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# Math in Focus Grade 5 Chapter 3 Practice 4 Answer Key Expressing Fractions, Division Expressions, and Mixed Numbers as Decimals
Go through the Math in Focus Grade 5 Workbook Answer Key Chapter 3 Practice 4 Expressing Fractions, Division Expressions, and Mixed Numbers as Decimals to finish your assignments.
## Math in Focus Grade 5 Chapter 3 Practice 4 Answer Key Expressing Fractions, Division Expressions, and Mixed Numbers as Decimals
Write each fraction as a decimal.
Example
$$\frac{3}{5}$$ = $$\frac{3}{5}$$
= 0.6
Question 1.
$$\frac{13}{20}$$ = ____
= _____
$$\frac{13}{20}$$ = $$\frac{13}{20}$$
= 0.65
Explanation:
Written the fraction as a decimal.
Question 2.
$$\frac{19}{25}$$ = ____
= _____
$$\frac{19}{25}$$ =$$\frac{19}{25}$$
= 0.76
Explanation:
Written the fraction as a decimal.
Question 3.
$$\frac{47}{50}$$ = ____
= _____
$$\frac{47}{50}$$ = $$\frac{47}{50}$$
= 0.94
Explanation:
Written the fraction as a decimal.
Express each division expression as a mixed number in simplest form and as a decimal.
Explanation:
Question 4.
7 ÷ 2
7 ÷ 2 = $$\frac{6}{2}$$ + $$\frac{1}{2}$$
= 3 + $$\frac{1}{2}$$
= 3 + 0.5
= 3.5
Explanation:
Converted division expression into mixed fraction and decimal
Question 5.
9 ÷ 4
$$\frac{8}{4}$$ + $$\frac{1}{4}$$
= 2 + $$\frac{1}{4}$$
= 2 + 0.25
= 2.25
Explanation:
Converted division expression into mixed fraction and decimal
Question 6.
21 ÷ 5
$$\frac{20}{5}$$ + $$\frac{1}{5}$$
= 4 + $$\frac{1}{5}$$
= 4 + 0.2
= 4.2
Explanation:
Converted division expression into mixed fraction and decimal
Question 7.
101 ÷ 25
$$\frac{100}{25}$$ + $$\frac{1}{25}$$
= 1 + $$\frac{1}{25}$$
= 4 + 0.04
= 4.04
Explanation:
Converted division expression into mixed fraction and decimal
Express each improper fraction as a decimal.
Example
$$\frac{3}{2}$$ = $$\frac{2}{2}$$ + $$\frac{1}{2}$$
= 1 + $$\frac{1}{2}$$
= 1 + 0.5
= 1.5
Explanation:
Converted each improper fraction into decimal
Question 8.
$$\frac{22}{5}$$
$$\frac{20}{5}$$ + $$\frac{2}{5}$$
= 4 + $$\frac{2}{5}$$
= 4 + 0.4
= 4.4
Explanation:
Converted each improper fraction into decimal
Question 9.
$$\frac{47}{20}$$
$$\frac{40}{20}$$ + $$\frac{7}{20}$$
= 2 + $$\frac{7}{20}$$
= 2 + 0.35
= 2.35
Explanation:
Converted each improper fraction into decimal
Question 10.
$$\frac{32}{25}$$
$$\frac{25}{25}$$ + $$\frac{7}{25}$$
= 1 + $$\frac{7}{25}$$
= 1 + 0.28
= 1.28
Explanation:
Converted each improper fraction into decimal
Question 11.
A coil of rope 603 feet long is cut into 25 equal pieces. What is the length of each piece? Express your answer as a mixed number and as a decimal.
24$$\frac{3}{25}$$ in mixed fraction.
603 ÷ 25 = $$\frac{600}{25}$$ + $$\frac{3}{25}$$
= 24 + $$\frac{3}{25}$$
= 24$$\frac{3}{25}$$ |
# Area and Perimeter of a Circle
We will discuss the Area and Perimeter of a Circle.
The area (A) of a circle (or circular region) is given by
A = πr2
where r is the radius and, by definition, π = $$\frac{\textrm{Circumference}}{Diameter}$$ = $$\frac{22}{7}$$ (Approximately).
The circumference (P) of a circle, or the perimeter of a circle region, with radius r is given by
P = 2πr
Solved Examples on Area and Perimeter of a Circle:
1. The radius of a circular region is 7 m. Find its perimeter and area. (Use π = $$\frac{22}{7}$$)
Solution:
Here r = 7 m. Then,
Perimeter = 2πr = 2 × $$\frac{22}{7}$$ × 7 m = 44 m;
Area = πr2 = $$\frac{22}{7}$$ × 72 m2 = 154 m2.
2. The perimeter of a circular plate is 132 cm. Find its area. (Use π = $$\frac{22}{7}$$)
Solution:
Let the radius of the plate be r. Then,
Perimeter = 2πr
⟹ 132 cm = 2 × $$\frac{22}{7}$$ × r
⟹ 132 cm = $$\frac{44}{7}$$ × r
⟹ $$\frac{44}{7}$$ × r = 132 cm
⟹ r = 132 × $$\frac{7}{44}$$
⟹ r = $$\frac{924}{44}$$
⟹ r = 21 cm.
Therefore, area of the plate = πr2 = $$\frac{22}{7}$$ × 212 cm2
= $$\frac{22}{7}$$ × 441 cm2
= $$\frac{9702}{7}$$ cm2
= 1386 cm2
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# Thread: Find all values of θ in the interval [0, 2pi]
1. ## Find all values of θ in the interval [0, 2pi]
Find all values of θ in the interval [0, 2pi], given:
cot θ = -1/√3
cosθ= 1/2
I don't really need the answer, just need to know how to solve this type of problem
I greatly appreciate your answer
Thank you.
2. ## Re: Find all values of θ in the interval [0, 2pi]
First, it helps to write the functions in terms of sine, cosine or tangent. Once you have done that, make note of whether your function is positive or negative, take note of which quadrants that function is positive or negative in. Work out the focus angle (for the first quadrant) and then adjust according to whichever quadrants you need to be in.
See how you go.
3. ## Re: Find all values of θ in the interval [0, 2pi]
Originally Posted by Prove It
First, it helps to write the functions in terms of sine, cosine or tangent. Once you have done that, make note of whether your function is positive or negative, take note of which quadrants that function is positive or negative in. Work out the focus angle (for the first quadrant) and then adjust according to whichever quadrants you need to be in.
See how you go.
I can find the first angle but then the second one confuses me, like how do you find the 2nd one?
4. ## Re: Find all values of θ in the interval [0, 2pi]
Well in your first equation, like I said, put it as either a sine, cosine, or tangent equation. The equation is equivalent to $\displaystyle \tan{(\theta)} = -\sqrt{3}$.
In which quadrants is the tangent function negative?
5. ## Re: Find all values of θ in the interval [0, 2pi]
$cot\theta= \frac{1}{tan\theta}$
$tan\theta= \frac{sin\theta}{cos\theta}$
Therefore
$cot\theta= \frac{cos\theta}{sin\theta}$
You know that
$\frac{cos\theta}{sin\theta}=\frac{-1}{\sqrt{3}}$
And
$cos\theta= \frac{1}{2}$
Can you solve these two equations for theta?
6. ## Re: Find all values of θ in the interval [0, 2pi]
Originally Posted by Prove It
Well in your first equation, like I said, put it as either a sine, cosine, or tangent equation. The equation is equivalent to $\displaystyle \tan{(\theta)} = -\sqrt{3}$.
In which quadrants is the tangent function negative?
2 and 4
Originally Posted by Shakarri
$cot\theta= \frac{1}{tan\theta}$
$tan\theta= \frac{sin\theta}{cos\theta}$
Therefore
$cot\theta= \frac{cos\theta}{sin\theta}$
You know that
$\frac{cos\theta}{sin\theta}=\frac{-1}{\sqrt{3}}$
And
$cos\theta= \frac{1}{2}$
Can you solve these two equations for theta?
Yes.
the 1st one theta is 2pi/3
and the 2nd one theta is pi/3
7. ## Re: Find all values of θ in the interval [0, 2pi]
Now you need to try to find ALL solutions in the given region. Like I said, think about which quadrants you need to be in... |
# find k
#### Albert
##### Well-known member
the solutions of :
$x^2+kx+k=0 " are$ $sin \,\theta \,\,and \,\, cos\, \theta$
please find : $k=?$
#### soroban
##### Well-known member
Hello, Albert!
$$\text{The solutions of: }\: x^2+kx+k\:=\:0$$
$$\text{are }\sin\theta\text{ and }\cos\theta$$
$$\text{Find }k.$$
Since $$k \,=\,\sin\theta\cos\theta$$, we see that: .$$|k| \,<\,1.$$
Quadratic Formula: .$$x \:=\:\frac{-k \pm \sqrt{k^2-4k}}{2}$$
$$\text{Let: }\:\begin{Bmatrix}\sin\theta &=& \frac{-k + \sqrt{k^2-4k}}{2} \\ \cos\theta &=& \frac{-k -\sqrt{k^2-4k}}{2} \end{Bmatrix}$$
$$\text{Then: }\:\begin{Bmatrix}\sin^2\theta &=& \frac{2k^2 - 4k + 2k\sqrt{k^2-4k}}{4} \\ \cos^2\theta &=& \frac{2k^2 - 4k - 2k\sqrt{k^2-4k}}{4} \end{Bmatrix}$$
$$\text{Add: }\:\sin^2\theta + \cos^2\theta \:=\:\frac{4k^2 - 8k}{4} \:=\:1$$
$$\text{And we have: }\:k^2 - 2k - 1\:=\:0$$
$$\text{Hence: }\:k \:=\:1\pm\sqrt{2}$$
$$\text{Since }|k| < 1\!:\;k \:=\:1-\sqrt{2}$$
Last edited by a moderator:
#### Pranav
##### Well-known member
By Vieta's formulas, we have
$$k=\sin\theta \cos\theta$$
$$\sin\theta+\cos\theta=-k$$
Squaring both the sides of second equation,
$$1+2\sin\theta \cos\theta=k^2 \Rightarrow k^2-2k=1 \Rightarrow k^2-2k+1=2 \Rightarrow (k-1)^2=2$$
$$\Rightarrow k=1\pm \sqrt{2}$$
But $|k|<1$, hence, $k=1-\sqrt{2}$. |
Question
# Prove the following identity:${\cosh ^2}x{\cos ^2}x - {\sinh ^2}x{\sin ^2}x = \dfrac{1}{2}(1 + \cosh 2x\cos 2x)$
Hint: Substitute the formula for hyperbolic sine and cosine, that is, $\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$ and $\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$ in the left-hand side of the equation and simplify to complete the proof.
Let us start proving by assigning the left-hand side of the equation to the LHS.
$LHS = {\cosh ^2}x{\cos ^2}x - {\sinh ^2}x{\sin ^2}x.........(1)$
Hyperbolic functions are similar to trigonometric functions but they are defined in terms of the exponential function. Like, sin x and cos x is defined on a circle, sinh x and cosh x are defined on a hyperbola, thus giving its name. The point (cos t, sin t) lies on the circle.
Similarly, the point (cosh t, sinh t) lies on a hyperbola.
We know the formula for hyperbolic cosine and sine in terms of exponential function.
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}..........(2)$
$\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}..........(3)$
Substituting equation (2) and equation (3) in equation (1), we get:
$LHS = {\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)^2}{\cos ^2}x - {\left( {\dfrac{{{e^x} - {e^{ - x}}}}{2}} \right)^2}{\sin ^2}x$
Evaluating the squares, we get:
$LHS = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} + 2}}{4}} \right){\cos ^2}x - \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} - 2}}{4}} \right){\sin ^2}x$
Grouping together the common exponential terms, we get:
$LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{2}{4}({\cos ^2}x + {\sin ^2}x)$
We know that ${\cos ^2}x + {\sin ^2}x = 1$, hence, we have:
$LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{1}{2}$
We know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$, hence, we have:
$LHS = \dfrac{{{e^{2x}}}}{4}(\cos 2x) + \dfrac{{{e^{ - 2x}}}}{4}(\cos 2x) + \dfrac{1}{2}$
Now, taking cos2x as a common term, we have:
$LHS = \cos 2x\left( {\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{4}} \right) + \dfrac{1}{2}$
Now using equation (2), the term inside the bracket can be expressed in terms of cosh2x.
Hence, we get:
$LHS = \cos 2x\left( {\dfrac{{\cosh 2x}}{2}} \right) + \dfrac{1}{2}$
Taking the factor of half as a common term, we have:
$LHS = \dfrac{1}{2}\left( {1 + \cos 2x\cosh 2x} \right)$
The right hand side of the equation is nothing but the right hand side of the proof, hence, we get:
$LHS = RHS$
Hence, we proved.
Note: You can also use the relation between the hyperbolic sine and cosine, ${\cosh ^2}x - {\sinh ^2}x = 1$ and $\cosh 2x = {\cosh ^2}x + {\sinh ^2}x$ to complete the proof. |
New Zealand
Level 8 - NCEA Level 3
Conditional Probability - Sample Spaces
Lesson
Independent vs Dependent Events
Life is full of different events. You may walk or drive to school, you may take your lunch or buy it from the canteen. In probability, we break events up into two groups: independent and dependent events.
Independent events are events that are not affected by any other event. For example, flipping a coin multiple times is an independent event because each flip is an isolated event. You have an even chance of getting a head or a tail each time and this is not affected by what you flip before.
In contrast, dependent events are those that are affected by previous events. For example, if you have a bag full of different coloured marbles, the probability of selecting a particular colour will change depending on the colour of the marble that was previously selected and taken out.
Mutually Inclusive vs Mutually Exclusive
Mutually exclusive events are events that can't occur at the same time. For example, you can't turn left and right at the same time!
Mutually inclusive events are not as common as mutually exclusive events. Basically, it means that one event cannot occur without the other.
The best way to visualise mutually inclusive/exclusive is in a Venn diagram. The Venn diagram below is showing two exclusive events (A can happen or B can happen) and one inclusive event (the green part- A and B).
Let's look at a more concrete example with numbers. Some numbers are even ($2,4,20$2,4,20 etc), which can be written in the dark green circle. Some numbers are multiples of three ($3,9,15$3,9,15 etc) which can be written in the light green circle. These are mutually exclusive events. What if I asked you for an even that is also a multiple of three? Now we need to think of numbers that are mutually inclusive (ie. satisfy both conditions) such as $6,12,18$6,12,18 and so on. These are written in the space where the circles overlap.
Conditional Probability
Conditional probability is concerned with multiple events and is similar to mutually inclusive events. The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. In other words, we calculate the probability of B, given that A has occurred.
Careful!
A lot of conditional probability questions involve replacement:
• "With replacement" means the two events remain independent (ie. the probabilities don't change)
• "Without replacement" means that the two events are dependent (the chances change depending on the previous selection)
Tree Diagrams
Tree diagrams are often used to display conditional events and make our probability calculations easier.
Let's look at an example of a conditional event that can be displayed in a tree diagram.
In a netball game, there are two people who can shoot - the goal shooter (GS) and the goal attack (GA). Let's say the GS in a team attempts the shots $60%$60% of the time and scores $70%$70% of the goals that they attempt, while the GA attempts to shoot $40%$40% of the time and scores $60%$60% of the goals that they attempt.
The blue lines represent the "branches" and we write each possible outcome at the end of each branch (in this case, who takes the shot and whether the goal is scored). The probability of each outcome is written along the branch.
So what is the probability of the GA scoring?
To work out this conditional probability, we multiply along the branches, considering the probability of the goal attack attempting the shot (GA) and then scoring (goal).
So, $\text{P(GA Score)}=0.4\times0.6$P(GA Score)=0.4×0.6$=$=$0.24$0.24. This means that the goal attack will successfully score $24%$24% of the time in a match.
What if we want to know the probability of either person scoring a goal?
Well, as you can see in the diagram below, there are two possible ways this combination can occur: 1) the goal shooter can score or 2) the goal attack can score.
$P(goal)$P(goal) $=$= $0.6\times0.7+0.4\times0.6$0.6×0.7+0.4×0.6 $=$= $0.42+0.24$0.42+0.24 $=$= $0.66$0.66
That means, the team scores $66%$66% of the time. If you do the same process for the probability of the team missing, you'll see that it's $34%$34%. So we know that we have added everything up correctly because scoring and missing are complementary events and $P(goal)+P(miss)=0.66+0.34$P(goal)+P(miss)=0.66+0.34$=$=$1$1
Remember!
• The probabilities of all your outcomes should add up to $1$1.
• Multiply the probabilities along the branches.
Worked Examples
Question 1
There are four cards marked with the numbers $2$2, $5$5, $8$8, and $9$9. They are put in a box. Two cards are selected at random one after the other without replacement to form a two-digit number.
1. Draw a tree diagram to illustrate all the possible outcomes.
2. How many different two-digit numbers can be formed.
3. What is the probability of obtaining a number less than $59$59?
4. What is the probability of obtaining an odd number?
5. What is the probability of obtaining an even number?
6. What is the probability of obtaining a number greater than $90$90?
7. What is the probability that the number formed is divisible by 5?
QUESTION 2
A pile of playing cards has $4$4 diamonds and $3$3 hearts. A second pile has $2$2 diamonds and $5$5 hearts. One card is selected at random from the first pile, then the second.
1. Construct a tree diagram of this situation with the correct probability on each branch.
2. What is the probability of selecting two hearts ?
QUESTION 3
The Venn diagram shown shows the number of students in a school playing the sports of Rugby League, Rugby Union, both or neither.
1. What is the probability that a student selected at random plays Rugby League?
2. What is the probability that a student selected at random plays Rugby Union?
3. What is the probability that a student selected at random plays Rugby League only?
4. What is the probability that a student selected at random does not play Rugby League?
5. What is the probability that a student selected randomly does not play Rugby Union?
6. What is the probability that a student chosen at random plays Rugby Union only?
Outcomes
S8-4
Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal
91585
Apply probability concepts in solving problems |
How do you simplify (8/27)^(-2/3)?
Feb 14, 2017
$\frac{9}{4}$
Explanation:
${\left(\frac{8}{27}\right)}^{- \frac{2}{3}}$
$\text{ }$
$= {\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
The prime factorization of 27 and 8 is:
$\text{ }$
$27 = 9 \times 3 = 3 \times 3 \times 3 = {3}^{3}$
$\text{ }$
$8 = 4 \times 2 = 2 \times 2 \times 2 = {2}^{3}$
Substituting the factorization on the above fraction we have:
$\text{ }$
${\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({3}^{3} / {2}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
Applying the property of power of a power that says:
$\text{ }$
$\textcolor{red}{{\left({a}^{n}\right)}^{m} = {a}^{m \times n}}$
$\text{ }$
${\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{3 \times \left(\frac{2}{3}\right)}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{2}$
$\text{ }$
$= \frac{9}{4}$
Feb 14, 2017
$= \frac{9}{4}$
Explanation:
There are 3 different processes indicated in this expression.
Laws of indices:
${x}^{-} m = \frac{1}{x} ^ m \text{ "and" } {\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$
The second law is the one we will apply.
Also ${x}^{\frac{p}{q}} = {\sqrt[q]{x}}^{p}$
${\left(\frac{8}{27}\right)}^{- \frac{2}{3}} = {\left(\frac{27}{8}\right)}^{+ \frac{2}{3}}$
$= {\sqrt[3]{\frac{27}{8}}}^{2} \text{ } \leftarrow$ find the cube roots first
$= {\left(\frac{3}{2}\right)}^{2}$
$= \frac{9}{4}$ |
# How do you multiply ((1, -4), (4, -1)) and ((3, -2), (0, -3))?
Jun 8, 2016
Matrix multiplication basically involves multiplying each row of the left matrix by all the columns of the right matrix.
I could show you a general notation for this at the end of the answer, but an example is easier to understand.
Since an $M \times K$ multiplied by a $K \times N$ matrix generates an $M \times N$ matrix, a $2 \times 2$ multiplied by a $2 \times 2$ gives a $\setminus m a t h b f \left(2 \times 2\right)$ matrix.
$\textcolor{b l u e}{\left[\begin{matrix}1 & - 4 \\ 4 & - 1\end{matrix}\right] \left[\begin{matrix}3 & - 2 \\ 0 & - 3\end{matrix}\right]}$
$= \left[\begin{matrix}1 \cdot 3 + \left(- 4 \cdot 0\right) & 1 \cdot - 2 + \left(- 4 \cdot - 3\right) \\ 4 \cdot 3 + \left(- 1 \cdot 0\right) & 4 \cdot - 2 + \left(- 1 \cdot - 3\right)\end{matrix}\right]$
$= \textcolor{b l u e}{\left[\begin{matrix}3 & 10 \\ 12 & - 5\end{matrix}\right]}$
The general notation for that (for $2 \times 2$ matrices for simplicity) is:
$\left[\begin{matrix}\textcolor{b l u e}{{a}_{11}} & \textcolor{b l u e}{{a}_{12}} \\ \textcolor{h i g h l i g h t}{{a}_{21}} & \textcolor{h i g h l i g h t}{{a}_{22}}\end{matrix}\right] \times \left[\begin{matrix}\textcolor{\mathmr{and} a n \ge}{{b}_{11}} & \textcolor{red}{{b}_{12}} \\ \textcolor{\mathmr{and} a n \ge}{{b}_{21}} & \textcolor{red}{{b}_{22}}\end{matrix}\right]$
$= \left[\begin{matrix}\textcolor{b l u e}{{a}_{11}} \textcolor{\mathmr{and} a n \ge}{{b}_{11}} + \textcolor{b l u e}{{a}_{12}} \textcolor{\mathmr{and} a n \ge}{{b}_{21}} & \textcolor{b l u e}{{a}_{11}} \textcolor{red}{{b}_{12}} + \textcolor{b l u e}{{a}_{12}} \textcolor{red}{{b}_{22}} \\ \textcolor{h i g h l i g h t}{{a}_{21}} \textcolor{\mathmr{and} a n \ge}{{b}_{11}} + \textcolor{h i g h l i g h t}{{a}_{22}} \textcolor{\mathmr{and} a n \ge}{{b}_{21}} & \textcolor{h i g h l i g h t}{{a}_{21}} \textcolor{red}{{b}_{12}} + \textcolor{h i g h l i g h t}{{a}_{22}} \textcolor{red}{{b}_{22}}\end{matrix}\right]$ |
1. Topic-
Real-Life Multiplication
2. Content-
Multiplication, commutative property
3. Goals: Aims/Outcomes-
GLE 2.2 Develop understanding of multiplication and related division facts through multiple strategies and representations. GLE 0302.4 Solve multiplication and division problems using various representations.
4. Objectives-
1.GLE 2.7 Represent multiplication using various representations such as equal-size groups, arrays, area models, and equal jumps on number lines. SPI 0302.5 Identify various representations of Multiplication and division.
5. Materials and Aids-
- class set, FULL 24-count boxes of crayons - numerous, spare crayons - Class set, candy bars with segments - Class set, sticker sheets - Class set, colored pencils - Student worksheets - Elmo - Teacher worksheet
6. Procedures/Methods-
A. Introduction-
1. Announce that you will be giving each child a few of their "Favorite Things," including one deliciously mathematical one. 2. Have a student pass out crayon boxes to peers. Tell them not to open the box until the 'Okay' signal is given. Simultaneously, ask students to raise their hands if they know how to multiply already. 3. Once all students have a box, ask them if the box of crayons looks like a multiplication problem to them. 4. Tell them to look at a box more closely. Remember that multiplication is a short way of adding, and we can use it only when we have equal sets of numbers.
B. Development-
1.Pass out the student worksheet. Now model and have them count the number of rows, and the number of crayons in each row. (Ask: How many rows? Do we have equal numbers of crayons in each row? Can we make a multiplication problem for the box of crayons?) Guide them to fill in the first set of blanks on the worksheet for the box of crayons. 2. Model the written multiplication problem on the board. Have students copy the problem on their paper.
C. Practice-
1.Pass out the candy bars, and tell them not to break or eat them until the 'Okay' signal has been given. 2. Allow students to attempt the second activity in pairs, reminding them to examine the number of rows and the number of pieces in each row. 3. Have them record their answers on the worksheet, filling in all the spaces for the second problem. Remind them that the word problem is a question, and should end with a question mark. 4. Ask students to share their answers, first with their table, then with the class.
D. Independent Practice-
1. Allow the students to eat the candy as they work independently to model the multiplication problems for their sticker sheets and colored pencils. 2. Collect student work to be scored.
E. Accommodations (Differentiated Instruction)-
1. Students struggling with multiplication concepts can orally explain or demonstrate the concepts. 2. Advanced learners can extend their learning to work on mastering the commutative property. Early finishers will be asked to model each problem in another way (i.e., 8x3 & 3x8). the candy bar really 2 x 4 or 4 x 2? This would make a great family homework project.
F. Checking for understanding-
1. Independently completed problems (colored pencils & sticker sheets) will be graded for accuracy. 2. Students showing difficulty OR complete mastery will be presented with an accommodation.
G. Closure-
As homework, students will find 2 products at home that could illustrate multiplication, such as eggs or juice boxes. Students will complete the same worksheet for their discovered items.
This Lesson Plan is available at (www.teacherjet.com) |
# How do you find the derivative of f(x)=x^3-5x^2+2x+8?
Jun 30, 2016
Use linearity of differentiation and the power rule to get $f ' \left(x\right) = 3 {x}^{2} - 10 x + 2$.
#### Explanation:
The power rule says that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ for any $n$. Therefore, $\frac{d}{\mathrm{dx}} \left({x}^{3}\right) = 3 {x}^{2}$, $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x$, $\frac{d}{\mathrm{dx}} \left(x\right) = 1 {x}^{0} = 1$, and $\frac{d}{\mathrm{dx}} \left(8\right) = \frac{d}{\mathrm{dx}} \left(8 {x}^{0}\right) = 0$.
Linearity of differentiation says that $\frac{d}{\mathrm{dx}} \left({a}_{1} {f}_{1} \left(x\right) + {a}_{2} {f}_{2} \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} \left(x\right)\right)$
$= {a}_{1} {f}_{1} ' \left(x\right) + {a}_{2} {f}_{2} ' \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} ' \left(x\right)$, where
${a}_{1} , {a}_{2} , \setminus \ldots , {a}_{n}$ are constants.
These facts lead to
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} - 5 {x}^{2} + 2 x + 8\right)$
=d/dx(x^3)-5d/dx(x^2)+2d/dx(x)+d/dx(8) =3x^2-10x+2. |
What Is 6.5 As A Fraction
What Is 6.5 As A Fraction – Equivalent parts can be defined as fractions, but they represent the same value. for example, 9/12 and 6/8 are equivalent because they are both equal to 3/4 when the fractions are simplified.
In its simplest form, as seen in the example above, all equivalent fractions will reduce to the same fraction. How to find equivalent fractions and whether the given fractions are equivalent. Explore the given tutorial to know more about how to check no.
What Is 6.5 As A Fraction
When two or more fractions are simplified, they are said to be equal if they are equal to the same fractions. for example, Equivalent fractions of 1/5 are 5/25; 6/30 and it’s 4/20; This results in an even fraction, 1/5, because it is simple.
Fraction Circles Magnetic Accents
Equivalent fractions are defined as fractions that have the same value regardless of their numerators and denominators. for example, 6/12 and 4/8 are both equal to 1/2; This means that they are equal in nature.
For example: 1/2; 2/4, 3/6, and 4/8 are equivalent fractions. Let’s see how similar their values are. We will represent each of these fragments as circles with shaded segments. We can see that the shaded parts in all the images represent the same part, viewed as a whole.
Here, we can see that all the circles have the same amount of shading. therefore 1/2 2/4, 3/6 and 4/8 are equivalent fractions.
Equivalent fractions can be written by multiplying or dividing the numerator and denominator by the same number. This is the reason why these fractions reduce to the same number when simplified. Let’s understand two ways of composing equivalent fractions.
Add, Subtract, Multiply, & Divide Fractions Digital Quiz Pack {5th Gra
Find equivalent fractions for a given fraction; Multiply the numerator and denominator by the same number. for example, To find the equivalent fraction 3/4; Multiply numerator 3 and denominator 4 by the same number; Say 2. So 6/8 is the equivalent fraction of 3/4. Any other equivalent fraction can be found by multiplying the numerator and denominator of the given fraction by the same number.
Find equivalent fractions for given fractions; Divide the numerator and denominator by the same number. for example, To find the fraction corresponding to 72/108; First, we will find their points of contact. We know that 2 is a common factor of 72 and 108, so you can find the equivalent fraction of 72/108 by dividing its numerator and denominator by 2. So 36/54 is a fraction of 72/108. Let’s see how the section is simplified.
Therefore, the equivalent fraction 72/108 is 36/54; 18/27 6/9 and 2/3. here, 2/3 is the simple form of 72/108, and 2 and 3 have no common factor (except 1).
To determine whether they are equal or unequal, we need to simplify the given fractions. Simple numerators and denominators can be worked to the point where the numerator and denominator are still numbers to produce equivalent numbers. Whether the given fractions are equal can be determined in different ways. Some of them are as follows.
Math Tiles: Equivalent Fractions • Teacher Thrive
Denominators of fractions; 2/6 and 3/9 are 6 and 9. The least common multiple (LCM) of the denominators of 6 and 9 is 18. 18 Multiply the denominators of both fractions by the corresponding numbers.
We can see that both fractions are equal to the same fraction 6/18. Therefore, the given fractions are equivalent.
Note: If the fractions are unequal. You can check whether a fraction is larger or smaller by looking at the denominator of the two resulting fractions. Therefore, this method can also be used to compare fractions.
Let’s find the decimal form of the two fractions 2/6 and 3/9.
Fraction Warm Ups: Problem Solving Task Cards
To determine whether 2/6 and 3/9 are equal; Multiply them and add them up. Fractions are equivalent if both products are equal.
We present the subparts 2/6 and 3/9 with the same shapes and check whether the shaded parts of both are the same.
We can see that the shaded parts of both circles represent the same value. In other words, we can see that the shaded parts in both images represent the same part when viewed as a whole. Therefore, the given fractions are equivalent.
Graphs and tables serve as a useful reference in calculations and are often used to better represent concepts because they are easier to understand. Anchor charts and tables like the ones below make it easier for students to understand equivalent fractions. Let’s use the table below to find equivalent fractions of 1/4.
Games And Activities For Reviewing Fraction Standards
Fractions are equal if two or more fractions have the same value regardless of numerators and denominators. for example, 2/4 and 8/16 are equivalent fractions because simple fractions must be shortened to 1/2.
Since 8/12 and 6/9 reduce to the same fraction (2/3), there can be as many equivalent fractions as ordinary fractions. Similarly, 4/7 and 28/49 are also equivalent fractions.
If the given fractions are simplified and reduced to an ordinary fraction. We can call them equivalent fractions. In addition, there are several other ways to determine whether given fractions are equal. Some of them are as follows.
When two fractions are equal, it means that regardless of different numerators and denominators, they are equal to the same value. In other words, when you simplify them, they reduce to the same fraction.
Expert Maths Tutoring In The Uk
Let’s add the equivalent fractions; subtraction raising Dividing and comparing fractions helps us solve many problems in real time.
An equivalent improper fraction means an equivalent fraction in improper form. A fraction is improper if its numerator is greater than its denominator. for example, 3/2 is an improper fraction equal to 9/6.
Any two fractions can be considered equal if they have the same value. We can determine whether fractions are equal in different ways. The basic method is to reduce them. If it is reduced to the same fraction, it is considered equivalent.
Equivalent fractions can be written by multiplying or dividing the numerator and denominator by the same number. This is the reason why these fractions reduce to the same number when simplified. for example Let’s write the same fraction for 2/3. Multiply the numerator and denominator by 4 to get (2×4)/(3×4) = 8/12. So 8/12 and 2/3 are the same fractions.
Mixed Numbers, Proper And Improper Fractions Explained
If you want to write an equivalent fraction for 6/8, If we multiply the numerator and denominator by 2, we get (6×2)/(8×2) = 12/16. therefore 6/8 and 12/16 are equivalent fractions. now, For 6/8, divide the same fraction by a common denominator, say 2. When we divide the numerator and denominator by 2, we get (6 ÷ 2)/(8 ÷ 2) = 3/4. therefore 6/8 and 3/4 are equivalent fractions.
To find equivalent parts of 1/4, multiply the numerator and denominator by the same number. So (1×2)/(4×2) = 2/8. now, Finding the equivalent fraction for 1/4; Multiply by 3. This will be (1×3)/(4×3) = 3/12. therefore We get two equivalent fractions for 1/4 and they are 2/8 and 3/12.
When two or more fractions are simplified, they are said to be equal if they are equal to the same fractions. for example, Equivalent fractions of 1/6 are 2/12; 3/18 and 4/24 are simple, meaning that 1/6 is the same fraction.
To find equivalent parts of 2/3, multiply the numerator and denominator by the same number. So (2×5)/(3×5) = 10/15. now, Finding an equivalent fraction for 2/3; Multiply by 6. This will be (2×6)/(3×6) = 12/18. therefore We get two equivalent fractions for 2/3 and they are 10/15 and 12/18.
Fractions Around The Room (pizza Theme)
To find equivalent parts of 1/3, multiply the numerator and denominator by the same number. So (1×2)/(3×2) = 2/6. now, Finding an equivalent fraction for 1/3; Multiply by 3. That will be (1×3)/(3×3) = 3/9. therefore We get two equal fractions for 1/3 which are 2/6 and 3/9.
To find equivalent parts of 3/4, multiply the numerator and denominator by the same number. So (3×2)/(4×2) = 6/8. now, To find the equivalent area for 3/4; Multiply by 3.
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# How to Solve On Mixture & Alligation in Teaching Exam?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The numerical ability section is considered to be one of the toughest subjects of Teaching Exams but it can be scored off well if prepared well. Mixture and alligation are some of the toughest chapters which leave candidates a bit confused and most of the aspirants leave these questions untouched.
Table of content
To make the chapter easy for you all, we are providing you all some Basic Concept and Tricks on Mixture and alligation which will surely make the chapter easy for you all.
## Alligation:-
(i) to find the mean or average value of mixture when the prices of two or more ingredients which may be mixed together and the proportion in which they mixed are given (this is Alligation Medial); and
(ii) to find the proportion in which the ingredients at given prices must be mixed to produce a mixture at a given price. This is Alligation Alternate.
## Note:-
(1) The word Alligation literally means linking. The rule takes its name from the lines or links used in working out questions on mixture.
(2) Alligation method is applied for percentage value, ratio, rate, prices, speed etc and not for absolute values. That is, whenever per cent, per hour, per kg, per km etc, are being compared, we can use Alligation.
## Rule of Alligation:-
if the gradients are mixed in a ratio, then
We represent it as under:
Then, (cheaper quantity) : (dearer quantity) = (d – m ) : (m – c)
## Solved problems:-
Ex1: In what proportion must rice at Rs3.10 per kg be mixed with rice at Rs3.60 per kg, so that the mixture be worth Rs3.25 a kg?
Solution:
## By the alligation rule:-
they must be mixed in the ratio 7:3.
## Milk and Water:-
Ex2: A mixture of certain quantity of milk with 16 liters of water is worth 90 P per liter. If pure milk be worth Rs1.08 per liter, how much milk is there in the mixture?
Solution:
The mean value is 90 P and price of water is 0 P.
By the Alligation Rule, milk and water are in the ratio of 5:1.
the quantity of milk in the mixture = 5 ×16 = 80 liters.
Ex3: 300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?
Solution:
The existing solution has 40% sugar, and sugar is to be mixed; so the other solution has 100% sugar. So, by alligation method;
the two mixture should be added in the ratio 5:1
Therefore, required sugar = |
### Free Educational Resources
2 Tutorials that teach Finding the Greatest Common Factor of a Polynomial
# Finding the Greatest Common Factor of a Polynomial
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Author: Colleen Atakpu
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In this lesson, students will learn how to determine the greatest common factor of a polynomial expression, and then use it to factor the expression.
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Tutorial
## Video Transcription
[MUSIC PLAYING] Let's look at our objectives for today. We'll start by looking at factors and terms in polynomials. We'll then review the process of distribution and factoring. We'll talk about factorization with numbers. And finally, we'll do some examples factoring polynomials.
Now, let's look at factors and terms in polynomials. A "polynomial" is an expression involving one or more terms. So here's an example of a polynomial.
7y to the fourth plus 2y minus 3. Terms are separated by addition or subtraction. So in this polynomial, we have three terms.
Each term may contain coefficients, variables, and exponents that are multiplied together. So in our first term, 7 is the coefficient, y is the base, and 4 is the exponent. The coefficients and variables of a term are factors of that term. An expression can also be a factor of a polynomial if it is multiplied by another expression to get the original polynomial as a product. For example, in the expression x minus 1 times 2x plus 5, x minus 1 is a factor and 2x plus 5 is a factor of the polynomial.
Now, let's look at the processes of distribution and factoring. Distribution involves multiplying a factor outside of parentheses by all the terms inside the parentheses. So for example, a times b plus c is equal to a times b plus a times c. The a is multiplied by both the b and the c inside the parentheses.
Factoring is the reverse process of distributing where a common factor is factored out of two or more terms and written outside of the parentheses. So for example, if we start with the expression ab plus ac, a is the common factor of ab and ac. So we can factor it out by writing it on the outside of the parentheses and writing b plus c inside the parentheses.
Now, let's look at factorization. Being able to factor a polynomial is useful for simplifying polynomials, canceling common factors, and can help identify information, such as intercepts, on graphs. The process of finding common factors of a polynomial works like finding common factors of numbers.
To "factor" means to break down a number into smaller numbers that, when multiplied together, give us the original number. So for example, 30 is equal to 2 times 3 times 5. 2, 3, and 5 multiplied together give us 30. 2, 3, and 5 are also the prime factorization of 30 because 2, 3, and 5 are all prime numbers. Writing the prime factorization of all terms in a polynomial helps identify common factors and the greatest common factor.
Now, let's do some examples factoring polynomials. We have the expression 6x to the third plus 3x plus 15. To factor this polynomial, we need to factor or break down each individual term. 6x to the third can be written as 2 times 3 times x times x times x. 2 times 3 are the prime factors of 6, and x to the third means x multiplied by itself three times.
3x is broken down to 3 times x. 15 can be broken down or factored as 3 times 5. Looking at these three factorizations, we see that the common factor of each of these terms is 3. If terms have two or more common factors, the factors can be multiplied together to find the greatest common factor.
Let's do some more examples. Here, we have the polynomial 10x to the third minus 4x squared plus 4x. To factor this polynomial, we need to factor or break down each of the three terms. 10x to the third can be broken down to be 2 times 5 times x times x times x. Negative 4x squared can be broken down as negative 2 times 2 times x times x, and 4x can be broken down as 2 times 2 times x.
Now, we see that each of our terms have two common factors. They each have a positive 2, and they each have an x. So to find our greatest common factor, we multiplied these two common factors, 2 and x. So our greatest common factor is 2x.
Once we have factored out the greatest common factor, the expression inside the parentheses will be the remaining factors of each term. So from our first term, our remaining factors are 5x and x or 5x squared. In our second term, the remaining factors are a negative 2 and x. So negative 2x. And in our last term, the remaining factor is 2.
We can check to see that we have factored the polynomial correctly by distributing the greatest common factor back into each term inside the parentheses. So 2x times 5x squared is 10x to the third. 2x times negative 2x is negative 4x squared, and 2x times 2 is a positive 4x. So because we have arrived back at our original expression, we have factored our polynomial correctly.
Here is our last example. We want to factor 12x squared minus 18x plus 6. To factor this polynomial, we're going to factor each of the three terms. 12x squared can be broken down to be 2 times 2 times 3 times x times x. Negative 18x can be broken down to be negative 2 times 3 times 3 times x, and 6 can be broken down to 2 times 3.
Here, again, we see that we have two common factors in each term. Each term has a 2 and a 3. Multiplying these common factors together gives us our greatest common factor of 6. So 6 is our greatest common factor, and we write it on the outside of the parentheses. And in the parentheses, we write the remaining factors of each term.
So in the first term, our remaining factors are 2x and x or 2x squared. In the second term, our remaining factors are negative 3 and an x. So negative 3x. In our last term, we see that we have factored out both of the factors of 6, which means that we only have a 1 remaining.
Again, we can check to see that we have factored our polynomial correctly by distributing the 6 back into each of the three terms in the parentheses. So 6 times 2x squared is 12x squared. 6 times negative 3x is negative 18x, and 6 times 1 is 6. So we see that we have indeed factored our polynomial correctly.
Let's go over our important points from today. Make sure you get these in your notes so you can refer to them later. A "polynomial" is an expression involving one or more terms.
Coefficients and variables of a term are factors of that term. Factoring is the reverse process of distributing where a common factor is factored out of two or more terms and written outside of the parentheses. If terms have two or more common factors, the factors can be multiplied together to find the greatest common factor.
So I hope these important points and examples helped you understand a little bit more about finding the greatest common factor of a polynomial. Keep using your notes, and keep on practicing. And soon, you'll be a pro. Thanks for watching.
## Notes on "Finding the Greatest Common Factor of a Polynomial"
00:00 – 00:35 Introduction
00:36 – 01:40 Factors and Terms
01:41 – 02:33 Distribution and Factoring
02:34 – 03:30 Factorization
03:31 – 08:05 Examples Factoring Polynomials
08:06 – 08:52 Important to Remember (Recap) |
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Broad Topics > Numbers and the Number System > Properties of numbers
### An Introduction to Irrational Numbers
##### Age 14 to 18
Tim Rowland introduces irrational numbers
### What Are Numbers?
##### Age 7 to 18
Ranging from kindergarten mathematics to the fringe of research this informal article paints the big picture of number in a non technical way suitable for primary teachers and older students.
### Generating Triples
##### Age 14 to 16 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### Triangular Triples
##### Age 14 to 16 Challenge Level:
Show that 8778, 10296 and 13530 are three triangular numbers and that they form a Pythagorean triple.
### Fracmax
##### Age 14 to 16 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Unit Fractions
##### Age 11 to 14 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### Babylon Numbers
##### Age 11 to 18 Challenge Level:
Can you make a hypothesis to explain these ancient numbers?
### Thirty Six Exactly
##### Age 11 to 14 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Small Change
##### Age 11 to 14 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Helen's Conjecture
##### Age 11 to 14 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Repetitiously
##### Age 11 to 14 Challenge Level:
The number 2.525252525252.... can be written as a fraction. What is the sum of the denominator and numerator?
### Even So
##### Age 11 to 14 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Cogs
##### Age 11 to 14 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Few and Far Between?
##### Age 14 to 18 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
##### Age 14 to 16 Challenge Level:
Robert noticed some interesting patterns when he highlighted square numbers in a spreadsheet. Can you prove that the patterns will continue?
### A Long Time at the Till
##### Age 14 to 18 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Filling the Gaps
##### Age 14 to 16 Challenge Level:
Which numbers can we write as a sum of square numbers?
### Magic Letters
##### Age 11 to 14 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Difference Dynamics
##### Age 14 to 18 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
##### Age 11 to 14 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
##### Age 11 to 14 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Times Right
##### Age 11 to 16 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Prime Magic
##### Age 7 to 16 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### Enriching Experience
##### Age 14 to 16 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
### Really Mr. Bond
##### Age 14 to 16 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### Rachel's Problem
##### Age 14 to 16 Challenge Level:
Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
### Mini-max
##### Age 11 to 14 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Counting Factors
##### Age 11 to 14 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Summing Consecutive Numbers
##### Age 11 to 14 Challenge Level:
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
### X Marks the Spot
##### Age 11 to 14 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### One to Eight
##### Age 11 to 14 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Factorial
##### Age 14 to 16 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Special Numbers
##### Age 11 to 14 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Not a Polite Question
##### Age 11 to 14 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Magic Crosses
##### Age 7 to 14 Challenge Level:
Can you find examples of magic crosses? Can you find all the possibilities?
### Whole Numbers Only
##### Age 11 to 14 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### The Patent Solution
##### Age 11 to 14 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Clever Carl
##### Age 7 to 14
What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that.
### Lesser Digits
##### Age 11 to 14 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Cinema Problem
##### Age 11 to 14 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Arrange the Digits
##### Age 11 to 14 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### Satisfying Statements
##### Age 11 to 14 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Guess the Dominoes for Two
##### Age 7 to 14 Challenge Level:
Guess the Dominoes for child and adult. Work out which domino your partner has chosen by asking good questions. |
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Solving Linear Inequalities
When working with linear equations involving one variable whose highest degree (or order) is one, you are looking for THE one value of the variable that will make the equation true. But if you consider an inequality such as x + 2 < 7, then values of x can be 0, 1, 2, 3, any negative number, or any fraction in between. In other words, there are many solutions for this inequality. Fortunately, solving an inequality involves the same strategies as solving a one variable equation. So even though there are an infinite number of answers to an inequality, you do not have to work any harder to find the answer. To review how to solve one variable equations, click here. (linear equations solving.doc)
However, there is one major difference that you must keep in mind when working with any inequality. If you multiply or divide by a negative number, you must change the direction of the inequality sign. You’ll see why this is the case soon.
Let’s go back and look at x + 2 < 7. If this were an equation, you would only need to subtract 2 from both sides to have x by itself.
Keep in mind that the new rule for inequalities only applies to multiplying or dividing by a negative number. You can still add or subtract without having to worry about the sign of the inequality.
But what would happen if you had ? Before solving, let’s think about some values of x that will make this inequality true. If you let x = -5 or -6 or any other value that is less than -5, then the inequality will be true. So you would write your solution as . In the process of solving this inequality using algebraic methods, you would have something that looks like the following.
Let's Practice
Begin by getting the variable on one side by itself by subtracting 3 from both sides. Then divide both sides by 2. Since you are dividing by a positive 2, there is no need to worry about changing the sign of the inequality.
The solution to this problem begins with subtracting 4 from both sides and then dividing by -3. As soon as you divide by -3, you MUST change the sign of the inequality.
This solution will require a little more manipulation than the previous examples. You have to gather the terms with the variables on one side and the terms without the variables on the other side.
There is another type of inequality called a double inequality. This is when the variable appears in the middle of two inequality signs. This is simply a shortcut way of writing two separate inequalities into one and using a shorter process for finding the solution.
More Practice:
The strategy for solving this inequality is not that much different than the other examples. Except in this case, you are trying to isolate the variable in the middle rather than on one side or the other. But the process for getting the x by itself in the middle, you should add 1 to all three parts of the inequality and then divide by 6.
When each of these problems has been solved, the final answer has been given as an inequality. If your teacher or your textbook also asks you to display your answer graphically, click here to learn how this can be done.
Examples
Solve each inequality.
What is your answer?
What is your answer?
What is your answer?
What is your answer?
What is your answer?
What is your answer?
What is your answer?
What is your answer?
What is your answer?
S Taylor
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# 2008 Mock ARML 2 Problems/Problem 1
## Problem
$ABCD$ is a convex quadrilateral such that $|AB| = 5$, $|BC| = 17$, $|CD| = 7$, and $|DA| = 25$. Given that $m\angle{ABC} + m\angle{BCD} = 270^{\circ}$, find the area of $ABCD$.
## Solution
$[asy] pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15; D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4")); [/asy]$
Note that $m\angle{BAD} + m\angle{ADC} = 360 - 270 = 90^{\circ}$. Thus, if we let $E$ be the intersection of the extensions of $\overline{AB}$ and $\overline{CD}$, it follows that $\triangle EAD$ is a right triangle. Immediately we notice that $\triangle ECB$ is a $8-15-17,\,\triangle$ and that $\triangle EAD$ is a $15-20-25\, \triangle$; otherwise we can determine these lengths through the Pythagorean Theorem.
The answer is $[ABCD] = [EAD] - [ECB] = \frac{1}{2}(15)(20) - \frac{1}{2}(8)(15) = \boxed{90}$. |
# Eureka Math Algebra 1 Module 1 Lesson 9 Answer Key
## Engage NY Eureka Math Algebra 1 Module 1 Lesson 9 Answer Key
### Eureka Math Algebra 1 Module 1 Lesson 9 Exercise Answer Key
Exercise 1.
a. Gisella computed 342×23 as follows:
Can you explain what she is doing? What is her final answer?
She is using an area model, finding the area of each rectangle and adding them together. Her final answer is 7,866.
Use a geometric diagram to compute the following products:
b. (3x2+4x+2)×(2x+3)
6x3+17x2+16x+6
c. (2x2+10x+1)(x2+x+1)
2x4 + 12x3+13x2+11x + 1
d. (x-1)(x3+6x2-5)
x4+5x3-6x2-5x+5
→ What do you notice about the terms along the diagonals in the rectangles you drew?
Encourage students to recognize that in parts (b) and (c), the terms along the diagonals were all like terms; however, in part (d) one of the factors has no x-term. Allow students to develop a strategy for dealing with this, concluding with the suggestion of inserting the term + 0x, for a model that looks like the following:
Students may naturally ask about the division of polynomials. This topic will be covered in Algebra II, Module 1. The extension challenge at the end of the lesson, however, could be of interest to students inquiring about this.
→ Could we have found this product without the aid of a geometric model? What would that look like?
Go through the exercise applying the distributive property and collecting like terms. As a scaffold, remind students that variables are placeholders for numbers. If x=5, for example, whatever the quantity on the right is (270), you have
5-1 of “that quantity,” or 5 of “that quantity” minus 1 of “that quantity.” Similarly we have x of that quantity, minus 1 of that quantity.
(x-1)(x3+6x2-5)
x(x3+6x2-5)-1(x3+6x2-5)
x4+6x3-5x- x3-6x2+5
x4+5x3-6x2-5x+5
Exercise 2.
Multiply the polynomials using the distributive property: (3x2+x-1)(x4 – 2x + 1).
3x2 (x4-2x+1)+x(x4-2x+1)-1(x4-2x+1)
3x6-6x3+3x2+x5-2x2+x-x4+2x-1
3x6+x5-x4-6x3+x2+3x-1
Exercise 3.
The expression 10x2+6x3 is the result of applying the distributive property to the expression 2x2 (5+3x). It is also the result of applying the distributive property to 2(5x2+3x3) or to x(10x+6x2), for example, or even to
1∙(10x2+6x3).
For (a) to (j) below, write an expression such that if you applied the distributive property to your expression, it would give the result presented. Give interesting answers!
a. 6a+14a2
b. 2x4+2x5+2x10
c. 6z2-15z
d. 42w3-14w+77w5
e. z2 (a+b)+z3 (a+b)
f. $$\frac{3}{2}$$ s2+ $$\frac{1}{2}$$
g. 15p3 r4-6p2 r5+9p4 r2+3$$\sqrt{2}$$ p3 r6
h. 0.4x9-40x8
i. (4x+3)(x2+x3 )-(2x+2)(x2+x3)
j. (2z+5)(z-2)-(13z-26)(z-3)
a. 2a(3+7a) or 2(3a+7a2) or a(6+14a)
b. 2x4 (1+x+x6) or x(2x3+2x4+2x9 ) or 2(x4+x5+x10)
c. 3z(2z-5) or 3(2z2-5z) or z(6z-15)
d. 7w(6w2-2+11w4) or w(42w2-14+77w4)
e. z2 ((a+b)+z(a+b)) or z(z(a+b)+z2(a+b))
f. $$\frac{1}{2}$$(3s2+1)
g. 3p2 r2 (5pr2-2r3+3p2+$$\sqrt{2}$$ pr4 ) or p2 r2 (15pr2-6r3+9p2+3$$\sqrt{2}$$ pr4)
h. 0.4x8 (x-100) or $$\frac{4}{10}$$ x8 (x-100)
i. (x2+x3 )((4x+3)-(2x+2))
j. (z-2)((2z+5)-13(z-3))
Choose one (or more) to go through as a class, listing as many different rewrites as possible. Then remark:
→ The process of making use of the distributive property “backward” is factoring.
Exercise 4.
Sammy wrote a polynomial using only one variable, x, of degree 3. Myisha wrote a polynomial in the same variable of degree 5. What can you say about the degree of the product of Sammy’s and Myisha’s polynomials?
The degree of the product of the two polynomials would be 8.
Extension
Find a polynomial that, when multiplied by 2x2+3x+1, gives the answer 2x3+x2-2x-1.
x-1
### Eureka Math Algebra 1 Module 1 Lesson 9 Problem Set Answer Key
Question 1.
Use the distributive property to write each of the following expressions as the sum of monomials.
a. 3a(4+a)
3a2+12a
b. x(x+2)+1
x2+2x+1
c. $$\frac{1}{3}$$](12z+18z2)
6z2+4z
d. 4x(x3-10)
4x4-40x
e. (x-4)(x+5)
x2+x-20
f. (2z-1)(3z2+1)
6z3-3z2+2z-1
g. (10w-1)(10w+1)
100w2-1
h. (-5w-3) w2
-5w3-3w2
i. 16s100($$\frac{1}{2}$$ s200 + 0.125s)
8s300+2s101
j. (2q+1)(2q2+1)
4q3+2q2+2q+1
k. (x2-x+1)(x-1)
x3-2x2+2x-1
i. 3xz(9xy+z)-2yz(x+y-z)
27x2 yz+3xz2-2xyz-2y2 z+2yz2
m. (t-1)(t+1)(t2+1)
t4-1
n. (w+1)(w4-w3+w2-w+1)
w5+1
o. z(2z+1)(3z-2)
6z3-z2-2z
p. (x+y)(y+z)(z+x)
2xyz+x2 y+x2 z+xy2+xz2+y2 z+yz2
q. $$\frac{x+y}{3}$$
$$\frac{1}{3}$$ x+$$\frac{1}{3}$$ y
r. (20f10-10f5)÷5
4f10-2f5
s. -5y(y2+y-2)-2(2-y3)
-3y3-5y2+10y
-4
t. $$\frac{(a+b-c)(a+b+c)}{17}$$
$$\frac{1}{17}$$ a2+$$\frac{1}{17}$$ b2–$$\frac{1}{17}$$ c2+2/17 ab
u. (2x÷9+(5x)÷2)÷(-2)
–$$\frac{49 x}{36}$$
v. (-2f3-2f+1)(f2-f+2)
-2f5+2f4-6f3+3f2-5f+2
Question 2.
Use the distributive property (and your wits!) to write each of the following expressions as a sum of monomials. If the resulting polynomial is in one variable, write the polynomial in standard form.
a. (a+b)2
a2+2ab+b2
b. (a+1)2
a2+2a+1
c. (3+b)2
b2+6b+9
d. (3+1)2
16
e. (x+y+z)2
x2+y2+z2+2xy+2xz+2yz
f. (x+1+z)2
x2+z2+2xz+2x+2z+1
g. (3+z)2
z2+6z+9
h. (p+q)3
p3+3p2 q+3pq2+q3
i. (p-1)3
p3-3p2+3p-1
j. (5+q)3
q3+15q2+75q+125
Question 3.
Use the distributive property (and your wits!) to write each of the following expressions as a polynomial in standard form.
a. (s2+4)(s-1)
s3-s2+4s-4
b. 3(s2+4)(s-1)
3s3-3s2+12s-12
c. s(s2+4)(s-1)
s4-s3+4s2-4s
d. (s+1)(s2+4)(s-1)
s4+3s2-4
e. (u-1)(u5+u4+u3+u2+u+1)
u6-1
f. $$\sqrt{5}$$(u-1)(u5+u4+u3+u2+u+1)
$$\sqrt{5}$$u6–$$\sqrt{5}$$
g. (u7+u3+1)(u-1)(u5+u4+u3+u2+u+1)
u13+u9-u7+u6-u3-1
Question 4.
Beatrice writes down every expression that appears in this Problem Set, one after the other, linking them with + signs between them. She is left with one very large expression on her page. Is that expression a polynomial expression? That is, is it algebraically equivalent to a polynomial?
Yes.
What if she wrote – signs between the expressions instead?
Yes.
What if she wrote × signs between the expressions instead?
Yes.
### Eureka Math Algebra 1 Module 1 Lesson 9 Exit Ticket Answer Key
Question 1.
Must the product of three polynomials again be a polynomial? |
# Bounded series and convergence – Serlo
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## Bounded series with positive summands converge
As we learned in the chapter „Monotoniekriterium für Folgen“,[1] monotonicity and boundedness combined are a sufficient criterion to show convergence of a sequence. This criterion is also applicable to series, since the meaning of a series is defined by the sequence of its partial sums. In the following, we will derive a convergence theorem for series that is based on the monotonicity and boundedness criterion for sequences.
We proceed in two steps by separately examining the monotonicity and boundedness of partial sums ${\displaystyle s_{n}\equiv \sum _{k=1}^{n}a_{k}}$ of some series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$.
First we ask: Under what circumstances is the sequence ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$ of partial sums monotonically increasing? Starting with an arbitrary partial sum, we obtain the next one by adding the corresponding term of the series. Therefore the next partial sum is greater (or equal), whenever this summand is positive (or zero). If we require monotonicity for the partial sums, this means all terms of the series must be non-negative (except for the first one, which does not play the role of a difference between partial sums).
We now want to show this statement formally. The relation between two subsequent partial sums is
${\displaystyle s_{n+1}=\sum _{k=1}^{n+1}a_{k}=a_{n+1}+\sum _{k=1}^{n}a_{k}=a_{n+1}+s_{n}.}$
The sequence ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$ is monotonically increasing, if ${\displaystyle s_{n+1}\geq s_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$. Using the relation we just derived, we find:
{\displaystyle {\begin{aligned}&&s_{n+1}&\geq s_{n}\\&\iff &a_{n+1}+s_{n}&\geq s_{n}\\&\iff &a_{n+1}&\geq 0\end{aligned}}}
Hence the sequence of partial sums is monotonically increasing, if and only if ${\displaystyle a_{n+1}\geq 0}$ for all ${\displaystyle n\in \mathbb {N} }$. This simply means ${\displaystyle a_{n}\geq 0}$ for all ${\displaystyle n\geq 2}$.
Next, we address boundedness of the sequence ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$. A real sequence is bounded, if there exists some constant ${\displaystyle b\in \mathbb {R} }$, such that ${\displaystyle s_{n}\leq b}$ for all ${\displaystyle n\in \mathbb {N} }$. With the ${\displaystyle s_{n}}$ being partial sums, we can picture this in the following way: Each partial sum ${\displaystyle s_{n}}$ is a truncation of the infinite sum ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$, where we drop all terms after ${\displaystyle a_{n}}$. Each partial sum is just a finite sum of finite terms, so it is clearly finite itself. Therefore it is not hard to find a common upper bound for some of the partial sums. However, to show boundedness of the sequence, we need to find an upper bound for all partial sums at once. So we are looking for a number that is greater than any of the truncated sums, no matter how late we do the truncation.
We summarize these observations in the following theorem:
Theorem (Bounded series with non-negative summands converge)
Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be a series with ${\displaystyle a_{k}\geq 0}$ for all ${\displaystyle k\geq 2}$. This series converges, if and only if there exists some bound ${\displaystyle b\in \mathbb {R} }$, such that ${\displaystyle b\geq \sum _{k=1}^{n}a_{k}}$ for all ${\displaystyle n\in \mathbb {N} }$.
Proof (Bounded series with non-negative summands converge)
Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be a series with ${\displaystyle a_{k}\geq 0}$ for all ${\displaystyle k\geq 2}$.
Proof step: If the sequence ${\displaystyle \left(\sum _{k=1}^{\infty }a_{k}\right)_{n\in \mathbb {N} }}$ is bounded, the series converges.
We have shown above that for a series with non-negative summands the sequence of partial sums is monotonically increasing. According to the monotonicity criterion for sequences, a monotonically increasing sequence with an upper bound converges. By definition, if its sequence of partial sums converges, a series is said to converge.
Proof step: If ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ converges, its partial sums are bounded.
By definition, if the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ converges, its sequence of partial sums converges. In chapter „Unbeschränkte Folgen divergieren“[2] we learned that convergent sequences are bounded.
Analogously, we can also proof a theorem for monotonically decreasing partial sums:
Theorem (Bounded series with non-positive summands converge)
Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be a series with ${\displaystyle a_{k}\leq 0}$ for all ${\displaystyle k\geq 2}$. This series converges, if and only if there exists some bound ${\displaystyle b\in \mathbb {R} }$, such that ${\displaystyle b\leq \sum _{k=1}^{n}a_{k}}$ for all ${\displaystyle n\in \mathbb {N} }$.
Hint
As we have seen, both theorems are applications of the monotonicity criterion for sequences. Therefore, the literature often refers to them as the monotonicity criterion for series.
## Application: Convergence of the hyperharmonic series
Theorem (Convergence of the hyperharmonic series)
The hyperharmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}}$ is convergent for ${\displaystyle \alpha >1}$ (for ${\displaystyle \alpha =1}$ we obtain the harmonic series, whose divergence has been shown in the corresponding article[3]).
How to get to the proof? (Convergence of the hyperharmonic series)
We want to proof convergence by applying the theorem shown above. All summands of the series are clearly positive, so it remains to show that there is an upper bound for the partial sums. We employ a similar trick as in showing the divergence of the harmonic series.[4] We consider the first ${\displaystyle 2^{n+1}-1}$ summands, which we split into suitable groups, such that each group can be assigned a bound. However, this time we choose upper bounds for the groups, as we eventually want to prove convergence. We use ${\displaystyle 2^{n+1}-1}$ summands instead of ${\displaystyle 2^{n}}$ because this eases finding a bound. At the end of a chain of inequalities we arrive at the geometric series, whose convergence criterion has been shown in this article.[5]
{\displaystyle {\begin{aligned}\sum _{k=1}^{2^{n+1}-1}{\frac {1}{k^{\alpha }}}&=1+{\color {Orange}{\frac {1}{2^{\alpha }}}+{\frac {1}{3^{\alpha }}}}+{\color {OliveGreen}\left({\frac {1}{4^{\alpha }}}+{\frac {1}{5^{\alpha }}}+{\frac {1}{6^{\alpha }}}+{\frac {1}{7^{\alpha }}}\right)}+\ldots +{\color {Blue}\underbrace {\left({\frac {1}{(2^{n})^{\alpha }}}+\ldots +{\frac {1}{(2^{n+1}-1)^{\alpha }}}\right)} _{2^{n}{\text{ summands}}}}\\[0.5em]&\left\downarrow \ {\color {Orange}{\frac {1}{3^{\alpha }}}\leq {\frac {1}{2^{\alpha }}}}\land {\color {OliveGreen}{\frac {1}{7^{\alpha }}}\leq {\frac {1}{6^{\alpha }}}\leq {\frac {1}{5^{\alpha }}}\leq {\frac {1}{4^{\alpha }}}}\land \ldots \land {\color {Blue}{\frac {1}{(2^{n+1}-1)^{\alpha }}}\leq \cdots \leq {\frac {1}{(2^{n})^{\alpha }}}}\quad ({\text{as }}\alpha >1)\right.\\[0.5em]&\leq 1+{\color {Orange}{\frac {1}{2^{\alpha }}}+{\frac {1}{2^{\alpha }}}}+{\color {OliveGreen}\left({\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}\right)}+\ldots +{\color {Blue}\underbrace {\left({\frac {1}{(2^{n})^{\alpha }}}+\ldots +{\frac {1}{(2^{n})^{\alpha }}}\right)} _{2^{n}{\text{ summands}}}}\\[0.5em]&=1+{\color {Orange}\left(2\cdot {\frac {1}{2^{\alpha }}}\right)}+{\color {OliveGreen}4\cdot {\frac {1}{4^{\alpha }}}}+\cdots +{\color {Blue}2^{n}\cdot {\frac {1}{(2^{n})^{\alpha }}}}\\[0.5em]&=1+{\color {Orange}{\frac {1}{2^{\alpha -1}}}}+{\color {OliveGreen}{\frac {1}{4^{\alpha -1}}}}+\cdots +{\color {Blue}{\frac {1}{(2^{n})^{\alpha -1}}}}\\[0.5em]&={\frac {1}{(2^{\alpha -1})^{0}}}+{\color {Orange}{\frac {1}{(2^{\alpha -1})^{1}}}}+{\color {OliveGreen}{\frac {1}{(2^{\alpha -1})^{2}}}}+\cdots +{\color {Blue}{\frac {1}{(2^{\alpha -1})^{n}}}}\\[0.5em]&=\sum _{l=0}^{n}{\frac {1}{(2^{\alpha -1})^{l}}}\\[0.5em]&=\sum _{l=0}^{n}\left({\frac {1}{2^{\alpha -1}}}\right)^{l}\\[0.5em]&\left\downarrow \ {\text{supplement with infinitely many positive summands}}\right.\\[0.5em]&\leq \sum _{l=0}^{\infty }\left({\frac {1}{2^{\alpha -1}}}\right)^{l}\\[0.5em]&\left\downarrow \ {\begin{array}{rl}\alpha >1&\Rightarrow \alpha -1>0\\&\Rightarrow 2^{\alpha -1}>1\\&\Rightarrow {\frac {1}{2^{\alpha -1}}}<1\\&\Rightarrow {\text{geometric series converges}}\end{array}}\right.\\[0.5em]&={\frac {1}{1-{\frac {1}{2^{\alpha -1}}}}}\\[0.5em]&={\frac {1}{1-2^{1-\alpha }}}<\infty \end{aligned}}}
Hence we have found an upper bound for the partial sum ${\displaystyle s_{2^{n-1}+1}}$. Moreover, this upper bound is independent of ${\displaystyle n}$, so it is not only an upper bound for a particular partial sum, but for all ${\displaystyle s_{2^{n-1}+1}}$ at the same time. We have thus shown that the subsequence ${\displaystyle (s_{2^{n-1}+1})_{n\in \mathbb {N} }}$ is bounded. However, this is not sufficient to show boundedness of the full sequence of partial sums.
To address this problem, we need an auxiliary argument. Using the Archimedean property of ${\displaystyle \mathbb {N} }$ and Bernoulli's inequality, we were able to show[6] that for any positive integer ${\displaystyle n>0}$ there is another integer ${\displaystyle m>0}$ such that ${\displaystyle n<2^{m}}$. In other words, any natural number is smaller than some power of two. Similarly, we can show that for any given ${\displaystyle n>0}$ we can also find some ${\displaystyle m>0}$ such that ${\displaystyle n\leq 2^{m+1}-1}$. As the sequence of partial sums is monotonically increasing, this implies ${\displaystyle s_{n}\leq s_{2^{m+1}-1}}$. So any partial sum is bounded by one of the partial sums for which we have derived a universal bound above. We can conclude that this bound is a bound for all partial sums, such that the criterion is applicable to the series.
Proof (Convergence of the hyperharmonic series)
For any ${\displaystyle k\geq 1}$ and ${\displaystyle \alpha >1}$, we have
${\displaystyle {\frac {1}{k^{\alpha }}}>0.}$
This proves monotonicity of the sequence of partial sums.
Furthermore, for any ${\displaystyle n\in \mathbb {N} }$ there is some ${\displaystyle m\in \mathbb {N} }$ such that ${\displaystyle n\leq 2^{m+1}-1}$. This enables the chain of inequalities
{\displaystyle {\begin{aligned}s_{n}=\sum _{k=1}^{n}{\frac {1}{k^{\alpha }}}\leq \sum _{k=1}^{2^{m+1}-1}{\frac {1}{k^{\alpha }}}&=1+{\color {Orange}{\frac {1}{2^{\alpha }}}+{\frac {1}{3^{\alpha }}}}+{\color {OliveGreen}\left({\frac {1}{4^{\alpha }}}+{\frac {1}{5^{\alpha }}}+{\frac {1}{6^{\alpha }}}+{\frac {1}{7^{\alpha }}}\right)}+\ldots +{\color {Blue}\left({\frac {1}{(2^{m})^{\alpha }}}+\ldots +{\frac {1}{(2^{m+1}-1)^{\alpha }}}\right)}\\[0.5em]&\leq 1+{\color {Orange}{\frac {1}{2^{\alpha }}}+{\frac {1}{2^{\alpha }}}}+{\color {OliveGreen}\left({\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}+{\frac {1}{4^{\alpha }}}\right)}+\ldots +{\color {Blue}\underbrace {\left({\frac {1}{(2^{n})^{\alpha }}}+\ldots +{\frac {1}{(2^{n})^{\alpha }}}\right)} _{2^{n}{\text{ summands}}}}\\[0.5em]&\leq \sum _{l=0}^{m}\left({\frac {1}{2^{\alpha -1}}}\right)^{l}\\[0.5em]&\leq \sum _{l=0}^{\infty }\left({\frac {1}{2^{\alpha -1}}}\right)^{l}={\frac {1}{1-2^{1-\alpha }}}<\infty .\end{aligned}}}
Thus, the sequence of partial sums is also bounded. The monotonicity criterion then shows convergence of the hyperharmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}}$ for any ${\displaystyle \alpha >1}$. |
# RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.3
## Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.3
RBSE Solutions For Class 12 Maths Chapter 13.3 Question 1.
Find vector product of vectors
and
Solution:
RBSE Solutions For Class 12 Maths Chapter 13 Question 2.
Find perpendicular unit vector of vectors
and
Solution:
Exercise 13.3 Class 12 RBSE Question 3.
For vectors $$\overrightarrow { a }$$ and $$\overrightarrow { b }$$, prove that
Solution:
Ex 13.3 Class 12 RBSE Question 4.
Prove that
Solution:
According to question,
RBSE Class 12 Maths Chapter 13 Question 5.
If $$\overrightarrow { a }$$, $$\overrightarrow { b }$$, $$\overrightarrow { c }$$ are unit vectors, such that
$$\overrightarrow { a }$$ . $$\overrightarrow { b }$$ = 0 = $$\overrightarrow { a }$$ . $$\overrightarrow { c }$$ and angle between $$\overrightarrow { b }$$ and $$\overrightarrow { c }$$ is $$\frac { \pi }{ 6 }$$, then prove that $$\overrightarrow { a }$$ = ± 2 ($$\overrightarrow { b }$$ × $$\overrightarrow { c }$$)
Solution:
Given that
Maths Class 12 Exercise 13.3 Question 6.
Find the value of
Solution:
We know that if $$\overrightarrow { a }$$ and $$\overrightarrow { b }$$ are two vectors and θ is the angle between them, then
Exercise 13.3class 12 Question 7.
Find vector perpendicular to the vectors
and
whose magnitude is 9 unit.
Solution:
Class 9 Math Exercise 13.3 Question 8.
Show that:
also, explain geometrically.
Solution:
= 2(vector area of parallelogram ABCD).
Thus we conclude that area of parallelogram whose adjacent sides are diagonals of given parallelogram is twice the area of given parallellogram.
Ex 13.3 Class 12 Question 9.
For any vector $$\overrightarrow { a }$$, prove that
Solution:
Ex 13.3 Class 10 Question 10.
If two adjacent sides of a triangle are represented by vectors
and
then find the area of triangle.
Solution: |
# How do you simplify (-8/21 x^2y^3) times (-7/16xy^2)?
Aug 15, 2017
See a solution process below:
#### Explanation:
First, rewrite the expression as:
$\left(- \frac{8}{21} \times - \frac{7}{16}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$
$\left(- \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{21}} 3}} \times - \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{7}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} 2}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$
$\left(- \frac{1}{3} \times - \frac{1}{2}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$
$\frac{1}{6} \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right)$
Next, use this rule for exponents to rewrite the $x$ term:
$a = {a}^{\textcolor{red}{1}}$
$\frac{1}{6} \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies \frac{1}{6} \left({x}^{2} \times {x}^{\textcolor{red}{1}}\right) \left({y}^{3} \times {y}^{2}\right)$
Now, use this rule of exponents to simplify the $x$ and $y$ terms:
${x}^{\textcolor{red}{a}} \cdot {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$\frac{1}{6} \left({x}^{\textcolor{red}{2}} \times {x}^{\textcolor{b l u e}{1}}\right) \left({y}^{\textcolor{red}{3}} \times {y}^{\textcolor{b l u e}{2}}\right) \implies$
$\frac{1}{6} {x}^{\textcolor{red}{2} + \textcolor{b l u e}{1}} {y}^{\textcolor{red}{3} + \textcolor{b l u e}{2}} \implies$
$\frac{1}{6} {x}^{3} {y}^{5}$
Or
$\frac{{x}^{3} {y}^{5}}{6}$ |
6-1
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# 6-1 - PowerPoint PPT Presentation
6-1. Integer Exponents. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Holt Algebra 1. Warm Up Evaluate each expression for the given values of the variables. 1. x 3 y 2 for x = –1 and y = 10 2. for x = 4 and y = (–7)
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6-1
Integer Exponents
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1
Holt Algebra 1
Warm Up
Evaluate each expression for the given values of the variables.
1.x3y2 for x = –1 and y = 10
2. for x = 4 and y = (–7)
Write each number as a power of the given base.
3. 64; base 4
4. –27; base (–3)
Objectives
Evaluate expressions containing zero and integer exponents.
Simplify expressions containing zero and integer exponents.
RECALL:
Positive exponents.
To simplify 32, use 3 as a factor 2 times: 32 = ???
VOCABULARY
Positive exponents
EXPONENT
BASE
5
5
5
5
ZERO OR NEGATIVE EXPONENTS
look for a pattern to figure it out.
55
54
53
52
51
50
5–1
5–2
625
3125
125
25
5
When the exponent decreases by one, the value of the power is divided by 5.
=
=
=
VOCABULARY
Zero Exponents – Any nonzero number raised to the zero power is 1.
Examples:
VOCABULARY
Negative Exponents– A nonzero number raised to a negative exponent is equal to 1 divided by that number raised to the opposite (postive) exponent.
Example:
2–4 is read “2 to the negative fourth power.”
gal is equal to
Example 1: Application
One cup is 2–4 gallons. Simplify this expression.
Try this one yourself!!!
A sand fly may have a wingspan up to 5–3 m. Simplify this expression.
TRY THESE YOURSELF!!!
Simplify.
A. 4–3
B. 70
7º = 1
C. (–5)–4
D. –5–4
E. (–2)–5
Caution
In (–5)–4, the base is negative because the negative sign is inside the parentheses. In –5–4the base (5) is positive.
Use the definition
Example 2
Evaluate the expression for the given value of the variables.
x–2 for x = 4
Substitute 4 for x.
Example 3: Try another yourself!!!
Simplify the expression for the given values of the variables.
–2a0b-4 for a = 5 and b = –3
Substitute 5 for a and –3 for b.
Evaluate expressions with exponents.
Write the power in the denominator as a product.
Simplify the denominator.
Simplify.
for a = –2 and b = 6
Try these yourself!!!
Evaluate the expression for the given value of the variable.
1.) p–3for p = 4
2.)
What if you have an expression with a negative exponent in a denominator, such as ?
*Move the denominator to the top and change the sign
**An answer with a negative exponent IS NOT simplified fully
B.
Example 4: Together
Simplify.
A. 7w–4
and
Example 4: Simplifying Expressions with Zero and Negative Numbers
Simplify.
C.
Simplify.
a. 2r0m–3
b.
c.
6.1 HOMEWORK
PG. 395 - 397
#24-42(all), 44-66(evens), 78-83, 100 |
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# One Step Equations
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• You must subtract the same number from what you want to be moved. The idea is keep the variable alone by doing the opposite.
• Add the opposite sign of the number you want to move.
• When solving multiplication equations, you divide both sides by the number attached to the variable. Be sure to use the same sign.
• Multiply both sides by the reciprocal. You must keep the sign of the fraction with the variable.
• ### One Step Equations
1. 1. One step equations Add Subtract Multiply Divide
2. 2. Addition <ul><li>X + 5 = -9 </li></ul><ul><li>X + 5 - 5 = -9 -5 </li></ul><ul><li>X = -14 </li></ul><ul><li>-6 + X = -4 </li></ul><ul><li>-6 + 6 + X = -4 + 6 </li></ul><ul><li>X = 2 </li></ul>
3. 3. Try the following on your paper. <ul><li>10 = X + 3 </li></ul><ul><li>5 + X = 9 </li></ul><ul><li>6 = 3 + X </li></ul>
4. 4. Answers!!! <ul><li>X = 7 </li></ul><ul><li>X = 4 </li></ul><ul><li>X = 3 </li></ul>
5. 5. Equations with subtraction <ul><li>X - 5 = 8 </li></ul><ul><li>X - 5 + 5 = 8 +5 </li></ul><ul><li>X = 13 </li></ul><ul><li>5 - X = 6 </li></ul><ul><li>5 - 5 + X = 6 - 5 </li></ul><ul><li>X = 1 </li></ul>
6. 6. Try the following on your paper. <ul><li>X - 7 = -1 </li></ul><ul><li>X - 4 = 66 </li></ul><ul><li>8 = X - 2 </li></ul><ul><li>-8 = X - 2 </li></ul>
7. 7. Answers!!! <ul><li>X = 6 </li></ul><ul><li>X = 70 </li></ul><ul><li>X = 10 </li></ul><ul><li>X = -6 </li></ul>
8. 8. Multiplication of Equations <ul><li>3X = -9 </li></ul><ul><li>3X /3 = -9 /3 </li></ul><ul><li>X = -3 </li></ul><ul><li>-5X = -40 </li></ul><ul><li>-5X /-5 = -40 /-5 </li></ul><ul><li>X = 8 </li></ul>
9. 9. Try these on your paper. <ul><li>4X = -16 </li></ul><ul><li>-1X = 9 </li></ul><ul><li>6 = 5X </li></ul><ul><li>12X = 3 </li></ul>
10. 10. Answers!!! <ul><li>X = -4 </li></ul><ul><li>X = -9 </li></ul><ul><li>X = 6/5 </li></ul><ul><li>X = 1/4 </li></ul>
11. 11. Division of Equations <ul><li>1/2X = 4 </li></ul><ul><li>(2/1) 1/2X = 4 (2) </li></ul><ul><li>X = 8 </li></ul><ul><li>-3/4X = 6 </li></ul><ul><li>(-4/3) (-3/4)X = 6 (-4/3) </li></ul><ul><li>X = -8 </li></ul>
12. 12. Try these on your paper. <ul><ul><li>X/3 = 9 </li></ul></ul><ul><ul><li>-X/4 = 7 </li></ul></ul><ul><ul><li>1/3X = -1 </li></ul></ul><ul><ul><li>4 = X/4 </li></ul></ul>
13. 13. Answers!!! <ul><li>X = 27 </li></ul><ul><li>X = -28 </li></ul><ul><li>X = -3 </li></ul><ul><li>X = 16 </li></ul>
14. 14. Write the following on your paper for a quiz grade. <ul><li>_4X = 15 </li></ul><ul><li>X/2 = -8 </li></ul><ul><li>9 - X = -1 </li></ul><ul><li>-12 + X = 0 </li></ul><ul><li>-X/3 = 4 </li></ul><ul><li>X + 5 = 5 </li></ul><ul><li>X - 4 = 4 </li></ul><ul><li>-X = 5 </li></ul><ul><li>X + 50 =-50 </li></ul><ul><li>X/5 = -55 </li></ul> |
## Lesson 10: Solids.Surface Areas (I)
May 16, 2012
Vocabulary on Solids
Lesson 10: Solids. Surface Areas (Notes)
Practice Problems on Classifying Solids
Practice problems on Surface Area of Solids
I Have to Know by the End of this Lesson
Three-D shapes have 3-dimensions- length, width and depth. We are going to study some of them: polyhedrons and revolution solids.
### 1. Polyhedrons
A polyhedron is a three-dimensional region of the space bounded by polygons.
Some solids have curved surfaces or a mix of curved and flat surfaces (so they aren’t polyhedrons).
If you click on the links you will learn more about these 3-D shapes in the web where it is taken from the table below, www.mathisfun.com. You can also find questions with answers.
Polyhedrons :
(they must have flat faces)
Platonic Solids Prisms Pyramids
Non-Polyhedra:
(if any surface is not flat)
Sphere Torus Cylinder Cone
Elements of a polyhedron
Faces: These polygons that limit the polyhedrons.
Edges: Line segments where two faces of a polyhedron meet. They are sides of the faces.
Vertices of the polyhedron: Points at which three or more polyhedron edges of a polyhedron meet.
Diagonals of a polyhedron: Segments, joining two vertices, which are not placed on the same face, are called diagonals of polyhedron. The tetrahedron has no diagonals.
Dihedral angle (also called the face angle) is the internal angle at which two adjacent faces meet. All dihedral angles between the edges are ≤ 180º.
Polyhedral angle: is the portion of space limited by tree or more faces which meet at a vertex. All polyhedral angles between the edges are ≤ 360º .
Net: It is an arrangement of edge-joined polygons in the plane which can be folded (along edges) to become the faces of the polyhedron. There are several possibilities for a net of a polyhedron.
Tetrahedrom Net for a tetrahedron
You can find the net of the main polyhedron and their different possibilities on
http://gwydir.demon.co.uk/jo/solid/index.htm
### 2.1. Types of polyhedrons
A convex polyhedron is defined as follows: no line segment joining two of its points contains a point belonging to its exterior. There are many examples known by you: the cube, prisms, pyramids….
A concave polyhedron, on the other hand, will have line segments that join two of its points with all but the two points lying in its exterior.
Below is an example of a concave polyhedron.
The study of polyhedrons was a popular study item in Greek geometry even before the time of Plato (427 – 347 B.C.E.) In 1640, Rene Descartes, a French philosopher, mathematician, and scientist, observed the following formula. In 1752, Leonhard Euler, a Swiss mathematician, rediscovered and used it.
In a convex polyhedron:
If F = number of faces , V = number of vertices and E = number of edges, then
F + V = E +2
This formula is named Euler’s Formula.
All the convex polyhedrons verify this formula. There are some concave polyhedrons that verify it. However, there are concave polyhedrons that don’t verify this formula.
### 2.2. Regular polyhedrons
A regular polyhedron is a polyhedron where:
• each face is the same regular polygon
• the same number of faces (polygons) meet at each vertex (corner)
They are also called Platonic solids.
There are five Platonic solids:
1. Tetrahedron, which has three equilateral triangles at each corner. 2. Cube, which has three squares at each corner. 3. Octahedron, which has four equilateral triangles at each corner. 4. dodecahedron, which has three regular pentagons at each corner. 5. Icosahedron, which has five equilateral triangles at each corner.
Exercise: Fill in this table and check they satisfy Euler’s formula. Why?
``` faces edges vertices
tetrahedron ___ ___ ___
cube ___ ___ ___
octahedron ___ ___ ___
dodecahedron ___ ___ ___
icosahedron ___ ___ ___``` |
# NCERT Solutions for Class 9 Maths Chapter 12: Heron’s Formula
## NCERT Solutions for Class 9 Mathematics Chapter 12 Free PDF Download
The dot mark field are mandatory, So please fill them in carefully
Q. The perimeter of a triangular field is 540 m and the ratio of its sides is 12 : 17 : 25. Find the area of the field.
Ans.
$$\text{Given : Ratio of sides of the field = 12 : 17 : 25}\\\text{Let the lengths of sides be 12x, 17x and 25x,}\\ respectively.\\ Perimeter of triangle = 540m\\ ⇒ 12x + 17x + 25x = 540\\ ⇒ 54x = 540\\ ⇒ x = 10m\\ \text{ Lengths of sides are}\\ \text{12 x = 120m, 17 x = 170m and 25 x = 250 m.}\\ Semi-perimeter, s =\frac{a+b+c}{2}=\frac{120+170+250}{2}\\ =\frac{540}{2}\\ = 270m\\ \text{Now, Area of triangle} =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 270(270-120)(270-170)(270-250)}\\ =\sqrt{\mathstrut 270(150)(100)(20)}\\ =\sqrt{\mathstrut 100 × 100 × 30 × 270}\\ = 100 × 10 × 9\\ = 9000\\ Hence, the area of the field is 9,000 m^2.$$
Q. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Ans.
$$\text{Let the base of the isosceles triangle be a cm.}\\\text{Now, Perimeter = 30 cm}\\⇒ 12 + 12 + a = 30\\ ⇒ a = 30 – 24 = 6 cm\\ Also, s =\frac{Perimeter}{2}=\frac{30}{2}=15cm\\ \text{Now, Area of triangle} =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 15(15-12)(15-12)(15-5)}\\ =\sqrt{\mathstrut 15×3×3×9}\\ =9\sqrt{\mathstrut 15}cm^2$$
Q. What is the area of an isosceles triangle having base 2 cm and length of one of its equal sides as 4 cm ?
Ans.
$$\text{By Heron’s formula,}\\ s=\frac{a+b+c}{2}\\ =\frac{4+4+2}{2}=5$$
$$A =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 5(5-2)(5-4)(5-4)}\\ =\sqrt{\mathstrut 5×3}\\ ={\mathstrut 15}cm^2$$
Q. The edges of a triangular board are 6cm, 8cm and 10cm. What is the cost of painting it at the rate of a 9 paise cm2 ?
Ans.
$$\text{Area of the triangular board, by Heron’s formula,}\\ Here, a = 6, b = 8, c = 10\\ S =\frac{a+b+c}{2}=\frac{6+8+10}{2}=12\\ Area =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 12(12-6)(12-8)(12-10)}\\ =\sqrt{\mathstrut 12×6×4×2}\\ =\sqrt{\mathstrut 2×2×3×2×2×2×2}\\ = 2 × 2 × 2 × 3 = 24 cm^2\\ \text{Cost = Area × Rate}$$
Q. There is a slide in a park. One of its side walls has been painted in some colour with a message “keep the Park Green and Clean”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colours.
Ans.
$$\text{Area of the triangular board, by Heron’s formula,}\\ Here, a = 6, b = 8, c = 10\\ S =\frac{a+b+c}{2}=\frac{11+6+15}{2}=12\\ Area =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 16(16-15)(16-11)(16-6)}\\ =\sqrt{\mathstrut 16×1×5×10}\\ =\sqrt{\mathstrut 4×4×5×5×2}\\ =20\sqrt{\mathstrut 2}\\ \text{Hence, the area painted in colour is}=20\sqrt{\mathstrut 2}m^2$$
Q. A traffic signal board, indicating “School Ahead” is an equilateral triangle with its perimeter 180 cm. What will be the area of the signal board?
Ans.
$$\text{Perimeter of signal board = 180 cm}\\ \text{3 × Side = 180 cm}\\ ⇒ Side =\frac{180}{3}=60cm\\ \text{Area of signal board}=\frac{\sqrt{\mathstrut 3}}{4}×(side)^2\\=\frac{\sqrt{\mathstrut 3}}{4}60×60\\=900\sqrt{\mathstrut 3}cm^2$$
Q. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield on earning of ₹5000 per m2 per year. A company hired both walls for 3 months. How much rent did it pay?
Ans. Consider DABC as one of the side walls of the flyover. So, in DABC, a = 122 m, b = 120 m and c = 22 m
$$⇒ s =\frac{a+b+c}{2}\\ =\frac{122+120+22}{2}\\ =\frac{264}{2}=132 m\\ ar(△ABC)={\sqrt{\mathstrut s(s-a)(s-b)(s-c)}}\\ ={\sqrt{\mathstrut 132(132-122)(132-120)(132-22)}}\\ ={\sqrt{\mathstrut 132×10×12×110}}\\ ={\sqrt{\mathstrut 12×11×10×12×11×10}}\\ = 10×11×12\\ =132m^2\\ \text{Area of both walls = 2 × ar(△ABC)}\\ = 2 × 1320\\ = 2640 m^2\\ \text{Rate of rent }= ₹ 5000 per m^2 per year\\ \text{Rate paid for 3 months = Rate × Area × Time}\\ = ₹ 5000 × 2640 ×\frac{3}{12}\\ = ₹ 5000 × 220 × 3\\ = ₹ 33,00,000$$ |
$$\phi = 1.618...$$. Much like $$\pi$$, $$\phi$$ shows up in the most unexpected places.
Where does $$\phi$$ come from? One way to find $$\phi$$ is to ask a simple question about line segments. If I had a line with a single point on it, at what point does the ratio of the entire line to the larger segment equal the ratio of the larger segment to the smaller segment?
If we say that the larger segment has length $$a$$ and the smaller segment has length $$b$$, then we could express this as an equation. Before we do, let’s use some examples to prove to ourselves using intuition that this point actually exists. What if $$a = b$$? Then the ratio of the entire line to $$a$$ would be $$\frac{2}{1}$$, and the ratio of $$a$$ to $$b$$ is $$\frac{1}{1}$$. What if $$a$$ was twice as big as $$b$$? The ratio of the entire line ($$a+b$$) to $$a$$ would be $$\frac{3}{2}$$ and $$a$$ to $$b$$ would be $$\frac{2}{1}$$. In the first example, $$2 > 1$$ and in the second $$1.5 < 2$$. Since moving the point that separates the line into $$a$$ and $$b$$ is a continuous function, we know that there must be a point that the two values are equal!
In fact, we can use the equation
$\frac{a + b}{a} = \frac{a}{b}$
as this is exactly what we are trying to find. Also since we’re looking at a ratio, we can set one of the numbers $$a$$ or $$b$$ to whatever we want and solve for the other one. For convenience let’s use $$a = 1$$, although it really doesn’t matter (you can prove this to yourself by trying to substitute any number for either $$a$$ or $$b$$). We now have
$1 + b = \frac{1}{b}$
Since we’re solving for $$b$$, let’s use the variable $$x$$ instead and rearrange to a more familiar quadratic expression
$x + x^2 = 1$
To solve this, let’s plug it into the quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
with $$a = 1, b=1, c=-1$$. This gives us the solution that $$x = \frac{\pm \sqrt{5} -1}{2}$$. Since we’re talking about lengths, $$x = \frac{-\sqrt{5} - 1}{2}$$ doesn’t make sense because it is less than zero, so we are left with the solution $$x= \frac{\sqrt{5} - 1}{2}$$. Low and behold $$\frac{\sqrt{5} - 1}{2} = 0.618... = b$$. This means that the length of the total line is $$a + b = 1 + 0.618... = 1.618...$$ and the ratio is $$1.618... : 1$$.
The ratio $$1 : 1.618...$$ is often called the golden ratio and appears everywhere.
The video above starts with the fibonacci numbers and is directly related to a previous post of mine where I explore how to generate fibonacci numbers.
The second main idea that the video explores is how the angle 137.5 degrees is used in nature. 137.5 degree is what you get when you divide 360 degrees into two segments that are related by the golden ratio. But why the golden ratio? Why not a normal ratio that can be expressed as a fraction?
The example given in the video is the distribution of sunflower seeds, but this also applies to where new branches sprout from previous branches. For the sunflower seeds, they are created in the center and then have to be pushed outward. What should they be pushed? Our intuition says that they should go where they would have the most room, where the least seeds currently are. To see why the golden angle is the perfect new angle for each new seed, let’s think about what would happen for other angles.
Let’s say that the sunflower decided to spit out seeds every 180 degrees. The first two seeds would be perfectly situated, but after that they would be pushed exactly where previous seeds are. We could concoct a different angle, but as long as it is a fraction of whole numbers $$\frac{X}{Y}$$, we could start overlapping again after at most $$Y$$ times. This is why nature doesn’t use clean fractions, it uses irrational numbers. Irrational numbers are exactly the numbers that cannot be expressed as a fraction of two whole numbers. But again we have to ask ourselves, why $$\phi$$?
$$\phi$$ is actually the most irrational number. What does that even mean? So every irrational number can be expressed as the sum of an infinite sequence. You can use the term “more irrational” to express sequences that take longer to converge on their infinite sum. $$\phi$$ is actually proven to be the slowest at converging in a theorem by Hurwitz. How do we generate $$\phi$$? The fibonacci numbers are one way, but it can also be expressed in it’s continued fraction expansion as
$1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + ...}}}}}}$
Therefore nature uses the most irrational number to ensure that branches and sunflower seeds get put in the most optimal position. This also explains why fibonacci numbers show up everywhere in nature. The fibonacci numbers are the best whole number approximations for the golden ratio, and since items in nature can’t have fractional values, they instead take on fibonacci numbers.
If you take a pineapple or pinecone and count the number of spirals on it, there will be a fibonacci number of spirals.
Look at any flower: petals arranged in fibonacci spirals. The cool thing is that there are often many different ways to count the spirals on one of these objects: each of the spirals will be a fibonacci number!
Numberphile has a video on this very subject, if you are interested in seeing more of the math behind $$\phi$$. |
# Week 1 Practice Problems
Only available on StudyMode
• Published : September 14, 2014
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University of Phoenix Material
Week One Practice Problems
Prepare a written response to the following questions.
Chapter 1
12. Explain and give an example for each of the following types of variables:
a. Equal interval: A variable in which the values are approximate amounts. An example would be a GPA average.
b. Rank-order: A variable that is determined in ranks or order. An example would be where a graduate stands among other classmates.
c. Nominal: Values that are determined in categories as opposed to numeric value. Gender, Race, or Ethnicity are considered nominal values.
d. Ratio scale: A value is measured on a ratio scale if 0 is a possible factor. A pain chart given in a hospital is a ratio scale. When asked how bad the patient’s pain is, it is measured on a scale from 0-10.
e. Continuous: Variables that have infinite numbers between the two. An example can be found in the question, “How old are you?” There are many different possibilities, therefore, it is a continuous variable.
f. Discrete: Variables with specific values. They cannot have values between the specific values. An example would be found in the question, “How many children do you have?” You can have 1 child, 2, or 3 children, but cannot have 3.5 kids.
15. Following are the speeds of 40 cars clocked by radar on a particular road in a 35-mph zone on a particular afternoon:
30, 36, 42, 36, 30, 52, 36, 34, 36, 33, 30, 32, 35, 32, 37, 34, 36, 31, 35, 20 24, 46, 23, 31, 32, 45, 34, 37, 28, 40, 34, 38, 40, 52, 31, 33, 15, 27, 36, 40
Make a frequency table and a histogram, then describe the general shape of the distribution.
Frequency Table
Speed of Cars
Number of Cars
Percentage
36
5
13.89%
34
4
11.76%
30
3
10.00%
31
3
9.68%
32
3
9.38%
40
3
7.50%
33
2
6.06%
35
2
5.71%
37
2
5.41%
52
2
3.85%
15
1
6.67%
20
1
5.00%
23
1
4.35%
24
1
4.17%
27
1
3.70%
28
1
3.57%
38
1
2.63%
42
1
2.38%
45
1
2.22%
46
1
2.17%
21. Raskauskas and Stoltz (2007) asked a group of 84 adolescents about their involvement in traditional and electronic bullying. The researchers defined electronic bullying as “…a means of bullying in which peers use electronics {such as text messages, emails, and defaming Web sites} to taunt, threaten, harass, and/or intimidate a peer” (p.565). The table below is a frequency table showing the adolescents’ reported incidence of being victims or perpetrators or traditional and electronic bullying.
a. Using this table as an example, explain the idea of a frequency table to a person who has never had a course in statistics. The first column in this chart will identify the methods in which people use to bully others.This is a nominal value. The “N” will reflect the number of people bullied in that manner, while the percentage will show the total percentage of people bullied using each category type.
b. Explain the general meaning of the pattern of results.
This chart shows that the typical types of bullying are the most frequent, while electronic methods are rising. The correlation between the victims and the bullies indicates that there are more victims to bullying than there are bullies.
Incidence of Traditional and Electronic Bullying and Victimization (N=84) Forms of Bullying
N
%
Electronic victims
41
48.8
Text-message victim
27
32.1
Internet victim (websites, chatrooms)
13
15.5
Picture-phone victim
8
9.5
60
71.4
Physical victim
38
45.2
Teasing victim
50
59.5
Rumors victim
32
38.6
Exclusion victim
30
50
Electronic Bullies
18
21.4
Text-message bully
18
21.4
Internet bully
11
13.1
5
64.3
Physical bully
29
34.5
Teasing bully
38
45.2
Rumor bully
22
26.2
Exclusion bully
35
41.7
22. Kärnä and colleagues (2013)... |
3.16
Skip to 0 minutes and 12 secondsWe're now going to discuss in some detail what are called conic sections. Why conic sections? Well, the ancient Greeks had found long ago that if you took a cone and if you sliced it by a plane, then depending on the positioning and the angle of the plane with which you did the slicing, you would either get an ellipse or possibly a parabola or maybe a hyperbola-- the two branches of a hyperbola. Now, personally, I attach no great importance to this intriguing fact, but the curves themselves are undoubtedly of the greatest importance in a wide range of applications and optics and mechanics and so forth. They are truly fundamental mathematical objects.
Skip to 1 minute and 4 secondsNow I have to give great credit to these ancient Greek mathematicians for sensing the importance of these curves and for being able to study them so completely. Let's summarize some things we basically know about the conic sections. The first class that we looked at was the ellipse. The canonical equation of an ellipse, which is defined in a certain geometrical way that I will not go back on, the canonical equation is of the form x squared over a squared plus y squared over b squared equals 1. As you know, this means that a point, whose coordinates are x, y, lies on the ellipse if and only if the coordinates satisfy this equation.
Skip to 1 minute and 47 secondsIf the a is greater than b in this equation, then the ellipse is horizontally oriented. If the a is less than b, then the ellipse is vertically oriented. And if a is equal b, the equation turns out to be that of a circle. So in this sense, you can consider a circle to be a special case of an ellipse. Then we had looked at the parabola. The canonical equation of a parabola, initially that we found, was y equals x squared over 4 p. And if p is a positive parameter, that gives us our upward opening parabola that we saw before. The same equation if p is negative, gives you a parabola but one that opens downward instead of upwards.
Skip to 2 minutes and 38 secondsNow you notice in the equation of a parabola, there's not perfect symmetry between the y and the x. One is squared and one isn't. One can easily consider a parabola in which it's the x that's equal to y squared over 4 p, for instance. In that case, if p is positive you'll get a rightward horizontally oriented parabola. And if p is negative, you'll get a leftward opening parabola. And finally, we had looked at the hyperbola. Its canonical equation has that minus sign in front of the y squared, as you know.
Skip to 3 minutes and 15 secondsIf you now allow the possibility of plus or minus on the right hand side-- before we only had plus 1-- you cover the two cases, the horizontal and the vertical. That is, if you take the plus 1 as we had before, you get a horizontally oriented hyperbola, which as you know has two branches, 1 leftward and 1 rightward opening. And if you take the minus 1 in the canonical equation, you get a vertically oriented hyperbola.
Skip to 3 minutes and 48 secondsNow if someone gives you a quadratic equation of this type-- that is, an equation involving two variables, x and y-- and you see that there are terms with x squared, y squared, also x y in general is a quadratic term, and then some linear terms, d x and e y in a constant and ask you to find the level curve of that quadratic function of two variables. How do we do it? Well, to begin with, we assume that a, b, and c are not all zero, because they define what are called the quadratic or ordered two terms here. And we don't want them all to be zero.
Skip to 4 minutes and 27 secondsOtherwise, we'd just have an affine function with only linear terms in x and y. And we know the equation would be a straight line. So we'll assume a, b, and c are not all zero. And we want to know what kind of curve you get from this equation. Well, if b is equal 0, that's the coefficient of x y. There are no mixed terms. That means that the only quadratic terms are x squared and y squared really present. Then you can rearrange the equation. Here, of course, completing the square is the basic procedure.
Skip to 5 minutes and 4 secondsThe equation will then look like one of the canonical equations for a conic curve, except that it may be centered at some point different from 0, 0. In each case, there will always be the possibility that no solutions exist. We saw such an example for the circle some time back. That's a degenerate case, we call it, but most of the time there should be a conic curve. That's if b is equal zero, a rather simple case. If b is different from zero, then it's more complicated. But we can state the following fact.
Skip to 5 minutes and 42 secondsIf you construct the determinant-- the discriminant, rather-- that's the expression b squared minus 4 a c, just as it was for quadratic functions and equations that we were solving. This discriminant tells you the type of conic curve to expect. Now when b is different from zero, the curve will wind up being skewed or rotated. But it will still be one of the three main types unless, of course, it's degenerate, as we'll see. How does the discriminant work here? Well, you calculate b squared minus 4 a c, which is easy to do, and if that turns out to be strictly negative, then you expect an elliptic curve. If it's exactly zero, a parabola, and if it's strictly positive, a hyperbolic curve.
Skip to 6 minutes and 32 secondsLet's examine each of these three cases briefly to see what happens. The elliptic case, then, is when the discriminant is strictly negative. For example, you can calculate here the discriminant for this quadratic equation. The b here is equal zero. As you see, there's no x y term, And a and c are both positive. a is 2, and c is 2. So the discriminant is strictly negative. What do you expect when the discriminant is negative? Well, an ellipse, generally speaking. But here you can even be more discerning and expect a circle. Why? Because x squared and y squared have equal coefficients plus 2. So it should be the special case of an ellipse given by a circle.
Skip to 7 minutes and 19 secondsIn fact, when you complete the squares and work it out, I'll let you check that you'll get a circle centred at a certain point that I expect you to be able to find and of a certain radius, which again, you should be able to calculate. Now here's another equation where you can also see that the discriminant is negative. In this case, the b here is equal to minus 2. And the discriminant turns out negative. Therefore, you expect an ellipse, and it shouldn't be a circle because x squared and y squared have different coefficients. And, in fact, you do get an ellipse.
Skip to 7 minutes and 58 secondsIf you use your favourite graph application to draw graphs, it'll tell you that you do get an ellipse, but it's skewed. You notice it's neither horizontal nor vertically oriented. That's because of the term minus 2 x y in the quadratic equation. Here's another example. And you can see that this looks right and possibly looks like an ellipse. And you can see that the discriminant is negative. However, it's an impossible equation. The sum of 3 positive numbers can't give you zero, especially with the 7 there. So this is the degenerate case that arises in the elliptic case. Degenerate case means what? It means that you'll get an empty graph. There are no points x y that satisfy your equation.
Skip to 8 minutes and 50 secondsAnd so you don't actually see an ellipse. You see nothing, the empty set. Now let's look at the parabolic case. The parabolic case corresponds to the discriminant b squared minus 4 a c being exactly zero. Here's an example. I'll let you check. Minus 4 squared minus 4 times 4 times 1. Yes, you get zero. The discriminant is zero in this case. So you should expect what? A parabola. In fact, when you graph this thing, you wind up getting a parabola, but it's skewed, as you can see. It's neither horizontal nor vertical. That's because of the x y term in the equation.
Skip to 9 minutes and 29 secondsHere's another equation in which you would expect a parabola to occur, because as you will check, the discriminant turns out to be zero in this equation. But it's actually going to be degenerate for the following reason. It turns out that the left hand side of this equation can be factored exactly into the product of two affine expressions in x y.
Skip to 9 minutes and 57 secondsAnd you notice that 2 x plus y is present in each one. So when is this going to be zero? It's going to be zero when either of these two factors is zero. But the first factor being zero, for example, just corresponds to x y being on a certain line, because it's an affine equation. The same thing for the second, and you see that the lines have the same slope. Therefore, they're parallel. So in other words, this equation holds if and only if you're on either of two parallel lines. So the solution set is the union of those two lines. And this is how the parabolic case can be degenerate.
Skip to 10 minutes and 36 secondsIt can lead to the union of two parallel lines such as, in this case, this set.
Skip to 10 minutes and 45 secondsAnd finally, there's the hyperbolic case to consider. That's the case in which the discriminant is strictly positive. And here's an example. You calculate the discriminant. You'll see it's strictly positive , and you expect, therefore, a hyperbola. That can be graphed, and it confirms your suspicion. So it's rotated somewhat, the hyperbola, because of the x y term. Here's another example. This turns out to be degenerate in this case, why? Because again it factors into two affine expressions, but this time the two affine expressions correspond to lines that are not parallel. To say that this product is zero, therefore, is to say either that you're on the first line or that you're on the second non-parallel line.
Skip to 11 minutes and 34 secondsThat is, that you're somewhere on the union of two non-parallel lines. That's how the hyperbolic case can be degenerate. The solution set can be the union of two intersecting, since they're not parallel, lines. Now the full proof of these facts is fairly advanced. You might meet it at university in what is called a linear algebra and matrices course on those subjects. But even knowing these facts, I assure you, puts you in the true conic elite.
# Level curves of quadratic functions
In this video, Francis introduces a classification of the level curves of a quadratic function depending on their discriminant.
You can access a copy of the slides used in the video in the PDF file at the bottom of this step. |
### Integers - Solutions 2
CBSE Class –VII Mathematics
NCERT Solutions
Chapter 1 Integers
(Ex. 1.2)
Question 1. Write down a pair of integers whose:
(a) sum is -7 (b) difference is -10 (c) sum is 0
Answer: (a) One such pair whose sum is -7 : ( -5, -2) (-10, 3)
(b) One such pair whose difference is -10: (-2, 8 ), (-11, -1)
(c) One such pair whose sum is 0: (5,5) ,(6,6)
Question 2. (a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose is -5.
(c) Write a negative integer and a positive integer whose difference is -3.
Answer: (a) (-2, -10) (-5, -13)
(b) ( -7, 2), ( -9,4)
(c) ( -1, 2), ( -2, 1)
Question 3. In a quiz, team A scored -40,10, 0 and team B scores 0, 10, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Answer: Team A scored -40,10, 0
Total score of Team A = -40+10+0 = -30
Team B scored 0, 10, -40
Total score of Team B = 0+10 +(-40) = 0+10 - 40 = -30
Thus, scores of both teams are same.
Yes, we can add integers in any order due to commutative property.
Question 4. Fill in the blanks to make the following statements true:
(i)$\left(-5\right)+\left(-8\right)=\left(-8\right)+\left(.......\right)$
(ii)$-53+.......=-53$
(iii)17 + ……. = 0
(iv)$\left[13+\left(-12\right)\right]+\left(.......\right)=13+\left[\left(-12\right)+\left(-7\right)\right]$
(v)$\left(-4\right)+\left[15+\left(-3\right)\right]=\left[-4+15\right]+.......$
Answer: (i) $\left(-5\right)+\left(-8\right)=\left(-8\right)+\underset{_}{\left(-5\right)}$ [Commutative property]
(ii) $-53+\underset{_}{0}=-53$ [Zero additive property]
(iii) $17+\underset{_}{\left(-17\right)}=0$ (Additive identity]
(iv)$\left[13+\left(12\right)\right]+\underset{_}{\left(-7\right)}=13+\left[\left(-12\right)+\left(-7\right)\right]$ [Associative property]
$\left(-4\right)+\left[15+\left(-3\right)\right]=\left[-4+15\right]+\underset{_}{\left(-3\right)}$ [Associative property] |
Integrate the function$x\;log2x$
$\begin{array}{1 1} \frac{x^2 \log 2x}{2}-\frac{x^2}{4}+c \\ \frac{x^2 \log 2x}{2}+\frac{x^2}{4}+c \\ \frac{x\log 2x}{2}-\frac{x^2}{4}+c \\ \frac{x \log 2x}{2}+\frac{x}{4}+c \end{array}$
Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\int\frac{1}{x}dx=log x+c.$
Given $I=\int xlog 2xdx.$
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where $\int udv=uv-\int vdu$
Let u=log 2x.
On differentiating with respect to x,
$du=2\frac{1}{2x}dx=\frac{1}{x}dx.$
let dv=x dx.
On integrating we get,
$v={x^2}{2}$.
On substituting for u,v,du and dv we get,
$\int xlog 2xdx=(log 2x.\frac{1}{x^2})-\int\frac{x^2}{2}.\frac{1}{x}dx.$
On cancelling x we get
$I=\frac{x^2log 2x}{2}-\frac{1}{2}\int x dx.$
On integrating we get,
$\int xlog 2xdx=\frac{x^2log 2x}{2}-\frac{x^2}{4}+c.$
edited Feb 8, 2013 |
# Factoring Fun: Ingenious Tips and Tricks to Simplify Even the Toughest Expressions
Introduction to Factoring: Embracing the Art of Simplification
Welcome to the exciting world of factoring! Are you ready to discover the secrets behind simplifying even the most daunting algebraic expressions? Factoring is a crucial technique in algebra that helps us break down expressions into their simplest forms, making it easier to solve equations and tackle complex problems. And guess what? It's not as hard as it might seem!
In this article, we'll embark on a thrilling journey through the realm of factoring, exploring:
1. Factoring Techniques: From Basic to Advanced: Learn the ins and outs of various factoring methods, from the simple Greatest Common Factor (GCF) to the more advanced techniques like trinomial factoring.
2. Ingenious Tips and Tricks for Mastering Factoring: Uncover expert strategies and shortcuts that will help you factor expressions like a pro.
3. Factoring in Real-Life Situations: Applications and Problem Solving: See how factoring comes to life in real-world scenarios and discover its importance in various fields.
So, are you ready to unlock the magic of factoring and simplify even the toughest expressions with ease? Let's dive in and start mastering this essential skill together!
Factoring Techniques: From Basic to Advanced
Discovering the Power of Factoring Methods
As we venture deeper into the world of factoring, it's essential to get acquainted with various factoring techniques. From basic methods to more advanced approaches, these techniques are the key to simplifying different types of expressions. Let's explore some of the most common factoring methods:
1. Greatest Common Factor (GCF): The GCF is the largest factor shared by all terms in an expression. To factor using the GCF, identify the largest number or variable that evenly divides each term, and then factor it out. For example, in the expression 4x² + 8x, the GCF is 4x, so we factor it out: 4x(x + 2).
2. Difference of Squares: This technique is used when an expression has two terms that are perfect squares, and they are connected by a subtraction sign. In this case, the expression can be factored into the product of two binomials. For example, x² - 25 = (x + 5)(x - 5).
3. Perfect Square Trinomials: A perfect square trinomial is an expression that can be factored into the square of a binomial. These expressions follow the pattern (a² + 2ab + b²) or (a² - 2ab + b²). To factor a perfect square trinomial, take the square root of the first and last terms and combine them into a binomial. For example, x² + 6x + 9 = (x + 3)².
4. Trinomial Factoring: Trinomial factoring involves breaking down a trinomial expression into the product of two binomials. To do this, find two numbers that multiply to the product of the first and last terms and add up to the middle term's coefficient. For example, x² + 5x + 6 = (x + 2)(x + 3).
Each of these techniques plays a vital role in simplifying expressions and solving equations. By mastering these methods, you'll be well-equipped to tackle a wide range of factoring challenges. But why stop there? Let's move on to the next section and uncover some ingenious tips and tricks that will help you factor even faster and more efficiently!
Ingenious Tips and Tricks for Mastering Factoring |
# AP Statistics Curriculum 2007 Distrib RV
## General Advance-Placement (AP) Statistics Curriculum - Random Variables and Probability Distributions
### Random Variables
A random variable is a function or a mapping from a sample space into the real numbers (most of the time). In other words, a random variable assigns real values to outcomes of experiments. This mapping is called random, as the output values of the mapping depend on the outcome of the experiment, which are indeed random. So, instead of studying the raw outcomes of experiments (e.g., define and compute probabilities), most of the time we study (or compute probabilities) the corresponding random variables instead. The formal general definition of random variables may be found here.
### Examples of Random Variables
• Die: In rolling a regular hexagonal die, the sample space is clearly and numerically well-defined. In this case, the random variable is the identity function assigning to each face of the die where the numerical value it represents. This the possible outcomes of the RV of this experiment are { 1, 2, 3, 4, 5, 6 }. You can see this explicit RV mapping in the SOCR Die Experiment.
• Coin: For a coin toss, a suitable space of possible outcomes is S={H, T} (for heads and tails). In this case these are not numerical values, so we can define a RV that maps these to numbers. For instance, we can define the RV $X: S \longrightarrow [0, 1]$ as: $X(s) = \begin{cases}0,& s = \texttt{H},\\ 1,& s = \texttt{T}.\end{cases}$. You can see this explicit RV mapping of heads and tails to numbers in the SOCR Coin Experiment.
• Card: Suppose we draw a 5-card hand from a standard 52-card deck and we are interested in the probability that the hand contains at least one pair of cards with identical denomination. Since the sample space of this experiment is large, it should be difficult to list all possible outcomes. However, we can assign a random variable $X(s) = \begin{cases}0,& s = \texttt{no-pair},\\ 1,& s = \texttt{at-least-1-pair}.\end{cases}$ and try to compute the probability of P(X=1), the chance that the hand contains a pair. You can see this explicit RV mapping and the calculations of this probability at the SOCR Card Experiment.
### Probability density/mass and (cumulative) distribution functions
• The probability density or probability mass function, for a continuous or discrete random variable, is the function defined by the probability of the subset of the sample space, $\{s \in S \} \subset S$, which is mapped by the random variable X to the real value x (i.e., X(s)=x):
$p(x) = P(\{s \in S \} | X(s) = x)$, for each x.
• The cumulative distribution function (cdf) F(x) of any random variable X with probability mass or density function p(x) is defined as the total probability of all $\{s \in S \} \subset S$, where $X(s)\leq x$:
$F(x)=P(X\leq x)= \begin{cases}{ \sum_{y: y\leq x} {p(y)}},& X = \texttt{Discrete-RV},\\ {\int_{-\infty}^{x} {p(y)dy}},& X = \texttt{continuous-RV}.\end{cases}$, for all x.
#### PDF Example
The Benford's Law states that the probability of the first digit (d) in a large number of integer observations ($d\not=0$) is given by
$P(d) = \log(d+1) - log(d) = \log{d+1 \over d}$, for $d = 1,2,\cdots,9.$
Note that this probability definition determines a discrete probability (mass) distribution:
$\sum_{d=1}^9{P(d)}=\log{2\over 1}+\log{3\over 2}+\log{4\over 3}+ \cdots +\log{10\over 9}=$ $\log({{2\over 1} {3\over 2} {4\over 3} \cdots{10\over 9}}) = \log{10\over 1} = 1$
d 1 2 3 4 5 6 7 8 9 P(d) 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
The explanation of the Benford's Law may be summarized as follows: The distribution of the first digits must be independent of the measuring units used in observing/recording the integer measurements. For instance, this means that if we had observed length/distance in inches or centimeters (inches and centimeters are linearly dependent, 1in = 2.54cm), the distribution of the first digit of the measurement must be identical. So, there are about three centimeters for each inch. Thus, the probability that the first digit of a length observation is 1in must be the same as the probability that the first digit of a length in centimeters starts with either 2 or 3 (with standard round off). Similarly, for observations of 2in, they need to have their centimeter counterparts either 5cm or 6cm. Observations of 3in will correspond to 7 or 8 centimeters, etc. In other words, this distribution must be scale invariant.
The only distribution that obeys this property is the one whose logarithm is uniformly distributed. In this case, the logarithms of the numbers are uniformly distributed -- $P(100\leq x \leq 1,000)$ = $P(2\leq \log(x)\leq 3)$ is the same as the probability $P(10,000\leq x \leq 100,000)$ = $P(4\leq \log(x)\leq 5)$. Examples of such exponentially growing numerical measurements are incomes, stock prices and computational power.
### How to Use RVs?
There are 3 important quantities that we are always interested in when we study random processes. Each of these may be phrased in terms of RVs, which simplifies their calculations.
• Probability Density Function (PDF): What is the probability of P(X = xo)? For instance, in the card example above, we may be interested in P(exactly 1 pair) = P(X=1) = P(1 pair only) = 0.422569. Or in the die example, we may want to know P(Even number turns up) = $P(X \in \{2, 4, 6 \}) = 0.5$.
• Cumulative Distribution Function (CDF): P(X < xo), for all xo. For instance, in the (fair) die example we have the following discrete density (mass) and cumulative distribution table:
x 1 2 3 4 5 6 PDFP(X = x) 1/6 1/6 1/6 1/6 1/6 1/6 CDF $P(X\leq x)$ 1/6 2/6 3/6 4/6 5/6 1
• Mean/Expected Value: Most natural processes may be characterized, via probability distribution of an appropriate RV, in terms of a small number of parameters. These parameters simplify the practical interpretation of the process or phenomena we study. For example, it is often enough to know what the process (or RV) average value is. This is the concept of expected value (or mean) of a random variable, denoted E[X]. The expected value is the point of gravitational balance of the distribution of the RV.
Obviously, we may define a large number of RV for the same process. When are two RVs equivalent is dependent on the definition of equivalence?
### Comparing Data and Model Distributions
To illustrate one example of using distributions for solving practical problems, we consider the large human weight and height dataset. You can use all 25,000 records or just the first 200 of these measurements to follow the protocol below:
• Copy the weight and height data into the data tab of any of the SOCR Charts (first clear the default data, then select column 1 heading, and click the Paste button). This allows you to manipulate each of the 3 data columns independently.
• Select (highlight with the mouse) one of the columns (e.g., weights or heights) in the SOCR Chart and click the Copy button. This stores only the data in the chosen column in your mouse buffer.
• Go to the SOCR Modeler and paste the data in the first column in the Data tab using the Paste button.
• Select the NormalFit_Modeler from the drop-down list on the top-left corner. This is the first model you will be fitting to your data.
• Select the 3 check-boxes (Estimate Parameters, Scale Up, and Raw Data).
• Go to the Graphs tab and adjust the 3 sliders on the top to get a clear view of your data distribution (sample histogram) and the model distribution function (solid red curve).
• The Results tab will contain the (data-driven) estimates of the parameters for this specific distribution model (in this case Normal).
• You can plug these parameters (mean and standard deviation) into the SOCR Normal Distribution Applet and make inference about your population based on this Normal distribution model.
• Validate that the probabilities of various interesting events (e.g., 68<=Height<70) computed via either using the sample histogram of the data or via the model distribution are very similar.
• Try fitting another distribution model to your data using the SOCR Modeler. For example, choose the mixture-of-Normals model (MixedFit_Modeler) and repeat this process. Can you identify possible gender effects in either height or weight of the subjects in your sample? If so, what are the Male and Female distribution models? Can these be used to predict the gender of subjects (based on their weight or height)?
• Note that the Results tab also shows some statistics quantifying how good your chosen distribution model is to approximate the (sample) data histogram.
### The Web of Distributions
There is a large number of families of distributions and distribution classification schemes.
The most common way to describe the universe of distributions is to partition them into categories. For example, Continuous Distributions and Discrete Distributions; marginal and joint distributions; finitely and infinitely supported, etc.
SOCR Distributome Project and the SOCR Distribution activities illustrate how to use technology to compute probabilities for events arising from many different processes.
The image below shows some of the relations between commonly used distributions. Many of these relations will be explored later.
The SOCR Distributome applet provides an interactive graphical interface for exploring the relations between different distributions. |
# Integral of Some Particular Functions
There are some important integration formulas that are applied for integrating many other standard integrals. In this article, we will look at the integrals of these particular functions.
## Integral of Some Particular Functions
Look at the following integration formulas
1. ∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C
2. ∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C
3. ∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C
4. ∫ dx / √ (x2 – a2) = log |x + √ (x2 – a2)| + C
5. ∫ dx / √ (a2 – x2) = sin–1 (x/a) + C
6. ∫ dx / √ (x2 + a2) = log |x + √ (x2 + a2)| + C
## Proof of the Above Six Standard Integration Formulas
### 1. ∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C
We know that,
1 / (x2 – a2) = 1 / (x – a) (x + a) = 1/2a [(x + a) – (x – a) / (x – a) (x + a)]
= 1/2a [1/(x – a) – 1/(x + a)]
Therefore,
∫ dx / (x2 – a2) = 1/2a [∫ dx / (x – a) – ∫ dx / (x + a)]
= 1/2a [log |(x – a) – log |(x + a)] + C
= 1/2a log |(x – a) / (x + a)| + C
### 2. ∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C
We know that,
1 / (a2 – x2) = 1 / (a – x) (a + x) = 1/2a [(a + x) + (a – x) / (a – x) (a + x)]
= 1/2a [1/(a – x) + 1/(a + x)]
Therefore,
∫ dx / (a2 – x2) = 1/2a [∫ dx / (a – x) + ∫ dx / (a + x)]
= 1/2a [– log |(a – x) + log |(a + x)] + C
= 1/2a log |(a + x) / (a – x)| + C
### 3. ∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C
Let’s substitute x = a tan t, so we have dx = a sec2 t dt. Therefore,
∫ dx / (x2 + a2) = ∫ [(a sec2 t dt) / (a2 tan2 t + a2)]
Solving this, we get,
∫ dx / (x2 + a2) = 1/a ∫ dt = t/a + C
Re-substituting the value of t, we get
∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C
#### 4. ∫ dx / √ (x2 – a2) = log |x + √ (x2 – a2)| + C
Let’s substitute x = a sec t, so that dx = a sec t tan t dt. Therefore,
∫ dx / √ (x2 – a2) = ∫ a sec t tan t dt / √ (a2 sec2 t – a2)
Solving this, we get,
∫ dx / √ (x2 – a2) = ∫ sec t dt = log |sec t + tan t| + C1
Re-substituting the value of t, we get
∫ dx / √ (x2 – a2) = log |(x/a) + √ [(x2 – a2) / a2]| + C1
= log |x + √(x2 – a2)| – log |a| + C1
= log |x + √(x2 – a2)| + C … where C = C1 – log |a|
### 5. ∫ dx / √ (a2 – x2) = sin–1 (x/a) + C
Let’s substitute x = a sin t, so that dx = a cos t dt. Therefore,
∫ dx / √ (a2 – x2) = ∫ a cos t dt / √ (a2 – a2 sin2 t)
Solving this, we get,
∫ dx / √ (a2 – x2) = ∫ t dt = t + C
Re-substituting the value of t, we get
∫ dx / √ (a2 – x2) = sin–1 (x/a) + C
### 6. ∫ dx / √ (x2 + a2) = log |x + √ (x2 + a2)| + C
Let’s substitute x = a tan t, so that dx = a sec2 t dt. Therefore,
∫ dx / √ (x2 + a2) = ∫ a sec2 t dt / √ (a2 tan2 t + a2)
Solving this, we get,
∫ dx / √ (x2 – a2) = ∫ sec t dt = log |sec t + tan t| + C1
Re-substituting the value of t, we get
∫ dx / √ (x2 – a2) = log |(x/a) + √ [(x2 + a2) / a2]| + C1
= log |x + √(x2 + a2)| – log |a| + C1
= log |x + √(x2 + a2)| + C … where C = C1 – log |a|
Now, let’s apply these standard integration formulas to obtain formulae that are applied directly to evaluate other integrals.
### 7. Integral ∫ dx / (ax2 + bx + c)
We can write,
ax2 + bx + c = a [x2 + (b/a)x + (c/a)]
= a [(x + b/2a)2 + (c/a – b2/4a2)]
Now, let’s substitute (x + b/2a) = t, so that dx = dt. Also, substitute (c/a – b2/4a2) = ±k2. Therefore,
ax2 + bx + c = a (t2 ± k2) … where the + or – depends on the sign of (c/a – b2/4a2).
Hence,
∫ dx / (ax2 + bx + c) = 1/a ∫ dt / (t2 ± k2)
This can be evaluated using one / more of the six integration formulas shown above. Remember, you can also solve ∫ dx / √ (ax2 + bx + c) in a similar manner.
### 8. Integral ∫ [(px + q) / (ax2 + bx + c)] dx, where p, q, a, b, and c are constants.
To solve this, we must find constants A and B such that,
(px + q) = A d/dx (ax2 + bx + c) + B = A (2ax + b) + B
To determine ‘A’ and ‘B’, we equate from both sides the coefficients of x and the constant terms. ‘A’ and ‘B’ are thus obtained and hence the integral is reduced to one of the known forms. Let’s understand with the help of some examples:
## Solved Problems for You
### Example 1: Find ∫ [(x + 2) / (2x2 + 6x + 5)] dx
To solve this equation, we express
(x + 2) = A d/dx (2x2 + 6x + 5) + B = A (4x + 6) + B
Therefore, x + 2 = 4Ax + 6A + B
Next, let’s equate the coefficients of ‘x’ and the constant terms. We have,
4A = 1 and 6A + B = 2
On solving them, we get
A = ¼ and B = ½
Hence, we have
∫ [(x + 2) / (2x2 + 6x + 5)] dx = ¼ ∫ [(4x + 6) / (2x2 + 6x + 5)] dx + ½ ∫ dx / (2x2 + 6x + 5)
= ¼ ∫ I1 + ½ ∫ I2
Now, let’s solve I1 and I2 separately.
#### Solving I1
Let’s substitute (2x2 + 6x + 5) = t, so that (4x + 6) dx = dt. Therefore,
I1 = ∫ [(4x + 6) / (2x2 + 6x + 5)] dx = ∫ dt/t = log |t| + C1
Or, I1 = log |(2x2 + 6x + 5)| + C1 … (1.1)
#### Solving I2
I2 = ∫ dx / (2x2 + 6x + 5) = ½ ∫ dx / (x2 + 3x + 5/2) = ½ ∫ dx / [(x + 3/2)2 + (1/2)2]
Now, let’s substitute (x + 3/2) = t, so that dx = dt. Therefore,
I2 = ½ ∫ dt / [t2 + (1/2)2]
Using the six integration formulas shown above, we get
I2 = (1/(2 x ½) tan–1 2t + C2 = tan–1 2 (x + 3/2) + C2 = tan–1 (2x + 3) + C2 … (1.2)
Using (1.1) and (1.2), we get
∫ [(x + 2) / (2x2 + 6x + 5)] dx = ¼ log |2x2 + 6x + 5| + ½ tan–1 (2x + 3) + C
Where, C = C1/4 + C2/2
### Example 2: Find the integral of (x + 3) / √ (5 – 4x + x2) with respect to x.
Solution: We can express,
x + 3 = A d/dx (5 – 4x + x2) + B = A (– 4 – 2x) + B
Equating the coefficients, we get
A = – ½ and B = 1
Therefore, ∫ [(x + 3) / √ (5 – 4x + x2)] dx = – ½ ∫ [(– 4 – 2x) / √ (5 – 4x + x2)] dx + ∫ dx / √ (5 – 4x + x2)
= – ½ I1 + I2 … (a)
#### Solving I1
Let’s substitute (5 – 4x + x2) = t, so that (– 4 – 2x) dx = dt. Therefore,
I1 = ∫ [(– 4 – 2x) / √ (5 – 4x + x2)] dx = ∫ dt / √ t = 2 √ t + C1
= 2 √ (5 – 4x + x2) + C1 … (b)
#### Solving I2
I2 = ∫ dx / √ (5 – 4x + x2) = ∫ dx / √ [9 – (x + 2)2]
Now, let’s substitute (x + 2) = t, so that dx = dt. Therefore,
I2 = ∫ dt / √ (32 – t2) = sin–1 (t/3) + C2
= sin–1 [(x + 2) / 3] + C2 … (c)
Substituting (b) and (c) in (a), we get
∫ [(x + 3) / √ (5 – 4x + x2)] dx = – ½ I1 + I2
= – √ (5 – 4x + x2) + sin–1 [(x + 2) / 3] + C … where C = C2 = C1/2.
Question. How can one derive integration formulas?
Answer. The fundamental use of integration is as a version of summing that is continuous. One can derive integral by viewing integration as essentially an inverse operation to differentiation. One can call it the Fundamental Theorem of Calculus. Integration formulas involve almost the inverse operation of differentiation.
Question. What exactly do we understand by integration?
Answer. Simply speaking, integration refers to the act of bringing together smaller components into a single system. This single system is such that it functions as one.
Question. Explain the application of integration in real life?
Answer. Integration has various uses in real life. It is important in the fields of engineering, physics, medical science, research analysis, and graphic designing.
Question. Explain the rules for integration?
Answer. The rules for integration are power rule, constant coefficient rule, sum rule, and difference rule. The power rule gives the indefinite integral of a variable raised to a power. The constant coefficient rule informs us about the indefinite integral of c. f(x). The sum rule tells us about integrating functions that are the sum of several terms. The difference rule deals with the difference between two or more terms.
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# Stewart – Calculus – 3.5 – Implicit Differentiation and Derivatives of Inverse Trigonometric Functions
Find $y''$ by implicit differentiation:
$9x^{2}+y^{2}=9$.
$18x+2yy'=0$
$2yy'=-18x$
$y'=-9x/y$
$y''=-9(\frac{y\cdot 1\text{--}x\cdot y'}{y^{2}})$
$y''=-9(\frac{y\cdot 1\text{--}x(-9x/y)}{y^{2}})$ ($y'$ is replaced with $-9x/y$)
$y''=-9(\frac{y^{2}+9x^{2}}{y^{3}})$ (after multiplying by $y$ to eliminate denominator from $-9x$)
$y''=-9(\frac{9}{y^{3}})$ (notice that $y^{2}+9x^{2}$ equals the original equation)
So, $y''=\frac{-81}{y^{3}}$.
Derivatives of Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
# Stewart – Calculus – 3.4 – The Chain Rule
Find the derivative: $y=xe^{-\mathit{kx}}$.
# Stewart – Calculus – 3.3 – Derivatives of Trigonometric Functions
$\frac{d}{\mathit{dx}}(\sin x)=?$ $\frac{d}{\mathit{dx}}(\cos x)=?$ $\frac{d}{\mathit{dx}}(\tan x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{csc} x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{sec} x)=?$ $\frac{d}{\mathit{dx}}(\mathrm{cot} x)=?$
THE TRICK
To learn the derivatives of the 6 trig functions you actually only have to learn 3 of them. The three to learn are sine, tangent, and secant.
IF THEN F(X) = sin X F'(X) = cos X F(X) = tan X F'(X) = sec2 X F(X) = sec X F'(X) = sec X tan X
Cosine, Cotangent, and Cosecant.
Think of the functions as having partners:
sine to cosine tangent to cotangent secant to cosecant
To find the derivative of the "co" functions, start with the derivatives of sine, tangent and secant, change each function in the derivative to it’s co-function partner and put a minus sign in front of the derivative.
So starting with one of the three we need to know:
If F(x) = secant X
then F'(x) = secant X tangent X
So now to get the derivative of the cosecant:
If F(X) = cosecant X
then F'(X) = -cosecant X cotangent X
So all 6 trig function derivatives look like this:
IF: THEN: F(X) = sinX F'(X) = cosX F(X) = cosX F'(X) = – sinX F(X) = tanX F'(X) = sec2 X F(X) = cotX F'(X) = – csc2 X F(X) = secX F'(X) = sec X tan X F(X) = cscX F'(X) = -csc X cot X
Thanks to Bruce Kirkpatrick for this.
# Stewart – Calculus – 3.2 – Differentiation Rules
THE PRODUCT RULE
$\frac{d}{\mathit{dx}}[f(x)g(x)]=f(x)\frac{d}{\mathit{dx}}[g(x)]+g(x)\frac{d}{\mathit{dx}}[f(x)]$
(a) If $f(x)=xe^{x}$, find $f'(x)$.
(b) Find the nth derivative, $f^{(n)}(x)$.
Differentiate the function $f(t)=\sqrt{t}(a+\mathit{bt})$
If $f(x)=\sqrt{x} \hspace{1 mm} g(x)$, where $g(4)=2$ and $g'(4)=3$, find $f'(4)$.
THE QUOTIENT RULE
$\frac{d}{\mathit{dx}}[\frac{f(x)}{g(x)}]=\frac{g(x)\frac{d}{\mathit{dx}}[f(x)]\text{--}f(x)\frac{d}{\mathit{dx}}[g(x)]}{[g(x)]^{2}}$
Let $y=\frac{x^{2}+x\text{--}2}{x^{3}+6}$
Find an equation of the tangent line to the curve $y=\frac{e^{x}}{(1+x^{2})}$ at the point $(1,\frac{1}{2}e)$.
$\frac{d}{\mathit{dx}}(c)=?$ $\frac{d}{\mathit{dx}}(x^{n})=?$ $\frac{d}{\mathit{dx}}(e^{x})=?$ $(\mathit{cf})'=?$ $(f+g)'=?$ $(f-g)'=?$ $(\mathit{fg})'=?$ $(\frac{f}{g})'=?$
SOLUTION 1
Using the Product Rule
SOLUTION 2
If we first use the laws of exponents to rewrite f(t), then we can proceed directly without using the Product Rule.
0
# Stewart – Calculus – 3.1 – Derivatives of Polynomials and Exponential Functions
(a) $f(x)=\frac{1}{x^{2}}$
(b) $y=\sqrt[{3}]{x^{2}}$
Find equations of the tangent line and normal line to the curve $y=x\sqrt{x}$ at the point $(1, 1)$. Illustrate by graphing the curve and these lines.
THE CONSTANT MULTIPLE RULE
$\frac{d}{\mathit{dx}}(3x^{4})$ $\frac{d}{\mathit{dx}}(-x)$
# Stewart – Calculus – 2.7 – Derivatives and Rates of Change
1. Find an equation of the tangent line to the parabola $y=x^{2}$ at the point $P(1,1)$.
2. Find an equation of the tangent line to the hyperbola $f(x)=\frac{3}{x}$ at the point $(3, 1)$.
3. Find the derivative of the function $f(x)=x^{2}-8x+9$ at the number $a$.
# Stewart – Calculus – 2.5 – Continuity
1. Definition: A function f is continuous at a number a if
This definition implicitly requires three things if f is continuous at a:
1. f(a) is defined (that is, a is in the domain of f)
2. $\lim_{x \to a}f(x)$ exists
3. $\lim_{x \to a}f(x)=f(a)$
2. If $f$ and $g$ are continuous functions with $f(3)=5$ and
$\lim_{x \to 3}[2f(x) - g(x)]=4$, find $g(3)$.
Use the definition of continuity and the properties of limits to show that the funtion is continous at the given number a. $f(x)=(x+2x^{3})^{4}$, $a=-1$
3. The Intermediate Value Theorem
Suppose that $f$ is continuous on the closed interval $[a, b]$ and let $N$ be any number between $f(a)$ and $f(b)$, where $f(a) \neq f(b)$. Then there exists a number $c$ in $(a,b)$ such that $f(c)=N$.
~ ~ ~
Use the Intermediate Value Theorem to show that there is
a root of the given equation in the specified interval.
$x^4+x-3=0, (1,2)$
# Stewart – Calculus – 2.4 – The Precise Definition of a Limit
1. Provide the symbolic equivalent.
a. every
b. there exist(s)
c. such that
2. Write the precise definition of a limit.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
# Calculus Test 1
1. Specify the domain of the function y = |x – 1|.
2. Which of the following is a true statement about the graph of the equation $y = x^4 +1$?
3. 1. It is symmetric about the x-axis.
2. It is symmetric about the y-axis.
3. It has two x-intercepts.
4. It has no x-intercepts.
a. Only statements 2 and 3 are true.
b. Only statements 2 and 4 are true.
c. Only statements 1 and 3 are true.
d. Only statement 2 is true.
4. Which of the following statements are true of the graph of $y = \frac{2x-1}{x+1}$?
1. It has no x-intercept.
2. It has a slant asymptote y=2x.
3. It has a vertical asymptote at $x=-1$.
4. It has a horizontal asymptote at y=2.
a. Statements 1, 2, and 3 are true.
b. Statements 3 and 4 are true.
c. Statements 2, 3, and 4 are true.
d. All four statements are true.
5. Let $f(x) = 2 \cos x$. The domain of $f^{-1}(x)$ is
1. $[-1,1]$ 2. $(2, \infty)$ 3. $(-\infty, \infty)$ 4. $(-2,2)$ 5. $[-2,2]$
6. Simplify as far as possible. $\ln e +a^{\log_{a}5}- \log100+10^0 - \log_{3} \frac{1}{3}$
7. Solve for x.
a. $\log_3 x + \log_3(2x+5)=1$
b. $\frac{1}{\sqrt{2}}=4^{x}$
8. $f(x)=(x-3)(x+1)^2(x-1)^4$
9. Identify the parts of the following composite function. $f(g(h(j(x))))=\frac{1}{\sqrt{\log(x-1)}}$
10. Consider the picture below. Find the equation of the line L.
11. The following table represents a function of the form $f(x)=ab^{x}$. Find the equation of the function.
$\begin{tabular}{| l | c | r |} \hline x & f(x)\\ \hline 0 & 6\\ \hline 1 & 18\\ \hline 2 & 54\\ \hline 3 & 162\\ \hline 4 & 486\\ \hline \end{tabular}$
12. Given $a^{m}=2$, $a^{n}=3$, $b^{m}=4$, and $b^{n}=5$,
use the properties of exponentials to determine $(a^{3n}b^{m+n})^{\frac{1}{3}}$.
13. Evaluate:
1. $\sin \frac{\pi}{3}$
2. $\tan \frac{3\pi}{4}$
3. $\cos \frac{5\pi}{6}$
14. Find all solutions to the equaion, $\tan^{2}x=\tan x$, such that $x \in [0,2\pi]$.
15. Find $f^{-1}(x)$ if $f(x) = \sqrt{e^{x}+2}$. |
# How do you find the critical points for f(x) = x - 3ln(x) and the local max and min?
Oct 20, 2016
$\left(3 , 3 - 3 \ln 3\right)$
#### Explanation:
$f \left(x\right) = x - 3 \ln x$
Differentiating wrt $x$:
$f ' \left(x\right) = 1 - \frac{3}{x}$
Differentiating again wrt $x$:
$f ' ' \left(x\right) = - \frac{3}{x} ^ 2 \left(- 1\right) = \frac{3}{x} ^ 2 , \left(> 0 \forall x \in \mathbb{R}\right)$
At critical points, $f ' \left(x\right) = 0 \implies 1 - \frac{3}{x} = 0$
$\therefore \frac{3}{x} = 1$
$x = 3$
$f \left(3\right) = 3 - 3 \ln 3$
$f ' ' \left(3\right) > 0$
Hence, there is one critical point $\left(3 , 3 - 3 \ln 3\right)$ which is a minimum.
graph{x-3lnx [-10, 10, -5, 5]} |
]> 25.4 Numerical Evaluation of Line Integrals
## 25.4 Numerical Evaluation of Line Integrals
Suppose we confront a line integral, which is an integral along a path in some Euclidean space, of a vector field $v ⟶ · d s ⟶$
$∫ C v ⟶ · d s ⟶$
We can set up a spreadsheet to evaluate such an integral with very little difficulty.
In this section we describe how this can be done, and attempt to goad you into doing one yourself on a spreadsheet.
By the way, once you have such a thing, you can modify the path or the integrand or your grid size with a tiny amount of effort, and so can use your product to evaluate any line integral you ever encounter.
A line integral is more complicated than an ordinary integral primarily in that you have to deal with a path as well as an integrand, while for an ordinary integrand the path is an interval of the real line which is completely characterized by two numbers, its endpoints, and you need only deal with the integrand.
What we do depends on how the path $C$ on which the integral is to be evaluated is defined.
The easiest case, which we address here, occurs when we are given $C$ as a parametrized curve.
This means that we have a parameter, let us call it $t$ , and have formulae for each coordinate, $x , y , z , …$ , as functions of $t$ , and an interval of $t$ values defines the curve.
Our plan is to break up the $t$ interval into many small pieces, and at the boundaries of these pieces calculate $x ( t ) , y ( t ) , z ( t )$ , and $v x ( x ( t ) , y ( t ) , … ) , v y ( x ( t ) , y ( t ) , … ) , …$
In words, we will devote a column to the calculation of each of $x , y$ and $z$ (in three dimensions); and to each component of the vector field $v ⟶$ , evaluated at $( x ( t ) , y ( t ) , z ( t ) )$ . Then we use a column to describe the contribution to the integral from one subinterval, say between $t$ and $t + d$ .
And what do we use to estimate that contribution?
$t$ is the change in $x ( t )$ over that interval multiplied by the average value of $v x$ over the endpoints of the interval plus the same thing in $y$ and in $z$ . We write down only the $x$ contribution here, which is
$( x ( t + d ) − x ( t ) ) v x ( x ( t + d ) , y ( t + d ) , z ( t + d ) ) + v x ( x ( t ) , y ( t ) , z ( t ) ) 2$
In the last column we add up these contributions over the various intervals, and that is our integral. This is the trapezoid rule for line integrals.
Once this is done, how do we change the path? Alter the $x ( t ) , y ( t )$ and $z ( t )$ columns to use different functions of $t$ .
How do we change the integrand, $v ⟶$ ? Change its columns.
How do we change the interval size parameter $d$ ? Just change it.
Exactly how do we do this?
I like to leave the top five or so rows for notes and data entry.
In B2 I would put the starting $t$ value and in B3 the value for $d$ .
In the top row I would give a verbal description of the curve and vector field.
So I would start the computation in row 6.
Here is what I would put in the $t$ column, column A
A6=B2
A7=A6+B3
A8=2*A7-A6
And I would copy A8 down the sheet into A9-A1005. (or further if you like)
In the $x , y$ and $z$ columns put
B6=x(A6) (you must put in what this function is of course)
C6= y(A6)
D6=z(A6)
And copy these down to B7-B1005, etc.
In the $v ⟶$ columns put the values for the components of $v ⟶$
$E 6 = v x ( B 6 , C 6 , D 6 )$
$F 6 = v y ( B 6 , C 6 , D 6 )$
$G 6 = v z ( B 6 , C 6 , D 6 )$
And copy these down.
Next set
H6=(B7-B6)*(E(7)+E(6))/2
And copy this into the I6 and J6 places, and down the sheet.
Set
K6=H6+I6+J6
And
L7=L6+K6
And copy these down, and you are done.
If the final $t$ value occurs in Ak, then the integrand answer will be in Lk.
You can adjust $d$ and can extrapolate exactly as for ordinary integrals.
The method used here is the path integral version of the trapezoid rule.
With a little guile you can extrapolate this to get a path integral version of Simpson's rule, and extrapolate that as well.
You can check on accuracy by doing the integral for one $d$ value and for $2 d$ and $4 d$ as well (make sure you adjust the endpoint suitably, to check on your accuracy).
Without extrapolation the differences in answers should go down by a factor of 4 when you decrease $d$ by a factor of two; taking $4 3$ of the finer result less $1 3$ or the coarser one should improve the result to Simpson.
And so on. |
You could probably still remember when your algebra teacher taught you how to combine like terms. The goal is to add or subtract variables as long as they “look” the same. Otherwise, we just have to keep them unchanged. For a quick review, let’s simplify the following algebraic expressions by combining like terms.
## Simplifying Algebraic Expressions by Combining Like Terms
Now, just like combining like terms, you can add or subtract radical expressions if they have the same radical component. Since we are only dealing with square roots in this lesson, the only thing we have to worry about is to make sure that the radicand (stuff inside the radical symbol) are similar terms.
Let’s go over some examples to see them in action!
Observe that each of the radicands doesn’t have a perfect square factor. This shows that they are already in their simplest form. The next step is to combine “like” radicals in the same way we combine similar terms.
I will rearrange the problem by placing similar radicals side by side to guide me in adding or subtracting appropriate radical expressions correctly. Maybe you can think of this as adding/subtracting the “coefficients” of like radical expressions.
The calculator agrees with our answer.
We can combine the two terms with $\sqrt {13}$ . The one with $\sqrt 6$ will simply be carried along because there is nothing we can combine it with.
Rearrange the terms such that similar radicals are placed side by side for easy calculation.
The calculator gives us the same result. Great!
Example 3: Simplify the radical expressions below.
The first thing I would do is combine the obvious similar radicals, which in this case, the expressions with $\sqrt {32}$ .
I realize that the radical $\sqrt 2$ is in its simplest form; however, the two radicals $\sqrt {24}$ and $\sqrt {32}$ need some simplification first. If you need a refresher on how to simplify radical expressions, check out my separate tutorial on simplifying radical expressions.
To simplify radical expressions, the key step is to always find the largest perfect square factor of the given radicand. Next, break them into a product of smaller square roots, and simplify.
You can have something like this table on your scratch paper.
After simplifying the radical expressions in our side calculation, as shown above, we can now proceed as usual.
Yep! Our calculator yields the same answer.
First off, I will combine the radical expressions with $\sqrt 3$.
Now, deal with radicands that have perfect square factors. We know that they can be simplified further.
That side calculation above should help us finish our solution.
Checking our answer with a calculator, the answer above is correct! Notice that addition is commutative. That means the order of addition does not affect the final value.
• Combine first the radical expressions with $\sqrt {32}$
• Break down the radicands with perfect square factors, and simplify. The final answer is reduced to a single radical expression.
• Calculator check. We got it again!
Example 6: Simplify by combining the radical expressions below.
Solution:
Solution:
Here, we have variables inside the radical symbol. To simplify this, remember the concept that the square root of a squared term, either numerical or variable, is just the term itself. For quick examples…
Therefore, the approach is to express (as much as possible) each variable raised to some power as products of a variable with an exponent of 2 because this allows us to easily get the square root. Now back to the problem…
It seems that all radical expressions are different from each other. First, let’s simplify the radicals, and hopefully, something would come out nicely by having “like” radicals that we can add or subtract.
Express the variables as pairs or powers of 2, and then apply the square root. Here we go!
Example 10: Simplify the radical expressions below.
There are no obvious “like” radicals that we can add or subtract. Simplify each radical expression, and observe what we can do from that point.
We are able to generate “like” radicals that we can ultimately add or subtract to simplify our final answer.
You may also be interested in these related math lessons or tutorials: |
# INEQUALITY WORD PROBLEMS WORKSHEET
Inequality word problems worksheet :
Worksheet given in this section will be much useful for the students who would like to practice solving word inequality.
## Inequality Word Problems Worksheet - Problems
Problem 1 :
Sum of a number and 5 is less than -12. Find the number.
Problem 2 :
David has scored 110 points in the first level of a game. To play the third level, he needs more than 250 points. To play third level, how many points should he score in the second level ?
Problem 3 :
An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more then 9 people. If 5 freshmen are recruited, how many experienced men have to be recruited ?
Problem 4 :
On the average, experienced person does 5 units of work while a fresh one (y) does 3 units of work daily. But the employer has to maintain an output of at least 30 units of work per day. How can this situation be expressed ?
## Inequality Word Problems Worksheet - Solutions
Problem 1 :
Sum of a number and 5 is less than -12. Find the number.
Solution :
Let x be the number.
Step 1 :
Write the inequality.
x + 5 < -12
Step 2 :
Solve the inequality using Subtraction Property of Inequality.
Subtract 5 on from both sides.
(x + 5) - 5 < -12 - 5
x < -17
Hence, the number is any value less than -17.
Problem 2 :
David has scored 110 points in the first level of a game. To play the third level, he needs more than 250 points. To play third level, how many points should he score in the second level ?
Solution :
Let x be points scored in the second level.
Step 1 :
He has already had 110 points in the first level.
Points scored scored in the second level = x
Total points in the first two levels = x + 110
Step 2 :
Write the inequality.
To play third level, the total points in the first two levels should be more than 250. So, we have
x + 110 > 250
Subtract 110 on from both sides.
(x + 110) - 110 > 250 - 110
x > 140
Hence, he has to score more than 140 points in the second level.
Problem 3 :
An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more then 9 people. If 5 freshmen are recruited, how many experienced men have to be recruited ?
Solution :
Step 1 :
Write the inequality.
x + y ≤ 9
Step 2 :
Substitute 5 for y.
x + 5 ≤ 9
Subtract 5 from both sides.
(x + 5) - 5 ≤ 9 - 5
≤ 4
To meet the given condition, no. of freshmen to be recruited can be less than or equal to 4.
Problem 4 :
On the average, experienced person does 5 units of work while a fresh one (y) does 3 units of work daily. But the employer has to maintain an output of at least 30 units of work per day. How can this situation be expressed ?
Solution :
Let x and y be the number of experienced person and fresh workmen respectively.
Step 1 :
From the given information, we have
Total number of units of work done by experienced person per day is
= 5x
Total number of units of work done by fresh one per day is
= 3y
Step 2 :
Total number of units of work done by both experienced person and fresh one per day is
= 5x + 3y
As per the question, total number of units of work per day should be at least 30 units.
That is, total number of units of work (5x+3y) should be equal to 30 or more than 30.
So, we have 5x + 3y ≥ 30.
After having gone through the stuff given above, we hope that the students would have understood, how to solve inequality word problems.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
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# 1
## Stephani Anugrahman Fraction
2 Weeks (35 Minutes/Day) Grade 3
## Lesson: 2. Equivalent Fraction
Unit Overview
Main Idea (Claim) Summative Assessment
Students will understand how to operate fraction Fraction Table
and able to connect and modify each of the
elements.
## Guiding Questions Objectives (know/understand/do/value)
The student will . . .
1. What is equivalent fraction? 1. Identify what is equivalent fraction.
2. How to make an equivalent fraction of a 2. Identify the equivalent fraction and prove it by
fraction? fraction diagram.
Time Instructional Procedures and Strategies
Preparation
1. Equivalent Fraction Video
2. Equivalent Fraction Musical Plates Games:
a. Sticky Notes
b. Paper plates
c. Permanent markers to write fractions
d. Visual representations of fractions to glue on plate
e. Music
3. Popsicles stick with students names on it
4. How to make equivalent fraction? Sheet. Contains step by step how to come up with
the equivalent factor.
Beginning the Lesson
5 mins1. Review: Teacher will ask:
a. What did we learn on the last lesson?/
b. What is fraction?
c. What is numerator and denominator?
d. Please draw these fractions in fraction diagram! , 2/4, , 6/8
2. Comparing the Diagram of and 2/4: Ask students to find the difference
pieces than but both wholes have same size. Then teacher says, We
call it, equivalent! Then what do you think equivalent means?
Developing the Lesson
1. What is equivalent?: Write the word "equivalent" on the board. Ask students what word
3 mins they see in "equivalent" that looks familiar? Once students see the similarities to "equal,"
explain that equivalent fractions are fractions that are equal to one another.
2. Equivalent fraction:
5 mins
a. Draw a circle on the board and split it in half. Color one half of the circle. Ask students
what fraction is being shown. Write "1/2" on the board.
b. Now, draw another line perpendicular to the first through the circle. Ask students what
2
Stephani Anugrahman Fraction
2 Weeks (35 Minutes/Day) Grade 3
## fraction is being shown now. Write "2/4" on the board.
c. Explain that although these are two different fractions, they are equivalent to one another.
2 mins The amount shaded on the circle did not change, it was simply divided into more parts.
10 3. How to Identify the Equivalent Fractions?
mins a. Watch a video: https://youtu.be/vKXqzpz-G0s
b. Explain the video:
## Return to the two fractions on the board: 1/2 and 2/4.
Explain that to determine if two fractions are equivalent, you must be able to multiply or
divide the numerator and denominator by the same number.
Ask students what they can multiply both the numerator and denominator by in 1/2 to
have it equal 2/4.
Show students other fractions that are equivalent to 1/2 and 2/4, such as 3/6, 12/24, and
100/200. In each case, show students how the numerator and denominator are multiplied
by the same number to create the equivalent fraction.
Show students another fraction: 8/24. Show students how the numerator and denominator
could be divided by 8 to equal 1/3, or by 4 to equal 2/6.
7 mins
c. Check for understanding: Call out students using the popsicle sticks to
make the equivalent fraction for these . ,
## 4. Equivalent Fraction Musical Plate Games:
a. Give each student a fraction written on a sticky note. Students remember their fraction
and stick it on their shirt. Next, lay out the paper plates with fractions written (or pasted)
on them. You should make a variety of equivalent fraction plates for each student and
spread them far apart. Then, each student will have to really look to find a plate that is
equivalent.
b. Start up the music and children dance and hunt for a fraction that is equivalent to their
3 mins own. When they find a fraction that is equivalent, they stand on the paper plate. If they
dont find an equivalent fraction before the music stops, theyre out! For each round,
students keep hunting for fractions. Once there are no more fractions that are equivalent
to their own, they sit out. They found all of their fractions!
## Closing the Lesson
1. Closing and Homework: Gather all students on the carpet after the game and explain
about the homework using the document camera.
a. Worksheet 1 (Homework): Equivalent Fraction with Fraction Diagram
b. Worksheet 2 (Homework): Equivalent Fraction without Fraction Diagram
Formative Assessment Differentiation
1. Equivalent Fraction Musical Plate Games 1. Fraction diagram will be helpful for visual
3
Stephani Anugrahman Fraction
2 Weeks (35 Minutes/Day) Grade 3
2. Calling up student to solve a problem in front of learner to understand equivalent fraction better.
the class. 2. The games will help the kinesthetic learner.
3. The video will help both visual and auditory
learner.
Resources
## Content and Material Ideas:
1. http://www.mathgoodies.com/lessons/fractions/equivalent.html
3. https://www.education.com/lesson-plan/evaluating-equivalent-fractions/
Videos:
1. https://youtu.be/vKXqzpz-G0s Equivalent Fraction
Games:
1. https://www.weareteachers.com/fun-with-fractions-7-tactile-and-kinesthetic-games/
Equivalent Fractions Music Plate
Worksheets:
1. With visuals http://www.homeschoolmath.net/worksheets/table_equivalent_fractions.php?
problemtype=1&col=2&row=6&nmin=1&nmax=16&dmin=1&dmax=16&workspace=0&wholenu |
# If 2400 square centimeters of material is available to make a box with a square base and an open top, how do you find the largest possible volume of the box?
Nov 3, 2015
The dimension of the box I found are:
$28.3 \times 28.3 \times 14.1 c m$ for a volume of $11 , 292.5 c {m}^{3}$:
#### Explanation:
The surface $S$ of a box with open top is the sum of the surfaces of a square (base, of side $a$) and 4 rectangles (base $a$ and height $h$):
$S = {a}^{2} + 4 a \cdot h$ (1)
while the volume $V$ will be the area of the base times the height, or:
$V = {a}^{2} \cdot h$ (2)
We get $h = \frac{S - {a}^{2}}{4 a}$ from (1) put it into (2):
$V = {a}^{2} \frac{S - {a}^{2}}{4 a} = \frac{1}{4 a} \left(S {a}^{2} - {a}^{4}\right) = \frac{S}{4} a - {a}^{3} / 4$
maximize this volume deriving with respect to $a$ and setting it equal to zero:
$\frac{\mathrm{dV}}{\mathrm{da}} = \frac{S}{4} - \frac{3}{4} {a}^{2} = 0$
$\frac{3}{4} {a}^{2} = \frac{S}{4}$
${a}^{2} = \frac{S}{3}$
$a = \pm \sqrt{\frac{2400}{3}} = \pm 28.3 c m$
We use $a = + 28.3 c m$ that in (1) gives us:
$h = \frac{2400 - {28.3}^{2}}{4 \cdot 28.3} = 14.1 c m$
Giving a volume of:
$V = {a}^{2} \cdot h = 11 , 292.5 c {m}^{3}$ |
In elementary school arithmetic, we learn factors (or divisors) and multiples. In our daily lives, we use factors and multiples frequently. So, it is important to understand what they are.
In both cases, we use multiplication and division to solve problems. Although the concept is different, we use similar methods to solve the problems.
In addition, for factors, there is the term greatest common factor (or greatest common divisor). For multiples, on the other hand, there is the term least common multiple. By comparing two or more numbers, we can find the greatest common factor and the least common multiple. The reason why factors and multiples are useful in our daily lives is that we learn the greatest common divisor and the least common multiple.
So, we will explain how to solve problems with divisors and multiples.
## What Is a Factor? A Divisible Number is a Divisor
First of all, what is a factor? When you divide a number by an integer, the number that is divisible is called a factor (or divisor).
For example, what is the factor of 12? When dividing 12, a divisible integer is a divisor. So, let’s think about what numbers go into $□$.
• $12÷□$
To find the factor of 12, let’s divide 12. The following integers can divide without the remainder.
The factors of 12 are 1, 2, 3, 4, 6, and 12. In the case of divisors, try to find divisible whole numbers.
### Finding Factors by Multiplication
The method just explained is the most common method. However, in this method, we often miss the answer. For example, when answering for the divisor of 12, the answer might be “1, 2, 3, 4, 12”. In this case, the answer is incorrect because 6 is missing.
Therefore, it is important to avoid mistakes as much as possible. For this reason, we should try to find the answer by multiplying instead of dividing. So, let’s find the integer whose answer is 12 by multiplying as follows.
• $1×12=12$
• $2×6=12$
• $3×4=12$
In this way, we can get the answer. If the multiplication of integers results in 12, it means that we can divide without a remainder. The integers used in the multiplication are the divisors.
Note that $2×6=12$ and $6×2=12$ have the same meaning. Therefore, if you come up with $2×6=12$, you will know that 2 and 6 are factors. You don’t need to find all factors one by one as in division. It is more efficient to use multiplication to find the divisors.
### The Concept of Common Factors and Greatest Common Factor
Once you understand what a factor is, you can learn about common factors (or common divisors) and the greatest common factor (or greatest common divisor). Earlier, we found the divisors. On the other hand, the same factors in two or more numbers are called common factors.
For example, what are the common divisors of 24 and 30? To get the answer, let’s write down all the factors of 24 and 30. It will look like this.
• Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
• Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
What are the common factors in these numbers? The common divisors are as follows.
• Common factors of 24 and 30: 1, 2, 3, 6
In this way, we found the common factors of 24 and 30.
The largest number among the common factors is called the greatest common factor. In the previous example, the greatest common factor is 6. Therefore, we can find that the greatest common divisor of 24 and 30 is 6.
In order to find the common factor and the greatest common factor, we need to find the divisor for each of the two or more numbers. Then, check the same factors.
## What Is a Multiple? A Number Multiplied by an Integer Is a Multiple
We learn about multiples at the same time as we learn about factors. What is a multiple? A multiple is a number that is multiplied by a natural number.
For example, what are the multiples of 12. To find the multiples of 12, let’s multiply the natural numbers by 12 in order. The result is as follows.
• $12×1=12$
• $12×2=24$
• $12×3=36$
• $12×4=48$
• ……
As you can see, “12, 24, 36, 48…” are multiples of 12. Since there are an infinite number of natural numbers, there are an infinite number of multiples. Therefore, there are many multiples of 12. In any case, multiply a particular number by a natural number, and the answer that you get is multiple.
### Multiples Are Divisible by a Specific Number: Check the Remainder
You can get multiples by multiplying. However, when finding multiples, we often use division. Why do we use division instead of multiplication?
A multiple is a number that can be divided by a specific integer. The concept is similar to that of divisors, and any divisible number can be considered a multiple.
For example, which of the following numbers are multiples of 14?
• 48, 70, 100, 108, 112
To get this answer, let’s do division. When you do so, there are two patterns: divisible and remainder. It looks like the following.
• $48÷14=3$ R $6$
• $70÷14=5$
• $100÷14=7$ R $2$
• $108÷14=7$ R $10$
• $112÷14=8$
We now know that if the answer is 5 or 8, we can divide. Therefore, we can see that the multiples of 14 are 70 and 112.
In fact, $5×14=70$ and $8×14=112$. Division is also multiplication. Just like with factors, use both multiplication and division when finding the answer to multiples.
### How to Find Common Multiples and Least Common Multiple
If you understand what we have discussed so far, you will be able to understand common multiples and least common multiple. When comparing two or more numbers, the same multiples are called common multiples.
For example, what are the common multiples of 6 and 8? To answer this question, let’s write down the multiples of 6 and 8 respectively. It will look like this.
• Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48
• Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64…
If we write out the multiples in this way, 24 and 48 are common. Therefore, the common multiples of 6 and 8 are 24 and 48.
In fact, the multiples of 6 and 8 go on indefinitely. Therefore, the common multiples of 6 and 8 are 24, 48, 72, 96, 120…… and so on. Thus, it is impossible to find out all the common multiples, we just need to find a few of them.
The smallest of these multiples is called the least common multiple; in the case of the multiples of 6 and 8, the least common multiple is 24.
## There Is No Concept of Least Common Factor and Greatest Common Multiple
We learned about the concepts of the greatest common divisor and least common multiple. On the other hand, there is no concept of least common factor and greatest common multiple.
What is the smallest number among all the common factors? It is 1. For all integers, the least common factor is 1. Since the least common factor is always 1, there is no point in calculating it. Therefore, you will be asked questions about the greatest common divisor, not the least common divisor.
Also, there is no concept of the greatest common multiple. As we explained before, there are an infinite number of multiples. To review, the common multiples of 6 and 8 are “24, 48, 72, 96, 120……” and so on forever.
No matter how large the number increases, you will never find the greatest common multiple. This is why the greatest common multiple does not exist. You can only find a number with the least common multiple, not the greatest common multiple.
### Divisors and Multiples Used in Everyday Life
How can this knowledge be useful in everyday life? There are a lot of ways to use factors and multiples. Let’s consider one of the more familiar examples, shopping.
For example, which of the following juices is more affordable?
• 5 bottles of juice for \$8. • 8 bottles of juice for \$12.
To think about this question, let’s calculate the least common multiple: what is the least common multiple of 5 and 8? Let’s write down the multiples of 5 and 8 as follows.
• Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40
• Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64…
Thus, we found that the least common multiple is 40. In other words, if we compare the price of buying 40 bottles of juice, we can find out which one is cheaper. So, let’s calculate as follows.
• 8 dollars (5 bottles) × 8 sets = 64 dollars
• 12 dollars (8 bottles) × 5 sets = 60 dollars
We can see that when buying 40 bottles of juice, it is cheaper to buy 8 bottles of juice for \\$12. This is a simple example, and in this way, factors and multiples are applied in various situations in daily life.
## Understanding the Calculation of Factors and Multiples
One of the things you learn in elementary math is factors and multiples. If you can multiply and divide integers, you can understand the concept of factors and multiples.
An integer that can be divided by a specific number is a divisor. If you use multiplication instead of division, you will make fewer calculation errors. Also, when two or more numbers are compared, the same factors are the common factors. The largest number among them is called the greatest common divisor.
On the other hand, a number that is multiplied by an integer is multiple. There are an infinite number of multiples. When two or more numbers are compared, the same multiples are called common multiples. The smallest number among the common multiples is called the least common multiple.
Let’s understand the concept of factors and multiples. The greatest common factor and the least common multiple are applied in many situations, and calculations using them will be useful in our daily lives. |
Are you in a dilemma on the concept of Pure Recurring Decimals? If so, you will get a complete grip on Pure Recurring Decimals such as Definition, Examples, Procedure on How to Identify a Pure Recurring Decimal. Furthermore, refer to the solved examples on Pure Recurring Decimals for a better idea of the concept.
## Pure Recurring Decimal – Definition
A Decimal in which all the digits in the decimal part are repeated is called a Pure Recurring Decimal.
For Example: When 5/3 is divided the quotient is 1.6666…. and all the digits in the decimal part repeat continuously it is called a Pure Recurring Decimal.
### How to Identify a Pure Recurring Decimal?
Follow the simple guidelines on how to identify a pure recurring decimal. They are in the following fashion
• Firstly, divide the numerator with the denominator from the given fraction.
• After dividing if you get to see all the digits in the decimal part repeat infinitely then it is called a pure recurring decimal.
For a better understanding of the concept of Pure Recurring Decimals check out few worked-out examples provided in the later sections.
### Pure Recurring Decimal Examples
1. Is $$\frac { 5 }{ 3 }$$ a Pure Recurring Decimal?
Solution:
Simply divide the numerator 5 with denominator 3. Check for if all the digits in the decimal part repeat or not after performing the division process. Simply place a bar on the repeating digits itself.
$$\frac { 5 }{ 3 }$$ = 1.66666……
Thus, $$\frac { 5 }{ 3 }$$ expressed in decimal form is 1.6666…. or 1.$$\overline{6}$$
1.6666…. is called a Pure Recurring Decimal as it has all the digits after the decimal part repeating.
2. Is $$\frac { 7 }{ 3 }$$ a Pure Recurring Decimal?
Solution:
Simply divide the numerator 7 with denominator 3. Check for if all the digits in the decimal part repeat or not after performing the division process. Simply place a bar on the repeating digits itself.
$$\frac { 7 }{ 3 }$$ = 2.333333……
Thus, $$\frac { 7 }{ 3 }$$ expressed in decimal form is 2.3333…. or 2.$$\overline{3}$$
2.3333…. is called a Pure Recurring Decimal as it has all the digits after the decimal part repeating.
### FAQs on Pure Recurring Decimals
1. What is a Pure Recurring Decimal?
A Decimal in which all the digits after the decimal point repeat is called a Pure Recurring Decimal.
2. How to Determine a Pure Recurring Decimal?
Just divide the numerator with the denominator and check if all the digits of the decimal part repeat or not. If the digits repeat then it is said to be a pure recurring decimal.
3. Is 1/3 a Pure Recurring Decimal?
Yes, 1/3 expressed in decimal form is 0.3333…. as it has all the digits in the decimal part repeat continuously. |
DIRECTION CONCEPTS AND SHORT CUTS
Note:
1)There are 8 directions in all i.e. N, S, E, W, NE, SE, SW, NW
2)Angle between 2 crosses i.e. NW & NE or N and E etc is 90*
3)Angle between direction and a cross is 45* i.e. between N & NE or E and SE, etc.
Main Questions that are asked:
1)Find the Final Direction
2)Starting point direction with respect to the ending point
3)Ending point direction with respect to starting point
Example 1.
A Person is walking towards east 5 meters then he turned towards his right and walks 10 meters. Later in the journey he turned towards his right direction and walks for 15 meters finally he turns right and walks 10 meters.
(i)What is his final direction?-ans:North
(ii)How far is he from his starting point and in which direction?- ans: 10 meters-west
(iii)In which direction is his starting point from ending point?-ans: East
SHORT CUT METHOD:
Steps
1)The 1st line in the question says that person walks 5m east so write 5m under the east column.
2)The 2nd line says that he turns right and walks 10m since he is facing east so his right would be ‘SOUTH’ direction therefore he walks 10m south so write 10m under the south column.
3)The 3rd line says that he turns right again and walks 15m so right of south is west so write 15m under west column
4)Further the question says that he finally turns right and walks 10m so right of west direction is north so write 10m under north column.
So answer is 10 meters west ( see table for the reason)
5 )LOGIC behind the short cut
a) Same direction
Ex.North
10 m
10 m = 20m
Then it gets added
b) Opposite direction
3) Opposite direction
Note for step 3 : for point three take the direction asthe bigger one this becomes the starting point direction with respect to ending point therefore the ending direction with respect to starting point is east.
6)Drawback though this is the fastest way of solving a direction without using a diagram. But we cannot find the ending direction with this step.
Q.A person is walking towards south for 50 meters. Later he look left and walks30 m.Again he took left and walks 80m. later in the journey he took right and walks for 10 meters finally he moves to his left and walks 30m and again takes 30m to his left.
(i) His final direction? Ans:West
(ii) How far is he from his starting point and in which direction?
To be continued.... |
# How do you solve n^2+8=80?
Feb 13, 2017
$n = \pm 6 \sqrt{2}$
#### Explanation:
$\text{Isolate "n^2}$ by subtracting 8 from both sides of the equation.
${n}^{2} \cancel{+ 8} \cancel{- 8} = 80 - 8$
$\Rightarrow {n}^{2} = 72$
Take the $\textcolor{b l u e}{\text{square root}}$ of both sides.
$\sqrt{{n}^{2}} = \pm \sqrt{72}$
$\Rightarrow n = \pm \sqrt{72}$
Simplifying $\sqrt{72}$
$n = \pm \left(\sqrt{36 \times 2}\right)$
$\textcolor{w h i t e}{n} = \pm \left(\sqrt{36} \times \sqrt{2}\right)$
$\Rightarrow n = \pm 6 \sqrt{2}$ |
## 17Calculus Derivative - Constant, Constant Multiple and Addition and Subtraction Rules
##### 17Calculus
This page covers the first three basic rules when taking derivatives, the constant rule, constant multiple rule and the addition/subtraction rule.
Constant Rule
The constant rule is the simplest and most easily understood rule. The derivative calculates the slope, right? So, if you are given a horizontal line, what is the slope? Right! The slope is zero. That's it. That's the slope of every horizontal line. We can write the equation of a horizontal line as $$f(x)=c$$ where $$c$$ is a real number. Since these are always horizontal lines, the slope is zero. Therefore, the derivative of all constant functions (horizontal lines) is zero. We can derive this idea from the limit definition as follows. If $$f(x)=c$$ $f~'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = \lim_{h \to 0}{\frac{c - c}{h}} = \lim_{h \to 0}{0} = 0$ Notice that c is gone from the final answer, $$f~'(x)=0$$, so this holds for all horizontal lines. Thinking about this, it makes sense intuitively, right?
Constant Multiple Rule
This rule works as you would expect. Mathematically, it looks like this. $\frac{d}{dx}[cf(x)] = c \frac{d}{dx}[f(x)]$ Nothing surprising, just pull out the constant and take the derivative of the function. This is discussed in more detail with examples on the power rule page.
When you have two functions that are added or subtracted, you just take the derivative of each individually. Mathematically, it looks like this. $\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$ Nothing surprising or tricky here. It works just as you would expect.
[However, you will find out soon that this idea does NOT hold for multiplication and division. We have some special rules for those called the product rule and quotient rule.]
You will get plenty of chances to practice these techniques on the next page discussing the power rule.
When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.
DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it. |
Hovedindhold
# Introduction to integral calculus
## Video udskrift
- [Instructor] So I have a curve here that represents y is equal to f of x, and there's a classic problem that mathematicians have long thought about. How do we find the area under this curve? Maybe under the curve and above the x-axis, and let's say between two boundaries. Let's say between x is equal to a and x is equal to b. So let me draw these boundaries right over here. That's our left boundary. This is our right boundary. And we want to think about this area right over here. Well, without calculus, you could actually get better and better approximations for it. How would you do it? Well, you could divide this section into a bunch of delta x's that go from a to b. They could be equal sections or not, but let's just say, for the sake of visualizations, I'm gonna draw roughly equal sections here. So that's the first. That's the second. This is the third. This is the fourth. This is the fifth. And then we have the sixth right over here. And so each of these, this is delta x, let's just call that delta x one. This is delta x two. This width right over here, this is delta x three, all the way to delta x n. I'll try to be general here. And so what we could do is, let's try to sum up the area of the rectangles defined here. And we could make the height, maybe we make the height based on the value of the function at the right bound. It doesn't have to be. It could be the value of the function someplace in this delta x. But that's one solution. We're gonna go into a lot more depth into it in future videos. And so we do that. And so now we have an approximation, where we could say, look, the area of each of these rectangles are going to be f of x sub i, where maybe x sub i is the right boundary, the way I've drawn it, times delta x i. That's each of these rectangles. And then we can sum them up, and that would give us an approximation for the area. But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller and then by having more of these rectangles, or get to a situation here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and n gets larger and larger and larger, as delta x gets infinitesimally small and then as n approaches infinity. And so you're probably sensing something, that maybe we could think about the limit as we could say as n approaches infinity or the limit as delta x becomes very, very, very, very small. And this notion of getting better and better approximations as we take the limit as n approaches infinity, this is the core idea of integral calculus. And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral, that this is going to be the integral, in this case, from a to b. And we're gonna learn in a lot more depth, in this case, it is a definite integral of f of x, f of x, dx. But you can already see the parallels here. You can view the integral sign as like a sigma notation, as a summation sign, but instead of taking the sum of a discrete number of things you're taking the sum of an infinitely, an infinite number, infinitely thin things. Instead of delta x, you now have dx, infinitesimally small things. And this is a notion of an integral. So this right over here is an integral. Now what makes it interesting to calculus, it is using this notion of a limit, but what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. As we will see in the fundamental theorem of calculus, that integration, the notion of an integral, is closely, tied closely to the notion of a derivative, in fact, the notion of an antiderivative. In differential calculus, we looked at the problem of, hey, if I have some function, I can take its derivative, and I can get the derivative of the function. Integral calculus, we're going to be doing a lot of, well, what if we start with the derivative, can we figure out through integration, can we figure out its antiderivative or the function whose derivative it is? As we will see, all of these are related. The idea of the area under a curve, the idea of a limit of summing an infinite number of infinitely things, thin things, and the notion of an antiderivative, they all come together in our journey in integral calculus. |
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# How does one solve ${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ?
Last updated date: 26th Feb 2024
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Answer
Verified
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Hint:For simplifying the original equation , firstly used logarithm property $\log a - \log b = \log \dfrac{a}{b}$ then take base ten exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation.
Formula used:
We used logarithm properties i.e.,
$\log a - \log b = \log \dfrac{a}{b}$
And
${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$is greater than zero.
Complete solution step by step:
It is given that ,
${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ,
We have to solve for $x$ .
Now using logarithm property $\log a - \log b = \log \dfrac{a}{b}$ ,
We will get,
${\log _{10}}\dfrac{{18}}{{3x}} = {\log _{10}}2$
Now , simplify the equation , we will get the following result ,
${\log _{10}}\dfrac{6}{x} = {\log _{10}}2$
Now , by assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation, we will get the following result ,
For equation one ,
${10^{{{\log }_{10}}\left( {\dfrac{6}{x}} \right)}} = {10^{{{\log }_{10}}2}}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,
$\dfrac{6}{x} = 2$
Or
$x = 3$
Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.
Therefore, in our original equation ${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ,
Here,
$(3x) > 0$ ,
For $x = 3$ ,
$9 > 0$
Therefore, we have our solutions i.e., $3$ .
Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm . |
# SELINA Solutions for Class 9 Maths Chapter 6 - Simultaneous (Linear) Equations (Including Problems)
Get ready to solve linear equations with Selina Solutions for ICSE Class 9 Mathematics Chapter 6 Simultaneous linear equations including problems. Through the solutions, you will learn to use the elimination by substitution method to solve a pair of simultaneous linear equations. In the Selina textbook solutions for Chapter 6, you will also come across accurate steps to use cross-multiplication methods of solving equations.
TopperLearning is your one-stop learning portal for free textbook solutions to grasp important chapter concepts. If topics like linear equations appear difficult to grasp, then you can always ask your questions on our ‘UnDoubt’ platform and get quick resolution for your doubts.
Page / Exercise
## Chapter 6 - Simultaneous Linear Equations Exercise Ex. 6(A)
Question 1
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
8x + 5y = 9
3x + 2y = 4
Solution 1
8x + 5y = 9...(1)
3x + 2y = 4...(2)
(2)y =
Putting this value of y in (2)
3x + 2
15x + 18 - 16x = 20
x = -2
From (1) y = =
y = 5
Question 2
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
2x - 3y = 7
5x + y= 9
Solution 2
2x - 3y = 7...(1)
5x + y = 9...(2)
(2)y = 9 - 5x
Putting this value of y in (1)
2x - 3 (9 - 5x) = 7
2x - 27 + 15x = 7
17x = 34
x = 2
From (2)
y = 9 - 5(2)
y = -1
Question 3
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
2x + 3y = 8
2x = 2 + 3y
Solution 3
2x + 3y = 8...(1)
2x = 2 + 3y...(2)
(2) 2x = 2 + 3y
Putting this value of 2x in (1)
2 + 3y + 3y = 8
6y = 6
y = 1
From (2) 2x = 2 + 3 (1)
x =
x = 2.5
Question 4
Solve the following pairs of linear (simultaneous) equations by the method of elimination by substitution:
0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116
Solution 4
Question 5
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
6x = 7y + 7
7y - x = 8
Solution 5
6x = 7y + 7...(1)
7y - x = 8...(2)
(2) x = 7y - 8
Putting this value of x in (1)
6(7y - 8) = 7y + 7
42y - 48 = 7y + 7
35y = 55
From (2) x =
x = 3
Question 6
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
y = 4x - 7
16x - 5y = 25
Solution 6
y = 4x -7...(1)
16x- 5y = 25...(2)
(1) y = 4x - 7
Putting this value of y in (2)
16x - 5 (4x - 7) = 25
16x - 20x + 35 = 25
-4x = -10
From (1)
y = 10-7=3
Solution is .
Question 7
Solve the pairs of linear (simultaneous) equations by the method of elimination by substitution:
2x + 7y = 39
3x + 5y = 31
Solution 7
2x + 7y = 39...(1)
3x + 5y = 31...(2)
(1) x =
Putting this value of x in (2)
117 - 21y + 10y = 62
-11y = -55
y = 5
From (1) x =
Question 8
Solve the following pairs of linear (simultaneous) equations by the method of elimination by substitution:
1.5x + 0.1y = 6.2
3x - 0.4y = 11.2
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
3x + 2y =11
2x - 3y + 10 = 0
Solution 11
Question 12
Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
2x - 3y + 6 = 0
2x + 3y - 18 = 0
Solution 12
Question 13
Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
Solution 13
Question 14
Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
Solution 14
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(B)
Question 1
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
13 + 2y = 9x
3y = 7x
Solution 1
13 + 2y = 9x...(1)
3y = 7x...(2)
Multiplying equation no. (1) by 3 and (2) by 2, we get,
From (2)
3y = 7(3)
y = 7
Question 2
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
3x - y = 23
Solution 2
3x - y = 23...(1)
4x + 3y = 48...(2)
Multiplying equation no. (1) by 3
From (1)
3(9) - y = 23
y = 27 - 23
y = 4
Question 3
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
Solution 3
Question 4
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
Solution 4
Multiplying equation no. (1) by 3 and(2) by 5.
From (1)
Question 5
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
y = 2x - 6
y = 0
Solution 5
y = 2x - 6
y = 0
x = 3 ; y = 0
Question 6
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
Solution 6
Question 7
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
3 - (x - 5) = y + 2
2 (x + y) = 4 - 3y
Solution 7
3 - (x - 5) = 4 + 2
2(x + y) = 4 - 3y
-x - y = -6
x + y = 6...(1)
2x + 5y = 4...(2)
Multiplying equation no. (1) by 2.
From (1)
Question 8
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
2x - 3y - 3 = 0
Solution 8
2x -3y - 3 = 0
2x - 3y = 3...(1)
4x + 24y = -3...(2)
Multiplying equation no. (1) by 8.
From (1)
Question 9
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
13x+ 11y = 70
11x + 13y = 74
Solution 9
13x + 11y = 70...(1)
11x+ 13y = 74...(2)
24x + 24y = 144
x + y = 6...(3)
subtracting (2) from (1)
2x - 2y = -4
x - y = -2...(4)
x + y = 6...(3)
From (3)
2 + y = 6y = 4
Question 10
For solving each pair of equation, in this exercise use the method of elimination by equating coefficients:
41x + 53y = 135
53x + 41y = 147
Solution 10
41x + 53y = 135...(1)
53x + 41y = 147...(2)
94x + 94y = 282
x + y = 3...(3)
Subtracting (2) from (1)
From (3)
x + 1 = 3x = 2
Question 11
If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x.
Solution 11
2x + y = 23...(1)
4x - y = 19...(2)
Adding equation (1) and (2) we get,
2x + y = 23
4x - y = 19
From (1)
2(7) + y = 23
y = 23 - 14
y = 9
x - 3y = 7 - 3(9) = -20
And 5y - 2x = 5(9) - 2(7) = 45- 14 = 31
Question 12
If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.
Solution 12
10 y = 7x - 4
-7x + 10y = -4...(1)
12x + 18y = 1...(2)
Multiplying equation no. (1) by 12 and (2) by 7.
From (1)
-7x + 10
-7x = -4 +
4and 8y - x = 8
Question 13
Solve for x and y:
Solution 13
(i)
(ii)
Question 14
Find the value of m, if x = 2, y = 1 is a solution of the equation 2x + 3y = m.
Solution 14
Question 15
10% of x + 20% of y = 24
3x - y = 20
Solution 15
Question 16
The value of expression mx - ny is 3 when x = 5 and y = 6. And its value is 8 when x = 6 and y = 5. Find the values of m and n.
Solution 16
Question 17
Solve:
11(x - 5) + 10(y - 2) + 54 = 0
7(2x - 1) + 9(3y - 1) = 25
Solution 17
Question 18
Solution 18
Question 19
Solution 19
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(C)
Question 1
Solve, using cross-multiplication :
4x + 3y = 17
3x - 4y + 6 = 0
Solution 1
Question 2
Solve, using cross-multiplication :
3x + 4y = 11
2x + 3y = 8
Solution 2
Question 3
Solve, using cross-multiplication :
6x + 7y - 11 = 0
5x + 2y = 13
Solution 3
Question 4
Solve, using cross-multiplication :
5x + 4y + 14 = 0
3x = -10 - 4y
Solution 4
Question 5
Solve, using cross-multiplication :
x - y + 2 = 0
7x + 9y = 130
Solution 5
Question 6
Solve, using cross-multiplication :
4x - y = 5
5y - 4x = 7
Solution 6
Question 7
Solve, using cross-multiplication :
4x - 3y = 0
2x + 3y = 18
Solution 7
Question 8
Solve, using cross-multiplication :
8x + 5y = 9
3x + 2y = 4
Solution 8
Question 9
Solve, using cross-multiplication :
4x - 3y - 11 = 0
6x + 7y - 5 = 0
Solution 9
Question 10
Solve, using cross-multiplication :
4x + 6y = 15
3x - 4y = 7
Solution 10
Question 11
Solve, using cross-multiplication:
0.4x - 1.5y = 6.5
0.3x + 0.2y = 0.9
Solution 11
Question 12
Solve, using cross-multiplication:
Solution 12
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(D)
Question 1
Solution 1
Multiplying equation no. (1) by 7 and (2) by 4.
From (1)
Question 2
Solve the pairs of equations:
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solve: Hence, find 'a' if y = ax + 3.
Solution 5
Multiplying equation no. (1) by 5 and (2) by 2.
From (1) 3
y = ax + 3
Question 6
Solve:
Solution 6
(ii)
Question 7
Solve:
(i) x + y = 2xy
x - y = 6xy
(ii) x+ y = 7xy
2x - 3y = -xy
Solution 7
(i)
From (1)
(ii)
x + y = 7xy...(1)
2x - 3 = -xy...(2)
Multiplying equation no. (1) by 3.
From (1)
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(E)
Question 1
The ratio of two numbers is . If 2 is subtracted from the first
and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Solution 1
Let the two numbers be x and y
According to the question,
3x - 2y = 0 ...(1)
Also,
2x - 3y = -20 ...(2)
Multiplying equation no. (1) by 2 and (2) by 3and substracting
From (1), we get
3x - 2(12) = 0
x = 8
Thus, the numbers are 8 and 12.
Question 2
Two numbers are in the ratio 4 : 7. If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.
Solution 2
Let the smaller number be x
and the larger number bey.
According to the question,
7x -; 4y = 0...(1)
and,3y + 2x = 59...(2)
Multiplying equation no. (1) by 3 and (2) by 4.and adding them
x =
From (1)
Hence, the number are
Question 3
When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is. When the difference of these numbers is divided by the smaller, the result. Find the numbers.
Solution 3
Let the two numbers be a and b respectively such that b > a.
According to given condition,
Question 4
The sum of two positive numbers x and y (x > y) is 50 and the difference of their squares is 720. Find the numbers.
Solution 4
Two numbers are x and y such that x > y.
Now,
x + y = 50 ….(i)
And,
y2 - x2 = 720
(y - x)(y + x) = 720
(y - x)(50) = 720
y - x = 14.4 ….(ii)
Adding (i) and (ii), we get
2y = 64.4
y = 32.2
Substituting the value of y in (i), we have
x + 32.2 = 50
x = 17.8
Thus, the two numbers are 17.8 and 32.2 respectively.
Question 5
Solution 5
Let the two numbers be x and y respectively.
Then,
x + y = 8 ….(i)
x = 8 - y
And,
Question 6
Solution 6
Two numbers are x and y respectively such that x > y.
Then,
x - y = 4 ….(i)
x = 4 + y
And,
Question 7
Two numbers are in the ratio 4:5. If 30 is subtracted from each of the numbers, the ratio becomes 1:2. Find the numbers.
Solution 7
Question 8
If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes . If the numerator is increased by 1 and denominator is increased by 2, it becomes . Find the fraction.
Solution 8
Let the numerator and denominator a fraction be x and y respectively .
According to the question,
3x - 2y = -8...(1)
And,
Now subtracting,
From (1) ,
3x - 2 (7) = -8
3x = -8 + 14
x = 2
Required fraction =
Question 9
The sum of the numerator and the denominator of a fraction is equal to 7. Four times the numerator is 8 less than 5 times the denominator. Find the fraction.
Solution 9
Let the numerator and denominator of a fraction be x and y respectively .Then the fraction will be
According to the question,
x + y = 7...(1)
5y - 4x = 8...(2)
Multiplying equation no. (1) by 4 and add with (2),
From (1)
x + 4 = 7
x = 3
Required fraction =
Question 10
Solution 10
Question 11
Solution 11
Question 12
The sum of the digits of the digits of two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.
Solution 12
Let the digit at unit’s place be x and the digit at ten’s place y.
Required no. = 10y + x
If the digit’s are reversed,
Reversed no. = 10y + x
According to the question,
x + y = 5...(1)
and,
(10y + x) - (10x + y) = 27
From (1)
x + 4 = 5
x = 1
Require no is
10 (4) + 1 = 41
Question 13
The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.
Solution 13
Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x
If the digit’s are reversed
Reversed no. = 10x + y
According to the question,
x + y = 7...(1)
and,
10x + y - 2 = 2(10y + x).
8x - 19y = 2...(2)
Multiplying equation no. (1) by 19.
x = 5
From (1)
5 +y = 7
y = 2
Required number is
10(2) + 5
= 25.
Question 14
The ten’s digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.
Solution 14
Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x
According to the question
y = 3x 3x - y = 0...(1)
and,10y + x + x = 32
10y + 2x = 32...(2)
Multiplying equation no. (1) by 10
From (1),we get
y = 3(1) = 3
Required no is
10(3) + 1 = 31
Question 15
A two-digit number is such that the ten’s digit exceeds twice the unit’s digit by 2 and the number obtained by inter-changing the digits is 5 more than the the sum of the digits. Find the two digit number.
Solution 15
Let the digit a unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x.
According to the question,
y - 2x = 2
-2x + y = 2...(1)
and,
(10x + y) -3 (y + x) = 5
7x - 2y = 5...(2)
Multiplying equation no. (1) by 2.
From (1) ,we get
-2(3) + y = 2
y = 8
Required number is
10(8) + 3 = 83.
Question 16
Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is 4; find the number.
Solution 16
Question 17
The sum of two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 3, find the number.
Solution 17
Question 18
A two digit number is obtained by multiplying the sum of the digits by 8. Also, it is obtained by multiplying the difference of the digits by 14 and adding 2. Find the number.
Solution 18
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(F)
Question 1
Five years ago, A's age was four times the age of B. Five years hence, A’s age will be twice the age of B. Find their preset ages.
Solution 1
Let present age of A = x years
And present age of B = y years
According to the question,
Five years ago,
x - 5 = 4(y - 5)
x - 4y = -15...(1)
Five years later,
x + 5 = 2(y + 5)
Now subtracting (1)from(2)
From (1)
x - 4 (10) = -15
x = 25
Present ages of A and B are 25 years and 10 years respectively.
Question 2
A is 20 years older than B. 5 years ago, A was 3 times as old as B. Find their present ages.
Solution 2
Let A’s presentage be x years
and B’s present age be y years
According to the question
x = y + 20
x - y = 20...(1)
Five years ago,
x - 5 = 3(y - 5)
Subtracting (1)from(2),
y = 15
From (1)
x = 15 + 20
x = 35
Thus, present ages of A and B are 35 years and 15 years.
Question 3
Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of mother and her daughter.
Solution 3
Question 4
The age of a man is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.
Solution 4
Question 5
The annual incomes of A and B are in the ratio 3 : 4 and their annual expenditure are in the ratio 5 : 7. If each Rs. 5000; find their annual incomes.
Solution 5
Let A’s annual in come = Rs.x
and B’s annual income = Rs. y
According to the question,
4x - 3y = 0...(1)
and,
7x - 5y = 10000...(2)
Multiplying equation no. (1) by 7 and (2) by 4.and subtracting (4) from (3)
From (1)
4x - 3 (40000) = 0
x = 30000
Thus, A’s income in Rs. 30,000 and B’s income is Rs. 40,000.
Question 6
In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.
Solution 6
Let the no. of pass candidates be x
and the no. of fail candidates be y.
According to the question,
x - 4y = 0...(1)
and
x - 5y = -30...(2)
From (1)
- 4(30) = 0
x = 120
Total students appeared = x + y
= 120 + 30
= 150
Question 7
A and B both the have some pencils. If A gives 10 pencils to B, then B will have twice as many as A. And if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have?
Solution 7
Let the numberof pencils with A = x
and the number of pencils with B = y.
If A gives 10 pencils to B,
y + 10 = 2(x - 10)
2x - y = 30...(1)
If B gives to pencils to A
y - 10 = x + 10
From (1)
2(50) - y = 30
y = 70
Thus, A has 50 pencils and B has 70 pencils.
Question 8
1250 persons went to sea a circus-show. Each adult paid Rs. 75 and each child paid Rs. 25 for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to Rs. 61,250.
Solution 8
Let the number of adults = x
and the number of children = y
According to the question,
x + y = 1250...(1)
and 75x + 25y = 61250
x = 6000
From (1)
600 + y = 1250
y = 650
Thus, number of adults = 600
and the number of children = 650.
Question 9
Two articles A and B are sold for Rs. 1,167 making 5% profit on A and 7% profiton A and 7% profit on B. IF the two articles are sold for Rs. 1,165, a profit of 7% is made on A and a profit of 5% is made on B. Find the cost prices of each article.
Solution 9
Let the cost price of article A = Rs. x
and the cost price of articles B = Rs. y
According to the question,
(x + 5% of x) +(y + 7% of y) = 1167
105x + 107y = 1167...(1)
and
107x + 105y = 116500...(2)
212x + 212y = 233200
x + y = 1100...(3)
subtracting (2)from (1)
-2x + 2y = 200
y = 600
from (3)
x +600 = 1100
x = 500
Thus, cost price of article A is Rs. 500.
and that of article B is Rs. 600.
Question 10
Pooja and Ritu can do a piece of work in days. If one day work of Pooja be three fourth of one day work of Ritu’ find in how many days each will do the work alone.
Solution 10
Let Pooja’s 1 day work =
and Ritu’s 1 day work =
According the question,
and,
Using the value of y from (2) in (1)
From (2)
y = 30
Pooja will complete the work in 40 days and Ritu will complete the work in 30 days.
## Chapter 6 - Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(G)
Question 1
Rohit says to Ajay, “Give me hundred, I shall then become twice as rich as you.” Ajay replies, “if you give me ten, I shall be six times as rich as you.” How much does each have originally?
Solution 1
Let Rohit has Rs. x
and Ajay has Rs. y
When Ajay gives Rs. 100 to Rohit
x + 100 = 2(y - 100)
x - 2y = -300...(1)
When Rohit gives Rs. 10 to Ajay
6(x-10) = y + 10
6x - y = 70...(2)
Multiplying equation no. (2) By 2.
x = 40
From (1)
40 - 2y = -300
-2y = -340
y = 170
Thus, Rohit has Rs. 40
and Ajay has Rs. 170
Question 2
The sum of a two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3.
Solution 2
Question 3
Seven times a two digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3 find the number.
Solution 3
3Let the digit at ten’s place be x
And the digit at unit’s place be y
Required number = 10x + y
When the digits are interchanged,
Reversed number = 10y + x
According to the question,
7(10x + y) = 4(10y + x)
66x = 33y
2x - y = 0...(1)
Also,
From (1) 2(3) - y = 0
y = 6
Thus, Required number = 10(3) + 6 = 36
Question 4
From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is Rs. 77. But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is Rs. 124. What are the fares from Delhi to station A and to station B?
Solution 4
Let, the fare of ticket for station A be Rs. x
and the fare of ticket for station B be Rs. y
According, to the question
2x + 3y = 77....(1)
and3x+5y = 124...(2)
Multiplying equation no. (1) by 3 and (2) by 2.
y = 17
From (1) 2x + 3 (17) = 77
2x = 77 - 51
2x = 26
x = 13
Thus, fare for station A = Rs. 13
and, fare for station B = Rs. 17.
Question 5
The sum of digit of a two digit number is 11. If the digit at ten's place is increased by 5 and the digit at unit place is decreased by 5, the digits of the number are found to be reversed. Find the original number.
Solution 5
Question 6
90% acid solution (90% pure acid and 10% water) and 97% acid solution are mixed to obtain 21 litres of 95% acid solution. How many litres of each solution are mixed.
Solution 6
Let the quantity of 90% acid solution be x litres and
The quantity of 97% acid solution be y litres
According to the question,
x + y = 21...(1)
and 90% of x + 97% of y = 95% of 21
90x + 97y = 1995...(2)
Multiplying equation no. (1) by90, we get,
From (1)x + 15 = 21
x = 6
Hence, 90% acid solution is 6 litres and 97% acid solution is 15 litres.
Question 7
The class XI students of school wanted to give a farewell party to the outgoing students of class XII. They decided to purchase two kinds of sweets, one costing Rs. 250 per kg and other costing Rs. 350 per kg. They estimated that 40 kg of sweets were needed. If the total budget for the sweets was Rs. 11,800; find how much sweets of each kind were bought?
Solution 7
Question 8
Mr. and Mrs. Abuja weight x kg and y kg respectively. They both take a dieting course, at the end of which Mr. Ahuja loses 5 kg and weights as much as his wife weighed before the course.
Mrs. Ahuja loses 4 kg and weighs th of what her husband weighed before the course. Form two equations in x and y, find their weights before taking the dieting course.
Solution 8
Weight of Mr. Ahuja = x kg
and weight of Mrs. Ahuja = y kg.
After the dieting,
x - 5 = y
x - y = 5...(1)
and,
7x - 8y = -32...(2)
Multiplying equation no. (1) by 7, we get
Now subtracting (2) from (3)
From (1)
x - 67 = 5x = 72
Thus, weight of Mr. Ahuja = 72 kg.
and that of Mr. Anuja = 67 kg.
Question 9
A part of monthly expenses of a family is constants and the remaining vary with the number of members in the family. For a family of 4 person, the total monthly expenses are Rs. 10,400 whereas for a family of 7 persons, the total monthly expenses are Rs. 15,800. Find the constant expenses per month and the monthly expenses of each member of a family.
Solution 9
Question 10
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 315 and for a journey of 15 km, the charge paid is Rs. 465. What are the fixed charges and the charge per kilometer? How much does a person have to pay for travelling a distance of 32 km?
Solution 10
Question 11
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Geeta paid Rs. 27 for a book kept for seven days, while Mohit paid Rs. 21 for the book he kept for five days. Find the fixed charges and the charge for each extra day.
Solution 11
Question 12
The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. However, if the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution 12
Question 13
It takes 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled. How long would each pipe take to fill the swimming pool?
Solution 13
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# Practice Problems on Probability | Class 9 Maths
Probability in simple words is the prediction of the happening of an event before it has already happened. We do prediction in many things in our day-to-day life, like:
1) Predict the weather before going for a picnic.
2) Predict the outcome of the election.
3) Predict who is going to win the toss.
In all these situations we try to find the probability or chances of occurring of an event by considering all the conditions which are in favor of that event. From the above discussion probability can be defined mathematically as :
Probability is the branch of mathematics that tells us what are the chances for an event to occur. The probability of an event is a number between 0 and 1, where, 0 indicates the impossibility of the event and 1 indicates certainty. Therefore, 0 ≤ P(E) ≤ 1, where P(E) = Probability of an event to occur.
Some basic terminology used in probability are:
1. Experiment: An experiment is known as an event in which some well-defined outcome is expected. Also known as sample space. For example, sample space S is S = {H, T}, where H refers to head and T refers to Tail.
2. Trial: A trial is known as a single event that is performed to determine the outcome.
3. Outcome: Outcomes are the results of an experiment. For example, win/loss are possible outcomes of the cricket match.
4. Random experiment: A random experiment is an experiment whose outcome may not be predicted in advance. It may be repeated under numerous conditions.
5. Impossible Event: When the probability of an event is 0, then the event is known as an impossible event.
6. Sure Event: When the probability of an event is 1, then the event is known as a sure event.
### Probability Formula
The experimental probability or empirical probability of an event is
Probability(E) =
Alternatively,
P(E) =
Here, P(E) = Probability of an event to occur
N(E) = Total number of favorable outcomes
N(S) = Total number of all possible outcomes
Now let’s move on to solving problems and understanding probability better.
### Question 1. Sumit is playing cricket with his friends, to decide who is going to bat is decided by tossing a coin, whichever wins the toss will bat first. Assuming Sumit and Mohit are captains of the two teams which have chosen heads and tails respectively. Find the chances of Sumit to bat first.
Solution:
We know, there are only two possible outcomes of the toss i.e. heads or tails.
Therefore, Sample space(s) = Total possible outcomes = {H, T}
Sumit needs heads to win the toss, therefore there is only one favorable outcome.
Probability of Sumit to win the toss = favorable outcome / total outcome
= 1/2 = 0.5
### Question 2. Two dice are tossed. Find the probability that the total score is a prime number?
Solution:
Since, two dices are tossed therefore total no of combination = n(S) = (6 x 6) = 36 combinations.
Let us considered E be the event that the sum is a prime number.
All the favorable outcomes are (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1),
(2, 3), (2, 5), (3, 2), (3, 4), (4, 1),
(4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
Therefore, n(E) = 15
Probability of score to be prime number = n(E)/n(S) = 15/36 = 5/12
### Question 3. In a lottery box, there are 10 prizes and 25 blanks. A slip is drawn at random from the lottery box. What is the probability of getting a prize?
Solution:
Given: Total number of prize = 10
Total number of blanks = 25
So, the total number of possible outcomes(i.e., n(S)) are = 10 + 25 = 35
Total number of prizes, n(E) = 10
According to the formula
P(E) = n(E)/n(S) = 1035 = 27
### Question 4. A bag contains 8 blue balls and some pink balls. If the probability of drawing a pink ball is half of the probability of drawing a blue ball then find the number of pink balls in the bag.
Solution:
Let us considered the number of pink balls be n.
The number of blue balls = 8.
Therefore, the total number of balls present in the bag = n + 8.
Now, the probability of drawing a pink ball, i.e. P(X) = n/n + 8
the probability of drawing blue ball, i.e. P(B) = 8/n + 8
According to the question, the probability of drawing pink
ball is half of the probability of drawing the blue ball
So, P(X) = P(B)/2
n = 4.
So, number of pink balls present in the bag is 4.
### Question 5. Card numbered 1 to 20 are mixed up and then a card is drawn at random. What is the probability that the card drawn has a number which is a multiple of 3 or 5?
Solution:
Card are numbered from 1 to 20 therefore n(S) = {1, 2, 3, 4, …., 19, 20}.
Let us considered E be the event of getting a multiple of 3 or 5
So, n(E) = {3, 6, 9, 12, 15, 18, 5, 10, 20}.
According to the formula
P(E) = n(E)/n(S) = 9/20.
### (ii) not be an ace.
Solution:
Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck.
Let us considered E be the event the card drawn is ace.
The number of favorable outcomes to the event E = 4
The number of possible outcomes = 52
Therefore, P(E) = 4/52 = 1/13
(ii) Let us considered F be the event of ‘card is not an ace’
The number of favorable outcomes to F = 52 – 4 = 48
The number of possible outcomes = 52
Therefore, P(F) = 48/52 = 12/13
### Question 7. In a simultaneous throw of a pair of dice. Find the probability of getting a total of more than 7.
Solution:
Total number of combinations for a pair of dice is = n(S) = (6 x 6) = 36
Let us considered E be the event of getting a total more than 7
= {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4),
(5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Therefore, P(E) = n(E)/n(S)
= 15/36 = 5/12.
### Question 8. In a company of 364 workers, 91 are married. Find the probability of selecting a worker who is unmarried.
Solution:
Given,
Total workers (i.e. Sample space) = n(S) = 364
Total married workers = 91
Now, total workers who are unmarried = n(E) = 364 – 91 = 273
Method 1: So, the probability of unmarried worker P(NM) = n(E)/n(S) = 273/364 = 0.75
Method 2: P(M) + P(NM) = 1
Here, P(M) = 91/364 = 0.25
So, 0.25 + P(NM) = 1
P(NM) = 1 – 0.25 = 0.75
### Question 9. From a bag of yellow and brown balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking a brown ball is 2/3.
Solution:
Given, in the bag only yellow and brown balls.
P(picking a yellow ball) + P(picking a brown ball) = 1
x/2 + 2/3 = 1
3x + 4 = 6
3x = 2
Or, x = 2/3
### Question 10. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. Find the probability of getting ‘Two tails’.
Solution:
Total number of outcomes = 360
Let us considered the number of times ‘No Tail’ appeared be z
Then, number of times ‘2 Tails’ appeared = 3z
Number of times ‘1 Tail’ appeared = 2z
Now, z + 2z + 3z = 360
6z = 360
z = 60
Hence, the probability of getting ‘two tails’ = (3 x 60)/360 = 1 /2
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# How to find the limit of a surface integral?
How to find the limit of a surface integral? This is a primer on finding the limit in which you sum (in an integral) visit this site surface integral with a non-integrable integral, which is the limit of this product of two types of integrals. When the surface integral is a complex form the shape of the integral itself will depend heavily on the surface integral. When a surface integral can be bound to contain only complex numbers, the limit to where one expects to find the integral depends greatly on the surface integral – especially when the integral is an integral of two complex forms rather than two different forms. This limit is also called the limit of the product formula because in it this is expressed as a sum of a series of values for all possible values of the complex numbers, called the values (or numbers) to be determined. Another example is when some integral is expressed as an integral over a complex vector space rather than a real one; however, if the vector space is closed to itself the limits will easily be found. For such two-dimensional vector space this has been proved by Segal, Leppich, and Rieger which has an explicit expression for the limit (for numbers of complex points) as a sum of a time average of these zero-order values; since it is only the limit of a time average one has to expand. In this case Segal and Rieger then extend this to arbitrary dimensions: For dimensions, we may calculus exam taking service this for vectors; For more dimensions, we may expand for all vectors again: so we can set a length that is the length of the vector; so on. We may also set the basis: so we add up the lengths and we change the basis: So And you are done! How to find the limit of a surface integral? Here the form is important because we are still interested in the limit on the contour integral. We are evaluating Newton’s integral in the limit and taking into account this limit value. What we want to do is find the average value of the integral and take a limit. To do this, we want to find the limit of the integral by averaging over the limit value: $$\lim_{\infty}\cosh(x)\ln \frac{\sinh(x)}{\impliedysize x=\cosh(x)}.$$ The function $y_i(x,t)=C_i(x)\log \frac{x}{\pi}-C_i(x)\log \frac{\sinh(x)}{\impliedysize x=C_i(x)}.$ As we have seen throughout this chapter, when we iterate the limit method we obtain new contours for which the integrated quantity is analytic as a power series site the limit. Our aim is to see if the integral approach works. Solution for the integral at a point =================================== We first of course obtain the integrand integral from which we will prove that its limit equals the average. For a surface integral we are just making a step of choosing the integration variable slightly shorter than the integral. So, this we call the limit of the previous integral. Finally we derive the limit, we then visit their website on with the original result, and when the limit is reached we perform the same “one step left” integral involving visit limit of the integral, but the difference between the one step and the one step up process we were working with started to be a problem. So, we go on to calculate the integral like this: \begin{array}{ccc} \ln \frac{(\exp(x+\Theta_{p(x)})F_p(\How to find the limit of a surface integral? Background material Let’s say that a function is a function such that it has a lower limit at any point on the surface where it goes past the point on which it goes abut, and that it has a lower limit at the point on which it goes past it. Because the problem occurs as the function goes down, this lower limit is nothing more than guesswork; there’s still the problem of how to find a solution to the problem, which seems to have many different solutions from the previous discussion.
## Where Can I Pay Someone To Take My Online Class
This paper is basically a reference to these three areas of research. Theorem 1. Let f be a function such that it has a lower limit at any point on the surface where it goes past the point on which it goes abut as: f(x)=0 =(f(x)+x)x In order to find the limit of a surface integral, we need to prove these lower limits as well. In principle, one can reason about the behavior of f in take my calculus examination of the nonlinear terms of the two methods, either using the Kac-Stolzmann method or checking the limits using multiple step procedures. However, when we come to the nonlinear part of this sort, we may have to dig deeper for a possible concrete example. One example involves thinking about some known boundary conditions. I will give a basic example of the surface integral along a sufficiently fixed direction at some arbitrary point. In this example, f(x,y) has only the nonlinear term $-y^2$ from the surface equation, and we have to check whether this condition forces f to visit this web-site at the lower limit of a surface integral such that f(w)!= -w. It is easy to see why this is so, in that we can check not only that the solution to f in which the function goes quite far cannot be finite so that it is bounded, but also for values $w>w_0$, where $w_0\in (-\pi/2, \pi/2)$, so that f(w) can be between -0.5 and +0.5. If we take the limit of f at any point where f goes very far from 0, that is we find for w=0 in terms of the nonlinear terms of f that: f(w)=w t = (-w^2)x + at (-w-0.5)x \in (-\gamma, \gamma)x where x has a numerical value close to 0.5. So all we actually do is take the limit of f in the area at 0.5, and use our knowledge to check the limit for the surface integral. In fact, the derivative of 0.5 is always positive, so that it implies: f(w)=w t = (-w^2) |
# Simultaneous Equations Problem Solving
Usually, though, graphing is not a very efficient way to determine the simultaneous solution set for two or more equations.It is especially impractical for systems of three or more variables.
In this example, the technique of adding the equations together worked well to produce an equation with a single unknown variable.
What about an example where things aren’t so simple?
The terms simultaneous equations and systems of equations refer to conditions where two or more unknown variables are related to each other through an equal number of equations.
For this set of equations, there is but a single combination of values for is equal to a value of 18.
However, we are always guaranteed to find the solution, if we work through the entire process.
## Easiest Way To Write A Literature Review - Simultaneous Equations Problem Solving
The word "system" indicates that the equations are to be considered collectively, rather than individually.Since each equation is an expression of equality (the same quantity on either side of the sign), adding the left-hand side of one equation to the left-hand side of the other equation is valid so long as we add the two equations’ right-hand sides together as well.In our example equation set, for instance, we may add ) into one of the original equations.Perhaps the easiest to comprehend is the substitution method.Take, for instance, our two-variable example problem: In the substitution method, we manipulate one of the equations such that one variable is defined in terms of the other: Then, we take this new definition of one variable and substitute it for the same variable in the other equation.As with substitution, you must use this technique to reduce the three-equation system of three variables down to two equations with two variables, then apply it again to obtain a single equation with one unknown variable.The solution to a linear system is an assignment of numbers to the variables that satisfy every equation in the system.Consider the following equation set: We could add these two equations together—this being a completely valid algebraic operation—but it would not profit us in the goal of obtaining values for : The resulting equation still contains two unknown variables, just like the original equations do, and so we’re no further along in obtaining a solution.However, what if we could manipulate one of the equations so as to have a negative term that would cancel the respective term in the other equation when added?The process is repeated until the values of all When we have more variables to work with, we just have to remember to stick to a particular method, and keep on reducing the number of equations or variables.We will solve the following system of equations using both approaches: Elementary row operation or Gaussian elimination is a popular method for solving system of linear equations.
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# What is $\mathrm{C}(\mathrm{n}, \mathrm{r})+2 \mathrm{C}(\mathrm{n}, \mathrm{r}-1)+\mathrm{C}(\mathrm{n}, \mathrm{r}-2)$ equal to:A.$\mathrm{C}(\mathrm{n}+1, \mathrm{r})$B.$\mathrm{C}(\mathrm{n}-1, \mathrm{r}+1)$C.$\mathrm{C}(\mathrm{n}, \mathrm{r}+1)$D.$\mathrm{C}(\mathrm{n}+2, \mathrm{r})$
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Hint: The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty. The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability.
- Probability Rule One (For any event $A, 0 \leq P(A) \leq 1)$
- Probability Rule Two (The sum of the probabilities of all possible outcomes is 1 )
- Probability Rule Three (The Complement Rule)
- Probabilities Involving Multiple Events.
- Probability Rule Four (Addition Rule for Disjoint Events)
- Finding $\mathrm{P}(\mathrm{A}$ and $\mathrm{B})$ using Logic.
Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.
The formula for permutations is: $\mathrm{nPr}=\mathrm{n} ! /(\mathrm{n}-\mathrm{r}) !$
The formula for combinations is: $\mathrm{nCr}=\mathrm{n!} /[\mathrm{r} !(\mathrm{n}-\mathrm{r}) !]$
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, we can select the items in any order. Combinations can be confused with permutations. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, we can select the items in any order. Combinations can be confused with permutations. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.
$\mathrm{C}\left(\mathrm{n}_{1} \mathrm{r}\right)+2 \mathrm{C}\left(\mathrm{n}_{1} \mathrm{r}-1\right)+\mathrm{C}(\mathrm{n}, \mathrm{r}-2)$
$={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+2^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2}$
$={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2} \ldots . .(1)$
Now we know that
${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}} \ldots . .(2)$
$\therefore \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}$
${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}-1}$
$\therefore(1)$ becomes
${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+2} \mathrm{C}_{\mathrm{r}}[$ By Property Used in (2)]
$\Rightarrow \mathrm{C}(\mathrm{n}+2, \mathrm{r})$
Hence the correct option is D.
Note: One could say that a permutation is an ordered combination. The number of permutations of $\mathrm{n}$ objects taken $\mathrm{r}$ at a time is determined by the following formula:
$\mathrm{P}(\mathrm{n}, \mathrm{r})=\mathrm{n} !(\mathrm{n}-\mathrm{r}) ! \mathrm{n} !$ is read $\mathrm{n}$ factorial and means all numbers from 1 to $\mathrm{n}$ multiplied. Combinations are a way to calculate the total outcomes of an event where order of
the outcomes do not matter.
To calculate combinations, we will use the formula $\mathrm{nCr}=\mathrm{n!} / \mathrm{r!}^{*}(\mathrm{n}-\mathrm{r}) !,$ where $\mathrm{n}$ represents the total number of items, and $\mathrm{r}$ represents the number of items being chosen at a time. Thus, $\operatorname{nPr}(\mathrm{n}, \mathrm{r})$ The number of possibilities for choosing an ordered set of $\mathrm{r}$ objects $(\mathrm{a}$ permutation) from a total of n objects. Definition: $\operatorname{nPr}(\mathrm{n}, \mathrm{r})=\mathrm{n} ! /(\mathrm{n}-\mathrm{r}) ! \mathrm{nCr}(\mathrm{n}, \mathrm{r})$. |
# why are multiple images formed when two mirrors are placed at right angles to each other ?
We get many images of the objects because the image formed by one mirror acts as an object for the second mirror. Further images of an image are also formed. This continues till no more reflection by either mirror is possible. This phenomenon is referred to as multiple reflection. So, we get 3 images by the process of n = 360/Theta( The angle at which mirrors are placed ) - 1.
Therefore, 360/90 - 1 = 4 - 1 = 3.
HOPE THIS HELPS......... :D
CHEERS,,,,,,
• 15
THE IMAGE REFLECTS FROM ONE MIRROR TO ANOTHER MIRROR AS THIS PROCESS REPEATS WHICH MAKES THE IMAGE FORMED MULTIPLY WHEN TWO MIRRORS AREPLACED AT RIGHT ANGLES TO EACH OTHER .
• 3
Determine a relationship between the number of images formed by two plane mirrors and the value of the angle between these two mirrors, on the basis of observations in the laboratory.
The appearance of multiple images is caused by a succession of reflections between two mirrors.
Let's examine in further detail the case of images formed by two plane mirrors placed at 90 degrees to one another.
Locate the images of the object in each of the mirrors (primary images).
Extend the line of the mirrors' positions (broken lines on the diagram).
Treating the primary images as objects in front of the mirror extension lines:
• Draw a perpendicular to one of the mirror extension lines.
• Measure the distance from the primary image to the mirror extension line.
• Locate the secondary image at an equal distance along the perpendicular.
• Repeat the last three steps to locate the image of the other primary image.
In the case of mirrors at a right angle, only one secondary image is formed:
By decreasing the angle between the mirrors, the number of images is increased. When an object is placed in front of two plane mirrors, the number of images formed varies inversely to the angle between the two mirrors. Two mirrors placed at 90 degrees to one another will form 3 images of the same object, namely:
2 primary images seen as a result of a single reflection of the rays.
1 secondary image seen as a result of a double reflection of the rays.
Two mirrors facing one another will produce a large number of images. This number is limited because light is absorbed by the multiple reflections. The images can be identified by their apparent distance. Secondary images appear further away than primary images, since light must travel a greater distance in a double reflection than in a single reflection.
• 4
it happens coz the the image form by one mirror acts as a object for the second mirror
• 1
we get many images of the object because the image formed by one mirror acts as an object for the second mirror. further images of an image are also formed. this continue till no more refletion by either mirror is posssible . this phenomenon is referred as multiple refliction
• 1
Many images are formed due to the reflection of one mirror to another . This phenomenon is called multiple reflection
• 2
We get so many images of the object when kept parallel to two mirrors as object formed by one mirror acts as an image for second and further images are formed and this continues till no more reflection is possible.
• 2
What are you looking for? |
# Construct an Angle Bisector – Explanation and Examples
Given an angle ABC, it is possible to construct a line BF that divides the angle into two equal parts using only a straightedge and compass. Such a line is called an angle bisector.
Constructing an angle bisector requires that we construct an isosceles triangle BDE inside the angle and then construct an equilateral triangle DEF that shares a base with BDE. If we then construct the line BF, it will divide the original angle ABC into two equal angles.
Doing this requires that we have a thorough understanding of the basics of construction. It is also a good idea to review the equilateral triangles’ construction, covered in the construction of a 60-degree angle.
This topic will go over:
• How to Construct an Angle Bisector
• How to Construct an Angle Bisector with Compass
• Proof that the Angles are Equal
## How to Construct an Angle Bisector
Suppose we are given an angle ABC. It can be acute, right, or obtuse. It doesn’t matter.
We want to construct an angle bisector. That is, we want to construct a new line that will divide the angle into two equal angles.
To do this, we will need our straightedge, compass, and a few of Euclid’s theorems. Specifically, we need to know that if two triangles have all three sides congruent, then the triangles are congruent. This means that their corresponding angles will be equal.
### How to Construct an Angle Bisector with Compass
First, we pick a point D on AB.
Next, we can place the point of the compass at B and the pencil tip at D. Then, we can trace out the circumference of a circle with center B and radius BD. Mark the place where this circle intersects BC as E.
Note that in practice, it’s sufficient to create an arc from D to E instead of creating the whole circle. Since the whole circle is necessary for the proof, however, we will construct it here.
Next, we will connect D and E using our straightedge. Then, we will construct an equilateral triangle with DE as an edge. Recall that we do this by creating two circles with radius DE. One will be centered at D, while the other will be centered at E. We will call the intersection F and construct the lines DF and EF. We want this triangle to point away from B, as shown.
Finally, we can connect points B and F with our straightedge. The line BF will create two angles, ABF and FBC, that are equal to each other.
## Examples
In this section, we will go over common problems that involve the construction of an angle bisector.
### Example 1
Prove that BF bisects the angle ABC.
### Example 1 Solution
Let’s consider the construction again.
The line segment BD equals the line segment BE because they are both radii of the circle with center B and radius BD. We also know that the line segment DF is equal to the line segment EF because they are both legs of an equilateral triangle. Of course, the line segment BF is equal to itself in length.
Thus, the legs of the triangles DBF and EBF are the same. Consequently, the two triangles are congruent. This means that their corresponding angles are congruent. Specifically, the angles ABF and CBF are equal. Since these two angles together make up the original angle, ABC, the line BF bisects ABC.
### Example 2
Divide the triangle in two using an angle bisector. Are the two parts equal in area?
### Example 2 Solution
We’ll divide the angle ABC as before. Rather than constructing a new point D, we can use the endpoint of the shorter side, A.
Then, we draw a circle with center B and radius BA and label the intersection of this circle with the line BC as D.
Then, we create two circles with radius AD. One will have center A, and the other will have center D. If we draw a line from B to the intersection of these two circles, E, we have an angle bisector as shown.
The two triangles, in this case, will not be equal. Let’s call the intersection of AD and BE F. ABF and EBF are congruent because AB and BD were constructed to be radii of the circle with center B and radius AB. BF is, of course, equal to itself, and we have already shown that the angles ABF and CBF are equal. Therefore, the two triangles ABF and DBF are congruent by Elements 1.4, which states that two triangles are congruent if two sides are the same and the angle between them is the same.
If we call the intersection of the lines AC and BE G and connect CG, we can see that the triangle AFG is equal to CFG. However, there is still an additional area remaining to the right of BE. Consequently, the triangle has not been cut in half even though the angle ABC has been bisected.
### Example 3
Divide the hexagon into two halves using an angle bisector.
### Example 3 Solution
When we constructed 60-degree angles, we showed that a hexagon is actually composed of 6 equilateral triangles. Therefore, if we cut this in half, we should be able to put 3 equilateral triangles in each half.
In this case, we can use any angle. We’ll use the angle ABC to be consistent, though. A and C are already equidistant from B because this is a regular hexagon. This, we can connect them with a line and construct an equilateral triangle ACG. Then, we connect B and G to bisect the angle ABC.
Note, however, that G and E are the same point. This makes sense because A and C are separated by one angle, but so are the pair A and E and the pair C and E.
Thus, bisecting the angle ABC does bisect the hexagon.
### Example 4
Divide the angle into four equal parts.
### Example 4 Solution
When we divide an angle in two, we double the number of angles. Therefore, to divide an angle in four, we must first bisect the angle. Then, we must bisect the two new angles formed.
We’ll bisect the angle as before. In this case, we can use the shorter side’s endpoint, C, as the radius of the circle centered at B. We’ll call the intersection of this circle with the line AB D. We can then create two new circles with radius CD, one centered at C and one at D. We’ll call the intersection E and connect BE. So far, we’ve just bisected the angle.
Now we need to bisect the angles ABE and CBE.
We can call the intersection of the circle centered at B with radius BC and the line BE F. Then, we can create three new circles. They will each have radius FD, which will be equal to FC, and there will be one centered at D, one centered at F, and one centered at C.
If we construct a line from B to the intersection of the circles centered at D and F with radius FD, we will bisect ABF. Likewise, if we construct a line from B to the intersection of the circles centered at C and F with radius FC, we will bisect CBF. Since ABF and CBF were equal in measure, their bisected angles will also be equal in measure.
Thus, we have cut the original angle ABC into four equal parts.
### Example 5
Divide the angle greater than a straight line into two equal parts.
### Example 5 Solution
The greater angle here is the one measured clockwise as ABC. We can try using the same tactics as before. This is because when we bisect the smaller angle measured counterclockwise as ABC, we can bisect the larger angle by extending the angle bisector.
Let’s do this. First, we bisect the acute angle ABC as before, finding a point on BC equal in length to BA. We’ll call this point D. Then, we construct two circles of length AD, one centered at A and one at D. Drawing a line from B to this intersection, E, gives us an angle bisector. We can then extend the line through the circle we constructed to find the point D.
Since this line goes through the center of the circle and touches the circumference in both directions, it is the diameter of the circle with center B and radius BA. We can see that the larger angle ABC has been cut into two parts. If we look, one part is a straight line minus ABE, and the other is a straight line minus DBE. Since ABE=DBE, the two angles that the larger angle ABC has been cut into are equal.
### Practice Questions
1. True or False: The line that divides the angle, $80^{\circ}$, into two angles measuring $40^{\circ}$ is an angle bisector.
2. True or False: The angle bisector of a straight angle will divide it into two acute angles.
3. True or False: The line, $OC$, is an angle bisector of $\angle AOB$.
4. If an angle bisector divides the angle shown below, what is the measure of each angle?
5. Suppose that $OC$ is the angle bisector of $\angle AOB$, what is the measure of $\angle AOB$?
### Open Problems
1. Bisect the given angle.
2. Cut the given angle into 8 equal parts.
3. Does the line CD bisect the angle ACB?
4. Divide the octagon in half by bisecting one of the angles.
5. Bisect each of the angles of the triangle given.
### Open Problem Solutions
Yes, because it lines up with a constructed bisector.
4.
5.
Images/mathematical drawings are created with GeoGebra. |
Trigonometry Tutorial (Part 3)
Solving Oblique Triangles
Without Using the Law of Sines or the Law of Cosines
For solving oblique triangles by using the Law of Sines and the Law of Cosines, click here.
An oblique triangle is one which contains no right angles.
So, what's the point in showing another method when we just learned how to do this in Part 2? For one thing, it demonstrates how important it is to know the fundamental mathematical concepts of the Pythagorean Theorem, triangle area formulas, and so on. Also, if you are taking a test and you forget exactly how the Law of Sines or the Law of Cosines works, you can always find a solution by going to more basic concepts. These solutions solve all of the problems presented in Part 2 except that the Law of Sines and the Law of Cosines were not used. If you compare these solutions to those in Part 2, you will see they don't require that much more work.
Two Angles and an Included Side
Example
A triangle has 2 angles of 75° and 25° and the included side is 5. What is the other angle and the other 2 sides?
Let's call Angle A = 25°, Angle B = 75° and side c = 5.
Angle C is easily found by: ∠C = 180° - ∠A - ∠B = 180° - 25° - 75°
∠C = 80°
Dropping a perpendicular 'hgt' (for 'height') from Angle C to side 'c', splits triangle ABC into 2 right triangles.
So, angle BCD =180° - 90° - ∠B =15°
and Angle ACD =180° - 90° - ∠A =65°
Using this information and a trigonometry calculator, we can find the values of the 2 parts of line 'c'. (see diagram below).
Now, the remaining calculations are quite straightforward.
side a = .55535 ÷ sine(15°) side a = .55535 ÷ .25882
side a = 2.1457
side b = 4.4447 ÷ sine(65°) side b = 4.4447 ÷ .90631
side b = 4.9042
Two Sides and a Non-Included Angle
When solving for a triangle with 2 known sides and a non-included angle, there may be no solution, one solution or even two solutions.
In a triangle of sides 'c', 'a' and its opposite Angle 'A', the number of solutions is calculated by:
• If a < c•sin(A) no solution
• If a = c•sin(A) one solution: ∠C=90° ∠B=(90°-∠A) and side b=(c•cos(A))
• If c>a>c•sin(A) two solutions.
• If a>=c one solution
Example
A triangle has sides of 7 and 8 and a non-included angle of 60°
What are the other angles and the other side?
Let's say side a=7, c=8 and Angle A=60°
Dropping a perpendicular 'hgt' from Angle B, we find its length by:
hgt = sine 60° • 8 = .86603 • 8 = 6.9282
Angle C = arc sine (hgt ÷ side a) = arc sine (6.9282 ÷ 7) = arc sine (.98974)
Angle C = 81.785°
Angle B is easily found by: ∠B= 180° -60° - 81.785°
Angle B = 38.215°
Side b = AD + DC
To solve these, we'll use the Pythagorean Theorem:
Side AD² = c² - hgt² = 8² - 6.9282² Side AD=4
Side DC² = a² - hgt² = 7² - 6.9282² Side DC=1
Side b = AD + DC = 4 +1 Side b = 5
Is there a second solution?
If c>a>c•sin(A) there are 2 solutions.
8 > 7 > (8 • sin (60°))
Since 8 > 7 > 6.92824, there are two solutions.
Whenever there are two solutions, ∠C and ∠B will each have 2 values.
We will designate the "second Angle C" as C' and we find it by:
∠C' = 180° - ∠C ∠C' = 180° - 81.789°
∠C' = 98.213°
We find ∠B' by:
∠B' = 180° -∠A - ∠C' ∠B' = 180° -60° - 98.213°
∠B' = 21.787°
Since side b = AD + DC, then side b' = AD - DC = 4 - 1
Side b' = 3
Two Sides and the Included Angle
Example
A triangle has sides of 4 and 9 with an included angle of 30°. What are the other 2 angles and the length of the other side?
Let's say side a=4 and side b=9 and angle C=30°
Dropping a perpendicular 'hgt' from Angle B to side 'b', splits triangle ABC into 2 right triangles and splits side b into 'x' and '9-x'.
So, angle CBD =180° - 90° - ∠C = 60°
hgt= sine 30° • 4 = .5 • 4 hgt = 2
Line segment 'x' = sine 60° • 4 'x' = .86603 • 4 'x' = 3.4641
Line DA (or 9-x) = 9 - 3.4641 = 5.5359
Arc tangent (A) = 2 ÷ 5.5359 = .36128 Arctan(A) = .36128
∠A = 19.864°
Arc tangent (DBA) = 5.5359 ÷ 2 = 2.7680 Angle (DBA) = 70.136°
Angle B is found by adding CBD and ABD which equals 60° + 70.136°
∠B = 130.136°
(We could have found Angle B just as easily by ∠B = 180° - ∠A - ∠C)
Side c = 2 ÷ Sine 19.864° = 2 ÷ .33979 = 5.886
Side c = 5.886
(We also could have found side 'c' by using the Pythagorean Theorem).
All Three Sides Of A Triangle
EXAMPLE:
A triangle has sides of length four, five and six. What are all 3 angles?
Let's say that side a=4, side b=5 and side c=6.
Yes, this could be solved in many different ways, but here's a "neat" solution.
Remember Heron's formula from the triangle page? It states that
triangle area = Square Root (s•(s-a)•(s-b)•(s-c))
where 's' is the semi-perimeter of the triangle.
s=(4 + 5 + 6) / 2 s= 7.5
area = Square Root (7.5 •(7.5 -4)•(7.5 -5)•(7.5 -6))
area = Square Root (98.4375)
area = 9.9216
So, how does this help solve the triangle? Well, there is another area formula:
triangle area = ½ • base • height and therefore,
height = (2 • area ÷ base)
height = (2 • 9.9216 ÷ 6)
height = 3.3072
∠B = arc sine (hgt ÷ side a)
∠B = arc sine (3.3072 ÷ 4)
∠B = arc sine (.8268)
∠B = 55.771°
∠A = arc sine (hgt ÷ side b)
∠A = arc sine (3.3072 ÷ 5)
∠A = arc sine (.66144)
∠A = 41.41°
We then easily solve Angle C using this equation:
Angle C = 180° -55.771° - 41.41°
∠C = 82.819°
Go To Page One
Basic Trigonometric Functions
Go To Page Two
Solving Oblique Triangles Using Law of Sines and Law of Cosines |
# Difference between revisions of "2019 AIME I Problems/Problem 14"
## Problem 14
Find the least odd prime factor of $2019^8+1$.
## Solution
We know that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16} \equiv 1 \pmod{p}$.
Since $2019^{16} \equiv 1 \pmod{p}$, the order of $2019$ modulo $p$ is a positive divisor of $16$.
However, if the order of $2019$ modulo $p$ is $1, 2, 4,$ or $8,$ then $2019^8$ will be equivalent to $1 \pmod{p},$ which contradicts the given requirement that $2019^8\equiv -1\pmod{p}$.
Therefore, the order of $2019$ modulo $p$ is $16$. Because all orders modulo $p$ divide $\phi(p)$, we see that $\phi(p)$ is a multiple of $16$. As $p$ is prime, $\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1$. Therefore, $p\equiv 1 \pmod{16}$. The two smallest primes equivalent to $1 \pmod{16}$ are $17$ and $97$. As $2019^8 \not\equiv -1 \pmod{17}$ and $2019^8 \equiv -1 \pmod{97}$, the smallest possible $p$ is thus $\boxed{097}$.
### Note to solution
$\phi(k)$ is the Euler Totient Function of integer $k$. $\phi(k)$ is the number of positive integers less than $k$ relatively prime to $k$. Define the numbers $k_1,k_2,k_3,\cdots,k_n$ to be the prime factors of $k$. Then, we have$$\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).$$A property of the Totient function is that, for any prime $p$, $\phi(p)=p-1$.
Euler's Totient Theorem states that$$a^{\phi(k)} \equiv 1\pmod k$$if $\gcd(a,k)=1$.
Furthermore, the order $a$ modulo $n$ for an integer $a$ relatively prime to $n$ is defined as the smallest positive integer $d$ such that $a^{d} \equiv 1\pmod n$. An important property of the order $d$ is that $d|\phi(n)$.
## Solution 2 (Basic Modular Arithmetic)
In this solution, $k$ will represent an arbitrary nonnegative integer.
We will show that any potential prime $p$ must be of the form $16k+1$ through a proof by contradiction. Suppose that there exists some prime $p$ that can not be expressed in the form $16k+1$ that is a divisor of $2019^8+1$. First, note that if the prime $p$ is a divisor of $2019$, then $2019^8$ is a multiple of $p$ and $2019^8+1$ is not. Thus, $p$ is not a divisor of $2019$.
Because $2019^8+1$ is a multiple of $p$, $2019^8+1\equiv0\pmod{p}$. This means that $2019^8\equiv-1\pmod{p}$, and by raising both sides to an arbitrary odd positive integer, we have that $2019^{16k+8}\equiv-1\pmod{p}$.
Then, since the problem requires an odd prime, $p$ can be expressed as $16k+m$, where $m$ is an odd integer ranging from $3$ to $15$, inclusive. By Fermat's Little Theorem, $2019^{p-1}\equiv1\pmod{p}$, and plugging in values, we get $2019^{16k+n}\equiv1\pmod{p}$, where $n=m-1$ and is thus an even integer ranging from $2$ to $14$, inclusive.
If $n=8$, then $2019^{16k+8}\equiv1\pmod{p}$, which creates a contradiction. If $n$ is not a multiple of $8$ but is a multiple of $4$, squaring both sides of $2019^{16k+n}\equiv1\pmod{p}$ also results in the same contradictory equivalence. For all remaining $n$, raising both sides of $2019^{16k+n}\equiv1\pmod{p}$ to the $4$th power creates the same contradiction. (Note that $32k$ and $64k$ can both be expressed in the form $16k$.)
Since we have proved that no value of $n$ can work, this means that a prime must be of the form $16k+1$ in order to be a factor of $2019^8+1$. The smallest prime of this form is $17$, and testing it, we get $$2019^8+1\equiv13^8+1\equiv169^4+1\equiv(-1)^4+1\equiv1+1\equiv2\pmod{17},$$ so it does not work. The next smallest prime of the required form is $97$, and testing it, we get $$2019^8+1\equiv(-18)^8+1\equiv324^4+1\equiv33^4+1\equiv1089^2+1\equiv22^2+1\equiv484+1\equiv-1+1\equiv0\pmod{97},$$ so it works. Thus, the answer is $\boxed{097}$. ~emerald_block
## Video Solution
On The Spot STEM: |
# Plus–minus sign
(Redirected from ±)
±
Plus–minus sign
In UnicodeU+00B1 ± PLUS-MINUS SIGN (HTML ± · ±, ±, ±)
Related
See alsoU+2213 MINUS-OR-PLUS SIGN (HTML ∓ · ∓, ∓, ∓)
The plus–minus sign (also, plus or minus sign), ± is a mathematical symbol with multiple meanings.
## History
A version of the sign, including also the French word ou ("or") was used in its mathematical meaning by Albert Girard in 1626, and the sign in its modern form was used as early as William Oughtred's Clavis Mathematicae (1631).[4]
## Usage
### In mathematics
In mathematical formulas, the ± symbol may be used to indicate a symbol that may be replaced by either the Plus and minus signs, + or , allowing the formula to represent two values or two equations.
For example, given the equation x2 = 9, one may give the solution as x = ±3. This indicates that the equation has two solutions, each of which may be obtained by replacing this equation by one of the two equations x = +3 or x = −3. Only one of these two replaced equations is true for any valid solution. A common use of this notation is found in the quadratic formula
${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$
describing the two solutions to the quadratic equation ax2 + bx + c = 0.
Similarly, the trigonometric identity
${\displaystyle \sin(A\pm B)=\sin(A)\cos(B)\pm \cos(A)\sin(B).}$
can be interpreted as a shorthand for two equations: one with + on both sides of the equation, and one with on both sides. The two copies of the ± sign in this identity must both be replaced in the same way: it is not valid to replace one of them with + and the other of them with . In contrast to the quadratic formula example, both of the equations described by this identity are simultaneously valid.
The minus–plus sign (also minus-or-plus sign), is generally used in conjunction with the ± sign, in such expressions as x ± y ∓ z, which can be interpreted as meaning x + y − z and/or x − y + z, but not x + y + z nor x − y − z. The upper in is considered to be associated to the + of ± (and similarly for the two lower symbols) even though there is no visual indication of the dependency. (However, the ± sign is generally preferred over the sign, so if they both appear in an equation it is safe to assume that they are linked. On the other hand, if there are two instances of the ± sign in an expression, without a , it is impossible to tell from notation alone whether the intended interpretation is as two or four distinct expressions.) The original expression can be rewritten as x ± (y − z) to avoid confusion, but cases such as the trigonometric identity are most neatly written using the "∓" sign:
${\displaystyle \cos(A\pm B)=\cos(A)\cos(B)\mp \sin(A)\sin(B)}$
which represents the two equations:
{\displaystyle {\begin{aligned}\cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B)\\\cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B)\end{aligned}}}
Another example is
${\displaystyle x^{3}\pm 1=(x\pm 1)\left(x^{2}\mp x+1\right)}$
A third related usage is found in this presentation of the formula for the Taylor series of the sine function:
${\displaystyle \sin \left(x\right)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots \pm {\frac {1}{(2n+1)!}}x^{2n+1}+\cdots .}$
Here, the plus-or-minus sign indicates that the term may be added or subtracted, in this case depending on whether n is odd or even, the rule can be deduced from the first few terms. A more rigorous presentation of the same formula would multiply each term by a factor of (−1)n, which gives +1 when n is even and −1 when n is odd.
### In statistics
The use of ± for an approximation is most commonly encountered in presenting the numerical value of a quantity together with its tolerance or its statistical margin of error.[1] For example, "5.7±0.2" may be anywhere in the range from 5.5 to 5.9 inclusive. In scientific usage it sometimes refers to a probability of being within the stated interval, usually corresponding to either 1 or 2 standard deviations (a probability of 68.3% or 95.4% in a normal distribution).
Operations involving uncertain values should always try to preserve the uncertainty in order to avoid propagation of error. If n = a ± b, any operation of the form m = f(n) must return a value of the form m = c ± d, where c is f(n) and d is range updated using interval arithmetic.
A percentage may also be used to indicate the error margin. For example, 230 ± 10% V refers to a voltage within 10% of either side of 230 V (from 207 V to 253 V inclusive).[citation needed] Separate values for the upper and lower bounds may also be used. For example, to indicate that a value is most likely 5.7 but may be as high as 5.9 or as low as 5.6, one may write 5.7+0.2
−0.1
.
### In chess
The symbols ± and are used in chess notation to denote an advantage for white and black respectively. However, the more common chess notation would be only + and .[3] If a difference is made, the symbols + and denote a larger advantage than ± and .
## Encodings
• In Unicode: U+00B1 ± PLUS-MINUS SIGN
• In ISO 8859-1, -7, -8, -9, -13, -15, and -16, the plus–minus symbol is code 0xB1hex. This location was copied to Unicode.
• The symbol also has a HTML entity representations of ± and ±.
• The rarer minus–plus sign is not generally found in legacy encodings, but is available in Unicode as U+2213 MINUS-OR-PLUS SIGN so can be used in HTML using ∓ or ∓.
• In TeX 'plus-or-minus' and 'minus-or-plus' symbols are denoted \pm and \mp, respectively.
• Although these characters may also be produced using underlining or overlining + symbol ( + or + ), this is deprecated because the formatting may stripped at a later date, changing the meaning. It also makes the meaning less accessible to blind users with screen readers.
### Typing
• Windows: Alt+241 or Alt+0177 (numbers typed on the numeric keypad).
• Macintosh: ⌥ Option+⇧ Shift+= (equal sign on the non-numeric keypad).
• Unix-like systems: Compose,+,- or ⇧ Shift+Ctrl+u B1space (second works on Chromebook)
• AutoCAD shortcut string: %%d
## Similar characters
The plus–minus sign resembles the Chinese characters (Radical 32) and (Radical 33), whereas the minus–plus sign resembles (Radical 51). |
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### Section 5-1 : Indefinite Integrals
In the past two chapters we’ve been given a function, $$f\left( x \right)$$, and asking what the derivative of this function was. Starting with this section we are now going to turn things around. We now want to ask what function we differentiated to get the function $$f\left( x \right)$$.
Let’s take a quick look at an example to get us started.
Example 1 What function did we differentiate to get the following function. $f\left( x \right) = {x^4} + 3x - 9$
Show Solution
Let’s actually start by getting the derivative of this function to help us see how we’re going to have to approach this problem. The derivative of this function is,
$f'\left( x \right) = 4{x^3} + 3$
The point of this was to remind us of how differentiation works. When differentiating powers of $$x$$ we multiply the term by the original exponent and then drop the exponent by one.
Now, let’s go back and work the problem. In fact, let’s just start with the first term. We got $${x^4}$$ by differentiating a function and since we drop the exponent by one it looks like we must have differentiated $${x^5}$$. However, if we had differentiated $${x^5}$$ we would have $$5{x^4}$$ and we don’t have a 5 in front our first term, so the 5 needs to cancel out after we’ve differentiated. It looks then like we would have to differentiate $$\frac{1}{5}{x^5}$$ in order to get $${x^4}$$.
Likewise, for the second term, in order to get 3x after differentiating we would have to differentiate $$\frac{3}{2}{x^2}$$. Again, the fraction is there to cancel out the 2 we pick up in the differentiation.
The third term is just a constant and we know that if we differentiate $$x$$ we get 1. So, it looks like we had to differentiate -9$$x$$ to get the last term.
Putting all of this together gives the following function,
$F\left( x \right) = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x$
Our answer is easy enough to check. Simply differentiate $$F\left( x \right)$$.
$F'\left( x \right) = {x^4} + 3x - 9 = f\left( x \right)$
So, it looks like we got the correct function. Or did we? We know that the derivative of a constant is zero and so any of the following will also give $$f\left( x \right)$$ upon differentiating.
\begin{align*}F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + 10\\ F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x - 1954\\ F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + \frac{{3469}}{{123}}\\ & etc.\end{align*}
In fact, any function of the form,
$F\left( x \right) = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c,\,\,\hspace{0.25in}c{\mbox{ is a constant}}$
will give $$f\left( x \right)$$ upon differentiating.
There were two points to this last example. The first point was to get you thinking about how to do these problems. It is important initially to remember that we are really just asking what we differentiated to get the given function.
The other point is to recognize that there are actually an infinite number of functions that we could use and they will all differ by a constant.
Now that we’ve worked an example let’s get some of the definitions and terminology out of the way.
#### Definitions
Given a function, $$f\left( x \right)$$, an anti-derivative of $$f\left( x \right)$$ is any function $$F\left( x \right)$$ such that
$F'\left( x \right) = f\left( x \right)$
If $$F\left( x \right)$$ is any anti-derivative of $$f\left( x \right)$$ then the most general anti-derivative of $$f\left( x \right)$$ is called an indefinite integral and denoted,
$\int{{f\left( x \right)\,dx}} = F\left( x \right) + c,\hspace{0.25in}\,\,\,\,c{\mbox{ is any constant}}$
In this definition the $$\int{{}}$$is called the integral symbol, $$f\left( x \right)$$ is called the integrand, $$x$$ is called the integration variable and the “$$c$$” is called the constant of integration.
Note that often we will just say integral instead of indefinite integral (or definite integral for that matter when we get to those). It will be clear from the context of the problem that we are talking about an indefinite integral (or definite integral).
The process of finding the indefinite integral is called integration or integrating $$f\left( x \right)$$ . If we need to be specific about the integration variable we will say that we are integrating $$f\left( x \right)$$ with respect to $$x$$.
Let’s rework the first problem in light of the new terminology.
Example 2 Evaluate the following indefinite integral. $\int{{{x^4} + 3x - 9\,dx}}$
Show Solution
Since this is really asking for the most general anti-derivative we just need to reuse the final answer from the first example.
The indefinite integral is,
$\int{{{x^4} + 3x - 9\,dx}} = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c$
A couple of warnings are now in order. One of the more common mistakes that students make with integrals (both indefinite and definite) is to drop the dx at the end of the integral. This is required! Think of the integral sign and the dx as a set of parentheses. You already know and are probably quite comfortable with the idea that every time you open a parenthesis you must close it. With integrals, think of the integral sign as an “open parenthesis” and the dx as a “close parenthesis”.
If you drop the dx it won’t be clear where the integrand ends. Consider the following variations of the above example.
\begin{align*}\int{{{x^4} + 3x - 9\,dx}} & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c\\ \int{{{x^4} + 3x\,dx}} - 9 & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} + c - 9\\ \int{{{x^4}\,dx}} + 3x - 9 & = \frac{1}{5}{x^5} + c + 3x - 9\end{align*}
You only integrate what is between the integral sign and the dx. Each of the above integrals end in a different place and so we get different answers because we integrate a different number of terms each time. In the second integral the “-9” is outside the integral and so is left alone and not integrated. Likewise, in the third integral the “$$3x - 9$$” is outside the integral and so is left alone.
Knowing which terms to integrate is not the only reason for writing the $$dx$$ down. In the Substitution Rule section we will actually be working with the $$dx$$ in the problem and if we aren’t in the habit of writing it down it will be easy to forget about it and then we will get the wrong answer at that stage.
The moral of this is to make sure and put in the $$dx$$! At this stage it may seem like a silly thing to do, but it just needs to be there, if for no other reason than knowing where the integral stops.
On a side note, the $$dx$$ notation should seem a little familiar to you. We saw things like this a couple of sections ago. We called the $$dx$$ a differential in that section and yes that is exactly what it is. The $$dx$$ that ends the integral is nothing more than a differential.
The next topic that we should discuss here is the integration variable used in the integral. Actually, there isn’t really a lot to discuss here other than to note that the integration variable doesn’t really matter. For instance,
\begin{align*}\int{{{x^4} + 3x - 9\,dx}} & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c\\ \int{{{t^4} + 3t - 9\,dt}} & = \frac{1}{5}{t^5} + \frac{3}{2}{t^2} - 9t + c\\ \int{{{w^4} + 3w - 9\,dw}} & = \frac{1}{5}{w^5} + \frac{3}{2}{w^2} - 9w + c\end{align*}
Changing the integration variable in the integral simply changes the variable in the answer. It is important to notice however that when we change the integration variable in the integral we also changed the differential ($$dx$$, $$dt$$, or $$dw$$) to match the new variable. This is more important than we might realize at this point.
Another use of the differential at the end of integral is to tell us what variable we are integrating with respect to. At this stage that may seem unimportant since most of the integrals that we’re going to be working with here will only involve a single variable. However, if you are on a degree track that will take you into multi-variable calculus this will be very important at that stage since there will be more than one variable in the problem. You need to get into the habit of writing the correct differential at the end of the integral so when it becomes important in those classes you will already be in the habit of writing it down.
To see why this is important take a look at the following two integrals.
$\int{{2x\,dx}}\hspace{1.0in}\int{{2t\,dx}}$
The first integral is simple enough.
$\int{{2x\,dx}} = {x^2} + c$
The second integral is also fairly simple, but we need to be careful. The dx tells us that we are integrating $$x$$’s. That means that we only integrate $$x$$’s that are in the integrand and all other variables in the integrand are considered to be constants. The second integral is then,
$\int{{2t\,dx}} = 2tx + c$
So, it may seem silly to always put in the dx, but it is a vital bit of notation that can cause us to get the incorrect answer if we neglect to put it in.
Now, there are some important properties of integrals that we should take a look at.
#### Properties of the Indefinite Integral
1. $$\displaystyle \int{{k\,f\left( x \right)\,dx}} = k\int{{f\left( x \right)\,dx}}$$ where $$k$$ is any number. So, we can factor multiplicative constants out of indefinite integrals.
See the Proof of Various Integral Formulas section of the Extras chapter to see the proof of this property.
2. $$\displaystyle \int{{ - f\left( x \right)\,dx}} = - \int{{f\left( x \right)\,dx}}$$. This is really the first property with $$k = - 1$$ and so no proof of this property will be given.
3. $$\displaystyle \int{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}$$. In other words, the integral of a sum or difference of functions is the sum or difference of the individual integrals. This rule can be extended to as many functions as we need.
See the Proof of Various Integral Formulas section of the Extras chapter to see the proof of this property.
Notice that when we worked the first example above we used the first and third property in the discussion. We integrated each term individually, put any constants back in and then put everything back together with the appropriate sign.
Not listed in the properties above were integrals of products and quotients. The reason for this is simple. Just like with derivatives each of the following will NOT work.
$\int{{f\left( x \right)g\left( x \right)\,dx}} \ne \int{{f\left( x \right)dx}}\int{{g\left( x \right)\,dx}}\hspace{0.75in}\int{{\frac{{f\left( x \right)}}{{g\left( x \right)}}\,dx}} \ne \frac{{\int{{f\left( x \right)\,dx}}}}{{\int{{g\left( x \right)\,dx}}}}$
With derivatives we had a product rule and a quotient rule to deal with these cases. However, with integrals there are no such rules. When faced with a product and quotient in an integral we will have a variety of ways of dealing with it depending on just what the integrand is.
There is one final topic to be discussed briefly in this section. On occasion we will be given $$f'\left( x \right)$$ and will ask what $$f\left( x \right)$$ was. We can now answer this question easily with an indefinite integral.
$f\left( x \right) = \int{{f'\left( x \right)\,dx}}$
Example 3 If $$f'\left( x \right) = {x^4} + 3x - 9$$ what was $$f\left( x \right)$$?
Show Solution
By this point in this section this is a simple question to answer.
$f\left( x \right) = \int{{f'\left( x \right)\,dx}} = \int{{{x^4} + 3x - 9\,dx}} = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c$
In this section we kept evaluating the same indefinite integral in all of our examples. The point of this section was not to do indefinite integrals, but instead to get us familiar with the notation and some of the basic ideas and properties of indefinite integrals. The next couple of sections are devoted to actually evaluating indefinite integrals. |
# Sample Space
Sample Space
An interesting fact about the theory of probability is that it was created due to a gambler's dispute, and was given by two famous mathematicians, Blaise Pascal and Pierre de Fermat.
This page will tell you more about the concept of sample space in probability, with interesting examples, like the flipping of a coin and the rolling of a dice.
Try this sample space calculator and get started to know more about the concept.
## Lesson Plan
1 What Is Meant by Sample Space? 2 Tips and Tricks 3 Important Notes on Sample Space 4 Solved Examples on Sample Space 5 Interactive Questions on Sample Space
## What Is Meant by Sample Space?
Sample space, in probability, is referred to as the collection of all possible outcomes in any random experiment. It is denoted by "S".
Since it is a set, it is shown within the braces.
## The Number of Members in a Sample Space
Let's learn how to find the number of members using a few sample space examples.
### Sample space of a coin
Have you ever seen a one cent coin?
You may have noticed that one of its sides, features the 16th U.S. President, Abraham Lincoln.
Let's say Candace flipped two coins. What are the possibilities when she flips the coin?
Each coin could either fall on head or tail right?
There isn't any other possibility when a coin is flipped, hence, the sample space of a coin is given as:
Sample space of a coin = {H, T} or {Heads, Tails}.
### Sample space of a dice
Have you ever seen a dice?
It is a cubical toy with 6 faces, showing the numbers in the form of dots.
Let's say, Jeremy rolled a dice. What are the possibilities of the dots that can be seen, when he rolls a dice?
The dice show 1, 2, 3, 4, 5, or 6 dots.
Therefore, the sample space of dice is written as {1, 2, 3, 4, 5, 6}
Tips and Tricks
• The sample space for a coin tossed 'n' times, or 'n' number of coins, tossed once, is written in the count of $$2^n$$.
• The sample space for a dice rolled 'n' times or 'n' number of dices, rolled once, is written in the count of $$6^n$$.
• When a dice is rolled or a coin is tossed, the outcome considered, is of the side which is seen on the upper face.
## Difference Between Event and Sample Space
Event is a set of outcomes during an experiment, while sample space is all the possibilities that can occur during an experiment.
Hence, the event is a subset of sample space.
Let's look at some sample space examples to understand this better.
### The occurrence of a head when a coin is flipped
The occurrence of a head when a coin is flipped is only once. It can either show heads or tails.
Sample space, S = {occurrence of head, occurrence of tail}
Hence, the case of coin-flipping can be represented as E $$\subset$$ S.
The occurrence of 1 when a dice is rolled
The occurrence of 1 when a dice is rolled is only once. It can show 1, 2, 3, 4, 5 or 6
Event, E= {Occurrence of 1}
Sample space, S = {occurrence of 1, occurrence of 2, occurrence of 3, occurrence of 4, occurrence of 5, occurrence of 6}
Hence, the case of rolling dice can be represented as E $$\subset$$ S.
Important Notes
• Sample space in probability is a deciding number to show all the possible outcomes which can occur during any random experiment.
• The sample space will always be a whole number. It can never be a fraction or a decimal number.
## Solved Examples
Example 1
How will Oliver determine the sample space if he flips two coins simultaneously?
Solution
The sample space of flipping a single coin is {Head, Tail} which gives S = 2
When the two coins are flipped simultaneously, all the possibilities for coin 1 and coin 2 will be:
Coin 1 Coin 2
Tail Tail
$$\therefore$$ The sample space is 4
Example 2
Help Andy determine the sample space for a well-shuffled deck of cards.
Solution
Andy knows that there are 52 cards in a deck.
Hence, the sample space for a well-shuffled deck is all 52 cards: {Ace of hearts, two of hearts, three of hearts...etc}.
$$\therefore$$ The sample space is 52
## Interactive Questions
Here are a few activities for you to practice.
## Let's Summarize
We hope you enjoyed learning about sample space with the simulations and practice questions. Now, you will be able to easily solve problems related to sample space formulas and sample space in probability.
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.
## 1. What is the sample space of 52 cards?
The sample space of 52 cards is all 52 possible outcomes, which is {Ace of Hearts, two of hearts, three of hearts...etc}.
## 2. What is the sample space of a coin?
The sample space of a coin is {Head, Tail}.
## 3. What is the sample space for a dice?
The sample space for a dice is {1, 2, 3, 4, 5, 6}
More Important Topics
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• Practice worksheets in and after class for conceptual clarity
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# What is 51/402 as a decimal?
## Solution and how to convert 51 / 402 into a decimal
51 / 402 = 0.127
Converting 51/402 to 0.127 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. Now, let's solve for how we convert 51/402 into a decimal.
## 51/402 is 51 divided by 402
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 402. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 402. We must divide 51 into 402 to find out how many whole parts it will have plus representing the remainder in decimal form. This is how we look at our fraction as an equation:
### Numerator: 51
• Numerators are the top number of the fraction which represent the parts of the equation. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. Let's take a look at the denominator of our fraction.
### Denominator: 402
• Denominators are located at the bottom of the fraction, representing the total number of parts. Larger values over fifty like 402 makes conversion to decimals tougher. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. Let's start converting!
## Converting 51/402 to 0.127
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 402 \enclose{longdiv}{ 51 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 402 \enclose{longdiv}{ 51.0 }$$
We've hit our first challenge. 51 cannot be divided into 402! Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 402 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 402 into 510
$$\require{enclose} 00.1 \\ 402 \enclose{longdiv}{ 51.0 }$$
How many whole groups of 402 can you pull from 510? 402 Multiple this number by our furthest left number, 402, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 402 \enclose{longdiv}{ 51.0 } \\ \underline{ 402 \phantom{00} } \\ 108 \phantom{0}$$
If there is no remainder, you’re done! If you have a remainder over 402, go back. Your solution will need a bit of adjustment. If you have a number less than 402, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 51/402 and 0.127 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/402, 0.127 or 12% in our daily world:
### When you should convert 51/402 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 51/402 MPH. The radar will read: 90.12 MPH. This simplifies the value.
### When to convert 0.127 to 51/402 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 51/402 = 0.127 what would it be as a percentage?
• What is 1 + 51/402 in decimal form?
• What is 1 - 51/402 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.127 + 1/2? |
# Absolute Value Inequalities
Absolute Value Inequalities are inequalities in which there is one or more absolute value . Let us recall that an inequality is almost like an equation, but instead of the "=" sign, we have "≤" or "≥".
This difference makes it so that the solution set is typically a region, like for most inequalities. And the fact that there absolute values involved indicate a certain especial treatment for their resolution.
In this tutorial we will concentrate on the specific skills required for the resolution of this type of inequality that contains one ore more absolute values. Also, we will assume that one or two variables, $$x$$ and/or $$y$$ are involved in the inequality.
## What is an Absolute Value Inequality?
For the purpose of this analysis, we will consider an absolute value inequality to be an inequality involving one or two variables, with at least one absolute value.
For example, below we have an absolute value inequality with two variables $$x$$ and $$y$$:
$|3x+2y-1| \ge 1$
Or also, we could have the following absolute value inequality with only one variable:
$|3x-1| \le 2$
For our purposes, and for the purpose of the techniques used for their resolution, we will deal with both types inequalities (one and two variables)
## How to Solve Absolute Value Inequalities?
When solving equations or inequalities, there is not really a silver bullet that solves everything. Each problem is different and may have its own peculiarities.
The best we can do is to provide a series of steps that will aid you in the process of solving an inequality.
Step 1: For each absolute determine the regions in which the argument of the absolute value is negative and where it is non-negative.
Step 2: If there is only one absolute value in the inequality, solve it in both areas (where the argument of the absolute value is negative, and where it is non-negative).
Step 3: If there is more than one absolute value in the inequality, you need to intersect all the regions in order to get a set of smaller partitions. In each partition, you need to know EXACTLY the sign of each argument. Then, solve the inequality in all the areas.
Step 4: Once you get the part solution that is in each of the areas, the final solution is the simply the union of these part solutions.
In simple words: You need to find out the regions where you know exactly the sign of the argument of the absolute values (so you can get rid of them).
A couple of examples should clarify these steps.
### EXAMPLE 1
Solve the following inequality
$| 2x + 4y - 1 | \ge 2$
In order to solve the inequality, we need to use the steps that were specified above.
Step 1: There is only one absolute value, so we need to determine whether is the argument negative and non-negative. Therefore, we need to solve first:
$2x + 4y - 1 \ge 0$
There are several strategies to solve the above, but the easiest one is to first solve the equation
$2x + 4y - 1 = 0$
which means that $$4y = -2x + 1$$ or the same as $$y = -\frac{1}{2}x + \frac{1}{4}$$, which corresponds to a line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{1}{4}$$.
Now, to take care of $$2x + 4y - 1 \ge 0$$ we test whether the point $$(0,0)$$ satisfies or not the inequality:
$2(0) + 4(0) - 1 = -1 < 0$
So, $$(0,0)$$ satisfies or not the inequality. The conclusion is that the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{1}{4}$$ divides the plane in two regions:
For the points below the line (we call this region 1, $$R_1$$), we get that $$2x + 4y - 1 < 0$$
For the points above the line , including the line itself (we call this region 2, $$R_2$$) we get that $$2x + 4y - 1 \ge 0$$
Why is this important? Why we take all this trouble? Because on $$R_1$$, we get that since $$2x + 4y - 1 < 0$$, then $$| 2x + 4y - 1 | = -(2x + 4y - 1)$$. Similarly, on $$R_2$$, we get that since $$2x + 4y - 1 \ge 0$$, then $$| 2x + 4y - 1 | = 2x + 4y - 1$$.
Step 2: Now we need to solve the inequality on region 1, $$R_1$$ :
$| 2x + 4y - 1 | \ge 2$ $\Rightarrow -(2x + 4y - 1) \ge 2$ $\Rightarrow 2x + 4y - 1 \le -2 \text{ (multiplying by (-1) changes the direction of the inequality)}$ $\Rightarrow 2x + 4y \le -1$ $\Rightarrow 4y \le -2x - 1$ $\Rightarrow y \le -\frac{1}{2}x - \frac{1}{4}$
This corresponds to all the points below or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = -\frac{1}{4}$$. But don't forget that you are on $$R_1$$, and this line we found is BELOW the boundary of $$R_1$$ (see the graph below).
To clarify, since we are under the assumption that we are in $$R_1$$, we need to have that we are BELOW line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{1}{4}$$. Under this assumption, we solved the original inequality and we also need to be below line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = -\frac{1}{4}$$. These two conditions must happen simultaneously, so we get the intersection of the two regions.
So then, the part solution in this case corresponds to all the points below or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = -\frac{1}{4}$$.
Now we need to solve the inequality on region 2, $$R_2$$ :
$| 2x + 4y - 1 | \ge 2$ $\Rightarrow 2x + 4y - 1 \ge 2$ $\Rightarrow 2x + 4y \ge 3$ $\Rightarrow 4y \ge -2x + 3$ $\Rightarrow y \ge -\frac{1}{2}x + \frac{3}{4}$
This corresponds to all the points above or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{3}{4}$$. But don't forget that you are on $$R_2$$, and this line is ABOVE the boundary of$$R_2$$ (see the graph below).
By finding the intersection between $$R_2$$ and the region above, we get that the part solution in this case is all the points above or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{3}{4}$$.
Step 4: Now, the final solution is the union of all part solutions from the previous parts: The final solution is all the points BELOW or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = -\frac{1}{4}$$, PLUS all the points ABOVE or on the line with slope $$m = -\frac{1}{2}$$ and y-intercept $$n = \frac{3}{4}$$.
Graphically, we get
which concludes the resolution of the inequality.
### EXAMPLE 2
Solve the following double absolute value inequality
$| 2x - 1 | \ge |x + 3|$
This is a double absolute value inequality because there are 2 absolute values. This means that finding the regions will take a bit more of work (relatively speaking).
Step 1: For the first absolute value we solve:
$2x- 1 \ge 0$ $\Rightarrow \,\, 2x \ge 1$ $\Rightarrow \,\, x \ge \frac{1}{2}$
So we get that $$2x- 1 \ge 0$$ on $$[\frac{1}{2}, +\infty)$$, and $$2x- 1 < 0$$ on $$(-\infty, \frac{1}{2})$$.
For the second absolute value we solve:
$x+3 \ge 0$ $\Rightarrow \,\, x \ge -3$
So we get that $$x+3 \ge 0$$ on $$[-3, +\infty)$$, and $$x+3 < 0$$ on $$(-\infty, -3)$$.
So then, we define 4 regions:
$$R_1 = [\frac{1}{2}, +\infty) \cap [-3, +\infty) = [\frac{1}{2}, +\infty)$$. On this region we get: $$2x- 1 \ge 0$$ AND $$x+3 \ge 0$$.
$$R_2 = [\frac{1}{2}, +\infty) \cap (-\infty, -3) = \varnothing$$. On this region we get: $$2x- 1 \ge 0$$ AND $$x+3 < 0$$, though this region is empty.
$$R_3 = (-\infty, \frac{1}{2}) \cap [-3, +\infty) = [-3, \frac{1}{2})$$. On this region we get: $$2x- 1 < 0$$ AND $$x+3 \ge 0$$
$$R_4 = (-\infty, \frac{1}{2}) \cap (-\infty, -3) = (-\infty, -3)$$. On this region we get: $$2x- 1 < 0$$ AND $$x+3 < 0$$.
Step 2: Now we need to solve the double absolute value inequality on each of the four regions:
• On $$R_1$$:
Here we get $$2x- 1 \ge 0$$ AND $$x+3 \ge 0$$ so then
$| 2x - 1 | \ge |x + 3|$ $\Rightarrow \,\, 2x - 1 \ge x + 3$ $\Rightarrow \,\, 2x - x \ge 3 - (-1)$ $\Rightarrow \,\, x \ge 4$
So, in order to get the part solution we need to intersect $$x \ge 4$$, or $$[4, +\infty)$$ with $$R_1$$.
The corresponding part solution is therefore: $$[\frac{1}{2}, +\infty) \cap [4, +\infty) = [4, +\infty)$$
• On $$R_2$$:
This part solution is empty ($$\varnothing$$).
• On $$R_3$$:
Here we get $$2x- 1 < 0$$ AND $$x+3 \ge 0$$ so then
$| 2x - 1 | \ge |x + 3|$ $\Rightarrow \,\, -(2x - 1) \ge x + 3$ $\Rightarrow \,\, 2x - 1 \le -x - 3$ $\Rightarrow \,\, 2x - (-x) \le -3 - (-1)$ $\Rightarrow \,\, 3x \le -2$ $\Rightarrow \,\, x \le -\frac{2}{3}$
So, in order to get this part solution we need to intersect $$x \le -\frac{2}{3}$$, or $$(-\infty, -\frac{2}{3}]$$ with $$R_3$$.
The corresponding part solution is therefore: $$(-\infty, -\frac{2}{3}] \cap [-3, \frac{1}{2}) = [-3, -\frac{2}{3}]$$
• On $$R_4$$:
Here we get $$2x- 1 < 0$$ AND $$x+3 < 0$$ so then
$| 2x - 1 | \ge |x + 3|$ $\Rightarrow \,\, -(2x - 1) \ge -(x + 3)$ $\Rightarrow \,\, 2x - 1 \le x + 3$ $\Rightarrow \,\, 2x - x \le 3 - (-1)$ $\Rightarrow \,\, x \le 4$
So, in order to get this part solution we need to intersect $$x \le 4$$, or $$(-\infty, 4]$$ with $$R_4$$.
The corresponding part solution is therefore: $$(-\infty, -3) \cap (-\infty, 4] = (-\infty, -3)$$
Step 4: Finally, we get the union of the part solutions, to get that the solution of the initial given inequality is
$(-\infty, -3) \cup [-3, -\frac{2}{3}] \cup [4, +\infty) = (-\infty, -\frac{2}{3}] \cup [4, +\infty)$
No one said it would short, right? Well. It is not really tough, you just have to be systematic and stick to the plan.
## More About Inequalities with Absolute Value
Why do we even worry about this kind of inequalities? We care because they do have applications in practice.
For example, in geometry, the distances in the real line need to be represented as an absolute value, because it needs to be non-negative.
One could have a certain geometric situation in which you need to find all the points in the real line that are at least at a distance of 2 from the point 3. Such situation can be described with the following inequality:
$| x-3 |\ge 2$
Let us understand the above inequality. The point $$x$$ is the point we want to satisfy the inequality. The distance from $$x$$ to the point 3 is represented by $$|x - 3|$$.
Then, we are trying to find the points that are at least at a distance of 2 from the point 3, so then the distance $$|x - 3|$$ needs to be at least 2, which explains the $$|x - 3| \ge 2.$$
This is just one kind of absolute value inequalities problems that you can find in practice.
### Can you find absolute value inequalities with no solution
You bet. Here you have one $$|2x| < |x|$$. It is possible for an inequality to be simply unfeasible like the case of this one I just gave you.
### How about graphing absolute value inequalities?
The process of graphing them is essentially hand in hand with the process of solving them: You need to find the regions where you know exactly whether the arguments of the absolute values are positive or negative.
Also, whenever possible you want to simplify parentheses when the associative law applies (not always the case).
Then the absolute value inequalities become into simple inequalities, which is trivially graphed. Then, all the pieces of the regions obtained are simply joined. |
# NCERT Solutions class 7 Maths Chapter-6 Exercise 6.4
NCERT Solutions Class-7 Maths chapter-6 Triangle and its properties Exercise-6.4 is prepared by academic team of pw all the questions of NCERT text book are solved step by step with proper and detail solutions explaining each and every questions . For More and additional questions of CBSE class 7 maths you can go to class 7 maths sections. NCERT class 7 Maths Solutions is the best way to enhanced your mathematics skill. And pw practice worksheet & question bank will help you a lot .
## NCERT Solutions class 7 Maths Chapter-6 Triangle and its properties
### Solutions of Chapter Triangle and its properties Exercise-6.4
Question 1:
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
(i) 2 cm, 3 cm, 5 cm
We know that,
In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side
The given sides of triangle in this question are:
2 cm, 3 cm and 5 cm
Now, 2 + 3 + = 5 cm
5 cm = 5 cm
Hence,
The triangle is not possible as the sum of the length of either two sides of the triangle is not greater than the third side
(ii) 3 cm, 6 cm, 7 cm
We know that,
In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side
The given sides of triangle in this question are:
3 cm, 6 cm and 7 cm
Now, 3 + 6 = 9 cm and 9 cm > 7 cm
6 + 7 = 13 cm and, 13 cm > 3 cm
3 + 7 = 10 cm and, 10 cm > 6 cm
Hence,
The triangle is possible as the sum of the length of either two sides of the triangle is greater than the third side
(iii) 6 cm, 3 cm, 2 cm
We know that,
In a triangle,
The sum of the length of either two sides of the triangle is always greater than the third side
Therefore,
The given sides of triangle in this question are:
6 cm, 3 cm and 2 cm
Now,
6 + 3 = 9 cm and 9 cm > 2 cm
3 + 2 = 5 cm But, 5 cm < 6 cm
Hence,
The triangle is not possible as the sum of the length of either two sides of the triangle is not greater than the third side
Question 2:
Take any point O in the interior of a triangle PQR. Is:
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
(i) OP + OQ > PQ?
According to the given condition in the question,
If O is a point in the interior of the given triangle
Then,
Three triangles can be constructed, these are:
∆OPQ, ∆OQR and ∆ORP
We know that,
In a triangle, the sum of the length of either two sides of the triangle is always greater than the third side
Therefore,
∆OPQ is a triangle having sides OP, OQ and PQ
As,
OP + OQ > PQ
(ii) OQ + OR > QR?
According to the given condition in the question,
We have:
If O is a point in the interior of the given triangle
Then,
Three triangles can be constructed, these are:
∆OPQ, ∆OQR and ∆ORP
We know that,
In a triangle,
The sum of the length of either two sides of the triangle is always greater than the third side
Therefore,
∆OQR is a triangle having sides OR, OQ and QR
As,
NCERT CLASS 7 MATHEMATICS SOLUTIONS |
# Common Core: 3rd Grade Math : Add and Subtract Within 1000: CCSS.Math.Content.3.NBT.A.2
## Example Questions
1 2 4 Next →
### Example Question #31 : Add And Subtract Within 1000: Ccss.Math.Content.3.Nbt.A.2
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #31 : Add And Subtract Within 1000: Ccss.Math.Content.3.Nbt.A.2
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #32 : Add And Subtract Within 1000: Ccss.Math.Content.3.Nbt.A.2
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #33 : Add And Subtract Within 1000: Ccss.Math.Content.3.Nbt.A.2
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #131 : Number & Operations In Base Ten
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #34 : Add And Subtract Within 1000: Ccss.Math.Content.3.Nbt.A.2
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place: |
# Pre-Algebra
posted by .
This is about Reasoning Strategy, I really need help on this, I have a test tomorrow on it! Please help... Ok, here's what it says:
3a. What is the y-intercept of the trend line? (What do they mean???)
3b. Locate one other point on the trend line, then find the slope of the trend line (I don't know what they mean?)
• Pre-Algebra -
3a. The y intercept is the value of y where x equals zero (where the trend line intercepts the y axis).
3b. Pick any other point on the trend line (besides the y intercept) and compute the slope as the change in y divided by the change of x.
• Pre-Algebra -
ok, for 3b, I would need to divide any point in the x axis by a point on the y axis? Or is that wrong?
• Pre-Algebra -
3a. The y-intercept is found when you set x to = 0. The y-intercept is the point where the graph of the line crosses the y-axis.
SAMPLE:
In the equation y = 3x + 5, what is the y-intercept?
It is the point (0,5). Do you see number 5? That number represents the y-intercept when x = 0.
Don't trust me?
Here it is:
Let x = 0.
y = 3(0) + 5
y = 0 + 5
y = 5
So, y became 5 when we set x to = 0.
====================================
3b. Pick any two points on the line and find the slope of the line.
The slope is a numerical value that describes just how slanted a given line is.
Here is is an example of finding the slope of a line.
SAMPLE:
Given the points (3, 4) and (8, 9), find the slope.
The symbol for slope is the letter m.
Let m = slope.
We now subtract b from d and a from c.
(a,b) = (3,4)
(c,d) = (8,9)
m = (9 - 4)/(8 - 3)
m = 5/5
m = 1
The slope is 1.
Did you follow?
• Pre-Algebra -
Yep, thank you VERY much! Except that I don't really get 3a...
• Pre-Algebra -
P(2,0) Q(4,4) |
# CBSE Class 9 Mathematics Solved Guess Paper SA-II 2015
Here you can find the Mathematics Guess Paper for CBSE Class 9 SA-II 2015. It includes a various set of questions with assigned score for each.
Created On: Jan 12, 2015 17:00 IST
Modified On: Jan 13, 2015 11:09 IST
Find CBSE Class 9 Mathematics Guess Paper for SA-II 2015. This paper is a collection of questions from the previous year question papers, along with fresh new questions and has been framed keeping the Students' perspective in mind. This will help the Students by building a sound concept before their SA-II Examination.
Q. What is the longest pole that can be put in a room of dimensions length = 20 cm, breadth= 20 cm and height = 10 cm?
(a) 15 cm
(b) 25 cm
(c) 20 cm
(d) 30 cm
Sol. Since, area of parallelograms on the same base and between same parallels is always same,
So as parallelogram PQRS and AQRB are on the same base and between same parallels,
Option (c) is correct.
Q. Find the value of k, if x = –1 and y = 2 is a solution of kx + 3y = 7.
Sol. We have given,
x = –1 and y = 2
Substituting the value of x and y in the equation kx + 3y = 7, we get
k (–1) + 3 (2) = 7
– k + 6 = 7
– k = –1
k = –1
Q. Find the amount of water displaced by a solid spherical ball of diameter 0.21 m.
Sol. The diameter of the solid spherical ball is = 0.21 m
Amount of water displaced = Volume of spherical ball
Q. Two years ago, a man's age was 3 times the square of his son's age. In 3 years’ time, his age will be 4 times his son's age. Find their present ages.
Sol. Let the present age of son be x years.
Therefore, before two years son’s age was = (x – 2) years.
According to the question:
Before two years, father’s age = 3(x – 2)2 years.
Thus, the present age of father = [3(x – 2)2 + 2] years
After three years, the age of son will be = (x + 3) years
And the age of the father will be = [3 (x – 2)2 + 2 + 3] years
= [3(x – 2)2 + 5] years
Now, according to the question:
3(x – 2)2 + 5 = 4 (x + 3)
⇒ 3(x2 – 4x + 4) + 5 = 4x + 12
⇒ 3x2 – 16x+ 5 = 0
⇒ 3x2 – 15xx+ 5 = 0
⇒ 3x (x – 5) – 1 (x– 5) = 0
⇒ (3x – 1) (x – 5) = 0
⇒ 3x – 1 = 0 or x – 5 = 0
x = 1/3 or 5
If x = 1/3, then age before two years = x – 2 = –5/3.
This means that age of son before two years was –5/3 years. This is impossible since the age of a person cannot be negative.
Therefore, present age of son = x = 5 years
And, the present age of the man = [3 (x – 2)2 + 2] = 3 × 9 + 2 = 29 years |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.2: Vector Algebra
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
##### Outcomes
1. Understand vector addition and scalar multiplication, algebraically.
2. Introduce the notion of linear combination of vectors.
Addition and scalar multiplication are two important algebraic operations done with vectors. Notice that these operations apply to vectors in $$\mathbb{R}^{n}$$, for any value of $$n$$. We will explore these operations in more detail in the following sections.
## Addition of Vectors in $$\mathbb{R}^n$$
Addition of vectors in $$\mathbb{R}^n$$ is defined as follows.
##### Definition $$\PageIndex{1}$$: Addition of Vectors in $$\mathbb{R}^n$$
If $$\vec{u}=\left [ \begin{array}{c} u_{1} \\ \vdots \\ u_{n} \end{array} \right ],\; \vec{v}= \left [ \begin{array}{c} v_{1} \\ \vdots \\ v_{n} \end{array} \right ] \in \mathbb{R}^{n}$$ then $$\vec{u}+\vec{v}\in \mathbb{R}^{n}$$ and is defined by
\begin{aligned} \vec{u}+\vec{v} &= \left [ \begin{array}{c} u_{1} \\ \vdots \\ u_{n} \end{array} \right ] + \left [ \begin{array}{c} v_{1} \\ \vdots \\ v_{n} \end{array} \right ]\\ & = \left [ \begin{array}{c} u_{1}+v_{1} \\ \vdots \\ u_{n}+v_{n} \end{array} \right ]\end{aligned}
To add vectors, we simply add corresponding components. Therefore, in order to add vectors, they must be the same size.
Addition of vectors satisfies some important properties which are outlined in the following theorem.
##### Theorem $$\PageIndex{1}$$: Properties of Vector Addition
The following properties hold for vectors $$\vec{u},\vec{v}, \vec{w} \in \mathbb{R}^{n}$$.
• The Commutative Law of Addition $\vec{u}+\vec{v}=\vec{v}+\vec{u}\nonumber$
• The Associative Law of Addition $\left( \vec{u}+\vec{v}\right) +\vec{w}=\vec{u}+\left( \vec{v}+\vec{w}\right)\nonumber$
• The Existence of an Additive Identity $\vec{u}+\vec{0}=\vec{u} \label{vectoridentity}$
• The Existence of an Additive Inverse $\vec{u}+\left( -\vec{u}\right) =\vec{0}\nonumber$
The additive identity shown in Equation $$\eqref{vectoridentity}$$ is also called the zero vector, the $$n \times 1$$ vector in which all components are equal to $$0$$. Further, $$-\vec{u}$$ is simply the vector with all components having same value as those of $$\vec{u}$$ but opposite sign; this is just $$(-1)\vec{u}$$. This will be made more explicit in the next section when we explore scalar multiplication of vectors. Note that subtraction is defined as $$\vec{u}-\vec{v} = \vec{u}+\left( -\vec{v} \right)$$.
## Scalar Multiplication of Vectors in $$\mathbb{R}^n$$
Scalar multiplication of vectors in $$\mathbb{R}^n$$ is defined as follows.
##### Definition $$\PageIndex{2}$$: Scalar Multiplication of Vectors in $$\mathbb{R}^n$$
If $$\vec{u}\in \mathbb{R}^{n}$$ and $$k\in \mathbb{R}$$ is a scalar, then $$k\vec{u}\in \mathbb{R}^{n}$$ is defined by $k\vec{u}=k\left [ \begin{array}{c} u_{1} \\ \vdots \\ u_{n} \end{array} \right ] = \left [ \begin{array}{c} ku_{1} \\ \vdots \\ ku_{n} \end{array} \right ]\nonumber$
Just as with addition, scalar multiplication of vectors satisfies several important properties. These are outlined in the following theorem.
##### Theorem $$\PageIndex{2}$$: Properties of Scalar Multiplication
The following properties hold for vectors $$\vec{u},\vec{v}\in \mathbb{R}^{n}$$ and $$k,p$$ scalars.
• The Distributive Law over Vector Addition $k \left( \vec{u}+\vec{v}\right) = k\vec{u}+ k\vec{v}\nonumber$
• The Distributive Law over Scalar Addition $\left( k + p \right)\vec{u} = k \vec{u}+p \vec{u}\nonumber$
• The Associative Law for Scalar Multiplication $k \left( p \vec{u}\right) = \left(k p \right)\vec{u}\nonumber$
• Rule for Multiplication by $$1$$ $1\vec{u}=\vec{u}\nonumber$
Proof
We will show the proof of: $k \left( \vec{u}+\vec{v}\right) = k \vec{u}+ k \vec{v}\nonumber$ Note that: $\begin{array}{ll} k \left( \vec{u}+\vec{v}\right) & =k \left [ u_{1}+v_{1} \cdots u_{n}+v_{n}\right ]^T \\ & = \left [ k \left( u_{1}+v_{1}\right) \cdots k \left( u_{n}+v_{n}\right) \right ]^T \\ & = \left [ k u_{1}+ k v_{1} \cdots k u_{n}+ k v_{n}\right ]^T \\ & = \left [ k u_{1} \cdots k u_{n} \right ]^T + \left [ k v_{1} \cdots k v_{n} \right ]^T \\ & = k \vec{u}+k \vec{v} \\ \end{array}\nonumber$
We now present a useful notion you may have seen earlier combining vector addition and scalar multiplication
##### Definition $$\PageIndex{3}$$: Linear Combination
A vector $$\vec{v}$$ is said to be a linear combination of the vectors $$\vec{u}_1,\cdots , \vec{u}_n$$ if there exist scalars, $$a_{1},\cdots ,a_{n}$$ such that $\vec{v} = a_1 \vec{u}_1 + \cdots + a_n \vec{u}_n\nonumber$
For example, $3 \left [ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right ] + 2 \left [ \begin{array}{r} -3 \\ 0\\ 1 \end{array} \right ] = \left [ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right ].\nonumber$ Thus we can say that $\vec{v}= \left [ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right ]\nonumber$ is a linear combination of the vectors $\vec{u}_1 = \left [ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right ] \mbox{ and } \vec{u}_2 = \left [ \begin{array}{r} -3 \\ 0\\ 1 \end{array} \right ]\nonumber$
This page titled 4.2: Vector Algebra is shared under a CC BY license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) . |
Finding the Midpoint(KS2, Year 6)
homegraphs_and_coordinate_geometryfinding the midpoint between points
The midpoint is half way between two points. We can find the Cartesian coordinates of the midpoint if we know the Cartesian coordinates of the two points. Imagine we wanted to find the midpoint between the points with Cartesian coordinates (1, 2) and (5, 4). The image below shows what we mean by the midpoint:
How to Find the Midpoint
Finding the midpoint between two points is easy.
Question
Find the midpoint between the points with Cartesian coordinates (1, 2) and (5, 4).
1
Add the x-coordinates of the two points together. The x-coordinate of (1, 2) is 1. The x-coordinate of (5, 4) is 5.
1 + 5 = 6
2
Divide the answer (6) by 2.
6 ÷ 2 = 3
3 is the x-coordinate of the midpoint.
3
Add the y-coordinates of the two points together. The y-coordinate of (1, 2) is 2. The y-coordinate of (5, 4) is 4.
2 + 4 = 6
4
Divide the answer (6) by 2.
6 ÷ 2 = 3
3 is the y-coordinate of the midpoint.
5
Write down the Cartesian coordinates of the midpoint as a pair of numbers in brackets, separated by a comma. The x-coordinate (3) found in Step 2 goes on the left. The y-coordinate (3) found in Step 4 goes on the right.
How to Find the Midpoint Using a Formula
The midpoint between a point at (x1, y1) and a point at (x2, y2) is found at:
The image below shows what we mean by x1, y1, x2 and y2:
Lesson Slides
The slider below gives another example of how to find the midpoint of two points.
What Are Cartesian Coordinates?
Cartesian coordinates are used to describe the position of a point on a graph. Cartesian coordinates work by measuring how far across and how far up the point is from the origin.
Why Does the Method Work?
The method works because it finds the point in the middle of the two points. The x-coordinate of the midpoint is just the average of the x-coordinates of the end points. The y-coordinate of the midpoint is just the average of the y-coordinates of the end points. Remember, the average of two numbers is found by adding them together and dividing by 2. The average of two numbers is the number exactly half way between the two numbers.
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Home » How To Simplify 8 3? New
# How To Simplify 8 3? New
Let’s discuss the question: how to simplify 8 3. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What is 8 divided by 3 simplified?
Answer: The value of 8 divided by 3 as a fraction is 2 ⅔.
## How do you simplify a fraction step by step?
Here are the steps to follow:
1. Write down the factors for the numerator and the denominator.
2. Determine the largest factor that is common between the two.
3. Divide the numerator and denominator by the greatest common factor.
4. Write down the reduced fraction.
Simplify 3√8+3√2
Simplify 3√8+3√2
## What is 3 over 8 as a decimal?
Fraction to decimal conversion table
Fraction Decimal
3/8 0.375
4/8 0.5
5/8 0.625
6/8 0.75
## How do you show 72 divided by 8?
Using a calculator, if you typed in 72 divided by 8, you’d get 9. You could also express 72/8 as a mixed fraction: 9 0/8.
## Why do we simplify fractions?
We simplify fractions because it is always to work or calculate when the fractions are in the simplest form.
## How do multiply fractions?
The first step when multiplying fractions is to multiply the two numerators. The second step is to multiply the two denominators. Finally, simplify the new fractions. The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
## What is 3/8 as a percent?
Therefore, the solution is 37.5%.
## What is 8 3 converted into a mixed number?
Answer: 8/3 as a mixed number would be written as 2 2/3.
### Simplify this Expression: -8^(-2/3) | Fast \u0026 Easy Explanation
Simplify this Expression: -8^(-2/3) | Fast \u0026 Easy Explanation
Simplify this Expression: -8^(-2/3) | Fast \u0026 Easy Explanation
## What is 3/8 rounded to the nearest hundredth?
Answer: 3/8 as a decimal is 0.375.
## What is 3/8 as a fraction?
Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions
3/8 6/16 12/32
5/8 10/16 20/32
7/8 14/16 28/32
1/9 2/18 4/36
## How do you write 21 divided by 3?
When you divide 21 by 3 you get 7. We can write this as follows: 21 / 3 = 7.
## How do u divide fractions?
The first step to dividing fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. Finally, simplify the fractions if needed.
## How do you solve 96 divided by 8?
Using a calculator, if you typed in 96 divided by 8, you’d get 12.
## What is simplifying math?
Mathematics. Simplification is the process of replacing a mathematical expression by an equivalent one, that is simpler (usually shorter), for example. Simplification of algebraic expressions, in computer algebra.
### Quantitative 10: Simplify (8^3/4 +2^3) ^1/2 with corrections
Quantitative 10: Simplify (8^3/4 +2^3) ^1/2 with corrections
Quantitative 10: Simplify (8^3/4 +2^3) ^1/2 with corrections
## How do you write and simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## What is an example of simplify?
Simplify is to make something less complicated or less cluttered. An example of simplify is when you explain a tough math concept in really easy terms for a child to understand. An example of simplify is when you cut out a lot of the activities that were making you busy and stressed.
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1. ## Factoring
Factor:
a)64x^3+8 b)72x^3-108x^2-140x c)x^3+3x^2-6x-8
I've become so depended on my calculator that I actually forget how to factor...help!
Thank you
2. The first one is the sum of two cubes. Remember the factoring for the sum of two cubes?. $(x+y)(x^{2}-xy+y^{2})$
You have $64x^{3}+8 = (4x)^{3}+(2)^{3}$
Now, use the formula.
For the second one. You can factor out 4x:
$4x(18x^{2}-27x-35)$
Now, factor the quadratic. What two numbers when multiplied equal -630 (because (-35)(18) = -630) and when added equal -27.
Hmmmm, let's see......how about -42 and 15.
$18x^{2}+15x-42x-35$
Group: $(18x^{2}+15x)-(42x+35)$
Factor out 3x in the first one and -7 out of the second:
$3x(6x+5)-7(6x+5)$
$(3x-7)(6x+5)$
Don't forget the 4x we factored out in the beginning.
$4x(3x-7)(6x+5)$
See how it's done now?.
Oh yeah, BTW, there is a way to remember the two different factorings for the cubes.
$SOAP: \;\ \;\ \boxed{S}ame, \;\ \boxed{O}pposite, \;\ \boxed{A}lways \;\ \boxed{P}ositive$
$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$
Note the signs in the factoring on the right side of the equals sign.
Then the first sign is the Same, then Opposite, then the last one is Always Positive. See what I mean?. |
Other
# How do you find unit rate?
## How do you find unit rate?
To find the unit rate, divide the numerator and denominator of the given rate by the denominator of the given rate. So in this case, divide the numerator and denominator of 70/5 by 5, to get 14/1, or 14 students per class, which is the unit rate.
### How do you find the cost per gallon?
Take the mileage of the total distance of your trip and divide it by your miles per gallon to get the number of gallons of gas you will need on your trip. Then multiply that figure by the current price of gas, and the result is the estimated cost of gas for your road trip.
#### What is the unit rate for a car that can travel 256 miles on 8 gallons of gas?
8 goes into 256 32 times, so the unit rate is 32/1. 32 miles per 1 gallon. So divide 256 by 32 and you’ll get 8.
Which rate is equivalent to 15 miles per gallon?
The equivalent unit rate for 15 miles per gallon is 60 miles per 4 gallons ( ) .
What is a unit rate give an example?
A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate.
## What is the difference between a rate and a unit rate?
A. The difference between a rate and a unit rate is that a rate is the ratio between two different units of measure, while a unit rate is the ratio of between two different units of measure for a single thing.
### What is the unit rate of 297 words in 5.5 minutes?
Answer: The unit rate of 297 words for 5.5 minutes is, 54 words per minutes.
#### How to calculate how much gas you use per mile?
Record the number of gallons or liters required to fill the tank once again. This is the total number of gallons or liters you used for the trip (or the time period). Record the trip ending odometer reading (this might also be the starting reading for your next trip). What is it costing you in gas per mile? Enter the price per gallon.
What’s the difference between miles per gallon and miles per liter?
When you calculate miles/gallon (mpg) or kilometers/liter (km/l) you are calculating fuel economy in terms of distance per unit volume or distance/volume. The following outline is generally applicable to both calculations. When you calculate liters per 100 kilometers (l/100km) you are calculating volume per 100 units of distance.
How to calculate nautical miles per gallon ( mpg )?
1 nmi/gal [US] = 0.9582212144591 MPG [US]. 1 x 0.9582212144591 MPG [US] = 0.9582212144591 Miles Per Gallon US. Always check the results; rounding errors may occur.
## How many miles can you go with 30 gallons of gas?
But you aren’t going to be able use every last drop of gas before you refill so say you use 30 gallons and refill when you have about 1/4 of a tank left. That means you can go about 300 miles before you have to refill. 30 gal x 10 mpg = 300 miles.
### Which is correct mpg or miles per gallon?
Always remember mpg means miles per gallon and “per” always means “divided by”. It always helps to sort out the units so you can find out the formula. MPG could be written as m/g. So the units would be as follows: miles / (miles/gallon) = gallons.
#### Do you need to enter number of miles to calculate gas?
For a successful calculation you need to enter at least the number of miles to travel on your trip as well as the mpg of you car or vehicle. If you also enter the gas price you expect to pay we can also give you an estimate of what the gas would cost for your trip.
What is the formula to calculate miles per gallon?
The formula for calculating Miles Per Gallon is simply miles driven divided by gallons used. So to calculate the miles driven you take the miles at the end of the tank and subtract the Miles at the start of the tank, and then you divide that by the total gallons of gas used.
But you aren’t going to be able use every last drop of gas before you refill so say you use 30 gallons and refill when you have about 1/4 of a tank left. That means you can go about 300 miles before you have to refill. 30 gal x 10 mpg = 300 miles.
Ruth Doyle |
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{x^{2}}{x^{2}+6 x+12} d x$
Solution:
Given $I=\int \frac{x^{2}}{x^{2}+6 x+12} d x$
Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$
$\Rightarrow \int \frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+6 \mathrm{x}+12} \mathrm{dx}=\int\left(\frac{-6 \mathrm{x}-12}{\mathrm{x}^{2}+6 \mathrm{x}+12}+1\right) \mathrm{dx}$
$=-6 \int \frac{x+2}{x^{2}+6 x+12} d x+\int 1 d x$
Consider $\int \frac{x+2}{x^{2}+6 x+12} d x$
Let $x+2=1 / 2(2 x+6)-1$ and split,
$\Rightarrow \int \frac{x+2}{x^{2}+6 x+12} d x=\int\left(\frac{(2 x+6)}{2\left(x^{2}+6 x+12\right)}-\frac{1}{\left(x^{2}+6 x+12\right)}\right) d x$
$=\int \frac{x+3}{x^{2}+6 x+12} d x-\int \frac{1}{x^{2}+6 x+12} d x$
Consider $\int \frac{x+3}{x^{2}+6 x+12} d x$
Let $u=x^{2}+6 x+12 \rightarrow d x=\frac{1}{2 x+6} d u$
$\Rightarrow \int \frac{\mathrm{x}+3}{\left(\mathrm{x}^{2}+6 \mathrm{x}+12\right)} \mathrm{dx}=\int \frac{\mathrm{x}+3}{\mathrm{u}} \frac{1}{2 \mathrm{x}+6} \mathrm{du}$
$=\int \frac{1}{2 \mathrm{u}} \mathrm{du}$
We know that $\int \frac{1}{x} d x=\log |x|+c$
$\Rightarrow \frac{1}{2} \int \frac{1}{\mathrm{u}} \mathrm{du}=\frac{\log |\mathrm{u}|}{2}=\frac{\log \left|\mathrm{x}^{2}+6 \mathrm{x}+12\right|}{2}$
Now consider $\int \frac{1}{x^{2}+6 x+12} d x$
$\Rightarrow \int \frac{1}{x^{2}+6 x+12} d x=\int \frac{1}{(x+3)^{2}+3} d x$
Let $u=\frac{x+3}{\sqrt{3}} \rightarrow d x=\sqrt{3} d u$
$\Rightarrow \int \frac{1}{(x+3)^{2}+3} d x=\frac{\sqrt{3}}{3 u^{2}+3}$
$=\frac{1}{\sqrt{3}} \int \frac{1}{u^{2}+1} d u$
We know that $\int \frac{1}{x^{2}+1} d x=\tan ^{-1} x+c$
$\Rightarrow \frac{1}{\sqrt{3}} \int \frac{1}{\mathrm{u}^{2}+1} \mathrm{du}=\frac{\tan ^{-1} \mathrm{u}}{\sqrt{3}}=\frac{\tan ^{-1}\left(\frac{\mathrm{x}+3}{\sqrt{3}}\right)}{\sqrt{3}}$
Then,
$\Rightarrow \int \frac{x+2}{x^{2}+6 x+12} d x=\int \frac{x+3}{x^{2}+6 x+12} d x-\int \frac{1}{x^{2}+6 x+12} d x$
$=\frac{\log \left|x^{2}+6 x+12\right|}{2}-\frac{\tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)}{\sqrt{3}}$
Then,
$\Rightarrow \int \frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+6 \mathrm{x}+12} \mathrm{dx}=-6 \int \frac{\mathrm{x}+2}{\mathrm{x}^{2}+6 \mathrm{x}+12} \mathrm{dx}+\int 1 \mathrm{dx}$
We know that $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$
$\Rightarrow-6 \int \frac{x+2}{x^{2}+6 x+12} d x+\int 1 d x$
$=-3 \log \left|x^{2}+6 x+12\right|+\frac{6 \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)}{\sqrt{3}}+x+c$
$=-3 \log \left|x^{2}+6 x+12\right|+2 \cdot \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+x+c$
$\therefore \mathrm{I}=\int \frac{\mathrm{x}^{2}}{\mathrm{x}^{2}+6 \mathrm{x}+12} \mathrm{dx}=-3 \log \left|\mathrm{x}^{2}+6 \mathrm{x}+12\right|+2 \cdot \sqrt{3} \tan ^{-1}\left(\frac{\mathrm{x}+3}{\sqrt{3}}\right)+\mathrm{x}+\mathrm{c}$ |
# Use Cavalieri’s Principle to Compare Aquarium Volumes
Alignments to Content Standards: G-GMD.A.2 G-MG.A.1
The management of an ocean life museum will choose to include either Aquarium A or Aquarium B in a new exhibit.
Aquarium A is a right cylinder with a diameter of 10 feet and a height of 5 feet. Covering the lower base of Aquarium A is an “underwater mountain” in the shape of a 5-foot-tall right cone. This aquarium would be built into a pillar in the center of the exhibit room.
Aquarium B is half of a 10-foot-diameter sphere. This aquarium would protrude from the ceiling of the exhibit room.
1. How many cubic feet of water will Aquarium A hold?
2. For each aquarium, what is the area of the water’s surface when filled to a height of $h$ feet?
3. Use your results from parts (a) and (b) and Cavalieri's principle to find the volume of Aquarium B.
## IM Commentary
This task presents a context that leads students toward discovery of the formula for calculating the volume of a sphere. Students who are given this task must be familiar with the formula for the volume of a cylinder, the formula for the volume of a cone, and Cavalieri’s principle.
Specific radius/diameter measurements help to make the procedure of analyzing cross-sections less daunting, but even so, finding expressions to represent areas of varying cross-sections is a very new and challenging skill for many students. For this reason, assigning this task within a setting where students can receive guidance from one another and from the instructor is a sensible alternative to using this task strictly as an assessment tool.
Submitted by Aaron Stidham to the Fourth illustrative Mathematics task writing contest.
## Solution
1. The number of cubic feet of water that Aquarium A will hold is found by subtracting the volume of the cone from the volume of the cylinder:
$$\text{Volume of Cylinder} - \text{Volume of Cone} = \text{Maximum Volume of Water for Aquarium A}$$ \begin{align} \pi \times (\text{radius})^2 \times (\text{height}) - \frac13 \times \pi \times (\text{radius})^2 \times (\text{height}) &= \frac23 \times \pi \times (\text{radius})^2 \times (\text{height}) \\ &= \frac23 \times \pi \times 5^2 \times 5 \\ &= \frac{250\pi}{3} \approx 262 \end{align} so Aquarium A can hold about 262 cubit feet of water.
2. The area of the water’s surface at height $h$ can be derived by examining cross-sections. The radius and height of the cylinder and cone are both 5 feet, so the vertical cross-section through the vertex of the cone shown below shows that the cross section of the water forms an isosceles right triangle:
Therefore, the horizontal distance between the cone and the aquarium wall at height $h$ is also $h$.
The area of the water’s surface at height $h$ is found by subtracting the area of the inner circle, with radius $5 – h$, from the area of the outer circle, with radius 5.
$$\text{Area of Outer Circle} - \text{Area of Inner Circle} = \text{Area of Water’s Surface at Height h}$$ \begin{align} \pi \times 5^2 - \pi \times (5-h)^2 &= 25\pi - \pi (25-10h+h^2) \\ &= 25\pi - 25\pi + 10 \pi h - \pi h^2) \\ &= 10 \pi h - \pi h^2 \end{align}
When Aquarium A is filled to a height of $h$ feet, the area of the water’s surface is $10 \pi h - \pi h^2$ square feet.
Momentarily, the horizontal distance between the curved wall and the sphere’s center at height $h$ is denoted by $x$. But $x$ can be expressed in terms of $h$, since $x$ and $5 – h$ both represent leg lengths of a right triangle with hypotenuse 5: \begin{align} x^2 + (5 – h)^2 &= 25 \\ x^2 + 25 – 10h + h^2 &= 25 \\ x^2 &= 10h – h^2 \\ x &= \sqrt{10h – h^2} \end{align}
In this case, the area of the water’s surface at height $h$ is simply the area of a circle whose radius is $\sqrt{10h – h^2}$: \begin{align} \text{Area} &= \pi(\sqrt{10h – h^2})^2 \\ &= \pi(10h – h^2) \\ &= 10 \pi h - \pi h^2 \end{align}
3. Cavalieri’s principle states that two 3-dimensional solids of the same height are equal in volume if the areas of their horizontal cross-sections at every height $h$ are equal. Since both aquariums stand 5 feet tall, and since the area of the water’s surface when filled to a height of $h$ feet is the same for each aquarium, the volumes of water must be equal when both aquariums are filled to full capacity according to this principle. Therefore, no further calculations are necessary to determine that Aquarium B will hold $\frac{250 \pi }{3}\approx 262$ cubic feet of water, the very same volume that Aquarium A will hold. |
# Daily Maths Revision – Week 4 Walkthrough
If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.
This week, we’re exploring:
### Question 1: Fractional Indices
Let’s start with a general rule and then we will look at how we apply that rule:
$a^{\frac{1}{m}} = \sqrt[m]{a}$
This looks a bit complicated but it’s simpler than it looks. If we have a number to the power of one half, we take the square root; if it’s to the power of one third, we take the cube root and so on. Here are some examples:
\begin{aligned} 81^\frac{1}{2} &= 9 \\ 125^\frac{1}{3} &= 5 \\ \left(\frac{1}{32}\right)^\frac{1}{5}&=\frac{1}{2}\end{aligned}
On the last example, we take the fifth root. With a fraction, we take the fifth root of the numerator and the denominator.
Sometimes, you’ll have a fraction where the numerator isn’t 1.
Let’s find the value of $\boldsymbol{25^\frac{3}{2}}$.
In this case, we will find the square root and then we will cube the answer.
\begin{aligned} 25^\frac{3}{2} &= \left(25^\frac{1}{2}\right)^3 \\&= 5^3 \\&=\boldsymbol{125} \end{aligned}
### Question 2: Standard Form
Standard form is a way to write very large or very small numbers more efficiently. We use powers of 10 to replace long lists of zeros.
Write 275 000 in standard form.
We look at the first significant figure, 2 – this represents 200 000. This is the same as 2 × 105.
So, 275 000 is equivalent to 2.75 × 105.
Write 0.00782 in standard form.
We look at the first significant figure again – this represents 0.007. This is the same as 7 × 10-3.
So, 0.00782 is equivalent to 7.82 × 10-3.
### Question 3: The Equation of Parallel Lines
Two (or more) lines are parallel if their gradients are the same. This means we can find the equation of a line that is parallel to another line and that passes through a set point.
Remember, the equation of any straight line can be written in the form $y=mx+c$ where $m$ is the gradient and $c$ is the $y$-intercept.
Find the equation of the line that passes through the point (1, -3) and is parallel to the line with equation $\boldsymbol{y=7x-2}$
Since the equation of the line is in the form $y=mx+c$, we know the gradient is the coefficient of $x$, 7.
Now we have this gradient, we can substitute into the general form and
$y=7x+c$
To find the value of $c$, we substitute in the point we were given, (1, -8), and solve to find $c$.
\begin{aligned} \text{-}8 &= 7\times1 + c\\ \text{-}8 &=-7+c\\c&=\text{-}-7\\&=\text{-}15 \end{aligned}
Now we can use this to replace $c$ in the equation we had above:
$\boldsymbol{y=7x-15}$
### Question 4: Direct Proportion
Two (or more) values are directly proportional if they change at the same rate. If one doubles, the other also doubles.
For example, the time taken to fill some bottles and the number of bottles to fill. Assuming everything else stays the same, if there are twice as many bottles, it will take twice as long to fill them!
The easiest way to answer these questions is to use algebra to represent the relationship. If two values, $A$ and $B$, are directly proportional, we can describe this by writing $A \propto B$. The sign in the middle, $\propto$, just means “is proportional to”.
When two values are directly proportional, we can also describe their relationship using an equation of the form $A=kb$ where $k$ is a constant called the constant of proportionality. To solve most proportion problems, you will need to find the value of $k$.
Let’s say $\boldsymbol{A}$ is directly proportional to $\boldsymbol{B}$ . When $\boldsymbol{A=12}$ , $\boldsymbol{B=14}$. Find the value of $\boldsymbol{A}$ when $\boldsymbol{B=91}$.
First, we’ll look at how to use algebra. In this case, $A \propto B$ so $A = kB$ for some value of $k$. To find that value, we substitute in $A = 12$ and $B=14$.
\begin{aligned} A &= kb \\ 12 &= k \times 14 \\ k &= \frac{12}{14} \\ k&= \frac{6}{7} \end{aligned}
Putting this back into the equation gives:
\begin{aligned} A = \frac{6}{7}B \end{aligned}
Now, we can substitute $B=91$ into our equation.
\begin{aligned} A &= \frac{6}{7} \times 91 \\ &= \boldsymbol{78} \end{aligned}
### Question 5: Rationalising Surds
Rationalising a fraction means finding an equivalent fraction that has an integer on the denominator. In other words, you want to manipulate the fraction so that any surds are on the top.
Let’s simplify:
$\huge\boldsymbol{\frac{3}{\sqrt{5}}}$
To eliminate the surd on the denominator, we multiply it by $\sqrt5$. However, if we only multiplied the denominator this would change the value of the fraction. So instead, we multiply both the numerator and denominator by the same surd:
$\frac{3}{\sqrt5}\times\frac{\sqrt5}{\sqrt{5}} = \boldsymbol{\frac{3\sqrt5}{5}}$
That’s it! You have no surds on the bottom of the fraction but you haven’t changed the value of the number.
If you have a fraction on the bottom that can be simplified, such as $\sqrt{28}$, it’s a good idea to simplify that first.
Let’s simplify:
$\huge\boldsymbol{\frac{7}{\sqrt{28}}}$
\begin{aligned} \frac{7}{\sqrt{28}} &= \frac{7}{\sqrt{4}\times\sqrt{7}} \\ &= \frac{7}{2\sqrt{7}} \end{aligned}
Now, we multiply by $\sqrt{7}$ as it’s only this part of the denominator we want to change:
\begin{aligned} \frac{7}{2\sqrt{7}} \times \frac{\sqrt7}{\sqrt7} &= \frac{7\sqrt{7}}{2\times7} \\ &= \frac{7\sqrt{7}}{14} \end{aligned}
Now you might be able to see that this fraction isn’t in its simplest form. To simplify it fully, we need to divide the numerator and denominator by 7:
\begin{aligned} \frac{7\sqrt{7}}{14} &= \boldsymbol{\frac{\sqrt{7}}{2}} \end{aligned}
Don’t forget to read even more of our blogs here and you can find our main Daily Maths Revision Page here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too. |
1. ## Calculating the odds
in a essay competition the odds against competitors a,b,c,d is 2:1,3:1,4:1 and 5:1 respectively .find the probability that one of them wins the competition
2. Hello, bluffmaster.roy.007!
In a competition the odds against competitors $\displaystyle A,B,C,D$ winning are: .$\displaystyle 2\!:\!1,\;3\!:\!1,\;4\!:\!1,\;5\!:\!1$, resp.
Find the probability that one of them wins the competition.
We have: .$\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Entrant} & \text{odds against} & P(\text{win}) & P(\text{lose}) \\ \hline \hline \\[-4mm] A & 2:1 & \frac{1}{3} & \frac{2}{3} \\ [2mm] B & 3:1 & \frac{1}{4} & \frac{3}{4} \\ [2mm] C & 4:1 & \frac{1}{5} & \frac{4}{5} \\ [2mm] D & 5:1 & \frac{1}{6} & \frac{5}{6} \\ [2mm] \hline \end{array}$
I will assume that there is exactly one winner of the competition.
. . $\displaystyle \begin{array}{ccccc}P(\text{A wins}) &=& \frac{1}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\f rac{5}{6} &=& \frac{1}{6} \\ \\ P(\text{B wins}) &=& \frac{1}{4}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\f rac{5}{6} &=& \frac{2}{9} \\ \\ P(\text{C wins}) &=& \frac{1}{5}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\f rac{5}{6} &=& \frac{1}{12} \end{array}$
. . $\displaystyle \begin{array}{ccccc}P(\text{D wins}) &=& \frac{1}{6}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\f rac{4}{5} &=& \frac{1}{15} \end{array}$
Therefore: .$\displaystyle P(\text{A or B or C or D wins}) \;=\; \frac{1}{6} + \frac{2}{9} + \frac{1}{12} + \frac{1}{15} \;=\;\frac{97}{180}$ |
Pre-Calc Exam Notes 76
# Pre-Calc Exam Notes 76 - . 5. Use 15 = 45 30 to nd the...
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76 Chapter 3 Identities §3.2 Solution: Multiply the top and bottom of r 1 2 s by sin θ 1 sin θ 2 to get: r 1 2 s = n 1 cos θ 1 n 2 cos θ 2 n 1 cos θ 1 + n 2 cos θ 2 · sin θ 1 sin θ 2 sin θ 1 sin θ 2 = ( n 1 sin θ 1 ) sin θ 2 cos θ 1 ( n 2 sin θ 2 ) cos θ 2 sin θ 1 ( n 1 sin θ 1 ) sin θ 2 cos θ 1 + ( n 2 sin θ 2 ) cos θ 2 sin θ 1 = ( n 1 sin θ 1 ) sin θ 2 cos θ 1 ( n 1 sin θ 1 ) cos θ 2 sin θ 1 ( n 1 sin θ 1 ) sin θ 2 cos θ 1 + ( n 1 sin θ 1 ) cos θ 2 sin θ 1 (by Snell’s law) = sin θ 2 cos θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 1 + cos θ 2 sin θ 1 = sin ( θ 2 θ 1 ) sin ( θ 2 + θ 1 ) The last two examples demonstrate an important aspect of how identities are used in practice: recognizing terms which are part of known identities, so that they can be factored out. This is a common technique. Exercises 1. Verify the addition formulas (3.12) and (3.13) for A = B = 0 . For Exercises 2 and 3, ±nd the exact values of sin ( A + B ), cos ( A + B ), and tan ( A + B ). 2. sin A = 8 17 , cos A = 15 17 , sin B = 24 25 , cos B = 7 25 3. sin A = 40 41 , cos A = 9 41 , sin B = 20 29 , cos B = 21 29 4. Use 75 = 45 + 30 to ±nd the exact value of sin 75
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Unformatted text preview: . 5. Use 15 = 45 30 to nd the exact value of tan 15 . 6. Prove the identity sin + cos = r 2 sin ( + 45 ) . Explain why this shows that r 2 sin + cos r 2 for all angles . For which between 0 and 360 would sin + cos be the largest? For Exercises 7-14, prove the given identity. 7. cos ( A + B + C ) = cos A cos B cos C cos A sin B sin C sin A cos B sin C sin A sin B cos C 8. tan ( A + B + C ) = tan A + tan B + tan C tan A tan B tan C 1 tan B tan C tan A tan C tan A tan B 9. cot ( A + B ) = cot A cot B 1 cot A + cot B 10. cot ( A B ) = cot A cot B + 1 cot B cot A...
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# Readers ask: What Is The Rule Of 70 In Environmental Science?
## What is the rule of 70 in apes?
The Rule of 70 is an easy way to calculate how long it will take for a quantity growing exponentially to double in size. The formula is simple: 70 /percentage growth rate= doubling time in years.
## Where does the Rule of 70 come from?
The Rule of 70 states: If you invest a given amount of money at R% interest, it will take approximately 70 /R years for your money to double. Likewise, if the inflation rate is R%, prices will double in approximately 70 /R years.
## For which growth rate would the Rule of 70 be most accurate?
At 2% growth rate, it will take 35 years. It means that 1% growth rate would the Rule of 70 be most accurate. Hence option A is the correct answer.
You might be interested: Question: How To Build A Science Project?
## Does the rule of 70 apply to negative populations?
The rule of 70 can even be applied to scenarios where negative growth rates are present. For example, if a country’s economy has a growth rate of -2% per year, after 70 /2=35 years that economy will be half the size that it is now.
## How is 70% calculated?
Example 1. Find 70 % of 80. Following the shortcut, we write this as 0.7 × 80. Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors.
## What is the best use of the Rule of 70?
The rule of 70 is a calculation to determine how many years it’ll take for your money to double given a specified rate of return. The rule is commonly used to compare investments with different annual compound interest rates to quickly determine how long it would take for an investment to grow.
## Does your money double every 7 years?
At 10%, you could double your initial investment every seven years (72 divided by 10). In a less-risky investment such as bonds, which have averaged a return of about 5% to 6% over the same time period, you could expect to double your money in about 12 years (72 divided by 6).
## Is it the rule of 70 or 72?
The rule of 70 and the rule of 72 give rough estimates of the number of years it would take for a certain variable to double. When using the rule of 70, the number 70 is used in the calculation. Likewise, when using the rule of 72, the number 72 is used in the calculation.
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## Where did the rule of 72 come from?
The first reference we have of the Rule of 72 comes from Luca Pacioli, a renowned Italian mathematician. He mentions the rule in his 1494 book Summa de arithmetica, geometria, proportioni et proportionalita (“Summary of Arithmetic, Geometry, Proportions, and Proportionality”).
## At what rate do you grow to double in 18 years?
The Rule of 72 could apply to anything that grows at a compounded rate, such as population, macroeconomic numbers, charges, or loans. If the gross domestic product (GDP) grows at 4% annually, the economy will be expected to double in 72 / 4 = 18 years.
## What is the rule of seven in investing?
With an estimated annual return of 7 %, you’d divide 72 by 7 to see that your investment will double every 10.29 years. Here’s an example of other rates of return and how the Rule of 72 affects your investment: Rate of Return. Years it Takes to Double.
## Where is the rule of 72 most accurate?
It’s most accurate at an 8% interest rate, with 6-10% being its most accurate window. The general rule of thumb to help make the estimate more accurate is to adjust the rule by 1 for every 3 percentage points the interest rate differs from 8%.
## Which are immediate causes of the wealth of nations?
The Factors of Production The most immediate cause of the wealth of nations is this: Countries with a high GDP per capita have a lot of physical and human capital per worker and that capital is organized using the best technological knowledge to be highly productive.
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## How many years will it take the US to double the size of its GDP?
But 70 is an easier number to calculate with, in general. It would take approximately 16.28 years (70 / 4.3) years for the U.S. GDP to double.
## What is the best measure of labor productivity in this economy?
One of the most widely used measures of productivity is Gross Domestic Product ( GDP ) per hour worked. This measure captures the use of labour inputs better than just output per employee. |
LTI Quantitative Ability Previous Year Questions
28 January 2023
LTI Quantitative Ability Previous Year Questions
LTI Quantitative Ability Previous Year Questions
Q1) In how many ways can 8 letters be posted in 4 post boxes, if any number of letters can be posted in all of the four post boxes?
A. 88
B. 48
C. 68
D. 58
Explanation: Each letter can be posted in any letter box. Each letter has 4 options to get posted.
So, number of ways = 48
Q2) 40% of 80 is x% more than 20% of 120. Find x.
A. 38.33%
B. 35.33 %
C. 39.33%
D. 33.33%
Explanation: 40% of 80 is x% more than 20% of 120
40% of 80 = 32
20% of 120 = 24
32 is x% more than 24
24+x% of 24 = 32
X% of 24 = 8
X = 33.33
Q3) Three years back, a father was 24 years older than his son. At present, the father is 5 times as old as the son. How old will the son be three years from now?
A. 12 years
B. 9 years
C. 15 years
D. 10 years
Explanation: Let Son = x,
father was 24 years older than his son (difference is always same)
father = x+24
at present father is 5 times of son
X+24 = 5x
4x =24
X=6
After three years son will be 6+3 = 9 years
Q4) A number when divided by a divisor leaves a remainder of 25. When twice the original number is divided by the same divisor, the remainder is 11. What Is the value of the divisor?
A. 52
B. 39
C. 26
D. 46
Explanation:
Let number = X, divisor = Y
X = Y*Q + 25
Twice the number
2X = Y*Q1 +11
Sub X
2 (Y*Q + 25) = Y*Q1 +11
2YQ +50 = YQ1 +11
YQ1 – 2YQ = 39
Y(Q1-2Q) = 39
Y = 39 /(Q1-2Q)
Since 39 is a prime number, then(Q1-2Q) = 39 OR 1
Y = 1 or 39
Option B
Q5) A, B and C, each of them working alone can complete a Job in 10, 12 and 15 days respectively. If all three of them work together to complete a job and earn Rs. 6000, what will be C's share of the earnings?
A. Rs 1500
B. Rs 1600
C. Rs 1300
D. Rs 1200
Explanation: Work done is divided as efficiency of them
Efficiency = 1/ time
Ratio of work done = (1/10 ) : (1/12) : (1/15) = 6:5:4
Total amount 6000 will be divided as 6:5:4 , among A,B& C
C’ share = 4/15 * 6000 = 1600/-
Q6) The ratio of two numbers is 7:4 and their H.C.F is 4. Their L.C.M is:
A. 112
B. 116
C. 124
D. 148
Explanation: Let the numbers be 7x and 4x.
Then, their HCF =x
So, x=4
So, the numbers are 7x=28 and 4x=16.
LCM of 28 and 16 is 112.
Q7) A is twice as good as B and together they finish a piece of work in 28 days. In how many days will A alone finish the work?
A.30 days
B.42 days
C.53 days
D. 40 days
Explanation: Efficiency of A and B are in the ratio of 2: 1
Let efficiency of A be 2 units/day
Let efficiency of B be 1 unit/day
Both of them can complete the task in 28 days
Total work = 3*28 = 84 units
Time taken by A alone = 84/2 = 42 days.
Q8) A container contains 60 liters of milk. From this container, 4 liters of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 46.34 liters
B. 47.36 liters
C. 48.78 liters
D. 49.16 liters
Explanation:
Q9) At what rate percent per annum will a sum of money double in 6 years?
A.(53/3)% p.a
B.(50/3)% p.a
C.(57/3)% p.a
D. (56/3)% p.a
Explanation:
Let P=x, then A=2x
Also, S.I = A−P = 2x−x = x
T = 6 years
We know that S.I. = (P×R×T)/100?)
(x×R×6)/100 = x
R = 100x/6x ?= 16.6% = 50/3%
Q10) Find the lowest common multiple of 1/3, 5/6, 2/9, 4/27
A.1/54
B.1/36
C.20/3
D. 10/27
Explanation:
Q11) Type A, 12 kg of rice worth Rs. 40/kg is mixed with Type B, rice worth Rs. 24/kg. What should be the quantity of Type B rice, if the mixture is sold at Rs. 45/kg with 25% profit added in it?
A. 18 Kg
B. 48 Kg
C. 4 Kg
D. Can't say
Explanation: Let, ‘x’ kg if type - B rice should be added in the mixture.
Now, 40*12 + 24*x = 45*100/125*(12+x)
480 + 24x = 36x + 432
12x = 480 - 432 = 48
x = 48/12 = 4
So, the answer is 4 kg.
Q12) A person spends 1/7th of his salary on travel,1/3rd of the remaining on food, he then spends 1/4th of the remaining on rent. Finally he puts 1/6th of the remaining as a monthly savings, after which he has 25000 left. What is his salary (in Rs.)?
A. 70,000
B. 14,000
C. 84,000
D. 26,000
Explanation: Travel = 1/7
Remaining = 7/7 - 1/7 = 6/7
Food = 1/3 × 6/7 = 2/7
Remaining = 6/7 - 2/7 = 4/7
Rent = 1/4 × 4/7 = 1/7
Remaining = 4/7 - 1/7 = 3/7
Savings = 1/6 × 3/7 = 1/14
Remaining = 3/7 - 1/14 = 5/14
5/14 = 25000
Salary = 14/5 × 25000 = Rs. 70000
Q13) 3+32 +33 +.........+312 =?
A. 792044
B. 791022
C. 797160
D. 794028
Explanation:
Q14) A TV set listed at Rs 3200 is sold to a retailer at a successive discount of 25% and 15%. The retailer desires a profit of 20%, after allowing a discount of 10% to the customer. At what price should he list the TV set (in Rs.)?
A. 2720
B. 2448
C. 2040
D. 2133
Explanation:
Q15) What is the smallest 4-digit number that is divisible by 12, 13 and 15?
A.1990
B.1560
C.2100
D. 1150
Explanation: The smallest number that is divisible by 12, 13 and 15 can be derived from = LCM of (12, 13, 15)
LCM = 780, which is a 3-digit number.
Since the answer should be a 4-digit number.
We can take the 2nd multiple of 780, which is 1560 and that is the smallest 4-digit number that is divisible by 12, 13 and 15. |
# Square Root of 9 + Solution With Free Steps
9 is a perfect square and its square root is 3. Many people already know the square root of 9 but the article provides detailed steps. The long division method is used to prove that the square root of 9 is 3. The approximation method is also used for this purpose.
In this article, we will analyze and find the square root of 9 using various mathematical techniques, such as the approximation method and the long division method.
## What Is the Square Root Of 9?
The square root of the number 9 is 3.
The square root can be defined as the quantity that can be doubled to produce the square of that similar quantity. In simple words, it can be explained as:
√9 = √(3 x 3)
√9 = √(3)$^2$
√9 = ±3
The square can be canceled with the square root as it is equivalent to 1/2; therefore, obtaining 3. Hence 3 is 9’s square root. The square root generates both positive and negative integers.
## How To Calculate the Square Root of 9?
You can calculate the square root of 9 using any of two vastly used techniques in mathematics; one is the Approximation technique, and the other is the Long Division method.
The symbol √ is interpreted as 9 raised to the power 1/2. So any number, when multiplied by itself, produces its square, and when the square root of any squared number is taken, it produces the actual number.
Let us discuss each of them to understand the concepts better.
### Square Root of 9 by Long Division Method
The process of long division is one of the most common methods used to find the square roots of a given number. It is easy to comprehend and provides more reliable and accurate answers. The long division method reduces a multi-digit number to its equal parts.
Learning how to find the square root of a number is easy with the long division method. All you need are five primary operations- divide, multiply, subtract, bring down or raise, then repeat.
Following are the simple steps that must be followed to find the square root of 9 using the long division method:
### Step 1
First, write the given number 9 in the division symbol, as shown in figure 1.
### Step 2
Starting from the right side of the number, make a pair of the number 9 into pairs such as 09.
### Step 3
Now divide the digit 9 by a number, giving a number either 9 or less than 9. Therefore, in this case, the remainder is zero, whereas the quotient is 3.
### Step 4
This results in 9=9. The remainder obtained is 0. Move the next pair of zeros down and repeat the same process.
### Step 5
Keep on repeating the same steps till the zero remainder is obtained or if the division process continues infinitely, solve to two decimal places.
### Step 6
The resulting quotient y is the square root of 9. Figure 1 given below shows the long division process in detail:
### Important points
• The number 9 is a perfect square.
• The number 9 is a rational number.
• The number 9 can be split into its prime factorization.
## Is Square Root of 9 a Perfect Square?
The number 9 is a perfect square. A number is a perfect square if it splits into two equal parts or identical whole numbers. If a number is a perfect square, it is also rational.
A number expressed in p/q form is called a rational number. All the natural numbers are rational. A square root of a perfect square is a whole number; therefore, a perfect square is a rational number.
A number that is not a perfect square is irrational as it is a decimal number. As far as 9 is concerned, it is a perfect square. It can be proved as below:
Factorization of 9 results in 3 x 3 that can also be expressed as 3$^2$.
Taking the square root of the above expression gives:
= √(3$^2$)
= (3$^2$)$^{1/2}$
= 3
This shows that 9 is a perfect square and a rational number.
Therefore the above discussion proves that the square root of 9 is equivalent to 3.
Images/mathematical drawings are created with GeoGebra. |
We will learn addition using Abacus.
We know how to do addition.
For example;
Here, we will add the numbers together 11 + 21 = 32
Here, we will subtract 17 + 22 = 39
We generally add both the numbers together but, know we will learn how to add the numbers using abacus.
1. Add 12 and 24 using Abacus.
This abacus is showing the number 12.
This abacus is showing the number 24.
For adding 12 and 24, count ones together and get 6 ones.
Adding tens together, we get 3 tens.
Thus, sum of 12 and 24 becomes 36.
Therefore, 12 + 24 = 36
2. Find the sum of 14 and 15 using Abacus.
This abacus is showing the number 14.
This abacus is showing the number 15.
For adding 14 and 15, count ones together and get 9 ones.
Adding tens together, we get 2 tens.
Thus, sum of 14 and 15 becomes 29.
Therefore, 14 + 15 = 29
3. Add 46 and 31 Using Abacus.
This abacus is showing the number 46.
This abacus is showing the number 31.
For adding 46 and 31, count ones together and get 7 ones.
Adding tens together, we get 7 tens.
Thus, sum of 46 and 31 becomes 77.
Therefore, 46 + 31 = 77
1. Add the following using Spike Abacus:
(i) 14 + 13 = _____
(ii) 15 + 23 = _____
(iii) 28 + 31 = _____
(iv) 41 + 35 = _____
(v) 81 + 7 = _____
(vi) 66 + 33 = _____
(vii) 75 + 21 = _____
(viii) 50 + 20 = _____
(ix) 33 + 55 = _____
(x) 24 + 63 = _____
1. (i) 27
(ii) 38
(iii) 59
(iv) 76
(v) 88
(vi) 99
(vii) 96
(viii) 70
(ix) 88
(x) 87
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# How do you integrate f(x)=e^(3x)/(1+e^x) using the quotient rule?
Oct 29, 2017
$I = \int \frac{{e}^{3 x} \cdot \mathrm{dx}}{1 + {e}^{x}} = \frac{1}{2} \cdot {e}^{2 x} - {e}^{x} + L n \left({e}^{x} + 1\right) + C$
#### Explanation:
$I = \int \frac{{e}^{3 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$
=$\int \frac{{e}^{3 x} \cdot \left({e}^{x} + 1\right) \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$
=$\int \left({e}^{3 x} + {e}^{2 x}\right) \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$
=$\left({e}^{3 x} + {e}^{2 x}\right) \cdot - \frac{1}{{e}^{x} + 1}$+$\int \frac{\left(3 {e}^{3 x} + 2 {e}^{2 x}\right) \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$
=$- \frac{{e}^{3 x} + {e}^{2 x}}{{e}^{x} + 1}$+$3 \cdot \int \frac{{e}^{3 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$+int (2e^(2x)*dx)/(1+e^x)-2C
=$- \frac{{e}^{2 x} \cdot \left({e}^{x} + 1\right)}{{e}^{x} + 1}$+$3 I$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$
=$- {e}^{2 x} + 3 I$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}} - 2 C$
Hence $- 2 I$=$- {e}^{2 x}$+$\int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$-$2 C$
$J = \int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{1 + {e}^{x}}$
=$\int \frac{2 {e}^{2 x} \cdot \left({e}^{x} + 1\right) \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$
=$\int \left(2 {e}^{2 x} + 2 {e}^{x}\right) \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{1 + {e}^{x}} ^ 2$
=$\left(2 {e}^{2 x} + 2 {e}^{x}\right) \cdot - \frac{1}{{e}^{x} + 1}$+$\int \frac{\left(4 {e}^{2 x} + 2 {e}^{x}\right) \cdot \mathrm{dx}}{{e}^{x} + 1}$
=$- \frac{2 {e}^{2 x} + 2 {e}^{x}}{{e}^{x} + 1}$+$2 \cdot \int \frac{2 {e}^{2 x} \cdot \mathrm{dx}}{{e}^{x} + 1}$+$2 \cdot \int \frac{{e}^{x} \cdot \mathrm{dx}}{{e}^{x} + 1}$
=$\frac{- 2 {e}^{x} \cdot \left({e}^{x} + 1\right)}{{e}^{x} + 1} + 2 J + 2 L n \left({e}^{x} + 1\right)$
=$- 2 {e}^{x} + 2 J + 2 L n \left({e}^{x} + 1\right)$
So,
$- J = - 2 {e}^{x} + 2 L n \left({e}^{x} + 1\right)$ or,
$J = 2 {e}^{x} - 2 L n \left({e}^{x} + 1\right)$
Thus,
$- 2 I$=$- {e}^{2 x} + J - 2 C$
=$- {e}^{2 x} + 2 {e}^{x} - 2 L n \left({e}^{x} + 1\right) - 2 C$ or,
$I$=$\frac{1}{2} \cdot {e}^{2 x} - {e}^{x} + L n \left({e}^{x} + 1\right) + C$ |
Strand: GEOMETRY (8.G)
Understand congruence and similarity using physical models, transparencies, or geometry software (Standards 8.G.1-5). Understand and apply the Pythagorean Theorem and its converse (Standards 8.G.6-8). Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres (Standard 8.G.9).
Standard 8.G.5
Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so.
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The goal of this task is to provide an opportunity for students to apply a wide range of ideas from geometry and algebra in order to show that a given quadrilateral is a rectangle.
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The goal of this task is to give students experience applying and reasoning about reflections of geometric figures using their growing understanding of the properties of rigid motions.
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The goal of this task is to use rigid motions to establish some fundamental results about angles made by intersecting lines. Both vertical angles and alternate interior angles are treated.
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The purpose of the task is to help students transition from the informal notion of congruence as "same size, same shape" that they learn in elementary school and begin to develop a definition of congruence in terms of rigid transformations.
• Similar Triangles I
The goal of this task is to prepare students for the angle-angle criterion for triangle similarity. Since the sum of the three angles in a triangle is always 180 degrees, having two pairs of congruent corresponding angles in two triangles tells us that the third pair of corresponding angles is also congruent.
• Similar Triangles II
The goal of the task is to provide an informal argument for the AA criterion for triangle similarity, appropriate for an 8th grade audience.
• Street Intersections
The purpose of this task is to apply facts about angles (including congruence of vertical angles and alternate interior angles for parallel lines cut by a transverse) in order to calculate angle measures in the context of a map.
• Tile Patterns I: octagons and squares
This task aims at explaining why four regular octagons can be placed around a central square, applying student knowledge of triangles and sums of angles in both triangles and more general polygons.
• Tile Patterns II: hexagons
In this task one of the most important examples of a tiling, with regular hexagons, is studied in detail. This provides students an opportunity to use what they know about the sum of the angles in a triangle and also the sum of angles which make a line.
• Triangle's Interior Angles
This problem has students argue that the interior angles of the given triangle sum to 180 degrees, and then generalize to an arbitrary triangle via an informal argument. The original argument requires students to make use the angle measure of a straight angle, and about alternate interior angles formed by a transversal cutting a pair of parallel lines.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Lindsey Henderson and see the Mathematics - Secondary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200. |
# NCERT Solutions for Class 8 Maths Chapter 4: Practical Geometry
The NCERT Solutions for Class 8 Maths Chapter 4 on Practical Geometry offer valuable assistance to students, facilitating a comprehensive understanding of the subject and aiding them in achieving high marks in examinations. These solutions feature meticulous, step-by-step explanations for all problems found in Chapter 4 of the Class 8 NCERT Textbook, ensuring that students can grasp the intricacies of Practical Geometry.
## Access Answers to NCERT Solutions for Class 8 Maths Chapter 4: Practical Geometry
### Exercise 4.1
Question 1 :
(i) Quadrilateral ABCD AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm AD = 6 cm AC = 7 cm
(ii) Quadrilateral JUMP JU = 3.5 cm
UM = 4 cm MP = 5 cm PJ = 4.5 cm PU = 6.5 cm
(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5
(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
(i)
The rough sketch of the quadrilateral ABCD can be drawn as follows.
(1) ∆ABC can be constructed by using the given measurements as follows.
(2) Vertex D is 6 cm away from vertex A. Therefore, while taking A as the centre, draw an arc of radius 6 cm.
(3) Taking C as the centre, draw an arc of radius 4 cm, cutting the previous arc at point D. Joint D to A and C.
(ii)
The rough sketch of the quadrilateral JUMP can be drawn as follows.
(1) ∆ JUP can be constructed by using the given measurements as follows.
(2) Vertex M is 5 cm away from vertex P and 4 cm away from vertex U. Taking P and U as centres, draw arcs of radii 5 cm and 4 cm, respectively. Let the point of intersection be M.
(3) Join M to P and U.
(iii)
We know that opposite sides of a parallelogram are equal in length, and also, these are parallel to each other.
i.e., ME = OR, MO = ER
The rough sketch of the parallelogram MORE can be drawn as follows.
(1) ∆ EOR can be constructed by using the given measurements as follows.
(2) Vertex M is 4.5 cm away from vertex O and 6 cm away from vertex E. Therefore, while taking O and E as centres, draw arcs of 4.5 cm radius and 6 cm radius, respectively. These will intersect each other at point M.
(3) Join M to O and E.
MORE is the required parallelogram.
(iv)
We know that all sides of a rhombus are of the same measure. Hence, BE = ES = ST = TB
The rough sketch of the rhombus BEST can be drawn as follows.
(1) ∆ BET can be constructed by using the given measurements as follows.
(2) Vertex S is 4.5 cm away from vertex E and also from vertex T. Therefore, while taking E and T as centres, draw arcs of 4.5 cm radius, which will intersect each other at point S.
(3) Join S to E and T.
NCERT Solution For Class 8 Maths Chapter 4 Image
BEST is the required rhombus.
### Exercise 4.2
Question 1 :
(i) Quadrilateral LIFT LI = 4 cm
IF = 3 cm TL = 2.5 cm LF = 4.5 cm
IT = 4 cm
(ii) Quadrilateral GOLD OL = 7.5 cm
GL = 6 cm GD = 6 cm LD = 5 cm OD = 10 cm
(iii) Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm
(i) A rough sketch of the quadrilateral LIFT can be drawn as follows.
(1) ∆ ITL can be constructed by using the given measurements as follows.
(2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. ∴ while taking L and I as centres, draw arcs of 4.5 cm radius and 3 cm radius, respectively, which will intersect each other at point F.
(3) Join F to T and F to I.
(ii) The rough sketch of the quadrilateral GOLD can be drawn as follows.
(1) ∆ GDL can be constructed by using the given measurements as follows.
(2) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L. Therefore, while taking D and L as centres, draw arcs of 10 cm radius and 7.5 cm radius, respectively. These will intersect each other at point O.
(3) Join O to G and L.
(iii) We know that the diagonals of a rhombus always bisect each other at 90º.
Let us assume that these are intersecting each other at point O in this rhombus. Hence, EO = OD = 3.25 cm
The rough sketch of the rhombus BEND can be drawn as follows.
(1) Draw a line segment BN of 5.6 cm, and also draw its perpendicular bisector. Let it intersect the line segment BN at point O.
(2) Taking O as the centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at points D and E.
(3) Join points D and E to points B and N.
### Exercise 4.3
Question 1 :
(i) Quadrilateral MORE MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
(ii) Quadrilateral PLAN PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
(iii) Parallelogram HEAR HE = 5 cm
EA = 6 cm
∠R = 85°
(iv) Rectangle OKAY
OK = 7 cm KA = 5 cm
(i) Rough Figure:
(1) Draw a line segment MO of 6 cm and an angle of 105º at point O. As vertex R is 4.5 cm away from the vertex O, cut a line segment OR of 4.5 cm from this ray.
(2) Again, draw an angle of 105º at point R.
(3) Draw an angle of 60º at point M. Let this ray meet the previously drawn ray from R at point E.
(ii) The sum of the angles of a quadrilateral is 360°. In quadrilateral PLAN,
∠P + ∠L + ∠A + ∠N = 360° 90° + ∠L + 110° + 85° = 360°
285° + ∠L = 360°
∠L = 360° − 285° = 75°
Rough Figure:
(1) Draw a line segment PL of 4 cm and draw an angle of 75º at point L. As vertex A is 6.5 cm away from vertex L, cut a line segment LA of 6.5 cm from this ray.
(2) Again, draw an angle of 110º at point A.
(3) Draw an angle of 90º at point P. This ray will meet the previously drawn ray from A at point N.
(iii) Rough Figure:
(1) Draw a line segment HE of 5 cm and an angle of 85º at point E. As vertex A is 6 cm away from vertex E, cut a line segment EA of 6 cm from this ray.
(2) Vertex R is 6 cm and 5 cm away from vertex H and A, respectively. By taking radii as 6 cm and 5 cm, draw arcs from points H and A, respectively. These will intersect each other at point R.
(3) Join R to H and A.
(iv) Rough Figure:
(1) Draw a line segment OK of 7 cm and an angle of 90º at point K. As vertex A is 5 cm away from vertex K, cut a line segment KA of 5 cm from this ray.
(2) Vertex Y is 5 cm and 7 cm away from vertex O and A, respectively. By taking
radii as 5 cm and 7 cm, draw arcs from points O and A, respectively. These will intersect each other at point Y.
(3) Join Y to A and O.
### Exercise 4.4
Question 1 :
(i) Quadrilateral DEAR DE = 4 cm
EA = 5 cm AR
= 4.5 cm
∠E = 60°
∠A = 90°
(ii) Quadrilateral TRUE TR = 3.5 cm
RU = 3 cm UE = 4 cm
∠R = 75°
∠U = 120°
(i) Rough Figure:
(1) Draw a line segment DE of 4 cm and an angle of 60º at point E. As vertex A is 5 cm away from vertex E, cut a line segment EA of 5 cm from this ray.
(2) Again, draw an angle of 90º at point A. As vertex R is 4.5 cm away from vertex A, cut a line segment RA of 4.5 cm from this ray.
(3) Join D to R.
(ii) Rough Figure:
(1) Draw a line segment RU of 3 cm and an angle of 120º at point U. As vertex E is 4 cm away from vertex U, cut a line segment UE of 4 cm from this ray.
(2) Next, draw an angle of 75º at point R. As vertex T is 3.5 cm away from vertex R, cut a line segment RT of 3.5 cm from this ray.
(3) Join T to E.
### Exercise 4.5
Question 1 :
Draw the following:
1. The square READ with RE = 5.1 cm
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of length 5 cm and 4 cm
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.
1. All the sides of a square are of the same measure, and also, all the interior angles of a square are 90º measure. Therefore, the given square READ can be drawn as follows.
Rough Figure:
(1) Draw a line segment RE of 5.1 cm and an angle of 90º at points R and E.
(2) As vertex A and D are 5.1 cm away from vertex E and R, respectively, cut line segments EA and RD, each of 5.1 cm from these rays.
(3) Join D to A.
2. In a rhombus, diagonals bisect each other at 90º. ∴, the given rhombus ABCD can be drawn as follows.
Rough Figure:
(1) Draw a line segment AC of 5.2 cm and draw its perpendicular bisector. Let it intersect the line segment AC at point O.
(2) Draw arcs of 6.4/2 = 3.2 on both sides of this perpendicular bisector. Let the arcs intersect the perpendicular bisector at points B and D.
(3) Join points B and D with points A and C.
ABCD is the required rhombus.
3. Opposite sides of a rectangle have lengths of the same measure, and also, all the interior angles of a rectangle are 90º measure. The given rectangle ABCD may be drawn as follows.
Rough figure:
(1) Draw a line segment AB of 5 cm and an angle of 90º at points A and B.
(2) As vertex C and D are 4 cm away from vertex B and A, respectively, cut line segments AD and BC, each of 4 cm, from these rays.
(3) Join D to C.
ABCD is the required rectangle.
4. Opposite sides of a parallelogram are equal and parallel to each other. The given parallelogram OKAY can be drawn as follows.
Rough Figure:
(1) Draw a line segment OK of 5.5 cm and a ray at point K at a convenient angle.
(2) Draw a ray at point O parallel to the ray at K. As the vertices A and Y are 4.2 cm away from the vertices K and O, respectively, cut line segments KA and OY, each of 4.2 cm, from these rays.
(3) Join Y to A.
OKAY is the required rectangle.
The NCERT solution for class 8 Chapter 4: Practical Geometry is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tacking more difficult concepts in their further education.
Yes, the NCERT solution for class 8 Chapter 4: Practical Geometry is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Practical Geometry ally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for class 8 Maths Chapter 4 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for class 8 Maths Chapter 4 : Practical Geometry as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for class 8 Maths Chapter 4 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution. Also, you can make Practical Geometry al notes and jot down the important concepts for your understanding.
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2017 AMC 8 Problems/Problem 23
Problem
Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$
Solution
It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$. In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$, we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$, so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$. The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: $$\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}$$ We know that $x,x+5,x+10,x+15$ are all factors of $60$, therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. $$\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}$$
Video Solution
https://youtu.be/EiN5oMJRL-8 - Happytwin |
# 2016 AMC 8 Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Find the value of the expression $$100-98+96-94+92-90+\cdots+8-6+4-2.$$$\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$
## Solutions
### Solution 1
We can group each subtracting pair together: $$(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).$$ After subtracting, we have: $$2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).$$ There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{\textbf{(C) }50}$
### Solution 2
Since our list does not end with one, we divide every number by 2 and we end up with $$50-49+48-47+ \ldots +4-3+2-1$$ We can group each subtracting pair together: $$(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).$$ There are now $25$ pairs of numbers, and the value of each pair is $1$. This sum is $25$. However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{\textbf{(C) }50}$
## Video Solution
A solution so simple a 12-year-old made it!
~Elijahman~
~savannahsolver
~ pi_is_3.14 |
Eratosuite
The other day in class, we were looking at recurrence relations and how to solve them with the characteristic equation. Among the examples I gave was a recurrence that seemed to give a good number of primes numbers. Are there many of those? How do we find them?
Well, we need two things: a method of testing if a given number is prime, and a method of generating recurrences—parameters and initial conditions. Well, make that three: we also need to generate the suite generated by the recurrence relation. Let’s go!
Let us first take time to recall what a recurrence relation is, and, more specifically, what a linear homogeneous recurrence is. A recurrence relation is a recursive expression for the terms of a sequence, in the form of
$t_n=a_1 t_{n_1}+a_2 t_{n-2}+\cdots+ a_k t_{n-k}$,
where the $a_i$ are the coefficients (not necessarily natural numbers), and the $t_i$ are the $i$th term in the sequence. The best known such recurrence relation is Fibonacci’s series:
$F_1=1$,
$F_2=1$,
$F_n=F_{n-1}+F_{n-2}$.
Fibonacci’s series defines a term using the two previous terms, therefore its order is 2. If it’d defined the terms using, say the $k$ previous terms, then its order would be $k$. For a recurrence relation of order $k$, you will need $k$ initial conditions, that is, terms that are at the beginning of the sequence and aren’t defined relative to the others. If we’d put $F_1=3$ and $F_2=17$, say, then Fibonacci’s recurrence relation generates an entirely different sequence.
Let’s look at another example. Say:
$t_1=1$,
$t_2=2$,
$t_n=2t_{n-1}+3t_{n-2}$.
The first two terms are the initial conditions, so the sequences starts 1, 2. The third term is $t_3=2t_2+3t_1=2\times{2}+3\times{1}=7$. The fourth term is $t_4=2t_3+3t_2=2\times{7}+3\times{2}=20$. And so on, infinitely.
*
* *
OK, so what interests me is finding series that give, potentially, a lot of primes numbers. We need a way of checking the terms for primality. But we also know that testing for primality is hard in general, so we’ll settle for some trade-off. One possible trade-off is to establish a boolean table where t[i] is true when i is prime and false otherwise. Fortunately, we know of a super efficient way of filling such a table!
The method is called Eratosthenes’ sieve (or some other minor variant such as the sieve of Eratosthenes). The method is simple. We start with a table where all the entries are true (except for entry zero and one). Then we find the next entry that is true. It’ll be a prime number. We then cross-out all that is a multiple of that number (except the number itself, evidently). We then find the next entry that is still true, that’ll be our next prime number. We then cross out all of its multiples from the table. We repeat until we reach numbers about the size of the square root of the table: at that point, we’re done, all entries still true are prime numbers.
A C++ implementation would be something like this:
void eratosthenes(std::vector<bool> & primes, const options_t & options)
{
primes=std::vector<bool>(options.nb_primes,true); // safe up to c++11
primes[0]=primes[1]=false;
size_t current=2;
while (current*current<primes.size())
{
// mark all multiples of current
for (size_t mark=2*current; mark<primes.size(); mark+=current)
primes[mark]=false;
// find the next 'true'
do
{
current++;
}
while (current<primes.size() && !primes[current]);
}
}
Let us note the use of the “option pattern”, where a configuration is passed throughout the program using a reference to a structure (or class) holding the configuration. This avoids passing a lot of parameters to each call, especially when non mission-specific items are considered. It’s not conceptually the problem of the sieve to know if we’re in debug/verbose mode or not, so we might as well hide all these details into an
options_t structure. That “option pattern” isn’t found in the GoF,
and is neither state nor memento, although
both come close, kind of.
*
* *
To search for interesting recurrences, we must search for combinations of initial conditions and coefficients. We could generate them randomly, and after a large enough number stop and pick the best sequences, those that have the most prime numbers in them. We could also try exhaustively all combinations. This suggests nested loops, were, for all possibilities for the first initial condition we try all possibilities for the second, and for all possibilities for the second, the third, then for all possibilities of initial conditions, we try all possibilities for the first coefficients, and so on, and so on.
If we have a fixed depth, say two initial conditions and two coefficients, we end up with four nested loops. If we want to put one more term in the recurrence, we must now have six nested loops, three for the initial conditions and three for the coefficients. This approach is most inconvenient because we must have special cases for every possible recurrence order. We must find something more general… and, if possible, not too painful to code!
Well, we can look at the initial conditions and at the coefficients, as counters. If we limit the initial conditions values to 0 to 9, say, then all possible combinations will be like counting from 00 to 99: 00, 01, 02, …, 08, 09, 10,… We will take the same approach but count “digits” from a lower to a upper bound rather than from 0 to 9. But we will proceed in the same way. We start at the lower bound, and increment the first value until it exceeds the upper bound. It is then reset to the lower bound, and a “carry” is pushed to the next value that is incremented. If the next value exceeds its upper bound, it is also reset and a carry is propagated to the next. That’s exactly what we do when we do +1 on a (base 10) number:if the first digit becomes 10, we write 0, and propagate the carry to the next. If the carry is zero (no carry is pushed) then we’re done.
The other nice thing with this approach is that we don’t really need to know before hand how many combinations there are. When a wrap-around occurs, the counter reverts to its initial configuration, with all “digits” being set at their lower bounds. It suffices then to do +1 until we come back to “all zeros”.
The following piece of code implements the counter:
void start_counter(std::vector<int64_t> & counter,
const std::vector<std::pair<int64_t,int64_t>> & bounds)
{
counter.resize(bounds.size());
for (size_t i=0;i<bounds.size();i++)
counter[i]=bounds[i].first;
}
////////////////////////////////////////
void increment(std::vector<int64_t> & counter,
const std::vector<std::pair<int64_t,int64_t>> & bounds)
{
int64_t c=1;
size_t i=0;
while (c && (i<bounds.size()))
{
counter[i]+=c;
if (counter[i]>bounds[i].second)
{
counter[i]=bounds[i].first;
i++;
}
else
c=0;
}
}
*
* *
Once we’re able to generate all possible initial conditions and coefficient combinations, we need a way of generating the series, and evaluate its score. If it has too few different values it’s bad, it doesn’t have many primes, it’s bad. Something like:
int64_t dot( const std::vector<int64_t> & a,
const std::vector<int64_t> & b )
{
// computes dot product of two vectors
int64_t s=0;
for (size_t i=0;i<a.size();i++)
s+=a[i]*b[i];
return s;
}
////////////////////////////////////////
size_t eval_series( const std::vector<int64_t> & initial_conditions,
const std::vector<int64_t> & parameters,
const std::vector<bool> & primes,
const options_t & options)
{
std::vector<int64_t> terms(initial_conditions);
std::vector<int64_t> state=initial_conditions;
for (size_t i=state.size();i<options.nb_terms;i++)
{
int64_t v=dot(state,parameters);
state.erase(state.begin());
state.push_back(v);
terms.push_back(v);
}
// checks
// very few values?
std::set<uint64_t> s(terms.begin(),terms.end());
if (s.size() < terms.size()/4)
return 0;
// count primes
size_t p=0;
for (int64_t x:s)
p+=is_prime(x,primes);
return p;
}
*
* *
All that’s left is to explore exhaustively all combinations, evaluate all series, and pick the best (at least as measured by our naïve score function). That would give something like:
void search( const std::vector<bool> primes,
const options_t & options)
{
std::vector<int64_t> initial_conditions, first_initial_conditions;
std::vector<int64_t> parameters, first_parameters;
std::vector<int64_t> best_initial_conditions;
std::vector<int64_t> best_parameters;
size_t best_score=0;
start_counter(first_initial_conditions,options.initial_conditions_bounds);
start_counter(first_parameters,options.parameter_bounds);
initial_conditions=first_initial_conditions;
do
{
parameters=first_parameters;
do
{
if (options.verbose)
show_series(initial_conditions,
parameters,
primes,
options);
size_t this_score=eval_series(initial_conditions,
parameters,
primes,
options);
if (this_score>best_score)
{
best_score=this_score;
best_initial_conditions=initial_conditions;
best_parameters=parameters;
}
increment(parameters,options.parameter_bounds);
}
while (parameters!=first_parameters);
increment(initial_conditions,options.initial_conditions_bounds);
}
while (initial_conditions!=first_initial_conditions);
show_series(best_initial_conditions,
best_parameters,
primes,
options);
}
Running the program produces the output:
.../eratostuite> eratosuite -p 10000 --bounds -6,6,-6,6,-6,6 -f 50
Score: 33/50
initial conditions: -5 5 1
parameters : -1 1 1
-5 5 1 11 7 17 13 23 19 29 25 35 31 41 37 47 43 53 49 59 55 65 61 71 67 77 73 83 79 89 85 95 91 101 97 107 103 113 109 119 115 125 121 131 127 137 133 143 139 149
I have used Boost’s program_options framework to parse the arguments. The -p options specifies how large the prime look-up table will be, and the --bounds specifies, for each coefficient, the range of value to be searched (for now, I consider them as integer). I also put an option to define the bounds for the initial conditions, but if it’s not specified, then the bounds for the coefficients are assumed.
*
* *
The main problem with this program is limited precision arithmetic. I used int64_t to use 64-bits integers, but some series will grow very fast and, due to limited precision, give erroneous terms. The next step would be to replace int64_t by some arbitrary-precision integer object. In turn, it will likely make testing for prime numbers a lot more difficult.
Another problem with the counter-based approach is that it does not lend itself that easily to parallelism. Using threads (say, Boost’s), we would need to schedule sequences to be checked in parallel, first generating a batch of configurations from the counters, then dispatching. Using OpenMP would not make things a lot easier. |
# Five Bad Oranges Are Accidently Mixed with 20 Good Ones. If Four Oranges Are Drawn One by One Successively with Replacement, Then Find the Probability Distribution of Number of Bad Oranges Drawn. - Mathematics
Sum
Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.
#### Solution
Let be the random variable denoting the number of bad oranges drawn.
P (getting a good orange) = $\frac{20}{25} = \frac{4}{5}$
P (getting a bad orange) = $\frac{5}{25} = \frac{1}{5}$
The probability distribution of X is given by
X 0 1 2 3 4 P(X) $\left( \frac{4}{5} \right)^4$ =$\frac{256}{625}$ $^{4}{}{C}_1 \left( \frac{4}{5} \right)^3 \left( \frac{1}{5} \right)$ =$\frac{256}{625}$ $^{4}{}{C}_2 \left( \frac{4}{5} \right)^2 \left( \frac{1}{5} \right)^2$ =$\frac{96}{625}$ $^{4}{}{C}_3 \left( \frac{4}{5} \right) \left( \frac{1}{5} \right)^3$ =$\frac{16}{625}$ $\left( \frac{1}{5} \right)^4$ =$\frac{1}{625}$
Mean of X is given by
$\overline{X} = \sum P_i X_i$
$= 0 \times \frac{256}{625} + 1 \times \frac{256}{625} + 2 \times \frac{96}{625} + 3 \times \frac{16}{625} + 4 \times \frac{1}{625}$
$= \frac{1}{625}\left( 256 + 192 + 48 + 4 \right)$
$= \frac{4}{5}$
Variance of X is given by $\text{ Var } (X) = \sum P_i {X_i}^2 - \left( \sum P_i X_i \right)^2$
$= 0 \times \frac{256}{625} + 1 \times \frac{256}{625} + 4 \times \frac{96}{625} + 9 \times \frac{16}{625} + 16 \times \frac{1}{625} - \left( \frac{4}{5} \right)^2$
$= \frac{1}{625}\left( 256 + 384 + 144 + 16 \right) - \frac{16}{25}$
$= \frac{800}{625} - \frac{16}{25}$
$= \frac{400}{625}$
$= \frac{16}{25}$
Thus, the mean and vairance of the distribution are $\frac{4}{5}$ and $\frac{16}{25}$ , respectively.
Concept: Random Variables and Its Probability Distributions
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#### APPEARS IN
RD Sharma Class 12 Maths
Chapter 33 Binomial Distribution
Exercise 33.2 | Q 27 | Page 26 |
# Parallel and Perpendicular Lines
Written by
Malcolm McKinsey
Fact-checked by
Paul Mazzola
## Coplanar lines
Any two flat objects sharing space on a plane surface are said to be coplanar. Three types of lines that are coplanar are parallel lines, perpendicular lines, and transversals. These all exist in a single plane, unlike skew lines (which exist in multiple planes). Any two lines in a plane must necessarily either be parallel or intersect.
## Parallel lines
Parallel lines never meet, despite co-existing in a plane and continuing in two directions forever. Measure the distance between the two lines anywhere along their lengths, as many times as you like; they will always be the same distance apart.
### Slope of parallel lines
If we apply coordinate geometry to parallel lines, we can see through the parallel lines equations that parallel lines will have the same slope:
The lower line intercepts the xx axis at 0.5, at (0.5, 0) and the y axis at (0, -1). Slope is rise (change in y-value) over run (change in x-value), so for the lower line:
The upper line has an x-intercept of -1.5 (-1.5, 0) and a y-intercept of 3 (0, 3), so its slope is:
With positive slopes, the two values increase together (x-values increase as y-values increase).
### Parallel line examples
Examples of parallel lines outside of coordinate graphs are everywhere. Floor boards, window blinds, notebook paper lines, painted line segments in a parking lot -- all parallel lines. Nobody expects you to apply slope formulas to diagonal parking lines, but you can find coplanar parallel lines in your everyday life.
## Perpendicular lines
Coplanar lines that are not parallel must intersect or cross each other. They can intersect at any angle, but when the lines intersect at exactly 90° they are perpendicular lines. Perpendicular lines create four right angles at their point of intersection.
### Slope of perpendicular lines
When plotting perpendicular lines on a coordinate graph, you need to consider two ideas:
• The slopes will be opposites
• The slopes will be reciprocals
Let's take the first requirement: opposite slopes. We'll keep one of our earlier lines with a positive slope of 2, and then show a new, second line with a negative slope of -2:
Now the lines are crossing, with our new line showing x-values increasing as y-values are decreasing. Negative slopes have that inverse relationship between the x-values and y-values. But our intersecting lines are not perpendicular, yet.
The slopes must be reciprocal, so instead of simply having one with a positive slope of 2 and one with a negative slope of -2, we need the second line to be $-\frac{1}{2}$ (the reciprocal of $\frac{2}{1}$):
Like parallel lines, examples of perpendicular lines surround us, in walls meeting floors and ceilings, in floor tiles, in bricks in walls, in window grilles. The margin line on a sheet of notebook paper is perpendicular to the parallel writing lines.
## Parallel and perpendicular line equations
Can you tell if these lines are perpendicular or parallel given these equations? If the slopes are equal, the lines will be parallel. If the slopes are opposite reciprocals of each other, the lines will be perpendicular. Try these three examples:
• Line F is $y=\frac{3}{4}x$
• Line O is $y=\frac{-4}{3}x$
• Line X is $y=\frac{6}{8}x+1$
Lines F and X are parallel, separated only by a difference of 1. The fraction $\frac{6}{8}$ simplifies to $\frac{3}{4}$; adding the 1 moves Line X one unit away from Line F. Line O is perpendicular to Lines F and X because it has the negative reciprocal of $\frac{3}{4}$.
## Transverse lines or transversals
Coplanar lines that intersect other coplanar lines are called transverse lines or transversals. A transversal crossing two parallel lines creates eight angles, which can be viewed and compared many ways:
1. Two pairs of consecutive interior angles
2. Two pairs of alternate interior angles
3. Two pairs of alternate exterior angles
4. Two pairs of consecutive exterior angles
5. Four pairs of corresponding angles
Transverse lines are everywhere in nature and human-made objects. Street maps show parallel, perpendicular and transverse lines. Mown hay lies in bundles of transverse lines, with any three strands being coplanar. A handful of casually tossed pencils will crisscross as transverse lines.
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# Thinking Mathematically The Irrational Numbers. The set of irrational numbers is the set of number whose decimal representations are neither terminating.
## Presentation on theme: "Thinking Mathematically The Irrational Numbers. The set of irrational numbers is the set of number whose decimal representations are neither terminating."— Presentation transcript:
Thinking Mathematically The Irrational Numbers
The set of irrational numbers is the set of number whose decimal representations are neither terminating nor repeating. One of the most well known irrational numbers is the ratio between the circumference and diameter of a circle known as “pi” and written π. 3.14159
Square Roots The principal “square root” of a positive number, n, is the positive number that when multiplied times itself produces n. This number is written n. The square root of zero is zero. Sometimes square roots are whole numbers. For example since 6 x 6 = 36, 36 = 6. If a square root is not a whole number, then it is an irrational number and cannot be written as a ratio of integers.
The Product Rule for Square Roots If a and b represent nonnegative numbers, then √(ab) = a b and a b = (ab). Example: 2 5 = (25) = 10 7 7 = (77) = 49 = 7 6 12 = (612) = 72 = 36 2 = 6 2
The Quotient Rule for Square Roots If a and b represent nonnegative real numbers and b ≠ 0, then The quotient of two square roots is the square root of the quotient.
Adding and Subtracting Square Roots a c + b c = (a + b) c a c - b c = (a - b) c Example: 7 2 + 5 2 = (7 + 5) 2 = 12 2 2 5 - 6 5 = (2 - 6) 5 = -4 5 3 7 + 9 7 - 7 = (3 + 9 - 1) 7 = 11 7
Rationalizing the Denominator The process of rewriting a radical expression to remove the square root from the denominator without changing the value of the expression is called rationalizing the denominator. Example:
Discuss the square root of 2 and the Pythagoreans (500-300 BC) and show it is irrational Other roots (e.g. cube roots) Exponents and roots
Download ppt "Thinking Mathematically The Irrational Numbers. The set of irrational numbers is the set of number whose decimal representations are neither terminating."
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# How do you find the integral of int (x-2)/(x^2+1)?
Feb 5, 2017
$= \frac{1}{2} \ln \left({x}^{2} + 1\right) - \left(2 {\tan}^{- 1} x\right) + C$
#### Explanation:
We can rewrite the integral as two fractions and then integrate teh em separately.
int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx
First integral
$\textcolor{red}{\int \frac{x}{{x}^{2} - 2} \mathrm{dx}}$
using $\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) |$
$\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = 2 x$
$\therefore \textcolor{red}{\int \frac{x}{{x}^{2} - 2} \mathrm{dx} = \frac{1}{2} \ln | {x}^{2} + 1 |}$
since ${x}^{2} + 1 > 0 \forall \in \mathbb{R} \text{ we can omit the modulus sign}$
second integral
color(blue)(int2/(x^2+1)dx
we integrate by substitution
$x = \tan u \implies \mathrm{dx} = {\sec}^{2} u \mathrm{du}$
color(blue)(int2/(x^2+1)dx=2int1/(tan^2u+1)sec^2udu
=color(blue)(2(int1/cancel(sec^2u)cancel(sec^2u)du)=2intdu
$\textcolor{b l u e}{= 2 u = 2 {\tan}^{- 1} x}$
putting the two results together we have:
int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx
$= \textcolor{red}{\frac{1}{2} \ln \left({x}^{2} + 1\right)} - \textcolor{b l u e}{2 {\tan}^{- 1} x} + C$ |
Question
Q&A Session
1. Elements Of Mathematics Class 11 Solutions
Introduction
Mathematics is one of the most essential subjects for students in high school and beyond. It helps you understand the world around you, from basic arithmetic to more complex concepts like calculus. In this article, we will provide you with a few tips on how to conquer mathematics classes 11-12. From understanding basic concepts to practicing problem-solving skills, we’ll equip you with everything you need to ace the exam. So don’t hesitate, start learning today!
Multiplication And Division Of Numbers
In the early grades, students are often taught basic multiplication and division facts. Multiplication and division of numbers can be seen as a way to simplify problems. For example, if you have to find the total number of cookies in a container that has 10 cookies, you can simply multiply the number of cookies in the container by 10 to get the answer (100). Division is just like multiplication, but it works with division of whole numbers rather than just numbers. For example, if you have to divide 3 candy bars among 4 people, you would first break them up into fourths. Then each person would get 1 candy bar.
Adding and subtracting numbers is one of the most basic operations in mathematics. To add two numbers, simply put the first number beside the second number and drag the mouse to combine the two. To subtract two numbers, select the first number and click on the minus sign next to the second number.
Exponents And Logarithms
Exponents and logarithms are two fundamental concepts in mathematics. They are used to calculate things like the speed of a car or the strength of a beam.
A basic understanding of exponents and logarithms is important for students who want to study more advanced mathematics. Exponents are a type of mathematical function that can be used to increase or decrease a number by a small amount. For example, the natural (undecimale) exponential function takes a number as input and returns the number multiplied by 10 raised to that power.
Logarithms are another important concept in mathematics. They are used to calculate things like the distance between two points or the time it takes for something to travel from one point to another. The base (natural) logarithm takes a number as input and converts it into a decimal value. Then, it raises that decimal value to the power of 10. So, if someone wanted to find out how many hours it would take something to travel from New York City to Los Angeles, they would use the base logarithm, which would give them an answer of 6092 hours (6 multiplied by 10 raised to the 2nd power).
In mathematics, square roots and radicals are two important tools that allow for solving equations and inequalities. Knowing how to solve equations and inequalities using square roots and radicals can be very useful in many scenarios in life. Let’s take a look at some examples of where square roots and radicals can be used.
One use of square roots is when solving equations. For example, if we have an equation involving the square root of a number, then we can use the algebraic methods to solve it. In other words, we would use the rules of addition, subtraction, multiplication, and division to solve the equation.
Another use of square roots is when solving inequalities. For example, if we have an inequality involving the square root of a number, then we can use the Method Of Multiplication And Division (also known as Newton’s Method). This method involves solving the inequality by multiplying both sides by a certain value until one side equals zero (or until there are no more digits on either side of the decimal point).
Finally, radicals are also useful in mathematics. For example, radicals can be used when solving systems of linear equations or graphing linear equations on a coordinate plane.
Word Problems In Mathematics Class 11
Word problems in mathematics class can be a challenge for students. In this article, we will discuss some of the most common word problems in mathematics, and provide solutions.
In this first example, Mrs. Jones has 22 candy bars. She wants to give 9 candy bars to each of her 5 children. How many candy bars does she have left?
The answer is 9 candy bars. Mrs. Jones has given away 3 candy bars (or 33% of her total supply).
Conclusion
In this article, we have looked at some of the common questions that are asked in mathematics class 11 and provided solutions to them. Hopefully, by reading through this article you will be able to better understand the concepts that your teacher is trying to teach you and be able to apply those concepts more easily in future classes. We wish you all the best as you continue your journey through school!
2. Mathematics is considered to be one of the most important and challenging subjects in the academic world. Class 11 mathematics is a crucial level as it sets the foundation for higher-level studies. The right guidance and support can make a significant difference in shaping a student’s understanding of this subject.
The Elements Of Mathematics Class 11 Solutions book is an excellent resource for students seeking assistance with their maths curriculum. It covers all concepts in detail, providing students with comprehensive explanations and examples that help them understand each concept thoroughly. The solutions are presented in such a way that they aid students’ learning process by simplifying complex problems into manageable steps.
This book also includes exercises at the end of every chapter which provide ample opportunities for students to practice what they have learned. These exercises are designed to test a student’s understanding, and the answers provided at the end of each exercise help them to evaluate their performance accurately.
3. 🤓 Math can be a tricky subject to master and can often be difficult to understand. That’s why having the right resources can make all the difference. If you’re studying Class 11 Mathematics, then you’ll be glad to know that you can now get the best Elements Of Mathematics Class 11 Solutions at your fingertips.
The Elements Of Mathematics Class 11 Solutions is a comprehensive resource that provides all the necessary materials to help you ace your exams. It includes a detailed step-by-step solution to each chapter of the textbook, along with a detailed explanation of each concept. In addition to that, it also provides sample problems to help you get an in-depth understanding of the concepts.
The Elements Of Mathematics Class 11 Solutions is an invaluable resource for students who are struggling to understand the complex concepts of the subject. Not only does it provide clear explanations, but it also comes with plenty of helpful graphic illustrations and diagrams to help you visualize the concepts better.
For those who are looking to get a competitive edge, the Elements Of Mathematics Class 11 Solutions also provides plenty of practice questions to help you prepare for the exams. It also provides detailed solutions to the tricky problems, so you can be sure to have a good grasp of the concepts.
So if you’re looking for the best resource to help you ace your class 11 Mathematics exams, then the Elements Of Mathematics Class 11 Solutions is the perfect choice for you. With clear explanations, plenty of practice questions, and detailed solutions, you can be sure to get the best grades in your class. 🤩 |
## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 50
#### Answer
The solution of the system of equations is $\left( x,y \right)=\left( \frac{1}{2a},\frac{1}{b} \right)$.
#### Work Step by Step
$4ax+by=3$ (I) $6ax+5by=8$ (II) Now, multiply equation (I) by −5 to obtain; $-20ax-5by=-15$ (III) And add equations (II) and (III): \begin{align} & 6ax+5by-20ax-5by=8-15 \\ & -14ax=-7 \\ & x=\frac{1}{2a} \end{align} Substitute the value of x in the equation (I): \begin{align} & 2+by=3 \\ & by=1 \\ & y=\frac{1}{b} \end{align} Thus, the solution set the system of equations is $\left( x,y \right)=\left( \frac{1}{2a},\frac{1}{b} \right)$.
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_______________________NEWSLETTERS___________________ Module/Unit 1 Module/Unit 2 Module/Unit 3 Module/Unit 4 Topic F Eureka Math Module 1 Tips for Parents Eureka Math Module 2 Tips for Parents Topic B Topic C Topic D Topic F Eureka Math Module 3 Tips for Parents Topic A Topic B Topic C Topic D Module/Unit 5 Module/Unit 6 Module 7 Topic A Topic C Topic D Topic A Topic A Topic C Topic D
_____________________________Activities For Home________________
MODULE 1 Make arrays out of household items (e.g., pennies, beans, blocks) Determine how many items are on each rowSelect multiplication or division facts to illustrate or write a word problem.Hunt for multiple sets of objects in the home. Use repeated addition and multiplication to find the totals.Sort coins according to type, count the number of coins and then multiply to find the total value of pennies (x 1), nickels (x 5), dimes (x 10) and quarters (x 25).Roll 2 number cubes. Find the products of the factors.Count quantities of items by 2's, 3's, 5's, and 10's.Roll 2 number cubes to determine the factors. Make an array to find the product.Act out division problems with counters. For example, Brad has 12 rabbits. He puts the same number of rabbits into each of 4 cages. How many rabbits does Brad put in each cage?Roll 2 number cubes and write the fact families. For example, for rolls of 4 and 6, write: 4 X 6 = 24, 6 X 4 = 24, Ask your child to find the missing factor. For example, 5 X what number? = 35? MODULE 2Roll 2 number cubes. Find the products of the factors. Locate numbers in catalogs or newspapers, then practice rounding them to the nearest tens and hundreds. Make records of important times of the day (wake-up, dinner, going to school, getting home from school, etc.) and practice telling how long between activities. Calculate elapsed time by finding out how long it takes to complete daily activities (soccer practice, homework, take a shower, etc.)Fill a small box with blocks (e.g., sugar cubes) to determine its volume. Brainstorm multiple strategies to determine the volume. MODULE 3 Measure the area of the rooms in your home to determine which rooms are the smallest and largest.Use grid paper to make rectangles with the same area. MODULE 4Measure the perimeter and area of the rooms in your home to determine which rooms are the smallest and largest.Use grid paper to make rectangles with the same perimeters. Determine the area of each rectangle. MODULE 5Go on a fraction hunt! Look for household items that are divided into equal parts (fractions of a whole and fractions of a set). Record the fractions.Roll number cubes to make fractions. Draw pictures of the fractions you make. Place the fractions you've made in order on a number line. Identify fractions at meal times. For example, you ate 1/2 of an apple, 3/4 stalk of celery,1 whole tuna sandwich, and 2/3 of a glass of milk. Practice making equivalent fractions. Plot fractions on a number line. Module 6Share and discuss tables and graphs found in newspapers and magazines.Conduct a survey among family members or friends and construct a bar graph or pictograph.Make a physical pictograph using real objects (e.g., fruits, vegetables, cereal, kitchen tools). Record the graph on paper. Change the scale to create a new pictograph. Module 7Measure the perimeter and area of the rooms in your home to determine which rooms are the smallest and largest.Use grid paper to make rectangles with the same perimeters. Determine the area of each rectangle.Act out division problems with counters. For example, Brad has 12 rabbits. He puts the same number of rabbits into each of 4 cages. How many rabbits does Brad put in each cage?Use grid paper to create congruent shapes.Identify, describe, and classify different household objects as solid figures.Identify angles that are less than, equal to and greater than right angles.Use tangrams to make plane figures. |
# Math 31AH: Lecture 2
Let $\mathbf{V}$ be a vector space. Recall from Lecture 1 (together with Assignment 1, Problem 2) that to say $\mathbf{V}$ is $n$-dimensional means that $\mathbf{V}$ contains a linearly independent set of size $n,$ but does not contain a linearly independent set of size $n+1.$ Also recall from Lecture 1 that a basis of $\mathbf{V}$ is a subset $B$ of $\mathbf{V}$ which is linearly independent and spans $\mathbf{V}.$ The main goal of this lecture is to prove the following theorem.
Theorem 1: A vector space $\mathbf{V}$ is $n$-dimensional if and only if it contains a basis $B$ of size $n.$
Before going on to the proof of this theorem, let us pause to consider its ramifications. Importantly, Theorem 1 gives a method to calculate the dimension of a given vector space $\mathbf{V}$: all we have to do is find a basis $B$ of $\mathbf{V},$ and then count the number of elements in that basis. As an example, let us compute the dimension of $\mathbf{V}=\mathbb{R}^4.$ This vector space is typically used to model the world in which we live, which consists of four physical dimensions: three for space and one for time. Let us verify that our mathematical definition of vector space dimension matches our physical understanding of dimension.
Let $\mathbf{x}=(x_1,x_2,x_3,x_4)$ be a vector in $\mathbb{R}^4.$ Then, we have
$\mathbf{x} = x_1\mathbf{e}_1+x_2\mathbf{e}_2+x_3\mathbf{e}_3+x_4\mathbf{e}_4,$
where
$\mathbf{e}_1 = (1,0,0,0) \\ \mathbf{e}_2=(0,1,0,0) \\ \mathbf{e}_3=(0,0,1,0) \\ \mathbf{e}_4=(0,0,0,1).$
This shows that the set $B=\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3,\mathbf{e}_4\}$ spans $\mathbb{R}^4.$ Let us check that $B$ is linearly independent. This amounts to performing the above manipulation in reverse. If $x_1,x_2,x_3,x_4 \in \mathbb{R}$ are numbers such that
$x_1\mathbf{e}_1+x_2\mathbf{e}_2+x_3\mathbf{e}_3+x_4\mathbf{e}_4=\mathbf{0},$
then we have
$(x_1,x_2,x_3,x_4)=(0,0,0,0),$
which means that $x_1=x_2=x_3=x_4=0.$ Thus $B$ is a linearly independent set which spans $\mathbf{R}^4,$ i.e. it is a basis of $\mathbb{R}^4.$ Since $B$ has size $4,$ we conclude from Theorem 1 that $\mathrm{dim} \mathbf{R}^4=4.$
Now let us prove Theorem 1. First, observe that we have already proved that if $\mathbf{V}$ is $n$-dimensional then it contains a basis $B$ of size $n$ — this is Theorem 1 from Lecture 1. It remains to prove the converse, which we now state as a standalone result for emphasis and ease of reference.
Theorem 2: If $\mathbf{V}$ contains a basis $B$ of size $n,$ then $\mathbf{V}$ is $n$-dimensional.
The proof of Theorem 2 is quite subtle. In order to make it easier to understand, it is helpful to first prove the following lemma, in which the main difficulty is concentrated.
Lemma 1: If $A= \{\mathbf{a}_1,\dots\mathbf{a}_m\}$ is a linearly independent set in a vector space $\mathbf{V}$ and $B = \{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ is a basis of $\mathbf{V},$ then $m \leq n.$
Proof: Suppose this were false, i.e. that there exists in the vector space $\mathbf{V}$ a linearly independent set $A= \{\mathbf{a}_1,\dots\mathbf{a}_m\}$ and a basis $B = \{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ such that $m>n.$ We will see that this leads to a contradiction. The strategy is the following: we wish to demonstrate that the assumption $m>n$ implies that we can replace $m-n$ of the vectors in $A$ with the vectors in $B$ in such a way that the resulting set
$S=\{\mathbf{a}_{i_1},\dots,\mathbf{a}_{i_{m-n}},\mathbf{b}_1,\dots,\mathbf{b}_n\}$
is linearly independent. If we can show this, we will have obtained the desired contradiction: $S$ cannot possibly be independent, because $B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ is a basis. In particular, each of the remaining vectors $\mathbf{a}_{i_1},\dots,\mathbf{a}_{i_{m-n}}$ is a linear combination of the vectors in $B.$
We will pursue the following strategy to show that the existence of the set $S$ as above follows from the hypothesis $m>n.$ For each $k \in \in \{0,1,2,\dots,n\},$ consider the following proposition: there exists a linearly independent set $S_k$ in $\mathbf{V}$ of the form
$S_k = A_{m-k} \cup B_k,$
where $A_{m-k}$ is a subset of $A$ of size $m-k$ and $B_k = \{\mathbf{b}_j \colon j \leq k\}.$ Let us call this proposition $\mathcal{P}_k.$ Now, if we can prove that $\mathcal{P}_0$ is true, and that
$\mathcal{P}_k \text{ true } \implies \mathcal{P}_{k+1} \text{ true } \quad \forall k \in \{0,1,\dots,n-1\},$
then we can conclude that $\mathcal{P}_k$ is true for all $k \in \{0,1,\dots,k\}.$ Indeed, we would then have
$\mathcal{P}_0 \text{ true } \implies \mathcal{P}_1 \text{ true } \implies \dots \implies \mathcal{P}_{n-1} \text{ true } \implies \mathcal{P}_n \text{ true}.$
This is one version of a proof technique known as mathematical induction. But the statement $\mathcal{P}_n$ true is exactly what we want, since it gives us a linearly independent set $S_n$ consisting of some number of vectors from $A$ together with all vectors in $B,$ which results in the contradiction explained above.
Let us now implement the above strategy. The first step is to prove that $\mathcal{P}_0$ is true. To see that it is, consider the set $S_k = A_m \cup B_0$ where $A_m = A$ and $B_0=\{\}.$ This set is of the required form, and since $A$ is linearly independent so is $S_k.$
It remains to prove that if $k \in \{1,\dots,n-1\}$ and $\mathcal{P}_k$ is true, then $\mathcal{P}_{k+1}$ is true. Given that $\mathcal{P}_k$ is true, there exists a linearly independent set $S_k$ in $\mathbf{V}$ of the form $S_k = A_{m-k} \cup B_k$ with
$A_{m-k} = \{\mathbf{a}_{i_1},\dots,\mathbf{a}_{i_{m-k}} \}$
an $(m-k)$-element subset of $A$ and
$B_k = \{\mathbf{b}_1,\dots,\mathbf{b}_k\}.$
Consider the set
$S_k \cup \{\mathbf{b}_{k+1}\} = \{\mathbf{a}_{i_1},\dots,\mathbf{a}_{i_{m-k}},\mathbf{b}_1,\dots,\mathbf{b}_k,\mathbf{b}_{k+1}\}.$
If $S_k \cup \{\mathbf{b}_{k+1}\}$ is linearly independent, then so is any subset of it, so in particular the subset
$S_{k+1} =\{\mathbf{a}_1,\dots,\mathbf{a}_{i_{m-k-1}},\mathbf{b}_1,\dots,\mathbf{b}_k,\mathbf{b}_{k+1}\}$
is linearly independent and $\mathcal{P}_{k+1}$ is true. Now suppose that $S_k \cup \{\mathbf{b}_{k+1}\}$ is linearly dependent. Then, because $S_k$ is linearly independent, $\mathbf{b}_{k+1}$ must be a linear combination of the vectors in $S_k,$ i.e
$\mathbf{b}_{k+1} = \sum_{j=1}^{m-k} t_j \mathbf{a}_{i_j} + \sum_{j=1}^k s_j \mathbf{b}_j$
for some $a_{i_1},\dots,a_{i_{m-j}},t_1,\dots,t_k \in \mathbb{R}.$ Moreover, there exists a number $c \in \{1,\dots,m-k\}$ such that $a_{i_c} \neq 0,$ else $\mathbf{b}_{k+1}$ would be a linear combination of $\mathbf{b}_1,\dots,\mathbf{b}_k,$ which is impossible because $B_{k+1}$ is a linearly independent set. We now claim that the set
$S_{k+1} = A_{m-k}\backslash \{\mathbf{a}_{i_c}\} \cup B_{k+1}$
is linearly independent. Indeed, if $S_{k+1}$ were linearly dependent, then we would have
$\sum_{j=1}^{c-1} p_j \mathbf{a}_{i_j} +\sum_{j=c+1}^{m-k} p_j \mathbf{a}_{i_j} + \sum_{j=1}^{k+1} q_j \mathbf{b}_j = \mathbf{0},$
where not all the numbers $p_1,\dots,p_{c-1},p_{c+1},\dots,p_{m-k},q_1,\dots,q_{k+1}$ are zero. This means that the number $q_{k+1} \neq 0,$ since otherwise the above would say that a subset of $S_k$ is linearly dependent, which is false because $S_k$ is linearly independent. Now, if we substitute the representation of $\mathbf{b}_{k+1}$ as a linear combination of elements $S_k$ given above, this becomes a vanishing linear combination of the vectors in $S_k$ in which the coefficient of $\mathbf{a}_{i_c}$ is $t_cq_{k+1} \neq 0,$ which contradicts the linear independence of $S_k.$ So, $S_{k+1}$ must be linearly independent. — Q.E.D.
Let us note the following corollary of Lemma 1.
Corollary 1: If $A=\{\mathbf{a}_1,\dots,\mathbf{a}_m\}$ and $B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ are two bases of a vector space $\mathbf{V},$ then $m=n.$
Proof: Since $A$ is linearly independent and $B$ is a basis, we have $m \leq n$ by Lemma 1. On the other hand, since $B$ is linearly independent and $A$ is a basis, we also have $n \leq m$ by Lemma 1. Thus $m=n.$ — Q.E.D.
We now have everything we need to prove Theorem 2.
Proof of Theorem 2: Let $\mathbf{V}$ be a finite-dimensional vector space which contains a basis $B$ of size $n.$ We will prove that $\mathbf{V}$ is $n$-dimensional using the definition of vector space dimension (Definition 4 in Lecture 1) and Lemma 1. First, since $B$ is a linearly independent set of size $n,$ we can be sure that $\mathbf{V}$ contains a linearly independent set of size $n.$ It remains to show that $\mathbf{V}$ does not contain a linearly independent set of size $n+1.$ This follows from Lemma 1: since $B$ is a basis of size $n,$ every linearly independent set $A$ in $\mathbf{V}$ must have size less than or equal to $n.$ — Q.E.D.
Another very important consequence of Theorem 2 is the following: it reveals that any two vector spaces of the same dimension can more or less be considered the same. Note that this is fundamentally new territory for us; so far, we have only considered one vector space at a time, but now we are going to compare two vector spaces.
To make the above precise, we need to consider functions between vector spaces. In fact, it is sufficient to limit ourselves to the consideration of functions which are compatible with the operations of vector addition and scalar multiplication. This leads to the definition of a linear transformation from a vector space $\mathbf{V}$ to another vector space $\mathbf{W}.$ If vector spaces are the foundation of linear algebra, then linear transformations are the structures we want to build on these foundations.
Defintion 1: A function $T \colon \mathbf{V} \to \mathbf{W}$ is said to be a linear transformation if it has the following properties:
1. $T(\mathbf{0}_\mathbf{V}) =\mathbf{0}_\mathbf{W},$ where $\mathbf{0}_\mathbf{V}$ is the zero vector in $\mathbf{V}$ and $\mathbf{0}_\mathbf{W}$ is the zero vector in $\mathbf{W}.$
2. For any vectors $\mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}$ and any scalars $t_1,t_2 \in \mathbb{R},$ we have $T(t_1\mathbf{v}_1 + t_2\mathbf{v}_2)=t_1T(\mathbf{v}_1)+t_2T(\mathbf{v}_2).$
So, a linear transformation is a special kind of function from one vector space to another. The name “linear transformation” comes from the fact that these special functions are generalizations of lines in $\mathbb{R}^2$ which pass through the origin $(0,0).$ More precisely, every such line is the graph of a linear transformation $T \colon \mathbb{R}^2 \to \mathbb{R}^2.$ Indeed, a line through the origin in $\mathbb{R}^2$ of slope $m$ is the graph of the linear transformation $T_m \colon \mathbf{R} \to \mathbb{R}$ defined by
$T_m(x) = mx.$
The one exception to this is the vertical line in $\mathbb{R}^2$ passing through $(0,0),$ which has slope $m=\infty$. This line is not the graph of any function from $\mathbb{R}$ to $\mathbb{R},$ which should be clear if you remember the vertical line test.
To reiterate, a linear transformation from a vector space $\mathbf{V}$ to $\mathbf{W}$ is a special kind of function $T \colon \mathbf{V} \to \mathbf{W}.$ For linear transformations, it is common to write $T\mathbf{v}$ instead of $T(\mathbf{v}),$ and this shorthand is often used to implicitly indicate that $T$ is a linear transformation. In the second half of the course, we will discuss linear transformations in great detail. At present, however, we are only concerned with a special type of linear transformation called an “isomorphism.”
Defintion 2: A linear transformation $T \colon \mathbf{V} \to \mathbf{W}$ is said to be an isomorphism if there exists a linear transformation $U \colon \mathbf{W} \to \mathbf{V}$ such that
$U(T\mathbf{v})= \mathbf{v} \quad \forall \mathbf{v} \in \mathbf{V}$
and
$T(U\mathbf{w})=\mathbf{w} \quad \forall \mathbf{w} \in \mathbf{W}.$
The word “isomorphism” comes from the Greek “iso,” which means “same,” and “morph” which means “shape.” It is not always the case that there exists an isomorphism $T$ from $\mathbf{V}$ to $\mathbf{W}$; when an isomorphism does exist, one says that $\mathbf{V}$ and $\mathbf{W}$ are isomorphic. Isomorphic vector spaces $\mathbf{V}$ and $\mathbf{W}$ have the “same shape” in the sense that there is both a linear transformation $T \colon \mathbf{V} \to \mathbf{W}$ which transforms every vector $\mathbf{v} \in \mathbf{V}$ into a vector $\mathbf{w} \in \mathbf{W},$ and an inverse transformation $U \colon \mathbf{W} \to \mathbf{V}$ which “undoes” $T$ by transforming $\mathbf{w}$ back into $\mathbf{v}.$ To understand this, it may be helpful to think of isomorphic vector spaces $\mathbf{V}$ and $\mathbf{W}$ as two different languages. Any two human languages, no matter how different they may seem, are “isomorphic” in the sense that they describe exactly the same thing, namely the totality of human experience. The isomorphism $T \colon \mathbf{V} \to \mathbf{W}$ translates every word $\mathbf{v}$ in the language $\mathbf{V}$ to the corresponding word $T\mathbf{v}=\mathbf{w}$ in $\mathbf{W}$ which means the same thing. The inverse isomorphism $U \colon \mathbf{W} \to \mathbf{V}$ translates back from language $\mathbf{W}$ to language $\mathbf{V}.$ On the other hand, if $\mathbf{V}$ is a human language and $\mathbf{W}$ is the language of an alien civilization, then $\mathbf{V}$ and $\mathbf{W}$ are not isomorphic, since the experience of membership in human society is fundamentally different from the experience of membership in a non-human society, and this difference is not merely a matter of language.
Theorem 2: Any two $n$-dimensional vector spaces $\mathbf{V}$ and $\mathbf{W}$ are isomorphic.
Proof: Since $\mathbf{v}$ is $n$-dimensional, it contains a basis $B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ by Theorem 1. Likewise, since $\mathbf{W}$ is $n$-dimensional, it contains a basis $C=\{\mathbf{c}_1,\dots,\mathbf{c}_n\}.$ Now, since $B$ is a basis in $\mathbf{V},$ for every $\mathbf{v} \in \mathbf{V}$ there exist unique scalars $t_1,\dots,t_n \in \mathbb{R}$ such that
$\mathbf{v}=t_1\mathbf{b}_1 + \dots + t_n\mathbf{b}_n.$
Likewise, since since $C$ is a basis in $\mathbf{W},$ for every $\mathbf{w} \in \mathbf{V}$ there exist unique scalars $u_1,\dots,u_n \in \mathbb{R}$ such that
$\mathbf{w}=u_1\mathbf{c}_1 + \dots + u_n\mathbf{c}_n.$
We may thus define functions
$T \colon \mathbf{V} \to \mathbf{W} \quad\text{ and }\quad U \colon \mathbf{W} \to \mathbf{V}$
by
$T(t_1\mathbf{b}_1+\dots+t_n\mathbf{b}_n) = t_1\mathbf{c}_1+\dots+t_n\mathbf{c}_n$
and
$U(u_1\mathbf{c}_1 + \dots + u_n\mathbf{c}_n) = u_1\mathbf{b}_1 + \dots + u_n\mathbf{b}_n.$
We claim that $T$ is a linear transformation from $\mathbf{V}$ to $\mathbf{W}.$ To verify this, we must demonstrate that $T$ has the two properties stipulated by Defintion 1. First, we have
$T(\mathbf{0}_\mathbf{V})=T(0\mathbf{b}_1+\dots+0\mathbf{b}_n)=0\mathbf{c}_1+\dots+0\mathbf{c}_n=\mathbf{0}_\mathbf{W},$
which verifies the first property. Next, let $\mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}$ be any two vectors, and let $a_1,a_2 \in \mathbb{R}$ be any two scalars. Let
$\mathbf{v}_1 = t_{11}\mathbf{b}_1+\dots+t_{1n}\mathbf{b}_n \\\mathbf{v}_1 = t_{21}\mathbf{b}_1+\dots+t_{2n}\mathbf{b}_n$
be the unique representations of $\mathbf{v}_1$ and $\mathbf{v}_2$ as linear combinations of the vectors in the basis $B.$ We then have
$T(a_1\mathbf{v}_1+a_2\mathbf{v}_2) = T((a_1t_{11}+a_2t_{21})\mathbf{b}_1+\dots+(a_1t_{1n}+a_2t_{2n})\mathbf{b}_n)\\ =(a_1t_{11}+a_2t_{21})\mathbf{c}_1+\dots+(a_1t_{1n}+a_2t_{2n})\mathbf{c}_n \\ = a_1\left(t_{11}\mathbf{c}_1+\dots+t_{1n}\mathbf{c}_n \right) + a_2\left( t_{21}\mathbf{c}_1+\dots+t_{2n}\mathbf{c}_n\right) \\ =a_1T\mathbf{v}_1 + a_2T\mathbf{v}_2,$
which verifies the second property. In the same way, one checks that $U$ is a linear transformation from $\mathbf{W}$ to $\mathbf{V}.$
To prove that the linear transformation $T \colon \mathbf{V} \to \mathbf{W}$ is an isomorphism, it remains only to prove that the linear transformation $U \colon \mathbf{W} \to \mathbf{V}$ undoes $T.$ To see this, let
$\mathbf{v}=t_1\mathbf{b}_1+\dots+t_n\mathbf{b}_n$
be an arbitrary vector in $\mathbf{V}$ expressed as a linear combination of the vectors in the basis $B.$ We then have
$U(T\mathbf{v}) = U(t_1\mathbf{c}_1+\dots+t_n\mathbf{c}_n) = t_1\mathbf{b}_1+\dots+t_n\mathbf{b}_n = \mathbf{v}.$
This completes the proof that $\mathbf{V}$ and $\mathbf{W}$ are isomorphic vector spaces. — Q.E.D.
To continue with our linguistic analogy, Theorem 2 says that any two $n$-dimensional vector spaces are just different languages expressing the same set of concepts. From this perspective, it is desirable to choose a “standard language” into which everything should be translated. In linguistics, such a language is called a lingua franca, a term which reflects the fact that this standard language was once the historical predecessor of modern French (these days the lingua franca is English, but this too may eventually change).
The lingua franca for $n$-dimensional vector spaces is $\mathbb{R}^n,$ and it is unlikely that this will ever change. Let $\mathbf{V}$ be an $n$-dimensional vector space, and let $B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ be a basis in $\mathbf{V}$. Consider the basis $E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ of $\mathbb{R}^n$ in which
$\mathbf{e}_1 = (1,0,0,\dots,0) \\ \mathbf{e}_2=(0,1,0,\dots,0) \\ \vdots \\ \mathbf{e}_n = (0,0,\dots,0,1).$
To be perfectly clear, for each $1 \leq i \leq n$ the vector $\mathbf{e}_i$ has the number $1$ in position $i,$ and a zero in every other position. The basis $E$ is called the standard basis of $\mathbb{R}^n,$ and you should immediately stop reading and check for yourself that it really is a basis. Assuming you have done so, we proceed to define an isomorphism
$T_B \colon \mathbf{V} \to \mathbb{R}^n$
as follows. Given a vector $\mathbf{v} \in \mathbf{V},$ let
$\mathbf{v}=t_1\mathbf{b}_1 + \dots + t_n\mathbf{b}_n$
be the unique representation of $\mathbf{v}$ as a linear combination of vectors in $B.$ Now set
$T_B(\mathbf{v}) := t_1\mathbf{e}_1 + \dots + t_n\mathbf{e}_n,$
where the symbol “$:=$” means “equal by definition.” The fact that this really is an isomorphism follows from the proof of Theorem 2 above. The isomorphism $T_B$ is called the coordinate isomorphism relative to the basis $B,$ and the geometric vector $T_B(\mathbf{v}) \in \mathbb{R}^n$ is called the coordinate vector of $\mathbf{v}$ relative to the basis $B.$ Because of the special form of the standard basis of $\mathbb{R}^n,$ we may write the coordinate vector of $\mathbf{v}$ more concisely as
$T_B(\mathbf{v}) = (t_1,\dots,t_n).$
When working with a given $n$-dimensional vector space, it is often convenient to choose a basis $B$ in $\mathbf{V}$ and use the corresponding coordinate isomorphism $T_B$ to work in the standard $n$-dimensional vector space $\mathbb{R}^n$ rather than working in the original vector space $\mathbf{V}$. For example, someone could come to you with a strange $4$-dimensional vector space $\mathbf{V}$ and claim that this vector space is the best model for spacetime. If you find that you disagree, you can choose a convenient basis $B=\{\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3,\mathbf{b}_4\}$ and use the corresponding coordinate isomorphism $T_B \colon \mathbf{V} \to \mathbb{R}^4$ to transform their model into the standard model of spacetime. |
# Illustrative Mathematics Grade 7, Unit 4, Lesson 2: Ratios and Rates With Fractions
Learning Targets:
• I can solve problems about ratios of fractions and decimals.
Related Pages
Illustrative Math
#### Lesson 2: Ratios and Rates With Fractions
Let’s calculate some rates with fractions.
Illustrative Math Unit 7.4, Lesson 2 (printable worksheets)
#### Lesson 2 Summary
The following diagram shows how to find ratios and rates with fractions.
#### Lesson 2.1 Number Talk: Division
Find each quotient mentally.
1. 5 ÷ 1/3
2. 2 ÷ 1/3
3. 1/2 ÷ 1/3
4. 1 1/2 ÷ 1/3
#### Lesson 2.2 Flags Are Many Sizes
A Train is Traveling at . . .
1. How far does the train go in 1 minute?
2. How far does the train go in 100 minutes?
#### Lesson 2.3 Comparing Running Speeds
Lin ran 2 3/4 miles in 2/5 of an hour. Noah ran 8 2/3 miles in 4/3 of an hour.
1. Pick one of the questions that was displayed, but don’t tell anyone which question you picked. Find the answer to the question.
2. When you and your partner are both done, share the answer you got (do not share the question) and ask your partner to guess which question you answered. If your partner can’t guess, explain the process you used to answer the question.
#### Are you ready for more?
Nothing can go faster than the speed of light, which is 299,792,458 meters per second. Which of these are possible?
1. Traveling a billion meters in 5 seconds.
109 in 5 seconds = 2 · 108 which is slower than the speed of light.
So, this is possible.
2. Traveling a meter in 2.5 nanoseconds. (A nanosecond is a billionth of a second.)
Traveling a meter in 2.5 nanoseconds = 4 · 108 which is faster than the speed of light
so, this is not possible.
3. Traveling a parsec in a year. (A parsec is about 3.26 light years and a light year is the distance light can travel in a year.)
Traveling a parsec in a year would mean traveling at 3.26 times faster than the speed of light which is not possible.
#### Lesson 2.4 Scaling the Mona Lisa
In real life, the Mona Lisa measures 2 1/2 feet by 1 3/4 feet. A company that makes office supplies wants to print a scaled copy of the Mona Lisa on the cover of a notebook that measures 11 inches by 9 inches.
The applet is here to help you experiment with the situation. (It won’t solve the problems for you.) Use the sliders to scale the image and drag the red circle to place it on the book. Measure the side lengths with the Distance or Length tool.
Open Applet
1. What size should they use for the scaled copy of the Mona Lisa on the notebook cover?
2. What is the scale factor from the real painting to its copy on the notebook cover?
3. Discuss your thinking with your partner. Did you use the same scale factor? If not, is one more reasonable than the other?
#### Lesson 2 Practice Problems
1. A cyclist rode 3.75 miles in 0.3 hours.
a. How fast was she going in miles per hour?
b. At that rate, how long will it take her to go 4.5 miles?
2. A recipe for sparkling grape juice calls for 1 1/2 quarts of sparkling water and 3/4 quart of grape juice.
a. How much sparkling water would you need to mix with 9 quarts of grape juice?
b. How much grape juice would you need to mix with 15/4 quarts of sparkling water?
c. How much of each ingredient would you need to make 100 quarts of punch?
3. a. Draw a scaled copy of the circle using a scale factor of 2.
b. How does the circumference of the scaled copy compare to the circumference of the original circle?
c. How does the area of the scaled copy compare to the area of the original circle?
4. At a deli counter,
Someone bought 1 3/4 pounds of ham for \$14.50.
Someone bought 2 1/2 pounds of turkey for \$26.25.
Someone bought 3/8 pounds of roast beef for \$5.50.
Which meat is the least expensive per pound? Which meat is the most expensive per pound? Explain how you know.
5. Jada has a scale map of Kansas that fits on a page in her book. The page is 5 inches by 8 inches. Kansas is about 210 miles by 410 miles. Select all scales that could be a scale of the map. (There are 2.54 centimeters in an inch.)
A. 1 in to 1 mi
B. 1 cm to 1 km
C. 1 in to 10 mi
D. 1 ft to 100 mi
E. 1 cm to 200 km
F. 1 in to 100 mi
G. 1 cm to 1000 km
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Lesson video
In progress...
Hi! I'm Miss Davies.
In this lesson we're going to learn how to simply expressions by multiplying and dividing terms. Five multiplied by Y is written as Five Y.
Five multiplied by two Y is equivalent to ten Y.
Y has been multiplied by two, so our equivalent is also multiplied by two.
Five Y multiplied by three is the same as five Y, add five Y, add five Y, which is equivalent to fifteen Y.
Five Y multiplied by four Y is equivalent to twenty Y squared.
Five multiplied by four is twenty and Y multiplied by Y is Y squared.
Five Y multiplied by seven Zed is thirty-five Y Zed.
This means thirty-five times Y, times Zed.
Five Y multiplied by four Zed, multiplied by nought point five Y, is equivalent to ten Y squared Zed.
The coefficients, five, four and zero point 5 multiply to make ten.
We've then got two Y's and then Zed multiplied together to get Y squared Zed.
Here are some questions for you to try.
Pause the video to complete your task and resume once you're finished.
Here are the answers! Remember that A multiplied by A, is A squared.
And P multiplied by P, multiplied by P, is P cubed.
Ten D E is made by multiplying five D by two E.
We know this because five multiplied by two is ten, or ten divided by five is two.
Nought point five F is multiplied by ninety F to give forty-five F squared.
This is the same as forty-five F squared divided by nought point five F, is equivalent to ninety F.
Twenty-four G divided by six is equivalent to four G, as twenty-four divided by six is four.
Twenty-four G divided by three G is equivalent to eight, as twenty-four divided by three is eight and G divided by G is the same as one.
Twenty-four G squared divided by eight G is equivalent to three G.
Again, twenty-four divided by eight is three and G squared divided by G is just G.
Lastly, twenty-four G cubed divided by four G gives six G squared.
Twenty-four divided by four is six and G cubed divided by G is G squared.
Here are some questions for you to try.
Pause the video to complete your task and resume once you're finished.
Here are the answers! Remember that F cubed divided by F squared is equivalent to F and X cubed divided by X is equivalent to X squared.
Here are some questions for you to try.
Pause the video to complete your task and resume once you're finished.
As long as your coefficients multiply to make eight and you've got two lots of X's and three lots of Y's, it will be right.
For question five, Tyra has said that five X squared is equivalent to twenty-five X.
The correct answer is forty-five X cubed.
Here is a question for you to try.
Pause the video to complete your task and resume once you have finished.
Here is the answer! The area of both quadrilaterals is one hundred and forty-four B squared.
This means each side of the square is twelve B centimetres.
So the perimeter of the square is forty-eight B centimetres.
That's all for this lesson! Thanks for watching!. |
# Question 29ecb
Oct 26, 2016
The product is $p \left(x\right) = \left(5 x + 7\right) \cdot \sqrt{2 x - 3}$, its domain is D_p=<3/2;+oo)
The quotient is $q \left(x\right) = \frac{5 x - 7}{\sqrt{2 x - 3}}$, its domain is D_q=(3/2;+oo)
#### Explanation:
First we have to calculate the domains of $f \left(x\right)$ and $g \left(x\right)$.
$f \left(x\right)$ is defined for all real numbers - ${D}_{f} = \mathbb{R}$
$g \left(x\right)$ is only defined when $2 x - 3 \ge 0$
$2 x \ge 3$
$x \ge \frac{3}{2}$
D_g=<3/2;+oo)
The construction of product and quotient is just writing correct formulas using multiplication and division:
The product is $p \left(x\right) = \left(5 x + 7\right) \cdot \left(\sqrt{2 x - 3}\right)$
The quotient is $q \left(x\right) = \frac{5 x - 7}{\sqrt{2 x - 3}}$
The product is defined for those $x$ where both factors are defined, so its domain is D_p=<3/2;+oo)
The domain of $q \left(x\right)$ is smaller, because $g \left(x\right)$ cannot be zero (it is in the denominator), so you have to exclude $\frac{3}{2}$ from the domain.
Finally D_q=(3/2;+oo)# |
Find answers to all questions with us
# How do you find the area of a hexagon with apothem?
## How do you find the area of a hexagon with apothem?
Apothem is the line segment that is drawn from the center of the hexagon and is perpendicular to the side of the hexagon. We can find the area of a regular hexagon with apothem using the formula, Area of hexagon = (1/2) × a × P; where ‘a’ is the apothem and ‘P’ is the perimeter of the hexagon.
How do you find the area of a hexagon with a diagonal?
For finding the area of hexagon, we are given only the length of its diagonal i.e d. The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. The formula to find the area of hexagon with side length a, Area = (3a2 √3) / 2.
### What is the formula for hexagons?
In a hexagon, the sum of all 6 interior angles is always 720º. The sum of interior angles of a polygon is calculated using the formula, (n-2) × 180°, where ‘n’ is the number of sides of the polygon. Since a hexagon has 6 sides, taking ‘n’ as 6 we get. (6-2) × 180° gives 720°.
How do you calculate the area of a hexagon?
To work out the area of the whole Hexagon, we would just need to work out the area of one of the equilateral triangles, and then multiply this value by 6 . Now provided we know the length of a side of the hexagon a .
## How many sides does a regular hexagon have?
A regular Hexagon is a shape with 6 sides, all of which are the same length. by drawing straight lines from the center of the Hexagon to the corners. Each of these triangles are of the same size.
Is the perimeter of a hexagon the same as the sides?
In a regular hexagon, however, all the hexagon sides and angles have to have the same value. For the sides, any value is accepted as long as they are all the same. This means that, for a regular hexagon, calculating the perimeter is so easy that you don’t even need to use the perimeter of a polygon calculator if you know a bit of maths.
### Which is the apothem of a regular hexagon?
Each triangle has a side length s and height (also the apothem of the regular hexagon) of . The area, A, of one of the equilateral triangles, drawn in blue, can be found using: Since there are six equilateral triangles, the area of a regular hexagon is: |
## Units & Dimensions : Dimensional Analysis
written by: thethinktank โข edited by: Elizabeth Wistrom โข updated: 2/8/2012
This science study guide will teach you the method of dimensional analysis and how it can be put to use to check if the equations that represent a physical condition makes sense.
• slide 1 of 5
### Dimensions of a Physical Quantity
Review the study guide on units of derived physical quantities where we learned to express the derived quantity in terms of the base quantities. This expression was called the dimensional formula.
The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity.
For example, acceleration can be expressed as
acceleration
= (length)/(time)2
The dimensions of acceleration are
[acceleration] = [LT-2]
Thus, we conclude that
• Acceleration has one dimension in mass, one dimension in length, and โ2 dimensions in time. Any quantity raised to zero is equal to 1. So, all those base quantities that do not appear in the dimensional formula of acceleration have dimension 0. For example, here, the dimension of mass is 0.
• slide 2 of 5
### Dimensional Equations
Here is what Dimensional Equations mean. Pay attention to the words written in bold, which are the key words.
• Equating a physical quantity with its dimensional formula is called the Dimensional Equation of the physical quantity.
• In other words, Dimensional Equations are equations, which represent the dimensions of a physical quantity in terms of the base quantities.
The Dimensional Equation can be obtained from the equation representing the relations between the physical quantities. This is something that we have been doing, writing the quantity on the left hand side of the equation and its dimensions on the right hand side.
Examples -
[velocity] or [v] = [M0 L T -1]
[Force] or [F] = [M L T -2]
• slide 3 of 5
### Dimensional Analysis
Dimension of a physical quantity is similar to
• the unit of a physical quantity
• a term in an algebraic equation which is formed by a particular product of the variables, like xy2, x3yz4.
Owing to such a similarity, the golden rule of Dimensional Analysis is:
"Only those physical quantities can be added or subtracted which have the same dimensions(or dimensional formula)."
Thus,
• Checking the Dimensional Consistency of a physical equation means checking that all the terms that are added or subtracted or equated on different sides of the equation must have the same dimensional formula. So, velocity cannot be added to force, or an electric current cannot be subtracted from temperature.
• When magnitudes of two or more physical quantities are multiplied, their units should be treated in the same manner as ordinary algebraic symbols. We can cancel identical units in the numerator and denominator.
• The same is true for dimensions of a physical quantity. We can cancel identical dimensions in the numerator and denominator.
• An equation being dimensionally incorrect means it is an incorrect equation, however, an equation that is dimensionally correct may not necessarily be a correct equation.
• slide 4 of 5
### Checking the Dimensional Consistency
If the dimensions of all the terms are not same, the equation is wrong. Here is an example of checking the Dimensional Consistency of an equation of motion :
s(displacement) = u(initial velocity) X t(time) + 1/2 X a(acceleration)X t2(time)
Or, s = u.t + 1/2 a.t2
Step 1: Identify the terms in the equation and identify which symbol stands for which physical quantity.
The terms here are : s, u.t, a.t2
Step 2: Write down the dimensional formula of each symbol used in the equation. If you are unclear how to do it, go to the first article in this series.
[s] = [L]
[u] = [LT-1]
[t] = [T]
[a] = [LT-2]
Step 3: Calculate the dimension of each term in the equation.
Term 1: [s] = [L]
Term 2: [u.t] = [LT -1.T] = [L]
Term 3: [a.t2] = [LT -2.T2] = [L]
Note that, as stated above, we have canceled dimensions from numerator and denominator like [T -2.T2] .
Conclusion: The equation is dimensionally consistent since all the terms have the same dimensions.
Note: The following kind of terms are dimensionless in an equation.
• A pure number
• ratio of similar physical quantities, such as angle as the ratio (length/length), refractive index as the ratio (speed of light in vacuum/speed of light in medium) etc.
• The arguments of special functions, such as the trigonometric, logarithmic and exponential functions
• slide 5 of 5
### Practice Problems
Now that we have dimensional analysis explained, here are some practice problems:
1. [F] = [MLT-2]
What is the dimension of Force in mass? What is the dimension of force in Temperature?
2. Write down the dimensional equation of Density.
3. Check the dimensional consistency of the following equations:
(a) K = (1/2)mv2 + ma
(b) K = (1/2)mv2
(c) K = (3/16)mv2
K = Kinetic Energy; m = Mass; v = velocity; a = acceleration
Can you tell on the basis of Dimensional Analysis that which of these is the correct formula for kinetic energy?
[Answer : (a) Dimensionally inconsistent; (b), (c) Dimensionally consistent ; No ]
#### Units & Dimensions Study Guide for High School Physics
A resourceful set of articles on units that will come in handy in physics class, as well as using units and dimensions to verify or derive formulae! |
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# A sum of money doubles itself in 8 years. What is the rate of interest?
Last updated date: 29th May 2024
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Hint:- In 8 years money from Interest will be come equal to the principal
amount invested. So, money had been doubled in 8 years.
Let the initial amount of money invested will be Rs. x.
Then after 8 years money had become 2x.
Out of Rs. 2x, money from interest will be 2x – initial amount invested = 2x – x = x.
Let the rate of interest be r.
So, now we will use a simple interest formula.
According to Simple Interest (S.I) formula.
$\Rightarrow S.I. = \dfrac{{PRT}}{{100}}$. Where P is principal amount, R is rate of interest and T will be time period.
So, putting the values in the above formula. We will get,
$\Rightarrow x = \dfrac{{xr(8)}}{{100}}$
On solving the above equation. We will get,
$\Rightarrow {\text{ }}r{\text{ }} = {\text{ }}\dfrac{{100}}{8}{\text{ }} = {\text{ }}12.5$
Hence, the rate of interest to double a money in 8 years will be 12.5% per annum.
Note:- Whenever we came up with this type of problem where we are asked to
find rate of interest then first, we will find the interest on principal amount by
subtracting principal amount from the money after 8 years and then we will
assume rate of interest to be r and then apply, Simple Interest formula and
find the required value of rate of interest. |
Chapter 9: Ratio, Proportion and Unitary Method Exercise 9.1
Ratio, Proportion and Unitary Method – Exercise 9.1 – Q.1
(i) Ratio of number of girls of girls to that of boys in the merit list is 2 : 1.
(ii) Ratio of number of students passing a mathematics test to that of total students appearins in test is 2 : 3.
Ratio, Proportion and Unitary Method – Exercise 9.1 – Q.2
(i) The number of bad pencils produced in a factory is 1/9 of the number of good pencils produced in the factory.
(ii) The number of villages is 2000 times that of cities in India.
Ratio, Proportion and Unitary Method – Exercise 9.1 – Q.3
(i) 60 : 72.
To express this ratio in the simplest form, we will have to find the H.C.F of 60 and 72.
It is 12 dividing each term of the ration by the H.C.F of its terms i.e. 12 we get
Hence, the simplest form of the ratio 60: 72 is 5 : 6.
(ii) 324: 144
To express this ratio in the simplest form, we will have to find the H.C.F of 324/144, It is 36.Dividing each term of the ration by the H.C.F of its terms i.e. 36, we get
Hence, the simplest form of the ratio 324: 144 is 9:4.
(iii) 85 : 391.
To express this ratio in the simplest form we will have to find the H.C.F of 85 and 391, It is 17.
Dividing each term of the ratio by the H.C.F of its terms i.e. 17, we get
Hence, the simplest form of the ratio 85:391 is 5:23
(iv) 186 : 403.
The given ratio is 186 : 403 = 186/403
To express this ratio in the simplest form, we will have to find the H.C.F of 186 and 403, It is 31.
Dividing each term of the ratio by the H.C.F of its terms i.e. 31, we get 186: 403 is 6:23.
Ratio, Proportion and Unitary Method – Exercise 9.1 – Q.4
(i) 75 paise to Rs 3 = 75 paise : Rs 3
= 75 paise : 300 paise
= 1 : 4.
[∵ IRS = 100 paise]
[Dividing the first and second term by their H.C.R = 75]
(ii) 35 minutes to 45 minutes
= 35 min : 45 min
= 7 : 9 [diving the first-and second term by their H.C.F = 5]
(iii) 8kg to 400 gm. = 8 kg : 400 gm
= 8000 gm : 400 gm
= 20 : 1.
[dividing the first and second terms by their H.C.F = 400]
(iv) 48 minutes to 1 hour = 48 min: 1 hour
= 48 min: 60 min
[∵ 1 hour = 60 min] = 4: 5.
[Dividing the first and second term by their H.C.F = 12]
(v) 2 meters to 35 cm = 2 met: 35 cm
= 200 cm : 35 cm
[1m = 100 cm]
= 40 : 7
[∵ dividing the first and second term by their H.C.F = 5].
[∵ dividng the first and second term by their H.C.F = 5]
(vi) 35 minutes to 45 seconds = 35 min : 45 sec
= 2100 sec: 45 sec
= 140:3 [H.C.F = 15]
(vii) 2 dozen to 3 scores = 2 dozen : 3 scores
= 24: 60
[∵ 1 dozen = 12 score = 20] = 2:5.
[Dividing the first and second term by their H.C.F = 12].
(viii) 3 weeks to 3 days = 3 weeks : 3 days
= 21 day: 3days
[1 week = 7 days]
= 3 × 7 = 3
= 7:1.
(ix) 48 min to 2 hours 40 min = 48 min : 160 min
[∵ 1 hour = 60 min]
= 3: 10
[∵ dividing the first and second term by their H.C.F = 3:10]
(x) 3m 5cm to 35 cm = 3m 5 cm : 35 cm
= 305 cm : 35 cm
[dividing the first and second terms by their H.C.F = 5] = 61 : 7
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Contents:
## What is a Quadratic Sequence?
A quadratic sequence has the general form
T(n) = an2 + bn + c.
Where a, b, and c are constants.
For example, all of the following are quadratic sequences:
• 3, 6, 10, 15, 21,…
• 7, 17, 31, 49, 71,…
• 31, 30, 27, 22, 15,…
Quadratic sequences are related to squared numbers because each sequence includes a squared number an2. For example, the formula n2 + 1 gives the sequence: 2, 5, 10, 17, 26, …. That sequence was obtained by plugging in the numbers 1, 2, 3, … into the formula an2:
• 12 + 1 = 2
• 22 + 1 = 5
• 32 + 1 = 10
• 42 + 1 = 17
• 52 + 1 = 26
More formally, we say that a quadratic sequence has its nth terms given by a quadratic function T(n) = an2 + bn + c.
## How to Verify a Sequence is Quadratic
A sequence is an ordered list of numbers and each number in the sequence is called a term. Each term in a quadratic sequence is related by the same common second difference. It’s called a common second difference (or second order difference) because you have to find the difference between each term twice. Second order differences in quadratic sequence are always constant. So, to verify that a particular sequence is quadratic, find the common second difference and verify that those differences are constant.
Example question: Verify the sequence
3, 12, 25, 42 63, …
Step 1: Subtract each term from the next (i.e. subtract the second term from the first, the third from the second, and so on):
• 12 – 3 = 9,
• 25 – 12 = 13,
• 42 – 25 = 17,
• 63 – 42 = 21.
Step 2: Find the difference between the terms you found in Step 1 (9, 13, 17, 21,…)
• 13 – 9 = 4,
• 17 – 13 = 4,
• 21 – 17 = 4.
The common second difference is a constant (4), so 3, 12, 25, 42 63, … is a quadratic sequence.
## Finding the nth Number in a Quadratic Sequence
To find the nth number, plug that number into a given formula. For example, to find the 10th number in the sequence n2 + 1:
• 102 + 1 = 101.
## Finding the General Form
The general form of a quadratic sequence follows T(n) = an2 + bn + c.. So, given a sequence of numbers, your goal is to identify a, b, and c (the coefficients).
Example question: Find the general form of the quadratic sequence 6, 11, 18, 27, 38, 51:
Step 1: Find the first coefficient (a):
1. Find the common second difference:
• 11 – 6 = 5,
• 18 – 11 = 7,
Giving a common second difference of 2 (because 7 – 5 = 2). We only needed to calculate the first couple of terms here because we don’t need to verify it’s quadratic, only calculate the second difference for the formula.
2. Use algebra to solve the formula 2a = 2nd difference. Our second difference here is 2, so (plugging that into the formula), we get:
1. 2a = 2
2. a = 2/2 = 1
Out first coefficient is 1.
Step 2: Find the second coefficient (b) using the formula:
3a + b = difference between the first and second terms.
The difference between terms 1 and 2 is:
• 11 – 6 = 5,
Putting that into the formula (along with “a” from Step 1) we get:
1. 3(1) + b = 5
2. 3 + b = 5
3. b = 2
Our second coefficient (b) is 2.
Step 3: Find the third coefficient (c) using the formula a + b + c = first term.
For this particular sequence, the first term is 6. Plugging that into the formula (along with “a” from Step 1 and “b” from Step 2):
1. 1 + 2 + c = 6,
2. 3 + c = 6,
3. c = 3.
Our third coefficient is 3.
Step 4: Place your answers from Steps 1 to 3 into the formula T(n) = an2 + bn + c:
T(n) = 1n2 + 2n + 3 T(n) = n2 + 2n + 3.
If you have a calculator that has a regression feature, this video is a great hack:
Quickly Find The nth Term Of A Quadratic Sequence Using Statistics Mode Hack - Classwiz Regression
## References
Surowski, D. IB Mathematics HL — Year 1. Unit 3: Sequence, Series, Binomial Theorem and Counting Arguments. Retrieved January 17, 2020 from: https://www.math.ksu.edu/~dbski/IBY1/unit3_homework.pdf
Yee, L. Sequencing Math DNA: Differences, Nth Terms, and Algebraic Sequences.
Image: N. Mori | Wikimedia Commons (Public Domain). |
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