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How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Arithmetic Sequence | AMC 10B, 2003 | Problem No. 24
## Problem - Arithmetic Sequence (AMC 10)
The first four terms in an arithmetic sequence are $x+y$ , $x-y$ xy , $\frac {x}{y}$ in that order. What is the fifth term?
• $\frac{12}{110}$
• $\frac {123}{40}$
• $\frac {16}{17}$
• 20
### Key Concepts
Arithmetic Sequence
Series and Sequence
Algebra
Answer: $\frac {123}{40}$
American Mathematics Competition
Challenges and Thrills - Pre - College Mathematics
## Try with Hints
Here is the first hint to start this sum:
There is a very easy method to do this sum
At first we can try to find the difference between two consecutive terms which is
$(x-y) - (x+y) = -2y$
So after that we can understand the third and forth terms in terms of x and y.
They can be : $( x-3y )$ and $( x - 5y )$
Now try to do rest of the sum......................................
If you got stuck after the first hint you can use this :
Though we from our solution we find the other two terms to be $(x-3y)$ and $(x-5y)$
but from the question we find that the other two terms are $xy$ and $\frac {x}{y}$
So both are equal.Thus ,
$xy = x - 3y$
$xy - x = - 3y$
$x (y - 1) = -3y$
$x = \frac {-3y}{ y - 1}$ .................................(1)
Again , similarly
$\frac {x}{y} = x -5y$
Now considering the equation (1) we can take the value of $\frac {x}{y}$
$\frac {-3}{y - 1}= \frac {-3y}{y -1} - 5y$ .........................(2)
$-3 = -3y - 5y(y-1)$
$0 = 5y^2 - 2y - 3$
$0 = ( 5y +3)(y-1)$
$y = - \frac {3}{5} , 1$
We are almost there with the answer. Try to find the answer.....
Now from the last hint we find the value of $y = - \frac {3}{5} , 1$
But we cannot consider the value of y to be 1 as the 1st and 2 nd terms would be $x+1$ and $x-1)$ but last two terms will be equal to x .
So the value of y be $- \frac {3}{5}$ and substituting the value of y in either $eq^n$ (1) or $eq^n$ (2) we get x = -$\frac {9}{8}$
so , $\frac {x}{y} -2y = \frac {9.5}{8.3} + \frac {6}{5}$
= $\frac {123}{40}$ (Answer ) |
# How do you solve y = x^2 -8x + 16 graphically and algebraically?
Feb 11, 2018
see below
#### Explanation:
Graphically
the roots are where the graph crosses the $x -$axis.
that is when $y = 0$
graph{x^2-8x+16 [-3.74, 14.04, -2.56, 6.33]}
As can be seen from the graph it touches the $x -$axis at one point only
$x = 4$
Algebraically
we could use factorising., completing the square or the formula.
look for factorising first
${x}^{2} - 8 x = 16 = 0$
${\left(x - 4\right)}^{2} = \left(x - 4\right) \left(x - 4\right) = 0$
$\therefore x - 4 = 0 \implies x = 4$
the repeated brackets show that we have repeated roots, and that the $x -$axis is a tangent to the graph.
Feb 11, 2018
$y = {x}^{2} - 8 x + 16$
Let's solve it algebraically first:
We solve this by finding the "zeros" or x-intercepts of the equation. To do this, we factor the equation, if possible.
We need to find 2 numbers (can be the same) that add up to $- 8$ and multiply to $16$.
$- 4$ works:
$- 4 + - 4 = - 8$
$- 4 \cdot - 4 = 16$
So our equation is:
$y = \left(x - 4\right) \left(x - 4\right)$ or $y = {\left(x - 4\right)}^{2}$
Since we are finding the x-intercept(s), let's set $y = 0$. So:
$0 = {\left(x - 4\right)}^{2}$
To simplify this, let's root both sides by $2$:
$\sqrt{0} = \sqrt{{\left(x - 4\right)}^{2}}$
$0 = x - 4$
$x = 4$
So the x-intercept is at $\left(4 , 0\right)$.
To graph this, we find the vertex and the slope.
In this case, since there is only one x-intercept, that means that it is the vertex.
The slope depends on the coefficient, or the number in front of the highest degree term.
${x}^{2}$ means $1 {x}^{2}$, so our slope is one.
Let's graph this now!
As you can see, the x-intercept is the same as the vertex.
Hope this helps! |
# Significant Digits Rules
There are three main significant digits rules that every student needs to keep in mind when performing calculations and determining the number of significant figures in a number:
1. Non-zero digits are always significant.
2. Zeros between non-zero digits are significant.
3. A zero or zeros at the end of a non-decimal or a decimal are significant.
Significant digits (figures) are the digits that contribute to the accuracy of the value and help round numbers correctly.
Example 1: in the number 45032 all digits are significant since 0 is between other non-zero digits.
In order to round this number to the first significant digit, we look at the first significant digit 4 and then at the value of the digit next to it 5. Since the value of the digit next to the first significant digit is 5 or higher, rounding to the first significant digit would be up: 5000.
Example 2: in the number 0.000745 zeros are not significant. The first significant digit here is 7 and the last is 5. Rounding this number to the first significant digit would mean taking 7, looking at the digit next to it, 4, and since 4 is less than 5, then the number is rounded down: 0.000700. Note that now the zeros following the 7 are significant, because they show the value of the number more accurately.
Example 3: when multiplying and dividing numbers, check how many significant digits each number has.
45.23 has 4 significant digits
2.1 has 2 significant digits.
The final answer (the product or the quotient of the two numbers) will have the least number of significant figures within any number in the problem.
So, if we multiplied 45.23 and 2.1 we would keep two significant digits in the answer.
Example 4: when adding and subtracting numbers, check how many significant digits exist in the decimal part of each number. Then add/subtract the numbers as usual and in the sum/difference keep the least number of significant figures from each number in the problem in the decimal portion of the answer.
67.03 has 2 significant digits in the decimal portion
6.483 has 3 significant digits in the decimal portion
When added together, the answer would contain 2 significant digits in the decimal portion.
The importance of significant figures was established in the 18th century A.D., when scientists realized the need to have more accurate solutions and made a connection between rounding and the incorrect final results. A German mathematician Carl Friedrich Gauss studied how significant digits rules were affecting the calculations. |
Question Video: Parallel and Perpendicular Lines | Nagwa Question Video: Parallel and Perpendicular Lines | Nagwa
Question Video: Parallel and Perpendicular Lines Mathematics • Third Year of Secondary School
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Consider the two lines π₯ = 4 + 2π‘, π¦ = 6 + π‘, π§ = 2 β 2π‘ and π« = β¨6, 7, 0β© + π‘β¨5, 4, 7β©. Determine whether they are parallel or perpendicular.
04:17
Video Transcript
Consider the two lines π₯ equals four plus two π‘, π¦ equals six plus π‘, π§ equals two minus two π‘, and π« equals six, seven, zero plus π‘ times five, four, seven. Determine whether they are parallel or perpendicular.
Now we can determine whether two lines are parallel or perpendicular by considering their direction vectors. And so what weβre going to do is begin by rewriting the equation of our first line in vector form so we can then extract its direction vector. And of course, thatβs of the form π₯ nought, π¦ nought, π§ nought plus π‘ times π, π, π, where π, π, π is the direction vector and π₯ nought, π¦ nought, π§ nought is the point which the line passes through. To achieve this, we begin by writing each expression as the component of our vector. Then we separate the components as shown. So we get four, six, two plus two π‘, π‘, negative two π‘. And then we take out that constant factor of π‘. So the vector form of our first line is four, six, two plus π‘ times two, one, negative two.
So letβs write down then the direction vector of each of our lines. Letβs call our first line πΏ sub one. Its direction vector is two, one, negative two. Then the direction vector of our second line, weβll call that πΏ sub two, is five, four, seven. And then we recall that if two vectors are parallel, specifically direction vectors, then one is a scalar multiple of the other. So for our lines to be parallel, we can say that the direction vector two, one, negative two can be written as some scalar multiple π of the vector five, four, seven.
Now, if we look at the second component one and four, we might deduce that for this to be true, π would have to be equal to one-quarter, since one-quarter of four is equal to one. But one-quarter of five is not equal to two, and a quarter of seven is not equal to negative two. And so this is not true. These vectors cannot be written as scalar multiples of one another, and so the lines cannot be parallel. We might deduce then that theyβre perpendicular. But weβre going to check. For two vectors to be perpendicular, of course, their dot product or scalar products have to be equal to zero. They also have to intersect at a point, but weβll deal with that in a moment.
Letβs just double-check that their dot product is indeed equal to zero. Well, their dot product or their scalar product is two times five plus one times four plus negative two times seven, which is indeed equal to zero. So this element so far is good. We do, of course, need to double-check that they intersect at a point. Now if the two lines intersect at a point, there must be some parameter values π‘ sub one in our first equation and π‘ sub two in our second that make the π₯-coordinates, π¦-coordinates, and π§ coordinates, respectively, equal.
For our π₯-coordinates, letβs say thatβs four plus two π‘ sub one is equal to six plus five π‘ sub two. For our π¦-coordinate, itβs six plus π‘ sub one equals seven plus four π‘ sub two. And then we have a corresponding equation for our π§-coordinate. Solving any two of these equations simultaneously, we obtain π‘ sub one equals one and π‘ sub two equals zero. Of course, we can check these by then substituting into the third equation, the third one that we hadnβt used. Substituting π‘ sub one equals one into the equation of our first line and we find that this point of intersection is four plus two, six plus one, two minus two, which is six, seven, zero.
Letβs double-check. We get the same values when we substitute into πΏ sub two. As expected, we do indeed get the value of six, seven, zero. And of course, if we had substituted π‘ sub one equals one and π‘ sub two equal zero into that third equation that we didnβt use, we would have expected this result. So since the direction vectors of the two lines are perpendicular and they intersect at a point, we can say that the lines themselves must also be perpendicular.
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# Lesson 10
Dilations on a Square Grid
### Lesson Narrative
In this lesson, students apply dilations to polygons on a square grid, both with and without coordinates. The grid offers a way of measuring distances between points, especially points that lie at the intersection of grid lines. If point $$Q$$ is three grid squares to the right and two grid squares up from $$P$$ then the dilation with center $$P$$ of $$Q$$ with scale factor 4 can be found by counting grid squares: it will be twelve grid squares to the right of $$P$$ and eight grid squares up from $$P$$. The coordinate grid gives a more concise way to describe this dilation. If the center $$P$$ is $$(0,0)$$ then $$Q$$ has coordinates $$(3,2)$$. The image of $$Q$$ after this dilation is $$(12,8)$$.
Students continue to find dilations of polygons, providing additional evidence that dilations map line segments to line segments and hence polygons to polygons. The scale factor of the dilation determines the factor by which the length of those segments increases or decreases. Using coordinates to describe points in the plane helps students develop language for precisely communicating figures in the plane and their images under dilations (MP6) and strategically use coordinates to perform and describe dilations (MP7). Student practice this type of language using the info gap structure. The student with the problem card needs to dilate a polygon on the coordinate grid. In order to do so, they need to request the coordinates of the polygon’s vertices and the center of dilation as well as the scale factor. After obtaining all of this information from the partner with the data card, the student performs the dilation. The focus here is on deciding what information is needed and communicating clearly to request the information and explain why it is needed.
### Learning Goals
Teacher Facing
• Create a dilation of a polygon on a square grid given a scale factor and center of dilation.
• Identify the image of a figure on a coordinate grid given a scale factor and center of dilation.
• Identify what information is needed to dilate a polygon on a coordinate grid. Ask questions to elicit that information.
### Student Facing
Let’s dilate figures on a square grid.
### Required Preparation
Two activities in this lesson call for pre-printed slips cut from copies of a blackline master. Organizing these copies to hand out before the start of class, possibly by using plastic bags and paperclips, will help the lesson run smoothly.
### Student Facing
• I can apply dilations to figures on a square grid.
• I can apply dilations to polygons on a rectangular grid if I know the coordinates of the vertices and of the center of dilation.
• If I know the angle measures and side lengths of a polygon, I know the angles measures and side lengths of the polygon if I apply a dilation with a certain scale factor. |
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# MEP Pupil Text 13-19, Additional Material
14 Loci and Transformations
Additional section to be used before Section 14.6.
14.5A Straight Edge and Compasses Constructions
In this section we consider how to draw some shapes and specific angles using only a straight edge and compasses. We begin by drawing the key constructions.
Construction of a Perpendicular at a Point on a Line Worked Example 1
Construct a perpendicular at a point, O, on a line.
O
Solution
Use compasses to mark points A and B on the line that are at equal distances from the given point O.
A O B
C
Draw arcs centred at A and B. These two arcs intersect at C and D. Then the required perpendicular line can be drawn through the points C and D.
A
O
B
D
Construction of the Perpendicular Bisector of a Line Segment Worked Example 2
Construct the perpendicular bisector of the line AB. Then label the midpoint of AB, M.
A B
1
Construction of the Perpendicular to a Line from a Point not on the Line C Worked Example 3 The diagram shows the line AB and the point C. This line is the perpendicular bisector of AB. A B 2 . We begin by drawing arcs of equal radius. Perpendicular bisector A M B The point where the perpendicular bisector intersects AB can then be labelled M. centred on the points A and B.14. as shown in the diagram. M is the midpoint of AB. 3 4 Then draw a line through the intersection points of the two arcs. The radius of these arcs should be A B 2 3 roughly to of the length AB. Draw a line through C that is perpendicular to AB.5A Solution MEP Pupil Text 13-19. Additional Material There are many lines that cut AB exactly in half. We have to construct the one that is perpendicular to AB.
14. using the points where the arc intersects the lines as the centres.5A Solution Using C as the centre. O 3 . C A B Construction of the Bisector of an Angle Worked Example 4 Bisect this angle. First draw an arc using O as the centre. O Solution To bisect an angle you need to draw a line that cuts the angle in half. draw two further arcs with radii of equal length. The perpendicular line can then be drawn from C through the point where these two new arcs cross. The bisector can then be drawn from O through the point where these two new arcs cross. O Then draw two further arcs of equal radius. draw an arc as shown. MEP Pupil Text 13-19. Additional Material C A B Then using the intersection points of this arc with the line AB as centres.
Then join O to B to complete the angle First draw an angle of 60 ° using the previous construction. as shown above.5A MEP Pupil Text 13-19. Construction 2: To draw an angle of 60 ° Join the ends of the line to the intersection of the arcs to complete the triangle B O A O A If the angle of 60 ° is to be drawn at O. Construction 3: To draw an angle of 30 ° Draw a second arc. with radius the same as the length of the first side of the triangle. of the same radius. one from each end of this side. Then draw two arcs. Additional Material Uses of these constructions Construction 1: To draw an equilateral triangle Draw one side. with the compass point at A.14. crossing the drawn side at A. with the compass point at O. then bisect that angle using the method described in Worked Example 4 earlier in this section. draw a line through O that will form one side of the angle. intersecting the first arc at B. Then draw an arc. A A C O B 4 O B .
5A MEP Pupil Text 13-19. A A C O B O B Worked Example 5 Construct a triangle ABC with AB = 8 cm. Additional Material Construction 4: To draw an angle of 45° Start by constructing a 90 ° angle using one of the methods described in Worked Examples 1.14. Then bisect that angle using the method described earlier in Worked Example 4. 5 . A B Extend the line AB and draw the line through B perpendicular to AB. Then construct an angle of 60 ° at the end A as shown below (see construction 2 above). Then measure (a) (b) (c) BC AC the angle ACB. Solution First draw a line of length 8 cm and label it AB. the angle BAC = 60 ° and ABC = 45 ° . 2 and 3. using the method described in Worked Example 1.
this right angle can now be bisected to give the 45° angle that is required.9 cm 6 (c) ∠ ACB = 75 ° . This is shown in the following diagram and completes the triangle.2 cm (b) AC = 5.14. Additional Material A B Using the method described in Worked Example 4. C A B The triangle is a shown below: C A 60˚ 8 cm 45˚ B (a) BC = 7.5A MEP Pupil Text 13-19.
and then construct perpendicular lines through P and Q using the method described in Worked Example 1. Measure 7 cm along the perpendicular through Q and mark in the point R.5A Worked Example 6 R 60˚ S MEP Pupil Text 13-19. using the method described in Construction 2. with length 5 cm. 7 cm Then measure (a) (b) RS PS the angle PSR. construct a 60 ° angle at R to complete the shape. R P Q Now.14. 7 . P 5 cm Q (c) Solution First draw the line PQ. Additional Material Use a ruler and compasses to construct the shape PQRS as shown opposite.
8 cm PS = 4.1 cm ∠ PSR = 120 ° 8 .5A MEP Pupil Text 13-19.14. Additional Material R S P Q The shape PQRS is shown below: R 60˚ S 7 cm P 5 cm Q (a) (b) (c) RS = 5.
(a) (b) (c) 5. (It is called the orthocentre of ∆ PQR . 6 cm 30˚ 10 cm 7. YZ and ZX.5A MEP Pupil Text 13-19. QT and RU all meet at a single point. and the perpendicular from R to PQ meeting it at U. Measure ∠ ACB . ∠ KLJ and ∠ LJK . Draw the triangle. Show that there are two possible solutions. Construct an angle of 135° . Construct the angle bisectors of ∠ JKL . centre I. Label the point of intersection. Additional Material Exercises 1. 9. Use a ruler and compasses to construct this parallelogram. Measure AC and BC.) Using straight edge and compasses: (a) construct a regular hexagon. W. such that this circle just touches all three sides of the triangle. 8. 3.14. (b) construct a regular octagon. D. radius WX. Show that this circle passes through all three vertices of the triangle. Show that these meet at a single point. Set your compasses to length BC and draw an arc centred on A. Take any triangle XYZ. Construct the perpendicular bisectors of XY. the perpendicular from Q to PR meeting it at T. 4. 10. Draw the circle. Explain why this construction produces the line through A parallel to BC and the line through C parallel to BA. Now set your compasses to length AB and draw an arc centred on C. Construct an angle of 75° .) 9 . ∠ LMN = 30 ° and LN = 7 cm. Draw the circle. I. Show that the lines PS. Construct the perpendicular from P to QR meeting it at S. A triangle ABC is such that AB = 10 cm. ∠ ABC = 30 ° and ∠ BAC = 60 ° . 6. Take any triangle ABC.) Take any triangle JKL. Join D to A and C. Draw triangle LMN with LM = 9 cm. (Hint: 135 ° = 90 ° + 45 ° ) Construct an angle of 15° . Show that these meet at a single point. (It is called the inscribed circle of ∆ JKL . 11. centre W. Take any triangle PQR. 2. (It is called the circumscribed circle of ∆ XYZ.
) 10 . BG and CH. 3 BM = 2 BG 3 CM = 2 CH 3 (M is called the centroid of ∆ ABC . by measuring. If you cut ∆ ABC out of card then M would be the centre of mass of ∆ ABC .5A 12. Additional Material Take any triangle ABC.14. M. Show. Construct the perpendicular bisector of BC to obtain the midpoint F of BC. that AM = 2 AF . Show that these meet at a single point. MEP Pupil Text 13-19. in the same way. Join AF. and the midpoint H of AB. Construct the midpoint G of CA.
Having done it this way. Additional Material Answers 14.MEP Pupil Text 13-19. 1. Teachers will need to check pupils' constructions on an individual basis.5A Straight Edge and Compasses Constructions Teachers will need to set appropriate accuracy levels for their pupils when they attempt these constructions. AC = 5 cm BC = 8. point out that the angle of 135° could also have been constructed using 180 ° − 45 ° . The hint leads pupils to construct and combine two separate angles of 90 ° and 45 ° . 3. Depending on the number of steps required to complete a given construction. accuracy levels of ± 1 mm or ± 2 mm . and ± 1° or ± 2 ° should be demanded. 11 . An angle of 15° can be constructed using 45 ° − 30 ° or as 60 ° − 45 ° . 4. An angle of 75° can be constructed using 45 ° + 30 ° or as 180 ° − 45 ° − 60 ° . which would require the construction of only one angle. (a) (b) (c) Diagram C 2.7 cm ∠ ACB = 90 ° B A N 5. N L M L M 6.
D C A B X 8. Additional Material Answers The figure drawn is a parallelogram. J I L 12 K .14. W Z Y 9.5A 7. MEP Pupil Text 13-19.
60˚ 60˚ 60˚ 60˚ 60˚ (b) To construct a regular octagon requires repetition 7 times of the following two steps: • measure the length of the edge • at one end of the edge. There are equally acceptable variations on these two constructions. 11.5A 10.B.14. measure an angle of 45° . 45˚ 45˚ 45˚ 45˚ 45˚ 45˚ N. P 45˚ 45˚ T U R S Q 13 . Additional Material Answers 60˚ To construct a regular hexagon requires repetition 5 times of the following two steps: • measure the length of the edge • at one end of the edge. measure an angle of 60 ° . (a) MEP Pupil Text 13-19.
Additional Material Answers A G M H C F B 14 .14.5A 12. MEP Pupil Text 13-19. |
# What are all the factors, the prime factorization, and factor pairs of 1787?
To find the factors of 1787, divide 1787 by each number starting with 1 and working up to 1787
## What is a factor in math ?
Factors are the numbers you multiply together to get another number. For example, the factors of 15 are 3 and 5 because 3 × 5 = 15.
The factors of a number can be positive or negative, but they cannot be zero.
The factors of a number can be used to find out if the number is prime or not.
A prime number is a number that has only two factors: itself and 1. For example, the number 7 is prime because its only factors are 7 and 1.
## List all of the factors of 1787 ?
To calculate the factors of 1787 , you can use the division method.
1. Begin by dividing 1787 by the smallest possible number, which is 2.
2. If the division is even, then 2 is a factor of 1787.
3. Continue dividing 1787 by larger numbers until you find an odd number that does not divide evenly into 1787 .
4. The numbers that divide evenly into 1787 are the factors of 1787 .
Now let us find how to calculate all the factors of One thousand seven hundred eighty-seven :
1787 ÷ 1 = 1787
1787 ÷ 1787 = 1
As you can see, the factors of 1787 are 1 and 1787 .
## How many factors of 1787 are there ?
The factors of 1787 are the numbers that can evenly divide 1787 . These numbers are 1 and 1787.
Thus, there are a total of 2 factors of 1787
## What are the factor pairs of 1787 ?
Factor Pairs of 1787 are combinations of two factors that when multiplied together equal 1787. There are many ways to calculate the factor pairs of 1787 .
One easy way is to list out the factors of 1787 :
1 , 1787
Then, pair up the factors:
and (1,1787) These are the factor pairs of 1787 .
## Prime Factorisation of 1787
There are a few different methods that can be used to calculate the prime factorization of a number. Two of the most common methods are listed below.
1) Use a factor tree :
1. Take the number you want to find the prime factorization of and write it at the top of the page
2. Find the smallest number that goes into the number you are finding the prime factorization of evenly and write it next to the number you are finding the prime factorization of
3. Draw a line under the number you just wrote and the number you are finding the prime factorization of
4. Repeat step 2 with the number you just wrote until that number can no longer be divided evenly
5. The numbers written on the lines will be the prime factors of the number you started with
For example, to calculate the prime factorization of 1787 using a factor tree, we would start by writing 1787 on a piece of paper. Then, we would draw a line under it and begin finding factors.
The final prime factorization of 1787 would be 1787.
2) Use a factorization method :
There are a few different factorization methods that can be used to calculate the prime factorization of a number.
One common method is to start by dividing the number by the smallest prime number that will divide evenly into it.
Then, continue dividing the number by successively larger prime numbers until the number has been fully factorised.
For example, to calculate the prime factorization of 1787 using this method, we keep dividing until it gives a non-zero remainder.
1787 ÷ 1787 = 1
So the prime factors of 1787 are 1787.
## Frequently Asked Questions on Factors
### What are all the factors of 1787 ?
The factors of 1787 are 1 and 1787.
### What is the prime factorization of 1787 ?
The prime factorization of 1787 is 1787 or 17871, where 1787 are the prime numbers .
### What are the prime factors of 1787 ?
The prime factors of 1787 are 1787 .
### Is 1787 a prime number ?
A prime number is a number that has only two factors 1 and itself.
1787 it is a prime number because it has the factors 1 and 1787. |
top
Chapter 5—Powers, Exponents, and Roots
5-7 Working with Scientific Notation
Recall Scientific notation is properly expressed (normalized) when there is only one non-zero digit to the left of the decimal point in the coefficient.
Multiplying and Dividing with Scientific Notation
The math procedures for multiplying and dividing terms expressed in scientific notation is no different from multiplying and dividing terms in any power-0f-ten format. The only thing unique about scientific notation is that the solution is given in the normalized form.
Procedure To multiply values expressed in scientific notation Multiply the coefficients Add the exponents Normalize for scientific notation, if necessary
Examples
Problem:(2 x 103)(4 x 102) = ______
1. Multiply coefficients and add the exponents:
(2 x 103)(4 x 102) = 2 4 x 103+2
2 4 x 103+2 = 8 x 105
2. Normalize for scientific notation
8 x 105 is normalized
Solution: (2 x 103)(4 x 102) = 8 x 105
Problem: (8 x 103)(4 x 104) = ______
1. Multiply the coefficients and add the exponents
(8 x 103) x (4 x 104) = 8 4 x 103+4 = 32 x 107
2. Normalize for scientific notation
32 x 107 = 3.2 x 108
Solution:(8 x 103)(4 x 104) = 1.28 x 1013
Examples & Exercises
Multiplying with Scientific Notation Multiply these terms and formalize the solution if necessary
Procedure To divide values expressed in scientific notation Divide the coefficients ( Subtract the exponents Normalize for scientific notation
Examples
Problem: (8 x 106) ¸ (4 x 104) = _____
1. Divide the coefficients and add the exponents:
(8 x 106) ¸ (4 x 104) = 8/4 x 106-4 = 2 x 102
2. Normalize for scientific notation:
2 x 102
Solution: (8 x 106) ¸ (4 x 104) = 2 x 102
Problem: (16 x 10-4) (0.5 x 102) = _____
1. Divide the coefficients and add the exponents:
16/0.5 x 10-4-2 = 32 x 10-6
2. Normalize for scientific notation:
32 x 10-6 = 3.2 x 10-5
Solution: (16 x 10-4) (0.5 x 102) = 3.2 x 10-5
Endless Examples and Exercises
Dividing with Scientific Notation Divide these terms and normalize the solution if necessary
Mixed Multiplication and Division
Examples
Problem: Perform the operations and show the results in normalized scientific notation.
(1.2 x 102)(4.5 x 103) 3 x 104
Procedure:
1. Complete the multiplication in the numerator:
(1.2 x 102)(4.5 x 103) A =A 1.2 + 4.5 x 102+3 A =A 5.4 x 105 3 x 104 3 x 104 3 x 104
2. Complete the division:
5.4 x 105 A =A 1.8 x 105-4 = 1.8 x 101 3 x 104
The result is already in normalized form.
Solution:
(1.2 x 102)(4.5 x 103) A =A 1.8 3 x 104
Endless Examples and Exercises
Mixed Multiplication and Division Complete these operations, presenting the solution in normalized scientific notation rounded to two decimal places.
Adding and Subtracting with Scientific Notation
The rules for adding and subtracting values in scientific notation are perhaps slightly more complicated than multiplication and division -- addition and subtraction requires that the exponents for the base are the same.
These terms can be added, because their exponents are equal:
The following terms can also be added,
but only after adjusting to make the exponents equal:
Notice that (3.45 x 105) was rewritten as (3450 x 102)
but the solution would have been the same by rewriting
(12.6 x 102) as (0.0126 x 105)
Procedure To add or subtract values expressed in scientific notation Adjust to produce identical exponents, if necessary Add or subtract the coefficients, as designated Attach the common power of ten Normalize for scientific notation, if necessary |
# Introduction to Functions
A function is a relationship between two variables where for every independent variable, there is only one dependent variable. This means that for every x value, there is only one y value.
A function is a special type of relation. It is like a machine where for every INPUT there is only one OUTPUT.
Notice below that there is only one ordered pair for each variable.
The ordered pairs are (A, 1), (B, 1), (C, 4), (D, 3) and (E, 2).
The first variable (INPUT) is called the independent variable and the second (OUTPUT) the dependent variable. The process is a rule or pattern.
For example, in $y=x+1$, we can use any number for x (the independent variable), say x = 3.
$x=3 \\ y=3+1 \\ =4$
As this value depends on the number we choose for x,y is called the dependent variable.
Below is an example of a relationship that is NOT a function. Can you see the difference between this example and the previous one?
In this example the ordered pairs are (A, 1), (A, 2), (B, 1), (C, 4), (D, 3) and (E, 2).
Notice that A has two dependent variables, 1 and 2. This means that it is NOT a function. Instead, a relation is a set of ordered points (xy) where the variables x and y are related according to some rule.
## Function Notation
If y depends on what value we give x in a function, then we can say that y is a function of x. We can write this as y = f(x).
For example, if f(x) = x + 1, evaluate f(3). This is the same as evaluating the function when x = 3.
f(x) = x + 1
f(3) = 3 + 1
f(3) = 4
If y = f(x) then f(a) is the value of y at the point on the function where x = a.
### Examples:
1. If $f\left( x \right) ={ x }^{ 3 }-{ x }^{ 2 }$, find the value of $f\left( 1 \right)$.
$f\left( x \right) ={ x }^{ 3 }-{ x }^{ 2 }\\ f\left( -1 \right) ={ (-1) }^{ 3 }-{ (-1) }^{ 2 }\\ f\left( -1 \right) =-1-1\\ f\left( -1 \right) =-2$
2. Find the values of x for which $f\left( x \right) =0$, given that $f\left( x \right) ={ x }^{ 2 }+3x-10$.
$f\left( x \right) =0\\ i.e.\quad f\left( x \right) ={ x }^{ 2 }+3x-10\\ (x+5)(x-2)=0\\ x+5=0\qquad or\qquad x-2=0\\ x=-5\qquad or\qquad x=2$
3. Find the value of $g\left( 1 \right) +g\left( -2 \right) -g\left( 3 \right)$ if:
$g\left( x \right) \begin{cases} { x }^{ 2 }\qquad \qquad \quad when\quad x>2 \\ 2x-1\quad \qquad when\quad -1\le x\le 2 \\ 5\qquad \qquad \quad when\quad x<-1 \end{cases}$
$g(1)=2(1)-1=1\qquad \qquad since\quad -1\le 1\le 2\\ g(-2)=5\qquad \qquad \qquad \quad \quad since\quad -1<-1\\ g(3)={ 3 }^{ 2 }=9\quad \qquad \qquad \quad since\quad 3>2\\ \\ \therefore \quad g(1)+g(-2)-g(3)\\ \qquad =1+5-9\\ \qquad =-3$
## Intercepts
One of the most useful techniques is to find the x- and y-intercepts.
For x-intercept, y=0.
For y-intercept, x=0.
### Example:
Find the x- and y-intercepts of the function: $f\left( x \right) ={ x }^{ 2 }+7x-8$.
For $y=0$
${ x }^{ 2 }+7x-8=0\\ (x+8)(x-1)=0\\ x+8=0\qquad or\qquad x-1=0\\ x=-8\qquad \quad or\qquad x=1$
For $x=0$
$y={ (0) }^{ 2 }+7(0)-8\\ y=8$
## Domain and Range
You have already seen that the x-coordinate is called the independent variable and the y-coordinate is the dependent variable.
The set of all real numbers x for which a function is define is called the domain.
The set of real values for y or f(x) as x varies is called the range (or image) of f.
For example, we can see the the domain and range of $f\left( x \right)={ x }^{ 2 }$from the graph, which is the parabola $y={ x }^{ 2 }$.
Notice that the parabola curves outwards gradually, and will take on any real value for x. However, it is always on or above the x-axis.
Domain: {all real x}
Range: {y:y$\ge$0}
We can also find the domain and range from the equation $y={ x }^{ 2 }$. Notice that we can substitute any value for x and you will find a value of y. However, all the y-values are positive or zero since squaring any number will give a positive answer (except zero).
## Odd and Even Functions
### Examples:
1. Show that $f(x)={ x }^{ 2 }+3$ is an even function.
$f(-x)={ (-x) }^{ 2 }+3\\ \qquad \quad ={ x }^{ 2 }+3\\ \qquad \quad =f(x)\\ \therefore \quad f(x)={ x }^{ 2 }+3$ is an even function
2. Show that $f(x)={ x }^{ 3 }-x$ is an odd function.
$f\left( -x \right) ={ (-x) }^{ 3 }-(-x)\\ \qquad \quad ={ -x }^{ 3 }+x\\ \qquad \quad =-({ x }^{ 3 }-x)\\ \qquad \quad =-f\left( x \right) \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }-x$ is an odd function
### Examples:
1. For the quadratic function $f\left( x \right) ={ x }^{ 2 }+x-6$
a) Find the x- and y-intercepts
b) Find the minimum value of the function
c) State the domain and range
d) For what values of x is the curve decreasing?
2. (a) Find the x- and y- intercepts and the maximum value of the quadratic function $f\left( x \right) =-{ x }^{ 2 }+4x+5$.
(b) Sketch the function and state the domain and range.
(c) For what values of x is the curve decreasing? |
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# If the radius of Earth somehow shrank by half without any change in Earth’s mass, what would be the value of g at the new surface? What would be the value of g above the new surface at a distance equal to the present radius?
Solutions
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Step 1
1 of 2
$\textbf{Given:}$
$M = 5.97 \times 10^{24}$ kg
$R = 6.37 \times 10^6$ m
$\textbf{g} = 9.8 \dfrac{\text{N}}{\text{kg}}$
The gravitational field strength can be calculated using the formula:
$\begin{gather} \textbf{g} = \dfrac{GM}{R^2} \end{gather}$
If the Earth shrank by half its original size, $R= \dfrac{R}{2}$, we calculate the $\textbf{g}$ at the surface using Eq (1) as follows:
\begin{align*} g_{S} &= \dfrac{GM}{\left(\dfrac{R}{2}\right)^2} \\ &= \dfrac{G \cdot (5.97 \times 10^{24})}{\left(\dfrac{6.37 \times 10^6}{2}\right)^2} \\ &= \boxed{39.28 \dfrac{\text{N}}{\text{kg}}} \end{align*}
This value is approximately four times its original $\textbf{g}$
Above the new surface, $R= R$, we calculate $\textbf{g}$ using Eq (1) as follows:
\begin{align*} g_{R} &= \dfrac{GM}{R^2} \\ &= \dfrac{G \cdot (5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \\ &= \boxed{9.82 \dfrac{\text{N}}{\text{kg}}} \end{align*}
This value is approximately the same its original $\textbf{g}$
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## What is variance and standard deviation with example?
Variance is the average squared deviations from the mean, while standard deviation is the square root of this number. Both measures reflect variability in a distribution, but their units differ: Standard deviation is expressed in the same units as the original values (e.g., minutes or meters).
What’s an example of variance?
Sample Variance Example Suppose a data set is given as 3, 21, 98, 17, and 9. The mean (29.6) of the data set is determined. The mean is subtracted from each data point and the summation of the square of the resulting values is taken. This gives 6043.2.
### What is difference between standard deviation and variance?
Standard deviation is the spread of a group of numbers from the mean. The variance measures the average degree to which each point differs from the mean. While standard deviation is the square root of the variance, variance is the average of all data points within a group.
How do you find the variance?
How to Calculate Variance
1. Find the mean of the data set. Add all data values and divide by the sample size n.
2. Find the squared difference from the mean for each data value. Subtract the mean from each data value and square the result.
3. Find the sum of all the squared differences.
4. Calculate the variance.
## How is variance used in real life?
Key Takeaways. Variance is a measurement of the spread between numbers in a data set. Investors use variance to see how much risk an investment carries and whether it will be profitable. Variance is also used to compare the relative performance of each asset in a portfolio to achieve the best asset allocation.
How do we calculate variance?
To calculate the variance follow these steps: Work out the Mean (the simple average of the numbers) Then for each number: subtract the Mean and square the result (the squared difference). Then work out the average of those squared differences.
### How do you calculate variance when given standard deviation?
s 2 {\\displaystyle s^{2}} = ∑[( x i {\\displaystyle x_{i}} – x̅) 2 {\\displaystyle^{2}}]/(n – 1)
• s 2 {\\displaystyle s^{2}} is the variance.
• x i {\\displaystyle x_{i}} represents a term in your data set.
• ∑,meaning “sum,” tells you to calculate the following terms for each value of x i {\\displaystyle x_{i}},then add them together.
• What is the difference between standard deviation and variation?
Standard Deviation. Standard deviation is a statistic that looks at how far from the mean a group of numbers is,by using the square root of the variance.
• Variance. The variance is the average of the squared differences from the mean.
• Standard Deviation and Variance in Investing.
• The Bottom Line.
• ## What are real life examples of standard deviation?
– Using an empirical concept, he finds 95% of student’s marks fluctuate between ( x + 2 σ ) e.15.5% and 100%. – On closely analyzing the marks, he found a very very low scoring student, roll n.6, who scored only 10%. – Roll no. – The teacher decides to remove roll no.
How to calculate standard deviation?
Add together all the cash flows you have put in the spreadsheet to calculate a total.
• Divide the total by the number of historical entries to calculate the mean average cash flow.
• Subtract the mean average cash flow from each recorded cash flow to calculate the difference.
• Square each cash flow difference by multiplying it against itself. |
# What Is 5/12 as a Decimal + Solution With Free Steps
The fraction 5/12 as a decimal is equal to 0.416.
A mathematical expression that tells the number of equal parts into which an object can be divided is known as a Fraction. There are two elements of a fraction that are separated by a slash or line. These are Numerator and Denominator, present above and below the slash, respectively.
Usually, fractions are solved by dividing the numerator with the denominator to get its equivalent decimal. In the fraction of 5/12, 12 is a denominator while 5 is a numerator.
Here, we will demonstrate the method of Long Division to simplify a fraction.
## Solution
To get the solution of a fraction, we begin by converting it into division. By doing so, the numerator of the fraction that is present above the slash becomes a Dividend, and the denominator present below the slash becomes a Divisor. Therefore, in this example, we get a dividend of 5 and a divisor of 12.
Dividend = 5
Divisor =12
Fraction 5/12 means to divide the number 5 into 12 equal parts and in the results, we get a numerical value of 1 part, also known as the Quotient. In some cases, fractions are not solved completely and we have a left-over value known as Remainder.
Quotient = Dividend $\div$ Divisor = 5 $\div$ 12
Now, let us solve a fraction of 5/12 as an example.
Figure 1
## 5/12 Long Division Method
An explanation of the Long Division method to solve a fraction is given below.
The fraction given to solve is:
5 $\div$ 12
We know that 5/12 is a Proper Fraction because 5 is less than 12. In a proper fraction, we must introduce a Decimal Point, which can be done by adding a zero to the right of the dividend. The dividend in our case is 5. By inserting a zero to its right we get 50. This 50 can now be divided by 12 as:
50 $\div$ 12 $\approx$ 4
Where:
12 x 4 = 48
As remainder 50 – 48 = 2 is a non-zero value, so we again put a zero to the right of the remainder i.e., 2, and make it 20. But here we don’t need another decimal point.
20 $\div$ 12 $\approx$ 1
Where:
12 x 1 = 12
Now, the remaining value is 8 as shown below:
20 – 12 = 8
When we plug in a zero to right of 8, it becomes 80, which can be divided by 12 as:
80 $\div$ 12 $\approx$ 6
Where:
12 x 2 = 72
This time, the Remainder 80 – 72 = 8 is the same as that obtained in the last step. This shows that it is a non-terminating and recurring fraction with a repeating decimal number. Thus, the Quotient of the given fraction is 0.416 and the remaining value is 8.
Images/mathematical drawings are created with GeoGebra. |
# 10.1 Key concepts (Page 3/4)
Page 3 / 4
## Project idea
Perform an experiment to show that as the number of trials increases, the relative frequency approaches the probability of a coin toss. Perform 10, 20, 50, 100, 200 trials of tossing a coin.
## Probability identities
The following results apply to probabilities, for the sample space $S$ and two events $A$ and $B$ , within $S$ .
$P\left(S\right)=1$
$P\left(A\cap B\right)=P\left(A\right)×P\left(B\right)$
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
What is the probability of selecting a black or red card from a pack of 52 cards
1. P(S)=n(E)/n(S)=52/52=1. because all cards are black or red!
What is the probability of drawing a club or an ace with one single pick from a pack of 52 cards
1. $P\left(\mathrm{club}\cup \mathrm{ace}\right)=\mathrm{P}\left(\mathrm{club}\right)+\mathrm{P}\left(\mathrm{ace}\right)-\mathrm{P}\left(\mathrm{club}\cap \mathrm{ace}\right)$
2. $\begin{array}{ccc}& =& \frac{1}{4}+\frac{1}{13}-\left(\frac{1}{4},×,\frac{1}{13}\right)\hfill \\ & =& \frac{1}{4}+\frac{1}{13}-\frac{1}{52}\hfill \\ & =& \frac{16}{52}\hfill \\ & =& \frac{4}{13}\hfill \end{array}$
Notice how we have used $P\left(C\cup A\right)=P\left(C\right)+P\left(A\right)-P\left(C\cap A\right)$ .
The following video provides a brief summary of some of the work covered so far.
## Probability identities
1. Rory is target shooting. His probability of hitting the target is $0,7$ . He fires five shots. What is the probability that all five shots miss the center?
2. An archer is shooting arrows at a bullseye. The probability that an arrow hits the bullseye is $0,4$ . If she fires three arrows, what is the probability that all the arrows hit the bullseye?
3. A dice with the numbers 1,3,5,7,9,11 on it is rolled. Also a fair coin is tossed. What is the probability that:
1. A tail is tossed and a 9 rolled?
2. A head is tossed and a 3 rolled?
4. Four children take a test. The probability of each one passing is as follows. Sarah: $0,8$ , Kosma: $0,5$ , Heather: $0,6$ , Wendy: $0,9$ . What is the probability that:
1. all four pass?
2. all four fail?
5. With a single pick from a pack of 52 cards what is the probability that the card will be an ace or a black card?
## Mutually exclusive events
Mutually exclusive events are events, which cannot be true at the same time.
Examples of mutually exclusive events are:
1. A die landing on an even number or landing on an odd number.
2. A student passing or failing an exam
3. A tossed coin landing on heads or landing on tails
This means that if we examine the elements of the sets that make up $A$ and $B$ there will be no elements in common. Therefore, $A\cap B=\varnothing$ (where $\varnothing$ refers to the empty set). Since, $P\left(A\cap B\right)=0$ , equation [link] becomes:
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$
for mutually exclusive events.
## Mutually exclusive events
1. A box contains coloured blocks. The number of each colour is given in the following table.
Colour Purple Orange White Pink Number of blocks 24 32 41 19
A block is selected randomly. What is the probability that the block will be:
1. purple
2. purple or white
3. pink and orange
4. not orange?
2. A small private school has a class with children of various ages. The table gies the number of pupils of each age in the class.
3 years female 3 years male 4 years female 4 years male 5 years female 5 years male 6 2 5 7 4 6
If a pupil is selceted at random what is the probability that the pupil will be:
1. a female
2. a 4 year old male
3. aged 3 or 4
4. aged 3 and 4
5. not 5
6. either 3 or female?
3. Fiona has 85 labeled discs, which are numbered from 1 to 85. If a disc is selected at random what is the probability that the disc number:
1. ends with 5
2. can be multiplied by 3
3. can be multiplied by 6
4. is number 65
5. is not a multiple of 5
6. is a multiple of 4 or 3
7. is a multiple of 2 and 6
8. is number 1?
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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# For what values of k, the line y = kx + 2 will be tangent to the conic 4x2 - 9y2 = 36
1. $$\pm\frac{2}{3}$$
2. $$\pm\frac{2\sqrt{2}}{3}$$
3. $$\pm\frac{8}{9}$$
4. $$\pm\frac{4\sqrt{2}}{3}$$
Option 2 : $$\pm\frac{2\sqrt{2}}{3}$$
## Detailed Solution
Concept:
The equation of a tangent to the horizontal hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$ whose slope is m is given by:
y = mx ± c where,
$$c = \sqrt{{a^2}{m^2} - {b^2}}$$
Calculation:
Given that,
4x2 - 9y2 = 36
⇒ $$\frac{x^2}{9}-\frac{y^2}{4}=1$$ ----(1)
We know that equation (1) is hyperbola and the tangent of a hyperbola is
y = mx ± $$\sqrt{{a^2}{m^2} - {b^2}}$$ ----(2)
According to the question, the tangent is
y = kx + 2 ---(3)
On comparing equation (2) & (3)
$$\sqrt{{a^2}{k^2} - {b^2}}$$ = 2
⇒ a2k2 - b2 = 4
From equation (1), a2 = 9, & b2 = 4
⇒ 9k2 - 4 = 4
⇒ k2 = 8/9
⇒ k = $$\pm\frac{2\sqrt{2}}{3}$$ |
# Quadrilaterals : Definition and its Types
In this chapter we will discuss basics of quadrilateral and its types.
Important properties of quadrilateral are also discussed in the chapter so make sure you remember it.
Any two dimensional closed shape with 4 sides are known as Quadrilaterals.
From the above examples we can note that:
(a) All the figure have four sides
(b) The figures can be of different shapes and sizes
Some of the common form of quadrilateral used in geometry are;
(a) Square
(b) Rectangle
(c) Parallelogram
(d) Rhombus
(e) Trapezium
Let us study their property one by one.
## Square
Its a 2D figure of four sides in which all sides are equal and each angle measures 90 degree
Important Points for Square
(a) It has 4 sides
(b) Opposite sides are parallel
(c) Each angle is 90 degree
Given above are two examples of square.
(a) Side Length: All sides are equal
(b) Side Orientation : Opposite sides are parallel
(c) Angles: All angles are 90 degree
∠ A = ∠ B = ∠ C = ∠ D = 90 degree
## Rectangle
Its a quadrilateral in which Opposite sides are equal and parallel and all angles measure 90 degree
Given above is the rectangle ABCD in which:
(a) Side Length: Opposite sides are equal
AB = CD and AD = BC
(b) Side Orientation : Opposite sides are parallel
(c) Angles: All angles are 90 degree
∠ A = ∠ B = ∠ C = ∠ D = 90 degree
## Parallelogram
Parallelogram is a quadrilateral in which Opposite sides are equal and parallel. Also in parallelogram opposite angles are equal.
The above parallelogram has following features;
(a) Side Length: Opposite sides are equal
AB = CD and AD = BC
(b) Side Orientation : Opposite sides are parallel
(c) Angles: Opposite angles are equal
∠ D = ∠B
∠ A = ∠ C
## Rhombus
Rhombus is a quadrilateral with following properties;
(a) All sides are equal
(b) Opposite sides are parallel
(c) Opposite angles are equal
Rhombus can also be imagined as tilted square in which sides has been tilted at certain angle.
The quadrilateral looks like a star.
The above rhombus ABCD has following features;
(a) Side Length: All sides are equal
(b) Side Orientation: Opposite sides are parallel
AB parallel to CD
CB parallel to DA
(c) Angles : Opposite angles are equal
∠ D = ∠B
∠ A = ∠ C
## Trapezium
Trapezium is a quadrilateral with following properties;
(a) one pair of opposite sides are parallel
(b) the other pairs are non parallel
Given above is figure of trapezium ABCD with following features;
Sides AB & CD are parallel pairs
Sides AD and BC are non parallel pairs
Solution
The Quadrilateral is a square as:
⟹ Its opposite sides are equal and parallel
⟹ All angles are 90 degree
(02) Identify the below image
Solution
The Quadrilateral is a square as:
⟹ all sides are equal
⟹ all angles are 90 degrees
Solution
The given quadrilateral is a trapezium because:
(a) One pair of opposite sides are parallel ( AD II BC)
(b) Other pairs of sides are not parallel
(04) Given below is the diagram of Rhombus.
The side AB = 6 cm, find the length of side CD
In Rhombus all sides are equal.
Hence, AB = BC = CD = DA = 6 cm
(05) Given below is the figure of parallelogram.
Here ∠B = 75 degree. Find value of ∠D
Solution
We know that in parallelogram opposite angles are equal to each other.
So ∠B = ∠D = 75 degree
(06) Identify the below figure
The above quadrilateral is Rhombus because:
(a) All sides are equal
(b) Opposite sides are parallel
(c) Opposite angles are equal
(07) Below is the image of rectangle
Find the value of all angles
In rectangle all the angles are 90 degree
Hence, ∠A = ∠B = ∠C = ∠D = 90 degree
(08) Two squares ABCD & EFGH are joined to form rectangle PQRS
Its given that AD = 2 cm, find the length of side PQ of the rectangle
Its given that ABCD is a square.
We know that in square, all sides are equal.
Hence, AD = CD = 2 cm
Similarly EFGH is a square with side 2 cm
From the image we can observe that:
⟹ DC + HG = SR
⟹ 2 + 2 = SR
⟹ SR = 4 cm
Since PQSR is a rectangle, the opposite sides are equal
PQ = SR
PQ = 4 cm
Hence, length of side PQ is 4 cm
(09) Identify the below shape
Solution
From the above image you can observe that:
Side AB = CD |
Question 1: $\displaystyle \text{In } \triangle ABC, \angle ABC = \angle DAC, AB= 8 cm, AC = 4 \text{ cm } \text{ and } AD = 5 \text{ cm}.$
(i) Prove that $\displaystyle \triangle ACD$ is similar to $\displaystyle \triangle BCA .$
(ii) Find $\displaystyle BC \text{ and } CD .$
(iii) Find $\displaystyle \text{ Area of } \triangle ACD : \text{ Area of } \triangle ABC .$ [2014]
$\displaystyle \text{(i) } \text{In } \triangle ACD \text{ and } \triangle BCA$
$\displaystyle \angle C = \angle C \text{ (common angle) }$
$\displaystyle \angle ABC = \angle CAD \text{ (given) }$
$\displaystyle \triangle ACD \sim \triangle BCA \text{ (AAA postulate) }$
$\displaystyle \text{ (ii) Since } \triangle ACD \sim \triangle BCA$
$\displaystyle \frac{AC}{ BC} = \frac{CD}{ CA} = \frac{AD}{ BA}$
$\displaystyle \frac{4 }{ BC} = \frac{CD}{ 4} = \frac{5}{ 8}$
$\displaystyle \Rightarrow BC = \frac{8}{ 5} \times 4 = 6.4 \text{ cm }$
$\displaystyle \Rightarrow CD = \frac{5}{ 8} \times 4 = 2.5 \text{ cm }$
$\displaystyle \text{ (iii) Since } \triangle ACD \sim \triangle ABC$
$\displaystyle \text{Therefore } \frac{ \text{ Area } \triangle ACD}{ \text{ Area } \triangle ABC} = \frac{AC^2}{AB^2} = \frac{4^2}{8^2} = \frac{1}{4}$
$\displaystyle \text{Hence } \text{ Area } \triangle ACD : \text{ Area } \triangle ABC= 1:4$
$\displaystyle \\$
Question 2: In the given triangle $\displaystyle P, Q \text{ and } R$ are mid points of sides $\displaystyle AB, BC \text{ and } AC$ respectively. Prove that $\displaystyle \triangle PQR \sim \triangle ABC .$
$\displaystyle \text{In } \triangle ABC, PR \parallel BC$
$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle PAR$
$\displaystyle \angle ABC = \angle APR \text{ (alternate angles) }$
$\displaystyle \angle ARP =\angle ACB$ alternate)
$\displaystyle \text{Therefore } \triangle ABC \sim \triangle ARP \text{ (AAA postulate) }$
$\displaystyle \frac{PR}{BC} = \frac{AP}{AB}$
Since $\displaystyle P$ is the mid point of $\displaystyle AB$
$\displaystyle AB = 2 AP$
$\displaystyle \frac{PR}{BC} = \frac{1}{2}$
$\displaystyle \text{Similarly, we can prove } \frac{PQ}{AC} = \frac{1}{2} \text{ and } \frac{RQ}{AB} = \frac{1}{2}$
$\displaystyle \text{Therefore } \frac{PR}{BC} = \frac{PQ}{AC} = \frac{RQ}{AB}$
$\displaystyle \text{Therefore } \triangle ABC \sim \triangle PQR \text{ (SSS postulate) }$
$\displaystyle \\$
Question 3: In the following figure $\displaystyle AD \text{ and } CE$ are medians of $\displaystyle \triangle ABC. DF \parallel CE .$ Prove that :
$\displaystyle \text{(i) } EF=FB$
$\displaystyle \text{(ii) } AG : GD = 2:1$
$\displaystyle \text{(i) } \text{Consider } \triangle BDF \text{ and } \triangle BCE$
$\displaystyle DF \parallel CE$
$\displaystyle \angle BDF = \angle BCE \text{ (alternate angles) }$
$\displaystyle \angle BFD = \angle BEC \text{ (alternate angles) }$
$\displaystyle \text{Therefore } \triangle BDF \sim \triangle BCE \text{ (AAA postulate) }$
$\displaystyle \frac{BD}{BC} = \frac{BF}{BE}$
$\displaystyle \frac{BD}{2BD} = \frac{BF}{BE}$
$\displaystyle \Rightarrow 2 BF = BE$
$\displaystyle \Rightarrow BF = FE$
$\displaystyle \text{(ii) } \text{Consider } \triangle AFD \text{ and } \triangle AEG$
$\displaystyle FD \parallel EG$
$\displaystyle \angle DFA = \angle GEA \text{ (alternate angles) }$
$\displaystyle \angle FDA =\angle EGA \text{ (alternate angles) }$
$\displaystyle \text{Therefore } \triangle AFD \sim \triangle AEG \text{ (AAA postulate) }$
$\displaystyle \frac{AG}{GD} = \frac{AE}{FE} \text{ By basic proportionality theorem }$
$\displaystyle \text{Given } AE = EB$
$\displaystyle AE = 2 FE \Rightarrow \frac{AE}{FE} = 2$
$\displaystyle \frac{AG}{GD} = \frac{2}{1}$
$\displaystyle \\$
Question 4: In the given figure $\displaystyle \triangle ABC \sim \triangle PQR .$ $\displaystyle AM \text{ and } PN$ are altitudes where as $\displaystyle AX \text{ and } PY$ are medians. Prove:
$\displaystyle \frac{AM}{PN} = \frac{AX}{PY}$
(Since $\displaystyle \triangle ABC \sim \triangle PQR$
$\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$
$\displaystyle \text{Given } AX \text{ and } PY$ are medians
$\displaystyle \text{Therefore } 2BX = BC \text{ and } 2 YR = QR$
$\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BX}{2QY} = \frac{BX}{QY} .$.. … … … (i)
$\displaystyle \text{Consider } \triangle ABM \text{ and } \triangle PQN$
$\displaystyle \angle ABN = \angle PQN$ Since $\displaystyle \triangle ABC \sim \triangle PQR$
$\displaystyle \angle AMB = \angle PNQ = 90^{\circ} \text{ (alternate angles) }$
$\displaystyle \text{Therefore } \triangle ABM \sim \triangle PQN \text{ (AAA postulate) }$
$\displaystyle \frac{AB}{PQ} = \frac{AM}{PN} .$.. … … … (ii)
From (i) and (ii) we get
$\displaystyle \frac{AB}{PQ} = \frac{AM}{PN}$
and $\displaystyle \angle ABX = \angle PQY$
$\displaystyle \text{Therefore } \triangle ABX \sim \triangle PQY$
$\displaystyle \Rightarrow \frac{AB}{PQ} = \frac{AX}{PY}$
$\displaystyle \text{Hence } \frac{AM}{PN} = \frac{AX}{PY}$
$\displaystyle \\$
Question 5: Two similar triangles are equal in area. Prove that the triangles are congruent.
Let the two triangles be $\displaystyle ABC \text{ and } PQR$
Since the two triangles are similar, We know
$\displaystyle \frac{Ar. \triangle ABC}{Ar. \triangle PQR} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}$
Since the are of the two triangles is equal
$\displaystyle \Rightarrow AB=PQ, BC=QR \text{ and } AC = PR$
$\displaystyle \text{Therefore } \triangle ABC \cong \triangle PQR$
$\displaystyle \\$
Question 6: The ratio between the altitudes of two similar triangles is $\displaystyle 3:5 .$ Write the ratios between their (i) medians (ii) perimeters (iii) areas.
The ratio of the altitude of two similar triangles is the same as the ratio of their sides. Given ratio $\displaystyle = 3:5$
(i) Ratio between their median $\displaystyle = 3:5$
(ii) Ratio between their perimeter $\displaystyle = 3:5$
(iii) Ratio between their areas $\displaystyle = 3^2:5^2 = 9:25$
$\displaystyle \\$
Question 7: The ratio between the altitudes of two similar triangles is $\displaystyle 16:25.$ Find the ratio between their: (i) perimeters (ii) altitudes (iii) medians
The ratio between the altitudes of two similar triangles is $\displaystyle 16:25 .$
This means that the ratio of the sides of the triangles = 4:5
(i) Ratio between their perimeter $\displaystyle = 4:5$
(ii) Ratio between their altitude $\displaystyle = 4:5$
(iii) Ratio between their median $\displaystyle = 4:5$
$\displaystyle \\$
Question 8: The following figure shows a $\displaystyle \triangle PQR$ in which $\displaystyle XY \parallel QR .$ If $\displaystyle PX:XQ=1:3 \text{ and } QR =9 \text{ cm }$ , find the length of $\displaystyle XY .$
$\displaystyle \text{Given } PX:XQ=1:3 \text{ and } QR =9 \text{ cm }$
$\displaystyle \text{Consider } \triangle PXY \text{ and } \triangle PQR$
$\displaystyle \angle PXY = \angle PQR \text{ (alternate angles) }$
$\displaystyle \angle PYX =\angle PRQ \text{ (alternate angles) }$
$\displaystyle \text{Therefore } \triangle PXY \sim \triangle PQR \text{ (AAA postulate) }$
$\displaystyle \frac{PX}{PQ} = \frac{XY}{QR}$
$\displaystyle \frac{PX}{PX+XQ} = \frac{XY}{QR}$
$\displaystyle \frac{PX/XQ}{PX/XQ+1} = \frac{XY}{QR}$
$\displaystyle \frac{1}{4} = \frac{XY}{9}$
$\displaystyle \Rightarrow XY = \frac{9}{P4}$
$\displaystyle \\$
Question 9: In the following figure, $\displaystyle AB, CD \text{ and } EF$ are parallel lines. $\displaystyle AB=6 \text{ cm }, CD= y \text{ cm }, EF=10 \text{ cm }, AC = 4 \text{ cm } \text{ and } CF= x \text{ cm}.$ Calculate: $\displaystyle x \text{ and } y .$ [1985]
$\displaystyle \text{Consider } \triangle FDC \text{ and } \triangle FBA$
$\displaystyle \angle FDC = \angle FDA \text{ (Corresponding angles) }$
$\displaystyle \angle DFC = \angle BFA \text{ (common angle) }$
$\displaystyle \triangle FDC \sim \triangle FBA \text{ (AAA postulate) }$
$\displaystyle \text{Therefore } \frac{CD}{ AB} = \frac{FC}{ FA}$
$\displaystyle \frac{y}{ 6} = \frac{x}{ x+4}$
Now $\displaystyle \text{Consider } \triangle FCE \text{ and } \triangle ACB$
$\displaystyle \angle FCE = \angle ACB \text{ (Vertically opposite angles) }$
$\displaystyle \angle CFE = \angle CAB \text{ (alternate angles) }$
$\displaystyle \triangle FCE \sim \triangle ACB \text{ (AAA postulate) }$
$\displaystyle \text{Therefore } \frac{FC}{ AC} = \frac{EF}{ AB}$
$\displaystyle x = \frac{10}{ 6} \times 4 = 6.67 \text{ cm }$
Also $\displaystyle y= \frac{6.67 }{ 6.67+4} \times 6 = 3.75 \text{ cm }$
$\displaystyle \\$
Question 10: On a map, drawn to a scale of $\displaystyle 1:20000$ , a rectangular plot of land $\displaystyle ABCD$ has $\displaystyle AB = 24 \text{ cm } \text{ and } BC = 32 \text{ cm}.$ Calculate:
(i) the diagonal distance of the plot in km
(ii) the area of the plot $\displaystyle \text{In } \text{ km }^2$
$\displaystyle k = \frac{1}{20000}$
Length of AB on map $\displaystyle = k \times$ actual length of AB
Actual length of $\displaystyle AB = 24 \times 20000 \text{ cm } = 4.8 \text{ km }$
Similarly Actual length of $\displaystyle BC = 32 \times 20000 \text{ cm } = 6.4 \text{ km }$
(i) Therefore the diagonal $\displaystyle = \sqrt{4.8^2+6.4^2} = 8 \text{ km }$
(ii) Area of the plot $\displaystyle = 4.8 \times 6.4 = 30.72 \text{ km }$
$\displaystyle \\$
Question 11: The dimension of a model of a multi storied building are $\displaystyle 1\text{ m }$ by $\displaystyle 60$ by $\displaystyle 1.20\text{ m } .$ If the scale factor is $\displaystyle 1:50$ , find the actual dimensions of the building. Also find:
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is $\displaystyle 50 cm^2$
(ii) the space inside the room of the model if the space inside the corresponding room if the building is $\displaystyle 90 m^3$
Dimension of model $\displaystyle = 100 \text{ cm } \times 60 \text{ cm } \times 120 \text{ cm }$
$\displaystyle \text{Scale factor } (k) = \frac{1}{50}$
$\displaystyle \text{Actual length } = 100 \times 50 = 50\text{ m }$
$\displaystyle \text{Actual breadth } = 60 \times 50 = 30\text{ m }$
$\displaystyle \text{Actual height } = 120 \times 50 = 60\text{ m }$
$\displaystyle \text{Therefore the actual dimension of the building } = 50\text{ m } \times 30\text{ m } \times 60\text{ m }$
$\displaystyle \text{(i) Floor area } = 50^2 \times 50 \text{ cm }^2 = 125000 \text{ cm }^2 = 12.5 \text{ m }^2$
$\displaystyle \text{(ii) Volume of the model } = \frac{90 m^3}{50^3} = 720 \text{ cm }^3$
$\displaystyle \\$
Question 12: $\displaystyle \text{In } \triangle PQR, L \text{ and } M$ are two points on the base $\displaystyle QR$ , such that $\displaystyle \angle LPQ = \angle QRP \text{ and } \angle RPM = \angle RQP .$ Prove that:
$\displaystyle \text{(i) } \triangle PQL \sim \triangle RPM$
$\displaystyle \text{(ii) } QL \times RM = PL \times PM$
$\displaystyle \text{(iii) } PQ^2= QR \times QL .$ [2003]
$\displaystyle \text{(i) } \text{Consider } \triangle PQL \text{ and } \triangle RMP$
$\displaystyle \angle LPQ = \angle QRP \text{ (given) }$
$\displaystyle \angle RQP = \angle RPM \text{ (given) }$
$\displaystyle \triangle PQL \sim \triangle RMP \text{ (AAA postulate) }$
$\displaystyle \text{ (ii) Since } \triangle PQL \sim \triangle RMP$
$\displaystyle \frac{PQ}{ RP} = \frac{QL}{ PM } = \frac{PL}{ RM}$
$\displaystyle \Rightarrow QL \times RM = PL \times PM$
$\displaystyle \text{(iii) } \text{Consider } \triangle PQL \text{ and } \triangle RQP$
$\displaystyle \angle LPQ = \angle QRP \text{ (given) }$
$\displaystyle \angle Q \text{ (common angle) }$
$\displaystyle \triangle PQL \sim \triangle RQP \text{ (AAA postulate) }$
$\displaystyle \text{Therefore } \frac{PQ}{ RQ} = \frac{QL}{ QP} = \frac{PL}{ PR}$
$\displaystyle \Rightarrow PQ^2 = QR \times QL$
$\displaystyle \\$
Question 13: $\displaystyle \text{In } \triangle ABC, \angle ACB = 90^{\circ} \text{ and } CD \perp AB .$
$latex \displaystyle \text{Prove that: } \frac{BC^2}{AC^2} = \frac{BD}{AD} .$
$\displaystyle \text{Consider } \triangle ACD \text{ and } \triangle ABC$
$\displaystyle \angle DAC = \angle BAC$ (Common)
$\displaystyle \angle CDA = \angle ACB = 90^{\circ} \text{ (given) }$
$\displaystyle \triangle ACD \sim \triangle ABC \text{ (AAA postulate) }$
$\displaystyle \frac{AC}{AB} = \frac{AD}{AC}$
$\displaystyle AC^2=AD \times AB .$.. … … … (ii)
$\displaystyle \text{Consider } \triangle BCD \text{ and } \triangle ABC$
$\displaystyle \angle CBD = \angle CBA$ (Common)
$\displaystyle \angle BDC = \angle ACB = 90^{\circ} \text{ (given) }$
$\displaystyle \triangle BCD \sim \triangle ABC \text{ (AAA postulate) }$
$\displaystyle \frac{BC}{AB} = \frac{BD}{BC}$
$\displaystyle BC^2=BD \times AB .$.. … … … (ii)
From (i) and (ii)
$\displaystyle \frac{BC^2}{AC^2} = \frac{BD \times AB}{AD \times AB}$
Hence Proved.
$\displaystyle \\$
Question 14: A $\displaystyle \triangle ABC$ with $\displaystyle AB=3 \text{ cm } , BC = 6 \text{ cm } \text{ and } AC = 4 \text{ cm }$ is enlarged to a $\displaystyle \triangle DEF$ such that the longest side of $\displaystyle \triangle DEF = 9 \text{ cm}.$ Find the scale factor and hence, the lengths of the other sides of $\displaystyle \triangle DEF .$
$\displaystyle \text{Scale factor } (k) = \frac{EF}{BC} = \frac{9}{6} = 1.5$
$\displaystyle \text{Therefore } \frac{ED}{AB} = 1.5 \Rightarrow ED = 3 \times 1.5 = 4.5 \text{ cm }$
$\displaystyle \frac{DF}{AC} = 1.5 \Rightarrow ED = 4 \times 1.5 = 6 \text{ cm }$
$\displaystyle \\$
Question 15: Two isosceles triangles have equal vertical angles. Show that the triangles are similar. If the ratio between the areas of these two triangles is $\displaystyle 16:25$, find the ratio between their corresponding altitudes.
$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle DEF$
$\displaystyle \angle BAC = \angle EDF$ (Common)
$\displaystyle AB=AC \text{ and } DE = DF \text{ (given) }$
$\displaystyle \triangle ABC \sim \triangle EDF \text{ (SAS postulate) }$
$\displaystyle \frac{\text{ Area } \triangle ABC}{\text{ Area } \triangle DEF} = \frac{AD^2}{PS^2}$
$\displaystyle \frac{16}{25} = ( \frac{AD}{PS} )^2$
$\displaystyle \Rightarrow \frac{AD}{PS}= \frac{4}{5}$
$\displaystyle \\$
Question 16: $\displaystyle \text{In } \triangle ABC, AP:PB=2:3. PQ \parallel BC$ and is extended to $\displaystyle Q$ so that $\displaystyle CQ \parallel BA .$ Find:
$\displaystyle \text{(i) } \text{ Area of } \triangle APO : \text{ Area of } \triangle ABC$
$\displaystyle \text{(ii) } \text{ Area of } \triangle APO : \text{ Area of } \triangle CQO$
$\displaystyle \text{(i) } \text{Consider } \triangle APO \text{ and } \triangle ABC$
$\displaystyle \angle APO = \angle ABC \text{ (alternate angles) }$
$\displaystyle \angle AOP = \angle ACB \text{ (alternate angles) }$
$\displaystyle \triangle APO \sim \triangle ABC \text{ (AAA postulate) }$
$\displaystyle \frac{AP}{PB} = \frac{2}{3}$
$\displaystyle \text{or } \frac{AP}{AB} = \frac{2}{5}$
$\displaystyle \frac{ \text{ Area of } \triangle APO}{ \text{ Area of } \triangle ABC} = \frac{2^2}{5^2} = \frac{4}{25}$
$\displaystyle \text{(ii) } \text{Consider } \triangle APO \text{ and } \triangle QOC$
$\displaystyle \angle AOP = \angle QOC \text{ (Vertically opposite angles) }$
$\displaystyle \angle PAO = \angle OCQ \text{ (alternate angles) }$
$\displaystyle \triangle APO \sim \triangle QOC \text{ (AAA postulate) }$
$\displaystyle \frac{ \text{ Area of } \triangle APO}{ \text{ Area of } \triangle QOC} = \frac{AP^2}{CQ^2} = \frac{AP^2}{PB^2} = \frac{2^2}{3^2} = \frac{4}{9}$
$\displaystyle \\$
Question 17:The following figure shows a $\displaystyle \triangle ABC$ in which $\displaystyle AD \perp BC \text{ and } BE \perp AC .$ Show that:
$\displaystyle \text{(i) } \triangle ADC \sim \triangle BEC \hspace{1.0cm} \text{(ii) } CA \times CE = CB \times CD \hspace{1.0cm} \\ \\ \text{(iii) } \triangle ABC \sim \triangle DEC \hspace{1.0cm} \text{(iv) } CD \times AB = CA \times DE$
$\displaystyle \text{(i) } \text{Consider } \triangle ADC \text{ and } \triangle BEC$
$\displaystyle \angle ADC = \angle BEC = 90^{\circ} \text{ (alternate angles) }$
$\displaystyle \angle ADC = \angle BCE \text{ (common angle) }$
$\displaystyle \triangle ADC \sim \triangle BEC \text{ (AAA postulate) }$
$\displaystyle \text{(ii) } \text{Therefore } \frac{CA}{CB} = \frac{CD}{CE}$
$\displaystyle CA \times CE = CB \times CD$
$\displaystyle \text{(iii) } \text{Consider } \triangle ABC \text{ and } \triangle DEC$
$\displaystyle \angle ADC = \angle BCE \text{ (common angle) }$
$\displaystyle \frac{CA}{CB} = \frac{CD}{CE}$
$\displaystyle \frac{CA}{CD} = \frac{CB}{CE}$
$\displaystyle \triangle ABC \sim \triangle DEC \text{ (SAS postulate) }$
$\displaystyle \text{(iv) } \text{Therefore } \frac{AC}{DC} = \frac{AB}{DE}$
$\displaystyle AC \times DE = CD \times AB$
$\displaystyle \\$
Question 18: In the given figure, $\displaystyle ABC$ is a triangle with $\displaystyle \angle EDB = \angle ACB .$ Prove that $\displaystyle \triangle ABC \sim \triangle EBD .$ If $\displaystyle BE=6 \text{ cm }, EC = 4 \text{ cm }, BD = 5 \text{ cm }$ and area of $\displaystyle \triangle BED = 9 \text{ cm }^2 .$ Calculate the:
(i) length of $\displaystyle AB$ (ii) area of $\displaystyle \triangle ABC$ [2010]
$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle EBD$
$\displaystyle \angle EDB = \angle ACB \text{ (given) }$
$\displaystyle \angle DBE = \angle ABC$ (common)
$\displaystyle \text{Therefore } \angle DEB = \angle BAC$
$\displaystyle \triangle ABC \sim \triangle EBD \text{ (AAA postulate) }$
$\displaystyle \text{(i) } \text{Given } BE=6 cm, EC=4 cm, BD=5 \text{ cm }$
$\displaystyle \frac{AB}{ EB} = \frac{BC}{ BD} = \frac{AC }{ED}$
$\displaystyle AB = \frac{BE+EC }{5} \times 6 = 2 \text{ cm }$
$\displaystyle \text{(ii) } \frac{ \text{ Area of } \triangle ABC}{ \text{ Area of } \triangle EBD} = \frac{AB^2}{EB^2} = \frac{144}{36}$
Area of $\displaystyle \triangle ABC = \frac{144}{ 36} \times 9 = 36 \text{ cm }^2$
$\displaystyle \\$
Question 19: In the given figure $\displaystyle \triangle ABC$ is a right angled triangle with $\displaystyle \angle BAC = 90^{\circ} .$
(i) Prove $\displaystyle \triangle ADB \sim \triangle CDA$
(ii) If $\displaystyle BD = 18 \text{ cm } \text{ and } CD = 8 \text{ cm }$ , find $\displaystyle AD$
(iii) Find the ratio of the area of $\displaystyle \triangle ADB$ is to area of $\displaystyle \triangle CDA .$ [2011]
$\displaystyle \text{(i) Let } \angle DAB = \theta$
$\displaystyle \text{Therefore } \angle DAC = 90^{\circ} - \theta$
$\displaystyle \angle DBA = 90^{\circ} - \theta$
$\displaystyle \angle DCA = \theta$
$\displaystyle \text{Therefore } \triangle ADB \sim \triangle CDA \text{ (AAA postulate) }$
$\displaystyle \text{(ii) } \frac{CD}{AD} = \frac{AD}{BD}$
$\displaystyle \Rightarrow AD^2 = CD \times BD = 8 \times 18 = 144$
$\displaystyle \text{Therefore } AD = \sqrt{144} = 12$
$\displaystyle \text{(iii) } \frac{ \text{ Area of } \triangle ADB}{ \text{ Area of } \triangle CDA} = \frac{\frac{1}{2} AD \times BD} {\frac{1}{2} AD \times CD} = \frac{BD}{CD} = \frac{18}{8} = \frac{9}{4}$
$\displaystyle \\$
Question 20: In the given figure $\displaystyle AB \text{ and } DE$ are perpendiculars to $\displaystyle BC .$
(i) Prove that $\displaystyle \triangle ABC \sim \triangle DEC$
(ii) If $\displaystyle AB = 6 cm, DE = 4 \text{ cm } \text{ and } AC = 15 \text{ cm }$ , calculate $\displaystyle CD$
(iii) Find the ratio of the $\displaystyle \text{ Area of } \triangle ABC : \text{ Area of } \triangle DEC .$ [2013]
$\displaystyle \text{(i) From } \triangle ABC \text{ and } \triangle DEC$
$\displaystyle \angle ABC = \angle DEC = 90^{\circ} \text{ (given) }$
$\displaystyle \angle ACB = \angle DCE \text{ (common angle) }$
$\displaystyle \triangle ABC \sim \triangle DEC \text{ (AAA postulate) }$
$\displaystyle \text{ (ii) Since } \triangle ABC \sim \triangle DEC$
$\displaystyle \text{In } \triangle ABC \text{ and } \triangle DEC$ ,
$\displaystyle \frac{AB}{ DE} = \frac{AC}{ CD}$
$\displaystyle AB = 6 cm, DE = 4 \text{ cm } \text{ and } AC = 15 \text{ cm }$
$\displaystyle \text{Therefore } CD = \frac{15}{6} \times 4 = 10 \text{ cm }$
$\displaystyle \text{ (iii) Since } \triangle ABC \sim \triangle DEC$
$\displaystyle \frac{ \text{ Area of } \triangle ABC}{ \text{ Area of } \triangle DEC} = \frac{AB^2}{ DE^2} = \frac{6^2}{4^2} = \frac{9}{4}$
Therefore the $\displaystyle \text{ Area of } \triangle ABC : \text{ Area of } \triangle DEC = 9:4$ |
# A closed box has a square base with side length l feet and height h feet. Given that the volume of the box is 35 cubic feet, express the surface area of the box in terms of l only. ??
Aug 23, 2017
$A = \frac{2 \left(70 + {I}^{3}\right)}{I}$
#### Explanation:
So we are told that the box has a square base, so all four sides are the same length I.
Volume = Length x Breadth x Height
So we know that Length and Breadth are both equal to I. Hence
$35 = I \cdot I \cdot h$
$\implies h = \frac{35}{I} ^ 2$
Now we need to calculate the surface are of the box. There are four side faces with area $I \cdot h$ as well as a top and bottom face of area ${I}^{2}$. Therefore
$A = 4 \left(I \cdot h\right) + 2 {I}^{2}$
$A = 4 I \cdot \frac{35}{I} ^ 2 + 2 {I}^{2}$
$A = \frac{140}{I} + 2 {I}^{2}$
$A = \frac{2 \left(70 + {I}^{3}\right)}{I}$ |
# What Are The Factors Of 26: Prime Factors Of 26
Knowing the factors of 26 will help you solve many maths problems. What are the factors? They are just the specific digits that divide the number equally leaving no remainders. Let us not waste the time and jump to learn the numbers that are factors of 26.
## What Are The Factors Of 26?
1, 2, 13, and 26 are factors of 26. The factor numbers are those specific numbers that divide the given digit in such a way that there will be no remainder left.
So all factors of 26 are 1, 2, 13, and 26. Now
• 1(26➗1=26)
• 2(26➗2=13)
• 13(26➗13=2)
• 26(26➗26=1)
1, 2, 13, and 26 are factors for 26.
After knowing what all the factors of 26 are let us know more about the 26 factors
## Prime Factors Of 26
To know the prime factors of 26, first, you should know the prime numbers up to 26. After that, we have to divide 26 by all prime numbers. Now, we will see the prime factors by factorization method.
• 26➗2 = 13
• 26➗3 = 8.66
• 26➗5 = 4.6
• 26➗7 = 3.71
• 26➗11 = 2.36
• 26➗13 = 2
• 26➗17 = 1.52
• 26➗19 = 1.36
• 26➗23 = 1.13
So the prime factor for the number 26 is 2, 5. and 7
## Factors Of 26 In Pairs
The product of the number whose result will be 26, that number is nothing but the pair factors of 26.
• 1 x 26 = 26
• 2 x 13 = 26
• 13 x 2 =26
• 26 x 1 = 26
So the pair factors of 26 are 1, 26, and 2, 13.
## Factors Of -26
The negative numbers of factors of 26 are called factors of -26.
Below You can see the numbers.
• -1(-26➗-1=-26)
• -2(-26➗-2=-13)
• -13(-26➗-13=-2)
• -26(-26➗-26=-1)
The Factors for -26 are -1, -2, -13, -26
## Factors Of 26 And Other Numbers
Here are the common factors of 26 and other numbers.
### The Factors Of 26 And 8
The factors for 26 and 8 are 1, 2, 13, 26, and 1, 2, 4, 8
• 26 factors – 1, 2, 13, 26
• 8 factors – 1, 2, 4, 8
HCF of 26 and 8 is 2
### The Factors Of 26 And 13
The factors for 26 and 6 are 1, 2, 13, 26, and 1, 13
• 26 factors – 1, 2, 13, 26
• 13 factors – 1, 13
HCF of 26 and 13 is 13
### The Factors Of 26 And 21
The factors for 26 and 21 are 1, 2, 13, 26, and 1, 3, 7, 21
• 26 factors – 1, 2, 13, 26
• 21 factors – 1, 3, 7, 21
HCF of 26 and 21 is 1
### The Factors Of 26 And 38
The factors for 26 and 38 are 1, 2, 13, 26, and 1, 2, 19, 38
• 26 factors – 1, 2, 13, 26
• 38 factors – 1, 2, 19, 38
HCF of 26 and 38 is 2
### The Factors Of 26 And 39
The factors for 26 and 39 are 1, 2, 13, 26, and 1, 3. 13, 39
• 26 factors – 1, 2, 13, 26
• 39 factors – 1, 3. 13, 39
HCF of 26 and 39 are 13
### The Factors Of 26 And 40
The factors for 26 and 40 are 1, 2, 13, 26, and 1, 2, 4, 5, 8, 10, 20, 40
• 26 factors – 1, 2, 13, 26
• 40 factors – 1, 2, 4, 5, 8, 10, 20, 40
HCF of 26 and 40 are 2
### The Factors Of 26 And 42
The factors for 26 and 42 are 1, 2, 13, 26, and 1, 2, 3, 6,7, 14, 21, 42
• 26 factors – 1, 2, 13, 26
• 42 factors – 1, 2, 3, 6,7, 14, 21, 42
HCF of 26 and 42 are 2
### The Factors Of 26 And 52
The factors for 26 and 52 are 1, 2, 13, 26, and 1, 2, 4, 13, 26, 52
• 26 factors – 1, 2, 13, 26
• 52 factors – 1, 2, 4, 13, 26, 52
HCF of 26 and 52 are 13
### The Factors Of 26 And 100
The factors for 26 and 100 are 1, 2, 13, 26, and 1, 2, 4, 5, 10, 20, 25, 50, 100
• 26 factors – 1, 2, 13, 26
• 100 factors – 1, 2, 4, 5, 10, 20, 25, 50, 100
HCF of 26 and 100 is 32
All factors for the numbers can be seen here on Factorsweb
## FAQ
### Which Factor Of 26 Is A Prime Number?
Prime numbers from 1 to 26 are 2, and 13
### How Many Primes Numbers Are There In 26?
There are 4 prime numbers from 1 to 26 which are 2, 3, 5, 7, 11, 13, 17, 19, and 23.
### What Are The Factors Of 26?
The factors of 26 are 1, 2, 13, and 26.
### What Are The Factors Pairs Of 26?
There are overall 4 factors of 26 among which 26 is the biggest factor and its positive factors are 1, 2, 13 and 26. The sum of all factors of 26 is 42 and its factors in Pairs are (1, 26) and (2, 13).
### What Are The Factors Of 26 GCF?
The greatest integer that can divide the numbers 26 and 91 exactly is the HCF value. The factors of 26 are 1, 2, 13, and 26, and the factors of 91 are 1, 7, 13, and 91.
### What Are The Prime Factors Of 24 And 26?
Prime factorization of 24 and 26 is (2 × 2 × 2 × 3) and (2 × 13) respectively. As visible, 24 and 26 have only one common prime factor i.e. 2.
## Conclusion
Today we gained knowledge of factors of 26 and also prime factors for 26, factors for 26 in pairs, factors for -26, and factor numbers of 26. As you have read the above article you can easily calculate and answer the question of what are factors of 26.
What are the prime factors of 26
What are the prime factors of 26? |
Question
Wed November 14, 2012 By:
# what should be done in splitting the middle term method?
Thu November 15, 2012
We will factorise the quadratic expression: 2x² + 3x - 5 to demonstrate the technique. Here are the steps:
(a) Split the first term into two factors with the variable in each one:
2x² = 2x , x
(b) Split the last term into two factors:
-5 = -1 , 5
(c) Write the factors like this:
term 1 term 3 2x -1 x 5
(d) Multiply the factors in opposite corners and add the two answers, these may be the two factors for splitting the middle term:
term 1 term 3 2x -1 x 5
2x , 5 + x , -1 = 10x - x
= 9x
If the two products add to give the middle term then the split is correct. But in this case, the middle term is 3x, not 9x.
We have not chosen the correct factors. TRY AGAIN ... BACK TO STEP (c)
(c) Try different term 3 factors, or the same factors in a different arrangement:
term 1 term 3 2x 5 x -1
(d) Multiply the factors in opposite corners:
term 1 term 3 2x 5 x -1
See if the two products add to give the middle term:
5x - 2x = 3x
Yes! So we have chosen the correct factors, and the quadratic expression can be written ready to be factorised, like this:
2x² - 2x + 5x - 5
The two factors from the top row of the table: ‘2x’ and ‘5’ and the two factors from the bottom row of the table: ‘x’ and ‘-1’ in step (c) now give the factorised quadratic immediately:
term 1 term 3 2x 5 x -1
(x - 1)(2x + 5)
Related Questions
Mon April 03, 2017
# Solve for x:. 1/2x-3 + 1/x-5 = 10/9
Fri March 31, 2017
Home Work Help |
# How to Measure Euler’s Number by Shuffling Cards.
Previously on this blog, I have talked about Buffon’s Needle, which is a way to measure $\pi$ using random numbers. In this entry, I will be talking about a way to measure another fundamental constant $e$, also by using random numbers.
To measure $e$, we will shuffle a deck of cards many times and count the number of shuffles that result in the deck being deranged. This means that no card remains in its original position in the deck. This type of arrangement is called a “derangement”.
We will start by computing the theoretical probability ($p$) getting a derangement, which turns out to be $1/e$ or about 36.7%, for a sufficiently large number of cards. We will then experimentally measure $e$ by shuffling cards.
To compute the theoretical probability $p$, we need to understand two numbers:
The first is the total number of arrangements that the cards can be in ($R$).
The second is how many of those arrangements are derangements ($D$).
The theoretical probability is the ratio of those two numbers:
$p = D/R$
and it approaches $1/e$.
(Or you could say $R/D$ approaches $e$.)
The first number is easy, it’s just the factorial of the number of cards in the deck ($n$).
$R_n = n!$
The second number $D_n$ is not as straightforward. We have to count the number of derangements a deck can have.
To show a simple example of counting derangements, let’s look at a deck of only 3 cards:
There are 6 ways to arrange these 3 cards ($3! = 6$)
$A,2,3$
$A,3,2$
$2,A,3$
$3,A,2$ (this one is a derangement)
$2,3,A$ (this one is a derangement)
$3,2,A$
Only 2 of these arrangements have every card moved from its original position. So the number of derangements for 3 cards is 2. So in this case, the probability of getting a derangement is about 33%.
$D_3 = 2$
$p_3 = D/R = 2/3! = 1/3$
That’s close to $1/e$, but not quite, because it turns out 3 cards are just not enough to produce that probability. We need more.
In general, with a deck of $n$ cards, we can arrange any number $k$ of those cards in $P_{nk}$ number of ways, where:
$P_{nk} = \frac{n!}{(n-k)!}$
We can use this formula to count the number of ways there are to rearrange a subset $k$ of the cards, while $n-k$ cards remain in a fixed position (we are counting the number of permutations).
Note that $n-k \neq 0$, then by definition, the corresponding arrangements will not be derangements. If we can count all of these, then we can subtract this number of arrangements from the total number $n!$ to get the number of derangements.
Let’s see this in action by going back to the deck of 3 cards. If we select all the cards to rearrange, we can get all 6 possible permutations:
$P_{3,3} = \frac{3!}{(3-3)!} = \frac{3!}{0!} = 3! = 6$
$A,2,3$
$A,3,2$
$2,A,3$
$3,A,2$
$2,3,A$
$3,2,A$
If we select only 2 cards out of the 6 to rearrange, we only get a subset of the arrangements. However if we count how many there are with the formula we get 6 permutations:
$P_{3,2} = \frac{3!}{(3-2)!} = \frac{3!}{1!} = 3! = 6$
So there are 6 ways to arrange 2 cards (6 ways to hold 1 card fixed). But why do we get 6 permutations here? Well, let’s look at all the ways to do this.
If we select 2 and 3 to rearrange, we get 2 permutations:
$A,2,3$
$A,3,2$
If we select A and 3 to rearrange, we get 2 permutations:
$A,2,3$
$3,2,A$
And if select A and 2 to rearrange, we get 2 permutations:
$A,2,3$
$2,A,3$
Notice however that $A,2,3$ occurs 3 times, and so there are actually 4 unique arrangements here, and they are not derangements by definition. So we have $6 - 4 = 2$ derangements for 6 cards.
But is there a way to count unique arrangements without doing it by hand like we did here? Let’s keep going with this pattern.
If we select only 1 card to rearrange, then that’s trivial, because you can’t rearrange one card.
$P_{3,1} = \frac{3!}{(3-1)!} = \frac{3!}{2!} = 3$
If selecting A:
$A,2,3$
If selecting 2:
$A,2,3$
If selecting 3:
$A,2,3$
All 3 permutations are the same arrangement!
Then if we select no cards to rearrange, this is also trivial.
$P_{3,0} = \frac{3!}{(3-0)!} = 1$
If selecting nothing:
$A,2,3$
That’s just 1 permutation.
All duplicate permutations, are accounted for here. The number of permutations for $P_{3,0}$ accounts for one of the duplicates in $P_{3,1}$:
$P_{3,1} - P_{3,0} = 3 - 1 = 2$
The permutations left in this difference account for 2 of the duplicates in $P_{3,2}$:
$P_{3,2} - (P_{3,1} - P_{3,0}) = 6 - 2 = 4$
And you are left with the unique 4 arrangements that are not derangements.
So to get the number of derangements, we just have to subtract this from the total number of arrangements $P_{3,3}$:
$P_{3,3} - (P_{3,2} - (P_{3,1} - P_{3,0})) = 6 - 4 = 2$
And so we see that $D_3 = 2$
We have a pattern here. We can see that by distributing the negative signs in the above equation we get:
$D_3 = P_{3,3} - P_{3,2} + P_{3,1} - P_{3,0}$
This can be expressed as a sum:
$D_3 = \sum\limits_{k = 0}^3 P_{3,k} (-1)^{3-k}$
Or more explicitly:
$D_3 = \sum\limits_{k = 0}^3 \frac{3!}{(3-k)!} (-1)^{3-k}$
It turns out that this pattern holds for decks of any number of cards $n$ (just replace the 3 here):
$D_n = \sum\limits_{k = 0}^n \frac{n!}{(n-k)!} (-1)^{n-k}$
For a deck of 3 cards, the number of derangements is: $D_3 = 2$
For a deck of 4 cards: $D_4 = 9$
For a deck of 5 cards: $D_5 = 44$
which then give us the probabilities of derangement:
$p_3 = 2/3! = 33.3\%$
$p_4 = 9/4! = 37.5\%$
$p_5 = 44/5! = 36.7\%$
So by 4 cards, we start seeing a probability closer to $1/e$.
Now, to get the probability $p$ of getting a derangement for $n$ cards:
$p_n = \frac{D_n}{R_n} = \frac{1}{n!}\,\sum\limits_{k = 0}^n \frac{n!}{(n-k)!} (-1)^{n-k}$
The $n!$ factors cancel out, and we are left with:
$p_n = \sum\limits_{k = 0}^n \frac{1}{(n-k)!} (-1)^{n-k}$
Since the order of addition doesn’t matter, we can change the direction of this sum to start at $n$ and end at 0:
$p_n = \sum\limits_{k = n}^0 \frac{1}{(n-k)!} (-1)^{n-k}$
If we do some relabeling, the relationship to $e$ will become more obvious. Let’s set $m = n-k$, so we can write:
$p_n = \sum\limits_{m=0}^n \frac{1}{m!} (-1)^{m}$
This looks like a truncated Taylor expansion of the exponential function $e^x$.
A Taylor expansion is a way of expressing a function as a polynomial.
The Taylor expansion of the exponential function is:
$e^x = \sum\limits_{m=0}^\infty \frac{1}{m!} x^m$
If we plug in $x = -1$ we get:
$e^{-1} = \sum\limits_{m=0}^\infty \frac{1}{m!} (-1)^m$
which is the limiting case of a deck with infinite cards:
$p_\infty = \frac{1}{e}$
This equivalence to the Taylor expansion is the reason why as you increase the number of cards, the probability of getting a derangement approaches $1/e$.
To put it another way:
$\frac{1}{p} = \frac{R}{D} \approx e$.
Now that we have seen the theoretical probability, let’s do the experiment.
Take a standard 52 card deck, note the original positions of the cards, then shuffle and see if you have a derangement. Keep shuffling and noting if the result is a derangement. Keep doing this about 40,000 times. Count the total number of shuffles $R$ and the number of derangements $D$ you get.
If you then take the ratio $\frac{R}{D}$ you will get something approaching $e$.
Just kidding. That would be extremely tedious. Instead, we can have a computer it for us.
For a deck of 52 cards, it seems to take about 40k shuffles for the ratio of shuffles to derangements to approach $e$. See the figure above.
The code used to compute $e$ from shuffles is shown below:
Here I used MATLAB code, to take advantage of vectorization and logical array operations. The inputs are cards (the number of cards in a deck) and games (the number of shuffles). Wins (a win is when you get a derangement) are then counted.
function eulersnum = eulerdeck(cards,games)
wins = 0;
series = 1:cards;
for ii=1:games
shuffled = randperm(cards);
wins = wins + ~any(series == shuffled);
end
eulersnum = games/wins;
end
The known value of $e$ is 2.7183…
To get an accurate measurement we should use a deck with a lot of cards, since that will include more terms of the Taylor expansion. We should also shuffle many times to be sure that we capture the probability.
So with a deck of 1000 cards, and 10 trials, each of 1000 shuffles, we measure $e$ to be:
$2.7265 \pm 0.1060$
# Fun with Repeating Decimals
So you may have noticed that when you divide an integer by 11, if it is not a multiple of 11, you get an alternating repeating decimal that looks like “.ABABABABABABAB…”. You may have also noticed that A and B always add up to 9. Let’s discuss why that is.
So let’s divide $n$ by 11, where $n$ is an integer less than 11. You only need to consider integers less than 11, because any improper fraction can be expressed as an integer plus a fraction where the numerator is less than the denominator.
For example: $\frac{48}{11} = 4 \frac{4}{11}$
Now if you convert that fraction into a decimal you get an alternating repeating decimal, $\frac{4}{11} = 0.363636363...$, where 3 and 6 add to 9.
This decimal can also be expressed in another form by turning the decimal places into a sum:
$\frac{4}{11} = 3 \cdot \frac{1}{10} + 6 \cdot \frac{1}{100} + 3 \cdot \frac{1}{1000} + 6 \frac{1}{10000} + ...$
We can then generalize this for any integer $n$ less than 11.
$\frac{n}{11} = A \cdot \frac{1}{10} + B \cdot \frac{1}{100} + A \cdot \frac{1}{1000} + B \frac{1}{10000} +....$
Where A and B are integers.
So now we want to prove that $A+B = 9$.
First let’s put the above equation into a summation form, since the sum is infinitely long, but can be expressed simply because A is multiplied by odd powers of 10, and B is multiplied by even powers of 10. This is done to simplify the expression so there is one instance of A and one instance of B.
$\frac{n}{11} = A \cdot \sum_{j=0}^{\infty} 10^{-(2j+1)} + B \cdot \sum_{j=0}^{\infty} 10^{-(2j+2)}$
Now we want to simplify this a bit more to get rid of the infinite sum.
First let’s pull out some factors of $\frac{1}{10}$ to make the sums in each term equivalent.
$\frac{n}{11} = A \frac{1}{10} \sum_{j=0}^{\infty} 10^{-2j} + B \frac{1}{100} \sum_{j=0}^{\infty} 10^{-2j}$
$= \left( A \frac{1}{10} + B \frac{1}{100} \right) \sum_{j=0}^{\infty} 10^{-2j}$
Now the sum can be reduced to a number: $\sum_{j=0}^{\infty} 10^{-2j} = 1.01010101... = \frac{100}{99}$
So now we have:
$\frac{n}{11} = \left( A \cdot \frac{1}{10} + B \cdot \frac{1}{100} \right) \frac{100}{99}$
which can be reduced to a nice expression:
$9n = 10A + B$
From this we want to prove that $A+B=9$, where $n, A, B$ are integers.
To prove this, let’s plug in a more general equation $A+B = \mu$ and solve for $\mu$. The result is:
$n = A + \frac{\mu}{9}$
Now since $n$ and $A$ are integers, $\frac{\mu}{9}$ must also be an integer, meaning $\mu$ must be a multiple of 9. Since $A$ and $B$ can only hold one digit, the only options are 9 and 18. But $\mu =18$ only occurs when $n=11$, which corresponds to $\frac{11}{11}=1$.
So for the case $n<11$, it must be that $A+B=9$. |
# RATIONALIZING THE DENOMINATOR
When a radical contains an expression that is not a perfect root, for example, the square root of 3 or cube root of 5, it is called an irrational number.
So, in order to rationalize the denominator, we have to get rid of all radicals that are in denominator.
The following steps are involved in rationalizing the denominator of rational expression.
Step 1 :
Multiply both numerator and denominator by a radical that will get rid of the radical in the denominator.
If the radical in the denominator is a square root, then we have to multiply by a square root that will give us a perfect square under the radical when multiplied by the denominator.
Step 2 :
Make sure all radicals are simplified.
Some radicals will already be in a simplified form, but we have to make sure that we simplify the ones that are not.
Step 3 :
Simplify the fraction if needed.
Be careful. We cannot cancel out a factor that is on the outside of a radical with one that is on the inside of the radical. In order to cancel out common factors, they have to be both inside the same radical or be both outside the radical.
Rationalize the denominator :
Example 1 :
¹²⁄
Solution :
¹²⁄
Multiply both numerator and denominator by √6 to get rid of the radical in the denominator.
= ⁽¹²ˣ⁶⁾⁄₍₆ₓ₆₎
= ⁽¹²ˣ⁶⁾⁄₆
= 2√6
Example 2 :
4√5/√10
Solution :
Simplify.
4√5/√10 = 4√5/√(2 ⋅ 5)
= 4√5/(√2 ⋅ √5)
On the right side, cancel out √5 in numerator and denominator.
= 4/√2
On the right side, multiply both numerator and denominator by √2 to get rid of the radical in the denominator.
= (4 ⋅ √2)/(√2 ⋅ √2)
= 4√2/2
= 2√2
Example 3 :
5/√7
Solution :
Multiply both numerator and denominator by √7 to get rid of the radical in the denominator.
5/√7 = (5 ⋅ √7)/(√7 ⋅ √7)
= 5√7 / 7
Example 4 :
12/√72
Solution :
Decompose 72 into prime factor using synthetic division.
√72 = √(2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3)
= 2 ⋅ 3 ⋅ √2
= 6√2
Then, we have
12/√72 = 12/6√2
Simplify.
= 2 / √2
On the right side, multiply both numerator and denominator by √2 to get rid of the radical in the denominator.
= (2 ⋅ √2) ⋅ (√2 ⋅ √2)
= 2√2/2
= √2
Example 5 :
1/(3 + √2)
Solution :
To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + √2), that is by (3 - √2).
1/(3 + √2) = [1 ⋅ (3 - √2)]/[(3 + √2) (3 - √2)]
(3 - √2)/[(3 + √2) (3 - √2)]
Using the algebraic identity a2 - b2 = (a + b)(a - b), simplify the denominator on the right side.
= (3 - √2)/[32 - (√2)2]
= (3 - √2)/(9 - 2)
= (3 - √2)/7
Example 6 :
(1 - √5)/(3 + √5)
Solution :
To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + √5), that is by (3 - √5).
(1 - √5)/(3 + √5) = [(1 - √5) ⋅ (3 - √5)]/[(3 + √5) (3 - √5)]
Simplify.
= [3 √5 - 3√5 + 5]/[32 - (√5)2]
= (8 - 4√5)/(9 - 5)
= 4(2 - √5)/4
= 2 - √5
Example 7 :
(√x + y)/(x - √y)
Solution :
To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (x - √y), that is by (x + √y).
(√x + y)/(x - √y) = [(√x + y) ⋅ (x + √y)]/[(x - y) (x + √y)]
Simplify.
= [x√x + √xy + xy + y√y]/[(x- (y)2]
= [x√x + √xy + xy + y√y]/(x- y2)
Example 8 :
3√(2/3a)
Solution :
3√(2/3a) = 3√2/33a
To rationalize the denominator in this case, multiply both numerator and denominator on the right side by the cube root of 9a2.
3√(2/3a) = [3√2 ⋅ 3√(9a2)]/[3√3a ⋅ 3√(9a2)]
Simplify.
3√(18a2)/3√(27a3)
3√(18a2)/3√(3 ⋅ 3 ⋅ 3 ⋅ a ⋅ a ⋅ a)
3√(18a2)/3a
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# How do you combine (4a)/(a^2-ab-2b^2) -( 6b)/(a^2+4ab+3b)?
##### 1 Answer
Mar 15, 2016
$\frac{4 a}{{a}^{2} - a b - 2 {b}^{2}} - \frac{6 b}{{a}^{2} + 4 a b + \textcolor{red}{3 {b}^{2}}}$
$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{{a}^{3} + 2 {a}^{2} b - 5 a {b}^{2} - 6 {b}^{3}}$
#### Explanation:
This problem makes more sense if the expression is:
$\frac{4 a}{{a}^{2} - a b - 2 {b}^{2}} - \frac{6 b}{{a}^{2} + 4 a b + \textcolor{red}{3 {b}^{2}}}$
$= \frac{4 a}{\left(a + b\right) \left(a - 2 b\right)} - \frac{6 b}{\left(a + b\right) \left(a + 3 b\right)}$
$= \frac{\left(4 a\right) \left(a + 3 b\right) - \left(6 b\right) \left(a - 2 b\right)}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$
$= \frac{4 {a}^{2} + 12 a b - 6 a b + 12 {b}^{2}}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$
$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{\left(a + b\right) \left(a - 2 b\right) \left(a + 3 b\right)}$
$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{\left({a}^{2} - a b - 2 {b}^{2}\right) \left(a + 3 b\right)}$
$= \frac{2 \left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)}{{a}^{3} + 2 {a}^{2} b - 5 a {b}^{2} - 6 {b}^{3}}$
Note that I did not try to factor $\left(2 {a}^{2} + 3 a b + 6 {b}^{2}\right)$ since it has a negative discriminant $\Delta = {3}^{2} - \left(4 \cdot 2 \cdot 6\right) = 9 - 48 = - 39$, so no linear factors with Real coefficients. |
# 1484 A Popsicle for a Hot Summer Day
### Today’s Puzzle:
It can be tricky to eat a melting popsicle on a hot summer day. Likewise, it can be tricky to solve a level 6 puzzle even if it looks like a popsicle! Making it was my granddaughter’s idea. We hope you enjoy it!
### Factors of 1484:
If you know that 7 x 12 is 84, it isn’t hard to recognize that 1484 is divisible by 7.
• 1484 is a composite number.
• Prime factorization: 1484 = 2 × 2 × 7 × 53, which can be written 1484 = 2² × 7 × 53
• 1484 has at least one exponent greater than 1 in its prime factorization so √1484 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1484 = (√4)(√371) = 2√371
• The exponents in the prime factorization are 2, 1, and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 1484 has exactly 12 factors.
• The factors of 1484 are outlined with their factor pair partners in the graphic below.
### Other facts about the Number 1484:
1484 is the difference of two squares in two different ways:
372² – 370² = 1484
60² – 46² = 1484
784-1260-1484 which is 28 times (28-45-53) |
# Squares And Perfect Squares
In these lessons, we will learn what are perfect squares and how to calculate the square root of a perfect square.
Related Pages
Perfect Cubes
Math Worksheets
### What Is The Square Of A Number?
The square of a number means to multiply the number by itself.
Example:
The square of 3 is 3 × 3 = 9
The square of a number can be written in exponent notation such as 32 where 3 is base and 2 is the exponent.
32 is read as “three to the second power” or “three squared”.
#### What Are Perfect Squares?
Perfect squares are the squares of whole numbers.
Example:
1, 4, 9, 16, 25 and 36 are the first 6 perfect squares because
12 = 1 × 1 = 1
22 = 2 × 2 = 4
32 = 3 × 3 = 9
42 = 4 × 4 = 16
52 = 5 × 5 = 25
62 = 6 × 6 = 36
#### How To Check For Perfect Squares?
We use repeated division by prime factors to check whether a given number is a perfect square.
Example:
Check whether 441 is a perfect square.
Solution:
441 = 3 × 3 × 7 × 7
= 3 × 7 × 3 × 7
= 21 × 21
= 212
So, 441 is a perfect square.
#### How To Determine If A Number Is A Perfect Square?
We can use prime factorization to check if a number is a perfect square.
#### How To Find Smallest Positive Whole Number That Is A Perfect Square And A Multiple Of A Specific Whole Number?
Examples:
1. What is the smallest positive whole number that is a perfect square and is a multiple of 24?
2. What is the smallest positive whole number that is a perfect square and is a multiple of 150?
#### How To Find The Square Of Negative Numbers, Decimals And Fractions?
We can also have the square of negative numbers, decimals and fractions.
When calculating the square of a number, take note of the following:
1. The square of a number is always positive.
Example:
(−5)2 = (−5) × (−5) = 25
(−7)2 = (−7) × (−7) = 49
Observe two important properties of a square in the last example above:
a) The square of a negative number becomes a positive number.
b) The square of a signed number is the same as the square of the unsigned number, i.e. (−7)2 = 49 = 72.
1. The square of a decimal will have twice the number of decimal places as the original decimal.
Example:
(0.3)2 = 0.3 × 0.3 = 0.09 ( 1 d.p. after squaring becomes 2 d.p.)
(0.03)2 = 0.03 × 0.03 = 0.0009 ( 2 d.p. after squaring become 4 d.p.)
10.22 = 10.2 × 10.2 = 104.04 ( 1 d.p. after squaring becomes 2 d.p.)
2. To square a fraction, multiply the numerator by itself and do the same for the denominator.
Example:
Take note that if a positive fraction which is less than 1 is squared, the result is always smaller than the original fraction.
1. To square a mixed number, change it to an improper fraction before calculating the square.
Example:
#### Squaring Negative Numbers
Look for parentheses to group negative numbers that are to be squared.
Example:
Simplify (−3)2 and −32
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. |
Class 9 Maths
# Polynomials
## Exercise 2.4 Part 2
Question 2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x^3 + x^2 – 2x – 1
g(x) = x + 1
Answer: Let g(x) = 0
Or, x + 1 = 0
Or, x = - 1
This means -1 is the zero of the given polynomial g(x) = x + 1
Now, according to Factor Theorem, if p(-1) is equal to zero, then g(x) is the factor of p(x) = 2x^3 + x^2 – 2x – 1
Or, p( - 1) = 2( - 1)^3 + ( - 1)^2 – 2( - 1) – 1
= 2 x ( - 1) + 1 + 2 – 1
= - 2 + 1 + 2 – 1 = 0
i.e. p( - 1) = 0
Since, p( - 1) = 0, therefore, g(x) = (x + 1) is a factor of p(x) = 2x^3 + x^2 – 2x - 1
(ii) p(x) = x^3 + 3x^2 + 3x + 1
g(x) = x + 2
Answer: Let g(x) = 0
Or, x + 2 = 0
Or, x = - 2, i.e. zero of (x + 2) = - 2
Now, (p – 2) = (- 2)^3 + 3( - 2)^2 + 3( - 2) + 1
= - 8 + 12 – 6 + 1 = - 1
Since, p(-2) ≠ 0, therefore, according to Factor Theorem g(x) = x + 2 is not the factor of given polynomial p(x) = x^3 + 3x^2 + 3x + 1
(iii) p(x) = x^3 – 4x^2 + x + 6
g(x) = x – 3
Answer: Let g(x) = 0
Or, x – 3 = 0
Or, x = 3
Now, p(3) = 3^3 – 4xx3^2 + 3 + 6
= 27 – 36 + 9 = 0
Since, p(3) = 0, therefore, according to Factor Theorem, g(x) = x + 3 is the factor of given polynomial p(x) = x^3 – 4x^2 + x + 6 |
# Full-Length GED Math Practice Test-Answers and Explanations
Did you take the GED Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.
## GED Mathematical Reasoning Practice Test Answers and Explanations
1- Choice A is correct
Plug in the value of $$x$$ and $$y$$: $$2x(y-2x)+(2y-x)^2$$ when $$x=-1$$ and $$y=2⇒2(-1)(2-2(-1))+(2(2)-(-1))^2=-8+25=17$$
2- Choice C is correct
The equation of a line in slope intercept form is: $$y=mx+b$$. Solve for $$y$$.
$$-4x-y=8⇒y=-8-4x$$. The slope of line is $$-4$$. Therefore, the slope of a line that is perpendicular to this line is $$m^{‘}=-\frac{1}{m}=\frac{1}{4}$$
3- Choice B is correct
To solve absolute values equations, write two equations.
$$2x-6$$ can equal positive 12, or negative 12. Therefore, $$\begin{cases}-2x-4=-8 \\ -2x-4=8 \end{cases}$$.
$$\begin{cases}-2x=-4 \\ -2x=12 \end{cases} ⇒𝑥=\frac{−8}{−4}=2, 𝑥=\frac{12}{−2}=−6$$
Find the product of solutions: $$2×(-6)=-12$$
4- Choice C is correct
A. $$(2, -8) ⇒ 2(2)-3(-8)=28≠13$$
B. $$(-2, -3) ⇒2(-2)-3(-3)=5≠13$$
C. $$(2, -3) ⇒2(2)-3(-3)=13=13$$
D. $$(2,3) ⇒2(2)-3(3)=-5≠13$$
Use PEMDAS (order of operation):$$-2×|-4|+3×(-2)-[10-6×(-7)]÷4=-8-6-13=-27$$
6- Choice D is correct
Use simple interest formula: I=prt
I = interest, p = principal, r = rate, t = time.
I=$$(25000)(0.029)(3)=2175$$
7- Choice A is correct
Let x be all expenses, then $$\frac{23}{100} x=575 →x=\frac{100×575}{23}=2500$$
He spent for his rent: $$\frac{34}{100}×2500=850$$
8- Choice B is correct
Write the numbers in order:$$4,12,15,19,21,25,34$$. Median is the number in the middle. So, the median is 19. The range is $$34-4=30$$
9- Choice D is correct
To get a sum of 8 for two dice, we can get 5 different options: $$(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)$$ To get a sum of 10 for two dice, we can get 3 different options: $$(6, 4), (5, 5), (4, 6)$$ Therefore, there are 8 options to get the sum of 8 or 10. Since, we have $$6 × 6 = 36$$ total options, the probability of getting a sum of 5 and 3 is 8 out of 36 or $$\frac{8}{36}=\frac{2}{9}$$.
The area of the square is 870.25. Therefore, the side of the square is the square root of the area. $$\sqrt{870.25}=29.5$$
Four times the side of the square is the perimeter: $$4 × 29.5 = 118$$
## The Absolute Best Book to Ace the GED Math Test
11- Choice A is correct
The perimeter of the trapezoid is 42 cm.
Therefore, the missing side (height) is $$= 45 – 15 – 9 – 5 = 13$$.
Area of a trapezoid: $$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2} (13) (5 + 9) = 91$$
Use formula of rectangle prism volume.
V = (length) (width) (height) $$⇒ 924 = (21) (11)$$ (height) ⇒ height $$= 924 ÷ 231 = 4$$
13- Choice D is correct
Change the numbers to decimal and then compare.
$$38\%=0.38 , 0.52=0.52, \frac{1}{10}=0.1, \frac{2}{5}=0.4$$ Therefore $$\frac{1}{10}<38\%<\frac{2}{5}<0.52$$.
14- Choice A is correct
Use the formula of areas of circles.
Area $$= πr^2 ⇒ 36 π> πr^2⇒ 36 > r^2⇒ r < 6$$
Radius of the circle is less than 6. Let’s put 6 for the radius. Now, use the circumference formula: Circumference $$=2πr=2π (6)=12 π$$
Since the radius of the circle is less than 6. Then, the circumference of the circle must be less than $$12π$$.
15- Choice A is correct
Volume of a box $$=$$ length $$×$$ width $$×$$ height $$= 5 × 5.5 × 9 = 247.5$$
16- Choice C and D is correct
(If you selected 3 choices and 2 of them are correct, then you get one point. If you answered 2 or 3 choices and one of them is correct, you receive one point. If you selected more than 3 choices, you won’t get any point for this question.) First, find the sum of five numbers.
average $$= \frac{sum of terms }{number of terms} ⇒ 32 = \frac{sum of 5 numbers}{5} ⇒$$sum of 5 numbers $$= 32 × 5 = 160$$
The sum of 5 numbers is 160. If a sixth number that is greater than 50 is added to these numbers, then the sum of 6 numbers must be greater than 210.
$$160 + 50 = 210$$
If the number was 50, then the average of the numbers is:
average $$=\frac{sum of terms }{number of terms}=\frac{210}{6}=35$$
Since the number is bigger than 50. Then, the average of six numbers must be greater than 35. Choices C and D are greater than 35.
17- Choice C is correct
If the length of the box is 45, then the width of the box is one fifth of it, 9, and the height of the box is 4.5 (one second of the width). The volume of the box is: V $$=$$ lwh $$= (45) (9) (4.5) = 1822.5$$
18- Choice B is correct
Write the equation and solve for $$B: 0.50 A = 0.80 B$$, divide both sides by 0.80, then: $$\frac{0.50}{0.80} A = B$$, therefore: $$B = 0.625 A$$, and $$B$$ is 0.625 times of $$A$$ or it’s $$62.5\%$$ of $$A$$.
19- Choice B is correct
Plug in 95 for F and then solve for C.
C $$= \frac{5}{9} (F – 32) ⇒$$ C $$= \frac{5}{9} (95 – 32) ⇒$$ C $$= \frac{5}{9} (63) = 35$$
20- Choice A is correct
Use simple interest formula: =prt (I = interest,p = principal,r = rate,t = time).
I=$$(75,00)(0.0325)(6)=1462.5$$
## Best GED Math Prep Resource for 2020
$$\frac{6^4}{3^2} =\frac{6×6×6×6}{3×3}=144$$
22- Choice B is correct
Use percent formula: part$$=\frac{percent}{100}×$$whole
$$57=\frac{percent}{100}×95 ⇒ 57=\frac{percent ×95}{100} ⇒$$ multiply both sides by 100.
$$5700=$$percent$$×95$$, divide both sides by 95. $$60=$$percent
23- Choice D is correct
Area of a trapezoid: $$A = \frac{1}{2} h (b_1 + b_2)$$. Therefore, the missing side is height(h). First, we should find $$x,h$$
$$x=\frac{32-20}{2}=4$$. And $$10^2=x^2+h^2⇒h=\sqrt{10^2-x^2}$$
$$h=\sqrt{10^2-4^2}=\sqrt{64}=8$$.
$$A=\frac{1}{2} h(b_1+b_2 )=\frac{1}{2}×8×(20+32)=208$$
24- Choice A is correct
Solving Systems of Equations by Elimination Add two equations.
$$\begin{cases}4x+2y=-9 \\ -2x-2y=-7 \end{cases} ⇒ 2x=-16 ⇒ x=-8$$
Plug in the value of x into one of the equations and solve for $$y$$.
$$4x+2y=-9⇒4(-8)+2y=-9⇒-32+2y=-9⇒2y=23⇒y=\frac{23}{2}$$
25- Choice B is correct
The equation of a line is in the form of $$y=mx+b$$, where m is the slope of the line and b is the y-intercept of the line.
Two points $$(1,0)$$ and $$(0,1)$$ are on line A. Therefore, the slope of the line A is:
slope of line A$$=\frac{y_2- y_1}{x_2 – x_1} = \frac{1-0}{0-1}=\frac{1}{-1}=-1$$
The slope of line A is $$-1$$. Thus, the formula of the line A is: $$y=mx+b=-x+b$$, choose a point and plug in the values of $$x$$ and $$y$$ in the equation to solve for b. Let’s choose point $$(0, 1)$$. Then:
$$y=x+b→1=0+b→b=1$$
The equation of line A is: $$y=-x+1$$
Now, let’s review the choices provided:
A.$$(1,2) ⇒2=-1+1$$. This is not true
B. $$(-1,2)⇒2=1+1=2$$. This is true
C. $$(2,1)⇒1=-2+1=-1$$. This is not true
D. $$(3,1)⇒1=-3+1=-2$$. This is not true
26- Choice A is correct
Use distance formula: Distance $$=$$ Rate $$×$$ time $$⇒ 540 = 75 ×$$ T, divide both sides by $$\frac{75. 540 }{75} =$$ T ⇒ T $$= 7.2$$ hours.
Change hours to minutes for the decimal part. 0.2 hours $$= 0.2 × 60 = 12$$ minutes
27- Choice D is correct
Let $$x$$ be the number. Write the equation and solve for $$x$$.
$$\frac{3}{5} ×22= \frac{3}{7}. x ⇒ \frac{3×22}{5}= \frac{3x}{7}$$, use cross multiplication to solve for $$x$$.
$$7×22=5x ⇒154=5x ⇒ x=30.8$$
28- Choice B is correct
To find the discount, multiply the number by ($$100\% –$$ rate of discount).
Therefore, for the first discount we get: $$(D) (100\% – 15\%) = (D) (0.85) = 0.85 D$$
For increase of $$10 \%: (0.85 D) (100\% + 10\%) = (0.85 D) (1.10) = 0.935 D = 93.5\%$$ of $$D$$
29- Choice C is correct
Use the formula for Percent of Change: $$\frac{New Value-Old Value}{Old Value} × 100 \%$$
$$\frac{34-45}{45} × 100 \% = –24.44 \%$$ (negative sign here means that the new price is less than old price).
30- Choices D and E are correct
(If you selected 3 choices and 2 of them are correct, then you get one point. If you answered 2 or 3 choices and one of them is correct, you receive one point. If you selected more than 3 choices, you won’t get any point for this question.) Some of prime numbers are: $$2, 3, 5, 7, 11, 13$$
Find the product of two consecutive prime numbers:
$$2 × 3 = 6$$ (not in the options)
$$3 × 5 = 15$$ (Choice D)
$$5 × 7 = 35$$ (Choice E)
$$7 × 11 = 77$$ (not in the options)
Choices D and E are correct.
The ratio of boy to girls is $$3:5$$. Therefore, there are 3 boys out of 8 students. To find the answer, first divide the total number of students by 8, then multiply the result by
$$3. 664 ÷ 8 = 83 ⇒ 83 × 3 = 249$$
32- Choice A is correct
The question is this: 440 is what percent of 550?
Use percent formula: part $$= \frac{percent}{100} ×$$ whole
$$440 = \frac{percent}{100} × 550 ⇒ 440= \frac{percent ×550}{100} ⇒44000 =$$ percent $$×550$$ percent $$= \frac{44000}{550} = 80. 440$$ is $$80 \%$$ of 550. Therefore, the discount is: $$100\% – 80\% = 20\%$$
33- Choice B is correct
If the score of Mia was 48, therefore the score of Ava is 16. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is 8.
34- Choice A is correct
The sample space S of the experiment described is as follows:
$$S={(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}$$
Let E be the event “the die shows an odd number and the coin shows a head”. Event E may be described as follows.
$$E={(1, H),(3,H),(5,H)}$$
The probability P(E) is given by P(E) $$= \frac{n(E)}{n(S)} = \frac{3 }{12} = \frac{1}{4}$$
35- Choice A is correct
Let $$x$$ be the smallest number. Then, these are the numbers:
$$x, x+1, x+2, x+3, x+4$$
average $$= \frac{sum of terms}{number of terms} ⇒ = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}⇒30=\frac{5x+10}{5} ⇒ 150=5x+10 ⇒ 140=5x ⇒ x=28$$
36- Choice D is correct
The area of the floor is: $$8$$ cm $$× 20$$ cm $$= 160$$ cm$$^2$$ The number of tiles needed $$= 160 ÷ 10 = 16$$
37- Choice B is correct
The weight of 9.6 meters of this rope is: $$9.6 × 500$$g $$= 4800$$ g $$1$$ kg $$= 1,000$$ g, therefore, $$4800$$ g $$÷ 1000 = 4.8$$kg
38- Choice B is correct
$$2.5\%$$ of the volume of the solution is alcohol. Let $$x$$ be the volume of the solution. Then:
$$2.5\%$$ of $$x = 30$$ ml $$⇒ 0.025 x = 30 ⇒ x = 30 ÷ 0.025 = 1200$$ ml
39- Choice C is correct
average $$= \frac{sum of terms}{number of terms}$$
The sum of the weight of all girls is: $$15 × 62 = 930$$ kg
The sum of the weight of all boys is: $$28 × 70 = 1960$$ kg
The sum of the weight of all students is: $$930 + 1960 = 2890$$ kg
average $$= \frac{2890}{43} = 67.21$$
40- Choice C is correct
Let $$x$$ be the original price. If the price of a laptop is decreased by $$12\%$$ to $$385$$, then:
$$88 \%$$ of $$x=385⇒ 0.88x=385 ⇒ x=385÷0.88=437.5$$
41- Choice A is correct
Write the numbers in order: $$14,14,15,16,18,22,62$$ Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 16.
42- Choice A is correct
Surface area of cone formula is : A$$=πr(r+\sqrt{h^2+r^2})$$. For r$$=17$$ and h$$=20$$ A$$=π(17)(17+\sqrt{20^2+17^2} )≈2309.79$$
43- Choice D is correct
Let $$x$$ be the number of years. Therefore, $$2,100$$ per year equals $$2100x$$.
starting from $$22,000$$ annual salary means you should add that amount to $$2100x$$.
Income more than that is: $$I > 2100x + 22000$$
44- Choice B is correct
The question is this: 1.65 is what percent of 1.05?
Use percent formula: part $$= \frac{percent}{100} ×$$ whole
$$1.65 = \frac{percent}{100} × 1.05 ⇒ 1.65=\frac{percent ×1.05}{100} ⇒165=$$percent $$×1.05 ⇒$$ percent$$=\frac{165}{1.05}=157\%$$
45- Choice D is correct
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$
$$20^2 + 15^2 = c^2 ⇒ 400 + 225 = c^2 ⇒ 625 = c^2 ⇒ c = 25$$
46- Choice A is correct
For each option, choose a point in the solution part and check it on both inequalities.
$$y>2x ,y≥-x+2$$
A. Point $$(0, 5)$$ is in the solution section. Let’s check the point in both inequalities.
$$5>0$$, It works
$$5≥0+2$$, it works (this point works in both)
B. Let’s choose this point $$(5, 0)$$
$$0>10$$ That’s not true!
$$0≥-5+2$$, it works
C. Let’s choose this point
$$(–5, 0) 0>-10$$, it works
$$0≥10+2$$ That’s not true!
D. Let’s choose this point
$$(0, 5) 5>0$$, That’s not true!
$$5≥0+2$$ it works
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# Case study Chapter 6 (Application of Derivative)
Case study 3:- Read the following and answer the question:(Case study application of derivative 3)
On the request of villagers, a construction agency designs a tank with the help of an architect. Tank consist of rectangular base with recangular sides, open at the top so that its depth is 2 m and volume is 8 cubic meter as shown below:
(i) If x and y represent the length and breadth of its rectangular base, then the relation between the variables is
(a) x + y = 8 (b) x . y = 4
(c) x + y = 4 (d) x/y = 4
(ii) If construction of tank cost ₹ 70 per sq. metre for the base and ₹ 45 per square metre for sides, then making cost ‘C’ expressed as a function of x is
(a) (b)
(c) (d)
(iii) The owner of a construction agency is interested in minimizing the cost ‘C’ of whole tank, for this to happen the value of x should be
(a) 4 m (b) 3 m
(c) 1 m (d) 2 m
(iv) For minimum cost ‘C’ the value of y should be
(a) 1 m (b) 3 m
(c) 2 m (d) 2 m
(v) The prsdan of village wants to know minimum cost. The minimum cost is
(a) ₹ 2000 (b) ₹ 4000
(c) ₹ 11,000 (d) ₹ 1000
8 = x.y.2
⇒ 2x.y = 8 ⇒ x.y = 4
Since ‘C’ is cost of making tank
∴ C = 70x.y + 45×2(2x + 2y)
⇒ C = 70x.y + 90(2x + 2y)
⇒ C = 70x.y +180(x + y)
⇒ C = 280 + 180(x + 4/x)
For maximum or minimum
⇒ x = 2(length can never be negative)
Now,
Hence to minimize C, x = 2 m
Since, x.y = 4
⇒ y = 4/x
⇒ y = 4/2 = 2 m
Since, C = 280+ 180(x + 4/x)
⇒ C = 280 + 180(2 + 2)
⇒ C = 280 + 180×4
⇒ C = 280 +720 = ₹1000
## Some other Case study problem
Case study 1: Read the following and answer the question(Case study application of derivative 1 )
A recangular hall is to be developed for a meeting of farmers in an agriculture college to aware them for new technique in cultivation. It is given that the floor has a fixed perimeter P as shown below.
Case study 2:- Read the following and answer the question.(Case study application of derivative 2)
Dr. Ritam residing to Delhi went to see an apartment of 3 BHK in Noida. The window of the house was in the form of a rectangle surmounted by a semicircular opening having a perimeter of the window 10 m as shown in figure.
Case study 4:- Read the following and answer the question:(Case study application of derivative 4 )
These days chinese and Indian troops are engaged in aggressive melee,face-off skirmishes at location near the disputed Pangong Lake in Ladakh.
One day a helicopterof enemy is flying along the curve represented by y = x² + 7. A soldier placed at (3, 7) wants to shoot down the helicopter when it is nearest to him. |
# The Modern GMAT’s Ancient Roots
Posted by on Sep 20th 2013
Mathematics credits its most famous formula to a legendary Greek, Pythagoras of Samos.
This gem of trigonometry has had far-reaching impact in every sphere of mathematics and appears quite frequently on the GMAT.
Pythagorean’s theorem applies to any right triangle (a triangle that contains a 90-degree angle).
In the equation, a and b represent the two legs (the shorter sides), and c represents the hypotenuse (the longer side, opposite the right angle).
Pythagorean’s theorem applies to any right triangle (a triangle that contains a 90-degree angle). In the equation, a and b represent the two legs (the shorter sides), and c represents the hypotenuse (the longer side, opposite the right angle).
Memorize these common Pythagorean triples.
Although every right-triangle must satisfy this equation, a few have the special distinction of having all their sides contain only integers. These special 3-integer combinations are known as Pythagorean triples. The most common example is the 3-4-5 right triangle.
New Pythagorean triples can be formed simply by multiplying an existing triple by a constant number. For example, if we multiply the 3-4-5 ratio by 2, we get the new Pythagorean triple, 6-8-10. Multiples of existing ratios can easily be calculated, so the only triples you need to memorize for the exam are these three unique ratios: 3-4-5, 5-12-13, 8-15-17.
Consider the following data sufficiency problem:
In the figure below, what is the value of c/a?
(1) a is 25% shorter than b
(2) b is 20% shorter than c.
To have sufficient data, we must have one and only one possible value for the ratio of c/a. We are given a right triangle, which means that the variables must obey the Pythagorean theorem. Perhaps if we can use the statements to replace the variable b, we might be able to express the equation in terms of only c and a alone. That will probably get us closer to finding the ratio.
If we translate statement 1), we have a = b – 0.25b = 3b/4. Rearranging, we have b=4a/3 . If we substitute this into Pythagorean’s theorem, we have:
Typically, when dealing with the squares of variables, we should consider both negative and positive solutions. In a geometry problem, however, lengths must always be positive numbers, so we know that both a and c can never take on negative values. c/a is therefore simply 5/3 — it is a single value, so we therefore have sufficient data.
We can apply the same technique to statement 2). Translating the statement and rearranging, we get b=4c/5. We substitute this back into Pythagorean’s theorem to find that:
Here, too, c/a = 5/3. Again, we have a single value, so we have sufficient data. Each statement alone provides sufficient data.
Ready to start your GMAT prep? Get the results you want, and the flexibility you need with The Economist GMAT Tutor. |
# Video: CBSE Class X • Pack 2 • 2017 • Question 20
CBSE Class X • Pack 2 • 2017 • Question 20
04:24
### Video Transcript
The dimensions of a solid iron cuboid are 4.4 metres by 2.6 metres by one metre. It is melted and recast into a hollow cylindrical pipe of inner radius 30 centimetres and thickness five centimetres. Find the length of the pipe.
In this scenario, we have two different shapes. First, we start out with a cuboid with given dimensions, which we’re told is made of metal and melted down into a totally different shape — a hollow cylinder. The hollow cylinder has an inner radius — we’ve called 𝑟 sub 𝑖 — which is given as 30 centimetres and a wall thickness of five centimetres, which we’ve called 𝑡.
Knowing all this, we want to solve for the length of the hollow cylinder 𝑙. If we label the volume of the cuboid 𝑉 sub 𝑐, we’re told that volume is 4.4 times 2.6 times one cubic metres. When it comes to the volume of the hollow cylinder, we can start calculating that by recalling that the area of a circle is equal to 𝜋 times the radius of this circle squared.
On the end of the cylinder, we want to solve for the shaded area, which is the difference between the area of two circles. If we call the volume of the hollow cylinder 𝑉 sub 𝑝 for the volume of the pipe, that volume is equal to the area of the larger circle on the end of the pipe 𝜋 times the inner radius plus the pipe thickness all squared minus the area of the smaller hollow circle on the end of the pipe 𝜋 times the inner radius squared all multiplied by 𝑙, the length.
We know that the volume of the cuboid is equal to the volume of the pipe. So we can say that the volume of the cuboid is equal to 𝜋 times the radius of the outer circle squared minus the radius of the inner circle squared all multiplied by the length of the pipe 𝑙.
Looking at the different units in this expression, we see that on the left-hand side we have length units of metres, while on the right we have length units of centimetres. Recalling that one metre is equal to 100 centimetres, we can rewrite our left-hand side as 440 times 260 times 100 cubic centimetres. And our right-hand side simplifies to 𝜋 times 35 squared square centimetres minus 30 squared square centimetres all multiplied by 𝑙. If we multiply 35 by itself, we find a result of 1225. And then, when we square 30, we find a result of 900.
When we then subtract the area of the inner circle from the area of the outer circle on the end of the pipe, we find a result of 325 square centimetres. Then, if we let the number 𝜋 equal exactly 3.14, we can multiply that value by the number of square centimetres on the right-hand side of our equation. When we do, we find a result of 1020.50.
And next, to convert this into an integer, we can multiply both sides of our equation by two. This leaves us with the equation 880 times 260 times 100 cubic centimetres is equal to 2041 square centimetres multiplied by 𝑙. We can rearrange the numbers on the left-hand side of our equation so that it reads 88 times 26 times 10000. Then, when we multiply 88 by 26, we find a result of 2288.
With that number on the left-hand side of our expression, we’re now ready to solve for the length 𝑙. To do that, we’ll divide both sides of our equation by 2041 square centimetres. This cancels this term out on the right-hand side. And considering the units on the left-hand side, we see that two factors of centimetres cancel out, leaving us with units of centimetres for our length.
We’re now ready to calculate the pipe length 𝑙. And we’ll do it by dividing 2041 into 22880000. To start off, 2041 goes into 2288 one time. Then, it divides into 2470 one time, it divides into 4290 two times, into 2080 one time, and finally into 390 zero times.
This means that 𝑙 is equal to 11210 centimetres or 112.10 metres. That’s the length of the pipe. |
# Divide Decimals by Decimals
Videos and solutions to help Grade 5 students learn how to divide decimal dividends by non-unit decimal divisors.
New York State Common Core Math Module 4, Grade 5, Lesson 31
Common Core Standards: 5.OA.1, 5.NBT.7, 5.NF.7
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 5
Common Core For Grade 5
Lesson 31 Problem Set
1. Estimate, then divide.
a. 53.2 ÷ 0.4 =
2. Estimate, then divide.
a. 9.42 ÷ 0.03 =
3. Solve using the standard algorithm. Use the thought bubble to show your thinking as you rename the divisor as a whole number.
b. 3.16 ÷ 0.04 = ______
4. The total distance of a race is 18.9 km.
a. If volunteers set up a water station every 0.7 km, including one at the finish line, how many stations will they have?
b. If volunteers set up a first aid station every 0.9 km, including one at the finish line, how many stations will they have?
(Homework) 4. Lucia is making a 21.6 centimeter beaded string to hang in the window. She decides to put a green bead every 0.4 centimeters and a purple bead every 0.6 centimeters. How many green beads and how many purple beads will she need?
5. In a laboratory, a technician combines a salt solution contained in 27 test tubes. Each test tube contains 0.06 liter of the solution. If he divides the total amount into test tubes that hold 0.3 liter each, how many test tubes will he need?
Lesson 31 Homework
This video shows how to divide decimal dividends by non-unit decimal divisors using the identity property of multiplication.
1. Estimate, then divide.
Example: 78.4 ÷ 0.7
a. 61.6 ÷ 0.8 =
2. Estimate, then divide.
a. 4.74 ÷ 0.06 =
Lesson 31 Homework
3. Solve using the standard algorithm. Use the thought bubble to show your thinking as you rename the divisor as a whole number.
b. 7.52 ÷ 0.08 = ______
5. A group of 14 friends collects 0.7 pound of blueberries and decides to make blueberry muffins. They put 0.05 pound of berries in each muffin. How many muffins can they make if they use all the blueberries they collected?
Lesson 30 Homework
1. Rewrite the division expression as a fraction, and divide.
2.4 ÷ 0.8 =
Lesson 31
Problem 2:
a. 21.56 ÷ 0.98 =
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# 1579 as a Product of Prime Factors
Do you want to express or show 1579 as a product of its prime factors? In this super quick tutorial we'll explain what the product of prime factors is, and list out the product form of 1579 to help you in your math homework journey!
## What is 1579 as a Product of Prime Factors?
Let's do a quick prime factor recap here to make sure you understand the terms we're using. When we refer to the word "product" in this, what we really mean is the result you get when you multiply numbers together to get to the number 1579.
In this tutorial we are looking specifically at the prime factors that can be multiplied together to give you the product, which is 1579.
Every number can be represented as a product of prime numbers. So when we talk aqbout prime factorization of 1579, we're talking about the building blocks of the number. A prime factor is a positive integer that can only be divided by 1 and itself. The prime factors of 1579 are all of the prime numbers in it that when multipled together will equal 1579.
In this case, the prime factors of 1579 are:
• 1579
We can now easily show 1579 as a product of the prime factors:
1 x 1579 = 1579
Note: 1579 and 1 are the only numbers that can be multiplied to equal 1579. Since 1 is not a prime number, this means that 1579 as a product of prime factors is not possible..
So there you have it. A complete guide to the factors of 1579. You should now have the knowledge and skills to go out and calculate your own factors and factor pairs for any number you like.
Feel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers!
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "1579 as a Product of Prime Factors". VisualFractions.com. Accessed on September 21, 2023. http://visualfractions.com/calculator/prime-factors/1579-as-a-product-of-prime-factors/.
• "1579 as a Product of Prime Factors". VisualFractions.com, http://visualfractions.com/calculator/prime-factors/1579-as-a-product-of-prime-factors/. Accessed 21 September, 2023.
• 1579 as a Product of Prime Factors. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/prime-factors/1579-as-a-product-of-prime-factors/.
## Product of Prime Factors Calculator
Want to find the prime factor for another number? Enter your number below and click calculate.
## Next Prime Factor Calculation
Eager to continue your learning of prime factorization? Why not try the next number on our list and see if you can calculate the product of prime factors for it for yourself? |
You are here:
Algebra/Maths
Question
Trying to work out:
8.75squared - 1.25 squared
I can do it by just multiplying out each side but wanted to know if there us a simpler formula. No Calculator allowed.
The problem is 8.75²-1.25². There are three ways to do it without a calculator.
1) As can be remembered from the equations in algebra, a²-b²=(a+b)(a-b).
Since a=8.75 and b=1.25, it can be seen that a+b = 10 and a-b = 7.5.
Thus, that answer is 10*7.5 = 75.
2) Note that if we have (n - m/4)² where n and m are integers,
the answer is n² - nm/2 + m²/16 (Well, n and m don't have to be integers, but it sure
makes for easy calculations if they are).
Here, we have n=9 and m=1, so the answer is 81 - 9/2 + 1/16.
Since 9/2 = 4.5 and 1/16 = .5/8 = .25/4 = .125/2 = .0625,
that gives 81 - 4.5 + 0.0625 = 76.5625.
It is known that 1.25 is 5/4, and the square of 5/4 is 25/16.
Now 25/16 = 1 9/16 = 1 + 8/16 + 1/16 = 1 + 1/2 + 1/16 = 1 + 0.5 + 0.0625 = 1.5625.
Thus, if we take 76.5625 - 1.5625, the decimals all cancel and the result is 75.
3) One last way to look at the problem it to note 8.75² - 1.25² = (35/4)² - (5/4)².
The 4² (4²=16)can be factored out, leaving (35² - 5²)/16.
Now 35² = 30² + 10*30 + 25 = 1,225 and 5²=25, so the difference is 1,200.
Since 12 is 3/4 of 16, it can be seen that 1200 = (3/4)1,600.
Using this, we get (3/4)1,600/16. Now 1,600/16 = 100, so we have 3/4 of 100,
which is known to be 75.
Note that (1) is probably the easiest way to do it, but I did it in my head by using (2) at first. After I thought of (2), I then realized the numbers could be expressed as fourths and the fractions combined, giving (3). Once I had done (2) and (3), I then realized (1) could be done, and I believe this is the easiest method to use.
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Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.
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# The Simplest Way to Find Factorial of A Number in Python
In this Python example, we will discuss the different ways to find the factorial of any given number
## 1. What is Factorial of a Number?
The factorial of a number is the function that multiplies the number by every natural number below it.
– Cuemath
In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n
Example:
Input : 4!
Output : 24
Explanation : 4 * 3 * 2 * 1
Input : 6!
Output : 720
Explanation : 6 * 5 * 4 * 3 * 2 * 1
Some of the topics which will be helpful for understanding the program implementation better are:
Let’s implement different ways and understand one by one.
## 2. Python Programs to Find Factorial
There are different ways by which we can find the factorial of a number. We will go through each of the methods and see how they can be implemented.
### 2.1. Using The Recursion
Using the recursive method, we will make one function and the function will be called recursively.
Let’s see how it works.
#Python Program for factorial using recursion
def factorial(n):
return 1 if (n==1 or n==0) else n * factorial(n - 1);
#Main
n = 7
print("Factorial of",n,"is",factorial(n))
Output
Factorial of 7 is 5040
In the above program, the function factorial calls itself under the condition if n is greater than 1
### 2.2. Using Iteration
In this method, we will iterate the loop and multiply the provided value till it reaches 1.
# Python Program to find factorial using reursion
def factorial(n):
if n < 0:
return "Not possible"
elif n == 0 or n == 1:
return 1
else:
f = 1
while(n > 1):
f *= n
n -= 1
return f
# Main
n = 5
print("Factorial of", n, "is", factorial(n))
Output
The Given Number is Odd Number
In the above program, we have made a function in which we have given the if…else… condition and if the certain condition is satisfied then the iteration will start and it will stop when the condition is false. Final value of f will be returned.
### 2.3. Using In-Built Function
Inside the math module, we have the in-built function factorial, which directly returns the factorial of the provided input number value.
# Python Program to find factorial using in-built function
import math
n = 8
print("Factorial of", n, "is", math.factorial(n))
In the above-provided program, we just call the factorial function from the math module and pass the value for which the factorial has to be calculated.
## 3. Conclusion
In this article, we discussed the different ways by which we can find the factorial of the given number. We have used a recursive function and inbuild function to find out the factorial of any given number in python.
Please follow the Python tutorial series or the menu in the sidebar for the complete tutorial series.
Also for examples in Python and practice please refer to Python Examples.
Complete code samples are present on Github project.
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In the figure PT is a tangent to the circle at T. If PA = 4 cm and AB = 5cm then PT =
In the figure PT is a tangent to the circle at T. If PA = 4 cm and AB = 5cm then PT =
1. A
4cm
2. B
5cm
3. C
6cm
4. D
cm
Fill Out the Form for Expert Academic Guidance!l
+91
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Verify OTP Code (required)
Solution:
From the given figure, we can see that OA, OB and OT are the radius of circles. Let us assume that the radius of the circle is r.
From the above figure let us consider the triangle OTP right angled at T. Here in triangle OTP, base is PT, altitude is OT and hypotenuse is OP.
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
Now let us consider the chord AB.
Now let us consider the property that any line perpendicular to a chord of the circle from the centre bisects the chord of the circle.
Here AB is a chord and OS is perpendicular to AB. So, from the above property we can say that S is the midpoint of AB.
So, as we have that AB=5cm, we get that
Now let us consider the triangle OSA right angled at S. Here in triangle OSA, base is SA, altitude is OS and hypotenuse is OA.
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
Now substituting the value of SA from above, we get,
Now, let us consider the triangle OPS right angled at S. Here in triangle OSP, base is SP, altitude is OS and hypotenuse is OP.
From the above figure we can see that the side SP is,
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
Now, from equations (2) and (3), we get,
Now, let us substitute this value in equation (1). Then we get,
So, the correct Ans is “Option 3”.
So, option 3 is correct.
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Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formula Perimeter of Triangle Formula Area Formulae Volume of Cone Formula Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics
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# Difference between revisions of "2007 AMC 12B Problems/Problem 17"
## Problem 17
If $a$ is a nonzero integer and $b$ is a positive number such that $ab^2=\log_{10}b$, what is the median of the set $\{0,1,a,b,1/b\}$?
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ a \qquad \mathrm{(D)}\ b \qquad \mathrm{(E)}\ \frac{1}{b}$
## Solution
Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $b=1$, $ab^2 > \log_{10} b$. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect.
This puts $a$ as the smallest in the set since it must be negative.
Checking the new equation: $-b^2 = \log_{10}b$
Near $b=0$, $-b^2 > \log_{10} b$ but at $b=1$, $-b^2 < \log_{10} b$
This implies that the solution occurs somewhere in between: $0 < b < 1$
This also implies that $\frac{1}{b} > 1$
This makes our set (ordered) $\{a,0,b,1,1/b\}$
The median is $b \Rightarrow \mathrm {(D)}$
## Cheap Solution that most people probably used
Led $b=0.1$. Then $a\cdot0.01 = -1,$ giving $a=-100$. Then the ordered set is $\{-100, 0, 0.1, 1, 10\}$ and the median is $0.1=b,$ so the answer is $\mathrm {(D)}$. |
# What is the percentage increase/decrease from 1 to 1094?
In this article we'll show you how to calculate the percentage increase/decrease from 1 to 1094. The calculation is very simple and knowing how to do this will help you calculate futurue increases or decreases in numbers as a percentage without having to search for it (or ask Siri).
In a rush and just need to know the answer? The percentage increase from 1 to 1094 is 109300%.
What is the % change from to
## Percentage increase/decrease from 1 to 1094?
Working out the percentage increase or decrease between two numbers is a common thing to do. For example, a shop owner that sold 1 t-shirts in October and then sold 1094 t-shirts in November. They might want to work out the percentage increase or decrease in sales to see how their store is performing and look for trends or reasons for that change.
Want to quickly learn or show students how to calculate percentage increase and decrease play this very quick and fun video now!
Like all types of percentage calculations, the way we work out a percentage increase or decrease between two numbers is pretty simple. The resulting number (the second input) is 1094 and what we need to do first is subtract the old number, 1, from it:
1094 - 1 = 1093
Once we've done that we need to divide the result, 1093, by the original number, 1. We do this because we need to compare the difference between the new number and the original:
1093 / 1 = 1093
We now have our answer in decimal format. How do we get this into percentage format? We multiply 1093 by 100:
1093 x 100 = 109300%
We're done! You just successfully calculated the percentage difference from 1 to 1094. You can now go forth and use this method to work out and calculate the increase/decrease in percentage of any numbers.
Hopefully this has given you the knowledge needed to go and calculate your own percentage increase or decrease between numbers without the need for Google searches. It's a very good skill to have and working with percentages regularly can help you in other aspects of math, like fractions.
Head back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand.
### Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is the percentage increase/decrease from 1 to 1094?". VisualFractions.com. Accessed on February 29, 2024. http://visualfractions.com/percentage-calculator/what-is-percentage-difference-from-1-to-1094/.
• "What is the percentage increase/decrease from 1 to 1094?". VisualFractions.com, http://visualfractions.com/percentage-calculator/what-is-percentage-difference-from-1-to-1094/. Accessed 29 February, 2024.
• What is the percentage increase/decrease from 1 to 1094?. VisualFractions.com. Retrieved from http://visualfractions.com/percentage-calculator/what-is-percentage-difference-from-1-to-1094/. |
### Mathematics Primary 4 (Basic Four) Second Term
Mathematics Primary 4 (Basic Four) Second Term
Subject Scheme & Timeline:
Please check through the topics down and be sure it conform with the scheme you are using.
WEEK ONE MULTIPLICATION OF NUMBERS BY 2-DIGIT NUMBERS Example 1 multiply 25 by 12 Method 1: column form Method 2: Expanded form
WEEK TWO SQUARES AND SQUARE ROOTS OF NUMBERS ( 1- digit and 2 – digit numbers)
WEEK THREE DIVISIONDivision of 2-digit and 3-digit numbers by numbers up to 9 without remainder Example 1: 78 ÷ 6
WEEK FOURLEAST COMMON MULTIPLES ( LCM)Example 1:Find the least common multiples of 2 and 3The multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36Thus the common multiples of 2 and three are 6, 12, 18 and 24 The smallest of these multiples(i.e. the least) is 6We say that the least common multiple of 2 and 3 is 6.
WEEK FIVE HIGHEST COMMON FACTOR (HCF) The product of 2 and 3 is; 2 x 3 = 62 and 3 are factors of 6 The factors of a number are numbers that divide the number without a remainder
WEEK SIXESTIMATIONRounding off decimals to the nearest whole numberRules for rounding off decimals to the nearest whole numberWhen rounding off decimals to the nearest whole number, look at the digit in the tenth place. If this digit is 5 or greater than 5, replace the digits after the decimal point with zero and add 1 to the digit in the units place If this digit is less than 5, replace the digits after the decimal point with zero.
WEEK SEVENMoney Addition of money Example: find the sum of N4.36, N3.79 and N4.82
WEEK EIGHT PROBLEM ON MULTIPLICATION OF MONEY EXAMPLE Find the cost of 3 books at N91.55 each.
WEEK TENPROFIT AND LOSSExample 1A man bought a leather bag for N350.00 and sold it for N360.00. will he have more money or less money with him?
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# GED Math : Numbers
## Example Questions
### Example Question #6 : Number Lines
Refer to the above number line. Which of the points is most likely the location of the number ?
Do not use a calculator.
Explanation:
, so .
Therefore, .
Of the four points, falls in this range, so it is the correct response.
### Example Question #7 : Number Lines
On a number line, how far apart are -2 and 7?
7 units
9 units
-9 units
5 units
2 units
9 units
Explanation:
On a number line, negative numbers lie to the left of zero and positive numbers lie to the right. To count the distance, count the slots between the two numbers:
There are TWO units between -2 and 0, and there are SEVEN units between 0 and 7. Together, there are NINE units between them.
### Example Question #41 : Numbers
On a number line, which of the following is the greatest distance?
The distance between -3 and 0.
The distance between -1 and 2.
The distance between -2 and 2.
The distance between -1 and 2.
The distance between -3 and 0.
The distance between -2 and 2.
All distances are equal.
The distance between -2 and 2.
Explanation:
On a number line, the negative numbers lie to the left of zero and positive numbers lie to the right. To visualize the distance between numbers, look at the units:
-3 and 0 are 3 units apart.
-1 and 2 are 3 units apart
-2 and 2 are 4 units apart. This is the greatest distance.
### Example Question #9 : Number Lines
Which distance is greatest?
The distance between -2 and 1.
The distance between 1 and 7.
The distance between -2 and 5.
The distance between -2 and 1.
All distances are equal.
The distance between -2 and 5.
The distance between 1 and 7.
The distance between -2 and 5.
Explanation:
Looking at a number line, you can visualize the distance between positive and negative numbers:
There are THREE units between -2 and 1
There are SIX units between 1 and 7.
There are SEVEN units between -2 and 5, so this is the greatest distance.
### Example Question #10 : Number Lines
A) The distance between -2 and 5 on a number line.
B) The distance between -1 and 3 on a number line.
4
3
5
6
7
3
Explanation:
The first step is to find out the values of A and B by counting the units between the given numbers.
A) The distance between -2 and 5 on a number line is 7.
B) The distance between -1 and 3 on a number line is 4.
### Example Question #42 : Numbers
A) The distance between -3 and 1 on a number line.
B) The distance between 2 and 5 on a number line.
C) The distance between -2 and 3 on a number line.
8
5
16
12
2
12
Explanation:
To find the distance between any two numbers on a number line, remember that negative numbers lie to the left of zero and positive numbers lie to the right. This is easier to do understand with a picture:
A) The distance between -3 and 1 on a number line is 4.
B) The distance between 2 and 5 on a number line is 3.
C) The distance between -2 and 3 on a number line is 5.
### Example Question #43 : Numbers
A) The distance between -3 and 0.
B) The distance between -1 and 2.
C) The distance between -2 and 2.
11
10
3
5
4
10
Explanation:
To find the sum of these values, you first need to find each individual difference. Use a number line to help visualize the distance between each pair of numbers:
A) The distance between -3 and 0 is 3.
B) The distance between -1 and 2 is 3.
C) The distance between -2 and 2 is 4.
### Example Question #44 : Numbers
Which of the following number lines represents the inequality ?
Explanation:
Start by solving the inequality.
Thus, the correct answer should show as between .
Recall that a sign means that there should be an open circle on the number line, and that a sign needs to have a closed circle on the number line.
### Example Question #1 : Decimals And Fractions
Convert the following decimal into a fraction:
Explanation:
The decimal, , is read as two-hundred forty-three thousandths, which translates to:
### Example Question #2 : Decimals And Fractions
Convert the following fraction to a decimal: |
## Ratios
4.1 Ratios
Definition Compares two or three quantities of the same kind that are measured in the same unit.
• The ratio of $$a$$ to $$b$$ is written as $$a:b$$.
Example The ratio of $$5\,000\text{ g}$$ to $$9\text{ kg}$$ can be represented as, \begin{aligned}&\space5\,000\text{ g}:9\text{ kg} \\\\&=5\text{ kg}:9\text{ kg} \\\\&=5:9. \end{aligned} Thus, the ratio is $$5:9$$.
The ratio of three quantities:
• Represent the relation between three quantities in the form of $$a:b:c$$.
Example Represent the ratio of $$0.02\text{ m}$$ to $$3\text{ cm}$$ to $$4.6\text{ cm}$$. \begin{aligned}&\space0.02\text{ m}:3\text{ cm}:4.6\text{ cm} \\\\&=2\text{ cm}:3\text{ cm}:4.6\text{ cm}\\\\&=2:3:4.6 \\\\&=20:30:46 \\\\&=10:15:23.\end{aligned}
Equivalent ratios:
Definition Two or more ratios that have the same value.
• Equivalent ratios can be found by writing the ratios as equivalent fractions.
Examples i) Multiplication Determine whether $$3:4$$ is the equivalent ratio of $$6:8$$. \begin{aligned} 3:4&=3\times2:4\times2 \\\\&=6:8. \end{aligned} Thus, $$3:4$$ is the equivalent ratio of $$6:8$$. ii) Division Determine whether $$7:28$$ is the equivalent ratio of $$1:4$$. \begin{aligned} 7:28&=7\div7:28\div7 \\\\&=1:4. \end{aligned} Thus, $$7:28$$ is the equivalent ratio of $$1:4$$.
Ratios in their simplest form:
• A ratio of $$a:b$$ is said to be in its simplest form if $$a$$ and $$b$$ are integers with no common factors other than $$1$$.
Example State $$800\text{ g}:1.8\text{ kg}$$ in its simplest form. \begin{aligned}&\space800\text{ g}:1.8 \text{ kg} \\\\&=800\text{ g}:1\,800\text{ g} \\\\&=800\div200:1\,800\div200 \\\\&=4:9. \end{aligned} Thus, $$4:9$$ is the simplest form of $$800\text{ g}:1.8\text{ kg}$$.
Examples i) Highest common factor (HCF) State the simplest form of $$32:24:20$$. Noted that $$4$$ is the HCF of $$32, 24$$ and $$20$$. Thus, \begin{aligned}&\space32:24:20 \\\\&= 32\div4:24\div4:20\div4 \\\\&=8:6:5. \end{aligned} ii) Lowest common multiple (LCM) State the simplest form of $$\dfrac{3}{5}:\dfrac{7}{10}$$. Noted that the LCM for $$5$$ and $$10$$ is $$10$$. Thus, \begin{aligned}&\space\dfrac{3}{5}:\dfrac{7}{10} \\\\&= \dfrac{3}{5} \times 10:\dfrac{7}{10}\times 10 \\\\&=6:7. \end{aligned} |
# Everyday Math Grade 3 Answers Unit 1 Math Tools, Time, and Multiplication
## Everyday Mathematics 3rd Grade Answer Key Unit 1 Math Tools, Time, and Multiplication
Finding Differences on a Number Grid
Family Note Today your child reviewed patterns on the number grid and used them to find differences between numbers. For example, one way to find the difference between 87 and 115 on the number grid is: Start at 87. Count the number of tens to 107. There are 2 tens, or 20. Count the number of ones from 107 to 115. There are 8 ones, or 8. The difference between 87 and 107 is 2 tens and 8 ones, or 28. Formal subtraction methods will be covered in the next unit.
Question 1.
The difference between 83 and 109 is _______ .
Question 2.
The difference between 97 and 125 is _______.
Question 3.
Explain how you solved Problem 2.
I solved the problem 2 by using the number grid.
I first counted the number of tens and then counted the number of ones.
After counting the tens and ones added the tens and ones and got the difference. You can look how i did in the image below.
Practice
Solve.
Question 4.
13 = 7 + _______
13 – 7 = 6
13 = 7 + 6
Question 5.
13 = 6 + _______
13 – 6 = 7
13 = 6 + 7
Question 6.
6 = _______ – 7
6 + 7 = 13
6 = 13 – 7
Question 7.
7 = _______ – 6
7 + 6 = 13
7 = 13 – 6 .
Telling Time
Family Note Today your child explored some of the math tools commonly used in third grade. We reviewed how to read a ruler to the nearest inch and centimeter, and how to tell time to the nearest hour, half hour, and 5 minutes. Help your child read each time by paying attention to the position of both the hour and the minute hands.
Question 1.
Draw the hour hand and the minute hand to show the time right now. Write the time.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed between 5 and 6 represents 5
and the Long hand is pointing to 8. 8 × 5 = 40.
So, the time on the clock is “5:40 o’clock”.
Write the time shown.
Question 2.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointing to 8
and the Long hand is pointing to 12.
So, the time on the clock is “8 o’clock”.
Question 3.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed in between 3 and 4
and the Long hand is pointing to 6.
So, the time on the clock is “3:30 o’clock”.
Question 4.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed between 6 past 7
and the Long hand is pointing to 3.
So, the time on the clock is “6:15 o’clock”.
Question 5.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed between 11 and 12
and the Long hand is pointing to 9.
So, the time on the clock is “11:45 o’clock”.
Question 6.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed to 7
and the Long hand is pointing to 2.
So, the time on the clock is “7:10 o’clock”.
Question 7.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed between 5 and 6
and the Long hand is pointing to 8.
So, the time on the clock is “5:40 o’clock”.
Question 8.
Explanation:
Clock is divided into 12 parts and each part is equal to 5 minutes.
Clock have 2 hands. Short hand is called the hour hand and the longer hand is called the minute hand.
As u can see in the image
Short hand is pointed between 5 and 6 represents 5
and the Long hand is pointing to 8. 8 × 5 = 40.
So, the time on the clock is “5:40 o’clock”.
Rounding Numbers
Family Note Today your child used open number lines ( see Example) to help round numbers to the nearest 10 and to the nearest 100. Rounding is one way to estimate calculations. For example, to estimate 83 − 37, your child can round 83 to 80 and 37 to 40, and then easily subtract 80 − 40 = 40, so an estimated answer for 83 − 37 is about 40. The actual answer, 46, is close to 40. Have your child explain how to use an open number line to round numbers.
Example: What is 72 rounded to the nearest 10? 70
Round each number. Show your work on an open number line.
Question 1.
What is 87 rounded to the nearest 10?
Explanation:
The nearest 10 to 87 are 80 and 90.
80 as the lowest 10 and 90 is the highest 10.
85 is the mid point of 80 and 90.
As 87 is above 85.
As the number is above the midpoint the largest is chosen.
87 is rounded to 90.
Question 2.
What is 283 rounded to the nearest 100?
Explanation:
Given 283
The nearest 100 to 283 are 200 and 300.
200 as the lowest 100 and 300 is the highest 100.
250 is the mid point of 200 and 300.
As 283 is above 250.
As the number is above the midpoint the largest is chosen.
283 is rounded to 300.
Question 3.
Round the numbers in the problem below to the nearest 10. You may sketch an open number line to help.
Use the rounded numbers to estimate the answer.
Then solve.
Estimate: ________ + __________ = __________
Telling Time to the Nearest Minute
Family Note Today your child practiced telling time to the nearest minute on analog clocks. Children used familiar times on the hour and half hour to help them read more precise times. For example, in Problem 1 the first clock shows 8:00. Children can use 8:00 as a familiar time to help them read the second clock as 8:06. They start at 8:00 and count by 5s to 8:05 and then 1 more to 8:06. As needed, help your child read and write each time.
Write each time shown. Use the first clock to help you read the time on the second clock.
Question 1.
Explanation:
The first clock represent 8 o’clock as long hand on 12 and short hand on 8.
In the second clock the short hand on 8 and the long hand is at 1 cross 1 that is 1 represents 5 minutes and plus 1 makes it 5 + 1 = 6 minutes.
It makes the time 8:06 o’clock.
Question 2.
Explanation:
In the first clock the long hand on 6 and short hand between 3 and 4 represent 3:30 o’clock .
In the second clock the short hand between 3 and 4 and the long hand is at 7 cross 4 that is 7 represents
7 × 5 minutes = 35 minutes and plus 4 makes it 35 + 4 = 39 minutes.
It makes the time 3:39 o’clock.
The time is 3:40 o’clock.
Question 3.
Explanation:
In the first clock the long hand on 9 and short hand between 1 and 2 represent 1:45 o’clock .
In the second clock the short hand between 1 and 2 and the long hand is at 10 cross 2 that is 10 represents
 10× 5 minutes = 50 minutes and plus 2 makes it 50 + 2 = 52 minutes.
It makes the time 1:52 o’clock.
The time is 1:50 o’clock.
Talk about when you may need to tell time to the nearest minute.
Finding Elapsed Time
Family Note Your child is learning how to use a model, such as a number line or clock, to determine elapsed time. Today we used an open number line like the one shown in the example below to figure out how long a morning class lasts. Have your child explain the example to you.
Example: A swim meet started at 3:45 P.M. and ended at 6:15 P.M. Fill in familiar times on the number line and use it to answer the question.
Ava solved the problem this way:
15 min + 1 hr + 1 hr + 15 min = 2 hrs and 30 min
How long was the swim meet? _ 2__ hours and __30__Â minutes
Question 1.
Explain Ava’s strategy to someone at home.
Ava was using number line with the clock values. The starting and ending time of the meet are 3:45 and 6:15.
The starting value is 3:45 and the ending value is 6:15
Ava rounded the starting time 3:45 to the nearest hour 4:00 and then added hours 4 + 1 = 5:00 and 5 + 1 = 6:00 and then add 15 minutes to reach the end point.
Question 2.
How much time do you usually have between the end of school and when you go to bed?
I leave school at ________. I go to bed at ________.
Solving Problems in Bar Graphs
Family Note Today your child collected and organized data about the number of letters in the class’s first and last names into tally charts. Then children represented the data in bar graphs. Help your child read the data in the tally chart below and then represent the data on the bar graph. Note that the scale on the bar graph shows intervals of 2, so each interval represents 2 children.
Look at the data in the tally chart.
Show the data in the tally chart on the bar graph. Look carefully at the scale
How Bay School 3rd Graders Get to School
Sharing Strategies for Multiplication
Family Note Today your child explored number stories that involved placing items in equal groups and organizing them into rows and columns, or arrays. (See examples below.) We used drawings and multiplication number models to help make sense of these stories. Help your child make sense of the number stories below. Note that each story can be represented by either an addition or a multiplication number model; one or the other is acceptable.
For each number story:
• Draw a picture to match.
• Solve the problem.
• Write a number model to represent the story and your answer.
Question 1.
Thaddeus buys 5 bags of apples for a picnic. There are 6 apples in each bag. How many apples does he have?
_________ apples
Number model: ___________
Question 2.
Elsa is planting a garden. She plants 3 rows of vegetables, with 8 plants in each row. How many plants in all are in Elsa’s garden?
___________ plants
Number model: ___________
Question 3.
Egg tray with 2 rows with 3 eggs each.
Ice tray with 2 rows with 10 ice cubes each.
Makeup kit
Cup cake tray.
Tv remote
Keyboard.
Question 4.
Write an equal-groups number story about one set of objects. Use the back of this page. Solve the number story.
Introducing Division
Family Note Today your child explored ways to solve number stories using division. In the stories below the total number of objects is given, so your child needs to find either the number in each group or the number of groups. If needed, help your child count out pennies or dried beans to match the total in each story and use them to act out the story.
Please send in an unopened, 1-liter bottle of water for use in an upcoming lesson on measuring mass
Question 1.
Connie has 18 toys to put away. She puts 6 toys in each basket. How many baskets does she use?
Question 2.
Jamal is bagging prizes for the school fair. There are 30 prizes and Jamal wants to put 3 prizes into each bag. How many bags did Jamal make?
_________ bags
Question 3.
Think of things at home that could be shared equally by your family. Record them on the back of this page.
Biscuits.
Car
Brinks
Question 4.
Write a number story about equally sharing one of the things you wrote for Problem 3. Use the back of this paper. Then solve your number story.
Their are 12 pieces of bread. Sam toasted 4 pieces of bread for each as breakfast . How many people had bread for breakfast?
Total number of bread pieces = 12
Number of bread pieces used by Sam = 4
Number of people had their breakfast = 12 ÷ 4 = 3
Foundational Multiplication Facts
Family Note Today your child worked on developing strategies for solving 2s, 5s, and 10s multiplication facts. These facts will be used later to help solve related multiplication facts. Fact Triangles are the Everyday Mathematics version of traditional flash cards. They are better tools for building fact fluency and mental-math reflexes, however, because they emphasize fact families.
A fact family is a group of facts made from the same three numbers. For 6, 5, and 30, the multiplication and division fact family is 5 × 6 = 30, 6 × 5 = 30, 30 ÷ 6 = 5, 30 ÷ 5 = 6.
Fact Triangles arrange the three numbers such that the product is below the dot at the top and the factors are in the other two corners.
Use Fact Triangles to practice basic facts with your child. Cut out the triangles from the three attached sheets. Cover either the number below the large dot (the product) or one of the numbers in a corner (a factor).
If your child misses a fact, flash the other two problems and then return to the fact that was missed.
Example: Ravi can’t answer 15 ÷ 3. Flash 3 × 5, and then 15 ÷ 5, and finally 15 ÷ 3 a second time.
Make this activity brief and fun. Spend about 10 minutes each night for the next few weeks, or until your child learns them all. The work you do at home will support the work we are doing at school.
×, ÷ Fact
Triangles 1: 2s, 5s, and 10s
Triangles 2: 2s, 5s, and 10s
Triangles 3: 2s, 5s, and 10s
Finding Elapsed Time
Family Note Today your child learned about elapsed time. Children use clocks and open number lines to figure out the total minutes and hours that pass from a start time to an end time. Throughout the year, they will practice calculating lengths of days using sunrise and sunset data.
Example: Ann starts swim practice at 4:05 P.M. and finishes at 4:55 P.M.
How long is Ann’s swim practice? Use the open number line to help . 50 minutes
Calculating elapsed time on an open number line:
5 + 5 + 30 + 10 = 50 minutes
Find the elapsed time. Use the open number line below to help.
Question 1.
Devin left for a bike ride at 10:15 A.M. He arrived at his friend’s house at 10:35 A.M. How long was his bike ride?
Devin’s bike ride was _________ minutes long
Estimating Mass
Family Note Today your child explored grams and kilograms by measuring the masses of different object s with a pan balance and standard masses. Help your child solve the number stories below.
Solve. Hint: 1 kilogram = 1,000 grams
Question 1.
If a bottle of water has a mass of about 1 kilogram, about how much mass will it have after someone drinks 500 grams of water from it?
Question 2.
Emmi’s bag has a mass of 2 kilograms. Marco’s bag has a mass of 1,000 grams. Whose bag has more mass? Explain
Practice
Fill in the unit box. Solve.
Question 3.
20 – 10 = _______
20 – 10 = 10
Explanation:
Subtracting any two number gives the difference.
20 – 10 = 10
20 is the minuend
10 is subtrahend
10 is the difference.
Question 4.
20 – 9 = ________
20 – 9 = 11
Explanation:
Subtracting any two number gives the difference.
20 – 9 = 11
20 is the minuend
9 is subtrahend
11 is the difference.
Question 5.
20 – 8 = _______
20 – 8 = 12
Explanation:
Subtracting any two number gives the difference.
20 – 8 = 12
20 is the minuend
8 is subtrahend
12 is the difference.
Question 6.
20 – 7 = _________ |
# Lesson 18
Applications of Expressions
• Let’s use expressions to solve problems.
### 18.1: Algebra Talk: Equivalent to $0.75t-21$
Decide whether each expression is equivalent to $$0.75t - 21$$. Be prepared to explain how you know.
$$\frac34t - 21$$
$$\frac34(t - 21)$$
$$0.75(t - 28)$$
$$t - 0.25t - 21$$
### 18.2: Two Ways to Calculate
Usually when you want to calculate something, there is more than one way to do it. For one or more of these situations, show how the two different ways of calculating are equivalent to each other.
1. Estimating the temperature in Fahrenheit when you know the temperature in Celsius
1. Double the temperature in Celsius, then add 30.
2. Add 15 to the temperature in Celsius, then double the result.
2. Calculating a 15% tip on a restaurant bill
1. Take 10% of the bill amount, take 5% of the bill amount, and add those two values together.
2. Multiply the bill amount by 3, divide the result by 2, and then take $$\frac1{10}$$ of that result.
3. Changing a distance in miles to a distance in kilometers
1. Take the number of miles, double it, then decrease the result by 20%.
2. Divide the number of miles by 5, then multiply the result by 8.
### 18.3: Which Way?
You have two coupons to the same store: one for 20% off and one for $30 off. The cashier will let you use them both, and will let you decide in which order to use them. • Mai says that it doesn’t matter in which order you use them. You will get the same discount either way. • Jada says that you should apply the 20% off coupon first, and then the$30 off coupon.
• Han says that you should apply the \$30 off coupon first, and then the 20% off coupon.
• Kiran says that it depends on how much you are spending.
Do you agree with any of them? Explain your reasoning. |
# How to teach fractions using the CPA approach
Editor’s Note:
This is an updated version of a blog post published on 12 October, 2020.
Many children and adults find the concept of fractions difficult, and with good reason. Maths mastery expert Dr Yeap Ban Har gives advice on how to teach this abstract convention by using the CPA approach.
A few years ago, I was fortunate enough to attend a professional development course taught by Dr Yeap Ban Har. He explained that one of the major reasons children struggle with conceptually understanding fractions simply comes down to the way we write them.
And yet, the importance of recording fractions in a way that’s easy for children to understand is fundamental to teaching maths for mastery.
## Why difficulties can arise from the way we record fractions
One of the major difficulties when it comes to understanding fractions is the convention of how we record the information.
To unpick this, one of the first things we need to understand is that, in mathematics, there are four kinds of numbers:
1. Cardinal Numbers
Cardinal numbers denote a quantity: 1, 2, 3, 4, 5.
2. Ordinal Numbers
Ordinal numbers define the position of something in a series: first, second, third.
3. Measurement Numbers
Measurement numbers have a unit attached to them. Saying “I weigh 50,” doesn’t make sense; we need to include a unit of measurement, like “I weigh 50 kgs.”
4. Nominal Numbers
A nominal number is a number used only as a name. For example, the 23 bus.
Why is knowing this important when teaching fractions?
If I asked you to write ‘3 quarters’ on your paper, most of us would likely write the symbolic notation 34 to represent the fraction. As you can see, this means we would write the numerical symbol 3 to represent the numerator, and the numerical symbol 4 to represent the denominator
However, the way we record the denominator so similarly to the numerator can cause issues for many learners.
When we record the numerator, we use a numerical symbol (so, a number) in a cardinal way to represent the quantity. When we record the denominator, we also use a numerical symbol, but we’re using it in a nominal way. By representing the denominator with a numerical symbol, many learners assume it is also a cardinal number like the numerator, when in fact it is not. The denominator is essentially a noun — it’s the name of the fraction.
For example, saying 34 is the same as saying ‘3 apples’. What we mean is ‘I have 3 of something. I have 3 of that thing. I have 3 quarters.
Children often struggle when we’re teaching them how to record fractions, because we don’t explicitly teach them that the denominator is not a cardinal number — it’s a nominal number.
## What’s the most common mistake children make when learning fractions?
This lack of understanding the difference between the numerator and denominator is illustrated when learners begin adding fractions. The most common mistake children make is that they will add the denominators together the same way they add the numerators. This reflects a lack of understanding that the denominator is a noun — it’s the name of the fraction — and not a quantity, like the numerator.
Key takeaway: We need to give learners an understanding that the denominator and the numerator are different, even though we are representing them both in the same way.
## Easy ways to help learners understand and record fractions
The symbolic notation of fractions can create problems for learners. And often, as teachers, we introduce it too early, which unintentionally creates further confusion. So how do we help children develop an understanding that the denominator and the numerator are different?
Here are some easy ways to teach fractions and give your learners a deep understanding of how they work.
1. Use concrete resources
Concrete resources are critical for giving learners a conceptual understanding of fractions. Using resources to support teaching the concept of adding fractions is an easy way to record fractions and help children see that only the numerator is cardinal and represents a quantity.
2. Don’t introduce the concept of fractions with the symbolic notation
Fractions follow the language convention of ‘number plus noun’, like place value. For example, 34 follows the same language convention as ‘3 tens.’ Both mean ‘we have 3 of that thing.’
When we introduce fractions to younger learners, explicitly following this convention will develop their understanding of the numerator and denominator. Rather than recording fractions with the symbolic notation, record the numerator with the numerical symbol and the denominator as a written word.
For example: 34 can be recorded as ‘3 quarters’.
Recording fractions initially as ‘the number plus the noun’ will help learners develop an understanding that the numerator is cardinal and the denominator is nominal.
3. Use pictorial and visual representations to support abstract maths.
Similar to using concrete resources, it’s important to support abstract maths with visual representations. Drawing diagrams to represent 34 added together with 24 will help learners develop an understanding that only the numerator represents a quantity.
Visually representing the abstract concept of fractions is important for developing your learners’ conceptual understanding of fractions throughout their primary school learning. Visual representations will help learners understand conceptually what the abstract notations and procedures mean.
Information in mathematics is highly encoded — really, we’re using a bunch of symbols to represent abstract concepts. So to ensure we give our learners a conceptual understanding of fractions, we need to carefully consider how we’re presenting the information to them.
## Book a free trial!
Get an in-depth introduction to Maths — No Problem! and see what this award-winning maths mastery programme can do for your school.
Alex Laurie |
### Completing the square
#### November 6, 2013
I was covering "completing the square" in class today and one of the examples caused a little difficulty. I thought I would use it as a chance to test a new plug-in for this site.
The question involved was: Use completing the square to solve $$2x^2 = 3x+6$$.
We begin by rearranging and setting equal to zero
$$2x^2-3x-6=0$$
and then remove the common factor of 2. I tend to remove it just from the first 2 terms to avoid fractions in the third term - this doesn't actually matter in this case but I will continue with the method I used in class
$$2x^2-3x-6 = 2\left(x^2-\frac{3}{2} \right) - 6.$$
Now we can completing the square:
$$2\left(x^2-\frac{3}{2} \right) - 6 = 2\left[\left(x-\frac{3}{4}\right)^2-\frac{9}{16}\right]-6.$$
In this step we, halved the $$\frac{3}{2}$$ to get the $$\frac{3}{4}$$ and the $$-\frac{9}{16}$$ comes from need to cancel the extra $$\frac{9}{16}$$ that is created from multiplying out
$$\left(x-\frac{3}{4}\right)^2 = \left(x-\frac{3}{4}\right)\left(x-\frac{3}{4}\right) = x^2-\frac{3}{4}x + \frac{9}{16}.$$ Finally we simplify to get $$2\left[\left(x-\frac{3}{4}\right)^2-\frac{9}{16}\right]-6 = 2\left(x-\frac{3}{4}\right)^2-\frac{18}{16}-6 = 2\left(x-\frac{3}{4}\right)^2-\frac{57}{8}.$$
We have now finished completing the square and know that
$$2x^2-3x-6= 2\left(x-\frac{3}{4}\right)^2-\frac{57}{8}.$$
Onto actually solving $$2x^2-3x-6=0$$. We use our work from above because it doesn't factorise neatly so
\begin{align} 2x^2-3x-6 & = 0 \\ 2\left(x-\frac{3}{4}\right)^2-\frac{57}{8} &= 0 \\ 2\left(x-\frac{3}{4}\right)^2 & = \frac{57}{8} \\ \left(x-\frac{3}{4}\right)^2 &= \frac{57}{16} \end{align}
Now we need to get rid of the $$^2$$ by taking the square root, remembering that we're now expecting to use both the postive and negative roots.
\begin{align} \left(x-\frac{3}{4}\right)^2 &= \frac{57}{16} \\ x-\frac{3}{4} &= \pm \sqrt{\frac{57}{16}} \\ x &= \pm\sqrt{\frac{57}{16}} + \frac{3}{4} \end{align}
Which gives the correct answer but we could simplify things - first notice that we can take the square root of 16 on the denominator and then we can make things look a little nicer.
$$x = \pm\sqrt{\frac{57}{16}} + \frac{3}{4} = \pm\frac{\sqrt{57}}{4}+\frac{3}{4} = \frac{\pm\sqrt{57}+3}{4}$$
$$x=\frac{\sqrt{57}+3}{4} \quad \text{or} \quad x=\frac{-\sqrt{57}+3}{4}$$ |
# How do you find the arithmetic means of the sequence 3, __, __, __, __, __, 27?
Feb 26, 2017
$\text{The Reqd. Means are, } 7 , 11 , 15 , 19 \mathmr{and} 23.$
#### Explanation:
Let the reqd. $5$ Arithmetic Means be ${A}_{1} , {A}_{2} , {A}_{3} , {A}_{4} , {A}_{5.}$
Then, $3 , {A}_{1} , {A}_{2} , {A}_{3} , {A}_{4} , {A}_{5} , 27$ are $7$ terms of an A.P., for
which, in the Usual Notation, $a = 3 \mathmr{and} {a}_{7} = 27.$
But, for an A.P., ${a}_{n} = a + \left(n - 1\right) d , n \in \mathbb{N} .$
$\therefore a = 3 , {a}_{7} = 27 \Rightarrow 27 = 3 + \left(7 - 1\right) d \Rightarrow 24 = 6 d \Rightarrow d = 4.$
Hence, ${A}_{1} = {a}_{2} = a + d = 3 = 4 = 7 , {A}_{2} = {a}_{3} = {a}_{2} + 4 = 11 ,$
${A}_{3} = {A}_{2} + 4 = 15 , {A}_{4} = 19 , {A}_{5} = 23.$
Thus, the Reqd. Means are, $7 , 11 , 15 , 19 \mathmr{and} 23.$
Enjoy Maths.! |
# A cylindrical bucket, 32 cm high and 18 cm of radius of the base,
`
Question:
A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap
Solution:
The height and radius of the cylindrical bucket are $h=32 \mathrm{~cm}$ and $r=18 \mathrm{~cm}$ respectively. Therefore, the volume of the cylindrical bucket is
$V=\pi r^{2} h$
$=\frac{22}{7} \times(18)^{2} \times 32$
The bucket is full of sand and is emptied in the ground to form a conical heap of sand of height $h_{1}=24 \mathrm{~cm}$. Let, the radius and slant height of the conical heap be $r_{1} \mathrm{~cm}$ and $l_{1} \mathrm{~cm}$ respectively. Then, we have
$l_{1}^{2}=r_{1}^{2}+h_{1}^{2}$
$\Rightarrow r_{1}^{2}=l_{1}^{2}-h_{1}^{2}$
$\Rightarrow r_{1}^{2}=l_{1}^{2}-(24)^{2}$
The volume of the conical heap is
$V_{1}=\frac{1}{3} \pi r_{1}^{2} h_{1}$
$=\frac{1}{3} \times \frac{22}{7} \times r_{1}^{2} \times 24$
$=\frac{22}{7} \times r_{1}^{2} \times 8$
Since, the volume of the cylindrical bucket and conical hear are same, we have
$V_{1}=V$
$\Rightarrow \frac{22}{7} \times r_{1}^{2} \times 8=\frac{22}{7} \times(18)^{2} \times 32$
$\Rightarrow \quad r_{1}^{2}=(18)^{2} \times 4$
$\Rightarrow \quad r_{1}=18 \times 2$
$\Rightarrow \quad r_{1}=36$
Then, we have
$l_{1}^{2}=r_{1}^{2}+h_{1}^{2}$
$\Rightarrow l_{1}^{2}=(36)^{2}+(24)^{2}$
$\Rightarrow l_{1}=43.27$
Therefore, the radius and the slant height of the conical heap are 36 cm and 43.27 cm respectively. |
# Sooraj’s father wants to construct a rectangular garden
Q:- Sooraj’s father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in the figure. He has 200 metres of fencing wire.
Based on the above information , answer the following question : [CBSE 2023]
(i) Let ‘x’ metres denotes the length of the side of the garden perpendicular to the brick wall and ‘y’ metres denote the length of the side parallel to the brick wall. Determine the relation representing the total length of feencing wire and also write A(x), the area of the garden. (2)
(ii) Detemine the maximum value of A(x). (2)
Solution:- Let ‘x’ metres denotes the length of the side of the garden perpendicular to the brick wall and ‘y’ metres denote the length of the side parallel to the brick wall
Length of fencing wire = 200
x + x + y = 200
⇒ 2x + y = 200
⇒ y = 200 – x
Relation between fencing of wire
y = 200 – x
Area of garden A(x) = x.y
A(x) = x(200 – x) . . . (i)
(ii) From equation (i)
A(x) = x(200 – x)
⇒ A(x) = 200x – x²
differentiate with respect to x
. . . (ii)
For maximaa and minima
200 – 2x = 0
⇒ x = 100
Again differentiate with respect to x of eq (ii)
Hence, area of garden is maximaum when x = 100
Now, maximum area
A(x) = x(200 – x)
⇒ A(100) = 100(200 – 100)
⇒ A(100) = 100 × 100 = 10000
Thus the maximum area of garden = 10,000 m²
Q:- A building contractor undertakes a job to construct 4 flats on a plot along with the parking area. Due to strike the probability of many construction workers not being present present for the job is 0.65. The probability that many are not present and still the work get completed on times is 0.35. The probability that work will be completed on time when all workers are present is 0.80. [CBSE 2023]
Solution:- See full solution |
Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
7.1 Prisms 7.2 Solids That Aren’t Prisms 7.3 Relating Length, Surface Area, and Volume 7.4 Inscribed Solids
7.5 Solids Produced by Rotating Polygons 7.6 Key Formulas 7.7 Review Questions 7.8 Explanations
Solids Produced by Rotating Polygons
Another type of Math IC question that you may come across involves a solid produced by the rotation of a polygon. The best way to explain how this type of problem works is to provide a sample question:
What is the surface area of the geometric solid produced by the triangle below when it is rotated 360 degrees about the axis AB?
When this triangle is rotated about AB, a cone is formed. To solve the problem, the first thing you should do is sketch the cone that the triangle will form.
The question asks you to figure out the surface area of the cone. The formula for surface area is πr2 + πrl, which means you need to know the lateral height of the cone and the radius of the circle. If you’ve drawn your cone correctly, you should see that the lateral height is equal to the hypotenuse of the triangle. The radius of the circle is equal to side BC of the triangle. You can easily calculate the length of BC since the triangle is a 30-60-90 triangle. If the hypotenuse is 2, then BC, being the side opposite the 30º angle, must be 1. Now plug both values of l and r into the surface area formula and then simplify:
Common Rotations
You don’t need to learn any new techniques or formulas for problems that deal with rotating figures. You just have to be able to visualize the rotation as it’s described and be aware of which parts of the polygons become which parts of the geometric solid. Below is a summary of which polygons, when rotated a specific way, produce which solids.
A rectangle rotated about its edge produces a cylinder. A semicircle rotated about its diameter produces a sphere. A right triangle rotated about one of its legs produces a cone. A rectangle rotated about a central axis (which must contain the midpoints of both of the sides that it intersects) produces a cylinder. A circle rotated about its diameter produces a sphere. An isosceles triangle rotated about its axis of symmetry (the altitude from the vertex of the non-congruent angle) produces a cone.
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# How Do You Find The Angle Of An Isosceles Triangle
## How Do You Find The Angle Of An Isosceles Triangle
Theorem: Angles opposite to equal sides of an isosceles triangle are equal.
Given: In ∆ABC, AB = AC
To Prove: ∠B = ∠C
Construction: Draw AD, bisector of ∠A
∴ ∠1 = ∠2
∠1 = ∠2 (by construction)
AB = AC
∴ ∠B = ∠C (c.p.c.t.) Proved.
and BD = DC (c.p.c.t.) ⇒ AD is median
∴ we can say AD is perpendicular bisector of BC or we can say in isosceles ∆, median is angle bisector and perpendicular to base also.
## Angle Of An Isosceles Triangle Example Problems With Solutions
Example 1: Find ∠BAC of an isosceles triangle in which AB = AC and ∠B = 1/3 of right angle.
Solution:
Example 2: In isosceles triangle DEF, DE = EF and ∠E = 70° then find other two angles.
Solution:
Example 3: ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P. Show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
Example 4: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see figure ). Show that:
(i) ∆ABM ≅ ∆PQN (ii) ∆ABC ≅ ∆PQR
Solution: |
# Two equations are equivalent if they have the same solutions. Solving a Linear Equation An equation is a statement in which two expressions are equal.
## Presentation on theme: "Two equations are equivalent if they have the same solutions. Solving a Linear Equation An equation is a statement in which two expressions are equal."— Presentation transcript:
Two equations are equivalent if they have the same solutions. Solving a Linear Equation An equation is a statement in which two expressions are equal. A linear equation in one variable is an equation that can be written in the form ax = b where a and b are constants and a ≠ 0. A number is a solution of an equation if the statement is true when the number is substituted for the variable. For instance, the equations x – 4 = 1 and x = 5 are equivalent because both have the number 5 as their only solution. The transformations, or changes, on the following slide produce equivalent equations and can be used to solve an equation.
Solving a Linear Equation TRANSFORMATIONS THAT PRODUCE EQUIVALENT EQUATIONS ADDITION PROPERTY OF EQUALITY SUBTRACTION PROPERTY OF EQUALITY MULTIPLICATION PROPERTY OF EQUALITY DIVISION PROPERTY OF EQUALITY Add the same number to both sides: If a = b, then a + c = b + c. Subtract the same number from both sides: If a = b, then a – c = b – c. Multiply both sides by the same nonzero number: If a = b and c ≠ 0, then ac = bc. Divide both sides by the same nonzero number: If a = b and c ≠ 0, then a c = b c.
Your goal is to isolate the variable on one side of the equation. Write original equation. Subtract 9 from each side. Simplify. SOLUTION Solving an Equation with a Variable on One Side Solve + 9 = 15. 3737 x + 9 = 15 3737 x = 6 3737 x Multiply each side by, the reciprocal of. 7373 3737 = 6 x 7373 (6) = 6 x 14 The solution is 14. Substitute 14 for x. Solution checks. + 9 = 15 3737 ( x ) C HECK ? 15 = 15 14 Check x = 14 in the original equation.
Solve 5n + 11 = 7n – 9. 5n + 11 = 7n – 9 5n + 11 = 2n – 9 5n + 20 = 2n 10 = n Divide each side by 2. Add 9 to each side. Subtract 5 n from each side. Write original equation. SOLUTION Solving an Equation with a Variable on Both Sides The solution is 10. Check this in the original equation.
Solve 4(3x – 5) = –2(–x + 8) – 6x. 4(3x – 5) = –2(–x + 8) – 6x 12x – 20 = 2x –16 – 6x 12x – 20 = –4x – 16 Divide each side by 16. Combine like terms. Distributive property Write original equation. SOLUTION Using the Distributive Property 16x – 20 = –16 Add 4 x to each side. 16x = 4 Add 20 to each side. x =x = 1414 The solution is. Check this in the original equation. 1414
4x + 3 = 12x – 2 Divide each side by 8. Distributive property Multiply each side by the LCD, 12. Write original equation. SOLUTION Solving an Equation with Fractions 3 = 8x – 2 Subtract 4x from each side. 5 = 8x Add 2 to each side. = x 5858 The solution is. Check this in the original equation. 5858 Solve x + = x – 1313 1414 1616 x + = x – 1313 1414 1616 12 x + = 12 x – 1313 1414 1616 ( )
REAL ESTATE A real estate broker’s base salary is \$18,000. She earns a 4% commission on total sales. How much must she sell to earn \$55,000 total? SOLUTION Verbal Model Labels Total income Commission rate Total income = 55,000 Base salary = 18,000 Commission rate = 0.04 (dollars) (percent in decimal form) Base salary + Using Linear Equations in Real Life = Total sales Total sales = x (dollars) Algebraic Model 18,000 + 0.04x55,000 =
REAL ESTATE A real estate broker’s base salary is \$18,000. She earns a 4% commission on total sales. How much must she sell to earn \$55,000 total? SOLUTION 18,000 + 0.04x55,000 = 37,000 = 0.04x 925,000 = x Write linear equation. Subtract 18,000 from each side. Divide each side by 0.04. The broker must sell real estate worth a total of \$925,000 to earn \$55,000. Writing and Using a Linear Equation
You have a 3 inch by 5 inch photo that you want to enlarge, mat, and frame. You want the width of the mat to be 2 inches on all sides. You want the perimeter of the framed photo to be 44 inches. By what percent should you enlarge the photo? SOLUTION Verbal Model Labels Perimeter + Perimeter = 44 (inches) 2 Width = 2 Length Width = 4 + 3x Length = 4 + 5x 44 = 2(4 + 3x) + 2(4 + 5x) Algebraic Model Writing and Using a Geometric Formula
44 = 2(4 + 3x) + 2(4 + 5x) Write linear equation. Distribute and combine like terms. 44 = 16 + 16x Subtract 16 from each side. 28 = 16x Divide each side by 16. 1.75 = x You have a 3 inch by 5 inch photo that you want to enlarge, mat, and frame. You want the width of the mat to be 2 inches on all sides. You want the perimeter of the framed photo to be 44 inches. By what percent should you enlarge the photo? SOLUTION You should enlarge the photo to 175% of its original size. Writing and Using a Geometric Formula
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# What are the extrema of f(x)=x^3-2x+5 on [-2,2]?
Feb 4, 2016
Minimum: $f \left(- 2\right) = 1$
Maximum: $f \left(+ 2\right) = 9$
#### Explanation:
Steps:
1. Evaluate the endpoints of the given Domain
$f \left(- 2\right) = {\left(- 2\right)}^{3} - 2 \left(- 2\right) + 5 = - 8 + 4 + 5 = \textcolor{red}{1}$
$f \left(+ 2\right) = {2}^{3} - 2 \left(2\right) + 5 = 8 - 4 + 5 = \textcolor{red}{9}$
2. Evaluate the function at any critical points within the Domain.
To do this find the point(s) within the Domain where $f ' \left(x\right) = 0$
$f ' \left(x\right) = 3 {x}^{2} - 2 = 0$
$\rightarrow {x}^{2} = \frac{2}{3}$
$\rightarrow x = \sqrt{\frac{2}{3}} \text{ or } x = - \sqrt{\frac{2}{3}}$
$f \left(\sqrt{\frac{2}{3}}\right) \approx \textcolor{red}{3.9}$ (and, no, I didn't figure this out by hand)
f(-sqrt(2/3))~color(red)(~6.1)#
Minimum of $\left\{\textcolor{red}{1 , 9 , 3.9 , 6.1}\right\} = 1$ at $x = - 2$
Maximum of $\left\{\textcolor{red}{1 , 9 , 3.9 , 6.1}\right\} = 9$ at $x = + 2$
Here is the graph for verification purposes:
graph{x^3-2x+5 [-6.084, 6.4, 1.095, 7.335]} |
# Multiplication with Money
When students are first asked to multiply with money they typically multiply an amount that contains dollars and cents with a whole number. This is a great introduction to multiplication with decimals. The main multiplication with decimal rules apply:
• Students do not have to line numbers up by the decimal. Lots of students like to do this anyway, but this introduces a lot of unnecessary decimals and space for careless errors. Try to explain that multiplication is different and does NOT require lining up decimals.
• Students count the number of digits behind decimals in the factors to determine how many digits should be behind the decimal in the answer.
But, because we are dealing with money (and only one amount with a decimal), there are always two digits behind the decimal point.
Example
$5.23 \times 7 =$
Write the problem vertically.
\eqalign{5.23\\\times 7\\\hline\qquad}
Multiply. Place the decimal two digits in in the answer.
\eqalign{\overset{1}{5}.\overset{2}{2}3\\\times 7\\\hline 36.61}
The process of multiplying with money does give students a start working with decimals (and, importantly, not forgetting to include the decimal in their answers), but be careful not to let students think that multiplying with decimals is always so automatic.
• ## Multiplication with Money
Find the products:
1. \$$4.51\times 3= 2. \$$20.31\times 4=$3. \$$2.22\times 9= 4. \$$2.52\times .5=$
5. If you buy 6 shirts for \$7.99, how much will it cost? 6. If you buy 4 hot dogs for 99 cents, how much will it cost? 7. If you buy 12 cupcakes for \$1.35 each, how much will it cost?
8. If you sell 5 hats for \\$5.05 each, how much will you make? |
## Geometry: Common Core (15th Edition)
$x = 6$
The perimeter of a triangle is the sum of the length of each of its sides. We are given expressions for the lengths of the legs and base of an isosceles triangle along with the value of its perimeter. If we add up the lengths of the legs and the base of the triangle, they should equal the perimeter. Let's set up this equation: $2(legs) + base$ = perimeter of the triangle Let's plug in what we know: $2(2x - 5) + x = 20$ Use the distributive property to get rid of the parentheses: $4x - 10 + x = 20$ Combine like terms: $5x - 10 = 20$ Add $10$ to both sides of the equation to isolate the $x$ term: $5x = 30$ Divide each side of the equation by $5$ to solve for $x$: $x = 6$ |
# Find the answer to the following problem (1.5 + 3.2i) – (-2.4 – 3.7i)
• Last Updated : 21 Mar, 2022
Complex number is the term that can be shown as a sum of a real number and an imaginary number. A complex number can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z.
For example:
6+2i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
9+5i is a complex number where 2 is a real number (Re) and 5i is an imaginary number (im)
The real number and imaginary number combination is called a Complex number.
Example: √-5, -8i, -13i are all imaginary numbers. here ‘i’ is an imaginary number called “iota’
### Find the answer to the following problem (1.5 + 3.2i) – (-2.4 – 3.7i)
Given: (1.5 + 3.2i) – (-2.4 – 3.7i)
By open bracket
= 1.5 + 3.2i + 2.4 + 3.7i
= 3.9 + 9.9i
### Similar Questions
Question 1: Simplify (-i)(2i)(-1/8i)3?
Solution:
Given: (-i)(2i)(-1/8i)3
= (-i)(2i) (-1/8i)3
= -2i2 (-1/512i3)
= -2i (-1/512 i3)
= -2(-1) {(-1/512)(- i)} {i2 = -1 and i3 = -i}
= 2(1/512i)
= (2/512)i
= 0 + 1/256i
Question 2: Express the (1-i) – (-1+6i)?
Solution:
Given : (1-i) – (-1+6i)
= 1 – i + 1 – 6i
= 2 – 7i
Question 3: Simplify ( 5 -3i)3?
Solution:
Given: (5 -3i)3
Here we will use identity (a-b)3 = a3 – b3 – 3a2b + 3ab2
= (5)3 – (3i)3 – 3(5)2(3i) + 3 (5) (3i)2
= 125 – (27i3) – 225i + 15(9i2)
= 125 -[27(-i)] – 225 i + 15(-9)
= 125 +27i -225i – 135
= -10 – 198i
Question 4: Express the {1/5 + 2/5i} – {4 + 5/2i}
Solution:
Given: {1/5 +2/5 i} – { 4 + 5/2 i}
= {(1+2i)/5} – {(8+5i)/ 2}
= [{2(1+2i) – 5(8+5i)} /10]
= {2+4i-40 -25i} /10
= (-38 -21i)/10
= -38/10 – 21/10 i
= -19/5 – 21/10 i
Question 5: Simplify (-6i)(7i)(-2)
Solution:
Given: (-6i)(7i)(-2)
= -6i x 7i x (-2)
= -42i2 x -2 {i2 = -1}
= -42 (-1) x -2
= 42 x -2
= -84 + 0i
Question 6: Prove that the {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} complex numbers are purely real?
Given : {(2+3i) / (3+4i) } { (2-3i)/(3-4i) }
= {(2+3i)(2-3i) } / {(3+4i)(3-4i)}
= {4 -6i +6i -9(i2)} / {9 -12i + 12i – 16(i2)}
= {4 +9} / {9 +16}
= 13/25 + 0i
Therefore {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} it is purely real
Question 7: Perform the indicated operation and write the answer in standard form (3-i)/(1+2i)?
Solution:
Given : (3-i)/(1+2i)
Multiply with the conjugate of denominator
= {(3-i)/(1+2i) x (1-2i)/(1-2i)}
= {(3-i)(1-2i)} / {(1)2 -(2i)2} {difference of squares formula . i.e (a+b)(a-b) = a2 – b2}
= {3 -6i -i + 2i2} / {1-4(-1)}
= {3-7i -2} / {1 + 4} { i2 = -1 }
= (1-7i)/5
= 1/5 -7/5 i
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Chapter 4 – Basic Geometrical Ideas
The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
Introduction
The study of geometry involves understanding of some basic geometric ideas like the concepts of shapes, sizes and dimensions. It means earth and its measurement marked by the words ‘geo’ and ‘metron’ in the word geometry. Geometry has application in architecture, engineering and other fields.
Basic Geometrical Ideas
The basic geometric ideas can be helpful in our daily application also. It is thus very important to understand the basic terminologies. Some of the basic geometric concepts include the concepts of points, line, intersecting lines, parallel lines, ray, curves, polygons, sides, vertices, diagonals, angles, triangles, quadrilaterals and circles. Let us understand them one by one.
Points
Point is nothing but a dot on the paper. It has no length, height or width. It can determine any location, and it is usually marked by just one alphabet. For instance, C is a point.
A Line Segment
The line is the shortest route towards two points.
A line moves to infinity in both directions. It is a unidirectional or one-directional geometric shape. A line lacks a width or a height. It can be intersecting or polygonal. As it can pass through various points, hence it can be named after picking the points.
Intersecting Lines
As the name suggests, they are lines that intersect each other at various common points.
Parallel Lines
When two lines or any pair of lines which never intersect each other at any points, they form parallel lines.
Example: CD II RS
Ray
A ray is nothing but a line with a single endpoint on one side and the other side it stretches to infinity. It is different from a line segment which moves to infinity on both sides.
Curves
A curve is a constant and a smooth flow of a line lacking any sharp turns. So it bends as well as changes its direction once at least. There are 2 types of curves. First is the Upward curve and the other is a downward curve.
Polygons
A polygon is a geometric figure which is enclosed by line segments on all its sides. The line segments are called sides and where two vertices meet are called its vertex. For example, a polygon has 4 sides PQ, QR, RS, ST. Now the sides QR and RS meet at R which is a vertex of polygon PQRST. The sides of the polygon are non-collinear.
Angles
A corner or angle is formed when two rays arise from common endpoints. Rays are called the sides and endpoints are called the vertex or are also called the angles. Vertex is always in the middle. There are different types of angles which include acute angle, right angle, straight angle and obtuse angle.
Triangles
A three-sided polygon is called a triangle. There are three vertices and three angles.
When a polygon has four sides, it is called a quadrilateral. It also has four angles and vertices. Quadrilaterals can be of various types like square, rectangle and others.
Circles
A geometric figure which neither has a start or an endpoint is a circle. It has a radius which is a connecting line from the centre of the circle to any of the points of the circle. A diameter is a line segment that connects the points of the circle passing through the centre. A chord is a line segment that connects any two points in the circle.
Conclusion
As we see above, the basic geometric idea is the study of shapes and sizes or the study of flat or different dimensional structure. They have very significant applications in real life. Look at the buildings around you. They are works of architecture. See the machines running and helping in our day to day lives like the mixer grinder to even aeroplanes. Everywhere there is the application of geometry.
FAQs
1. What are the basic geometrical ideas?
They are as follows:-
• Point is any dot on the paper and has no length, height or width.
• The line is the shortest route between two points.
• The line can be an intersecting and parallel line.
• A ray is a line where one end continues till infinity, and another point is limited.
• A geometric figure enclosed by a line on each side is a polygon.
• A polygon with three sides is a triangle and with four sides is a quadrilateral.
• The circle has no start or endpoint.
1. What are the basic concepts of geometry?
The basic concept of geometry revolves around point, line circle and enclosed figure. As discussed above, it can be of various types.
1. What are adjacent sides Class 6?
Adjacent Sides is with reference to a polygon where any two sides have a common end. For example, say that sides AB and BC are adjacent as they terminate at a common end C.
1. What is geometry math?
The branch of mathematics that studies the sizes, positions, shapes, angles and various dimensions is called geometry. It can be the study dealing with flat figures like the circle or line or even the study of the complicated 2 or 3-dimensional structures.
1. What are 10 geometric concepts?
As explained above, they involve the concept of points, line, intersecting lines, parallel lines, ray, curves, polygons, sides vertices, diagonals, angles, triangles, quadrilaterals, circles.
1. What are 10 geometric terms?
Points, line, intersecting lines, parallel lines, ray, curves, polygons, sides vertices, diagonals, angles, triangles, quadrilaterals, circles.
You can learn the basic concepts of geometry better with MSVgo. Make your concepts clear for all Math & Science topics with MSVgo’s expansive 6,000+ videos library that provides explanatory visualisations and real-life examples. Download MSVgo Now!
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• Money
• Multiplication and Factors
• Multiply and Divide
• Numbers
• Parts and Wholes
• Pattern Recognition
• Patterns
• Play With Patterns
• Rupees And Paise
• Shapes And Angles
• Shapes And Designs
• Shapes and Space
• Similarity
• Smart Charts
• Squares
• Subtraction
• Tables And Shares
• Tenths and Hundredths
• Time |
# How do you solve the equation x^2-4x+1=0 by completing the square?
Mar 4, 2018
$x = 2 \pm \sqrt{3}$
#### Explanation:
The difference of squares identity can be written:
${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$
We can complete the square and use this with $A = \left(x - 2\right)$ and $B = \sqrt{3}$ as follows:
$0 = {x}^{2} - 4 x + 1$
$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 - 3$
$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} - {\left(\sqrt{3}\right)}^{2}$
$\textcolor{w h i t e}{0} = \left(\left(x - 2\right) - \sqrt{3}\right) \left(\left(x - 2\right) + \sqrt{3}\right)$
$\textcolor{w h i t e}{0} = \left(x - 2 - \sqrt{3}\right) \left(x - 2 + \sqrt{3}\right)$
Hence:
$x = 2 \pm \sqrt{3}$
Mar 4, 2018
In this question you are asked to solve the given equation by the completing the square method. The whole purpose of using this method is to create a perfect square that can be factored "perfectly".
#### Explanation:
Right now, if you tried factoring this question it wouldn't work. So, let's complete the square!
The first step is identifying your terms, in which your current equation looks like this: $a {x}^{2} - b x + c$ = 0
For completing the square method, you will need to your 'a' and 'b' terms, and this is done by moving the 'c' term over to the other side of the equation to look like this: $a {x}^{2} - b x = c$
Now, we add in component that will lead us to completing the square, which is ${\left(\frac{b}{2}\right)}^{2}$. This means that you will add ${\left(\frac{b}{2}\right)}^{2}$ (which in this case is equal to 4), and you will have to add it to both sides of the equation to keep them balanced.
Then, once you have added that part in, solve the equation by factoring:
${x}^{2} - 4 x + 4 = 1 + 4$
${x}^{2} - 4 x + 4 = 5$
$\left(x - 2\right) \left(x - 2\right) = 5$
${\left(x - 2\right)}^{2} = 5$
Now, to find 'x' you will apply the Null Factor Law. This means you will now let 'x' = 0, and simply move the -2 over to the other side to find your answer. Note: since we are using a perfect square, x will only have one value since there is only one answer (x-2) in the brackets, meaning in this case we can ignore the exponent (as the answers will be the same). If there was more than one answer in the bracket (e.g. (x-4) and (x-6)), then we would apply Null Factor Law to both brackets and solve for x.
$\left(x - 2\right) = 5$
$x = 5 + 2$
$x = 7$ |
# System of two equations solver
Math can be a challenging subject for many learners. But there is support available in the form of System of two equations solver. Our website can solve math problems for you.
## The Best System of two equations solver
This System of two equations solver helps to quickly and easily solve any math problems. There are many ways to solve quadratic functions, but one of the most popular methods is known as the quadratic formula. This formula is based on the fact that any quadratic equation can be rewritten in the form of ax^2 + bx + c = 0. The quadratic formula then states that the roots of the equation are given by: x = (-b +/- sqrt(b^2 - 4ac)) / (2a). In other words, the roots of a quadratic equation are always symmetrical around the axis of symmetry, which is given by x = -b/(2a). To use the quadratic formula, simply plug in the values of a, b, and c into the formula and solve for x. Keep in mind that there may be more than one root, so be sure to check all possible values of x. If you're struggling to remember the quadratic formula, simply Google it or look it up in a math textbook. With a little practice, you'll be solvingquadratics like a pro!
Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning.
There are a variety of methods that can be used to solve mathematical equations. One of the most common is known as elimination. This method involves adding or subtracting terms from both sides of the equation in order to cancel out one or more variables. For example, consider the equation 2x + 3y = 10. To solve for x, we can add 3y to both sides of the equation, which cancels out y and leaves us with 2x = 10. We can then divide both sides by 2 in order to solve for x, giving us a final answer of x = 5. While elimination may not always be the easiest method, it can be very effective when used correctly.
In this case, we are looking for the distance travelled by the second train when it overtakes the first. We can rearrange the formula to solve for T: T = D/R. We know that the second train is travelling at 70 mph, so R = 70. We also know that the distance between the two trains when they meet will be the same as the distance travelled by the first train in one hour, which we can calculate by multiplying 60 by 1 hour (60 x 1 = 60). So, plugging these values into our equation gives us: T = 60/70. This simplifies to 0.857 hours, or 51.4 minutes. So, after 51 minutes of travel, the second train will overtake the first.
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# As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A.
Question:
As shown in Fig.7.40, the two sides of a step ladder BA and CA are $1.6 \mathrm{~m}$ long and hinged at A A rope DE, $0.5 \mathrm{~m}$ is tied half way up. A weight $40 \mathrm{~kg}$ is suspended from a point $F, 1.2 \mathrm{~m}$ from $B$ along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ )
(Hint: Consider the equilibrium of each side of the ladder separately.)
Solution:
The given situation can be shown as:
NB = Force exerted on the ladder by the floor point B
NC = Force exerted on the ladder by the floor point C
= Tension in the rope
BA = CA = 1.6 m
DE = 0. 5 m
BF = 1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
$\triangle \mathrm{ABI}$ and $\triangle \mathrm{AlC}$ are similar
$\therefore \mathrm{BI}=\mathrm{IC}$
Hence, $I$ is the mid-point of $B C$.
$D E \| B C$
$B C=2 \times D E=1 \mathrm{~m}$
$\mathrm{AF}=\mathrm{BA}-\mathrm{BF}=0.4 \mathrm{~m} \ldots(i)$
$D$ is the mid-point of $A B$.
Hence, we can write:
$\mathrm{AD}=\frac{1}{2} \times \mathrm{BA}=0.8 \mathrm{~m}$ $\ldots(i i)$
Using equations (i) and (ii), we get:
FE = 0.4 m
Hence, F is the mid-point of AD.
$\mathrm{FG} \| \mathrm{DH}$ and $\mathrm{F}$ is the mid-point of $\mathrm{AD}$. Hence, $\mathrm{G}$ will also be the mid-point of $\mathrm{AH}$.
$\triangle \mathrm{AFG}$ and $\triangle \mathrm{ADH}$ are similar
$\therefore \frac{\mathrm{FG}}{\mathrm{DH}}=\frac{\mathrm{AF}}{\mathrm{AD}}$
$\frac{\mathrm{FG}}{\mathrm{DH}}=\frac{0.4}{0.8}=\frac{1}{2}$
$\mathrm{FG}=\frac{1}{2} \mathrm{DH}$
$=\frac{1}{2} \times 0.25=0.125 \mathrm{~m}$
In $\triangle A D H$ :
$\mathrm{AH}=\sqrt{\mathrm{AD}^{2}-\mathrm{DH}^{2}}$
$=\sqrt{(0.8)^{2}-(0.25)^{2}}=0.76 \mathrm{~m}$
For translational equilibrium of the ladder, the upward force should be equal to the downward force.
$N_{\mathrm{C}}+N_{\mathrm{B}}=m g=392 \ldots$ (iii)
For rotational equilibrium of the ladder, the net moment about $A$ is:
$-N_{\mathrm{B}} \times \mathrm{BI}+m \mathrm{~g} \times \mathrm{FG}+N_{\mathrm{C}} \times \mathrm{Cl}+T \times \mathrm{AG}-T \times \mathrm{AG}=0$
$-N_{\mathrm{B}} \times 0.5+40 \times 9.8 \times 0.125+N_{\mathrm{C}} \times(0.5)=0$
$\left(N_{\mathrm{C}}-N_{\mathrm{B}}\right) \times 0.5=49$
$N_{\mathrm{C}}-N_{\mathrm{B}}=98$ $\ldots(i v)$
Adding equations (iii) and (iv), we get:
$N_{\mathrm{C}}=245 \mathrm{~N}$
$N_{\mathrm{B}}=147 \mathrm{~N}$
For rotational equilibrium of the side $A B$, consider the moment about $A$.
$-N_{\mathrm{B}} \times \mathrm{BI}+m \mathrm{~g} \times \mathrm{FG}+T \times \mathrm{AG}=0$
$-245 \times 0.5+40+9.8 \times 0.125+T \times 0.76=0$
$0.76 T=122.5-49$
$\therefore T=96.7 \mathrm{~N}$ |
# Unleashing the Fun in Fractions: The Fraction Line Up Game
Hello math teachers! Welcome back to our exciting world of mathematics where we transform complex concepts into engaging activities. Today, we’re going to delve into the realm of fractions. Fractions may sometimes feel like a daunting topic for students, but with the right activities, we can make this topic an enjoyable learning journey.
One of the wonderful ways we’re going to do this is through a highly engaging game called “Fraction Line Up.” This hands-on activity enables students to explore the concept of ordering fractions in a fun, competitive setting. So, whether you’re looking for a new activity to shake up your routine, or need a fresh strategy to help students grasp ordering fractions, we’ve got you covered.
## What is the Fraction Line Up Game?
The Fraction Line Up game is a competitive, problem-solving activity that focuses on teaching students how to order fractions from least to greatest. It is a board game that can be played individually or in teams of two. It incorporates problem-solving skills and strategy, encouraging students to think critically about where to place their fractions.
The game is designed for 1 vs 1 gameplay, but you can also facilitate 2 vs 2 games to encourage teamwork and communication among students. The game continues until a player (or team) manages to arrange all their dominoes from least to greatest, hence the name “Fraction Line Up.”
## Equipment Needed for the Fraction Line Up Game
You’ll need only two types of materials for this activity, making it an easy game to set up in any classroom:
1. Dominoes Fraction Cards: These special dominoes have fractions instead of numbers. You’ll need one set of these cards per game.
2. Game Sheet: Each player (or team) requires one game sheet, where they will line up their fraction cards.
## How to Play the Fraction Line Up Game
After you have the necessary materials, the next step is understanding how to play the game. Here is the sequence of gameplay:
1. Start by placing all the dominoes face down in a pile.
2. Players take turns picking a domino. When a player picks a domino, they must place it on their game sheet.
3. The objective is to line up all the fractions from least to greatest on their game board.
4. It’s important to remember that once a fraction is placed, it cannot be moved. This is where strategy comes in! Players need to think about where they’re placing their fractions.
5. Players are not allowed to place equivalent fractions next to each other. This helps to diversify the fractions used and encourages a broader understanding of different fractions.
6. If a player picks a domino that can’t be placed on their game board, it’s placed in a reject pile.
7. The first player to successfully line up all the dominoes from least to greatest (and with an equal number of turns as the other player) wins the round.
8. If a player runs out of rejects before they finish their lineup, they “bust” and can’t win the round.
9. It’s possible for the game to end with one player winning, both players tying, or both players busting.
## Gameplay Scenario
Let’s consider an example gameplay scenario. Let’s say two students, Jack and Jill, are playing the game.
Jack picks a domino with the fraction 3/4. He chooses to put this fraction on the right side of his game board, predicting that he will pick smaller fractions later. Jill picks a domino with the fraction 1/2 and places it in the middle of her game board.
The game continues, and they both manage to place a few more fractions. However, Jack picks a fraction that’s larger than 3/4 but his board only has room for smaller fractions. He can’t place this domino, so it goes into his reject pile.
Jill, on the other hand, successfully places all her fractions in order from least to greatest. She wins the round, and Jack learns a valuable lesson about strategy and probability.
## Accommodations and Modifications
We understand that every classroom is unique, and every student learns differently. Hence, we’re providing some ways to modify the game to better suit your students’ needs.
1. Level of Difficulty: Adjust the level of difficulty by using fractions that have the same denominators, or are parts of the same whole (like halves, quarters, etc.). As students become more comfortable with the game, start introducing fractions with different denominators.
2. Gameplay Modification: To accommodate slower learners, you can allow players to rearrange their fractions once or twice during the game. This modification can make the game less stressful and more enjoyable for these students.
3. Peer Assistance: For students who might be struggling, consider pairing them up with students who understand the concept well. This way, they can help each other during the game, enhancing both their understanding and communication skills.
## Fraction Line Up Game
The Fraction Line Up game is a dynamic, exciting way to teach students about ordering fractions. Not only does it help improve their understanding of fractions, but it also enhances their problem-solving and strategic thinking skills.
Remember, fractions don’t have to be scary or complicated. With interactive games like Fraction Line Up, your students will be engaged and excited to learn. And who knows, they might even start asking for more fractions activities!
Do try out the Fraction Line Up game in your classroom and let us know how it went. We’re always excited to hear about your experiences and the creative ways you’re making math more fun for your students.
## Common Core State Standards (CCSS)
The Fraction Line Up game aligns with the following Common Core State Standards:
1. CCSS.MATH.CONTENT.3.NF.A.3.D: Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole.
2. CCSS.MATH.CONTENT.4.NF.A.2: Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2.
Remember, the goal isn’t just to have fun; it’s also to develop our students’ love for learning, especially for math. Here’s to making fractions a favorite topic in your classroom!
Have fun with fractions, and until next time, keep making math magical!
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# GED Math
##### General Study Tips: If you haven't found them already, Chris has a great series of videos on HOW to study including how to counter math anxiety prior to a big test and the best ways to prepare. Study Tips
Content Category Skills Example Lessons
Operations with Integers Order of operations, adding, subtracting, multiplying, and dividing basic integers \$72+4÷2+5\$ Intro to PEMDAS
Operations with Fractions Adding, subtracting, multiplying, and dividing fractions \$2/3+(3/4×1/2)+(3/4-1/3)\$ Adding & Subtracting Fractions Multiplying & Dividing Fractions
Operations with Decimals Adding,subtracting,multiplying,and dividing decimals \$7.46+1.85+2.5×3\$ Decimals Decimals rounding
Operations between mixed numbers and improper fractions Adding, subtracting, multiplying,and dividing using mixed numbers and improper fractions \$7(1/2)+ 18.85+4(2/3)\$ Improper Fractions and Mixed Numbers
Exponents Raising numbers to "a power" (how many times a number is multiplied by itself) \$2^4+3^2\$ Intro to Exponents Exponent Rules, Like Bases, & Negative Exponents Scientific Notation
Square Roots A number that produces a specified quantity when multiplied by itself. \$√4+√81\$ Simplifying Roots & Radicals
Ratios and Proportions "Part over whole" or fractions that equals a similar fraction \$X/2=12/8\$ Ratios & Proportions
Percentages "A part" of a certain value which is used in operations. 20% of 100 + 30% of 60 Percents as Ratios Percent Increase & Decrease Percents As Decimals Hard Multi-Step Percent Problems
Averages Another word for the "mean" which is a sum of all numbers in a set of data divided by the amount of numbers Steve tried to compute the average of his 5 test scores. He scored 86 on one, 55 on the next one, then 99, 77, and 90. What is Steve's average test score? Calculating The Mean (Average)
Geometry Knowing both the area and the perimeter of triangles, quadrilaterals, and circles If one angle of a triangle is 95 degrees and another is 35 degrees, what is the third angle of the triangle? 180 Triangle Rule Problems Areas of Rectangles & Squares Rectangles & Squares Basic Circle Area Problems
Substituting values for x Being given a value for x and plugging in What is the value at \$(x^2-2)/(x-1)\$ for x=2? Intro to Variables & Expressions
Setting up an Equation Translating an equation from words into numbers and variables A car's distance is determined by the velocity multiplied by the amount of time the car is at that speed.If a car can go at a maximum velocity of 225 meters per second, what equation shows how far the car has travelled at its maximum speed? How to Translate Words Into Equations
Operations with Polynomials Adding,subtracting,multiplying,and dividing equations with multiple variables Simplify: \$2a+3b-(-6a+4b)\$ Adding Polynomials (a.k.a. Combining Like Terms) Distribution (a.k.a. "multiplying stuff out") Solving Equations With Distribution Basic FOIL on 2-by-2 parentheses
Factoring Undoing a polynomial and simplifying it into its most basic equations as well as using it to find certain input values What is one possible factor of \$x^2+4x+3\$ ? Factoring Stuff Out (a.k.a. "distrubution in reverse") Easier Factoring Problems Factoring Difference of Squares Hard Factoring Problems
Linear Equations to One Variable An input which leads to an output and using the output to solve for the input Solve for x when \$2(x-2)=-10\$ Solving Equations With Addition & Subtraction Only] Solving Equations With Multiplication & Division] Solving Two Step Equations
Exponent Properties Multiplying, dviding, or taking the power of a number or variable to a power \$(x^2)((x^5)^2)/ x^9\$ =? Basic Exponent Rules & Rational Expressions Negative Exponents Common Algebraic Canceling Mistakes
Irrational functions and multiplying by the Complex Conjugate Simplifying a polynomial over a polynomial and being able to simplify roots in a denominator. \$ √x/(3√x- √y)\$ Dividing Roots & Rationalizing Denominators Solving Rational Equations (x's in denominators)
System of Equations Solving for an intersection point of two equation at a designated x or y value. Solve for x and y: \$2x+3y=6\$ \$8x-4y=-12\$ Solving Systems of Simultaneous Equations with Elimination Solving Systems of Simultaneous Equations with Substitution Solving 3 Simultaneous Equations with 3 Unknowns
Inequalities Solving for the range of a functions x or y value using manipulation of a number line What is the range of x for the function \$2x+5<6\$ ? Solving "Linear" Inequalities |
Question Video: Velocity Graphs | Nagwa Question Video: Velocity Graphs | Nagwa
# Question Video: Velocity Graphs
The diagram shows a graph of the variation of the position 𝑥 of an object with time 𝑡. What is the velocity of the object in the time interval 𝑡 > 0 s to 𝑡 = 0.5 s? What is the velocity of the object in the time interval 𝑡 > 0.5 s to 𝑡 = 1.0 s? What is the velocity of the object in the time interval 𝑡 > 1.0 s to 𝑡 = 2.0 s?
03:04
### Video Transcript
The diagram shows a graph of the variation of the position 𝑥 of an object with time 𝑡. What is the velocity of the object in the time interval 𝑡 greater than zero seconds to 𝑡 equal 0.5 seconds? What is the velocity of the object in the time interval 𝑡 greater than 0.5 seconds to 𝑡 equals 1.0 seconds? What is the velocity of the object in the time interval 𝑡 greater than 1.0 seconds to 𝑡 equals 2.0 seconds?
In this exercise, we want to solve for three average velocities. We’ll call them 𝑣 one, 𝑣 two, and 𝑣 three. These average velocities are based on our position versus time graph, where 𝑣 one applies to the first leg of the journey, 𝑣 two applies to the second leg, and 𝑣 three applies to the third and final leg of this movement. To get started, we can recall that average velocity 𝑣 sub avg is equal to displacement divided by time 𝑡. Let’s apply this relationship to solve first for 𝑣 sub one.
𝑣 sub one is equal to the position of our object 𝑝 when time equals 0.5 seconds minus our object’s position when time equals zero seconds. And this difference in position, called the displacement, is then divided by the time it takes to move that distance. Looking at our diagram, we can mark out the position of the object at 𝑡 equals 0.5 seconds and 𝑡 equals zero seconds. And if we drop a vertical line down from our second point, we see that the time interval Δ𝑡 for all this to happen is equal to 0.5 seconds. This means that 𝑣 sub one is 1.0 meters per second. That’s the average velocity of the object over this initial time interval.
Next, we move on to calculating 𝑣 sub two. 𝑣 sub two is the average velocity of the object between time equals 1.0 seconds and 0.5 seconds. Looking at the diagram of the object’s position versus time, we see that its position at both these time values is the same. It’s 0.5 meters. That means 𝑣 sub two is 0.0 meters per second. That’s the average velocity of our object over a time interval where its position doesn’t change.
Finally, we move on to calculating 𝑣 three, the average velocity of our object over the third time interval from time 𝑡 equals 2.0 seconds to 1.0 seconds. Looking on our diagram, we see that at 2.0 seconds our object has a position of 0.0 meters and at 1.0 seconds it has a position of 0.5 meters. Calculating this fraction, we find 𝑣 sub three is negative 0.50 meters per second. That’s the object’s average velocity on the last leg of its movement.
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# What are the types of irregular shapes?
## What are the types of irregular shapes?
Some of the examples of irregular polygons are scalene triangle, rectangle, kite, etc. When the angles and sides of a pentagon and hexagon are not equal, these two shapes are considered irregular polygons.
## What are the examples of irregular solid?
Examples of irregular solids include rocks, prisms, beans and screws. Each of these objects feature an irregular shape that is not composed of other recognized shapes. An irregular solid is distinguished by the inability to determine its area, volume or mass by measuring its length, width and height.
Are rectangles irregular shapes?
Shape (a), for example, is a rectangle (and therefore by definition also a parallelogram and a quadrilateral). It is irregular by virtue of the fact that, although opposite sides are equal in length, adjacent sides are not. It is irregular because adjacent sides are not equal, and adjacent angles are not equal.
What are irregular 3D shapes?
Irregular shapes are shapes of which the sides and interior angles are not all the same. They can be harder for children to identify as they don’t look like the conventional shape they are used to seeing. You can find examples of both irregular 2D and irregular 3D shapes.
### What are regular and irregular objects?
Regular shapes have sides that are all equal and interior (inside) angles that are all equal. Irregular shapes have sides and angles of any length and size.
### Is a semi circle an irregular shape?
Depending on the shape and curves, a part of the figure can be a circle, semicircle or quadrant as well. The following figure is an irregular shape with 8 sides, including one curve. Decompose the figure into two rectangles and a semicircle.
Which is an example of an irregular shape?
Irregular shapes are polygons with five or more sides of varying lengths. These shapes or figures can be decomposed further into known shapes like triangles, squares, and quadrilaterals to evaluate the area.
How is the area of an irregular shape determined?
A regular shape has equal sides and angles, whereas, an irregular shape can be of any size and length. Thus, the area of regular shapes can be determined by directly applying suitable formulas for it whereas the area of an irregular shape can be found by decomposing an irregular shape into several regular shapes.
## Which is an example of a regular object?
Those substances that have fixed geometrical shapes are called regular objects. Some of the examples of regular objects are books, pencils, chalk box, basketball, etc. Those substances which do not have a fixed geometrical shape are called irregular objects.
## Which is a regular shape a square or a rectangle?
A square, by definition, is a regular shape: it is, in fact, a regular rectangle. A regular triangle is called an equilateral triangle. To browse our entire collection of free and premium maths resources for teachers and parents, register to join the Third Space Learning maths hub. |
Calculate volume of a cube
If you need calculating the volume of a cubeOur online tool allows you to find the volume of this regular geometric figure by simply entering the length of one of its sides.
When you press the calculate button you will get the result you are looking for.
Formula to calculate the volume of a cube
The cube is a regular geometric figure. which is composed of six faces whose shape is that of a square. This implies that all edges measure exactly the same.
This makes the calculations much easier, because if you want to calculate the area of a square we have to square the length of a side, to know the volume we have to cubing the edge valueThis is represented in the following mathematical formula:
cube volume = a3
Let's imagine that we have a cube whose side measures 3 centimeters, in this case, the volume of the figure will be:
volume = 3 x 3 x 3 x 3 = 27 cm3
It is important to remember that since we are working with volumes, the unit also rises cubed. In the previous example we start with centimeters and end with cubic centimeters.
If you want to calculate the volume of a rectangular cube, then we recommend that you learn how to calculate the volume of a rectangular cube. volume of a prismis practically the same but taking into account the different measurements of each side.
How to calculate the area of a cube without knowing how long the edge is
It may be that in the mathematics exercise for calculate the volume of a cubedo not give us how long the edge is, but give us how much is the surface of one of its faces.
In this case, we know that the surface of a square is equal to the base squared (b2). Therefore, if we know the surface area, we can find out how long one of the sides is of the figure and from there, find its volume.
To do this, the first thing we have to do is to calculate how long the side of the face of the cube measures from its surface. Simply we make the square root of the value of the surface area and the result will be the length of the edge.
Finally, we cube the value obtained as we have seen in the previous point and that's it. We have found the volume of this regular figure from the surface of one of its faces. |
# Linear Transformation and Matrices
I have been studying linear algebra for a while now, and I still can't understand the basic concept of linear transformation and the easy ''translation'' of them the matrices.
I understand that every proof we make can be done with both concepts, and that they are connected pretty close, and there is a basis which ''moves'' us from one concept to another.
If anyone can write a brief explanation, or some reading material, It would be a big step for me for further understanding of the subject.
Thanks,
Alan
Consider the linear transformation $T$ that takes a vector in $(x,y)\in\mathbb{R}^2$ and maps it to the vector $(x+2y,3x+4y)\in\mathbb{R}^2$. That is,
$$\begin{bmatrix} x\\ y \end{bmatrix} \overset{T}{\longmapsto} \begin{bmatrix} x+2y\\ 3x+4y \end{bmatrix}. \tag{1}$$
We want to express this transformation in the form of $A\mathbf{x}$ where $A$ is a matrix and $\mathbf{x}=(x,y)$ is the input vector. Hopefully you can see here that the matrix of the linear transformation described in $(1)$ is $$A= \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$$ since $$A\mathbf{x}=\underbrace{\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}}_{\text{matrix of }T}\ \ \underbrace{\begin{bmatrix}x\\ y\end{bmatrix}}_\text{input vector} = \underbrace{\begin{bmatrix} x+2y\\ 3x+4y \end{bmatrix} }_\text{output vector}$$
Note that $A$ is not the linear transformation per se; rather $A$ is a computationally efficient way to write down the action taking place in $(1)$ in the form of the matrix-vector multiplication $A\mathbf{x}$.
Implicit in this conversation is that we are writing down the matrix of the linear transformation in $(1)$ with respect to some fixed basis. If changing the basis is your emphasis rather than the general idea of the connection between a linear transformation and its associated matrix, you might study this. |
Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths
# Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download
INTRODUCTION
We have studied in detail about the properties of a triangle. We also know that the triangle is a figure obtained
by joining three non collinear points in pair. In this chapter we shall discuss about four non-collinear points such that no three of them are collinear.
We know that the figure obtained on joining three non-collinear points in pairs is a triangle. If we mark four
points and join them in some order, then there are three possibilities for the figure obtained:
(i) If all the points are collinear (in the same line), we obtain a line segment.
(ii) If three out of four points are collinear, we get a triangle.
(iii) If no three points out of four are collinear, we obtain a closed figure with four sides.
Each of the figure obtained by joining four points in order is called a quadrilateral. (quad means four and lateral for sides).
A quadrilateral has four sides, four angles and four vertices.
In quadrilateral ABCD, AB, BC, CD and DA are the four sides; A, B, C and D are the four vertices and A,
B, C and D are the four angles formed at the vertices.
If we join the opposite vertices A to C and B to D, then AC and BD are the two diagonals of the quadrilateral ABCD.
Question for Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths
Try yourself:
Which of the following figures is obtained when four non-collinear points are joined in order?
We find so many objects around us which are of the shape of a quadrilateral the floor, walls, ceiling, windows
of our classroom, the blackboard, each face of the duster, each page of our mathematics book, the top of our
study table, etc. Some of these are given below.
4. SOME RELATED TERMS TO QUADRILATERALS
In a quadrilateral ABCD, we have
(i) VERTICES :- The points A, B, C and D are called the vertices of quadrilateral ABCD.
(ii) SIDES :- The line segments AB, BC, CD and DA are called the sides of quadrilateral ABCD.
(iii) DIAGONALS :- The line segments AC and BD are called the diagonals of quadrilateral ABCD.
(iv) ADJACENT SIDES :- The sides of a quadrilateral are said to be adjacent sides if they have a common end point.
Here, in the above figure, (AB,BC),(BC,CD),(CD,DA)and(DA,AB) are four pairs of adjacent sides or consecutive sides of quadrilateral ABCD.
(v) OPPOSITE SIDES :- Two sides of a quadrilateral are said to be opposite sides if they have no common end point.
Here, in the above figure, (AB, DC) and (BC, AD) are two pairs of opposite sides of quadrilateral ABCD.
(vi) CONSECUTIVE ANGLES :- Two angles of a quadrilateral are said to be consecutive angles if they have a common arm.
Here, in the above figure, (A,B), (B,C), (C, D) and (D, A) are four pairs of consecutive angles.
(vii) OPPOSITE ANGLES :- Two angles of a quadrilateral are said to be opposite angles if they have no common arm.
Here, in the given figure, (A, C) and (B, D) are two pairs of opposite angles of quadrilateral ABCD.
(i) PARALLELOGRAM :- A quadrilateral in which Dboth pair of opposite Csides are parallel is called a parallelogram.
In figure, ABCD is a quadrilateral in which AB || DC, BC || AD.
(ii) RHOMBUS :- A parallelogram whose all sides are equal is called rhombus.
In figure, ABCD is a parallelogram in which AB = BC = CD = DA, AB | DC and BC || AD .
parallelogram ABCD is a rhombus.
(iii) RECTANGLE :- A parallelogram whose each angle is equal to 90°, is called a rectangle.
In figure, ABCD is a parallelogram in which
A = B = C = D = 90° ,
AB || DC and BC || AD.
Parallelogram ABCD is a rectangle.
(iv) SQUARE :- A rectangle in which a pair of adjacent sides are equal is said to be a square.
In figure, ABCD is a rectangle in which A = B = C = D = 90° ,
AB = BC, BC = CD, CD = DA, DA= AB.
i.e., AB = BC = CD = DA .
rectangle ABCD is a square.
(v) TRAPEZIUM :- A quadrilateral in which exactly one pair of opposite sides is parallel, is called a trapezium.
A In figure, ABCD is a quadrilateral in which ABDC.
ABCD is trapezium.
(vi) ISOSCELES TRAPEZIUM :- A trapezium whose non-parallel sides are equal is called an isosceles trapezium.
In figure, ABCD is a trapezium in which AB || DC and BC = AD .
trapezium ABCD is isosceles trapezium.
(vii) KITE :- A quadrilateral in which two pairs of adjacent sides are equal is called a kite.
In figure, ABCD is a quadrilateral in which AB = AD and BC = CD.
Question for Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths
Try yourself:
Which type of quadrilateral has all sides equal and opposite sides parallel?
6. ANGLE SUM PROPERTY OF A QUADRILATERAL
THEOREM-I : The sum of the four angles of a quadrilateral is 360°.
To Prove :A + B + C + D = 360°
Construction : Join AC. 2 1
Proof :
Hence, proved.
Ex.1 Three angles of a quadrilateral measure 56°, 100° and 88°. Find the measure of the fourth angle.
Sol. Let the measure of the fourth angle be x°.
56° + 100° + 88° + x° = 360° [Sum of all the angles of quadrileteral is 360°]
⇒ 244+x=360
⇒ x = 360 – 244 = 116
Hence, the measure of the fourth angle is 116°.
Ex.2 In figure, ABCD is a trapezium in which AB || CD. If D = 45° and C = 75°, find A and B.
Sol.
We have, AB || CD and AD is a transversal.
so, A + D = 180° [Interior angles on the same side of the transversal]
Ex. 3 In the given figure, sides AB and CD of the quadrilateral ABCD are produced. Find the value of x.
Sol.
Ex.4 In the given figure, ABCD is a quadrilateral in which AE and BE are the angle bisectors of A and B. Prove that C + D = 2AEB.
Sol. Given : ABCD is a quadrilateral in which AE and BE are the
Hence proved.
Ex.5 In figure, ABCD is a quadrilateral in which AB = AD and BC = CD . Prove that
(i) AC bisects
A and C
(ii) BE = DE.
Sol. Given : ABCD is a quadrilateral in which AB = AD and BC = CD
To Prove :
(i) AC bisects A and C
(ii) BE = DE.
Proof :
Hence proved
Ex.6 In a quadrilateral ABCD, AO and BO are the bisectors of A and B respectively. Prove that
Sol. Given : In a quadrilateal ABCD, AO and BO are the bisectors of A and B
Hence proved
Ex.7 In fig. bisectors of B and D of quadrilateral ABCD meet CD and AB produced at P and
Q respectively. Prove that
Sol. Given : bisectors of B and D of quadrilateral ABCD meet CD and AB produced at P and Q
To Prove :
Proof :
STATEMENT REASON
[Sum of the angles of a quadrilateral equals is 360°]
Hence proved
Ex.8 In quadrilateral ABCD B = 90°, C – D = 60° and A – C – D = 10°. Find A, C and D.
Sol. A + B + C + D = 360° (Sum of the four angles of a quadrilateral is 360°)
A + C = 140°, A : C = 1 : 3 and B : D = 5 : 6. Find the A, B, C and D.
Sol.
Hence, A = 35°, B = 100°, C = 105° and D = 120°
Ex.10 In figure, ABCD is a parallelogram in which D = 72°. Find A, B and C.
Sol. We have D = 72°
But B = D [Opposite angles of the parallelogram]
B = 72°
Now, AB || CD and AD and BC are two transversals.
So, A + D = 180° [Interior angles on the same side of the transversal AD]
The document Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 is a part of the Class 9 Course Extra Documents & Tests for Class 9.
All you need of Class 9 at this link: Class 9
## Extra Documents & Tests for Class 9
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## FAQs on Chapter 8 - Quadrilaterals, Solved Examples, Class 9, Maths - Extra Documents & Tests for Class 9
Ans. Quadrilaterals are polygons that have four sides, four vertices, and four angles. They are two-dimensional shapes that can be convex or concave. Examples of quadrilaterals include squares, rectangles, parallelograms, rhombuses, kites, and trapezoids.
2. What is the difference between convex and concave quadrilaterals?
Ans. Convex quadrilaterals have all their interior angles measuring less than 180 degrees and all their vertices pointing outwards. Concave quadrilaterals, on the other hand, have at least one interior angle greater than 180 degrees, and at least one vertex pointing inwards.
3. How do you find the perimeter of a quadrilateral?
Ans. To find the perimeter of a quadrilateral, you need to add up the lengths of all four sides. For example, if a quadrilateral has sides measuring 5 cm, 7 cm, 9 cm, and 6 cm, then its perimeter would be 27 cm (i.e., 5 + 7 + 9 + 6).
4. What is the sum of interior angles of a quadrilateral?
Ans. The sum of interior angles of a quadrilateral is always 360 degrees. This means that if you add up all the angles inside a quadrilateral, the total will always be 360 degrees.
5. How do you determine if a quadrilateral is a parallelogram?
Ans. A quadrilateral is a parallelogram if it has two pairs of parallel sides. Additionally, if one angle of a parallelogram is a right angle, then all four angles are right angles. If the diagonals of a quadrilateral bisect each other, then it is also a parallelogram.
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# Maharashtra Board Practice Set 2 Class 7 Maths Solutions Chapter 1 Geometrical Constructions
Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 2 Answers Solutions Chapter 1 Geometrical Constructions.
## Geometrical Constructions Class 7 Practice Set 2 Answers Solutions Chapter 1
Question 1.
Draw triangles with the measures given below:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.
ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.
iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.
Solution:
i. In ∆ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.
ii. In ∆STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.
iii. In ∆PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.
Question 2.
Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.
Solution:
Question 3.
Draw an equilateral triangle with side 6.5 cm.
Solution:
Question 4.
Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene triangle.
Solution:
i. Equilateral triangle LMN, l(LM) = l(MN) = l(LN) = 4 cm.
ii. Isosceles triangle STU, l(ST) = l(TU) = 4cm, l(SU) = 6 cm
iii. Scalene triangle XYZ, l(XY) = 4.5 cm, l(XY) = 6.5 cm, l(XZ) = 5.5 cm
Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 2 Intext Questions and Activities
Question 1.
Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm. (Textbook pg. no. 3)
1. Can this triangle be drawn?
2. A number of triangles can be drawn to fulfill these conditions. Try it out.
3. Which further condition must be placed if we are to draw a unique triangle using the above information?
Solution:
1. ∆ABC triangle cannot be drawn as length of third side is not given.
2. For ∆ABC to draw l(AC) > l(AB) + l(BC)
i.e., l(AC) > 4 + 3
i.e., l(AC) > 7 cm
∴ number of triangles can be drawn if l(AC) > 7 cm
3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle. |
# Multiplying Expressions – Methods & Examples
The operation of rational expressions might seem difficult to a few students, but the rules for multiplying expressions are just the same with integers. In Mathematics, a rational number is defined as a number in the form p/q, where p and q are integers and q is not equal to zero.
Examples of rational numbers are: 2/3, 5/8, -3/14, -11/-5, 7/-9, 7/-15 and -6/-11 etc.
An algebraic expression is a mathematical phrase where variables and constants are combined using the operational (+, -, × & ÷) symbols.
For example, 10x + 63 and 5x – 3 are examples of algebraic expressions. Similarly, a rational expression is in the form p/q, and either or both p and q are algebraic expressions.
Examples of rational expression include: 3/ (x – 3), 2/ (x + 5), (4x – 1)/3, (x2 + 7x)/6, (2x + 5)/ (x2 + 3x – 10), (x + 3)/(x + 6) etc.
## How to Multiply Rational Expressions?
In this article, we will learn how to multiply rational expressions, but before that, let us remind ourselves two fractions are multiplied.
Multiplication of two fractions entails finding the numerator of the first and the second fractions and the product of the denominator. In other words, the multiplication of two rational numbers is equal to the product of the numerators/product of their denominators.
Similarly, the multiplication of rational numbers is equal to the product of their numerators/product of their denominators. For instance, if a/b and c/d are two rational expressions, then multiplication of a/b by c/d is given by; a/b × c/d = (a × c)/ (b × d).
Alternatively, you can perform multiplication of rational expressions by; first factoring and canceling out the numerator and denominator and then multiplying the remaining factors.
Below are the steps required for multiplying rational expressions:
• Factor out both the denominator and numerator of each expression.
• Reduce the expressions to the lowest terms possible only if the numerators and denominators’ factors are common or similar.
• Multiply together the remaining expressions.
Example 1
Multiply 3/5y * 4/3y
Solution
Separately multiply the numerators and denominators;
3/5y * 4/3y = (3 * 4)/ (5y * 3y)
= 12/15y 2
Reduce the fraction by cancelling out by 3;
12/15y 2 = 4/5y2
Example 2
Multiply {(12x – 4x 2)/ (x 2 + x -12)} * {(x 2 + 2x -8)/ (x 3-4x)}
Solution
Factor out both the numerators and denominators of each expression;
= {- 4x (x – 3)/(x-3) (x + 4)} * {(x – 2) (x + 4)/x (x + 2) (x – 2)}
Reduce or cancel the expressions and rewrite the remaining fraction;
= -4/ x + 2
Example 3
Multiply (x 2 – 3x – 4/x 2 -x -2) * (x 2 – 4/ x2 + x – 20).
Solution
Factor the numerators and denominators of all expressions;
= (x – 4) (x + 1)/ (x + 1) (x – 2) * (x + 2) (x – 2)/ (x – 4) (x + 5)
Cancel out and rewrite the remaining factors;
= x + 2/ x + 5
Example 4
Multiply
(9 – x 2/x 2 + 6x + 9) * (3x + 9/3x – 9)
Solution
Factor the numerators and denominators and cancel out common factors;
= – 1 (x + 3) (x – 3)/ (x + 3)2 * 3(x + 3)/3(x – 30
= -1
Example 5
Simplify: (x2+5x+4) * (x+5)/(x2-1)
Solution
By factoring the numerator and denominator, we get;
=>(x+1) (x+4) (x+5)/(x+1) (x-1)
On canceling the common terms, we get;
=>(x+4) (x+5)/x-1
Example 6
Multiply ((x + 5) / (x – 4)) * (x / x + 1)
Solution
= ((x + 5) * x) / ((x – 4) * (x + 1))
= (x2 + 5x) / (x2 – 4x + x – 4)
= (x2 + 5x) / (x2 – 3x– 4)
When you multiply a whole number by an algebraic expression, you multiply the number by the expression’s numerator.
This is possible because any whole number always has a denominator of 1. And therefore, the multiplication rules between an expression and a whole does not change.
Consider the example 7 below:
Example 7
Multiply ((x + 5) / (x2 – 4)) * x
Solution
= ((x + 5) / (x2 – 4)) * x / 1
= (x + 5) * x / (x2 – 4) × 1
= (x2 + 5x) / (x2 – 4)
### Practice Questions
1. Simplify the expression: 4xy2/3y * 2x/4y.
2. Simplify the expression: (8x 2 – 6x)/ (4 – x) * (x 2 -16)/(4x 2 – 7x – 3) * (-5x -5)/(2x + 8).
3. Simplify the expression: (8x 2 – 6x)/ (4 – x) * (x 2 -16)/(4x 2 – 7x – 3) * (-5x -5)/(2x + 8).
4. Simplify the expression: [(x2 – 7x + 10)/ (x 2 – 9x + 14)] * [(x 2 – 6x -7)/(x 2 + 6x + 5)].
5. Simplify the expression: [(2x + 1)/(x2 – 1)] * [(x + 1)/(2x 2 + x)].
6. Simplify the expression: [(-3x^2 +27)/(x^3 – 1)] *[ (7x^3 + 7x^2 + 7x)/(x – 3x)] * (x – 1)/21.
7. Simplify the expression: [(x2 – 5x – 14)/ (x2 – 3x + 2)] * [(x 2 – 4)/(x2 – 14x + 49)].
8. The product of the two numbers’ sum and the difference is equal to 17. If the product of the two numbers is 72, what are the two numbers? |
# Understanding Binary Numbering
## Understanding Binary Numbering
All computers function using a system of switches that can be in one of two positions, on or off. This is called a binary system, with “off” being represented by the digit 0 and “on” being represented by the digit 1. A binary number will include only the digits 0 and 1. Network device addresses also use this binary system to define their location on the network. The IP address is based on a dotted decimal notation of a binary number. You must have a basic understanding of the mathematical properties of a binary system to understand networking. The following sections describe the mathematics involved in the binary numbering system and explain how to convert a decimal (base 10) number to a binary (base 2) number and vice versa.
#### Decimal and Binary Systems
The decimal (base 10) system is the numbering system used in everyday mathematics, and the binary (base 2) system is the foundation of computer operations. In the decimal system, the digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. When quantities higher than 9 are required, the decimal system begins with 10 and continues all the way to 99. Then the decimal system begins again with 100, and so on, with each column to the left raising the exponent by 1.
The binary system uses only the digits 0 and 1. Therefore, the first digit is 0, followed by 1. If a quantity higher than 1 is required, the binary system goes to 10, followed by 11. The binary system continues with 100, 101, 110, 111, 1000, and so on. Table 4-1 shows the binary equivalents of the decimal numbers 0 through 19.
Table 4-1 Decimal Versus Binary Numbers
Decimal Number Binary Number 0 0 1 1 2 10 3 11 4 100 5 101 6 110 7 111 8 1000 9 1001 10 1010 11 1011 12 1100 13 1101 14 1110 15 1111 16 10000 17 10001 18 10010 19 10011
#### Least Significant Bit and Most Significant Bit
Most people are accustomed to the decimal numbering system. While the base number is important in any numbering system, it is the position of a digit that confers value. The number 10 is represented by a 1 in the tens position and a 0 in the ones position. The number 100 is represented by a 1 in the hundreds position, a 0 in the tens position, and a 0 in the ones position.
In a binary number, the digit on the rightmost side is the least significant bit (LSB), and the digit on the leftmost side is the most significant bit (MSB). The significance of any digits in between these sides is based on their proximity to either the LSB or the MSB. Figure 4-7 shows the relationships of bit significance to the values for base 10 and base 2 numbering systems.
Figure 4-7 Bit Significance
Base-10 Decimal Conversion – 63204829
Base-2 Binary Conversion – 1110100 (233)
#### Base 2 Conversion System
Understanding the base 2 system is important because an IP version 4 (IPv4) address consists of 32 binary bits. Each digit is 1 bit. The 32 bits are divided into four sets of 8 bits, called octets. A dot (period) is placed between each set to separate them. (A byte is another name for 8 bits; however, for the purposes of this module, 8 bits will be referred to as an octet.)
The various classes of addresses are based on the octet boundaries, so it is helpful to get used to such groupings. It is also an ease-of-use issue, because 8-bit binary numbers are easier to convert than 32-bit binary numbers. When converting a binary IP address, you only convert one octet at a time. The highest possible binary octet is 11111111, which converts to the decimal number 255. The lowest possible binary octet is 00000000, which is the decimal number 0. That means that with 8 bits, you can have 256 different number combinations, 0 to 255 inclusive.
#### Powers of 2
To understand how binary numbers are used in IP addressing, you must understand the mathematical process of converting a decimal number to a binary number and vice versa. Calculator batteries run down and charts can be misplaced, but if you know the mathematical principles, a piece of paper and pencil are all that you need to convert binary numbers to decimal numbers and to convert decimal numbers to binary numbers. There are charts available to help with decimal-to-binary conversion, showing, for example, 20 = decimal 1, 21 = decimal 2, 22 = decimal 4, and so on. Table 4-2 illustrates which decimal numbers are produced for powers of 2.
Table 4-2 Powers of 2
Decimal-to-Binary Conversion
Decimal numbers can be converted to binary numbers through a specific process, as shown in Figure 4-8.
Figure 4-8 Decimal-to-Binary Conversion
This example shows a simple binary conversion of the decimal number 35. The base exponent line shows base 2 numbers and their exponents (2 * 2 = 4 * 2 = 8, and so on). The decimal value of the base exponent number is listed in the second row, and the binary number is displayed in the third row. The table describes the steps to determine the binary number. Notice that the first 2 bits of the binary number are 0s; these are known as leading 0s. In reality, the decimal number 35 would only be a 6-bit binary number. Because IP addresses are laid out as four sets of octets, the binary number is made into an octet by placing 0s to the left of the 6-bit number.
The steps used in converting the number 35 to a binary number are as follows:
Step 1 Looking at Figure 4-8, what is the greatest power of 2 that is less than or equal to 35? 128 does not go into 35, so place a 0 in that column.
Step 2 64 does not go into 35, so place a 0 in that column.
Step 3 25 (32) is smaller than 35. 32 goes into 35 one time. Place a 1 in that column.
Step 4 Calculate how much is left over by subtracting 32 from 35. The result is 3.
Step 5 Check to see whether 16 (the next lower power of 2) fits into 3. Because
it does not, a 0 is placed in that column.
Step 6 The value of the next number is 8, which is larger than 3, so a 0 is placed in that column, too.
Step 7 The next value is 4, which is still larger than 3, so it too receives a 0.
Step 8 The next value is 2, which is smaller than 3. Because 2 fits into 3 one time, place a 1 in that column.
Step 9 Subtract 2 from 3, and the result is 1.
Step 10 The decimal value of the last bit is 1, which fits in the remaining number. Therefore, place a 1 in the last column. The binary equivalent of the decimal number 35 is 00100011.
#### Binary-to-Decimal Conversion
As with decimal-to-binary conversion, there is usually more than one way to convert binary numbers to decimal numbers. You can convert binary numbers to decimal numbers using the positional values based on the powers of 2 and identifying the columns with nonzero values, which contribute to the final numerical value. Figure 4-9 illustrates this process.
Figure 4-9 Binary-to-Decimal Conversion
The steps used for converting the binary number 10111001 to a decimal number are as follows:
Step 1 Find the place value that corresponds to any 1 bit in the binary number, according to its position. For example, as shown in Figure 4-9, the binary bit in the 27 column is 1, so the decimal total is 128.
Step 2 There is a 0 in the 26 (64) column. The decimal equation is 128 + 0 = 128.
Step 3 There is now a 1 in the 25 (32) column. The decimal equation becomes 128 + 32 = 160.
Step 4 There is a 1 in the 24 (16) column. Adding the value to the decimal total gives 160 + 16 = 176.
Step 5 The next column, 23, has a 1, so add the value 8 to the decimal total, giving 176 + 8 = 184.
Step 6 There are 0s in the 22 and 21 columns. Add 0s to the decimal total: 184 + 0 + 0 = 184.
Step 7 Finally, there is a 1 in the 20 (1) column. Now, add 1 to 184. The result is 185. The decimal equivalent of the binary number 10111001 is 185. |
# Basic Things About Corresponding Angles Which You Should be Clear About?
Being clear about the corresponding angles is considered to be the best possible way of ensuring that kids can have a good command of the subject of mathematics. Corresponding angles are the angles that can be formed by matching corners or corresponding corners with the transversal when two parallel lines will be intersected by any other line. The very basic example is the angles that are found on the opposite side of the transversal. It is very much important for people to note that transversal can intersect to parallel or non-parallel lines because of which the corresponding angles can be of two types. These are categoriszed into two categories which are explained as follows:
• Corresponding angles formed by parallel lines and transversal:
If a line of transversal crosses any kind of two parallel lines then the corresponding angles which have been formed will always have equal measure and a set of two parallel lines along with transversal will be forming eight angles with the transversal among which corresponding angles will be equal.
• Corresponding angles formed by non-parallel lines and transversal:
For the non-parallel lines if a transversal intersects them because such angles formed by them will not have any kind of relationship with each other they will never be equal which was possible in the cases of parallel lines. In such cases, there will be no relationship between the interior angles, exterior angles, vertically opposite angles and consecutive angles.
## The corresponding angles postulate has been perfectly explained as follows:
According to this particular concept, the corresponding angles will always be congruent if the transversal is intersecting two parallel lines. In other words, if the transversal will intersect two parallel lines the corresponding angles will always be equal without any kind of doubt and the best part is that kids will be able to solve the questions very easily and efficiently.
The corresponding angles are also available in the cases of the triangle and these will be the ones that are contained by the concurrent pair of sides of two similar or congruent to the triangle. The corresponding angles into the triangle will always have the same measure without any kind of doubt. The corresponding angles can also be supplementary if the transversal intersects two parallel lines perpendicular because every angle will be having a 90° measure. In such cases, the sum will come out to be 180°. Corresponding angles will not always be equal and they will always be formed by transversal and parallel lines. The angle rule of corresponding angles is also very much available in this particular area which further makes sure that corresponding angles will be equal if the transversal will cut two parallel lines. Corresponding angles can be both complementary as well as supplementary without any kind of issue.
## Following are some of the very basic terms associated with the whole process of parallel lines and transversal:
1. Corresponding angles will always be formed at the same relative position at every intersection.
2. Vertically opposite angles can be formed by transversal and parallel lines and will always be opposite and equal to each other.
3. The angles which are formed on the interior side or inside of two parallel lines with the transversal will be termed as the alternate interior angles.
4. The angle formed on the outside exterior side of the parallel lines and transversal will be the alternate exterior angles.
5. The angles are found inside two parallel lines but one side of the transversal is consecutive interior angles and the angles which are supplementary to each other will always have the sum of 180°.
Hence, being clear about all the above-mentioned points is very much important for the people and apart from this people also need to be clear about the concept of the complementary angles so that they can solve the questions very easily. Mastering this particular subject is only possible if people register themselves on platforms like Cuemath so that overall goals are efficiently achieved. |
# What is the Associative Property?
Tricia Christensen
Tricia Christensen
The associative property of mathematics refers to the ability to group certain numbers together in specific mathematical operations, in any type of order without changing the answer. Most commonly, children begin to study the associative property of addition and then move on to study the associative property of multiplication. With both these operations, changing the order of the numbers being added or numbers being multiplied won’t result in a changed sum or product.
Some confuse the associative property with the commutative property, but the commutative property tends to apply to two numbers only. In contrast, the associative property is often used to express the unchanging nature of sums or products when three or more numbers are used. The property may also be discussed in relationship to how parentheses are used in math. Placing parentheses around some of the numbers that will all be added together doesn’t change the results.
Consider the following examples:
1 + 2 + 3 +4 = 10. This will remain true even if the numbers are grouped differently.
(1 + 3) + (2 + 4) and (1 + 2 + 3) + 4 both equal ten. You don’t have to consider the order of these numbers or their grouping since the act of adding means they will still have the same total sum.
In the associative property of multiplication, the same basic idea holds true. A X B X C = (AB)C or (AC)B. No matter how you group these numbers together, product remains constant.
Especially in multiplication, the associative property can prove very helpful. Take for instance the basic formula for calculating the area of a triangle: 1/2bh or half of the base times the height. Now consider that the height is 4 inches and the base is 13 inches. It is simpler to take half of height (4/2 = 2) than it is to take half of the base (13/2 = 6.5). It’s a lot easier to solve the resultant problem 2 X 13 than it is to solve 6.5 X 4.
We can do this when we understand the associative property because we will know that it doesn’t matter what order we multiply these numbers in. This can take the work out of some complicated calculations and make math work just a little bit easier. Note that this property does not work when you use division or subtraction. Changing order and grouping with these operations will impact results.
Tricia Christensen
Tricia has a Literature degree from Sonoma State University and has been a frequent InfoBloom contributor for many years. She is especially passionate about reading and writing, although her other interests include medicine, art, film, history, politics, ethics, and religion. Tricia lives in Northern California and is currently working on her first novel.
Tricia Christensen
Tricia has a Literature degree from Sonoma State University and has been a frequent InfoBloom contributor for many years. She is especially passionate about reading and writing, although her other interests include medicine, art, film, history, politics, ethics, and religion. Tricia lives in Northern California and is currently working on her first novel. |
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• 1.)luis had 8/10 meter of wood for the frame that...
# 1.)luis had 8/10 meter of wood for the frame that he is doing. his father gave him 1/4 meter more. how many meters of wood does he have now? 2)angelica needs 4/5 kilogram of chicken for her recipe in epp-h. e. subject. she already has 2/4 kilogram. how many more kilograms does she need? 3)ella had a piece of ribbon. after using 3/8 meter for her headband, she had 1/4 meter left. how many meters of ribbon did she have at first?
• Réponse publiée par: 09389706948
1) 42/40 or 1 & 2/40 or 1 & 1/20m of wood
2) 3/10 kg more
3) 5/8 ribbon
Step-by-step explanation:
1) 8/10 + 1/4
MAKE FRACTIONS SIMILAR
1. get the GCF or greatest common factor of the denominators, in this case, 10 and 4
GCF of 4&10= 40
2. change the old denominators(10 &4) to the new denominators(40)
1/4= change 4 to 40 8/10= change 10 to 40
3. divide the GCF (20) by the old denominators then multiply to the old numerator; these numbers will be the new numerators
1/4- 40 divided by 4= 10, 10x 1 8/10- 40 divided by 10= 4, 4x8
1/4= 10/40 8/10= 32/40
10/40 + 32/40= 42/40
10+ 32= 42
YOU MAY CONVERT TO IMPROPER FRACTION
42/40= 1 & 2/40
divide the numerator by the denominator. the quotient (answer) is the whole number(1) and the remainder is the new numerator (2). the denominator stays the same (40)
since the improper fraction is 2/40, you may simplify the fraction and get 1/20 just look for the GCF of the numerator(2) and denominator(40) and divide to denominator and numerator.
2/40 --- 2 divided by 2 and 40 divided by 2
= 1/20m
2) 4/5- 2/4 SUBTRACT
MAKE FRACTIONS SIMILAR
4/5 and 2/4 = 16/20 and 10/20
SUBTRACT
subtract the numerator while the denominator stays the same
16/20- 10/20= 6/20
YOU MAY SIMPLIFY
6/20= 3/10
find the GCF or greatest common factor of the numerator and the denominator (2)
divide them by their GCF ( 6- 3 and 20- 10)
=3/10 kg more
SIMPLIFY 3/8= 3/8 and 1/4 = 2/8
=5/8 ribbon
• Réponse publiée par: HaHannah
didal
Explanation:
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• Réponse publiée par: tayis
dko mkita yong picture
• Réponse publiée par: abyzwlye
ANSWER:SAAN POHH BAHH YUNG TANONG NYO POHH
Connaissez-vous la bonne réponse?
1.)luis had 8/10 meter of wood for the frame that he is doing. his father gave him 1/4 meter more. h... |
Q1. What is the difference between a left neighborhood and a right neighborhood of a number? How does this concept become relevant in determining a limit of a function? Answer: Left neighborhood of a number ‘a’ represents numbers lesser than the number ‘a’ and is denoted by ‘a-’ or ‘a-d’, where d is infinitesimally small. Similarly, right neighborhood of a number ‘a’ represents numbers greater than the number ‘a’ and is denoted by ‘a+’ or or ‘a+d’, where d is infinitesimally small. This concept is very important in determining limit of a function. A function f(x) of ‘x’ will have a limit at x = a; if and only if f(a-d) = f(a+d) = f(a); where d is infinitesimally small. Q2. A limit of a function at a point of discontinuity does not exist. Why? Give an example. Answer: For existence of limit of function f(x) of ‘x’; at x = a; the necessary and sufficient condition is f(a-d) = f(a+d) = f(a); where d is infinitesimally small. At a point of discontinuity, f(a-d) ≠ f(a+d). Therefore, limit of a function does not exist at a point of discontinuity. The following example will make it clear. Let us take example of integer function. This function is defined in the following manner: f(x) = a; where ‘a’ is an integer less than or equal to x. Let us check if limit exists for this function at x = ‘a’, where ‘a’ is an integer. Now left hand side limit = f(a-d) = a-1 And right hand side limit = f(a+d) = a Thus, f(a-d) ≠ f(a+d); and hence limit does not exists for this function. If this function is plotted, there is discontinuity at all integer points. Thus it can be seen that limit of a function does not exist at a point of discontinuity. 3. What is the difference between a derivative of a function and its slope? Give a detailed explanation. Answer: Derivative of a function is another function, which remains same throughout the domain of the function at all the points. Slope of a function on the other hand is the value of the derivative. This value may change from point to point depending on the nature of the function. Let us take an example. Derivative of Sin(x) is Cos(x) for all values of ‘x’. If one looks at the slope of Sin(x), its value keeps changing in [-1, +1] range from point to point. Slope of Sin(x) is -1 for x = odd integral multiples of p; +1 for x = even multiples of p and 0 for x = odd multiples of p/2. Thus, it can be seen that while derivative of a function remains the same while its slope could be changing from point to point. |
# Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers
Here we are providing Pair of Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.
## Extra Questions for Class 10 Maths Arithmetic Progressions with Answers Solutions
Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions with Solutions Answers
### Arithmetic Progressions Class 10 Extra Questions Very Short Answer Type
Question 1.
Which of the following can be the nth term of an AP?
4n + 3, 3n2 + 5, n2 + 1 give reason.
Solution:
4n + 3 because nth term of an AP can only be a linear relation in n as an = a + (n – 1)d.
Question 2.
Is 144 a term of the AP: 3, 7, 11, …? Justify your answer.
Solution:
No, because here a = 3 an odd number and d = 4 which is even. so, sum of odd and even must be odd whereas 144 is an even number.
Question 3.
The first term of an AP is p and its common difference is q. Find its 10th term.
Solution:
210 = a + 9d = p + 99.
Question 4.
For what value of k: 2k, k + 10 and 3k + 2 are in AP?
Solution:
Given numbers are in AP
∴ (k + 10) – 2k = (3k + 2) – (k + 10)
⇒ -k + 10 = 2k – 8 or 3k = 18 or k = 6.
Question 5.
If an = 5 – 11n, find the common difference.
Solution:
We have an = 5 – 11n
Let d be the common difference
d = an+1 – an
= 5 – 11(n + 1) – (5 – 11n)
= 5 – 11n – 11 -5 + 11n = -11
Question 6.
If nth term of an AP is $$\frac{3+n}{4}$$ find its 8th term.
Solution:
Question 7.
For what value of p are 2p + 1, 13, 5p – 3, three consecutive terms of AP?
Solution:
since 20 + 1, 13, 5p – 3 are in AP.
∴ second term – First term = Third term – second term
⇒ 13 – (2p + 1) = 5p – 3 – 13
⇒ 13 – 2p – 1 = 5p – 16
⇒ 12 – 2p = 5p – 16
⇒ -7p = – 28
⇒ p = 4
Question 8.
In an AP, if d = -4, n = 7, a, = 4 then find a.
Solution:
We know, an = a + (n – 1)d
Putting the values given, we get
⇒ 4 = a + (7 – 1)(-4) or a = 4 + 24
⇒ a = 28
Question 9.
Find the 25th term of the AP: -5, $$\frac{-5}{2}$$ , 0, $$\frac{-5}{2}$$ ………
Solution:
Here, a = -5, b = –$$\frac{5}{2}$$ – (-5) = $$\frac{5}{2}$$
We know,
a25 = a + (25 – 1 )d
= (-5) + 24($$\frac{5}{2}$$) = -5 + 60 = 55
Question 10.
Find the common difference of an AP in which a18 – a14 = 32.
Solution:
Given, a18 – a14 = 32
⇒ (a + 17d) – (a + 13d) = 32
⇒ 17d – 13d = 32 or d = $$\frac{32}{4}$$
Question 11.
If 7 times the 7th term of an AP is equal to 11 times its 11th term, then find its 18th term.
Solution:
Given, 7a7= 11a11
⇒ 7(a + 6d) = 11(a + 100) or 7a + 42d = 11a + 110d
⇒ 4a + 68d = 0 or a + 17d = 0
Now, a18 = a + 17d = 0
Question 12.
In an AP, if a = 1, an = 20, and sn = 399, then find n.
Solution:
Given, An = 20
= 1 + (n – 1)d = 20
⇒ (n – 1) d = 19
sn = $$\frac{n}{2}$$ {2a + (n – 1)d}
⇒ 399 = $$\frac{n}{2}$${2 × 1 + 19}
⇒ $$\frac{399 \times 2}{21}$$ = n
⇒ n = 38
Question 13.
Find the 9th term from the end (towards the first term) of the AP 5, 9, 13, …, 185.
Solution:
l = 185, d = 4
l9 = l – (n – 1) d
= 185 – 8 × 4 = 153
### Arithmetic Progressions Class 10 Extra Questions Short Answer Type 1
Question 1.
In which of the following situations, does the list of numbers involved to make an AP? If yes, give a reason.
(i) The cost of digging a well after every meter of digging, when it costs 150 for the first meter and rises by 50 for each subsequent meter.
(ii) The amount of money in the account every year, when 10,000 is deposited at simple interest at 8% per annum.
Solution:
(i) The numbers involved are 150, 200, 250, 300, …
Here 200 -150 = 250 – 200 = 300 – 250 and so on
∴ It forms an AP with a = 150, d = 50
(ii) The numbers involved are 10,800, 11,600, 12,400, …
which forms an AP with a = 10,800 and d = 800.
Question 2.
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
We have, last term = 1 = 253
And, common difference d = 2nd term – 1st term = 8 – 3 = 5
Therefore, 20th term from end = 1 -(20 – 1) × d = 253 – 19 × 5 = 253 – 95 = 158.
Question 3.
If the sum of the first p terms of an AP is ap2 + bp, find its common difference.
Solution:
ap = sp – sp-1 = (ap2 + bp) -[a(p – 1)2 + b(p – 1)]
= ap2 + bp – (ap2 + a – 2ap + bp – b)
= ap2 + bp – ap2 – a + 2ap – bp + b = 2ap + b-a .
= a1 = 2a + b – a = a + b and a2 = 4a + b – a = 3a + b
⇒ d = a2 – a1 = (3a + b) – (a + b) = 2a
Question 4.
The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Solution:
Let the first term be ‘a’ and common difference be ‘d’.
Given, a = 5, Tn = 45, sn = 400 .
Tn = a + (m – 1)d
⇒ 45 = 5 + (m – 1)d
⇒ (n – 1) d = 40 ………(i)
sn = $$\frac{n}{2}$$ (a + Tn)
⇒ 400 = $$\frac{n}{2}$$ (5 + 45)
⇒ n = 2 × 8 = 16 substituting the value of n in (i)
⇒ (16 – 1)d = 40
⇒ d = $$\frac{40}{15}$$ = $$\frac{8}{3}$$
Question 5.
Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Solution:
Natural numbers between 101 and 999 divisible by both 2 and 5 are 110, 120, … 990.
so, a1 = 110, d = 10, an = 990
We know, an = a1 + (n – 1)d
990 = 110 + (n – 1) 10
(n – 1) = $$\frac{990-110}{10}$$
⇒ n = 88 + 1 = 89
Question 6.
Find how many integers between 200 and 500 are divisible by 8.
Solution:
AP formed is 208, 216, 224, …, 496
Here, an = 496, a = 208, d = 8
an = a + (n – 1) d
⇒ 208 + (n – 1) x 8 = 496
⇒ 8 (n – 1) = 288
⇒ n – 1 = 36
⇒ n = 37
Question 7.
The sum of the first n terms of an AP is 3n2 + 6n. Find the nth term of this
Solution:
Given: sn = 3n2 +6n
sn-1 = 3(n – 1)2 + 6(n – 1)
⇒ 3(n2 + 1 – 2n) + 6n – 6
⇒ 3m2 + 3 – 6n + 6n – 6 = 3n2 – 3
The nth term will be an
sn = sn-1 + an
an = sn – sn-1
⇒ 3n2 + 6n – 3n2 + 3
⇒ 6n + 3
Question 8.
How many terms of the AP 18, 16, 14, …. be taken so that their sum is zero?
Solution:
Here, a = 18, d = -2, sn = 0
Therefore, $$\frac{n}{2}$$ [36 + (n – 1) (- 2)] = 0
⇒ n(36 – 2n + 2) = 0
⇒ n(38 – 2n) = 0
⇒ n = 19
Question 9.
The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term.
Solution:
∵ a4 = 0 (Given)
⇒ a + 3d = 0
⇒ a = -3d
a25 = a + 24d = – 3d + 24d = 21d
3a11 = 3(a + 100) = 3(70) = 21d
∴ a25 = 3011
Hence proved.
Question 10.
If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
Solution:
Question 11.
What is the common difference of an AP in which a21 – a7 = 84?
Solution:
Given: a21 – a7 = 84
⇒ (a + 20d) – (a + 6d) = 84
⇒ 14d = 84
⇒ d = 6
Question 12.
For what value of n, are the nth terms of two APs 63, 65, 67,… and 3, 10, 17,… equal?
Solution:
Let nth terms for two given series be an and a’n
According to questions
an = a’n
⇒ a + (n – 1)d = a + (n – 1)d’
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 5n = 65
⇒ n = 13.
### Arithmetic Progressions Class 10 Extra Questions Short Answer Type 2
Question 1.
Which term of the AP: 3, 8, 13, 18, … , is 78?
Solution:
Let an be the required term and we have given AP
3, 8, 13, 18, …..
Here, a = 3, d = 8 – 3 = 5 and an = 78
Now, an = a + (n – 1)d
⇒ 78 = 3 + (n – 1) 5
⇒ 78 – 3 = (n – 1) × 5
⇒ 75 = (n – 1) × 5
⇒ $$\frac{75}{5}$$ = n – 1
⇒ 15 = n – 1
⇒ n = 15 + 1 = 16
Hence, 16th term of given AP is 78.
Question 2.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Let the first term be a and common difference be d.
Now, we have
Putting the value of d in equation (i), we have
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70
⇒ a = – 32
We have, a = -32 and d = 7
Therefore, a31 = a + (31 – 1)d
⇒ a31 = a + 30d
⇒ (-32) + 30 × 7
⇒ – 32 + 210
= a31 = 178
Question 3.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Let a be the first term and d be the common difference.
since, given AP consists of 50 terms, so n = 50
a3 = 12
⇒ a + 2d = 12 …(i)
Also, a50 = 106
⇒ a + 490 = 106 … (ii)
subtracting (i) from (ii), we have
47d = 94
⇒ d = $$\frac{94}{47}$$ = 2
Putting the value of d in equation (i), we have
a + 2 × 2 = 12
⇒ a = 12 – 4 = 8
Here, a = 8, d = 2
∴ 29th term is given by
a29 = a + (29 – 1)d = 8 + 28 × 2
⇒ a29 = 8 + 56
⇒ a29 = 64
Question 4.
If the 8th term of an AP is 31 and the 15th term is 16 more than the 11th term, find the AP.
Solution:
Let a be the first term and d be the common difference of the AP.
We have, a8 = 31 and a15 = 16 + a11
⇒ a + 7d = 31 and a + 14d = 16 + a + 10d
⇒ a + 7d = 31 and 4d = 16
⇒ a + 7d = 31 and d = 4
⇒ a + 7 × 4 = 31
⇒ a + 28 = 31
⇒ a= 3
Hence, the AP is a, a + d, a + 2d, a + 3d…..
i.e., 3, 7, 11, 15, 19, …
Question 5.
Which term of the arithmetic progression 5, 15, 25, …. will be 130 more than its 31st term?
Solution:
We have, a = 5 and d = 10
∴ a31 = a + 30d = 5 + 30 × 10 = 305
Let nth term of the given AP be 130 more than its 31st term. Then,
an = 130 + a31
∴ a + (n – 1)d = 130 + 305
⇒ 5 + 10(n – 1) = 435
⇒ 10(n – 1) = 430
⇒ n – 1 = 43
⇒ n = 44
Hence, 44th term of the given AP is 130 more than its 31st term.
Question 6.
Which term of the progression $$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}$$, is the first negative term?
Solution:
Question 7.
Find the sum given below:
7 + 10 $$\frac{1}{2}$$ + 14 +…+ 84
Solution:
Let a be the first term, d be the common difference and an be the last term of given AP.
Question 8.
In an AP: given l = 28, s = 144, and there are total 9 terms. Find a.
Solution:
We have, l = 28, s = 144 and n = 9
Now, l = an = 28
28 = a + (n – 1) d 28 = a + (9 – 1)d
⇒ 28 = a + 8d ……(i)
and S = 144
⇒ 141 = $$\frac{1}{2}$$ [2a + (n – 1)d]
⇒ 144 = $$\frac{9}{2}$$ [12a +(9 – 1) d]
$$\frac{144 \times 2}{9}$$ = 2a + 8d
⇒ 32 = 2a + 8d
⇒ 16 = a + 4d … (ii)
Now, subtracting equation (ii) from (i), we get
4d = 12 or d = 3
Putting the value of d in equation (i), we have
a + 8 × 3 = 28
⇒ a + 24 = 28
⇒ a = 28 – 24
∴ a = 4.
Question 9.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Let sum of n terms be 636.
sn = 636, a = 9, d = 17 – 9 = 8
⇒ $$\frac{n}{2}$$[2a + (n – 1) d] = 636
⇒ $$\frac{n}{2}$$[2 x 9 + (n – 1) × 8] = 636
⇒ $$\frac{n}{2}$$ × 2[9+ (n – 1) 4] = 636
⇒ n[9 + 4n – 4] = 636
⇒ n[5 + 4n] = 636
⇒ 5n + 4n2 = 636
⇒ 4n2 + 5n – 636 = 0
Thus, the sum of 12 terms of given AP is 636.
Question 10.
How many terms of the series 54, 51, 48 ……. be taken so that, their sum is 513? Explain the
Solution:
Clearly, the given sequence is an AP with first term a = 54 and common difference d = -3. Let
the sum of n terms be 513. Then,
sn = 513
⇒ $$\frac{n}{2}$$ {2a + (n – 1) d} = 513
⇒ $$\frac{n}{2}$$ (108 + (n – 1) – 3) = 513
⇒ n[108 – 3n + 3) = 1026
⇒ -3n2 + 111n = 1026
⇒ n2 – 37n + 342 = 0
⇒ (n – 18) (n – 19) = 0
⇒ n = 18 or 19
Here, the common difference is negative. so, 19th term is given by
a19 = 54 + (19 – 1) × – 3 = 0
Thus, the sum of 18 terms as well as that of 19 terms is 513.
Question 11.
The first term, common difference and last term of an AP are 12, 6 and 252 respectively. Find the sum of all terms of this AP.
Solution:
We have, a = 12, d = 6 and l = 252
Now, l = 252
⇒ an = 252
= l = a + (n – 1)d
⇒ 252 = 12 + (n – 1) × 6
⇒ 240 = (n – 1) × 6
⇒ n – 1 = 40 or n = 41
Thus, Sn = $$\frac{n}{2}$$(a + l)
⇒ S41 = $$\frac{41}{2}$$(12 + 252) = $$\frac{41}{2}$$ (264) = 41 × 132 = 5412
Question 12.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
We have, S7 = 49
⇒ 49 = $$\frac{7}{2}$$[2a + (7 – 1) × d]
⇒ 49 × $$\frac{2}{7}$$ = 2a + 6d
⇒ 14 = 2a + 6d
⇒ a + 3d = 7
and S17 = 289
⇒ 289 = $$\frac{17}{2}$$ [2a + (17 – 1)d]
⇒ 2a + 16d = $$\frac{289 \times 2}{17}$$ = 34
⇒ a + 8d = 17
Now, subtracting equation (i) from (ii), we have
5d = 10 ⇒ d = 2
Putting the value of d in equation (i), we have
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
Here, a = 1 and d = 2
Now, sn = $$\frac{n}{2}$$ [2a + (n – 1) d]
= $$\frac{n}{2}$$ [2 × 1 + (n – 1) × 2]
= $$\frac{n}{2}$$[2 + 2n – 2]
= $$\frac{n}{2}$$ × 2n = n2
Question 13.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Question 14.
If the seventh term of an AP is a and its ninth term is, find its 63rd term.
Solution:
Question 15.
The sum of the 5th and the 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.
Solution:
a5 + a9 = 30
⇒ (a + 4d) + (a + 8d) = 30 = 2a + 12d = 30
⇒ a + 6d = 15
a = 15 – 6d …(i)
a25 = 3a8
⇒ a + 24d = 3(a + 7d)
á + 24d = 3a + 21d
⇒ 2a = 3d
Putting the value of a form (i), we have
2(15 – 6d) = 3d
⇒ 30 – 12d = 3d
⇒ 15d = 30
⇒ d = 2
So, a = 15 – 6 × 2 = 15 – 12 [From equation (i)]
⇒ a = 3
The AP will be 3, 5, 7, 9….
Question 16.
The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term of this AP.
Solution:
sum of first seven terms,
Question 17.
If the ratio of the sum of first n terms of two AP’s is (7n + 1): (4n + 27), find the ratio of their mth terms.
Solution:
Question 18.
Find the sum of the following series :
5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)
Solution:
The series can be rewritten as,
(5 + 9 + 13 + … +81) + (41 + (-39) + (-37) + … + (-5) + (-3))
For the series 5 + 9 + 13 + … 81
a = 5, d = 4 and an = 81
nth term = a + (n – 1)d = an
⇒ 5+ (n – 1)4 = 81
⇒ 4n = 80
⇒ n =
20 sum of 20 terms for this series.
sn = $$\frac{20}{2}$$ (5 + 81) = 860 …(i)
[∵ sn = $$\frac{n}{2}$$ (a + an)]
For the series (41) + (-39) + (-37)… + (-5) + (-3)
a = 41, d = 2 and and = -3
nth term = a + (n – 1)d = an.
⇒ -41 + (n – 1)2 = -3
⇒ 2n = 40
⇒ n = 20
sum of 20 terms for this series
sn = $$\frac{20}{2}$$ (-41 – 3) = – 440 …(ii)
By adding (i) and (ii), we get
sum of series = 860 – 440
= 420
Question 19.
Find the sum of all two digit natural numbers which are divisible by 4.
Solution:
Here a = 12, d = 4, an = 96
The formula is an = a + (n – 1)d
Therefore 96 = 12 + (n – 1) × 4
⇒ 96 = 8 + 4n
⇒ n = $$\frac{88}{4}$$
⇒ n = 22
Apply the formula for sum,
sn = $$\frac{n}{2}$$ [2a + (n – 1)d]
Hence, s22 = 11/24 + 21 × 4] = 11[24 + 84]
= 11 × 108 = 1188.
### Arithmetic Progressions Class 10 Extra Questions Long Answer Type
Question 1.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Solution:
We have, a4 + a8 = 24
⇒ a + (4 – 1)d + a + (8 – 1) d = 24
⇒ 2a + 3d + 7d = 24
⇒ 2a + 10d = 24
⇒ 2(a + 5d) = 24
∴ a + 5d = 12
and, a6 + a10 = 44
⇒ a + (6 – 1)d + a + (10 – 1) d = 44
⇒ 2a + 5d + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22
subtracting (i) from (ii), we have
2d = 10
∴d = $$\frac{10}{2}$$ = 5
Putting the value of d in equation (i), we have
a + 5 × 5 = 12
⇒ a = 12 – 25 = -13
Here, a = -13, d = 5
Hence, first three terms are
-13, -13, + 5, -13 + 2 × 5 i.e., -13, -8, -3
Question 2.
The sum of the first n terms of an AP is given by sn = 3n2 – 4n. Determine the AP and the 12th term.
Solution:
We have, sn = 3n2 – 4n …(i)
Replacing n by (n – 1), we get
sn-1 = 3(n – 1)2 – 4(n – 1) ….(ii)
We know, .
an = sn – sn-1 = {3n2 – 4n} – {3(n – 1)2 – 4(n – 1)}.
= {3n2 – 4n} – {3n2 + 3 – 6n – 4n + 4}
= 3n2 – 4n – 3n2 – 3 + 6n + 4n – 4 = 6n – 7
so, nth term an = 6n – 7
To get the AP, substituting n = 1, 2, 3… respectively in (iii), we get
a1 = 6 × 1-7 = -1,
a2 = 6 × 2 – 7 = 5
a3 = 6 × 3 – 7 = 11,…
Hence, AP is – 1,5, 1:1, …
Also, to get 12th term, substituting n = 12 in (iii), we get
a12 = 6 × 12 – 7 = 72 – 7 = 65
Question 3.
Divide 56 into four parts which are in AP such that the ratio of product of extremes to the product of means is 5 : 6.
Solution:
Let the four parts be a – 3d, a-d, a + d, a + 3d.
Given, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56
Question 4.
In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.
Solution:
Let ‘a’ be the first term and ‘d be the common difference.
Question 5.
If s, denotes the sum of the first n terms of an AP, prove that s30 = 3 (s20 – s10).
Solution:
Question 6.
A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief?
Solution:
Let total time be n minutes
Total distance covered by thief = 100 n metres
Total distance covered by policeman = 100 + 110 + 120 + … + (n – 1) terms
∴ 100m = $$\frac{n-1}{2}$$ [100(2) + (n – 2)10]
⇒ 200n = (n – 1)(180 + 10n)
⇒ 102– 30n – 180 = 0
⇒ n2 – 3n – 18 = 0
⇒ (n-6) (n + 3) = 0
⇒ n = 6
Policeman took (n – 1) = (6 – 1) = 5 minutes to catch the thief.
Question 7.
The houses in a row are numbered consecutively from 1 to 49. show that there exists a value of X such that sum of numbers of houses preceeding the house numbered X is equal to sum of the numbers of houses following X. Find value of X.
Solution:
The numbers of houses are 1, 2, 3, 4……….49.
The numbers of the houses are in AP, where a = 1 and d = 1
sum of n terms of an AP = $$\frac{n}{2}$$[2a + (n – 1)d]
Let Xth number house be the required house.
sum of number of houses preceding Xth house is equal to sx-1 i.e.,
since number of houses is positive integer,
∴ X = 35
Question 8.
If the ratio of the 11th term of an AP to its 18th term is 2:3, find the ratio of the sum of the first five terms to the sum of its first 10 terms.
Solution:
### Arithmetic Progressions Class 10 Extra Questions HOTS
Question 1.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are
8, 16, 24, … 120
Clearly, these numbers are in AP with first term a = 8 and common difference, d = 16 – 8 = 8
Question 2.
Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder.
Solution:
Two digit natural numbers which when divided by 3 yield 1 as remainder are:
10, 13, 16, 19, …, 97, which forms an AP.
with a = 10, d = 3, an = 97
an = 97 = a + (n – 1) d = 97
or 10 + (n – 1)3 = 97
⇒ (n – 1) = $$\frac{87}{3}$$ = 29
⇒ n = 30
Now, s30 = [2 × 10 + 29 × 3) = 15(20 + 87) = 15 × 107 = 1605
Question 3.
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
Therefore, the sum of prizes is
a + 60 + a + 40 + a + 20 + a ta – 20 + a – 40 +a – 60 = 700
= 7a = 700
⇒ a = $$\frac{700}{7}$$ = 100
Thus, the value of seven prizes are
100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60
i.e., ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹340
Question 4.
If the mth term of an AP is $$\frac{1}{n}$$ and nth term is $$\frac{1}{m}$$, then show that its (mn)th term is 1.
Solution:
Let a and d be the first term and common difference respectively of the given AP. Then
am = a + (m – 1) d
⇒ a + (m – 1)d = $$\frac{1}{n}$$ …… (i)
Question 5.
If the sum of m terms of an AP is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero.
Solution:
Let a be the first term and d be the common difference of the given AP.
Then, sm = sn
Question 6.
The ratio of the sums of m and n terms of an AP is m2 : n2. show that the ratio of the mth and nth terms is (2m – 1): (2n – 1).
Solution:
Let a be the first term and d the common difference of the given AP. Then, the sums of m and n terms are given by
Question 7.
The sum of n, 2n, 3n terms of an AP are s1, s2, and s3 respectively. Prove that s3 = 3(s2 – s1).
Solution:
Let a be the first term and d be the common difference of the AP
Question 8.
If a2, b2, c2, are in AP, prove that $$\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$$ are in AP.
Solution:
since a2, b2, c2, are in AP …(i) |
High School Physics
# Series Circuits, Resistors in series & equivalent resistance formula & numerical
Last updated on April 20th, 2023 at 04:07 pm
In this post, we will study simple series circuits with the help of a series circuit diagram. We will also go through related topics like resistance in series or resistors in series, series circuit formula of equivalent resistance, etc. This post also covers series circuit numerical problems with solutions.
## Series circuit | resistance in series | resistors in series
What is a series circuit? A series circuit consists of several resistors connected in sequence, one after the other. This increases the length of the conductor. Therefore total resistance increases as the number of resistors in series are increased in the circuit.
A series circuit is also referred to as resistance in series or resistors in series.
### Series Circuit Diagram (a simple one)
In the above, series circuit diagram three resistors R1, R2, and R3 are connected in sequence, one after the other. In other words, these are connected in series.
### Characteristics of Series Circuit as found from experiments
Experiments show that the current in the series circuit made up of one loop is the same everywhere.
As there are potential drops across each resistor, the total voltage across the battery is equal to the sum of the potential drops across each resistor.
If one resistor goes out, then the whole circuit goes off. (like when a light bulb in a series circuit goes out, the entire circuit becomes inoperative)
### Equivalent Resistance of a series circuit | series circuit formula of equivalent resistance
Using Ohm’s Law and our experimental results, we find that: (see the circuit diagram above)
The current flowing through the circuit = I = current through each of the resistors in series
I = I1 = I2 = I3
VT = Total voltage or Potential difference
Req = VT/I = equivalent resistance
VT = V1 + V2 + V3 = sum of potential drops across each resistor
Req = VT/I = (V1 + V2 + V3) /I = V1/I + V2/I + V3/I = V1/I1 + V2/I2 + V3/I3 = R1 + R2 + R3
Series circuit formula of equivalent resistance Req = R1 + R2 + R3
### Series circuit numerical problems with solutions | Numerical Problems (based on series circuit and its equivalent resistance)
1] Resistors of 10 Ω, 15 Ω, and 5 Ω are all connected in series to a 90 V source of potential difference. Calculate the total equivalent resistance of the circuit, the total current in the circuit, and the potential drop across each resistor.
Solution:
Equivalent Resistance = Req = (10 + 15 + 5) Ω = 30 Ω
Total current in the circuit I = V/Req = 90 V/ 30 Ω = 3 Ampere
The potential drop across each resistor:
Potential drop across 10 Ω resistor = I x resistance = 3 x 10 V = 30 V
Potential drop across 15 Ω resistor = I x resistance = 3 x 15 V = 45 V
Potential drop across 5 Ω resistor = I x resistance = 3 x 5 V = 15 V
2] Three resistors of 50 Ω, 30 Ω, and 40 Ω are connected in series to a 60 V battery. Calculate the equivalent resistance of the circuit, the total current in the circuit, the total power generated by the circuit, and the voltage drop across the resistors.
Solution:
Equivalent Resistance = (50 + 30 + 40) Ω = 120 Ω
Total current = I = V/R = 60/120 Ampere = 0.5 A [ we used Ohm’s law here]
Total power = P = VI = 60 x 0.5 Watt = 30 W
Voltage drop across 50 Ω = IR =0.5 x 50 V= 25 V
Voltage drop across 30 Ω = IR = 0.5 x 30 V = 15 V
Voltage drop across 40 Ω = IR = 0.5 x 40 V = 20 V
3] Fifteen resistors with equal resistance are connected in a series circuit. if the voltage and current are equal to 120 v and 1.5 amp, respectively, determine the individual resistance of each resistor.
Solution:
Say, the individual resistance of each resistor is R.
Hence, the equivalent resistance of 15 such resistors in a series circuit is Req=15R
V = 120 V
I = 1.5 A
Using Ohm’s law:
V = IReq
=> Req = V/I = 120/1.5 = 80 ohm
15R = 80
=>R = 80/15 = 5.33 ohm
#### Formula Used to solve the series circuit numerical problems
Related Posts
Ohm’s Law
Ohm’s Law Numericals
See also How to find resistance using Resistor Colour Codes
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error: physicsTeacher.in |
## Numerical Measures of Central Tendency and Variability
Read these sections and complete the questions at the end of each section. First, we will define central tendency and introduce mean, median, and mode. We will then elaborate on median and mean and discusses their strengths and weaknesses in measuring central tendency. Finally, we'll address variability, range, interquartile range, variance, and the standard deviation.
### Measures of Variability
#### Variance
Variability can also be defined in terms of how close the scores in the distribution are to the middle of the distribution. Using the mean as the measure of the middle of the distribution, the variance is defined as the average squared difference of the scores from the mean. The data from Quiz 1 are shown in Table 1. The mean score is $\mathrm{7.0}$. Therefore, the column "Deviation from Mean" contains the score minus $\mathrm{7}$. The column "Squared Deviation" is simply the previous column squared.
Table 1. Calculation of Variance for Quiz 1 scores.
Scores Deviation from Mean Squared Deviation
9 2 4
9 2 4
9 2 4
8 1 1
8 1 1
8 1 1
8 1 1
7 0 0
7 0 0
7 0 0
7 0 0
7 0 0
6 -1 1
6 -1 1
6 -1 1
6 -1 1
6 -1 1
6 -1 1
5 -2 4
5 -2 4
Means
7 0 1.5
One thing that is important to notice is that the mean deviation from the mean is 0. This will always be the case. The mean of the squared deviations is $\mathrm{1.5}$. Therefore, the variance is $\mathrm{1.5}$. Analogous calculations with Quiz 2 show that its variance is $\mathrm{6.7}$. The formula for the variance is:
$\sigma^{2}=\frac{\sum(X-\mu)^{2}}{N}$
where $\sigma^{2}$ is the variance, $\mu$ is the mean, and $N$ is the number of numbers. For Quiz $1, \mu=7$ and $N=20$
If the variance in a sample is used to estimate the variance in a population, then the previous formula underestimates the variance and the following formula? should be used:
$s^{2}=\frac{\sum(X-M)^{2}}{N-1}$
where $s^{2}$ is the estimate of the variance and $M$ is the sample mean. Note that $M$ is the mean of a sample taken from a population with a mean of $\mu$. Since, in practice, the variance is usually computed in a sample, this formula is most often used. The simulation "estimating variance" illustrates the bias in the formula with $N$ in the denominator.
Let's take a concrete example. Assume the scores $1,2,4$, and 5 were sampled from a larger population. To estimate the variance in the population you would compute $s^{2}$ as follows:
\begin{aligned} M &=(1+2+4+5) / 4=12 / 4=3 \\ S^{2} &=\left[(1-3)^{2}+(2-3)^{2}+(4-3)^{2}+(5-3)^{2}\right] /(4-1) \\ &=(4+1+1+4) / 3=10 / 3=3.333 \end{aligned}
There are alternate formulas that can be easier to use if you are doing your calculations with a hand calculator. You should note that these formulas are subject to rounding error if your values are very large and/or you have an extremely large number of observations.
$\sigma^{2}=\frac{\sum X^{2}-\frac{\left(\sum X\right)^{2}}{N}}{N}$
and
$s^{2}=\frac{\sum X^{2}-\frac{\left(\sum X\right)^{2}}{N}}{N-1}$
For this example,
\begin{aligned} &\sum X^{2}=1^{2}+2^{2}+4^{2}+5^{2}=46 \\ &\frac{\left(\sum X\right)^{2}}{N}=\frac{(1+2+4+5)^{2}}{4}=\frac{144}{4}=36 \\ &\sigma^{2}=\frac{(46-36)}{4}=2.5 \\ &s^{2}=\frac{(46-36)}{3}=3.333 \text { as with the other formula } \end{aligned} |
# Texas Go Math Grade 3 Lesson 7.3 Answer Key Multiply with 3 and 6
Refer to our Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Lesson 7.3 Answer Key Multiply with 3 and 6.
## Texas Go Math Grade 3 Lesson 7.3 Answer Key Multiply with 3 and 6
Essential Question
what are some ways to multiply with 3 and 6?
Unlock the Problem
Sabrina is making triangles with toothpicks. She uses 3 toothpicks for each triangle. She makes 4 triangles.
How many toothpicks does Sabrina use?
She uses 3 toothpicks for each triangle.
Explanation:
Number of toothpicks she uses = 3.
• Why does Sabrina need 3 toothpicks for each triangle?
Sabrina need 3 toothpicks for each triangle because to make 4 triangles.
Explanation:
To make 4 triangles Sabrina needs 3 toothpicks for each triangle.
Draw a picture.
Step 1
Complete the 4 triangles.
Step 2
Skip count by the number of sides. ___, ___, ___, ___,
How many craft sticks will she use in all? ____
How many toothpicks are there in all?
4 × __ = ___
4 triangles have ___ toothpicks.
So, Sabrina uses ___ toothpicks.
Math Talk
Mathematical Processes
How can you use what you know about the number of toothpicks needed for 4 triangles to find the number of toothpicks needed for 8 triangles? Explain.
Jessica is using craft sticks to make 6 octagons. How many craft sticks will she use in all?
Use doubles.
When at least one factor is an even number, you can use doubles.
First multiply with half of an even number.
After you multiply, double the product.
Another Way
Use a multiplication table.
Find the product 6 × 8 where row 6 and column 8 meet.
6 × 8 = ___
• Shade the row for 3 in the table. Then, compare the rows for 3 and 6. What do you notice about their products?
6 × 8 = 48.
It is noticed that 3 products are half of 6. 6 products are double of 3 products.
Explanation:
Question 1.
Use 5s facts and addition to find 6 × 4 =___
6 × 4 = 5 × 4 + 4 = 24
6 × 4 = ___
6 × 4 = 24.
Explanation:
6 × 4 = 5 × 4 + 4
=> 6 × 4 = 20 + 4
=> 6 × 4 = 24.
Math talk
Mathematical Processes
Explain how you would use 5s facts and addition to find 6 × 3.
6 × 3 = 5 × 3 + 3 = 18.
= 15 + 3
= 18.
Explanation:
6 × 3 = 5 × 3 + 3
= 15 + 3
= 18.
Find the product.
Question 2.
6 × 1 = ___
6 × 1 = 6.
Explanation:
6 × 1 = 5 × 1 + 1
= 5 + 1
= 6.
Go Math Lesson 7.3 Answer Key Grade 3 Question 3.
___ = 3 × 7
21 = 3 × 7.
Explanation:
3 × 7 = 2 × 7 + 7
= 14 + 7
= 21.
Question 4.
___ = 6 × 5
30 = 6 × 5.
Explanation:
6 × 5 = 5 × 5 + 5
= 25 + 5
= 30.
Question 5.
3 × 9 = ___
3 × 9 = 27.
Explanation:
3 × 9 = 2 × 9 + 9
= 18 + 9
= 27.
H.O.T. Algebra Complete the table.
Explanation:
6. 4 × 3 = 3 × 3 + 3
= 9 + 3 = 12.
7. ?? × 3 = 18.
=> ?? = 18 ÷ 3
=> ?? = 6.
8. ?? × 6 = 30.
=> ?? = 30 ÷ 6
=> ?? = 5.
9. 7 × 6 = 6 × 6 + 6
=> 36 + 6 = 42.
Problem Solving
Use the table for 12-13.
Question 12.
Use Tools The table tells about quilt pieces Jenna has made. How many squares are there in 6 of Jenna’s quilt pieces?
Number of squares in 6 of Jenna’s quilt pieces = 36.
Explanation:
Number of squares in one quilt pieces = 6.
Number of quilt pieces Jenna’s having = 6.
Number of squares in 6 of Jenna’s quilt pieces = Number of squares in one quilt pieces × Number of quilt pieces Jenna’s having
= 6 × 6
= 5 × 6 + 6
= 30 + 6
= 36.
Go Math Grade 3 Lesson 7.3 Answer Key Question 13.
Compare How many more squares than triangles are in 3 of Jenna’s quilt pieces?
3 more squares than triangles are in 3 of Jenna’s quilt pieces.
Explanation:
Number of squares in one quilt pieces = 6.
Number of quilt pieces Jenna’s having = 3.
Number of squares in 3 of Jenna’s quilt pieces = Number of squares in one quilt pieces × Number of quilt pieces Jenna’s having
= 6 × 3
= 5 × 3 + 3
= 15 + 3
= 18.
Number of triangles in one quilt pieces = 4.
Number of quilt pieces Jenna’s having = 3.
Number of triangles in 3 of Jenna’s quilt pieces = Number of triangles in one quilt pieces × Number of quilt pieces Jenna’s having
= 4 × 3
= 3 × 3 + 3
= 9 + 3
= 12.
Difference:
Number of squares in 3 of Jenna’s quilt pieces – Number of triangles in 3 of Jenna’s quilt pieces
= 18 – 15
= 3.
Question 14.
H.O.T. Multi-Step Alii had some craft sticks. She decided to make shapes. If she used one craft stick for each side of the shape, would Alii use more craft sticks for 5 squares or 6 triangles? Explain.
Alii use 2 more craft sticks for 5 squares than 6 triangles.
Explanation:
Number of craft stick for each side of the shape she used = 1.
Number of sides for a square = 4.
Number of craft sticks for making 5 squares = 5 × (Number of craft stick for each side of the shape she used × Number of sides for a square)
= 5 × (1 × 4)
= 5 × 4
= 4 × 4 + 4
= 16 + 4
= 20.
Number of sides for a triangle = 3.
Number of craft sticks for making 6 triangle = 6 × (Number of craft stick for each side of the shape she used × Number of sides for a triangle)
= 6 × (1 × 3)
= 6 × 3
= 18.
Difference:
Number of craft sticks for making 5 squares – Number of craft sticks for making 6 triangle
= 20 – 18
= 2.
Go Math 3rd Grade Lesson 7.3 Answer Key Question 15.
H.O.T. Multi-Step Draw a picture and use words to explain the Commutative Property of Multiplication with factors 3 and 4.
Explanation:
The commutative property states that the change in the order of numbers in an addition or multiplication operation does not change the sum or the product. The commutative property of addition is written as A + B = B + A. The commutative property of multiplication is written as A × B = B × A.
Suppose: 3 × 4 = 12.
4 × 3 = 12.
=> 3 × 4 = 4 × 3.
Fill in the bubble for the correct answer choice.
Use multiplication strategies to solve.
Question 16.
Each day a hornworm eats 8 leaves. How many leaves will the hornworm eat in 3 days?
(A) 21
(B) 11
(C) 24
(D) 16
Number of leaves will the hornworm eat in 3 days = 24.
(C) 24.
Explanation:
Number of leaves each day a hornworm eats = 8.
Number of leaves will the hornworm eat in 3 days = 3 × Number of leaves each day a hornworm eats
= 3 × 8
= 2 × 8 + 8
= 16 + 8
= 24.
Question 17.
Representation Which shows a way to use a 5s fact to find 6 × 9?
(A) 5 × 9 + 9
(B) 5 × 9 + 5
(C) 5 × 5 + 5
(D) 5 × 9 + 6
5s fact to find 6 × 9: 6 × 9 = 5 × 9 + 9 = 54.
(A) 5 × 9 + 9
Explanation:
5s fact to find 6 × 9:
6 × 9 = 5 × 9 + 9
= 45 + 9
= 54.
Lesson 7.3 Answer Key Go Math 3rd Grade Question 18.
Multi-Step Bobbie has 4 packs of trading cards. Each pack holds 6 cards. Bobbie gives 6 of the cards to his sister. How many cards does Bobbie have left?
(A) 2
(B) 24
(C) 18
(D) 36
Number of cards Bobbie have left = 18.
(C) 18.
Explanation:
Number of packs of trading cards Bobbie has = 4.
Number of cards each pack holds = 6.
Number of cards Bobbie gives to his sister = 6.
Number of cards Bobbie have left = (Number of packs of trading cards Bobbie has × Number of cards each pack holds) – Number of cards Bobbie gives to his sister
= (4 × 6) – 6
= 24 – 6
= 18.
Texas Test Prep
Question 19.
There are 8 crayons in each of 6 packages. How many crayons are there?
(A) 24
(B) 14
(C) 2
(D) 48
Number of crayons are there = 48.
(D) 48.
Explanation:
Number of crayons in each pack = 8.
Number of packages of crayons = 6.
Number of crayons are = Number of crayons in each pack × Number of packages of crayons
= 8 × 6
= 7 × 6 + 6
= 42 + 6
= 48.
### Texas Go Math Grade 3 Lesson 7.3 Homework and Practice Answer Key
Algebra Complete the table.
Explanation:
Multiple by 3:
1. ?? × 3 = 15.
=> ?? = 15 ÷ 3
=> ?? = 5.
2. 7 × 3 = 21.
Multiple by 6:
3. 6 × 6 = 36.
4. ?? × 6 = 48.
=> ?? = 48 ÷ 6
=> ?? = 8.
Use the table for 5-8.
Craft Stick Animals
Question 5.
Joella makes 6 craft stick birds. How many craft sticks does Joella use?
Number of craft sticks Joella uses = 36.
Explanation:
Number of craft stick birds Joella makes = 6.
Number of birds craft sticks = 6.
Number of craft sticks Joella uses = Number of craft stick birds Joella makes × Number of birds craft sticks
= 6 × 6
= 5 × 6 + 6
= 30 + 6
= 36.
Go Math Practice and Homework Lesson 7.3 Answer Key Question 6.
Lebraun makes 3 dogs and 2 birds. Does he use more craft sticks for the birds or for the dogs?
He uses 3 more craft sticks for the birds than for the dogs.
Explanation:
Number of dogs Lebraun makes = 3.
Number of dog craft sticks = 5.
Number of craft stick dogs Lebraun makes = Number of dogs Lebraun makes × Number of dog craft sticks
= 3 × 5
= 2 × 5 + 5
= 10 + 5
= 15.
Number of birds Lebraun makes = 2.
Number of birds craft sticks = 6.
Number of craft stick birds Lebraun makes = Number of birds Lebraun makes × Number of birds craft sticks
= 2 × 6
= 1 × 6 + 6
= 6 + 6
= 12.
Difference:
Number of craft stick dogs Lebraun makes – Number of craft stick birds Lebraun makes
= 15 – 12
= 3.
Question 7.
Ariel makes 2 cats and 2 dogs. How many more craft sticks does she need to make the cats?
6 more craft sticks she needs to make the cats.
Explanation:
Number of dogs Ariel makes = 2.
Number of dog craft sticks = 5.
Number of craft stick dogs Ariel makes = Number of dogs Ariel makes × Number of dog craft sticks
= 2 × 5
= 1 × 5 + 5
= 5 + 5
= 10.
Number of cats Ariel makes = 2.
Number of cats craft sticks = 8.
Number of craft stick cats Ariel makes = Number of cats Ariel makes × Number of cats craft sticks
= 2 × 8
= 1 × 8 + 8
= 8 + 8
= 16.
Difference:
Number of craft stick cats Ariel makes – Number of craft stick dogs Ariel makes
= 16 – 10
= 6.
Problem Solving
Question 8.
Sam makes 3 dogs and 2 cats. Does he need more craft sticks for the dogs or for the cats? Explain.
He need 1 more craft sticks for the cats than for the dogs.
Explanation:
Number of dogs Sam makes = 3.
Number of dog craft sticks = 5.
Number of craft stick dogs Sam makes = Number of dogs Sam makes × Number of dog craft sticks
= 3 × 5
= 2 × 5 + 5
= 10 + 5
= 15.
Number of cats Sam makes = 2.
Number of cats craft sticks = 8.
Number of craft stick cats Sam makes = Number of cats Sam makes × Number of cats craft sticks
= 2 × 8
= 1 × 8 + 8
= 8 + 8
= 16.
Difference:
Number of craft stick cats Sam makes – Number of craft stick dogs Sam makes
= 16 – 15
= 1.
Lesson Check
Go Math Answer Key Grade 3 Lesson 7.3 Homework Answers Question 9.
Each day Mr. Mosell drives a total of 8 miles to work and back. How many miles does Mr. Mosell drive in 5 days?
(A) 45 miles
(B) 40 miles
(C) 13 miles
(D) 58 miles
Number of miles Mr. Mosell drives in 5 days = 40.
(B) 40 miles.
Explanation:
Number of miles each day Mr. Mosell drives to work and back = 8.
Number of miles Mr. Mosell drives in 5 days = 5 × Number of miles each day Mr. Mosell drives to work and back
= 5 × 8
= 4 × 8 + 8
= 32 + 8
= 40.
Question 10.
Tanya walks 3 miles every day for one week. How many miles does Tanya walk in all?
(A) 7 miles
(B) 15 miles
(C) 3 miles
(D) 21 miles
Number of miles Tanya walk in all = 21.
(D) 21 miles.
Explanation:
Number of miles every day for one week Tanya walks = 3.
Number of days in a week = 7.
Number of miles Tanya walk in all = Number of miles every day for one week Tanya walks × Number of days in a week
= 3 × 7
= 2 × 7 + 7
= 14 + 7
= 21.
Question 11.
Terrel wants to use a 5s fact to multiply 6 × 8. Which expression can Terrel use?
(A) 5 × 5 + 8
(B) 5 × 8 + 5
(C) 5 × 8 + 8
(D) 6 × 8 + 5
Expression can Terrel use is 6 × 8 = 5 × 8 + 8 = 48.
(C) 5 × 8 + 8.
Explanation:
Terrel wants to use a 5s fact to multiply 6 × 8:
=> 6 × 8 = 5 × 8 + 8
=> 6 × 8 = 40 + 8
=> 6 × 8 = 48.
Question 12.
Anita wants to use a 5s fact to multiply 6 × 3. Which expression can Anita use?
(A) 5 × 3 + 3
(B) 5 × 3 + 5
(C) 5 × 8 + 5
(D) 5 × 3 + 6
Expression can Anita use is 6 × 3 = 5 × 3 + 3 = 18.
(A) 5 × 3 + 3.
Explanation:
Anita wants to use a 5s fact to multiply 6 × 3:
=> 6 × 3 = 5 × 3 + 3
=> 6 × 3 = 15 + 3
=> 6 × 3 = 18.
Go Math Grade 3 Lesson 7.3 Homework Answer Key Question 13.
Multi-Step Milo collects 6 boxes of shells. Each box holds 7 shells. Milo gives 2 boxes of shells to his friend. How many shells does Milo have left?
(A) 30
(B) 42
(C) 14
(D) 28
Number of shells Milo has left = 28.
(D) 28.
Explanation:
Number of boxes of shells Milo collects = 6.
Number of shells each box holds = 7.
Number of boxes of shells Milo gives to his friend = 2.
Number of shells Milo have left = (Number of boxes of shells Milo collects × Number of shells each box holds) – (Number of boxes of shells Milo gives to his friend × Number of shells each box holds)
= (6 × 7) – (2 × 7)
= 42 – 14
= 28.
3rd Grade Go Math Lesson 7.3 Answer Key Question 14.
Multi-Step Henley collects 3 eggs from each of the nests of 9 chickens. Henley uses 6 eggs to make breakfast. How many eggs does Henley have left?
(A) 21
(B) 27
(C) 18
(D) 54 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Scientific Notation Values
## Decimals written as a power of ten
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Scientific Notation Values
David is working on his science homework. The assignment is to find the mass of each planet in the solar system and write it using scientific notation. While researching the assignment, he finds that the mass of the Earth is 5,973,600,000,000,000,000,000,000 kg. How can David express the mass of the Earth in scientific notation?
In this concept, you will learn how to write numbers using scientific notation.
### Scientific Notation
Scientific notation is a shortcut for writing very small and very large numbers. It is very useful for scientists, mathematicians, and engineers. It is useful in careers where people work with very large or very small numbers. For example, the distance from the Earth to the Sun is 96,960,000 miles. Instead of writing out the number every time, you can use scientific notation. Numbers in scientific notation follow the form
a×10n,\begin{align*}a \times 10^n,\end{align*}
where a\begin{align*}a\end{align*}, a number between 1 and 10, is multiplied by a power of ten.
Remember that a power of ten is written as 10n\begin{align*}10^n\end{align*}, where n\begin{align*}n\end{align*} is the exponent that tells you how many times 10 is multiplied by itself.
Here is 96,960,000 written in scientific notation.
96,960,000=9.696×107\begin{align*}96,960,000=9.696 \times 10^7\end{align*}
To change 96,960,000 into a number between 1 and 10, you must move the decimal point 7 spaces to the left. So to get 96,960,000 from 9.696, you must do the opposite and move the decimal point 7 spaces to the right. Moving the decimal point to the right requires multiplying by 10 a total of 7 times or 107\begin{align*}10^7\end{align*}. Note that large numbers written in scientific notation will use positive exponents.
Here is a very small number written in scientific notation.
0.000000023=2.3×108\begin{align*}0.000000023 = 2.3 \times 10^{-8}\end{align*}
To change 0.000000023 into a number between 1 and 10, you must move the decimal point 8 spaces to the right. So to convert 2.3 to 0.000000023, move the decimal 8 spaces to the left. Moving the decimal to the left requires multiplying by a negative power of 10 a total of 8 times or 108\begin{align*}10^{-8}\end{align*}. Remember that multiplying by a negative power of ten is the same as dividing by a power of ten.
108=1108\begin{align*}10^{-8} = \frac{1}{10^8}\end{align*}
Note that decimal numbers less than 1 will use negative powers of ten when written in scientific notation.
Here is a small decimal number. Write the number using scientific notation.
0.00056\begin{align*}0.00056\end{align*}
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 5.6.
5.6×10n\begin{align*}5.6 \times 10^{n}\end{align*}
Then, identify the power of ten, n\begin{align*}n\end{align*}. Since 0.00056 is a number less than 1, the power of ten will be negative. Notice that to go from 0.00056 to 5.6, you must move the decimal point four places to the right. This means the exponent will be −4.
5.6×104\begin{align*}5.6 \times 10^{-4}\end{align*}
To check if this is correct, multiply 5.6 times 104\begin{align*}10^{-4}\end{align*}. When multiplying by a negative power of ten, move the decimal point to the left 4 times.
0.00056\begin{align*}0.\underleftarrow{0005}6\end{align*}
Here are some charts that might help you remember how to convert numbers to scientific notation and scientific notation to numbers.
Converting a Number to Scientific Notation
Large Numbers → Positive Power of Ten
Small Decimal Numbers → Negative Power of Ten
Converting Scientific Notation to a Number
Positive Power of Ten → Move the Decimal to the Right
Negative Power of Ten → Move the Decimal to the Left
### Examples
#### Example 1
Earlier, you were given a problem about David’s science homework.
David needs to express the mass on the Earth, 5,973,600,000,000,000,000,000,000 kg, using scientific notation.
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 5.9736.
5.9736×10n\begin{align*}5.9736 \times 10^n\end{align*}
Then, identify the power of ten, n\begin{align*}n\end{align*}. Since the mass of the Earth is a large number, the power of ten will be positive. The decimal moves 24 places to get from 5,973,600,000,000,000,000,000,000 to 5.9736, so n\begin{align*}n\end{align*} is 24.
5.9736×1024\begin{align*}5.9736 \times 10^{24}\end{align*}
The scientific notation form of 5,973,600,000,000,000,000,000,000 kg is 5.9736×1024 kg\begin{align*}5.9736 \times 10^{24} \ kg\end{align*}.
#### Example 2
Write the number in scientific notation.
0.0000000034
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 3.4.
\begin{align*}3.4 \times 10^n\end{align*}
Then, identify the power of ten, \begin{align*}n\end{align*}. Since 0.0000000034 is a small number, the power of ten will be negative. The decimal moves 9 places to get from 0.0000000034 to 3.4, so \begin{align*}n\end{align*} is -9.
\begin{align*}3.4 \times 10^{-9}\end{align*}
The scientific notation form of 0.0000000034 is \begin{align*}3.4 \times 10^{-9}\end{align*}.
Write the numbers in scientific notation.
#### Example 3
\begin{align*}0.0012\end{align*}
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 1.2.
\begin{align*}1.2 \times 10^n\end{align*}
Then, identify the power of ten, \begin{align*}n\end{align*}. Since 0.0012 is a small number, the power of ten will be negative. The decimal moves 3 places to get from 0.0012 to 1.2, so \begin{align*}n\end{align*} is -3.
\begin{align*}1.2 \times 10^{-3}\end{align*}
The scientific notation form of 0.0012 is \begin{align*}1.2 \times 10^{-3}\end{align*}.
#### Example 4
\begin{align*}78,000,000\end{align*}
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 7.8.
\begin{align*}7.8 \times 10^n\end{align*}
Then, identify the power of ten, \begin{align*}n\end{align*}. Since 78,000,000 is a large number, the power of ten will be positive. The decimal moves 7 places to get from 78,000,000 to 7.8, so \begin{align*}n\end{align*} is 7.
\begin{align*}7.8 \times 10^7\end{align*}
The scientific notation form of 78,000,000 is \begin{align*}7.8 \times 10^7\end{align*}.
#### Example 5
\begin{align*}345,102,000,000\end{align*}
First, use the scientific notation form. Change the number to be a number between 1 and 10. This number is 3.45102.
\begin{align*}3.45102 \times 10^n\end{align*}
Then, identify the power of ten, \begin{align*}n\end{align*}. Since 345,102,000,000 is a large number, the power of ten will be positive. The decimal moves 11 places to get from 345,102,000,000 to 3.45102, so \begin{align*}n\end{align*} is 11.
\begin{align*}3.45102 \times 10^{11}\end{align*}
The scientific notation form of 345,102,000,000 is \begin{align*}3.45102 \times 10^{11}\end{align*}.
### Review
Write each decimal in scientific notation.
1. 0.00045
2. 0.098
3. 30,000,000
4. 0.000987
5. 3,400,000
6. 0.0000021
7. 1,230,000,000,000
8. 0.00000000345
9. 0.00056
10. 0.0098
11. 0.024
12. 0.000023
13. 4,300
14. 0.0000000000128
15. 980
16. 0.00000045
To see the Review answers, open this PDF file and look for section 4.15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# MCQ Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers
Here you will find NCERT MCQ Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers PDF Free Download based on the important concepts and topics given in the textbook as per CBSE new exam pattern. This may assist you to understand and check your knowledge about the Chapter 3 Pair of Linear Equations in Two Variables. Students also can take a free test of the Multiple Choice Questions of Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables. Each question has four options followed by the right answer. These MCQ Questions are selected supported by the newest exam pattern as announced by CBSE.
## Q1. 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys who took part in the quiz.
(a) 3
(b) 7
(c) 2
(d) 8
(a) 3 No. of boys = x and No. of girls = y x + y =10 y = x + 4 x = 3 y = 7 => Number of boys = 3
## Q2. Area of figure OABC is (in square units):
(a) 1
(b) 2
(c) 3
(d) 4
(d) 4 Area of figure OABC = 4 square units => Correct option is (d).
## Q3. If x + 5y = 34 and x – 5y = – 6, find the value of 5y – 2x.
(a) – 8
(b) 14
(c) 8
(d) 20
(a) – 8 x + 5y = 34 x – 5y = – 6 x=14 Solving these, x = 14; y = 4 => Correct option is (a).
## Q4. 5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens cost Rs. 46. Find the cost of 2 pencils and 3 pens.
(a) Rs. 42
(b) Rs. 39
(c) Rs. 21
(d) Rs. 16
(c) Rs. 21 Let, cost of 1 pencil = x and cost of 1 pen = y – 2 Cost of 2 pencil and 3 pens = 6 +15 = Rs. 21 => Correct option is (c)
## Q5. An equation ax + by + c = 0 is a linear equation in 2 variables, where a, b, c are :
(a) natural numbers
(b) whole numbers
(c) integers
(d) real numbers
(d) real numbers
## Q6. A linear equation in two variables has:
(a) 1 solution
(b) 2 solutions
(c) no solution
(d) infinitely many
(d) infinitely many A linear equation in two variables has infinite number of solutions. => Correct option is (d)
## Q7. In a AABC, ?A = x°, ?B = (3a – 2)° and ?C = y°. Also, ?C – ?B = 9°. Find the value of ?B.
(a) 73°
(b) 82°
(c) 25°
(d) 49°
(a) 73° ?A + ?B + ?C = 180° => x + (3x – 2) + y = 180 4x + y = 182 …(1) y – 3x + 2 = 9 …(2) Solving (1) & (2), y = 82, x = 25° => ?B = 73°
## Q8. The perimeter of a rectangle is 44 cm. Its length exceeds twice its breadth by 4 cm. Find the area of the rectangle.
(a) 46 cm²
(b) 49 cm²
(c) 96 cm²
(d) 69 cm²
(c) 96 cm² Let, Length = x cm Breadth = y cm => 2(x + y) = 44 => x + y =221, y = 6 cm ,x = 16 cm x = 2y+4 => Area = 16 x 6 = 96 cm²
(a) k = 12
(b) k = 4
(c) k = 36
(d) k = 2
(b) k = 4
## Q10. A pair of linear equations is not consistent if:
(a) if has one solution
(b) it has many solutions
(c) graph intersect or coincide
(d) graph is parallel
(d) graph is parallel 37. Correct option is (d)
## Q11. The pair of equations y = 9 and y = – 7 has:
(a) one solution
(b) two solutions
(c) infinitely many
(d) no solution solutions
(d) no solution solutions y = 9 and y = – 7 are lines parallel to X-axis, thus these are parallel lines and hence no solution
## Q12. In a triangle, the sum of two angles is equal to the third angle. If the difference between two angles is 30°, find the angles.
(a) 15°, 45°, 75°
(b) 20°, 50°, 80°
(c) 30°, 60°, 90°
(d) 45°, 45°, 90°
(c) 30°, 60°, 90° (x) + (y) + (x + y) 180° x + y = 90° x – y = 30° => x = 60°, y = 30° => Angles are 30°, 60° and 90°
## Q13. The line x – y + 1 = 0 meets Y-axis at:
(a) (0,1)
(b) (4,0)
(c) (2,3)
(d) (0,6)
(a) (0,1) Correct option is (a).
## Q14. The solutions of the system of equations, x – y + 1 = 0, 3x + 2y -12 = 0 is :
(a) (0,1)
(b) (4,0)
(c) (2,3)
(d) (0,6)
(c) (2,3) Since the two lines meet at (2, 3) Correct option is (c).
## Q15. The sum of a two digit number is 8. The number obtained by reversing the digits exceeds the number by 18. Then the given number is :
(a) 53
(b) 35
(c) 26
(d) 62
(b) 35 Let, the digit at units’ place = x and digit at ten’s place = y => x + y = 8 and lOx + y = 18 + (10y + x) No. = 35.
## Q16. Equation of line m is :
(a) x = – 2
(b) x = 2
(c) y = 2
(d) y = – 2
(c) y = 2 y = 2 => Correct option is (c).
## Q17. The figure obtained by lines l, m, X-axis and Y-axis, can not be the :
(a) square
(b) rectangle
(c) trapezium
(d) triangle
(d) triangle The figure obtained is square. Since every square is a rectangle and is a trapezium also. Thus, the wrong name is triangle => Correct option is (d).
## Q18. Coordinates of A, B, C (in order) are:
(a) (2,0), (2,-2) (0,-2)
(b) (0,2), (-2,2), (-2,0)
(c) (-2,0), (2,2), (0,-2)
(d) (0,2), (2,2), (0,-2)
(b) (0,2), (-2,2), (-2,0) A(0,2), B(-2,2), C(-2,0). => Correct option is (b).
## Q19. Find the values of x and y if ABCD is a cyclic quadrilateral, ?A = 6x + 10°, ?B = (5x)°, ?C = (x + y)° (?D)=(3y-10)°.
(a) x = 30°, y = 20°
(b) x = 20°, y = 30°
(c) x = 40°, y = 10°
(d) x = 10°, y = 40°
(b) x = 20°, y = 30° ABCD is cyclic => ?A + ?C = 180° and ?B + ?D = 180° => 6x + 10 + x + y = 180° => 7x + y = 170° 5x + 3y – 10 = 180° => 5x + 3y = 190° x = 20°, y = 30° => Correct option is (b)
Practicing NCERT Maths MCQs Chapter 3 Pair of Linear Equations in Two Variables With Answers Pdf Class 10 is one of the best ways to prepare for the CBSE Class 10 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or one wants to score better. If you have any queries regarding The Chapter 3 Pair of Linear Equations in Two Variables CBSE Class 10 Maths MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.
## MCQs with Solutions for Class 10 Maths
The multiple-choice questions for all the chapters of Class 10 Maths are available here online for students. Below are the links for respective chapters starting from 1 to 15. These objective questions cover all the concepts explained in the individual chapter so that students can test themselves as to their level of preparation for exams.
These materials will help students score in the upcoming board exam and perform better in competitive exams, such as Maths olympiads. Hence, it is suggested to students to solve all these MCQs with their problem-solving skills and cross-check the answers. Also, try to maintain a time limit for each question.
Below is the list of all the chapters for multiple-choice questions. Click on the particular link to get the MCQs with answers and also download PDFs provided for each chapter for extra MCQs to practise more.
### Benefits of CBSE NCERT MCQs for Class 10 Mathematics
You can find various perks of referring to 10th-grade maths cbse MCQs with answers pdf provided on our page. They are as such:
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## CBSE Class 10 Maths Sample Paper & MCQs
CBSE Class 10 Sample Paper & MCQs for Term-1: CBSE has scheduled Mathematics Basic & Standard Exam for Class 10 on 04th December 2021. As now just 1 day left so you should practice with as many as Multiple Choice Questions to know what is your level of preparation for the final exam. In this article, we have provided you with 60 MCQs for Class 10 Maths for your practice, just keep solving and scroll the page to test your preparation for the Class 10 Maths Term-1 Exam with these Maths MCQ Class 10 Term-1 with Solutions.
## Maths Sample Paper for Class 10 Term-1
Mathematics Subject of Class 10 is divided into two parts- Basic Mathematics & Standard Mathematics and we have provided Class 10 MCQs for both sections in the below section. If you are going to appear in CBSE Class 10 Maths Term-1 on 04th December 2021, then you must not miss these questions as you never know which question can be asked in the final exam or questions can be of the same type as provided below.
## Maths Class 10 MCQ Term-1 with Solution
If you are looking for Class 10th Maths Sample Paper with Solution for your Term-1 Exam, then you are at the right place, as we have listed a few important MCQs for your help.
# MCQs for Class 10 Mathematics
## Class 10 Maths MCQs Multiple Choice Questions with Answers
Practicing CBSE NCERT Objective MCQ Questions of Class 10 Maths with Answers Pdf is one of the best ways to prepare for the CBSE Class 10 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or you want to score better. By practicing more Class 10th Maths Objective Questions Pdf, students can improve their speed and accuracy which can help them during their board exam.
# MCQ Questions for Class 10 Mathematics
Get Class 10 Math MCQ with answer PDF Download, which is based on the latest pattern of CBSE and NCERT. It involves every one of the points given in NCERT class 10 Math book. You can easily download these MCQ Questions for Class 10 Maths with Answers PDF with answer given below.
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Multiple choice Questions for Class 10 Mathematics with solutions have been provided below for each chapter in NCERT Book for Class 10 Commerce Mathematics Book. Students can click on the link below for any chapter and refer to the questions and answers provided below. all MCQs for Class 10 Mathematics have been provided with solutions so that the students can read the questions and assess their performance.
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Free MCQs in PDF of CBSE Class 10 Mathematics have been designed by our highly expert class 10th teachers as per CBSE NCERT guidelines and examination pattern issued this year.
## MCQ Questions for ICSE Class 10 Math With Answers
“JustTutors” brings you topic wise MCQ Questions for Class 10 Maths with all the subtopics from every chapter to test your knowledge. Multiple Choice Questions for ICSE Class 10 Maths are available for free, you can test your knowledge anytime and share the links with your friends to help them check their knowledge.
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– Students who are preparing for various school tests such as unit-tests, term-tests, semester tests, half-yearly and yearly tests and examinations on Mathematics – Class 10.
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– Students who wish to sharpen their knowledge of Mathematics.
– Students who are studying in CBSE, ICSE, IGCSE schools as well as various State Board schools.
– Students who are preparing for competitive examinations such as NTSE, Olympiad, etc for class 10
# MCQ Questions for Class 10 Maths with Answers
MCQ Questions is vital for students who need to score great marks in their CBSE board tests. Students who can Practice Multiple Choice Questions Class 10 Maths with Answers to improve your score in Board Exams. Practice is the solitary key for accomplishment in the CBSE Exam. You can begin with CBSE class 10 sample papers Board Exams.
The Multiple Choice Questions for every one of the chapters of Class 10 Maths are accessible here online for students. The following are the connections for separate chapters beginning from 1 to 15. These objective questions cover every one of the ideas clarified in the individual chapters so students can test themselves concerning their degree of the groundwork for tests.
Practicing NCERT Maths MCQ for Class 10 CBSE with Answers Pdf is one of the best ways to prepare for the CBSE Class 10 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or one wants to score better. By practicing more Maths Quiz Questions with Answers for Class 10 Pdf, students can improve their speed and accuracy which can help them during their board exam.
### How can Important questions for class 10 maths can help you
Scoring good marks in class 10 maths required extensive practice.The fundamental and key points which is required to score good marks in class 10 maths are :
1. Concept clarity in the chapter.Read theory given in NCERT text book and Entrancei class 10 maths section.Solve as many as Maths Questions as you can.
2. Must remember the formula use in question.Note down all importnat formulas of chapter.
3. Application of concept in numerical can be learned through solved examples.
4. NCERT text book theory and questions given in exercise is must.
5. Numerical practice from entrancei MCQ section and online quiz will be highly helpful.
One must cover all the above points before moving ahead to Important questions for class 10 maths. Start with revision of NCERT text book of your class 10 .Try to make two notes one for the detail theory part and second is for important formula of class 10 maths.Refer the maths formula section of entrancei to get all important formula in one page.Notes and formula help you to complete revision of the class 10 maths.After that follow the application of concept and formulas in numericals before writing the maths solutions of Important questions for class 10 maths.
## Class 10 Maths MCQs Multiple Choice Questions with Answers
Prepare effectively for the Class 10 Maths Exam taking the help of CBSE NCERT Objective MCQ Questions of Class 10 Maths with Answers PDF provided here. Refer to the Chapter Wise 10th Standard Maths during your preparation and attempt the exam with confidence.
Use the above-provided Chapter Wise NCERT MCQ Questions for Class 10 Maths with Answers Pdf free download. We hope the given NCERT MCQ Questions for Class 10 Maths PDF Free Download will definitely yield fruitful results. If you have any queries regarding The Solid State CBSE Class 10 Maths MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.
## Q1. What is the best strategy while answering the Multiple Choice Questions for exams?
The best strategy to follow while answering the MCQ Questions for exams is to go through the question as many times as possible and understand it clearly.
## Q2. Where to download CBSE MCQs for all Classes and Subjects?
CBSE MCQs for all the classes and subjects are provided online or you can view them on our site. Detailed description of the questions is given by subject experts and you can use them as a reference during your preparation.
## Q3. What is NCERT Exemplar and where can we get its Solutions?
NCERT Exemplars are basically practice-books that contain additional questions of higher level and give you comprehensive and thorough learning. The NCERT Solutions for the questions are present on our site or you can find them online too.
## Q4. Which Book is best for CBSE MCQs?
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## Q5. What are the benefits of CBSE MCQs?
CBSE MCQs lay a stronger foundation of concepts in students that can be useful for them in their later stages to prepare for any competitive exams.
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# Convert Mixed Numbers to Improper Fractions
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## Mixed Numbers
A mixed number is a combination of a whole number and a fraction, for example: 2
1
2
.
## Improper Fractions
A fraction in which the numerator is larger than or equal to the denominator, like
5
2
,
17
3
, or
6
6
is called an improper fraction.
## Convert Mixed Numbers to Improper Fractions
A mixed number can be expressed as a fraction.
• Multiply the whole number by the denominator.
• Then add the result to the numerator.
• This result becomes new numerator.
• The denominator remains the same.
Let's illustrate through examples.
### Example 1
Write 2
1
2
as an improper fraction.
### Solution
To write 2
1
2
as an improper fraction,
Multiply the whole number by the denominator.
The whole number is 2.
The denominator is 2.
2 x 2 = 4
Add the result to the numerator:
The numerator is 1.
4 + 1 = 5
The numerator is 5. The denominator remains 2.
2
1
2
=
5
2
### Example 2
Write 5
2
3
as an improper fraction.
### Solution
To write 5
2
3
as an improper fraction,
Multiply the whole number by the denominator.
The whole number is 5.
The denominator is 3.
5 x 3 = 15
Add the result to the numerator:
The numerator is 2.
15 + 2 = 17
The numerator is 17. The denominator remains 3.
5
2
3
=
17
3
## Convert Mixed Numbers to Improper Fractions
• A mixed number is a combination of a whole number and a fraction.
• A fraction in which the numerator is larger than or equal to the denominator, like
5
2
,
17
3
, or
6
6
is called an improper fraction.
• A mixed number can be expressed as a fraction.
• Multiply the whole number by the denominator.
• Then add the result to the numerator.
• This result becomes new numerator.
• The denominator remains the same.
## Similar Lessons
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# Integration by parts for Definite integral problems with limits or bounds
Home > Math Article > Definite integral using integration by parts formula
Page Contents
## Integration by parts for definite integral with limits, UV formulas, and rules
In this article, you will learn how to evaluate the definite integral using integration by parts UV formula. Generally, most of the students are confused about how to use the limit of the integral function after applying the integration by parts UV formula.
Integration by parts formula for solving definite integral limits problems-
### Rules for solving integration by parts for definite integral limits
1. The first one is that you can apply limits after the end of your integrating result as you did in indefinite integration but make sure your variable is the same.
Let’s take an example of $\int _ { a } ^ { b } f ( y ) dx$
⇒ First, solve the integration of this function
$\int _ { a } ^ { b } f ( y ) dx = F ( y ) + C$
⇒ then you put a limit in F(y)
$\int _ { a } ^ { b } f ( y ) dx = F ( b ) – F ( a ) + C$
2. The second one is that you can use the limit as you are going with integrals of functions. I recommended to you to use the second method when two or more functions are given as it minimizes your chance of error in Definite integral problems.
$\int _ { a } ^ { b } udv = u.v \Big| _ a ^ b – \int _ { a } ^ { b } v.du$
3. If you are using the first method then make sure first you solve integration as you do in indefinite integration then put the limit at the end of your result.
4. In the case of substitution integration, limits will keep changing depending on the substitution value.
So, you understand the rules and formulas of integration by parts for definite integral problems. If not, then we will understand it better by solving its examples. Let’s go-
## Examples of Integration by parts for Definite integral limits/ Changing limits/ infinite limits
### Example 1: Evaluate Definite Integral limits using integration by parts with both Rule 1 and 2.
$\int _ { 1 } ^ { 2 } x.lnx dx$
Solution: For solving the above definite integral problem with integration by parts using Rule 1, we have to apply limits after the end of our result
First solve it, according to this: $\int _ { } ^ { } u.dv = u.v – \int _ { } ^ { } v.du$
So, we have u = lnx and $v = \frac { x ^ { 2 } } { 2 }$ [ ∴ dv = x ]
$\int _ { } ^ { } lnx.xdx = lnx . \frac { x ^ { 2 } } { 2 } – \int _ { } ^ { } \frac { x ^ { 2 } } { 2 } . \frac { 1 } { x }$
= $\int _ { } ^ { } lnx.xdx = lnx . \frac { x ^ { 2 } } { 2 } – \frac { x ^ { 2 } } { 4 }$
Now apply limits to evaluate definite integral
$\int _ { 1 } ^ { 2 } lnx.xdx = \left [ lnx . \frac { x ^ { 2 } } { 2 } – \frac { x ^ { 2 } }{ 4 } \right ] \Big| _ 1 ^ 2$
$= \left [ ln2.2 – 1 \right ] – \left [ ln1. \frac { 1 } { 2 } – \frac { 1 } { 4 } \right ]$ [ ∴ $\int _ { a } ^ { b } f ( y ) dx = F ( b ) – F ( a ) + C$ ]
$= ln4 – 1 + \frac { 1 } { 4 }$
$= ln4 – \frac { 3 } { 4 }$ + C Answer
Rule 2: The second rule said that you can use the limit as you are going with integrals of functions to solve integration by parts Definite integral with limits or bounds.
According to this ⇒ $\int _ { a } ^ { b } udv = u.v \Big| _ a ^ b – \int _ { a } ^ { b } v.du$
So, we have u = lnx and $v = \frac { x ^ { 2 } } { 2 }$ [ ∴ dv = x ]
$\int _ { 1 } ^ { 2 } lnx.x dx = lnx. \frac { x ^ { 2 } } { 2 } \Big| _ 1 ^ 2 – \int _ { 1 } ^ { 2 } \frac { x ^ { 2 } } { 2 } . \frac { 1 } { x }$
= $\left [ 2ln2 – \frac { 1ln1 } { 2 } \right ] – \frac { 1 } { 2 } \int _ { 1 } ^ { 2 } x ^ { 2 } dx$
= $\left [ ln4 – 0 \right ] – \left [ \frac { x ^ { 2 } } { 4 } \right ] \Big| _ 1 ^ 2$ [ ∴ $\int _ { a } ^ { b } f(y)dx = F ( b ) – F ( a ) + C$ ]
= $\left [ ln4 – 0 \right ] – \left [ 1 – \frac { 1 } { 4 } \right ]$
= $ln4 – \frac { 3 } { 4 } + C Answer$
### Example 2: Evaluate the definite integral using integration by parts with limits Rule 1
$\int _ { 0 } ^ { \sqrt { 3 } } t ^ { 3 } ( 1 + t ^ { 2 } ) ^ { – 3 } dt$
Solution:
Rule 1: First solve it by integration by parts as indefinite integral then use the limits
So we have integration by parts uv formula
After solving this we get-
$\int _ { } ^ { } t ^ { 3 } ( 1 + t ^ { 2 } ) ^ { -3 } dt = – \frac { 1 } { 4 } t ^ { 2 } ( 1 + t ^ { 2 } ) ^ { -2 } – \frac { 1 } { 4 } ( 1 + t ^ { 2 } ) ^ { -1 } + C$
Now solve this using definite integral limits
$\int _ { 0 } ^ { \sqrt { 3 } } t ^ { 3 } ( 1 + t ^ { 2 } ) ^ { -3 } dt = \left [ – \frac { 1 } { 4 } t ^ { 2 } ( 1 + t ^ { 2 } ) ^ { – 2 } – \frac { 1 } { 4 } ( 1 + t ^ { 2 } ) ^ { – 1 } \right ] \Big| _ 0 ^ 3$
$= \left [ – \frac { 1 } { 4 } ( \sqrt { 3 } ) ^ { 2 } ( 1+ ( \sqrt { 3 } ) ^ { 2 } ) ^ { -2 } – \frac { 1 } { 4 } ( 1 + ( \sqrt { 3 } ) ^ { 2 } ) ^ { -1 } \right ] – \left [ – \frac { 1 } { 4 } ( 0 ) ^ { 2 } ( 1+ ( 0 ) ^ { 2 } ) ^ { – 2 } – \frac { 1 } { 4 } ( 1 + ( 0 ) ^ { 2 } ) ^ { -1 } \right ]$
$= – \frac { 3 } { 64 } – \frac { 1 } { 16 } + \frac { 1 } { 4 }$
$= \frac { 9 } { 64 } Answer$
### Example 3: Evaluate definite integral by Rule 2 of integration by parts
$\int _ { 0 } ^ { 2 } ( x + 1 ) e ^ { 4x } dx$
Solution:
Rule 2: The second rule said you can use the limit as you are going with integrals of functions to solve integration by parts with limits or bounds.
So we have, u = x + 1 dv = e4x
According to this ⇒ $\int _ { a } ^ { b } udv = u.v \Big| _ a ^ b – \int _ { a } ^ { b } v.du$
$= \left [ ( x + 1 ) \frac { e ^ { 4x } } { 4 } \right ] \Big| _ 0 ^ 2 – \int _ { 0 } ^ { 2 } \frac { e ^ { 4x } } { 4 } dx$
$= \left [ ( 2 + 1 ) \frac { e ^ { 8 } } { 4 } – \frac { 1 } { 4 } \right ] – \left [ \frac { e ^ { 4x } } { 16 } \right ] \Big| _ 0 ^ 2$
$= \left [ ( 2 + 1 ) \frac { e ^ { 8 } } { 4 } – \frac { 1 } { 4 } \right ] – \left [ \frac { e ^ { 8 } } { 16 } – \frac { 1 } { 16 } \right ]$
$= \frac { 11 } { 16 } e ^ { 8 } – \frac { 3 } { 16 }$ Answer
### Example 4: Evaluate definite integral with an infinite limit by integration by parts second rule
$\int _ { 0 } ^ { \infty } x.e ^ { -x } dx$
Solution:
Rule 2: $\int _ { a } ^ { b } udv = u.v \Big| _ a ^ b – \int _ { a } ^ { b } v.du$
Let u = x and dv = e-x
$\int _ { 0 } ^ { \infty } x.e ^ { -x } dx = \left [ -xe ^ { -x } \right ] \Big| _ 0 ^ \infty + \int _ { 0 } ^ { \infty } -e ^ { -x } dx$ (∴ v = -e-x )
= $\left [ -xe ^ { -x } \right ] \Big| _ 0 ^ \infty + \left [ -e ^ { -x } \right ] \Big| _ 0 ^ \infty$
= $\left [ -xe ^ { -x } -e ^ { -x } \right ] \Big| _ 0 ^ \infty$
Take limit extent to infinity
= $\lim _ { x \rightarrow \infty } \left [ -xe ^ { -x } -e ^ { -x } \right ] – \left [ -0.e ^ { -0 } – e ^ { -0 } \right ]$
= $\left [ 0 -0 \right ] – \left [ 0 – 1 \right ]$
### Example 5: Evaluate Definite integral by changing limit of integration by parts using Rule 2
$\int _ { 0 } ^ { 1 } t \sqrt { t + 2 } dt$
Solution:
According to the Rule 2$\int _ { a } ^ { b } udv = u.v \Big|_a ^ b – \int _ { a } ^ { b } v.du$
Let u = t and dv = $\sqrt { t + 2 } dt$
v = $\frac { 2 } { 3 } ( t + 2 ) ^ { 3/2 }$
Now put value in integration by part with limit formula
= $\frac { 2 } { 3 } t( t + 2 ) ^ { 3/2 } \Big|_ 0 ^ 1 – \int _ { 0 } ^ { 1 } \frac { 2 } { 3 } ( t + 2 ) ^ { 3/2 } dt$
= $\frac { 2 } { 3 } t ( t + 2 ) ^ { 3/2 } \Big| _ 0 ^ 1 – [ \frac { 4 } { 15 } ( t + 2 ) ^ { 5/2 } ] \Big| _ 0 ^ 1$
= $\frac { 2 } { 3 } ( 3 ) ^ { 3/2 } – \frac { 4 } { 15 } ( 3 ^ { 5/2 } – 2 ^ { 5/2 } )$ Answer
Hope you understand that how can we solve the integration by parts for Definite integrals with limits or bounds problems, its substitution rules, formulas, etc.
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# Represent the following data with the help of a pie-diagram:
Question:
Represent the following data with the help of a pie-diagram:
Items Wheat Rice Tea Production (in metric tons) 3260 1840 900
Solution:
We know:
Central angle of a component $=\left(\right.$ component value $/$ sum of component values $\left.\times 360^{\circ}\right)$
Here, total production = 6000 (in metric tons)
Thus, the central angle for each component can be calculated as follows:
Item Production(in metric tons) Sector angle Wheat 3260 3260/6000 x 360 = 195.6 Rice 1840 1840/6000 x 360 =1 10.4 Tea 900 900/6000 x 360 = 54
Total = 6000 (in metric tons)
Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here, the largest central angle is 195.6o. Draw a sector with the central angle 195.6 o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct the other sectors representing the other items in the clockwise direction in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them as shown in the figure below. |
# DOMAIN AND RANGE OF TRIGONOMETRIC FUNCTIONS
To make the students to understand domain and range of a trigonometric function, we have given a table which clearly says the domain and range of trigonometric functions.
## Domain of sin x and cos x
In any right angle triangle, we can define the following six trigonometric ratios.
sin x, cos x, csc x, sec x, tan x, cot x
In the above six trigonometric ratios, the first two trigonometric ratios sin x and cos x are defined for all real values of x.
The two trigonometric ratios sin x and cos x are defined for all real values of x.
So, the domain for sin x and cos x is all real numbers.
## Range of sin x and cos x
The diagrams given below clearly explains the range of sin x and cos x.
Range of sin x
Range of cos x
From the pictures above, it is very clear that the range of y = sin x and y = cos x is
{y | -1 ≤ y ≤ 1}
## Domain of csc x and sec x
We know that sin (kπ) = 0, cos [(2k+1)π] /2 = 0, here "k" is an integer.
Then,
k = ...........-2, -1, 0, 1, 2, ..........
For k = -2,
sin (-2π) = 0 and cos (-3π/2) = 0
For k = -1
sin (-π) = 0 and cos (-π/2) = 0
For k = 0,
sin (0) = 0 and cos (π/2) = 0
For k = 1,
sin (π) = 0 and cos (3π/2) = 0
For k = 2,
sin (2π) = 0 and cos (5π/2) = 0
Stuff 1 :
We know that csc x and sec x are the reciprocals of sin x and cos x respectively.
Let us see the values of csc x for
x = .......-2π, -π, 0, π, 2π, .........
csc(-2π) = 1/sin(-2π) = 1/0 = Undefined
csc(-π) = 1/sin(-π) = 1/0 = Undefined
csc(0) = 1/sin(0) = 1/0 = Undefined
csc(π) = 1/sin(π) = 1/0 = Undefined
csc(2π) = 1/sin(2π) = 1/0 = Undefined
From the above examples, it is very clear, that csc x is defined for all real values of x except
x = .......-2π, -π, 0, π, 2π, .........
So the domain of csc x is
{x | x ≠ ...-2π, -π, 0, π, 2π, ..}
In the same way, domain of sec x is
{x | x≠ ...-3π/2, -π/2, π/2, 3π/2, 5π/2 ...}
## Range of csc x and sec x
Let y = csc x.
In the trigonometric function y = csc x, when plug values for x such that
x ∈ R - {.......-2π, -π, 0, π, 2π,.......},
we will get values for "y" which are out of the interval
(-1, 1)
So the range of csc x is
{y | y ≤ -1 or y ≥ 1}
In the same way, for the function y = sec x, when plug values for x such that
x ∈ R - {.......-3π/2, -π/2, π/2, 3π/2, 5π/2.......},
we will get values for y which are out of the interval
(-1, 1)
So the range of sec(x) is
{ y | y ≤ -1 or y ≥ 1}
## Domain of tan x and cot x
The trigonometric function tan x will become undefined for
x = [(2k + 1)π] / 2
here k is an integer.
Substituting k = ...........-2, -1, 0, 1, 2, .......... we get
x = ..........-3π/2, -π/2, π/2, 3π/2, 5π/2........
For the above values of x, tan x becomes undefined and tan x is defined for all other real values.
Therefore, domain of tan x is
{x | x ≠......-3π/2, -π/2, π/2, 3π/2, 5π/2.....}
The trigonometric function cot x will become undefined for
x = kπ
here k" is an integer.
Substituting k = ...........-2, -1, 0, 1, 2, .......... we get
x = ..........-2π, -π, 0, π, 2π.......
For the above values of x, cot x becomes undefined and cot(x) is defined for all other real values.
So, the domain of cot x is
{x | x ≠......-2π, -π, 0, π, 2π.......}
## Range of tan x and cot x
In the trigonometric function y = tan x, if we substitute values for x such that
x ∈ R - {.......-3π/2, -π/2, π/2, 3π/2, 5π/2.....},
we will get all real values for "y" .
So the range of tan x is
All Real Values
In the same way, for cot x, if we substitute values for x such that
x ∈ R - {.......-2π, -π, 0, π, 2π......},
we will get all values for y.
So the range of cot x is
All Real Values
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A Parallelogram is a 2-D figure which is used in the Mensuration. The Area and Perimeter of the Parallelogram are similar to the area and perimeter of the rectangle. Learn the basic formulas of Parallelogram with the help of this article. Scroll down this page to know the formulas of Area and Perimeter of Parallelogram with solved problems.
Before learning the formulas we suggest the students to know what is a parallelogram. A parallelogram is a regular polygon with four sides and four angles. Get the properties, formulas from the below section.
## Area and Perimeter of Parallelogram
Area of Parallelogram: The area of the parallelogram is the product of base and height. It is used to measure the region occupied by the parallelogram. The units to measure the area of the parallelogram is square units.
Perimeter of Parallelogram: The perimeter of the parallelogram is defined as the sum of all four sides. The units to measure the perimeter of the parallelogram is units.
### Properties of Parallelogram
• The diagonals of the parallelogram bisect each other.
• Opposite sides of the parallelogram are congruent.
• If one of the angles is the right angle then all the angles are right.
• Opposite angles are congruent.
### Formulas for Area and Perimeter of Parallelogram
The area of the parallelogram is the product of base and height.
Area = b × h
Where,
b = base
h = height
The perimeter of the parallelogram is the sum of the four sides.
P = a + a + b + b
P = 2a + 2b
P = 2(a + b)
Where,
a and b are the lengths of the parallelogram.
### Solved Examples on Area of Parallelogram
1. Find the area of the parallelogram if the base and height are 12m and 10m?
Solution:
Given,
b = 12m
h = 10m
We know that,
Area of the Parallelogram = b×h
A = (12m)(10m)
A = 120 sq. meters
Therefore the area of the parallelogram is 120 sq.m
2. Find the perimeter of the parallelogram if the sides are 5 cm and 4 cm.
Solution:
Given,
a = 5 cm
b = 4 cm
We know that,
The perimeter of the parallelogram is 2(a + b)
P = 2(5 cm + 4 cm)
P = 2(9 cm)
P = 18 cm
Thus the perimeter of the parallelogram is 18 cm.
3. Find the height of the parallelogram if the base is 14m and Area is 112 sq. meters.
Solution: Given,
b = 14m
A = 112 sq. m
Area of the Parallelogram = b×h
112 sq. m = 14 × h
h = 112/14
h = 8 meters
Thus the height of the parallelogram is 8 meters.
4. Find the length of the parallelogram whose base length is 6m and the perimeter is 16m.
Solution:
Given,
a = 6m
b =?
P = 16m
Perimeter of the Parallelogram = 2(a + b)
16m = 2(6 + b)
16/2 = 6 + b
8m = 6m + b
b = 8m – 6m
b = 2m
Thus the length of the parallelogram is 2 meters.
### FAQs on Area and Perimeter of the Parallelogram
1. What is the formula for area and perimeter of the parallelogram?
Area of the parallelogram = bh
Perimeter of the parallelogram = 2(a + b)
2. How to find the perimeter of the parallelogram?
The perimeter of the parallelogram can be found by adding the sides of the parallelogram i.e, a + a + b + b.
3. What is the height of the parallelogram?
The height of the parallelogram is the distance between the opposite sides of the parallelogram. |
# How to Use the Acceleration Formula
By Kathy Foust
Having some trouble understanding the acceleration formula? This study guide will walk you through all the steps.
Acceleration is known as the time rate of change in velocity as stated in the study guide on motion.
Acceleration is caused by a change in speed (which also means a change in velocity since speed is part of velocity.), a change in direction or both. Average acceleration is found using the average acceleration formula which is described below.
## Acceleration Formula
To find average acceleration, you must have constant acceleration from the time that the change in velocity occurs until the change in velocity stops. For these equations we are are going to assume constant acceleration or the uniform change in velocity. This is always necessary when working with mathematical formulas--such as the formula for acceleration--and is what most textbooks use as well. Let's try the acceleration formula below.
• To begin with, the acceleration formula is expressed as average acceleration=change in velocity/time for change to occur.
• A car at rest accelerates uniformly on a straight track and reaches a speed of 70 km/h in 5.0 seconds. What is the acceleration?
• To begin with we must change the km/h to m/s using the conversion formula 70 x .278= 19.46 m/s
• We need to find the change in velocity by subtracting the end velocity from the beginning velocity.
• We know both of the amounts because the problem states the car was "at rest" and therefore had a 0 change in velocity at that moment.
• We also know that the car was traveling on a straight track which also tells us there was no change in direction.
• The speed attained was 19.46 m/s, so 19.46 m/s - 0 m/s= 19.46 m/s in 5.0 seconds was the change in velocity from beginning to end.
• Use the acceleration formula (19.46 m/s) / 5.0 s = 3.892 m/s2 where 19.46 m/s=change in velocity and 5.0 s is the time for the change to occur.
• Following the steps for average acceleration, we find the answer to be 3.892 m/s2.
• Round the answer down to significant numbers to get the answer of 3.9 m/s2.
The main things to remember when dealing with the acceleration formula is that the changes and movements in the formula must be constant and must be changes in velocity. Each change needs to be measured and the formula done for each one. |
# 2022 AMC 8 Problems/Problem 8
## Problem
What is the value of $$\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?$$
$\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}$
## Solution 1
Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes $$\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.$$
~MRENTHUSIASM
## Solution 2
The original expression becomes $$\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.$$ ~hh99754539
~Math-X
## Video Solution (CREATIVE THINKING!!!)
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Home | Math | What is Odd Number Mean in Math?-Definition, And Properties
What is Odd Number Mean in Math?-Definition, And Properties
August 23, 2022
written by Azhar Ejaz
Odd numbers are numbers that cannot be divided into two groups equally. In other words, all numbers except the completely divided of 2 are odd numbers. In this article learn more about odd numbers. Even numbers are those numbers that can be divided into two equal groups and are exactly divisible by 2.
What are the odd numbers?
Odd numbers are those numbers that cannot be divided into two groups or pairs equally. In other words, odd numbers are that cannot be categorized into groups of two.
For example:
3,5,7,9
Properties of an odd number
The following properties of odd numbers. Each of these can be explained in a detailed way property as given below.
The sum of two odd numbers always answers an even number
For example:
7 + 9 = 16 .
• Subtraction of odd Numbers:
Subtraction of two odd numbers always answers in an even number.
For example:
5 – 3 = 2
• Multiplication of odd numbers:
Multiplication of two odd numbers always results in an odd number.
For example:
11 × 3 = 33
• Division of odd numbers
The division of two odd numbers always answers an odd number.
For example:
9 ÷ 3 = 3
Types of odd numbers
Odd numbers are a list of all the numbers that are not completely divided by 2. So we can have many types of odd numbers starting from whether the odd numbers have factors or not, what is the difference between two odd numbers, what is their position on the number line, etc.
Composite Odd Numbers
Composite odd numbers are said to be odd numbers that are not prime numbers. These types of odd numbers are made by the product of two smaller positive odd integers. The composite odd numbers from 1 to 100 are 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, and 99.
Consecutive odd number
Consecutive odd numbers are those odd numbers that are listed in sequence of their order.
For example:
if n is an odd number, then the numbers n and n + 2 are grouped under the consecutive odd numbers. They always have a difference of 2 between them and are consecutive in nature, hence the name consecutive odd numbers.
For example:
3 and 5, 11 and 13, 25 and 27, 37 and 39, 49 and 51, and so on.
Summary
• The sum of two odd numbers is always an even number.
• A simple method to differentiate whether a number is odd or even is to divide it by 2.
• If the number is not divided by 2 completely.
• Odd numbers always have 1, 3, 5, 7, or 9 in their unit place. Even numbers always have 0, 2, 4, 6, or 8 in their unit place.
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## Objective
SWBAT add mixed numbers with like denominators by changing them to improper fractions. They validate their answers with a visual model.
#### Big Idea
A mixed number can be solved by changing it to an improper fraction.
## Whole Class Discussion
15 minutes
In today's lesson, the students learn to add mixed numbers with like denominators. This relates to 4.NF.B3c because the students add mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction.
Before we begin the lesson, I give the students a review by asking a series of questions. First, I ask, What is a mixed number? Student response: A mixed numbers is when you have a whole number and a fraction. I let the students know that today, we are adding mixed numbers with like denominators. What are like denominators? Student response: When the denominators are the same. If the denominators are the same number, what must we do to solve the problem? Student response: Add the numerators. I explain to the students that we are not working with unlike denominators today, so we do not have to find a common denominator.
The Adding Mixed Numbers with Like denominators.pptx power point is displayed on the Smart board. The students are sitting at their desk with paper and pencil. I let the students know that we are learning two ways to add mixed numbers. I let the students know that they can solve the problem in either way that is easiest for them. I instruct them to write down the sample problem from the power point, 5 1/5 + 3 2/5. I want the students to work with me as we discuss the power point. (I find that this method works well for my students because it keeps them focused on what is being said in class.) Be sure to line up the fractions under the fractions and the whole numbers under the whole numbers. We learned in addition that we line numbers up according to place value. This also applies to fractions because fractions can be written as decimals.
Back to the problem, next we add the whole number. What is 5 + 3? 8. We have 8 3/5 as our answer. I instruct the students to draw 5 1/5. What did we learn about the wholes? Student response: They have to be the same size. As the students draw their models, I draw models on the board as well. (A teacher modeling for the students can help give those lost students a starting point if they are completely lost.) Together we shade 5 whole and 1 out of 5 for our 5 1/5. Next, the students draw a model of 3 2/5, then shade 3 whole and 2 out of 5 to show 3 2/5. Now, let's see if our model connects to our answer. We said that the answer to the problem was 8 3/5. Let's count to see how many "wholes" are shaded. The students count to get 8. Next, count the boxes that represent the fraction. The students count 3/5 shaded. Therefore, our model and answers connect.
Another Way:
## Skill Building/Exploration
20 minutes
For this activity, I let the students work independently. I give each student an Adding Mixed Numbers with Like Denominators.docx activity sheet. The students must add mixed numbers with like denominators in one of two ways: 1) by adding the mixed numbers, then simplifying the fraction, or 2) by changing the mixed numbers to improper fractions. The students must draw a model of their mixed numbers (MP4).
The students work on real-world scenarios to add mixed numbers with like denominators. The students are guided to the conceptual understanding through questioning by me. As I walk around the classroom, I am questioning the students and looking for common misconceptions among the students. Any misconceptions are addressed at that point, as well as whole class at the end of the activity.
Any student that finishes the assignment early, can go to the computer to practice the skill at the following site until we are ready for the whole group sharing:
My Findings:
In today's lesson, the students learned to add mixed numbers in two ways. As I monitored and answered questions from the students, I noticed that some of the students still had a hard time drawing the models. I encouraged those students to solve the problem first, then draw a model last. Most of the students could solve the problems, but as stated earlier, the models were giving a few of them a problem. After I kept driving home the point that the denominator tells you how to divide the whole, it became better for most of the ones having problems.
I was surprised that several of the students used equivalent fractions to add the mixed numbers. I expected more of the students to just add the fractions and whole numbers. Overall, I was pleased with the lesson.
## Closure
15 minutes
To close the lesson, we review the answers to the problems as a whole class. This gives those students who still do not understand another opportunity to learn it. I like to use my document camera to show the students' work during this time. Some students do not understand what is being said, but understand clearly when the work is put up for them to see.
I feel that by closing each of my lessons by having students share their work is very important to the success of the lesson. Students need to see good work samples, as well as work that may have incorrect information. From the Video - Adding Mixed Numbers, you can hear a student explain how she solved a problem. More than one student may have had the same misconception. During the closing of the lesson, all misconceptions that were spotted during the activity will be addressed whole class.
Misconception(s):
-Drawing models correctly (as stated earlier) |
# 2020 AMC 10B Problems/Problem 12
## Problem
The decimal representation of$$\dfrac{1}{20^{20}}$$consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$
## Solution 1
$$\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}$$
Now we do some estimation. Notice that $2^{20} = 1024^2$, which means that $2^{20}$ is a little more than $1000^2=1,000,000$. Multiplying it with $10^{20}$, we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$. Notice that when we divide $1$ by an $n$ digit number, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$, there are $\boxed{\textbf{(D) } \text{26}}$ zeros in the initial string after the decimal point. -PCChess
## Solution 2
First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$.
$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits.
Our answer is $\boxed{\textbf{(D) } \text{26}}$.
## Solution 3 (Brute Force)
Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of$\boxed{\textbf{(D) } \text{26}}$ zeroes to be formed. -OreoChocolate
## Solution 4 (Smarter Brute Force)
Just as in Solutions $2$ and $3,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We can then look at the number of digits in powers of $5$. $5^1=5$, $5^2=25$, $5^3=125$, $5^4=625$, $5^5=3125$, $5^6=15625$, $5^7=78125$ and so on. We notice after a few iterations that every power of five with an exponent of $1 \mod 3$, the number of digits doesn't increase. This means $5^{20}$ should have $20 - 6$ digits since there are $6$ numbers which are $1 \mod 3$ from $0$ to $20$, or $14$ digits total. This means our expression can be written as $\dfrac{k\cdot10^{14}}{10^{40}}$, where $k$ is in the range $[1,10)$. Canceling gives $\dfrac{k}{10^{26}}$, or $26$ zeroes before the $k$ since the number $k$ should start on where the one would be in $10^{26}$. ~aop2014
## Solution 5 (Logarithms)
$$\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}$$
Let $$x = \dfrac{1}{10^{20}\cdot2^{20}}$$
Remember: $$log_{a}b + log_{a}c = log_{a}bc$$ And $$log_{a}bc = log_{a}b + log_{a}c$$ (We will use the second property) Like this
$$\dfrac{1}{log_{10}(10^{20}\cdot2^{20})} = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}$$
Taking $log_{10}$ on both the sides results in
$$log_{10}x = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}$$
$$= \dfrac{1}{20 \cdot log_{10}10 + 20 \cdot log_{10}2}$$
Remember that $$log_{10}10 = 1$$ and $$log_{10}2 = 0.3010$$ Note that this value is an approximate. This makes our equation
$$log_{10}x = \dfrac{1}{20 + 20 \cdot {0.3010}}$$ $$= \dfrac{1}{20 + 6.020}$$ $$= \dfrac{1}{26.02}$$
Removing the log we get $$x = 10^{-26.02}$$
Number of zero's after the decimal point is $$| [-26.02] + 1 |= | -26 | = 26$$
And thus our answer is $\boxed{\textbf{(D) } \text{26}}$.
~phoenixfire
~IceMatrix |
# Rijesh Augustine
Tree Climber / Photographer / Peafowl Enthusiast
A blog where I detail some of the fun projects I am working on.
## Objectives
One of my favorite problems is calculating pi based on probability. This blog post will show you how to do that with python, c#, and ocaml. Basic knowledge of computer programming, in any one of those languages, will definitely be an asset. This post can also show you how to translate from a language that you know to a new one.
## PI
pi is an irrational constant that is the ratio of a circle’s circumference to its diameter
It is also used to calculate a circle’s area.
If we overlay a circle on a square as shown below we can compare the differences in the area of the two.
So if we were to randomly choose any location on the square what would be the probability that the location would also be inside the circle. The probability is just the ratio of areas.
So the chance that we hit the inside of the circle is $\frac{\pi}{4}$.
## Making it computationally easier.
Let’s just look at $\frac{1}{4}$ of the drawing. This will make a bit more sense later on.
In this diagram, the radius of the circle is 1. The probability of being in the $\frac{1}{4}$ circle when choosing a random position inside the smaller square is still the same probability as in eq 5. This is because we have divided the area of both by 4. This diagram also shows an x and y-axis with an origin at the centre of the original circle.
Ok. So if we can generate two random numbers between 0 and 1, we can use the first random number to represent an x coordinate and the second number to represent a y coordinate. We just need some equation to govern whether the randomly chosen x and y coordinate are inside the $\frac{1}{4}$ circle.
We can use the Pythagorean theorem to help us. We know that the radius of the circle is always 1. So…
## Calculating with Python
We will be using a few premade python libraries. So the first step is to import these libraries in.
this allows us to use math functions like square root and power as well as a function to generate random numbers. We can perform those operations like this.
It is a little tedious to keep writing the words math and random, so instead, we can import all functions of these two libraries and call the functions directly.
After importing the libraries we need to generate a variable to hold the total number of iterations we will perform and the number of times those iterations succeed in being inside the $\frac{1}{4}$ circle.
Next, we can start looping through all of the iterations. The easiest way to make a for loop in python is using the range keyword. This block of code will run multiple times while changing the value of i from 0 to total_iterations.
With Python, we have to be careful about our indentations and note that there is a colon after a function, loop, or if statement.
Finally, we can then print out our answer for pi.
You can test this online at https://repl.it/ or https://www.python.org/shell/.
The full file should look like this.
## Calculating with C#
Ok, so this starts off very similar to python. Start off with our imports. In c# we use the using keyword.
Next, we construct the main body of our program. We need to make a class and the main function. This is how the compiler knows where the program starts. You need to do something similar in python when you have larger projects, but python interprets everything at runtime line by line, so smaller projects are fine.
In c# we need to define variables and the type of data associated with them. We will be using integer variables and doubles. Doubles are numbers that can hold decimal values.
We can generate random numbers using the Random class.
With for-loops, we may pass three different operations. The first may be a variable initializer, the second is a condition that will stop the loop, and finally, we have an operation that will be done at the end of each loop. Any or all of these may be blank.
Our full loop will llok like this.
And finally, print out the answer. Note we are casting to a double. Otherwise, the arithmetic will return an integer value.
You can test this online at https://repl.it/ or http://rextester.com/.
The full file should look like this. C, C++, C#, will all have very similar implementations as what is shown in the following file.
## Calculating with Ocaml
Ocaml is a functional programming language so it is quite different than the last two examples. There are no type declarations like in C# but the type is inferred based on context. Arithmetic operators must be specifically chosen for the type being used.
Basic function:
The name of this function is f and takes in one parameter x. The return of this function is the return of the function float_of_int x. Basically, you can think of this function as an alias. We don’t want to write float_of_int x everywhere. Instead, we would rather write f x. The in keyword is defining function f in the all of the code leading up to a ;;.
This function called hyp takes in two parameters a and b. You will notice that the expression on the right-hand side has *. and +. operators. These are operators used on floating point numbers. By using these the ocaml compiler knows that a and b must be floats and will issue an error if we try to pass in an integer.
The randcheck function takes in an empty parameter. This makes sure that each time the function is called it will be reevaluated. This functions passes 2 random floats to the hypotenuse function and returns a logical true and false depending on if the hypotenuse is less than or greater than 1. Note that the Random.float function takes a float value as an argument. This argument limits the maximum return value and also note that it is 1.0 (a float) and not 1 (an int).
This function is only evaluated once since no arguments are passed into it.
This function defines a reference to a variable. This way we can change the value of the variable. By default, it is set to 0.
Finally lets deal with the for loop.
the ! and := operators are how we derefence the value inside count_inside_quarter_circle and how we then put a new value in.
Finally let’s print.
You can try this online at tutorialspoint.
The full file should look like this. |
## Equation of a straight line
The general equation for a straight line in its gradient form is:
$y = mx + c$
where = the gradient of the line
c = a constant
If we let x = 0, then it is found that y = c. Therefore is equal to the y-intercept.
#### Application
Sketch the graph of the equation y = 2x + 3.
Firstly, c = 3 and therefore the y-intercept is at the point (0, 3).
To find the x-intercept, let y = 0. The x-intercept is therefore at (-3/2, 0).
Plot both of these points and join them to create the line.
### Intercept form
Another way to express a linear relation is in the form:
$ax + by = c$
where a, b, and c are all constants
This is referred to as the intercept form as it is convenient in determining the x and y intercepts.
#### Application
Sketch the graph 8x + 11y = 88.
When x = 0, y = 8. Therefore the y-intercept is at (0, 8).
When y = 0x = 11. Therefore the x-intercept is at (11, 0).
## Finding the equation of a straight line
The gradient, m, of the line can be rearranged using a general point (x, y) on the same line
$y - y_1 = m(x - x_1)$
This form is most convenient when determining the equation of a straight line.
### Given two points
When two points of a line are found, they can be substituted into the gradient formula in order to find the slope, m. After that, the re-arranged gradient formula can be used for convenience.
Find the equation of the straight line passing through the points (6, 4) and (2, -4).
The gradient must first be found:
$m = \frac{-4 - 4}{2 - 6}$
$m = 2$
This value, as well as any of the given points can be substituted into the formula:
$y - 4 = 2(x - (-4))$
$y = 2x + 12$
### Given a point, and the gradient
Using the same re-arranged formula, the equation can be quickly deduced.
Find the equation of the line that passes through the point (2,5) and has a gradient of -3.
Using $y - y_1 = m(x - x_1)$
$y - 5 = -3(x - 2)$
$y = -3x + 11$
### Given a graph
In a graph, the value of c reads as the y-intercept. m is the gradient and can be calculated using two points on the line. The equation can therefore be written in the form y = mx + c |
# A truck pulls boxes up an incline plane. The truck can exert a maximum force of 2,500 N. If the plane's incline is (3 pi )/8 and the coefficient of friction is 1/3 , what is the maximum mass that can be pulled up at one time?
Aug 5, 2017
$m = 242$ $\text{kg}$
#### Explanation:
We're asked to find the maximum mass that the truck is able to pull up the ramp, given the truck's maximum force, the ramp's angle of inclination, and the coefficient of friction.
The forces acting on the boxes are
• the gravitational acceleration (acting down the incline), equal to $m g \sin \theta$
• the friction force ${f}_{k}$ (acting down the incline because it opposes motion)
• the truck's upward pulling force
In the situation where the mass $m$ is maximum, the boxes will be in equilibrium; i.e. the net force acting on them will be $0$.
We thus have our net force equation
sumF_x = overbrace(F_"truck")^"upward"- overbrace(f_k - mgsintheta)^"downward" = 0
The expression for the frictional force ${f}_{k}$ is given by
ul(f_k = mu_kn
And since the normal force color(green)(n = mgcostheta, we can plug that in above:
$\sum {F}_{x} = {F}_{\text{truck}} - {\overbrace{{\mu}_{k} \textcolor{g r e e n}{m g \cos \theta}}}^{{f}_{k} = {\mu}_{k} \textcolor{g r e e n}{n}} - m g \sin \theta = 0$
Or
$\sum {F}_{x} = {F}_{\text{truck}} - m g \left({\mu}_{k} \cos \theta + \sin \theta\right) = 0$
Therefore,
$\underline{{F}_{\text{truck}} = m g \left({\mu}_{k} \cos \theta + \sin \theta\right)} \textcolor{w h i t e}{a a a}$ (net upward force$=$net downward force)
And if we rearrange this equation to solve for the maximum mass $m$, we have
color(red)(ulbar(|stackrel(" ")(" "m = (F_"truck")/(g(mu_kcostheta + sintheta))" ")|)
The problem gives us
• ${F}_{\text{truck}} = 2500$ $\text{N}$
• ${\mu}_{k} = \frac{1}{3}$
• $\theta = \frac{3 \pi}{8}$
• and $g = 9.81$ ${\text{m/s}}^{2}$
Plugging in these values:
m = (2500color(white)(l)"N")/((9.81color(white)(l)"m/s"^2)(1/3cos[(3pi)/8] + sin[(3pi)/8])) = color(blue)(ulbar(|stackrel(" ")(" "242color(white)(l)"kg"" ")|) |
## Topic H. Linear Formulas—Word Problems
### Objectives:
1. Solve applications problems using linear models and interpret the results.
2. Re-define the input variable if needed to make the model easier to interpret in a useful manner.
### Terminology:
We use the following instructions interchangeably:
Write a formula to express the linear relationship.
Write an equation to express the linear relationship.
Write an algebraic linear model to express the relationship.
### Overview:
To solve the applications problems in this topic, follow these steps.
1. Variables:
1. What are the names of the two variables?
2. What are the units of each variable?
2. Prediction:
1. Which variable will we predict in the main question? (That’s the output variable, i.e. y-variable.)
2. What letters will we use for each of the variables in our linear model?
3. Are there any limits on the values of either variable? If so, what? Write them in mathematical notation.
4. What are some points, that is, values of both variables for more than one point? Write these in appropriate ordered pair notation, with the x-variable first and the y-variable second.
3. Is a linear model appropriate? How do you know? (Use one of these three methods: (1) the problem says that a linear model should be used, (2) graph the data given and see that it forms a straight line, (3) the problem says that a certain increase in the input variable will always give the same increase in the output variable.)
4. Slope:
1. Find the slope.
2. Write a sentence to interpret the slope, using the units of each variable in the sentence.
5. Find the formula for the line.
6. Write a sentence to interpret the y-intercept, using the units of each variable in the sentence. If this value is out of the range of the acceptable values for the y-variable, comment on that.
7. Use the formula for the line to answer the each prediction question. Do the algebra and write a sentence, naming the variable and the units, interpreting your answer.
8. Graphing and checking your algebraic work.
1. Make a graph appropriate to answer the prediction questions.
(If you make the graph from only the information in the original problem, then you can check your work in developing the model by determining whether the results of the model are consistent with the graph. If you graph the formula you obtained, then you should check to be sure that the graph is consistent with the information given in the original problem.)
2. Use the graph to answer the prediction questions. Did you find the same answers as when you used the formula?
3. Do your answers to the prediction questions make sense? Explain something you looked at to see if they are reasonable. |
# Section 3.2 Calculus 2
Volumes by Cross-Sectioning
## 3.2 Volumes by Cross-Sectioning
### 3.2.1 Defining Volume with Integrals
• The volume of a solid defined between $$x=a$$ to $$x=b$$ with a cross-sectional area of $$A(x)$$ at each $$x$$-value is defined to be $$V=\int_a^b A(x)\,dx$$.
• Steps for solving such problems:
1. Sketch the solid along the $$x$$-axis with a typical cross-section at some $$x$$ value.
2. Find the formula for $$A(x)$$, and the minimal/maximal $$x$$ values $$a,b$$.
3. Evaluate $$V=\int_a^b A(x)\,dx$$.
• Example Show that the volume of a pyramid with a square base of sidelength $$2$$ and height $$3$$ is $$4$$ cubic units.
### 3.2.2 Circular Cross-Sections
• In the case that all cross-sections are circular, we may replace $$A(x)$$ with $$\pi[R(x)]^2$$, where $$R(x)$$ is the radius of the circular cross-section at that $$x$$ value.
• Example Prove that a cone of radius $$r$$ and height $$h$$ has volume $$V=\frac{1}{3}\pi r^2 h$$.
### Review Exercises
1. Find the volume of a solid located between $$x=-1$$ and $$x=2$$ with cross-sectional area $$A(x)=x^2+1$$ for all $$-1\leq x\leq 2$$.
2. Find the volume of a solid located between $$x=0$$ and $$x=1$$ whose cross-sections are parallelograms with base length $$b(x)=x+1$$ and height $$h(x)=x^2+1$$ for all $$0\leq x\leq 1$$.
3. Find the volume of a wedge cut from a circular cylinder with radius $$2$$, sliced out at a $$45^\circ$$ angle from the diameter of its base. (Hint: Sketch the diameter of the cylinder along the $$x$$-axis from $$-2$$ to $$2$$, and use the equation $$x^2+y^2=2^2$$. The cross-sections will be isosceles triangles.)
4. Prove that the volume of a sphere with radius $$r$$ is $$V=\frac{4}{3}\pi r^3$$. (Hint: Draw a diameter of the sphere on the $$x$$-axis from $$-r$$ to $$r$$, and use the equation $$x^2+y^2=r^2$$.)
5. Find the volume of the solid whose base is the region $$0\leq y\leq 4-x^2$$ and whose cross-sections are equilateral triangles perpendicular to the $$x$$-axis.
Solutions
### Textbook References
• University Calculus: Early Transcendentals (3rd Ed)
• 6.1 |
Categories
# Inverse Uniform Distribution | ISI MStat 2007 PSB Problem 4
This problem is an intersting application of the inverse uniform distribution family, which has infinite mean. This problem is from ISI MStat 2007. The problem is verified by simulation.
This problem is an interesting application of the inverse uniform distribution family, which has infinite mean. This problem is from ISI MStat 2007. The problem is verified by simulation.
## Problem
The unit interval (0,1) is divided into two sub-intervals by picking a point at random from inside the interval. Denoting by $Y$ and $Z$ the
lengths of the long and the shorter sub-intervals respectively show that $\frac{Y}{Z}$ does not have a finite expectation.
This is the 4th Problem ISI MStat 2008. Enjoy it.
### Prerequisites
• Distribution Function
• Distribution Function of $\frac{1}{X}$ in terms of the Distribution Function of $X$.
## Solution
$\frac{Y}{Z} + 1 = \frac{Y+Z}{Z} = \frac{1}{Z}$, where $Z$ is the shorter length of the broken stick.
So, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) – 1$.
Let’s try to find the distribution of $\frac{1}{Z}$.
Let $U$ ~ Unif $(0,1)$ whcih denotes the random uniform cut.
The shorter stick of length smaller than $x$ can be achieved if the stick is cut either before $x$ or it is cut after $1-x$.
Observe that $P( Z \leq x) = P ( U \leq x ) + P ( U \geq 1 – x) = x + 1 – (1-x) = 2x$. This answer is natural since, the total valid length is $2x$.
$P( \frac{1}{Z} \leq z) = P ( Z \geq \frac{1}{z} ) = 1 – \frac{2}{z} \Rightarrow F_{\frac{1}{Z}}(z) = 1 – \frac{2}{z}$ if $2 \leq z < \infty$.
Therefore, $f_{\frac{1}{Z}}(z) = \frac{2}{z^2}$ if $2 \leq z < \infty$.
Hence, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) – 1 = (\int_{2}^{\infty} \frac{2}{z} dz) – 1 = \infty$
## Simulation and Verification
Exercise: Prove that $F_{\frac{Y}{Z}}(x) = \frac{(x-1)}{(x+1)}$ if $1 \leq x < \infty$.
u = runif(1000,0,1)
w = 1 - u
Z = pmin(u,w)
Y = pmax(u,w)
YbyZ = Y/Z
plot(ecdf(YbyZ), xlim = c(0,50))
x = seq(0, 50, 0.01)
curve((x - 1)/(x+1), from = 0, col = "red", add = TRUE)
The Mean moves really slowly to infinity ~ logx. Hence it is really hard to show it is going to $\infty$. Also, the probability of occurrence of high value is almost 0. Hence, it really hard to show my simulation that the mean is $\infty$. But, we can show that the mean of the maximum values is really large.
v = rep(0,200)
m = NULL
for ( i in 1:200)
{
#v[i] = 100*i
u = runif(10000,0,1)
w = 1 - u
Z = pmin(u,w)
Y = pmax(u,w)
YbyZ = Y/Z
m = c(m, max(YbyZ))
}
mean(m) = 79079.43
Beware of the simulation, it can be totally counterintuitive. This is really enjoyable though.
Stay Tuned!
## By Srijit Mukherjee
I Learn. I Dream. I Enjoy. I Share.
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# Easy Time, Speed & Distance Solved QuestionAptitude Discussion
Q. A tower is 61.25 m high. A rigid body is dropped from its top and at the same instant another body is thrown up-wards from the bottom of the tower with such a velocity that they meet in the middle of the tower. The velocity of projection of the second body is:
✔ A. 24.5 m/s ✖ B. 20 m/s ✖ C. 25 m/s ✖ D. 22 m/s
Solution:
Option(A) is correct
Let the body moving down-wards take '$t$' sec to reach half the height $\left(= \dfrac{61.25}{2} \right)$.
As we know,
$s=ut+\dfrac{1}{2}at^2$
Since, body is DROPPED from the top, initial velocity $u=0$. and $s=\dfrac{61.25}{2}$
$\Rightarrow \dfrac{61.25}{2}=0+\dfrac{1}{2}\times 9.8\times t^2$
$\Rightarrow t=\dfrac{5}{2}$ sec
Again, assume that the second body is projected minimum velocity '$u$' up-wards, using the same formula,
$s=ut-\dfrac{1}{2}at^2$
$\dfrac{61.25}{2}=u\times \dfrac{5}{2}- \left[\dfrac{1}{2} \times 9.8 \times \left( \dfrac{5}{2}\right)^2 \right]$
$\Rightarrow 61.25=u\times \dfrac{5}{2}- \left[\dfrac{245}{8} \right]$
$\Rightarrow 61.25=u\times \dfrac{5}{2}- \dfrac{61.25}{2}$
$\Rightarrow u=\dfrac{49}{2}$
$= \textbf{24.5 m/sec}$
Edit: Taking note of the comments, solution has been updated.
## (9) Comment(s)
Jazzi
()
From where they got 9.8? i can't understand plz guide me
Deepali
()
When the body falls from the top or moves towards the ground then gravitational force occurs on the body as an acceleration(9.8m/s/s).
When the body thrown upward from the ground then gravitational force occurs on the body as a deceleration.(- 9.8 m/s/s )
so, by default g=9.8 m/s/s
i.e the formula is changed over here for both the situation
Vejayanantham TR
()
How come you take both takes same "t" ?
Rishi
()
A rigid body is dropped from the tower's top and at the same instant another body is thrown upwards from the bottom of the tower with such a velocity that they meet in the middle of the tower.
So both the bodies spend the same time in the travel, making 't' same for both of the bodies.
Suhail Ahmed
()
It is a Physics problem. The formula stated is derived of Netwon's Law of Gravity. And secondly it states free fall. I had forgotten the formula. Couldn't solve it.
Fara
()
i also don't understand from where 245 came? n second thing, when a body is moving upward, 9.8 should b negative...
Shahid
()
To your second query, Formula is used as $s=ut-\dfrac{1}{2}at^2$. So no need to take acceleration a negative value.
Swathik
()
How come does this 245 took place.......?
Krish
()
How is this calculation done.. can some one explain how 245 came to picture... |
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# Tutor profile: Dana H.
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Dana H.
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## Questions
### Subject:Pre-Algebra
TutorMe
Question:
Solve for X 3X= 15
Inactive
Dana H.
When solving for X you want to get X alone on one side of the equal sign. 3 is multiplied to X so use the inverse operation division (division is opposite to multiplication) to cancel out the 3 and get X alone. Divide both sides by 3. 3X\3 = 15\3 X = 5
### Subject:Basic Math
TutorMe
Question:
5 + 7=
Inactive
Dana H.
When adding it is helpful to use some objects from around the house. Gather 5 objects such as coins, teddy bears, pencils... then gather 7 of the same or different objects then add them together. XXXXX + XXXXXXX= XXXXXXXXXXXX 5+ 7 =12
### Subject:Algebra
TutorMe
Question:
Simplify then solve for X 3X+ 7 -5X -2= 11 + 2X
Inactive
Dana H.
First simplify by adding like terms. A like term is a term that has the same letter to the same power or like terms can be numbers that have no letter. In the problem 3X, -5X, 2X are like terms and can be combined. 7, -2, and 11 are like terms and can be combined. When solving for X you want to get X alone on one side of the equal side. So bring 2X over to the left side of the equal side by doing the inverse operation, subtracting 2X from both sides then combine the terms with X. 3X- 5X- 2X+ 7- 2 = 11 +2X- 2X -4X +7 -2= 11 Continue to work on getting X alone by bringing the terms with just a number no X to the right side and combine them. -4X +7 -7 -2 +2= 11 -7+2 -4X=6 Then divide by -4 to get X alone -4X/-4 =6/-4 Reduce the fraction and the answer is X = -3/2
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# CLASS-10RATIO & PROPORTION - AN IMPORTANT PROPERTY
An Important Property
a c e
If, ------ = ------- = ------- = k,
b d f a + c + e
prove that each ratio is equal to ------------
b + d + f
sum of antecedent
i.e., each ratio = --------------------
sum of consequent
a c e
Proof – Let, ------- = -------- = -------- = k.
b d f
Then, a = bk, c = dk, and e = fk
a + c + e bk + dk + fk
So, ------------ = ---------------
b + d + f b + d + f
k (b + d + f)
= --------------- = k
(b + d + f)
a + c + e
Hence, each ratio = ------------
b + d + f
x + y y + z z + x
Example.1) If, --------- = --------- = ---------, prove that each of
ax + by ay + bz az + bx
2
these ratios is equal to -------- unless x + y + z = 0
(a + b)
Sum of Antecedents
Ans.) As we know, each Ratio = --------------------
Sum of Consequent
x + y y + z z + x (x + y) + (y + z) + (z + x)
-------- = --------- = -------- = ---------------------------
ax + by ay + bz az + bx (ax + by) + (ay + bz) + (az + bx)
(2x + 2y + 2z)
= -----------------------------
(ax + ay + az) + (by + bz + bx)
2 (x + y + z)
= ---------------------------
a (x + y + z) + b (x + y + z)
2 (x + y + z)
= --------------------
(x + y + z) (a + b)
2
= --------- (proven)
(a + b)
a b c
Example.2) If ------- = -------- = --------,
(b + c) (c + a) (a + b)
Sum of Antecedents
Ans.) As we know that, each ratio = ----------------------
Sum of Consequents
a b c a + b + c
So, -------- = -------- = --------- = ------------------------
(b + c) (c + a) (a + b) (b + c) + (c + a) + (a + b)
a + b + c
= ---------------
(2a + 2b + 2c)
(a + b + c) 1
= --------------- = ------
2 (a + b + c) 2
1
= ------- [where (a + b + c) ≠ 0]
2
Now, when a + b + c = 0, we have
b + c = - a, a + b = - c, and a + c = - b
a a
so, -------- = ------ = - 1
b + c - a
b b
now, -------- = ------- = - 1
a + c - b
c c
and, -------- = ------- = - 1
a + b - c
Hence, in this case, each ratio is – 1 (proven) |
Miscellaneous
Chapter 9 Class 11 Straight Lines
Serial order wise
### Transcript
Misc 6 Find the equation of a line drawn perpendicular to the line 𝑥/4 + 𝑦/6 = 1 through the point, where it meets the y-axis. Let equation of line AB be 𝑥/4 + 𝑦/6 = 1 Above equation is of the form 𝑥/𝑎 + 𝑦/b = 1 Where a = x-intercept of line = 4 b = y-intercept of line = 6 Since line AB makes y-intercept 6 ∴ Line AB meets y-axis at point (0, 6) Let line CD be drawn perpendicular to the line AB through the point where AB meet at the y-axes So, line CD is perpendicular to the line AB & passes through the point (0,6) We need to find the equation of line CD Now, Given that line CD is perpendicular to line AB And we know that if two lines are perpendicular, the product of their slopes is equal to -1 So, slope of line CD × slope of line AB = − 1 Slope of CD = ( − 1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) Finding slope of line AB Equation of line AB is 𝑥/4 + 𝑦/6 = 1 (3𝑥 + 2𝑦 )/12 = 1 3x + 2y = 12 2y = − 3x + 12 y = ( − 3𝑥 + 12)/2 y = (( − 3)/2)x + (12/2) The above equation is of the form y = mx + c where m = slope of line ∴ Slope of line AB = m = ( − 3)/2 From (1) Slope of CD = ( − 1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) = ( − 1)/(( − 3)/2) = (−1 × 2)/(−3) = 2/3 We know that equation of line passing through (x1, y1) & having slope m is (y − y1) = m(x − x1) Equation of line CD passing through point (0, 6) & having slope 2/3 is (y − 6) = 2/3 (x − 0) y − 6 = 2/3 x 3(y − 6) = 2x 3y − 18 = 2x 3y – 2x – 18 = 0 − 3y + 2x + 18 = 0 2x − 3y + 18 = 0 Which is required equation
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. |
# How do you find all the zeros of x^3 + 2x^2 + 5x +1?
Mar 12, 2016
Use Cardano's method...
#### Explanation:
$f \left(x\right) = {x}^{3} + 2 {x}^{2} + 5 x + 1$
To cut down on the number of fractions we need to work with, multiply by ${3}^{3} = 27$ first...
$0 = 27 f \left(x\right)$
$= 27 {x}^{3} + 54 {x}^{2} + 135 x + 27$
$= {\left(3 x + 2\right)}^{3} + 33 \left(3 x + 2\right) - 47$
Substitute $t = 3 x + 2$ ...
$= {t}^{3} + 33 t - 47$
Using Cardano's method, substitute $t = u + v$ ...
$= {u}^{3} + {v}^{3} + 3 \left(u v + 11\right) \left(u + v\right) - 47$
Add the constraint $v = - \frac{11}{u}$ to eliminate the $\left(u + v\right)$ term ...
$= {u}^{3} - \left({11}^{3} / {u}^{3}\right) - 47$
$= {u}^{3} - \frac{1331}{u} ^ 3 - 47$
Multiply through by ${u}^{3}$ to get this quadratic in ${u}^{3}$ ...
${\left({u}^{3}\right)}^{2} - 47 \left({u}^{3}\right) - 1331 = 0$
Solve using the quadratic formula to get:
${u}^{3} = \frac{47 \pm \sqrt{{47}^{2} + 4 \cdot 1331}}{2}$
$= \frac{47 \pm \sqrt{7533}}{2}$
$= \frac{47 \pm 9 \sqrt{93}}{2}$
Since the derivation was symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to derive the Real root:
${t}_{1} = \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$
and the Complex roots:
${t}_{2} = \omega \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$
${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$
where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$
Then $x = \frac{t - 2}{3}$ hence zeros of the original function:
x_1 = 1/3(-2+root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2) )
${x}_{2} = \frac{1}{3} \left(- 2 + \omega \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}\right)$
${x}_{3} = \frac{1}{3} \left(- 2 + {\omega}^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}\right)$ |
# What is the limit of ( x^3 - 8 )/ (x-2) as x approaches 2?
Jun 17, 2015
The limit is $12.$
#### Explanation:
Notice how you have a difference of two cubes.
${x}^{3} - {y}^{3} = {x}^{3} + {x}^{2} y + x {y}^{2} - {x}^{2} y - x {y}^{2} - {y}^{3}$
$= \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$
In our expression, we have $y = 2$:
$\frac{{\left(x\right)}^{3} - {\left(2\right)}^{3}}{x - 2} = \frac{\cancel{\left(x - 2\right)} \left({x}^{2} + 2 x + 4\right)}{\cancel{x - 2}} = {x}^{2} + 2 x + 4$
At this point, the limit can be evaluated:
$\implies \textcolor{b l u e}{{\lim}_{x \to 2} \frac{{x}^{3} - 8}{x - 2}}$
$= {\lim}_{x \to 2} {x}^{2} + 2 x + 4$
$= {\left(2\right)}^{2} + 2 \left(2\right) + 4$
$= 4 + 4 + 4$
$= \textcolor{b l u e}{12}$
Sep 24, 2017
$12.$
#### Explanation:
We can have another soln., if we use the following useful Standard Limit :
${\lim}_{x \to a} \frac{{x}^{n} - {a}^{n}}{x - a} = n \cdot {a}^{n - 1} \ldots \ldots \ldots \ldots . . \left(\star\right) .$
Accordingly,
${\lim}_{x \to 2} \frac{{x}^{3} - 8}{x - 2} ,$
$= {\lim}_{x \to 2} \frac{{x}^{3} - {2}^{3}}{x - 2} ,$
$= 3 \cdot {2}^{3 - 1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left[\because , \left(\star\right)\right] ,$
$= 3 \cdot {2}^{2} ,$
$= 12 ,$ as Respected EEt-AP has already obtained! |
## Basic College Mathematics (9th Edition)
Published by Pearson
# Chapter 7 - Measurement - 7.5 Metric - U.S. Measurement Conversions and Temperature - 7.5 Exercises: 2
#### Answer
a.) $\frac{0.95 liter}{1 quart}$, $\frac{1 quart}{0.95 liter}$ b.) qt; denominator
#### Work Step by Step
A.) As we know 1 quart = 0.95 liter The two unit fractions using the information for converting between quart and liter To convert quart to liter Unit for the answer (liter) is in numerator; unit being changed (quart) is in denominator so it will divide out. The unit fraction is (0.95 liter)/(1 quart). To convert liter to quart Unit for the answer (quart) is in numerator; unit being changed (liter) is in denominator so it will divide out. The unit fraction is (1 quart)/(0.95 liter). b.) 3 qt to liters To convert quart to liter Unit for the answer (liter) is in numerator; unit being changed (quart) is in denominator so it will divide out. The unit fraction is (0.95 liter)/(1 quart) When you will multiply 3 qt to the unit fraction qt unit will be divide out. In the unit fraction the unit being changed will be written in the denominator.
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