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Section 2.2: Complex Numbers
Learning Outcomes
• Express square roots of negative numbers as multiples of i.
• Plot complex numbers on the complex plane.
• Add and subtract complex numbers.
• Multiply and divide complex numbers.
The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as
${x}^{2}+4=0$
Our best guesses might be +2 or –2. But if we test +2 in this equation, it does not work. If we test –2, it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it.
Express square roots of negative numbers as multiples of i
We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number $i$ is defined as the square root of negative 1.
$\sqrt{-1}=i$
So, using properties of radicals,
${i}^{2}={\left(\sqrt{-1}\right)}^{2}=-1$
We can write the square root of any negative number as a multiple of i. Consider the square root of –25.
\begin{align} \sqrt{-25}&=\sqrt{25\cdot \left(-1\right)} \\ &=\sqrt{25}\sqrt{-1} \\ &=5i \end{align}
We use 5and not $-\text{5}i$ because the principal root of 25 is the positive root.
Figure 1
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written + bi where a is the real part and bi is the imaginary part. For example, $5+2i$ is a complex number. So, too, is $3+4\sqrt{3}i$.
Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.
A General Note: Imaginary and Complex Numbers
A complex number is a number of the form $a+bi$ where
• a is the real part of the complex number.
• bi is the imaginary part of the complex number.
If $b=0$, then $a+bi$ is a real number. If $a=0$ and b is not equal to 0, the complex number is called an imaginary number. An imaginary number is an even root of a negative number.
How To: Given an imaginary number, express it in standard form.
1. Write $\sqrt{-a}$ as $\sqrt{a}\sqrt{-1}$.
2. Express $\sqrt{-1}$ as i.
3. Write $\sqrt{a}\cdot i$ in simplest form.
Example 1: Expressing an Imaginary Number in Standard Form
Express $\sqrt{-9}$ in standard form.
Try It
Express $\sqrt{-24}$ in standard form.
Plot complex numbers on the complex plane
We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis.
Let’s consider the number $-2+3i$. The real part of the complex number is –2 and the imaginary part is 3i. We plot the ordered pair $\left(-2,3\right)$ to represent the complex number $-2+3i$.
Figure 2
A General Note: Complex Plane
Figure 3
In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis.
How To: Given a complex number, represent its components on the complex plane.
1. Determine the real part and the imaginary part of the complex number.
2. Move along the horizontal axis to show the real part of the number.
3. Move parallel to the vertical axis to show the imaginary part of the number.
4. Plot the point.
Example 2: Plotting a Complex Number on the Complex Plane
Plot the complex number $3 - 4i$ on the complex plane.
Try It
Plot the complex number $-4-i$ on the complex plane.
Add and subtract complex numbers
Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.
A General Note: Addition and Subtraction of Complex Numbers
Adding complex numbers:
$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$
Subtracting complex numbers:
$\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i$
How To: Given two complex numbers, find the sum or difference.
1. Identify the real and imaginary parts of each number.
2. Add or subtract the real parts.
3. Add or subtract the imaginary parts.
Example 3: Adding Complex Numbers
Add $3 - 4i$ and $2+5i$.
Try It
Subtract $2+5i$ from $3 - 4i$.
Multiplying Complex Numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
Multiplying a Complex Number by a Real Number
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,
Figure 5
How To: Given a complex number and a real number, multiply to find the product.
1. Use the distributive property.
2. Simplify.
Example 4: Multiplying a Complex Number by a Real Number
Find the product $4\left(2+5i\right)$.
Try It
Find the product $-4\left(2+6i\right)$.
Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get
$\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bd{i}^{2}$
Because ${i}^{2}=-1$, we have
$\left(a+bi\right)\left(c+di\right)=ac+adi+bci-bd$
To simplify, we combine the real parts, and we combine the imaginary parts.
$\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i$
How To: Given two complex numbers, multiply to find the product.
1. Use the distributive property or the FOIL method.
2. Simplify.
Example 5: Multiplying a Complex Number by a Complex Number
Multiply $\left(4+3i\right)\left(2 - 5i\right)$.
Try It
Multiply $\left(3 - 4i\right)\left(2+3i\right)$.
Dividing Complex Numbers
Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of $a+bi$ is $a-bi$.
Note that complex conjugates have a reciprocal relationship: The complex conjugate of $a+bi$ is $a-bi$, and the complex conjugate of $a-bi$ is $a+bi$. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide $c+di$ by $a+bi$, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
\begin{align} \frac{c+di}{a+bi} &= \frac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}&&\text{Multiply the numerator and denominator by the complex conjugate of the denominator.} \\ &=\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}} && \text{Apply the distributive property.} \\&=\frac{ca-cbi+adi-bd\left(-1\right)}{{a}^{2}-abi+abi-{b}^{2}\left(-1\right)} && \text{Simplify, remembering that } i^2 = -1 \\ &= \frac{\left(ca+bd\right)+\left(ad-cb\right)i}{{a}^{2}+{b}^{2}} \end{align}
A General Note: The Complex Conjugate
The complex conjugate of a complex number $a+bi$ is $a-bi$. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
• When a complex number is multiplied by its complex conjugate, the result is a real number.
$(a+bi)(a-bi)=a^2-abi+abi-b^2i^2=a^2-b^2*(-1)=a^2+b^2$
• When a complex number is added to its complex conjugate, the result is a real number.
$(a+bi)+(a-bi)=2a$
Example 6: Finding Complex Conjugates
Find the complex conjugate of each number.
1. $2+i\sqrt{5}$
2. $-\frac{1}{2}i$
How To: Given two complex numbers, divide one by the other.
1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.
Example 7: Dividing Complex Numbers
Divide $\left(2+5i\right)$ by $\left(4-i\right)$.
Example 8: Substituting a Complex Number into a Polynomial Function
Let $f\left(x\right)={x}^{2}-5x+2$. Evaluate $f\left(3+i\right)$.
Try It
Let $f\left(x\right)=2{x}^{2}-3x$. Evaluate $f\left(8-i\right)$.
Example 9: Substituting an Imaginary Number in a Rational Function
Let $f\left(x\right)=\frac{2+x}{x+3}$. Evaluate $f\left(10i\right)$.
Try It
Let $f\left(x\right)=\frac{x+1}{x - 4}$. Evaluate $f\left(-i\right)$.
Simplifying Powers of i
The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.
\begin{align}&{i}^{1}=i\\ &{i}^{2}=-1\\ &{i}^{3}={i}^{2}\cdot i=-1\cdot i=-i\\ &{i}^{4}={i}^{3}\cdot i=-i\cdot i=-{i}^{2}=-\left(-1\right)=1\\ &{i}^{5}={i}^{4}\cdot i=1\cdot i=i\end{align}
We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.
\begin{align}&{i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1\\ &{i}^{7}={i}^{6}\cdot i={i}^{2}\cdot i={i}^{3}=-i\\ &{i}^{8}={i}^{7}\cdot i={i}^{3}\cdot i={i}^{4}=1\\ &{i}^{9}={i}^{8}\cdot i={i}^{4}\cdot i={i}^{5}=i\end{align}
Example 10: Simplifying Powers of i
Evaluate ${i}^{35}$.
Key Concepts
• The square root of any negative number can be written as a multiple of i.
• To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.
• Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts.
• Complex numbers can be multiplied and divided.
• To multiply complex numbers, distribute just as with polynomials.
• To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator.
• The powers of i are cyclic, repeating every fourth one.
Glossary
complex conjugate
the complex number in which the sign of the imaginary part is changed and the real part of the number is left unchanged; when added to or multiplied by the original complex number, the result is a real number
complex number
the sum of a real number and an imaginary number, written in the standard form abi, where a is the real part, and bi is the imaginary part
complex plane
a coordinate system in which the horizontal axis is used to represent the real part of a complex number and the vertical axis is used to represent the imaginary part of a complex number
imaginary number
a number in the form bi where $i=\sqrt{-1}$
Section 2.2 Homework Exercises
1. Explain how to add complex numbers.
2. What is the basic principle in multiplication of complex numbers?
3. Give an example to show the product of two imaginary numbers is not always imaginary.
4. What is a characteristic of the plot of a real number in the complex plane?
For the following exercises, evaluate the algebraic expressions.
5. $\text{If }f\left(x\right)={x}^{2}+x - 4$, evaluate $f\left(2i\right)$.
6. $\text{If }f\left(x\right)={x}^{3}-2$, evaluate $f\left(i\right)$.
7. $\text{If }f\left(x\right)={x}^{2}+3x+5$, evaluate $f\left(2+i\right)$.
8. $\text{If }f\left(x\right)=2{x}^{2}+x - 3$, evaluate $f\left(2 - 3i\right)$.
9. $\text{If }f\left(x\right)=\frac{x+1}{2-x}$, evaluate $f\left(5i\right)$.
10. $\text{If }f\left(x\right)=\frac{1+2x}{x+3}$, evaluate $f\left(4i\right)$.
For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown.
11.
12.
For the following exercises, plot the complex numbers on the complex plane.
13. $1 - 2i$
14. $-2+3i$
15. i
16. $-3 - 4i$
For the following exercises, perform the indicated operation and express the result as a simplified complex number.
17. $\left(3+2i\right)+\left(5 - 3i\right)$
18. $\left(-2 - 4i\right)+\left(1+6i\right)$
19. $\left(-5+3i\right)-\left(6-i\right)$
20. $\left(2 - 3i\right)-\left(3+2i\right)$
21. $\left(-4+4i\right)-\left(-6+9i\right)$
22. $\left(2+3i\right)\left(4i\right)$
23. $\left(5 - 2i\right)\left(3i\right)$
24. $\left(6 - 2i\right)\left(5\right)$
25. $\left(-2+4i\right)\left(8\right)$
26. $\left(2+3i\right)\left(4-i\right)$
27. $\left(-1+2i\right)\left(-2+3i\right)$
28. $\left(4 - 2i\right)\left(4+2i\right)$
29. $\left(3+4i\right)\left(3 - 4i\right)$
30. $\frac{3+4i}{2}$
31. $\frac{6 - 2i}{3}$
32. $\frac{-5+3i}{2i}$
33. $\frac{6+4i}{i}$
34. $\frac{2 - 3i}{4+3i}$
35. $\frac{3+4i}{2-i}$
36. $\frac{2+3i}{2 - 3i}$
37. $\sqrt{-9}+3\sqrt{-16}$
38. $-\sqrt{-4}-4\sqrt{-25}$
39. $\frac{2+\sqrt{-12}}{2}$
40. $\frac{4+\sqrt{-20}}{2}$
41. ${i}^{8}$
42. ${i}^{15}$
43. ${i}^{22}$
For the following exercises, use a calculator to help answer the questions.
44. Evaluate ${\left(1+i\right)}^{k}$ for $k=\text{4, 8, and 12}\text{.}$ Predict the value if $k=16$.
45. Evaluate ${\left(1-i\right)}^{k}$ for $k=\text{2, 6, and 10}\text{.}$ Predict the value if $k=14$.
46. Evaluate $\left(1+i\right)^{k}-\left(1-i\right)^{k}$ for $k=\text{4, 8, and 12}$. Predict the value for $k=16$.
47. Show that a solution of ${x}^{6}+1=0$ is $\frac{\sqrt{3}}{2}+\frac{1}{2}i$.
48. Show that a solution of ${x}^{8}-1=0$ is $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$.
For the following exercises, evaluate the expressions, writing the result as a simplified complex number.
49. $\frac{1}{i}+\frac{4}{{i}^{3}}$
50. $\frac{1}{{i}^{11}}-\frac{1}{{i}^{21}}$
51. ${i}^{7}\left(1+{i}^{2}\right)$
52. ${i}^{-3}+5{i}^{7}$
53. $\frac{\left(2+i\right)\left(4 - 2i\right)}{\left(1+i\right)}$
54. $\frac{\left(1+3i\right)\left(2 - 4i\right)}{\left(1+2i\right)}$
55. $\frac{{\left(3+i\right)}^{2}}{{\left(1+2i\right)}^{2}}$
56. $\frac{3+2i}{2+i}+\left(4+3i\right)$
57. $\frac{4+i}{i}+\frac{3 - 4i}{1-i}$
58. $\frac{3+2i}{1+2i}-\frac{2 - 3i}{3+i}$
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### Building Tetrahedra
Can you make a tetrahedron whose faces all have the same perimeter?
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.
# Nested Square Roots
##### Age 14 to 16 ShortChallenge Level
Squaring both sides
Let $x=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}$
\begin{align}x^2&=\left(\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}\right)^2\\ &=3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}\\ &=3+2x\end{align}
So $x^2-2x-3=0\Rightarrow (x-3)(x+1)=0$, so $x=3$ or $x=-1$.
Since $x$ is positive, $x=3$.
Check: $\sqrt{3+2\times(3)}=3$
'Working it out' using a sequence
Let $x=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}$
We can approximate $x$ by imagining that we are going to find that large square root as a calculation.
We can start by finding $\sqrt{3}$, and then double it, add $3$, and find the square root of that, and so on.
So if we let $a=\sqrt{3}$, and then $b=\sqrt{3+2\times a}=\sqrt{3+2\sqrt{3}}$, then $c=\sqrt{3+2\times b}=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3}}}$, and $d=\sqrt{3+2\times c}=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3}}}}$, then the sequence $a, b, c, d, ...$ will get closer and closer to $x$.
Using a calculator, or a spreadsheet, $a=1.73205$, $b=2.54246$, $c=2.84340$, $d=2.94734$, $e=2.98239$, $f=2.99413$, ...
It looks like they are getting closer and closer to $3$, so it looks like $x=3$. If we got to $3$ in our sequence, then the next number in the sequence would be $\sqrt{3+2\times 3}$, which is $\sqrt{3+6}=\sqrt{9}=3$. So if the sequence ever gets to $3$, it will stay at $3$ forever, which strongly suggests that $x=3$.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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# True Discount Questions
FACTS AND FORMULAE FOR TRUE DISCOUNT QUESTIONS
Suppose a man has to pay Rs.156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of Rs. 100 now will clear off the debt of Rs.156 due 4 years. Hence, we say that :
Sum due = Rs.156 due 4 years hence;
Present Worth (P.W) = Rs. 100;
True Discount (T.D) = (Sum due) - (P.W)=Rs. (156 - 100) = Rs. 56
We define :
T.D = Interest on P.W
Amount = (P.W) + (T.D)
Interest is reckoned on P.W and true discount is reckoned on the amount.
IMPORTANT FORMULAE
Let rate = R% per annum and Time = T years. Then,
1.$\inline P.W = \frac{100\times Amount}{100+(R\times T)}=\frac{100\times T.D}{R\times T}$
2. $\inline T.D = \frac{(P.W)\times R\times T}{100}=\frac{Amount\times R\times T}{100+(R\times T)}$
3. $\inline Sum = \frac{(S.I)\times (T.D)}{(S.I)-(T.D)}$
4. (S.I) - (T.D )= S.I on T.D
5. When the sum is put at compound interest, then
$\inline P.W=\frac{Amount}{(1+\frac{R}{100})^T}$
Q:
The profit earned by selling an article for Rs 900 is double the loss incurred when the same article is sold for Rs.490. At what price should the article be sold to make 25% profit?
A) 715 B) 469 C) 400 D) 750
Explanation:
Let C.P be Rs. x
900 - x = 2(x - 450)
=> x = Rs.600
C.P = 600 gain required is 25%
S.P = [(100+25)*600]/100
= Rs.750
18 12034
Q:
A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:
A) gains Rs. 55 B) gains Rs. 50 C) loses Rs. 30 D) gains Rs. 30
Explanation:
12 10380
Q:
The difference between Simple Interest and True Discount on a certain sum of money for 6 months at $12\frac{1}{2}$% per annum is Rs 25. Find the sum.
A) Rs.6800 B) Rs.6500 C) Rs.6000 D) Rs.6200
Explanation:
13 10353
Q:
A man wants to sell his scooter .There are two offers one at Rs12000 cash and other at a credit of Rs12880 to be paid after 8 months ,money being at 18% per annum which is better offer?
A) Rs.12880 B) Rs.12000 C) Both are equally good D) None of the above
Explanation:
PW of Rs.12,880 due 8 months hence
= Rs[(12880 x 100)/(100+(18 x 8/12))] =Rs.11500
Clearly 12000 in cash is a better offer.
5 9601
Q:
A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:
A) 0% B) 5% C) 7.5% D) 10%
Explanation:
C.P =Rs.3000
S.P =Rs. [3600*10]/[100+(10*2)] = Rs.3000
Gain =0%
12 7032
Q:
The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is?
A) 12% B) 13% C) 15% D) 14%
Explanation:
P.W = 2562-122 =Rs.2440
Rate = (100 x 122)/(2440 x 1/3) =15%
7 4200
Q:
The present worth of Rs.1404 due in two equal half-yearly instalments at 8% per annum. Simple interest is:
A) 1320 B) 1300 C) 1325 D) 1200
Explanation:
Required Sum = PW of Rs.702 due 6 months hence + PW of Rs.702 due 1 year hence
= Rs.[(100 x 702)/(100+(8 x 1/2))] + [(100 x 702)/(100+(8 x 1))]
=Rs.1325
7 4094
Q:
Find the present worth of Rs.930 due 3 years hence at 8% per annum.Aso find the discount?
A) 186 B) 180 C) 185 D) 189
Explanation:
P.W = [(100 x Amount)/(100+(R x T))]
=[(100 x 930)/(100+(8x3))] = Rs. 750
T.D = Amount - P.W = 930 - 750 = Rs.180
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# Discrete Mathematics Propositional Logic - Discrete Mathematics
## What is Propositional Logic in Discrete Mathematics?
Theoretical base for many areas of mathematics and computer science is provided by logical reasoning. Some of the areas such as artificial intelligence, programming languages etc.
By propositional logic, the statements are analyzed and the truth vales are assigned. The analysis is done either for individual statement or as a composite of statements.
### Prepositional Logic – Definition
The declarative statement, which has either of the truth values, is termed as a proposition. Capital letters are used for representing propositional variables, and the variables are linked by a connective.
Some illustrations for propositions are:
• "Apple is a fruit", for this, the truth value “TRUE” is returned.
• "9+8+3=3 – 2", for this, the truth value “FALSE” is returned.
## What are Connectives in Propositional Logic?
Propositional logic provides five different types of connectives -
• OR (∨)
• AND (∧)
• Negation/ NOT (¬)
• Implication / if-then (→)
• If and only if (⇔).
### OR (∨)
If any one of the variables, A or B is true, then the OR operation is true. The OR operation is represented as A∨B. The following is the truth table for OR operation -
A B A ∨B True True True True False True False True True False False False
### AND (∧)
If both the variables, A and B is true, then the AND operation is true. The AND operation is represented as A∧B. The following is the truth table for AND operation -
A B A ∧B True True True True False False False True False False False False
### Negation (¬)
For a true variable A, the negation is false. The Negation operation is represented as ¬A. The following is the truth table for this operation -
A ¬ A True False False True
### Implication / if-then (→)
It is a conditional statement and If the variable A is true and the variable B is False, then this conditional statement would be false. The following is the truth table for this operation -
A B A →B True True True True False False False True True False False True
### If and only if (⇔)
If both variables A and B are same, that is either True or False, then the result is True. If both the Variables are not same, then the result is false. The following is the truth table for this operation -
A B A ⇔B True True True True False False False True False False False True
## What are Tautologies?
For all the values of the propositional variables, tautology is always true.
Example − Prove [(A→B)∧A]→B is a tautology
The following is the truth table −
A B A →B (A → B) ∧A [( A → B ) ∧ A] →B True True True True True True False False False True False True True False True False False True False True
It is observed that it is a tautology for every value is True.
For all the values of the propositional variables, contradiction is always false.
Example − Prove (A∨B)∧[(¬A)∧(¬B)] is a contradiction
The following is the truth table −
A B A ∨B ¬ A ¬ B (¬ A) ∧ ( ¬ B) (A ∨ B) ∧ [( ¬ A) ∧ (¬ B)] True True True False False False False True False True False True False False False True True True False False False False False False True True True False
It is observed that it is a contradiction for every value is False.
## What is a Contingency?
For each of the value of the propositional variables, contingency is either true or false.
Example − Prove (A∨B)∧(¬A) a contingency
The following is the truth table −
A B A ∨B ¬ A (A ∨ B) ∧ (¬ A) True True True False False True False True False False False True True True True False False False True False
It is observed that it is a contingency is either true or False.
## What are Propositional Equivalences?
Two statements are said to be equivalent logically if they satisfy the following conditions -
• •The truth values for each of the statement are same.
• •The bi-conditional statement X⇔Y is a tautology.
Example − Prove ¬(A∨B)and[(¬A)∧(¬B)] are equivalent
According to Matching truth table method –
A B A ∨B ¬ (A ∨ B) ¬ A ¬ B [(¬ A) ∧ (¬ B)] True True True False False False False True False True False False True False False True True False True False False False False False True True True True
The statements are said to be equivalent as it is observed that the truth values of both the statements ¬(A∨B)and[(¬A)∧(¬B)] are true.
According to Bi-Conditional Method -
A B ¬ (A ∨ B ) [(¬ A) ∧ (¬ B)] [¬ (A ∨ B)] ⇔ [(¬ A ) ∧ (¬ B)] True True False False True True False False False True False True False False True False False True True True
The statements are said to be equivalent as it is observed that [¬(A∨B)]⇔[(¬A)∧(¬B)] is a tautology.
## Inverse, Converse, and Contra-positive
The conditional statement has two parts -
• Hypothesis, p
• Conclusion, q
It is denoted as p→q.
Example of Conditional Statement − “If you eat health food, you will not be sick.” For this statement, hypothesis is – eat health food and conclusion is – will not be sick.
### Inverse
The negation of both hypothesis and conclusion is known as Inverse of the conditional statement.
Example – The inverse for the above discussed example is “If you do not eat health food, you will be sick.”
### Converse
The interchange of hypothesis and conclusion is known as Converse of the conditional statement.
Example – The converse for the example is “You will not be sick if you eat health food”.
### Contra-positive
The interchange of hypothesis and conclusion of inverse statement is Contra-positive.
Example − The Contra-positive for the example is “You will be sich if you do not eat health food”.
## What is Duality Principle?
According to Duality principle, if the statement and true and if unions are interchanged into intersections and if universal set is interchanged into the Null set, the result of the interchange in both the cases is true.
Example − The dual of (A∩B)∪C is (A∪B)∩C
## What are Normal Forms?
The propositions can be converted into any of the norm forms, which are of two types -
• Conjunctive normal form
• Disjunctive normal form
### Conjunctive Normal Form
When AND operation is executed on the variables connected with OR operation, the compound statement formulated is said to be in conjunctive normal form.
Examples
• (A∨B)∧(A∨C)∧(B∨C∨D)
• (P∪Q)∩(Q∪R)
### Disjunctive Normal Form
When OR operation is executed on the variables connected with AND operation, the compound statement formulated is said to be in disjunctive normal form.
Examples
• (A∧B)∨(A∧C)∨(B∧C∧D)
• (P∩Q)∪(Q∩R)
|
Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II PhysicsStrategies for Taking SAT II PhysicsVectorsKinematicsDynamicsWork, Energy, and PowerSpecial Problems in MechanicsLinear MomentumRotational MotionCircular Motion and GravitationThermal PhysicsElectric Forces, Fields, and PotentialDC CircuitsMagnetismElectromagnetic InductionWavesOpticsModern PhysicsPhysics GlossaryPractice Tests Are Your Best Friends
4.1 What’s a Vector? 4.2 Vector Addition 4.3 Vector Subtraction 4.4 Multiplication by a Scalar 4.5 Vector Components
4.6 Vector Multiplication 4.7 Key Formulas 4.8 Practice Questions 4.9 Explanations
Vector Components
As we have seen, vector addition and scalar multiplication can produce new vectors out of old ones. For instance, we produce the vector A + B by adding the two vectors A and B. Of course, there is nothing that makes A + B at all distinct as a vector from A or B: all three have magnitudes and directions. And just as A + B can be construed as the sum of two other vectors, so can A and B. In problems involving vector addition, it’s often convenient to break a vector down into two components, that is, two vectors whose sum is the vector in question.
Basis Vectors
We often graph vectors in an xy-coordinate system, where we can talk about vectors in purely numerical terms. For instance, the vector (3,4) is the vector whose tail is at the origin and whose tip is at the point (3,4) on the coordinate plane. From this coordinate, you can use the Pythagorean Theorem to calculate that the vector’s magnitude is 5 and trigonometry to calculate that its direction is about 53.1º above the x-axis.
Two vectors of particular note are (1,0), the vector of magnitude 1 that points along the x-axis, and (0,1), the vector of magnitude 1 that points along the y-axis. These are called the basis vectors and are written with the special hat notation: and respectively.
The basis vectors are important because you can express any vector in terms of the sum of multiples of the two basis vectors. For instance, the vector (3,4) that we discussed above—call it A—can be expressed as the vector sum .
The vector is called the “x-component” of A and the is called the “y-component” of A. In this book, we will use subscripts to denote vector components. For example, the x-component of A is and the y-component of vector A is .
The direction of a vector can be expressed in terms of the angle by which it is rotated counterclockwise from the x-axis.
Vector Decomposition
The process of finding a vector’s components is known as “resolving,” “decomposing,” or “breaking down” a vector. Let’s take the example, illustrated above, of a vector, A, with a magnitude of A and a direction above the x-axis. Because , , and A form a right triangle, we can use trigonometry to solve this problem. Applying the trigonometric definitions of cosine and sine,
we find:
|
Working With the Arrow Representation for Vectors
# Working With the Arrow Representation for Vectors
PRACTICE (online exercises and printable worksheets)
Whenever you get a new mathematical object, you need to learn what you can do with it.
So ... what operations can be performed with vectors?
• vectors can be added and subtracted
• vectors can be multiplied by a scalar (a real number)
In this section, you learn how to perform these operations with the arrow representation for vectors.
In the next section, you learn how to perform these operations with the analytic representation for vectors.
## Multiplying a Vector by a Scalar
Let $\,k\,$ be a real number (a scalar).
Then, $\,k\vec v\,$ is a vector.
That is, a scalar times a vector produces a vector.
In other words, a real number times a vector produces a vector.
The image below illustrates the relationship between an original vector $\,\vec v\,$ and various scaled versions:
Multiplying a vector by a positive number $\,k\,$: doesn't change the direction of the vector changes the length by a factor of $\,k\,$ For example: if $\,k = 2\,$, the length is doubled if $\,k=\frac 12\,$, the length is halved Multiplying a vector by a negative number $\,k\,$: produces a vector pointing in the opposite direction changes the length by a factor of $\,|k|\,$ For example: if $\,k = -2\,$, the length is doubled,and the vector points in the opposite direction if $\,k=-\frac 12\,$, the length is halved,and the vector points in the opposite direction Multiplying any vector by the real number $\,0\,$ produces the zero vector. For all vectors $\,\vec v\,$, $$-\vec v\, = -1\cdot\vec v$$ is the opposite of $\,\vec v\,$.
### Finding the Size of a Scaled Vector
To find the size of a scaled vector, you multiply together two numbers:
• the absolute value (size) of the scaling constant
• the size of the original vector
Precisely, we have:
For all real numbers $\,k\,$, and for all vectors $\,\vec v\,$, $$\| k \vec v\| = |k|\,\cdot\,\|\vec v\|$$
Keep in mind:
• Whenever you see $\,|\cdot|\,$, there must be a real number inside.
• Whenever you see $\,\|\cdot\|\,$, there must be a vector inside.
For example:
$\|3\vec v\|$ $=$ $|3|\cdot\|\vec v\|$ $=$ $3\|\vec v\|$ and $\|-3\vec v\|$ $=$ $|-3|\cdot\|\vec v\|$ $=$ $3\|\vec v\|$
So, $\|-3\vec v\| = \|3\vec v\|\,$.
A vector and its opposite have the same length.
Adding the arrow representations of vectors is done using the ‘head-of-first to tail-of-second’ rule.
Here's how to add $\,\vec u\,$ to $\,\vec v\,$:
• draw an arrow representing $\,\vec u\,$
• to the head of $\,\vec u\,$, attach the tail of vector $\,\vec v\,$
• the sum $\,\vec u+\vec v\,$ goes from the tail of $\,\vec u\,$ to the head of $\,\vec v\,$
It sounds complicated when written out.
The diagram below shows how simple it really is:
• the configuration to find $\vec u + \vec v\,$ is shown in red:
head of $\,\vec u\,$ is attached to tail of $\,\vec v\,$
• the configuration to find $\vec v + \vec u\,$ is shown in blue:
head of $\,\vec v\,$ is attached to tail of $\,\vec u\,$
• in both cases, going from the tail of the first to the head of the second
gives the same vector (shown in black)
• So, $\,\vec u + \vec v = \vec v + \vec u\,$.
This always works!
• the configuration to find $(\vec u + \vec v) + \vec w\,$ is shown in red:
head of $\,\vec u + \vec v\,$ is attached to tail of $\,\vec w\,$
• the configuration to find $\vec u + (\vec v + \vec w)\,$ is shown in blue:
head of $\,\vec u\,$ is attached to tail of $\,\vec v + \vec w\,$
• in both cases, going from the tail of the first to the head of the last
gives the same vector (shown in green)
• So, $\,(\vec u + \vec v) + \vec w = \vec u + (\vec v + \vec w)\,$.
This always works!
Therefore, we can write $\,\vec u + \vec v + \vec w\,$ (no parentheses) without ambiguity.
The vector from the tail of the first to the head of the last is the vector sum.
## Subtracting Vectors
To subtract a vector, just add its opposite: $$\,\vec u - \vec v := \vec u + (-\vec v)\,$$ (Remember that ‘$\,:=\,$’ means ‘equals, by definition’).
## Other Operations with Vectors
• There are different types of vectors.
The vectors we're talking about here are two-dimensional vectors—vectors in a plane.
There are also three-dimensional vectors—vectors in space—and lots more.
When you add vectors, they have to be vectors of the same type.
• Depending on what type of vectors you're working with, there may be other operations defined.
For example, there is a ‘dot product’ and a ‘cross product’ that you'll likely come across if you study Calculus.
Master the ideas from this section
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# Fundamentals of Physics/Vectors
A vector is a two-element value that represents both magnitude and direction.
Vectors are normally represented by the ordered pair ${\displaystyle {v}=(v_{x}\,v_{y})}$ or, when dealing with three dimentions, the tuple ${\displaystyle {v}=(v_{x}\,v_{y}\,v_{z})}$. When written in this fashion, they represent a quantity along a given axis.
The following formulas are important with vectors:
${\displaystyle \left\|\mathbf {v} \right\|={\sqrt {{v_{x}}^{2}+{v_{y}}^{2}+{v_{z}}^{2}}}}$
${\displaystyle v_{x}=\left\|\mathbf {v} \right\|\cos {\theta }}$
${\displaystyle v_{y}=\left\|\mathbf {v} \right\|\sin {\theta }}$
${\displaystyle \theta =\tan ^{-1}({\frac {v_{y}}{v_{x}}})\,\!}$
Addition is performed by adding the components of the vector. For example, c = a + b is seen as:
${\displaystyle {c}=(a_{x}+b_{x}\,a_{y}+b_{y})}$
With subtraction, invert the sign of the second vector's components.
${\displaystyle {c}=(a_{x}-b_{x}\,a_{y}-b_{y})}$
## Multiplication (Scalar)
The components of the vector are multiplied by the scalar:
${\displaystyle s*{v}=(s*v_{x}\,s*v_{y})}$
## Division
While some domains may permit division of vectors by vectors, such operations in physics are undefined. It is only possible to divide a vector by a scalar.
As with multiplication, the components of the vector are divided by the scalar:
${\displaystyle s*{v}=({\frac {s_{x}}{v_{x}}}\,{\frac {s_{y}}{v_{y}}})}$
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# Distance, Rate, and Time Learning Log
Distance rate and time are all units of measuring somethings rate, where rate is the actual rate. In equation, the rate is distance over time, or r = d/t, where r is rate, d is distance, and t is time. For example, say you have a mouse that can run 10 feet every 7 seconds, the rate would be 10/7, so the rate would be about 1.428 feet per 7 seconds, so the unit rate would be 0.204 ft every hour.
# Multiplication Equations Learning Log
Solving an equation using multiplication is really easy, depending on the equation. For example, say you have the problem
5/7d = 15
You know you need to multiply d with 5/7 to get 15, so you would use multiplication. You multiply the denominator (7), with 15 in this problem, and your result is 105. So that is your answer, d = 105. Lets Check, 105/7 = 15, you can go check on a calculator. Basically, you use multiplication to solve a problem when you have a numeral less than the final result, and you need to multiply the less numeral with a fraction. However, there is another way to solve the 5/7d = 15 problem. It’s addition. 5/7 + q = 15, s if you calculate, q is 14 2/7.
# Representations, Center, Shape, and Spread of Data Learning Log
Different types of graphs exist, and each of them shows things better than others. Maximum, Minimum, Mode, Range, Median, and Mean, all of them are shown clearly in some and not s clearly on others. For example, a box plot shows the exact Median always no matter what because it is based around it, while a steam and leaf plot can show the maximum minimum and range clearly, and still a histogram can show a sort of idea of where all the data in centralised, the sort of “mode”. For showing shape, instantly throw the box plot out, because it doesn’t show shape. Anyways, a histogram and a steam and leaf plot both show the shape of the information, where the stem and leaf plot is basically a histogram with bins of 10.
# Mean Absolute Deviation Learning Log
The Absolute Deviation is the Mean of the distance each information piece is from the informations mean. I mean, say you have the information 11, 12, 14, 15. The mean is 13, so we need to find the distances. 11 and 15 are 2 numerals away from 13, and 12 and 14 are online 1 numeral away, so we add that up. 1+1+2+2 = 6/4 = 1.5. So 1.5 is the Absolute Deviation for that information. This can help to tell how dynamically far away each information piece can be, like if we added the information 1 and 25, which are both 12 away from 13, it would become 30/4 which would become 5, because the information is a lot more dynamic.
# Mean and Median Learning Log
There is something called the Mean, and something called the Median. And I don’t mean Mean, in math, the Mean means average. The Median, however, is just the center of all the information, it’s the information that stands in the middle of all of the information. Both, can show the middle of the information, but there are times where it is better to use one than the other and times where both are extremely close to each other. For example, say you have the information :
11,12,13,13,13,17,17,19,21
In the set of information above, you should be using the median, which is 13, because if you used the Mean, it would be lower than 16 because 11,12,13,13 and 13 of the question would partially lean it towards one direction, making the Mean lower. However, in other informations, like these :
20,20,21,21,21,22,22,23,23
It’s better to use the mean because the median would be 21, which you know is not the average of all the information. So using the mean would be a better substitute to find the “typical” information.
# Distributive Property
Distributive property is showing an expression without a parentheses, and instead multiply the factor with each expression inside the parentheses. For example in the expression 4(7x+6) the distributive property would multiply 7 by 4 and 6 by 4, so the new expression would be 28x + 24. The expressions are equivalent, just shown in different “formats”.
# Inverse operations
Inverse operations are basically the opposite of the original operation. For example, if you add add “X”, the inverse operations is to subtract “X”. Similarly, it also works vice versa. Multiplication and division are also inverses. Inverse operations undo the previous operations, making both operations cancelled from the question. Follow the examples below:
10 +4 = 14 – 4 = 10
See above how the subtraction undid what the addition did to the original number. That is called inverse operations. A multiplication and division inverse operation is:
2*4 = 8 ÷ 4 = 2
See how the division undo’s the multiplication? This works the other way round too. You subtract then add, or divide then multiply. That, is inverse operations.
# Using a Super Giant One Learning Log
Using a SUPER GIANT ONE is really quite simple. It’s easier to teach while showing an example than to explain it.
Say you want to divide 5/6 by 11/17. You would confusingly represent the problem as 5/6/11/17 (The red line is the divide line, while the black line is part of representing the fraction). You would want to make the denominator 1, so you find the reciprocal of the denominator fraction, which in this case is 17/11. Then you would show it confusingly, just try to follow:
5/6 x 17/11 = 85/66 11/17 x 17/11 = 1
If you have followed above, the denominator should always equal 1, and the numerator is the answer to the problem. This is because any number divided by 1 has a result of the same as the original number. So our answer is 85/66, or 1 and 19/66. You always find the reciprocal of the denominator because when you multiply a fraction by it’s reciprocal, it will always have a result of one, meaning all you have to do is multiply the numerator with the reciprocal of the denominator.
# Random Graph
Take the graph below, we have a large figure in a graph, hover your mouse over it, and you see the side lengths, where “X = 4”. COOL!
# Algebra Tiles Learning Log
Today in math we used algebra tiles to learn about variables and showing algebraic expressions. We used algebra tiles. We had small unit tiles, tiles that equal 5 units, an x tile, and x2 tiles. We used these tiles to show algebraic expressions, like 1*3+5+3x+2x2 . Our feeble learnings let us understand how algebraic equations worked.
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### AlxaEGCS5_11_05_04
```Copyright © Cengage Learning. All rights reserved.
Chapter 5
Similar Triangles
5.4 The Pythagorean Theorem
The Pythagorean Theorem
The following theorem will enable us to prove the wellknown Pythagorean Theorem.
Theorem 5.4.1
The altitude drawn to the hypotenuse of a right triangle
separates the right triangle into two right triangles that are
similar to each other and to the original right triangle.
3
The Pythagorean Theorem
Theorem 5.4.1 is illustrated by Figure 5.18, in which the
right triangle ABC has its right angle at vertex C so that
is the altitude to hypotenuse
.
The smaller triangles are shown in Figures 5.18(b) and (c),
and the original triangle is shown in Figure 5.18(d).
Figure 5.18
Note the matching arcs indicating congruent angles.
4
The Pythagorean Theorem
In Figure 5.18(a),
and
(parts) of the hypotenuse
are known as segments
.
Furthermore,
is the segment of the hypotenuse
and
is the segment of the
5
The Pythagorean Theorem
Theorem 5.4.2
The length of the altitude to the hypotenuse of a right
triangle is the geometric mean of the lengths of the
segments of the hypotenuse.
Lemma 5.4.3
The length of each leg of a right triangle is the geometric
mean of the length of the hypotenuse and the length of the
segment of the hypotenuse adjacent to that leg.
6
The Pythagorean Theorem
Theorem 5.4.4 (Pythagorean Theorem)
The square of the length of the hypotenuse of a right
triangle is equal to the sum of the squares of the lengths of
the legs.
7
Example 1
Given
RST with right
S in Figure 5.23, find:
Figure 5.23
a) RT if RS = 3 and ST = 4
b) RT if RS = 4 and ST = 6
c) RS if RT = 13 and ST = 12
d) ST if RS = 6 and RT = 9
8
Example 1 – Solution
With right S, the hypotenuse is
and ST = b.
Then RT = c, RS = a,
a) 3² + 4² = c² → 9 + 16 = c²
c² = 25
c = 5;
RT = 5
b) 4² + 6² = c² → 16 + 36 = c²
c² = 52
9
Example 1 – Solution
cont’d
c=
c) a² + 12² = 13² → a² + 144 = 169
a² = 25
10
Example 1 – Solution
cont’d
a = 5;
RS = 5
d) 6² + b² = 9² → 36 + b² = 81
b² = 45
b=
11
Example 1 – Solution
cont’d
The converse of the Pythagorean Theorem is also true.
12
The Pythagorean Theorem
Theorem 5.4.5 (Converse of Pythagorean Theorem)
If a, b, and c are the lengths of the three sides of a triangle,
with c the length of the longest side, and if c² = a² + b²,
then the triangle is a right triangle with the right angle
opposite the side of length c.
Theorem 5.4.6
If the hypotenuse and a leg of one right triangle are
congruent to the hypotenuse and a leg of a second right
triangle, then the triangles are congruent (HL).
13
The Pythagorean Theorem
Our work with the Pythagorean Theorem would be
incomplete if we did not address two issues.
The first, Pythagorean triples, involves natural (or counting)
numbers as possible choices of a, b, and c.
The second leads to the classification of triangles
according to the lengths of their sides.
14
PYTHAGOREAN TRIPLES
15
Pythagorean Triples
Definition
A Pythagorean triple is a set of three natural numbers
(a, b, c) for which a² + b² = c².
Three sets of Pythagorean triples encountered in this
section are (3, 4, 5), (5, 12, 13),and (8, 15, 17).
These numbers will always fit the sides of a right triangle.
16
Pythagorean Triples
Natural-number multiples of any of these triples will also
constitute Pythagorean triples.
For example, doubling (3, 4, 5) yields (6, 8, 10), which is
also a Pythagorean triple. In Figure 5.29, the triangles are
similar by SSS~.
Figure 5.29
17
Pythagorean Triples
The Pythagorean triple (3, 4, 5) also leads to (9, 12, 15),
(12, 16, 20), and (15, 20, 25). The Pythagorean triple
(5, 12, 13) leads to triples such as (10, 24, 26) and (15, 36,
39).
Basic Pythagorean triples that are used less frequently
include (7, 24, 25), (9, 40, 41), and (20, 21, 29).
18
Pythagorean Triples
Pythagorean triples can be generated by using select
formulas.
Where p and q are natural numbers and p > q,
one formula uses 2pq for the length of one leg, p² – q² for
the length of other leg, and p² + q² for the length of the
hypotenuse (See Figure 5.30.).
Figure 5.30
19
Pythagorean Triples
Table 5.1 lists some Pythagorean triples corresponding to
choices for p and q.
20
Pythagorean Triples
The triples printed in boldface type are basic triples, also
known as primitive triples.
In application, knowledge of the primitive triples and their
multiples will save you considerable time and effort.
In the final column, the resulting triple is provided in the
order from a (small) to c (large).
21
THE CONVERSE OF THE
PYTHAGOREAN THEOREM
22
The Converse of the Pythagorean Theorem
The Converse of the Pythagorean Theorem allows us to
recognize a right triangle by knowing the lengths of its
sides.
A variation on the converse allows us to determine
whether a triangle is acute or obtuse.
23
The Converse of the Pythagorean Theorem
Theorem 5.4.7
Let a, b, and c represent the lengths of the three sides of a
triangle, with c the length of the longest side.
1. If c² > a² + b², then the triangle is obtuse and the obtuse
angle lies opposite the side of length c.
2. If c² < a² + b², then the triangle is acute.
24
Example 6
Determine the type of triangle represented if the lengths of
its sides are as follows:
a) 4, 5, 7
b) 6, 7, 8
c) 9, 12, 15
d) 3, 4, 9
25
Example 6 – Solution
a) Choosing c = 7, we have 7² > 4² + 5², or 49 > 16 + 25;
the triangle is obtuse.
b) Choosing c = 8, we have 8² < 6² + 7², or 64 < 36 + 49;
the triangle is acute.
c) Choosing c = 15, we have 15² = 9² + 12², or
225 = 81 + 144; the triangle is a right triangle.
d) Because 9 > 3 + 4, no triangle is possible. (Remember
that the sum of the lengths of two sides of a triangle
must be greater than the length of the third side.)
26
```
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# Time and Work Practice Set 1
5 Steps - 3 Clicks
# Time and Work Practice Set 1
### Introduction
Work is defined as the amount of job assigned or the amount of job actually done. Work is always considered as a whole or 1. Work based problems are more or less related to time speed and distance. The article Time and Work Practice Set 1 provides information about Time and Work, a important topic of Quantitative Aptitude section. Consists of different types questions with solutions useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.
### Quiz
1. A laborer was appointed by a contractor on the condition he would be paid Rs 150 for each day of his work but would be, fined a the rate of Rs 30 per day for his absent. After 20 days, the contractor paid the laborer’s 2820. Find the number of days he worked:
A. 13 days
B. 19 days
C. 5 days
D. 12 days
E. None of these
Explanation: Let the required number of days = x days
So, 150x- (20-x)30 = 2820
x=19 days
2. In the beginning, Ram works at a rate such that he can finish a piece of work in 24 hrs, but he only works at this rate for 16 hrs. After that, he works at a rate such that he can do the whole work in 18 hrs. If Ram is to finish this work at a stretch, how many hours will he take to finish this work?
A. 12 hrs
B. 18 hrs
C. 11$$\frac{50}{3}$$
D. 15 hrs
E. 22 hrs
Explanation: Ram’s 16 hr work = $$\frac{16}{24}$$ =$$\frac{60 × 70}{(60 + 70)}$$. Remaining work = 1 – $$\frac{2}{3}$$ = $$\frac{1}{3}$$.
Using work and time formula: This will be completed in $$\frac{2}{3}$$ × 18 i.e. 6 hrs.
So, total time taken to complete work = 16 + 6= 22 hrs.
3. A can do a piece of work in 10 days, and B can do the same work in 20 days. With the help of C, they finished the work in 4 days. C can do the work in how many days, working alone?
A. 5 days
B. 10 days
C. 15 days
D. 20 days
E. 25 days
Explanation: Their combined 4 day work = 4($$\frac{1}{10}$$ + $$\frac{1}{15}$$) = $$\frac{12}{20}$$ = $$\frac{3}{5}$$.
Remaining work = 1 – $$\frac{3}{5}$$ = $$\frac{2}{5}$$.
This means C did $$\frac{2}{5}$$ work in 4 days, hence he can finish the complete work in $$\frac{5}{2}$$ × 4 = 10 days.
4. Ajay and Vijay undertake to do a piece of work for Rs. 480. Ajay alone can do it in 75 days while Vijay alone can do it in 40 days. With the help of Pradeep, they finish the work in 25 days. How much should Pradeep get for his work?
A. Rs. 40
B. Rs. 20
C. Rs. 360
D. Rs. 100
E. Rs. 60
Explanation: In 24 days, they would have done $$\frac{1}{3}$$ and $$\frac{5}{8}$$ of the work.
The remaining work is 1 – ($$\frac{1}{3}$$ + $$\frac{5}{8}$$) = $$\frac{1}{24}$$.
This means Pradeep has done $$\frac{1}{24}$$th of the work, so he should be paid $$\frac{1}{24}$$th of the amount i.e. 480 × $$\frac{1}{24}$$ = Rs. 20 is the answer.
5. Daku and Tamatar can do a piece of work in 70 and 60 days respectively. They began the work together, but Daku leaves after some days and Tamatar finished the remaining work in 47 days. After how many days did Daku leave?
A. 14 days
B. 16 days
C. 18 days
D. 10 days
E. 7 days
Explanation: Tamatar would have done $$\frac{47}{60}$$ work in 47 days. The remaining work i.e. $$\frac{13}{60}$$ must have been done by Daku and Tamatar together. They can do the whole work in $$\frac{60 × 70}{(60 + 70)}$$ = 60 × $$\frac{70}{130}$$ = $$\frac{420}{13}$$ days.
So, they would have done $$\frac{13}{60}$$ work in $$\frac{420}{13}$$ × $$\frac{13}{60}$$ = 7 days. Therefore, Daku left work after 7 days.
1. A contractor takes a road construction project to finish it in 40 days and for that he engaged 200 men. After 30 days he employed 100 more men in this project, then the work finished on time. Find if the 100 more men would not worked then how many more days required to finish the work ?
A. 8 days
B. 10 days
C. 12 days
D. 7 days
E. None of these
Explanation: 100 * 10 days = 1000
Now 1000/200 = 5 days (Initial total no. of men engaged in the project)
Hence, 5 more days required to finish the work if 100 more men would not joined
2. A woman has her three daughters. First and second can take 24 and 30 days resp. to complete a work. In how many days third one takes to complete the work. If woman can complete the whole work alone in 3(3/11) days. The efficiency of woman is double than her three daughters.
A. 22 days
B. 12 days
C. 13 days
D. 21 days
E. 19 days
Explanation: LCM = 72 (if we are taking woman and her two daughters )
Here it is given
Woman Three daughters
Time taken = 1 : 2
Efficiency= 2 : 1Three daughters, let P+Q+R = 11
3+2+R = 11
R=6 days
Her third daughter complete the work in = $$\frac{72}{6}$$= 12 Days
3. Ramesh and Ram can do a piece of work in 24 and 30 days respectively. They both started and worked for 6 days. Ram then leaves the work and another their friend Rohit joins the work and completed the remaining work with Ramesh in 11 days. Find how many days are taken by Rohit alone to finish the work
A. 110 days
B. 132 days
C. 150 days
D. 120 days
E. None of these.
Explanation: ($$\frac{1}{24}$$ + $$\frac{1}{30}$$) *6 +($$\frac{1}{24}$$ + $$\frac{1}{Rohit}$$ ) *11 = 1
Therefore, Rohit takes 120 days to finish the work
4. A printer A can print one thousand books in 15 hours ,printer B can print the same number of books in 10 hours and printer C can print the same number of books in 12 hours . If all the printers are started to print the books at 8 A.M, After sometime printer A is closed at 9 A.M and printer B and printer C remains working. Find at what time the printing will be completed?
A. 4$$\frac{3}{11}$$ hours
B. 3$$\frac{1}{11}$$ hours
C. 5$$\frac{3}{11}$$ hours
D. 3$$\frac{5}{11}$$ hours
E. None of these.
Explanation: Let printing completed in be T hours
Then A ‘s 1 hour work, B’s T hours work and C;s T hours work =Total work
$$\frac{1}{15}$$ +$$\frac{T}{10}$$+$$\frac{T}{12}$$ =1
T= 5($$\frac{1}{11}$$)
Hence, the printing of books will be completed at 5($$\frac{1}{11}$$)hours
5. A group of men decided to do a job in 4 days but 20 men dropped out everyday, the job was completed at the end of the 7th day. Find the men who are in the work initially?
A. 155
B. 135
C. 120
D. 140
E. 160
Explanation: Total work = M * 4 = 4M
M + (M+20) +…….
$$\frac{7}{2}$$ [2M +6(-20)] = 4M
M=140
1. Raj can build a house alone in 16 days but Suraj alone can build it in 12 days. Raj and Suraj work on alternate days. If Raj works on first day, the house will be built in how many days?
A. 12.5 days
B. 13$$\frac{3}{4}$$ days
C. $$\frac{48}{7}$$ days
D. $$\frac{24}{7}$$ days
E. None of these
2. 12 Men or 18 women can reap a field in 14 days. The number of days that 8 men and 16 women will take to reap it?
A. 5
B. 7
C. 8
D. 9
E. 10
Explanation: 12 men = 18 women or 1 man = $$\frac{3}{2}$$ women
Therefore 8 men + 16 women = (8 × $$\frac{3}{2}$$ + 16) women i.e.., 28 women
Now 18 women can reap the field in 14 days
Therefore, 28 women can reap it in = 9 days
3. 8 Children and 12 men complete a certain piece of work in 9 days. If each child takes twice the time taken by man to finish the work. In how many days will 12 men finish the same work?
A. 8
B. 15
C. 9
D. 12
E. None of these.
Explanation: 2 children = 1 man
Therefore (8 children +12 men) = 16 men
Now less men more days 12: 16:: 9: x
Therefore, x = ($$\frac{144}{12}$$) = 12 days
4. A can do a certain job in 12 days. B is 60% more efficient than A. Then Number of days it takes B to do the same piece of work is?
A. 6
B. 6$$\frac{1}{4}$$ hours
C. 7$$\frac{1}{2}$$ hours
D. 8
E. 4
Explanation:Ratio of times taken by A and B = 160 : 100 = 8 : 5
If A takes 8 days B takes 5 days
If A takes 12 days, B takes =($$\frac{5}{8}$$ x 12) = 7 ½ days
5. A, B and C together earn Rs.150 per day while A and C together earn Rs. 94 and B and C together earn Rs. 76. The daily earning of C is?
A. Rs. 75
B. Rs.56
C. Rs.34
D. Rs.20
E. Rs.160
Explanation: B’s daily earning = RS.(150 -94) = Rs. 56
A’s daily earning = Rs.(150 -76) = Rs.74
C’s daily earning = Rs.[150 – (56+74)] = Rs. 20
### Exams
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# Rational Roots and Synthetic Division
by Justin Skycak on
The rational roots theorem can help us find zeros of polynomials without blindly guessing.
This post is a chapter in the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Rational Roots and Synthetic Division. Justin Math: Algebra. https://justinmath.com/rational-roots-and-synthetic-division/
In the previous post, we learned how to find the remaining zeros of a polynomial if we are given some zeros to start with. But how do we get those initial zeros in the first place, if they’re not given to us and aren’t obvious from the equation?
## Rational Roots Theorem
The rational roots theorem can help us find some initial zeros without blindly guessing. It states that for a polynomial with integer coefficients, any rational number (i.e. any integer or fraction) that is a root (i.e. zero) of the polynomial can be written as some factor of the constant coefficient, divided by some factor of the leading coefficient.
For example, if the polynomial $p(x)=2x^4+x^3-7x^2-3x+3$ has a rational root, then it is some positive or negative fraction having numerator $1$ or $3$ and denominator $1$ or $2$.
The possible roots are then $\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, or $\pm 3$. We test each of them below.
\begin{align*} &\mathbf{p\left( \frac{1}{2} \right) = 0} \hspace{.5cm} && p\left( -\frac{1}{2} \right) = \frac{11}{4} \\ & p(1)=-4 \hspace{.5cm} && \mathbf{p(-1)=0} \\ & p\left( \frac{3}{2} \right) = -\frac{15}{4} && p\left( -\frac{3}{2} \right) = -\frac{3}{2} \\ & p(3)=120 && p(-3)=84 \end{align*}
We see that $x=\frac{1}{2}$ and $x=-1$ are indeed zeros of the polynomial.
Therefore, the polynomial can be written as
\begin{align*} \left( x-\frac{1}{2} \right) (x+1)(2x^2+bx+c) \end{align*}
for some constants $b$ and $c$, which we can find by expanding and matching up coefficients.
\begin{align*} 2x^4+x^3-7x^2-3x+3 &= \left( x-\frac{1}{2} \right) (x+1)(2x^2+bx+c) \\ &= 2x^4+(1+b)x^3+\left( c+\frac{b}{2}-1 \right)x^2 + \left( \frac{c}{2} - \frac{b}{2} \right)x - \frac{c}{2} \end{align*}
We find that $b=0$ and $c=-6$.
The remaining quadratic factor becomes $2x^2-6$, which has zeros $x=\pm \sqrt{3}$.
Thus, the zeros of the polynomial are $-\sqrt{3}$, $-1$, $\frac{1}{2}$, and $\sqrt{3}$.
## Synthetic Division
To speed up the process of finding the zeros of a polynomial, we can use synthetic division to test possible zeros and update the polynomial’s factored form and rational roots possibilities each time we find a new zero.
Given the polynomial $x^4+3x^3-5x^2-21x-14$, the rational roots possibilities are $\pm 1$, $\pm 2$, $\pm 7$, and $\pm 14$.
To test whether, say, $2$ is a zero, we can start by setting up a synthetic division template which includes $2$ at the far left, followed by the coefficients of the polynomial (in the order that they appear in standard form).
We put a $0$ under the first coefficient (in this case, $1$) and add down the column.
Then, we multiply the result by the leftmost number (in this case, $2$) and put it under the next coefficient (in this case, $3$).
We repeat the same process over and over until we finish the final column.
The bottom-right number is the remainder when we divide the polynomial by the factor corresponding to the zero being tested. Therefore, if the bottom-right number is $0$, then the top-left number is indeed a zero of the polynomial, because its corresponding factor is indeed a factor of the polynomial.
In this case, though, the bottom-right number is not $0$ but $-36$, so $2$ is NOT a zero of the polynomial.
However, when we repeat synthetic division with $-2$, the bottom-right number comes out to $0$ and we conclude that $-2$ is a zero of the polynomial.
Then $x+2$ is a factor of the polynomial, and the bottom row gives us the coefficients in the sub-polynomial that multiplies to yield the original polynomial.
\begin{align*} x^4+3x^3-5x^2-21x-14 &= (x+2)(1x^3+1x^2-7x-7) \\ &= (x+2)(x^3+x^2-7x-7) \end{align*}
The next factor will come from $x^3+x^2-7x-7$, so the rational roots possibilities are just $\pm 1$ and $\pm 7$.
We use synthetic division to test whether $x=1$ is a zero of $x^3+x^2-7x-7$.
Since the bottom-right number is $-12$ rather than $0$, we see that $x=1$ is not a zero of $x^3+x^2-7x-7$. However, $x=-1$ is!
Using the bottom row as coefficients, we update the factored form of our polynomial.
\begin{align*} x^4+3x^3-5x^2-21x-14 &= (x+2)(x^3+x^2-7x-7) \\ &= (x+2)(x+1)(1x^2+0x-7) \\ &= (x+2)(x+1)(x^2-7) \end{align*}
Now that we’re down to a quadratic, we can solve it directly.
\begin{align*} x^2-7 &= 0 \\ x^2 &= 7 \\ x &= \pm \sqrt{7} \end{align*}
Thus, the zeros of the polynomial are $-2$, $-1$, $\sqrt{7}$, and $-\sqrt{7}$, and the factored form of the polynomial is
\begin{align*} (x+2)(x+1)(x+\sqrt{7})(x-\sqrt{7}). \end{align*}
## Final Remarks
In this example, the polynomial factored fully into linear factors. However, if the last factor were $x^2+7$, which does not have any zeros, we would leave it in quadratic form. The zeros of the polynomial would be just $-2$ and $-1$, and the fully factored form of the polynomial would be $(x+2)(x+1)(x^2+7)$.
One last thing about synthetic division: be sure to include ALL coefficients of the original polynomial in the top row of the synthetic division setup, even if they are $0$. For example, the polynomial $3x^4+2x$ is really $3x^4+0x^3+0x^2+2x+0$, so the top row in the synthetic division setup should read $\hspace{.5cm} 3 \hspace{.5cm} 0 \hspace{.5cm} 0 \hspace{.5cm} 2 \hspace{.5cm} 0$.
## Exercises
For each polynomial, find all the zeros and write the polynomial in factored form. (You can view the solution by clicking on the problem.)
$1) \hspace{.5cm} 3x^3+18x^2+33x+18$
Solution:
$3(x+1)(x+2)(x+3)$
$\text{zeros: } -1, -2, -3$
$2) \hspace{.5cm} 2x^3-5x^2-4x+3$
Solution:
$(2x-1)(x+1)(x-3)$
$\text{zeros: } \frac{1}{2}, -1, 3$
$3) \hspace{.5cm} x^4-4x^3+3x^2+4x-4$
Solution:
$(x+1)(x-1)(x-2)^2$
$\text{zeros: } \pm 1, 2$
$4) \hspace{.5cm} x^4-3x^3+5x^2-9x+6$
Solution:
$(x^2+3)(x-2)(x-2)$
$\text{zeros: } 1, 2$
$5) \hspace{.5cm} x^4-2x^3-x^2+4x-2$
Solution:
$(x-1)^2(x+\sqrt{2})(x-\sqrt{2})$
$\text{zeros: } 1, \pm \sqrt{2}$
$6) \hspace{.5cm} 2x^4+7x^3+6x^2-x-2$
Solution:
$(2x-1)(x+2)(x+1)^2$
$\text{zeros: } \frac{1}{2}, -2, -1$
$7) \hspace{.5cm} 2x^4-8x^3+10x^2-16x+12$
Solution:
$2(x-1)(x-3)(x^2+2)$
$\text{zeros: } 1, 3$
$8) \hspace{.5cm} 21x^5+16x^4-74x^3-61x^2-40x-12$
Solution:
$(x+2)(x-2)(7x+3)(3x^2+x+1)$
$\text{zeros: } \pm 2, -\frac{3}{7}$
This post is a chapter in the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Rational Roots and Synthetic Division. Justin Math: Algebra. https://justinmath.com/rational-roots-and-synthetic-division/
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How to do basic math - Mathematic
# How to do basic math
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To solve a slope intercept form equation, you need to first isolate the y variable on one side of the equation. Once you have done that, you can then use algebra to solve for the y variable. After you have solved for the y variable, you can then plug that back into the original equation to solve for the x variable.
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The three main branches of trigonometry are Plane Trigonometry, Spherical Trigonometry, and Hyperbolic Trigonometry. Plane Trigonometry is concerned with angles and sides in two dimensions, while Spherical Trigonometry deals with angles and sides on the surface of a sphere. Hyperbolic Trigonometry is concerned with angles and sides in three dimensions. The applications of trigonometry are endless, making it a vital tool for anyone who wants to pursue a career in mathematics or science.
There are a lot of great math apps out there that can help you with your math homework. However, my personal favorite is called Photomath. This app is simple to use and can quickly give you the answer to any math problem. Just take a picture of the problem and the app will give you the answer. You can also use the app to see step-by-step solutions for more complex problems.
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# Teach me math app
There are a lot of Teach me math app that are available online. Math can be a challenging subject for many students.
## The Best Teach me math app
Here, we will be discussing about Teach me math app. The factor calculator will then return a list of all of the factors for that number. In some cases, the factor calculator may also return additional information such as the prime factors or the greatest common divisor. Factor calculators can be very useful when working with large numbers or when trying to solve complex mathematical problems.
How to solve for roots. There are multiple ways to solve for the roots of a polynomial equation. One way is to use the Quadratic Formula. The Quadratic Formula is: x = -b ± √b² - 4ac/2a. You can use the Quadratic Formula when the highest exponent of your variable is 2. Another way you can solve for the roots is by factoring. You would want to factor the equation so that it is equal to 0. Once you have done that, you can set each factor equal to 0 and solve for your variable. For example, if you had the equation x² + 5x + 6 = 0, you would first want to factor it. It would then become (x + 2)(x + 3) = 0. You would then set each factor equal to zero and solve for x. In this case, x = -2 and x = -3. These are your roots. If you are given a cubic equation, where the highest exponent of your variable is 3, you can use the method of solving by factoring or by using the Cubic Formula. The Cubic Formula is: x = -b/3a ± √(b/3a)³ + (ac-((b) ²)/(9a ²))/(2a). To use this formula, you need to know the values of a, b, and c in your equation. You also need to be able to take cube roots, which can be done by using a graphing calculator or online calculator. Once you have plugged in the values for a, b, and c, this formula will give you two complex numbers that represent your two roots. In some cases, you will be able to see from your original equation that one of your roots is a real number and the other root is a complex number. In other cases, both of your roots will be complex numbers.
This will usually result in a quadratic equation which can be solved using standard methods. In some cases, it may be possible to solve the equations directly without first solving for one of the variables. However, this is usually more difficult and it is often easier to use substitution. Whether or not to use substitution depends on the form of the equations and the preference of the person solving them.
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## Math checker you can trust
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Website that answers math problems College algebra math answers Helping with math Help me with this math problem How to solve for x with logarithms
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Chapter 3 Practice Test
# Chapter 3 Practice Test - Directions : Consider y = f(x)...
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Chapter 3 Practice Test Directions : Determine whether each function is even, odd, or neither. Explain. 1. y = 3x 4 - 6x 2 + 2 2. y = 10x 12 - 7x 3 + 1 Directions : The graphs below are portions of completed graphs. Sketch a completed graph showing the symmetry requested. 3. 4. symmetry about the x-axis symmetry about the origin Directions : State whether each of the following functions is even, odd, or neither. Explain. 5. 6. 7. 8. Directions : Solve the following problems. Show your work. 9. Where does the graph of y = x - 17 x 2 - 6x + 4 touch the x-axis? Name________________ Period_______
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10. What is the horizontal asymptote of y = -2x 2 - 5x + 4 3x 12 - 14x ? Directions : Find the inverse of each function below. 11. 3 23 8 14 5 2 y x f(x) 12. y = x x - 3 13.
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Unformatted text preview: Directions : Consider y = f(x) shown at the right. On the grids provided, sketch the shift requested. 14. 15. y =-f(x) y = f(x) 16. 17. y = w(-x) y = w(x ) 18. Find the inverse of y = x-2 ( ) 3 + 4. Directions: On a separate sheet of graph paper sketch each function. Make sure you number and label each graph. Plot at least 2 points for accuracy. 19. y ≥-4x+ 2 20. y < 3x + 2-1 21. y = x 3 + 2 22. y =- x +1 ( ) 2 + 6 23. y = 3x-2 x-1 24. y = x 2-25 x-5 25. y =-3 x 2-8x + 16 26. y =-x 2-4 27. Let f(x) = 1 x if x < 3 if 0 < x < 4 x-4 if x ≥ 4 Graph f(x), -f(x), f(-x), f(x) , and f x ( ) ....
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## This note was uploaded on 04/11/2011 for the course MATH 1400 taught by Professor Grether during the Spring '08 term at North Texas.
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Chapter 3 Practice Test - Directions : Consider y = f(x)...
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SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024
### Chapter : 6 REGRESSION ANALYSIS
Regression analysis is a method of predicting or estimating one variable knowing the value of the other variable. Estimation is required in different fields in every day life. A businessman wants to know the effect of increase in advertising expenditure on sales or a doctor wishes to observe the effect of a new drug on patients. An economist is interested in finding the effect of change in demand pattern of some commodities on prices.
We observe, different pairs of variables related to each other like saving depends upon income, cost of production depends upon the number of units produced, the production depends on the number of workers present on a particular day etc. The relationship between two variables can be established with the help of any measure of correlation, as we have seen in the previous chapter. When it is observed that two variables are highly correlated, it leads to interdependence of the variables. We can study the cause and effect relationship between them and then we can apply the regression analysis. The analysis helps in finding a mathematical model of the relationship. In this chapter we will discuss linear regression only.
There are two types of regression lines viz., regression equation of x on y and regression equation of y on x. The first one is used to estimate the value of x when the corresponding value of y is given and the second used to find the value of y when the corresponding value of x is given. They are given by
Regression equation of x on y is and
Regression equation of y on x is where and are known as regression coefficients.
The regression coefficients are given by
Thus, when the regression coefficients are known to us we can determine the co-efficient of correlation.
The regression coefficients can be found independently as
Problems:
1. From the following data find the two regression equations and hence estimate y when x = 13 and estimate x when y = 10.
x: 14 10 15 11 9 12 6 y: 8 6 4 3 7 5 9
[ Answer: 5.2858 & 8.1428 ]
2. Find the two regression equations and also estimate y when x = 13 and estimate x when y = 10
x: 11 7 9 5 8 6 10 y: 16 14 12 11 15 14 17
[ Answer: 17.5359 & 5.0693 ]
3. The following data represents the marks in Algebra (x) and Geometry (y) of a group of 10 students. Find both regression equations and hence estimate y if x = 78 and x if y = 94.
x: 75 80 93 65 87 71 98 68 89 77 y: 82 78 86 72 91 80 95 72 89 74
[ Answer: 80.394 ~ 80 and 94.9337 ~ 95 ]
4. Find the regression equations for the following data and hence estimate y when x = 15 and x when y = 18.
x: 10 12 14 19 8 11 17 y: 20 24 25 21 16 22 20
[ Answer: 21.64 & 11.54 ]
5. From the following data, find the regression equations and further estimate y if x = 16 and x if y = 18.
x: 3 4 6 10 12 13 y: 12 11 15 16 19 17
[ Answer: 20.32 & 11.8 ]
6. The heights in cms of a group of mothers and daughters are given below. Find the regression equations and hence find the most likely height of a mother when the daughter’s height is 164 cms. Also obtain the estimate of the height of a daughter when mother’s height is 162 cms.
Height of mother (x): 157 160 163 165 167 168 170 164 Height of daughter (y): 162 159 165 167 172 170 168 166
[ Answer: 164.2942 & 162.4556 ]
7. The following data gives the marks obtained at the preliminary examination (x) and the final examination (y) for a group of 10 students. Obtain the regression equation of y on x. Hence find the most probable marks at the final examination of a student who has scored 70 marks at the preliminary examination.
x: 54 65 75 82 57 59 60 64 58 62 y: 58 67 76 80 60 64 65 65 60 70
[ Answer: 71.21 ]
8. The following data gives the expenditure on advertising, expressed in hundreds of Rs. (x) and the expenditure on office staff, expressed in thousands of Rs. (y) for 7 different companies. Find the two regression equations and hence estimate the expenditure on advertising when expenditure on office staff is Rs.97,000. Also obtain the most likely expenditure on office staff when the advertising expenditure is Rs.13,200.
x: 129 137 138 135 139 134 145 y: 98 94 93 91 96 95 100
[ Answer: Rs.13746 and Rs.94780 ]
9. The following data represents the figures in kgs. for demands of two commodities A and B. Find (i) the most probable demand for A when demand for B is 107 kgs. (ii) the most probable demand for B when demand for A is 115 kgs.
x: 107 113 109 103 110 117 114 y: 105 110 103 100 110 111 108
[ Answer: 110.7069 & 110.0996 ]
10. From the following data, estimate y when x = 142 and x when y = 126.
x: 140 155 163 167 145 150 148 y: 122 135 140 139 125 130 125
[ Answer: 123.25 & 146.3 ]
11. For a bivariate distribution, the following results are obtained.
Mean value of x = 65 Mean value of y = 53 Standard deviation = 4.7 Standard deviation = 5.2 Coefficient of correlation = 0.78
Find the two regression equations and hence obtain
1. The most probable value of y when x = 63
2. The most probable value of x when y = 50
[ Answer: 51.274 & 62.885 ]
12. Given that for 10 pair of observations.
Find the regression equation of y on x and then estimate y when x = 78.
[ Answer: 81.4495 ]
13. The averages for rainfall and yield of a crop are 42.7 cms and 850 kgs respectively. The corresponding standard deviations are 3.2 cms and 14.1 kgs. The coefficient of correlation is 0.65. Estimate the yield when the rainfall is 39.2 cms. [ Estimated yield is 839.99 kgs. ]
14. The regression equation of supply in thousands of Rs.(y) on price in thousands of Rs.(x) is . The average supply is Rs.18,000. The ratio of standard deviation of supply and price is . Find the average price and the coefficient of correlation between supply and price. [ Average price is Rs.15,000 & r = 0.6 ]
15. Find the regression equation and hence estimate y when x = 56 and x when y = 45.
[ 40.9 & 62.04 ]
16. Given the two regression equations, find
(i) mean values of x and y
(ii) coefficient of correlation
where the equation are
[ 1, 1 and 0.9682 ]
17. Given the regression equations
. Find the means of x and y and the coefficient of correlation. [1, 1 and -0.8165 ]
18. Find the mean values of x and y and correlation coefficient, if the regression equations are [ 1, 1 and -0.5774 ]
19. Given the regression equations , find
1. Mean values of x and y
2. Coefficient of correlation
3. Estimate of y when x = 17
4. Estimate of x when y = 25
[ 15, 20, 0.4714, 21.33 and 16.67 ]
20. Find the two regression equations for the following:
[ ]
21. The regression equation of income (x) on expenditure (y) is . The ratio of the standard deviation of income and expenditure is 4 : 3. Find the coefficient of correlation between income and expenditure. Also find the average income if the average expenditure is Rs.1800. [2500, 0.5 ]
22. From the following regression equations , find
1. Mean values of x and y
2. Coefficient of correlation
3. Most probable value of y when x = 28
4. Most probable value of x when y = 35. [ 25, 33, 0.8164, 37, 26 ]
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##### Determine, algebraically, the vertices of the triangle formed by the lines3x-y = 22x-3y = 2and x + 2y = 8
Given equation of lines are
3x - y = 2 …(i)
2x -3y = 2 …(ii)
and x + 2y = 8 …(iii)
Let lines (i), (ii) and (iii) represent the side of a ∆ABC i.e., AB, BC and CA respectively.
On solving lines (i) and (ii), we will get the intersecting point B.
On multiplying Eq. (i) by 3 in Eq. (i) and then subtracting, we get
(9x-3y)-(2x-3y) = 9-2
7x = 7
x = 1
On putting the value of x in Eq. (i), we get
3×1-y = 3
y = 0
So, the coordinate of point or vertex B is (1, 0)
On solving lines (ii) and (iii), we will get the intersecting point C.
On multiplying Eq. (iii) by 2 and then subtracting, we get
(2x + 4y)-(2x-3y) = 16-2
7y = 14
y = 2
On putting the value of y in Eq. (iii), we get
Hence, the coordinate of point or vertex C is (4, 2).
On solving lines (iii) and (i), we will get the intersecting point A.
On multiplying in Eq. (i) by 2 and then adding Eq. (iii), we get
(6x-2y) + (x + 2y) = 6 + 8
7x = 14
x = 2
On putting the value of x in Eq. (i), we get
3×2 - y = 3
y = 3
So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ∆ABC formed by the given lines are A (2, 3), B(1, 0) and C (4, 2).
6
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# Lesson 14
Alternate Interior Angles
Let’s explore why some angles are always equal.
### 14.1: Angle Pairs
1. Find the measure of angle $$JGH$$. Explain or show your reasoning.
2. Find and label a second $$30^\circ$$ degree angle in the diagram. Find and label an angle congruent to angle $$JGH$$.
### 14.2: Cutting Parallel Lines with a Transversal
Lines $$AC$$ and $$DF$$ are parallel. They are cut by transversal $$HJ$$.
1. With your partner, find the seven unknown angle measures in the diagram. Explain your reasoning.
2. What do you notice about the angles with vertex $$B$$ and the angles with vertex $$E$$?
3. Using what you noticed, find the measures of the four angles at point $$B$$ in the second diagram. Lines $$AC$$ and $$DF$$ are parallel.
4. The next diagram resembles the first one, but the lines form slightly different angles. Work with your partner to find the six unknown angles with vertices at points $$B$$ and $$E$$.
5. What do you notice about the angles in this diagram as compared to the earlier diagram? How are the two diagrams different? How are they the same?
Parallel lines $$\ell$$ and $$m$$ are cut by two transversals which intersect $$\ell$$ in the same point. Two angles are marked in the figure. Find the measure $$x$$ of the third angle.
### 14.3: Alternate Interior Angles Are Congruent
1. Lines $$\ell$$ and $$k$$ are parallel and $$t$$ is a transversal. Point $$M$$ is the midpoint of segment $$PQ$$.
Find a rigid transformation showing that angles $$MPA$$ and $$MQB$$ are congruent.
2. In this picture, lines $$\ell$$ and $$k$$ are no longer parallel. $$M$$ is still the midpoint of segment $$PQ$$.
Does your argument in the earlier problem apply in this situation? Explain.
### Summary
When two lines intersect, vertical angles are equal and adjacent angles are supplementary, that is, their measures sum to 180$$^\circ$$. For example, in this figure angles 1 and 3 are equal, angles 2 and 4 are equal, angles 1 and 4 are supplementary, and angles 2 and 3 are supplementary.
When two parallel lines are cut by another line, called a transversal, two pairs of alternate interior angles are created. (“Interior” means on the inside, or between, the two parallel lines.) For example, in this figure angles 3 and 5 are alternate interior angles and angles 4 and 6 are also alternate interior angles.
Alternate interior angles are equal because a $$180^\circ$$ rotation around the midpoint of the segment that joins their vertices takes each angle to the other. Imagine a point $$M$$ halfway between the two intersections—can you see how rotating $$180^\circ$$ about $$M$$ takes angle 3 to angle 5?
Using what we know about vertical angles, adjacent angles, and alternate interior angles, we can find the measures of any of the eight angles created by a transversal if we know just one of them. For example, starting with the fact that angle 1 is $$70^\circ$$ we use vertical angles to see that angle 3 is $$70^\circ$$, then we use alternate interior angles to see that angle 5 is $$70^\circ$$, then we use the fact that angle 5 is supplementary to angle 8 to see that angle 8 is $$110^\circ$$ since $$180 -70 = 110$$. It turns out that there are only two different measures. In this example, angles 1, 3, 5, and 7 measure $$70^\circ$$, and angles 2, 4, 6, and 8 measure $$110^\circ$$.
### Glossary Entries
• alternate interior angles
Alternate interior angles are created when two parallel lines are crossed by another line called a transversal. Alternate interior angles are inside the parallel lines and on opposite sides of the transversal.
This diagram shows two pairs of alternate interior angles. Angles $$a$$ and $$d$$ are one pair and angles $$b$$ and $$c$$ are another pair.
• transversal
A transversal is a line that crosses parallel lines.
This diagram shows a transversal line $$k$$ intersecting parallel lines $$m$$ and $$\ell$$.
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It is quite difficult to describe the significance of a Quadratic Transformations Worksheet. A small preview of the importance can be seen in the fact that many of the students are frequently asked about quadratic equations when they visit us for a long stay in the classroom.
275 best Education Trigonometry images on Pinterest in 2018 from quadratic transformations worksheet , source:pinterest.com
An ordinary two by two matrices has four rows and two columns, so a quadratic equation in this form has a normal form. The quadratic equation can be represented in three different ways – using a quadratic formula or using the quadratic formula. The quadratic formulas of the form x^3+x^2= 1 can also be written as x^.
In the fourth place, it is quite possible to transform a quadratic formula by means of a quadratic formula. For this purpose, it is necessary to find out the four possible factors of a quadratic formula to do this transformation.
We can get these four factors by means of a quadratic formula or by means of the quadratic formula, and so the quadratic formula worksheet can be used for this purpose. There are different quadratic formulas which can be used for the purpose of transforming the quadratic equations.
As a matter of fact, the quadratic formula has been named a quadratic transformation of an ordinary matrix. A quadratic formula is actually a quadratic formula as well. The quadratic formula is then a quadratic transformation of a quadratic matrix in two-dimensional.
Algebra 2 Worksheets from quadratic transformations worksheet , source:math-aids.com
With a quadratic formula, the quadratic formula transformation can be applied to multiply the matrix by the quadratic formula. This will produce an array which will be called a matrix factorization, and we can transform it by means of a quadratic formula.
The important fact here is that the transformation matrix can be transformed with a quadratic formula by means of a quadratic formula. The transform function can be found using a quadratic formula and then we can transform it by means of a quadratic formula. It can be seen that the quadratic formula worksheet can be used for the purpose of creating a matrix factorization.
It is quite easy to transform a quadratic formula by means of a quadratic formula. The quadratic formula worksheet is really very useful for transforming the quadratic formula.
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algebra : brackets
Multiplying Out(expanding) - a pair of brackets with a single term infront
The term outside the brackets multiplies each of the terms in turn inside the brackets.
example:
further examples:
Multiplying Out(expanding) - two pairs of brackets
Think of the two terms in the first bracket as separate single terms infront of a pair of brackets.
example:
Multiply the contents of the 2nd bracket by the 1st term in the 1st bracket.
Multiply the contents of the 2nd bracket by the 2nd term in the 1st bracket.
Example #1
Example #2
Example #3
Squared Brackets
note: a common mistake
Difference of Two Squares
Brackets - Simple Factorising - This involves taking out a common term from each expression and placing it infront of the brackets.
examples:
This is best illustrated with an example:
You must first ask yourself which two factors when multiplied will give 12 ?
The factors of 12 are : ........1 x 12, ..............2 x 6, ...............3 x 4
Now which numbers in a group added or subtracted will give 7 ?
1 : 12 gives 13, 11 ..............2 : 6 gives 8, 4 .....................3 : 4 gives 7, 1
so
which of the '+' & '-' terms makes +12? ........and when added gives -7?
these are the choices: (+3)(+4), (-3)(+4), (+3)(-4) or (-3)(-4)
clearly, (-3)(-4) are the two factors we want
therefore
Example #1
Example #2
Example #3
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# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12, the angle between sides B and C is (5pi)/12, and the length of B is 10, what is the area of the triangle?
Aug 17, 2017
The area of the triangle is $12.50$ sq.unit .
#### Explanation:
The angle between sides $A \mathmr{and} B$ is $\angle c = \frac{\pi}{12} = \frac{180}{12} = {15}^{0}$
The angle between sides $B \mathmr{and} C$ is /_a =(5 pi)/12=(5*180)/12= 75^0
The angle between sides $C \mathmr{and} A$ is $\angle b = 180 - \left(75 + 15\right) = {90}^{0}$
We can find side $A$ by aplying sine law A/sina = B/sinb ; B=10
$A = B \cdot \sin \frac{a}{\sin} b = 10 \cdot \sin \frac{75}{\sin} 90 \approx 9.66$
Now we have Sides $A \left(9.66\right) , B \left(10\right)$ and their included angle
$\angle c = {15}^{0}$. The area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$
or ${A}_{t} = \frac{10 \cdot 9.66 \cdot \sin 15}{2} = 12.50$ sq.unit [Ans]
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# What Is A Pure Imaginary Number? (3 Key Ideas To Know)
Imaginary numbers are useful throughout mathematics, including algebra (for solving some quadratic equations) calculus, and differential equations.
So, what is a pure imaginary number? A pure imaginary number is a complex number whose real part is zero (that is, Re(z) = 0). So, if a complex number z has the form a + bi and a = 0, then z = bi is a pure imaginary number. A pure imaginary number bi is a square root of the negative number –b2 (the other square root of –b2 is –bi).
Of course, the word “pure” is not necessary – a pure imaginary number is the same as an imaginary number, since both have a real part that is zero.
In this article, we’ll talk about (pure) imaginary numbers and answer some questions about them. We’ll also take a look at how to graph imaginary and complex numbers.
Let’s get started.
## What Is A Pure Imaginary Number?
A pure imaginary number is a specific type of complex number. A complex number a + bi is the sum of a real part “a” and an imaginary part “bi”.
A pure imaginary number has a real part that is zero – that is, a = 0. So, a pure imaginary number has the form 0 + bi, or just bi.
Another notation you may see is Re(z) = 0. Re(z) just means “the real part of the complex number z”. So Re(z) = 0 means that the real part of the complex number z is zero.
A pure imaginary number is one square root of a negative number (the other square root is the negative of that number).
For example, 2i is a square root of -4, and -2i is another square root of -4, since:
• (2i)2 = 4i2 = 4(-1) = -4
• (-2i)2 = 4i2 = 4(-1) = -4
### How Imaginary Numbers Work
Imaginary numbers work by giving us a way to talk about the “square root of a negative number”. We start out by defining the imaginary unit i = √-1.
So, i is the square root of -1, and so i2 = -1. We can also say that i3 = -i and i4 = 1, since:
• i3 = i2*i = (-1)*i = -i
• i4 = i2*i2 = (-1)*(-1) = 1
We can use these four basic powers of i (along with the rules of exponents) to simplify any power of i. For example:
• i2022 = (i2)1011 = (-1)1011 = -1 (since 1011 is odd)
Once we have defined the imaginary unit i = √-1, we can use it to define other imaginary numbers. For example:
• √-4 = √4√-1 = 2i
Note that -2i is also a square root of -4.
More generally, the two square roots of a negative number –b2 are bi and –bi.
So, the set of imaginary number is the set:
• {bi | b is a real number}
We can also write this as:
• {a + bi | a = 0 and b is a real number}
This is a subset of the set of complex numbers, which are defined by:
• {a + bi | a and b are real numbers}
So, a complex number is just a sum of a real number (the “a” part) and an imaginary number (the “bi” part).
### What Is The Square Root Of The Imaginary Unit? (Square Root Of i)
First, remember that the square root of i is a complex number, so it has the form a + bi.
So, we want a number a + bi such that:
• (a + bi)*(a + bi) = i
First, let’s use FOIL on the left side, then we will combine like terms and use the powers of i:
• a2 + abi + abi + b2i2 =i [used FOIL on the left side]
• a2 + 2abi + b2i2 = i [combined like terms]
• a2 + 2abi + b2(-1) = i [used i2 = -1]
• (a2 – b2) + (2ab)I = 0 + 1i [grouped real and imaginary parts]
When we set the real terms on both sides of the equation equal to each other, we get:
• a2 – b2 = 0
This implies that (a + b)(a – b) = 0 [factored by a difference of squares!]
So a + b = 0, or a – b = 0, meaning a = -b, or a = b.
Now, when we set the imaginary terms on both sides of the equation equal to each other, we get:
• 2abi = 1i
• 2ab = 1
• ab = 1/2
Combining this with the equations above, we get two possibilities.
For a = -b:
• ab = 1/2
• -b*b = 1/2
• -b2 = 1/2
• b2 = -1/2
This is impossible, since b is a real number (not imaginary).
For a = b:
• ab = 1/2
• b*b = 1/2
• b2 = 1/2
• b = √(1/2)
• b = √2 / 2
Since a = b, this means a = √2 / 2 also. So one of the square roots of i is:
• a + bi = (√2 / 2) + (√2 / 2)i = (√2 / 2)(1 + i)
The negative of this number is the other square root of i:
• (-√2 / 2) – (√2 / 2)i = (√2 / 2)(1 – i)
### Is Zero A Pure Imaginary Number?
Zero is a pure imaginary number, since its real part is zero. That is, we can write the number zero as follows:
• 0 = 0 + 0i
which has the form a + bi, meaning 0 = 0 + 0i is a complex number (a = 0, b = 0). The real part is equal to zero (a = 0), so 0 = 0 + 0i is a pure imaginary number.
Note that zero is also a real number, since its imaginary part is zero (b = 0).
### Can Imaginary Numbers Be Negative?
Imaginary numbers cannot be negative. The concepts of positive and negative only apply to nonzero real numbers.
Remember that:
• a positive number is to the right of zero on the real number line
• a negative number is to the left of zero on the real number line
On the other hand, an imaginary number is not on the real number line. So, we cannot say which side of zero an imaginary number can be found.
Thus, an imaginary number is neither positive nor negative.
However, we can take the coefficient of the imaginary unit i and look at its sign. That is, in the imaginary number bi, we can tell if the value of b is positive or negative.
For example:
• 2i is an imaginary number that is neither positive nor negative. However, b = 2 is positive.
• -5i is an imaginary number that is neither positive nor negative. However, b = -5 is negative.
### Can Imaginary Numbers Be Rational?
Imaginary numbers cannot be rational, since a rational number is a real number, and an imaginary number is not a real number.
However, we can take the coefficient of the imaginary unit i and determine if it is rational. That is, in the imaginary number bi, we can tell if the value of b is rational or not.
For example:
• 0.4i is an imaginary number that is not rational. However, b = 0.4 can be written as the fraction 2/5, which means b is rational.
• (√2)*i is an imaginary number that is not rational. However, b = √2 cannot be written as a ratio of two integers, so it is irrational.
### Can Imaginary Numbers Be Irrational?
Imaginary numbers cannot be irrational, since an irrational number is a real number, and an imaginary number is not a real number.
However, we can take the coefficient of the imaginary unit i and determine if it is irrational or not (see the examples above). That is, in the imaginary number bi, we can tell if the value of b is irrational or not.
### Can You Have An Imaginary Number In The Denominator?
You can have an imaginary number in the denominator of a fraction. This is a perfectly valid expression.
#### Example 1: A Fraction With An Imaginary Number In The Denominator
Let’s say we have the fraction 1 / i. This is a fraction with an imaginary denominator.
We can simplify this fraction as follows:
• 1 / i
• =(1 / i)*(i / i) [multiply by i / i]
• = 1*i / i*i
• =i / i2
• =i / -1 [since i2 = -1]
• =-i
So 1 / i is equal to –i (which is also equal to i3).
#### Example 2: A Fraction With A Complex Number In The Denominator
Let’s say we have the fraction 1 / (2 + i). This is a fraction with a complex denominator.
We can simplify this fraction as follows:
• 1 / (2 + i)
• =(1 / (2 + i)*((2 – i) / (2 – i)) [multiply by (2 – i) / (2 – i), since 2 – i is the complex conjugate of 2 + i]
• = 1*(2 – i) / (2 + i)*(2 – i)
• =(2 – i) / (4 + 2i – 2i – i2) [FOIL the denominator]
• =(2 – i) / (4 – i2) [combine like terms in the denominator]
• =(2 – i) / (4 – (-1)) [since i2 = -1]
• =(2 – i) / 5
So 1 / (2 + i) is equal to (2 – i) / 5 or (2/5) – (2/5)i.
### Can You Square An Imaginary Number?
You can square an imaginary number, and the result will always be a negative number (or zero when the imaginary number is zero). We can prove this as follows.
Let bi be an imaginary number (so b is a real number).
When we square bi, we get:
• (bi)2
• =b2i2
• =b2(-1) [since i2 = -1]
• =-b2
Since b is real, then b2 >= 0, and so –b2 <= 0. That is, (bi)2 is negative for nonzero b, and zero for b = 0.
### Can Imaginary Numbers Be Graphed?
You can graph imaginary numbers. However, you need to use a real axis (x-axis) and an imaginary axis (y-axis) to graph complex or imaginary numbers.
Here are some examples.
#### Example 1: Graphing The Imaginary Number 2i
The number 2i is really 0 + 2i, which is a complex number of the form a + bi (a = 0, b = 2). So, we would graph this as the point where a = 0 (real axis) and b = 2 (imaginary axis).
This point is 2 units above the origin in the complex plane. You can see this illustrated below.
#### Example 1: Graphing The Complex Number 3 + 4i
The number 3 + 4i is really a complex number of the form a + bi (a = 3, b = 4). So, we would graph this as the point where a = 3 (real axis) and b = 4 (imaginary axis).
This point is 5 units away from the origin in the complex plane. You can see this illustrated below.
## Conclusion
Now you know what a pure imaginary number is and how they are connected to square roots of negative numbers.
You can learn all about the uses of imaginary numbers here.
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Logarithms Exponents and Trees
Base
Logarithms and exponents have a base which is a number. Similarly, as a tree grows the tree will sprout new branches. This “branching factor” can be thought of the base of a tree, exponent, or logarithm. Take for example trees of base two, three and four.
At each level of the tree, new branches sprout equal to the base. If you were to try and draw trees with higher bases, you might be surprised by how quickly they grow. This consequence of branching is one of the fundamental ideas that separates exponents and logarithms from “linear thinking”. Instead, exponents are like thinking in a “multiplicative-way”. At each level the number of nodes is multiplied instead of having some constant added. Inversely, taking the logarithm of a number is like thinking in a “division-like-way”.
Trees
Play around with the variables below to see how the number of leaves change based on the number of levels in the tree and the branching factor at each level.
Exponents
The above image describes the mathematical form of exponentiation. The base corresponds to the branching factor, y corresponds with the number of levels, and finally x corresponds to the number of leaves in the tree.
Example:
The expression 23 can be represented by the tree to the left. In this example, the base is two, so at each level of the tree the node will split into two. If we draw the tree and count the number of leaves then we quickly come up with the answer eight: .
Example:
The expression 32 can be represented by the tree to the left. In this example, the base is three, so at each level of the tree the node will split into three. If we draw the tree and count the number of leaves then we quickly come up with the answer nine: .
Logarithms
The above interactive also describes the mathematical form of taking the logarithm of a number, which is the inverse operation of exponentiation. Here the roles are reversed, instead of solving for the number of leaves, we solve for the number of levels in the tree.
Example:
The expression can be represented by the tree to the left. Note that the base is two, which corresponds to a branching factor of two. Instead of counting the number of leaves, we count the number of levels in the tree.
Example:
The expression can be visualzed by the tree to the left. Note that the base is three, which corresponds to a branching factor of three. To evaluate this expression we count the number of levels in the tree: . Another way to think of this is how many times do we divide nine by three and end up with one?
Conclusion
Try and formulate some simple problems, such as 42 or log3 27, and draw a little tree to solve them. Hopefully, this gives you a simple tool to visualize and reason about problems concerning exponents and logarithms. Maybe the next time you look at a tree, bush, or plant, you will see it a little bit more like a mathematician.
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# Equation of a Circle when the circle passes through the origin
Expressing a circle in mathematical equation form when the circle passes through the origin is defined equation of a circle when the circle passes through the origin.
A circle can be passed through the origin of the Cartesian coordinate system. It can be expressed in algebraic form, based on the geometrical relation between the circle and Cartesian coordinate system. Here, you learn how to develop an equation of a circle in algebraic form expression when the circle passes through the origin of the Cartesian coordinate geometric system.
## Derivation
Assume, $O$ is an origin of the Cartesian coordinate system.
Assume, a circle having $r$ units radius, passes through the origin of the Cartesian coordinate geometric system.
Assume, the centre of the circle is located at $a$ units distance in horizontal direction and $b$ units distance in vertical direction from the origin. The centre of the circle is named $P$. So, the coordinates of the centre is represented as $P\left(a,b\right)$ in coordinate geometric system.
Join the origin $O$ and centre of the circle $P$. Actually, the circle is passing through the origin. Therefore, the distance from origin to centre of the circle is equal to the radius of the circle. So, $\stackrel{‾}{OP}=r$.
Draw a perpendicular line to horizontal axis from the centre of the circle and assume, it meets horizontal axis at a point, named $Q$. According to figure, the point $Q$ is located at $a$ units distance from the origin in horizontal axis direction and zero units distance from the same origin in vertical axis direction. Therefore, the distance between origin and point $Q$ is $\stackrel{‾}{OQ}=a$.
The distance from point $Q$ to centre of the circle is $\stackrel{‾}{PQ}=b$.
A right angle triangle, known $\Delta POQ$ is constructed inside the circle. The sides of the right angle triangle are line segments $\stackrel{‾}{OP},\stackrel{‾}{PQ}$ and $\stackrel{‾}{OQ}$.
According to the Pythagorean Theorem,
${\left(Hypotenuse\right)}^{2}={\left(Opposite side\right)}^{2}+{\left(Adjacent side\right)}^{2}$
${OP}^{2}={PQ}^{2}+{OQ}^{2}⇒{r}^{2}={b}^{2}+{a}^{2}$
Later, this mathematical relation will be used in developing the equation of the circle.
It will be used later in developing the equation of the circle. So, look away from this mathematical relation temporarily.
Consider a point on the circle and assume the point represents each and every point on the circle. It means, it is a variable point, named $S$. Assume, it is located at $x$ units distance in horizontal axis direction and $y$ units distance in vertical axis direction from the origin. Therefore, the coordinates of the point $S$ is written $S\left(x,y\right)$.
Join the point $P$ and $S$ and the distance between point $P$ and $S$ is equal to radius of the circle. Therefore, $\stackrel{‾}{PS}=r$.
Draw a perpendicular line towards horizontal axis from the point $S$ and also draw a line, parallel to the horizontal axis from point $P$. Assume, both lines meet at a point $R$. Thus, a right angled triangle, known $\Delta SPR$ is constructed.
Apply Pythagorean Theorem to right angled triangle $\Delta SPR$.
${PS}^{2}={SR}^{2}+{PR}^{2}$
As per the right angled triangle, and $\stackrel{‾}{PS}=r$. Substitute these values in the Pythagorean Theorem to get the required equation.
It is already proved above that ${r}^{2}={b}^{2}+{a}^{2}$. Substitute it in the above expression to start simplifying the expression.
It can be written as follows
$⇒{x}^{2}+{y}^{2}=2\left(ax+by\right)$
It is an algebraic equation which represents a circle when the circle passes through the origin of the Cartesian coordinate system.
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# Congruence of Triangles
Two triangles are congruent if their corresponding sides and angles are equal. They can be flipped, rotated, or moved, but they will still look the same. The symbol for congruence is ‘‘.
In mathematics, congruence means two shapes are the same in both size and shape. We can tell if two triangles are congruent by following four main rules, and we only need to know three out of six measurements to prove it. The matching sides and angles of congruent triangles are always equal.
## Congruence in Math
Congruence in math refers to shapes that can be moved or flipped to fit exactly on top of each other. Two shapes are congruent if they have the same shape and size, like a shape and its mirror image.
## Congruent Triangles
A triangle is a three-sided polygon. Two triangles are congruent if their sides are the same length and their angles are the same measure. For example, in the figure, triangles Δ ABC and Δ PQR are congruent if:
• Vertices match: A with P, B with Q, C with R
• Sides match: AB = PQ, BC = QR, AC = PR
• Angles match: ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
Congruent triangles have the same area and perimeter.
## CPCT Full Form
CPCT stands for “Corresponding Parts of Congruent Triangles.” This means that the matching parts of congruent triangles are equal. When solving problems about triangles, we use CPCT to quickly refer to this concept.
## CPCT Rules in Math
After proving triangles congruent, we can predict the other dimensions without measuring. Here are the rules:
• SSS (Side-Side-Side): If all three sides of one triangle match the three sides of another, they are congruent.
• SAS (Side-Angle-Side): If two sides and the included angle of one triangle match the corresponding parts of another, they are congruent.
• ASA (Angle-Side-Angle): If two angles and the included side of one triangle match the corresponding parts of another, they are congruent.
• AAS (Angle-Angle-Side): If two angles and a non-included side of one triangle match the corresponding parts of another, they are congruent.
• RHS (Right angle-Hypotenuse-Side): If the hypotenuse and one side of a right-angled triangle match the corresponding parts of another, they are congruent.
## FAQ’s
How can congruence be used in real-life problems?2024-08-09T10:26:51+05:30
#### How can congruence be used in real-life problems?
Congruence can help in solving problems involving shapes and structures in geometry, ensuring designs are accurate and matching parts fit together perfectly.
What does CPCT stand for?2024-08-09T10:26:20+05:30
#### What does CPCT stand for?
CPCT stands for “Corresponding Parts of Congruent Triangles”
What is the symbol for congruence?2024-08-09T10:25:54+05:30
#### What is the symbol for congruence?
The symbol for congruence is ‘‘.
What are the rules to prove triangle congruence?2024-08-09T10:22:42+05:30
#### What are the rules to prove triangle congruence?
The main rules are SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), AAS (Angle-Angle-Side), and RHS (Right angle-Hypotenuse-Side).
What does it mean for two triangles to be congruent?2024-08-09T10:21:45+05:30
#### What does it mean for two triangles to be congruent?
Two triangles are congruent if their corresponding sides and angles are equal, meaning they have the same shape and size.
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# Probability and Cumulative Distribution Functions
• Apr 20, 2008
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#### Transcript
1 Lesson 20 Probability and Cumulative Distribution Functions
2 Recall If p(x) is a density function for some characteristic of a population, then
3 Recall If p(x) is a density function for some characteristic of a population, then We also know that for any density function,
4 Recall We also interpret density functions as probabilities: If p(x) is a probability density function (pdf), then
5 Cumulative Distribution Function Suppose p(x) is a density function for a quantity. The cumulative distribution function (cdf) for the quantity is defined as Gives: The proportion of population with value less than x The probability of having a value less than x.
6 Example: A Spinner Last class: A spinner that could take on any value 0x 360. Density function: p(x) = 1/360 if 0x 360, and 0 everywhere else.
7 Example: A Spinner Last class: A spinner that could take on any value 0x 360. Density function: p(x) = 1/360 if 0x 360, and 0 everywhere else. CDF:
8 Example: A Spinner Cumulative Density Function: Distribution Function:
9 Properties of CDFs P(x) is the probability of values less than x If P(x) is the cdf for the age in months of fish in a lake, then P(10) is the probability a random fish is 10 months or younger. P(x) goes to 0 as x gets smaller: (In many cases, we may reach 0.)
10 Properties of CDFs Conversely, P(x) is non-decreasing. The derivative is a density function, which cannot be negative. Also, P(4) cant be less than P(3), for example.
11 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
12 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
13 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
14 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
15 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
16 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
17 Practice Life expectancy (in days) of electronic component has density function , for x 1, and p(x) =0 for x
18 Example Someone claims this is the CDF for grades on the 2015 final exam. Probability a random student scored. 25 or lower? 1, or 100%. Conclusion? 50 or lower? Theyre lying! This cannot be a cumulative distribution 0.5, or 50%. function! (It decreases.)
19 Relating the CDF and DF According the FTC version 2, since , then P'(x)=p(x).
20 Relating the CDF and DF According the FTC version 2, since , then P'(x)=p(x). So the density function is the derivative, or rate of change, of the cumulative distribution function.
21 Example Sketch the density function for the cdf shown.
22 Example Sketch the density function for the cdf shown. Slope?
23 Example Sketch the density function for the cdf shown. Slope? 0
24 Example Sketch the density function for the cdf shown. Slope? 0 Slope 1/2
25 Example Sketch the density function for Slope 0 the cdf shown. Slope? 0 Slope 1/2
26 Example Sketch the density function for Slope 0 the cdf shown. Slope? Slope 1/2 0 Slope 1/2
27 Example Slope 0 Sketch the density function for Slope 0 the cdf shown. Slope? Slope 1/2 0 Slope 1/2
28 Example Slope 0 Sketch the density function for Slope 0 the cdf shown. Slope? Slope 1/2 0 Slope 1/2 Density Function
29 DF vs. CDF You must know which is which. We work differently with density functions than with cumulative distribution functions.
30 Another Example Suppose the cumulative distribution function for the height of trees in a forest (in feet) is given by Find the height, x for which exactly half the trees are taller than x feet, and half the trees are shorter than x. In case - Quadratic Formula: !b b 2 ! 4ac x= 2a
31 Another Example Find the height, x for which exactly half the trees are taller than x feet, and half the trees are shorter than x. 0.1x ! 0.0025x 2 = 0.5 !0.0025x 2 + 0.1x ! 0.5 = 0 !0.0025(x 2 ! 40x + 200) = 0 using quadratic : x = 20 10 2 = 20 !10 2 " 5.86 ft
32 Example Assume p(x) is the density function "1 \$ ,!!!if !1! x ! 3 p(x) = # 2 \$%0,!!!if !x 3 Let P(x) represent the corresponding (1) Cumulative Distribution function. Then (a) For the density function given above, what is the cumulative distribution function P(x) for values of x between 1 and 3? (2) (b) P(5) = ? (c) Choose the correct pair of graphs for p(x)!and!P(x) . (3) (4)
33 In-class Assignment The time to conduct a routine maintenance check on a machine has a cumulative distribution function P(t), which gives the fraction of maintenance checks completed in time less than or equal to t minutes. Values of P(t) are given in the table. t, minutes 0 5 10 15 20 25 30 P(t), fraction completed 0 .03 .08 .21 .38 .80 .98 a. What fraction of maintenance checks are completed in 15 minutes or less? b. What fraction of maintenance checks take longer than 30 minutes? c. What fraction take between 10 and 15 minutes?
34 In-class Assignment The time to conduct a routine maintenance check on a machine has a cumulative distribution function P(t), which gives the fraction of maintenance checks completed in time less than or equal to t minutes. Values of P(t) are given in the table. t, minutes 0 5 10 15 20 25 30 P(t), fraction completed 0 .03 .08 .21 .38 .80 .98 a. What fraction of maintenance checks are completed in 15 minutes or less? 21% b. What fraction of maintenance checks take longer than 30 minutes? 2% c. What fraction take between 10 and 15 minutes? 13%
35 Lesson 21 Mean and Median
36 The Median The median is the halfway point: Half the population has a lower value, and half a higher value. If p(x) is a density function, then the median of the distribution is the point M such that M #!" p(x)dx = 0.5
37 Example Suppose an insect life span of months, and 0 elsewhere. Whats the median life span? We need Then solve for M: months
38 Median From CDF If we have a CDF P(x), the median M is the point where P(M)=0.5 (This of course says 1/2 of the population has a value less than M!)
39 Example CDF: P(t) = t2, 0t 1. Whats the median of this distribution?
40 Example CDF: P(t) = t2, 0t 1. Whats the median of this distribution?
41 The Mean What most people think of as the average: Add up n values and divide by n. How should we represent this for something represented by a density function p(x)?
42 The Mean We define the mean of a distribution given by density function p(x) to be
43 Example Previous insect population example: and p(x) =0 elsewhere. Whats the mean lifespan?
44 Example Previous insect population example: and p(x) =0 elsewhere. Whats the mean lifespan? 8 months. The mean is slightly below the median. This means
45 Example Previous insect population example: and p(x) =0 elsewhere. Whats the mean lifespan? 8 months. The mean is slightly below the median. This means more than half the insects live longer than the mean lifespan.
46 Example 2 From Another Example, p(x) = 0.1! 0.005x # # " 20 mean = x p(x)dx = x(0.1! 0.005x)dx = !" 0 # 3 20 20 0.1x ! 0.005x dx = 0.05x ! 0.0016 x 2 2 0 0 = 6.64 ft
47 Geometry Turns out: the mean is the point where the graph of the distribution would balance if we cut it out.
48 Example Find the mean and median for and zero elsewhere. Median: Solve for M
49 Example Find the mean and median for and zero elsewhere. Mean: Evaluate
50 Example Find the mean and median for and zero elsewhere. median mean
51 The Normal Distribution A special distribution:
52 The Normal Distribution A special distribution:
53 The Normal Distribution Heres an example with =0 and =1: The distribution is symmetric about x= . So the mean is , and so is the median. is called the standard deviation, and describes how spread out the curve is.
54 The Normal Distribution Unfortunately, has no elementary antiderivative. So we must use numerical means to evaluate integrals involving the normal distribution.
55 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 What are the mean and median?
56 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 What are the mean and median? Here, =10, so both the mean and median are 10million people.
57 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 What is the standard deviation?
58 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 What is the standard deviation? Here, we see that =2, so 2million people.
59 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 What is the standard deviation? Here, we see that =2, so 2million people.
60 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 Write an integral for the fraction of time between 8 and 12 million people are on the internet.
61 Example Suppose a density function for number of people (in millions) on the internet at one time were 2 Write an integral for the fraction of time between 8 and 12 million people are on the internet. 2
62 Evaluating the Integral How to calculate the required integral? 2
63 Evaluating the Integral How to calculate the required integral? 2 We could use our Simpsons rule calculator:
64 Evaluating the Integral How to calculate the required integral? 2 We could use our Simpsons rule calculator: n !x Result 2 2 0.693237 4 1 0.683058 8 0.5 0.682711 16 0.25 0.682691 32 0.125 0.68269
65 Evaluating the Integral How to calculate the required integral? 2 We could use our Simpsons rule calculator: n !x Result Seems to converge 2 2 0.693237 4 1 0.683058 to about 0.68. 8 0.5 0.682711 16 0.25 0.682691 32 0.125 0.68269
66 In Class Assignment Find the mean and median of the distribution with density function: (calculations should be to 2 decimal places) 1 3 p(x) = x for 0 < x < 2!and!0!elsewhere 4
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# How To Work Out Ratio
Here we will learn about how to work out ratio, including how to write and simplify ratios.
There are also working out ratio worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is working out ratio?
Working out ratio is a way of showing the comparative size between two or more quantities. Ratios are usually written in the form a:b where a and b are integers (whole numbers).
Example \bf{1}
In a class of students there are 13 boys and 17 girls.
The ratio of boys to girls can be written as 13:17.
The ratio of girls to boys can be written as 17:13.
Example \bf{2}
In a recipe there are three eggs, one onion and two tomatoes. The ratio of eggs to onions to tomatoes is 3:1:2.
Note: You can also write a ratio as a fraction in the form \frac{a}{b} (where a and b are integers) but this is less commonly used in the GCSE syllabus.
The ratio of boys to girls from example 1 would be \frac{13}{17}.
The ratio of girls to boys from example 1 would be \frac{17}{13}.
The subject mentioned first is the numerator of the fraction and the subject mentioned second is the denominator.
Example \bf{3} : Here are some coloured squares.
Rearranging them into coloured bars, we can easily see the number of purple squares in comparison to the number of pink squares.
There are 7 purple squares and 5 pink squares. There are 12 squares in total.
The ratio of purple to pink squares is 7:5.
The ratio of pink to purple squares is 5:7.
The ratio of purple squares to the total number of squares is 7:12.
The ratio of pink squares to the total number of squares is 5:12.
Note: After you have written a ratio from a word problem, like in the examples above, you must then check if the ratio simplifies. When a ratio is fully simplified the parts of the ratio are all integers (whole numbers) with no common factors. This means a ratio must not include any decimals or fractions.
Step-by-step guide: Simplifying ratios
## How to work out ratio
In order to work out ratio:
1. Identify the different quantities in the question.
2. Identify the order in which the quantities are to be represented.
3. Write the ratio using a colon.
4. Check if the ratio can be simplified*.
*Remember that a simplified ratio does not include any fractions or decimals.
## Related lessons on ratio
How to work out ratio is part of our series of lessons to support revision on ratio. You may find it helpful to start with the main ratio lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
## How to work out ratio examples
### Example 1: writing a simple ratio about an everyday life situation
Ms Holly is looking after 7 children. She only has 5 toys in her nursery. Write the ratio of children to toys.
1. Identify the different quantities in the question.
There are 7 children.
There are 5 toys.
2Identify the order in which the quantities are to be represented.
Children to toys.
3Write the ratio using a colon.
children : toys
7:5
4Check if the ratio can be simplified.
This ratio is already in its simplest form.
### Example 2: writing a three part ratio about a real life situation
On a farm there are 20 pigs, 8 cows and 124 chickens. Write the ratio of cows to pigs to chickens in simplest form.
Identify the different quantities in the question.
Identify the order in which the quantities are to be represented.
Write the ratio using a colon.
Check if the ratio can be simplified.
### Example 3: writing a ratio from a picture
Use the diagram below to write the ratio of pink counters to the total number of counters in simplest form.
Identify the different quantities in the question.
Identify the order in which the quantities are to be represented.
Write the ratio using a colon.
Check if the ratio can be simplified.
### Example 4: writing a ratio from a graph
The comparative bar chart shows the number of left handed and right handed people in a group.
Write the ratio of left handed people to the total number of people in its simplest form.
Identify the different quantities in the question.
Identify the order in which the quantities are to be represented.
Write the ratio using a colon.
Check if the ratio can be simplified.
### Example 5: writing a ratio from a word problem involving fractions
In a fridge there are three different types of drinks in bottles: water, coke and lemonade. \frac{3}{8} of the drinks are coke. \frac{1}{2} of the drinks are water. Write the ratio of water to coke to lemonade.
Identify the different quantities in the question.
Identify the order in which the quantities are to be represented.
Write the ratio using a colon.
Check if the ratio can be simplified.
### Example 6: writing an algebraic ratio from a word problem
Amy is \textbf{a} years old. Imran is 12 years older than Amy. Write the ratio of Imran’s age to Amy’s age.
Identify the different quantities in the question.
Identify the order in which the quantities are to be represented.
Write the ratio using a colon.
Check if the ratio can be simplified.
### Common misconceptions
• Writing the parts of the ratio in the wrong order
When a ratio is written, the order is very important. Remember to read the word problem carefully to ensure you get it right.
For example, there are 12 dogs and 13 cats. What is the ratio of cats to dogs?
A common mistake would be to write 12:13 because this is the order the numbers are presented in the information. However the question asks for the ratio of ‘cats to dogs’ so the correct answer is 13:12 because there are 13 cats and 12 dogs.
### Practice working out ratio questions
1. In a ball pool there are 258 red balls, 300 yellow balls and 546 blue balls. Write the ratio of blue balls to yellow balls to red balls in simplest form.
91:50:43
258:300:546
273:150:129
31:25:21
The parts of the ratio should be in the order blue : yellow : red.
First write 546:300:258.
By dividing through by 6 this simplifies to 91:50:43.
The correct ratio must have its parts in the correct order and be simplified.
2. A café sells 21 cups of coffee and 14 cups of tea during one lunch break. Write the ratio of coffee to tea sales in its simplest form.
21:14
3:5
3:2
21:35
21 cups of coffee, 14 cups of tea.
Ratio of coffee to tea is 21:14.
By dividing through by 7 this simplifies to 3:2.
The correct ratio must have it’s part in the correct order and be simplified.
3. Matio is investigating how many people have the Facebook application on their phone. He stands outside a shop and conducts a survey on people he sees using a smartphone. 75 people report that they have the Facebook application.
25 people report that they do not have the Facebook application. Write the ratio of people who don’t have the Facebook application on their phone to the total number of people surveyed.
1:3
3:1
3:4
1:4
The parts of the ratio should be in the order ‘no Facebook to total’.
Calculate the total 75+25 = 100.
Write the ratio 25:100.
By dividing through by 25 this simplifies to 1:4.
4. The pie chart shows the variety of chocolate bars in a tin. Write the ratio of Leon to KittyKats to Twisles in simplest form.
1:2
90:90:180
1:1:2
\frac{1}{4}:\frac{1}{4}:\frac{1}{2}
Leon has 90^o of the pie chart, KittyKat has 90^o of the pie chart and Twisles has 180^o of the pie chart.
Ratio of Leon to KittyKat to Twisles is 90:90:180 .
By dividing through by 90 this simplifies to 1:1:2.
5. On a bookcase there are F number of fiction books and N number of non-fiction books. Write the ratio of fiction books to the total number of books.
F:N
F:FN
F:N-F
F:F+N
There are F+N books in total.
Ratio of fiction books to total number of books is F:F+N .
6. At a school \frac{1}{3} of students study French. The rest of the students study German. Write the ratio of students who study German to students who study French in simplest form.
2:1
\frac{1}{3}:\frac{2}{3}
1:3
3:4
\frac{1}{3} of the students study French. The rest of the students study German which as a fraction is \frac{2}{3}.
Ratio of students who study German to students who study French is \frac{2}{3}:\frac{1}{3} (be careful of the order here).
By multiplying through by 3 this simplifies to 2:1.
### How to work out ratio GCSE questions
1. Robbie earns £8.40 per hour and Joseph earns £9.20 per hour. Robbie works for 5 hours each day, and Joseph works for 4.5 hours each day.
What is the ratio of Robbie’s daily pay to Joseph’s daily pay? Write your answer in its simplest form.
(2 marks)
Robbie’s earning for one day = 8.40 \times 5 = 42 .
Joseph’s earning for one day = 9.2 \times 4.5 =41.4 .
42 and 41.4
(1)
\begin{aligned} &42:41.4 \\\\ &420:414 \\\\ &70:69 \end{aligned}
(1)
2. An ice cream van sells three flavours of ice cream; vanilla, strawberry and chocolate. The pie chart shows the proportion of each flavour that was sold one hot summer’s day.
(a) Write the ratio of chocolate to strawberry to vanilla for the ice cream sales shown in this pie chart.
(b) If 42 ice creams were sold in total, how many of them were chocolate?
(3 marks)
(a)
Chocolate = \frac{1}{6}, Strawberry = \frac{1}{3}, Vanilla = \frac{1}{2} / Chocolate 60^{o}, Strawberry 120^{o}, Vanilla 180^o
(1)
1:2:3
(1)
(b)
42 \div 6 = 7
(1)
3. Sammy, Yip and Harleen go to a sweet shop. Sammy buys twice as many sweets as Yip. Harleen buys 5 more sweets than Sammy. If Yip buys \textbf{n} sweets, write the ratio of Yip’s sweets to the total number of sweets.
(3 Marks)
Yip = n, Sammy = 2n, Harleen = 2n + 5
(1)
Total = n + 2n + 2n + 5 = 5n + 5
(1)
n:5n+5
(1)
## Learning checklist
You have now learned how to:
• Use ratio notation, including reduction to simplest form
• Express a multiplicative relationship between 2 quantities as a ratio or a fraction
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.
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Introduction, Basic Concepts & Formulae: Mensuration- 1
# Important Formulas for CAT Mensuration- 1
## General Test Preparation for CUET
154 videos|396 docs|683 tests
## FAQs on Important Formulas for CAT Mensuration- 1
1. What is mensuration and why is it important in mathematics?
Ans. Mensuration is a branch of mathematics that deals with the measurement of geometric figures and their parameters such as area, volume, and perimeter. It is important in mathematics as it helps in understanding and solving real-life problems related to measurement, construction, and design.
2. What are the basic concepts of mensuration?
Ans. The basic concepts of mensuration include understanding the concepts of length, area, volume, and perimeter. Length refers to the distance between two points, area refers to the measure of surface covered by a figure, volume refers to the measure of space occupied by a figure, and perimeter refers to the distance around a closed figure.
3. What are the formulas for finding the area of different geometric figures in mensuration?
Ans. The formulas for finding the area of different geometric figures in mensuration are: - Rectangle: Area = length × width - Square: Area = side × side - Triangle: Area = 1/2 × base × height - Circle: Area = π × radius^2
4. How do you find the volume of different 3D shapes in mensuration?
Ans. The volume of different 3D shapes in mensuration can be found using the following formulas: - Cube: Volume = side × side × side - Cylinder: Volume = π × radius^2 × height - Cone: Volume = 1/3 × π × radius^2 × height - Sphere: Volume = 4/3 × π × radius^3
5. How is mensuration used in real-life applications?
Ans. Mensuration is used in various real-life applications such as construction, architecture, engineering, and design. It helps in calculating the materials required, estimating costs, designing structures, and measuring quantities accurately. For example, mensuration is used in calculating the area of a field, determining the volume of a container, or designing the layout of a room.
## General Test Preparation for CUET
154 videos|396 docs|683 tests
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## What you need to know
Things to remember:
• We can extend the number line so that we have numbers on the left of 0.
• We can flip a subtraction, and it will change the answer from positive to negative, or negative to positive.
Usually our number line starts at 0 and goes up in 1s like this:
And we can use our number line to help us with subtractions.
What is $5 -4$?
Using our number line we’d start at 5, and count back 4.
So, we can see that $5-4=1$. But, what would happen if we subtracted a bigger number from a smaller number?
What is $3-6$?
So, we can only subtract 3 before we don’t have anywhere to go, so let’s add our negative numbers.
Now that we have negative numbers, we can start counting back again.
So, we can see that $3-6=-3$. Interestingly, $6-3=3$. Let’s compare these.
$$3-6=-3$$
$$6-3=3$$
If we change the order of the subtraction, then we either add or remove the subtraction.
Earlier we saw that $5-4=1$, so $4-5=-1$. Let’s see some more.
$$8-5=3$$
$$5-8=-3$$
$$12-11=1$$
$$11-12=-1$$
$$25-8=17$$
$$8-25=-17$$
## Example Questions
#### Question 1: What is $5-7$?
So, $5-7=-2$.
Alternatively, we can flip it!
$$7-5=2$$
so
$$5-7=-2$$
#### Question 2: What is $3-8$?
So, $3-8=-5$.
Alternatively, we can flip it!
$$8-3=5$$
so
$$3-8=-5$$
|
# divide 2 numbers
Discover divide 2 numbers, include the articles, news, trends, analysis and practical advice about divide 2 numbers on alibabacloud.com
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### Detailed explanation of divide and conquer Algorithms
Detailed explanation of divide and conquer Algorithms I. Basic Concepts In computer science, divide and conquer is an important algorithm. The literal explanation is "divide and conquer". It is to divide a complicated problem into two or more
### Divide and Conquer law (II.)
Refer to the fourth chapter of algorithmic design and analysis Anany levitin translation version of Tsinghua University PressIn the previous article, we introduced the idea of divide-and-conquer strategy, the main theorem, and some classical cases
### Divide and Conquer
Divide and Conquer is an algorithmic paradigm based on multi-branched recursion.A typical Divide and Conquer algorithm solves a problem using following three steps: 1. Divide: Break the given problem into subproblems of same type.2. Conquer:
### Divide and Conquer algorithm (i.)
When we solve some problems, because these problems to deal with a considerable amount of data, or the process is quite complex, so that the direct solution in time is quite long, or simply can not be directly obtained. For this kind of problem, we
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### C language division: Calculate the minimum common divisor of two numbers, and divide the common divisor.
C language division: Calculate the minimum common divisor of two numbers, and divide the common divisor. The maximum division of the moving phase is used to calculate the maximum common divisor of two numbers.(A, B) is used to represent the maximum
### Lightoj 1214 (divide large numbers)
Large DivisionDescriptionGiven integers, a and b, you should check whether a is divisible by b or not. We know A integer a is divisible by an integer b if and only if there exists an integer C such that a = b * C.InputInput starts with an integer T
### 59. Summary: N (n = 1, 2, 3) numbers that only appear once in the array [find N numbers which appear only once in array]
[Link to this article] Http://www.cnblogs.com/hellogiser/p/find-n-numbers-which-appear-only-once-in-array.html 【Question] Three numbers a, B, and c appear only once in an array, and the other numbers appear twice. Find three numbers that appear only
### Using divide-and-conquer method to realize multiplication, addition and subtraction of large numbers (Java implementation)
Large number multiplication is the problem of polynomial multiplication, the product C (x) of a (x) and B (x), the complexity of the Naïve solution O (n^2), the basic idea is to write the polynomial a (x) and B (x)A (x) =a*x^m+bB (x) =c*x^m+dWhere a,
### Numbers, cardinality and representations
Numbers, cardinality and representations integerIntegers are these familiar numbers ...,-1, 0, +1, +2, .... Integer values are also referred to as ' complete ' and are divided into positive numbers (1 to infinity), negative numbers (-1 to
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## Characteristics of Graphs of Logarithmic Functions
### Learning Outcomes
• Determine the domain and range of a logarithmic function.
• Determine the x-intercept and vertical asymptote of a logarithmic function.
• Identify whether a logarithmic function is increasing or decreasing and give the interval.
• Identify the features of a logarithmic function that make it an inverse of an exponential function.
Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.
Recall that the exponential function is defined as $y={b}^{x}$ for any real number x and constant $b>0$, $b\ne 1$, where
• The domain of $y$ is $\left(-\infty ,\infty \right)$.
• The range of $y$ is $\left(0,\infty \right)$.
In the last section we learned that the logarithmic function $y={\mathrm{log}}_{b}\left(x\right)$ is the inverse of the exponential function $y={b}^{x}$. So, as inverse functions:
• The domain of $y={\mathrm{log}}_{b}\left(x\right)$ is the range of $y={b}^{x}$: $\left(0,\infty \right)$.
• The range of $y={\mathrm{log}}_{b}\left(x\right)$ is the domain of $y={b}^{x}$: $\left(-\infty ,\infty \right)$.
Transformations of the parent function $y={\mathrm{log}}_{b}\left(x\right)$ behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of shape.
Previously we saw that certain transformations can change the range of $y={b}^{x}$. Similarly, applying transformations to the parent function $y={\mathrm{log}}_{b}\left(x\right)$ can change the domain. Therefore, when finding the domain of a logarithmic function, it is important to remember that the domain consists only of positive real numbers. That is, the value you are applying the logarithmic function to, also known as the argument of the logarithmic function, must be greater than zero.
For example, consider $f\left(x\right)={\mathrm{log}}_{4}\left(2x - 3\right)$. This function is defined for any values of x such that the argument, in this case $2x - 3$, is greater than zero. To find the domain, we set up an inequality and solve for $x$:
$\begin{array}{l}2x - 3>0\hfill & \text{Show the argument greater than zero}.\hfill \\ 2x>3\hfill & \text{Add 3}.\hfill \\ x>\dfrac{3}{2}\hfill & \text{Divide by 2}.\hfill \end{array}$
In interval notation, the domain of $f\left(x\right)={\mathrm{log}}_{4}\left(2x - 3\right)$ is $\left(\dfrac{3}{2},\infty \right)$.
### Tip for success
Finding the domain of the logarithm function is similar to finding the domain of the square root function.
Recall that for $y = \sqrt{u}, u\geq0$, so the domain of $y=\sqrt{2x-3}$ is all $x$ for which $2x-3\geq0$, or all $x\geq\dfrac{3}{2}$.
Since the logarithm function takes input only strictly greater than zero, we see in the example above that the domain of $f\left(x\right)={\mathrm{log}}_{4}\left(2x - 3\right)$ is $\left(\dfrac{3}{2},\infty \right)$.
### How To: Given a logarithmic function, identify the domain
1. Set up an inequality showing the argument greater than zero.
2. Solve for $x$.
3. Write the domain in interval notation.
### Example: Identifying the Domain Resulting From a Logarithmic Shift
What is the domain of $f\left(x\right)={\mathrm{log}}_{2}\left(x+3\right)$?
### Try It
What is the domain of $f\left(x\right)={\mathrm{log}}_{5}\left(x - 2\right)+1$?
### Example: Identifying the Domain Resulting From a Logarithmic Shift and Reflection
What is the domain of $f\left(x\right)=\mathrm{log}\left(5 - 2x\right)$?
### Try It
What is the domain of $f\left(x\right)=\mathrm{log}\left(x - 5\right)+2$?
## Graphing a Logarithmic Function Using a Table of Values
Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function $y={\mathrm{log}}_{b}\left(x\right)$ along with all of its transformations: shifts, stretches, compressions, and reflections.
We begin with the parent function $y={\mathrm{log}}_{b}\left(x\right)$. Because every logarithmic function of this form is the inverse of an exponential function of the form $y={b}^{x}$, their graphs will be reflections of each other across the line $y=x$. To illustrate this, we can observe the relationship between the input and output values of $y={2}^{x}$ and its equivalent logarithmic form $x={\mathrm{log}}_{2}\left(y\right)$ in the table below.
### recall inverse relationships
Recall that if an invertible function $f(x)$ contains a point, $\left(a, b\right)$, then the inverse function $f^{-1}\left(x\right)$ must contain the point $\left(b, a\right)$.
x –3 –2 –1 0 1 2 3 ${2}^{x}=y$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{2}$ 1 2 4 8 ${\mathrm{log}}_{2}\left(y\right)=x$ –3 –2 –1 0 1 2 3
Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions $f\left(x\right)={2}^{x}$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
$f\left(x\right)={2}^{x}$ $\left(-3,\frac{1}{8}\right)$ $\left(-2,\frac{1}{4}\right)$ $\left(-1,\frac{1}{2}\right)$ $\left(0,1\right)$ $\left(1,2\right)$ $\left(2,4\right)$ $\left(3,8\right)$ $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ $\left(\frac{1}{8},-3\right)$ $\left(\frac{1}{4},-2\right)$ $\left(\frac{1}{2},-1\right)$ $\left(1,0\right)$ $\left(2,1\right)$ $\left(4,2\right)$ $\left(8,3\right)$
As we would expect, the and y-coordinates are reversed for the inverse functions. The figure below shows the graphs of f and g.
Notice that the graphs of $f\left(x\right)={2}^{x}$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ are reflections about the line y = x since they are inverses of each other.
Observe the following from the graph:
• $f\left(x\right)={2}^{x}$ has a y-intercept at $\left(0,1\right)$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ has an x-intercept at $\left(1,0\right)$.
• The domain of $f\left(x\right)={2}^{x}$, $\left(-\infty ,\infty \right)$, is the same as the range of $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
• The range of $f\left(x\right)={2}^{x}$, $\left(0,\infty \right)$, is the same as the domain of $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
### A General Note: Characteristics of the Graph of the Parent Function $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$
For any real number x and constant > 0, $b\ne 1$, we can see the following characteristics in the graph of $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$:
• one-to-one function
• vertical asymptote: = 0
• domain: $\left(0,\infty \right)$
• range: $\left(-\infty ,\infty \right)$
• x-intercept: $\left(1,0\right)$ and key point $\left(b,1\right)$
• y-intercept: none
• increasing if $b>1$
• decreasing if 0 < < 1
### Try it
Use an online graphing tool to graph the function $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$
Adjust the $b$ value to investigate how changes in the base of the logarithmic function affect the graph of that function. Which function is an increasing function? Which is decreasing? Does the x-intercept change when you change the base? Make sure to use values between $0$ and $1$ as well as values greater than $1$.
The graphs above show how changing the base b in $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$ can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function $\mathrm{ln}\left(x\right)$ has base $e\approx \text{2}.\text{718.)}$
The graphs of three logarithmic functions with different bases all greater than 1.
### How To: Given a logarithmic function Of the form $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$, graph the function
1. Draw and label the vertical asymptote, x = 0.
2. Plot the x-intercept, $\left(1,0\right)$.
3. Plot the key point $\left(b,1\right)$.
4. Draw a smooth curve through the points.
5. State the domain, $\left(0,\infty \right)$, the range, $\left(-\infty ,\infty \right)$, and the vertical asymptote, x = 0.
### Example: Graphing a Logarithmic Function Of the Form $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$
Graph $f\left(x\right)={\mathrm{log}}_{5}\left(x\right)$. State the domain, range, and asymptote.
### Try It
Graph $f\left(x\right)={\mathrm{log}}_{\frac{1}{5}}\left(x\right)$. State the domain, range, and asymptote.
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# How do you find the perimeter of a triangle given (2, 2), (-2, 2), and (-2, -3)?
Apr 21, 2018
Perimeter is $15.4$
#### Explanation:
To find perimeter, we should find sum of all the three sides of the triangle formed by $A \left(2 , 2\right)$, $B \left(- 2 , 2\right)$ and $C \left(- 2 , - 3\right)$
Before we do that observe that between $A$ and $B$, ordinate is same and hence line joining them is parallel to $x$-axis. Further as difference between abscissa is $4$, length of $A B$ is $4$.
Also observe that between $B$ and $C$, abscissa is same and hence line joining them is parallel to $y$-axis. Further as difference between ordinate is $5$, length of $B C$ is $5$.
As $A B$ is parallel to $x$-axis and $B C$ is parallel to $y$-axis, $\Delta A B C$ is right angled triangle and hence from Pythagoras theorem, hypotenuse $A C = \sqrt{{4}^{2} + {5}^{2}} = \sqrt{16 + 25} = \sqrt{41} = 6.403 \cong 6.4$
and perimeter is $4 + 5 + 6.4 = 15.4$
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# Unravel the equation: Discover the relation between x and y.
Welcome to Warren Institute! In this article, we will delve into the fascinating world of Mathematics education. Today, we will focus on understanding the relationship between variables x and y by completing the equation that describes their connection. By applying critical thinking and problem-solving skills, we will unlock the secrets behind this mathematical puzzle. Join us as we explore the intricacies of equations and empower ourselves with knowledge. Let's unleash the power of mathematics and uncover the hidden truths within these mathematical expressions. Stay tuned for an enlightening journey into the world of Mathematics education!
## Identifying the Relationship between x and y
In this section, we will explore how to identify the relationship between the variables *x* and *y* in a given equation. Understanding this relationship is crucial in mathematics education as it helps students solve problems and make connections between different mathematical concepts.
To identify the relationship between *x* and *y*, we need to analyze the equation and determine the pattern or rule governing their connection. This can be done by examining the coefficients, exponents, and arithmetic operations involved in the equation.
For example, if we have an equation like *y = 2x + 3*, we can observe that for every value of *x*, *y* is equal to twice that value plus three. This indicates a linear relationship between *x* and *y*, where *x* increases or decreases in a constant ratio with respect to *y*.
Understanding the relationship between *x* and *y* allows students to make predictions, solve equations, and graph functions more effectively.
## Solving Equations with x and y
Once we have identified the relationship between *x* and *y*, the next step is to solve equations involving these variables. Solving equations is a fundamental skill in mathematics education as it helps students find the values of unknown variables and validate their solutions.
To solve an equation with *x* and *y*, we typically isolate one variable and express it in terms of the other. This process allows us to substitute known values and find the corresponding values for the other variable.
For instance, let's consider the equation *3x - 2y = 10*. To solve for *y*, we can start by isolating *y* by moving the *3x* term to the other side of the equation:
*-2y = 10 - 3x*
*y = (10 - 3x)/(-2)*
Solving equations involving *x* and *y* empowers students to find solutions, verify their accuracy, and apply their mathematical knowledge in real-world scenarios.
## Graphing the Relationship between x and y
In mathematics education, graphing the relationship between *x* and *y* is a visual representation that helps students understand the connection between the variables. Graphs provide valuable insights into the behavior, patterns, and trends exhibited by mathematical equations.
To graph the relationship between *x* and *y*, we plot the values of *x* on the horizontal axis (x-axis) and the corresponding values of *y* on the vertical axis (y-axis). By connecting these points, we can visualize the relationship and identify any notable characteristics.
For example, if we have the equation *y = x^2*, we can plot a graph where each point represents an ordered pair (*x*, *y*) satisfying the equation. The resulting graph will be a parabola, indicating a quadratic relationship between *x* and *y*.
Graphing the relationship between *x* and *y* enables students to interpret data, analyze trends, and make informed mathematical conclusions.
## Applying the Relationship to Real-World Scenarios
The ability to apply the relationship between *x* and *y* to real-world scenarios is a crucial aspect of mathematics education. It allows students to connect abstract mathematical concepts with practical situations, fostering critical thinking and problem-solving skills.
By understanding the relationship between *x* and *y*, students can analyze real-world problems, formulate equations, and find solutions. This skill is particularly valuable in fields such as physics, economics, and engineering, where mathematical models are used to describe and predict various phenomena.
Applying the relationship between *x* and *y* to real-world scenarios equips students with the ability to make connections between mathematics and the world around them, leading to a deeper understanding and appreciation of the subject.
### What is the equation that represents the relationship between x and y?
The equation that represents the relationship between x and y is y = mx + b, where m represents the slope of the line and b represents the y-intercept.
### Can you provide an example of a real-life scenario where x and y are related through an equation?
One example of a real-life scenario where x and y are related through an equation is the relationship between distance and time in physics. According to the equation y = mx + b, where x represents time and y represents distance, if an object is moving at a constant speed, the distance it covers is directly proportional to the time it takes. As time increases, the distance covered also increases proportionally.
### How can I determine the slope of the line that represents the relationship between x and y in the equation?
To determine the slope of the line representing the relationship between x and y in an equation, use the formula y = mx + b. The coefficient m represents the slope of the line.
### Are there any restrictions or limitations on the values of x and y in the equation?
Yes, there may be restrictions or limitations on the values of x and y in an equation. It depends on the specific equation and context. Some equations may have restrictions based on the domain of the function, such as division by zero or square roots of negative numbers. Other equations may have limitations based on the problem's constraints or real-world applications.
### Is it possible to solve for either x or y in the equation if the other variable is known?
Yes, it is possible to solve for either x or y in an equation if the other variable is known, provided that the equation is solvable and not dependent or inconsistent.
In conclusion, understanding the relationship between x and y is crucial in Mathematics education. By completing the equation that describes their connection, students can develop a deeper comprehension of mathematical concepts and enhance their problem-solving skills. This knowledge empowers learners to tackle more complex equations and equations in real-life scenarios, fostering critical thinking and analytical reasoning abilities. Encouraging students to explore and discover the relationships between variables leads to a more engaging and interactive learning experience, enabling them to apply these principles across various mathematical domains. Ultimately, mastering the art of completing equations provides a solid foundation for future mathematical endeavors and equips students with the necessary skills to succeed in higher-level mathematics courses.
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Find the equation of the lines through the point (3, 2) which make an angle of $45^{\circ}$ with the line $x-2y=3.$
$\begin {array} {1 1} (A)\;3x+y=7 ; x-3y=9 & \quad (B)\;3x-y=7 ; x+3y=9 \\ (C)\;3y-x=7 ; 3x+y=9 & \quad (D)\;x-3y=7 ; 3x-y=9 \end {array}$
Toolbox:
• Slope $m = \tan \theta$
• Angle between two lines whose slopes are $m_1$ and $m_2$ is $\tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Let the slope of the required line be $m_1$.
Equation of the given line is $x-2y=3$
$\qquad y = \large\frac{1}{2}$$x- \large\frac{3}{2} which is of the form y=mx+c Hence the slope of the given line is m_2 = \large\frac{1}{2} given angle between the above line and the required line is 45^{\circ} Hence \tan 45^{\circ} = \large\frac{|m_1-m_2|}{1+m_1m_2} But \tan 45 = 1 \therefore 1 = \large\frac{|\Large\frac{1}{2}-m_1|}{1+\Large\frac{m_1}{2}} \Rightarrow 1 = \large\frac{\bigg|\bigg( \Large\frac{1-2m_1}{2} \bigg) \bigg|}{\Large\frac{2+m_1}{2}} \Rightarrow 1= \pm \bigg( \large\frac{1-2m_1}{2+m_1} \bigg) (i.e.), 1= \large\frac{1-2m_1}{2+m_1}$$ or 1= \large\frac{-(1-2m_1)}{2+m_1}$
If $1= \large\frac{1-2m_1}{2+m_1}$
then $2+m_1=1-2m_1$
$\Rightarrow m_1= -\large\frac{1}{m}$
If $1= \large\frac{-(1-2m_1)}{2+m_1}$
(i.e.,) $2+m_1=-1+2m_1$
$\Rightarrow m_1=3$
When $m_1=3$
The equation of the line passing through (3, 2) and slope 3 is
$y-2=3(x-3)$
$\Rightarrow y-2=3x-9$
$3x-y=7$
When $m_1= -\large\frac{1}{3}$
The equation of the line passing through (3,2) and having a slope $-\large\frac{1}{3}$ is
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# 51000 in words
51000 in words is written as Fifty One Thousand. In 51000, 5 has a place value of ten thousand and 1 has the place value of thousand. The article on Place Value gives more information. The number 51000 is used in expressions that relate to money, distance, length, population and many more. For example, “A new township has Fifty One Thousand people.” We can also use 51000 as “A Mango farm area was 5.1 hectares. It means the area was 51000 sq meters.”
51000 in words Fifty One Thousand Fifty One Thousand in Numbers 51000
## How to Write 51000 in Words?
We can convert 51000 to words using a place value chart. The number 51000 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 5 1 0 0 0
Thus, we can write the expanded form as:
5 × Ten thousand + 1 × Thousand + 0 × Hundred + 0 × Ten + 0 × One
= 5 × 10000 + 1 × 1000 + 0 × 100 + 0 × 10 + 0 × 1
= 51000
= Fifty One Thousand
51000 is the natural number that is succeeded by 50999 and preceded by 51001.
51000 in words – Fifty One Thousand.
Is 51000 an odd number? – No.
Is 51000 an even number? – Yes.
Is 51000 a perfect square number? – No.
Is 51000 a perfect cube number? – No.
Is 51000 a prime number? – No.
Is 51000 a composite number? – Yes.
## Solved Example
1. Write the number 51000 in expanded form
Solution: 5 x 10000 + 1 x 1000 + 0 x 100 + 0 x 10 + 0 x 1
Or Just 5 x 10000 + 1 x 1000
We can write 51000 = 50000 + 1000 + 0 + 0 + 0
= 5 x 10000 + 1 x 1000 + 0 x 100 + 0 x 10 + 0 x 1
## Frequently Asked Questions on 51000 in words
Q1
### How to write 51000 in words?
51000 in words is written as Fifty One Thousand.
Q2
### State True or False. 51000 is divisible by 4?
True. 51000 is divisible by 4.
Q3
### Is 51000 divisible by 10?
Yes. 51000 is divisible by 10. It is also divisible by 2, 3, 4, 5, 6, 10, 100 and 1000.
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# How do you integrate int x^3*(1-x^4)^6 dx ?
Apr 5, 2018
$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$
#### Explanation:
$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} - - - \left(1\right)$
method 1: inspection
we note that the function outside the bracket is a multiple of the bracket differentiated, so we can approach this integral by inspection.
We will 'guess' the bracket to the power +1, differentiate and compare it with the required integral.
$\frac{d}{\mathrm{dx}} {\left(1 - {x}^{4}\right)}^{7} = 7 {\left(1 - {x}^{4}\right)}^{6} \times \left(- 4 {x}^{3}\right)$
$= - 28 {x}^{3} {\left(1 - {x}^{4}\right)}^{6} - - \left(2\right)$
comparing $\left(1\right) \text{ & } \left(2\right)$ we can conclude
$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$
method 2 : substitution
$\int {x}^{3} {\left(1 - {x}^{4}\right)}^{6} \mathrm{dx} - - - \left(1\right)$
$u = 1 - {x}^{4} \implies \mathrm{du} = - 4 {x}^{3} \mathrm{dx}$
$\left(1\right) \rightarrow \int \cancel{{x}^{3}} {u}^{6} \times - \frac{1}{4 \cancel{{x}^{3}}} \mathrm{du}$
$= - \frac{1}{4} \int {u}^{6} \mathrm{du} = - \frac{1}{4} \times {u}^{7} / 7 + c$
$= - \frac{1}{28} {u}^{7} + c = - \frac{1}{28} {\left(1 - {x}^{4}\right)}^{7} + c$
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# PPT on Linear Equations in two variables
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### PPT on Linear Equations in two variables
1. 1. Created By SAGAR Class-10 Roll No-23
2. 2. Linear Equations Definition of a Linear Equation A linear equation in two variable x is an equation that can be written in the form ax + by + c = 0, where a ,b and c are real numbers and a and b is not equal to 0. An example of a linear equation in x is 2x – 3y + 4 = 0.
3. 3. A pair of linear equations in two variables can be solved by the: (i) Graphically method (ii) Algebraic method
4. 4. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION Let us consider the following system of two simultaneous linear equations in two variable. 2x – y = -1 3x + 2y = 9 Here we assign any value to one of the two variables and then determine the value of the other variable from the given equation.
5. 5. xxx 00
6. 6. Types of Solutions of Systems of Equations • One solution – the lines cross at one point • No solution – the lines do not cross • Infinitely many solutions – the lines coincide
7. 7. Algebraic method TO SOLVE A PAIR OF LINEAR EQUATION IN TWO VARIABLE
8. 8. To solve a pair of linear equations in two variables algebraically, we have following methods:(i) Substitution method (ii) Elimination method (iii) Cross-multiplication method
9. 9. SUBSTITUTION METHOD STEPS Obtain the two equations. Let the equations be a1x + b1y + c1 = 0 ----------- (i) a2x + b2y + c2 = 0 ----------- (ii) Choose either of the two equations, say (i) and find the value of one variable , say ‘y’ in terms of x Substitute the value of y, obtained in the previous step in equation (ii) to get an equation in x
10. 10. Solve the equation obtained in the previous step to get the value of x. Substitute the value of x and get the value of y. Let us take an example x + 2y = -1 ------------------ (i) 2x – 3y = 12 -----------------(ii)
11. 11. x + 2y = -1 x = -2y -1 ------- (iii) Substituting the value of x in equation (ii), we get 2x – 3y = 12 2 ( -2y – 1) – 3y = 12 - 4y – 2 – 3y = 12 - 7y = 14 ; y = -2 ,
12. 12. Putting the value of y in eq. (iii), we get x = - 2y -1 x = - 2 x (-2) – 1 =4–1 =3 Hence the solution of the equation is ( 3, - 2 )
13. 13. ELIMINATION METHOD • In this method, we eliminate one of the two variables to obtain an equation in one variable which can easily be solved. Putting the value of this variable in any of the given equations, the value of the other variable can be obtained. • For example: we want to solve, 3x + 2y = 11 2x + 3y = 4
14. 14. Let 3x + 2y = 11 --------- (i) 2x + 3y = 4 ---------(ii) Multiply 3 in equation (i) and 2 in equation (ii) and subtracting eq iv from iii, we get 9x + 6y = 33 ------ (iii) 4x + 6y = 8 ------- (iv) 5x = 25 x=5
15. 15. • putting the value of X in equation (ii) we get, 2x + 3y = 4 2 x 5 + 3y = 4 10 + 3y = 4 3y = 4 – 10 3y = - 6 y=-2 Hence, x = 5 and y = -2
16. 16. CROSS MULTIPLICATION METHOD
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# Using the balance model with negative terms in an equation
It is not so easy to portray negatives in an equation with the balance model because in the natural we don't have any 'negative weight', but you can think of the negatives as holes or empty spots that when filled with positives, they become nothing or zero.
x - 4 = 3 Add 4 balls to both sides to 'fill' the negative balls. x - 4 + 4 = 7 Now we have added 4 balls to both sides of the balance/equation. x = 7 The 4 balls 'fill' the negative balls, or cancel the negatives. So in the end there is nothing - or zero - left.
### Another example 2x + 3 = -5
2x + 3 = -5 We want the left hand side to contain ONLY blocks therefore we have to take away three balls from both sides. 2x = - 8 But since there are no balls from which to take away on the right side, we have to 'add' negative balls. x = - 4 In the end we take half of each side to arrive to the solution x = -4.
The same thing can happen with the blocks or x's - they can be negative too! Try the following example:
2x - 3 = -x + 3 To have ONLY x's on the left side we get rid of the -x on the right side. Since it is negative, we add one x to both sides. 2x - 3 + x = -x + 3 + x. Now -x and x 'cancel' each other. 2x - 3 + x = 3 Then we can combine 2x and x on the left side - together they are 3x. 3x - 3 = 3 The next step is to get rid of the number - 3 on the left hand side. For that end we have to add 3 on both sides. 3x - 3 + 3 = 3 + 3 And again - 3 and + 3 cancel each other (or produce a zero). 3x = 6 As the last step, since there are three x's on the left side and we want to know how much one is worth, we have to take a third part of both sides, or divide both sides by three. x = 2.
Checking the solution: we substitute x = -4 to the equation 2x + 3 = -5:
2(-4) + 3 = -5
-8 + 3 = -5
-5 = -5
So it was the right solution.
### More exercises - you can solve these equations with the help of the balance model or just with pencil and paper
1. x - 5 = 5
2. 2x - 5 = 5
3. 3x - 4 = 2x + 4
4. 5x - 3 = -x + 3
5. 5x + 3 = - 2x + 4
A Balanced Equation Model from Absorb Mathematics
An interactive animation illustrating solving the equation 4x + 6 = x - 3. Drag the green handles to balance each side. Click the arrow button to reset the animation. On the right side, you'll see links to similar animations of equation solving using a balance.
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# Class 7 Maths NCERT Solutions for Chapter 1 Integers Chapter 1 Integers EX – 1.2
## Integers
Question 1.
Write down a pair of integers whose:
(a)
The sum is – 7
(b) Difference is -10
(c) Sum is 0
Solution.
(a) A pair of integers whose sum is – 7 can be (- 1) and (- 6).
∵ (- 1) + (- 6) = – 7
(b) A pair of integers whose difference is -10 can be (- 11) and (- 1)
∵ – 11 – (- 1) = – 11 + 1 = – 10
(c) A pair of integers whose sum is 0 can be 1 and (- 1).
∵ (- 1) + (1) = 0.
Question 2.
(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is – 5.
(c) Write a negative integer and a positive integer whose difference is – 3.
Solution.
(a) A pair of negative integers whose difference gives 8 can be – 12 and – 20.
∵ (-12) – (- 20) = -12 + 20 = 8 .
(b) A negative integer and a positive integer whose sum is -5 can be – 13 and 8.
∵ (- 13) + 8 = -13 + 8 = – 5
(c) A negative integer and a positive integer whose difference is -3 can be – 1 and 2.
∵ (- 1) – 2 = – 1 – 2 = – 3
Question 3.
In a quiz, team A scored – 40, 10, 0, and Team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Solution.
Total scores of team A = (- 40) + 10 + 0
= – 40 + 10 + 0 = – 30
and, total scores of team B = 10 + 0 + (- 40)
= 10 + 0 – 40 = – 30
Since the total scores of each team are equal.
∴ No team scored more than the other but each has an equal score.
Yes, integers can be added in any order and the result remains unaltered. For example, 10 + 0 + (- 40) = – 30 = – 40 + 0 + 10
Question 4.
Fill in the blanks to make the following statements true:
(i)
(- 5) + (- 8) = (- 8) + (………)
(ii) – 53 + ……. = – 53
(iii) 17 + …… = 0
(iv) [13 + (- 12)] + (……) = 13 + [(- 12) + (- 7)]
(v) (- 4) + [15 + (- 3)] = [- 4 + 15] + ……
Solution.
(i) (- 5) + (- 8) = (- 8) + (- 5)
(ii) – 53 + 0 = – 53
(iii) 17 + (-17) = 0
(iv) [13 + (- 12)] + (- 7) = (13) + [(- 12) + (- 7)]
(v) (- 4) + [15 + (- 3)] = [(- 4 ) + 15] + (- 3)
|
# Conditional Probability and It’s Examples
## Conditional Probability
Probability is a branch of Mathematics which deals with the study of occurrence of an event. There are several approaches to understand the concept of probability which include empirical, classical and theoretical approaches. The conditional probability of an event is when the probability of one event depends on the probability of occurrence of the other event. When two events are mutually dependent or when an event is dependent on another independent event, the concept of conditional probability comes into existence.
### Conditional Probability Definition:
Conditional probability of occurrence of two events A and B is defined as the probability of occurrence of event ‘A’ when event B has already occurred and event B is in relation with event A.
The above picture gives a clear understanding of conditional probability. In this picture, ‘S’ is the sample space. The circles A and B are events A and B respectively. The sample space S is restricted to the region enclosed by B when event B has already occurred. So, the probability of occurrence of event A lies within the region of B. This probability of occurrence of event A when event be has already existed lies within the region common to both the circles A and B. So, it can be denoted as the region of A ∩ B.
### Conditional Probability Examples:
• The man travelling in a bus reaches his destination on time if there is no traffic. The probability of the man reaching on time depends on the traffic jam. Hence, it is a conditional probability.
• Pawan goes to a cafeteria. He would prefer to order tea. However, he would be fine with a cup of coffee if the tea is not being served. So, the probability that he would order a cup of coffee depends on whether tea is available in the cafeteria or not. So, it is a conditional probability.
• It will rain at the end of the hottest day. Here, the probability of occurrence of rainfall is depending on the temperature throughout the day. So, it is a conditional probability.
• In a practical record book, the diagrams are written with a pencil and the explanation is written in black ink. Here, the theory part is written in black ink irrespective of whether the diagrams are drawn with a pencil or not. So, the two events are independent and hence the probabilities of occurrence of these two events are unconditional.
### Conditional Probability Formula:
The formula for conditional probability is given as:
P(A/B) = $\frac{N(A\cap B)}{N(B)}$
In the above equation,
P (A | B) represents the probability of occurrence of event A when event B has already occurred
N (A ∩ B) is the number of favorable outcomes of the event common to both A and B
N (B) is the number of favorable outcomes of event B alone.
If ‘N’ is the total number of outcomes of both the events in a sample space S, then the probability of event B is given as:
P(B) = $\frac{N(B)}{N}$ → (1)
Similarly, the probability of occurrence of event A and B simultaneously is given as:
P(A ∩ B) = $\frac{N(A\cap B)}{N}$→ (2)
Now, in the formula for conditional probability, if both numerator and denominator are divided by ‘N’, we get
P(A/B) = $\frac{\frac{N(A\cap B)}{N}}{\frac{N(B)}{N}}$
Substituting equations (1) and (2) in the above equation, we get
P(A/B) = $\frac{P(A\cap B)}{P(B)}$
### Conditional Property Problems:
Question 1) When a fair die is rolled, find the probability of getting an odd number. Also find the probability of getting an odd number given that the number is less than or equal to 4.
Solution:
In the given questions there are two events. Let A and B represent the 2 events.
A = Getting an odd number when a fair die is rolled
B= Getting a number less than 4 when a fair die is rolled
The possible outcomes when a die is rolled are {1, 2, 3, 4, 5, 6}
The total number of possible outcomes in this event of rolling a die: N = 6
For the event A, the number of favorable outcomes: N (A) = 3
For the event B, the number of favorable outcomes: N (B) = 4
The number of outcomes common for both the events: N (A ∩ B) = 2
The probability of event A is given as:
P(A) = $\frac{N(A)}{N} = \frac{3}{6}$ = 0.5
The probability of occurrence of event A given event B is
P(A/B) = $\frac{N(A\cap B)}{N(B)} = \frac{2}{4}$ = 0.5.
### Fun Facts:
• The conditional probability of two events A and B when B has already occurred is represented as P (A | B) and is read as “the probability of A given B”.
• The probability of occurrence of an event when the other event has already occurred is always greater than or equal to zero.
• If the probability of occurrence of an event when the other event has already occurred is equal to 1, then both the events are identical.
1. What are the Fundamental Rules of Probability?
The branch of Mathematics which deals with the computation of likelihood of an event being true is called the probability. There are a few important facts that should be known before solving any problem related to probability. They are:
• The probability of an event ranges from 0 to 1, 0 being the lowest range and 1 being the highest range.
• If the probability of an event is zero, then it is called an impossible event.
• If the probability of an event is one, then it is called a sure or a certain event.
• The sum of the probabilities of occurrence and nonoccurrence of an event is equal to unity. So, the probability that the event does not occur can be found by subtracting the probability of occurrence of an event from 1.
2. How are Conditional Probabilities Different from Unconditional Probabilities?
In an unconditional probability, the occurrence of a number of events are independent of each other. Each event occurs individually and does not depend on any of the other events occurring in a sample space. However, in case of conditional probability, the probability of occurrence of an event is dependent on the occurrence of the other event. The probability of getting a head when a coin is tossed and that of getting an even number when dice are rolled are unconditional probability events. The bank remains closed on government holidays. When this event is considered, the probability that the bank is open depends on whether the day is a government holiday or not.
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# We Try To Predict Eddie Cibrian’s Future (03/26/2020)
How will Eddie Cibrian fare on 03/26/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first work out the destiny number for Eddie Cibrian, and then something similar to the life path number, which we will calculate for today (03/26/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 03/26/2020: We will take the month (03), the day (26) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 26 we do 2 + 6 = 8. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 8 + 4 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 03/26/2020.
DESTINY NUMBER FOR Eddie Cibrian: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Eddie Cibrian we have the letters E (5), d (4), d (4), i (9), e (5), C (3), i (9), b (2), r (9), i (9), a (1) and n (5). Adding all of that up (yes, this can get tedious) gives 65. This still isn’t a single-digit number, so we will add its digits together again: 6 + 5 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Eddie Cibrian.
CONCLUSION: The difference between the path number for today (6) and destiny number for Eddie Cibrian (2) is 4. That is higher than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t fear! As mentioned earlier, this is for entertainment purposes only. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
|
CMIIh
2020-12-17
Solve differential equation $y={e}^{\left(}6x\right)-3x-2$
Luvottoq
${y}^{\prime }-6y=7x+3$
$y={e}^{6x}-3x-2$
${y}^{\prime }\left(x\right)+p\left(x\right)y=q\left(x\right)$
$p\left(x\right)=6$
In order to solve the ordinary first order differential equation, we use integration factor that is:
$\mu \left(x\right)={e}^{\int p\left(x\right)dx}$
$\mu \left(x\right)={e}^{\int \left(-6\right)dx}$
$\mu \left(x\right)={e}^{-6\int 1dx}$
$\mu \left(x\right)={e}^{-6x}$
Write the equation in the form
$\left(\mu \left(x\right)y{\right)}^{\prime }=\mu \left(x\right)q\left(x\right)$
So
$\left({e}^{-6x}y{\right)}^{\prime }={e}^{-6x}\left(7x+3\right)$
$\left({e}^{-6x}y{\right)}^{\prime }=6x{e}^{-6x}+3{e}^{-6x}$
Solve
$\left({e}^{-6x}y{\right)}^{\prime }=7x{e}^{-6x}+3{e}^{-6x}$
${e}^{-6x}y=7/36\left(-6{e}^{-6x}x-{e}^{-6x}\right)+3\left(-1/6{e}^{-6x}\right)+c$
${e}^{-6x}y=42/36{e}^{-6x}x-7/36{e}^{-6x}-3/6{e}^{-6x}+c$
${e}^{-6x}y=-42/36{e}^{-6x}x-7/36{e}^{-6x}-18/36{e}^{-6x}+c$
${e}^{-6x}y=\left(-42{e}^{-6x}-25{e}^{-6x}\right)/36+c$
$y=-42{e}^{-6x}x-25{e}^{-6x}/\left(36{e}^{-6x}\right)+c$
Do you have a similar question?
|
# Estimating Products
Estimating products by rounding numbers to the nearest ten, hundred, thousand etc., we know how to estimate the sum, difference, etc., to the nearest 10, 100, 1000, etc. Now we have to estimate the products.
In order to estimate products, we round the given factors to the required place value. Estimating products help us to check the reasonableness of an answer.
To estimate the product, we first round off the multiplier and the multiplicand to the nearest tens, hundreds, or thousands and then multiply the rounded numbers.
I. Estimate the products of 34 and 86.
34 ⟶ 30 34 is rounded down to 30
86 ⟶ 90 86 rounded up to 90
34 ⟶ 30 86 ⟶ 90 34 is rounded down to 3086 rounded up to 90
Calculate mentally 30 × 90 = 2700
The estimated product is 2700.
II. Estimate the products of 331 and 267 by rounding to the nearest hundred.
331 ⟶ 300
267 ⟶ 300
300 × 300 = 90000
The estimated product is 90000.
III. Find the estimated product of 34 and 17 by rounding off to the nearest tens.
Rounding off 34 to nearest tens.
34 is nearer to 30 than 40.
So, 34 is rounded down to 30.
17 is nearer to 20 than 10.
So, 17 is rounded up to 20.
Now 30 × 20 = 600
So, the estimated product of 34 × 17 = 600
IV. Find the estimated product of 148 and 14 by rounding off to the nearest tens.
Rounding off 148 to nearest tens.
148 is nearer to 150 than 140.
So, 148 is rounded up to 150.
14 is nearer to 10 than 20.
So, 14 is rounded down to 10.
Now 150 × 10 = 1500
So, the estimated product of 148 × 14 = 1500
Consider the following examples on estimating products:
1. Estimate the product: 42 × 57
Solution:
4 2 → 4 0
× 5 7 → × 6 0
2 4 0 0
Thus estimated product = 2400
2. Estimate the product: 47 × 84
Solution:
4 7 × 8 4 47 is rounded up to 50
↓ ↓
5 0 × 8 0 84 is rounded down to 80
= 4 0 0 0
The estimated product = 4000
3. Estimate the product: 74 × 85
Solution:
7 4 × 8 5 74 is rounded up to 80
↓ ↓
7 0 × 9 0 85 is rounded down to 90
= 6 3 0 0
6300 is the estimated product
4. Estimate the product: 358 × 326
Solution:
One number is rounded up and the other is rounded down. Here 358 is rounded up to the nearest hundred, i.e., 400 while 326 is rounded down to the nearest hundred, i.e., 300.
3 5 8 × 3 2 6 358 is rounded up to 400
↓ ↓
4 0 0 × 3 0 0 326 is rounded down to 300
= 1 2 0 0 0 0
120000 is the estimated product
5. Estimate the following product by rounding numbers to the nearest;
(i) Hundred
(ii) Ten ‘341 × 267’
Solution:
(i) 3 4 1 × 2 6 7 341 is rounded up to 300
↓ ↓
3 0 0 × 3 0 0 267 is rounded down to 300
= 90,000 & The estimated product is 90,000
(ii) 3 4 1 × 2 6 7 341 is rounded down to 340, the nearest ten
↓ ↓
3 4 0 × 2 7 0 267 is rounded up to 270, the nearest ten
= 91,800 The estimated product is 91,800
Word Problems on Estimating Products:
To solve any word problem, we need to read each word carefully and select numbers carefully from the question to solve it.
1. Ron has a crayon box containing 12 colors. He has 7 such boxes. How many crayons does he have in all?
Solution:Colors in a crayon box = 15Total number of boxes = 7Total number of crayons in. 7 such boxes = 12 × 7
TRon has 84 crayons in all.
Questions and Answers on Estimation in Multiplication:
1. Find the estimated product for each of the given. First one is done for you.
(b) 70 × 40, 28000
(c) 300 × 100, 30000
(d) 200 × 200, 40000
(e) 700 × 200, 140000
(f) 500 × 400, 200000
(g) 900 × 500, 450000
(h) 3000 × 200, 600000
2. Find the estimated product of the following by rounding off the 2 and 3-digit numbers to the nearest tens.
(i) 121 × 17
(ii) 17 × 14
(iii) 93 × 12
(iv) 189 × 15
(v) 164 × 16
(vi) 335 × 13
2. (i) 2400
(ii) 200
(iii) 900
(iv) 3800
(v) 3200
(vi) 3400
3. Choose the best estimate and tick the right answer.
I. A shopkeeper has 92 packets of chocolates. If each packet has 37 chocolates, then how many chocolates are there in the shop.
(i) 3600 (ii) 4000
II. A museum has 219 marble jars. Each jar has 178 marbles. What is the total number of marbles in the museum?
(i) 40000 (ii) 50000
III. There are 76 houses in a locality. Each house uses 278 units of electricity each day. How many units of electricity is used each day in the locality?
(i) 24000 (ii) 30000
IV. A hotel has 18 water tanks and each tank has the capacity of 4089 litres of water. What is the total quantity of water that can be stored by the hotel?
(i) 80000 (ii) 16000
3. I. (ii)
II. (i)
III. (ii)
IV. (i)
Worksheet on Word Problems on Estimating Products:
4. Solve the given word problems.
(i)There are 157 pages in a book. How many pages are there in 4 such books?
(ii) Mary eat 6 nuts daily. How many nuts does she eat in 120 days?
(iii) Tina bought a pencil box which has 24 pencils in it. If she gets 5 such boxes then how many pencils will she have in all?
(iv) A packet of nuts contains 233 nuts. There are 3 such packets. How many nuts are there in all?
(v) There are 120 students in one class. How many students are there in 8 classes?
(vi) Shelly and her brother filled a basket with 57 fruits. They both together filled 6 such baskets. How many fruits they have in all the baskets?
4. (i) 628
(ii) 720
(iii) 120
(iv) 699
(v) 960
(vi) 342
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# How to calculate gallons of water in a rectangular tank
Asked by wiki @ in Mathematics viewed by 536 People
A steady stream of water flows into a partially-filled rectangular tank. After 6 minutes, there are 87 gallons of water in the tank. After 21 minutes, there are 222 gallons. Write an equation to represent the volume of water in the tank y after x minutes. How much water was in the tank to begin?
33 gallons of water to begin with.
Step-by-step explanation:
So we essentially are given two coordinates: (6,87) and (21,222). To find an equation, we simply need to find the slope and y-intercept. We know it's a linear equation because it's a steady stream, meaning a constant slope.
The slope is:
So, the rate at which the stream flows is 9 gallons per minute.
Now, let's find the initial amount of water. To do this, we can use point-slope form. Pick either of the two points. I'm going to use (6,87).
So, there were 33 gallons of water in the tank to begin with.
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High School: Number and Quantity
The Complex Number System HSN-CN.A.2
2. Use the relation i2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
Addition and subtraction of complex numbers is easy. Just because it says "complex" doesn't mean it is. Some ice cream cartons say, "nonfat," but do you really think it doesn't have the same effect on your waistline as regular ice cream? Please.
The i looks just like a variable and in this context, it acts just like one too. The basic rule for adding and subtracting is the one your students might imagine: we add and subtract like terms. Just make sure they remember to express their answers in a + bi form. In other words, we want the i term to come after the "non-i" term.
So, for example, if we want to add 4 + 10i and 7 – 2i, we combine the like terms. That means the 4 and the 7 go together, as do the 10i and -2i. (We can do this thanks to a generous donation from the Commutative Property. Remember to send it a thank you card.) The answer, then, is 11 + 8i.
If we want to subtract, we simply remember to distribute the negative sign (à la Distributive Property).
(21 + 4i) – (16 – i) becomes 21 + 4i – 16 + i, or even more simply, 5 + 5i.
Multiplication also follows the rules of algebra. Many people use the "FOIL" method to multiply binomials (First, Outer, Inner, Last). We can also think of it as the "double distributive" property. First, distribute the real portion of the first complex number, then the imaginary part.
When multiplying, it's worth noting that i2 = -1. This will help a lot when we simplify. Don't be surprised when a real number results from us fiddling around with these imaginary numbers. It happens sometimes. We just gotta roll with it.
Drills
1. Find the sum of 4 + 6i and 11 – 2i.
Correct Answer:
15 + 4i
Answer Explanation:
All we have to do is combine the like terms. For the real numbers, 4 + 11 = 15. On the imaginary side, 6i – 2i = 4i. Put 'em together and what do you get? Uh, well (B). It's just like second-grade math, isn't it?
2. What does (8 + 4i) – (3 + i) equal?
Correct Answer:
5 + 3i
Answer Explanation:
We add and subtract numbers just like we always did—along with distributing the negative sign. When we do so, (8 + 4i) – (3 + i) becomes 8 + 4i – 3 – i. If we combine the like terms and express our answer in a + bi form, we should get (A).
3. Find the product of 6 and 3 + i.
Correct Answer:
18 + 6i
Answer Explanation:
The product of 6 and 3 + i is 6(3 + i). First, we have to distribute the 6. It gets around anyway. We then have 6 × 3 = 18 and 6 × i = 6i. If we put our answer in a + bi form, like we will for now and forever, we'll get (C) as the correct answer.
4. What does (5i)2 equal?
Correct Answer:
-25
Answer Explanation:
You know that to square a monomial, you square each part of it. That means we have to consider the 5 and the i separately. We know that 52 = 25 and i2 = -1. Since the two are multiplied together, 25 × -1 = -25. In other words, (A).
5. What does (2 + i) × (3 + 4i) equal?
Correct Answer:
2 + 11i
Answer Explanation:
All right! It's time for some double distributive action (or "FOIL," if that's what you prefer). First, let's distribute that 2 to the second binomial. That gives us 6 + 8i so far. Now, let's distribute that i to the second binomial. That gives us 3i + 4i2. Now, we combine them and make 6 + 8i + 3i + 4i2. But wait a minute—remember the definition of i. If i2 = -1, then 4i2 = 4(-1) or -4. That means we have 6 + 8i + 3i – 4 instead. If we simplify, our answer will be (B).
6. What is (3 – i)2?
Correct Answer:
8 – 6i
Answer Explanation:
Asking us to square (3 – i) is the same as asking us to multiply (3 – i) by (3 – i). Now it's time for more double distributive. First, we distribute the 3 to the second set of parentheses to get 9 – 3i. Then, we distribute the -i to get -3i + i2. What do we know about i2, again? So that part is actually -3i – 1. If we combine the two, we get 9 – 3i – 3i – 1 and after we combine like terms (rational goes with rational, imaginary goes with imaginary) and express it in a + bi form, we get (D).
7. What does i(6 – 2i) equal?
Correct Answer:
2 + 6i
Answer Explanation:
Those negative signs can get tricky, huh? Well, given i (6 – 2i), the first thing we want to do is distribute the i. That means we get 6i – 2i2. Since i2 = -1, we actually have 6i – (-2), or 6i + 2. Just switch those around so that the real number is in front, and we have (A).
8. What does 7i(7i – 5) equal?
Correct Answer:
-49 – 35i
Answer Explanation:
The first thing to notice is that the stuff inside the parenthesis isn't in a + bi form. Don't let that confuse you. Proceed as normal and distribute the 7i. If we do that, we end up with 49i2 – 35i. Since we know i2 = -1, that turns into -49 -35i. That's (C)—and look! It's already in a + bi form. What could be better?
9. Solve (8 – 2i)(8 + 2i).
Correct Answer:
68
Answer Explanation:
Distribute things double-time! First we distribute the 8, which gives us 64 + 16i, and then the -2i, which gives us -16i – 4i2. If we put them together, we get: 64 + 16i – 16i – 4i2. Since 16i – 16i = 0 (even in the imaginary world), we really just have 64 – 4i2. You're probably sick of hearing it, but i2 = -1, so we really have 64 + 4, which is just (D).
10. Solve (-3 + 5i)(3 – 5i).
Correct Answer:
16 + 30i
Answer Explanation:
The biggest challenge with imaginary numbers and distribution is the negative signs. Here, distributing -3 gives us -9 + 15i. Then, distributing the 5i makes 15i – 25i2. That means we have -9 + 15i + 15i -25i2. Sometimes people would rather not deal with imaginary numbers (even though they're perfectly nice), so they try and cancel them out. In this case, the two 15i terms are added, not subtracted. That means we have -9 + 30i – 25i2. We don't have to tell you this is the same as -9 + 30i + 25, do we? We should get (B) as our answer.
|
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
You make known to me the path of life; you will fill me with joy in your presence, with eternal pleasures at your right hand. Psalms 16:11 NIV
9-03 Matrix Operations
Summary: In this section, you will:
• Multiply a scalar with a matrix.
• Multiply a matrix with a matrix.
SDA NAD Content Standards (2018): PC.6.4
A digital picture can be described as a matrix of colored pixels. Each element is the position and color of the pixel. If you wanted to enlarge the picture so it would be 4 times bigger, you could multiply the position elements of the photo matrix by 4. Many things in computer programming and graphics are simplified by using matrix operations. The difference is that computer programming names them arrays instead of matrices.
To add or subtract matrices, add or subtract the corresponding elements. Because each element is combined with its corresponding element, the matrices have to be the same size.
In order to add or subtract matrices, they must be the same size.
Add or subtract matrices by adding or subtracting all the corresponding elements.
$$\left[\begin{matrix} 1 & 3 \\ -2 & 4 \end{matrix}\right] + \left[\begin{matrix} -3 & 0 \\ -1 & 2 \end{matrix}\right]$$
Solution
The matrices are the same size, so we can add them. Add the corresponding elements as indicated by the color.
$$\left[\begin{matrix} \color{blue}{1} & \color{red}{3} \\ \color{green}{-2} & \color{purple}{4} \end{matrix}\right] + \left[\begin{matrix} \color{blue}{-3} & \color{red}{0} \\ \color{green}{-1} & \color{purple}{2} \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{1 + (-3)} & \color{red}{3 + 0} \\ \color{green}{-2 + (-1)} & \color{purple}{4 + 2} \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{-2} & \color{red}{3} \\ \color{green}{-3} & \color{purple}{6} \end{matrix}\right]$$
Example 2: Subtracting Two Matrices
$$\left[\begin{matrix} 1 & 3 \\ -2 & 4 \end{matrix}\right] - \left[\begin{matrix} -3 & 0 \\ -1 & 2 \end{matrix}\right]$$
Solution
The matrices are the same size, so we can subtract them. Subtract the corresponding elements as indicated by the color.
$$\left[\begin{matrix} \color{blue}{1} & \color{red}{3} \\ \color{green}{-2} & \color{purple}{4} \end{matrix}\right] - \left[\begin{matrix} \color{blue}{-3} & \color{red}{0} \\ \color{green}{-1} & \color{purple}{2} \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{1 - (-3)} & \color{red}{3 - 0} \\ \color{green}{-2 - (-1)} & \color{purple}{4 - 2} \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{4} & \color{red}{3} \\ \color{green}{-1} & \color{purple}{2} \end{matrix}\right]$$
Try It 1
$$\left[\begin{matrix} 2 & -7 & 1 \\ 0 & 3 & -5 \end{matrix}\right] + \left[\begin{matrix} 0 & -9 & 3 \\ -1 & 8 & 4 \end{matrix}\right]$$
$$\left[\begin{matrix} 2 & -16 & 4 \\ -1 & 11 & -1 \end{matrix}\right]$$
Scalar Multiplication
A scalar is a single number as opposed to a matrix. To multiply a scalar with a matrix, multiply each element by the scalar. This is similar to distributing.
Multiply a Scalar and a Matrix
Multiply each element in the matrix by the scalar.
This is similar to distribution.
Example 3: Multiply a Scalar with a Matrix
$$3 \left[\begin{matrix} 3 & -2 & 1 \\ -4 & 5 & -1 \end{matrix}\right]$$
Solution
Multiply the scalar, 3, with each of the elements in the matrix
$$3 \left[\begin{matrix} 3 & -2 & 1 \\ -4 & 5 & -1 \end{matrix}\right]$$
$$\left[\begin{matrix} 3(3) & 3(-2) & 3(1) \\ 3(-4) & 3(5) & 3(-1) \end{matrix}\right]$$
$$\left[\begin{matrix} 9 & -6 & 3 \\ -12 & 15 & -3 \end{matrix}\right]$$
Example 4: Combined Operations
$$2 \left[\begin{matrix} 1 & 3 \\ -2 & 4 \end{matrix}\right] - 3 \left[\begin{matrix} 0 & -1 \\ 2 & 5 \end{matrix}\right]$$
Solution
Follow the order of operations. Multiplication comes before subtraction.
$$2 \left[\begin{matrix} 1 & 3 \\ -2 & 4 \end{matrix}\right] - 3 \left[\begin{matrix} 0 & -1 \\ 2 & 5 \end{matrix}\right]$$
$$\left[\begin{matrix} 2 & 6 \\ -4 & 8 \end{matrix}\right] - \left[\begin{matrix} 0 & -3 \\ 6 & 15 \end{matrix}\right]$$
Now that the multiplication is done, subtract the matrices.
$$\left[\begin{matrix} 2-0 & 6-(-3) \\ -4-6 & 8-15 \end{matrix}\right]$$
$$\left[\begin{matrix} 2 & 9 \\ -10 & -7 \end{matrix}\right]$$
Try It 2
$$\left[\begin{matrix} 2 \\ -4 \end{matrix}\right] + 4 \left[\begin{matrix} -1 \\ 3 \end{matrix}\right]$$
$$\left[\begin{matrix} -2 \\ 8 \end{matrix}\right]$$
Matrix Multiplication
In order to multiply 2 matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. If the orders of the two matrices are written in order, then the two middle numbers must be the same. The two outer numbers give the order of the product.
$$(\color{red}{3} × \color{blue}{2}) \cdot (\color{blue}{2} × \color{red}{4})$$
Notice the middle numbers, the blue 2's, are the same. The product will be the outside red numbers, 2 × 4. Because of this, order is important in matrix multiplication. It is NOT commutative! A 2 × 3 can be multiplied with a 3 × 1. It is not possible if the order is switched. A 3 × 1 cannot be multiplied by a 2 × 3.
Matrix Multiplication
Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix.
Matrix multiplication is not commutative.
To multiply two matrices,
1. Verify that the number of columns in the first matrix equals the number of rows in the second matrix.
2. Choose a row in the first matrix and a column in the second matrix.
3. Multiply the first entries in that row and column.
4. Put a plus sign and multiply the second entries in that row and column.
5. Continue adding plus signs and the products of the entries in that row and column until the end of the row and column is reached.
6. Repeat steps 2-5 for all combinations of rows and columns. The results go in the answer matrix in the same location as the chosen row and column. For example, multiplying the 2nd row and 1st column gives a result for the element in the 2nd row 1st column of the answer matrix.
7. Simplify each element of the answer.
Example 5: Matrix Multiplication
$$\left[\begin{matrix} 1 & 3 & -2 \\ 2 & 0 & -1 \end{matrix}\right] \left[\begin{matrix} 4 \\ 0 \\ -3 \end{matrix}\right]$$
Solution
First check to see if the matrices can be multiplied. Are the number of columns in the first matrix equal to the number of rows in the second matrix? Write the orders of the matrices and check to see if the middle numbers are the same.
$$(2 × \color{blue}{3}) · (\color{blue}{3} × 1)$$
The middle numbers are the same, so it is possible to multiply the matrices. The product will be a 2 × 1 matrix, the first and last number from above.
Pick a row in the first matrix and a column in the second matrix such as 1st row, 1st column. Multiply the left and top element, plus, multiply the next element in the row and column, plus, multiply the next element in the row and column, plus, and continue until reaching the end of the row and column. The result goes in the 1st row, 1st column of the product matrix.
$$\left[\begin{matrix} \color{blue}{1} & \color{blue}{3} & \color{blue}{-2} \\ 2 & 0 & -1 \end{matrix}\right] \left[\begin{matrix} \color{blue}{4} \\ \color{blue}{0} \\ \color{blue}{-3} \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{1} & \rightarrow & \color{purple}{3} & \rightarrow & \color{red}{-2} \\ 2 & \quad & 0 & \quad & -1 \end{matrix}\right] \left[\begin{matrix} \color{blue}{4} \\ \downarrow \\ \color{purple}{0} \\ \downarrow \\ \color{red}{-3} \end{matrix}\right] = \left[\begin{matrix} \color{blue}{1(4)} + \color{purple}{3(0)} + \color{red}{-2(-3)} \\ \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. The only other combination in this problem is 2nd row, 1st column. The result will go in the 2nd row, 1st column of the product.
$$\left[\begin{matrix} 1 & \quad & 3 & \quad & -2 \\ \color{blue}{2} & \rightarrow & \color{purple}{0} & \rightarrow & \color{red}{-1} \end{matrix}\right] \left[\begin{matrix} \color{blue}{4} \\ \downarrow \\ \color{purple}{0} \\ \downarrow \\ \color{red}{-3} \end{matrix}\right] = \left[\begin{matrix} 1(4) + 3(0) + -2(-3) \\ \color{blue}{2(4)} + \color{purple}{0(0)} + \color{red}{-1(-3)} \end{matrix}\right]$$
Simplify the result.
$$= \left[\begin{matrix} 10 \\ 11 \end{matrix}\right]$$
Example 6: Matrix Multiplication
$$\left[\begin{matrix} 1 & -1 \\ 0 & 2 \end{matrix}\right] \left[\begin{matrix} 5 & 1 & 3 \\ -3 & -2 & 0 \end{matrix}\right]$$
Solution
First check to see if the matrices can be multiplied. Are the number of columns in the first matrix equal to the number of rows in the second matrix. Write the orders of the matrices and check to see if the middle numbers are the same.
$$(2 × \color{blue}{2}) · (\color{blue}{2} × 3)$$
The middle numbers are the same, so it is possible to multiply the matrices. The product will be a 2 × 3 matrix, the first and last number from above.
Pick a row in the first matrix and a column in the second matrix such as 1st row, 1st column. Multiply the left and top element, plus, multiply the next element in the row and column, plus, multiply the next element in the row and column, plus, and continue until reaching the end of the row and column. The result goes in the 1st row, 1st column of the product matrix.
$$\left[\begin{matrix} \color{blue}{1} & \color{blue}{-1} \\ 0 & 2 \end{matrix}\right] \left[\begin{matrix} \color{blue}{5} & 1 & 3 \\ \color{blue}{-3} & -2 & 0 \end{matrix}\right]$$
$$\left[\begin{matrix} \color{blue}{1} & \rightarrow & \color{purple}{-1} \\ 0 & \quad & 2 \end{matrix}\right] \left[\begin{matrix} \color{blue}{5} & 1 & 3 \\ \downarrow & \quad & \quad \\ \color{purple}{-3} & -2 & 0 \end{matrix}\right] = \left[\begin{matrix} \color{blue}{1(5)} + \color{purple}{-1(-3)} & \quad & \quad \\ \quad & \quad & \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. perhaps the 1st row, 2nd column. The result will go in the 1st row, 2nd column of the product.
$$\left[\begin{matrix} \color{blue}{1} & \rightarrow & \color{purple}{-1} \\ 0 & \quad & 2 \end{matrix}\right] \left[\begin{matrix} 5 & \color{blue}{1} & 3 \\ \quad & \downarrow & \quad \\ -3 & \color{purple}{-2} & 0 \end{matrix}\right] = \left[\begin{matrix} 1(5) + -1(-3) & \color{blue}{1(1)} + \color{purple}{-1(-2)} & \quad \\ \quad & \quad & \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. perhaps the 1st row, 3rd column. The result will go in the 1st row, 3rd column of the product.
$$\left[\begin{matrix} \color{blue}{1} & \rightarrow & \color{purple}{-1} \\ 0 & \quad & 2 \end{matrix}\right] \left[\begin{matrix} 5 & 1 & \color{blue}{3} \\ \quad & \quad & \downarrow \\ -3 & -2 & \color{purple}{0} \end{matrix}\right] = \left[\begin{matrix} 1(5) + -1(-3) & 1(1) + -1(-2) & \color{blue}{1(3)} + \color{purple}{-1(0)} \\ \quad & \quad & \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. perhaps the 2nd row, 1st column. The result will go in the 2nd row, 1st column of the product.
$$\left[\begin{matrix} 1 & \quad & -1 \\ \color{blue}{0} & \rightarrow & \color{purple}{2} \end{matrix}\right] \left[\begin{matrix} \color{blue}{5} & 1 & 3 \\ \downarrow & \quad & \quad \\ \color{purple}{-3} & -2 & 0 \end{matrix}\right] = \left[\begin{matrix} 1(5) + -1(-3) & 1(1) + -1(-2) & 1(3) + -1(0) \\ \color{blue}{0(5)} + \color{purple}{2(-3)} & \quad & \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. perhaps the 2nd row, 2nd column. The result will go in the 2nd row, 2nd column of the product.
$$\left[\begin{matrix} 1 & \quad & -1 \\ \color{blue}{0} & \rightarrow & \color{purple}{2} \end{matrix}\right] \left[\begin{matrix} 5 & \color{blue}{1} & 3 \\ \quad & \downarrow & \quad \\ -3 & \color{purple}{-2} & 0 \end{matrix}\right] = \left[\begin{matrix} 1(5) + -1(-3) & 1(1) + -1(-2) & 1(3) + -1(0) \\ 0(5) + 2(-3) & \color{blue}{0(1)} + \color{purple}{2(-2)} & \quad \end{matrix}\right]$$
Pick another row in the first matrix and a column in the second matrix. perhaps the 2nd row, 3rd column. The result will go in the 2nd row, 3rd column of the product.
$$\left[\begin{matrix} 1 & \quad & -1 \\ \color{blue}{0} & \rightarrow & \color{purple}{2} \end{matrix}\right] \left[\begin{matrix} 5 & 1 & \color{blue}{3} \\ \quad & \quad & \downarrow \\ -3 & -2 & \color{purple}{0} \end{matrix}\right] = \left[\begin{matrix} 1(5) + -1(-3) & 1(1) + -1(-2) & 1(3) + -1(0) \\ 0(5) + 2(-3) & 0(1) + 2(-2) & \color{blue}{0(3)} + \color{purple}{2(0)} \end{matrix}\right]$$
There are no more combinations of rows and columns, so now simplify the result.
$$= \left[\begin{matrix} 8 & 3 & 3 \\ -6 & -4 & 0 \end{matrix}\right]$$
Try It 3
Multiply.
$$\left[\begin{matrix} 2 & 5 \end{matrix}\right] \left[\begin{matrix} 1 & -3 \\ 0 & -2 \end{matrix}\right]$$
$$\left[\begin{matrix} 2 & -16 \end{matrix}\right]$$
Lesson Summary
In order to add or subtract matrices, they must be the same size.
Add or subtract matrices by adding or subtracting all the corresponding elements.
Multiply a Scalar and a Matrix
Multiply each element in the matrix by the scalar.
This is similar to distribution.
Matrix Multiplication
Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix.
Matrix multiplication is not commutative.
To multiply two matrices,
1. Verify that the number of columns in the first matrix equals the number of rows in the second matrix.
2. Choose a row in the first matrix and a column in the second matrix.
3. Multiply the first entries in that row and column.
4. Put a plus sign and multiply the second entries in that row and column.
5. Continue adding plus signs and the products of the entries in that row and column until the end of the row and column is reached.
6. Repeat steps 2-5 for all combinations of rows and columns. The results go in the answer matrix in the same location as the chosen row and column. For example, multiplying the 2nd row and 1st column gives a result for the element in the 2nd row 1st column of the answer matrix.
7. Simplify each element of the answer.
Practice Exercises
1. Describe a scalar and give an example.
2. Add or subtract the matrices.
3. $$\left[\begin{matrix} 1 & 9 \\ -3 & -2 \end{matrix}\right] + \left[\begin{matrix} 0 & -4 \\ -1 & -7 \end{matrix}\right]$$
4. $$\left[\begin{matrix} -4 & 2 & 4 \\ 1 & -2 & -7 \end{matrix}\right] + \left[\begin{matrix} 8 & 2 \\ 4 & 5 \end{matrix}\right]$$
5. $$\left[\begin{matrix} -2 & -3 \\ 1 & -2 \\ 4 & 7 \end{matrix}\right] - \left[\begin{matrix} 2 & 1 \\ 0 & 4 \\ -5 & 6 \end{matrix}\right]$$
6. Perform the indicated operations.
7. $$\left[\begin{matrix} 2 & -1 \end{matrix}\right] + 2 \left[\begin{matrix} -5 & 5 \end{matrix}\right]$$
8. $$-1 \left[\begin{matrix} -3 & 2 & 6 \\ 1 & 4 & -7 \end{matrix}\right] + 3 \left[\begin{matrix} -1 & 2 & -5 \\ 4 & -3 & 0 \end{matrix}\right]$$
9. $$2 \left[\begin{matrix} 3 \\ 2 \\ 0 \end{matrix}\right] - 3 \left[\begin{matrix} 0 \\ -3 \\ 1 \end{matrix}\right] + \left[\begin{matrix} 4 \\ -2 \\ -1 \end{matrix}\right]$$
10. Multiply the matrices.
11. $$\left[\begin{matrix} 2 & 1 \end{matrix}\right] \left[\begin{matrix} 0 & -2 \\ -1 & 3 \end{matrix}\right]$$
12. $$\left[\begin{matrix} 3 & -2 \end{matrix}\right] \left[\begin{matrix} 7 & 9 \end{matrix}\right]$$
13. $$\left[\begin{matrix} -1 & 0 \\ 2 & -2 \end{matrix}\right] \left[\begin{matrix} 4 & 3 \\ -2 & 0 \end{matrix}\right]$$
14. $$\left[\begin{matrix} 5 & 1 \\ -2 & -4 \\ 0 & 3 \end{matrix}\right] \left[\begin{matrix} -1 \\ 6 \end{matrix}\right]$$
15. $$\left[\begin{matrix} 1 & 1 & 0 \\ -2 & 1 & 5 \end{matrix}\right] \left[\begin{matrix} 2 & 1 \\ 0 & 4 \\ -1 & 0 \end{matrix}\right]$$
16. Use the matrices $$A = \left[\begin{matrix} 2 & -1 \\ 0 & -2 \end{matrix}\right]$$ and $$B = \left[\begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix}\right]$$ to verify that matrix multiplication has no commutative property by comparing the products AB and BA.
17. Problem Solving
18. The corners of a figure are at the coordinates A(0, 0), B(3, 1), C(4, 5), and D(1, 5). These can be written as the matrix $$\begin{matrix} \begin{matrix} A & B & C & D \end{matrix} \\ \left[\begin{matrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 5 & 5 \end{matrix}\right] \end{matrix}$$ where each column is a point. Jane wants to enlarge the figure by a factor of 5. Use scalar multiplication to find the coordinates of the enlargement.
19. A student is buying supplies for two different classes at school. Math class requires 20 pencils, 2 paper packs, and 1 textbook. English class requires 15 pencils, 2 paper packs, and 5 textbooks. Pencils are 25¢ each, paper packs are $2 each, and textbooks are$20 each. Write the supply requirements as a matrix and the costs as another matrix. Show how to use matrix multiplication to find the total cost for each class.
20. Mixed Review
21. (9-02) Use Gauss-Jordan Elimination to solve \left\{\begin{align} x + 3y - 2z &= -8 \\ y + 5z &= 16 \\ -x + z &= 8 \end{align}\right.
22. (8-02) Use elimination to solve \left\{\begin{align} 10x + 5y &= 6 \\ 30x + 20y &= 17 \end{align}\right.
23. (6-05) Find the dot product $$(2\hat{i} - 3\hat{j}) · (4\hat{i} + \hat{j})$$
24. (5-04) Solve $$1 = 2 \sin (2θ)$$ on the interval $$0 ≤ θ < 2π$$.
25. (5-03) Verify the trigonometric identity $$\sin θ = \frac{\sin 2θ}{2 \cos θ}$$
1. A single number, not a matrix or vector, such as 3.
2. $$\left[\begin{matrix} 1 & 5 \\ -4 & -9 \end{matrix}\right]$$
3. Not possible
4. $$\left[\begin{matrix} -4 & -4 \\ 1 & -6 \\ 9 & 1 \end{matrix}\right]$$
5. $$\left[\begin{matrix} -8 & 9 \end{matrix}\right]$$
6. $$\left[\begin{matrix} 0 & 4 & -21 \\ 11 & -13 & 7 \end{matrix}\right]$$
7. $$\left[\begin{matrix} 10 \\ 11 \\ -4 \end{matrix}\right]$$
8. $$\left[\begin{matrix} -1 & -1 \end{matrix}\right]$$
9. Not possible
10. $$\left[\begin{matrix} -4 & -3 \\ 12 & 6 \end{matrix}\right]$$
11. $$\left[\begin{matrix} 1 \\ -22 \\ 18 \end{matrix}\right]$$
12. $$\left[\begin{matrix} 2 & 5 \\ -9 & 2 \end{matrix}\right]$$
13. $$AB = \left[\begin{matrix} 0 & 6 \\ -4 & 0 \end{matrix}\right]$$, $$BA = \left[\begin{matrix} 2 & -7 \\ 4 & -2 \end{matrix}\right]$$
14. A′ = (0, 0), B′ = (15, 5), C′ = (20, 25), D′ = (5, 25)
15. $$\begin{matrix} \begin{matrix} \quad \end{matrix} & \begin{matrix} Pencil & Paper & Book \end{matrix} \\ \begin{matrix} Math \\ Eng \end{matrix} & \left[\begin{matrix} 20 & 2 & 1 \\ 15 & 2 & 5 \end{matrix}\right] \end{matrix} \begin{matrix} \begin{matrix} Pencil \\ Paper \\ Book \end{matrix} \left[\begin{matrix} 0.25 \\ 2 \\ 20 \end{matrix}\right] \end{matrix} = \left[\begin{matrix} 29 \\ 107.75 \end{matrix}\right]$$
16. (−5, 1, 3)
17. $$\left(\frac{7}{10}, -\frac{1}{5}\right)$$
18. 5
19. $$\frac{π}{12}, \frac{5π}{12}, \frac{13π}{12}, \frac{17π}{12}$$
20. Show work
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## Sum Of Two Abundant Numbers
#### Problem
The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of $28$ are $1, 2, 4, 7, 14,$ and $28,$ so the sum of proper divisors is $1 + 2 + 4 + 7 + 14 = 28$.
Similarly, if the sum of the proper divisors exceeds the number we call the number abundant. For example, $12$ is abundant because the divisors of $12$ are $1, 2, 3, 4, 6, 12,$ and the sum of proper divisors, $1 + 2 + 3 + 4 + 6 = 14 \gt 12$.
By first showing that $315p$ is abundant for all primes, $p \le 103,$ prove that all integers greater than $28123$ can be written as the sum of two abundant numbers.
#### Solution
Let $S(n)$ represent the sum of proper divisors of $n$.
$S(315) = S(3^2 \times 5 \times 7) = 1 + 3 + 5 + 7 + 9 + 15 + 21 + 35 + 45 + 63 + 105 = 309$.
When considering $S(315p)$ we must deal with two cases.
Case 1: $p$ is coprime with $315$ ($p \ne 3, 5, 7$)
$\therefore S(315p) = p(1 + 3 + ... + 105) + (1 + 3 + ... + 105) + 315 = 309p + 624$
For $315p$ to be abundant, $315p \lt 309p + 624 \Rightarrow p \lt 104$.
Case 2: $p = 3, 5, 7$
$S(315p) = p(1 + 3 + ... + 105) + q,$ where $q$ is the sum of divisors not containing a factor of $3, 5,$ or $7$ respectively.
So for $315p$ to be abundant it is sufficient to show $315p \lt 309p + q \Rightarrow 6p \lt q$.
When $p = 3, 6p = 18, q = 1 + 5 + 7 + 35 = 48 \Rightarrow 6p \lt 48$
When $p = 5, 6p = 30, q = 1 + 3 + 7 + 9 + 21 + 63 = 104 \Rightarrow 6p \lt 104$
When $p = 7, 6p = 42, q = 1 + 3 + 5 + 9 + 15 + 45 = 48 \Rightarrow 6p \lt 78$
Hence $315p$ is abundant for all primes, $p \le 103$.
It can be shown that multiples of abundant numbers are also abundant (see Even Sum Of Two Abundant Numbers). Thus all values of $m$ from $2$ to $103$ will either be prime or contain a prime in that domain. So although $315$ is deficient, $315m$ is guaranteed to be abundant for $2 \le m \le 103$.
We now search for the smallest abundant number that is coprime with $315$. Considering numbers of the form $2^k \times 11$ we find that $88$ is the smallest such example. Therefore the expression $315a + 88b$ will be the sum of two abundant numbers for $2 \le a \le 103$ and $b \ge 1$.
Clearly the expression produces integers congruent with $315a \pmod{88},$ and although $0 le a le 87$ will produce all possible congruences we require $a \ge 2$ to ensure $315a$ is abundant; that is, $2 \le a \le 89$ will produce all possible congruences.
It is necessary to have at least one multiple of $88$, so $315 \times 89 + 88$ represents that last integer before the congruences repeat. Hence every integer $n \gt 315 \times 89 + 88 = 28123$ can be written in the form $315a + 88b,$ which will be the sum of two abundant numbers. Q.E.D.
Problem ID: 348 (08 Nov 2008) Difficulty: 4 Star
Only Show Problem
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# Conversion of a Decimal Fraction into a Fractional Number
Now this topic will deal with conversion of decimal fraction into fractional number. For converting decimal fraction into fractional number we will have to write the given decimal number in the numerator leaving the decimal point and in the denominator we will write 1 and as many zeros as there are number of places after the decimal point.
Solved Examples on Conversion of a Decimal Fraction into a Fractional Number:
Change 150.05 into fractional number.
For changing 150.05 into fractional number we will first write 15005 in the numerator (leaving the decimal point).
Then 100 in the denominator as there are two places after the decimal point.
Therefore we will get,
15005/100
Now the next step is to change it into lowest terms
15005/100 ÷ 5/5
= 3001/20
Now 3001/20 cannot be further reduced into lowest terms any more hence this is the answer
Here are few more examples showing conversion of decimal fraction into a fractional number.
1. Change 34.20 into fractional number.
First write 3420 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
3420/100
Then changing it into lowest terms
3420/100 ÷ 20/20
= 171/5
2. Change 42.50 into fractional number.
First write 4250 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
4250/100
Then changing it into lowest terms
4250/100 ÷ 50/50
= 85/2
3. Change 78.639 into fractional number.
First write 78639 in the numerator (leaving the decimal point) and 1000 in the denominator as there are three places after the decimal point.
78639/1000
This cannot be further changed into lowest terms hence this will remain as answer
4. Change 56.675 into fractional number
First write 56675 in the numerator (leaving the decimal point) and 1000 in the denominator as there are three places after the decimal point.
56675/1000
Then changing it into lowest terms
56675/1000 ÷ 25/25
2267/40
5. Change 78.50 into fractional number.
First write 7850 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
7850/100
Then changing it into lowest terms
7850/100 ÷ 25/25
314/4
6. Change 845.75 into fractional number.
First write 84575 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
84575/100
Then changing it into lowest terms
84575/100 ÷ 25/25
3383/4
7. Change 52.32 into fractional number.
First write 5232 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
5232/100
Then changing it into lowest terms
5232/100 ÷ 4/4
= 1308/25
8. Change 105.32 into fractional number.
First write 10532 in the numerator (leaving the decimal point) and 100 in the denominator as there are two places after the decimal point.
10532/100
Then changing it into lowest terms
10532/100 ÷ 4/4
2633/25
This is how we need to convert decimal number into a fractional number.
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Q:
# How do I calculate cumulative averages?
A:
The cumulative average is calculated by weighting each grade by the number of credits the course was worth. The calculation of a cumulative average is how you determine your grade point average in school. This requires you to have access to your transcript so that both the credits earned and the grade earned for each course are visible.
## Keep Learning
1. Record the credit hours and grades
Record the credit hours and the grades from your transcript for each course. Typically, an A is worth four points, a B is worth three points, a C is worth two points, a D is worth one point and an F isn't worth any points. In some systems, an F is worth negative points, so make sure to check the values your school uses.
2. Calculate the total
Add the total number of points from the grades you received. It may be best to calculate the totals separately and add them together. For example, if you received 8 credits worth of an A grade and six credits for a B, you would perform the following calculation: 8*4 + 6*3 = 50.
3. Divide the total by credit hours
Add up the total number of credit hours that you have taken and divide the calculation from the previous step by this number. From the previous step, you would have 14 credit hours and perform the following calculation: 50/14 = 3.57. The number that you get is your cumulative average.
Sources:
## Related Questions
• A:
Introductory algebra is a fundamental mathematics course. It is essential to master this course before moving on to more advanced material. Key concepts in introductory algebra involve the study of variables, expressions and equations.
Filed Under:
• A:
An experimental variable is something that a scientist changes during the course of an experiment. It is distinguished from a controlled variable, which could theoretically change, but the scientists keep constant.
Filed Under:
• A:
In a ninth-grade Algebra I course, students learn to solve problems using the foundational rules of mathematical computations, and they learn to solve algebraic expressions to find the values of variables. Some Algebra I problems require students to graph equations and inequalities. The study of equations in Algebra I helps prepare students for the study of functions in Precalculus and Calculus courses.
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# 3.2.1 - Expected Value and Variance of a Discrete Random Variable
3.2.1 - Expected Value and Variance of a Discrete Random Variable
By continuing with example 3-1, what value should we expect to get? What would be the average value?
We can answer this question by finding the expected value (or mean).
Expected Value (or mean) of a Discrete Random Variable
For a discrete random variable, the expected value, usually denoted as $\mu$ or $E(X)$, is calculated using:
$\mu=E(X)=\sum x_if(x_i)$
The formula means that we multiply each value, $x$, in the support by its respective probability, $f(x)$, and then add them all together. It can be seen as an average value but weighted by the likelihood of the value.
## Example 3-2: Expected Value
In Example 3-1 we were given the following discrete probability distribution:
$x$ $f(x)$ 0 1 2 3 4 1/5 1/5 1/5 1/5 1/5
What is the expected value?
\begin{align} \mu=E(X)=\sum xf(x)&=0\left(\frac{1}{5}\right)+1\left(\frac{1}{5}\right)+2\left(\frac{1}{5}\right)+3\left(\frac{1}{5}\right)+4\left(\frac{1}{5}\right)\\&=2\end{align}
For this example, the expected value was equal to a possible value of X. This may not always be the case. For example, if we flip a fair coin 9 times, how many heads should we expect? We will explain how to find this later but we should expect 4.5 heads. The expected value in this case is not a valid number of heads.
Now that we can find what value we should expect, (i.e. the expected value), it is also of interest to give a measure of the variability.
Variance of a Discrete Random Variable
The variance of a discrete random variable is given by:
$\sigma^2=\text{Var}(X)=\sum (x_i-\mu)^2f(x_i)$
The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. Then sum all of those values.
There is an easier form of this formula we can use.
$\sigma^2=\text{Var}(X)=\sum x_i^2f(x_i)-E(X)^2=\sum x_i^2f(x_i)-\mu^2$
The formula means that first, we sum the square of each value times its probability then subtract the square of the mean. We will use this form of the formula in all of our examples.
Standard Deviation of a Discrete Random Variable
The standard deviation of a random variable, $X$, is the square root of the variance.
$\sigma=\text{SD}(X)=\sqrt{\text{Var}}(X)=\sqrt{\sigma^2}$
## Example 3-3: Standard Deviation
Consider the first example where we had the values 0, 1, 2, 3, 4. The PMF in tabular form was:
$x$ $f(x)$ 0 1 2 3 4 1/5 1/5 1/5 1/5 1/5
Find the variance and the standard deviation of X.
$\text{Var}(X)=\left[0^2\left(\dfrac{1}{5}\right)+1^2\left(\dfrac{1}{5}\right)+2^2\left(\dfrac{1}{5}\right)+3^2\left(\dfrac{1}{5}\right)+4^2\left(\dfrac{1}{5}\right)\right]-2^2=6-4=2$
$\text{SD}(X)=\sqrt{2}\approx 1.4142$
## Example 3-4: Prior Convictions
Click on the tab headings to see how to find the expected value, standard deviation, and variance. The last tab is a chance for you to try it.
Let X = number of prior convictions for prisoners at a state prison at which there are 500 prisoners. ($x = 0,1,2,3,4$)
$X=x$ $Number\ of\ Prisoners$ $f(x) = P(X=x)$ $f(x)=P(X=x)$ 0 1 2 3 4 80 265 100 40 15 80/500 265/500 100/500 40/500 15/500 0.16 0.53 0.2 0.08 0.03
What is the expected value for number of prior convictions?
$X=x$ $Number\ of\ Prisoners$ $f(x) = P(X=x)$ $f(x)=P(X=x)$ 0 1 2 3 4 80 265 100 40 15 80/500 265/500 100/500 40/500 15/500 0.16 0.53 0.2 0.08 0.03
For this we need a weighted average since not all the outcomes have equal chance of happening (i.e. they are not equally weighted). So, we need to find our expected value of $X$, or mean of $X$, or $E(X) = \Sigma f(x_i)(x_i)$. When we write this out it follows:
$=(0.16)(0)+(0.53)(1)+(0.2)(2)+(0.08)(3)+(0.03)(4)=1.29$
Stop and think! Does the expected value of 1.29 make sense?
Calculate the variance and the standard deviation for the Prior Convictions example:
$X=x$ $Number\ of\ Prisoners$ $f(x) = P(X=x)$ $f(x)=P(X=x)$ 0 1 2 3 4 80 265 100 40 15 80/500 265/500 100/500 40/500 15/500 0.16 0.53 0.2 0.08 0.03
Using the data in our example we find that...
\begin{align} \text{Var}(X) &=[0^2(0.16)+1^2(0.53)+2^2(0.2)+3^2(0.08)+4^2(0.03)]–(1.29)^2\\ &=2.53–1.66\\ &=0.87\\ \text{SD}(X) &=\sqrt(0.87)\\ &=0.93 \end{align}
Note! If variance falls between 0 and 1, the SD will be larger than the variance.
What is the expected number of prior convictions? Below is the probability distribution table for the prior conviction data. Use this table to answer the questions that follow.
$X=x$ $f(x)=P(X=x)$ 0 1 2 3 4 0.16 0.53 0.2 0.08 0.03
a. What is the probability a randomly selected inmate has exactly 2 priors?
$P(X=2) = 100/500 = 0.2$
b. What is the probability a randomly selected inmate has < 2 priors?
$P(X<2)=P(X=0\ or\ 1)=P(X=0)+P(X=1)=0.16+0.53=0.69$
c. What is the probability a randomly selected inmate has 2 or fewer priors?
$P(X≤2)=(X=0)+P(X=1)+P(X=2)=0.16+0.53+0.2=0.89$
d. What is the probability a randomly selected inmate has more than 2 priors?
$P(X>2)=P(X=3\ or\ 4)=P(X=3)+P(X=4)\ or\ 1−P(X≤2)=0.11$
e. Finally, which of a, b, c, and d above are complements?
c. and d. are complements
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# Lesson 10
Using Data Displays to Find Associations
## 10.1: Sports and Musical Instruments (5 minutes)
### Warm-up
The purpose of this warm-up is for students to answer questions about relative frequency of items after finding missing information in a two-way table.
Monitor for students who find the percentages for the final two questions using different strategies to share during the whole-class discussion.
### Launch
Give students 2 minutes of quiet work time followed by a whole-class discussion.
### Student Facing
For a survey, students in a class answered these questions:
• Do you play a sport?
• Do you play a musical instrument?
1. Here is a two-way table that gives some results from the survey. Complete the table, assuming that all students answered both questions.
plays instrument does not play instrument total
plays sport 5 16
does not play sport
total 15 25
2. To the nearest percentage point, what percentage of students who play a sport don’t play a musical instrument?
3. To the nearest percentage point, what percentage of students who don’t play a sport also don’t play a musical instrument?
### Activity Synthesis
Ask students to share the missing information they found for the table. Record and display their responses for all to see.
Select students previously identified to explain how they found the percentages for the final two questions and what that percentage represents.
1. Students who find a percentage using the values given (for example 31% since $$\frac{5}{16} \approx 0.31$$), then subtract from 100% (for example 69% since $$100 - 31 = 69$$) to answer the question.
2. Students who find the actual values first by subtracting (for example $$16 - 5 = 11$$) then compute the percentage (for example 69% because $$\frac{11}{16}=0.6875$$).
Ask the rest of the class if they agree or disagree with the strategies and give time for any questions they have.
## 10.2: Sports and Music Association (20 minutes)
### Activity
Now that students are more familiar with two-way tables showing relative frequency, they are ready to create their own segmented bar graphs. In this activity, students create two segmented bar graphs based on the same two-way table by considering percentages of the rows and columns separately. After creating the segmented bar graphs, they are analyzed to determine if there is an association present in the data.
### Launch
Arrange students in groups of 2. After a brief introduction, give 5–10 minutes of quiet work time. Ask students to compare their answers with their partner and try to resolve any differences. Finish with a whole-class discussion.
Display the two-way table from the previous lesson's cool-down activity containing the data collected about the class's playing sports and musical instruments. If the data is unavailable, the data from this lesson's warm-up can be used.
Tell students they should work with their partners to each work on one of the graphs. One student should work on problems 1 and 2 while their partner should work on 3 and 4. After they have completed their graphs, they should work together to understand their partners graphs and complete the last problem together.
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organization and problem solving. For example, present one question at a time. Some students may benefit from a checklist on how to create a segmented bar graph.
Supports accessibility for: Organization; Attention
### Student Facing
Your teacher will give you a two-way table with information about the number of people in your class who play sports or musical instruments.
1. Complete this table to make a two-way table for the data from earlier. The table will show relative frequencies by row.
plays instrument does not play instrument row total
plays sport 100%
does not play sport 100%
2. Make a segmented bar graph for the table. Use one bar of the graph for each row of the table.
3. Complete the table to make a two-way table for the data from earlier. The table will show relative frequencies by column.
plays instrument does not play instrument
plays sport
does not play sport
column total 100% 100%
4. Using the values in the table, make a segmented bar graph. Use one bar of the graph for each column of the table.
5. Based on the two-way tables and segmented bar graphs, do you think there is an association between playing a sport and playing a musical instrument? Explain how you know.
### Anticipated Misconceptions
Students may draw the segmented bar graph incorrectly. Most likely, they will accidentally graph frequency instead of relative frequency. They may also graph relative frequencies, but without stacking them. Both segmented bars should go from 0 to 100.
### Activity Synthesis
To clarify how to create and interpret segmented bar graphs, ask:
• “What different information can be seen by the two segmented bar graphs?”
• “Why are the numbers in the top left box in the two tables different? What do they mean?” (In the first table it represents the percentage who also play musical instruments out of all the people who play sports. In the second table it represents the percentage of people who also play sports out of all the people who play musical instruments.)
• “Is there an association between the two variables? Explain or show your reasoning.” (The answer will depend on class data, but the reasoning should include an analysis of the relative frequencies within categories. There is an association if the percentages within one category are very different from the percentages in another category.)
If there is an association, ask what the segmented bar graphs would look like if there was no association. If there is not an association, ask what the segmented bar graphs would look like if there was one.
Writing, Speaking: MLR1 Stronger and Clearer Each Time. Use this routine to give students a structured opportunity to revise and refine their response to the last question. Ask each student to meet with 2–3 other partners in a row for feedback. Provide students with prompts for feedback that will help them strengthen their ideas and clarify their language (e.g., “Why do you think there is a (positive/negative) association?”, “How do the relative frequencies help to answer this question?”, “How could you say that another way?”, etc.). Students can borrow ideas and language from each partner to strengthen the final product. They can return to the first partner and revise and refine their initial response.
Design Principle(s): Optimize output (for explanation)
## 10.3: Colored Erasers (15 minutes)
### Activity
This activity provides students less structure for their work in creating segmented bar graphs to determine an association (MP4). In addition, the data in this activity is split into more than two options. Students work individually to create a segmented bar graph based on either columns or rows and then share their information with a partner who has created the other segmented bar graph. Together, partners discuss the segmented bar graphs to determine if there is an association between the variables (MP3). In particular, students should notice that there is evidence of an association is the relative frequencies within a category are very different from the relative frequencies in another category.
As students work, identify groups that use the different segmented bar graphs to explain why there is an association between the color of the eraser and flaws.
### Launch
Keep students in groups of 2. Give 5 minutes quiet work time followed by 5 minutes of partner discussion and then a whole-class discussion.
Provide students access to colored pencils. Either assign or have partners choose which will make a graph for each row and which will make a graph for each column.
Representation: Access for Perception. Read the directions aloud. Students who both listen to and read the information will benefit from extra processing time. Check for understanding by inviting students to rephrase directions in their own words.
Supports accessibility for: Language
### Student Facing
An eraser factory has five machines. One machine makes the eraser shapes. Then each shape goes through the red machine, blue machine, yellow machine, or green machine to have a side colored.
The manager notices that an uncolored side of some erasers is flawed at the end of the process and wants to know which machine needs to be fixed: the shape machine or some of the color machines. The manager collected data on the number of flawed and unflawed erasers of each color.
unflawed flawed total
red 285 15 300
blue 223 17 240
yellow 120 80 200
green 195 65 260
total 823 177 1000
1. Work with a partner. Each of you should make one segmented bar graph for the data in the table. One segmented bar graph should have a bar for each row of the table. The other segmented bar graph should have one bar for each column of the table.
2. Are the flawed erasers associated with certain colors? If so, which colors? Explain your reasoning.
### Student Facing
#### Are you ready for more?
Based on the federal budgets for 2009, the table shows where some of the federal money was expected to go. The values are in billions of U.S. Dollars.
United States Japan United Kingdom
defense 718.4 42.8 49.2
education 44.9 47.5 113.9
1. Why would a segmented bar graph be more useful than the table of data to see any associations between the country and where the money is spent?
2. Create a segmented bar graph that represents the data from the table.
3. Is there an association between the country’s budget and their spending in these areas? Explain your reasoning.
### Activity Synthesis
The purpose of this discussion is to identify strategies for creating segmented bar graphs and for analyzing them to determine if there is an association among variables.
Ask, “What strategies did you use to create the segmented bar graphs?” (First, we created a new table of the relative frequencies. Then we approximated the heights of the segments based on the percentages from the table.)
Select previously identified groups to share their explanation for noticing an association.
1. Groups that use the segmented bar graph based on rows.
2. Groups that use the segmented bar graph based on columns.
After both explanations are shared, ask students, "Do you think that noticing the association was easier with one of the graphs?" (Likely the segmented bar graph based on rows is easier since there are only 2 segments and it is easier to see that the yellow and green erasers are more flawed.)
Finally, ask students, "If there was not an association between color and flaws, what might the segmented bar graph based on the rows look like? What might the segmented bar graph based on the columns look like?” (The segmented bar graph based on the rows would have each segmented bar look about the same. That is, the line dividing the two segments would be at about the same height in each bar. The segmented bar graph based on the columns would have segments that are all approximately equal. That is, each segment should represent about 25% of the entire bar.)
## Lesson Synthesis
### Lesson Synthesis
Remind students that we have been looking for associations in categorical data, and that there is evidence of an association if the relative frequencies of some characteristic are very different from each other in the different groups. Ask:
• “Is it easier to see evidence of an association in a frequency table or a relative frequency table?” (It depends on the data. If the two groups are approximately the same size, it doesn't matter very much, but when they are different sizes, it is usually easier to compare using relative frequencies.)
• “How can we see evidence of an association in a two-way table of either kind?” (By numerically comparing the proportions between the two groups.)
• “How can we see evidence of an association in a bar graph or segmented bar graph?” (By visually comparing the proportions between the two groups.)
## Student Lesson Summary
### Student Facing
In an earlier lesson, we looked at data on meditation and state of mind in athletes.
Is there an association between meditation and state of mind?
The bar graph shows that more athletes were calm than agitated among the group that meditated, and more athletes were agitated than calm among the group that did not. We can see the proportions of calm meditators and calm non-meditators from the segmented bar graph, which shows that about 66% of athletes who meditated were calm, whereas only about 27% of those who did not meditate were calm.
This does not necessarily mean that meditation causes calm; it could be the other way around, that calm athletes are more inclined to meditate. But it does suggest that there is an association between meditating and calmness.
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### Removing Brackets Practice Question
#### Question 2
Remove the brackets and simplify the following expression.
A. -3y + 1
B. -3x - 1
C. 3x - 1
D. 3x + 1
### Step by Step Solution
• #### Step 1
First, let's remove the brackets in -(5x -3).
• #### Step 2
Multiply -1 and -3 together. You will get -5x. Note that "-" is the same as "-1"
• #### Step 3
Now, multiply -1 and -3 together. You will get +3.
• #### Step 4
Put back -5x +3 into the expression. Next, we remove the brackets in 2(4x -2).
• #### Step 5
Now, multiply 2 and 4x together. You will get +8x.
• #### Step 6
Now, multiply 2 and -2 together. You will get -4.
• #### Step 7
Put back 8x -4 into the expression. The expression now becomes:
Notice that this expression has like terms. Therefore, let's arrange the like terms together. The expression now becomes:
• #### Step 8
Now, simplify the expression by adding +3 and -4. You will get -1.
• #### Step 9
Now, simplify the expression by adding +8x and -5x. You will get 3x.
• #### Step 10
The expression becomes 3x-1. There is no more like terms, therefore this is the simplest expression. Clearly, the answer is C.
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Number Systems: A number system is a format for representing different numerical values Any number system calculates the value of the ...
06. Aptitude Questions - Numbers Theory - Number Systems - Conversions
Number Systems:
A number system is a format for representing different numerical values
Any number system calculates the value of the number being represented by
1) The digit used to represent and
2) The place where the digits are present in the actual number.
How the value of the number is calculated also depends on the base of the number system.
Definition of Base of a number system: The base of a number system is the number of digits used to represent the number in that number system. The base is also called radix
We use 10 digits to represent the numbers in the decimal number system, so the base of the decimal number system is 10.
In case a number XYZ *where X ,Y, and Z are digits are represented in some number system, which has a base “B”, its value would be calculated by the expression: X x B2 + Y x B + Z
Decimal Number System:
The system generally used by the convention is the decimal number system. The base of the decimal number system is 10(ten), and the digits used are from 0(zero) to 9(nine)
In the decimal number system, a number can be represented as
47218= 4 x 104 + 7 x 103 + 2 x 102 + 1 x 101 + 8 x 100
= 40000 + 7000 + 200 +10 +8
Binary Number System:
The digits used for the binary system would be 0 and 1. We use only 2 digits in the binary number system, so the base of the binary number system is 2.
The binary digits o and 1 are called bits.
In the binary number system, zero is represented by 0, and 1 is represented by 1. After 1, there is no any digit to represent two. Therefore, two is written is as 10, three written as 11. Again the number four is represented as 100.
For a number represented as 10101 in the binary system would have a value of
(1 x 20) + (0 x 21) + ( 1 x 22) + (0 x 23 ) + 1 x 24) which amounts to 1 +0 + 4 + +0 + 16 =21
Octal Number System:
The digits use for the octal system would be 0 through 7. The number of digits used in this system is 8, so base of the octal number system is 8.
The number 8 of the decimal number system is represented by 10, 9 by 11, and so on.
For example , a number represented as 123 in the octal number system would have a value of 1 x 82 + 2 x 81 + 3 x 80 which amounts to 64 + 16 + 3=83 as represented in the decimal system.
The base of the hexadecimal number system is 16. The ten digits from 0 to 9 and letters from A to F are used to represent numbers in this system.
The digits 0 to 9 are same as those for the decimal number system, and digits from 10 to 15 are represented by alphabets A to F respectively.
For example, a number represented as 6A9 in the hexadecimal number system would have a value of 6 x 162 + A x 161 + 9 x 160 which amounts to 6 x 256 + 10 x 16 + 9 = 1705 as represented in the decimal number system.
Conversion of Base 10 to Base 2 :
Conversion of Decimal Number to Binary Number:
In case a number ABC (with A, B, and C as the digits) represented in the base 10 system, is to be converted to a binary system, then it has to be repeatedly divided by 2. The remainders obtained on each division provide the digits that would appear in the binary system.
Thus a number 25 can be repeatedly divided by 2 as follows.
Number / Base Quotient Remainder 25 / 2 12 1 12 / 2 6 0 6 / 2 3 0 3 / 2 1 1 1 / 2 0 1
Repeated division until we get the remainder 0, gives the remainders .1,0,0,1 and 1 in that order. On writing these in the reverse order, we get the binary number 11001.
Example 2: Convert 125(10) to equivalent binary number.
2 | 125
2 | 62 - 1
2 | 31- 0
2 | 15 - 1
2 | 7 - 1
2 | 3 - 1
2 | 1 - 1
0 - 1
The binary equivalent of 125(10) is 1111101(2)
Conversion of a Decimal Fraction to Binary Fraction:
The conversion of a decimal fraction to a binary fraction can be done using the successive multiplication by 2 techniques.
Following example will explain the procedure to convert a decimal fraction to a binary fraction:
Fraction Fraction x2 Remainder Fraction Integer 0.25 0.5 0.5 0 0.5 1 0 1
0.25(10) = 0.01
Conversion of A Binary Number System To Decimal Number System
To convert binary number to decimal number, follow the procedure:
Example : Convert 11001(2) to equivalent decimal number
Solution : 11001 = 1 x 24 + 1 x 23 + 0 x 22 + 0 x21 + 1
=16 + 8 + 0 + 0 + 1
=25
Example 2: Convert 10.101 to equivalent decimal number
Solution : 10.101(2) = 1 x 2 + 0 x 2 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3
= 1 x 2 + 0 x 2+ 1/2 + 0/4 + 1/8
= 2 + 0 +0.5 +0 +0.125
= 3.625
Conversion of a decimal number to octal number :
How to convert a number from base 10 to base 8 :
In case a number ABC (with A, B, and C as the digits) represented in the base 10 system, is to be converted to an octal system, then it has to be divided by 8 repeatedly. The remainders obtained on each division provide the digits that would appear in the octal number.
Number / Base Quotient Remainder 205 / 8 25 5 25 / 8 3 1 3 / 2 1 1
We have to repeat the division until we get the remainder zero. The remainders obtained are 5,1, and 1. On writing these in the reverse order, we get the octal number 115.
Example 2: Convert 463(10) to equivalent octal number.
8 | 463
8 | 57- 7
8 | 7- 1
0 - 7
The octal equivalent of 463(10) is 717(8)
Conversion of octal number to decimal number :
Example 1:Convert 716(8) to equivalent decimal number
Solution: 716(8) = 7 x 82 + 1 x 81 + 6 x 80
= 7 x 64 + 1 x 8 + 6
= 448 + 8 + 6 = 462
Example 2 : Convert 65.24 to equivalent decimal number
Solution : 65.24(8) = 6 x 8 + 5 + 2/8 + 4/82
=48 + 5 +0.25+0.0625
= 53.3125
Conversion of a decimal number to Hexadecimal Number:
In case a number ABC (with A, B, and C as the digits) represented in the base 10 system, is to be converted to an octal system, then it has to be divided by 16 repeatedly. The remainders obtained on each division provide the digits that would appear in the octal number.
Number / Base Quotient Remainder 410 / 16 25 10 25 / 16 1 9 9 / 16 0 9
We have to repeat the division until we get the remainder zero. The remainders obtained are 10, 9, and 9.
In the hexadecimal number system, the equivalent number for 10 is A.
On writing these in the reverse order, we get the octal number 99A.
Example 2: Convert 8655(10) to equivalent decimal number.
16 | 8655
16 | 540- 15
16 | 33- 12
16 | 2 - 1
0 - 2
The hexadecimal equivalents to 12 and 15 are C, F
The octal equivalent of 8655(10) is 21CF(8)
Conversion of the Hexadecimal number to Decimal Number:
To convert a hexadecimal number to a decimal number, multiply each digit of the given hexadecimal number by powers of 16 as follows :
Example: Convert 90BA(16) to equivalent decimal number
Solution : 90BA(16) = 9 x 163 + 0 x 162 + B x 161 + A x 160
= 9 x 4096 + 0 x 256 + 11 x 16 +10 x 1
=36864 + 0 + 176 +10 =37050
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# The sum of the angles of a regular n-gon is 1440 degrees. What is the sum of the angles of another regular
The sum of the angles of a regular n-gon is 1440 degrees. What is the sum of the angles of another regular polygon if it is known that the vertices of the first polygon taken through one serve as the vertices of the second.
To solve the problem, we use the theorem on the sum of the angles of a convex polygon:
N is a gon, the sum of the angles is:
180 ° * (n – 2).
By the condition of the problem, the sum of the angles of the n-gon is known, we compose the equation:
180 ° * (n – 2) = 1440 °
n – 2 = 8
n = 10 – the number of angles given by the condition of the polygon.
The number of corners of the second polygon is:
10/2 = 5 (taken through one by condition).
Find the sum of the angles of the pentagon:
180 ° * (n – 2) = 180 ° * 3 = 180 ° * (5 – 2) = 540 °.
Answer: the sum of the angles of the second polygon is 540 °.
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# Triangles, NCERT Solutions, Class 9, Maths Class 9 Notes | EduRev
## Class 9 : Triangles, NCERT Solutions, Class 9, Maths Class 9 Notes | EduRev
The document Triangles, NCERT Solutions, Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.
All you need of Class 9 at this link: Class 9
### Theorems with logical proofs
Theorem 1. : If a ray stands on a line, then the sum of the adjacent angles so formed is 180°.
Given: The ray PQ stands on the line XY.
To Prove: m∠QPX + m∠YPQ = 180°.
Construction: Draw PE perpendicular to XY.
Proof: m∠QPX = m∠QPE + m∠EPX
m∠QPE + 90° (1)
m∠YPQ = m∠YPE − m∠QPE
90° − m∠QPE (2)
(1) + (2) ⇒ m∠QPX + m∠YPQ = (m∠QPE + 90°) + (90° − m ∠QPE) = 180°.
Thus the theorem is proved.
Theorem 2: If two lines intersect, then the vertically opposite angles are of equal measure.
Given: Two lines AB and CD intersect at the point O
To prove: m∠AOC = m∠BOD, m∠BOC = m∠AOD.
Proof: The ray OB stands on the line CD.
∴ m∠BOD + m∠BOC = 180° (1)
The ray OC stands on the line AB.
∴ m∠BOC + m∠AOC = 180° (2)
From (1) and (2),
m∠BOD + m∠BOC = m∠BOC + m∠AOC
∴ m∠BOD = m∠AOC.
Since the ray OA stands on the line CD,
m∠AOC + m∠AOD = 180° (3)
From (2) and (3), we get
m∠BOC + m∠AOC = m∠AOC + m∠AOD
∴ m∠BOC = m∠AOD.
Hence the theorem is proved.
Theorem 3: The sum of the three angles of a triangle is 180°.
Given: ABC is a triangle (see Figure 6.54).
To prove: + ∠+ ∠= 180°.
Construction: Through the vertex A, draw the line
XY parallel to the side BC.
Proof: XY || BC ∴ m∠XAB = m∠ABC (alternate angles).
= ∠B. (1)
Next, AC is a transversal to the parallel lines XY and BC.
∴ m∠YAC = m∠ACB (alternate angles) = ∠C. (2)
We also have m∠BAC = m∠A. (3)
(1) + (2) + (3) ⇒ m∠XAB + m∠YAC + m∠BAC = m∠+ m∠+ m∠A
⇒ (m∠XAB + m∠BAC) + m∠CAY = m∠+ m∠+ m∠C
⇒ m∠XAC + m∠CAY = m∠+ m∠+ m∠C
⇒ 180° = m∠+ m∠+ m∠C.
Hence the theorem is proved.
Theorem4. : The angles opposite to equal sides of a triangle are equal.
Given: ABC is a triangle where AB AC).
To prove: = ∠C.
Construction: Mark the mid point of BC as and join AM.
Proof: In the triangles AMB and AMC
(i) BM CM (ii) AB AC (iii) AM is common.
∴ By the SSS criterion, ΔAMB ≡ ΔAMC.
∴ Corresponding angles are equal. In particular, ∠= ∠C.
Hence the theorem is proved.
Theorem 5. : The side opposite to the larger of two angles in a triangle is longer than the side opposite to the smaller angle.
Given: ABC is a triangle, where ∠is larger than ∠C, that is m∠> m∠C.
To prove: The length of the side AC is longer than the length of the side AB.
i.e., AC AB (see Figure 6.56).
Proof: The lengths of AB and AC are positive numbers. So three cases arise
(i) AC AB
(ii) AC AB
(iii) AC AB
Case (i) Suppose that AC AB. Then the side AB has longer length than the side AC. So the angle ∠which is opposite to AB is larger measure than that of ∠which is opposite to the shorter side AC. That is, m∠> m∠B. This contradicts the given fact that m∠> m∠C. Hence the assumption that AC AB is wrong.
AC AB.
Case (ii) Suppose that AC AB. Then the two sides AB and AC are equal. So the angles opposite to these sides are equal. That is ∠= ∠C. This is again a contradiction to the given fact that ∠> ∠C. Hence AC AB is impossible. Now Case (iii) remains alone to be true.
Hence the theorem is proved.
Theorem 6. : A parallelogram is a rhombus if its diagonals are perpendicular.
Given: ABCD is a parallelogram where the diagonals AC and BD are perpendicular.
To prove: ABCD is a rhombus.
Construction: Draw the diagonals ACand BD. Let be the point of intersection of AC and BD (see Proof: In triangles AMB andBMC,
(i) ∠AMB = ∠BMC = 90°
(ii) AM MC
(iii) BM is common.
∴ By SSA criterion, ΔAMB ≡ ΔBMC.
∴Corresponding sides are equal.
In particular, AB BC.
Since ABCD is a parallelogram, AB CDBC AD.
∴ AB BC CD AD.
Hence ABCD is a rhombus. The theorem is proved.
Theorem 7:Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base.
Solution: Let ABC be an isosceles triangle where AB AC. Let AD be the bisector of the vertex angle ∠A. We have to prove thatAD is the median of the base BC. That is, we have to prove that is the mid point of BC. In the triangles ADB and ADC,
we have AB AC, m∠BAD = m∠DAC AD is an angle (bisector), AD is common.
∴ By SAS criterion, ∠ABD ≡ ΔACD.
∴ The corresponding sides are equal.
∴ BD DC.
i.e., is the mid point of BC.
Theorem 8 Prove that the sum of the four angles of a quadrilateral is 360°.
Solution: Let ABCD be the given quadrilateral. We have to prove that ∠+ ∠+ ∠+ ∠= 360°. Draw the diagonal AC. From the trianglesACD and ABC, we get
DAC + ∠+ ∠ACD = 180° (1)
CAB + ∠+ ∠ACB = 180° (2)
(1) (2) ⇒ ∠DAC + ∠+ ∠ACD + ∠CAB
+ ∠+ ∠ACB = 360°
⇒ (∠DAC + ∠CAB) + ∠+ (∠ACD + ∠ACB)+ ∠= 360°
⇒ ∠+ ∠+ ∠+ ∠= 360°.
Theorem 8: In a rhombus, prove that the diagonals bisect each other at right angles.
Solution: Let ABCD be a rhombus, Draw the diagonals ACand BD. Let them meet at O. We have to prove that is the mid point of both AC and BD and that AC is perpendicular (⊥) to BD.
Since a rhombus is a parallelogram, the diagonals AC and BD bisect each other.
∴ OA OCOB OD.
In triangles AOB and BOC, we have
(i) AB BC
(ii) OB is common
(iii) OA OC
∴ ΔAOB = ΔBOC, by SSS criterion.
∴ ∠AOB = ∠BOC.
Similarly, we can get ∠BOC = ∠COD, ∠COD =∠DOA.
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA (say)
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°
∴ += 360°
∴ 4= 360° or = 4360° = 90°.
∴ The diagonals bisect each other at right angles.
Theorem 9: Prove that a diagonal of a rhombus bisects each vertex angles through which it passes.
Solution: Let ABCD be the given rhombus. Draw the diagonals AC and BD. Since AB || CDand AC is a transversal to AB and CD. We get
BAC = ∠ACD (alternate angles are equal) (1)
But AD CD (since ABCD is a rhombus)
∴∠ACD = ∠DAC
(angles opposite to the equal sides are equal)(2)
From (1) and (2), we get
BAC = ∠DAC
i.e., AC bisects the angle ∠A.
Similarly we can prove that AC bisects ∠CBD bisects ∠and BD bisects ∠D
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# What is the sum of all positive even divisors of 1000?
I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade students.Thank you very much!
What is the sum of all positive even divisors of 1000?
• @ColinMcLarty What about, say, 8? Sep 12, 2016 at 17:40
• @GoodDeeds The divisors correspond, but are not identical. Each number 2*k that divides 1000 corresponds to the number k that divides 500. Sep 12, 2016 at 17:55
• @chepner I had misunderstood, thank you. Sep 12, 2016 at 17:56
First consider the prime factorization of $1000$. We have:
$$1000=2^3\times 5^3$$
Now, how can we list all the factors of $1000$? We see that we can try listing them in a table:
$$\begin{array}{c|c|c|} & \text{5^0} & \text{5^1} & \text{5^2} & \text{5^3} \\ \hline \text{2^0} & 1 & 5 & 25 & 125 \\ \hline \text{2^1} & 2 & 10 & 50 & 250 \\ \hline \text{2^2} & 4 & 20 & 100 & 500 \\ \hline \text{2^3} & 8 & 40 & 200 & 1000 \\ \hline \end{array}$$
We see that we can take $(2^1+2^2+2^3) \times (5^0 + 5^1 + 5^2 + 5^3) = 2184$. To get the sum of all factors, we would also include $2^0$ on the left side of the multiplication. We exclude $2^0$ because those would be odd factors.
• Interesting visual approach, but this would only work for smaller numbers, yes? If I take i.e. $1778700=2^2\cdot3^1\cdot5^2\cdot7^2\cdot11^2$, you would have 5 "dimensions" for your table, which would not be easy to get the even factors out? Or is there another way? Sep 13, 2016 at 11:44
• @hamena314 All odd factors will lie in the 4-dimensional 'row' which corresponds to the factor $2^0$ and all even ones will lie in the row corresponding to $2^1$ and $2^2$. So again, the sum of positive even divisors will be $(2^1 + 2^2) \times (3^0 + 3^1) \times (5^0 + 5^1 + 5^2) \times (7^0 + 7^1 + 7^2) \times (11^0 + 11^1 + 11^2).$ Sep 13, 2016 at 16:04
Since $$1000=2^3\cdot5^3$$, the even divisors of $$1000$$ have the form $$2^i5^j$$, where $$1\leq i\leq 3$$ and $$0\leq j\leq 3$$. There are only 12 of them, so you can do this calculation directly.
Alternatively, it is $$\sum_{i=1}^3\sum_{j=0}^32^i5^j=(\sum_{i=1}^3 2^i)(\sum_{j=0}^3 5^j)=\frac{2^4-2}{2-1}\cdot\frac{5^4-1}{5-1}=14\cdot156=2184$$.
First, factor $1000=2^3\cdot 5^3$. A divisor of $1000$ has to be of the form $2^a\cdot 5^b$. If you want it to be even, you need $a \ge 1$. How many choices do you have? You can simplify the calculation (though it is not worth it for this small a case) by making it the product of two geometric series. If you wanted the sum of even divisors of $10^{32}$ it would be worthwhile, and is worth understanding.
We know that product of two odds is always odd and since $1000=2^3\cdot5^3$, the only odd terms are $1$, $5^1$, $5^2$, $5^3$, and their sum is 1 + 5 + 25 + 125 = 156.
Also sum of divisors of 1000 = σ($2^3$.$5^3$) = [($2^4$-1)/ (2-1)].[($5^4$-1)/(5-1)] = 15.156 = 2340.
Subtracting the sum of odd divisors gives the sum of even divisors, 2340-156 = 2184.
I know the function for the summation of divisors of a number, σ ,maybe a bit new for the 8th grade but it is easy to grasp and worthwhile to know.
• The product of two evens is even and also product of an odd and even is also even. The only odd terms are product of odd and an odd .
– user356774
Sep 13, 2016 at 8:58
$n$ is a positive even divisor of $1000$ if and only if $n = 2m$ where $m$ is a divisor of $500$. Since $500 = 2^2 \times 5^3$, there are $(2+1)(3+1) = 12$ divisors of $500$. Those divisors are
\begin{array}{rr} 1, & 500, \\ 2, & 250, \\ 4, & 125, \\ 5, & 100, \\ 10, & 50, \\ 20, & 25 \\ \end{array}
so there are $12$ positive even divisors of $1000$.
Those divisors are
\begin{array}{cc} 2, & 1000, \\ 4, & 500, \\ 8, & 250, \\ 10, & 200, \\ 20, & 100, \\ 40, & 50 \\ \end{array}
The sum of the positive divisors of $500 = 2^2 \times 5^3$ equals $\dfrac{2^3 - 1}{2 - 1} \times \dfrac{5^4 - 1}{5 - 1} = 1092$
So the sum of the even divisors of $1000$ is $2 \times 1092 = 2184$
We want to exclude the odd divisors, and the odd divisors of $1000$ are exactly the divisors of $125$. Therefore your answer is the sum of divisors of $1000$, minus the sum of divisors of $125$, or using $\sigma$ for the sum of divisors, $$\sigma(1000) - \sigma(125).$$ Now, to compute $\sigma(n)$, you split $n$ into its relatively prime parts and then sum the divisors of each part. So we get \begin{align*} \sigma(1000) - \sigma(125) &= \sigma(125) \sigma(8) - \sigma(125) \\ &= (1 + 5 + 25 + 125)(1 + 2 + 4 + 8) - (1 + 5 + 25 + 125) \\ &= 156 \cdot15 - 156 \\ &= 156 \cdot 14 \\ &= 2184. \end{align*}
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# Interactive video lesson plan for: What is average rate of change? (KristaKingMath)
#### Activity overview:
► My Derivatives course: https://www.kristakingmath.com/derivatives-course
We already know that the slope of a function at a particular point is given by the derivative of that function, evaluated at that point. So we can easily find the slope, or the rate of change, in one particular location, and so we could call this the instantaneous rate of change, because it’s the rate of change at that particular instant.
If instead we want to find the rate of change over a larger interval, then we’d need to use the average rate of change formula. After all, we’re looking for the average rate at which the function changes over time, or over this particular interval. To do that, all we need is an equation for the function, and the endpoints of the interval we’re interested in. We can plug those things into the average rate of change formula, and we’ll have the average rate of change over the interval.
So in the same way that the derivative at a point, which we can also call instantaneous rate of change, is equal to the slope of the tangent line at that point, the average rate of change over an interval is equal to the slope of the secant line that connects the endpoints of the interval.
0:27 // Formula for the average rate of change
0:49 // Average rate of change is equal to the slope of the line
1:17 // When average rate of change is negative, positive, and zero
2:30 // Average rate of change vs. instantaneous rate of change
3:12 // Average rate of change from a table
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● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: http://www.kristakingmath.com
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Tagged under: calculus, average rate change formula,average rate change table,negative, average rate change,positive,Krista King, average rate change represent,average rate change,instantaneous rate change,educational,secant line
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# A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm.
Question:
A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres.
Solution:
We have,
height of cone, $h=60 \mathrm{~cm}$,
the base radius of cone, $r=30 \mathrm{~cm}$,
the height of cylinder, $H=180 \mathrm{~cm}$ and
the base radius of the cylinder, $R=60 \mathrm{~cm}$
Now,
Volume of water left in the cylinder = Volume of cylinder-Volume of cone
$=\pi R^{2} H-\frac{1}{3} \pi r^{2} h$
$=\frac{22}{7} \times 60 \times 60 \times 180-\frac{1}{3} \times \frac{22}{7} \times 30 \times 30 \times 60$
$=\frac{22}{7} \times 30 \times 30 \times 60\left(2 \times 2 \times 3-\frac{1}{3}\right)$
$=\frac{22}{7} \times 54000\left(12-\frac{1}{3}\right)$
$=\frac{22}{7} \times 54000 \times \frac{35}{3}$
$=1980000 \mathrm{~cm}^{3}$
$=\frac{1980000}{1000000} \mathrm{~m}^{3}$
$=1.98 \mathrm{~m}^{3}$
So, the volume of water left in the cylinder is 1.98 m3.
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## Thursday, April 23, 2015
### Find The Area Of The Equilateral Triangle
Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.
My solution:
The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:
$2x^3+3x^2-1=(x+1)^2(2x-1)=0$
Thus, we know the points:
$(x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)$
are on the given curve. Next, if we begin with the line:
$y=1-x$
and cube both sides, we obtain:
$y^3=1-3x+3x^2-x^3$
We may arrange this as:
$x^3+3x(1-x)+y^3=1$
Since $y=1-x$, we may now write
$x^3+3xy+y^3=1$
And since the point $\left(\frac{1}{2},\frac{1}{2} \right)$ is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.
Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:
$h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}$
Using the Pythagorean theorem, we find that the side lengths of the triangle must be:
$s=\frac{2}{\sqrt{3}}h=\sqrt{6}$
And so the area of the triangle is:
$A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}$
#### 1 comment:
1. Great job on how you did this problem Mark! Thanks for sharing it on your Google+. -M
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## Related Articles
• CBSE Class 9 Syllabus (All Subjects)
• CBSE Class 9 Maths Notes
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# Surface Area of a Sphere
• Last Updated : 25 Aug, 2022
The surface area of a sphere is the area occupied by the curved surface of that sphere. In geometry, a sphere is a three-dimensional solid figure that is round in shape, with every point on its surface equidistant from its center. For example, the globe, ball, and so on. The distance between any point on the surface of a sphere and its center is called the “radius of a sphere.” A sphere is a three-dimensional figure that is defined in three dimensions, i.e., the x-axis, the y-axis, and the z-axis, whereas a circle is a two-dimensional shape that is defined in a plane.
## What is the Surface Area of a Sphere?
The surface area of a sphere is the region covered by the outer surface in the 3-dimensional space. It can be said that a sphere is the 3-dimensional form of a circle. The surface area of a sphere formula is given in terms of pi (π) and radius.
Surface Area of a Sphere = 4πr2 square units
Where, r = radius of the sphere.
## Surface Area of a Sphere Formula
The surface area of a sphere is calculated using the radius of the sphere. If the surface area of the sphere is “S” and the radius is “r”, then the formula for surface area of a sphere is:
Surface area of a sphere = 4πr2
If the diameter of the sphere is given instead of the radius, the formula will become:
Surface area of a sphere = πd2
## Derivation of Surface Area of a Sphere
The area occupied by the surface of a sphere in space is the surface area of a sphere. We know that a sphere is round in shape, so to calculate its surface area, we can connect it to a curved shape, such as a cylinder. A cylinder is a three-dimensional figure that has a curved surface with two flat surfaces on either side. Let’s consider that the radius of the sphere and the radius of a cylinder is the same. So the sphere can perfectly fit into a cylinder. Therefore, the height of the sphere is equal to the height of a cylinder, i.e., the diameter of a sphere. This fact was proved by the mathematician Archimedes that the surface area of a sphere of radius “r” is equal to the lateral surface area of a cylinder of radius “r”. Therefore,
The Surface area of a sphere = The Lateral surface area of a cylinder
We know that,
The lateral surface area of a cylinder = 2πrh,
Where r is the radius of the cylinder and h is its height.
We have assumed that the sphere perfectly fits into the cylinder. So, the height of the cylinder is equal to the diameter of the sphere.
Height of the cylinder (h) = Diameter of the sphere (d) = 2r (where r is the radius)
Therefore,
The Surface area of a sphere = The Lateral surface area of a cylinder = 2πrh
Surface area of the sphere = 2πr × (2r) = 4πr2
Hence,
The surface area of the sphere = 4πr2 square units
Where r = radius of the sphere
## How to Find the Surface Area of a Sphere
The surface area of a sphere is simply the area occupied by its surface. Let’s consider an example to see how to determine the surface area of a sphere using its formula.
Example: Find the surface area of a sphere of radius 7 cm.
Step 1: Note the radius of the given sphere. Here, the radius of the sphere is 47 cm.
Step 2: We know that the surface area of a sphere = 4πr2. So, substitute the value of the given radius in the equation = 4 × (3.14) × (7)2 = 616 cm2.
Step 3: Hence, the surface area of the sphere is 616 square cm.
## Curved Surface Area of a Sphere
The sphere has only one curved surface. Therefore, the curved surface area of the sphere is equal to the total surface area of the sphere, which is equal to the surface area of the sphere in general. Therefore, it is safe to say that,
Curved Surface Area of a Sphere = 4πr2
## Solved Examples on Surface Area of Sphere
Example 1: Calculate the total surface area of a sphere with a radius of 15 cm. (Take π = 3.14)
Solution:
Given, the radius of the sphere = 15 cm
We know that the total surface area of a sphere = 4 π r2 square units
= 4 × (3.14) × (15)2
= 2826 cm2
Hence, the total surface area of the sphere is 2826 cm2.
Example 2: Calculate the diameter of a sphere whose surface area is 616 square inches. (Take π = 22/7)
Solution:
Given, the curved surface area of the sphere = 616 sq. in
We know,
The total surface area of a sphere = 4 π r2 square units
⇒ 4 π r2 = 616
⇒ 4 × (22/7) × r2 = 616
⇒ r2 = (616 × 7)/(4 × 22) = 49
⇒ r = √49 = 7 in
We know, diameter = 2 × radius = 2 × 7 = 14 inches
Hence, the diameter of the sphere is 14 inches.
Example 3: Find the cost required to paint a ball that is in the shape of a sphere with a radius of 10 cm. The painting cost of the ball is ₨ 4 per square cm. (Take π = 3.14)
Solution:
Given, the radius of the ball = 10 cm
We know that,
The surface area of a sphere = 4 π r2 square units
= 4 × (3.14) × (10)2
= 1256 square cm
Hence, the total cost to paint the ball = 4 × 1256 = ₨ 5024/-
Example 4: Find the surface area of a sphere whose diameter is 21 cm. (Take π = 22/7)
Solution:
Given, the diameter of a sphere is 21 cm
We know,
⇒ 21 = 2 × r ⇒ r = 10.5 cm
Now, the surface area of a sphere = 4 π r2 square units
= 4 × (22/7) × (10.5)
= 1386 sq. cm
Hence, the total surface area of the sphere = 1386 sq. cm
Example 5: Find the ratio between the surface areas of two spheres whose radii are in the ratio of 4:3. (Take π = 22/7)
Solution:
Given, the ratio between the radii of two spheres = 4:3
We know that,
The surface area of a sphere = 4 π r2
From the equation, we can say that the surface area of a sphere is directly proportional to the square of its radius.
⇒ A1/A2 = (r1)2/(r2)2
⇒ A1/A2 = (4)2/(3)2 = 16/9
Therefore, the ratio between the total surface areas of the given two spheres is 16:9.
Example 6: Find the ratio between the radii of two spheres when their surface areas are in the ratio of 25:121. (Take π = 22/7)
Solution:
Given, the ratio between the total surface areas of two spheres = 25:121
We know that,
The total surface area of a sphere = 4 π r2
From the equation, we can say that the surface area of a sphere is directly proportional to the square of its radius.
⇒ r1/r2 = √A1/√A2
⇒ r1/r2 = √25/√121 = 5/11
Therefore, the ratio between the radii of the given two spheres is 5:11.
## FAQs on Surface Area of Sphere
Question 1: How to find the surface area of a sphere?
The surface area of a sphere is given by the formula,
Surface area = 4πr2
Where r is the radius of the sphere.
Question 2: What is the surface area of a hemisphere?
The surface area of a hemisphere is given by the sum of half of the sphere’s surface area and the base area, that is,
S.A. = 2πr2 + πr2
The surface area of a hemisphere = 3πr2
Question 3: What is the lateral surface area of a sphere?
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## Thursday, 17 November 2011
### How to tackle 8th grade Polynomial Functions
In my opinion, Polynomial functions are one of the interesting areas of studies but sometimes it comes with so much complexities that it becomes quite a bit difficult to solve them. So the most necessary thing to understand is that we need to practice it a lot and it requires full concentration while solving it. Polynomial is basically a term which deals in almost every type of mathematical equations or statements. The most common terminologies used in polynomial expressions comes in eighth standard mathematics are monomial, binomial and trinomials. Algebraic equation with all variables having whole number, exponents or powers are called polynomials. The expressions in which the power of variables are negative and which include rational numbers are not polynomials. Algebraic expression having single term is known as Monomial and expression with two terms are known as Binomial whereas expressions with more than two terms or having three terms are known as Trinomials.
Now lets talk about Polynomial Functions. A polynomial functions p is basically a function or an expression that can be formed by combining the variable and some constants by a finite number of additions, subtractions, and multiplications.
A polynomial equation comes with the sum of the power of same derivatives and includes different integer constants, while the derivatives used are finite in numbers. The standard form of any polynomial equation is as:
bn yn + bn-1 yn-1 + ….......................+. b2 y2 + ….... + b0 y0
Lets take some examples of Polynomial Equations to understand it better.
Function p (x) = x2 – 3 is a polynomial of degree 2. in standard form it can be represented as : a2 = 1, a1= 0 and a0 = -3.
An example to show a non polynomial function is P (d) = 1/d. This function becomes arbitrarily large for values of d close to zero and no polynomial does that.
## Friday, 11 November 2011
### Some examples of polynomial equations
A polynomial function f, is a function of the form f(x) = anxn + an-1xn-1 + ... + a2x2 + a1x1 + a0 where a0, a1...an are real numbers. It will soon become more understandable with some more examples. If "n" is not zero, then f is said to have degree n. A polynomial f(x) with real coefficients and of degree n has n zeros (not necessarily all different). Some or all are real zeros and appear as x-intercepts when f(x) is graphed. To make it more simple lets start by explaining the word polynomial, it is the word which comes made of two terms "poly" states many and nominal states "terms". Nomenclature of different polynomial functions, depending upon the terms present in the equation is done such as, if it has only one term it is called as monomial, if two terms it is called as binomial and if three terms it is called as trinomial, and so on with increasing variable terms.
Lets make it more elaborate with the help of examples, X + X>2 = 4 is an example of monomial, x + y = 5 is an example of binomial, X + Y + Z = 7 is an example of trinomial. When there are equations involving polynomial it is known as a Polynomial equation. For solution of a polynomial equation different values for variables in the equation satisfying the equation along with the given constant coefficient values used in the polynomial.
Lets see it practically with the help of some example, 5 x + 6 y = 0 is a polynomial equation, for a point A in a plane having coordinates (0, 0). Co-ordinates states value of x = 0, and y = 0 for this equation. Substituting values of x and y in the Polynomial equations we have 5 (0) + 6 (0) = 0 + 0 = 0. As the values on both sides of the equation are equal this is a solution of the equation. Hence point A is the solution for this Polynomial equations.
## Thursday, 10 November 2011
### TutorVista help on Polynomials
Polynomial is one of the most important term used in mathematical world which plays an important role in almost every type of mathematical equations or statements. Terminologies or concepts used in Polynomial equations are are monomial, binomial and trinomials. Algebraic equation with all variables having whole number, exponents or powers are called polynomials. Monomials are the Algebraic expression consist of single term and those algebraic expression comes equipped with two terms are known as Binomial Whereas the expressions with more than two terms or having three terms are known as Trinomials.
Lets talk about Polynomial functions used in mathematics. Polynomial function includes various things like terms, factors, variables, and constants. Let us talk about all the above terminologies in detail which are required to form a polynomial function. Terms can be explained as when numbers are implemented with addition or subtraction are known as terms. Terms can be further divided in to two sections that are Like terms or Unlike terms. Terms that has the same power of the same variables are called Like terms. The terms used in an expression that do not contain the same power of the same variables are called unlike terms. In an expression if the product of the numbers are used then the expression is called as factors. Variables are just representing a symbol which uses different values under it whereas constant is a single value symbol.
A polynomial equations comes with the sum of the power of same derivatives and includes different integer constants, while the derivatives used are finite in numbers.
Lets take some examples of Polynomial equations to understand it better.
7xyz : Monomial
x + 7y : Binomial
x + 3y – c : Trinomial
## Monday, 7 November 2011
### Learn Polynomial functions by taking help of Tutorvista
In mathematics, problems are represented in expression form on which standard principles of math are applicable. The major part of mathematical problems is represented in form of any polynomial function. A polynomial is a mathematical expression which may consist of various derivatives of various orders which are related with each other by arithmetical operators to form an math expression.
The standard form of any polynomial is as:
cnyn + cn-1 yn-1 + .......................+c2 y2 + ….... + c0y0
The standard form consists of derivatives of only one variable, but as the name suggests “poly” means many , and “nomial” means terms, it means polynomial may consist of derivatives of different variables in it. According to this property polynomial equations are further categorizes as: monomial, binomial, trinomial and so on. Every polynomial function consist of different terms, factors, variables and constant integers coefficients. If two numbers or variables are combined through addition or subtraction operator then they form a term and if the degree of all those terms is same then they are said to be like terms otherwise unlike terms. The numbers are multiplied or product of two numbers is used in a polynomial than that part of it is known as factor. Variable is a symbol which represents the value which may change and constant is a fixed numerical value.
Let's see few examples of polynomial equations to explore more about it:
y – y2 = 2 (a Monomial equation)
x + y = 2 (Binomial equation)
x + y - z = 1 (Trinomial equation)
All the algebraic equations like linear equations, quadratic equations, etc and they are different kind of polynomial equations also and for solving these polynomial equations the variables of the functions should be replaced by appropriate numerical value which will satisfy the equation in the end. To learn more about this topic switch to tutorvista.
## Friday, 4 November 2011
### Solving polynomial equations of mathematics
Polynomial is basically a term which deals in almost every type of mathematical equations or statements. The most common terminologies used in polynomial expressions are monomial, binomial and trinomials. Algebraic equation with all variables having whole number, exponents or powers is called polynomial. The expressions in which the power of variables are negative and rational numbers are not polynomials. Algebraic expression having single term known as monomial and expression with two terms are known as binomial whereas expressions with more than two terms or having three terms are known as trinomials.
Polynomial functions come equipped with terms, factors, variables, and constants. Let us explore about all these required objects to form a polynomial function. When numbers are implemented with addition or subtraction then they are said to be terms.Terms are of two types Like terms or Unlike terms. Terms that have the same power of the same variable are called like terms. The terms used in an expression that do not contain the same power of the same variables are called unlike terms. In an expression if the product of the numbers are used then the expression is called as factors. Variables are just representing a symbol which uses different values under it whereas constant is a single value symbol.
Polynomial equations, come with the sum of the power of same derivatives and includes different integer constants, while the derivatives used are finite in numbers. The standard form of any polynomial equation is as:
bnyn + bn-1 yn-1 + ….......................+. b2 y2 + ….... + b0y0
In the above equation, y is the variable and b is the integer coefficient used.
Lets see some of the examples of polynomial equations to understand it better.
10xyz : Monomial
3x + 7y : Binomial
3x + 7y – c : Trinomial
## Wednesday, 2 November 2011
### How to Solve Polynomials
Polynomial is a term which implies with every kind of mathematical expression. Polynomial function consists of terms, factors, variables, and constants. Let us explore about all these required objects to form a polynomial function, When numbers are implemented with addition or subtraction than they are said to be terms, when product of the numbers are used than that form is called factors. Variables are just representing a symbol which uses different values under it whereas constant is a single value symbol.
A polynomial equation includes sum of the power of same derivatives with different integer coefficients and these all derivatives are finite in numbers. The standard form of any polynomial equation is as:
bn yn + bn-1 yn-1 + ….......................+. b2 y2 + ….... + b0 y0
Here y is the variable with n types of derivatives, 'b' is an integer co-efficient and 'n' represents the finite number of derivatives in polynomial equation.
Sometimes any polynomial equation may consist of number of different derivatives or variables. So according to this property polynomial is categorized in various types which are binomial, monomial, trinomial and so on.
A polynomial is said to be a monomial when it only have one single variable derivatives and if equations consist of derivatives of two variables than that is a form of binomial equation. Similarly a trinomial will include derivatives of 3 different variables. Let us take some examples of polynomial equations:
x – x2 = 2 (a Monomial)
x + y = 1 (Binomial)
x + y + z = 3 (Trinomial)
In mathematics most of the equations are in form of polynomial equation like every algebraic equation is a type of polynomial equation. So it is clear that polynomial functions are important in mathematics equation formation so for enhancing your knowledge in this topic and various other math topics you can take online math help on a math tutoring website “ TutorVista”.
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# RD Sharma Solutions for Class 10 Mathematics Chapter 2: Polynomials
This chapter explains in detail polynomials and their various types, namely, linear polynomial, quadratic polynomial and cubic polynomial, followed by the concept of the zeroes of a polynomial. Finally, a brief mention of the degrees of a polynomial helps the students understand the basic essence of this chapter.
A linear polynomial is a polynomial of degree 1. For example, 2x-3 is a linear polynomial. Similarly, a polynomial of degree 2 is called a quadratic polynomial, for example, 2x2-3x+2; and a polynomial with degree 3 is called a cubic polynomial. For example, x3-2x2+3x-4 is a cubic polynomial.
Chapter 2: Polynomials also talks about the geometrical meaning of the zeroes of a polynomial. This is done by explaining the concept of the zeroes of a polynomial with the help of graphs, as pictorial representations are often easier to grasp. The zeros of the polynomial mean the values of x for which the value of the polynomial converts to zero.
It then sheds light on the relationship between the roots of a polynomial and its coefficients. This is a direct method to find out the zeroes of a given polynomial based on its coefficients.
If α and β are the roots of a polynomial, ax2+bx+c, where a, b, c are real numbers with a≠0,
α + β = -b/a; αβ = c/a.
An important topic discussed in this chapter is the 'Division Algorithm for Polynomials. This Algorithm is similar to Euclid's Division Algorithm, which we encountered in the first chapter. It states that "if p(x) and g(x) are two given polynomials such that g(x)≠0, then we can find polynomials, q(x) and r(x) by the following-
P(x) = g(x) * q(x)+r(x)”.
Where, r(x)=0 or the degree of r(x)
The above-given method is also known as the remainder theorem
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Cartesian Coordinate System and Quadrants
Introduction
Cartesian Coordinate System is used to describe the position of a point in a plane using coordinates in coordinate geometry. Coordinates are the way of telling the position of a point using numbers.
Rene Discartes invented the Coordinate System and the term Cartesian Coordinate System is named after the name of Discartes mathematician to honour the inventor.
Cartesian geometry system introduces coordinate axes, coordinates and quadrants to describe the position of a point in a plane and its distances from other points.
Let’s learn about these new terms next.
Coordinate axes
Cartesian coordinate system uses a plane which has two mutually perpendicular lines which are called as coordinate axes. Out of the two perpendicular lines, one line is in horizontal position and another line is in vertical position. The figure below shows how the two coordinate axes look like.
Note
Mind the words axis and axes, how they are used. Axis is used to represent one axis, it is singular. On the other hand, axes is used to represent more than one axis, which is plural.
Horizontal axis in coordinate geometry
Vertical axis in coordinate geometry
The horizontal line marked with XX is called as x-axis and vertical line marked as YY is called as y-axis. Both axes, x-axis and y-axis intersect each other at point O and are perpendicular to each other. The point O is called as origin.
Directions of axes
So, x-axis has two sides, one side lies on right side of y-axis and can be read as OX. The second side lies on left side of y-axis and is read as OX. Similarly, y-axis has two sides top and bottom of x-axis. The top side of y-axis is read as OY and bottom side of y-axis is OY’.
The OX on x-axis and OY on y-axis called as positive directions of x-axis and y-axis respectively. Similarly, the OX on x-axis and OY on y-axis called as negative directions of x-axis and y-axis respectively.
Directions of axis in coordinate geometry
Scaling of axes
Scaling of axes is meant by marking the x-axis and y-axis with numbers which are placed at equal distances. The ray OX on x-axis has positive numbers (1, 2, 3, 4, ….) and ray OX has negative numbers (-1, -2, -3, -4, ….) only. Similarly,
The ray OY on y-axis has positive numbers (1, 2, 3, 4, ….) and ray OY has negative numbers (-1, -2, -3, -4, ….) on it. In other words, we can say that positive numbers lie in the directions of ray OX and ray OY and the negative numbers lie in the direction of ray OX and ray OY.
The two coordinate axes x-axis and y-axis in coordinate geometry divide the plane into four parts. These four parts are called as quadrants.
Quadrants in coordinate geometry
In above figure, x-axis and y-axis is dividing the plane into four parts which are XOY, XOY, XOY and XOY.
XOY is called as first quadrant.
XOY is called as second quadrant.
XOY is called as third quadrant.
XOY is called as fourth quadrant.
Also, the four quadrants are numbered as I, II, III and IV anticlockwise starting from first quadrant XOY and the last quadrant as XOY.
Coordinates of a point
Any point in a plane can lie in any of the four quadrants I, II, III or IV. The position of a point is in the plane is called as coordinates of that point. Coordinates of the point in a quadrant is determined by knowing its perpendicular distances from its nearest x-axis and y-axis.
Coordinates of a point is written by writing its perpendicular distance in brackets in the format of (x,y), where x is the perpendicular distance of the point from x-axis and y is the perpendicular distance of the point from y-axis.
The x-axis coordinate of a point is called as abscissa.
The y-axis coordinate of a point is called as ordinate.
Coordinates of a point can be written only in specific order i.e (x,y). First we write abscissa followed by ordinate and seperated by a comma.
Note
Abscissa and ordinate of a point cannot be interchanged.
i.e. (x,y) ≠ (y,x)
Moreover, (x,y) = (y,x) if x = y
Let’s take an example to understand it precisely
Coordinates of a point in coordinate geometry
Example
From the above figure, we can see
P is the point that lie in first quadrant and is marked has P.
The perpendicular distance PN of point P from y-axis is 3 units, measured along the positive direction of x-axis as OM which is 3 units.
The perpendicular distance PM of point P from x-axis is 4 units, measured along the positive direction of y-axis as ON which is 4 units.
So, the perpendicular distance of point P from y-axis is 3 and x-axis is 4, so the coordinate of the point P can be written as (3,4).
Note
Coordinates of origin are always written as (0,0) because perpendicular distance of origin from x-axis is zero and from y-axis is also zero.
Note
The ordinate of any point on x-axis is 0 i.e coordinate of any point on x-axis is (x,0).
The abscissa of any point on y-axis is 0 i.e coordinate of any point on y-axis is (0,y).
Sign conventions in quadrants
As we have read above, the x-axis XOX’ and y-axis YOY’ divide the coordinate plane into four quadrants.
The ray OX is regarded as positive x-axis.
The ray OY is regarded as positive y-axis.
The ray OX is regarded as negative x-axis.
The ray OY is regarded as negative y-axis.
So, any point that lies in first quadrant will always have positive abscissa and positive ordinate.
Any point that lie in second quadrant has negative abscissa and positive ordinate.
Any point that lie in third quadrant has negative abscissa and negative ordinate.
Any point that lie in fourth quadrant has positive abscissa and negative ordinate.
Please refer to the following figure, how the signs of abscissa and ordinate looks like in the four quadrants.
Sign conventions in quadrants in coordinate geometry
So, In the figure:
first quadrant has points with sign (+,+).
second quadrant has points with sign (-,+).
third quadrant has points with sign (-,-).
fourth quadrant has points with sign (+,-).
1) What is coordinate geometry?
The branch of mathematics in which coordinate system is used to solve geometric problems is known as coordinate geometry.
2) What else is a horizontal line passing through origin called as?
Horizontal line is called as x-axis.
3) What else is a vertical line passing through origin called as?
Vertical line is called as y-axis.
4) What is called the x coordinate of a point as?
The x coordinate of a point is called as abscissa.
5) What is called the y coordinate of a point as?
The y coordinate of a point is called as ordinate.
Solved Examples
1) Plot the points P(3,4), Q(-5,3), R(6,0) and S(-5,0) on the graph paper.
Solution
Point P(3,4) lies in the first quadrant. So we move 3 units along OX and 4 units in upward direction i.e along OY.
Point Q(-5,3) lies in second quadrant. So we move 5 units along OX' and 3 units in upward direction i.e along OY.
Point R(-5,0) lies on negative direction of x-axis at the distance of 5 units from origin.
Point S(6,0) lies on positive direction of x-axis at the distance of 6 units from origin.
2) Plot the points A(3,-4), B(-3,-5), C(0,5) and D(-0,-2) on the graph paper.
Solution
Point A(3,-4) lies in fourth quadrant. So we move 3 units along OX and 4 units in downward direction i.e along OY'.
Point B(-3,-5) lies in third quadrant. So we move 3 units along OX' and 5 units in downward direction i.e along OY'.
Point C(0,5) lies on positive direction of y-axis at the distance of 5 units from origin.
Point D(0,-2) lies on negative direction of y-axis at the distance of 2 units from origin.
3) Locate the points A(3,3), B(-3,3), C(-3,-3) and D(3,-3) on graph paper. Name the figure by joining these points. Also find its perimeter.
Solution
The figure joined by these points is a square.
AB = BC = CD = DA = 6 units
side of square = 6 units
Perimeter of square= 4 × 6 = 24 sq. units
4) Plot the points A(3,2), B(-4,2), C(-4,-2) and D(3,-2) on graph paper. Name the figure by joining these points. Also find its area.
Solution
AB = CD = 7 units
BC = AD = 4 units
So, ABCD is a rectangle
Area of rectangle = length × breadth
= 7 × 4 = 28 sq. units
5) Plot the points M(3,0), N(3,4), and O(0,0) on graph paper. Find area of the figure by joining these points.
Solution
The figure is right angled triangle when we join these points.
OM = 3 units
MN = 4 units
∠OMN = 900
Area of δOMN = $\frac{1}{2}$ × base × height
= $\frac{1}{2}$ × OM × MN
= $\frac{1}{2}$ × 3 × 4
= 6 sq. units
Multiple Choice Questions
1. abscissa
2. ordinate
3. origin
4. coordinates
2) Which mathematician did develop the cartesian system of coordinates?
1. Thales
2. Euclid
3. Pythagoras
4. Rene descartes
3) A Horizontal line and a vertical line in a coordinate plane are called as
1. x-axis and y-axis
2. y-axis and x-axis
3. abscissa and ordinate
4. ordinate and abscissa
4) Points (2,5) and (5,2) represent a same point in a cartesian plane.
1. True
2. False
3. Maybe
4. None of these
1. (-4,0)
2. (0,-4)
3. (2,-1)
4. (-1,-2)
1. (+,+)
2. (+,-)
3. (-,+)
4. (-,-)
8) Points (x,y) and (y,x) represent a same point in a plane if
1. x < y
2. x > y
3. x = y
4. not possible
9) If in cartesian plane, the points (x+3,6) and (-1,y+2) represent a same point, then value of x and y will be
1. x = -4 and y = 4
2. x = -4 and y = -4
3. x = 3 and y = 6
4. x = -1 and y = 6
1. (4,0)
2. (0,4)
3. (-3,1)
4. (-3,-2)
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ASVAB Mathematics: Divisibility Rules for the Numbers 1, 2 and 3
Examining techniques that can be made use of to determine whether a number is evenly divisible by other numbers, is a crucial topic in primary number theory.
These are faster ways for testing a number’s factors without resorting to division computations.
The regulations transform a provided number’s divisibility by a divisor to a smaller sized number’s divisibilty by the exact same divisor.
If the result is not evident after applying it as soon as, the guideline should be used again to the smaller number.
In kids’ math text books, we will normally locate the divisibility guidelines for 2, 3, 4, 5, 6, 8, 9, 11.
Also locating the divisibility guideline for 7, in those publications is a rarity.
In this short article, we offer the divisibility policies for prime numbers in general and also apply it to particular situations, for prime numbers, below 50.
We offer the regulations with instances, in a basic method, to follow, understand and apply.
Divisibility Policy for any prime divisor ‘p’:.
Think about multiples of ‘p’ till (least numerous of ‘p’ + 1) is a multiple of 10, to ensure that one tenth of (least several of ‘p’ + 1) is an all-natural number.
Allow us say this natural number is ‘n’.
Thus, n = one tenth of (the very least multiple of ‘p’ + 1).
Locate (p – n) likewise.
Example (i):.
Let the prime divisor be 7.
Multiples of 7 are 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,.
7×7 (Got it. 7×7 = 49 and also 49 +1= 50 is a several of 10).
So ‘n’ for 7 is one tenth of (the very least multiple of ‘p’ + 1) = (1/10) 50 = 5.
‘ p-n’ = 7 – 5 = 2.
Instance (ii):.
Let the prime divisor be 13.
Multiples of 13 are 1×13, 2×13,.
3×13 (Got it. 3×13 = 39 as well as 39 +1= 40 is a several of 10).
So ‘n’ for 13 is one tenth of (the very least several of ‘p’ + 1) = (1/10) 40 = 4.
‘ p-n’ = 13 – 4 = 9.
The values of ‘n’ as well as ‘p-n’ for other prime numbers listed below 50 are given below.
p n p-n.
7 5 2.
13 4 9.
17 12 5.
19 2 17.
23 7 16.
29 3 26.
31 28 3.
37 26 11.
41 37 4.
43 13 30.
47 33 14.
After discovering ‘n’ and also ‘p-n’, the divisibility regulation is as adheres to:.
To discover, if a number is divisible by ‘p’, take the last number of the number, multiply it by ‘n’, and include it to the remainder of the number.
or increase it by ‘( p – n)’ as well as deduct it from the remainder of the number.
If you get an answer divisible by ‘p’ (consisting of no), after that the initial number is divisible by ‘p’.
If you do not know the new number’s divisibility, you can use the guideline once again.
So to create the regulation, we need to select either ‘n’ or ‘p-n’.
Usually, we choose the lower of both.
With this knlowledge, allow us state the divisibilty policy for 7.
For 7, p-n (= 2) is less than n (= 5).
Divisibility Policy for 7:.
To find out, if a number is divisible by 7, take the last digit, Increase it by 2, and deduct it from the rest of the number.
If you get a solution divisible by 7 (consisting of zero), then the initial number is divisible by 7.
If you don’t understand the brand-new number’s divisibility, you can use the regulation once again.
Instance 1:.
Locate whether 49875 is divisible by 7 or not.
Solution:.
To inspect whether 49875 is divisible by 7:.
Twice the last number = 2 x 5 = 10; Rest of the number = 4987.
Deducting, 4987 – 10 = 4977.
To examine whether 4977 is divisible by 7:.
Two times the last number = 2 x 7 = 14; Rest of the number = 497.
Subtracting, 497 – 14 = 483.
To check whether 483 is divisible by 7:.
Twice the last digit = 2 x 3 = 6; Rest of the number = 48.
Deducting, 48 – 6 = 42 is divisible by 7. (42 = 6 x 7 ).
So, 49875 is divisible by 7. Ans.
Now, let us mention the divisibilty regulation for 13.
For 13, n (= 4) is lower than p-n (= 9).
Divisibility Regulation for 13:.
To find out, if a number is divisible by 13, take the last figure, Increase it with 4, and also add it to the remainder of the number.
If you get a solution divisible by 13 (including zero), then the initial number is divisible by 13.
If you don’t recognize the new number’s divisibility, you can apply the guideline again.
Example 2:.
Find whether 46371 is divisible by 13 or otherwise.
Service:.
To inspect whether 46371 is divisible by 13:.
4 x last figure = 4 x 1 = 4; Rest of the number = 4637.
Including, 4637 + 4 = 4641.
To examine whether 4641 is divisible by 13:.
4 x last figure = 4 x 1 = 4; Remainder of the number = 464.
Including, 464 + 4 = 468.
To inspect whether 468 is divisible by 13:.
4 x last digit = 4 x 8 = 32; Remainder of the number = 46.
Adding, 46 + 32 = 78 is divisible by 13. (78 = 6 x 13 ).
( if you desire, you can apply the guideline once again, below. 4×8 + 7 = 39 = 3 x 13).
So, 46371 is divisible by 13. Ans.
Now Number Place Value let us mention the divisibility guidelines for 19 as well as 31.
for 19, n = 2 is more convenient than (p – n) = 17.
So, the divisibility policy for 19 is as follows.
To figure out, whether a number is divisible by 19, take the last number, multiply it by 2, as well as add it to the remainder of the number.
If you obtain an answer divisible by 19 (including absolutely no), after that the initial number is divisible by 19.
If you do not know the new number’s divisibility, you can apply the rule again.
For 31, (p – n) = 3 is easier than n = 28.
So, the divisibility guideline for 31 is as complies with.
To figure out, whether a number is divisible by 31, take the last digit, increase it by 3, as well as deduct it from the rest of the number.
If you get a response divisible by 31 (including no), then the initial number is divisible by 31.
If you do not understand the new number’s divisibility, you can apply the regulation again.
Like this, we can define the divisibility policy for any kind of prime divisor.
The technique of locating ‘n’ given over can be encompassed prime numbers above 50 likewise.
Prior to, we close the write-up, allow us see the proof of Divisibility Policy for 7.
Proof of Divisibility Guideline for 7:.
Let ‘D’ (> 10) be the returns.
Allow D1 be the devices’ digit and also D2 be the rest of the variety of D.
i.e. D = D1 + 10D2.
We need to show.
( i) if D2 – 2D1 is divisible by 7, after that D is also divisible by 7.
and also (ii) if D is divisible by 7, then D2 – 2D1 is additionally divisible by 7.
Evidence of (i):.
D2 – 2D1 is divisible by 7.
So, D2 – 2D1 = 7k where k is any kind of natural number.
Increasing both sides by 10, we get.
10D2 – 20D1 = 70k.
Including D1 to both sides, we get.
( 10D2 + D1) – 20D1 = 70k + D1.
or (10D2 + D1) = 70k + D1 + 20D1.
or D = 70k + 21D1 = 7( 10k + 3D1) = a multiple of 7.
So, D is divisible by 7. (verified.).
Evidence of (ii):.
D is divisible by 7.
So, D1 + 10D2 is divisible by 7.
D1 + 10D2 = 7k where k is any type of all-natural number.
Subtracting 21D1 from both sides, we obtain.
10D2 – 20D1 = 7k – 21D1.
or 10( D2 – 2D1) = 7( k – 3D1).
or 10( D2 – 2D1) is divisible by 7.
Considering that 10 is not divisible by 7, (D2 – 2D1) is divisible by 7. (proved.).
In a similar style, we can show the divisibility policy for any type of prime divisor.
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Question Video: Finding a Side Length in a Triangle given the Corresponding Side in a Similar Triangle and the Ratio of Similarity between Them | Nagwa Question Video: Finding a Side Length in a Triangle given the Corresponding Side in a Similar Triangle and the Ratio of Similarity between Them | Nagwa
# Question Video: Finding a Side Length in a Triangle given the Corresponding Side in a Similar Triangle and the Ratio of Similarity between Them Mathematics • First Year of Secondary School
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If πΆπΈ = (π₯ + 2) cm, what is π₯?
02:08
### Video Transcript
If πΆπΈ equals π₯ plus two centimeters, what is π₯?
First of all, on our figure, we can label πΆπΈ as π₯ plus two centimeters. And then we should think about what else we know based on the figure. First of all, we see that line segment πΈπ· is parallel to line segment πΆπ΅. And then we can say that line segment π΄π΅ and line segment π΄πΆ are transversals of these two parallel lines. Based on these two facts, we can draw some conclusions. We can say that the parallel lines πΈπ· and πΆπ΅ cut this triangle proportionally. So we can say line segment π΄πΈ over line segment π΄π· will be equal to line segment πΆπΈ over line segment π·π΅ by parallel lines and transversal properties.
To solve then, we can just plug in the values that we know for these line segments. Six over π₯ plus two is equal to four over eight. The first way we could solve this is by using cross multiplication. We can say six times eight is equal to four times π₯ plus two. Therefore, 48 equals four times π₯ plus two. And if we divide both sides of the equation by four, we see that 12 is equal to π₯ plus two. So we subtract two from both sides, and we see that π₯ equals 10. Now, I said this is one way to solve. And thatβs because if we think about proportionality, and we know that the parallel lines cut these line segments proportionally, we notice that line segment π·π΅ is two times line segment π΄π·.
And in order for things to be proportional, that would mean that the same thing would have to be true on the other side. This means that π₯ plus two must be equal to six times two, which again shows us that side length πΆπΈ must be equal to 12 and therefore π₯ plus two must be equal to 12. So again, π₯ equals 10.
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Lesson video
In progress...
Hello, and welcome to this lesson on similar triangles with me, Miss Oreyomi.
For today's lesson you'll be needing your pencil, your protractor, so you'll be needing this, you'll also be needing a ruler, a rubber, and, of course, you will be needing your book.
So, pause the video now if you need to go get this equipment, also, try to minimise distraction by putting your phone on silent, get it into a space where there's less distraction and being ready to learn, because today's lesson is so fun, as you'll be using this tool to draw triangles.
So, pause the video now and go and get your equipment, when you're ready, press play to begin the lesson.
You have four triangles on your screen and your job is to find the value of the missing angles in each triangle, and it's also very important that you explain your reasoning clearly.
So, pause the video now and attempt these four tasks, once you're done, press play to come back to find out the answers.
Okay, I hope you found that task straightforward, we're just going to go through the answers very quickly.
So over here, I know that the sum of angles in a triangle add up to 180 degrees, so my a must be 130 degrees.
Here, this is an isosceles triangle, so therefore this angle here is 34 degrees, and 34 plus 34 is 68, so 180 takeaway 68 is b, which is 112 degrees.
If I move on to this, this we've already been given this angle as 112 degrees, well, 180 takeaway 112 is 64, but because it's an isosceles triangle I'm going to split it between my two base angles, so this will be 34 degrees and my d would be 34 degrees too.
Well here, again, the sum of angles in a triangle add up to 180, I have 150 at the moment, so therefore e must be 30 degrees.
So I hope you got that, if you didn't just check your work and try to see where you went wrong and correct yourself as well.
Whilst you were doing that task.
I wonder if you noticed something about these triangles, I wonder if you thought, hmm, I can see some things are the same, whereas I can see some things are different, because that links in very nicely to what we're doing in today's lesson, did you notice similarities and differences? Well, I noticed that these two angles here have the same, they have the same interior angles, however, this triangle is smaller than this triangle.
And also the same for this, these two have the same interior angles, however, the position has been different, so these two are the same, whereas the two base angles have been put in different positions, and that links into similar triangles.
So our connect task, two students were asked to draw a triangle where two of the angles were 70 degrees and 55 degrees.
So firstly, what's the first thought you're thinking? I have 70 degrees and, I've got one angle is 70 degrees and another angle is 55 degrees, so my third angle must be? Well, it must be 55 degrees as well, because I've got 70 and 55, and I need to add something that would give me 180, and therefore my missing angle is 55 degrees, so what type of triangle are these students trying to draw? Well yes, they're trying to draw an isosceles triangle.
First student started by drawing a base of six centimetre, so she started by drawing a base of six centimetre and then she's gone ahead and measured 70 degrees using her protractor, and then our second student started by drawing a base of three centimetre and then he's measured 55 degrees first.
How can these students complete their drawing? So what can our first student do over here to complete her drawing and what can our second student do over here to complete his drawing? Well, she could put her protractor over here and measure 55 degrees, and he could place the protractor over here and measure 70 degrees.
We are going to have a go at doing this.
Can you, in your book, draw a triangle of 70 degrees and 55 degrees? If you're confident using a protractor, you can pause the video now and try to draw a triangle where the base is 70 degrees, and where two of the angles in your triangle are 70 degrees and 55 degrees.
If you're not so sure how to do this, then carry on watching the video and I'll be providing a tutorial video of how to use a protractor to draw a triangle.
This is hurting.
Okay, before we proceed with you drawing your triangles, I thought a video and me explaining what is happening in the video would give you the support and the foundation you need to be able to use your own protractor.
So before we start I'm going to, is I'm going to talk over in the videos but before we start, let's just do some labelling very quickly.
This part of your protractor is called the centre, so this is called the centre, the zero going from the outside is called the outer scale because it's reading from the outside.
If you see here, we also have a zero and this is the inner scale, so outer scale, inner scale, and then this straight line here is called the base line, so you're going to be needing this, you'll be needing to know this for this task, okay? Let's start with the video, so I'm going to press play.
If at any point you don't quite understand what I'm doing, please pause the video and rewind to watch again.
So first thing I'm doing is I am drawing my base line, or I'm drawing my base length for my triangle and I have chosen five centimetre, because I wasn't given a base line, I chose any value and I chose five centimetre.
Now, I have taken my protractor, I have put, so that's the outer scale, that's the inner scale, and that's the centre, so I'm taking my centre now and putting it on my line.
Now, because the base line of my protractor is on the straight line that I've drawn to this way, I am reading from the outer scale, so I'm always reading from zero.
My zero is on the outer scale on this side, so I am reading from zero, and I'm looking for 70 degrees.
I have found 70 degrees so I am marking that on my book.
I am then going to draw a line of any length, just so I've marked that this is my 70 degree angle.
I like to measure again just to make sure I haven't made a mistake, and yes, it is 70 degrees, so I am going to mark that in my book.
Okay, notice how my centre is now at the other end of my line, my center's now at the other end of my line, and my base line for my protractor is now on this side of the line that I've drawn.
So previously my base line was here, so I was reading the outer value, now my base line is this way, so I am reading, I have to read the inner value because that is where my zero is starting from on this line, okay? So I am now looking for 55 degrees, I have found 55 degrees so I am going to mark it on my book.
I am going to take my ruler and connect it up, notice that I have excess lines at the top but I'm not going to rub those out because, well, it just shows that I have drawn this using a protractor.
So that's a 70 degrees, I am measuring making sure it is roughly around 55, it is 55 degrees, and then the top one, I am just going to measure to make sure that is roughly around 55 degrees, and it is roughly, it's about 57, if your angle is two degrees more or two degrees less, that is fine, so I'm going to write it's approximately 55 degree.
Because this is a new skill to learn, I am going to do another example, so if you need to play this at a slower rate then please do so, okay? Now this time around, I have chosen my base length for my triangle to be four centimetre, so I'm going to label that as four centimetre.
Then again, I'm going to take my protractor, put the centre of my protector at the end of one line and read from the zero value, this time around, I'm starting with 55 degrees.
I'm going to take my ruler and connect it up from where I put the centre of my protractor.
Going to take my protractor, or just measure it again, like I said, I like to measure it, make sure that I've got in roughly the right value.
Now, I'm going to put the centre of my protractor and read the inner value this time for 70 degrees.
Again, I'm going to connect it up, and measure it again, just to make sure that I have the, I have drawn the correct angle, and that should also be roughly around 55 degrees.
You can use this video to help you, using your protractor and your ruler, can you draw a triangle where two of the angles are 70 degrees and 55 degrees? So you can choose your base length for your triangle to be of any length, your choice, so pause the video now, attempt this and then come back when you're ready and we can go through what you've drawn as well.
Okay, I hope you managed to construct your own triangle.
We're now going to compare what is the same and different about the triangles our students on our screen have drawn and the triangle that you managed to draw as well.
So if we link this to our try this task, well, we can see that this is 55 degrees and this is 70 degrees, the same as this triangle here is 55 degrees for this student and it's 70 degrees for this student.
If we complete, if we fill in this missing angle, this is going to also be 55 degrees, just as this is going to be 55 degrees.
So, for both students, we can say the interior angles are the same, exactly, the interior angles for both triangles are the same.
Well, what is different? Well, the lengths are different, this is three centimetre and this is six centimetre, so we could say that these are different lengths.
What else could we say? Both isosceles triangle? Yes, they are, they're different orientation though, so one, the base length of the triangle is at the bottom whereas the base line of the triangle is at the top for this smaller triangle over here.
And we can say one has sides that are twice as long as the other, so this is six centimetre, this is three centimetre, this is 3.
4 centimetre, this is 6.
8 centimetre, and this is three centimetre again, and this is six centimetre.
So, we could say these two triangles are similar, they are similar because they have the same interior angle, and they're also, there's a link between the lengths of their sides.
Can you draw different triangles where at least one of the sides is 10 centimetre long? So, using your ruler, you're going to measure one of the sides of the triangle to be 10 centimetre long, and then you're going to measure two for the first one where two of the angles are 60 degrees, so you want to draw a triangle where at least one of the sides is 10 centimetre long and where two of the angles are 60 degrees.
Secondly, how many different triangles can you draw where one of the side length is 10 centimetre, and for the second one, two of the angles are 20 degrees, okay, so you want two angles to be 20 degrees.
And then for C, you want one angle to be 20 degree, the other angle to be 60 degree, and I've given you a hint here, work out the third angle.
So just to recap your instruction, when I tell you to, you're going to pause your screen, you're going to draw, A, you're going to draw a triangle where at least one of the side length is 10 centimetre and two of your angles in your triangle for A are 60 degrees.
For B, you're going to draw a triangle where at least one of the side length is 10 centimetre and two of your angles are 20 degrees.
For C, you are going to draw a triangle where again, you have one side that is 10 centimetre, one angle is 20 degrees, another angle is 60 degrees, and what is going to be your third angle? Pause your screen now and attempt this, and when you finish, come back and we can discuss some of the possible triangles that you came up with.
How did you get on, did you manage to draw your different triangles? We're going to go over the possibilities of what you could have drawn.
So for the first one, one of the side length is 10 centimetre, well, it is an equilateral triangle, isn't it, because if I've got two angles as 60 degrees, well, the third angle has got to be 60 degrees, so there's only one way I could have drawn this triangle because however way I draw it, all my angles will always be 60 degrees, and because it's an equilateral triangle, my side lengths are the same.
Now for this one, where two angles are 20 degrees and the side length is 10 centimetre, I could have started by drawing my side length to be here, so if I'd drawn my side length here of 10 centimetre, then this is one possible way I could have drawn my triangle.
Another way would be to draw my base length to be 10 centimetre, if I had done that, this would be a way I could have drawn my triangle.
Notice how, when my side length is opposite a smaller angle, my triangle is bigger.
There are three ways I could have drawn the last one.
I could have started from here, drawing this side as 10 centimetre first, then measuring my 20 degrees, my 60 degrees, and therefore my last angle would be 100 degrees.
I could have drawn my base length, I could have drawn my base length to be here as 10 centimetre, and the last possible way to have done it would be to draw my base length here.
We're now moving on to our independent task, I want you to pause the video now and attempt all the questions on your worksheet.
It's probably better if you look at the questions on the worksheet rather than on the video, and attempt the questions and once you're done, come back and we'll go through the answers together.
Okay, let's go over our answers together.
So we've got 180, because this is an isoceles triangle, 180 take away 54 is 126, so we divide that by two, I get 63 degrees here and I get 63 degrees here.
90 plus 27, that takes us to 117, so a will be 63 degrees.
Okay, for the third one I've already written 63 degrees so we have 63 degrees because base line, isosceles triangle have equal angle, so therefore this would be 54 degrees, and d here would be 90 degrees.
For the next one I have drawn this and I've inserted the picture so checking that yours looks similar to this.
Okay, and then for c and d, I started by drawing a base length for my triangle of seven centimetre for each, and then I measured, well, I didn't need to measure the right angle for this one because it just came up to a straight line.
Okay, now for 3, I started by drawing a base line at the bottom here, and then I thought I would change things up a bit and then I drew the base line over here, I drew my 3.
5 centimetre over here.
Checking your work, making sure they have drawn two different angles of interior angles of 35 degrees, 45 degrees and 100 degrees for both.
Okay, let's move on to our explore task.
Your job is to draw an accurate sketch of this triangle and then draw three different isosceles triangle that have at least one angle that is 40 degrees, and at least one side that is eight centimetre long.
So draw three different isosceles triangle that have at least one angle that is 90 degree and at least one side that is eight centimetre long.
So pause your screen now, I'm not going to provide you with support this time around because I think, I believe, that you should know how to do this, so pause your screen now, attempt this, and then when you finish, come back and we'll go over the answer.
Okay, how did you get on with yours? Well for this one, we were told that our triangle must have at least one angle that is 40 degrees.
Well, I've got one angle that's 40 degrees and then my bottom angles are 70 degrees each, so that's one way of doing that.
And then another way of doing it is, my base angle are 40 degrees, both 40 degrees, and then I've got 100 degrees at the top as well, and then I've got two lengths here of eight centimetre.
Over here I've got my base length of eight centimetre, and then again, the same as this, this is 40 degrees and this is 40 degrees, but this is 100 degrees.
Notice how in this one, my two lengths here are equal, whereas here this would be eight centimetre, and then this length here would be equal to this length here.
Did you draw the same ones as I did on the board? Okay, we have now reached the end of today's lesson and a very big well done for completing the task, you know, learning how to use a protractor, and again, keep practising.
Before you go though, show off your knowledge by completing the quiz, so just ensure that you complete the quiz to consolidate your knowledge and just to show yourself how much you've learned from today's lesson.
And I will see you at the next lesson.
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# Find the value of x in the figure?
Aug 6, 2016
In Fig.(a), $x = 8.75$
In Fig.(b), $x \cong 8.57$
#### Explanation:
In Similar Triangles, the corresponding sides are in proportion.
In Fig.(a), the small triangle ling inside the big one is similar to each other.
Hence, $\frac{x}{7} = \frac{8 + 2}{8} \Rightarrow x = \frac{70}{8} = 8.75$
In Fig.(b), x/15=12/21 rArr x=180/21~=8.57
Aug 6, 2016
In Figure (a): $\textcolor{g r e e n}{x = 8 \frac{3}{4}}$
In Figure (b): $\textcolor{g r e e n}{x = 8 \frac{4}{7}}$
#### Explanation:
Figure (a)
I have assumed that the lines labelled with $7$ and $x$ are parallel (otherwise this question can not be solved).
Reproducing the figure (a) with labelled vertices for reference purposes:
Notice the similar triangles:
color(white)("XXX")triangleABC~triangleADE
rarrcolor(white)("XXX")abs(BC)/abs(AB)=abs(DE)/(abs(AD)
$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{7}{8} = \frac{x}{8 + 2}$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} 8 x = 70$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = 8 \frac{6}{8} = 8 \frac{3}{4}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Figure (b)
Similarly in figure (b) I have had to assume that sides with lengths $15$ and $x$ are parallel.
Again, reproducing the image with labelled vertices:
trianglePQR ~ trianglePST#
$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{\left\mid Q R \right\mid}{\left\mid P Q \right\mid} = \frac{\left\mid S T \right\mid}{\left\mid P S \right\mid}$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{x}{12} = \frac{15}{9 + 12}$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} 21 x = 15 \times 12$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} x = \frac{180}{21} = 8 \frac{4}{7}$
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# 6th Class Mathematics Related to Competitive Exam Reasoning Aptitude
Reasoning Aptitude
Category : 6th Class
Reasoning and Aptitude
Reasoning and logic skills are an integral part of subjects like Mathematics. In this chapter, we will learn various problems related to reasoning and aptitude.
Problems Based on Missing Numbers
In these types of problems, we find out a missing number from a given set of numbers, which is appropriate and follow a certain pattern.
• Example:
Complete the series given below:
4, 10, 26, 72, 208, 614?
(a) 1815 (b) 1820
(c) 1830 (d) 1836
(e) None of these
Explanation: Here the pattern is given below:
$4\times 3-2\times 1=\text{ }10;\text{ }10\times 3\text{ }-\text{ }2\times 2\text{ }=\text{ }26$
$26\times 3\text{ }-\text{ }2\times 3\text{ }=\,\,\,72;\text{ }72\times 3\text{ }-\text{ }2\times 4\text{ }=\text{ }208$
$208\times 3\text{ }-\text{ }2\times 5\text{ }=614;\text{ }614\times 3\text{ }-\text{ }2\times 6\text{ }=1830$
Problems Based on Coding-decoding
In these types of problems, you will learn to code a word by using a certain pattern or rule.
• Example:
If 'SAMPLE' is coded as 'FMNOBT' then how would you code 'CLIMAX'?
(a) YBJLMD (b) YJBMLD
(c) XJMNLD (d) DBJLMY
(e) None of these
Explanation: Here we have the following pattern.
Problems Based on Puzzle
In these types of problems, the given information is summarized by a table.
• Example:
Joseph and John are good in hockey and cricket. Ketan and Yash are good in cricket and football. Arjun and Tuffey are good in baseball and volleybalL Ketan and Tapan are good in hockey and baseball.
Based on above information answer the given question.
Who is good in Hockey, Cricket, Football and baseball?
(a) Joseph (b) John
(c) Ketan (d) Yash
(e) None of these
Explanation:
Joseph John Ketan Yash Arjun Tuffey Tapan Hockey √ √ √ √ Cricket √ √ √ √ Football √ √ Baseball √ √ √ √ Volleyball √ √
Problems Based on Figure
In some of these types of problems two pairs of figures are given. You have to identify a certain pattern or rule used in the first pair of the figure and then apply the rule in another pair.
• Example:
(a) (b)
(c) (d)
(e) None of these
Explanation: Clearly, in second figure the shape moves in $90{}^\circ$ clockwise direction and circle disappeared. Hence by following this pattern correct answer is (c).
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of the form
by
Brock F. Miller
In this investigation we will be looking at various quadratic functions and their graphs by keeping the quadratic of the form:
,
a, b, and c are elements of the Reals,
and a not equal to zero.
Through investigation we will show how quadratics travel around the Cartesian Plane, find the vertex, compare x-intercepts, and compare y-intercepts.
Before we begin let us review the basic concepts and rules for vertex, x-intercept, and y-intercept
I. Vertex of a Parabola
The vertex of a parabola is the point on the graph that either contains the maximum or the minimum for the parabola (discussed later). It also is the point where the line of symmetry will travel vertically (perpendicular to the x-axis).
To find the vertex: Like all points on the Cartesian Plane the vertex is of the form (x , y). Using idea of quadratic equation and the midpoint rule we can show that:
To find the y coordinate of the vertex we can just substitute the x coordinate into the equation and solve for y.
EXAMPLE: Find the vertex of the quadratic
Solution:
The x coordinate of the vertex: a = 3 and b = -12 therefore:
By substituting x = 2 into our equation we get
We can conclude that the vertex is at the coordinate (2 , -6) and the line of symmetry is x = 2.
II. Y-Intercept
The y-intercept (abbreviated y-int) is the point(s) on a graph where the curve intercepts the y-axis. The one constant of all points that cross the y-axis is that x = 0.
To find the y-int , substitute x = 0 into your equation and solve for y. For quadratics we can come up with an easy proof for finding y-intercepts.
Remember the general equation of a quadratic is ## To find the y-int for this form all we do is substitute x = 0 and solve for y:
Therefore a quadratic function of the form:
has a y-intercept of "c" and is at the coordinate (0 , c).
III. X- Intercept
The x-intercept (abbreviated x-int) is the point(s) on a graph where the curve intercepts the x-axis. The one constant of all points that cross the y-axis is that y = 0.
To find the x-int, substitute y = 0 into your equation and solve for x. The formula for the quadratic equation is:
For any quadratic there can be from zero to two real x-intercepts. Though intercepts can be thought of in terms of imaginary numbers, for this investigation into quadratics we will focus on the real number set.
Now that the basic rules for this investigation have been reviewed to you let's look at a few different quadratics. Note for all of the tested quadratics we will show graphically the vertex, y-intercept, and the x-intercept(s) (if any). It is up to you the reader to use the rules stated above to calculate the exact values.
A) Vertex
B) y-int
C) x-int
To begin with lets look at a quadratic with a = 1 and b = c = 0
.
It is not difficult to visually (or algebraically) to show that the vertex of this parabola is (0 , 0) and x-int = y-int = 0.
Now we will look at what happens with change in a, b, and c individually with preset values of 1 for the other two parameters.
Notice that the y-intercept for all of these graphs is at the coordinate (0 , 1). Also that no real x-intercepts exist. Wish preset b and c at one we began with the first equation a =1, second equation a = 2, third equation a = 3, and fourth equation a = 1/2. By investigation we can see that as a grows from 1 to 3 the parabola "narrow" while as a goes less that one we get a wider parabola. All parabolas have a minimum value at the vertex as there are no other y values less that it. Now let's look when a < 0.
Still looking at each of the five graphs a few similarities are there. First all have a y-intercept at the coordinate (0 , 1) when a < -1 we see a narrowing parabola and when -1< a < 0 the parabola seems to widen. There are two major differences between the first graphs and the second set of it's negative a counterparts: y-values of vertex are now maximums and the direction of the curve.
Because the graphs are directed down when a < 0 we see that the y value of the vertex is the largest y value in the range of the quadratic, hence it is a maximum. Secondly and probably the most telling is that when a < 0 we see that the parabolas are directed downward.
Also you should notice that the curves for the negative value of "a" have two x-ints.
To see what happens when we vary "a" on the interval [-5 , 5]: click here
We see that the change in "a" has a bevy of characteristics for a quadratic function. The most obvious of such is the control of the direction of the parabola but also remember it aids in the placement of the vertex as well as the possible zeroes for the curve.
Now we will look at some different quadratics with different values put for the parameter "b"
The value of "b" has many traits that will define a specific parabola the one would notice if they looked at the rules/properties listed at the beginning of the investigation that "b" was a mainstay in vertex selection as well as the quadratic formula. With that in mind one could decide that the value of "b" has a strong emphasis on where the parabola will be placed on the Cartesian Plane.
Now lets look at negative values for "b". For the sake of argument the positive value of b=1 will remain on the sketch.
Upon looking at the curves all still have a y-int at the point (0 , 1) but the curves when b<0 have moved primarily in Quadrants I and IV. As "b" becomes more negative we would assume that the curve drops further to the down and to the right.
If we look at the curves when b = 1 and b= -1 we should see two curves that are symmetric round the line x = 0. It seems as if jumping to a negative counterpart (or positive if you started with a b<0) would give this idea with fixed values for "a" and "c"
From the beginning of the investigation we showed the "c" has a major role in the y-int of the curve. It also has control in other facets but as far as looking at "c" with fixed "a" and "b" we should see a curve that strongly resembles the others but has different y-int. Let's see.
When looking at the curves we can see that as the value of "c" grows the resulting parabolas are contained within the lesser value of "c". we can also see that when c =1 and less we have two real zeroes and when c is 1/2 and higher we have no real zeroes. But there must be a value of "c" that will give the curve 1 real zero. Thus setting parameters on the two real zeroes as well as the no real zeroes for a specific equation.
Now that we have looked at the changed individually for each "a", "b" and "c" one can see that the values are the essence of the parabola. They all have aid in the control of the zeroes of the curve as well as the placement of the curve. The "c" value is the only value that has a lone specific element as it is the value of the y-int though it aids in all else.
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In conclusion, due to the vast possible changes one can do with the values of "a", "b", and "c" one cannot fully see all the possibilities but hopefully this has helped you see ideas that make up a parabola. If there are any questions please contact me back by email back on my web page.
BFM
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# Decrease Percentage
How to find the decrease percentage?
It can easily be understood if it is expressed as per cent. We will follow the following steps to convert the decrease into per cent
Step I: first find the decrease
Step II: divide it by the original quantity
Step III: multiply the fraction by 100%
Formula for finding the Decrease% is = Decrease in value/Original value × 100%
Note: decrease per cent is calculated on the original value.
For example:
The price of sugar decreases from $8 per kg to$6.40 per kg.
Decrease in price = $8 -$6.40 = $1.60 and decrease % = decrease in price/Original price × 100 % = 1.60/8 × 100 % = 160/8 % = 20 % We will apply the concept of solving some real-life problems by using the formula for finding the decrease percent. Solved examples: 1. The cost of an article is decreased by 15%. If the original cost is$80, find the decrease cost.
Solution:
Original cost = $80 Decrease in it = 15% of$80
= 15/100 × 80
= 1200/100
= $12 Therefore, decrease cost =$80 - $12 =$68
2. A television manufacturing company declares that a television is now available for $5600 as against$8400 one year before. Find the percentage reduction in the price of television offered by the company.
Solution:
Price of the television a year before = $8400 Price of the television after a year =$5600
Decrease in price = $(8400 - 5600) =$2800
Therefore, decrease % = 2800/8400 × 100 % = 100/3 = 33 1/3%
3. Find the decrease value if 300 decreased by 30%
Solution:
Decrease 300 by 30%
= 30/100 × 300
= 90
Therefore, decrease value = 300 – 90 = 210
4. Find the number which when decreased by 12 % becomes 198.
Solution:
Let the number be m.
Decrease = 12 % of m
= 12/100 × m = 3m /25
Therefore, decrease number = m – 3m/25 = (25m – 3m)/25 = 22m/25
According to the question 22m/25 = 198
22m = 198 × 25
m = 4950/22
m = 225
5. A number 42 was misread as 24. Find the reading error in per cent.
Solution:
Error = 42 – 24 = 18
Therefore, % error = 18/42 = 100%;
[Since, we know decrease% = decrease in value/original value × 100 %]
= 300/7 %
= 42.8 %
Fraction into Percentage
Percentage into Fraction
Percentage into Ratio
Ratio into Percentage
Percentage into Decimal
Decimal into Percentage
Percentage of the given Quantity
How much Percentage One Quantity is of Another?
Percentage of a Number
Increase Percentage
Decrease Percentage
Basic Problems on Percentage
Solved Examples on Percentage
Problems on Percentage
Real Life Problems on Percentage
Word Problems on Percentage
Application of Percentage
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# How to solve 9 − 2 + 5 (5−4) × [(5+2)+1] ÷ 2 of 4
Welcome to my article How to solve 9 − 2 + 5 (5−4) × [(5+2)+1] ÷ 2 of 4. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 9 − 2 + 5 (5−4) × [(5+2)+1] ÷ 2 of 4, read and understand it carefully till the end.
Let us know how to solve this question How to solve 9 − 2 + 5 (5−4) × [(5+2)+1] ÷ 2 of 4.
First write the question on the page of the notebook.
## How to solve 9 − 2 + 5 (5−4) × [(5+2)+1] ÷ 2 of 4
Write this question in this way and solve it in simple way,
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( {5-4} \right)\text{ }\times \text{ }\left[ {\left( {5+2} \right)+1} \right]\text{ }\div \text{ }2\text{ }of\text{ }4
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( {5-4} \right)\text{ }\times \text{ }\left[ {7+1} \right]\text{ }\div \text{ }2\text{ }of\text{ }4
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( {5-4} \right)\text{ }\times \text{ }\left[ 8 \right]\text{ }\div \text{ }2\text{ }of\text{ }4
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( 1 \right)\text{ }\times \text{ 8 }\div \text{ }2\text{ }\times \text{ }4
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( 1 \right)\text{ }\times \text{ 8 }\div \text{ 8}
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }\left( 1 \right)\text{ }\times \text{ 1}
\displaystyle 9\text{ }-\text{ }2\text{ }+\text{ }5\text{ }
\displaystyle \text{14 }-\text{ }2\text{ }
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# Algebra II: Graphs and Functions
On this page we hope to clear up problems that you might have with graphs and functions. Graphs are used to give graphical representation of equations, usually functions. Read on or follow any of the links below to start understanding graphs and functions!
Graphing points
Graphing lines
Finding the slope of a line
Functions
Parallel and perpendicular lines
Quiz on Graphs and Functions
## Graphing Points
Graphing single ordered pairs is usually covered in most pre-algebra classes and that custom has been followed on this site. You can follow this link to learn about graphing points.
## Graphing Lines
Graphing simple equations such as y = 2x - 3 is a topic usually covered in most elementary algebra (Algebra I) classes and that custom has been followed on this site. You can follow this link to learn about graphing simple equations.
## Finding the Slope of a Line
When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
This can be explained with a formula: ((y2) - (y1))/((x2) - (x1)). (The varialbes would be subscripted if text only browsers allowed for subscripted characters.) To find the slope, you pick any two points on the line and find the change in y, and then divide it by the change in x.
Example:
```1. Problem: The points (1, 2) and (3, 6) are on a line. Find the
line's slope.
Solution: Plug the given points into the slope formula.
(y2) - (y1) The variables would be sub-
m = ----------- scripted if text only browsers
(x2) - (x1) allowed for subscripted
characters.
6 - 2
m = -----
3 - 1
After simplification, m = 2.
```
## Functions
A function is a relation (usually an equation) in which no two ordered pairs have the same x-coordinate when graphed.
One way to tell if a graph is a function is the vertical line test, which says if it is possible for a vertical line to meet a graph more than once, the graph is not a function. The accompanying figure is an example of a function.
Functions are usually denoted by letters such as f or g. If the first coordinate of an ordered pair is represented by x, the second coordinate (the y-coordinate) can be represented by f(x). In the accompanying figure, f(1) = -1 and f(3) = 2.
When a function is an equation, the domain is the set of numbers that are replacements for x that give a value for f(x) that is on the graph. Sometimes, certain replacements do not work, such as 0 in the following function: f(x) = 4/x (you cannot divide by 0). In that case, the domain is said to be x <> 0.
There are a couple of special functions whose graphs you should have memorized because they are sometimes hard to graph. They are the absolute value function (example figure) and the greatest integer function (example figure).
The greatest integer function, y = [x] is defined as follows: [x] is the greatest integer that is less than or equal to x.
## Parallel and Perpendicular Lines
If nonvertical lines have the same slope but different y-intercepts, they are parallel.
Example:
```1. Problem: Determine whether the graphs of y = -3x + 5 and
4y = -12x + 20 are parallel lines.
Solution: Use the Multiplication Principle to
get the second equation in slope-intercept form.
y = -3x + 5
y = -3x + 5
The slope-intercept equations are the same. The two
equations have the same graph.
2. Problem: Determine whether the graphs of 3x - y = -5 and
y - 3x = -2 are parallel.
Solution: By solving each equation for y, you get the
equations in slope-intercept form.
y = 3x + 5
y = 3x - 2
The slopes are the same, and the y-intercepts are
different, so the lines are parallel.
```
Sometimes, you will be asked to find the equation of a line parallel to another line. Not all the information to put the equation in slope-intercept form will always be given.
Example:
```1. Problem: Write an equation of the line parallel to the line
2x + y - 10 = 0 and containing the point (-1, 3).
Solution: First, rewrite the given equation in slope-intercept
form.
y = -2x + 10
This tells us the parallel line must have a slope of -2.
Plug the given point and the slope into the
slope-intercept formula to find the y-intercept
of the parallel line.
3 = -2(-1) + b
Solve for b.
1 = b
The parallel line's equation is y = -2x + 1.
```
If two nonvertical lines have slopes whose product is -1, the lines are perpendicular.
Example:
```1. Problem: Determine whether the lines 5y = 4x + 10 and
4y = -5x + 4 are perpendicular.
Solution: Find the slope-intercept equations by solving for
y.
y = (4/5)x + 2
y = -(5/4)x + 1
The product of the slopes is -1, so the lines are
perpendicular.
```
Sometimes, you will be asked to find the equation of a line perpendicular to another line. Not all the information to put the equation in slope-intercept form will always be given.
Example:
```1. Problem: Write an equation of the line perpendicular to
4y - x = 20 and containing the point (2, -3).
Solution: Rewrite the equation in slope-intercept form.
y = .25x + 5
We know the slope of the perpendicular line is -4
because .25 * -4 = -1. (Notice that the slope of the
perpendicular line is the reciprocal of the other line's
slope.)
Now plug the given point and the slope into a slope-
intercept equation to find the y-intercept.
-3 = (-4)2 + b
Solve for b.
b = 5
Now you have the information you need to write an
equation for a line perpendicular to 4y - x = 20.
The answer is the following equation: y = -4x + 5.
```
Take the quiz on graphs and functions. The quiz is very useful for either review of to see if you've really got the topic down.
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Math for Morons Like Us -- Algebra II: Graphs and Functions
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Transformations of Functions
Transformations of Functions and their Graphs
Before we begin our discussion on transformations of function, it will be helpful to compile a list of some common functions and their graphs. Armed with those, we shall consider various transformations and what effect that has on the graph of the function.
Function: y = x Function: y = |x| Function: y = x2 Function: y = x3
Domain: (-, ) Domain: (-, ) Domain: (-, ) Domain: (-, )
Range: (-, ) Range: [0, ) Range: [0, ) Range: (-, )
Function: Function: Function: y = 1/x Function: y = 1/x2
Domain: [0, ) Domain: [-1, 1] Domain: x 0 Domain: x 0
Range: [0, ) Range: [0, 1] Range: y 0 Range: (0, )
Figure 1: Eight Common Functions and their Graphs
There are two types of transformations that shall discuss. First are translations. By a translation of a graph, we mean a shift in its location such that every point of the graph is moved the same distance and in the same direction. Essentially, think of lifting the graph out of the paper, moving it around, and then placing it down at a new location.
There are four ways to move the graph: left, right, up and down. The effect this has on the graph is summarized in the following table:
#### Translations
Suppose that c is a positive constant.
Equation Effect on the Graph
1. y = f(x) + c Translate c units upward
2. y = f(x) c Translate c units downward
3. y = f(x + c) Translate c units to the left
4. y = f(x c) Translate c units to the right
Example 1:
Write out the function of and graph the translation the graph of y = x2 to the left by 1.
Solution:
From the table above, we see that a translation to the left by 1 can be accomplished by replacing x with x + 1. That is, our function is y = (x + 1)2. Its graph is the following:
Figure 2: The graph y = (x + 1)2
Example 2:
Translate the graph of y = |x| to the right by 2 and down by 1.
Solution:
First, we translate to the right by 2, and then we translate down by 1.
Figure 3: The graph of y = |x 2| 1
The second type of transformation we are interested in is a reflection. There are two types of reflections that we will be concerned about. A reflection about the x-axis is where each point (x, y) is mapped to the point (x, -y). That is, we think of the x-axis as fixed and we spin our graph 180°.
Similarly, a reflection about the y-axis is where each point (x, y) is mapped to the point (-x, y). This time we think of the y-axis as fixed and we spin our graph 180°. We record this in the following:
#### Reflections
Equation Effect on the Graph
1. y = f(x) Reflect about the x-axis
2. y = f(x) Reflect about the y-axis
Example 3:
Graph the functions and .
Solution:
Notice that the first graph is a translation of 1/x2 to the right by 1. The second graph is a reflection of the first graph about the x-axis. Their graphs appear below.
Figure 4: The graphs of and
Example 4:
Graph the functions and .
Solution:
Notice that the first graph is a reflection of about the y-axis. The second graph is a translation of the first graph up by 1. Their graphs appear below.
Figure 5: The graphs of and
What if instead we wanted to graph the function . There will be a reflection involved because of the x. But there will also be a translation, because of the +1. So, which do we do first?
A common mistake would be to apply the translation rule y = f(x + c). The problem is that there is no “x” in the rule. Instead, we have to massage the function.
Notice that . We read this to say “translate the graph to the right by 1, then reflect about the y-axis”.
In general, when presented with both reflections and translations, factor out the negative signs first. Then perform the translations, and finally apply the reflections.
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movie movie
movie movie
If there really is a unifying formula that explains the creation of everything, then perhaps we can use our understanding of it to support our health. To explore this idea further, we need to look at the basic mathematics behind these principles. Our starting point must be a suitable fractal equation, one that is easy to use but is clear enough for our body's intelligence to understand the intent behind its formation.
Some fractals are based on the square root of negative numbers, which takes us further into the conceptual world than we need to travel. For this reason, the Fibonacci sequence is the perfect equation. It uses 'real' numbers, which are easier for our bodies to interpret.
This is the core design behind the equations in this manual. I have occasionally used a Julia Set sequence in some areas, but only when working with concepts outside the basic functions of the human body, like blocking electromagnetic stress, or undoing the trauma of dental work.
So, if we are going to use the Fibonacci sequence as our primary building block, we need to start looking at how it works.
Take any two numbers you wish, say 2 and 5. Add them together and you get 7. Add the 5 and the 7 together and you get 12. 12 and 7 makes 19. Keep adding the last two numbers together and before long you have the following sequence:
2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804 etc
As the numbers grow, you move increasingly closer to the 'golden ratio' of approximately 1 to 1.618. In other words, every subsequent number is 1.618 times the previous one. I'll explain the significance of this in a minute.
Lets work through another sequence first - the true Fibonacci progression which starts with 0 and 1. (I like the undertones here; first there was nothing, then there was the first act of creation...)
0 + 1 = 1 1 + 1 = 2 1 + 2 = 3 2 + 3 = 5 3 + 5 = 8 ... and off it runs... ...0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597 etc
The mathematical representation of this sequence of additions looks like this:
All this is saying is that the third number in the sequence (n+2) is the addition of the previous two, (n+1) and (n). I have added infinity signs - <-> + to show that the fractal should be computed in both directions, as in nature where there is no beginning or end to the depth of the calculations.
- - - - - - - - - - - - - - - - - - - - - - - - - -
The term "fractal" was coined by Benoit Mandelbrot in 1975. It comes from the Latin fractus, meaning an irregular surface like that of a broken stone. Fractals are non-regular geometric shapes that have the same degree of non-regularity on all scales. Just as a stone at the base of a foothill can resemble in miniature the mountain from which it originally tumbled down, so are fractals self-similar whether you view them from close up or very far away.
Fractals are the kind of shapes we see in nature. We can describe a right triangle by the Pythagorean theorem, but finding a right triangle in nature is a different matter altogether. We find trees, mountains, rocks and cloud formations in nature, but what is the geometrical formula for a cloud? How can we determine the shape of a dollop of cream in a cup of coffee? Fractal geometry, chaos theory, and complex mathematics attempt to answer questions like these. Science continues to discover an amazingly consistent order behind the universe's most seemingly chaotic phenomena.
Mathematicians have attempted to describe fractal shapes for over one hundred years, but with the processing power and imaging abilities of modern computers, fractals have enjoyed a new popularity because they can be digitally rendered and explored in all of their fascinating beauty. Fractals are being used in schools as a visual aid to teaching math, and also in our popular culture as computer-generated surfaces for landscapes and planetary surfaces in the movie industry.
- - - - - - - - - - - - - - - - - - - - - - - - - -
Mira's Model
The coordinates of the points on the Mira curve are generated iteratively through the following system of nonlinear difference equations:
where:
Here, a=–0.99, and we consider the cases b=1 and b=0.98. The starting point coordinates are (4, 0). This case can be viewed by editing and executing the following script M-file (MATLAB):
for n=1:12000
a=-0.99;b1=1;b2=0.98;
x1(1)=4;y1(1)=0;x2(1)=4;y2(1)=0;
x1(n+1)=b1*y1(n)+a*x1(n)+2*(1-a)*(x1(n))^2/(1+(x1(n)^2));
y1(n+1)=-x1(n)+a*x1(n+1)+2*(1-a)*(x1(n+1)^2)/(1+(x1(n+1)^2));
x2(n+1)=b2*y2(n)+a*x2(n)+2*(1-a)*(x2(n))^2/(1+(x2(n)^2));
y2(n+1)=-x2(n)+a*x2(n+1)+2*(1-a)*(x2(n+1)^2)/(1+(x2(n+1)^2));
end
subplot(2,1,1); plot(x1,y1,'.')
title('a=-0.99 b=1')
subplot(2,1,2); plot(x2,y2,'.')
title('a=-0.99 b=0.98')
Manifest the computer artist inside yourself. Generate new geometrical morphologies, in Mira's model, by new choices of the parameters (-1 a 1 and b=1) and of the starting point. You can start with:
a b1 b2 (x1,y1) -0,48 1 0,93 (4,0) -0,25 1 0,99 (3,0) 0,1 1 0,99 (3,0) 0,5 1 0,9998 (3,0) 0,99 1 0,9998 (0,12)
Henon’s Model
The coordinates of the Henon's orbits are generated iteratively through the following system of nonlinear difference equations:
x1(1)=0.5696;y1(1)=0.1622;
x2(1)=0.5650;y2(1)=0.1650;
for n=1:120
x1(n+1)=a*x1(n)-b*(y1(n)-(x1(n))^2);
y1(n+1)=b*x1(n)+a*(y1(n)-(x1(n))^2);
x2(n+1)=a*x2(n)-b*(y2(n)-(x2(n))^2);
y2(n+1)=b*x2(n)+a*(y2(n)-(x2(n))^2);
end
plot(x1,y1,'ro',x2,y2,'bx')
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Exponential Terms Raised to an Exponent
## Multiply to raise exponents to other exponents
0%
Progress
Practice Exponential Terms Raised to an Exponent
Progress
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Recognize and Apply the Power of a Power Property
Have you ever tried to multiply a power by a power when there is a monomial? Take a look at this dilemma.
(x2y3z3)3\begin{align*}(x^2y^3z^3)^3\end{align*}
This is a monomial expression that is being raised to the third power. Do you know how to simplify this expression?
Pay attention and you will know how to complete this dilemma by the end of the Concept.
### Guidance
We have raised monomials to a power, products to a power, and quotients to a power. You can see that exponents are a useful tool in simplifying expressions. If you follow the rules of exponents, the patterns become clear. We have already seen powers taken to a power. For example, look at the quotient:
(x7y9)4=(x7)4(y9)4=(x7)(x7)(x7)(x7)(y9)(y9)(y9)(y9)=x7+7+7+7y9+9+9+9=x28y36
If you focus on just the numerator, you can see that (x7)4=x28\begin{align*}(x^7)^4=x^{28}\end{align*}. You can get the exponent 28 by multiplying 7 and 4. This is an example of the Power of a Power Property which says for any nonzero numbers a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} and any integer n\begin{align*}n\end{align*}:
(am)n=amn
Here is one.
(x5)3=x5.3=x15\begin{align*}(x^5)^3=x^{5.3}=x^{15}\end{align*}
Take a look at this one.
(x6y3)7=x67y37=x42y21\begin{align*}(x^6 y^3)^7=x^{6 \cdot 7} y^{3 \cdot 7}=x^{42} y^{21}\end{align*}
Apply the Power of a Power Property to each example.
#### Example A
(x7)3\begin{align*}(x^7)^3\end{align*}
Solution: x21\begin{align*}x^{21}\end{align*}
#### Example B
(x3y4)3\begin{align*}(x^3y^4)^3\end{align*}
Solution: x9y12\begin{align*}x^9y^{12}\end{align*}
#### Example C
(a7)8\begin{align*}(a^7)^8\end{align*}
Solution: a56\begin{align*}a^{56}\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
(x2y3z3)3\begin{align*}(x^2y^3z^3)^3\end{align*}
Next, we have to take each part of the monomial and raise it to the third power.
(x2)3=x(2×3)=x6\begin{align*}(x^2)^3 = x(2 \times 3) = x^6\end{align*}
(y3)3=y(3×3)=y9\begin{align*}(y^3)^3 = y(3 \times 3) = y^9\end{align*}
(z3)3=z(3×3)=z9\begin{align*}(z^3)^3 = z(3 \times 3) = z^9\end{align*}
Now we can put it altogether.
x6y9z9\begin{align*}x^6y^9z^9\end{align*}
This is our solution.
### Vocabulary
Monomial
a single term of variables, coefficients and powers.
Coefficient
the number part of a monomial or term.
Variable
the letter part of a term
Exponent
the little number, the power, that tells you how many times to multiply the base by itself.
Base
the number that is impacted by the exponent.
Expanded Form
write out all of the multiplication without an exponent.
Power of a Power Property
the exponent is applied to all the terms inside the parentheses by multiplying the powers together.
### Guided Practice
Here is one for you to try on your own.
(x2y4z3)4\begin{align*}(x^2y^4z^3)^4\end{align*}
Solution
First, we are going to separate each part of the monomial and raise it to the fourth power.
(x2)4=x8\begin{align*}(x^2)^4 = x^8\end{align*}
(y4)4=y16\begin{align*}(y^4)^4 = y^{16}\end{align*}
(z3)4=z12\begin{align*}(z^3)^4 = z^{12}\end{align*}
Our final answer is x8y16z12\begin{align*}x^8y^{16}z^{12}\end{align*}.
### Practice
Directions: Simplify each monomial expression by applying the Power of a Power Property.
1. (x2)2\begin{align*}(x^2)^2\end{align*}
2. (y4)3\begin{align*}(y^4)^3\end{align*}
3. (x2y4)3\begin{align*}(x^2y^4)^3\end{align*}
4. (x3y3)4\begin{align*}(x^3y^3)^4\end{align*}
5. (y6z2)6\begin{align*}(y^6z^2)^6\end{align*}
6. (x3y4)5\begin{align*}(x^3y^4)^5\end{align*}
7. (a5b3)3\begin{align*}(a^5b^3)^3\end{align*}
8. (a4b4)5\begin{align*}(a^4b^4)^5\end{align*}
9. (a3b6c7)3\begin{align*}(a^3b^6c^7)^3\end{align*}
10. (x12)3\begin{align*}(x^{12})^3\end{align*}
11. (y9)6\begin{align*}(y^9)^6\end{align*}
12. (a2b8c9)4\begin{align*}(a^2b^8c^9)^4\end{align*}
13. (x4b3c3)5\begin{align*}(x^4b^3c^3)^5\end{align*}
14. (a4b3c7d8)6\begin{align*}(a^4b^3c^7d^8)^6\end{align*}
15. (a3b11)5\begin{align*}(a^3b^{11})^5\end{align*}
16. (x6y10z12)5\begin{align*}(x^6y^{10}z^{12})^5\end{align*}
### Vocabulary Language: English
Base
Base
When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Coefficient
Coefficient
A coefficient is the number in front of a variable.
Expanded Form
Expanded Form
Expanded form refers to a base and an exponent written as repeated multiplication.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Monomial
Monomial
A monomial is an expression made up of only one term.
Power of a Power Property
Power of a Power Property
The power of a power property states that $(a^m)^n = a^{mn}$.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# What are the steps to solve this simple algebraic equation?
This is the equation that I use to calculate a percentage margin between cost and sales prices, where x = sales price and y = cost price:
$$z=\frac{x-y}{x}*100$$
This can be solved for x to give the following equation, which calculates sales price based on cost price and margin percentage:
$$x=\frac{y}{1-(\frac{z}{100})}$$
My question is, what are the steps involved in solving the first equation for x? It's been 11 years since I last did algebra at school and I can can't seem to figure it out.
I'm guessing the first step is to divide both sides by 100 like so:
$$\frac{z}{100}=\frac{x-y}{x}$$
Then what? Do I multiply both sides by x? If so how to I reduce the equation down to a single x?
-
I think your problem is you don't remember that $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$. You can use that on $\frac{x-y}{x}$ and you'll find $\frac{x}{x}-\frac{y}{x}$. Unless $x=0$ this simplifies to $1 - \frac{y}{x}$. (In fact, it does too when $x=0$, then both equations are simply not defined. But you shouldn't worry about that too much.) – Myself Feb 17 '11 at 21:40
First, clear the denominator by multiplying both sides by $x$: \begin{align*} z &= \frac{100(x-y)}{x}\\ zx &= 100(x-y) \end{align*} Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$: \begin{align*} zx &= 100x - 100y\\ zx - 100x &= -100y\\ x(z-100) &= -100y \end{align*} Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get: $$x = \frac{-100y}{z-100}.$$ Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have: \begin{align*} x & = \frac{-100y}{z-100}\\ x &= \frac{100y}{100-z}\\ x &= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\ x &= \frac{y}{1 - \frac{z}{100}}. \end{align*}
Added: Alternatively, following Myself's very good point, you can go "unsimplify" $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from $$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$ to $$\frac{y}{x} = 1 - \frac{z}{100}.$$ Taking reciprocals and multiplying through by $y$ gives \begin{align*} \frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\ x = \frac{y}{1-\frac{z}{100}} \end{align*} which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place.
-
Great answer, really well explained thanks! – Adam J. Forster Feb 18 '11 at 9:33
$$z = 100 \cdot \frac{x-y}{x}$$
$$zx = 100(x-y)$$
$$zx - 100x = -100y$$
$$x(z-100) = -100y$$
$$x = -\frac{100y}{z-100}$$
Then divide both numerator and denominator by $-100$ to get $$x = \frac{y}{1-(\frac{z}{100})}$$
-
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# 1.5 Elementary Matrices and a Method for Finding - PowerPoint PPT Presentation
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## 1.5 Elementary Matrices and a Method for Finding
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### 1.5 Elementary Matrices and a Method for Finding An elementary row operation on a matrix A is any one of the following three types of operations: – PowerPoint PPT presentation
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Title: 1.5 Elementary Matrices and a Method for Finding
1
1.5 Elementary Matrices and a Method for Finding
• An elementary row operation on a matrix A is any
one of the following three
• types of operations
• Interchange of two rows of A.
• Replacement of a row r of A by c r for some
number c ? 0.
• Replacement of a row r1 of A by the sum r1 c
r2 of that row and a
• multiple of another row r2 of A.
An nn elementary matrix is a matrix produced by
applying exactly one elementary row operation to
In
Examples
2
When a matrix A is multiplied on the left by an
elementary matrix E, the effect is To perform an
elementary row operation on A.
Theorem (Row Operations by Matrix
Multiplication) Suppose that E is an mm
elementary matrix produced by applying a
particular elementary row operation to Im, and
that A is an mn matrix. Then EA is the matrix
that results from applying that same elementary
row operation to A
Theorem Every elementary matrix is invertible,
and the inverse is also an elementary matrix.
Remark The above theorem is primarily of
theoretical interest. Computationally, it is
preferable to perform row operations directly
rather than multiplying on the left by an
elementary matrix.
3
Theorem
Theorem (Equivalent Statements)
• If A is an nn matrix, then the following
statements are equivalent, that is, all true or
all false.
• A is invertible.
• Ax 0 has only the trivial solution.
• The reduced row-echelon form of A is In.
• A is expressible as a product of elementary
matrices.
4
A Method for Inverting Matrices
• By previous Theorem, if A is invertible, then the
reduced row-echelon form of A is In. That is, we
can find elementary matrices E1, E2, , Ek such
that
• Ek E2E1A In.
• Multiplying it on the right by A-1 yields
• Ek E2E1In A-1
• That is,
• A-1 Ek E2E1In
• To find the inverse of an invertible matrix A, we
must find a sequence of elementary row operations
that reduces A to the identity and then perform
this same sequence of operations on In to obtain
5
Using Row Operations to Find A-1
Example Find the inverse of
• Solution
• To accomplish this we shall adjoin the identity
matrix to the right side of A, thereby producing
a matrix of the form A I
• We shall apply row operations to this matrix
until the left side is reduced to I these
operations will convert the right side to ,
so that the final matrix will have the form I
6
Row operations
rref
Thus
7
If and n X n matrix A is not invertible, then it
cannot be reduced to In by elementary row
operations, i.e, the computation can be stopped.
Example
8
1.6 Further Results on Systems of Equations and
Invertibility
Theorem 1.6.1 Every system of linear equations
has either no solutions, exactly one solution, or
in finitely many solutions.
Theorem 1.6.2 If A is an invertible nn matrix,
then for each n1 matrix b, the system of
equations Ax b has exactly one solution,
namely, x b.
Remark this method is less efficient,
computationally, than Gaussian elimination, But
it is important in the analysis of equations
involving matrices.
9
Example Solve the system by using
10
Linear Systems with a Common Coefficient Matrix
To solve a sequence of linear systems, Ax b1,
Ax b2, , Ax bk, with common coefficient
matrix A
• If A is invertible, then the solutions x1
b1, x2 b2 , , xk bk
• A more efficient method is to form the matrix
A b1 b2 bk , then
• reduce it to reduced row-echelon form we can
solve all k systems at
• once by Gauss-Jordan elimination (Here A may
not be invertible)
11
Example
Solve the system
Solution
12
Theorem 1.6.3 Let A be a square matrix (a) If B
is a square matrix satisfying BA I, then B
(b) If B is a square matrix satisfying AB I,
then B
Theorem 1.6.5 Let A and B be square matrices of
the same size. If AB is invertible, then A and B
must also be invertible
13
Theorem 1.6.4 (Equivalent Statements) If A is an
nn matrix, then the following statements are
equivalent
• A is invertible
• Ax 0 has only the trivial solution
• The reduced row-echelon form of A is In
• A is expressible as a product of elementary
matrices
• Ax b is consistent for every n1 matrix b
• Ax b has exactly one solution for every n1
matrix b
14
A Fundamental Problem Let A be a fixed mXn
matrix. Find all mX1 matrices b such Such that
the system of equations Axb is consistent.
If A is an invertible matrix, then for every mXn
matrix b, the linear system Axb has The unique
solution x b.
If A is not square, or if A is a square but not
invertible, then theorem 1.6.2 does not Apply. In
these cases the matrix b must satisfy certain
conditions in order for Axb To be consistent.
15
Determine Consistency by Elimination
Example What conditions must b1, b2, and b3
satisfy in order for the system of equations
To be consistent?
Solution
16
Example What conditions must b1, b2, and b3
satisfy in order for the system of equations
To be consistent?
Solution
17
Section 1.7 Diagonal, Triangular, and Symmetric
matrices
• A square matrix in which all the entries off the
main diagonal are zero is called a diagonal
matrix.
• For example
• A general nxn diagonal matrix
(1)
• A diagonal matrix is invertible if and only if
all its diagonal entries are nonzero in this
case the inverse of (1) is
18
Diagonal Matrices
• Powers of diagonal matrices are easy to compute
if D is the diagonal matrix (1) and k is a
positive integer, then
• In words, to multiply a matrix A on the left by a
diagonal matrix D, one can multiply successive
rows of A by the successive diagonal entries of
D, and to multiply A on the right by D, one can
multiply successive columns of A by the
successive diagonal entries of D.
19
Triangular Matrices
• A square matrix in which all the entries above
the main diagonal are zero is called low
triangular, and a square matrix in which all the
entries below the main diagonal are zero is
called upper triangular. A matrix that is either
upper triangular or lower triangular is called
triangular.
• Theorem 1.7.1
• The transpose of a lower triangular matrix is
upper triangular, and the transpose of an upper
triangular matrix is lower triangular.
• The product of lower triangular matrices is lower
triangular, and the product of upper triangular
is upper triangular.
• A triangular matrix is invertible if and only if
its diagonal entries are all nonzero.
• The inverse of an invertible lower triangular
matrix is lower triangular, and the inverse of an
invertible upper triangular matrix is upper
triangular.
20
Symmetric matrices
• A square matrix A is called symmetric if AAT.
• A matrix Aaij is symmetric if and only if
aijaji for all values of I and j.
• Theorem 1.7.2
• If A and B are symmetric matrices with the same
size, and if k is any scalar, then
• AT is symmetric.
• AB and A-B are symmetric.
• kA is symmetric.
• Note in general, the product of symmetric
matrices is not symmetric.
• If A and B are matrices such that ABBA, then we
say A and B commute.
• The product of two symmetric matrices is
symmetric if and only if the matrices commute.
21
Theorems
• Theorem 1.7.3
• If A is an invertible symmetric matrix, then A-1
is symmetric.
• Theorem 1.7.4
• If A is an invertible matrix, then AAT and ATA
are also invertible.
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# Introduction to Integers, Fractions and the Order of Operations
Integers are positive whole numbers and negative numbers. They do not contain any fractions or decimals. Positive integers are numbers that are greater than zero, while negative integers are numbers that are smaller than zero. Some very basic examples include, the numbers one, two, three, negative four, negative seven and so much more.
Order of Operations for Integers
Here are the basic order of operations you can take note of when dealing with integers and all the four operations.
• Always look for brackets and simplify expressions within the brackets first.
• Starting from left to right, perform multiplication or division next (whichever comes earlier)
• Again from left to right, perform addition or subtraction next (whichever comes earlier).
• Simplify the expression.
Introducing Fractions
Now, let’s look into fractions. A fraction consists of a numerator (top number) and a denominator (bottom number). A proper fraction is one where the numerator is smaller than the denominator. For example, ½ and ¼ are proper fractions. An improper fraction, on the other hand, is one where the numerator is greater than the denominator. Some examples of improper fractions include 4/3 and 8/7. Equivalent fractions have the same value. For example ½ and 2/4 are equivalent fractions since both their values equal to 0.5. Lastly, a mixed number is the sum of a whole number and a fraction.
The basic methods for adding and subtracting fractions can be summed up below.
• If the fractions have different denominators, you need to make them the same first.
• Add or subtract the numerators but just copy back the denominator. Write down the solution of the adding or subtracting of the numerator on the top while the denominator is written at the bottom.
• If need be, simplify the fraction into the lowest terms.
Multiplication of Fractions
For multiplication of fractions, there are a few rules to follow which may seem complicated at first. But once you have mastered them, it’s actually quite easy to handle multiplication of fractions.
The following summarises the basic steps when dealing with multiplication of fractions or mixed numbers:
Change all mixed numbers (if there are any) to improper fractions first.
Simplify the fractions by canceling from the top and bottom or sideways.
Multiply the numerators and then the denominators.
Simplify the answer to the simplest form (if required).
Division of Fractions
Now that you are clear about multiplying fractions, let’s delve into the division of fractions instead. The following summarises the basic steps when dealing with division of fractions or mixed numbers.
• Change all mixed numbers (if there are any) to improper fractions first.
• Turn the second fraction upside down and change the divide sign to multiply.
• Simplify the fractions by canceling from the top and bottom or sideways.
• Multiply the numerators and then the denominators.
• Simplify the answer to the simplest form (if required).
Getting Help in Maths for your Child
If you are confused about how you can help your child in numbers, fractions, the order of operations and maths in general, then you should consider enrolling them in Edufront. You do not have to waste your time looking for the best maths tuition Singapore has to offer, once your child is enrolled in Edufront’s tuition programmes.
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# 16.11 Energy in waves: intensity (Page 2/3)
Page 2 / 3
## Determine the combined intensity of two waves: perfect constructive interference
If two identical waves, each having an intensity of $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , interfere perfectly constructively, what is the intensity of the resulting wave?
Strategy
We know from Superposition and Interference that when two identical waves, which have equal amplitudes $X$ , interfere perfectly constructively, the resulting wave has an amplitude of $2X$ . Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.
Solution
1. Recall that intensity is proportional to amplitude squared.
2. Calculate the new amplitude:
$I\prime \propto {\left(X\prime \right)}^{2}={\left(2X\right)}^{2}={4X}^{2}.$
3. Recall that the intensity of the old amplitude was:
${I}^{}\propto {X}^{2}.$
4. Take the ratio of new intensity to the old intensity. This gives:
$\frac{I\prime }{I}=4.$
5. Calculate to find $I\prime$ :
$I\prime =4I=4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}.$
Discussion
The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , yet their sum has an intensity of $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in Superposition and Interference suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ each, there will be places in the room where the intensity is $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ , other places where the intensity is zero, and others in between. [link] shows what this interference might look like. We will pursue interference patterns elsewhere in this text.
Which measurement of a wave is most important when determining the wave's intensity?
Amplitude, because a wave’s energy is directly proportional to its amplitude squared.
## Section summary
Intensity is defined to be the power per unit area:
$I=\frac{P}{A}$ and has units of ${\text{W/m}}^{2}$ .
## Conceptual questions
Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.
Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.
## Problems&Exercises
Medical Application
Ultrasound of intensity $1\text{.}\text{50}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output?
0.225 W
The low-frequency speaker of a stereo set has a surface area of $0\text{.}\text{05}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity $0\text{.}1\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ?
To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased?
7.07
Engineering Application
A device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm 2 and registers 6.50 W. What is the intensity in ${\text{W/m}}^{2}$ ?
Astronomy Application
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of $1.30\phantom{\rule{0.25em}{0ex}}{\text{kW/m}}^{2}.$ How long does it take for $1.8×{10}^{9}\phantom{\rule{0.25em}{0ex}}\text{J}$ to arrive on an area of $1\text{.}00\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ ?
16.0 d
Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?
2.50 kW
Engineering Application
(a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is $700\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2},$ what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 ¢ per kilowatt-hour.
A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally $\text{2.00}×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2},$ but is turned up until the amplitude increases by 30.0%, what is the new intensity?
$\text{3.38}×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$
Medical Application
(a) What is the intensity in ${\text{W/m}}^{2}$ of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about $700\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.
Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
what is angular velocity
Why does earth exert only a tiny downward pull?
hello
Islam
Why is light bright?
an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
thanks so much. i undersooth well
what is physics
is the study of matter in relation to energy
Kintu
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
what is power?
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
hi
Godfred
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
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# What is the standard form of y= (x – 1) (x + 2) (x + 3) ?
May 11, 2017
$y = {x}^{3} + 4 {x}^{2} - 9$
#### Explanation:
Expand the formula and ensure the power and coefficient go first.
$y = \left(x - 1\right) \left(x + 2\right) \left(x + 3\right)$
$= \left({x}^{2} + 2 x - x - 3\right) \left(x + 3\right)$ (Use FOIL on the first two terms)
$= \left({x}^{2} + x - 3\right) \left(x + 3\right)$ (Simplify)
$= {x}^{2} \left(x + 3\right) + x \left(x + 3\right) - 3 \left(x + 3\right)$ (Distribute the $\left(x + 3\right)$)
$= {x}^{3} + 3 {x}^{2} + {x}^{2} + 3 x - 3 x - 9$
$= {x}^{3} + 4 {x}^{2} - 9$ (Simplify)
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# Roots and Powers of Algebraic Expressions
### Subtraction Property and Limits: Definition & Examples
What are limits of functions? What happens to limits if we subtract two functions that have limits at the same value? In this lesson, we will look at the subtraction property for limits.
### Definition of a Limit
Most of the time, a function f(x) has a value for every possible value of x, but not always. For example, f(x) = (x - 4)(x - 6)/2(x - 6) is undefined at the value x = 6 because dividing by 2(6- 6) = 0 is just not feasible. Even then, however, we can look at what the function is like when we get closer and closer to the limit: in this case, the limit when x = 6.
As we can see below, the closer x gets to 6, the closer y gets to 1, so we say that the limit of f(x) as x approaches 6 is 1. We can define a limit if we can get as close as we want to a particular value of x (call it x sub 0) and if it consistently gets closer and closer to one particular value of y when we do. If we can't get close to the x value with a definition of a function, or if the values of y jump around when we get closer and closer to x sub 0, or if the values of y go up or down without ever settling near a number, we cannot define a limit.
### Subtraction Property
The subtraction property for limits says that, if two functions, say f(x) and g(x), have a limit at the same point (let's call it a), the limit as x approaches a of (f - g)(x) is the difference, as x approaches a, of the two limits at a. Note that both limits need to be approaching the same number.
Now, suppose we have two functions, and both have limits at a certain value. For example, suppose f(x) = (x^2 - 1)/(x + 1) and g(x) = x + 3. Each of them has a limit at x = -1: the limit of f(x) as x approaches -1 is -2, and the limit of g(x) as x approaches -1 is 2. Let's look at the graphs:
So far, so good. Notice that g(x) has a defined value at x = -1 and f(x) does not. The limits still work out just fine.
Now, suppose we look at (f - g)(x). We can subtract most functions by subtracting the expressions that define them, so in our example,
(f - g)(x) = (x^2 - 1)/(x + 1) - (x + 3) = (x^2 - 1)/(x + 1) - x - 3.
Let's graph the new function and see what happens to the limit as x approaches -1:
Look! The function turned out to be almost a constant function, with every value except (f - g)(-1) coming out a -4. Of course, (f - g)(x) is not defined for x= -1.
So the limit as x approaches -1 for f(x) is -2, the limit as x approaches -1 for g(x) is 2, and the limit as xapproaches -1 of (f - g)(x) is -4, which is -2 - 2.
We can subtract the two limits to get the limit of the difference function and save some work. This is the Subtraction Property for limits:
The limit as x approaches some value a of (f - g)(x) is equal to the limit as x approaches a of f(x) - the limit as x approaches a of g(x), providing that both limits are defined.
### Summary
When we have a limit defined at a certain value a for two functions f and g, the limit at that value for the difference function (f - g)(x) is the difference of the limits for f(x) and g(x) at that value. Instead of defining the difference function and finding the new limit, we can just subtract the known limits. Note that, if the value of either function is undefined at a, the value of the difference function is also undefined at a.
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# Table of Contents Polynomials: Algebraically finding exact zeros Example 1:Find all exact zeros of the polynomial, P(x) = 3x 3 – 16x 2 + 19x – 4. First,
## Presentation on theme: "Table of Contents Polynomials: Algebraically finding exact zeros Example 1:Find all exact zeros of the polynomial, P(x) = 3x 3 – 16x 2 + 19x – 4. First,"— Presentation transcript:
Table of Contents Polynomials: Algebraically finding exact zeros Example 1:Find all exact zeros of the polynomial, P(x) = 3x 3 – 16x 2 + 19x – 4. First, use the Rational Zero Test to make up a list of possible rational zeros. (see separate slideshow) Second, graph the function in the decimal viewing window by pressing, entering the function, and pressing. Y = ZOOM 4 Looking at the graph and the list of possible rational zeros, suggests 4/3 might be a rational zero of P(x).
Table of Contents Polynomials: Algebraically finding exact zeros Slide 2 The rational zero can be confirmed by pressing 2ndTRACEENTER and entering 4/3, then. ENTER Since 4/3 is a zero, (x – 4/3) is a factor, so next, divide P(x) by (x – 4/3). 3 - 16 19 - 4 4 - 16 4 4/3 3 - 12 3 0 This means P(x) can be written as: P(x) = (x – 4/3)(3x 2 – 12x + 3).
Table of Contents Polynomials: Algebraically finding exact zeros Slide 3 Remember, finding zeros of a polynomial, P(x) means solving the equation P(x) = 0. Therefore, to find the remaining two zeros, solve 3x 2 – 12x + 3 = 0, or x 2 – 4x + 1 = 0. Using the quadratic formula, The exact zeros are 4/3, and Note also P(x) factors as: P(x) =
Table of Contents Polynomials: Algebraically finding exact zeros Slide 4 Example 2:Find all exact zeros of the polynomial, P(x) = 2x 4 – x 3 + 7x 2 – 4x – 4. First, use the Rational Zero Test to make up a list of possible rational zeros. (see separate slideshow) Second, graph the function in the decimal viewing window by pressing, entering the function, and pressing. Y = ZOOM 4 Looking at the graph and the list of possible rational zeros, suggests - 1/2 and 1 are rational zeros of P(x).
Table of Contents Polynomials: Algebraically finding exact zeros Slide 5 Since 1 is a zero, (x – 1) is a factor, so next, divide P(x) by (x – 1). 2 - 1 7 - 4 - 4 2 1 8 4 1 2 1 8 4 0 This means P(x) can be written as: P(x) = (x – 1)(2x 3 + x 2 + 8x + 4). The rational zeros can be confirmed by pressing and pressing the and keys. TRACE Note, - 1/2 is also a zero, so P(x) can furthermore be written as: P(x) = (x – 1)(x + 1/2)(quadratic factor).
Table of Contents Polynomials: Algebraically finding exact zeros Slide 6 To find the unknown quadratic factor, divide 2x 3 + x 2 + 8x + 4 by x + 1/2. 2 1 8 4 - 1 0 - 4 - 1/2 2 0 8 0 This means P(x) can be written as: P(x) = (x – 1)(x + 1/2)(2x 2 + 8). Last, find the remaining two zeros by solving 2x 2 + 8 = 0, or x 2 + 4 = 0. x 2 + 4 = 0,x 2 = - 4, x = 2i The exact zeros are - 1/2, 1, - 2i, and 2i.
Table of Contents Polynomials: Algebraically finding exact zeros Slide 7 Try:Find all exact zeros of the polynomial, P(x) = 4x 3 – 5x 2 + 14x + 15. The exact zeros are - 3/4,1 + 2i, and 1 – 2i. Try:Find all exact zeros of the polynomial, P(x) = x 4 – 11x 2 – 2x + 12. The exact zeros are -3, 1, and
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# How do you make a clock in math?
Contents
## What is a clock in mathematical terms?
In math, time can be defined as the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future. … We measure and define what time of the day it is using clocks. A clock in general has 12 numbers written on it, from 1 to 12.
## Why is the number 13 always in a clock?
Harry from the Beacon School solved this using an algebraic method. The sum of all the numbers is 78. If we call x the total on the small side, then 5x+x=78, so 6x=78 and x=13. So the sum on the small side is 13.
## Why clock sum is 13?
In this case it is about the sum of the numbers on a clock. If we want t divide the clock in 6 parts with each part an equal sum, we now know that the sum of each part must be 13. If we want t divide the clock in 6 parts with each part an equal sum, we now know that the sum of each part must be 13.
## Is time man made or natural?
Time as we think of it isn’t innate to the natural world; it’s a manmade construct intended to describe, monitor, and control industry and individual production.
IT IS AMAZING: Are Timex watches accurate?
## What is a clock called?
A clock or watch is called “analog” when it has moving hands and (usually) hours marked from 1 to 12 to show you the time.
## Is am or am proper?
The first and most common way to write them is with lowercase “a.m.” and “p.m.” This way requires periods, and both Chicago Style and AP Style recommend this way of writing the abbreviations. This subway train will leave daily at 10:05 a.m. After 10:00 p.m. I really need to sleep.
## What are two ways to write time?
There are two ways of telling the time in English – the 12-hour clock and the 24-hour clock. In the 24-hour clock, we use the numbers from 0 – 23 to indicate the hours. In the 12-hour clock, we use 1 – 12.
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# 2019 AIME II Problems/Problem 2
## Problem 2
Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$. From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
## Solution
Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$. Then $P_7 = 1$, $P_6 = \frac12$, and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$. Working our way down, we find $$P_5 = \frac{3}{4}$$ $$P_4 = \frac{5}{8}$$ $$P_3 = \frac{11}{16}$$ $$P_2 = \frac{21}{32}$$ $$P_1 = \frac{43}{64}$$ $43 + 64 = \boxed{107}$.
## Solution 2(Casework)
Define a one jump to be a jump from k to K + 1 and a two jump to be a jump from k to k + 2.
Case 1: (6 one jumps) (1/2)^6 = 1/64
Case 2: (4 one jumps and 1 two jumps) 5C1 x (1/2)^5 = 5/32
Case 3: (2 one jumps and 2 two jumps) 4C2 x (1/2)^4 = 3/8
Case 4: (3 two jumps) (1/2)^3 = 1/8
Summing the probabilities gives us 43/64 so the answer is 107.
- pi_is_3.14
## Solution 3 (easiest)
Let $P_n$ be the probability that the frog lands on lily pad $n$. The probability that the frog never lands on pad $n$ is $\frac{1}{2}P_{n-1}$, so $1-P_n=\frac{1}{2}P_{n-1}$. This rearranges to $P_n=1-\frac{1}{2}P_{n-1}$, and we know that $P_1=1$, so we can compute $P_7$ to be $\frac{43}{64}$, meaning that our answer is $\boxed{107}$
-Stormersyle
## Solution 4
For any point $n$, let the probability that the frog lands on lily pad $n$ be $P_n$. If the frog is at lily pad $n-2$, it can either double jump with probability $\frac{1}{2}$ or single jump twice with probability $\frac{1}{4}$ to get to lily pad $n$. Now consider if the frog is at lily pad $n-3$. It has a probability of landing on lily pad $n$ without landing on lily pad $n-2$ with probability $\frac{1}{4}$, double jump then single jump. Therefore the recursion is $P_n = \frac{3}{4}P_{n-2} + \frac{1}{4}P_{n-3}$. Note that all instances of the frog landing on lily pad $n-1$ has been covered. After calculating a few values of $P_n$ using the fact that $P_1 = 1$, $P_2 = \frac{1}{2}$, and $P_3 = \frac{3}{4}P_1 = \frac{3}{4}$ we find that $P_7 = \frac{43}{64}$. $43 + 63 = \boxed{107}$
2019 AIME II (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# 2001 AIME I Problems/Problem 13
## Problem
In a certain circle, the chord of a $d$-degree arc is $22$ centimeters long, and the chord of a $2d$-degree arc is $20$ centimeters longer than the chord of a $3d$-degree arc, where $d < 120.$ The length of the chord of a $3d$-degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$
## Solution
### Solution 1
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree arcs, and let $CD$ be the chord of the $3d$-degree arc. Also let $x$ be equal to the chord length of the $3d$-degree arc. Hence, the length of the chords, $AD$ and $BC$, of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem.
Using Ptolemy's theorem,
$$AB(CD) + AC(BD) = AD(BC)$$ $$22x + 22(22) = (x + 20)^2$$ $$22x + 484 = x^2 + 40x + 400$$ $$0 = x^2 + 18x - 84$$
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. $$x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}$$ $$x = \frac{-18 + \sqrt{660}}{2}$$
$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$.
### Solution 2
Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, $$2R\sin z=22$$ $$2R(\sin 2z-\sin 3z)=20$$ Dividing the latter by the former, $$\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}$$ $$4\cos^2z-2\cos z-\frac{1}{11}=0 (1)$$ We want to find $$\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).$$ From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174.}$
### Solution 3
Let $z=\frac{d}{2},$ R$be the circumradius, and$a$be the length of 3d degree chord. Using the extended sine law, we obtain: 22=2Rsin(z) 20+a=2Rsin(2z) a=2Rsin(3z) Dividing the second from the first we get$cos(z)=\frac{20+a}{44}$By the triple angle formula we can manipulate the third equation as follows:$a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))\$ Solving it gives the answer.
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## How do you find the slope of the best fit line?
The line’s slope equals the difference between points’ y-coordinates divided by the difference between their x-coordinates. Select any two points on the line of best fit. These points may or may not be actual scatter points on the graph. Subtract the first point’s y-coordinate from the second point’s y-coordinate.
## What does the line of best fit tell you?
Line of best fit refers to a line through a scatter plot of data points that best expresses the relationship between those points. A regression involving multiple related variables can produce a curved line in some cases.
## How do you determine the equation of a line?
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## How do you predict a line of best fit?
A line of best fit is drawn through a scatterplot to find the direction of an association between two variables. This line of best fit can then be used to make predictions. To draw a line of best fit, balance the number of points above the line with the number of points below the line.
## How do I find the slope of the line?
To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
## How do you find the slope and y intercept?
Slope-intercept form is y=mx+b, where m represents the slope and b represents the y-intercept.
## How do you interpret the slope of a regression line?
Interpreting the slope of a regression line The slope is interpreted in algebra as rise over run. If, for example, the slope is 2, you can write this as 2/1 and say that as you move along the line, as the value of the X variable increases by 1, the value of the Y variable increases by 2.
## What two things make a best fit line?
A line of best fit is a straight line drawn through the maximum number of points on a scatter plot balancing about an equal number of points above and below the line.
## Why all the points do not lie on the line of best fit?
Student: The line of best fit will touch all of those points because those points make a straight line. The line will go upwards and it will be pretty steep.
## Is a line of best fit always straight?
Lines of best fit can be straight or curved. Some will pass through all of the points, while others will have an even spread of points on either side. There is usually no right or wrong line, but the guidelines below will help you to draw the best one you can.
## What is the Y intercept formula?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## How do you graph a line from an equation?
To graph a linear equation, we can use the slope and y-intercept.Locate the y-intercept on the graph and plot the point.From this point, use the slope to find a second point and plot it.Draw the line that connects the two points.
### Releated
#### Equation for
How do you find the equation of a function? It is relatively easy to determine whether an equation is a function by solving for y. When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is […]
#### Factoring an equation
How do you factor algebraic equations? So, if, in your equation, your b value is twice the square root of your c value, your equation can be factored to (x + (sqrt(c)))2. For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we […]
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# How do you find the VERTEX of a parabola y= x^2 + 6x + 5?
Jul 17, 2015
Complete the square to get the equation into vertex form:
$y = {\left(x - \left(- 3\right)\right)}^{2} + \left(- 4\right)$
Then the vertex can be read as $\left(- 3 , - 4\right)$
#### Explanation:
Given any quadratic: $y = a {x}^{2} + b x + c$, we can complete the square to get:
$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$
In our example, $a = 1$, $b = 6$ and $c = 5$, so we find:
${x}^{2} + 6 x + 5 = {\left(x + 3\right)}^{2} + \left(5 - {3}^{2}\right) = {\left(x + 3\right)}^{2} + \left(5 - 9\right)$
$= {\left(x + 3\right)}^{2} - 4$
So $y = {\left(x + 3\right)}^{2} - 4$
Strictly speaking, vertex form is $y = a {\left(x - h\right)}^{2} + k$, from which you can read off the vertex $\left(h , k\right)$. So let's replace the $\left(x + 3\right)$ with $\left(x - \left(- 3\right)\right)$ and the $- 4$ with $+ \left(- 4\right)$ to get:
$y = {\left(x - \left(- 3\right)\right)}^{2} + \left(- 4\right)$
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# How to solve logarithms
College algebra students learn How to solve logarithms, and manipulate different types of functions. We can solve math problems for you.
## How can we solve logarithms
In this blog post, we will provide you with a step-by-step guide on How to solve logarithms. It is available for free on both iOS and Android devices. Photomath is able to solve simple mathematical problems by taking a picture of the problem. It can also provide step-by-step instructions on how to solve the problem.
Solving equations is one of the most basic skills you can have as a mathematician. It's also one of the most important, because without it you can't do much in math. Solving equations is all about grouping numbers together and finding the relationship between them. You do that by using addition, subtraction, multiplication, or division to combine the numbers. You can also use inverse operations (like dividing by negative 1) to undo the effects of addition and subtraction. Once you know how to solve equations, you can use them for almost anything! They may seem easy at first, but if you practice solving equations every day, you'll soon be a pro! Here are some tips for solving equations: Group like terms together (like 2 + 5 = 7). Add or subtract one number at a time until you reach your target answer. If you're not sure what to do next, try multiplying both sides by each other (like 12 × 5 = 60). If that doesn't work, try dividing both sides by each other (like 12 ÷ 5 = 4). If none of these works, just look at your answer choices and pick the correct one.
solving equations is a process that involves isolating the variable on one side of the equation. This can be done using inverse operations, which are operations that undo each other. For example, addition and subtraction are inverse operations, as are multiplication and division. When solving an equation, you will use these inverse operations to move everything except for the variable to one side of the equal sign. Once the variable is isolated, you can then solve for its value by performing the inverse operation on both sides of the equation. For example, if you are solving for x in the equation 3x + 5 = 28, you would first subtract 5 from both sides of the equation to isolate x: 3x + 5 - 5 = 28 - 5. This results in 3x = 23. Then, you would divide both sides of the equation by 3 to solve for x: 3x/3 = 23/3. This gives you x = 23/3, or x = 7 1/3. Solving equations is a matter of isolating the variable using inverse operations and then using those same operations to solve for its value. By following these steps, you can solve any multi-step equation.
There are two methods that can be used to solve quadratic functions: factoring and using the quadratic equation. Factoring is often the simplest method, and it can be used when the equation can be factored into two linear factors. For example, the equation x2+5x+6 can be rewritten as (x+3)(x+2). To solve the equation, set each factor equal to zero and solve for x. In this case, you would get x=-3 and x=-2. The quadratic equation can be used when factoring is not possible or when you need a more precise answer. The quadratic equation is written as ax²+bx+c=0, and it can be solved by using the formula x=−b±√(b²−4ac)/2a. In this equation, a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, if you were given the equation 2x²-5x+3=0, you would plug in the values for a, b, and c to get x=(5±√(25-24))/4. This would give you two answers: x=1-½√7 and x=1+½√7. You can use either method to solve quadratic functions; however, factoring is often simpler when it is possible.
There are many math answers websites available online. These websites allow students to submit questions and receive answers from other students or from tutors. This can be a great resource for students who are struggling with a particular concept or who need extra help outside of class. However, it is important to note that not all of these websites are created equal. Some are more reliable than others, and some may even charge for their services. Before using any math answers website, be sure to do your research to
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# Chapter 2.5 – Compound Inequalities
## Presentation on theme: "Chapter 2.5 – Compound Inequalities"— Presentation transcript:
Chapter 2.5 – Compound Inequalities
Objectives I will find the intersection of two sets.
I will solve compound inequalities containing and. I will find the union of two sets. I will solve compound inequalities containing or.
What’s the difference? You get a discount if you are at least 18 years old and no more than 60 years old. 18 < x < 60 You get a discount if you are less than 18 years old or at least 60 years old. 18 > x > 60
Compound Inequalities
Two inequalities joined by the words and or or are called compound inequalities. Compound Inequalities x + 3 < 8 and x > 2
Intersection of Two Sets
The intersection of two sets. A and B, is the set of all elements common to both sets. A intersect B is denoted by, B A
The numbers 4 and 6 are in both sets.
Example 1 If A = { x / x is an even number greater than 0 and less than 10} and B = {3, 4, 5, 6} find A ∩ B. List the elements in sets A and B A = {2, 4, 6, 8} B = {3, 4, 5, 6} The numbers 4 and 6 are in both sets. The intersection is {4, 6}
You try it! Find the intersection: {1, 2, 3, 4, 5} ∩ {3,4,5,6}
{3,4,5}
Solutions to compound inequalities
A value is a solution of a compound inequality formed by the word and if it is a solution of both inequalities. For example, the solution set of the compound inequality x ≤ 5 and x ≥ 3 contains all values of x that make both inequalities true.
Compound Inequalities
A compound inequality such as x ≥ 3 and x ≤ 5 can be written more compactly. 3 ≤ x ≤ 5
Graphing compound inequalities
{ x / x ≤ 5} ] 1 5 3 4 2 6 {x / x ≥ 3} [ 1 5 3 4 2 6 { x / 3 ≤ x ≤ 5 1 5 3 4 2 6 [ ]
Example 2 Solve x – 7 < 2 and 2x + 1 < 9
Step 1: Solve each inequality separately x – 7 < 2 and 2 x + 1 < 9 x < 9 and x < 8 x < 9 and x < 4
Example 2 Step 2: Graph the two intervals on two number lines to find their intersection. x < 9 x < 4 4 8 6 7 5 9 ) 4 8 6 7 5 9 )
Example 2 The solution set is ( - ∞, 4)
Step 3: Graph the compound inequality { x / x < 9 and x < 4} = { x / x < 4} 3 7 5 6 4 8 ) The solution set is ( - ∞, 4)
Solve: x + 5 < 9 and 3x -1 < 2
Give it a try! Solve: x + 5 < 9 and 3x -1 < 2
Example 3 Solve 2 x ≥ 0 and 4 x – 1 ≤ -9
First Step: Solve each inequality separately.
Example 3 Step 2: Graph the intervals to find their intersection.
Example 3 Step 3: Graph the intersection
You try it! Solve: 4x ≥ 0 and 2x + 4 ≤ 2
Example 4 Solve 2 < 4 – x < 7
Step 1: Isolate the x in the middle 2 < 4 – x < 7 2 – 4 < 4 – x – 4 < 7 – 4 -2 < -x < 3 -2 < - x < 3 2 > x > 3 Subtract 4 from all 3 parts Divide all three parts by -1 Must reverse symbol because dividing by a negative.
2 > x > -3 is equivalent to -3 < x < 2
Example 4 2 > x > -3 is equivalent to -3 < x < 2 Interval Notation (-3, 2) Graph the inequality -3 1 -1 -2 2 ( )
Give it a try! Solve: 5 < 1 – x < 9
Example 5 Solve
Example 5 Graph the solution Interval Notation
Give it a try!
Union of two sets The solution set of a compound inequality formed by the word or is the union of the solution set of two inequalities. The union of two sets, A and B, is the set of elements that belong to either of the sets. A union B is denoted by A U B
Example 6 If A = {x / x is an even number greater than 0 and less than 10} and B = {3, 4, 5, 6} Find A U B. List the elements in Set A and Set B A: {2, 4, 6, 8} B: {3, 4, 5, 6} Union : {2,3, 4, 5, 6, 8}
Union A value is a solution of a compound inequality formed by the word or if it is a solution of either inequality. Graph of {x / x ≤ 1} Graph of { x / x ≥ 3}
Unions Graph of { x/ x ≤ 1 or x ≥ 3}
Give it a try! Find the union: {1, 2, 3, 4, 5} U {3, 4, 5, 6}
{1,2,3,4,5,6}
Give it a try! Solve: 5 x – 3 ≤ 10 or x + 1 ≥ 5
First solve each inequality separately.
Example 7 Now we can graph each interval and find their union.
Example 7: Solution set:
Give it a try! Solve: 3x – 2 ≥ 10 or x – 6 ≤ -4
Example 8 Solve: – 2x – 5 < -3 or 6x < 0
Step 1: Solve each inequality separately
Example 8 Graph each interval and find their union
Give it a try! Solve: x – 7 ≤ -1 or 2x – 6 ≥ 2
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### Section 5-1 : Dividing Polynomials
For problems 1 – 6 use long division to perform the indicated division.
1. Divide $$7{x^2} + 4x - 9$$ by $$x - 1$$
2. Divide $$8{x^3} - 4x + 1$$ by $$x + 6$$
3. Divide $${x^4} - 2{x^2} + 7x$$ by $$x - 4$$
4. Divide $$2{x^4} - 9{x^3} + 2x + 8$$ by $$x + 3$$
5. Divide $$8{x^4} + {x^3} - 3{x^2} + 1$$ by $${x^2} - 2$$
6. Divide $$4{x^5} - 7{x^3} + {x^2} - 4x + 2$$ by $$2{x^2} - 3x - 6$$
For problems 7 – 11 use synthetic division to perform the indicated division.
1. Divide $$- {x^3} - 8{x^2} + x + 10$$ by $$x + 2$$
2. Divide $$10{x^3} - 9x$$ by $$x - 10$$
3. Divide $$3{x^4} + 5{x^3} + x - 2$$ by $$x + 7$$
4. Divide $${x^4} + 2{x^3} - 9x + 11$$ by $$x + 3$$
5. Divide $$5{x^4} - 4{x^3} + 3{x^2} - 2x + 1$$ by $$x - 1$$
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# USING TABLES TO COMPARE RATIOS
## About "Using tables to compare ratios"
Using tables to compare ratios :
When two ratios are given and if want to compare them, we will be converting them into like fractions (having same denominator) and compare numerator.
To make the above method easier, we use tables to compare the given two ratios.
Let us see how to use tables to compare the given two ratios with some examples.
## Using tables to compare ratios
Example 1 :
Anna’s recipe for lemonade calls for 2 cups of lemonade concentrate and 3 cups of water. Bailey’s recipe calls for 3 cups of lemonade concentrate and 5 cups of water. Whose recipe makes stronger lemonade? How do you know ?
Solution :
Anna’s recipe :
2 cups of lemonade and 3 cups of water.
So, the ratio is 2 : 3.
Let us write equivalent ratios to the ratio 2 : 3.
Bailey’s recipe :
3 cups of lemonade and 5 cups of water.
So, the ratio is 3 : 5.
Let us write equivalent ratios to the ratio 3 : 5.
Find two columns, one in each table, in which the amount of water is the same. Circle those two columns.
From the circled columns, we get two ratios.
They are,
10 : 15 and 9 : 15
In these two ratios, the second quantity (water) is same.
So, we have to compare the first quantity (Lemonade).
The first quantity (10) in the first ratio is more than the first quantity (9) in the second ratio.
When the quantity of water is same (15) in both recipes, Anna's recipe has more quantity of lemonade concentrate.
Therefore, Anna's recipe has stronger lemonade.
Example 2 :
Find which ratio is greater :
2 1/3 : 3 1/3 and 3.6 : 4.8
Solution :
If we want to compare two ratios using table, both the terms of the ratio must be integers.
Let us convert the terms of the first ratio into integers.
2 1/3 : 3 1/3 = (7/3) : (10/3)
2 1/3 : 3 1/3 = 7 : 10 ------> multiplied by 3
Let us write equivalent ratios to the ratio 7 : 10
Let us convert the terms of the second ratio into integers.
3.6 : 4.8 = 36 : 48 ------> multiplied by 10
3.6 : 4.8 = 3 : 4 ------> divided by 12
Let us write equivalent ratios to the ratio 3 : 4
Find two columns, one in each table, in which the the second term is same. Circle those two columns.
From the circled columns, we get two ratios.
They are,
14 : 20 and 15 : 20
In these two ratios, the second term is same.
So, we have to compare the first terms 14 and 15.
The first term in the second ratio (15) is more than the first term in the first ratio (14).
Therefore, the second ratio is greater than the first ratio.
That is, 3.6 : 4.8 is greater than 2 1/3 : 3 1/3.
After having gone through the stuff given above, we hope that the students would have understood "Using tables to compare ratios".
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## Book: RD Sharma - Mathematics (Volume 2)
### Chapter: 21. Areas of Bounded Regions
#### Subject: Maths - Class 12th
##### Q. No. 25 of Exercise 21.1
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
25
##### Compare the areas under the curves y = cos2x and y = sin2 x between x = 0 and x = π.
Given equations are:
y = cos2x …..(i)
y = sin2x …..(ii)
x = 0 ……(iii)
x = …..(iv)
A table for values of y = cos2x and y = sin2x is: -
A rough sketch of the curves is given below: -
The area under the curve y = cos2x, x = 0 and x = is
A1 = (area of the region OABCO + area of the region CEFGC)
A1 = 2(area of the region CEFGC)
(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)
(As x is between and the value of y varies)
Apply reduction formula:
On integrating we get,
On applying the limits we get
The area under the curve y = cos2x, x = 0 and x = is
A2 = (area of the region OBDGEO)
(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)
(As x is between and the value of y varies)
Apply reduction formula:
On integrating we get,
On applying the limits we get
Hence A1 = A2
Therefore the areas under the curves y = cos2x and y = sin2 x between x = 0 and x = π are equal.
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Rd Sharma 2019 2020 Solutions for Class 7 Math Chapter 4 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 7 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 7 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.
#### Question 1:
Determine whether the following rational numbers are in the lowest form or not:
(i) $\frac{65}{84}$
(ii) $\frac{-15}{32}$
(iii) $\frac{24}{128}$
(iv) $\frac{-56}{-32}$
#### Answer:
(i) We observe that 65 and 84 have no common factor i..e., their HCF is 1.
Thus, $\frac{65}{84}$ is in its lowest form.
(ii) We observe that $-$15 and 32 have no common factor i..e., their HCF is 1.
Thus, $\frac{-15}{32}$is in its lowest form.
(iii) HCF of 24 and 128 is not 1.
Thus, given rational number is not in its simplest form.
(iv) HCF of 56 and 32 is 8.
Thus, given rational number is not in its simplest form.
#### Question 2:
Express each of the following rational numbers to the lowest form:
(i) $\frac{4}{22}$
(ii) $\frac{-36}{180}$
(iii) $\frac{132}{-428}$
(iv) $\frac{-32}{-56}$
Lowest form of:
#### Question 3:
Fill in the blanks:
(i) $\frac{-5}{7}=\frac{\dots }{35}=\frac{\dots }{49}$
(ii)
(iii)
(iv)
#### Question 1:
Write each of the following rational numbers in the standard form:
(i) $\frac{2}{10}$
(ii) $\frac{-8}{36}$
(iii) $\frac{4}{-16}$
(iv) $\frac{-15}{-35}$
(v) $\frac{299}{-161}$
(vi) $\frac{-63}{-210}$
(vii) $\frac{68}{-119}$
(viii) $\frac{-195}{275}$
#### Answer:
(i) The denominator is positive and HCF of 2 and 10 is 2.
$\therefore$ Dividing the numerator and denominator by 2, we get:
(ii) The denominator is positive and HCF of 8 and 36 is 4.
$\therefore$ Dividing the numerator and denominator by 4, we get:
(iii)
The denominator is negative.
HCF of 4 and 16 is 4.
$\therefore$ Dividing the numerator and denominator by 4, we get:
(iv) The denominator is negative.
HCF of 15 and 35 is 5.
$\therefore$ Dividing the numerator and denominator by 5, we get:
(v) The denominator is negative.
HCF of 299 and 161 is 23.
$\therefore$ Dividing the numerator and denominator by 23, we get:
(vi) The denominator is negative.
HCF of 63 and 210 is 21.
$\therefore$ Dividing the numerator and denominator by 21, we get:
(vii) The denominator is negative.
HCF of 68 and 119 is 17.
$\therefore$ Dividing the numerator and denominator by 17, we get:
(viii) The denominator is positive and HCF of 195 and 275 is 5.
$\therefore$ Dividing divide the numerator and denominator by 5, we get:
#### Question 1:
Which of the following rational numbers are equal?
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv)
#### Question 2:
If each of the following pairs represents a pair of equivalent rational numbers, find the values of x:
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv)
#### Question 3:
In each of the following, fill in the blanks so as to make the statement true:
(i) A number which can be expressed in the form $\frac{p}{q}$, where p and q are integers and q is not equal to zero, is called a .....
(ii) If the integers p and q have no common divisor other than 1 and q is positive, then the rational number $\frac{p}{q}$ is said to be in the ....
(iii) Two rational numbers are said to be equal, if they have the same .... form.
$\frac{a}{b}=\frac{a÷m}{....}$
(v) If p and q are positive integers, then $\frac{p}{q}$ is a ..... rational number and $\frac{p}{-q}$ is a ..... rational number.
(vi) The standard form of −1 is ...
(vii) If $\frac{p}{q}$ is a rational number, then q cannot be ....
(viii) Two rational numbers with different numerators are equal, if their numerators are in the same .... as their denominators.
#### Answer:
(i) rational number
(ii) standard rational number
(iii) standard form
(iv) $\frac{a}{b}=\frac{a÷m}{b÷m}$
(v) positive rational number, negative rational number
(vi) $\frac{-1}{1}$
(vii) zero
(viii) ratio
#### Question 4:
In each of the following state if the statement is true (T) or false (F):
(i) The quotient of two integers is always an integer.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every fraction is a rational number.
(v) Every rational number is a fraction
(vi) If $\frac{a}{b}$ is a rational number and m any integer, then $\frac{a}{b}=\frac{a×m}{b×m}$
(vii) Two rational numbers with different numerators cannot be equal.
(viii) 8 can be written as a rational number with any integer as denominator.
(ix) 8 can be written as a rational number with any integer as numerator.
(x)
#### Answer:
(i) False; not necessary
(ii) True; every integer can be expressed in the form of p/q, where q is not zero.
(iii) False; not necessary
(iv) True; every fraction can be expressed in the form of p/q, where q is not zero.
(v) False; not necessary
(vi) True
(vii) False; they can be equal, when simplified further.
(viii) False
(ix) False
(x) True; in the standard form, they are equal.
#### Question 1:
Draw the number line and represent the following rational numbers on it:
(i) $\frac{2}{3}$
(ii) $\frac{3}{4}$
(iii) $\frac{3}{8}$
(vi) $\frac{-5}{8}$
(v) $\frac{-3}{16}$
(vi) $\frac{-7}{3}$
(vii) $\frac{22}{-7}$
(viii) $\frac{-31}{3}$
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
#### Question 2:
Which of the two rational numbers in each of the following pairs of rational numbers is greater?
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
#### Answer:
(i) We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus,
$\frac{-3}{8}>0$
(ii) $\frac{5}{2}>0.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iii) $\frac{-4}{8}<\frac{3}{11}.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iv)
(v)
(vi)
(vii)
(viii)
#### Question 3:
Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
(i)
(ii)
(iii)
(iv)
#### Answer:
(i) $\frac{-6}{-13}=\frac{6}{13}<\frac{7}{13}$
(ii) $\frac{16}{-5}<3$
(iii)
(iv)
#### Question 4:
Fill in the blanks by the correct symbol out of >, =, or <:
(i) $\frac{-6}{7}.....\frac{7}{13}$
(ii) $\frac{-3}{5}.....\frac{-5}{6}$
(iii) $\frac{-2}{3}.....\frac{5}{-8}$
(iv)
#### Question 5:
Arrange the following rational numbers in ascending order:
(i)
(ii)
#### Answer:
(i) Ascending order:
(ii)
#### Question 6:
Arrange the following rational numbers in descending order:
(i)
(ii)
#### Answer:
We have to arrange them in descending order.
(i)
(ii)
#### Question 7:
Which of the following statements are true:
(i) The rational number $\frac{29}{23}$ lies to the left of zero on the number line.
(ii) The rational number $\frac{-12}{-17}$ lies to the left of zero on the number line.
(iii) The rational number $\frac{3}{4}$ lies to the right of zero on the number line.
(iv) The rational numbers are on the opposite side of zero on the number line.
(v) The rational numbers are on the opposite side of zero on the number line.
(vi) The rational number $\frac{-3}{-5}$ is one the right of $\frac{-4}{7}$ on the number line.
#### Answer:
(i) False; it lies to the right of zero because it is a positive number.
(ii) False; it lies to the right of zero because it is a positive number.
(iii) True
(iv) True; they are of opposite signs.
(v) False; they both are of same signs.
(v) True; they both are of opposite signs and positive number is greater than the negative number. Thus, it is on the right of the negative number.
#### Question 1:
Mark the correct alternative in each of the following:
$\frac{44}{-77}$ in standard form is
(a) $\frac{4}{-7}$ (b) $-\frac{4}{7}$ (c) $-\frac{44}{77}$ (d) None of these
#### Answer:
The denominator of $\frac{44}{-77}$ is negative.
Firstly, multiply the numerator and denominator by −1 to make it positive.
$\frac{44}{-77}=\frac{44×\left(-1\right)}{-77×\left(-1\right)}=\frac{-44}{77}$
Now,
HCF of 44 and 77 = 11
Dividing the numerator and denominator of $\frac{-44}{77}$ by 11, we have
$\frac{-44}{77}=\frac{-44÷11}{77÷11}=\frac{-4}{7}=-\frac{4}{7}$
Thus, the standard form of $\frac{44}{-77}$ is $-\frac{4}{7}$.
Hence, the correct answer is option (b).
#### Question 2:
Mark the correct alternative in each of the following:
$-\frac{102}{119}$ in standard form is
(a) $-\frac{6}{7}$ (b) $\frac{6}{7}$ (c) $-\frac{6}{17}$ (d) None of these
#### Answer:
The denominator of the rational number $-\frac{102}{119}$ is positive.
In order to write the rational number in standard form, divide its numerator and denominator by the HCF of 102 and 119.
HCF of 102 and 119 = 17
Dividing the numerator and denominator of $-\frac{102}{119}$ by 17, we have
$-\frac{102}{119}=-\frac{102÷17}{119÷17}=-\frac{6}{7}$
Thus, the standard form of $-\frac{102}{119}$ is $-\frac{6}{7}$.
Hence, the correct answer is option (a).
#### Question 3:
Mark the correct alternative in each of the following:
A rational number equal to $\frac{-2}{3}$ is
(a) $\frac{-10}{25}$ (b) $\frac{10}{-15}$ (c) $\frac{-9}{6}$ (d) None of these
#### Answer:
We know that two rational numbers are equal if they have the same standard form.
The rational number $\frac{-2}{3}$ is in its standard form.
Consider the rational number $\frac{10}{-15}$.
This rational number can be expressed in standard form as follows:
$\frac{10}{-15}=\frac{10×\left(-1\right)}{-15×\left(-1\right)}=\frac{-10}{15}$ (Multiplying numerator and denominator by −1 to make denominator positive)
HCF of 10 and 15 = 5
Dividing the numerator and denominator of $\frac{-10}{15}$ by 5, we have
$\frac{-10}{15}=\frac{-10÷5}{15÷5}=\frac{-2}{3}$
Thus, the standard form of $\frac{-10}{15}$ is $\frac{-2}{3}$, which is same as the given rational number.
So, the rational number equal to $\frac{-2}{3}$ is $\frac{10}{-15}$.
Let us check why options (a) and (c) are not correct.
The standard form of $\frac{-10}{25}$ is $\frac{-2}{5}$.
HCF of 10 and 25 = 5
Dividing the numerator and denominator of $\frac{-10}{25}$ by 5, we have
$\frac{-10}{25}=\frac{-10÷5}{25÷5}=\frac{-2}{5}$
The standard form of $\frac{-9}{6}$ is $\frac{-3}{2}$.
HCF of 6 and 9 = 3
Dividing the numerator and denominator of $\frac{-9}{6}$ by 3, we have
$\frac{-9}{6}=\frac{-9÷3}{6÷2}=\frac{-3}{2}$
Hence, the correct answer is option (b).
#### Question 4:
Mark the correct alternative in each of the following:
If $\frac{-3}{7}=\frac{x}{35}$, then x =
(a) 15 (b) 21 (c) −15 (d) −21
#### Answer:
Firstly, write $\frac{-3}{7}$ as a rational number with denominator 35.
Multiplying the numerator and denominator of $\frac{-3}{7}$ by 5, we have
$\frac{-3}{7}=\frac{-3×5}{7×5}=\frac{-15}{35}$
$\therefore \frac{-3}{7}=\frac{x}{35}\phantom{\rule{0ex}{0ex}}⇒\frac{-15}{35}=\frac{x}{35}\phantom{\rule{0ex}{0ex}}⇒x=-15$
Hence, the correct answer is option (c).
#### Question 5:
Mark the correct alternative in each of the following:
Which of the following is correct?
(a) $\frac{5}{9}>\frac{-3}{-8}$ (b) $\frac{5}{9}<\frac{-3}{-8}$ (c) $\frac{2}{-3}<\frac{-8}{7}$ (d) $\frac{4}{-3}>\frac{-8}{7}$
#### Answer:
Consider the rational numbers $\frac{5}{9}$ and $\frac{-3}{-8}$.
We write the rational number $\frac{-3}{-8}$ with positive denominator.
$\frac{-3}{-8}=\frac{-3×\left(-1\right)}{-8×\left(-1\right)}=\frac{3}{8}$
Now, we write the rational numbers so that they have a common denominator.
LCM of 8 and 9 = 72
So, $\frac{5}{9}=\frac{5×8}{9×8}=\frac{40}{72}$ and $\frac{3}{8}=\frac{3×9}{8×9}=\frac{27}{72}$
Now,
$40>27\phantom{\rule{0ex}{0ex}}⇒\frac{40}{72}>\frac{27}{72}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{9}>\frac{3}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{9}>\frac{-3}{-8}$
It can also be checked that $\frac{2}{-3}>\frac{-8}{7}$ and $\frac{4}{-3}<\frac{-8}{7}$.
Hence, the correct answer is option (a).
#### Question 6:
Mark the correct alternative in each of the following:
If the rational numbers $\frac{-2}{3}$ and $\frac{4}{x}$ represent a pair of equivalent rational numbers, then x =
(a) 6 (b) −6 (c) 3 (d) −3
#### Answer:
It is given that the rational numbers $\frac{-2}{3}$ and $\frac{4}{x}$ represent a pair of equivalent rational numbers.
We know that the values of two equivalent rational numbers is equal.
Hence, the correct answer is option (b).
#### Question 7:
Mark the correct alternative in each of the following:
What is the additive identity element in the set of whole numbers?
(a) 0 (b) 1 (c) −1 (d) None of these
#### Answer:
If a is a whole number then a + 0 = a = 0 + a.
Therefore, 0 is the additive identity element for addition of whole number because it does not change the identity or value of the whole number during the operation of addition.
Hence, the correct answer is option (a).
#### Question 8:
Mark the correct alternative in each of the following:
What is the multiplicative identity element in the set of whole numbers?
(a) 0 (b) 1 (c) −1 (d) None of these
#### Answer:
We know that if a is a whole number, then a × 1 = a = 1 × a.
Therefore, 1 is the multiplicative identity element for multiplication of whole numbers because it does not change the identity or value of the whole number during the operation of multiplication.
Hence, the correct answer is option (b).
#### Question 9:
Mark the correct alternative in each of the following:
Which of the following is not zero?
(a) 0 × 0 (b) $\frac{0}{3}$ (c) $\frac{7-7}{3}$ (d) 9 ÷ 0
#### Answer:
If any number is multiplied by 0, the product is 0.
∴ 0 × 0 = 0
If 0 is divided by any number (≠ 0), the quotient is always 0.
∴ $\frac{0}{3}=0$ and $\frac{7-7}{3}=\frac{0}{3}=0$
Division of any number by 0 is meaningless and is not defined.
∴ 9 ÷ 0 is not defined.
Hence, the correct answer is option (d).
#### Question 10:
Mark the correct alternative in each of the following:
The whole number nearest to 457 and divisible by 11 is
(a) 450 (b) 451 (c) 460 (d) 462
#### Answer:
The numbers 450 and 460 are not divisible by 11.
Now, both the numbers 451 and 462 are divisible by 11.
Distance between 457 and 451 on the number line = 457 − 451 = 6
Distance between 457 and 462 on the number line = 462 − 457 = 5
Thus, the whole number nearest to 457 and divisible by 11 is 462.
Hence, the correct answer is option (d).
#### Question 11:
Mark the correct alternative in each of the following:
If $-\frac{3}{8}$ and $\frac{x}{-24}$ are equivalent rational numbers, then x =
(a) 3 (b) 6 (c) 9 (d) 12
#### Answer:
It is given that the rational numbers $-\frac{3}{8}$ and $\frac{x}{-24}$ are equivalent rational numbers.
We know that the values of two equivalent rational numbers is equal.
Hence, the correct answer is option (c).
#### Question 12:
Mark the correct alternative in each of the following:
If $\frac{27}{-45}$ is expressed as a rational number with denominator 5, then the numerator is
(a) 3 (b) −3 (c) 6 (d) −6
#### Answer:
In order to express $\frac{27}{-45}$ as a rational number with denominator 5, firstly find a number which gives 5 when −45 is divided by it.
This number is −45 ÷ 5 = −9.
Dividing the numerator and denominator of $\frac{27}{-45}$ by −9, we have
$\frac{27}{-45}=\frac{27÷\left(-9\right)}{-45÷\left(-9\right)}=\frac{-3}{5}$
Thus, the numerator is −3.
Hence, the correct answer is option (b).
#### Question 13:
Mark the correct alternative in each of the following:
Which of the following pairs of rational numbers are on the opposite side of the zero on the number line?
(a) $\frac{3}{7}$ and $\frac{5}{12}$ (b) $-\frac{3}{7}$ and $\frac{-5}{12}$ (c) $\frac{3}{7}$ and $\frac{-5}{12}$ (d) None of these
#### Answer:
The rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are positive rational numbers. We know that every positive rational number is greater than 0, so both the rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are represented by points on the right of the zero on the number line.
The rational numbers $-\frac{3}{7}$ and $\frac{-5}{12}$ are negative rational numbers. We know that every negative rational number is less than 0, so both the rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are represented by points on the left of the zero on the number line.
The rational numbers $\frac{3}{7}$ is a positive rational number whereas the rational number $\frac{-5}{12}$ is a negative rational numbers. We know that every negative rational number is less than 0 and every positive rational number is greater than 0, so the rational number $\frac{3}{7}$ is represented by point on the right of the zero and $\frac{-5}{12}$ is represented by point on the left of the zero on the number line.
Thus, the rational numbers $\frac{3}{7}$ and $\frac{-5}{12}$ are on the opposite side of the zero on the number line.
Hence, the correct answer is option (c).
#### Question 14:
Mark the correct alternative in each of the following:
The rational number equal to $\frac{2}{-3}$ is
(a) $\frac{14}{-18}$ (b) $\frac{-6}{9}$ (c) $\frac{-8}{-12}$ (d) $\frac{3}{-2}$
#### Answer:
We know that two rational numbers are equal if they have the same standard form.
$\frac{2}{-3}=\frac{2×\left(-1\right)}{-3×\left(-1\right)}=\frac{-2}{3}$
The standard form of $\frac{2}{-3}$ is $\frac{-2}{3}$.
Consider the rational number $\frac{-6}{9}$.
HCF of 6 and 9 = 3
Dividing the numerator and denominator of $\frac{-6}{9}$ by 3, we have
$\frac{-6}{9}=\frac{-6÷3}{9÷3}=\frac{-2}{3}$
Thus, the standard form of $\frac{-6}{9}$ is $\frac{-2}{3}$.
So, the rational number $\frac{-6}{9}$ is equal to $\frac{2}{-3}$.
It can be checked that
Standard form of $\frac{14}{-18}$ = $\frac{-7}{9}$
Standard form of $\frac{-8}{-12}$ = $\frac{2}{3}$
Standard form of $\frac{3}{-2}$ = $\frac{-3}{2}$
Hence, the correct answer is option (b).
#### Question 15:
Mark the correct alternative in each of the following:
If $-\frac{3}{4}=\frac{6}{x}$, then x =
(a) −8 (b) 4 (c) −4 (d) 8
#### Answer:
Hence, the correct answer is option (a).
#### Question 1:
Write down the numerator of each of the following rational numbers:
(i) $\frac{-7}{5}$
(ii) $\frac{15}{-4}$
(iii) $\frac{-17}{-21}$
(iv) $\frac{8}{9}$
(v) 5
#### Answer:
Numerators are:
(i) $-$7
(ii) 15
(iii) $-$17
(iv) 8
(v) 5
#### Question 2:
Write down the denominator of each of the following rational numbers:
(i) $\frac{-4}{5}$
(ii) $\frac{11}{-34}$
(iii) $\frac{-15}{-82}$
(iv) 15
(v) 0
#### Answer:
Denominators are:
(i) 5
(ii) $-$34
(iii) $-$82
(iv) 1
(v) 1
#### Question 3:
Write down the rational number whose numerator is (−3) × 4, and whose denominator is (34 − 23) × (7 − 4).
#### Answer:
According to the question:
Numerator = ($-$3) × 4 = $-$12
Denominator = (34 $-$ 23) × (7 $-$ ​4) = 11 × 3 = 33
∴ Rational number = $\frac{-12}{33}$
#### Question 4:
Write the following rational numbers as integers:
#### Answer:
Integers are 7, $-$12, 34, $-$73 and 95.
#### Question 5:
Write the following integers as rational numbers with denominator 1:
#### Answer:
Rational numbers of given integers with denominator 1 are:
#### Question 6:
Write down the rational number whose numerator is the smallest three digit number and denominator is the largest four digit number.
#### Answer:
Smallest three-digit number = 100
Largest four-digit number = 9999
∴ Required rational number = $\frac{100}{9999}$
#### Question 7:
Separate positive and negative rational numbers from the following rational numbers:
#### Answer:
Given rational numbers can be rewritten as:
Thus, positive rational numbers are:
or,
Negative rational numbers are:
or,
#### Question 8:
Which of the following rational numbers are positive:
(i) $\frac{-8}{7}$
(ii) $\frac{9}{8}$
(iii) $\frac{-19}{-13}$
(iv) $\frac{-21}{13}$
#### Answer:
The numbers can be rewritten as:
Positive rational numbers are (ii) and (iii), i.e., .
#### Question 9:
Which of the following rational numbers are negative?
(i) $\frac{-3}{7}$
(ii) $\frac{-5}{-8}$
(iii) $\frac{9}{-83}$
(iv) $\frac{-115}{-197}$
#### Answer:
The numbers can be rewritten as:
Negative rational numbers are (i) and (iii).
#### Question 1:
Express each of the following as a rational number with positive denominator:
(i) $\frac{-15}{-28}$
(ii) $\frac{6}{-9}$
(iii) $\frac{-28}{-11}$
(iv) $\frac{19}{-7}$
#### Answer:
Rational number with positive denominators:
(i) Multiplying the number by $-$1, we get:
$\frac{-15}{-28}=\frac{-15×-1}{-28×-1}=\frac{15}{28}$
(ii) Multiplying the number by $-$1, we get:
$\frac{6}{-9}=\frac{6×-1}{-9×-1}=\frac{-6}{9}$
(iii) Multiplying the number by $-$1, we get:
$\frac{-28}{-11}=\frac{-28×-1}{-11×-1}=\frac{28}{11}$
(iv) Multiplying the number by $-$1, we get:
$\frac{19}{-7}=\frac{19×-1}{-7×-1}=\frac{-19}{7}$
#### Question 2:
Express $\frac{3}{5}$ as a rational number with numerator:
(i) 6
(ii) −15
(iii) 21
(iv) −27
#### Answer:
Rational number with numerator:
(i) 6 is:
(ii)
(iii)
(iv)
#### Question 3:
Express $\frac{5}{7}$ as a rational number with denominator:
(i) −14
(ii) 70
(iii) −28
(iv) −84
#### Answer:
$\frac{5}{7}$ as a rational number with denominator:
(i) −14 is:
(ii) 70 is:
(iii) −28 is:
(iv) −84 is:
#### Question 4:
Express $\frac{3}{4}$ as a rational number with denominator:
(i) 20
(ii) 36
(iii) 44
(iv) −80
#### Answer:
3/4 as rational number with denominator:
(i)
(ii)
(iii)
(iv)
#### Question 5:
Express $\frac{2}{5}$ as a rational number with numerator:
(i) −56
(ii) 154
(iii) −750
(iv) 500
#### Answer:
2/5 as a rational number with numerator:
(i)
(ii)
(iii)
(iv)
#### Question 6:
Express $\frac{-192}{108}$ as a rational number with numerator:
(i) 64
(ii) −16
(iii) 32
(iv) −48
#### Answer:
Rational number with numerator:
#### Question 7:
Express $\frac{168}{-294}$ as a rational number with denominator:
(i) 14
(ii) −7
(iii) −49
(iv) 1470
#### Answer:
Rational number with denominator:
#### Question 8:
Write $\frac{-14}{42}$ in a form so that the numerator is equal to:
(i) −2
(ii) 7
(iii) 42
(iv) −70
#### Answer:
Rational number with numerator:
#### Question 9:
Select those rational numbers which can be written as a rational number with numerator 6:
#### Answer:
Given rational numbers that can be written as a rational number with numerator 6 are:
#### Question 10:
Select those rational numbers which can be written as a rational number with denominator 4:
#### Answer:
Given rational numbers that can be written as a rational number with denominator 4 are:
122 (On multiplying by 6) = 613223 (On multiplying by 3) = 6934 (On multiplying by 2) = 6867 (On multiplying by 1) = 67
#### Question 11:
In each of the following, find an equivalent form of the rational number having a common denominator:
(i)
(ii)
(iii)
#### Answer:
Equivalent forms of the rational number having common denominator are:
(i) .
(ii)
(iii)
View NCERT Solutions for all chapters of Class 7
|
Education Technology
# Activities
• ##### Subject Area
• Math: Algebra I: Quadratic Functions
9-12
90 Minutes
• ##### Device
• TI-83 Plus Family
• TI-84 Plus
• TI-84 Plus Silver Edition
TI Connect™
• ##### Accessories
TI Connectivity Cable
• ##### Other Materials
This is Activity 3 from the EXPLORATIONS Book:
A Hands-On Look at Algebra Functions: Activities for Transformation Graphing .
## Exploring the Vertex Form of the Quadratic Function (Algebra Application)
#### Activity Overview
Students explore the vertex form of the parabola and discover how the vertex, direction, and width of the parabola can be determined by studying the parameters. They predict the location of the vertex of a parabola expressed in vertex form.
#### Before the Activity
• Set up the Transformation Graphing Application on the calculator using the TI Connect™ software
• See the attached PDF file for detailed instructions for this activity
• Print pages 20 - 26 from the attached PDF file for your class
• #### During the Activity
Distribute the pages to your class.
• Graph a quadratic equation y = (x - 2)2 + 1 on the calculator and record its vertex and the direction in which the curve opens
• Use the Transformation Graphing application and enter the general vertex form of a quadratic equation Y = A(X - B)2 + C
• Identify the relationship between the values of A, B, and C (the coefficients) and the vertex and magnitude of the graph
• Value of B gives the x-coordinate of the vertex;
for the equation Y = (X - 3)2, B = 3 and the vertex is at X = 3;
for the equation Y = (X + 1)2, B = -1 and the vertex at X = -1
• Changes in C create a vertical translation of the curve; when C increases the curve moves up; when C decreases the curve moves down; Value of C is the y-coordinate of the vertex
• Value of A determines the direction of the parabola and its width; larger the magnitude of A, the narrower the curve; smaller the magnitude of A, the wider the curve; a positive sign means the parabola is opening up; a negative sign means the parabola is opening down
• Observe the direction in which the graph opens and determine the maximum/minimum values
• #### After the Activity
Students will complete the Student Worksheet and answer questions listed on it.
• Review student results
• As a class, discuss questions that appeared to be more challenging
• Re-teach concepts as necessary
|
# Circles III - The A- Z of Tangent Formula to Solve Circle Questions
## In this post, we will discuss all the tangent formulas that will help you solve circle questions and save time in SSC CGL Exam.
Questions from geometry section, especially circle questions always appear in SSC Exams. These Circle Questions can be easily solved with the help of formulas.
This is the third post in the series of Circles. In this post, we will discuss tangent formulas that will help you solve circle questions easily and thus help you save time during SSC Exams.
Before moving further, let’s quickly review the concept of circles.
### What Is A Tangent?
A tangent is a line that touches the circle exactly at one point on the circumference of a circle.
### What Is A Secant?
Secant is a line that touches the circumference or cuts the circumference of the circle at two points.
List of Tangent Formulas to Solve Circle Questions
### Tangent Formula 1:
The angle between the tangent and the radius of the circle at the point of contact is 90 degrees.
### Tangent Formula 2:
This property is based on the Length of a tangent:
1. The length of two tangents from an external point is equal in length.
2. Tangent Formula to measure the length of a chord
Let's take the distance between the centre and the point P as ‘d’ At the point of contact, where the radius and tangent meet 90 degree - right angle is formed. Now we get two right angled triangles.
OAP and OBP are right angled triangles. Hence, by using Pythagoras theorem, we know that OA is the radius, OP is the distance and AP is the hypotenuse. Therefore, hypotenuse AP which is the tangent can be found by:
√d² - r²
### Tangent Formula 3:
This tangent formula is based on a tangent and a secant.
1.There is a circle, from external point P we draw a tangent and a secant.
PT is the length of the tangents and TBA is the length of the secant. Therefore,
PT² = PA x PB
Hence, When we draw a tangent and secant from an external point then, the length of the tangent square will be equal to PA and PB
2. When we draw two secant lines, suppose PAB and PCD then,
PA x PB = PC x PD
Have a look at chord formulas too, to have a firm grip over circle questions.
Hope these tangent formulas have been of help to you while preparing for SSC CGL 2017 as circle questions often appear in these competitive exams.
Do write in the comment section below on how these Tangent Formulas have helped you solve circle questions asked in mock tests for SSC CGL. Also, download free questionnaire for practice!
|
# Class 9 – Co-Ordinate Geometry
Take practice tests in Co-Ordinate Geometry
## Online Tests
Topic Sub Topic Online Practice Test
Co-Ordinate Geometry
• Distance formula
• Section formula
Take Test See More Questions
Co-Ordinate Geometry
• Slope and intersection of two lines
• Equation of a line in various forms
• Area of a triangle
Take Test See More Questions
## Study Material
Introduction: Analytical geometry was invented by the French philosopher Rene Descartes (1596-1650) and the year of invention is generally set as 1637, when he published his La Geometrica, which provided a new tool for unifying the two branches of mathematics, algebra and geometry. Prior to this, mathematicians confined themselves to Euclidean geometry and did not know how to apply algebra advantageously to the study of geometrical relationships. Co-ordinate geometry: The study of that branch of Mathematics which deals with the interrelationship between geometrical and algebraic concepts is called Co-ordinate geometry (or) Cartesian geometry in honour of the French mathematician Rene Descartes. Co-ordinate geometry is that branch of geometry in which two numbers, called coordinates. These are used to indicate the position of a point in a plane and which makes use of algebraic methods in the study of geometric figures. Co-ordinate axes: In graphs, two number lines at right angles to each other are used as reference lines. These lines are called axes. The direction to the right of the vertical axis is denoted by a positive sign, and to the left by a negative sign, while direction upward from the horizontal axis is positive and downward is negative. The horizontal line is usually termed as the x-axis and the vertical as the y-axis. Cartesian Co-ordinates of a point: Let and be the co-ordinate’s axes, and let P be any point in the plane. Draw perpendiculars PM and PN on x and y-axis respectively. The length of the directed line segment OM in the units of scale chosen is called the x-coordinate (or) abscissa of point P. Similarly, the length of the directed line segment ON on the same scale is called the y-coordinate (or) ordinate of point P.
Let OM = x and ON = y. Then the position of the point P is the plane with respect to the co-ordinate axes is represented by the ordered pair .The ordered pair is called the coordinates of point P.
i) is called x – axis.
is positive x - axis and is negative x – axis.
ii) is called y – axis
is positive y – axis and is negative y – axis
Abscissa: The perpendicular distance of any pointfrom the y-axis is called the abscissa of the point.
Ordinate: The perpendicular distance of any pointfrom the x-axis is called the ordinate of the point.
Co-ordinates: The abscissa and ordinate of a pointare together called its co-ordinates.
Example: The co-ordinates of the point A in the figure above are (4, 5), where 4 is abscissa and 5 is ordinate and it is written as A (4, 5). Similarly, the other points are
B (3, 2), C (-4, 7), D (-3, -4).
Quadrants: The two axes namely x and y axes divide the plane into four equal parts called quadrants, numbered I, II, III and IV as shown in the figure.
Origin: The intersection of the two axes is called the origin. It is usually denoted by O. The co-ordinates of the originare (0, 0).
I Quadrant (Q1): x >0, y > 0 II Quadrant (Q2): x <0, y > 0
III Quadrant (Q3): x <0, y < 0 IV Quadrant (Q4): x >0, y < 0
The coordinates of the origin are taken as (0, 0). The coordinates of any point on x-axis are of the form (x, 0) and the coordinates of any point on y-axis are of the form (0, y). Thus, if the abscissa of a point is zero, it would lie some where on the y-axis and if its ordinate is zero it would lie on x-axis.
It follows from the above discussion that by simply looking at the coordinates of a point we can tell in which quadrant it would lie.
Example:.
Distance formula: The distance d between the points and is given by the formula
Proof: Let and be the two points. From P, Q draw PL,QM perpendiculars on the x-axis and PR perpendicular on MQ.
Then,
units
Distance from origin: Let O (0, 0) are two points then
Collinearity: If A, B, C are three points, then
i) are collinear.
ii) are collinear.
iii) are collinear.
iv) are collinear.
Example: Show that the points A(1,1), B(-2,7)and C(3,-3) are collinear .
AB =
BC =
AC =
Clearly, BC = AB + AC
Hence C, A, B are collinear.
Important points:
a) Let A, B, C are three non-collinear points:
i) ABC forms a scalene triangle if any two sides are not equal .
ii) ABC forms an isosceles triangle if any two sides are equal.
iii) ABC forms an equilateral triangle i fall the three sides are equal .
iv) ABC forms a right angle triangle if square of longest side is equal to sum of squares of other two sides.
v) ABC forms an acute angled triangle if (or) (or) .
vi) ABC forms an obtuse angled triangle if (or) (or).
vii) ABC forms an isosceles right angled triangle if any two sides are equal and square of the unequal side = sum of the squares of equal sides.
Example: Find the distance between the following pairs of points:
i) A(14, 3) andB(10, 6) ii) M(-1, 2) and N(0, -6)
Solution:
i) Distance between two points units
ii) Here
Midpoint formula: Let P, Q be the points with co-ordinates respectively and (x, y) is there quired coordinates of M, which is the mid-point of PQ. Draw PD and QF and ME perpendiculars to OX. Through M, draw NML parallel to OX and meet DP producedat N and QF at L.
Then, from congruent triangles PMN and QML, we get
NM = ML DE= EF
OE– OD = OF – OE
Again, from the same congruent triangles, we get
PN = LQ DN- DP = FQ – FL
EM EM
Hence, the coordinates of the mid-point of the join of P and Q is
Example: Find the mid-point of the following:
i) A (2, 3) ,B (-3, 5) ii) P (-5, -3) , Q (3, -4)
Solution:
i) A (2, 3) B (-3, 5)
Mid-point of
Mid-point of.
ii) P (-5,-3) Q (3, -4)
Mid-point of
Mid-point of
Important points:
Let A, B, C, D are four non-collinear points taken in order:
i) ABCD forms a Parallelogram, if AB = CD;AD = BC and AC ≠ BD. ( Opposite sides are equal and diagonal sare not equal).
ii) ABCD forms a Rectangle, if AB = CD; AD =BC and AC = BD. ( Opposite sides are equal and diagonal sare equal).
iii) ABCD forms a Rhombus, if AB = BC = CD =AD and AC ≠ BD.
( All sides are equal and diagonals are not equal).
iv) To prove ABCD forms a Square, AB = BC =CD = AD and AC = BD.
( All sides are equal and diagonals are equal).
1. Show that the pointsP (–1, –1), Q (2, 3) and R (–2, 6) are the vertices of a right – angled triangle.
Solution:
2. Show that the points A (1, 2), B (4, 5) and C (-1, 0) lie on a straight line.
Solution: Here,
i.e.BA + AC = BC
Hence, B, A, C lie on a straight line. In other words B, A, C are collinear.
3. Prove that the points (2a, 4a),(2a, 6a) and are the vertices of an equilateral triangle whose side is 2a.
Solution:Let the points be A(2a,4a), B(2a, 6a) and C
4. Show that four points (0,-1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle.
Solution: Let A (0,-1), B (6, 7),C (-2, 3) and D (8, 3) be the given points.
units
BC units
units
AC = units
BD =units
By observation AB = CD, BC = AD and AC = BD (Opposite sides and diagonals are equal).
ABCDis a rectangle.
Find the mid-points of i) A (5, 0) , B (9, 0) ii) P(0, -3) , Q (0, 5)
Solution:
i) A (5, 0) B(9, 0)
Mid-point of
Mid-point of
ii) P(0, -3) Q(0, 5)
Mid-point of
Mid-point of
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Chords in Circles
## Line segments whose endpoints are on a circle.
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Practice Chords in Circles
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Chords in Circles
What if you were asked to geometrically consider the Gran Teatro Falla, in Cadiz, Andalucía, Spain, pictured below? This theater was built in 1905 and hosts several plays and concerts. It is an excellent example of circles in architecture. Notice the five windows, \begin{align*}A-E\end{align*}. \begin{align*}\bigodot A \cong \bigodot E\end{align*} and \begin{align*}\bigodot B \cong \bigodot C \cong \bigodot D\end{align*}. Each window is topped with a \begin{align*}240^\circ\end{align*} arc. The gold chord in each circle connects the rectangular portion of the window to the circle. Which chords are congruent? How do you know? After completing this Concept, you'll be able to use properties of chords to answer questions like these.
### Watch This
CK-12 Foundation: Chapter9ChordsinCirclesA
Learn more about chords and a circle's center by watching the video at this link.
### Guidance
A chord is a line segment whose endpoints are on a circle. A diameter is the longest chord in a circle. There are several theorems that explore the properties of chords.
Chord Theorem #1: In the same circle or congruent circles, minor arcs are congruent if and only if their corresponding chords are congruent.
Notice the “if and only if” in the middle of the theorem. This means that Chord Theorem #1 is a biconditional statement. Taking this theorem one step further, any time two central angles are congruent, the chords and arcs from the endpoints of the sides of the central angles are also congruent. In both of these pictures, \begin{align*}\overline{BE} \cong \overline{CD}\end{align*} and \begin{align*}\widehat{BE} \cong \widehat{CD}\end{align*}. In the second picture, we have \begin{align*}\triangle BAE \cong \triangle CAD\end{align*} because the central angles are congruent and \begin{align*}\overline{BA} \cong \overline{AC} \cong \overline{AD} \cong \overline{AE}\end{align*} because they are all radii (SAS). By CPCTC, \begin{align*}\overline{BE} \cong \overline{CD}\end{align*}.
##### Investigation: Perpendicular Bisector of a Chord
Tools Needed: paper, pencil, compass, ruler
1. Draw a circle. Label the center \begin{align*}A\end{align*}.
2. Draw a chord in \begin{align*}\bigodot A\end{align*}. Label it \begin{align*}\overline{BC}\end{align*}.
3. Find the midpoint of \begin{align*}\overline{BC}\end{align*} by using a ruler. Label it \begin{align*}D\end{align*}.
4. Connect \begin{align*}A\end{align*} and \begin{align*}D\end{align*} to form a diameter. How does \begin{align*}\overline{AD}\end{align*} relate to the chord, \begin{align*}\overline{BC}\end{align*}?
Chord Theorem #2: The perpendicular bisector of a chord is also a diameter.
In the picture to the left, \begin{align*}\overline{AD} \bot \overline{BC}\end{align*} and \begin{align*}\overline{BD} \cong \overline{DC}\end{align*}. From this theorem, we also notice that \begin{align*}\overline{AD}\end{align*} also bisects the corresponding arc at \begin{align*}E\end{align*}, so \begin{align*}\widehat{BE} \cong \widehat{EC}\end{align*}.
Chord Theorem #3: If a diameter is perpendicular to a chord, then the diameter bisects the chord and its corresponding arc.
##### Investigation: Properties of Congruent Chords
Tools Needed: pencil, paper, compass, ruler
1. Draw a circle with a radius of 2 inches and two chords that are both 3 inches. Label as in the picture to the right. This diagram is drawn to scale.
2. From the center, draw the perpendicular segment to \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CD}\end{align*}.
3. Erase the arc marks and lines beyond the points of intersection, leaving \begin{align*}\overline{FE}\end{align*} and \begin{align*}\overline{EG}\end{align*}. Find the measure of these segments. What do you notice?
Chord Theorem #4: In the same circle or congruent circles, two chords are congruent if and only if they are equidistant from the center.
Recall that two lines are equidistant from the same point if and only if the shortest distance from the point to the line is congruent. The shortest distance from any point to a line is the perpendicular line between them. In this theorem, the fact that \begin{align*}FE = EG\end{align*} means that \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CD}\end{align*} are equidistant to the center and \begin{align*}\overline{AB} \cong \overline{CD}\end{align*}.
#### Example A
Use \begin{align*}\bigodot A\end{align*} to answer the following.
a) If \begin{align*}m \widehat{BD}= 125^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*}.
b) If \begin{align*}m \widehat{BC}= 80^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*}.
Solutions:
a) From the picture, we know \begin{align*}BD = CD\end{align*}. Because the chords are equal, the arcs are too. \begin{align*}m \widehat{CD}= 125^\circ\end{align*}.
b) To find \begin{align*}m \widehat{CD}\end{align*}, subtract \begin{align*}80^\circ\end{align*} from \begin{align*}360^\circ\end{align*} and divide by 2. \begin{align*}m \widehat{CD}=\frac{360^\circ - 80^\circ}{2}=\frac{280^\circ}{2}=140^\circ\end{align*}
#### Example B
Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
The diameter here is also perpendicular to the chord. From Chord Theorem #3, \begin{align*}x = 6\end{align*} and \begin{align*}y = 75^\circ\end{align*}.
#### Example C
Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
Because the diameter is perpendicular to the chord, it also bisects the chord and the arc. Set up an equation for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
\begin{align*}(3x-4)^\circ &= (5x-18)^\circ && \ y+4=2y+1\\ 14^\circ &= 2x && \qquad 3=y\\ 7^\circ &= x\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter9ChordsinCirclesB
#### Concept Problem Revisited
In the picture, the chords from \begin{align*}\bigodot A\end{align*} and \begin{align*}\bigodot E\end{align*} are congruent and the chords from \begin{align*}\bigodot B, \bigodot C\end{align*}, and \begin{align*}\bigodot D\end{align*} are also congruent. We know this from Chord Theorem #1. All five chords are not congruent because all five circles are not congruent, even though the central angle for the circles is the same.
### Guided Practice
1. Is the converse of Chord Theorem #2 true?
2. Find the value of \begin{align*}x\end{align*}.
3. \begin{align*}BD = 12\end{align*} and \begin{align*}AC = 3\end{align*} in \begin{align*}\bigodot A\end{align*}. Find the radius and \begin{align*}m \widehat{BD}\end{align*}.
1. The converse of Chord Theorem #2 would be: A diameter is also the perpendicular bisector of a chord. This is not a true statement, see the counterexample to the right.
2. Because the distance from the center to the chords is congruent and perpendicular to the chords, then the chords are equal.
\begin{align*}6x-7 &= 35\\ 6x &= 42\\ x &= 7\end{align*}
3. First find the radius. In the picture, \begin{align*}\overline{AB}\end{align*} is a radius, so we can use the right triangle \begin{align*}\triangle ABC\end{align*}, such that \begin{align*}\overline{AB}\end{align*} is the hypotenuse. From Chord Theorem #3, \begin{align*}BC = 6\end{align*}.
\begin{align*}3^2+6^2 &= AB^2\\ 9+36 &= AB^2\\ AB &= \sqrt{45}=3 \sqrt{5}\end{align*}
In order to find \begin{align*}m \widehat{BD}\end{align*}, we need the corresponding central angle, \begin{align*}\angle BAD\end{align*}. We can find half of \begin{align*}\angle BAD\end{align*} because it is an acute angle in \begin{align*}\triangle ABC\end{align*}. Then, multiply the measure by 2 for \begin{align*}m \widehat{BD}\end{align*}.
\begin{align*}\tan^{-1} \left( \frac{6}{3} \right) &= m \angle BAC\\ m \angle BAC & \approx 63.43^\circ\end{align*}
This means that \begin{align*}m \angle BAD \approx 126.9^\circ\end{align*} and \begin{align*}m \widehat{BD} \approx 126.9^\circ\end{align*} as well.
### Explore More
Find the value of the indicated arc in \begin{align*}\bigodot A\end{align*}.
1. \begin{align*}m \widehat{BC}\end{align*}
2. \begin{align*}m \widehat{BD}\end{align*}
3. \begin{align*}m \widehat{BC}\end{align*}
4. \begin{align*}m \widehat{BD}\end{align*}
5. \begin{align*}m \widehat{BD}\end{align*}
6. \begin{align*}m \widehat{BD}\end{align*}
Algebra Connection Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*}.
1. \begin{align*}AB = 32\end{align*}
2. Find \begin{align*}m \widehat{AB}\end{align*} in Question 10. Round your answer to the nearest tenth of a degree.
3. Find \begin{align*}m \widehat{AB}\end{align*} in Question 15. Round your answer to the nearest tenth of a degree.
In problems 18-20, what can you conclude about the picture? State a theorem that justifies your answer. You may assume that \begin{align*}A\end{align*} is the center of the circle.
1. Find the measure of \begin{align*}\widehat{AB}\end{align*}in each diagram below.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 9.4.
### Vocabulary Language: English Spanish
chord
A line segment whose endpoints are on a circle.
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
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# Algebra & Graphs
## Polynomial Equations
A polynomial function $p(x)$ is not an equation. An equation implies equality, preferably to zero and must contain an $=$ sign.
## Linear Equations
$ax+b=0 \qquad a \neq 0$
$x = {-b\over a}$
$ax^2+bx+c=0$
There are three main methods to solve quadratic equations.
1. factorising (always try this first)
2. using 'the formula'
3. completing the square
### Factorisation
To solve $x^2-4x-21=0$:
$(x+3)(x-7)=0 \quad \Rightarrow \quad x=-3, \ \mbox{or}\ x = 7$
### Formula
$\displaystyle x=\displaystyle \frac{-b\pm \sqrt {b^2-4ac}}{2a}$
$b^2-4ac$ is called the discriminant.
• If the descriminant is zero then there is one real solution, sometimes referred to as two equal solutions.
• If it is greater than zero, there are two different (distinct) real solutions.
• If it is less than zero, there are no real solutions, but two complex conjugate solutions. (A number is complex if it contains $\sqrt -1$, represented by $i$ or $j$)
### Completing the square
To solve
$\begin{array}{lrcl}& x^{2} - 4x + 3 & =& 0\\ \Rightarrow & (x - 2)^{2} - 4 + 3& =& 0\\ \Rightarrow & (x - 2)^{2}& =& 1\\ \Rightarrow & x - 2& =& \pm \sqrt {1}\\ \Rightarrow & x& =& 2\pm \sqrt {1}\\ \Rightarrow & x& =& 3 \hbox{ or } 1\\ \end{array}$
Note: In this case the quadratic expression would factorise and it would have been easier to factorise than to complete the square.
$y=ax^2+bx+c$
$\matrix { & a>0 & & & a<0 & \qquad \cr (1) & b^2-4ac<0 & \qquad & (1) & b^2-4ac>0& \cr (2) & b^2-4ac=0 & & (2) & b^2-4ac=0& \cr (3) & b^2-4ac>0 & & (3) & b^2-4ac<0& \cr }$
To help sketch graphs the process of completing the square can be very useful.
### Graphs of other common functions
Graphs of $y=x^2$, $y=x^3$ and $\displaystyle y=\frac{1}{x}, \ y=a^ x\ (a>1)$
### The modulus function
$|x| = \left\{ \matrix {x & {\rm if } & x\geqslant 0 \cr -x & {\rm if } & x<0\cr }\right.$
## Basic Transformations
### Graphs of Inverse Functions
The graphs of a function and its inverse have reflection symmetry about the line $y = x$
Example: consider the function $f(x)=3x-1$
Its inverse is $f^{-1}(x)=\displaystyle \frac{x+1}{3}$
These can be shown graphically as
## Polynomials
A polynomial function, $p(x)$, is made by summing terms which contain only positive integer powers of x together, perhaps with a constant.
Note: The polynomial may contain just one term.
### Examples
$2x^{3}+5x, \quad 3x+2 ,\quad 7$
In general,
$p\left(x\right) = a_{0}+a_{1}x+a_{2}x^{2}+\ldots +a_{n}x^{n} = \displaystyle \sum _{i=0}^{n}a_{i}x^{i}$
$a_{i}$ are the coefficients of $x^{i}$ and may be zero.
($a_{0}x^{0}=a_{0}$ since $x^{0}=1$ and $a_{1}x^{1}=a_{1}x$ since $x^{1}=x$)
### Degree of a polynomial
The degree of a polynomial is the value of the highest power of $x$.
A polynomial of degree 1 is called linear, one of degree 2 is quadratic, one of degree 3 is cubic and one of degree 4 is quartic.
## Factorisation & Expansion
### Factorisation
Difference of two squares:
$x^2-a^2= (x-a)(x+a)$
Difference of two cubes:
$x^3-a^3=(x-a)(x^2+ax+a^2)$
Sum of two cubes:
$x^3+a^3=(x+a)(x^2-ax+a^2)$
### Expansions
$\begin{array}{rcl} (x+a)^2& =& x^2+2ax+a^2 \\ (x-a)^2& =& x^2-2ax+a^2 \\ (x+a)^3& =& x^3+3ax^2+3a^2x+a^3 \\ (x-a)^3& =& x^3-3ax^2+3a^2x-a^3 \end{array}$
## The Factor Theorem
For a given polynomial $p(x)$, if $p(a) = 0$ then $(x - a)$ is a factor of $p(x)$.
Following on from this, if $(x - a)$ is a factor of $p(x)$ then $p(a) = 0$.
### The Remainder Theorem
If a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$.
## Partial Fractions
For proper fractions $\displaystyle {P(x)\over Q(x)}$ where $P$ and $Q$ are polynomials with the degree of $P$ less than the degree of $Q$:
a linear factor $ax+b$ in the denominator produces a partial fraction of the form $\displaystyle {A\over ax+b}$
repeated linear factors $\displaystyle (ax+b)^2$ in the denominator produce partial fractions of the form $\displaystyle {A\over ax+b}+\displaystyle {B\over (ax+b)^2}$
a quadratic factor $ax^2+bx+c$ in the denominator produces a partial fraction of the form $\displaystyle {Ax+B\over ax^2+bx+c}$
Improper fractions require an additional term. This is a polynomial of degree $n-d$ where $n$ is the degree of the numerator and $d$ is the degree of the denominator.
## Inequalities
$a>b \hbox{ means } a \hbox{ is greater than } b$
$a<b \hbox{ means } a \hbox{ is less than } b$
$a\geqslant b \hbox{ means } a \hbox{ is greater than or equal to } b$
$a\leqslant b \hbox{ means } a \hbox{ is less than or equal to } b$
$|x|<k \hbox{ means } -k<x<k$
$|x|>k \hbox{ means } x>k \hbox{ or } x<-k$
## The Binomial Theorem
### Pascal's triangle
$\matrix {& & & & & & 1& & & & & & \cr & & & & & 1& & 1& & & & & \cr & & & & 1& & 2& & 1& & & \cr & & & 1& & 3& & 3& & 1& \cr & & 1& & 4& & 6& & 4& & 1\cr & 1& & 5& & 10& & 10& & 5& & 1\cr \vdots & & \vdots & & \vdots & & \vdots & & \vdots & & \vdots & & \vdots \cr }$
Note that the rows of Pascal's triangle give the coefficients in the expansion of $(a+b)^ n$:
$(a+b)^0=1$
$(a+b)^1= 1a+1b$
$(a+b)^2 = 1a^2+2ab + 1b^2$
$(a+b)^3=1a^3+3a^2b+3ab^2+1b^3$
$\vdots$
### Combinations
The number of ways of making a selection of $r$ objects from a set of $n$ different objects when the order of selection is not important (e.g. selecting 6 numbers on a lottery ticket) is denoted $^ nC_ r$ or $({n\atop r})$, given by
$^ nC_ r = \displaystyle \frac{n!}{r!(n-r)!}\quad \hbox{sometimes written as} \left({n\atop r}\right)$
### Factorials
$n!=n\times (n-1)\times (n-2)\times \ldots \times 3\times 2\times 1$
Here, $n$ is a positive integer. $0!$ is defined as 1.
### Binomial Theorem
If $n$ is a positive integer
$\begin{array}{rcl} (a+b)^ n & =& a^ n + \left({n\atop 1}\right)a^{n-1}b^1+\left({n \atop 2}\right)a^{n-2}b^2 \\ & & \kern -30pt{+\ldots + \left(n\atop r\right)a^{n-r}b^ r + \ldots + b^ n \ (n\in \mathbb N)}\\ \end{array}$
### Binomial Series
If $n$ is not a positive integer
$\begin{array}{rcl} (1+x)^ n& =& 1+nx + {n(n-1)\over 1\times 2}x^2\\ & & +\ldots + {n(n-1)\ldots (n-r+1)x^ r\over 1\times 2\times 3 \ldots \times r}\\ & & +\ldots \end{array}$
which is valid for any real number $n$ providing that $|x|<1$.
## Indices, Exponentials & Logarithms
### Laws of Indices
$a^ ma^ n=a^{m+n} ,\quad {a^ m\over a^ n}=a^{m-n}, \quad (a^ m)^ n=a^{mn}$
$a^0=1,\ a^{-m}={1\over a^ m},\ a^{1/n}=\sqrt [n]{a}, a^{m\over n}=(\sqrt [n]{a})^ m$
### Laws of Logarithms
For any positive base $b$ (with $b\ne 1$)
$\log _ bA=c \quad \hbox{ means }\quad A=b^ c$
$\log _ b A + \log _ b B = \log _ b AB$
$\log _ b A-\log _ b B=\log _ b{A\over B},$
$n\, \log _ b A=\log _ b A^ n,\quad \log _ b 1=0, \quad \log _ bb=1$
### Formula for change of base
$\log _ ax={\log _ bx\over \log _ b a}$
Logarithms to base e, denoted $\log _{\rm e}$ or alternatively $\ln$ are called natural logarithms. The letter e stands for the exponential constant which is approximately 2.718.
### Solving equations with an unknown power
To solve $a^ x=b$ where $a>0$, $b>0$:
\begin{eqnarray*} & a^ x=b \\ \hbox{where} & a>0 \\ \hbox{and} & b>0 \\ \end{eqnarray*}
Take logs of both sides to give
\begin{eqnarray*} \log (a^ x) & =& \log b \\ x\log a & =& \log b \\ x & =& \displaystyle \frac{\log b}{\log a} \end{eqnarray*}
Logs of any base may be used including base 10 ($\log$) and base e ($\ln$).
### Exponential functions
A pair of graphs showing exponential growth and decay, respectively:
A graph of powers of different bases:
### Logarithmic functions
Graphs of $y=\ln x$ and $y=\log _{10} x$
$y=e^ x\Leftrightarrow x=\ln y$
|
# What Is 69 percent of 35 + Solution with Free Steps?
The 69 percent of 35 is equal to 24.15. This solution is estimated by multiplication of 0.35 by 69.
The calculation of 69 percent of 35 may be used in many examples. For example lets say that you want to purchase apples at the rate of 35 dollars per dozen. Now you find out that Walmart is offering a discount of 69%. Since you know that 69% of 35 is equal to 24.15 dollars, you can easily know that you can purchase the same amount of apples for a discounted price of 24.15 dollars.
The main objective of this question is to find the value which is 69 percent of 35.
## What is 69 percent of 35?
The 69 percent of 35 is 24.15. This answer is calculated by multiplication of 0.35 by 69
The percentage is the part of 100 that can be written as either a number or ratio. It doesn’t have a unit and the only symbol for it is “%.”. We can rearrange the percentage formula in order to find the percent value which is 69 percent of 35. Thus:
a = ( Percentage x b ) x 100
Where a is the actual value and b is the total value.
## How to calculate 69% of 50?
Following are the steps used to represent that which value is 69 percent of 35.
## Step 1
We are given that:
The total value is 35.
The percentage is 69.
While we have to find the actual value.
## Step 2
Let represent the actual value by “a” which is an unknown.
## Step 3
We know that the percentage is calculated as:
Percentage = ( a / b ) x 100
Where a is the part of the whole number while b represents the whole number.
Rearranging the percentage formula according to our requirement results in:
a = ( b x Percentage ) / 100
By putting the values, we get”
a = ( 35 x 69 ) / 100
a = ( 35 x 69 ) / 100
a = 2415 / 100
Thus dividing by 100 results in:
a = 24.15.
Hence, 69 % of 35 is 24.15
## Step 4
To visualize what 69% of 35 looks like, we may look at the corresponding Pie Chart presented above. If we split the whole value into 100 equal parts, then 69% of 35 is 24.15, which is in the blue area, and 31% of 100 is in the yellow area.
A percentage is a number and maybe even a ratio written as just a fraction of 100. The word “percentage” comes from the Latin word “per centum,” which means “by a hundred.” A symbol represents the percentage as ” % “.
With the percentage formula, you can figure out how much of a whole something is in terms of 100. You can show a number as just a fraction of 100 by using the concept of percentage. Mathematically the percentage can be calculated as:
Percentage = ( a / b ) x 100
Where a is the part of the whole number while b represents the whole number.
We know that one percent, which is written as 1″%“, is a hundredth of the whole, so 100% is whole while 200% is twice the given amount.
All Images/Mathematical drawings are created with Geogebra.
#### What Is 65 Percent Of 69 | Percentage of a Number List | What Is 69 Percent Of 42
A percentage is a number and maybe even a ratio written as just a fraction of 100. The word “percentage” comes from the Latin word “per centum,” which means “by a hundred.” A symbol represents the percentage as ” % “.
With the percentage formula, you can figure out how much of a whole something is in terms of 100. You can show a number as just a fraction of 100 by using the concept of percentage. Mathematically the percentage can be calculated as:
Percentage = ( a / b ) x 100
Where a is the part of the whole number while b represents the whole number.
We know that one percent, which is written as 1″%“, is a hundredth of the whole, so 100% is whole while 200% is twice the given amount.
The main objective of this question is to find the value which is 69 percent of 35.
## Solution: What is 69 percent of 35?
The 69 percent of 35 is 24.15.
The percentage is the part of 100 that can be written as either a number or ratio. It doesn’t have a unit and the only symbol for it is “%.”. We can rearrange the percentage formula in order to find the percent value which is 69 percent of 35. Thus:
a = ( Percentage x b ) x 100
Where a is the actual value and b is the total value.
## How to calculate 69% of 35?
Following are the steps used to represent that which value is 69 percent of 35.
## Step 1
We are given that:
The total value is 35.
The percentage is 69.
While we have to find the actual value.
## Step 2
Let represent the actual value by “a” which is an unknown.
## Step 3
We know that the percentage is calculated as:
Percentage = ( a / b ) x 100
Where a is the part of the whole number while b represents the whole number.
Rearranging the percentage formula according to our requirement results in:
a = ( b x Percentage ) / 100
By putting the values, we get”
a = ( 35 x 69 ) / 100
a = ( 35 x 69 ) / 100
a = 2415 / 100
Thus dividing by 100 results in:
a = 24.15.
Hence, 69 % of 35 is 24.15
## Step 4
To visualize what 69% of 35 looks like, we may look at the corresponding Pie Chart presented above. If we split the whole value into 100 equal parts, then 69% of 35 is 24.15, which is in the blue area, and 31% of 100 is in the orange area.
All Images/Mathematical drawings are created with Geogebra.
|
# When Will We Reach One Half?
### How many digits are needed before the hundreds chart is half covered?
First grade students became deeply engaged in this open problem. They explored patterns, made conjectures (predictions) as to what digit they would be on when half the 100 board would be covered. They discussed and defined what one half would mean on the hundred chart. In the process of solving this interesting problem, they discovered other numeration concepts along the way. They learned the meaning of a “digit” versus a number. They also gained reinforcement and practice in recognizing, sequencing, writing, and saying numbers up to 100.
## Is The Longest Block Area Also The Biggest?
Blog entry by Ariane Stern & Julie Kim
# First Graders Explore Area and Perimeter
Earlier in the year, we introduced the concept of area—that you can measure how small or large a space is. After measuring various shapes with non-standard units (e.g. tiles, beans, paperclips), first graders developed an understanding that with bigger units like popsicle sticks, you’d need fewer of them to cover a certain area and that you would need many more small units like beans to fill that same area. Later in the year, we began a unit on linear measurement. Again, we used manipulatives to measure the length of objects. This time, we focused on ways to accurately measure length, such as figuring out which side is the length, and measuring by starting at the edge. After learning how to measure length accurately with one unit, we moved again to the idea of measuring the same object with different sized units. Once again, it takes less popsicle sticks to measure the length of a book, and more cubes, if you were measuring the same length.
This year, to culminate our study on linear measurement, we challenged first graders to measure how long each Lower School classroom’s block area is. In small groups, first graders went to each classroom, calculated which side was the longest side and measured it with string. They brought their strings back to our classroom and proceeded to measure how long they were with popsicle sticks, tiles, and double unit wooden blocks. Afterwards, we collected each group’s data and compared the number of double unit blocks it took to measure the length of each string. We put the lengths in order to see which block area was the longest.
Along the way, we ran into some problems, just as mathematicians do in real life! What would we do if the double units were too long or too short for the end of the string? That led us into a conversation about fractions. We looked at halves, quarters, and thirds of rectangles and also looked at circles to deepen our understanding of fractions. This helped some groups to make more accurate measurements.
Then, Julie showed the class some pictures. Tasha’s block area was the longest, but it was very narrow and skinny. Diane’s block area was the shortest, but it was very wide. We wondered, just because a block area is the longest, does that mean it’s the biggest?
What a big question!
How would we figure that out? We knew what all the lengths of the block areas were, but how would we figure out how big they were? What were we even looking for? One student shouted, “The area! How were we going to figure out the area? We drew a rectangle on the board, showing how we knew the lengths. What else would we need to know to find the area? This was a puzzle. But calling upon what we had already learned about area, we realized that we needed to know the width of each block area. Wow!
Each group then went back to their assigned classroom, this time measuring the width of each block area. When they came back, they measured their new strings again with double unit wooden blocks. Then they used graph paper to replicate their block area. Each square of the graph paper represented one double unit block. Then they counted all the squares to figure out the area. This was hard work! Many of the shapes they drew had over 100 squares! Each group had to count the squares in their shape multiple times to try and get the most accurate count.
# First Graders Cook, Question, and Count
By Julie Kim, First Grade Associate teacher
In the real world, we confront daily math problems through a process of noticing and wondering. After our mind has determined a question about a scenario, whether it is counting how many more blocks you need to walk or how many servings to cook for dinner, we proceed to the next step: plan, search, and gather. We plan for what steps we are going to take in order to answer the question. We search for the separate variables and pieces of information that we need in order to solve the question. We gather these pieces of information then puzzle them up in a way that will help us solve the problem. Will we, as mathematicians, get the answer we are looking for the first time around? Not always. Will we get an answer immediately? Not guaranteed. Will we persevere and try over and over again until we do? We should. Through real world work, first graders develop stamina and perseverance as they attack the challenges and questions that they are eager to solve. This is where the real work happens.
# Teachers posed this question to first graders during their study of geometry.
First graders went beyond naming basic two-dimensional shapes to exploring specific characteristics, or attributes, that define them. Their conversations evolved from seeing a shape as a “whole” such as a triangle, to analyzing and deciding the specific features that prove that it is a triangle. Students decided that a shape can only be called a triangle if it has three straight lines that make up the sides, three corners, or vertices, and does not have any openings (it needs to be closed). Through this specific definition, they discovered that there are many types of triangles. They used this critical thinking foundation to explore a variety of two-dimensional shapes.
# “Does a Circle Have Sides?”
This was an interesting question to ponder. First grade teachers, Sarah and Ariane, posed it to their students to see if they could apply the skills they had learned about defining geometric attributes to this question. It turns out that it wasn’t an easy question to answer! Continue reading
## Family Math Night at LREI
### Students and Parents look forward to Family Math Night every year.
Students in grades one through four celebrate mathematics, as well as continue to hone their fluency in combination facts by playing fun games. Fourth grade students create their own math games as a capstone experience, and then teach them to family and friends during Family Math Night. Continue reading
## Security Camera Equations & Algebra in First Grade
Students in first grade have been working with the equal symbol, and the greater than/less than symbol. They’ve created number stories and equations using the data they collected from counting the number of security cameras the stores in the neighborhood have. Some of these equations are simply true statements, and some have missing addends, or missing sums, depending on the story they created. Continue reading
## Introducing the Equal Symbol as a Relationship Model
Students in first grade are learning that the equal symbol doesn’t necessarily mean to “do” something. It can just mean that a mathematical statement is “true”. Continue reading
## First Grader’s Data Representations of “Safe” and “Unsafe” Lead Them to Social Responsibility.
The students in first grade are learning how to collect data and communicate the results of their data in a representation that makes sense to them.
Both classes spent time outside observing and recording “safe” and “unsafe” events in the neighborhood before each class decided on a topic to collect data on. Safety is also the larger topic they are learning about in social studies. Sarah’s class collected data on bicyclists and whether or not they wore helmets. Ariane’s class collected data on broken benches in the nearby parks. Continue reading
## A rich student-to-student discussion by first graders on the meaning of the equal symbol.
What does: 1 + 7 = ___ + 6
have to do with: 3x + 9 = 5x + 5
…and why are first graders arguing with each other over the meaning of the equal symbol?
|
# Solve the Mystery of Missing Dollar
### Missing Dollar Riddle
Three friends check into a hotel. They pay \$30 to the manager and go to their room. The manager suddenly remembers that the room rate is \$25 and gives \$5 to the bellboy to return to the people. On the way to the room the bellboy reasons that \$5 would be difficult to share among three people so he pockets \$2 and gives \$1 to each person. Now each person paid \$10 and got back \$1. So they paid £9 each, totalling \$27. The bellboy has \$2, totalling \$29.
Where is the missing \$1?
The above missing dollar riddle is a famous problem that illustrates problems of confusion and misdirection in conversation. It illustrates how misdirection and irrelevant facts and questions can foil a person’s clear understanding of a problem. Apply your mind and explain the logic behind the missing dollar.
Follow-up to the above problem – An interesting resolution is often cited as the solution to the above problem. Just read the following paragraph and post a comment whether you agree or not the given resolution to the above mystery.
A few months later, two of the original three guests check into a hotel room in the same hotel. The clerk says the bill is \$20, so each guest pays \$10. Later the clerk realizes the bill should only be \$15. To rectify this, he gives the bellhop \$5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn’t know the total of the revised bill, the bellhop decides to just give each guest \$1 and keep \$3 for himself.
Now that each of the guests has been given \$1 back, each has paid \$9, bringing the total paid to \$18. The bellhop has \$3, so \$18 + \$3 = \$21, and the guests originally handed over \$20, so that’s where the missing dollar from the original problem is!
#### Vineet Patawari
Hi, I'm Vineet Patawari. I fell in love with numbers after being scared of them for quite some time. Now, I'm here to make you feel comfortable with numbers and help you get rid of Math Phobia!
## 16 thoughts to “Solve the Mystery of Missing Dollar”
1. miracle says:
on their return — the same
they paid \$18 (\$9 each person). \$15 went to the clerk and \$3 went to the bellhop. so no need to add another \$3 from the bellhop. its already there in the \$18 (\$15+\$3). so to get \$20, you just add the \$2 (\$1 each-the 2 person took back). so no excess there…. 🙂
2. miracle says:
they paid \$27 (\$9 for each person) –\$25 went to the clerk and \$2 to the bell boy. so the \$2 dollars is already included in the \$27 they paid. so the only number your going to add is the \$3 (\$1 each person got back) to get \$30. so nothing is missing.
3. Room rent=25, so each has paid \$8.33
Among remaining \$5 , bellboy had kept \$2 and return \$1 to each one.
so each one had pay \$9.33 not \$9.Which totals \$ 30(28+2)
4. Anish says:
On the one hand, the question is about the savings and on the other hand, comparison is being done with the amount paid. If we compare same things, there would not be any difference.
5. Xpenditure+balance in hand = total amount. 27(9*3) + 3(1+1+1)=30 xpenditure=roomrent+bellboycharge. 27=25 + 2
6. Daemion says:
They paid \$9 each totaling \$27 of which the manager has \$25 and the bellboy \$2.
7. NANDEESH says:
I think this puzzle is one of the most glorified puzzles but a stupid one.
8. Avijit says:
As You Know Room Rent Is \$30. So three Friend’s are paid \$10 each.And exact cost of room is \$25.Manager send rest of money to the hand of Bell boy \$5.but he take \$2 for easy to distribute them.so they return \$1 each now problem is that if they paid \$9 each then total money is \$9*3=\$27.
And bell boy take \$2 so total amount =\$27+\$2=\$29.
Missing \$1. ok.
suppose they paid \$8*3=\$24.
So remainder is =\$6.
We add \$1 in side of room rent=\$24+\$1=\$25.
now remainder = \$5
bell boy take \$2.
so now remainder =\$3
They equally distribute=\$1 each.
Now Problem Is solve
They Paid=\$9 each.
\$2 have in bell boy pocket.
\$1 problem is also solved.
9. Vijayendra Bose says:
Simple explanation:- Since \$1 has been returned to each of the three friends, the total paid by them is no longer \$30. The total paid by them is now \$27. So the room rate of \$25 and the money pocketed by the bell boy i.e \$2, also add up the the same total of \$27.
The problem statement tries to confuse us into comparing with the original total of \$30 instead of the new total of \$27
10. GVS says:
Expenditure by some person(s) = Income of some other person(s)
3 people expended \$27 = \$25 income of hotelier + \$2 income of bellboy
\${[3*(10-1)]} = \$25 + \$2
\$3*9 = \$27
\$27 = \$27.
11. GVS says:
Expenditure made by some person(s) = Income of some other person(s)
Here the 3 people has expended \$25 income of hotelier + \$2 income of
{[3*(10-1)] = 27} dollars = bellboy
Therefore, \$27 = \$27.
Thats it.
12. Ajay says:
well its simple and nothing complex that appears in it..
they paid \$25+\$5. out of which \$3 were given back to them amounting to \$28 and \$2 lies with the waiter which totals to \$30.
Its the language which is actually confusing here, but the calculation part is pretty simple and all this confusion only appears because neither \$25 not \$5 and also not \$2 happens to be the multiple of 3.
13. 2\$ should be not be added at the end of statement rather these 2\$’s should be subtracted..to get 25\$
14. priyank says:
in 1st case out of 25\$ eash paid 8\$ and remainind 1\$ is 0.33\$=25\$.they received 1\$ back therefore 25\$+3\$=28\$ and bellboy keep the 2\$. there fore 25\$+3\$+2\$=30\$.
in 2nd case out of 15\$ each share 7.5\$=15\$. they get back 1\$=2\$ and bellshop keep 3\$. therefore 15\$+2\$+3\$=20\$.
15. Combining Daemion’s and Anish’s we get the complete answer. Kudos guys 🙂
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