, to have an average of $85$ minutes, Gage need to skate for $85+35$ minutes, which is $120$ minutes, or $\boxed{\text{(E)}\ 2\ \text{hr}}$.
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# RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D
In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20D for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20D Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D Maths book pdf download. Now you will get step by step solution to each question.
### RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20D Download PDF
Question 1.
Solution:
(i) Base of the triangle = 42 cm
Height = 25 cm
Area = 12 x base x height
= 12 x 42 x 25 = 525 cm²
(ii) Base of the triangle = 16.8 m
and height = 75 cm = 0.75 m
Area = 12 x Base x height
= 12 x 16.8 x 0.75 m2 = 6.3 m²
(iii) Base of a triangle (b) = 8 m = 80 cm
and height (h) = 35 cm
Area = 12 bh = 12 x 80 x 35 = 1400 cm²
Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area = 12 bh = 12 x 3x x x
Question 6.
Solution:
Area of the right angled triangle = 129.5 cm²
Question 7.
Solution:
In right angled ∆ABC,
Base BC = 1.2 m
and hypotenuse AC = 3.7 m
But AC² = AB² + BC² (Pythagoras Theorem)
⇒ (3.7)² = AB² + (1.2)²
⇒ 13.69 = AB² + 1.44
⇒ AB² = 13.69 – 1.44
⇒ AB² = 12.25 = (3.5)²
⇒ AB = 3.5 m
Now, area of ∆ABC = 12 x base x altitude
= 12 x 1.2 x 3.5 m² = 2.1 m²
Question 8.
Solution:
Legs of a right angled triangle = 3 : 4
Let one leg (base) = 3x
Then second leg (altitude) = 4x
Area = 12 x base x altitude
= 12 x 3x x 4x = 6x²
6x² = 1014
⇒ x² = 10146 = 169 = (13)²
x = 13
one leg'(Base) = 3x = 3 x 13 = 39 cm
and second leg (altitude) = 4x = 4 x 13 = 52 cm
Question 9.
Solution:
One side BC of a right triangular scarf = 80 cm
and longest side AC = 1 m = 100 cm
By Pythagoras Theorem,
AC² = AB² + BC²
⇒ (100)² = AB² + (80)²
⇒ 10000 = AB² + 6400
⇒ AB² = 10000 – 6400
⇒ AB² = 3600 = (60)²
⇒ AB = 60
Second side = 60 cm
Area of the scarf = 12 x b x h
= 12 x 80 x 60 cm2 = 2400 cm²
Rate of cost = Rs. 250 per m²
Total cost =2400100×100 x 250 = Rs. 60
Question 10.
Solution:
(i) Side of the equilateral triangle (a) = 18 cm
Question 11.
Solution:
Area of equilateral triangle = 16√3 cm²
Let each side = a
then √34 a² = 16√3
⇒ a² = 16√3×4√3
⇒ a² = 64 = (8)²
a = 8 cm
Each side = 8 cm
Question 12.
Solution:
Each side of an equilateral triangle = 24cm
Length of altitude = √32 a = √32 x 24
= 12√3 cm = 12 (1.73) = 20.76 cm
Question 13.
Solution:
(i) a = 13 m, b = 14 m, c = 15 m
= 2 x 2 x 3 x 7 x 7 x 7 = 4116 m²
Question 14.
Solution:
Let a = 33 cm, b = 44 cm, c = 55 cm
Question 15.
Solution:
Perimeter of the triangle = 84 cm
Ratio in side = 13 : 14 : 15
Sum of ratios =13 + 14 + 15 = 42
Let then first side = 84×1342 = 26 cm
Question 16.
Solution:
Let a = 42 cm, b = 34 cm, c = 20 cm
Question 17.
Solution:
In isosceles ∆ABC
Base BC = 48 cm.
and AB = AC = 30cm.
Question 18.
Solution:
Perimeter of an isosceles triangle = 32 cm
Base = 12 cm
Sum of two equal sides = 32 – 12 = 20 cm
Length of each equal side = 202 = 10cm
BD = DC = 122 = 6 cm
Question 19.
Solution:
diagonal AC = 26 cm.
and perpendiculars DL = 12.8cm, BM = 11.2 cm
= 12 (Sum of perpendicular) x diagonal
= 12 (12.8 + 11.2) x 26 cm²
= 12 x 24 x 26 = 312 cm²
Question 20.
Solution:
AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm
and diagonal AC = 30 cm
In ∆ABC,
Question 21.
Solution:
ABCD is a rectangle in which AB = 36 m
and BC = 24m
In ∆AED,
EF = 15 m
AD = BC = 24 m.
Now area of rectangle ABCD = l x b = 36 x 24 cm² = 864 cm²
Area of ∆AED = 12 x AD x EF
= 12 x 24 x 15 cm² = 180 cm²
Area of shaded portion = 864 – 180 = 684 m²
Question 22.
Solution:
In the fig. ABCD is a rectangle in which AB = 40 cm, BC = 25 cm.
P, Q, R and S and the mid points of sides, PQ, QR, RS and SP respectively
Then PQRS is a rhombus.
Now, join PR and QS.
PR = BC = 25cm and QS = AB = 40cm
Area of PQRS = 12 x PR x QS
= 12 x 25 x 40 = 500 cm²
Question 23.
Solution:
(i) Length of rectangle (l) = 18 cm
and breadth (b) = 10 cm
Area = l x b = 18 x 10 = 180 cm²
Area of right ∆EBC = 12 x 10 x 8 = 40 cm²
and area of right ∆EDF = 12 x 10 x 6 = 30 cm²
Area of shaded region = 180 – (40 + 30) = 180 – 70 = 110 cm²
(ii) Side of square = 20 cm
Area of square = a² = (20)² = 400 cm²
Area of right ∆LPM = 12 x 10 x 10 cm² = 50 cm²
Area of right ∆RMQ = 12 x 10 x 20 = 100 cm²
and area of right ∆RSL = 12 x 20 x 10 = 100 cm²
Area of shaded region = 400 – (50 + 100 + 100) cm2 = 400 – 250 = 150 cm²
Question 24.
Solution:
BD = 24 cm
AL ⊥ BD and CM ⊥ BD
AL = 5 cm and CM = 8 cm
All Chapter RS Aggarwal Solutions For Class 7 Maths
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## Related Rates: Defining Variables
By Tutor GuyNo Comments
Related rate problems will no longer seem so complicated once you master the art of defining rates as variables. And not just any variables, but variables in the form of derivatives! In other words, express the known and unknown rates as derivatives such as $dV/dt$ or $dA/dt$. Remember that $dA/dt$ means “the rate of change of the Area”. Your task is to find expressions for rates in the problem and “translate” them from English into Calculus. Here are some examples:
• “The radius is increasing at a rate of 3.0 cm/s.”
$\dfrac{dr}{dt}=3.0$
• “If the volume is decreasing at a rate of 1.5 liters per minute…”
$\dfrac{dV}{dt}=-1.5$
• “How fast is the ladder falling down the side of the building?”
$\dfrac{dy}{dt}= \, ?$
Remember that speed at which something moves is the rate of change of its position. Also, this derivative expresses the unknown we are going to find in the problem.
• “The temperature is dropping by 2.3° per hour.”
$\dfrac{dT}{dt}=-2.3$
Note that when a value is falling or decreasing, the value of the derivative is negative.
Once you have defined all your rates as variables, your next step is to write the equation that expresses one variable in terms of the others. However, the equation has to be in terms of the quantities in the numerators, not in terms of the derivatives. Yes, this seems all wrong, but you will get the derivatives when you differentiate each side of the equation. For example, if your two rates are dV/dt and dr/dt, then you need to write an equation that expresses V in terms of r. When you differentiate this equation, you get the related rates equation and can plug in all the values you have.
Example: A hot air balloon in the shape of a sphere is being filled with gas at the rate of 20 cubic meters per minute. How fast is the radius changing when the volume is 288π cubic meters?
Solution: First, find all the rates mentioned in the problem:
$\dfrac{dV}{dt}=20; \; \dfrac{dr}{dt}= \, ?$
Now, you need an equation that relates volume (V) to radius (r). Because the balloon is a sphere, use the formula for the volume of a sphere:
$V= \dfrac{4}{3} \pi r^3$
Next, differentiate with respect to time:
$\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt}$
This equation includes r. You need to plug in the value of V in the original equation to solve for r:
$V= \dfrac{4}{3} \pi r^3 \rightarrow 288 \pi= \dfrac{4}{3} \pi r^3 \rightarrow 216=r^3 \rightarrow r=6 \, m$
Now plug in all the values to solve for the unknown rate:
$\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt} \rightarrow 20=4 \pi (6)^2 \dfrac{dr}{dt} \rightarrow \dfrac{dr}{dt}= \dfrac{20}{4 \pi (36)} \approx 0.044 \, m/min$
(Look for other tips on solving related rate problems on this website if you need help with other aspects of related rate problems.)
Calculus
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# Solving $y' -3y = x \cdot e^x$
I'm struggling with this differential equation: $y' -3y = x \cdot e^x$
Step one: Solve the homogenous equation $y' -3y = 0$ $$y_0 = e^{3x} \cdot C$$
The approach for the particular solution $y_p$ should be $(c_1x+c_0)\cdot Ce^x$
This leads me to the following term: $$y_p = Ce^x(ax+b)$$ $$y'_p= Ce^x\cdot(ax+b)+Ce^x\cdot a$$ $$Ce^x\cdot(ax+b)+Ce^x\cdot a - 3Ce^x(ax+b) = x\cdot e^x$$ $$C\cdot(ax+b)+C\cdot a - 3C(ax+b) = x$$
The C is a real problem here. After taking a look in my textbook I saw that the approach for $y_p$ is: $$y_p = (ax + b)\cdot e^x$$ $$y'_p = e^x \cdot (ax +b)+e^x \cdot a$$ Plugging this into $y' - 3y = x\cdot e^x$ and simplify gives:
$$x(-2a)+a-2b=x \Rightarrow a=-\frac{1}{2}; b=-\frac{1}{4}$$ And finally $$y= e^{3x}\cdot C - \frac{1}{4}(e^x+2xe^x)$$
When I omit the C I'm able to solve this but I don't know why this is correct as the approach for $g(x)$ would be $C\cdot e^{bx}$
• The particular solution is supposed to be any solution to the whole equation. $C$ is extraneous as you can absorb it into $a$ and $b$. You can see that by looking at your solution where it always multiplies one or the other. If you are going to use $C$ you should use a different variable than the one you used for the general solution. That is a good way to get confused. – Ross Millikan Sep 19 '18 at 0:16
Just note that as $c_0 , c_1 , C$ vary all over $\mathbb{R}$ you have that $a=c_1C$ and $b=c_0C$ vary all over $\mathbb{R}$ AND $(c_1x+c_0)Ce^x = (ax+b)e^x$.
So considering $3$ variables is useless (they are too many) since you only want a constant for $xe^x$ and a constant for $e^x$.
$$C\cdot(ax+b)+C\cdot a - 3C(ax+b) = x$$ $$-2Cax+(Ca-2Cb)=x$$ $$\implies aC=-1/2,$$ $$\implies 2Cb=Ca \implies Cb=-1/4$$ $$y_p = Ce^x(ax+b) \implies y_p = e^x(-\frac12x-\frac 14)$$ $$y_p = -\frac 12e^x(x+\frac 12)$$
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# Lesson 9Guess My ParallelogramPractice Understanding
Use the given scale drawings to find the desired actual lengths and areas.
### 1.
Jordan’s living room is a rectangle represented in the scaled drawing with a scale factor of .
Find the perimeter and the area of the real-life size living room.
### 2.
The figures shown are scale drawings of a quarter sector of a circle. Use them to determine the scale factor for both enlarging the smaller figure to create the larger one and also for shrinking the larger figure to create the smaller one. Then find the areas of the sectors and explain how they are related.
Scale Factor from small to big:
Scale Factor from big to small:
Area of small sector:
Area of large sector:
Explanation of relationship between the areas:
### 3.
The scale drawing of two pentagons has some corresponding sides labeled with the number of units they measure.
Use these given sides to find the scale factor and determine each of the missing side lengths.
Reproduce each of the drawings using the given scale factor.
Scale Factor
Scale Factor
## Set
Use the given information and the diagram to prove each statement. Create a proof idea and be sure your reasoning is logical. Justify your statements.
(Hint: Consider using congruent triangles or transformations.)
### 6.
Given: is a Rhombus
Prove: bisects
### 7.
Given: is a Rhombus
Prove:
### 8.
Given: is a Rhombus
Prove: bisects
## Go
Use the given information and the diagram to prove each statement. Create a proof idea and be sure your reasoning is logical. Justify your statements. (Hint: Consider using congruent triangles or transformations.)
### 9.
Given: is a rectangle
Prove: bisects
### 10.
Given: is a rectangle
Prove:
Given: and
Prove:
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Perimeter of a Rhombus – Explanation & Examples
The perimeter of a rhombus is the total length measured across its boundaries.
All sides of a rhombus are equal to each other. If the length of any single side is equal to $x$, as shown in the figure above, then the perimeter is given as
Perimeter $=4x$
We get the perimeter of a rhombus by adding the value of all of its sides. This topic will help you understand the properties of a rhombus and how to calculate its perimeter.
Before we jump to the topic, you must know the difference between a rhombus, a square, and a parallelogram, as all of them are quadrilaterals (i.e., four-sided geometrical figures) and share some commonalities. The differences between them is presented in the table below.
Parallelogram Square Rhombus The opposite sides of a parallelogram are equal All sides of a square are equal All sides of a rhombus are equal The opposite angles of a parallelogram are equal, while the adjacent angles supplement each other. All angles ( interior & adjacent) are equal. All angles are right angles, i.e., 90 degrees. The sum of two interior angles of a rhombus equals 180 degrees. Therefore, if all the angles of a rhombus are equal, they will be $90^o$ each, making it a square. The diagonals of a parallelogram bisect each other. The diagonals of the square are equal in length. The diagonals of the rhombus bisect each other and are equal in length. Every Parallelogram is not a rhombus. Every rhombus is a parallelogram. All four sides of a square are perpendicular to each other. The sides of a rhombus are not necessarily perpendicular.
What Is the Perimeter of a Rhombus?
The perimeter of a rhombus is the total distance covered around its boundaries. A rhombus is a flat geometrical figure with four sides, and if we add the length of all four sides, it will give us the perimeter of the rhombus.
All sides of a rhombus are equal, similar to a square, and the perimeter is calculated by multiplying 4 with the length of a single side.
Take note that unlike a square, the four angles of a rhombus are not necessarily equal to $90^{o}$. A rhombus is a mixture of a rectangle and a square, and the properties of a rhombus are given below.
1. All four sides of a rhombus are equal to each other.
2. Opposite sides of a rhombus are parallel to each other.
3. The diagonals of a rhombus bisect each other at $90^{0}$.
4. The opposite angles of a rhombus are equal to each other.
5. Just like a rectangle, the sum of two adjacent angles of a rhombus is $180^{o}$.
The perimeter is a linear measure, so the units of the perimeter are the same as the units of the lengths of each side, i.e., centimeters, meters, inches, ft, etc.
How To Find the Perimeter of a Rhombus
The perimeter of a rhombus is defined as the sum of all the sides of a rhombus. If we add all sides, it will give us the perimeter of the rhombus. This method is only applicable if we are given the length of any one side of a rhombus.
Sometimes, we are given the diagonals of a rhombus and we are asked to find the perimeter. Thus, the given data determines which method we should use to calculate the perimeter of a rhombus.
Perimeter of a Rhombus Using the Side Method
This method is used when we are given the length of any one side of a rhombus. As discussed earlier, all the sides of a rhombus are equal. Therefore, if one side of a rhombus is “x,” then we can calculate the perimeter of the rhombus by multiplying “x” with 4.
Perimeter of a Rhombus Using the Diagonal Method
This method is used when we are given the length of the diagonals of a rhombus and no data regarding the lengths of the sides of the rhombus is available. However, we know that the diagonals of a rhombus bisect each other at a right angle, so when we draw the diagonals of a rhombus, it provides us with four congruent right-angled triangles, as shown in the picture below.
To calculate the perimeter by using this method, we follow the steps listed below:
1. First, write down the measurements of the diagonals of the rhombus.
2. Then, apply the Pythagorean theorem to get the value of any one side of the rhombus.
3. Finally, multiply the calculated value in step 2 by “4”.
Perimeter of a Rhombus Formula
We can derive the formula for the perimeter of a rhombus by multiplying the length of any of the sides by “4”. We know that all the sides of a rhombus are equal, and we can write the formula for the perimeter of a rhombus as:
Perimeter of a rhombus $= x + x + x + x$
Perimeter of a rhombus $= 4\times x$
Perimeter of a Rhombus When Two Diagonals Are Given
Let us derive the formula of the perimeter of a rhombus when we are provided with the length of the diagonals. Consider this picture of a rhombus with the values of both diagonals available.
We can take any of the four triangles to solve for the formula. Lets us take the triangle ABP. We know the diagonals of the rhombus bisect each other at $90^{o}$, so we can write AP and BP as $\dfrac{a}{2}$ and $\dfrac{b}{2}$ respectively. Now, if we apply the Pythagorean theorem on the triangle ABP:
$c^{2} = (\dfrac{a}{2})^{2} + (\dfrac{b}{2})^{2}$
$c^{2} = (\dfrac{a^{2}}{4}) + (\dfrac{b^{2}}{4})$
$c = \dfrac{\sqrt{(a^{2}+ b^{2})}}{2}$
We know that we can write the formula for the perimeter of the rhombus when one side (in this instance, the side “c”) is given as:
Perimeter of a rhombus $= 4 \times c$
Plugging the value of “c” in the above formula:
Perimeter of a rhombus $= 4 \times \dfrac{\sqrt{(a^{2}+ b^{2})}}{2}$
Perimeter of a rhombus $= 2 \times \sqrt{(a^{2}+ b^{2})}$
Note: You can also use the above formula to calculate the perimeter of the rhombus if you are provided with the length of one diagonal along with the area of the rhombus. Formula for the area of the rhombus $= \dfrac{diagonal\hspace{1mm} 1\times diagonal \hspace{1mm} 2}{2}$. So, we can calculate the length of the second diagonal using the area formula and then use the perimeter formula given above to calculate the perimeter of the rhombus.
Real-life Applications of the Perimeter of a Rhombus
The word perimeter is a combination of two Greek words: “Peri,” which means surrounding or boundaries of a surface or an object, and “Meter,” which means measurement of the surface or object, so perimeter means the total measurement of boundaries of a given surface.
With this information, we can use the perimeter of a rhombus in numerous real-life applications. Various examples are given below:
• For example, we can use the perimeter of a rhombus to calculate the distance of a pitcher’s spot from the striker in baseball if the whole pitch is shaped like a rhombus.
• The perimeter formula is also helpful in designing tables and cupboards having a rhombus shape.
• It is also helpful in the construction of rhombus-shaped offices and rooms.
Example 1:
If the length of one side of a rhombus is 11 cm, what will be the length of the rest of the sides?
Solution:
We know that all the sides of a rhombus are equal in length, so the length of the rest of the three sides is also 11 cm each.
Example 2:
Calculate the perimeter of a rhombus for the figure given below.
Solution:
We are given the length of one side of a rhombus, and we know that all the sides are equal in length.
Perimeter of the rhombus $= 4\times 8$
Perimeter of the rhombus $= 32 cm$
Example 3:
If the perimeter of a rhombus is 80cm, what will be the length of all the sides of the rhombus?
Solution:
We are given the perimeter of the rhombus. We can calculate the length of each side of a rhombus by using the perimeter formula:
Perimeter of a rhombus $= 4\times side$
$80 = 4\times side$
Side $= \frac{80}{4}$
Side $= \frac{80}{4}$
Side $= 20 cm$
All sides of the rhombus are 20 cm.
Example 4:
If the length of the diagonals of a rhombus is 9 cm and 11cm, what will be the perimeter of the rhombus?
Solution:
We are given the length of the two diagonals of the rhombus: let “a” and “b” be the two diagonals of the rhombus. Then, we can calculate the perimeter of the rhombus by using the formula given below.
Perimeter of the rhombus $= 2 \times \sqrt{(a^{2}+ b^{2})}$
Perimeter of the rhombus $= 2 \times \sqrt{(9^{2}+ 11^{2})}$
Perimeter of the rhombus $= 2 \times \sqrt{99 + 121}$
Perimeter of the rhombus $= 2 \times \sqrt{220}$
Perimeter of the rhombus $= 2 \times 14.83$
Perimeter of the rhombus $= 29.67 cm$ approx.
Example 5:
A rhombus has an area of $64 cm^{2}$, and the length of one diagonal of the rhombus is $8 cm$. What will be the perimeter of the rhombus?
Solution:
Let diagonal “a” = 8cm and we have to find “b”
Area of the rhombus $= \dfrac{a\times b}{2}$
$64 = \dfrac{8\times b}{2}$
$128 = 8 \times b$
$b = \dfrac{128}{8}$
$b = 16 cm$
Perimeter of a rhombus $= 2 \times \sqrt{(a^{2}+ b^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{(8^{2}+ 16^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{64 + 256}$
Perimeter of a rhombus $= 2 \times \sqrt{320}$
Perimeter of a rhombus $= 2 \times 17.89$
Perimeter of a rhombus $= 35.78 cm$ approx.
Practice Questions
1. If one side of a rhombus is $20 cm$, what is the length of the remaining sides and the perimeter of the rhombus?
2. If the perimeter of a rhombus is $100 cm$, what is the length of the sides of the rhombus?
3. If the length of the diagonals of a rhombus is $9 cm$ and $12cm$, what will be the rhombus’ perimeter and area?
4. Consider a rhombus having an area of $36 cm ^{2}$ while the length of one of the diagonal is $4 cm$. What will be the perimeter of the rhombus?
1. We know that all sides of a rhombus are equal in length. If the length of one side of the rhombus is 20 cm, then the length of the remaining three sides will also be the same, i.e., 20 cm.
Perimeter of the rhombus $= 4\times side$
Perimeter of the rhombus $= 4\times 20$
Perimeter of the rhombus $= 80 cm$
2. We are given the perimeter of the rhombus. We can calculate the length of each side of the rhombus by using the perimeter formula:
Perimeter of a rhombus $= 4\times side$
$100 = 4\times side$
Side $= \frac{100}{4}$
Side $= 25 cm$
We know that all sides of a rhombus are equal in length, so all sides of the rhombus are $25 cm$ long.
3. We are given the lengths of the two diagonals of the rhombus. Let “a” and “b” be the two diagonals. Then, we can calculate the perimeter and area of the rhombus by using the values of the diagonals.
Area of the rhombus $= \dfrac{a\times b}{2}$
Area of the rhombus $= \dfrac{9\times 12}{2}$
Area of the rhombus $= 9\times 6 = 54 cm^{2}$
Now let us calculate the perimeter of the rhombus.
Perimeter of a rhombus $= 2 \times \sqrt{(a^{2}+ b^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{(9^{2}+ 12^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{81 + 144}$
Perimeter of a rhombus $= 2 \times \sqrt{225}$
Perimeter of a rhombus $= 2 \times 15$
Perimeter of a rhombus $= 30 cm$ approx.
4. Let diagonal “a” $= 4 cm$ and we have to find “b”
Area of the rhombus $= \dfrac{a\times b}{2}$
$36 = \dfrac{4 \times b}{2}$
$72 = 4 \times b$
$b = \dfrac{72}{4}$
$b = 18 cm$
Perimeter of a rhombus $= 2 \times \sqrt{(a^{2}+ b^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{(4^{2}+ 18^{2})}$
Perimeter of a rhombus $= 2 \times \sqrt{16 + 324}$
Perimeter of a rhombus $= 2 \times \sqrt{340}$
Perimeter of a rhombus $= 2 \times 18.44$
Perimeter of a rhombus $= 36.88 cm$ approx.
Images/Mathematical drawings are created using GeoGebra.
|
Courses
# Trees Part 1 Notes | EduRev
## : Trees Part 1 Notes | EduRev
``` Page 1
1
Trees : Part 1
Section 4.1
(1) Theory and Terminology
(2) Preorder, Postorder and
Levelorder Traversals
Theory and Terminology
?Definition: A tree is a connected graph
with no cycles
?Consequences:
?Between any two vertices, there is exactly one
unique path
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
Theory and Terminology
?Definition: A rooted tree is a graph G
such that:
?G is connected
?G has no cycles
?G has exactly one vertex called the root of the
tree
Theory and Terminology
?Consequences
?The depth of a vertex v is the length of the
unique path from root to v
?G can be arranged so that the root is at the top,
its neighboring vertices are vertices of depth 1,
and so on…
?The set of all vertices of depth k is called level k
of the tree
Page 2
1
Trees : Part 1
Section 4.1
(1) Theory and Terminology
(2) Preorder, Postorder and
Levelorder Traversals
Theory and Terminology
?Definition: A tree is a connected graph
with no cycles
?Consequences:
?Between any two vertices, there is exactly one
unique path
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
Theory and Terminology
?Definition: A rooted tree is a graph G
such that:
?G is connected
?G has no cycles
?G has exactly one vertex called the root of the
tree
Theory and Terminology
?Consequences
?The depth of a vertex v is the length of the
unique path from root to v
?G can be arranged so that the root is at the top,
its neighboring vertices are vertices of depth 1,
and so on…
?The set of all vertices of depth k is called level k
of the tree
2
A Rooted Tree
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height = 2
height = 3
height = 0
height = 1
Rooted Tree: Recursive definition
? A graph with N nodes and N - 1 edges
? Graph has
?one root r
?Zero or more non-empty sub-trees, each of whose root
is connected to r by an edge.
?Every node except the root has one parent
Theory and Terminology
? Definition: A descending path in a rooted tree is
a path, whose edges go from a vertex to a
deeper vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
8
12
Theory and Terminology
? Consequences:
?A unique path from the root to any vertex is a descending
path
?The length of this path is the depth of the vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
3
8
11
Theory and Terminology
?Definition: If there is a descending path
from v
1
to v
2
, v
1
is an ancestor of v
2
,
and v
2
is a descendant of v
1
.
Theory and Terminology
?Suppose v is a vertex of depth k:
?Any vertex that is adjacent to v must have
depth k - 1 or k + 1.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
7 8
Page 3
1
Trees : Part 1
Section 4.1
(1) Theory and Terminology
(2) Preorder, Postorder and
Levelorder Traversals
Theory and Terminology
?Definition: A tree is a connected graph
with no cycles
?Consequences:
?Between any two vertices, there is exactly one
unique path
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
Theory and Terminology
?Definition: A rooted tree is a graph G
such that:
?G is connected
?G has no cycles
?G has exactly one vertex called the root of the
tree
Theory and Terminology
?Consequences
?The depth of a vertex v is the length of the
unique path from root to v
?G can be arranged so that the root is at the top,
its neighboring vertices are vertices of depth 1,
and so on…
?The set of all vertices of depth k is called level k
of the tree
2
A Rooted Tree
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height = 2
height = 3
height = 0
height = 1
Rooted Tree: Recursive definition
? A graph with N nodes and N - 1 edges
? Graph has
?one root r
?Zero or more non-empty sub-trees, each of whose root
is connected to r by an edge.
?Every node except the root has one parent
Theory and Terminology
? Definition: A descending path in a rooted tree is
a path, whose edges go from a vertex to a
deeper vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
8
12
Theory and Terminology
? Consequences:
?A unique path from the root to any vertex is a descending
path
?The length of this path is the depth of the vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
3
8
11
Theory and Terminology
?Definition: If there is a descending path
from v
1
to v
2
, v
1
is an ancestor of v
2
,
and v
2
is a descendant of v
1
.
Theory and Terminology
?Suppose v is a vertex of depth k:
?Any vertex that is adjacent to v must have
depth k - 1 or k + 1.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
7 8
3
Theory and Terminology
?Suppose v is a vertex of depth k:
?Vertices adjacent to v of depth k + 1 are
called children of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
7 8
Theory and Terminology
? Suppose v is a vertex of depth k:
?If k > 0, there is exactly one vertex of depth k – 1 that
is adjacent to v in the graph.
?This vertex is called the parent of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
Theory and Terminology
?Definitions
?A vertex with no children is called a leaf
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
Theory and Terminology
? Definitions
?Depth of a vertex v is its distance from the root.
?Height of a vertex v is the distance of the longest path from
v to one of its descendant leaves.
?The height of a tree is the maximum depth of its vertices
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height
Theory and Terminology
? Definitions
?The root is the only vertex of depth 0. The root has no
parent.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
Example of rooted tree
? Which are the parent nodes?
? Which are the child nodes?
? Which are the leaves?
? What is the height and depth of the tree?
? What is the height and depth of node E? Node F?
Page 4
1
Trees : Part 1
Section 4.1
(1) Theory and Terminology
(2) Preorder, Postorder and
Levelorder Traversals
Theory and Terminology
?Definition: A tree is a connected graph
with no cycles
?Consequences:
?Between any two vertices, there is exactly one
unique path
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
Theory and Terminology
?Definition: A rooted tree is a graph G
such that:
?G is connected
?G has no cycles
?G has exactly one vertex called the root of the
tree
Theory and Terminology
?Consequences
?The depth of a vertex v is the length of the
unique path from root to v
?G can be arranged so that the root is at the top,
its neighboring vertices are vertices of depth 1,
and so on…
?The set of all vertices of depth k is called level k
of the tree
2
A Rooted Tree
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height = 2
height = 3
height = 0
height = 1
Rooted Tree: Recursive definition
? A graph with N nodes and N - 1 edges
? Graph has
?one root r
?Zero or more non-empty sub-trees, each of whose root
is connected to r by an edge.
?Every node except the root has one parent
Theory and Terminology
? Definition: A descending path in a rooted tree is
a path, whose edges go from a vertex to a
deeper vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
8
12
Theory and Terminology
? Consequences:
?A unique path from the root to any vertex is a descending
path
?The length of this path is the depth of the vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
3
8
11
Theory and Terminology
?Definition: If there is a descending path
from v
1
to v
2
, v
1
is an ancestor of v
2
,
and v
2
is a descendant of v
1
.
Theory and Terminology
?Suppose v is a vertex of depth k:
?Any vertex that is adjacent to v must have
depth k - 1 or k + 1.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
7 8
3
Theory and Terminology
?Suppose v is a vertex of depth k:
?Vertices adjacent to v of depth k + 1 are
called children of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
7 8
Theory and Terminology
? Suppose v is a vertex of depth k:
?If k > 0, there is exactly one vertex of depth k – 1 that
is adjacent to v in the graph.
?This vertex is called the parent of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
Theory and Terminology
?Definitions
?A vertex with no children is called a leaf
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
Theory and Terminology
? Definitions
?Depth of a vertex v is its distance from the root.
?Height of a vertex v is the distance of the longest path from
v to one of its descendant leaves.
?The height of a tree is the maximum depth of its vertices
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height
Theory and Terminology
? Definitions
?The root is the only vertex of depth 0. The root has no
parent.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
Example of rooted tree
? Which are the parent nodes?
? Which are the child nodes?
? Which are the leaves?
? What is the height and depth of the tree?
? What is the height and depth of node E? Node F?
4
Overview of Tree Implementation
? Each node points to
?Its first child
?Its next sibling
?Back to its parent (optional)
? What could be an alternate representation?
Tree Traversals
?Definition: A traversal is the process for
“visiting” all of the vertices in a tree
?Often defined recursively
?Each kind corresponds to an iterator type
?Iterators are implemented non-recursively
Preorder Traversal
?Visit vertex, then visit child vertices
(recursive definition)
?Depth-first search
?Begin at root
?Visit vertex on arrival arrival
?Implementation may be recursive, stack-
based, or nested loop
Preorder Traversal
1
2 3 4
5 6 8 7
root
2
5 6
3
7 8
4
Preorder Traversal of
UNIX Directory Tree
Postorder Traversal
?Visit child vertices, then visit vertex
(recursive definition)
?Depth-first search
?Begin at root
?Visit vertex on departure departure
?Implementation may be recursive, stack-
based, or nested loop
Page 5
1
Trees : Part 1
Section 4.1
(1) Theory and Terminology
(2) Preorder, Postorder and
Levelorder Traversals
Theory and Terminology
?Definition: A tree is a connected graph
with no cycles
?Consequences:
?Between any two vertices, there is exactly one
unique path
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
A Tree?
1
2 3 4
5 6 8 7
9 10 12 11
Theory and Terminology
?Definition: A rooted tree is a graph G
such that:
?G is connected
?G has no cycles
?G has exactly one vertex called the root of the
tree
Theory and Terminology
?Consequences
?The depth of a vertex v is the length of the
unique path from root to v
?G can be arranged so that the root is at the top,
its neighboring vertices are vertices of depth 1,
and so on…
?The set of all vertices of depth k is called level k
of the tree
2
A Rooted Tree
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height = 2
height = 3
height = 0
height = 1
Rooted Tree: Recursive definition
? A graph with N nodes and N - 1 edges
? Graph has
?one root r
?Zero or more non-empty sub-trees, each of whose root
is connected to r by an edge.
?Every node except the root has one parent
Theory and Terminology
? Definition: A descending path in a rooted tree is
a path, whose edges go from a vertex to a
deeper vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
8
12
Theory and Terminology
? Consequences:
?A unique path from the root to any vertex is a descending
path
?The length of this path is the depth of the vertex
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
3
8
11
Theory and Terminology
?Definition: If there is a descending path
from v
1
to v
2
, v
1
is an ancestor of v
2
,
and v
2
is a descendant of v
1
.
Theory and Terminology
?Suppose v is a vertex of depth k:
?Any vertex that is adjacent to v must have
depth k - 1 or k + 1.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
7 8
3
Theory and Terminology
?Suppose v is a vertex of depth k:
?Vertices adjacent to v of depth k + 1 are
called children of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
7 8
Theory and Terminology
? Suppose v is a vertex of depth k:
?If k > 0, there is exactly one vertex of depth k – 1 that
is adjacent to v in the graph.
?This vertex is called the parent of v.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
3
1
Theory and Terminology
?Definitions
?A vertex with no children is called a leaf
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
Theory and Terminology
? Definitions
?Depth of a vertex v is its distance from the root.
?Height of a vertex v is the distance of the longest path from
v to one of its descendant leaves.
?The height of a tree is the maximum depth of its vertices
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
height
Theory and Terminology
? Definitions
?The root is the only vertex of depth 0. The root has no
parent.
1
2 3 4
5 6 8 7
9 10 12 11
root
depth = 1
depth = 0
depth = 3
depth = 2
1
Example of rooted tree
? Which are the parent nodes?
? Which are the child nodes?
? Which are the leaves?
? What is the height and depth of the tree?
? What is the height and depth of node E? Node F?
4
Overview of Tree Implementation
? Each node points to
?Its first child
?Its next sibling
?Back to its parent (optional)
? What could be an alternate representation?
Tree Traversals
?Definition: A traversal is the process for
“visiting” all of the vertices in a tree
?Often defined recursively
?Each kind corresponds to an iterator type
?Iterators are implemented non-recursively
Preorder Traversal
?Visit vertex, then visit child vertices
(recursive definition)
?Depth-first search
?Begin at root
?Visit vertex on arrival arrival
?Implementation may be recursive, stack-
based, or nested loop
Preorder Traversal
1
2 3 4
5 6 8 7
root
2
5 6
3
7 8
4
Preorder Traversal of
UNIX Directory Tree
Postorder Traversal
?Visit child vertices, then visit vertex
(recursive definition)
?Depth-first search
?Begin at root
?Visit vertex on departure departure
?Implementation may be recursive, stack-
based, or nested loop
5
Postorder Traversal
1 root
2 3 4
5 6 7 8
2
5 5 6 6
2 3
7 7 8 8
3 4 4
1
Postorder Traversal
Calculating Size of Directory
Levelorder Traversal
?Visit all vertices in level, starting with level
0 and increasing
?Begin at root
?Visit vertex on departure
?Only practical implementation is queue-
based
Levelorder Traversal
1 root
2 3 4
5 6 7 8
2
5 5 6 6
2 3
7 7 8 8
3 4 4
1
Tree Traversals
?Preorder: depth-first search (possibly
stack-based), visit on arrival
?Postorder: depth-first search (possibly
stack-based), visit on departure
|
# Percentage Problems With Solutions
Percentage Problems With Solutions, deals with various concepts which are as under:-
• Finding a number, whose percentage value is given
• Finding what percentage of a given term is another term
• Net Increase or Decrease Percentage in a Number
• Finding a number when its decreased percentage is given
• Finding a number when its increased percentage is given
• Salary saved
Percentage Problems With Solutions – Finding a number, whose percentage value is given
Question 1:
If 30 % of a number is 60, find the number?
Solution:
We can form a linear equation to express the given statement and then arrive at the number
Let the required number be x
We are given that
30 % of x is 60
or in other words
30/100 × x = 60
On cross multiplication, we get
x = (60 × 100)/30
x = 200
Hence, the number is 200
#### Percentage Problems With Solutions – Finding what percentage of a given term is another term
Question 2:
What percent of 20 litres is 10 litres ?
Solution:
Let a % of 20 litres = 10 litres
Step 1 – Convert the Percentage into Fraction
i.e a % = a/100
Step 2 – Multiply the Fraction with given quantity.
]
a = 50
Hence, 10 litres is 50 % of 20 litres
Question 3:
What percent of 100 m is 80 m?
Solution:
Let a % of 100 m = 80 m
Step 1 – Convert the Percentage into Fraction
a = 80
Hence, 80 m is 80 % of 100 m
Question 4:
What percent of 70 kg is 49 kg?
Solution:
Let a % of 70 kg = 49 kg
Step 1 – Convert the Percentage into Fraction
a = 70
Hence, 49 kg is 70 % of 70 kg
#### Percentage Problems With Solutions – Net Increase or Decrease Percentage in a Number
Question 5:
A number is increased by 25 % and then decreased by 8 %. The net increase or decrease percentage is?
Solution:
We can form a linear equation to express the given statement and then arrive at the number
Let the required number be “a”
Step I:
If the number is increased by 25 % it would be equal to ( Given Number + 25 % of Given Number )
Step II:
If the number at Step 1 is decreased by 8 % it would be equal to ( Increased Number – 8 % of Increased Number )
= 15 %
Hence, Net Increase Percentage is = 15 %
#### Percentage Problems With Solutions – Finding a number when its decreased percentage is given
Question 6:
A number when decreased by 20% gives 160. The number is?
Solution:
Here we take the help of linear equation. First we will form the equation.
Let the number be ‘a’
20 % of ‘a’ = 20a/100
Since, the equation is:
Hence, the number is 200
#### Percentage Problems With Solutions – Finding a number when its increased percentage is given
Question 7:
A number, when increased by 25% gives 375. The number is?
Solution:
The problem can be solved using linear equation.
Let the number be ‘a’
a = 300
Hence, the number is 300
#### Percentage Problems With Solutions – Salary saved
Question 8:
Shruti spends 60 % of her Salary and saves the rest If she saves ₹ 5960 per month.
Solution:
Let Shruti’s monthly salary be ₹ a
Expenditure = 60 % of her Salary
Expenditure = 60 % of ₹ a
Given,
Savings = 5960 = 40 % of a
a = ₹ 14900
Shruti’s monthly salary is ₹ 14900
|
# Solving Exponential Equations using Logarithms
To solve an exponential equation:
$1\right)$ Isolate the exponential expression.
$2\right)$ Take the logarithms of both sides.
$3\right)$ Solve for the variable .
Example 1:
Solve for $x$ : ${2}^{x}=12$
$\begin{array}{l}\mathrm{log}{2}^{x}=\mathrm{log}12\\ x\mathrm{log}2=\mathrm{log}12\\ x=\frac{\mathrm{log}12}{\mathrm{log}2}\approx 3.585\end{array}$
Example 2:
Solve for $x$ : $8\left({10}^{x}\right)=12$
$\begin{array}{l}{10}^{x}=\frac{12}{8}=\frac{3}{2}\\ \mathrm{log}{10}^{x}=\mathrm{log}\frac{3}{2}\\ x\mathrm{log}10=\mathrm{log}\frac{3}{2}\\ x=\mathrm{log}\frac{3}{2}\approx 0.176\end{array}$
Example 3:
Solve for $x$ : ${e}^{5x}=30$
$\begin{array}{l}\mathrm{ln}{e}^{5x}=\mathrm{ln}30\\ 5x\mathrm{ln}e=\mathrm{ln}30\\ 5x=\mathrm{ln}30\\ x=\frac{\mathrm{ln}30}{5}\approx 0.680\end{array}$
|
### Welcome to our community
#### Prove It
##### Well-known member
MHB Math Helper
5. To start with, we should work out the x intercepts, they are x = -2 and x = 2. That means your region in the first quadrant will be integrated over \displaystyle \begin{align*} x \in [0,2] \end{align*}.
You should note that rotating the function \displaystyle \begin{align*} y = 4 - x^2 \end{align*} about the line \displaystyle \begin{align*} y = -\frac{1}{2} \end{align*} will give the exact same volume as rotating \displaystyle \begin{align*} y = \frac{9}{2} - x^2 \end{align*} around the x axis. We would then subtract the volume of the region bounded by the line \displaystyle \begin{align*} y = \frac{1}{2} \end{align*} rotated about the x axis.
So our required volume is
\displaystyle \begin{align*} V &= \int_0^2{ \pi\,\left( \frac{9}{2} - x^2 \right) ^2 \,\mathrm{d}x } - \int_0^2{ \pi\,\left( \frac{1}{2} \right) ^2\,\mathrm{d}x } \\ &= \pi \int_0^2{ \left[ \left( \frac{9}{2} - x^2 \right) ^2 - \left( \frac{1}{2} \right) ^2 \right] \,\mathrm{d}x } \\ &= \pi \int_0^2{ \left( \frac{81}{4} - 9\,x^2 + x^4 - \frac{1}{4} \right) \,\mathrm{d}x } \\ &= \pi \int_0^2{ \left( 20 - 9\,x^2 + x^4 \right) \,\mathrm{d}x } \\ &= \pi \,\left[ 20\,x - 3\,x^3 + \frac{x^5}{5} \right] _0^2 \\ &= \pi \,\left[ \left( 20 \cdot 2 - 3 \cdot 2^3 + \frac{2^5}{5} \right) - \left( 20 \cdot 0 - 3 \cdot 0^3 + \frac{0^5}{5} \right) \right] \\ &= \pi \, \left( 40 - 24 + \frac{32}{5} - 0 \right) \\ &= \pi \, \left( 16 + \frac{32}{5} \right) \\ &= \frac{112\,\pi}{5}\,\textrm{units}^3 \\ &\approx 70.371\,68 \, \textrm{units}^3 \end{align*}
I will do the second question when I have a spare moment.
#### Prove It
##### Well-known member
MHB Math Helper
To evaluate the second volume, you need to imagine the region being made up of a very large number of vertically oriented cylinders. The areas of the curved surfaces of the cylinders together build up to the volume of your solid.
In each cylinder, the radius is the x value, and the height is the y value. So each cylinder has area \displaystyle \begin{align*} 2\,\pi\,x\,y \end{align*}, where \displaystyle \begin{align*} y = 3 + \frac{1}{4}\,\sqrt{x} \end{align*} and \displaystyle \begin{align*} x \in \left[ 1, \frac{3}{2} \right] \end{align*}. Thus the volume is
\displaystyle \begin{align*} V &= \int_1^{\frac{3}{2}}{ 2\,\pi\,x\,\left( 3 + \frac{1}{4}\,\sqrt{x} \right) \,\mathrm{d}x } \\ &= 2\,\pi\int_1^{\frac{3}{2}}{ \left( 3\,x + \frac{1}{4}\,x^{\frac{3}{2}} \right) \,\mathrm{d}x } \\ &= 2\,\pi\,\left[ \frac{3\,x^2}{2} + \frac{1}{4}\,\left( \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right) \right]_1^{\frac{3}{2}} \\ &= 2\,\pi\,\left[ \frac{3\,x^2}{2} + \frac{x^{\frac{5}{2}}}{10} \right] _1^{\frac{3}{2}} \\ &= 2\,\pi\,\left\{ \left[ \frac{3\,\left( \frac{3}{2} \right) ^2}{2} + \frac{\left( \frac{3}{2} \right) ^{\frac{5}{2}}}{10} \right] - \left[ \frac{3\,\left( 1 \right) ^2}{2} + \frac{1^{\frac{5}{2}}}{10} \right] \right\} \\ &= 2\,\pi \,\left( \frac{27}{8} + \frac{9\,\sqrt{6}}{80} - \frac{3}{2} - \frac{1}{10} \right) \\ &= \pi\,\left( \frac{27}{4} + \frac{9\,\sqrt{6}}{40} - 3 - \frac{1}{5} \right) \\ &= \pi \,\left( \frac{270}{40} + \frac{9\,\sqrt{6}}{40} - \frac{120}{40} - \frac{8}{40} \right) \\ &= \frac{\left( 142 + 9\,\sqrt{6} \right)\,\pi }{40}\,\textrm{units}^3 \\ &\approx 12.884\,10\,\textrm{units}^3 \end{align*}
|
APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
EARTHQUAKE WORD PROBLEMS:
As with any word problem, the trick is convert a narrative statement or question to a mathematical statement.
Before we start, let's talk about earthquakes and how we measure their intensity.
In 1935 Charles Richter defined the magnitude of an earthquake to be
where I is the intensity of the earthquake (measured by the amplitude of a seismograph reading taken 100 km from the epicenter of the earthquake) and S is the intensity of a ''standard earthquake'' (whose amplitude is 1 micron =10-4 cm).
The magnitude of a standard earthquake is
Richter studied many earthquakes that occurred between 1900 and 1950. The largest had magnitude of 8.9 on the Richter scale, and the smallest had magnitude 0. This corresponds to a ratio of intensities of 800,000,000, so the Richter scale provides more manageable numbers to work with.
Each number increase on the Richter scale indicates an intensity ten times stronger. For example, an earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is times strong than an earthquake of magnitude 5. An earthquake of magnitude 8 is times stronger than an earthquake of magnitude 5.
Example 3: If one earthquake is 25 times as intense as another, how much larger is its magnitude on the Richter sclae?
Solution: Another of saying this is that one earthquake is 25 times as intense as another
where I1 is the intensity of the larger earthquake and I2 is the intensity of the smaller earthquake.
We are trying to determine the ratio of the larger magnitude M1 to the smaller magnitude I2 or M1-M2. The reason we are subtracting the magnitudes instead of dividing them is the question asked how much larger, not how many times larger.
Solve for I1 by multiplying both sides of the equation by I2.
We can write M1-M2 as and we can write
The larger earthquake had a magnitude 1.4 more on the Richter scale than the smaller earthquake.
Let's check our answer: Suppose the larger earthquake had a magnitude of 8.6 and the smaller earthquake had a magnitude of 8.6-1.4=7.2).
Convert both of these equations to exponential equations.
Example 4: How much more intense is an earthquake of magnitude 6.5 on the Richter scale as one with a magnitude of 4.9?
Solution: The intensity (I) of each earthquake is different. Let I1 represent the intensity of the stronger earthquake and I2 represent the intensity of the weaker earthquake.
What you are looking for is the ratio of the intensities: So our task is to isolate this ratio from the above given information using the rules of logarithms.
Convert the logarithmic equation to an exponential equation.
The stronger earthquake was 40 times as intense as the weaker earthquake.
If you would like to work another example, click on example.
If you would like to test your knowledge by working some problems, click on problem.
[Exponential Rules] [Logarithms]
[Algebra] [Trigonometry ] [Complex Variables]
|
Read- Identify Arithmetic Patterns Video for Kids | 3th, 4th & 5th Grade
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# Read About Identifying Number Patterns (In Arithmetic and the Multiplication Table)
WHAT IS IDENTIFYING NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE)
Having explored patterns with shapes, you will now discover that patterns can be found in numbers as well! Specifically, you will explore outcomes when even and odd numbers are added, subtracted, and multiplied.
To better understand identifying number patterns…
WHAT IS IDENTIFYING NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE). Having explored patterns with shapes, you will now discover that patterns can be found in numbers as well! Specifically, you will explore outcomes when even and odd numbers are added, subtracted, and multiplied. To better understand identifying number patterns…
## LET’S BREAK IT DOWN!
### Odd and Even Socks
Adesina, April, and Marcos are folding laundry. Each person has several socks, and they want to know how they can tell if it’s an even or odd number. If we can put all the socks into pairs, then there is an even number. April has 6 socks, and they can be put into 3 pairs, so 6 is even. Marcos has 8 socks, and they can all be paired up as well, so 8 is even. Combined, they have 14 pairs of socks, and 14 socks can all be paired up as well, so that number is even. They have just discovered a pattern: Even + even = even, since there are no left overs. Next, they fold socks that are a different color. This time, Marcos has 5 socks and April has 9 socks. 5 is an odd number because one of the socks is not paired. 9 is also an odd number because one of the socks is unpaired. But, if Marcos and Adesina put their socks together, now their unpaired socks make a pair together! Therefore, odd + odd = even. Next, they work on a new pile of socks. This time, Marcos has 6 socks and April has 5 socks. 6 is even, since Marcos' socks can all be paired. 5 is odd, since April has one sock left unpaired. If they combine their socks, one sock is still unpaired, and the answer is still odd. Therefore, odd + even = odd! Now you try: Without calculating, can you tell if 7 + 12 is odd or even?
Odd and Even Socks Adesina, April, and Marcos are folding laundry. Each person has several socks, and they want to know how they can tell if it’s an even or odd number. If we can put all the socks into pairs, then there is an even number. April has 6 socks, and they can be put into 3 pairs, so 6 is even. Marcos has 8 socks, and they can all be paired up as well, so 8 is even. Combined, they have 14 pairs of socks, and 14 socks can all be paired up as well, so that number is even. They have just discovered a pattern: Even + even = even, since there are no left overs. Next, they fold socks that are a different color. This time, Marcos has 5 socks and April has 9 socks. 5 is an odd number because one of the socks is not paired. 9 is also an odd number because one of the socks is unpaired. But, if Marcos and Adesina put their socks together, now their unpaired socks make a pair together! Therefore, odd + odd = even. Next, they work on a new pile of socks. This time, Marcos has 6 socks and April has 5 socks. 6 is even, since Marcos' socks can all be paired. 5 is odd, since April has one sock left unpaired. If they combine their socks, one sock is still unpaired, and the answer is still odd. Therefore, odd + even = odd! Now you try: Without calculating, can you tell if 7 + 12 is odd or even?
### Googly Eyes
Adesina starts by putting 12 googly eyes into pairs. 12 can be put into 6 pairs, so 12 is even. If we subtract 4, which is an even number, then we have 8 googly eyes left, which is also even. Subtracting an even number from an even number doesn't leave any googly eyes unpaired. Even – even = even. If we try subtracting an odd number from an odd number, like 11 – 5 = 6, we end up with an even number! The unpaired googly eye in 11 is taken away by the unpaired eye in 5. Odd – odd = even. If we subtract an odd number from an even number, like 14 – 3 = 11, we get an odd number. Likewise, if we subtract an even number from an odd number, like 13 – 4 = 9, we also get an odd number. Then, even – odd = odd, and odd – even = odd. Now you try: Without calculating, can you tell if 21 – 9 is even or odd?
Googly Eyes Adesina starts by putting 12 googly eyes into pairs. 12 can be put into 6 pairs, so 12 is even. If we subtract 4, which is an even number, then we have 8 googly eyes left, which is also even. Subtracting an even number from an even number doesn't leave any googly eyes unpaired. Even – even = even. If we try subtracting an odd number from an odd number, like 11 – 5 = 6, we end up with an even number! The unpaired googly eye in 11 is taken away by the unpaired eye in 5. Odd – odd = even. If we subtract an odd number from an even number, like 14 – 3 = 11, we get an odd number. Likewise, if we subtract an even number from an odd number, like 13 – 4 = 9, we also get an odd number. Then, even – odd = odd, and odd – even = odd. Now you try: Without calculating, can you tell if 21 – 9 is even or odd?
### Multiplication Table
Adesina, Marcos, and April are looking at a multiplication table. They notice some interesting number patterns: The row showing multiples of 2 contains all even numbers. All multiples of 2 are even! Looking at the row with multiples of 5, they notice that you add 5 each time you move along the row. All multiples of 5 end in 5 or 0. This is how you know you cannot divide 23 by 5: it doesn’t end in 5 or 0. Looking at the row with multiples of 10, we can see that all multiples of 10 end in 0. Discovering these patterns can help us check our work. Now you try: Is 73 divisible by 5? Is 90 divisible by 10? Is 43 divisible by 2?
Multiplication Table Adesina, Marcos, and April are looking at a multiplication table. They notice some interesting number patterns: The row showing multiples of 2 contains all even numbers. All multiples of 2 are even! Looking at the row with multiples of 5, they notice that you add 5 each time you move along the row. All multiples of 5 end in 5 or 0. This is how you know you cannot divide 23 by 5: it doesn’t end in 5 or 0. Looking at the row with multiples of 10, we can see that all multiples of 10 end in 0. Discovering these patterns can help us check our work. Now you try: Is 73 divisible by 5? Is 90 divisible by 10? Is 43 divisible by 2?
### Multiplying Evens and Odds
Let’s see what happens when we multiply using even and odd numbers. To find the product of two numbers, we can look up the two numbers on the multiplication table to see what they multiply to. 2 and 4 are even, and when we multiply them, we get 8, which is also an even number. Then, even × even = even. 1 and 3 are odd, and if we multiply them, we get 3, an odd number. Same with 5 × 7 = 35. Then, odd × odd = odd. 2 is even and 3 is odd, and if we multiply them, we get 6, an even number. Same with 6 × 9 = 54. Then, even × odd = even. Also, odd × even = even. Now you try: Without calculating, can you tell if 11 × 16 is odd or even?
Multiplying Evens and Odds Let’s see what happens when we multiply using even and odd numbers. To find the product of two numbers, we can look up the two numbers on the multiplication table to see what they multiply to. 2 and 4 are even, and when we multiply them, we get 8, which is also an even number. Then, even × even = even. 1 and 3 are odd, and if we multiply them, we get 3, an odd number. Same with 5 × 7 = 35. Then, odd × odd = odd. 2 is even and 3 is odd, and if we multiply them, we get 6, an even number. Same with 6 × 9 = 54. Then, even × odd = even. Also, odd × even = even. Now you try: Without calculating, can you tell if 11 × 16 is odd or even?
## IDENTIFY NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE) VOCABULARY
Pattern
Something that follows a rule.
Combined
Pair
Two of the same thing.
Sequence
When one thing follows another according to a rule.
Equation
Mathematical expressions that are equal to the same number.
Multiplication table
A tool that helps us quickly see the product of two numbers.
Can be put into two equal groups, or into groups of two.
Cannot be put into two equal groups, or into groups of two.
## IDENTIFY NUMBER PATTERNS (IN ARITHMETIC AND THE MULTIPLICATION TABLE) DISCUSSION QUESTIONS
### If I have 4 socks and you have 8 socks, can we pair them all up if we put them together?
Yes! 4 can be put into pairs, and 8 can be put into pairs, so if we put them together, we are just adding pairs.
### If I have 5 socks and you have 11 socks, can we pair them all up if we put them together?
Yes. All my socks can be put into pairs except one. All your socks can be put into pairs except one. Together, we have several pairs of socks and two unpaired socks. Those unpaired socks become a pair together!
### If you start with 12 socks and take away 3 socks, you have 9 socks left. Which rule can you learn from this story?
I started with an even number of socks that can be put into pairs. Then I took away an odd number of socks, which means that I take away some pairs and also break up a pair to take one sock. That means that I have an unpaired sock left over, so the amount of leftover socks is an odd number. The rule is: even – odd = odd.
### What happens if you multiply an even number by an odd number? Explain.
I get an even number! I can think of it as an odd number of groups of even numbers (which can be paired).
### What happens if you multiply an odd number by an odd number? Explain.
I get an odd number. For example, if I have 3 groups that each contain 5 socks, the socks in each group make 2 pairs with 1 sock unpaired. When I combine the groups, there are 3 unpaired socks, which still leaves 1 unpaired.
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Lesson Two
Triangle inside a Rectange
Two sixth grade students, Nate and Matthew, excited with the extensions and solutions to the triangle inside a rectangle problem.
Goals/Objectives:
To solve the triangle inside a rectangle presented below:
A triangle has two shared vertices and one shared side with a rectangle.
The third vertex is anywhere on the side opposite of the shared side (see
figures above).
How does the area of the triangle compare with the area of the rectangle?
Why do you think this relationship holds?
Extensions
Where would you move the third vertex of the triangle to have a minimum
perimeter? Explain why this position results in a minimum perimeter.
Conjecture/Hypothesis/Prediction:
We felt the area of the large triangle would be half the area of the rectangle.
Procedures/Materials
Using Geosketchpad we constructed a rectangle with a triangle inside that had two shared vertices and one shared side with the rectangle. We found the area the large triangle and found its ratio to the total area of the rectangle. See below:
Conclusion (analysis using technology and understanding why it works)
We concluded our hypothesis was true that the area of the triangle is half the area of the rectangle.
To have the maximum perimeter, you would have to move the vertex to the midpoint of side CD of the large rectangle.
Comments and reactions (what did I learn)
I learned this would be an excellent activity for sixth grade students to reinforce this concept in Geometry.
Extensions (what other questions can I ask or explore?)
Ask the students to move point E over to the edge CA of the rectangle and hypothesize what the measure of the perimeter and area would be. Describe the relationship of isosceles and equilateral triangles.
Ask the students to measure the area of two small triangles in relationship to the larger triangle.
NCTM Process Standard(s)
Standard 12:
1. identify, describe, compare, and classify geometric figures
2. visualize and represent geometric figures with special attention to developing spatial sense
3. explore transformations of geometric figures
4. understand and apply geometric properties and relationships
Link to my Twotriangle GSP file
Author and Contact
Peggy Reigle
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# 2016 AMC 12A Problems/Problem 12
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Solution
By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$
$\frac{6}{AE} = \frac{7}{8 - AE}$, so $AE = \frac{48}{13}$
Similarly, $CD = 4$
Now, we use mass points.
Assign point $C$ a mass of $1$.
Because $\frac{AE}/{EC} = \frac{6}{7}, A$ will have a mass of $\frac{7}{6}$.
Similarly, $B$ will have a mass of $\frac{4}{3}$.
$mE = mA + mC = \frac{13}{6}$.
Similarly, $mD = mC + mB = \frac{7}{3}$.
The mass of $F$ is the sum of the masses of $E$ and $B$.
$mF = mE + mB = \frac{7}{2}$.
This can be checked with $mD + mA$, which is also $\frac{7}{2}$.
So $\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$
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### About Linear Inequalities in two Variables:
To graph a linear inequality in two variables we solve our inequality for y and replace the inequality symbol with an equality symbol. We then graph this equality as our boundary line. The boundary line separates the solution region from the non-solution region. We can then shade above the boundary line for a greater than, or below the line for a less than.
Test Objectives
• Demonstrate the ability to place an equation in slope-intercept form
• Demonstrate the ability to graph a boundary line
• Demonstrate the ability to graph a linear inequality in two variables
Linear Inequalities in two Variables Practice Test:
#1:
Instructions: Graph each linear inequality.
a) $$7x - 3y ≥ -15$$
#2:
Instructions: Graph each linear inequality.
a) $$y ≥ 7$$
#3:
Instructions: Graph each linear inequality.
a) $$5x + 3y < 12$$
#4:
Instructions: Graph each compound inequality.
a) $$6x + y < 3$$ or $$6x + y ≥ 6$$
#5:
Instructions: Graph each compound inequality.
a) $$x ≤ -3$$ and $$2x + 3y ≥ -3$$
Written Solutions:
#1:
Solutions:
a) $$y ≤ \frac{7}{3}x + 5$$
#2:
Solutions:
a) $$y ≥ 7$$
#3:
Solutions:
a) $$y < -\frac{5}{3}x + 4$$
#4:
Solutions:
a) $$y < -6x + 3$$ or $$y≥ -6x + 6$$
#5:
Solutions:
a) $$x ≤ -3$$ and $$y ≥ -\frac{2}{3}x - 1$$
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Launch-o-Rocket
School, from our house as the crow flies, is 5.73 km. If we neglect air resistance and deal strictly with ballistic flight then we can materialize a wonderful fantasy. Starting in the backyard, extending over the top of the house, is a launch-o-rocket, a rail-like launcher that accelerates the school-bound student until he or she can cruise over the city and arrive without bother of traffic. Our charter is to find the acceleration of the student from the launch-o-rocket.
Finding the Initial Velocity
We rely on the well-known fact that the maximum distance in a throw occurs when the departure angle is 45°. The vertical speed and the horizontal speed are equal. We denote these two identical speeds as $s$. Since distance is time multiplied by speed, the distance from home to school $d$ is
$$d = t\cdot s.$$
We know the distance $d =$ 5.73 km.
Turning to the vertical speed, the student departs the launch-o-rocket with vertical speed $s$, but is immediately subject to gravitational acceleration. Since the student’s upward flight is exactly matched by his or her downward flight. Because the flight is matched, the student spends $t/2$ time rising and $t/2$ time descending. Since the student has no vertical speed at the top, we know that his or her speed is
$$s = g\frac{t}{2},$$
where $g$ is the gravitational acceleration 9.8 m/s2.
Now, we have a system of equations
$$d = t\cdot s$$
$$s = g\frac{t}{2}.$$
The system looks like it has a many variables, but really there are only two, $s$ and $t$. We know $g$ and $d$. To solve the system we substitute for $s$ in the first equation with the second to get
$$d = tg\frac{t}{2} = g\frac{t^2}{2}$$
Solve for $t$
$$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2\cdot 5.73\,\text{m}}{9.8\,\text{m/s}^2}} \approx 34.2\,\text{s}.$$
Not a bad commute, a little over half a minute.
With $t$ in hand, we can find the magnitude of the initial velocity. Remember that the initial velocity is $s$ in the horizontal direction and $s$ in the vertical direction, so the speed when leaving the launcher is
$$\left| \mathbf{v}_0\right| = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}.$$
The initial speed the student must attain is given by the very first equation, $d = s\cdot t$. Solving for $s$ with the value of $t$ we found, we get
$$s = \frac{5.73\,\text{km}}{34.2\,\text{s}} = 168\,\text{m/s}.$$
Finding the Acceleration
The ramp lives on a footprint that is about 80 ft, or 24.4 m. It is also 24.4 m tall, so special zoning is surely required! The rail of the launch-o-rocket is the hypotenuse of a triangle, and that triangle has sides 24.4 m, and a total length of $\sqrt{2}\cdot 24.4\,\text{m} = 34.5\,\text{m}$.
The formula for position after a period of acceleration is
$$p = \frac{1}{2}a\tau^2.$$
For our system, we also know that the acceleration is the change in speed divided by the change in time. Our speed goes from zero to 168 m/s in $\tau$. Again, we have a system of equations,
$$34.5\, \text{m} = \frac{1}{2}a\tau^2$$
$$a = \frac{168\,\text{m/s}}{\tau}.$$
Solve for $a$ by first solving the second equation for $\tau$, and then substituting that result into the first equation to get
$$34.5\, \text{m} = \frac{1}{2}a\left(\frac{168\,\text{m/s}}{a}\right)^2$$
$$a = \frac{\left( 168\, \text{m/s}\right)^2}{2 \cdot 34.5\,\text{m}} = 407\, \text{m/s}^2 = 41.5\, g.$$
The typical onset of death occurs when acceleration exceeds about $10g$, so unfortunately, the launch-o-rocket is a single try system.
|
# A-level Mathematics/OCR/C3/Numerical Methods
< A-level Mathematics | OCR | C3
## Finding Roots of an Equation
In this module we will explore approximating where an equation has a root. Below we have two graphs.
### Finding A Root Using a Graph
In the graph we have two functions. If we want to approximate where the roots are we have to look at where the function cross the x-axis. The lilac function crosses the x-axis somewhere between 2.05 and 2.25. The green function has roots around 1 and 3.5. If we want to know where the green function is equivalent to the lilac function we need to look at the graph. When the two graphs cross they are equivalent. This number will be the root. In this case it is around 1.75. We can also say that on this domain, the functions will only cross once.
### Finding A Root By Searching for A Sign Change
When a function has a root the value of the function will change from a positive value to a negative value or vice versa. If below is the table of values for the lilac function we can say that the root occurs between 2.05 and 2.10:
x f(x) 2 -1 2.05 -.4849 2.1 .061 2.15 .63838 2.2 1.248 2.25 1.08906
## Iterative Formulae
An iterative formula is a formula that is composed of itself. That is the output of the function is the next input of the function. ${\displaystyle x_{n+1}=F(n)}$. These functions are a sequence of approximations that usually converge to the value of a function. You can tell if a function converges by the fact that the outputs become closer and closer to each other, if this does not happen then the function diverges and there is no value. The output of an iterative formula is written as:
${\displaystyle x_{2}\,}$
${\displaystyle x_{3}\,}$
${\displaystyle x_{4}\,}$
The first x is provided for you. The output of the iterative formula is written as the Greek letter alpha: α. α can be found to the required degree of accuracy when ${\displaystyle x_{n}=x_{n+1}}$ to the required number of decimal places. Given an iterative function you can find the root of an equation. Also you find the function from a given iterative function by setting x = iterative function, then solving so that you have a zero on one side. The aforementioned process can be done in reverse.
### Example
Given the sequence that is defined by the iterative formula ${\displaystyle \left(2x+5\right)^{\frac {1}{3}}}$ with ${\displaystyle x_{1}=2}$ converges to ${\displaystyle \alpha }$
1. Find ${\displaystyle \alpha }$ correct to 4 decimal places.
2. Find equation that has ${\displaystyle \alpha }$ as a root.
3. Does the equation have any more roots?
With an iterative formula we have:
1. We plug in ${\displaystyle x_{1}}$ to get ${\displaystyle x_{2}}$ and so on
1. ${\displaystyle x_{2}=\left(2\times (2)+5\right)^{\frac {1}{3}}=2.0801}$
2. ${\displaystyle x_{3}=\left(2\times (2.0801)+5\right)^{\frac {1}{3}}=2.0924}$
3. ${\displaystyle x_{4}=\left(2\times (2.0924)+5\right)^{\frac {1}{3}}=2.0942}$
4. ${\displaystyle x_{5}=\left(2\times (2.0942)+5\right)^{\frac {1}{3}}=2.0945}$
5. ${\displaystyle x_{6}=\left(2\times (2.0945)+5\right)^{\frac {1}{3}}=2.0945}$
6. ${\displaystyle \alpha =2.0945\,}$
2. ${\displaystyle x=\left(2x+5\right)^{\frac {1}{3}}}$
1. ${\displaystyle x^{3}=2x+5\,}$
2. ${\displaystyle x^{3}-2x-5=0\,}$
3. To determine if the function has any roots just graph the graph of the highest power of x and then the rest. The number of times they cross is the number of roots you have.
1. If we plot ${\displaystyle y=x^{3}}$ and ${\displaystyle y=2x+5}$ we can see that they only cross once.
2. The function only has one root.
## Simpson's Rule for Area
To find the area beneath a curve we have already learned the trapezium rule. The trapezium rule is not very accurate it takes a very large number of trapezoids to get an very accurate area. The Simpson rule is much more accurate to find the area underneath a curve. Simpson's Rule states:
${\displaystyle \int _{a}^{b}ydx\approx {\frac {1}{3}}h\left\{\left(y_{0}+y_{n}\right)+4\left(y_{1}+y_{3}+\ldots +y_{n-1}\right)+2\left(y_{2}+y_{4}+\ldots +y_{n-2}\right)\right\}}$
where${\displaystyle h={\frac {b-a}{n}}}$ and n is even
#### Example
Use Simpson's Rule to evaluate ${\displaystyle \int _{1}^{5}x^{2}+2x\,dx}$ with h = 1.
${\displaystyle \int _{1}^{5}x^{2}+2x\,dx\approx {\frac {1}{3}}\left[\left(1^{2}+2\times 1\right)+4\left(2^{2}+2\times 2\right)+2\left(3^{2}+2\times 3\right)+4\left(4^{2}+2\times 4\right)+\left(5^{2}+2\times 5\right)\right]}$
${\displaystyle \int _{1}^{5}x^{2}+2x\,dx=65{\frac {1}{3}}}$
The area under the curve ${\displaystyle x^{2}+2x\,}$is equal to ${\displaystyle 65{\frac {1}{3}}}$.
This is the true value. If we compare it to the value of 66 that we trapezium rule and to the value of 65 that we got from the midpoint rule we can see that Simpson's rule is the most accurate.
This is part of the C3 (Core Mathematics 3) module of the A-level Mathematics text.
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Lesson 2
Representations of Growth and Decay
• Let’s revisit ways to represent exponential change.
Priya borrowed $160 from her grandmother. Each month, she pays off one fourth of the remaining balance that she owes. 1. What amount will Priya pay her grandmother in the third month? 2. Discuss with a partner why the expression $$160 \boldcdot \left(\frac34\right)^3$$ represents the balance Priya owes her grandmother at the end of the third month. 2.2: Climbing Cost The tuition at a college was$30,000 in 2012, $31,200 in 2013, and$32,448 in 2014. The tuition has been increasing by the same percentage since the year 2000.
1. The equation $$c(t)=30,\!000 \boldcdot (1.04)^t$$ represents the cost of tuition, in dollars, as a function of $$t$$, the number of years since 2012. Explain what the 30,000 and 1.04 tell us about this situation.
2. What is the percent increase in tuition from year to year?
3. What does $$c(3)$$ mean in this situation? Find its value and show your reasoning.
1. Write an expression to represent the cost of tuition in 2007.
2. How much did tuition cost that year?
Jada thinks that the college tuition will increase by 40% each decade. Do you agree with Jada? Explain your reasoning.
2.3: Two Vans and Their Values
A small business bought a van for $40,000 in 2008. The van depreciates by 15% every year after its purchase. 1. Which graph correctly represents the value of the van as a function of years since its purchase? Be prepared to explain why each of the other graphs could not represent the function. 2. Find the value of the van 8 years after its purchase. Show your reasoning. 3. In the same year (2008), the business bought a second van that cost$10,000 less than the first van and depreciates at 10% per year. Would the second van be worth more or less than the first van 8 years after the purchase? Explain or show your reasoning.
4. On the same coordinate plane as the graph you chose in the first question, sketch a graph that shows the value of the second van, in dollars, as a function of years since its purchase.
Summary
There are lots of ways to represent an exponential function. Suppose the population of a city was 20,000 in 1990 and that it increased by 10% each year.
We can represent this situation with a table of values and show, for instance, that the population increased by a factor of 1.1 each year.
year population
1990 20,000
1991 22,000
1992 24,200
1993 26,620
We can also use a graph to show how the population was changing. While the graph looks almost linear, it has a slight upward curve since the population is increasing by a factor of 1.1 and not a constant value each year.
An equation is another useful representation. In this case, if $$t$$ is the number of years since 1990, then the population is a function $$f$$ of $$t$$ where $$f(t) = 20,\!000 \boldcdot (1.1)^t$$. Here we can see the 20,000 in the expression represents the population in 1990, while 1.1 represents the growth factor due to the 10% annual increase each year. We can even use the equation to calculate the population predicted by the model in 1985. Since 1985 is 5 years before 1990, we use an input of -5 to get $$f(\text-5)=20,\!000 \boldcdot (1.1)^{\text-5}$$, which is about 12,418 people.
Throughout this unit, we will examine many exponential functions. All four representations—descriptions, tables, graphs, and equations—will be useful for determining different information about the function and the situation the function models.
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# Shell integration
A volume is approximated by a collection of hollow cylinders. As the cylinder walls get thinner the approximation gets better. The limit of this approximation is the shell integral.
Shell integration (the shell method in integral calculus) is a means of calculating the volume of a solid of revolution, when integrating along an axis perpendicular to the axis of revolution. This is in contrast to disc integration which integrates along the axis parallel to the axis of revolution.
## Definition
The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the xy-plane around the y-axis. Suppose the cross-section is defined by the graph of the positive function f(x) on the interval [a, b]. Then the formula for the volume will be:
${\displaystyle 2\pi \int _{a}^{b}xf(x)\,dx}$
If the function is of the y coordinate and the axis of rotation is the x-axis then the formula becomes:
${\displaystyle 2\pi \int _{a}^{b}yf(y)\,dy}$
If the function is rotating around the line x = h or y = k, the formulas become:[1]
${\displaystyle {\begin{cases}\displaystyle 2\pi \int _{a}^{b}(x-h)f(x)\,dx,&{\text{if}}\ h\leq a
and
${\displaystyle {\begin{cases}\displaystyle 2\pi \int _{a}^{b}(y-k)f(y)\,dy,&{\text{if}}\ k\leq a
The formula is derived by computing the double integral in polar coordinates.
## Example
Consider the volume, depicted below, whose cross section on the interval [1, 2] is defined by:
${\displaystyle y=(x-1)^{2}(x-2)^{2}}$
Cross-section
3D volume
In the case of disk integration we would need to solve for x given y. Because the volume is hollow in the middle we will find two functions, one that defines the inner solid and one that defines the outer solid. After integrating these two functions with the disk method we subtract them to yield the desired volume.
With the shell method all we need is the following formula:
${\displaystyle 2\pi \int _{1}^{2}x(x-1)^{2}(x-2)^{2}\,dx}$
By expanding the polynomial the integral becomes very simple. In the end we find the volume is π/10 cubic units.
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# 7.11: Factorization of Special Cubics
Difficulty Level: Advanced Created by: CK-12
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Practice Sum and Difference of Cubes
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Factor the following cubic polynomial: 375x3+648\begin{align*}375x^3+648\end{align*}.
### Guidance
While many cubics cannot easily be factored, there are two special cases that can be factored quickly. These special cases are the sum of perfect cubes and the difference of perfect cubes.
• Factoring the sum of two cubes follows this pattern: x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}
• Factoring the difference of two cubes follows this pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}
#### Example A
Factor: x3+27\begin{align*}x^3+27\end{align*}.
Solution: This is the sum of two cubes and uses the factoring pattern: x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
x3+33=(x+3)(x23x+9)\begin{align*}x^3+3^3=(x+3)(x^2-3x+9)\end{align*}.
#### Example B
Factor: x3343\begin{align*}x^3-343\end{align*}.
Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
x373=(x7)(x2+7x+49)\begin{align*}x^3-7^3=(x-7)(x^2+7x+49)\end{align*}.
#### Example C
Factor: 64x31\begin{align*}64x^3-1\end{align*}.
Solution: This is the difference of two cubes and uses the factoring pattern: x3y3=(xy)(x2+xy+y2)\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
(4x)313=(4x1)(16x2+4x+1)\begin{align*}(4x)^3-1^3=(4x-1)(16x^2+4x+1)\end{align*}.
#### Concept Problem Revisited
Factor the following cubic polynomial: 375x3+648\begin{align*}375x^3+648\end{align*}.
First you need to recognize that there is a common factor of 3\begin{align*}3\end{align*}. 375x3+648=3(125x3+216)\begin{align*}375x^3+648=3(125x^3+216)\end{align*}
Notice that the result is the sum of two cubes. Therefore, the factoring pattern is x3+y3=(x+y)(x2xy+y2)\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
375x3+648=3(5x+6)(25x230x+36)\begin{align*}375x^3 +648 = 3(5x+6)(25x^2-30x+36)\end{align*}
### Vocabulary
Difference of Two Cubes
The difference of two cubes is a special polynomial in the form of x3y3\begin{align*}x^3-y^3\end{align*}. This type of polynomial can be quickly factored using the pattern: (x3y3)=(xy)(x2+xy+y2)\begin{align*}(x^3-y^3)=(x-y)(x^2+xy+y^2)\end{align*}
Sum of Two Cubes
The sum of two cubes is a special polynomial in the form of x3+y3\begin{align*}x^3+y^3\end{align*}. This type of polynomial can be quickly factored using the pattern: (x3+y3)=(x+y)(x2xy+y2)\begin{align*}(x^3+y^3)=(x+y)(x^2-xy+y^2)\end{align*}
### Guided Practice
Factor each of the following cubics.
1. x3+512\begin{align*}x^3+512\end{align*}
2. 8x3+125\begin{align*}8x^3+125\end{align*}
3. x3216\begin{align*}x^3-216\end{align*}
1. x3+83=(x+8)(x28x+64)\begin{align*}x^3+8^3=(x+8)(x^2-8x+64)\end{align*}.
2. (2x)3+53=(2x+5)(4x210x+25)\begin{align*}(2x)^3+5^3=(2x+5)(4x^2-10x+25)\end{align*}.
3. x363=(x6)(x2+6x+36)\begin{align*}x^3-6^3=(x-6)(x^2+6x+36)\end{align*}.
### Practice
Factor each of the following cubics.
1. x3+h3\begin{align*}x^3+h^3\end{align*}
2. a3+125\begin{align*}a^3+125\end{align*}
3. 8x3+64\begin{align*}8x^3+64\end{align*}
4. x3+1728\begin{align*}x^3+1728\end{align*}
5. 2x3+6750\begin{align*}2x^3+6750\end{align*}
6. h364\begin{align*}h^3-64\end{align*}
7. s3216\begin{align*}s^3-216\end{align*}
8. p3512\begin{align*}p^3-512\end{align*}
9. 4e332\begin{align*}4e^3-32\end{align*}
10. 2w3250\begin{align*}2w^3-250\end{align*}
11. x3+8\begin{align*}x^3+8\end{align*}
12. y31\begin{align*}y^3-1\end{align*}
13. 125e38\begin{align*}125e^3-8\end{align*}
14. 64a3+2197\begin{align*}64a^3+2197\end{align*}
15. 54z3+3456\begin{align*}54z^3+3456\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Cubed
The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
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# 2.7: Derivatives of Inverse Functions
Recall that a function $$y=f(x)$$ is said to be one to one if it passes the horizontal line test; that is, for two different $$x$$ values $$x_1$$ and $$x_2$$, we do not have $$f(x_1)=f(x_2)$$. In some cases the domain of $$f$$ must be restricted so that it is one to one. For instance, consider $$f(x)=x^2$$. Clearly, $$f(-1)= f(1)$$, so $$f$$ is not one to one on its regular domain, but by restricting $$f$$ to $$(0,\infty)$$, $$f$$ is one to one.
Now recall that one to one functions have inverses. That is, if $$f$$ is one to one, it has an inverse function, denoted by $$f^{-1}$$, such that if $$f(a)=b$$, then $$f^{-1}(b) = a$$. The domain of $$f^{-1}$$ is the range of $$f$$, and vice-versa. For ease of notation, we set $$g=f^{-1}$$ and treat $$g$$ as a function of $$x$$.
Since $$f(a)=b$$ implies $$g(b)=a$$, when we compose $$f$$ and $$g$$ we get a nice result: $f\big(g(b)\big) = f(a) = b.$ In general, $$f\big(g(x)\big) =x$$ and $$g\big(f(x)\big) = x$$. This gives us a convenient way to check if two functions are inverses of each other: compose them and if the result is $$x$$, then they are inverses (on the appropriate domains.)
When the point $$(a,b)$$ lies on the graph of $$f$$, the point $$(b,a)$$ lies on the graph of $$g$$. This leads us to discover that the graph of $$g$$ is the reflection of $$f$$ across the line $$y=x$$. In Figure 2.29 we see a function graphed along with its inverse. See how the point $$(1,1.5)$$ lies on one graph, whereas $$(1.5,1)$$ lies on the other. Because of this relationship, whatever we know about $$f$$ can quickly be transferred into knowledge about $$g$$.
For example, consider Figure 2.30 where the tangent line to $$f$$ at the point $$(a,b)$$ is drawn. That line has slope $$f^\prime(a)$$. Through reflection across $$y=x$$, we can see that the tangent line to $$g$$ at the point $$(b,a)$$ should have slope $$\frac{1}{f^\prime(a)}$$. This then tells us that $$g^\prime(b) = \frac{1}{f^\prime(a)}.$$
Consider:
FIX
$\begin{center} \begin{tabular}{ccc} Information about $$f$$ & & Information about $$g=f\primeskip^{-1}$$ \\ \hline $$(-0.5,0.375)$$ lies on $$f$$ & \hskip 40pt & \parbox{100pt}{\centering $$(0.375,-0.5)$$ lies on $$g}\\ \rule{0pt}{20pt}\parbox{100pt}{\centering Slope of tangent line to \(f$$ at $$x=-0.5$$ is $$3/4} & & \parbox{100pt}{\centering Slope of tangent line to \(g$$ at $$x=0.375$$ is $$4/3}\rule{0pt}{17pt} \\ \rule{0pt}{15pt}f^\prime(-0.5) = 3/4$$ & & $$g\primeskip'(0.375) = 4/3\rule{0pt}{12pt} \end{tabular} \end{center}$ We have discovered a relationship between \(f^\prime$$ and $$g^\prime$$ in a mostly graphical way. We can realize this relationship analytically as well. Let $$y = g(x)$$, where again $$g = f^{-1}$$. We want to find $$y^\prime$$. Since $$y = g(x)$$, we know that $$f(y) = x$$. Using the Chain Rule and Implicit Differentiation, take the derivative of both sides of this last equality.
\begin{align*}\frac{d}{dx}\Big(f(y)\Big) &= \frac{d}{dx}\Big(x\Big) \\ f^\prime(y)\cdot y^\prime &= 1\\ y^\prime &= \frac{1}{f^\prime(y)}\\ y^\prime &= \frac{1}{f^\prime(g(x))} \end{align*}
This leads us to the following theorem.
Theorem 22: Derivatives of Inverse Functions
Let $$f$$ be differentiable and one to one on an open interval $$I$$, where $$f^\prime(x) \neq 0$$ for all $$x$$ in $$I$$, let $$J$$ be the range of $$f$$ on $$I$$, let $$g$$ be the inverse function of $$f$$, and let $$f(a) = b$$ for some $$a$$ in $$I$$. Then $$g$$ is a differentiable function on $$J$$, and in particular,
$$1. \left(f^{-1}\right)^\prime (b)=g^\prime (b) = \frac{1}{f^\prime(a)}$$\quad \text{ and }\quad 2. \left(f^{-1}\right)^\prime (x)=g^\prime (x) = \frac{1}{f^\prime(g(x))}\)
The results of Theorem 22 are not trivial; the notation may seem confusing at first. Careful consideration, along with examples, should earn understanding.
In the next example we apply Theorem 22 to the arcsine function.
Example 75: Finding the derivative of an inverse trigonometric function
Let $$y = \arcsin x = \sin^{-1} x$$. Find $$y^\prime$$ using Theorem 22.
Solution:
Adopting our previously defined notation, let $$g(x) = \arcsin x$$ and $$f(x) = \sin x$$. Thus $$f^\prime(x) = \cos x$$. Applying the theorem, we have
\begin{align*} g^\prime (x) &= \frac{1}{f^\prime(g(x))} \\ &= \frac{1}{\cos(\arcsin x)}. \end{align*}
This last expression is not immediately illuminating. Drawing a figure will help, as shown in Figure 2.32. Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, specifically, the ratio "opposite over hypotenuse.'' This means that the arcsine function takes as input a ratio of sides and returns an angle. The equation $$y=\arcsin x$$ can be rewritten as $$y=\arcsin (x/1)$$; that is, consider a right triangle where the hypotenuse has length 1 and the side opposite of the angle with measure $$y$$ has length $$x$$. This means the final side has length $$\sqrt{1-x^2}$$, using the Pythagorean Theorem.
$\text{Therefore }\cos (\sin^{-1} x) = \cos y = \sqrt{1-x^2}/1 = \sqrt{1-x^2},\text{ resulting in }$$\frac{d}{dx}\big(\arcsin x\big) = g^\prime (x) = \frac{1}{\sqrt{1-x^2}}.$
Remember that the input $$x$$ of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; the absolute value of this ratio will never be greater than 1. Therefore the inside of the square root will never be negative.
In order to make $$y=\sin x$$ one to one, we restrict its domain to $$[-\pi/2,\pi/2]$$; on this domain, the range is $$[-1,1]$$. Therefore the domain of $$y=\arcsin x$$ is $$[-1,1]$$ and the range is $$[-\pi/2,\pi/2]$$. When $$x=\pm 1$$, note how the derivative of the arcsine function is undefined; this corresponds to the fact that as $$x\to \pm1$$, the tangent lines to arcsine approach vertical lines with undefined slopes.
In Figure 2.33 we see $$f(x) = \sin x$$ and $$f^{-1} = \sin^{-1} x$$ graphed on their respective domains. The line tangent to $$\sin x$$ at the point $$(\pi/3, \sqrt{3}/2)$$ has slope $$\cos \pi/3 = 1/2$$. The slope of the corresponding point on $$\sin^{-1}x$$, the point $$(\sqrt{3}/2,\pi/3)$$, is $\frac{1}{\sqrt{1-(\sqrt{3}/2)^2}} = \frac{1}{\sqrt{1-3/4}} = \frac{1}{\sqrt{1/4}} = \frac{1}{1/2}=2,$verifying yet again that at corresponding points, a function and its inverse have reciprocal slopes.
Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. In Figure 2.31 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible.
FIX TABLE
$\begin{tabular}{cccccc} Function & Domain & Range &\parbox[b]{40pt}{\centering Inverse Function} & Domain & Range\\ \hline \rule{0pt}{12pt} $$\sin x$$ & $$[-\pi/2, \pi/2]$$ & $$[-1,1]&\sin^{-1} x$$ & $$[-1,1]$$ & $$[-\pi/2, \pi/2]$$ \\ \rule{0pt}{12pt}\cos x\) & $$[0,\pi]$$ & $$[-1,1]&\cos^{-1}(x)$$ & $$[-1,1]$$ & $$[0,\pi]$$ \\ \rule{0pt}{12pt}\tan x\) & $$(-\pi/2,\pi/2)$$ & $$(-\infty,\infty)&\tan^{-1}(x)$$ & $$(-\infty,\infty)$$ & $$(-\pi/2,\pi/2)$$ \\ \rule{0pt}{12pt} $$\csc x$$ & $$[-\pi/2,0)\cup (0, \pi/2]$$ & $$(-\infty,-1]\cup [1,\infty)&\csc^{-1} x$$ & $$(-\infty,-1]\cup [1,\infty)$$ & $$[-\pi/2,0)\cup (0, \pi/2]$$ \\ \rule{0pt}{12pt}\sec x\) & $$[0,\pi/2)\cup (\pi/2,\pi]$$ & $$(-\infty,-1]\cup [1,\infty)&\sec^{-1}(x)$$ & $$(-\infty,-1]\cup [1,\infty)$$ & $$[0,\pi/2)\cup (\pi/2,\pi]$$ \\ \rule{0pt}{12pt}\cot x\) & $$(0,\pi)$$ & $$(-\infty,\infty)&\cot^{-1}(x)$$ & $$(-\infty,\infty)$$ & (0,\pi) \end{tabular} \captionsetup{type=figure} \caption{Domains and ranges of the trigonometric and inverse trigonometric functions.}\label{fig:domain_trig} %\end{center} %\normalsize \end{minipage} %\captionsetup{type=figure}% %\caption{Domains and ranges of the trigonometric and inverse trigonometric functions.}\label{fig:domain_trig} %\end{minipage} %}$ Theorem 23: Derivatives of Inverse Trigonometric Functions The inverse trigonometric functions are differentiable on all open sets contained in their domains (as listed in Figure 2.31) and their derivatives are as follows: \begin{align} &1. \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} \qquad &&4.\frac{d}{dx}\big(\cos^{-1}(x)\big) = -\frac{1}{\sqrt{1-x^2}} \\ &2.\frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{|x|\sqrt{x^2-1}} &&5.\frac{d}{dx}\big(\csc^{-1}(x)\big) = -\frac{1}{|x|\sqrt{x^2-1}} \\ &3.\frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} &&6.\frac{d}{dx}\big(\cot^{-1}(x)\big) = -\frac{1}{1+x^2} \end{align} Note how the last three derivatives are merely the opposites of the first three, respectively. Because of this, the first three are used almost exclusively throughout this text. In Section 2.3, we stated without proof or explanation that \( \frac{d}{dx}\big(\ln x\big) = \frac{1}{x}. We can justify that now using Theorem 22, as shown in the example.
Example 76: Finding the derivative of y = ln x
Use Theorem 22 to compute $$\frac{d}{dx}\big(\ln x\big)$$.
Solution:
View $$y= \ln x$$ as the inverse of $$y = e^x$$. Therefore, using our standard notation, let $$f(x) = e^x$$ and $$g(x) = \ln x$$. We wish to find $$g^\prime (x)$$. Theorem 22 gives:
\begin{align*} g^\prime (x) &= \frac{1}{f^\prime(g(x))} \\ &= \frac{1}{e^{\ln x}} \\ &= \frac{1}{x}. \end{align*}
In this chapter we have defined the derivative, given rules to facilitate its computation, and given the derivatives of a number of standard functions. We restate the most important of these in the following theorem, intended to be a reference for further work.
Theorem 24: Glossary of Derivatives of Elementary Functions
FIX THIS
{Let $$u$$ and $$v$$ be differentiable functions, and let $$a$$, $$c$$ and $$n$$ be real numbers, $$a>0$$, $$n\neq 0$$. \\
\noindent%
\begin{minipage}{.5\specialboxlength}
\begin{enumerate}
\item \(\frac{d}{dx}\big(cu\big) = cu'$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(u\cdot v\big) = uv'+u'v$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(u(v)\big) = u'(v)v'$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(x\big) = 1$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(e^x\big) = e^x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\ln x\big) = \frac{1}{x}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\sin x\big) = \cos x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\csc x\big) = -\csc x\cot x$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\tan x\big) = \sec^2x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\sin^{-1}x\big) = \frac{1}{\sqrt{1-x^2}}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\csc^{-1}x\big) = -\frac{1}{|x|\sqrt{x^2-1}}$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\tan^{-1}x\big) = \frac{1}{1+x^2}$\addtocounter{enumi}{1}
\end{enumerate}
\normalsize
\end{minipage}
\begin{minipage}{.5\specialboxlength}
\item \(\frac{d}{dx}\big(u\pm v\big) = u'\pm v'$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\frac uv\big) = \frac{u'v-uv'}{v^2}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(c\big) = 0$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(x^n\big) = nx^{n-1}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(a^x\big) = \ln a\cdot a^x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\log_a x\big) = \frac{1}{\ln a}\cdot\frac{1}{x}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\cos x\big) = -\sin x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\sec x\big) = \sec x\tan x$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\cot x\big) = -\csc^2x$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\cos^{-1}x\big) = -\frac{1}{\sqrt{1-x^2}}$\addtocounter{enumi}{1}
\item \(\frac{d}{dx}\big(\sec^{-1}x\big) = \frac{1}{|x|\sqrt{x^2-1}}$\addtocounter{enumi}{1} \item \(\frac{d}{dx}\big(\cot^{-1}x\big) = -\frac{1}{1+x^2}$
\end{enumerate}
\normalsize
\end{minipage}
}
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Dear Students,
### Quantitative Aptitude Quiz for IBPS CLERK
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Q1. In how many different ways can 5 boys and 4 girls be arranged in a row such that all the boys stand together and all the girls stand together?
7,500
5,760
5,670
7,560
8,500
Solution:
Total ways = 5! × 4! × 2!
= 5,760
Q2. The average of the ages of 4 friends is 24. Even if the age of the 5th friends is added the average remains 24. What is the age of the 5th friend ?
23 yr
21 yr
24 yr
Cannot be determined
None of these
Solution:
Age of 5th friend = 5 × 24 – 4 × 24
= 24 years
Q3. 9 men working for 8 h a day complete a piece of work in 15 days. In how many days can 6 men working for 15 h a day complete the same piece of work ?
21
12
14
20
31
Solution:
Q4. The total area of a circle and a rectangle is equal to 586 sq cm. The diameter of the circle is 14 cm. What is the sum of the circumference of the circle and the perimeter of the rectangle if the length of the rectangle is 24 cm ?
128 cm
130 cm
134 cm
132 cm
138 cm
Solution:
Q5. The simple interest accrued on a sum of certain principal is Rs. 5,400 in 3 yr at the rate of 18% per annum. What would be the compound interest accrued on that principal at the rate of 10% per annum in 2 yrs. ?
Rs. 2,000
Rs. 2,150
Rs. 1,800
Rs. 2,100
None of these
Solution:
Directions (6-10): Study the following graph carefully to answer the questions that follow:
Q6. The number of non-graduates from Q is approximately what percent of the total number of non-graduates from all the companies together ?
32
21
37
43
27
Solution:
Q7. What is the difference between the total number of graduates and the total number of non-graduates from all the companies together ?
440
520
580
460
560
Solution:
Required difference=(140+300+180+250+240)-(80+200+100+150+120)
=1110-650 =460
Q8. What is the average number of graduates from all the companies together ?
122
126
130
134
144
Solution:
Q9. The number of graduates from company R is approximately what per cent of the total number of graduates from all the companies together ?
15
23
31
44
56
Solution:
Required percentage=100/650×100 =15.38 ≈15%
Q10. What is the respective ratio of number of non-graduates from company P to the number of non-graduates from T?
1 : 2
7 : 12
5 : 12
3 : 4
12: 7
Solution:
Required ratio=140/240=7 :12
Directions (11-15): What should come in place of question mark (?) in the following questions?
Q11. 205% of 3850 – 105% of 2640 = ?
4218.5
5120.5
4448.5
4628.5
6120.5
Solution:
Q12. 4527+ 785 + 968 – ? = 4560
1,720
1,700
1,675
1,800
1,520
Solution:
? = 1720
Q13. 27.4 × 3 + 5.4 – 57.6 = ? + 14.8
15.7
12.6
14.8
15.2
13.2
Solution:
? = 30 – 14.8 = 15.2
Q14. 7.53 + 6.32 + 0.54 + 41.47 + 345 = ?
397.27
419.57
400.86
406.67
500.86
Solution:
? = 400.86
Q15.
1,250
1,290
1,270
1,300
1,450
Solution:
|
July 16, 2024
Algebra Formulas
Numbers and letters are both used in algebra. The algebraic formula’s unknown quantities are represented by letters or alphabets. Now, an equation or formula is created using a mix of integers, letters, factorials, fractions, decimals, log etc. There are a few highly crucial algebraic formulae and equations that students must memorise while they prepare for their exams. The foundation of basic or elementary algebra is these formulas. Knowing the formulas alone is insufficient. you need to know how to use these formulas to solve problems.
Here, we’ll list all the significant algebraic formulas. The extensive list will make it possible for the students to look over it quickly before tests or to refer to it at any time they like.
Important Algebra Formulas (List of all Algebraic Identities)
An algebraic identity is an algebraic equation which is true for all values of the variable (s).
• $(a+b)^{2}=a^{2}&space;+b^{2}+2ab$
• $(a-b)^{2}=a^{2}&space;+b^{2}-2ab$
• $(a+b)^{2}=(a-b)^{2}+4ab$
• $(a-b)^{2}=(a+b)^{2}-4ab$
• $a^{2}+b^{2}=&space;\frac{1}{2}[(a+b)^{2}+(a-b)^{2}]$
• $a^{2}-b^{2}=&space;(a-b)(a+b)$
• $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
• $(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)$
• $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$
• $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
• $(a+b+c)^{2}=a^{2}&space;+b^{2}+c^{2}+2ab+2bc+2ca$
• $(a-b-c)^{2}=a^{2}&space;+b^{2}+c^{2}-2ab-2bc-2ca$
• $(a+b+c)^{3}=a^{3}&space;+b^{3}+c^{3}+3(a+b)(b+c)(c+a)$
• $a^{3}&space;+b^{3}+c^{3}-3abc=&space;(a+b+c)(a^{2}&space;+b^{2}+c^{2}-ab-bc-ca)$
• If $a+b+c=0&space;\,&space;\,&space;\mathrm{then&space;}\,&space;\,&space;a^{3}&space;+b^{3}+c^{3}=3abc$
• $a^{3}+b^{3}+c^{3}-3abc=\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$
• $\mathrm{If&space;\,&space;\,&space;a^{3}+b^{3}+c^{3}=3abc&space;\,&space;then&space;\,&space;\,&space;either&space;\,&space;\,&space;a+b+c=o&space;\,&space;\,&space;or\,&space;\,&space;a=b=c&space;}$
• $\mathrm{If&space;\,&space;\,&space;a^{3}+b^{3}+c^{3}=ab+bc+ca&space;\,&space;\,&space;then\,&space;\,&space;a=b=c}$
• $a^{4}+b^{4}+a^{2}b^{2}=(a^{2}+b^{2}+ab)(a^{2}+b^{2}-ab)$
• $a^{4}-b^{4}=(a-b)(a+b)(a^{2}+b^{2})$
• $a^{8}-b^{8}=(a-b)(a+b)(a^{2}+b^{2})(a^{4}+b^{4})$
• $If&space;\,&space;\,&space;a+\frac{1}{a}=&space;x&space;\,&space;\,&space;then&space;\,&space;\,&space;a^{3}+\frac{1}{a^{3}}&space;=x^{2}-3x$
• $If&space;\,&space;\,&space;a-\frac{1}{a}=&space;x&space;\,&space;\,&space;then&space;\,&space;\,&space;a^{3}-\frac{1}{a^{3}}&space;=x^{2}+3x$
• If “$n$” is a natural number then $a^{n}&space;-&space;b^{n}&space;=&space;(a-b)&space;(a^{(n-1)}&space;+&space;a^{(n-2)}b&space;+....+b^{(n-2)}a&space;+&space;b^{n-1})$
• If “$n$” is a even number then $a^{n}&space;+&space;b^{n}&space;=&space;(a+b)&space;(a^{(n-1)}&space;-&space;a^{(n-2)}b&space;+.....+&space;b^{(n-2)}a&space;-&space;b^{(n-1))})$
• If “$n$” is an odd number then $a^{n}&space;+&space;b^{n}&space;=&space;(a-b)&space;(a^{(n-1)}&space;-a^{(n-2)}b&space;+....&space;-&space;b^{(n-2)}a&space;+&space;b^{(n-1)})$
• $If&space;\,&space;\,&space;\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+........\infty&space;}}}}&space;\,&space;\,&space;where&space;\,&space;\,&space;x=&space;n(n+1)$ then $If&space;\,&space;\,&space;\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+........\infty&space;}}}}&space;=n+1$
• $If&space;\,&space;\,&space;\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-........\infty&space;}}}}&space;\,&space;\,&space;where&space;\,&space;\,&space;x=&space;n(n+1)$ then $If&space;\,&space;\,&space;\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-........\infty&space;}}}}&space;=n$
Remainder Theorem:
Let $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+..........+a_{n}x^{n}$
be a polynomial of degree $n\geq&space;1$, and let $a$ be any real number. When is $p(x)$ divided by $(x-a)$, then the remainder is $p(a)$.
Factor theorem :
Let $p(x)$ be a polynomial of degree greater than or equal to 1 and a be a real number such that $p(a)$ = 0 , then $(x-a)$ is a factor of $p(x)$.
Conversely, if $(x-a)$ is a factor of $p(x)$ , then $p(a)$ = 0 .
Componendo and Dividendo rule :
• $If&space;\,&space;\,&space;\frac{a}{b}=\frac{b}{c}&space;\,&space;\,&space;then&space;\,&space;\,&space;\frac{a+b}{a-b}=\frac{c+d}{c-d}$
• $If&space;\,&space;\,&space;\frac{a+b}{a-b}=\frac{c}{d}&space;\,&space;\,&space;then&space;\,&space;\,&space;\frac{a}{b}=\frac{c+d}{c-d}$
Binomial theorem:
$(a+b)^{n}=&space;^{n}C_{0}a^{n}b^{0}+&space;^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}+&space;.....+^{n}C_{n-1}a^{1}b^{n-1}+^{n}C_{n}a^{0}b^{n}$
where n is a positive number and $^{n}C_{r}=&space;\frac{n!}{r!(n-r)!}$ .
You might also be interested in:
Some tips for mathematics students SSC CHSL 2024 Exam Date ssc chsl 2023 tier 1 cut off NIRF Rankings 2023 : Top 10 Engineering colleges in India CBSE Compartment Exam 2023 Application Form
|
## Engage NY Eureka Math 5th Grade Module 2 Lesson 29 Answer Key
### Eureka Math Grade 5 Module 2 Lesson 29 Problem Set Answer Key
Solve.
Question 1.
Lamar has 1,354.5 kilograms of potatoes to deliver equally to 18 stores. 12 of the stores are in the Bronx. How many kilograms of potatoes will be delivered to stores in the Bronx?
The kilograms of potatoes will be delivered to stores in the Bronx = 225.75 kg.
Explanation:
In the above-given question,
given that,
Lamar has 1,354.5 kilograms of potatoes to deliver equally to 18 stores.
12 of the stores are in the Bronx.
18 – 12 = 6.
1354.5/6.
225.75 kgs.
the kilograms of potatoes will be delivered to stores in the Bronx = 225.75 kg.
Question 2.
Valerie uses 12 fluid oz of detergent each week for her laundry. If there are 75 fluid oz of detergent in the bottle, in how many weeks will she need to buy a new bottle of detergent? Explain how you know.
The number of weeks will she need to buy a new bottle of detergent = 150.
Explanation:
In the above-given question,
given that,
Valerie uses 12 fluid oz of detergent each week for her laundry.
if there are 75 fluid oz of detergent in the bottle.
12 x 7 = 54.
75 x 12 = 150.
Question 3.
The area of a rectangle is 56.96 m2. If the length is 16 m, what is its perimeter?
The perimeter of rectangle = 39.12 m.
Explanation:
In the above-given question,
given that,
area of the rectangle = 56.96 sq m.
length = 16m
56.96/16 = 3.56 m
perimeter = 2 x (l + w).
perimeter = 2 x (16 + 3.56)
perimeter = 2 x 19.56.
p = 39.12 m.
Question 4.
A city block is 3 times as long as it is wide. If the distance around the block is 0.48 kilometers, what is the area of the block in square meters?
The area of the block in sq meters = 1.44 sq m.
Explanation:
In the above-given question,
given that,
A city block is 3 times as long as it is wide.
if the distance around the block is 0.48 kilometers.
area = l x b.
area = 0.48 x 3.
area = 1.44 sq m.
### Eureka Math Grade 5 Module 2 Lesson 29 Exit Ticket Answer Key
Solve.
Hayley borrowed $1,854 from her parents. She agreed to repay them in equal installments throughout the next 18 months. How much will Hayley still owe her parents after a year? Answer: Hayley still owes her parents after a year =$103.
Explanation:
In the above-given question,
given that,
Hayley borrowed $1854 from her parents. she agreed to repay them in equal installments throughout the next 18 months.$1854/18.
103$. ### Eureka Math Grade 5 Module 2 Lesson 29 Homework Answer Key Solve. Question 1. Michelle wants to save$150 for a trip to the Six Flags amusement park. If she saves $12 each week, how many weeks will it take her to save enough money for the trip? Answer: The number of weeks will it take her to save enough money for the trip =$13.
Explanation:
In the above-given question,
given that,
Michelle wants to save $150 for a trip to the Six Flags amusement park. if she saves$12 each week.
13 weeks.
12 x 13.
$156. Question 2. Karen works for 85 hours throughout a two-week period. She earns$1,891.25 throughout this period. How much does Karen earn for 8 hours of work?
The Karen earn for 8 hours of work = $236.40. Explanation: In the above-given question, given that, Karen works for 85 hours throughout a two-week period. she earns$1,891.25 throughout this period.
1891.25/8.
236.40$. Question 3. The area of a rectangle is 256.5 m2. If the length is 18 m, what is the perimeter of the rectangle? Answer: The perimeter of rectangle = 64.5 m. Explanation: In the above-given question, given that, area of the rectangle = 56.96 sq m. length = 18m 256.5/18 = 14.25 m perimeter = 2 x (l + w). perimeter = 2 x (18 + 14.25) perimeter = 2 x 32.25. p = 64.5 m. Question 4. Tyler baked 702 cookies. He sold them in boxes of 18. After selling all of the boxes of cookies for the same amount each, he earned$136.50. What was the cost of one box of cookies?
The cost of one box of cookies = $7.58. Explanation: In the above-given question, given that, Tyler baked 702 cookies. he sold them in boxes of 18. After selling all of the boxes of cookies for the same amount each he earned$136.50.
136.50/18.
7.58.
the cost of one box of cookies = \$7.58.
Question 5.
A park is 4 times as long as it is wide. If the distance around the park is 12.5 kilometers, what is the area of the park?
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# 2.2 Bernoulli’s equation (Page 4/7)
Page 4 / 7
## Making connections: take-home investigation with two strips of paper
For a good illustration of Bernoulli’s principle, make two strips of paper, each about 15 cm long and 4 cm wide. Hold the small end of one strip up to your lips and let it drape over your finger. Blow across the paper. What happens? Now hold two strips of paper up to your lips, separated by your fingers. Blow between the strips. What happens?
## Velocity measurement
[link] shows two devices that measure fluid velocity based on Bernoulli’s principle. The manometer in [link] (a) is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity ( ${v}_{1}=0$ ) in front of it, while fluid passing the other tube has velocity ${v}_{2}$ . This means that Bernoulli’s principle as stated in ${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}$ becomes
${P}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}$
Thus pressure ${P}_{2}$ over the second opening is reduced by $\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}$ , and so the fluid in the manometer rises by $h$ on the side connected to the second opening, where
$h\propto \frac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}$
(Recall that the symbol $\text{∝}$ means “proportional to.”) Solving for ${v}_{2}$ , we see that
${v}_{2}\propto \sqrt{h}\text{.}$
[link] (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air speed indicators in aircraft.
## Summary
• Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\mathrm{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}.$
• Bernoulli’s principle is Bernoulli’s equation applied to situations in which depth is constant. The terms involving depth (or height h ) subtract out, yielding
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}.$
• Bernoulli’s principle has many applications, including entrainment, wings and sails, and velocity measurement.
## Conceptual questions
You can squirt water a considerably greater distance by placing your thumb over the end of a garden hose and then releasing, than by leaving it completely uncovered. Explain how this works.
Water is shot nearly vertically upward in a decorative fountain and the stream is observed to broaden as it rises. Conversely, a stream of water falling straight down from a faucet narrows. Explain why, and discuss whether surface tension enhances or reduces the effect in each case.
Look back to [link] . Answer the following two questions. Why is ${P}_{\text{o}}$ less than atmospheric? Why is ${P}_{\text{o}}$ greater than ${P}_{\text{i}}$ ?
how environment affect demand and supply of commodity ?
Wht at the criteria for market ?
Amos
what is difference between monitory policy and fiscal policy?
monetary policy is a policy thrust by National Govt(CBN) to influence government spending, purchase &taxes
Frank
necessity of economics
I will say want,choice,opportunity cost,scarcity,scale of preference
Alao
what is monopoly market.How price output are determined under monopoly market
bisham
b) Monopoly market is an impecfect market where s single firm having the innovation to produce a particular commodity.Prices are determined through output since there are no other competitive.
Frank
Monopoly market:firm has market power & does not respond to market price
Frank
Explain the process of price determination under perfect competition market with suitable diagram
Price determination under perfect competition via this process :firms have no market power to influence price rather firms respond to market price.
Frank
price is different from demand- demand is amount of commodity
demand is amount /quantity of commodity a potential buyer is willing to buy at a given price at market
Frank
demand is a desire of customer on commodity with the ability to pay it and willing to buy it at given price of commodity
Harika
demand is price of what
show that shortrun average cost
what is economics
what is money
Mbah
what is money
Mbah
Difine macro economics
agaba
money is a medium of exchange between goods and services,maybe inform of currency.
Wesonga
Economics is study of how human beings strive to satisfy numerous wants using limited available resources.
Wesonga
how do you find the maximum number of workers the firms should employ order to produce where there are increasing returns
Jane
what are implications of computing national income?.
agaba
pl
MUDASIRU
what is the formulae for calculating national income
MUDASIRU
it calculated by value added method
Praveen
classify the production units like agriculture, banking, transport etc
Praveen
money is anything that is generally acceptetable for human
Ogbaji
Estimate the net value added(NVA) at fixed cost by each industrial structure
Praveen
definition of unemployment
what are the causes of unemployment?
The main causes of unemployment are listed below. 1. Frictional unemployment 2. Cyclical unemployment 3. Structural unemployment
assani
We can also categorize the causes on a broader sense as: 1. Political and 2. Social cause As unemployeement root causes are embaded in this two.
Yonathan
would opportunity cost exist if there was no scarcity?
assani
yes just because the opportunity cost arose when there is Alternative to choose among the alternatives.
I am thinking that, if our resources were unlimited, then there wouldn't be any need to forgo some wants. Hence the inexistence if opportunity cost
assani
Politics
Job
politics has done what?
assani
consider time assani
Mary
I'm Emmanuel,...I taught the main cause is the change in gov't.
Emmanuel
...Lack of capital to set up a firm respectively
Emmanuel
🙈
Emmanuel
I would like to bring in Educational levels can also be the cause the cause of the problem respectively
Emmanuel
I think the main causes of unemployment is lack of INFRASTRUCTURAL DEVELOPMENT OVER POPULATION OVER DEPENDENT ON GOVERNMENT LACK OF SELF EMPOWERMENT...
ananti
lack of skills among the new generation is the serious issue.
Vishal
Where I come from , I don't see why education or personal aspects seem to do with unimployment, technically the motivation and eigerness in all works of live is there , dispite the cultural influence and physical bearriors;the thing we lacking is Government Support and open market ethics.
Joe
sorry about that-(repation). We have a over powering ethical political system that's displacing the marketing asspects of economy and causing large scale unemployment right across the board...
Joe
can someone Explain Expansionary Monetary Policy and Contractionary Monetary Policy Using one of the instrument of Monetary Policy? Please am kinda lost here?. ta
using a graph show the case of substitute and compliment goods
can anyone give me a simple explanation to Five Sector Macroeconomics?
Emmanuel
Can someone please define what economics is
economics simply is a social science subject that study human behavior.
dajan
economics is a social science which studies human behaviour as a relationship between ends and scarce means that has alternative uses
Alao
Can someone please tell me how to calculate GDP
Emmanuel
emmanual kapal to calculate GDP (Gross Domestic Product) has three method in calculating it (1)income approach (2) expenditure approach (3) value added method
Alao
thanks Alae
Emmanuel
u are welcome
Alao
in basic terms economics is revered to as battery system, it date back to when Men sees the need to exchange sapless goods and produce to gain , either wealth , basic necessities or to establish trading ties for personal benefit or social asspects in terms of coexistence and continuity, future .
Joe
what is the law of demand
keep other thing constant, when the price increases demand decrease when the price decreases demand increases of the commodity.
sj
all things being equal,quantity demanded decrease as price increase and increase as price decrease
Seth
there's practial joke to it ..." the higher the demand ; scarcity, increase in production and drop in quality"... quite the controversy - for example China vs Europe, United States and we are all boxed up in between somewhere...
Joe
Other thing remain constant the low price of commodity the high quantity of commodity and vice versa is true
Baraka
Explain Effective demand
What is effective demand
Anita
like Modi is in demand...best example of effective demand
Pranav
Don't get you
Anita
Anita you mean you don't get me or who?
Onyeking
level of demand that represents a real intention to purchase by people with the means to pay
Pranav
Difference between extinct and extici spicies
While the American heart association suggests that meditation might be used in conjunction with more traditional treatments as a way to manage hypertension
in a comparison of the stages of meiosis to the stage of mitosis, which stages are unique to meiosis and which stages have the same event in botg meiosis and mitosis
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
Got questions? Join the online conversation and get instant answers!
|
### 1): Solving Equations with Variables on Both Sides
```Objectives
The student will be able to:
1. solve equations with variables on
both sides.
2. solve equations containing
grouping symbols.
SOL: A.4df
Designed by Skip Tyler, Varina High School
1) Solve. 3x + 2 = 4x - 1
You need to get the variables on one
side of the equation. It does not
matter which variable you move.
Try to move the one that will keep
1.
2.
3.
4.
5.
6.
1) Solve 3x + 2 = 4x - 1
Draw “the river”
- 3x
- 3x
Subtract 3x from
2 = x-1
both sides
Simplify
+1
+1
sides
3
=
x
Simplify
Check your
3(3)
+
2
=
4(3)
1
9 + 2 = 12 - 1
1.
2.
3.
4.
5.
6.
7.
8.
2) Solve 8y - 9 = -3y + 2
+ 3y
+ 3y
Draw “the river”
11y – 9 =
2
Simplify
+9
+9
Simplify
11y = 11
Divide both sides by
11
Simplify
11
11
y=1
8(1) - 9 = -3(1) + 2
What is the value of x if 3 - 4x = 18 + x?
1. -3
2.
3.
1
3
1
3
4. 3
1.
3) Solve 4 = 7x - 3x
Draw “the river”
4 = 4x
– Notice the variables
4 4
are on the same side!
Combine like terms
1=x
Divide both sides by 4
2.
3.
4. Simplify
4 = 7(1) - 3(1)
4) Solve -7(x - 3) = -7
1. Draw “the river”
2. Distribute
3. Subtract 21 from both
sides
4. Simplify
5. Divide both sides by -7
6. Simplify
-7x + 21 = -7
- 21 - 21
-7x
= -28
-7
-7
x=4
-7(4 - 3) = -7
-7(1) = -7
What is the value of x if
3(x + 4) = 2(x - 1)?
1.
2.
3.
4.
-14
-13
13
14
3 1
1
3
5) Solve x x
8 4
2
4
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Draw “the river”
Clear the fraction – multiply
each term by the LCD
Simplify
Simplify
Simplify
Divide both sides by 6
Simplify
3
1
1
3
(8) (8) x (8) x (8)
8
4
2
4
3 - 2x = 4x – 6
+ 2x +2x
3
= 6x – 6
+6
+6
9
= 6x
6
6
3
or 1.5 = x
2
3 1
1
3
1.5 1.5
8 4
2
4
Special Case #1
6) 2x + 5 = 2x - 3
1. Draw “the river”
2. Subtract 2x from both
sides
3. Simplify
-2x
-2x
5 = -3
This is never true!
No solutions
Special Case #2
7) 3(x + 1) - 5 = 3x - 2
Draw “the river”
Distribute
Combine like terms
Subtract 3x from both
sides
5. Simplify
1.
2.
3.
4.
3x + 3 – 5 = 3x - 2
3x - 2 = 3x – 2
-3x
-3x
-2 = -2
This is always true!
Infinite solutions
or identity
What is the value of x if
-3 + 12x = 12x - 3?
1.
2.
3.
4.
0
4
No solutions
Infinite solutions
Challenge! What is the value of x if
-8(x + 1) + 3(x - 2) = -3x + 2?
1.
2.
3.
4.
-8
-2
2
8
|
# Complex Numbers If we wish to work with , we need to extend the set of real numbers Definitions i is a number such that i2 = -1 C is the set of.
## Presentation on theme: "Complex Numbers If we wish to work with , we need to extend the set of real numbers Definitions i is a number such that i2 = -1 C is the set of."— Presentation transcript:
Complex Numbers If we wish to work with , we need to extend the set of real numbers Definitions i is a number such that i2 = -1 C is the set of numbers Z, of the form where a and b are real numbers. a is called the real part of Z and we write a = R(z) of a = Re(z) b is called the imaginary part of Z and we write b = i(z) or b = Im(z)
Given and Addition is defined by: Multiplication is defined by: We may write a + bi or a + ib, whichever we find more convenient
a) Given find (i) (ii) (iii) (i) (ii) (iii)
b) Solve the equation Using the quadratic formula
Page 90 Exercise 1 Questions 1, 2, 3, 6, 7, 8
Complex Conjugate When , then its complex conjugate is denoted by
This is useful when we wish to carry out a division.
a) Calculate
b) Calculate Let where Then Equating parts we get: also
Since
Page 91 Ex 2 Questions 1(a), (b), (c), 2(c), (e) 3(a), (b), (f), 5(a), (b) TJ Exercise 2. TJ Exercise 1 - if needed.
Argand Diagrams The complex number is represented on the plane by the point P(x,y). The plane is referred to as “The Complex Plane”, and diagrams of this sort are called Argand Diagrams. r y x p Any point on the x-axis represents a purely Real Number Any point on the y-axis represents a purely imaginary number
The size of the rotation is called the amplitude or argument of z.
It is often denoted Arg z. This angle could be We refer to the value of Arg z which lies in the range -< as the principal argument. It is denoted arg z, lower case ‘a’. r y x p By simple trigonometry: This is referred to as the Polar form of z.
a) Find the modulus and argument of the complex number
Since (3,4) lies in the first quadrant, n = 0 b) Find the modulus and argument of the complex number Since (-3,-4) lies in the third quadrant, n = -1
(2,2) is in Q1
Page 94 Exercise 3 Questions 3a, b, d, e, i 6a, b, f 7a, b, c
Loci-Set of points on the complex plane
This is a circle, centre the origin radius 4 4 -4 y x (i) 4 -4 y x (ii)
y x This is a circle centre (2, 0) radius 3 units.
This is a straight line through the origin gradient
Page 96 Exercise 4 Questions 1a, b, d, f, j
4a, c TJ Exercise 7
Polar Form and Multiplication
Note arg(z1z2) lies in the range (-, ) and adjustments have to be made by adding or subtracting 2 as appropriate if Arg(z1z2) goes outside that range during the calculation.
Note:
Now turn to page 96 Exercise 5 Questions 1 and 2.
Let us now look at question 3 on page 99.
This leads to the pattern:
De Moivre’s Theorem
a) Given find
b) Given find Round your answer to the nearest integer
Page 101 Exercise 6 questions 1 to 3, 4g, h, i, j.
Roots of a complex number
It would appear that if then
The solutions are radians apart,
By De Moivre’s theorem, when finding the nth root of a complex number we are effectively dividing the argument by n. We should therefore study arguments in the range (-n, n) so that we have all the solutions in the range (-, ) after division. The position vectors of the solution will divide the circle of radius r, centre the origin, into n equal sectors.
For k = 0 For k = 1
For k = 2
Page 106 Exercise 7: Question 2 plus a selection from 1
Polynomials In 1799 Gauss proved that every polynomial equation with complex coefficients, f(z) = 0, where z C, has at least one root in the set of complex numbers. He later called this theorem the fundamental theorem of algebra. In this course we restrict ourselves to real coefficients but the fundamental theorem still applies since real numbers are also complex.
We need to find z2, z3 and z4 And substitute them into the Original equation.
Using Division Hence the complimentary real factor is Hence all four roots are:
Page 108 Exercise 8 Questions 2, 3, 4, 5 and 6
Review on Page 110
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A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to th | MathCelebrity Forum
# A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to th
#### math_celebrity
##### Administrator
Staff member
A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to the reciprocal of the original fraction. Find the original fraction.
Let the fraction be x/y. We're given two equations:
1. x/y = 3/4
2. (x + 7)/y = 4/3. (The reciprocal of 3/4 is found by 1/(3/4)
Cross multiply equation 1 and equation 2:
1. 4x = 3y
2. 3(x + 7) = 4y
Simplifying, we get:
1. 4x = 3y
2. 3x + 21 = 4y
If we divide equation 1 by 4, we get:
1. x = 3y/4
2. 3x + 21 = 4y
Substitute equation (1) into equation (2) for x:
3(3y/4) + 21 = 4y
9y/4 + 21 = 4y
Multiply the equation by 4 on both sides to eliminate the denominator:
9y + 84 = 16y
To solve this equation for y, we type it in our math engine and we get:
y = 12
We then substitute y = 12 into equation 1 above:
x = 3 * 12/4
x = 36/4
x = 9
So our original fraction x/y = 9/12
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# How do you convert $0.8333$ to fraction?
Last updated date: 22nd Jul 2024
Total views: 384k
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Hint: To convert .8333 into fraction count the numbers after the point then, the denominator can be calculated by ${10^n}$ where $n$ is the number of digits after the decimal. Then, we replace the decimal with ${10^n}$ in the denominator and put its value.
Complete step by step solution:
A decimal fraction can be defined as the fraction in which the denominator of any rational number is in the power of 10 such as 10,100,1000 etc. depending on the number of digits after the decimal. The denominator can be calculated by ${10^n}$ where $n$ is the number of digits after the decimal.
As it is given in the question, we have to convert .8333 into a fraction. We can also write it as 0.8333 So, there are four digits after the decimal in 0.8333.
Therefore, the denominator can be calculated as –
$\Rightarrow {10^4} = 10000$
Hence, the number in the denominator will be 10000. So, we have the decimal with 1 in the denominator and write zero for each number after the decimal.
Therefore, the fraction can be written as –
$\Rightarrow \dfrac{{8333}}{{10000}}$
which is in the fractional form but it should be in the simplest form. So, we will convert the above fraction into its simplest form.
So, find the common factor of 8333 and 10000. Since, there is no common factor for 8333 and 10000 therefore, the fraction, we get –
$\Rightarrow \dfrac{{8333}}{{10000}}$
Hence, this is the required fraction.
Note:
It is not always necessary that we get the simplest form of the fraction as if in any fraction, there is no common factor between numerator and denominator then, the fraction is already in its simplest form and it cannot be further changed.
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What Is 10/55 as a Decimal + Solution With Free Steps
The fraction 10/55 as a decimal is equal to 0.181.
A fractional form is the way of representation of a decimal operation. This is expressed as a fractional number o/p, where o is the numerator (upper value) and p is the denominator (lower value).
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction-to-decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 10/55.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 10
Divisor = 55
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 10 $\div$ 55
This is when we go through the Long Division solution to our problem. Given is the long division process in Figure 1:
Figure 1
10/55 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 10 and 55, we can see how 10 is Smaller than 55, and to solve this division, we require that 10 be Bigger than 55.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 10, which after getting multiplied by 10 becomes 100.
We take this 100 and divide it by 55; this can be done as follows:
100 $\div$ 55 $\approx$ 1
Where:
55 x 1 = 55
This will lead to the generation of a Remainder equal to 100 – 55 = 45. Now this means we have to repeat the process by Converting the 45 into 450 and solving for that:
450 $\div$ 55 $\approx$ 8
Where:
55 x 8 = 440
This, therefore, produces another Remainder which is equal to 450 – 440 = 10. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 100.
100 $\div$ 55 $\approx$ 1
Where:
55 x 1 = 55
Finally, we have a Quotient generated after combining the three pieces of it as 0.181, with a Remainder equal to 45.
Images/mathematical drawings are created with GeoGebra.
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# Describing Division
Standard
Division can be a tricky, but understanding some basic terms can help. There are three basic terms to describe a division problem; they are dividend, divisor, and quotient. An example of this is 24/6=4. In this example the 24 is the dividend, 6 is the divisor, and 4 is the quotient.
There are two different concepts to find the solution to a division problem. They are the sharing concept and the measurement concept. The sharing concept uses the divisor to divide the problem into number of groups, and uses the quotient to determine how many parts are in each group. A great way to visualize this is by using fruit. For example, we have a total of 24 apples, and we want to divide them equally between 6 groups. How many apples will each group receive? According to the sharing concept the divisor, 6, is used to find the number of groups, and the quotient, 4, is used to find the number of apples in each group. This means we will have 6 groups with 4 apples in each group. On the other hand, the measurement concept uses the quotient to divide the problem into number of groups, and the divisor to determine how many parts are in each group. If we use the apple problem again, there would be 4 groups with 6 apples in each group. The picture provided may help to clear up any confusion about the concepts.
Drawing pictures is a great way to practice division, but I have also found a fun game that is great for practice. The game is called Alien Munchtime; in the game it allows you to choose the fact families that you need help with. The choices include 2 through 12. After you have chosen your fact families, an alien pops up and says she needs help serving lunch. Once you click the “lets go” button the game begins. Hungry alien students come through the lunch line with a division question like 88/8 and you have to serve them the answer like 11. This is a great way to get practice with division and have fun at the same time!
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2001 AMC 12 Problems/Problem 3
The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page.
Problem
The state income tax where Kristin lives is levied at the rate of $p\%$ of the first $\textdollar 28000$ of annual income plus $(p + 2)\%$ of any amount above $\textdollar 28000$. Kristin noticed that the state income tax she paid amounted to $(p + 0.25)\%$ of her annual income. What was her annual income?
$\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000$
Solution
Solution 1
Let the income amount be denoted by $A$.
We know that $\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}$.
We can now try to solve for $A$:
$(p+.25)A=28000p+Ap+2A-28000p-56000$
$.25A=2A-56000$
$A=32000$
So the answer is $\boxed{B}$
Solution 2
Let $A$, $T$ be Kristin's annual income and the income tax total, respectively. Notice that \begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*} We are also given that $$T = (p + 0.25)\%\cdot A = p\%\cot A + 0.25\%\cdot A$$ Thus, $$p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cot A + 0.25\%\cdot A$$ $$2\%\cdot(A - 28000) = 0.25\%\cdot A$$ Solve for $A$ to obtain $A = 32000$. $\boxed{B}$
~ Nafer
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# Fields and Fences/Class 4
Chapter 13 of NCERT/CBSE Class 4 Mathematics is about Fields and Fences. Here in this lesson you can find practice problems about perimeter.
NCERT SOLUTIONS FOR CLASS 4 MATHEMATICS
Fields and Fences – Chapter 13.
1. The length of the boundary of a field is 9m, 15m, 21m and 9m.
a)How much wire need to fence the field?
b) If Rahmath bought a roll of 70m wire for the fence, how much wire left with her?
2. If the length of the boundary of Ganapat’s field is 15m, 15m, 9m ,9m and 18m.
a) How long is the boundary of Ganapat’s field?
b) How much more wire will Ganapat need for his field, if Rahmath gave 16m wire to Ganpat?
3. Which one has the longest boundary?
a) 6m, 15m, 15m, 24m
b) 6m, 3m, 6m, 6m, 12m, 9m
c) 9m, 12m, 15m
4. The length of the boundary of a field is given by 100m, 150m, 100m; 150m.Chandu’s father goes for a walk around the field every morning. Every day he takes four rounds of this field. What is the total distance he covers?
5. Ganapat’s wife works in a tailor’s shop. She has to fix lace around a table cloth. The length of the boundary of the table is 1m 50 cm length and 50 cm breadth. She bought a 100m roll of lace.
a) How much lace is used for one table cloth?
b) How much lace will be used in 3 such table cloths?
c) How much lace will be left in the roll?
6. A hockey field is 91m 40cm long and 55m wide. How long is the boundary of the field?
7. Usha and Valsamma are running a race. Usha is running on the inner circle. Valsamma is running on the outer circle. Valsamma runs faster than Usha. But still she loses the race. Can you guess why?
8. Which of the following has the longest boundary?
9. Find the perimeter of a square garden of length 11m?
10. Find the distance around a rectangular field of length 12m and breadth 5m?
1. a) The wire needed to fence the field = 9 + 15 + 21+9 = 54m.
b) Length of the wire that Rahmath bought = 70m
Length of wire left with her = 70 – 54 = 16m
2. a) Boundary of Ganapat’s field = 15 + 15 + 9 + 9 + 18 = 66m
b) If Rahmath gave 16m wire to Ganapath, wire required = 66 – 16 = 50m more wire needed.
3. a) Length of the boundary = 6 + 15 + 15 + 24 = 60m
b) Length of the boundary = 6 + 3 + 6 + 6 + 12+ 9 = 42m
c) Length of the boundary = 9 + 12 + 15 = 36m
d) Length of the boundary = 15 + 9 + 15 + 15 + 9 + 15 = 78m
Therefore a field with length of boundary 15m, 9m, 15m, 15m, 9m, 15m has the longest boundary.
4. Total distance he covers = 4 x (100 + 150 + 100 + 150)
= 4 x 500 = 2000m = 2 km.
5. Length of the boundary of the table is 50 cm and 1m 50cm.
a) Lace used for one table cloth = 50 cm + 50 cm + 1m 50 cm + 1m 50 cm
= 50 + 50 + 150 + 150 = 400 cm = 4m.
b) Lace required for 3 such table cloths = 3 x 4 = 12m
c) Lace left in the roll = 100 – 12 = 88m.
6. Length of the hockey field = 91m 40cm = 9140 cm
Breadth of the hockey field = 55m = 5500 cm
Length of the boundary = 9140 + 5500 = 14640 cm = 146m 40cm.
7. Outer circle has longer boundary and inner circle has shorter boundary. So Valsamma has to cover more length. Therefore Valsamma loses the race.
8. a) Length of the boundary = 8 + 3 + 8 + 3 = 22m
b) Length of the boundary = 6 + 4 + 6 + 4 = 20m
Therefore, the field with length 8m, breadth 3m has the longest boundary.
9. Perimeter of a square garden = 11 + 11 + 11+ 11 = 44m
10. Length of the boundary = 12 + 5+ 12 + 5 = 34m.
### 5 Responses
1. Deepti Kumari says:
Because I want CBSE maths worksheet 3
• Sanjusha says:
CBSE Class 3 Worksheet is available in this blog. Please visit
• STHUTHI says:
YEAH ONE OF THE QUESTION WAS WORING
IT WAS THE 6 TH QUESTION
• Sanjusha says:
Thank you for your valuable comment….Updated
2. Lekha says:
Good one…was useful as it included all category of perimeter problems. Thank you!!!!!!!!1
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# Box & Whisker Plot: Lesson for Kids
I have a Bachelor's degree in Elementary Education and Spanish. I have taught for 5 years in bilingual classrooms of various elementary grade levels.
What do you do when you have a lot of numbers that are spread out, and you want to group them into sets and graph them? In this lesson, you will learn how to create and read a box and whisker plot which is perfect to use with that type of data.
## Uses of a Box and Whisker Plot
Cookies, cupcakes, fruit pies! Oh my! Your school just had its annual bake sale, and each class had to make and sell their sweets. Your teacher wants to see how well the 16 students in her class did and suggests using a box and whisker plot, but what exactly is that?
A box and whisker plot is a special type of graph that is used to show groups of number data and how they are spread. It shows the median, which is the middle value of the numbers in your data, the lowest number, the highest number and the quartiles, which divides the data into four equal groups. Quartiles are like quarters. There are 4 quarters in a dollar, each of which is worth 25 cents, so there are 4 quartiles in a box and whisker plot, each showing 25% of the number data. Let's take a look at our bake sale numbers and put this information to use.
## Creating a Box and Whisker Plot
Take a look at the image which shows the numbers of goodies each classmate sold. Now it's time to create your own box and whisker plot! Your teacher tells you to:
1. Order the numbers from least to greatest:
• 0, 3, 7, 9, 10, 15, 22, 27, 28, 30, 32, 35, 36, 40, 50, 55
2. Find the median and record it on the line. If there is an odd amount of number data, this will be the middle number. If there is an even amount of number data, take the average of the two numbers in the middle. This will create halves.
• We have 16 numbers, so our two middle numbers are 27 and 28. Add them together and divide by two to find the average = 27.5. This is our median.
3. Plot the lowest number at the beginning and the highest number at the end of the number line.
• 0 is our lower extreme and 55 is our upper extreme
4. Find the median for each half of the data - this is dividing the data into quartiles. To do this, you must repeat the process for finding the median, but do it for each half. Record both numbers on the line.
• Our first half of the data are the first eight numbers. The two middle numbers are 9 and 10. So, the average is 9.5. Our second half of data are the last eight numbers. The middle numbers are 35 and 36, so the average is 35.5.
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Saturday, July 20, 2013
Exact Differential Equations (Problem Solutions)
The Problems Again:
(1) State whether the DE: dy/dx = (2x + y2)/ -2xy is exact
(2) Show that the DE below is exact and find the solution
(3xy4 + x)dx + (6x2y3 – 2y2 + 7)dy = 0
(3) Consider the differential equation:
(3x2 y + 2) dx + (x3 + y) dy = 0
Determine whether it is an exact DE or not. If it is, find the general solution and then the particular soln. for an initial condition such that : y(1) = 3.
Solutions:
1) We first re-arrange to obtain:
(-2xy)dy = (2x + y2) dx
And: (2x + y2) dx +(2xy)dy = 0
Then: M = (2x + y2) and N = (2xy)
Take partial derivatives: M/y = 2y and N/x = 2y
Since the two partials are equal, the DE is exact.
2) First, take the partials to make sure it’s exact:
M = (3xy4 + x) and M/y = 12xy3
N = (6x2y3 – 2y2 + 7) and N/x = 12xy3
So, it’s exact. Now, let:
f(x,y) = òx (3xy4 + x) dx + C(y) = 3x2y4/ 2 + x2/ 2 + C(y)
Then: dC/dy = 6x2y3 -2y2 + 7 – 6x2y3 = -2y2 + 7
And: ò dC(y) = C(y) = ò (-2y2 + 7) dy = -2y3/ 3 + 7y
So the general solution is: f(x,y) = 3x2y4/ 2 + x2/ 2 -2y3/ 3 + 7y
3) This is straightforward, having solved the previous example.
We have: M = (3x2 y + 2) so M/y = 3x2
And: N = (x3 + y) so N/x= 3x2
So, the DE is exact. We have then:
f(x,y) = òx (3x2 y + 2) dx + C(y) = (x3 y + 2x) + C’(y)
whence:
òy dC(y)dy = C(y) = òy (x3 + y) dy + c1= y2/2 + c1
This solution satisfies: f(x,y) = c2
Or: x3y + 2x + y2/ 2 + c1 = c2 = òx (3x2 y + 2) dx + C(y)
So the final general soln. is: : x3y + 2x + y2/ 2 + c = 0
To satisfy the condition y(1) = 3: (1)3 + 2(1) + (3)2/ 2 + c = 0
= 3 + 9/2 + c = 0
So: c = -7 ½
Therefore: : x3y + 2x + y2/ 2 - 15/2 = 0
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# Find Averages from Frequency Charts
In this worksheet, students will read grouped data from frequency charts in order to work out the mean, median and mode.
Key stage: KS 3
Curriculum topic: Statistics
Curriculum subtopic: Understand Variables, Representation, Measures and Spread
Difficulty level:
#### Worksheet Overview
How tall are you?
The following frequency chart shows the heights of children in a Year 8 class at the EdPlace school.
Let's remember that frequency is just 'how many times' something occurs and when data is grouped (like in this chart!), the exact value is not known.
For example, here we have:
0 children have a height between 110 cm and 115 cm.
1 child has a height between 115 cm and 120 cm.
3 children have a height between 120 cm and 125 cm ..... etc.
We can use a frequency chart to work out our three averages: mean, median and mode!
Mean
We can only estimate the mean by using the midpoint value in each class.
These are 112.5, 117.5, 122.5, 127.5, 132.5, 137.5, 142.5, 147.5 and 152.5 cm
We multiply these midpoint values by the frequency that they occur. eg. 3 x 122.5 because three children are in that height group.
The estimated sum of all the heights is:
(1 x 117.5) + (3 x 122.5) + (4 x 127.5) + (7 x 132.5) + (4 x 137.5) + (2 x 142.5) + (1 x 152.5) = 2,910 cm
We add all the frequencies to find the number of children in the class:
1 + 3 + 4 + 7 + 4 + 2 + 1 = 22 children in total
The estimate for the mean height is therefore:
2,910 ÷ 22 = 132.3 cm (1 dp)
Median
We have found that there are 22 children in total.
So the 'middle person' will be the (22 + 1) ÷ 2 = 11.5th person, i.e. the value between the height of the 11th and 12th person.
The 11th and 12th people both fall into the 130 cm to 135 cm class.
So the median class is 130 cm to 135 cm.
Mode
This is the class with the highest frequency.
We can see that the tallest bar (i.e. the one with the highest frequency) is the one for 130 to 135 cm.
So the modal class is 130 cm to 135 cm.
Easy!
That's a lot of information to take in - don't worry if you can't remember it all, you can check back to this introduction at any point during the activity by clicking on the red button on the side of the screen!
Grab your calculator and let's get started!
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
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# Number Sentence Roll and Write: Multiplication and Division
Last year when I was focusing on number bonds and fact families for addition and subtraction, I created this super simple, but really fun activity. Seriously, there’s just something about rolling the giant foam dice that my kids love! Today I’m sharing a similar activity for multiplication and division! It’s hard to believe that I’m already introducing my daughter to multiplication and division, but she has shown that she gets it, and enjoys the new challenge. This fun multiplication and division game is such a simple activity, but helps form a strong foundation!
*Please Note: Some of the links in this post are affiliate links and help support the work of this site. Read our full disclosure policy here.*
It’s incredibly important that math concepts not be taught in isolation. Mathematics is a rich, interconnected subject, and we should take every opportunity to help kids see and form those connections.
Multiplication and division are what are called inverse operations. This means they are opposite each other and “undo” one another.
For instance, if take the number 3 and multiply it by 5, you get 15. To “undo” that and get back to your original value, you would simply divide by 5.
Related: Solving Problems by Working Backwards (using inverse operations).
## Multiplication and Division Fact Families:
A fact family is a set of three numbers that can be related to each other using multiplication and division. (Or as in this post, with addition and subtraction).
For example, the fact family above includes the numbers 3, 5, and 15. This creates 4 different equations:
• 3 x 5 = 15
• 5 x 3 = 15
• 15 / 3 = 5
• 15 / 5 = 3
This Roll and Write multiplication and division game gives kids the opportunity to write out the fact families.
Please note, though, that when you are dealing with doubles, there will only be two equations, not four (i.e. 5 x 5 = 25 and 25 / 5 = 5).Â
By creating and comparing fact families with multiplication and division, kids learn (or reinforce) several important math concepts:
1. The commutative property of multiplication
2. That multiplication and division are inverse operations
3. How to correctly write equations involving multiplication and division
4. That knowing multiplication facts means they also know division facts
Plus, they’re able to connect a new math concept (division) with a known math concept (multiplication).
## How to use the Number Sentence Roll and Write Activity:
As I’ve said, this activity is SUPER simple. All you need is the student handout (free in my shop) and a set of dice (or just one die, rolled twice)!
Personally, whenever I create or use an activity involving dice, I like to get out our giant foam dice. My kids just love these, and it seems to make it even more fun (but regular dice will totally do the trick, too!)
Once your kids are ready, they start by rolling the dice and recording the numbers on the answer sheet.
They then use those two values to write two multiplication problems (thus demonstrating the commutative property).
After finding the solution to those equations, they can then start with the solution to write two division equations.
They then complete the chart by rolling to get two new values.
And that’s it! Simple, right? And yet, so many important math concepts are being explored and connections made! 🙂
UPDATE: This download has been updated to include a page that focuses only on multiplication. Students roll two numbers and write a multiplication sentence, then rewrite it using the commutative property.
If your students are not quite ready to work on the complete fact family, hopefully this will provide a helpful bridge to get there! 🙂
I hope you’re able to use this multiplication and division game with your child or in your classroom. It will reinforce known facts or start to form connections between the math they’re learning.
And most of all, have fun forming multiplication and division fact families!
### {Click HERE to go to my shop and grab the Roll and Write Multiplication and Division Game!}
Looking for more multiplication and division ideas? Try one of these!
Looking for a fun and hands on way to teach multiplication to your kids? This huge lesson bundle covers 5 different representations of multiplication, as well as the commutative property, printable games, word problems and more!
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# Question: What Is The First Rule Of Algebra?
## What are the 3 rules of algebra?
There are many laws which govern the order in which you perform operations in arithmetic and in algebra.
The three most widely discussed are the Commutative, Associative, and Distributive Laws.
Over the years, people have found that when we add or multiply, the order of the numbers will not affect the outcome..
## What is the golden rule of algebra?
The Golden Rule of Algebra. Do unto one side of the equation what you do to the other.
## What is a 1 in algebra?
Algebra 1 is the second math course in high school and will guide you through among other things expressions, systems of equations, functions, real numbers, inequalities, exponents, polynomials, radical and rational expressions.
## What are the four rules of maths?
The four basic Mathematical rules are addition, subtraction, multiplication, and division.
## Do you add first or multiply first?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right. Next, add and subtract from left to right.
## What is algebra formula?
Algebra includes both numbers and letters. Numbers are fixed, i.e. their value is known. Letters or alphabets are used to represent the unknown quantities in the algebra formula.
## What is the rule for solving equations?
The following steps provide a good method to use when solving linear equations. Simplify each side of the equation by removing parentheses and combining like terms. Use addition or subtraction to isolate the variable term on one side of the equation. Use multiplication or division to solve for the variable.
## What is formula of a2 b2?
a2 + b2 = (a+b)2 – 2ab nd a2 – b2 = (a + b)(a – b)
## Is AB the same as BA in algebra?
The like terms are abc and acb; ab and ba. Note, abc and acb are identical and ab and ba are also identical. It does not matter in what order we multiply – for example, 3 x 2 x 4 is the same as 3 x 4 x 2. (c) –3a; –3ab; –3b There are no like terms in this example.
## What are the basic rules of algebra?
The Basic Laws of Algebra are the associative, commutative and distributive laws. They help explain the relationship between number operations and lend towards simplifying equations or solving them. The arrangement of addends does not affect the sum.
## What is basic algebra?
What Is Basic Algebra? Basic algebra is the field of mathematics that it one step more abstract than arithmetic. Remember that arithmetic is the manipulation of numbers through basic math functions. Algebra introduces a variable, which stands for an unknown number or can be substituted for an entire group of numbers.
## What are the 3 types of equations?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## What does AB in algebra mean?
Algebraic Terms 2a means 2 × a ab means a × b a means a × a a means a × a × a means a ÷ b means a × a × b ÷ c Adding a.
## What is the golden rule for solving equations?
How do we apply the Golden Rule? First it should be stated, that when solving for an unknown variable in an equation, you must try to get 0 on the side with the unknown variable in addition/subtraction (and get 1 in multiplication/division).
## What are the four steps for solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal. If we subtract the same number from both sides of an equation, both sides will remain equal.
## What is the order of algebra?
The order of operations is Parenthesis, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). This can be remembered in two ways: “Please Excuse My Dear Aunt Sally” or PEMDAS.
## Are equations algebra?
Algebraic equation, statement of the equality of two expressions formulated by applying to a set of variables the algebraic operations, namely, addition, subtraction, multiplication, division, raising to a power, and extraction of a root. … Examples are x3 + 1 and (y4x2 + 2xy – y)/(x – 1) = 12.
## What is formula of a3 b3?
a3 – b3 = (a – b) (a2 + ab + b2 ). 9. a3 + b3 = (a + b) (a2 – ab + b2 ).
## Is Bodmas wrong?
BODMAS is wrong. … It contains no brackets, powers, division, or multiplication so we’ll follow BODMAS and do the addition followed by the subtraction: This is erroneous.
## What is M in algebra?
In the equation y = mx + b for a straight line, the number m is called the slope of the line. Definition 2. In the equation y = mx + b for a straight line, the. number b is called the y-intercept of the line. Page 9.
## What are the two basic rules for solving algebraic equations?
In algebra 1 we are taught that the two rules for solving equations are the addition rule and the multiplication/division rule. The addition rule for equations tells us that the same quantity can be added to both sides of an equation without changing the solution set of the equation.
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# Equation: Even More Examples
Example 1 Find the equation of the circle, whose diameter’s end points are (1, 3) and (4, 2).
Solution Pretty simple. Using the form we derived in the previous lesson, the required equation is (x – 1)(x – 4) + (y – 3)(y – 2) = 0, or x2 + y2 – 5x – 5y + 10 = 0.
Example 2 Find the equation of the circle passing through the points (0, 0), (a, 0) and (0, b).
Solution I’ve covered a similar example in a previous lesson (see Example 9). But now we have a better method to obtain the equation.
Observe that the points (a, 0) and (0, b) lie on the X and the Y axes respectively. That means angle AOB = 90°, which implies that AB must be the diameter of the circle.
Therefore the required equation is (x – a)(x – 0) + (y – 0)(y – b) = 0, or x2 + y2 – ax – by = 0.
Note that the center of the circle is (a/2, b/2) (midpoint of AB, the diameter), and the radius is $$\sqrt{a^2+b^2}/2$$ (half of AB, the diameter)
Example 3 Find the equation of the circle, whose chord with end-points as (1, 2) and (0, 4) subtends an angle of 45° at its circumference.
Solution This equation type was also covered in the previous lesson.
I’ll use that equation directly: (x – 1)(x – 0) + (y – 2)(y – 4) = ± cot45° [(y – 2)(x – 0) – (y – 4)(x – 1)].
As I mentioned earlier, there’ll be two different circles satisfying the given conditions: x2 + y2 – 3x – 7y + 12 = 0 and x2 + y2 + x – 5y + 4 = 0.
And that’s it for this lesson. See you in the next with the parametric equation of the circle.
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# NCERT Solutions Class 12 Maths Chapter 12 Linear Programming
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming prepared by the subject experts at CoolGyan’S have been provided here. We can say that linear programming is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. In this chapter students learn about Linear Programming in detail. Solving all the questions from this chapter of the NCERT book of Class 12 maths, repeatedly, will help the students in understanding the methods of solving the questions.
The Class 12 NCERT solutions of the chapter Linear Programming are very easy to understand as the subject experts at CoolGyan’S ensure that the solutions are given in simplest form. These NCERT Solutions cover all the exercise questions included in the book and are in accordance with the latest CBSE guidelines.
## Download PDF of NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming
### Access All Exercises of Class 12 Maths Chapter 12
Exercise 12.1 Solutions 10 Questions
Exercise 12.2 Solutions 11 Questions
Miscellaneous Exercise On Chapter 12 Solutions 10 Questions
#### NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming
Exercise 12.1 page no: 513
1. Maximise Z = 3x + 4y
Subject to the constraints:
Solution:
The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is given below
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given below
Corner point Z = 3x + 4y O (0, 0) 0 A (4, 0) 12 B (0, 4) 16 Maximum
Hence, the maximum value of Z is 16 at the point B (0, 4)
2. Minimise Z = −3x + 4y
subject to.
Solution:
The feasible region determined by the system of constraints,
is given below
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region
The values of Z at these corner points are given below
Corner point Z = – 3x + 4y O (0, 0) 0 A (4, 0) -12 Minimum B (2, 3) 6 C (0, 4) 16
Hence, the minimum value of Z is – 12 at the point (4, 0)
3. Maximise Z = 5x + 3y
subject to.
Solution:
The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are given below
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these corner points are given below
Corner point Z = 5x + 3y O (0, 0) 0 A (2, 0) 10 B (0, 3) 9 C (20 / 19, 45 / 19) 235 / 19 Maximum
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19)
4. Minimise Z = 3x + 5y
such that.
Solution:
The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2, and x, y ≥ 0 is given below
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2)
The values of Z at these corner points are given below
Corner point Z = 3x + 5y A (3, 0) 9 B (3 / 2, 1 / 2) 7 Smallest C (0, 2) 10
7 may or may not be the minimum value of Z because the feasible region is unbounded
For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check the resulting half plane have common points with the feasible region or not
Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7
Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2)
5. Maximise Z = 3x + 2y
subject to.
Solution:
The feasible region determined by the constraints, x + 2y ≤ 10, 3x + ≤ 15, x ≥ 0, and y ≥ 0, is given below
A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the feasible region.
The values of Z at these corner points are given below
Corner point Z = 3x + 2y A (5, 0) 15 B (4, 3) 18 Maximum C (0, 5) 10
Hence, the maximum value of Z is 18 at the point (4, 3)
6. Minimise Z = x + 2y
subject to
.
Solution:
The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y ≥ 0, is given below
A (6, 0) and B (0, 3) are the corner points of the feasible region
The values of Z at the corner points are given below
Corner point Z = x + 2y A (6, 0) 6 B (0, 3) 6
Here, the values of Z at points A and B are same. If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6
Hence, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line x + 2y = 6
7. Minimise and Maximise Z = 5x + 10y
subject to.
Solution:
The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and y ≥ 0, is given below
A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the corner points of the feasible region. The values of Z at these corner points are given
Corner point Z = 5x + 10y A (60, 0) 300 Minimum B (120, 0) 600 Maximum C (60, 30) 600 Maximum D (40, 20) 400
The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30)
8. Minimise and Maximise Z = x + 2y
subject to.
Solution:
The feasible region determined by the constraints, x + 2y ≥ 100, 2x − y ≤ 0, 2x + ≤ 200, x ≥ 0, and y ≥ 0, is given below
A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the corner points of the feasible region. The values of Z at these corner points are given below
Corner point Z = x + 2y A (0, 50) 100 Minimum B (20, 40) 100 Minimum C (50, 100) 250 D (0, 200) 400 Maximum
The maximum value of Z is 400 at point (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40)
9. Maximise Z = − x + 2y, subject to the constraints:
.
Solution:
The feasible region determined by the constraints,
is given below
Here, it can be seen that the feasible region is unbounded.
The values of Z at corner points A (6, 0), B (4, 1) and C (3, 2) are given below
Corner point Z = – x + 2y A (6, 0) Z = – 6 B (4, 1) Z = – 2 C (3, 2) Z = 1
Since the feasible region is unbounded, hence, z = 1 may or may not be the maximum value
For this purpose, we graph the inequality, – x + 2y > 1, and check whether the resulting half plane has points in common with the feasible region or not.
Here, the resulting feasible region has points in common with the feasible region
Hence, z = 1 is not the maximum value.
Z has no maximum value.
10. Maximise Z = x + y, subject to.
Solution:
The region determined by the constraints, is given below
There is no feasible region and therefore, z has no maximum value.
Exercise 12.2 page no: 519
1.Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture?
Solution:
Let the mixture contain x kg of food P and y kg of food Q.
Hence, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given
Vitamin A (units / kg) Vitamin B (units / kg Cost (Rs / kg) Food P 3 5 60 Food Q 4 2 80 Requirement (units / kg) 8 11
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Hence, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost of purchasing food is, Z = 60x + 80y
So, the mathematical formulation of the given problem can be written as
Minimise Z = 60x + 80y (i)
Now, subject to the constraints,
3x + 4y ≥ 8 … (2)
5+ 2y ≥ 11 … (3)
xy ≥ 0 … (4)
The feasible region determined by the system of constraints is given below
Clearly, we can see that the feasible region is unbounded
A (8 / 3, 0), B (2, 1 / 2) and C (0, 11 / 2)
The values of Z at these corner points are given below
Corner point Z = 60x + 80 y A (8 / 3, 0) 160 Minimum B (2, 1 / 2) 160 Minimum C (0, 11 / 2) 440
Here the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this purpose, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the resulting half plane has points in common with the feasible region or not
Here, it can be seen that the feasible region has no common point with 3x + 4y < 8
Hence, at the line segment joining the points (8 / 3, 0) and (2, 1 / 2), the minimum cost of the mixture will be Rs 160
2. One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes?
Solution:
Let the first kind of cakes be x and second kind of cakes be y. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as shown below
Flour (g) Fat (g) Cakes of first kind, x 200 25 Cakes of second kind, y 100 50 Availability 5000 1000
So, 200x + 100y ≤ 5000
2x + y ≤ 50
25x + 50y ≤ 1000
x + 2y ≤ 40
Total number of cakes Z that can be made are
Z = x + y
The mathematical formulation of the given problem can be written as
Maximize Z = x + y (i)
Here, subject to the constraints,
2x + y ≤ 50 (ii)
x + 2y ≤ 40 (iii)
x, y ≥ 0 (iv)
The feasible region determined by the system of constraints is given as below
A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the corner points
The values of Z at these corner points are as given below
Corner point Z = x + y A (25, 0) 25 B (20, 10) 30 Maximum C (0, 20) 20 O (0,0) 0
Hence, the maximum numbers of cakes that can be made are 30 (20 cakes of one kind and 10 cakes of other kind)
3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
(ii) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution:
Let x and y be the number of rackets and the number of bats to be made.
Given that the machine time is not available for more than 42 hours
Hence, 1.5x + 3y ≤ 42 ……………. (i)
Also, given that the craftsman’s time is not available for more than 24 hours
Hence, 3x + y ≤ 24 ………… (ii)
The factory is to work at full capacity. Hence,
1.5x + 3y = 42
3x + y = 24
On solving these equations, we get
x = 4 and y = 12
Therefore, 4 rackets and 12 bats must be made.
(i) The given information can be compiled in a table as give below
Tennis Racket Cricket Bat Availability Machine Time (h) 1.5 3 42 Craftsman’s Time (h) 3 1 24
1.5x + 3y ≤ 42
3x + y ≤ 24
x, y ≥ 0
Since, the profit on a racket is Rs 20 and Rs 10
Hence, Z = 20x + 10y
The mathematical formulation of the given problem can be written as
Maximize Z = 20x + 10y ………….. (i)
Subject to the constraints,
1.5x + 3y ≤ 42 …………. (ii)
3x + y ≤ 24 …………….. (iii)
x, y ≥ 0 ………………… (iv)
The feasible region determined by the system of constraints is given below
A (8, 0), B (4, 12), C (0, 14) and O (0, 0) are the corner points respectively.
The values of Z at these corner points are given below
Corner point Z = 20x + 10y A (8, 0) 160 B (4, 12) 200 Maximum C (0, 14) 140
Therefore, the maximum profit of the factory when it works to its full capacity is Rs 200
4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 hours a day?
Solution:
Let the manufacturer produce x package of nuts and y package of bolts. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given below
Nuts Bolts Availability Machine A (h) 1 3 12 Machine B (h) 3 1 12
The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7
Hence, the constraints are
x + 3y ≤ 12
3x + y ≤ 12
Then, total profit, Z = 17.5x + 7y
The mathematical formulation of the given problem can be written as follows
Maximize Z = 17.5x + 7y …………. (1)
Subject to the constraints,
x + 3y ≤ 12 …………. (2)
3x + y ≤ 12 ………… (3)
x, y ≥ 0 …………….. (4)
The feasible region determined by the system of constraints is given below
A (4, 0), B (3, 3) and C (0, 4) are the corner points
The values of Z at these corner points are given below
Corner point Z = 17.5x + 7y O (0, 0) 0 A (4, 0) 70 B (3, 3) 73.5 Maximum C (0, 4) 28
Therefore, Rs 73.50 at (3, 3) is the maximum value of Z
Hence, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs 73.50
5. A factory manufacturers two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Solution:
On each day, let the factory manufacture x screws of type A and y screws of type B.
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as given below
Screw A Screw B Availability Automatic Machine (min) 4 6 4 × 60 = 240 Hand Operated Machine (min) 6 3 4 × 60 = 240
The profit on a package of screws A is Rs 7 and on the package screws B is Rs 10
Hence, the constraints are
4x + 6y ≤ 240
6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem can be written as
Maximize Z = 7x + 10y …………. (i)
Subject to the constraints,
4x + 6y ≤ 240 …………. (ii)
6x + 3y ≤ 240 …………. (iii)
x, y ≥ 0 ……………… (iv)
The feasible region determined by the system of constraints is given below
A (40, 0), B (30, 20) and C (0, 40) are the corner points
The value of Z at these corner points are given below
Corner point Z = 7x + 10y A (40, 0) 280 B (30, 20) 410 Maximum C (0, 40) 400
The maximum value of Z is 410 at (30, 20)
Hence, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410
6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding / cutting machine and a sprayer. It takes 2 hours on grinding / cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding / cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding / cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shade that he produces, how should he schedule his daily production in order to maximize his profit?
Solution:
Let the cottage industry manufacture x pedestal lamps and y wooden shades respectively
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
Lamps Shades Availability Grinding / Cutting Machine (h) 2 1 12 Sprayer (h) 3 2 20
The profit on a lamp is Rs 5 and on the shades is Rs 3. Hence, the constraints are
2x + y ≤ 12
3x + 2y ≤ 20
Total profit, Z = 5x + 3y …………….. (i)
Subject to the constraints,
2x + y ≤ 12 …………. (ii)
3x + 2y ≤ 20 ………… (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints is given below
A (6, 0), B (4, 4) and C (0, 10) are the corner points
The value of Z at these corner points are given below
Corner point Z = 5x + 3y A (6, 0) 30 B (4, 4) 32 Maximum C (0, 10) 30
The maximum value of Z is 32 at point (4, 4)
Therefore, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.
7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?
Solution:
Let the company manufacture x souvenirs of type A and y souvenirs of type B respectively
Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
Type A Type B Availability Cutting (min) 5 8 3 × 60 + 20 = 200 Assembling (min) 10 8 4 × 60 = 240
The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Hence, the constraints are
5x + 8y ≤ 200
10x + 8y ≤ 240 i.e.,
5x + 4y ≤ 120
Total profit, Z = 5x + 6y
The mathematical formulation of the given problem can be written as
Maximize Z = 5x + 6y …………… (i)
Subject to the constraints,
5x + 8y ≤ 200 ……………. (ii)
5x + 4y ≤ 120 ………….. (iii)
x, y ≥ 0 ………….. (iv)
The feasible region determined by the system of constraints is given below
A (24, 0), B (8, 20) and C (0, 25) are the corner points
The values of Z at these corner points are given below
Corner point Z = 5x + 6y A (24, 0) 120 B (8, 20) 160 Maximum C (0, 25) 150
The maximum value of Z is 200 at (8, 20)
Hence, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 160
8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
Solution:
Let the merchant stock x desktop models and y portable models respectively.
Hence,
x ≥ 0 and y ≥ 0
Given that the cost of desktop model is Rs 25000 and of a portable model is Rs 40000.
However, the merchant can invest a maximum of Rs 70 lakhs
Hence, 25000x + 40000y ≤ 7000000
5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
Hence, x + y ≤ 250
The profit on a desktop model is 4500 and the profit on a portable model is Rs 5000
Total profit, Z = 4500x + 5000y
Therefore, the mathematical formulation of the given problem is
Maximum Z = 4500x + 5000y ………… (i)
Subject to the constraints,
5x + 8y ≤ 1400 ………… (ii)
x + y ≤ 250 ………….. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints is given below
A (250, 0), B (200, 50) and C (0, 175) are the corner points.
The values of Z at these corner points are given below
Corner point Z = 4500x + 5000y A (250, 0) 1125000 B (200, 50) 1150000 Maximum C (0, 175) 875000
The maximum value of Z is 1150000 at (200, 50)
Therefore, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs 1150000.
9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution:
Let the diet contain x units of food F1 and y units of food F2. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
Vitamin A (units) Mineral (units) Cost per unit (Rs) Food F1 (x) 3 4 4 Food F2 (y) 6 3 6 Requirement 80 100
The cost of food F1 is Rs 4 per unit and of food F2 is Rs 6 per unit
Hence, the constraints are
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem can be written as
Minimise Z = 4x + 6y …………… (i)
Subject to the constraints,
3x + 6y ≥ 80 ………… (ii)
4x + 3y ≥ 100 ………. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the constraints is given below
We can see that the feasible region is unbounded.
A (80 / 3, 0), B (24, 4 / 3), and C (0, 100 / 3) are the corner points
The values of Z at these corner points are given below
Corner point Z = 4x + 6y A (80 / 3, 0) 320 / 3 = 106.67 B (24, 4 / 3) 104 Minimum C (0, 100 / 3) 200
Here, the feasible region is unbounded, so 104 may or not be the minimum value of Z.
For this purpose, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half plane has points in common with the feasible region or not
Here, it can be seen that the feasible region has no common point with 2x + 3y < 52
Hence, the minimum cost of the mixture will be Rs 104
10. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6 / kg and F2 costs Rs 5 / kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution:
Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2. Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
Nitrogen (%) Phosphoric Acid (%) Cost (Rs / kg) F1 (x) 10 6 6 F2 (y) 5 10 5 Requirement (kg) 14 14
F1 consists of 10% nitrogen and F2 consists of 5% nitrogen.
However, the farmer requires at least 14 kg of nitrogen
So, 10% of x + 5% of y ≥ 14
x / 10 + y / 20 ≥ 14
By L.C.M we get
2x + y ≥ 280
F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid.
However, the farmer requires at least 14 kg of phosphoric acid
So, 6% of x + 10 % of y ≥ 14
6x / 100 + 10y / 100 ≥ 14
3x + 5y ≥ 700
Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem can be written as
Minimize Z = 6x + 5y ………….. (i)
Subject to the constraints,
2x + y ≥ 280 ……… (ii)
3x + 5y ≥ 700 ………. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints is given below
Here, we can see that the feasible region is unbounded.
A (700 / 3, 0), B (100, 80) and C (0, 280) are the corner points
The values of Z at these points are given below
Corner point Z = 6x + 5y A (700 / 3, 0) 1400 B (100, 80) 1000 Minimum C (0, 280) 1400
Here, the feasible region is unbounded, hence, 1000 may or may not be the minimum value of Z.
For this purpose, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 6x + 5y < 1000
Hence, 100 kg of fertilizer F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is Rs 1000
11. The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
(A) p = q
(B) p = 2q
(C) p = 3q
(D) q = 3p
Solution:
The maximum value of Z is unique
Here, it is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5)
Value of Z at (3, 4) = Value of Z at (0, 5)
p (3) + q (4) = p (0) + q (5)
3p + 4q = 5q
3p = 5q – 4q
3p = q or q = 3p
Therefore, the correct answer is option (D)
Miscellaneous Exercise page no: 607
1. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solution:
Let the diet contain x and y packets of foods P and Q respectively. Hence,
x ≥ 0 and y ≥ 0
The mathematical formulation of the given problem is given below
Maximize z = 6x + 3y ………….. (i)
Subject to the constraints,
4x + y ≥ 80 …………. (ii)
x + 5y ≥ 115 ………. (iii)
3x + 2y ≤ 150 ………… (iv)
x, y ≥ 0 …………… (v)
The feasible region determined by the system of constraints is given below
A (15, 20), B (40, 15) and C (2, 72) are the corner points of the feasible region
The values of z at these corner points are given below
Corner point z = 6x + 3y A (15, 20) 150 B (40, 15) 285 Maximum C (2, 72) 228
So, the maximum value of z is 285 at (40, 15)
Hence, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used.
The maximum amount of vitamin A in the diet is 285 units.
2. A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution:
Let the farmer mix x bags of brand P and y bags of brand Q respectively
The given information can be compiled in a table is given below
Vitamin A (units / kg) Vitamin B (units / kg) Vitamin C (units / kg) Cost (Rs / kg) Food P 3 2.5 2 250 Food Q 1.5 11.25 3 200 Requirement (units / kg) 18 45 24
The given problem can be formulated as given below
Minimize z = 250x + 200y ………… (i)
3x + 1.5y ≥ 18 ………….. (ii)
2.5x + 11.25y ≥ 45 ……….. (iii)
2x + 3y ≥ 24 ………….. (iv)
x, y ≥ 0 ………. (v)
The feasible region determined by the system of constraints is given below
A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region
The values of z at these corner points is given below
Corner point z = 250x + 200y A (18, 0) 4500 B (9, 2) 2650 C (3, 6) 1950 Minimum D (0, 12) 2400
Here, the feasible region is unbounded, hence, 1950 may or may not be the minimum value of z
For this purpose, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, and check whether the resulting half plane has points in common with the feasible region or not
Here, it can be seen that the feasible region has no common point with 5x + 4y < 39
Hence, at point (3, 6) the minimum value of z is 1950
Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950
3. A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
Solution:
Let the mixture contain x kg of food X and y kg of food Y respectively
The mathematical formulation of the given problem can be written as given below
Minimize z = 16x + 20y …….. (i)
Subject to the constraints,
x + 2y ≥ 10 …………. (ii)
x + y ≥ 6 ………… (iii)
3x + y ≥ 8 …………. (iv)
x, y ≥ 0 ………… (v)
The feasible region determined by the system of constraints is given below
A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region
The values of z at these corner points are given below
Corner point z = 16x + 20y A (10, 0) 160 B (2, 4) 112 Minimum C (1, 5) 116 D (0, 8) 160
Since the feasible region is unbounded, hence, 112 may or may not be the minimum value of z
For this purpose, we draw a graph of the inequality, 16x + 20y < 112 or 4x + 5y < 28, and check whether the resulting half plane has points in common with the feasible region or not.
Here, it can be seen that the feasible region has no common point with 4x + 5y < 28
Hence, the minimum value of z is 112 at (2, 4)
Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.
The minimum cost of the mixture is Rs 112.
4. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is give below:
Types of Toys Machines I II III A 12 18 6 B 6 0 9
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution:
Let x and y toys of type A and type B be manufactured in a day respectively.
The given problem can be formulated as given below
Maximize z = 7.5x + 5y ………….. (i)
Subject to the constraints,
2x + y ≤ 60 …………. (ii)
x ≤ 20 ……….. (iii)
2x + 3y ≤ 120 ……….. (iv)
x, y ≥ 0 ……………. (v)
The feasible region determined by the constraints is given below
A (20, 0), B (20, 20), C (15, 30) and D (0, 40) are the corner points of the feasible region.
The values of z at these corner points are given below
Corner point z = 7.5x + 5y A (20, 0) 150 B (20, 20) 250 C (15, 30) 262.5 Maximum D (0, 40) 200
262.5 at (15, 30) is the maximum value of z
Hence, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.
5. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
Solution:
Let the airline sell x tickets of executive class and y tickets of economy class respectively.
The mathematical formulation of the given problem can be written as given below
Maximize z = 1000x + 600y ………… (i)
Subject to the constraints,
x + y ≤ 200 …………….. (ii)
x ≥ 20 ………… (iii)
y – 4x ≥ 0 …………… (iv)
x, y ≥ 0 ……………(v)
The feasible region determined by the constraints is given below
A (20, 80), B (40, 160) and C (20, 180) are the corner points of the feasible region
The values of z at these corner points are given below
Corner point z = 1000x + 600y A (20, 80) 68000 B (40, 160) 136000 Maximum C (20, 180) 128000
136000 at (40, 160) is the maximum value of z
Therefore, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.
6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs) From / To A B D 6 4 E 3 2 F 2.50 3
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let godown A supply x and y quintals of grain to the shops D and E
So, (100 – x – y) will be supplied to shop F.
Since, x quintals are transported from godown A, so the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.
Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F
The given problem can be represented diagrammatically as given below
x ≥ 0, y ≥ 0, and 100 – x – y ≥ 0
Then, x ≥ 0, y ≥ 0, and x + y ≤ 100
60 – x ≥ 0, 50 – y ≥ 0, and x + y – 60 ≥ 0
Then, x ≤ 60, y ≤ 50, and x + y ≥ 60
Total transportation cost z is given by,
z = 6x + 3y + 2.5 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)
= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180
= 2.5x + 1.5y + 410
The given problem can be formulated as given below
Minimize z = 2.5x + 1.5y + 410 …………. (i)
Subject to the constraints,
x + y ≤ 100 ……….. (ii)
x ≤ 60 ……….. (iii)
y ≤ 50 ………. (iv)
x + y ≥ 60 ……… (v)
x, y ≥ 0 …………..(vi)
The feasible region determined by the system of constraints is given below
A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points
The values of z at these corner points are given below
Corner point z = 2.5x + 1.5y + 410 A (60, 0) 560 B (60, 40) 620 C (50, 50) 610 D (10, 50) 510 Minimum
The minimum value of z is 510 at (10, 50)
Hence, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals and 40 quintals respectively and from B to D, E and F is 50 quintals, 0 quintals, 0 quintals respectively
Thus, the minimum cost is Rs 510
7. An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Distance in (km) From / To A B D 7 3 E 6 4 F 3 2
Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution:
Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, (7000 – x – y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.
Similarly, (3000 – y)L and 3500 – (7000 – x – y) = (x + y – 3500) L will be transported from depot B to petrol pump E and F respectively.
The given problem can be represented diagrammatically as given below
x ≥ 0, y ≥ 0, and (7000 – x – y) ≥ 0
Then, x ≥ 0, y ≥ 0, and x + y ≤ 7000
4500 – x ≥ 0, 3000 – y ≥ 0, and x + y – 3500 ≥ 0
Then, x ≤ 4500, y ≤ 3000, and x + y ≥ 3500
Cost of transporting 10L of petrol = Rs 1
Cost of transporting 1L of petrol = Rs 1 / 10
Hence, total transportation cost is given by,
z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 – x – y) + 3 / 10 (4500 – x) + 4 / 10 (3000 – y) + 2 / 10 (x + y – 3500)
= 0.3x + 0.1y + 3950
The problem can be formulated as given below
Minimize z = 0.3x + 0.1y + 3950 ………. (i)
Subject to constraints,
x + y ≤ 7000 ………. (ii)
x ≤ 4500 ……….. (iii)
y ≤ 3000 ……. (iv)
x + y ≥ 3500 ……… (v)
x, y ≥ 0 ………… (vi)
The feasible region determined by the constraints is given below
A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.
The values of z at these corner points are given below
Corner point z = 0.3x + 0.1y + 3950 A (3500, 0) 5000 B (4500, 0) 5300 C (4500, 2500) 5550 D (4000, 3000) 5450 E (500, 3000) 4400 Minimum
The minimum value of z is 4400 at (500, 3000)
Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F respectively.
Therefore, the minimum transportation cost is Rs 4400.
8. A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table, Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric acid 1 2 Potash 3 1.5 Chlorine 1.5 2
Solution:
Let the fruit grower use x bags of brand P and y bags of brand Q respectively
The problem can be formulated as given below
Minimize z = 3x + 3.5y …………. (i)
Subject to the constraints,
x + 2y ≥ 240 ……….. (ii)
x + 0.5y ≥ 90 …….. (iii)
1.5x + 2y ≤ 310 ……….. (iv)
x, y ≥ 0 …………. (v)
The feasible region determined by the system of constraints is given below
A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region
The value of z at these corner points are given below
Corner point z = 3x + 3.5y A (140, 50) 595 B (20, 140) 550 C (40, 100) 470 Minimum
The maximum value of z is 470 at (40, 100)
Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimize the amount of nitrogen
Hence, the minimum amount of nitrogen added to the garden is 470 kg.
9. Refer to Question 8. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution:
Let the fruit grower use x bags of brand P and y bags of brand Q respectively
The problem can be formulated as given below
Maximize z = 3x + 3.5y ……….. (i)
Subject to the constraints,
x + 2y ≥ 240 ……….. (ii)
x + 0.5y ≥ 90 ……….. (iii)
1.5x + 2y ≤ 310 ……….. (iv)
x, y ≥ 0 …………. (v)
The feasible region determined by the system of constraints is given below
A (140, 50), B (20, 140) and C (40, 100) are the corner points of the feasible region
The values of z at these corner points are given below
Corner point z = 3x + 3.5y A (140, 50) 595 Maximum B (20, 140) 550 C (40, 100) 470
The maximum value of z is 595 at (140, 50)
Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximize the amount of nitrogen.
Thus, the maximum amount of nitrogen added to the garden is 595 kg.
10. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximize the profit?
Solution:
Let x and y be the number of dolls of type A and B respectively that are produced in a week.
The given problem can be formulated as given below
Maximize z = 12x + 16y ……….. (i)
Subject to the constraints,
x + y ≤ 1200 ………… (ii)
y ≤ x / 2 or x ≥ 2y ………. (iii)
x – 3y ≤ 600 …………. (iv)
x, y ≥ 0 …………… (v)
The feasible region determined by the system of constraints is given below
A (600, 0), B (1050, 150) and C (800, 400) are the corner points of the feasible region
The values of z at these corner points are given below
Corner point z = 12x + 16y A (600, 0) 7200 B (1050, 150) 15000 C (800, 400) 16000 Maximum
The maximum value of z is 16000 at (800, 400)
Hence, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of Rs 16000.
Also Access NCERT Exemplar for Class 12 Maths Chapter 12 CBSE Notes for Class 12 Maths Chapter 12
Beneath are the theoretical concepts covered in Class 12 Maths Chapter 12 Linear Programming:
12.1 Introduction
This section recollects the discussion of linear equations, linear inequalities, applications of linear inequalities in previous grades. It introduces the concept of optimisation problems and a special case of optimisation problem called linear programming problem, using an example. An ideal example of optimisation would be maximising the profit and minimising the cost of a production unit.
12.2 Linear Programming Problem and its Mathematical Formulation
This section consists of an example of a furniture dealer who is trying to maximise the profits by choosing and experimenting with different combinations of buying chairs and tables.
12.2.1 Mathematical formulation of the problem
This section further explains the formulation of the above mentioned mathematical problem. It defines the non-negative constraints, objective function, decision variables.
The optimal value of a linear function can be defined as an objective function.
A linear programming problem is finding the optimal value [maximum or minimum] of a linear function of variables, which are subjected to certain conditions and satisfying a set of linear constraints.
The variables involved in the objective function are called decision variables.
The constraints are the restrictions on the variables.
12.2.2 Graphical method of solving linear programming problems
This section comprises the definition of the feasible region, feasible solution and infeasible solution, optimal solution, bounded and unbounded region of feasible solution. It briefs about the Corner Point Method, which is used to solve linear programming problems with solved examples.
The region obtained by the constraints [including non-negative constraints] can be termed as the feasible region.
The points within the feasible region are called feasible solutions.
The points outside the feasible region are called infeasible solutions.
The point in the feasible region which gives the optimal value of the objective function is called an optimal solution.
12.3 Different Types of Linear Programming Problems
In this section, the different types of linear programming problems are discussed.
12.3.1 Manufacturing problems
These problems can be seen in the manufacturing sector in order to optimise production by maximising profits. The profits can be a function of the number of workers, working hours, materials required, the value of the product in the market, the demand for the product, the supply of the product etc.
12.3.2 Diet problems
Such problems involve the optimisation of the amount of intake of different types of foods, which are required by the body to obtain necessary nutrients. The agenda of a diet problem will be selecting those foods with the required nutrient at a lesser cost.
12.3.3 Transportation problems
These problems involve the transportation of manufactured goods efficiently to different places such that the transportation cost is minimum. For big companies, the analysis of transportation cost is very much important as it caters to a widespread area.
Exercise 12.1 Solutions 10 Questions
Exercise 12.2 Solutions 11 Questions
Miscellaneous Exercise On Chapter 12 Solutions 10 Questions
## A few points on Chapter 12 Linear Programming
The chapter Linear Programming itself makes up a whole unit that carries five marks of the total eighty marks. There are two exercises along with a miscellaneous exercise in this chapter to help the students understand the concepts related to Linear Programming thoroughly. Some of the topics discussed in Chapter 12 of NCERT Solutions for Class 12 Maths are as follows:
1. A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables (called objective function), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative.
2. A few important linear programming problems are:
(i) Diet problems
(ii) Manufacturing problems
(iii) Transportation problems
3. The common region determined by all the constraints including the non-negative constraints x ≥ 0, y ≥ 0 of a linear programming problem is called the feasible region (or solution region) for the problem.
4. Points within and on the boundary of the feasible region represent feasible solutions of the constraints.
1. Any point outside the feasible region is an infeasible solution.
2. Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.
5. If the feasible region is unbounded, then a maximum or a minimum may not exist. However, if it exists, it must occur at a corner point of R.
6. The corner point method is used in solving a linear programming problem.
7. If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maximum or minimum, then any point on the line segment joining these two points is also an optimal solution of the same type.
Studying the Linear Programming of Class 12 enables the students to understand the following:
Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints). Furthermore, it helps the student while facing other exams in the future.
## Frequently Asked Questions on NCERT Solutions for Class 12 Maths Chapter 12
### How many problems are there in each exercise of the NCERT Solutions for Class 12 Maths Chapter 12?
Chapter 12 of the NCERT Solutions for Class 12 Maths has three exercises. Exercise 12.1 has 10 questions, Exercise 12.2 has 11 questions whereas the miscellaneous exercise which covers all the topics in this chapter has 10 questions. Each and every problem is solved with utmost care to provide the students with the accurate solutions as per the CBSE guidelines.
### What do you mean by linear programming in Chapter 12 of NCERT Solutions for Class 12 Maths?
Linear programming is a concept in which the mathematical functions are subjected to constraints and are minimized or maximized accordingly. The objective functions and the constraints covered in this chapter are completely linear. For further deductions, the minimum and maximum values of the linear functions are plotted on the xy coordinate. The main objective of linear programming is to optimize the mathematical functions exposed to linear constraints.
### Where I get NCERT Solutions for Class 12 Maths Chapter 12 online?
Yes, you can download the NCERT Solutions for Class 12 Maths Chapter 12 on CoolGyan’S. The solutions to all the problems are curated by the subject experts to provide the students with a best reference material to rely on. You can download both chapter wise or exercise wise solutions and get a clear idea about the problem solving techniques which are implemented while answering the textbook problems.
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## Numerical Summaries of Data
### Learning Outcomes
• Calculate the mean, median, and mode of a set of data
• Calculate the range of a data set, and recognize it’s limitations in fully describing the behavior of a data set
• Calculate the standard deviation for a data set, and determine it’s units
• Identify the difference between population variance and sample variance
• Identify the quartiles for a data set, and the calculations used to define them
• Identify the parts of a five number summary for a set of data, and create a box plot using it
It is often desirable to use a few numbers to summarize a data set. One important aspect of a set of data is where its center is located. In this lesson, measures of central tendency are discussed first. A second aspect of a distribution is how spread out it is. In other words, how much the data in the distribution vary from one another. The second section of this lesson describes measures of variability.
## Measures of Central Tendency
### Mean, Median, and Mode
Let’s begin by trying to find the most “typical” value of a data set.
Note that we just used the word “typical” although in many cases you might think of using the word “average.” We need to be careful with the word “average” as it means different things to different people in different contexts. One of the most common uses of the word “average” is what mathematicians and statisticians call the arithmetic mean, or just plain old mean for short. “Arithmetic mean” sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word “average.”
### Mean
The mean of a set of data is the sum of the data values divided by the number of values.
### examples
Marci’s exam scores for her last math class were 79, 86, 82, and 94. What would the mean of these values would be?
The number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season are shown below.
37 33 33 32 29 28 28 23 22 22 22 21 21 21 20
20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6
What is the mean number of TD passes?
Both examples are described further in the following video.
The price of a jar of peanut butter at 5 stores was $3.29,$3.59, $3.79,$3.75, and $3.99. Find the mean price. ### examples The one hundred families in a particular neighborhood are asked their annual household income, to the nearest$5 thousand dollars. The results are summarized in a frequency table below.
Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7
What is the mean average income in this neighborhood?
Extending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand).
What is the new mean of this neighborhood’s income?
Both situations are explained further in this video.
While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a “typical” value. Imagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below. If we graph our household data, the$5 million data value is so far out to the right that the mean has to adjust up to keep things in balance.
For this reason, when working with data that have outliers – values far outside the primary grouping – it is common to use a different measure of center, the median.
### Median
The median of a set of data is the value in the middle when the data is in order.
• To find the median, begin by listing the data in order from smallest to largest, or largest to smallest.
• If the number of data values, N, is odd, then the median is the middle data value. This value can be found by rounding N/2 up to the next whole number.
• If the number of data values is even, there is no one middle value, so we find the mean of the two middle values (values N/2 and N/2 + 1)
### example
Returning to the football touchdown data, we would start by listing the data in order. Luckily, it was already in decreasing order, so we can work with it without needing to reorder it first.
37 33 33 32 29 28 28 23 22 22 22 21 21 21 20
20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6
What is the median TD value?
Find the median of these quiz scores: 5 10 8 6 4 8 2 5 7 7
The price of a jar of peanut butter at 5 stores was $3.29,$3.59, $3.79,$3.75, and $3.99. Find the median price. ### Example Let us return now to our original household income data Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7 What is the mean of this neighborhood’s household income? If we add in the new neighbor with a$5 million household income, then there will be 101 data values, and the 51st value will be the median. As we discovered in the last example, the 51st value is 35 thousand. Notice that the new neighbor did not affect the median in this case. The median is not swayed as much by outliers as the mean is. View more about the median of this neighborhood’s household incomes here. In addition to the mean and the median, there is one other common measurement of the “typical” value of a data set: the mode. ### Mode The mode is the element of the data set that occurs most frequently. The mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed. ### Example In our vehicle color survey earlier in this section, we collected the data Color Frequency Blue 3 Green 5 Red 4 White 3 Black 2 Grey 3 Which color is the mode? Mode in this example is explained by the video here. It is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once. ### Try It Reviewers were asked to rate a product on a scale of 1 to 5. Find 1. The mean rating 2. The median rating 3. The mode rating Rating Frequency 1 4 2 8 3 7 4 3 5 1 ## Measures of Variation ### Range and Standard Deviation Consider these three sets of quiz scores: Section A: 5 5 5 5 5 5 5 5 5 5 Section B: 0 0 0 0 0 10 10 10 10 10 Section C: 4 4 4 5 5 5 5 6 6 6 All three of these sets of data have a mean of 5 and median of 5, yet the sets of scores are clearly quite different. In section A, everyone had the same score; in section B half the class got no points and the other half got a perfect score, assuming this was a 10-point quiz. Section C was not as consistent as section A, but not as widely varied as section B. In addition to the mean and median, which are measures of the “typical” or “middle” value, we also need a measure of how “spread out” or varied each data set is. There are several ways to measure this “spread” of the data. The first is the simplest and is called the range. ### Range The range is the difference between the maximum value and the minimum value of the data set. ### example Using the quiz scores from above, For section A, the range is 0 since both maximum and minimum are 5 and 5 – 5 = 0 For section B, the range is 10 since 10 – 0 = 10 For section C, the range is 2 since 6 – 4 = 2 In the last example, the range seems to be revealing how spread out the data is. However, suppose we add a fourth section, Section D, with scores 0 5 5 5 5 5 5 5 5 10. This section also has a mean and median of 5. The range is 10, yet this data set is quite different than Section B. To better illuminate the differences, we’ll have to turn to more sophisticated measures of variation. The range of this example is explained in the following video. ### Standard deviation The standard deviation is a measure of variation based on measuring how far each data value deviates, or is different, from the mean. A few important characteristics: • Standard deviation is always positive. Standard deviation will be zero if all the data values are equal, and will get larger as the data spreads out. • Standard deviation has the same units as the original data. • Standard deviation, like the mean, can be highly influenced by outliers. Using the data from section D, we could compute for each data value the difference between the data value and the mean: data value deviation: data value – mean 0 0-5 = -5 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 5 5-5 = 0 10 10-5 = 5 We would like to get an idea of the “average” deviation from the mean, but if we find the average of the values in the second column the negative and positive values cancel each other out (this will always happen), so to prevent this we square every value in the second column: data value deviation: data value – mean deviation squared 0 0-5 = -5 (-5)2 = 25 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 5 5-5 = 0 02 = 0 10 10-5 = 5 (5)2 = 25 We then add the squared deviations up to get 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 25 = 50. Ordinarily we would then divide by the number of scores, n, (in this case, 10) to find the mean of the deviations. But we only do this if the data set represents a population; if the data set represents a sample (as it almost always does), we instead divide by n – 1 (in this case, 10 – 1 = 9).[1] So in our example, we would have 50/10 = 5 if section D represents a population and 50/9 = about 5.56 if section D represents a sample. These values (5 and 5.56) are called, respectively, the population variance and the sample variance for section D. Variance can be a useful statistical concept, but note that the units of variance in this instance would be points-squared since we squared all of the deviations. What are points-squared? Good question. We would rather deal with the units we started with (points in this case), so to convert back we take the square root and get: \begin{align}&\text{populationstandarddeviation}=\sqrt{\frac{50}{10}}=\sqrt{5}\approx2.2\\&\text{or}\\&\text{samplestandarddeviation}=\sqrt{\frac{50}{9}}\approx2.4\\\end{align} If we are unsure whether the data set is a sample or a population, we will usually assume it is a sample, and we will round answers to one more decimal place than the original data, as we have done above. ### To compute standard deviation 1. Find the deviation of each data from the mean. In other words, subtract the mean from the data value. 2. Square each deviation. 3. Add the squared deviations. 4. Divide by n, the number of data values, if the data represents a whole population; divide by n – 1 if the data is from a sample. 5. Compute the square root of the result. ### example Computing the standard deviation for Section B above, we first calculate that the mean is 5. Using a table can help keep track of your computations for the standard deviation: data value deviation: data value – mean deviation squared 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 0 0-5 = -5 (-5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 10 10-5 = 5 (5)2 = 25 Assuming this data represents a population, we will add the squared deviations, divide by 10, the number of data values, and compute the square root: $\sqrt{\frac{25+25+25+25+25+25+25+25+25+25}{10}}=\sqrt{\frac{250}{10}}=5$ Notice that the standard deviation of this data set is much larger than that of section D since the data in this set is more spread out. For comparison, the standard deviations of all four sections are: Section A: 5 5 5 5 5 5 5 5 5 5 Standard deviation: 0 Section B: 0 0 0 0 0 10 10 10 10 10 Standard deviation: 5 Section C: 4 4 4 5 5 5 5 6 6 6 Standard deviation: 0.8 Section D: 0 5 5 5 5 5 5 5 5 10 Standard deviation: 2.2 See the following video for more about calculating the standard deviation in this example. ### Try It The price of a jar of peanut butter at 5 stores was3.29, $3.59,$3.79, $3.75, and$3.99. Find the standard deviation of the prices.
Where standard deviation is a measure of variation based on the mean, quartiles are based on the median.
### Quartiles
Quartiles are values that divide the data in quarters.
The first quartile (Q1) is the value so that 25% of the data values are below it; the third quartile (Q3) is the value so that 75% of the data values are below it. You may have guessed that the second quartile is the same as the median, since the median is the value so that 50% of the data values are below it.
This divides the data into quarters; 25% of the data is between the minimum and Q1, 25% is between Q1 and the median, 25% is between the median and Q3, and 25% is between Q3 and the maximum value.
While quartiles are not a 1-number summary of variation like standard deviation, the quartiles are used with the median, minimum, and maximum values to form a 5 number summary of the data.
### Five number summary
The five number summary takes this form:
Minimum, Q1, Median, Q3, Maximum
To find the first quartile, we need to find the data value so that 25% of the data is below it. If n is the number of data values, we compute a locator by finding 25% of n. If this locator is a decimal value, we round up, and find the data value in that position. If the locator is a whole number, we find the mean of the data value in that position and the next data value. This is identical to the process we used to find the median, except we use 25% of the data values rather than half the data values as the locator.
### To find the first quartile, Q1
1. Begin by ordering the data from smallest to largest
2. Compute the locator: L = 0.25n
3. If L is a decimal value:
• Round up to L+
• Use the data value in the L+th position
4. If L is a whole number:
• Find the mean of the data values in the Lth and L+1th positions.
### To find the third quartile, Q3
Use the same procedure as for Q1, but with locator: L = 0.75n
Examples should help make this clearer.
### examples
Suppose we have measured 9 females, and their heights (in inches) sorted from smallest to largest are:
59 60 62 64 66 67 69 70 72
What are the first and third quartiles?
Suppose we had measured 8 females, and their heights (in inches) sorted from smallest to largest are:
59 60 62 64 66 67 69 70
What are the first and third quartiles? What is the 5 number summary?
The 5-number summary combines the first and third quartile with the minimum, median, and maximum values.
What are the 5-number summaries for each of the previous 2 examples?
More about each set of women’s heights is in the following videos.
Returning to our quiz score data: in each case, the first quartile locator is 0.25(10) = 2.5, so the first quartile will be the 3rd data value, and the third quartile will be the 8th data value. Creating the five-number summaries:
Section and data 5-number summary Section A: 5 5 5 5 5 5 5 5 5 5 5, 5, 5, 5, 5 Section B: 0 0 0 0 0 10 10 10 10 10 0, 0, 5, 10, 10 Section C: 4 4 4 5 5 5 5 6 6 6 4, 4, 5, 6, 6 Section D: 0 5 5 5 5 5 5 5 5 10 0, 5, 5, 5, 10
Of course, with a relatively small data set, finding a five-number summary is a bit silly, since the summary contains almost as many values as the original data.
A video walkthrough of this example is available below.
### Try It
The total cost of textbooks for the term was collected from 36 students. Find the 5 number summary of this data.
$140$160 $160$165 $180$220 $235$240 $250$260 $280$285
$285$285 $290$300 $300$305 $310$310 $315$315 $320$320
$330$340 $345$350 $355$360 $360$380 $395$420 $460$460
### Example
Returning to the household income data from earlier in the section, create the five-number summary.
Income (thousands of dollars) Frequency 15 6 20 8 25 11 30 17 35 19 40 20 45 12 50 7
This example is demonstrated in this video.
Note that the 5 number summary divides the data into four intervals, each of which will contain about 25% of the data. In the previous example, that means about 25% of households have income between $40 thousand and$50 thousand.
For visualizing data, there is a graphical representation of a 5-number summary called a box plot, or box and whisker graph.
### Box plot
A box plot is a graphical representation of a five-number summary.
To create a box plot, a number line is first drawn. A box is drawn from the first quartile to the third quartile, and a line is drawn through the box at the median. “Whiskers” are extended out to the minimum and maximum values.
### examples
The box plot below is based on the 9 female height data with 5 number summary:
59, 62, 66, 69, 72.
The box plot below is based on the household income data with 5 number summary:
15, 27.5, 35, 40, 50
Box plot creation is described further here.
### Try It
Create a box plot based on the textbook price data from the last Try It.
Box plots are particularly useful for comparing data from two populations.
### examples
The box plot of service times for two fast-food restaurants is shown below.
While store 2 had a slightly shorter median service time (2.1 minutes vs. 2.3 minutes), store 2 is less consistent, with a wider spread of the data.
At store 1, 75% of customers were served within 2.9 minutes, while at store 2, 75% of customers were served within 5.7 minutes.
Which store should you go to in a hurry?
The box plot below is based on the birth weights of infants with severe idiopathic respiratory distress syndrome (SIRDS)[2]. The box plot is separated to show the birth weights of infants who survived and those that did not.
Comparing the two groups, the box plot reveals that the birth weights of the infants that died appear to be, overall, smaller than the weights of infants that survived. In fact, we can see that the median birth weight of infants that survived is the same as the third quartile of the infants that died.
Similarly, we can see that the first quartile of the survivors is larger than the median weight of those that died, meaning that over 75% of the survivors had a birth weight larger than the median birth weight of those that died.
Looking at the maximum value for those that died and the third quartile of the survivors, we can see that over 25% of the survivors had birth weights higher than the heaviest infant that died.
The box plot gives us a quick, albeit informal, way to determine that birth weight is quite likely linked to survival of infants with SIRDS.
The following video analyzes the examples above.
1. The reason we do this is highly technical, but we can see how it might be useful by considering the case of a small sample from a population that contains an outlier, which would increase the average deviation: the outlier very likely won't be included in the sample, so the mean deviation of the sample would underestimate the mean deviation of the population; thus we divide by a slightly smaller number to get a slightly bigger average deviation.
2. van Vliet, P.K. and Gupta, J.M. (1973) Sodium bicarbonate in idiopathic respiratory distress syndrome. Arch. Disease in Childhood, 48, 249–255. As quoted on http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398296§ion=1.1.3
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Students can download Maths Chapter 7 Mensuration Ex 7.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.4
Question 1.
The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is ……..
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
Solution:
(c) 30 cm
Hint:
l = 15 cm, b = 20 cm, h = 25 cm
Semi-perimeter = $$\frac{a+b+c}{2}$$
= $$\frac{15+20+25}{2}$$
= 30 cm
Question 2.
If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is ………
(a) 3 cm²
(b) 6 cm²
(c) 9 cm²
(d) 12 cm²
Solution:
(b) 6 cm²
Hint:
a- 3 cm, b = 4 cm, c = 5 cm
s = $$\frac{a+b+c}{2}$$
= $$\frac{3+4+5}{2}$$
= 6 cm
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{6×3×2×1}$$
= $$\sqrt{36}$$
= 6 cm²
Question 3.
The perimeter of an equilateral triangle is 30 cm. The area is ……..
(a) 10 √3 cm²
(b) 12 √3 cm²
(c) 15 √3 cm²
(d) 25 √3 cm²
Solution:
(d) 25 √3 cm²
Hint:
Perimeter of an equilateral triangle = 30 cm
3a = 30 cm
a = $$\frac{30}{3}$$
= 10 cm
Area of an equilateral triangle = $$\frac{√3}{4}$$ a² sq.units
= $$\frac{√3}{4}$$ × 10 × 10
= 25 √3 cm²
Question 4.
The lateral surface area of a cube of side 12 cm is ……..
(a) 144 cm²
(b) 196 cm²
(c) 576 cm²
(d) 664 cm²
Solution:
(c) 576 cm²
Hint:
Side of a cube (a) = 12 cm
L.S.A. of a cube = 4a² sq.units
= 4 × 12 × 12 cm²
= 576 cm²
Question 5.
If the lateral surface area of a cube is 600 cm², then the total surface area is ………
(a) 150 cm²
(b) 400 cm²
(c) 900 cm²
(d) 1350 cm²
Solution:
(c) 900 cm²
Hint:
L.S.A. of a cube = 600 cm²
4a² = 600
a² = $$\frac{600}{4}$$
= 150
Total surface area of a cube = 6a² sq.units
= 6 × 150 cm²
= 900 cm²
Question 6.
The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is ………
(a) 280 cm²
(b) 300 cm²
(c) 360 cm²
(d) 600 cm²
Solution:
(a) 280 cm²
Hint:
T.S.A. of a cuboid = 2(lb + bh + lh) sq.units
= 2(10 × 6 + 6 × 5 + 10 × 5) cm²
= 2(60 + 30 + 50) cm²
= 2 × 140 cm²
= 280 cm²
Question 7.
If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be ………
(a) 4 : 6
(b) 4 : 9
(c) 6 : 9
(d) 16 : 36
Solution:
(b) 4 : 9
Hint:
Ratio of the surface area of cubes = 4a12 : 4a22
= a12 : a22
= 4² : 9²
= 4 : 9
Question 8.
The volume of a cuboid is 660 cm and the area of the base is 33 cm². Its height is ………
(a) 10 cm
(b) 12 cm
(c) 20 cm
(d) 22 cm
Solution:
(c) 20 cm
Hint:
Volume of a cuboid = 660 cm³
l × b × h = 660
33 × h = 660 (Area of the base = l × b)
h = $$\frac{660}{33}$$
= 20 cm
Question 9.
The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is ………
(a) 75 litres
(b) 750 litres
(c) 7500 litres
(d) 75000 litres
Solution:
(d) 75000 litres
Hint:
The capacity of a tank = l × b × h cu.units
= (10 × 5 × 1.5) m³
= 75 m³
= 75 × 1000 litres [1m³ = 1000 lit]
= 75000 litres
Question 10.
The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m x 3 m x 2 m is ………
Solution:
(a) 1000
(b) 2000
(c) 3000
(d) 5000
Solution:
(a) 1000
Hint:
Volume of one brick = 50 × 30 × 20 cm³
Volume of the wall = l × b × h
[l = 5m = 500 cm]
[b = 3m = 300 cm]
[h = 2m = 200 cm]
= 500 × 300 × 200 cm³
No. of bricks
= 10 × 10 × 10
= 1000 bricks
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# Circle A has a radius of 3 and a center at (1 ,2 ). Circle B has a radius of 5 and a center at (3 ,7 ). If circle B is translated by <2 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Jun 3, 2018
$\text{no overlap } \approx 2.82$
#### Explanation:
$\text{what we have to do here is compare the distance (d)}$
$\text{between the centres to the sum of the radii}$
• " if sum of radii">d" then circles overlap"
• " if sum of radii"< d" then no overlap"
$\text{before calculating d we require to find the new centre}$
$\text{of B under the given translation}$
$\text{under the translation } < 2 , 4 >$
$\left(3 , 7\right) \to \left(3 + 2 , 7 + 4\right) \to \left(5 , 11\right) \leftarrow \textcolor{red}{\text{new centre of B}}$
$\text{to calculate d use the "color(blue)"distance formula}$
•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
$\text{let "(x_1,y_1)=(1,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 11\right)$
$d = \sqrt{{\left(5 - 1\right)}^{2} + {\left(11 - 2\right)}^{2}} = \sqrt{16 + 81} = \sqrt{117} \approx 10.82$
$\text{sum of radii } = 3 + 5 = 8$
$\text{since sum of radii"< d" then no overlap}$
$\text{minimum distance "=d-" sum of radii}$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} = 10.82 - 8 = 2.82$
graph{((x-1)^2+(y-2)^2-9)((x-5)^2+(y-11)^2-25)=0 [-20, 20, -10, 10]}
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# 2014 AMC 10A Problems/Problem 22
## Problem
In rectangle $ABCD$, $\overline{AB}=20$ and $\overline{BC}=10$. Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$. What is $\overline{AE}$?
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$
## Solution (Trigonometry)
Note that $\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$. (If you do not know the tangent half-angle formula, it is $\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}$). Therefore, we have $DE=10\sqrt 3$. Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$
## Solution 2 (Without Trigonometry)
Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$. By the angle bisector theorem, $\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}$. Since $\triangle BFC$ is a $30-60-90$ right triangle, $\overline{CF} = \frac{10\sqrt{3}}{3}$ and $\overline{BF} = \frac{20\sqrt{3}}{3}$. Additionally, $$\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}$$Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}$ and $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$. Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}$, so $\overline{DE} = 10\sqrt{3}$. Because $\triangle ADE$ is a $30-60-90$ triangle, $\overline{AE} = \boxed{\textbf{(E)}~20}$.
## Solution 3 (Trigonometry)
By Law of Sines$$\frac{BC}{\sin 75^\circ}=\frac{EC}{\sin15^\circ}\rightarrow\frac{10}{\frac{\sqrt{2}+\sqrt{6}}{4}}=\frac{EC}{\frac{\sqrt{6}-\sqrt{2}}{4}}\rightarrow\frac{10}{EC}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\rightarrow EC=\frac{10}{2+\sqrt{3}}=20-10\sqrt{3}.$$Thus, $DE=20-(20-10\sqrt{3})=10\sqrt{3}.$
We see that $\triangle{ADE}$ is a $30-60-90$ triangle, leaving $\overline{AE}=\boxed{\textbf{(E)}~20}.$
## Solution 4 (Measuring)
If we draw rectangle $ABCD$ and whip out a protractor, we can draw a perfect $\overline{BE}$, almost perfectly $15^\circ$ degrees off of $\overline{BC}$. Then we can draw $\overline{AE}$, and use a ruler to measure it. We can clearly see that the $\overline{AE}$ is $\boxed{\textbf{(E)}~20}$.
NOTE: this method is a last resort, and is pretty risky. Answer choice $\textbf{(D)}~11\sqrt{3}$ is also very close to $\textbf{(E)}~20$, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of $\triangle ADE$, we can clearly see that it is a $30-60-90$ triangle, which verifies our answer of $\boxed{20}$.
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# Practice questions on Height balanced/AVL Tree
• Difficulty Level : Easy
• Last Updated : 03 Aug, 2022
AVL tree is binary search tree with additional property that difference between height of left sub-tree and right sub-tree of any node can’t be more than 1. Here are some key points about AVL trees:
• If there are n nodes in AVL tree, minimum height of AVL tree is floor(log2n).
• If there are n nodes in AVL tree, maximum height can’t exceed 1.44*log2n.
• If height of AVL tree is h, maximum number of nodes can be 2h+1 – 1.
• Minimum number of nodes in a tree with height h can be represented as: N(h) = N(h-1) + N(h-2) + 1 for n>2 where N(0) = 1 and N(1) = 2.
• The complexity of searching, inserting and deletion in AVL tree is O(log n).
We have discussed types of questions based on AVL trees.
Type 1: Relationship between number of nodes and height of AVL tree –
Given number of nodes, the question can be asked to find minimum and maximum height of AVL tree. Also, given the height, maximum or minimum number of nodes can be asked.
Que – 1. What is the maximum height of any AVL-tree with 7 nodes? Assume that the height of a tree with a single node is 0.
• (A)
• (B)
• (C)
• (D)
Solution: For finding maximum height, the nodes should be minimum at each level. Assuming height as 2, minimum number of nodes required: N(h) = N(h-1) + N(h-2) + 1 N(2) = N(1) + N(0) + 1 = 2 + 1 + 1 = 4. It means, height 2 is achieved using minimum 4 nodes. Assuming height as 3, minimum number of nodes required: N(h) = N(h-1) + N(h-2) + 1 N(3) = N(2) + N(1) + 1 = 4 + 2 + 1 = 7. It means, height 3 is achieved using minimum 7 nodes. Therefore, using 7 nodes, we can achieve maximum height as 3. Following is the AVL tree with 7 nodes and height 3.
Que – 2. What is the worst case possible height of AVL tree?
• (A) 2*logn
• (B) 1.44*log n
• (C) Depends upon implementation
• (D) θ(n)
Solution: The worst case possible height of AVL tree with n nodes is 1.44*logn. This can be verified using AVL tree having 7 nodes and maximum height.
Checking for option (A), 2*log7 = 5.6, however height of tree is 3.
Checking for option (B), 1.44*log7 = 4, which is near to 3.
Checking for option (D), n = 7, however height of tree is 3.
Out of these, option (B) is the best possible answer.
Type 2: Based on complexity of insertion, deletion and searching in AVL tree –
Que – 3. Which of the following is TRUE?
• (A) The cost of searching an AVL tree is θ(log n) but that of a binary search tree is O(n)
• (B) The cost of searching an AVL tree is θ(log n) but that of a complete binary tree is θ(n log n)
• (C) The cost of searching a binary search tree is O(log n ) but that of an AVL tree is θ(n)
• (D) The cost of searching an AVL tree is θ(n log n) but that of a binary search tree is O(n)
Solution: AVL tree’s time complexity of searching, insertion and deletion = O(logn). But a binary search tree, may be skewed tree, so in worst case BST searching, insertion and deletion complexity = O(n).
Que – 4. The worst case running time to search for an element in a balanced in a binary search tree with n*2^n elements is
Solution: Time taken to search an element is Θ(logn) where n is number of elements in AVL tree. As number of elements given is n*2^n, the searching complexity will be Θ(log(n*2^n)) which can be written as:
```= Θ(log(n*2^n))
= Θ(log(n)) + Θ(log(2^n))
= Θ(log(n)) + Θ(nlog(2))
= Θ(log(n)) + Θ(n)```
As logn is asymptotically smaller than n, Θ(log(n)) + Θ(n) can be written as Θ(n) which matches option C.
Type 3: Insertion and Deletion in AVL tree – The question can be asked on the resultant tree when keys are inserted or deleted from AVL tree. Appropriate rotations need to be made if balance factor is disturbed.
Que – 5. Consider the following AVL tree.
Which of the following is updated AVL tree after insertion of 70?
• (A)
• (B)
• (C)
• (D) None
Solution: The element is first inserted in the same way as BST. Therefore after insertion of 70, BST can be shown as:
However, balance factor is disturbed requiring RL rotation. To remove RL rotation, it is first converted into RR rotation as:
After removal of RR rotation, AVL tree generated is same as option (C).
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Related Articles
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Current voltage power and resistance relationship
Current, Voltage and Resistance - Humane Slaughter Association
They are Power (P) or (W), measured in Watts, Voltage (V) or (E), measured in Volts, Current or Amperage (I), measured in Amps (Amperes), and Resistance (R ). Did you know that electrical current is affected by the voltage and resistance in a circuit? In this lesson, we'll use Ohm's law, which tells us. Current, Resistance, Voltage, and Power. Current Current is a measure of the flow of electric charge . This relation can be found from the formula for power.
Ohm's Law Calculations With Power
We saw these concepts in action with the garden hose. Increasing the pressure caused the flow to increase, but getting a kink in the hose increased the resistance, which caused the flow to decrease. Using this diagram is an easy way to solve equations. The way the equation is written here, it would be easy to use Ohm's law to figure out the current if we know the voltage and the resistance.
But, what if we wanted to solve for the voltage or the resistance instead? One way to do this would be to rearrange the terms of the equation to solve for the other parameters, but there's an easier way.
The diagram above will give us the appropriate equation to solve for any unknown parameter without using any algebra.
To use this diagram, we simply cover up the parameter we're trying to find to get the proper equation. This will make more sense once we start using it, so let's do some examples.
Ohm's Law in Action Below is a simple electric circuit that we'll use to do our examples. Our voltage source is a battery that is connected to a light bulb, which provides resistance to the electric current.
To start off with, let's say our battery has a voltage of 10 volts, the light bulb has a resistance of 20 ohms, and we need to figure out the current flowing through the circuit. Using our diagram, we cover up the parameter that we're trying to find, which is current, or i, and that leaves us with the voltage, v, over the resistance, r.
In other words, to find the current, we need to divide the voltage by the resistance.
Doing the math, 10 volts divided by 20 ohms results in one half ampere of current flowing in the circuit. To find the current, divide the voltage 20 volts by the resistance 20 ohms. Next, let's increase the voltage to see what happens to the current. We'll use the same light bulb but switch to a volt battery.
The Basics of Voltage, Current and Resistance
Using the same equation as before, we divide 20 volts by 20 ohms and we get 1 amp of current. The formulas related to circuits are true for "Ohmic" materials, and "non-Ohmic" materials are not discussed in this course. The resistivity of an Ohmic conductor depends on the temperature of the material. One Ohm is equal to one Volt per Ampere, Resistance depends on temperature in the same way as resistivity, This formula requires R0, the resistance at a reference temperature T0.
Current, Voltage and Resistance
A resistor is a device that is used in electric circuits, and has a certain fixed resistance. Resistors are made by choosing a piece of material with a certain resistivity, length, and area, and wrapping it in an insulator with wires leading out of each end. In circuit diagrams, it is represented with the symbol, Voltage Voltage is a difference in electric potential between two points.
A voltage source is a device used in electric circuits that has a fixed potential difference between its ends. A voltage source can be a battery, or another source of direct current with a fixed potential difference.
Current, Resistance, Voltage, and Power
In circuit diagrams, it is represented with the symbol, If the ends of a voltage source are connected through a circuit with any number of resistors or other components, a complete circuit is formed, and current can flow from one terminal to the other.
If current is flowing, it will be the same on both terminals of the voltage source. For an ideal source, the electromotive force is equal to the voltage difference, Real sources like batteries are not ideal, and so there is some amount of internal resistance.
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# Linear Algebra
## Determinants by Cofactors
We came across the determinant of a $$2\times 2$$ matrix in a previous lesson. That determinant can be useful when to find determinants of larger square matrices. This is known as the cofactor method.
The determinant of a $$2\times 2$$ matrix $$A=\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]$$ is defined by $detA=\left|\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right|=det\left[\begin{array}{cc} a_{11} & a_{12}\\a_{21}&a_{22}\end{array}\right]=a_{11}a_{22}-a_{12}a_{21}$
Example: Find the determinant of $$\left[\begin{array}{cc} 10&-3\\4&-8\end{array}\right]$$
Solution: $$\left|\begin{array}{cc} 10&-3\\4&-8\end{array}\right|=10(-8)-4(-3)=-68$$
Cofactors of a $$3\times 3$$ matrix Let $$A\in M_{3\times 3}(\mathbb{R})$$ and let $$A(i,j)$$ denote the $$2\times 2$$ submatrix obtained from $$A$$ by deleting the i-th row and the j-th column. Define the $$(i,j)-cofactor of A to be $C_{ij}=(-1)^{i+j}|A(i,j)|$ Example: Let \(A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]$$
Cofactor $$C_{11}$$ is defined by $$C_{11}=(-1)^{1+1}|A(1,1)|$$
This means to remove row 1 and column 1 from the matrix and calculate the determinant of the remaining submatrix $$A(1,1)$$.
$A(1,1)=\left[\begin{array}{cc} -1&3\\-2&3\end{array}\right]$
The determinant $$|A(1,1)|=\left|\begin{array}{cc} -1&3\\-2&3\end{array}\right|=3$$, therefore, the cofactor $$C_{11}=+3$$
Let's calculate $$C_{12}$$, which requires us to remove row 1 and column 2, with submatrix
$A(1,2)=\left[\begin{array}{cc} 0&3\\1&3\end{array}\right]$
The determinant $$|A(1,2)|=\left|\begin{array}{cc} 0&3\\1&3\end{array}\right|=-3$$, therefore, the cofactor $$C_{12}=(-1)^{1+2}(-3)=-(-3)=3$$
Continuing with this process, the determinant $$|A(1,3)|=\left|\begin{array}{cc} 0&-1\\1&-2\end{array}\right|=1$$, therefore, the cofactor $$C_{13}=(-1)^{1+3}(-3)=+(1)=1$$.
There are 9 possible cofactors of matrix $$A$$. Can you find them all?
Determinant of $$n\times n$$ matrix using Cofactors Let $$A\in M_{n\times n}(\mathbb{R})$$ with $$n>2$$. Let $$A(i,j)$$ denote the $$(n-1)\times (n-1)$$ submatrix obtained from $$A$$ by deleting the i-th row and j-th column. The determinant of $$A\in M_{n\times n}(\mathbb{R})$$ is defined by $|A|=a_{k1}C_{k1}+a_{k2}C_{k2}+\ldots+a_{kn}C_{kn}$ where $$k$$ is any row and the $$(i,j)$$-cofactor of $$A$$ is defined to be $C_{ij}=(-1)^{i+j}|A(i,j)|$ The determinant can also be defined along any column $$l$$ $|A|=a_{1l}C_{1l}+a_{2l}C_{2l}+\ldots+a_{nl}C_{nl}$
Example: Find the determinant of $$A=\left[\begin{array}{ccc} 1&0&3\\ 0&-1&3\\1&-2&3\end{array}\right]$$
Solution
We can find the determinant using cofactors of row 1.
\begin{align} |A|&=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ &=1(3)+0(3)+3(1)\\&=6\end{align}
Tips:
• Since any row or column can be chosen. Choosing the row or column with the most 0 elements will decrease steps in calculation.
• You can perform adding or subtracting a row by another first then calculate the determinant.
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Solving Linear Programming Via Big M method
$$\text{Max } x_1 +3x_2$$ $$\text{s.t } 3x_1+x_2\leq 3$$ $$x_1-x_2\geq 2$$ $$x_1,..,x_2\geq 0$$
So we first transform to the standard form:
$$\text{Min } -x_1 -3x_2$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4= 2$$ $$x_1,..,x_4\geq 0$$
Because there is no basic feasible solution we will add and artificial variable to move to a canonical form:
$$\text{Min } -x_1 -3x_2$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$
Using Big M we will add $$mx_5$$ to the objective function:
$$\text{Min } -x_1 -3x_2+mx_5$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$
But now, it is is not in basic form as the objective function need to be written in the non basic variables, so we will add -M times equation 2 to the objective function:
$$\text{Min } (-m-1)x_1 +(m-3)x_2+mx_4=-m-1$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$
Now we can start the simplex:
moving $$x_1$$ to the basis we get:
But here all the coefficients of the objective function is positive as $$m$$ is a big positive number, how can we continue? or we can not and we still have $$x_5$$ is the basis so there is no feasible solution?
• The latter: there is no feasible solution. – LinAlg Jul 8 at 12:39
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# Complex Numbers (7 Common Questions Answered)
Complex numbers come up in algebra, calculus, and other areas of mathematics. This set of numbers often raises some questions about its nature in relation to other sets.
Complex numbers contain the set of real numbers, rational numbers, and integers. So, some complex numbers are real, rational, or integers. The conjugate of a complex number is often used to simplify fractions or factor polynomials that are irreducible in the real numbers. The modulus of a complex number gives us information about where a complex number lies in the coordinate plane.
Of course, complex numbers cannot be ordered like real numbers, since they are not on a line, but rather in an entire plane. This is because the imaginary part adds a 2nd dimension when we graph complex numbers.
Let’s get started.
## Can A Complex Number Be A Real Number?
A complex number can be a real number in some cases, when its imaginary part is zero (b = 0). The set of real numbers is a subset of the set of complex numbers.
Remember that a complex number has the form
a + bi
where a and be are real numbers (a is the real part and b is the imaginary part of the complex number).
There are two special cases for complex numbers:
• Pure Imaginary Numbers (a = 0): this gives us a number of the form 0 + bi or just bi, which is a pure imaginary number for any nonzero b. Some examples are i (a = 0, b = 1) and -2i (a = 0, b = -2).
• Real Numbers (b = 0): this gives us a number of the form a + 0i or just a, which is a real number for any a. Some examples are 7 (a = 7, b = 0) and -5 (a = -5, b = 0).
From the 2nd special case, we can see that a complex number is a real number if its imaginary part is zero (that is, b = 0). These numbers have the form a + 0i.
Alternatively, we can write any real number a as a complex number by writing it as a + 0i.
The diagram below illustrates the relationship between complex, real, and pure imaginary numbers.
## Can A Complex Number Be Rational?
A complex number can be rational if its complex part is zero (b = 0) and its real part is rational (a is rational). This is due to the fact that the set of rational numbers is a subset of the set of real numbers.
One example is the rational number ½, which can be written as ½ + 0i (a = ½, b = 0).
Another example is the rational number -13/2, which can be written as -13/2 + 0i (a = -13/2, b = 0).
Note that a complex number can have both a and b rational, but these complex numbers are not rational numbers if b is nonzero.
For example, ½ + (2/3)i has a = ½ and b = 2/3, which are both rational. However, the complex number ½ + (2/3)i itself is not rational.
The number (4/5)i has a = 0 and b = 4/5, which are both rational. However, (4/5)i is not a rational number.
The diagram below adds rational numbers to the diagram showing the relationship between complex, real, and pure imaginary numbers.
## Can A Complex Number Be An Integer?
A complex number can be an integer if its complex part is zero (b = 0) and its real part is an integer (a is an integer). This is due to the fact that the set of integers is a subset of the set of rational numbers (which is a subset of the set of real numbers).
One example is the integer 3, which can be written as 3 + 0i (a = 3, b = 0).
Another example is the integer -8, which can be written as -8 + 0i (a = -8, b = 0).
Note that a complex number can have both a and b as integers, but these complex numbers are not integers if b is nonzero.
For example, 2 + 5i has a = 2 and b = 5, which are both integers. However, the complex number 2 + 5i itself is not an integer.
The number 6i has a = 0 and b = 6, which are both integers. However, 6i is also not an integer.
The diagram below adds integers to the diagram showing the relationship between complex, real, and pure imaginary numbers.
## Can A Complex Number Be Even Or Odd?
A complex number can be even if its real part is an even integer and its imaginary part is zero (a = 2n for some integer n and b = 0).
A complex number can be odd if its real part is an odd integer and its imaginary part is zero (a = 2n + 1 for some integer n and b = 0).
Any complex number with nonzero b cannot be even or odd (since it is not an integer).
## What Is The Conjugate Of A Complex Number?
For a complex number a + bi, its conjugate (or complex conjugate) is a – bi.
In other words, find the conjugate of a complex number, keep the real part the same, but take the opposite (negative) of the imaginary part.
For example, the conjugate of 2 + 3i is 2 – 3i.
The conjugate of 8 – 6i is 8 + 6i.
The conjugate of -4 – i is -4 + i.
The conjugate of 7i is -7i (The conjugate of any pure imaginary number is the opposite (negative) of the number.)
The conjugate of 5 is 5 (The conjugate of any real number is the number itself.)
[table of sample complex numbers and conjugates]
The product of any complex number and its conjugate is a real number, since:
• (a + bi)(a – bi) [product of a complex number and its conjugate]
• a2 – abi + abi – b2i2 [FOIL]
• a2 – b2i2 [combine like terms]
• a2 + b2 [since i2 = -1, by definition of the imaginary unit i]
Since a and b are simply real numbers, then so are a2 and b2, and so is their sum, a2 + b2.
The key is that the imaginary parts cancel each other out when we combine like terms after we FOIL.
Complex conjugates are often used to convert the denominator of a complex fraction to a real number.
For example, if we have 1 / 2i, we can multiply by the complex conjugate -2i on the top and bottom of the fraction to get:
• 1 / 2i [original fraction]
• (1*-2i) / (2i*-2i) [multiply by -2i on the top and bottom]
• -2i / -4i2
• -2i / 4 [since i2 = -1, by definition of the imaginary unit i]
• -i / 2
This fraction has a real number in the denominator.
We can also use complex conjugates to factor polynomials that are irreducible over the real numbers.
For example, f(x) = x2 + 1 is irreducible in the real numbers.
However, we can factor it as f(x) = (x + i)(x – i) in the complex numbers.
Note that i and –i are complex conjugates.
## What Is The Modulus Of A Complex Number?
The modulus of a complex number is the square root of the sum of the squares of its real and imaginary parts. So, for a complex number z = a + bi, the modulus is |z| = (a2 + b2)1/2 which is a nonnegative real number.
Due to the notation |z|, which denotes the modulus of z, we also use the phrase “absolute value of z”, since |z| tells us the distance from z to the origin in the complex plane.
For example, the complex number z = 3 + 4i has a modulus of |z| = (32 + 42)1/2 = (9 + 16)1/2 = (25)1/2 = 5. This means that if we graph z = 3 + 4i in the complex plane, it is a distance of 5 units from the origin (the point 0 + 0i).
Note that the modulus of a complex number is the same as the square root of the product of the complex number and its conjugate. So, for the complex number z = a + bi:
• (a + bi)(a – bi) [product of a + bi and its complex conjugate]
• a2 – abi + abi – b2i2 [FOIL]
• a2 – b2i2 [combine like terms]
• a2 + b2 [since i2 = -1, by definition of the imaginary unit i]
• (a2 + b2)1/2 [take the square root]
• |z| [by definition of the modulus of z = a + bi]
The modulus of a complex number is similar to absolute value in that it is never negative, and it is only zero for the complex number 0 + 0i (a = 0 and b = 0).
## Can A Complex Number Be Negative?
A complex number cannot be negative – this idea of negative or positive does not make sense for complex numbers.
A real number can be negative, since we can compare its position to zero on the number line:
• If a real number is to the left of zero on the number line, it is negative.
• If a real number is to the right of zero on the number line, it is positive.
However, we cannot graph complex numbers on a line as we can the real numbers. This is because there is an extra dimension: the imaginary part of the complex number, or bi.
As a result, we must graph a complex number on a plane, rather than on a line. As such, the concept of “left or right of zero” loses its meaning in two dimensions.
So, there is no concept of “negative” or “positive” complex numbers as there is for the real numbers. Another way of saying this is that we cannot order the complex numbers in the way we can order the real numbers.
With that being said, we can still have complex numbers with a negative real part (a < 0), a negative imaginary part (b < 0), or both (a < 0 and b < 0). However, we could not say that a + bi is negative in any of these cases.
We can also take the negative (opposite) of any complex number by changing the sign of both a and b. For example, the negative (opposite) of 2 – 3i is -2 + 3i.
## Conclusion
Now you know more about complex numbers and how they relate to other number sets. You also know about the conjugate, argument, and modulus of a complex number and what they can tell us.
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# Teach Equivalent Fractions Using This Simple Paper and Crayon Activity
Page content
## Needed Materials
rectangular pieces of paper
crayons
## Description of the Activity
• Provide each student with a piece of rectangular paper. Fold the paper in half. After you have folded the paper in half, instruct students to do the same. Explain that a fraction is a part of a whole. You have divided a whole piece of paper into two equal parts.
• Instruct students to color one of the two equal parts. Ask a student to write 1/2 on the board to show that one out of the two equal parts is now shaded.
• Introduce the vocabulary words numerator and denominator. The numerator is the number of parts shaded and the denominator is the total number of equal parts. (For those students who have difficulty remembering which is the numerator and which is the denominator, try this memory association technique - In a fraction, one number is UP above the line and one is DOWN below the line. Numerator has a “u” in it and so does up; denominator begins with “d” and so does down.)
• Demonstrate and have students fold additional pieces of paper (one for each fractional amount) to represent 1/4, 3/4, 1/3, 2/3, and 1/8. Each time, a student should write the fraction on the board and identify its numerator and denominator. If you prefer, project a rectangle onto the board and divide the rectangle into the same fractions as those in the paper-folding demonstration.
• Equivalent Fractions: Ask students to fold a rectangular sheet of paper in half and color one of the two equal parts. Ask what fraction of the paper is colored. (1/2) Now have them refold the same paper and then fold it in half once again. Unfold. How many equal parts now? (4) What fraction is shaded? (2/4 or 1/2) Since the amount of shading has not changed, this means that 1/2=2/4. Tell students that 1/2 and 2/4 are two names for the same amount. Therefore, they are equivalent. Now have students refold the papers and then fold in half a third time. Unfold. What new fraction have they found that is equivalent to 1/2 and 2/4? (4/8) These three fractions (1/2, 2/4, 4/8) name the same amount.
## Assessment
• Students can demonstrate fractions and equivalent fractions using paper strips.
• Students can write the fractions for the amounts demonstrated using the paper strips.
## Integration
Fractions can be integrated with music by connecting whole, half, quarter, eighth, and sixteenth notes with the appropriate fractions.
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# Question Video: Solving Word Problems Involving Subtraction of Two Multidigit Numbers Mathematics
The number of houses built in one state was 33,726, while 47,612 houses were built in another. What is the difference between the number of houses built in the two states?
03:34
### Video Transcript
The number of houses built in one state was 33726, while 47612 houses were built in another. What is the difference between the number of houses built in the two states?
When a question asks us to find the difference between two amounts, we’re to count on. Or we can subtract to find the answer. We’ve been given two five-digit numbers that represent the number of houses built in two different states, 33726 and 47612. These numbers don’t look like they’re going to be easy to count on from one to the other. So perhaps, the best way to find the answer is to subtract one from the other. So let’s find the difference using columns subtraction.
First, we need to write one number on top of the other and make sure that we put the larger number first. The second thing to make sure is that all of our digits are lined up, so they’re in the correct columns when it comes to subtract them. What’s two ones take away six ones? But we can’t do this. Six is greater than two. But we know we can take six ones away from the number because we’ve got 47612.
What we need to do is to write the number slightly differently. We can take one ten from the tens columns — so instead of one ten, we’ve now got zero tens — and exchange it for ten ones. The whole number is still worth the same. But we’ve just juggled it around slightly. We now can subtract the ones. 12 ones take away six ones leaves us with six ones. On to the tens column. Zero tens take away two tens, but we can’t do this. So okay, we’re going to have to exchange. We’re gonna take one hundred from the hundreds column instead of six hundreds. We’ve now got a five hundreds. And we’re gonna exchange it for 10 tens.
So we now have 10 tens in the tens column. 10 tens take away two tens leaves us with eight tens. We’re going to have to exchange again in the hundreds column because seven is larger than five. So we’ll take one thousand from the thousands columns instead of seven thousands. We now have six thousands. And we’re gonna exchange it for 10 hundreds. We now have 15 hundreds. 15 hundreds take away seven hundreds leaves us with eight hundreds.
On to the thousands, six thousands take away three thousands leaves us with three thousands. And finally, four 10 thousands take away three 10 thousands leaves us with one lot of 10 thousands. We used column subtraction to find the answer. If the number of houses built in one state is 33726, while 47612 houses are built in another state, the difference between these two numbers of houses built is 13886.
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## Do you round up or down with percentages?
Level 1: For tables with percentages, the general rule is to round to one decimal.
## Does 2.5 get rounded up or down?
Both 1.5 and 2.5 are rounded to 2 . 3.5 and 4.5 are rounded to 4 .
How do you round up and down decimals?
There are certain rules to follow when rounding a decimal number. Put simply, if the last digit is less than 5, round the previous digit down. However, if it’s 5 or more than you should round the previous digit up. So, if the number you are about to round is followed by 5, 6, 7, 8, 9 round the number up.
### How do you round percentages to the nearest hundredth?
To round a number to the nearest hundredth , look at the next place value to the right (the thousandths this time). Same deal: If it’s 4 or less, just remove all the digits to the right. If it’s 5 or greater, add 1 to the digit in the hundredths place, and then remove all the digits to the right.
### Can percents have decimals?
Percents can be written in decimal form. Per-cent means per-100. So, we divide the percent by 100 to get an equivalent decimal. For example, 65% can be converted to decimal form by solving 65÷100.
Does 0.45 round up or down?
For example, 0.25 gets rounded up to 0.3, but 0.45 gets rounded down to 0.4. These are both correct, since rounding a 5 up or down is an arbitrary decision.
#### How do you round mathly?
Here’s the general rule for rounding:
1. If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40.
2. If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down. Example: 33 rounded to the nearest ten is 30.
#### Does 3.5 round up or down?
3.5 or more rounds up to 4. 3.49 or less rounds down to 3. 3.57 rounded to 1 decimal place is 3.6 .
How do you round decimals to the hundredths?
## Where is the hundredths place in a decimal?
The first digit after the decimal represents the tenths place. The next digit after the decimal represents the hundredths place.
## How do you convert a decimal into a percentage?
Multiply by 100 to convert a number from decimal to percent then add a percent sign %. Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %. The shortcut to convert from decimal to percent is to move the decimal point 2 places to the right and add a percent sign.
How do you write percent as a decimal?
A percent can always be written as a decimal, and a decimal can be written as a percent, like this: 0.85 = 85%. We can find any percent of a given number by changing the percent to a decimal and multiplying. One hundred percent of a number is just the number itself. Two hundred percent of a number is twice that number.
### What is 70 percent as a decimal?
To convert 70% to decimal, simply divide 70 by 100 as follows: 70 / 100 = 0.7. 70% means 70 per every 100. We find it useful to convert 70% to decimal, because if you need to find 70% of any number, you can simply multiply that number with 0.7.
### How do you round percents to the nearest tenth?
To change any decimal to a percent, you multiply the decimal by #100#.. In order to round to the nearest tenth, you examine the value in the hundredths place to determine whether or not the number in the tenths place will be kept or rounded up. However, there is no hundredths place in this case, so you are not able to round.
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# SAT II Math I : Finding Angles
## Example Questions
### Example Question #1 : Finding Angles
What angle do the minute and hour hands of a clock form at 4:45?
Explanation:
There are twelve numbers on a clock; from one to the next, a hand rotates . At 4:45, the minute hand is exactly on the "9" - that is, at the . The hour hand is three-fourths of the way from the "4" to the "5" - that is, on the position. Therefore, the difference is the angle they make:
### Example Question #2 : Finding Angles
What angle do the minute and hour hands of a clock form at 8:50?
Explanation:
There are twelve numbers on a clock; from one to the next, a hand rotates . At 8:50, the minute hand is exactly on the "10" - that is, on the position. The hour hand is five-sixth of the way from the "8" to the "9" - that is, on the position. Therefore, the difference is the angle they make:
.
### Example Question #3 : Finding Angles
Note: Figure NOT drawn to scale.
The above hexagon is regular. What is ?
Explanation:
Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures
.
The four angles of the quadrilateral are . Their sum is , so we can set up, and solve for in, the equation:
### Example Question #4 : Finding Angles
Can a triangle have a set of angles that are and degrees?
No
Yes
No
Explanation:
A triangle's angles must add up to degrees.
The angles given add up to 181,
.
That means that this cannot be an actual triangle.
### Example Question #5 : Finding Angles
If two angles of a triangle are and , find the measurement of the third angle.
Explanation:
Step 1: Recall the sum of the angles of a triangle...
The sum of the internal angles of a triangle is .
Step 2: To find the missing angle, subtract the given angles from ...
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# Question Video: Finding the Position of the Centre of Mass of a Uniform Equilateral Triangular Lamina Mathematics
Find the position of the center of mass of the uniform lamina π΄π΅πΆ, which is in the shape of an equilateral triangle.
03:42
### Video Transcript
Find the position of the center of mass of the uniform lamina π΄π΅πΆ, which is in the shape of an equilateral triangle.
Okay, in our sketch, we see this equilateral triangle oriented on an π₯π¦-plane. Vertex π΄ is at the origin, π΅ is up here, and πΆ is here along the π₯-axis. This shape is filled up by a uniform lamina. We can think of it as a very thin sheet of mass of material. Mass is evenly distributed throughout this lamina, and we want to solve for its center of mass. This is the location where all of this laminaβs mass is effectively concentrated.
Now, since the lamina π΄π΅πΆ is uniform, that means if we can find the geometric center of this shape, then weβll also have found its center of mass. For any triangle, whether equilateral or not, its geometric center can be found from the coordinates of its three vertices. That is, for our triangle, if we know the coordinates of vertex π΄, π΅, and πΆ, then we can use that information to solve for the geometric center of our triangle. And as we said, thatβs located at the same point as its center of mass.
Our first task then is to solve for the coordinates of these three vertices. Considering first vertex π΄, since itβs located at the origin, we know that its π₯- and π¦-coordinates are zero, zero. Then letβs consider vertex π΅. Its π₯- and π¦-coordinates are indicated here, and we can see that its π₯-coordinate will be one-half the base of our triangle, where that base is seven π. But what about the π¦-coordinate of this vertex? At this point, we can recall that this is an equilateral triangle. That means that all of the interior angles are the same. And so they must be 60 degrees.
Knowing that, we can say that this height of the triangle, what weβre really trying to solve for here, is equal to the hypotenuse of this right triangle multiplied by the sin of 60 degrees. That hypotenuse length is the length of all the sides of the triangle, seven times π. So therefore, the π¦-coordinate of vertex π΅ is seven π times the sin of 60 degrees. We can then recall that the sin of 60 degrees equals exactly the square root of three over two. So we now have the π₯- and π¦-coordinates of vertex π΅. And if we then consider those coordinates of vertex πΆ, we see that the π₯-coordinate of this point is seven times π, while the π¦-coordinate is zero.
Letβs now remind ourselves of what we noted earlier that the π₯- and π¦-coordinates of the center of mass of our triangle are equal, respectively, to the average π₯- and π¦-coordinates of our vertices. In other words, if we average out these three values, then weβll have found the π₯-coordinate of the center of mass of our triangle. And the same is true for the π¦-coordinate of our triangle center of mass. Plugging in the π₯-coordinates of our three vertices, their average will be zero plus seven π over two plus seven π all divided by three. This equals three-halves times seven π divided by three or in simplified form seven π divided by two.
Now letβs move on to calculate the π¦-coordinate of our triangle center of mass. Similarly to the π₯-coordinate, the π¦-coordinate of the center of mass equals the average π¦-value of the three vertices. This simplifies to the square root of three over six times seven π. We can now write out the coordinates of our center of mass. For this equilateral triangle, its center of mass is located at seven π over two, root three over six times seven π or seven π over two, seven root three π over six.
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# CHAPTER 1: Picturing Distributions with Graphs
## Presentation on theme: "CHAPTER 1: Picturing Distributions with Graphs"— Presentation transcript:
CHAPTER 1: Picturing Distributions with Graphs
Basic Practice of Statistics - 3rd Edition CHAPTER 1: Picturing Distributions with Graphs Chapter 5
Chapter 1 Concepts Individuals and Variables
Categorical Variables: Pie Charts and Bar Graphs Quantitative Variables: Histograms Interpreting Histograms Quantitative Variables: Stemplots Time Plots
Chapter 1 Objectives Define statistics.
Define individuals and variables. Categorize variables as categorical or quantitative. Describe the distribution of a variable. Construct and interpret pie charts and bar graphs. Construct and interpret histograms and stemplots. Construct and interpret time plots.
Quantitative Variable
Statistics Statistics is the science of data. The first step in dealing with data is to organize your thinking about the data: Categorical Variable Places individual into one of several groups or categories. Individual An object described by data Variable Characteristic of the individual Quantitative Variable Takes numerical values for which arithmetic operations make sense.
Exploratory Data Analysis
An exploratory data analysis is the process of using statistical tools and ideas to examine data in order to describe their main features. Exploring Data Begin by examining each variable by itself. Then move on to study the relationships among the variables. Begin with a graph or graphs. Then add numerical summaries of specific aspects of the data.
Distribution of a Variable
To examine a single variable, we want to graphically display its distribution. The distribution of a variable tells us what values it takes and how often it takes these values. Distributions can be displayed using a variety of graphical tools. The proper choice of graph depends on the nature of the variable. Categorical Variable Pie chart Bar graph Quantitative Variable Histogram Stemplot
Categorical Data The distribution of a categorical variable lists the categories and gives the count or percent of individuals who fall into that category. Pie Charts show the distribution of a categorical variable as a “pie” whose slices are sized by the counts or percents for the categories. Bar Graphs represent each category as a bar whose heights show the category counts or percents.
Pie Charts and Bar Graphs
US Solid Waste (2000) Material Weight (million tons) Percent of total Food scraps 25.9 11.2% Glass 12.8 5.5 % Metals 18.0 7.8 % Paper, paperboard 86.7 37.4 % Plastics 24.7 10.7 % Rubber, leather, textiles 15.8 6.8 % Wood 12.7 Yard trimmings 27.7 11.9 % Other 7.5 3.2 % Total 231.9 100.0 %
Quantitative Data The distribution of a quantitative variable tells us what values the variable takes on and how often it takes those values. Histograms show the distribution of a quantitative variable by using bars whose height represents the number of individuals who take on a value within a particular class. Stemplots separate each observation into a stem and a leaf that are then plotted to display the distribution while maintaining the original values of the variable.
Histograms For quantitative variables that take many values and/or large datasets. Divide the possible values into classes (equal widths). Count how many observations fall into each interval (may change to percents). Draw picture representing the distribution―bar heights are equivalent to the number (percent) of observations in each interval.
Histograms Example: Weight Data―Introductory Statistics Class
Stemplots (Stem-and-Leaf Plots)
For quantitative variables. Separate each observation into a stem (first part of the number) and a leaf (the remaining part of the number). Write the stems in a vertical column; draw a vertical line to the right of the stems. Write each leaf in the row to the right of its stem; order leaves if desired.
Stemplots Example: Weight Data – Introductory Statistics Class 5 2 2
Key 20|3 means 203 pounds Stems = 10’s Leaves = 1’s 11 009 14 08 16 555 19 245 20 3 21 025 22 0 23 24 25 26 0 2 2 5 Stems Leaves
Stemplots (Stem-and-Leaf Plots)
If there are very few stems (when the data cover only a very small range of values), then we may want to create more stems by splitting the original stems. Example: If all of the data values were between 150 and 179, then we may choose to use the following stems: Leaves 0-4 would go on each upper stem (first “15”), and leaves 5-9 would go on each lower stem (second “15”).
Describing Distributions
In any graph of data, look for the overall pattern and for striking deviations from that pattern. You can describe the overall pattern by its shape, center, and spread. An important kind of deviation is an outlier, an individual that falls outside the overall pattern.
Describing Distributions
A distribution is symmetric if the right and left sides of the graph are approximately mirror images of each other. A distribution is skewed to the right (right-skewed) if the right side of the graph (containing the half of the observations with larger values) is much longer than the left side. It is skewed to the left (left-skewed) if the left side of the graph is much longer than the right side. Symmetric Skewed-left Skewed-right
Basic Practice of Statistics - 3rd Edition
Time Plots A time plot shows behavior over time. Time is always on the horizontal axis, and the variable being measured is on the vertical axis. Look for an overall pattern (trend), and deviations from this trend. Connecting the data points by lines may emphasize this trend. Look for patterns that repeat at known regular intervals (seasonal variations). Chapter 5
Chapter 1 Objectives Review
Define statistics. Define individuals and variables. Categorize variables as categorical or quantitative. Describe the distribution of a variable. Construct and interpret pie charts and bar graphs. Construct and interpret histograms and stemplots. Construct and interpret time plots.
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11.4 - Fractional equations
Before reading this section you may want to review the following topics:
A fractional equation is one that contains fraction terms. In section 4.2 we saw how to solve a linear equation that contains fractions. The steps for solving any fractional equation are exactly the same:
• Look at the denominators of all the fraction terms and find their lowest common multiple (LCM) (this is also called the lowest common denominator (LCD) of the fractions).
• Multiply both sides of the equation by the LCM.
• Distribute the LCM over both sides of the equation.
• The equation no longer contains fraction terms and you can continue solving it by using the basic procedures for solving equations.
• Check the solution. This is especially important with fractional equations. There are two possible problems:
• If the denominator of any fraction term contains x, then the LCM will also contain x, and multiplying both sides of the equation by the LCM will increase the degree of x in the equation. This often leads to extraneous solutions.
• When substituting the solutions back into the original equation to check them, any solution that causes any fraction term to have a denominator of zero must be dropped because division by zero is forbidden in mathematics.
Example 1: Solve this fractional equation for x:
Solution: The fraction terms have denominators of 3, 2 and 6. The LCM of these numbers is 6. Multiply both sides of the equation by 6. (Don’t forget to put brackets around both sides of the equation.)
Distribute on both sides of the equation:
x − 3 = 6 x + 7.
The fractions are now cleared so this is no longer a fractional equation. Finish solving the equation by collecting linear terms on the left-hand-side and constant terms on the right-hand-side. This gives:
−2 x = 10.
Divide both sides by −2. This gives the solution:
x = −5.
Check it by substituting it back into the original equation. This gives −23 / 6 = −23 / 6, so the solution checks out.
Example 2: Solve this fractional equation for x:
Solution: The fraction terms have denominators of x 2 + x − 2, x + 2, and x − 1. It might appear that the LCM is just the product of all three, but because x 2 + x − 2 can be factored as (x + 2)(x − 1), the LCM is actually just (x + 2)(x − 1). Multiply both sides of the equation by it. (Don’t forget to put brackets around both sides of the equation.)
Distribute on both sides of the equation:
9 = 3 (x − 1) + 7 (x + 2).
The fractions are now cleared so this is no longer a fractional equation; it is a linear equation. Solve it using the usual techniques. Distribute once more on the right-hand-side:
9 = 10 x + 11.
Collect constant terms on the left-hand-side:
−2 = 10 x.
Divide both sides by 10. This gives the solution:
x = −1/5.
Check it by substituting it back into the original equation. This gives −25 / 6 = −25 / 6, so the solution checks out.
Example 3: The purpose of this example is to illustrate a solution that must be rejected because it causes a division by zero. The equation is identical to the one in the previous example except that it differs in the sign of one term. Solve this fractional equation for x:
Solution: Compare each step here with the corresponding step in the example above. Multiply both sides of the equation by the LCM, which again is (x + 2)(x − 1):
Distribute on both sides of the equation:
9 = −3 (x − 1) + 7 (x + 2).
Distribute once more on the right-hand-side:
9 = 4 x + 17.
This time the solution is x = −2. If we try to substitute it back into the original equation we get divisions by zero in two of the fractions. Therefore we must reject this solution and state that the equation has no solution.
Algebra Coach Exercises
If you found this page in a web search you won’t see the
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# Linear Programming?!?! Sec. 7.5. Linear Programming In management science, it is often required to maximize or minimize a linear function called an objective.
## Presentation on theme: "Linear Programming?!?! Sec. 7.5. Linear Programming In management science, it is often required to maximize or minimize a linear function called an objective."— Presentation transcript:
Linear Programming?!?! Sec. 7.5
Linear Programming In management science, it is often required to maximize or minimize a linear function called an objective function. This is a linear programming problem. In two dimensions, the objective function takes the form f = ax + by, and is used with a system of inequalities, called constraints. The solution to a linear programming problem occurs at one of the vertex points, or corner points, along the boundary of the region.
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 5x + 3y, subject to the constraints given by the system of inequalities. Start with a graph! (0,8) (0,3) (4,0) (9,0)
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 5x + 3y, subject to the constraints given by the system of inequalities. Start with a graph! Next, find the corner points! Then evaluate f at the corner points! (9,0), (0,8), (3,2) (x, y) f (9, 0) 45 (0, 8) 24 (3, 2) 21 at (3, 2)none! (unbounded region!)
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 5x + 8y, subject to the constraints given by the system of inequalities. Where’s the graph? (0,10) (0,14/3) (5,0) (7,0)
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 5x + 8y, subject to the constraints given by the system of inequalities. Where’s the graph? Corner points? Evaluation of f ? (0,0), (0,14/3), (5,0), (4,2) (x, y) f (0, 0) 0 (0, 14/3) 112/3 (5, 0) 25 (4, 2) 36 at (0, 0) at (0, 14/3)
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 3x – 2y, subject to the constraints given by the system of inequalities. Where’s the graph? (0,8) (0,1) (0,10) (2,0)(3,0) (10,0)
Linear Programming Practice Problems Find the maximum and minimum values of the objective function f = 3x – 2y, subject to the constraints given by the system of inequalities. Where’s the graph? Corner points? Evaluation of f ? (2/5,8), (7,8), (90/49,40/49), (10/3,2/3) (x, y) f (2/5, 8) –14.8 (7, 8) 5 (90/49, 40/49)(10/3, 2/3) at (2/5,8) at (10/3,2/3) 190/4926/3
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## The Ratio Word Problems Tutorial Series
This is a series of tutorials regarding ratio word problems. Ratio is defined as the relationship between two numbers where the second number is how many times the first number is contained. In this series of problems, we will learn about the different types of ratio word problems.
How to Solve Word Problems Involving Ratio Part 1 details the intuitive meaning of ratio. It uses arithmetic calculations in order to explain its meaning. After the explanation, the algebraic solution to the problem is also discussed.
How to Solve Word Problems Involving Ratio Part 2 is a continuation of the first part. In this part, the ratio of three quantities are described. Algebraic methods is used as a solution to solve the problem.
How to Solve Word Problems Involving Ratio Part 3 in this post, the ratio of two quantities are given. Then, both quantities are increased resulting to another ratio.
How to Solve Word Problems Involving Ratio Part 4 involves the difference of two numbers whose ratio is given.
If you have more math word problems involving ratio that are different from the ones mention above, feel free to comment below and let us see if we can solve them.
## How to Solve Word Problems Involving Ratio Part 4
This is the fourth and the last part of the solving problems involving ratio series. In this post, we are going to solve another ratio word problem.
Problem
The ratio of two numbers 1:3. Their difference is 36. What is the larger number?
Solution and Explanation
Let x be the smaller number and 3x be the larger number.
3x – x = 36
2x = 36
x = 18
So, the smaller number is 18 and the larger number is 3(18) = 54.
Check:
The ratio of 18:54 is 1:3? Yes, 3 times 18 equals 54.
Is their difference 36? Yes, 54 – 18 = 36.
Therefore, we are correct.
## How to Solve Word Problems Involving Ratio Part 3
In the previous two posts, we have learned how to solve word problems involving ratio with two and three quantities. In posts, we are going to learn how to solve a slightly different problem where both numbers are increased.
Problem
The ratio of two numbers is 3:5 and their sum is 48. What must be added to both numbers so that the ratio becomes 3:4?
Solution and Explanation
First, let us solve the first sentence. We need to find the two numbers whose ratio is 3:5 and whose sum is 48.
Now, let x be the number of sets of 3 and 5.
3x + 5x = 48
8x = 48
x = 6
Now, this means that the numbers are 3(6) = 18 and 5(6) = 30.
Now if the same number is added to both numbers, then the ratio becomes 3:4.
Recall that in the previous posts, we have discussed that ratio can also be represented by fraction. So, we can represent 18:30 as $\frac{18}{30}$. Now, if we add the same number to both numbers (the numerator and the denominator), we get $\frac{3}{4}$. If we let that number y, then
$\dfrac{18 + y}{30 + y} = \dfrac{3}{4}$.
Cross multiplying, we have
$4(18 + y) = 3(30 + y)$.
By the distributive property,
$72 + 4y = 90 + 3y$
$4y - 3y = 90 - 72$
$y = 18$.
So, we add 18 to both the numerator and denominator of $\frac{18}{30}$. That is,
$\dfrac{18 + 18}{30 + 18} = \dfrac{36}{48}$.
Now, to check, is $\dfrac{36}{48} = \frac{3}{4}$? Yes, it is. Divide both the numerator and the denominator by 12 to reduce the fraction to lowest terms.
## How to Solve Word Problems Involving Ratio Part 2
This is the second part of a series of post on Solving Ratio Problems. In the first part, we have learned how to solve intuitively and algebraically problems involving ratio of two quantities. In this post, we are going to learn how to solve a ratio problem involving 3 quantities.
Problem 2
The ratio of the red, green, and blue balls in a box is 2:3:1. If there are 36 balls in the box, how many green balls are there?
Solution and Explanation
From the previous, post we have already learned the algebraic solutions of problems like the one shown above. So, we can have the following:
Let $x$ be the number of grous of balls per color.
$2x + 3x + x = 36$
$6x = 36$
$x = 6$
So, there are 6 groups. Now, since we are looking for the number of green balls, we multiply x by 3.
So, there are 6 groups (3 green balls per group) = 18 green balls.
Check:
From above, $x = 6(1)$ is the number of blue balls. The expression 2x represent the number of red balls, so we have 2x = 2(6) = 12 balls. Therefore, we have 12 red balls, 18 green balls, and 6 blue balls.
We can check by adding them: 12 + 18 + 6 = 36.
This satisfies the condition above that there are 36 balls in all. Therefore, we are correct.
## How to Solve Word Problems Involving Ratio Part 1
In a dance school, 18 girls and 8 boys are enrolled. We can say that the ratio of girls to boys is 18:8 (read as 18 is to 8). Ratio can also be expressed as fraction so we can say that the ratio is 18/8. Since we can reduce fractions to lowest terms, we can also say that the ratio is 9/4 or 9:4. So, ratio can be a relationship between two quantities. It can also be ratio between two numbers like 4:3 which is the ratio of the width and height of a television screen.
Problem 1
The ratio of boys and girls in a dance club is 4:5. The total number of students is 63. How many girls and boys are there in the club?
Solution and Explanation
The ratio of boys is 4:5 means that for every 4 boys, there are 5 girls. That means that if there are 2 groups of 4 boys, there are also 2 groups of 5 girls. So by calculating them and adding, we have
4 + 5 = 9
4(2) +5(2) =18
4(3) +5(3) =27
4(4) +5(4) = 36
4(5) +5(5) = 45
4(6) +5(6) =54
4(7) +5(7) =63
As we can see, we are looking for the number of groups of 4 and, and the answer is 7 groups of each. So there are 4(7) = 28 boys and 5(7) = 35 girls.
As you can observe, the number of groups of 4 is the same as the number of groups of 5. Therefore, the question above is equivalent to finding the number of groups (of 4 and 5), whose total number of persons add up to 63.
Algebraically, if we let x be the number of groups of 4, then it is also the number of groups of 5. So, we can make the following equation.
4 x number of groups + 5 x number of groups of 5 = 63
Or
4x + 5x = 63.
Simplifying, we have
9x = 63
x = 7.
So there are 4(7) = 28 boys and 5(7) = 35 girls. As we can see, we confirmed the answer above using algebraic methods.
## How to Solve Investment Word Problems in Algebra
Investment word problems in Algebra is one of the types of problems that usually come out in the Civil Service Exam. In solving investment word problems, you should know the basic terms used. Some of these terms are principal (P) or the money invested, the rate (R) or the percent of interest, the interest (I) or the return of investment (profit), and the time or how long the money is invested. Investment is the product of the principal, the rate, and the time, and therefore, we have the formula
I = PRT.
This tutorial series discusses the different types of problems in investment and discussed the method and strategies used in solving them.
How to Solve Investment Problems Part 1 discusses the common terminology used in investment problems. It also discusses an investment problem where the principal is invested at two different interest rates.
How to Solve Investment Problems Part 2 is a discussion of another investment problem just like in part 1. In the problem, the principal is invested at two different interest rates and the interest in one investment is larger than the other.
How to Solve Investment Problems Part 3 is very similar to part 2, only that the smaller interest amount is described.
How to Solve Investment Problems Part 4 discusses an investment problem with a given interest in one investment and an unknown amount of investment at another rate to satisfy a percentage of interest for the entire investment.
## How to Solve Investment Problems Part 4
This is the fourth part of the Solving Investment Problems Series. In this part, we discuss a problem which is very similar to the third part. We discuss an investment at two different interest rates.
Problem
Mr. Garett invested a part of $20 000 at a bank at 4% yearly interest. How much does he have to invest at another bank at a 8% yearly interest so that the total interest of the money is 7%. Solution and Explanation Let x be the money invested at 8% (1) We know that the interest of 20,000 invested at 4% yearly interest is 20,000(0.04) (2) We also know that the interest of the money invested at 8% is (0.08)(x) (3) The interest of total amount of money invested is 7%. So, (20,000 + x)(0.07) Now, the interest in (1) added to the interest in (2) is equal to the interest in (3). Therefore, 20,000(0.04) + (0.08)(x) = (20,000 + x)(0.07) Simplifying, we have 800 + 0.08x = 1400 + 0.07x To eliminate the decimal point, we multiply both sides by 100. That is 80000 + 8x = 140000 + 7x 8x – 7x = 140000 – 80000 x = 60000 This means that he has to invest$60,000 at 8% interest in order for the total to be 7% of the entire investment.
Check:
$20,000 x 0.04 =$800
$60,000 x 0.08 = 4800 Adding the two interest, we have$5600. We check if this is really 7% of the total investment.
Our total investment is $80,000. Now,$80,000 x 0.07 = \$5600.
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# Bisector of the Angle which Contains the Origin
We will learn how to find the equation of the bisector of the angle which contains the origin.
Algorithm to determine whether the origin lines in the obtuse angle or acute angle between the lines
Let the equation of the two lines be a$$_{1}$$x + b$$_{1}$$y + c$$_{1}$$ = 0 and a$$_{2}$$x + b$$_{2}$$y + c$$_{2}$$ = 0.
To determine whether the origin lines in the acute angles or obtuse angle between the lines we proceed as follows:
Step I: Obtain whether the constant terms c$$_{1}$$ and c$$_{2}$$ in the equations of the two lines are positive or not. Suppose not, make them positive by multiplying both sides of the equations by negative sign.
Step II: Determine the sign of a$$_{1}$$a$$_{2}$$ + b$$_{1}$$b$$_{2}$$.
Step III: If a$$_{1}$$a$$_{2}$$ + b$$_{1}$$b$$_{2}$$ > 0, then the origin lies in the obtuse angle and the “ + “ symbol gives the bisector of the obtuse angle. If a$$_{1}$$a$$_{2}$$ + b$$_{1}$$b$$_{2}$$ < 0, then the origin lies in the acute angle and the “ Positive (+) “ symbol gives the bisector of the acute angle i.e.,
$$\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}$$ = + $$\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$$
Solved examples on the equation of the bisector of the angle which contains the origin:
1. Find the equations of the two bisectors of the angles between the straight lines 3x + 4y + 1 = 0 and 8x - 6y - 3 = 0. Which of the two bisectors bisects the angle containing the origin?
Solution:
3x + 4y + 1 = 0 ……….. (i)
8x - 6y - 3 = 0 ……….. (ii)
The equations of the two bisectors of the angles between the lines (i) and (ii)
$$\frac{3x + 4y + 1}{\sqrt{3^{2} + 4^{2}}}$$ = + $$\frac{8x - 6y - 3}{\sqrt{8^{2} + (-6)^{2}}}$$
⇒ 2 (3x + 4y + 1) = (8x - 6y - 3)
Therefore, the required two bisectors are given by,
6x + 8y + 2 = 8x+ 6y - 3 (taking +' sign)
⇒ 2x - 14y = 5
And 6x+ 8y + 2 = - 8x + 6y + 3 (taking -' sign)
⇒ 14x + 2y = 1
Since the constant terms in (i) and (ii) are of opposite signs, hence the bisector which bisects the angle containing the origin is
2 (3x + 4y + 1) = - (8x - 6y - 3)
⇒ 14x + 2y= 1.
2. For the straight lines 4x + 3y - 6 = 0 and 5x + 12y + 9 = 0 find the equation of the bisector of the angle which contains the origin.
Solution:
To find the bisector of the angle between the lines which contains the origin, we first write down the equations of the given lines in such a form that the constant terms in the equations of the lines are positive. The equations of the given lines are
4x + 3y - 6 = 0 ⇒ -4x - 3y + 6 = 0 ……………………. (i)
5x + 12y + 9 = 0 ……………………. (ii)
Now the equation of the bisector of the angle between the lines which contains the origin is the bisector corresponding to the positive symbol i.e.,
$$\frac{-4x - 3y + 6}{\sqrt{(-4)^{2} + (-3)^{2}}}$$ = + $$\frac{5x + 12y + 9}{\sqrt{5^{2} + 12^{2}}}$$
⇒ -52x – 39 y + 78 = 25x + 60y + 45
⇒ 7x + 9y – 3 = 0
Form (i) and (ii), we have a1a2 + b1b2 = -20 – 36 = -56 <0.
Therefore, the origin is situated in an acute angle region and the bisector of this angle is 7x + 9y – 3 = 0.
The Straight Line
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Home > CCA2 > Chapter Ch12 > Lesson 12.1.1 > Problem12-12
12-12.
If you remember what n! means, you can do some messy calculations quickly or compute problems that are too large for your calculator’s memory. For instance, if you wanted to calculate $\frac { 9 ! } { 6 ! }$, you could use the $n!$ button on your calculator and find that $9! = 362,880$ and $6! = 720$, so $\frac { 9 ! } { 6 ! } = \frac { 362,880 } { 720 } = 504$. You could also use a simplification technique. Since $9! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1$ and $6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1$, you can rewrite $\frac { 9 ! } { 6 ! } = \frac { 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } { 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }$ $= 9 ⋅ 8 ⋅ 7 = 504$.
Use this simplification technique to simplify each of the following problems before computing the result.
1. $\frac { 10 ! } { 8 ! }$
$90$
1. $\frac { 20 ! } { 18 ! 2 ! }$
$\frac{20\cdot19}{2\cdot1}$
1. $\frac { 7 ! } { 4 ! 3 ! }$
Refer to part (b). Remember to write factors for both $4!$ and $3!$.
1. $\frac { 75 ! } { 72 ! }$
In this case, it would be very tedious to write out all the factors.
Think of all the Giant Ones you would create if you did write it all out. What is left?
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# Related rates
• In the problem type of related rates, based on a verbal description, you have to set up an equation, which you then have to derivate implicitly to get the solution.
• Example. 3.9.2.
• Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?
• You can see that this is a related rates problem from that it's asking about the rate of change of the area of the spill.
• Since the rate of change of the radius is given in m/s, you can see that the variable with respect to which you'll have to derivate is the time, measured in seconds.
• Let's set up notation: time is $$t$$, radius is $$r$$, area is $$A$$.
• This problem type is called related rates, because the rate of change of $$r$$ is related to the rate of change of $$A$$.
• We have to find the relation between $$r$$ and $$A$$.
• It's the area formula for a circular disk: $A=r^2\pi$
• Now we have to use implicit differentiation on this, with respect to $$t$$: $A'=2rr'\pi$
• Now we can substitute the numerical values given in the problem: $A'=2\cdot30\cdot1\cdot\pi=60\pi\,\mathrm m/\mathrm s.$
## Example 2
• 3.9.12.
• A particle is moving along a hyperbola $$xy=8$$. As it reaches the point $$(4,2)$$, the $$y$$-coordinate is decreasing at a rate of 3 cm/s. How fast is the $$x$$-coordinate of the point changing at that instant?
• Here, the expression relating $$x$$ and $$y$$ is already given to us, so we'll have to derivate it implicitly.
• What you need to look out for is that the variable with respect to which we're differentiating is $$t$$, the time.
• Therefore implicit differentiation yields the following. $x'y+xy'=0$
• Substituting the known numerical values, we get the following. $x'\cdot2+4\cdot(-3)=0.$
• This yields $$x'=6$$ cm/s.
## Example 3
• 3.9.14.
• If a snowball melts so that its surface area decreases at a rate of $$1\,\mathrm{cm}^2/\mathrm{min}$$, find the rate at which the diameter decreases when the diameter is 10 cm.
• The area of a sphere is $$A=4r^2\pi$$. But we need an equation with the area $$A$$, and the diameter $$d=2r$$.
• Therefore, we want to use $A=4(d/2)^2\pi=d^2\pi.$
• Implicit differentiation gives $A'=2dd'\pi$
• Substituting the given numerical values, we get $-1=2\cdot10\cdot d'\cdot\pi$
• Therefore, we get $$d'=-\frac{1}{20\pi}$$ cm/min.
• Exercises. 3.9: 4, 6, 8, 16, 18, 22, 24, 30, 38
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# L. Riddler 6th Grade Mathematics
## Presentation on theme: "L. Riddler 6th Grade Mathematics"— Presentation transcript:
Place Value L. Riddler 6th Grade Mathematics
What is a “digit?” A digit is just one number that is a part of a number A digit can be one of these numbers: Sometimes people call their fingers “digits” “Digits” is also a slang word that means telephone number.
Let’s count the digits in these numbers:
21 375 7 6,503,211 50 F) 30.7 G) 7,180 H) 70, I) 911 J) 23, 004
Let’s add the digits in these numbers:
23 407 1,234 2,200,256 96 F) 67 G) 593 H) 1,000,000 I) 96,301 J) 81
Let’s count the number of whole number digits:
31.5 1,341 F) 54,461.0 G) 67.89 H) I) J)
Let’s count the number of decimal digits:
31.5 1,341 F) 54,461.0 G) 67.89 H) I) J)
What is “place value?” Place value is the amount assigned to a digit.
Some important place values that we will use are: millions, hundred thousands, ten thousands, thousands, hundreds, tens, ones, tenths, and hundredths.
But there are way, way, way more than that.
Undecillion Duodecillion Tredecillion Quattuordecillion Quindecillion Sexdecillion Septendecillion Octodecillion Novemdecillion Vigintillion And on and on and on… Googol Googolplex And there is even bigger numbers than that! Million Billion Trillion Quadrillion Quintillion Sextillion Septillion Octillion Nonillion Decillion
We will need to memorize the names of the important place values.
Take out a piece of unlined paper. Fold it in half horizontally. Use scissors to cut 10 flaps half way through the paper. Look at the example to label your notes.
What is the place value of the digit that is 5? 345.7 85461 34.5
Name the place value of the digit that is a zero (0). A) B) C) D) E) 34.08 F) G) 10789 H) I) J) 50 If you are having trouble (there are two or more you CANNOT do on your own), raise your hand to let your teacher know!
Activity Break Work on activities that your teacher has selected for you independently or in groups. If you need help with place value, work with the teacher in small group. Do not interrupt the teacher unless you have a really, really, really good reason during small group time.
What can I do with place value?
You can use it put numbers in order from least to greatest or greatest to least. You can use it to put numbers into word form. You can use it to put numbers into expanded form.
How do I put numbers in order?
Put the following numbers in order from least to greatest: ; ; 511; 307.7;
Step One: Find the number with the largest number of digits.
345.04 307.04 511 307.7 417.7 Both of these numbers have 5 digits!
Step Two: Make a Chart This chart has 5 columns because the largest number of digits is 5. It has 5 rows because we are comparing 5 numbers.
Step Three: Line up the Decimals
3 4 5 7 1
Step Four: Compare the Digits, Going from the Biggest (left side) to the Smallest (right side)
You always read the digits on a number in the same order you read words– from left to right. 3 4 5 7 1 Middle Smallest Biggest Next to Smallest Next to Biggest
Step Five: Write your numbers in the order that is asked for in the problem.
Least to Greatest = 307.04 307.7 345.04 417.7 511
Here are some practice problems that you can use your notes to help with:
Put these in order from least to greatest: 717 710 707 700.17 700.07 Put these in order from greatest to least: 1001 1001.5 100 100.5 5001.5
Here are some problems to do without your notes:
Put these in order from greatest to least: 5601 5061 5160 5161 5610 Put these in order least to greatest: 60.1 60.02 61.02 60.01 61 If you are having trouble (there are two or more you CANNOT do on your own), raise your hand to let your teacher know!
Activity Break Work on activities that your teacher has selected for you independently or in groups. If you need help with ordering numbers, work with the teacher in small group. Do not interrupt the teacher unless you have a really, really, really good reason during small group time.
What is “word form?” It is the way you say the number out loud with all the numbers spelled out. People write numbers in word form when they write a check.
Practice writing these numbers in word form:
561= five hundred sixty one 1,342= one thousand three hundred forty two 1,789,453= one million seven hundred eighty nine thousand four hundred fifty three 56.45= fifty six and forty five hundredths 10.6= ten and six tenths
What is “expanded form?”
It looks like an addition problem with each digit represented as its own number. Expanded form helps explain why you can carry and borrow when you add and subtract large numbers.
Practice writing these numbers in expanded form:
617= 7,846,567= 6001= 7801= 6.12=
Now without your notes, write each number in expanded form and in word form:
511 50 613 46.2 If you are having trouble (there is one or more you CANNOT do on your own), raise your hand to let your teacher know!
Activity Break Work on activities that your teacher has selected for you independently or in groups. If you need help with ordering numbers, work with the teacher in small group. Do not interrupt the teacher unless you have a really, really, really good reason during small group time.
Vocabulary Study Guide
Digit Place value Decimal Least Greatest Order Increasing Decreasing Expanded form Word form Standard form
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Question
# The value of $\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ }$is equal to:A.$\frac{2}{3}$B.$\frac{1}{2}$C.$\frac{1}{8}$D.$\frac{1}{3}$
Use $\sin a\sin b$ formula in the first pair and $\sin ({90^ \circ } - \theta )$formula In the third try and try to solve.
Consider the given expression: $\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ }$.We know the formula:
$\sin a\sin b = \frac{1}{2}[\cos (a - b) - \cos (a + b)]$, where consider,$a = {48^ \circ },b = {12^ \circ }$. Putting the values in the given expression will give us,
$(\sin {12^ \circ }\sin {48^ \circ })\sin {54^ \circ } \\ \Rightarrow \frac{1}{2}(\cos ({48^ \circ } - {12^ \circ }) - \cos ({48^ \circ } + {12^ \circ }))\sin ({90^ \circ } - {36^ \circ })\;{\text{ [Using }}\sin a\sin b{\text{ and sin(}}{90^ \circ } - \theta {\text{) formula]}} \\ \Rightarrow \frac{1}{2}(\cos {36^ \circ } - \cos {60^ \circ })\cos {36^ \circ }{\text{ [}}\cos ( - x) = \cos x{\text{ and }}\sin ({90^ \circ } - x) = \cos x{\text{]}} \\ \Rightarrow \frac{1}{2}(\frac{{\sqrt 5 + 1}}{4} - \frac{1}{2})(\frac{{\sqrt 5 + 1}}{4}){\text{ [}}\cos {36^ \circ } = \frac{{\sqrt 5 + 1}}{4}{\text{]}} \\ \Rightarrow \frac{1}{{2 \times 4 \times 4}}(\sqrt 5 + 1 - 2)(\sqrt 5 + 1) \\ \Rightarrow \frac{1}{{32}}(\sqrt 5 - 1)(\sqrt 5 + 1) \\ \Rightarrow \frac{1}{{32}}({(\sqrt 5 )^2} - {1^2}){\text{ [}}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\ \Rightarrow \frac{1}{{32}}(5 - 1) \\ \Rightarrow \frac{1}{{32}} \times 4 \\ \Rightarrow \frac{1}{8} \\$
And hence,$\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ } = \frac{1}{8}$
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# Beta Distribution
Last updated date: 28th May 2024
Total views: 429k
Views today: 4.29k
## What is Beta Distribution?
The Beta Distribution is the type of the probability distribution related to probabilities that typically models the ancestry of probabilities. The Beta curve distribution is a versatile and resourceful way of describing outcomes for the percentages or the proportions. As the Beta Distribution basically represents the probability, its domain is restricted between 0 and 1. For instance, what is the possibility of Vladmir Putin winning the next presidential election in Russia? While some might think the probability for that is 0.2, others might think it is 0.25. The Beta Distribution is a concept that provides a way of explaining this.
### The Examples of Beta Distribution
The Beta Distribution can be used for representing the different probabilities as follows.
1. The likelihood of the audience rating the new movie release.
2. The click-through rate of the website, which is the proportion of visitors.
4. What is the survival chance of a person having blood cancer.
### The Formula for the Beta Distribution
The standard formula for Beta Distribution pdf is as follows.
$f(x)=\frac{(x−a)^{p−1}(b−x)^{q−1}}{B(p,q)(b−a)^{p+q−1}}$ a≤x≤b;p,q>0
Here, p and q represent the shape parameters. ‘A’ and ‘b’ are used for representing lower and the upper bounds respectively for the distribution. B(p, q) is the beta function. The beta function has this formula:
$B(\alpha,\beta) = \int_{1}^{0}t^{(α−1)}(1−t)^{(\beta−1)}dt.$
An event where the value of a = 0, and b = 1, is known as the standard Beta Distribution. Mathematical equation or formula related to standard Beta Distribution can be described as: $F (x) = \frac{x^{p−1} (1−x)^{q−1}}{ B (p,q)}$ 0≤x≤1;p,q>0.
Generally the usual form of the distribution is described with regards to scale and location parameters. The beta is a little different in the sense the usual distribution regarding upper and lower bounds is described. However, the scale and location parameters are specified in the form of lower and the upper limits as mentioned below.
Location = a
Scale = b - a
### The Application of Beta Density Function
Beta Distribution is implemented and integrated in a wide range of applications like Bayesian hypothesis testing, task duration modelling, and Rule of Succession. The Beta Distribution is particularly the right project and planning control systems such as CPM and PERT primarily due to the fact that function is contrived by the interval with the max value of 1 and min value of 0.
### Solved Examples
There are various examples of Beta Distribution probability and solving them can help the students to understand them well and prepare for their exams. For instance you can find out about the probability of someone going out on a movie with you using the method of Beta Distribution. You can refer to the Vedantu notes for numerous solved examples and explanations behind it.
## FAQs on Beta Distribution
1. How is Beta Distribution useful for probabilities?
If we only consider probability distribution for representing the probability, then any arbitrary distribution across (0,1) would work in that order. And making one of these should be easy. You need to lay hold on the function which doesn’t boom at any point between 0 and 1 and also remains positive. Now you can integrate it from 0 to 1, and then divide the function using that outcome. You obtain the probability distribution which can be employed directly to represent the probability.
2. Why is beta distribution considered special?
The Beta Distribution is considered the conjugate before Bernoulli, binomial, geometric distributions, and negative binomial in the Bayesian hypotesizing.As the machine learning scientist, you specific is hardly ever complete and you must keep updating the model as new data flows in and this is why there is an insistence on usage of the Bayesian Inference. The computation involved in the Bayesian inference can be quite difficult or unimaginable. However, if we can execute closed-form beta formulas with conjugate prior, computation then becomes a walkover.
3. What are the common uses of beta distribution?
The Beta Distribution model is commonly used to document uncertainty regarding probability of the success of random experiments. In project management, the 3 point technique known as Beta Distribution is deployed that recognizes the uncertainty in estimation of project time. It also provides quantitative tools that are powerful combined with basic stats for computing the confidence levels of expected completion time. The Beta Distribution is also widely used in PERT for producing the bell-shaped curve.
4. What can i expect from the vedantu notes on “beta distribution”?
You can expect accurate and extremely insightful notes on “Beta Distribution” from Vedantu. These notes are drafted by the expert professionals with high proficiency in the subject. These notes cover all aspects related to the Beta Distribution including definition, function, formula, applications, logic, and solved examples. These notes are specially created keeping the students’ comprehension capabilities in mind to provide them with the best resource for learning.
5. How can i download the notes on “beta distribution” from vedantu?
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# Definition, Angles, Formulas, and Properties of a Rhombus
Are you finding difficulties in understanding the concept of a Rhombus? If yes, then you landed at the right place! The rhombus is a type of quadrilateral and usually has a diamond shape. It is one of the most commonly used Quadrilateral shapes in Mathematics and Geometry. Moreover, all the diagonals in a rhombus intersect with each other at a 90 degrees angle.
In this post, we are going to discuss and explain the definition, angles, formulas, and properties of a rhombus. Trust me; you will feel like you are having the easiest lesson of your life from a professional Math tutor while reading this post! But before going into the details of a rhombus, let us take a quick look at what is a quadrilateral!
A quadrilateral is a closed shape that has four angles enclosed with four verticals and four sides. Moreover, when you sum up all the interior angles of the quadrilateral, the answer would be equal to 360 degrees. There are further 6 types of quadrilateral:
1. Rhombus
2. Rectangle
3. Square
4. Trapezium
5. Kite
6. Parallelogram
## Definition of Rhombus
A rhombus is defined as a type of quadrilateral, and a special type of parallelogram as well. All the opposite sides in a rhombus are parallel, and all the opposite angles are equal. Also, all four sides in a rhombus have equal and same length. Additionally, all the diagonals of a rhombus bisect each other at right angles.
Moreover, a Rhombus has three additional names; since a rhombus has a diamond shape, it is also called a ‘rhombus diamond’, Lozenge, or simply a ‘diamond’. In plural form, it is called rhombuses or rhombi.
### Is Square a Rhombus?
All the sides of a rhombus are equal, right? Well, so does a square. Not only this, but all the diagonals bisect the opposite angles of the square. Also, a square has four right angles as well. Hence, a square can definitely be a type of Rhombus.
## Angles of a Rhombus
We hope that you already know a rhombus has four interior angles, right? Now let us go through some significant facts about Rhombus angles:
1. When we add up all the interior angles of a rhombus, we will get a sum of 360 degrees.
2. In a rhombus, all the angles that are opposite to each other are the same.
3. All the diagonals in a rhombus will bisect each other at the right angles.
### Rhombus Formulas
There are two basic Rhombus formulas to find out these two things:
1. Area of a rhombus
2. The perimeter of a rhombus
Here is how you can find these both things:
### Area of a Rhombus
The region that a rhombus covers in a two-dimensional plane is known as the area of a rhombus. Here is the formula to find it:
Area of a Rhombus = A = (d1 x d2)/2 square units
In this formula, d1 and d2 are known as the diagonals.
### The perimeter of a Rhombus
The total length of the boundaries of a Rhombus shape is its perimeter. In simpler words, the sum of all the four sides of a rhombus is known as its perimeter. Here is the formula to find the perimeter:
The perimeter of a Rhombus = P = 4a units
In this formula, ‘a’ is the side.
## Properties of a Rhombus
Now that we have gone through the definition, angles, and formulas, let’s move to the properties of a Rhombus! All these following properties are extremely important and you should know them by heart to fully absorb the concept of a Rhombus. Read on!
1. All the sides of a Rhombus will always be equal.
2. All the opposite sides of a Rhombus are parallel to each other.
3. All the opposite angles of a rhombus will be equal.
4. All the diagonals of a Rhombus will always bisect each other at right angles.
5. All the diagonals bisect the angles.
6. After summing up two adjacent angles, you will get a sum of 180 degrees.
7. In a Rhombus, two diagonals can make four right-angled triangles.
8. When you join a midpoint of the sides in a Rhombus, you can form a rectangle.
9. If you join the midpoints from the half of the diagonal, you can form another Rhombus!
10. You can not form any sort of circumscribing circle around a Rhombus.
11. You can not form any inscribing circle inside a Rhombus.
12. When you join the midpoints of all the four sides of a rhombus, you will form a rectangle. However, the width and length of that rectangle will be half of the diagonal. Eventually, the area of that rectangle will be half of the area of the rhombus.
## Rhombus Solved Problems and Examples
Here are a few solved problems and examples related to the area and the perimeter of a Rhombus to help you have a better understanding!
Question no. 1
The diagonal lengths of a Rhombus (d1 and d2) are 7 cm and 15 cm. What is the area of this Rhombus?
Solution:
d1 = 7 cm
d2 = 15 cm
Now, we apply the formula:
A = (d1 x d2)/2 square units
A = ( 7 x 15)/2
A = 105/2
A = 52.5 cm2
Question no. 2
If the area of a Rhombus is 90 cm2, and the length of its longest diagonal is 15 cm. What is the diagonal of this Rhombus?
Solution:
Here, area of Rhombus = 90 cm2 and supposedly d1 = 15 cm.
Now, we apply the formula:
A = (d1 x d2)/2 square units
90 = (15 x d2)/2
121 = 7.5 x d2
or 7.5 = d2
Thus, the diagonal of this Rhombus is 7.5.
Question no. 3
If all the sides of a Rhombus are 8 cm, then what will be its perimeter?
Solution:
Side of the Rhombus = 8 cm
Now, we know that all sides are equal. Hence, we apply the formula:
Perimeter = 4 x side
P = 4 x 8
P = 32 cm
Hence, the perimeter of the Rhombus is 32 cm.
### Is square a type of Rhombus?
Not a type, but yes, a square can be a rhombus.
### Can a Rhombus have 4 right angles?
No, a Rhombus can never have 4 right angles.
### In a Rhombus, are all the angles equal?
No, only the opposite angles of a rhombus are equal.
## Final Words
Now that you have gone through the entire post, we are sure your concepts about the definition, angles, formulas, and properties of a Rhombus are cleared. For more easy and detailed lessons, keep following and checking our blog or book one of our professional math tutors. You will surely find the answers to everything!
### Find Top Tutors in Your Area
With over 3 years of experience in teaching, Chloe is very deeply connected with the topics that talk about the educational and general aspects of a student's life. Her writing has been very helpful for students to gain a better understanding of their academics and personal well-being. I’m also open to any suggestions that you might have! Please reach out to me at chloedaniel402 [at] gmail.com
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How to Find Remainder of Large Powers Quickly
Find Remainders of Large Powers Quickly
Ever wondered how you can calculate remainder of large powers quickly ? Like –
252627when divided by 3 what would be result?
You would need to learn the following –
• Basic Rules
• Common Factor
• Co Primes
• Fermat Theorem
• Euler Theorem
• Wilson Theorem
Basic Rules
Rule 0
Rem[(xy)/d] = Rem[ry/d]
where r is remainder when x divided by d
Example
• Rem[(3132)/3]
• = Rem[132/3] (as remainder is 1 when 31 is divided by 3)
• = 1
or
• Rem[(2929)/3] = Rem[(-1)29/3] = -1 = +2
The above -1 can be written as 2 as -1 remained in case of 3 is nothing but + 2 remainder.
We didn’t do 2 power 29 as it would have taken additional steps and would’ve been lengthy
or
• Rem[(2930)/3] = Rem[(-1)30/3] = 1
Rule 1
Rem[(a*b*c)/d] = Rem[a/d] * Rem[b/d] * Rem[c/d]
Example
• Rem[(30*31*32)/7] = Rem[30/7] * Rem[31/7] * Rem[32/7]
• Rem[2] * Rem[3] * Rem[4]
• 3
Rule 2
Rem[(a+b+c)/d] = Rem[a/d] + Rem[b/d] + Rem[c/d]
Example
• Rem[(30+31+32)/7] = Rem[30/7] + Rem[31/7] + Rem[32/7]
• Rem[2] + Rem[3] + Rem[4]
• Rem[2+3+4] => Rem[9/7]
• 2
Applications of above rules
Rem[(3030)/7]
Solution
• Rem[(3030)/7] = Rem[(30)/7] * Rem[(30)/7] * Rem[(30)/7] * Rem[(30)/7] …….. (30 times)
• Rem[(3030)/7] = 2 * 2 * 2 ….. (30 times) = 230
• Rem[(3030)/7] = 230 = (26)5= (64)5
• Rem[(3030)/7] = Rem[64/7])5 => 15
• Rem[(3030)/7] = 1
Rem[(303132)/7]
Solution
• Rem[(303132)/7] = Rem[(23132)/7] (as Rem[30/7] = 2)
Hypothesis
1. We know, 23 will give us remainder 1 when divided by 7 (Rem[8/7] = 1)
2. Thus, for any K > 0, 23K will also give us remainder 1 when divided by 7
Futhemore,
1. 23K+1 will give us remainder 2
1. 23K+1 can be written as (23K)*2 => Rem[23K/7] * 2
2. 1* 2 = 2
Similarly,
1. 23K+2 will give us remainder 4
1. 23K+2 can be written as (23K)*2*2 => Rem[23K/7] * 2 * 2
2. 1* 2 * 2 = 4
Now, we need to write 3132 in 3K, 3K + 1, 3K + 2 format by doing follows –
• Rem[3132/3] = Rem[132/3] = 1
Thus, can be written in format 3k+1
• Rem[(303132)/7] = Rem[(23132)/7] = Rem[(23K+1)/7] = 2
Thus, the remainder will be 2
Common Factor
Rem[X/Y] = Rem[kx/ky] = k *resultOf(Rem[x/y])
Example
• Rem[(415)/28] = Rem[(4*414)/(4*7)] = 4* Rem[(414)/(7)]
• 4 * resultOf(Rem[(4*4*412)/(7)])
• 4 * resultOf(Rem[(4)/(7)] * Rem[(4)/(7)] * Rem[(412)/(7)])
• 4 * resultOf(4 *4 * Rem[(644)/(7)])
• 4 * resultOf(4 *4 * Rem[(14)/(7)])
• 4 * resultOf(Rem[(4*4*1)/7])
• 4 * resultOf(Rem[(16)/7])
• 4 * resultOf(2])
• 8
Example
• Rem[(516)/35] = Rem[(5*515)/(5*7)] = 5* Rem[(515)/(7)]
• 5 * resultOf(Rem[(1255)/(7)])
• 5 * resultOf(Rem[(65)/(7)])
• 5 * resultOf(Rem[(6*36*36)/(7)])
• 5 * resultOf(Rem[(6)/(7)] * Rem[(36)/(7)] * Rem[(36)/(7)])
• 5 * resultOf(6 *1 * 1)
• 30
Co Primes
When trying to find out the remainder, if the divisor can be broken down into smaller co-prime factors; then
Rem[M/N] = Rem[M/(a*b)]
HCF(a, b) = 1 (then only these are co-primes)
Let, Rem[M/a] = r1 & Rem[M/b] = r2
Rem[M/N] = axr2 + byr1
Such that ax + bx = 1
Example
• Rem[(715)/15]
• Rem[(715)/(3*5)]
a = 3 & b = 5
• Rem[(715)/(3)] = 1 &
• Rem[(715)/(5)] = Rem[(215)/(5)] = 3
(Trick) How is Rem[(215)/(5)] = 3 ?
We can observe a pattern for remainders when divided by 5.
• (21)/5 , Rem is 2
• (22)/5 , Rem is 4
• (23)/5 , Rem is 3
• (24)/5 , Rem is 1
• (25)/5 , Rem is 2
• (26)/5 , Rem is 4
• (27)/5 , Rem is 3
• (28)/5 , Rem is 1
So remainder for (215)/5 is 3.
So we now have,
r2 = 3, r1= 1
• axr2 + byr1 = 3*x*3 + 5*y*1
Such that, ax + by = 1 | 3x + 5y = 1
Valid values are x = -3 and y = 2
Thus final answer will be: 3*(-3)*3 + 5*2*1 = – 27 + 10 = 17
Fermat Theorem
If p is a prime, and HCF (a, p) = 1 (a and p are co-primes), then Rem[ap-1/p] = 1
Example
• Rem[2345/11]
• Rem[((210)34(2))/11]
• Rem[((210)34)/11] * Rem[(2)5)/11]
• p = 11 and p – 1 = 10 so using fermats theorem
• Rem[((1)34)/11] * Rem[(2)5)/11]
• Rem[((1)34)/11] * Rem[(2)5)/11]
• Rem[1/11] * Rem[32)/11]
• 1 * 10
• 10
Euler’s Remainder Theorem
For a number of the form Px/Q , where P & Q are co-primes, then Rem[Pϕ(Q)/Q] =1, where
ϕ(Q) is called the Euler’s Number.
Let us first learn How to find Euler Number ϕ(Q) ?
ϕ(Q) = Q (1 – 1/a) (1 – 1/b) (1 – 1/c)……………. ,
where Q = { al x bm x cn } and a, b & c are prime factors of Q.
Example : Euler’s Number for 36, i.e ϕ(36)
• 33 can be prime factorised as { 22 x 32 } which means a = 2 and b = 3
• ϕ(36) = 36 (1 – 1/a) (1 – 1/b) (Using Formula)
• ϕ(36) = 36 (1 – 1/2) (1 – 1/3) , since a = 2 and b = 3
• ϕ(36) = 12
Example 1 Using Euler’s Theorem
Remainer of 267/33
• Here P = 2 , Q = 33 and x = 67
• P and Q are co-prime i.e 2 and 33 are co-prime to each other
• Q = 33 can be prime factorised as { 111 x 31 } which means a = 11 and b = 3
• ϕ(Q) = ϕ(33) = 33 (1 – 1/11) (1 – 1/3) = 20
• So ϕ(Q) = 20
• Now Divide x by ϕ(Q) and find the remainder ‘y’
• y = Rem[x / ϕ(Q)]
• y = Rem[67 / 20] = 7
• Now find Rem[Py/Q]
• i.e Rem[27/33] = 29
• So, Rem[267/33] = 29
Example 2 Using Euler’s Theorem
Remainder of 353/63
• Here P = 3 , Q = 63 and x = 53
• P and Q are not co-prime i.e 3 and 63 are not co-prime to each other
• 353/63 = { 32 x 351 } / { 32 x 7 }
• 353/63 = { 351 } / { 7 }
• New Values P=3 and Q=7 are co-prime to each other. And New value of x = 51
• ϕ(7) = 6
• Now Divide x by ϕ(Q) and find the remainder ‘y’
• y = Rem[51 / 6] = 3
• Now find Rem[Py/Q]
• i.e Rem[33/7] = 6
• So, Rem[353/63] = 6 * 9 = 54 (as we eliminated 9 as common factor initially )
Wilson’s Theorem
• When (P-1)! is divided by P, the remainder is (P-1), where P must be a prime number.
• When (P-2)! is divided by P, the remainder is 1, where P must be a prime number.
Example 1
Find the remainder when 40! is divided by 41?
• Rem[40!/41]
• Rem[ (41-1)! / 41]
• Here P = 41 is a prime number.
• So Rem[ (41-1)! / 41] = (41 – 1) = 40
Example 2
Find the remainder when 45! is divided by 47?
• Rem[45!/47]
• Rem[ (47-2)! / 47]
• Here P = 47 is a prime number.
• So Rem[ (47-2)! / 47] = 1
Example 3
Find the remainder when 21! is divided by 361?
• Rem[21!/361]
• Rem[ (21*20*19*18!) / (19 * 19)]
• Rem[ (21*20*18!) / 19] , removing common terms
• Rem[(21*20*18!)/19] = Rem[21/19] * Rem[20/19] * Rem[18!/19]
• Rem[2] * Rem[1] * Rem[18!/19]
• Now, P = 19 is a prime number.
• Rem[2] * Rem[1] * Rem[18!/19]
• Rem[2] * Rem[1] * Rem[(19-1)!/19]
• Rem[2] * Rem[1] * Rem[19-1], by wilson’s theorem
• Rem[(2*1*18)/19]
• Rem[36/19]
• 17
• Now we removed common term 19 initially
• So final answer would be 17*19 = 323
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# Solving the Mystery of the Cube Root of 125
As children, we are often introduced to the concept of mathematical roots. We learn how to find the square root and cube root of numbers, but these are often limited to small numbers. However, as we progress in our mathematical journey, we begin to encounter more complex problems that demand a deeper understanding of the concept of roots. One such problem is finding the cube root of 125.
At first glance, the task seems straightforward. To find the cube root of 125, we need to find a number that, when multiplied by itself three times, gives us 125. However, as we begin to think about this problem more deeply, we realize that it is not as simple as it seems.
To understand why finding the cube root of 125 is not trivial, we need to delve into the properties of roots. When we take the square root of a number, we are essentially trying to find a number that, when multiplied by itself, gives us the original number. The same logic applies to cube roots. The cube root of a number is the number that, when multiplied by itself three times, gives us the original number.
However, there is one important difference between square roots and cube roots. When we take the square root of a positive number, we get two answers – one is positive and the other is negative. For example, the square root of 4 is 2, but -2 is also a valid answer because (-2)^2 = 4. However, when we take the cube root of a positive number, we only get one answer. This means that if we want to find the cube root of 125, we need to find a single number that satisfies the equation x^3 = 125.
So, how do we go about finding this number? One approach is to use trial and error. We can start with a small number, say 2, and see if 2^3 = 125. If not, we move on to the next number and repeat the process until we find the right answer. However, this approach is not practical for larger numbers because it would take us a long time to try every possible number.
Fortunately, there is a more efficient way to find the cube root of 125. This involves using a mathematical formula called the cube root formula. The formula states that the cube root of any number can be expressed as the product of the number’s prime factors, each raised to the power of 1/3. Let’s apply this formula to 125.
The prime factorization of 125 is 5 x 5 x 5. Using the cube root formula, we can write the cube root of 125 as (5 x 5 x 5)^(1/3) = 5^(3/3) = 5. Therefore, the cube root of 125 is 5.
The cube root formula works because of the fundamental theorem of arithmetic, which states that every positive integer can be uniquely represented as a product of prime numbers. This means that every number has a prime factorization that is unique to it. By using the cube root formula, we can extract the cube root of any number by taking the cube root of each of its prime factors and multiplying them together.
In conclusion, finding the cube root of 125 may seem like a daunting task at first, but with the right approach, it can be done quickly and efficiently. By understanding the properties of roots and using the cube root formula, we can find the cube root of any number with ease. So, the next time you encounter a cube root problem, don’t let it intimidate you – just remember the cube root formula and you’ll be able to solve it in no time!
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RBSE Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2
RBSE Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2 is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2.
Board RBSE Class Class 9 Subject Maths Chapter Chapter 2 Chapter Name Number System Exercise Ex 2.2 Number of Questions Solved 3 Category RBSE Solutions
RBSE Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2
Question 1.
Classify the following numbers (RBSESolutions.com) as rational or irrational.
Solution:
(i) 2 – √5 is an irrational number.
(ii) (3 + √23) – √23 = 3 + √23 – √23 = 3
Hence, it is (RBSESolutions.com) a rational number.
$$\frac { 2\sqrt { 11 } }{ 7\sqrt { 11 } } =\frac { 2 }{ 7 }$$
Hence, it a rational number.
(iv) $$\frac { 1 }{ \sqrt { 3 } }$$ is an irrational number.
(v) 2π is an irrational number.
Question 2.
Rationalist the (RBSESolutions.com) denominator of the following:
Solution:
Question 3.
$$\frac { 3+2\sqrt { 2 } }{ 3-\sqrt { 2 } } =a+b\sqrt { 2 }$$, where a and b are rational, then (RBSESolutions.com) find the values of a and b.
Solution:
$$\frac { 3+2\sqrt { 2 } }{ 3-\sqrt { 2 } } =a+b\sqrt { 2 }$$
Rationalizing the denominator of L.H.S. by multiplying and dividing by 3 + √2
We hope the given RBSE Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 2 Number System Ex 2.2, drop a comment below and we will get back to you at the earliest.
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# Golden Triangle
The Golden Triangle, often known as the sublime triangle, is an isosceles triangle. The ratio of the side ‘a’ to base ‘b’ is equal to the golden ratio, $$\frac{a}{b} = \varphi$$.
Considering the above figure, the vertex angle will be:.
$$\theta = 2sin^{-1}(\frac{b}{2a}) = 2sin^{-1}(\frac{1}{2\varphi}) = \frac{1}{5}\pi = 36^{\circ}$$
And then the height(h) to base(b) of the traingle will be related as,
$$4h^2 = b^2 (5+ 2 \sqrt{5})$$
Golden Ratio
Two quantities are in the golden ratio, if the ratio of the quantities is same as the ratio of their sum to the larger of the two quantities. Algebraically expressed, for quantities ‘a’ and ‘b’ with a > b > 0,
$$\frac {a+b}{a} = \frac {a}{b} = \varphi$$
Where, Greek letter phi($$\varphi$$ or $$\phi$$) represents the golden ratio. It is an irrational number that is a solution to the quadratic equation $$x^2 − x − 1 = 0$$ , with a value of,
$$\varphi = \frac {1+{\sqrt {5}}}{2}=1.6180339887\ldots$$.
Because of its frequent appearance in geometry, Ancient Greek mathematicians first studied what we now call the golden ratio.
## Applications of Golden Triangle
Some points of a logarithmic spiral are formed using golden triangles.
This triangle is used in determining the dimensions of the layout in Architecture.
The golden triangle is classically used in paintings and photography to visual presentation for paintings and photographs, especially those that have elements with diagonal lines.
## FAQs
How is the golden triangle related to the golden ratio?
Golden triangle is such that the ratio of the hypotenuse ‘a’ to base ‘b’ is equal to the golden ratio, $$\frac{a}{b} = \varphi$$.
How do you make a golden triangle?
The Golden Triangle is an isosceles triangle with a vertex angle of 36° and base angles of 72°. When a base angle is bisected, the angle bisector divides the opposite side in a golden ratio and forms two smaller isosceles triangles.
Know about more interesting triangles herre
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Worksheet of Algebraic Expressions - CREST Olympiads
# Worksheet of Algebraic Expressions
## Algebraic Expressions - Sub Topics
• What are Algebraic Expressions?
• Types of Algebraic Expressions
• Algebraic Expressions
• Algebraic Identities
• How to Multiply Algebraic Expressions?
• ## What are Algebraic Expressions?
An algebraic expression is a combination of constants and variables connected by four fundamental operations such as addition, subtraction, multiplication and division. Algebraic expressions are used to represent real-life situations concisely and clearly, allowing us to analyze and predict outcomes.
Example: 2x + 3 is an algebraic expression,
Where;
x is a variable
2 and 3 are constants
These expressions can be evaluated by substituting a specific value for the variable and performing the operations in the correct order. Algebraic expressions are often used in solving equations, graphing functions and simplifying expressions.
The addition sign (+) indicates that the two values should be added together. If we were to substitute a specific value for x (such as x = 5), we could simplify the expression to
2x + 3
= 2(5) + 3
= 10 + 3
= 13
## Types of Algebraic Expressions
### Monomial
Algebraic expression with a single term.
For example: 5x, -7xy, 7/5
### Binomial
Algebraic expression with two terms.
For example: 7x - 3y, 6x2 - 4xy
### Trinomial
Algebraic expression with three terms.
For example: -6x + 7y -4xy, 5p - 106p2q + 2
### Polynomial
An algebraic expression with one or more terms, each with a non-zero coefficient and variables with only positive exponents.
For example: 3x2 + 2x + 1 or 4y3 - 2y2 + y + 6
## Algebraic Expressions
Variables: In algebraic expressions, variables are used to represent values that are not yet known or can vary based on the context of the problem. These variables are typically represented by letters such as x, y or z.
For example, in the expression 7x - 3, x is the variable and its value can change depending on the context.
Constants: Constants, on the other hand, are values that do not change and are represented by numbers that have a specific value.
For example, in the expression 7x - 3 the number 3 is constant. They have specific values and do not change.
Terms: Terms are the parts of an algebraic expression that are separated by addition or subtraction.
For example, in the expression 7x - 3, the term 7x represents the product of 7 and x and the term 3 represents the constant value.
Like and Unlike Terms: Terms having the same variables with the same exponents are called like terms, otherwise unlike terms.
Coefficient: The coefficient of an algebraic expression is the numerical value that is multiplied by the variable.
For example, in the expression 7x - 3, the coefficient of the variable x is 7. This means that x is multiplied by 7 in the term 7x.
Degree: Highest power of the variable in that expression is known as a degree.
For example, in the expression 7x3 + 2x2 - 1, the degree is 3, because that is the highest power of x that appears in the expression.
All of the above are important concepts in algebraic expressions. Variables represent unknown values, constants represent specific values, terms are the parts of an expression separated by addition or subtraction and the coefficient is the numerical value that is multiplied by the variable. Understanding these concepts is crucial for solving and manipulating algebraic expressions.
## Algebraic Identities
Algebraic identities are mathematical statements that are always true for specific values of the variables involved.
Algebraic expressions serve the purpose of simplifying and manipulating equations, and they have a crucial role in solving equations and dealing with problems in algebra and other mathematical disciplines. Additionally, they are used in the factorization of polynomials. To find algebraic expressions and solve various polynomials, algebraic identities are used.
### Standard Algebraic Identities
The standard Algebraic Identities are formed from the Binomial Theorem.
Some Standard Algebraic Identities list are listed below:
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b) (a – b)
Identity IV: (x + a) (x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 = (a + b) (a2 – ab + b2)
Identity IX: a3 - b3 = (a - b) (a2 + ab + b2)
Identity X: a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
## How to Multiply Algebraic Expressions?
### Multiplication of Monomials
To multiply the monomials put the constants together and multiply the like variables together.
Therefore, 3z2 x 5z3y2 = (3 x 5) (z2 x z3y2)
= 15 z5y2
### Multiplication of Binomials
To multiply the binomials (x + 2) and (x - 3) first take term of (x + 2) as x with (x - 3) than 2 with (x - 3) and add them.
Therefore, (x + 2) (x - 3) = x (x - 3) + 2(x - 3)
= x2 - 3x + 2x - 6
= x2 - x – 6
NOTE: It's important to remember the order of operations (PEMDAS) when multiplying algebraic expressions.
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## Allen/Cone
Home | Routine of the Month | Homework | Spelling & Vocabulary | Science | Math | Social Science | Reading | Classroom Forms | Links | Scholastic Books | Mrs. Cone's Class Schedule | Mrs. Allen's Class Schedule
## Math
To Access the Math Book Online:
Click on the "Links" icon. You will see "Math Book Online."
All 5th graders are invited every Thursday to Math Lab. The students bring their lunches and stay through lunch recess.
Divisibility
Rules
A number is
divisible by 2 if it is an even number.
A number is
divisible by 3 if the sum of its digits is a
multiple of 3.
A number is
divisible by 5 if it ends in a 5 or a 0.
A number is
divisible by 9 if the sum of its digits is a
multiple of 9.
A number is
divisible by 10 if it ends in a 0.
Prime Factorization
Prime factorization
Prime factorization is finding the factors of a number that are all prime. Here's how you do it: Find 2 factors of your number. Then look at your 2 factors and determine if one or both of them is not prime. If it is not a prime factor it. Repeat this process until all your factors are prime. Here's an example:
Find the prime factors of the number 84:
```
84
/ \
42 x 2 (84 is 42 times 2)
/ \
21 x 2 (42 is 21 times 2)
/ \
7 x 3 (21 is 7 times 3)
(7 and 3 are both prime, so we stop!)
```
So the prime factors of 84 are 7 x 3 x 2 x 2.
## Decimals
To understand decimal numbers you must first know about Place Value.
When we write numbers, the position (or "place") of each number is important.
In the number 327:
• the "7" is in the Ones position, meaning just 7 (or 7 "1"s),
• the "2" is in the Tens position meaning 2 tens (or twenty),
• and the "3" is in the Hundreds position, meaning 3 hundreds.
"Three Hundred Twenty Seven"
As we move left, each position is 10 times bigger! From Units, to Tens, to Hundreds
... and ...
As we move right, each position is 10 times smaller. From Hundreds, to Tens, to Units
But what if we continue past Units?What is 10 times smaller than Units?1/10 ths (Tenths) are!
But we must first write a decimal point,so we know exactly where the Units position is: "three hundred twenty seven and four tenths"but we usually just say "three hundred twenty seven point four"
## Decimal Point
The decimal point is the most important part of a Decimal Number. It is exactly to the right of the Units position. Without it, we would be lost ... and not know what each position meant.
Now we can continue with smaller and smaller values, from tenths, to hundredths, and so on, like in this example:
## Large and Small
So, our Decimal System lets us write numbers as large or as small as we want, using the decimal point. Numbers can be placed to the left or right of a decimal point, to indicate values greater than one or less than one.
17.591 The number to the left of the decimal point is a whole number (17 for example) As we move further left, every number place gets 10 times bigger. The first digit on the right means tenths (1/10). As we move further right, every number place gets 10 times smaller (one tenth as big).
## Ways to think about Decimal Numbers ...
### ... as a Whole Number Plus Tenths, Hundredths, etc
You could think of a decimal number as a whole number plus tenths, hundredths, etc:
### Example 1: What is 2.3 ?
• On the left side is "2", that is the whole number part.
• The 3 is in the "tenths" position, meaning "3 tenths", or 3/10
• So, 2.3 is "2 and 3 tenths"
### Example 2: What is 13.76 ?
• On the left side is "13", that is the whole number part.
• There are two digits on the right side, the 7 is in the "tenths" position, and the 6 is the "hundredths" position
• So, 13.76 is "13 and 7 tenths and 6 hundredths"
### ... as a Decimal Fraction
Or, you could think of a decimal number as a Decimal Fraction.
A Decimal Fraction is a fraction where the denominator (the bottom number) is a number such as 10, 100, 1000, etc (in other words a power of ten)
So "2.3" would look like this:
23 10
And "13.76" would look like this:
1376 100
### ... as a Whole Number and Decimal Fraction
Or, you could think of a decimal number as a Whole Number plus a Decimal Fraction.
So "2.3" would look like this:
2 and
3 10
And "13.76" would look like this:
13 and
76 100
Fractions
Finding Common Denominators
Step 1
See if one of the denominators is a multiple or factor of the other denominator.
Example: 1 + 1 1 + 2
6 3 = 6 6
Step 2
If Step 1 doesn’t work, find a common multiple of the two denominators.
Example: 1 + 1 5 + 2
4 10 = 20 20
Because 20 is a multiple of both 4 and 10
Step 3
If Step 1 doesn’t work, and you can’t easily find a common multiple of the two denominators as in Step 2, multiply the two denominators together for a common denominator.
Example: 1 + 1 7 + 5
5 7 = 35 35
Simplifying Fractions
Simplifying a fraction means to rewrite a fraction as an equivalent fraction with a smaller numerator and denominator. To do this, you need to find a common factor of the numerator and denominator. For example:
3
9 can be simplified because both the numerator (3) and
denominator (9) can be divided by 3. So,
3 ÷ 3 1
9 ÷ 3 = 3
PERCENTAGES
You can think of a percentage as the numerator of a fraction with 100 as the denominator or a hundredths decimal. You can write percents as fractions and decimals. For example:
35
100 = 0.35 = 35%
As long as a fraction has a denominator of 100, it can easily be written as a percent. If a fraction is not written with a denominator of 100, it needs to be converted into either an equivalent fraction with a denominator of 100, or written as a decimal and then converted to a percent. For example:
2
5 can easily be written as an equivalent fraction with a denominator of 100.
2 x 20 40
5 x 20 = 100, so this translates to 40%
However,
5
8 cannot easily be written as an equivalent fraction with a denominator of 100. In this case, we divide 8 by 5 and get 0.625.
This is 625/1000, which equals
62.5
100 = 62.5%
VOLUME AND SURFACE AREA
To find the volume of a rectangular prism, simply multiply the width by the length by the height.
V = w x l x h
To find the surface area, you need to calculate the area of each face of the prism.
3 in. q 5 in.
2 in
.
You will have:
2 faces that are 2in. x 3 in. (front and back)
2 faces that are 3 in. x 5 in. (sides)
2 faces that are 2 in. x 5 in. (top and bottom)
2 x 2 x 3 = 12 sq. in.
2 x 3 x 5 = 30 sq. in.
2 x 2 x 5 = 20 sq. in.
Total 62 sq. in.
ROUNDING
When rounding to the nearest tenth, look at the number in the hundredths place. If the number in the hundredths place is 5 or above, round the number in the tenths place up. If the number in the hundredths place is 4 or less, leave the number in the tenths place the same.
Example: Round 9.654 to the nearest tenth.
The easiest way to do this is to place 9.654 between two numbers one-tenth apart.
9.600
9.654
9.700
Look at the number in the hundreths place: 9.654
Since the "5" is 5 or above, you will round 9.654 to 9.7. You are saying that 9.654 is closer to 9.7 than it is to 9.6.
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Succeed with maths – Part 2
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# 1.4 Describing shapes
When you are describing a geometrical figure or shape, you often need to refer to a particular line or angle on the diagram, so others know what you are referring to. This can be done by labelling the diagram with letters. For example, Figure 10 shows a triangle labelled clockwise at the corners with A,B,C. This is then known as triangle ABC, in which the longest side is AB and the angle is a right angle. is the angle formed by the lines AC and CB. The point where two lines meet is known as a vertex (the plural is vertices). So A, B and C are vertices of the triangle.
Figure _unit7.1.10 Figure 10 A triangle labelled ABC
Note that you can use the shorthand notation ‘’ for ‘the triangle ABC’ if you wish. There is a lot of new maths vocabulary in these last few sections, so you might find it useful to make a note of these to refer back to when completing this next activity, or for this week’s quiz and the badged quiz in Week 8.
## Activity _unit7.1.1 Activity 1 What can you see?
Timing: Allow approximately 10 minutes
Look at the image below and then answer the following questions using the letters shown.
Figure _unit7.1.11 Figure 11
• a.Which sets of lines appear to be parellel?
• a.
• b.AB is parallel to DC.
AD is parallel to GI and BC.
AI is parallel to EH.
BD is parallel to HJ
• b.Which lines are perpendicular?
• b.AB is perpendicular to AD and to BC.
DC is perpendicular to AD and BC.
EH is perpendicular to DB and HJ.
AI is perpendicular to HJ and DB.
• c.How many triangles can you see?
• c.
• d.What other shapes can you see?
• d.The parallelogram, GBJI.
The squares, ABCD and EFIH.
The trapeziums, DHIF, EHIG, DBJH, BJHE and DGIH.
This completes your work on defining shapes and how to label them in order to describe them clearly to others.
The next section looks at the different ways for measuring shapes.
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# How do you factor z^3 + 7z + 7^2 + 7?
Jun 13, 2017
It can be factored in two different ways. Check out the Explanation! :)
#### Explanation:
Apply the power function when it appears on a constant value:
${z}^{3} + 7 z + {7}^{2} + 7 = {z}^{3} + 7 z + 49 + 7 = {z}^{3} + 7 z + 56$.
We can now do two different factorings:
$\to$ by $z$:
$z \left({z}^{2} + 7\right) + 56$
or
$\to$ by $7$:
${z}^{3} + 7 \left(z + 8\right)$
Jul 21, 2017
${z}^{3} + 7 z + {z}^{2} + 7 = \left(z + 1\right) \left({z}^{2} + 7\right)$
$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left(z - \sqrt{7} i\right) \left(z + \sqrt{7} i\right)$
#### Explanation:
I suspect the question has been mistranscribed somewhere along the line. A more plausible cubic that we can factor by grouping would be:
${z}^{3} + 7 z + {z}^{2} + 7 = \left({z}^{3} + 7 x\right) + \left({z}^{2} + 7\right)$
$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = z \left({z}^{2} + 7\right) + 1 \left({z}^{2} + 7\right)$
$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left({z}^{2} + 7\right)$
This can only be factored further using complex coefficients, since ${z}^{2} + 7 > 0$ for any real values of $z$ ...
$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left({z}^{2} - {\left(\sqrt{7} i\right)}^{2}\right)$
$\textcolor{w h i t e}{{z}^{3} + 7 z + {z}^{2} + 7} = \left(z + 1\right) \left(z - \sqrt{7} i\right) \left(z + \sqrt{7} i\right)$
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# Addition Strategies Review
Contributor: Samantha Penna. Lesson ID: 11618
If you completed the four lessons in the Addition Strategies series, you've met pictures, tally marks, hopping frogs, and your fingers! Now it's time to add them all together - you're ready to review!
categories
## Elementary
subject
Math
learning style
Visual
personality style
Beaver
Primary (K-2)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
If you have worked through the first four Related Lessons in the Addition Strategies series, you have learned four strategies to solve addition problems.
• What are the four strategies?
Can you remember all four addition strategies you learned in this series? Did you share your answer with your parent or teacher?
If you cannot remember all four, catch up with the Related Lessons in the right-hand sidebar.
Great job! Now, take a look at the list below to make sure you remembered all the fun and helpful strategies you have learned:
• drawing pictures
• writing tally marks
• using a number line
• counting on
These are all the strategies you have learned that can help you solve addition problems.
Now, it's time to review each strategy so you can use them in a snap!
First, you will review the picture drawing strategy. If you ever come across an addition problem that is too tricky to solve using mental math, you can draw pictures to help find the solution. Take a look at the problem below:
### 4 + 3 =
• What are the two numbers you see in this problem?
That's right! There is a 4 and a 3 in this problem. Since this is not a word problem that asks for a certain object, we can draw circles to represent the numbers in the problem.
Take out a sheet of paper and a pencil. Once your materials are ready, solve the problem above by using pictures. Remember to draw above the number that matches the amount of pictures you drew.
Draw and solve the problem, then share your answers with your parent or teacher.
• Does your answer look like the one below?
The second strategy you used to solve addition problems was writing tally marks. This is very similar to the picture drawing strategy, except you draw tally marks instead. Try drawing tally marks to solve the problem below:
### 3 + 3 =
• Does your answer look like the one below?
Fantastic!
The third strategy you learned involved using a number line to solve addition problems. Draw a number line like the one below:
Look at the example below of the addition problem being solved.
• Do you remember how to use this strategy?
Don't forget your hopping frogs!
### 5 + 3 =
Use this strategy to solve the problem below.
### 6 + 3 =
Solve the problem on the number line you drew, then share your answer with your parent or teacher.
Excellent work!
The fourth strategy you learned was the counting on strategy. You will need your fingers for this strategy; hopefully, you didn't lose them!
This strategy is meant to be used with numbers smaller than ten. For example, if you were solving the problem, 2 + 2, first you would hold up two fingers, then you would hold up two more fingers. You would count all of the fingers you held up, and the number of all your fingers together is your answer.
Try counting on with your fingers to solve the problem below. Share your answer with your parent or teacher:
### 4 + 3 =
• Did you solve the problem and find the answer below?
### 4 + 3 = 7
Great work!
• Was it encouraging to review all of the addition strategies you learned?
Move on to the Got It? section to practice using all of the strategies you have learned.
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future.
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## Summary and examples
The general equation of a quadratic function is:
, and are called parameters. , because otherwise it is no quadratic function.
The graph of a quadratic function is a parabola. There are two types, one that opens up and one that opens down, depending on the parameter .
: opens up
: opens down
A parabola intersects the -axis in the point . You can easily see this, because for any point on the -axis we have . When is substituted in the equation we get .
When a parabola intersects the -axis we can calculate the intersection points in various ways. The -formula, explained in Quadratic equations (abc-formula), always gives a solution. Another way is based on factorizing the equation, see the topic Quadratic equations (factorizing). This is faster, though not always possible.
The intersection points can be calculated by applying the -formula:
In this formula the discrriminant plays an important role:
We have:
• : the parabola has two different intersection points with the -axis;
• : the parabola has two coinciding intersection points with the -axis (usually it is said that the parabola has just one intersection point); and
• : the parabola has no intersection points with the -axis.
A parabola is a symmetric graph and has a symmetry axis, namely the line:
Then it is clear that the top of the parabola is on this symmetry axis:
This can also be deduced from the -formula. This optimum can also be found by differentiating the function and make the derivative equal to :
#### Example 1
Given the function:
1. Determine the intersection point of the graph with the -axis;
2. Does this graph have an intersection point with the -axis, and if so, how many and determine their coordinates.
3. Determine the symmetry axis.
1. Because in the formula is equal to , the intersection point with the -axis is equal to .
2. We have to compute the discriminant:
is greater than en thus the graph has two intersection points with the -axis. The -coordinates of these intersection points are:
so:
and the coordinates are:
and
3. The symmetry axis can be calculated in two way. First by using the formula above, but also by taking the mean value of the two intersection points, namely:
#### Example 2
Given the function:
1. Determine the intersection point of the graph with the -axis;
2. Does this graph have an intersection point with the -axis, and if so, how many and determine their coordinates.
3. Determine the symmetry axis.
1. Because in the formula is equal to , the intersection point with the -axis is equal to .
2. We can compute the discriminant to find out whether there are one or more intersection points, but another way is faster. The equation can be factorized:
and thus we may conclude that the graph has two intersection points with the -axis: en .
3. The symmetry axis lies just in the middle of these two points, so the symmetriy-axis is .
#### Example 3
For which value of does the parabola:
have two coinciding intersection points with the -axis.
These intersection points coincide if the discriminant , so:
0
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Expansion of (a ± b)$$^{2}$$
A binomial is an algebraic expression which has exactly two terms, for example, a ± b. Its power is indicated by a superscript. For example, (a ± b)2 is a power of the binomial a ± b, the index being 2.
A trinomial is an algebraic expression which has exactly three terms, for example, a ± b ± c. Its power is also indicated by a superscript. For example, (a ± b ± c)3 is a power of the trinomial a ± b ± c, whose index is 3.
Expansion of (a ± b)2
(a +b)$$^{2}$$
= (a + b)(a + b)
= a(a + b) + b(a+ b)
= a$$^{2}$$ + ab + ab + b$$^{2}$$
= a$$^{2}$$ + 2ab + b$$^{2}$$.
(a - b)$$^{2}$$
= (a - b)(a - b)
= a(a - b) - b(a - b)
= a$$^{2}$$ - ab - ab + b$$^{2}$$
= a$$^{2}$$ - 2ab + b$$^{2}$$.
Therefore, (a + b)$$^{2}$$ + (a - b)$$^{2}$$
= a$$^{2}$$ + 2ab + b$$^{2}$$ + a$$^{2}$$ - 2ab + b$$^{2}$$
= 2(a$$^{2}$$ + b$$^{2}$$), and
(a + b)$$^{2}$$ - (a - b)$$^{2}$$
= a$$^{2}$$ + 2ab + b$$^{2}$$ - {a$$^{2}$$ - 2ab + b$$^{2}$$}
= a$$^{2}$$ + 2ab + b$$^{2}$$ - a$$^{2}$$ + 2ab - b$$^{2}$$
= 4ab.
Corollaries:
(i) (a + b)$$^{2}$$ - 2ab = a$$^{2}$$ + b$$^{2}$$
(ii) (a - b)$$^{2}$$ + 2ab = a$$^{2}$$ + b$$^{2}$$
(iii) (a + b)$$^{2}$$ - (a$$^{2}$$ + b$$^{2}$$) = 2ab
(iv) a$$^{2}$$ + b$$^{2}$$ - (a - b)$$^{2}$$ = 2ab
(v) (a - b)$$^{2}$$ = (a + b)$$^{2}$$ - 4ab
(vi) (a + b)$$^{2}$$ = (a - b)$$^{2}$$ + 4ab
(vii) (a + $$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ + 2a ∙ $$\frac{1}{a}$$ + ($$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ + $$\frac{1}{a^{2}}$$ + 2
(viii) (a - $$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ - 2a ∙ $$\frac{1}{a}$$ + ($$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ + $$\frac{1}{a^{2}}$$ - 2
Thus, we have
1. (a +b)$$^{2}$$ = a$$^{2}$$ + 2ab + b$$^{2}$$.
2. (a - b)$$^{2}$$ = a$$^{2}$$ - 2ab + b$$^{2}$$.
3. (a + b)$$^{2}$$ + (a - b)$$^{2}$$ = 2(a$$^{2}$$ + b$$^{2}$$)
4. (a + b)$$^{2}$$ - (a - b)$$^{2}$$ = 4ab.
5. (a + $$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ + $$\frac{1}{a^{2}}$$ + 2
6. (a - $$\frac{1}{a}$$)$$^{2}$$ = a$$^{2}$$ + $$\frac{1}{a^{2}}$$ - 2
Solved Example on Expansion of (a ± b)2
1. Expand (2a + 5b)$$^{2}$$.
Solution:
(2a + 5b)$$^{2}$$
= (2a)$$^{2}$$ + 2 ∙ 2a ∙ 5b + (5b)$$^{2}$$
= 4a$$^{2}$$ + 20ab + 25b$$^{2}$$
2. Expand (3m - n)$$^{2}$$
Solution:
(3m - n)$$^{2}$$
= (3m)$$^{2}$$ - 2 ∙ 3m ∙ n + n$$^{2}$$
= 9m$$^{2}$$ - 6mn + n$$^{2}$$
3. Expand (2p + $$\frac{1}{2p}$$)$$^{2}$$
Solution:
(2p + $$\frac{1}{2p}$$)$$^{2}$$
= (2p)$$^{2}$$ + 2 ∙ 2p ∙ $$\frac{1}{2p}$$ + ($$\frac{1}{2p}$$)$$^{2}$$
= 4p$$^{2}$$ + 2 + $$\frac{1}{4p^{2}}$$
4. Expand (a - $$\frac{1}{3a}$$)$$^{2}$$
Solution:
(a - $$\frac{1}{3a}$$)$$^{2}$$
= a$$^{2}$$ - 2 ∙ a ∙ $$\frac{1}{3a}$$ + ($$\frac{1}{3a}$$)$$^{2}$$
= a$$^{2}$$ - $$\frac{2}{3}$$ + $$\frac{1}{9a^{2}}$$.
5. If a + $$\frac{1}{a}$$ = 3, find (i) a$$^{2}$$ + $$\frac{1}{a^{2}}$$ and (ii) a$$^{4}$$ + $$\frac{1}{a^{4}}$$
Solution:
We know, x$$^{2}$$ + y$$^{2}$$ = (x + y)$$^{2}$$ – 2xy.
Therefore, a$$^{2}$$ + $$\frac{1}{a^{2}}$$
= (a + $$\frac{1}{a}$$)$$^{2}$$ – 2 ∙ a ∙ $$\frac{1}{a}$$
= 3$$^{2}$$ – 2
= 9 – 2
= 7.
Again, Therefore, a$$^{4}$$ + $$\frac{1}{a^{4}}$$
= (a$$^{2}$$ + $$\frac{1}{a^{2}}$$)$$^{2}$$ – 2 ∙ a$$^{2}$$ ∙ $$\frac{1}{a^{2}}$$
= 7$$^{2}$$ – 2
= 49 – 2
= 47.
6. If a - $$\frac{1}{a}$$ = 2, find a$$^{2}$$ + $$\frac{1}{a^{2}}$$
Solution:
We know, x$$^{2}$$ + y$$^{2}$$ = (x - y)$$^{2}$$ + 2xy.
Therefore, a$$^{2}$$ + $$\frac{1}{a^{2}}$$
= (a - $$\frac{1}{a}$$)$$^{2}$$ + 2 ∙ a ∙ $$\frac{1}{a}$$
= 2$$^{2}$$ + 2
= 4 + 2
= 6.
7. Find ab if a + b = 6 and a – b = 4.
Solution:
We know, 4ab = (a + b)$$^{2}$$ – (a – b)$$^{2}$$
= 6$$^{2}$$ – 4$$^{2}$$
= 36 – 16
= 20
Therefore, 4ab = 20
So, ab = $$\frac{20}{4}$$ = 5.
8. Simplify: (7m + 4n)$$^{2}$$ + (7m - 4n)$$^{2}$$
Solution:
(7m + 4n)$$^{2}$$ + (7m - 4n)$$^{2}$$
= 2{(7m)$$^{2}$$ + (4n)$$^{2}$$}, [Since (a + b)$$^{2}$$ + (a – b)$$^{2}$$ = 2(a$$^{2}$$ + b$$^{2}$$)]
= 2(49m$$^{2}$$+ 16n$$^{2}$$)
= 98m$$^{2}$$ + 32n$$^{2}$$.
9. Simplify: (3u + 5v)$$^{2}$$ - (3u - 5v)$$^{2}$$
Solution:
(3u + 5v)$$^{2}$$ - (3u - 5v)$$^{2}$$
= 4(3u)(5v), [Since (a + b)$$^{2}$$ - (a – b)$$^{2}$$ = 4ab]
= 60uv.
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# How do you solve 4 sqrt (x) = 8 + 2 sqrt (x)?
Sep 13, 2015
color(green)(x = 16
#### Explanation:
We are given that $4 \sqrt{x} = 8 + 2 \sqrt{x}$
Transposing $2 \sqrt{x}$ to the left hand side, we get:
$4 \sqrt{x} - 2 \sqrt{x} = 8$
$\left(4 - 2\right) \sqrt{x} = 8$
$2 \sqrt{x} = 8$
Divising both sides by 2, we get:
$\frac{\cancel{2} \sqrt{x}}{\cancel{2}} = \frac{8}{2}$
$\sqrt{x} = 4$
Squaring both sides we get:
${\left(\sqrt{x}\right)}^{2} = {4}^{2}$
In exponents, color(blue)(sqrta*sqrta = a
Hence we get color(green)(x = 16
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## - Interval Bisection
With this method we repeatedly bisect the interval containing the root in order to improve the approximation.
LESSON 1
Given that $$x^3-3x^2=1-x$$ has a root between 2 and 3, find this root to 1 decimal place using the bisection method.
SOLUTION
$x^3-3x^2=1-x$
Rewrite the equation in the form $$f(x)=0$$
$x^3-3x^2+x-1=0$
Let $$f(x)=x^3-3x^2+x-1$$
$f(2)=2^3-3(2)^2+2-1=-3$
$f(3)=3^3-3(3)^2+3-1=2$
• Determine the midpoint of this interval in which the root lies.
Mid – point of interval is 2.5
$f(2.5)=2.5^3-3(2.5)^2+2.5-1=-1.625$
• Since there is a sign change between 2.5 and 3 the root occurs within this interval. We therefore now bisect this interval. Mid – point is 2.75
$f(2.75)=2.75^3-3(2.75)^2+2.75-1$
$=-0.140625$
• Due to sign changes the root must be in the interval 2.75 and 3.
Mid – point is 2.875
$f(2.875)=2.875^3-3(2.875)^2+2.875-1=0.8418$
• Due to sign changes the root is between 2.75 and 2.875
Mid – point is 2.8125
$f(2.8125)=2.8125^3-3(2.8125)^2+2.8125-1=0.33$
Therefore, root lies between 2.75 and 2.8125. When these 2 values are rounded to 1 decimal place the answers are both 2.8. Consequently, the root is approximately 2.8.
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