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# How to Solve Systems of Linear Inequalities?
Linear inequalities are expressions in which two linear expressions are compared using the inequality symbols. In this step-by-step guide, you will learn about solving systems of linear inequalities.
The solution to a system of a linear inequality is the region where the graphs of all linear inequalities in the system overlap.
## Astep-by-step guide to solving systems of linear inequalities
The system of linear inequalities is a set of equations of linear inequality containing the same variables. Several methods of solving systems of linear equations translate to the system of linear inequalities. However, solving a system of linear inequalities is somewhat different from linear equations because the signs of inequality prevent us from solving by the substitution or elimination method. Perhaps the best way to solve systems of linear inequalities is by graphing the inequalities.
To solve a system of inequalities, graph each linear inequality in the system on the same $$x-y$$ axis by following the steps below:
1. Solve the inequality for $$y$$.
2. Treat the inequality as a linear equation and graph the line as either a solid line or a dashed line depending on the inequality sign. If the inequality sign does not contain an equals sign $$(< or >)$$ then draw the line as a dashed line. If the inequality sign does have an equals sign $$(≤ or ≥)$$ then draw the line as a solid line.
3. Shade the region that satisfies the inequality.
4. Repeat steps $$1 – 3$$ for each inequality.
5. The solution set will be the overlapped region of all the inequalities.
### Solving Systems of Linear Inequalities – Example 1:
Solve the following system of inequalities.
$$\begin{cases}x\:-\:5y\ge \:6\\ \:3x\:+\:2y>1\end{cases}$$
Solution:
First, isolate the variable $$y$$ to the left in each inequality:
$$x -5y≥ 6$$
$$x≥6 + 5y$$
$$5y≤ x- 6$$
$$y≤0.2 x -1.2$$
And:
$$3x+ 2y> 1$$
$$2y>1-3x$$
$$y> 0.5-1.5x$$
Now, graph $$y≤ 0.2x-1.2$$ and $$y > 0.5 -1.5x$$ using a solid line and a broken one, respectively.
The solution of the system of inequality is the darker shaded area which is the overlap of the two individual solution regions.
## Exercises for Solving Systems of Linear Inequalities
### Determine the solution to the following system of inequalities.
• $$\color{blue}{\begin{cases}5x-2y\le 10 \\ \:3x+2y>\:6\end{cases}}$$
• $$\color{blue}{\begin{cases}-2x-y<\:-1 \\ \:4x+\:2y\:\le -6\end{cases}}$$
• $$\color{blue}{\begin{cases}5x-2y\le 10 \\ \:3x+2y>\:6\end{cases}}$$
• $$\color{blue}{\begin{cases}-2x-y<\:-1 \\ \:4x+\:2y\:\le -6\end{cases}}$$
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Sequence and Series | kullabs.com
Notes, Exercises, Videos, Tests and Things to Remember on Sequence and Series
Please scroll down to get to the study materials.
## Note on Sequence and Series
• Note
• Things to remember
• Videos
• Exercise
• Quiz
### Sequence
A Sequence is a list of things (usually numbers) that are in order. let us consider the following of numbers.
(i) 1, 4, 7, 10,.....
(ii) 20, 18, 16, 14,....
(iii) 1, 3, 9, 27, 81,....
(iv) 1, 2, 3, 4, ...
We observe that each term after the first term
(i) is formed by adding 3 to the preceding term;
(ii) is formed by subtracting 2 from the preceding term;
(iii) is formed by multiplying the preceding term by 3; each term in
(iv) is formed by squaring the natural numbers 1, 2, 3, 4,.....
In all the above case, we see that set of number follow a certain rule and we can easily say what number will come next to given number. thus, the numbers come in succession in accordance with a certain rule or low. A succession of numbers formed and arranged in a definite order according to a certain definite rule is called a sequence. the successive number in a sequence are called its terms.
### Series
A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.
eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.
The successive numbers forming the series are called the terms of the series and the successive terms are denoted by t1, t2, t3,....., t,which denotes the 1st, 2nd, 3rd, ...... nth term respectively. The nth term, tn, of a series, is called its general term. Thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.
### Progression
A sequence of number is said to be a progression if the difference or ratio between its two successive terms is constant throughout the whole sequence. An example of progression is as follows.
(i) 1, 3, 5, 7,..... (ii) 1, 3, 9, 27,.....
In (i), the difference between two successive terms is equal to 2.
In (ii), the ratio of two successive terms is equal to 3.
### Types of Progression
Progression is divided into following two types.
(i) Arithmetic progression
(ii) Geometric progression
#### Arithmetic progression or Sequence
A sequence is called an arithmetic progression if the difference between its two successive terms is constant throughout the whole sequence. An arithmetic progression can be denoted by A.P. The constant number obtained by subtracting succeeding term from its preceding term is called the common difference.
For example:-
(i) 1, 3, 5, 7, 9,....
(ii) 15, 12, 9, 6,......
From (i), we find that
second term - first term = 3 -1 = 2,
third term - second term = 5 - 3 = 2,
fourth term - third term = 7 - 5 = 2 and so on.
From (ii), we find that
second term - first term = 12-15 = -3,
third term - second term = 9 - 12 = -3,
fourth term - third term = 6 - 9 = -3 and so on.
Hence, the common difference 'd' is calculated by
d = succeeding term - proceeding term = tn - tn-1
Here, we find that the difference between two successive terms, in both sequences, are same or constant. So, such sequence is called arithmetic progressions. The C.D. of the two progressions are 2 and -3 respectively. Thus, arithmetic progressions is a series in which the successive terms increase or decrease by the common difference.
##### General term or nth term of an A.P.
To find the nth term of an A.P.
Let, t1 be the first term, n be the number of terms and 'd' the common difference of an A.P. respectively. Then,
t= a = a + (1-1)d
t= a + d = a + (2-1)d
t= a + 2d = a + (3-1)d
t= a + 3d = a + (4-1)d
In general, tn = a + (n-1)d
Formula: If tdenotes the nth term, of the arithmetic progression whose first term, common term and number of terms are a, d and n respectively.
With this term, arithmetic sequence and series can be written as:
Arithmetic sequence: a, a+d, a+2d, a+3d, ............
Arithmetic series: a+ (a+d) + (a+2d) + (a+3d), ..........
#### Arithmetic Mean
The terms between the arithmetic progression are known as arithmetic mean. Such as the three numbers 2, 4, 6 are in arithmetic progression with the common difference d = 2, then 4 is the arithmetic mean between 2 and 6.
For example:
Let a, b,c are in arithmetic progression
b-a = c-b
or, b+b = a+c
or, 2b = a+c
or, b = $$\frac{a+c}{2}$$
Hence the arithmetic mean between a and c is ($$\frac{a+c}{2}$$)
n Arithmetic Means between two numbers a and b
Let m1, m2, m3, .........mn be the arithmetic means between the given term a and b. Then, a, m1, m2, m3, .........mn, b are in A.P.
Here, numbers of arithmetic means = n
So, numbers of terms of A.P. = n+2
It means,
b = (n+2)th term of AP
or, b = a + (n+2-1)d, where d is common difference
or, b =a + (n+1)d
or, (n+1)d = b-a
∴ d= $$\frac{b-a}{n+1}$$
Now, m1 = a+d = a + $$\frac{b-a}{n+1}$$
m2 = a + 2d = a + $$\frac{2(b-a)}{n+1}$$
m3 = a + $$\frac{3(b-a)}{n+1}$$
.............................................
mn = a + $$\frac{n(b-a)}{n+1}$$
#### Sum of n terms of series in A.P.
Let us consider an arithmetic series
a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l
Here, the first term = a,
first term = a,
common difference = d,
number of terms= n,
last term (tn) = l
the term before last term = l-d
if the sum of n terms is denoted by Sn, then
Sn = a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l .... (i)
Writing term in the reverse order,
Sn = l + (l-d) + (l-2d) + ...... + (a+3d) + (a+2d) + (a+d) + a .... (ii)
Adding the corresponding terms of (i) and (ii)
$$\frac{S_n \;= \;a \;+\; (a+d)\; +\; (a+2d)\; +\; (a+3d)\; + \;......\; + \;(l-2d) \;+\; (l-d)\; + \;l\\S_n\;=\;l\; +\;(l-d)\;+\;(l-2d)\;+\;......\;+\;(a+3d)\;+\;(a+2d)\;+\;(a+d)\;+a\:}{2S_n\;= \;(a+l) \;+ \;(a+l)\; + \;(a+l)\; +\; ............ \;+\; (a+l)\; + \;(a+l)\; +\; (a+l)}$$
= n times (a+l)
= n (a+l)
= $$\frac{n}{2}$$(a+l)
But, the last term l = a + (n-1)d
So, Sn = $$\frac{n}{2}$$(a+l) = $$\frac{n}{2}$$[a+a+(n-1)d] = $$\frac{n}{2}$$[2a+(n-1)d]
∴ Sn = $$\frac{n}{2}$$[2a+(n-1)d]
Thus, if d is unknown, Sn = $$\frac{n}{2}$$(a+l)
And, if l is unknown, Sn = $$\frac{n}{2}$$[2a+(n-1)d]
1. Sum of first n natural numbers
the numbers 1, 2, 3, 4, ......, n are called the first n natural numbers.
Here, first term (a) = 1
Common difference (d) = 2-1 = 1
Number of terms (n) = n
If Sn denotes the sum of these first n natural numbers, then
Sn = $$\frac{n}{2}$$[2a+(n-1)d] = $$\frac{n}{2}$$[2.1+(n-1).1] = $$\frac{n}{2}$$[2+n-1] = $$\frac{n}{2}$$(n+1)
2. Sum of first n odd numbers
1, 3, 5, 7, ......., (2n-1) are the first n odd numbers.
Here, first term (a) = 1
Common difference (d) = 3-1 = 2
Number of terms (n) = n
If Sn denotes the sum of these first n odd numbers, then
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
3. Sum of first n even numbers
2, 4, 6, 8, ......., 2n are the first n even numbers.
Here, first term (a) = 2
Common difference (d) = 4-2 = 2
Number of terms (n) = n
If Sn denotes the sum of these first n even numbers, then
Sn = $$\frac{n}{2}$$[2a+(n-1)d] = $$\frac{n}{2}$$[2.2+(n-1).2] = $$\frac{n}{2}$$(4+2n-2) = $$\frac{n}{2}$$(2n-2) = n(n+1)
##### Geometric Progression or Sequence
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly,10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.
Examples of a geometric sequence are powers rk of a fixed number r, such as 2and 3k. The general form of a geometric sequence is
a, ar, ar2, ar3, ............arn
where r ≠ 0 is the common ratio and a is a scale factor equal to the sequence's start value.
General Term or nth term of G.P.
We use the following notations for terms and expression involved in a geometrical progression:
The first term = a the nth term = toor b
The number of terms = n Common ratio = r.
The expression arn-1 gives us the nth term or the last term of the geometric progression whose first term, common ratio and a number of terms are a, r and n respectively.
∴tn = arn-1
With the help of this general term, geometric sequence and series can be written in the following ways:
Geometric Sequence: a, ar, ar2, ar$$^3$$, .....
Geometric series: a + ar + ar2 + ar$$^3$$ + ........
#### Geometric Mean
If the three numbers are in G.P., then the middle term is called the geometric mean of the other two terms. In other words, the geometric mean of two non-zero numbers is defined as the square root of their product.
Let a, G, b be three numbers in G. P., then the common ratio is the same i.e.
$$\frac{G}{a}$$ =$$\frac{b}{G}$$
or, G2 = ab
or, G =$$\sqrt{a}{b}$$
Hence, the geometric mean of two numbers a and b is the square root of their product i.e. $$\sqrt{a}{b}$$.
So, the geometric mean between two number 2 and 8 is G =$$\sqrt ab$$ = $$\sqrt2*8$$ = $$\sqrt16$$ = 4.When
When any number of quantities are in G. P., all the terms in between the first and last terms are called the geometric means between these two quantities.
Here, Gn = arn = a $$\begin{pmatrix}b\\a\\ \end{pmatrix}$$$$\frac{n}{n + 1}$$
#### Relation between arithmetic mean and geometric mean
"Arithmetic mean (A. M) is always greater than Geometric mean (G. M.) between two position real unequal numbers".
Let us consider two numbers 2 and 8
Here, AM between 2 and 8 =$$\frac{2 + 8}{2}$$ = 5
GM between 2 and 8 = $$\sqrt 2 * 8$$) = 4
∴ AM > GM.
The sum of n terms of a series in G. P.
Let us consider geometric series a + ar + ar2 + ar$$^3$$ + .......+ arn -3+ arni2+ arn-1
Here, first = a common ratio = r number of terms = n last term (l) = arn-1
∴ Sn = $$\frac{lr - a}{r - 1}$$
If the number of terms is odd, we take the middle term as aand the common ratio as r. If the number of terms is even, we take $$\frac{a}{r}$$ and ar as the middle terms and r2 as the common ratio.
A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.
eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.
The successive numbers forming the series are called the terms of the series and the successive terms are denoted by$$t_1, t_2, t_3,....., t_n,$$ where$$t_1, t-2, t-3, ........ tn$$ denote the$$1^st, 2^nd, 3^rd ,...... .n^{nt}$$ term respectively. The n^th term, t_n, of a series, is called its general term. thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.
The terms between the first term and last term of an A.P are called arithmetic mean.,
.
### Very Short Questions
$$\sum_{n=2}^5 (n^2- 3)$$
Putting n = 2, 3, 4, 5
= (22 - 3) + (32 - 3) + (42 - 3) + (52 - 3)
= (4 - 3) + (9 - 3) + (16 - 3) + (25 - 3)
= 1 + 6 +13 +22
= 42 Ans
un + 1 = 1 - $$\frac {1}{u_n}$$ and u1 = 3, u2 and u3 = ?
If n = 1,un + 1 = 1 - $$\frac {1}{u_n}$$
u1 + 1 =1 - $$\frac {1}{u_1}$$
or, u2 = 1 - $$\frac 13$$ = $$\frac {3 - 1}{3}$$ = $$\frac 23$$
If n = 2, un + 1 =1 - $$\frac {1}{u_n}$$
u2 + 1 =1 - $$\frac {1}{u_2}$$
or, u3 = 1 - $$\cfrac{1}{\cfrac{2}{3}}$$ = 1 - $$\frac 32$$ = $$\frac {2 - 3}{2}$$ = -$$\frac 12$$
∴ u2 = $$\frac 23$$ and u3 = -$$\frac 12$$ Ans
Sequence: A sequence is a set of numbers or quantities, which are formed according to some governed laws.
i.e. 1, 4, 16 ............................ and 2, 3, 4, 5...................
Series: The sum of the term of a sequence is called a series.
i.e. 1 + 4 + 16 + ....................... and 2 + 3 + 4 + 5 + ......................
Here,
$$\sum_{n=4}^7 (3n- 2)$$
where: n = 4, 5, 6, 7
$$\sum_{n=4}^7 (3n- 2)$$
= (3×4-2) + (3×5-2) +(3×6-2) +(3×7-2)
= (12-2) + (15-2) + (18-2) + (21-2)
= 10 + 13 + 16 + 19
= 58 Ans
$$\sum_{k=3}^7 (k^2 + 1)$$
= (32 + 1) +(42 + 1) +(52 + 1) +(62 + 1) +(72 + 1)
= (9 + 1) +(16 + 1) +(25 + 1) +(36 + 1) +(49 + 1)
= 10 + 17 + 26 + 37 + 50
= 140 Ans
Here,
First term (a) = 1
Second term (b) = 4
Common Difference (d) = b - a = 4 - 1 = 3
Last term (l) = 34
We know,
last term (l) = a + (n - 1)d
or, 34 = 1 + (n - 1) 3
or, 34 = 1 + 3n - 3
or, 3n - 2 = 34
or, 3n = 34 + 2
or, n = $$\frac {36}{3}$$
∴ n = 12
Again,
Sn = $$\frac n2$$ [a + l]
Sn = $$\frac{12}{2}$$ [1 + 34] = 6× 35 = 210
∴ Sn = 210 Ans
Here,
first term (a) = 2
second term (b) = 7
common difference (d) = b - a = 7 - 2 = 5
number of items (n) = 16
sum (S16) = ?
We know that,
Sn = $$\frac n2$$ [2a + (n - 1) d]
S16 = $$\frac {16}{2}$$ [2 × 2 + (16 - 1) 5] = 8 [4 + 15× 5] = 8 [4 + 75] = 8 × 79 = 632 Ans
Here,
first term (a) =3$$\frac12$$ = $$\frac 72$$
second terrn (b) = 1
common difference (d) = b - a = 1 - $$\frac 72$$ = $$\frac {2 - 7}{2}$$ = -$$\frac 52$$
no. of items (n) = ?
last term (l) =-21$$\frac 12$$ = -$$\frac {43}{2}$$
We know that,
l = a + (n - 1) d
or,-$$\frac {43}{2}$$ = $$\frac 72$$ + (n - 1) (-$$\frac 52$$)
or, - 43 = 7 - 5n + 5
or, 5n = 12 + 43
or, n = $$\frac {55}{5}$$ = 11
∴-21$$\frac 12$$ is the 11th term. Ans
Here,
first term (a) = 20
common difference (d) = -3
numbers of term (n) = 7
seventh term (t7) = ?
We know that,
tn= a + (n - 1) d
t7 = 20 + (7 - 1) (-3) = 20 + 6 (-3) = 20 - 18 = 2 Ans
Here,
first term (a) = 2
second term (b) = 4
common difference (d) = b - a = 4 - 2 = 2
no. of terms (n) = 20
sum (S20) = ?
We know that,
Sn = $$\frac n2$$ [2a + (n - 1) d]
S20 = $$\frac {20}{2}$$ [2× 2 + (20 - 1)× 2] = 10 [4 + 38] = 10× 42 = 420 Ans
Here,
first term (a) = 1
second term (b) = 3
common difference (d) = b - a = 3 - 1 = 2
number of items (n) = 20
twenty term (t20) = ?
We know,
tn= a + (n - 1) d
t20 = 1 + (20 - 1) 2
t20 = 1 + 19× 2 = 1 + 38 = 39 Ans
Here,
first term (a) = 25
second term (b) = $$\frac {45}{2}$$
common difference (d) = b - a = $$\frac {45}{2}$$ - 25 = $$\frac {45 - 50}{2}$$ = -$$\frac 52$$
last term (l) = - 15
number of terms (n) = ?
We know that,
l = a + (n - 1) d
or, - 15 = 25 + (n - 1) - $$\frac 52$$
or, - 15 - 25 = -$$\frac 52$$n + $$\frac 52$$
or, - 40 - $$\frac 52$$ = -$$\frac 52$$n
or - $$\frac {80 - 5}{2}$$ = -$$\frac {5n}{2}$$
or, -$$\frac {85}{2}$$× -$$\frac {2}{5}$$ = n
∴ n = 17
∴ The number of term = 17. Ans
Here,
first term (a) = 3
last term (l) = -9
number of terms (n) = 4 [including a and b]
common difference (d) = $$\frac {b - a}{n - 1}$$ = $$\frac {-9 - 3}{4 - 1}$$ = -$$\frac {12}{3}$$ = - 4
x = a + d = 3 - 4 = - 1
y = a + 2d = 3 + 2× -4 = 3 - 8 = - 5
∴ x = -1 and y = -5 Ans
Here,
first term (a) = 2
second term (b) = - 9
common difference (d) = b - a = -9 - 2 = -11
last term (l) = -130
number of terms (n) = ?
We know,
l = a + (n - 1) d
or, -130 = 2 + (n - 1) (-11)
or, -130 - 2 = - 11n + 11
or, -132 - 11 = -11n
or, -11n = -143
or, n = $$\frac {-143}{-11}$$
∴ n = 13
Again, Sn = $$\frac n2$$ (a + 1) = $$\frac {13}{2}$$ (2 - 130) = 6.5× -128
∴ Sn = -832 Ans
Here,
common difference (d) = -3
number of terms (n) = 7
sum (S7) = 0
first term (a) = ?
We know,
Sn = $$\frac n2$$ [2a + (n - 1) d]
or, 0 = $$\frac 72$$ [2a + (7 - 1) (-3)]
or, 0× $$\frac 27$$ = 2a + 6× (-3)
or, 0 = 2a - 18
or, 2a = 18
or, a = $$\frac {18}{2}$$ = 9
∴ first term (a) = 9 Ans
Here,
first term (a) = $$\frac 14$$
second term (b) = $$\frac 12$$
common ratio (r) = $$\frac ba$$ =$$\cfrac{\frac{1}{2}}{\cfrac{1}{4}}$$ = $$\frac 12$$× $$\frac 41$$ = 2
last term (l) = 128
number of terms (n) = ?
We know that,
l = arn-1
or, 128 = $$\frac 14$$ (2)n-1
or, 27× 22 = 2n-1
or, 2n-1 = 29
or, n -1 = 9
or, n = 9 + 1
∴ n = 10
∴ The number of term = 10 Ans
Here,
first term (a) = 16
last term (l) = 36
We know that,
geometric mean (G.M.) =$$\sqrt {ab}$$
G.M. =$$\sqrt{({16}×{36})}$$
= $$\sqrt {({4^2}×{6^2})}$$
= $$\sqrt {({4×6})^2}$$
= $$\sqrt {({24})^2}$$
∴ Geometric mean = 24 Ans
Here,
first term (a) = $$\frac 13$$
second term (b) = 1
common ratio (r) = $$\frac ba$$ = $$\cfrac {1}{\cfrac 13}$$ = 3number of terms (n) = 8
number of terms (n) = 8
eighth term (t8) = ?
Using formula,
tn = arn-1
t8 = ($$\frac 13$$) (3)8-1 = $$\frac 13$$× 3× 36 = 729 Ans
Here,
first term (a) = 3
second term (b) = 1
common ratio (r) = $$\frac ba$$ = -$$\frac 13$$
last term (l) = -$$\frac {1}{81}$$
number of terms (n) = ?
We know that,
l = arn-1
or, -$$\frac {1}{81}$$ = 3 (-$$\frac 13$$)n-1
or, - $$\frac {1}{3^4× 3}$$ = (-$$\frac 13$$)n-1
or, (-$$\frac 13$$)5 =(-$$\frac 13$$)n-1
or, n - 1 = 5
or, n = 5 + 1
∴ n = 6
∴ The number of term = 6 Ans
Here,
first term (a) = $$\frac 23$$
second term (b) = -$$\frac 16$$
common ratio (r) =$$\frac ba$$ = $$\cfrac{\frac{-1}{6}}{\cfrac{2}{3}}$$= -$$\frac 16$$ × $$\frac 32$$ = -$$\frac 14$$
number of items (n) = 5
5th term (t5) = ?
We know that,
tn= arn-1
or, t5 = $$\frac 23$$(-$$\frac 14$$)5-1
or, t5 = $$\frac 23$$×(-$$\frac 14$$)4
or, t5 = $$\frac 23$$× $$\frac {1}{16}$$×$$\frac {1}{16}$$ = $$\frac {1}{384}$$
∴ fifth term = $$\frac {1}{384}$$ Ans
Here,
first term (a) = $$\frac 23$$
second term (b) = x
third term (c) = $$\frac 32$$
We know that,
G.M. = $$\sqrt {ab}$$
x = $$\sqrt {(\frac 23) × (\frac 32)}$$
x = 1
r = $$\frac ba$$ =$$\cfrac{1}{\cfrac{2}{3}}$$ = $$\frac 32$$
y = ar = $$\frac 32$$× $$\frac 32$$ = $$\frac 94$$
∴ x = 1 and y = $$\frac 94$$ Ans
Here,
first term (a) = 12
second term (b) = 4 common ratio (r) = $$\frac ba$$ = $$\frac 42$$ = 2
number of terms (n) = 10
sum of the ten terms (S10) = ?
We know that,
Sn = $$\frac {a(r^n - 1)}{r - 1}$$ = $$\frac {2(2^10 - 1)}{2 - 1}$$ =$$\frac {2(1024 - 1)}{1}$$ = 2× 1023 = 2046
∴ The sum of tenth terms = 2046 Ans
Here,
first term (a) = 81
second term (b) = 27
common ratio (r) = $$\frac ba$$ = $$\frac {27}{81}$$ = $$\frac 13$$
number of terms (n) = 5
sum of five terms (S5) = ?
We know,
Sn = $$\frac {a(r^n - 1)}{r - 1}$$
S5 = $$\frac {81[(\frac 13)^5 - 1]}{\frac 13 - 1}$$
=$$\frac {81[(\frac {1}{243}) - 1]}{\frac {1 - 3}{3}}$$
=$$\frac {81 × (\frac {1 - 243}{243})}{\frac {-2}{3}}$$
= 81× -$$\frac {242}{243}$$× -$$\frac 32$$
= 121
∴ The sum of five terms = 121 Ans
Let,
first term = a
common ratio = r
From equation,
S6 = 9 S3
or,$$\frac {a(r^6 - 1)}{r - 1}$$ = 9 [$$\frac {a(r^3 - 1)}{r - 1}$$] [$$\because$$Sn =$$\frac {a(r^n - 1)}{r - 1}$$]
or, $$\frac {a[(r^3)^2 - (1)^2]}{r - 1}$$×$$\frac {r - 1}{a(r^3 - 1)}$$ = 9
or, $$\frac{(r^3 + 1)(r^3 - 1)}{r^3 - 1}$$ = 9
or, r3 = 9 - 1
or, r3 = 8
or, r3 = 23
∴ r = 2
∴ The common ratio = 2 Ans
Let: the three terms of an G.P. are $$\frac ar$$, a and ar
From Question,
$$\frac ar$$× a× r = 729
or, a3 = 93
∴ a = 9
∴ The second term = 9 Ans
Here,
first term (a) = 6
third term (b) = x
A.M. = 30
A.M. = $$\frac {a + b}{2}$$
or, 30 = $$\frac {6 + x}{2}$$
or, 60 - 6 = x
∴ x = 54
G.M. = $$\sqrt {ab}$$
G.M. = $$\sqrt {6 × 54}$$
G.M. = $$\sqrt {2^3 × 3^3 ×3^3}$$
G.M. = 18
∴ The G.M. = 18 Ans
Here,
first term (a) = $$\frac 19$$
last term (l) = 9
number of Geometric mean (n) = 3
total number of term (n) = 5
3 Geometric mean (m1), (m2), (m3) = ?
l = arn-1
or, 9 = $$\frac 19$$r5-1
or, 81 = r4
or, 34 = r4
∴ r = 3
m1 = ar = $$\frac 19$$× 3 = $$\frac 13$$
m2 = ar2 = $$\frac 19$$× 32 = $$\frac 19$$× 9 = 1
m3 = ar3 = $$\frac 19$$× 33 = $$\frac 19$$× 27 = 3
∴ m1 = $$\frac 13$$, m2 = 1, m3 = 3 Ans
If M and G be the arithmetic mean and geometric mean between two positive numbers a and b, then
m = $$\frac {a + b}{2}$$ .................... (1)
G = $$\sqrt {ab}$$ ................................(2)
Equation (1) - (2) we get
M - G = $$\frac {a + b}{2}$$ - $$\sqrt {ab}$$
= $$\frac {a + b - 2\sqrt{ab}}{2}$$
= $$\frac {(\sqrt a)^2 - 2\sqrt a ⋅ \sqrt b + (\sqrt b)^2}{2}$$
= $$\frac 12$$ ($$\sqrt a$$ - $$\sqrt b$$)2
$$\sqrt a$$ - $$\sqrt b$$ gives positive value
∴ M - G = $$\frac {(\sqrt a - \sqrt b)^2}{2}$$≥ 0
M - G≥ 0 or, M ≥G
∴ A.M.≥ G.M. Ans
Let:
Sn = 1, 3, 5, ..................... n
First term (a) = 1
Second term (b) = 3
Common difference (d) = b - a = 3 - 1 =2
number of items = n
We know,
Sn = $$\frac n2$$[2a + (n - 1) d]
or, Sn = $$\frac n2$$[ 2× 1 + (n - 1) 2]
or, Sn = $$\frac n2$$[2 + 2n - 2]
or, Sn = $$\frac n2$$× 2n
∴ Sn = n2Ans
Here,
first term (a) = 1
common ratio (r) = 2
sum (Sn) = 255
We know,S
Sn = $$\frac{a(r^n - 1)}{r - 1}$$
or, Sn =$$\frac{1(2^n - 1)}{2 - 1}$$
or,$$\frac{2^n - 1}{1}$$ = 255
or, 2n = 255 + 1
or, 2n = 256
or, 2n = 28
∴ n = 8
∴ The no. of terms = 8 Ans
Let:
first term (a) = 3
last term (b) = 243
no. of geometric means (n) = 3
common ratio (r) = ?
3 G.M are X1, X2, X3 = ?
r = ($$\frac ba$$)$$\frac {1}{n+1}$$= ($$\frac {243}{3}$$)$$\frac {1}{3+1}$$= (81)$$\frac 14$$= (3)$$\frac {4×1}{4}$$ = 3
X1 = ar = 3× 3 = 9
X2 = ar2 =3× (3)2 = 27
x3 = ar3 =3× (3)3 = 81
∴ 9, 27, 81 Ans
Let:
first term (a) = 27
last term (l) = $$\frac {32}{9}$$
number of terms (n) = 4
common ratio (r) = ($$\frac ba$$)$$\frac {1}{n+1}$$
= ($$\cfrac{\frac{32}{9}}{27}$$)$$\frac {1}{4+1}$$
=( $$\frac {32}{9×27}$$)$$\frac 15$$
= ($$\frac 23$$)$$\frac {5×1}{5}$$
=$$\frac 23$$
p = ar = 27×$$\frac 23$$ = 18
q = ar2=27×$$\frac 23$$×$$\frac 23$$ = 12
r = ar3 =27×$$\frac 23$$×$$\frac 23$$×$$\frac 23$$ = 8
s = ar4 =27×$$\frac 23$$×$$\frac 23$$×$$\frac 23$$×$$\frac 23$$ = $$\frac {16}{3}$$
∴ p = 18, q = 12, r = 8 and s = $$\frac {16}{3}$$ Ans
Here,
first term (a) = 2
second term (b) = -6
common ratio (r) = $$\frac ba$$ = $$\frac {-6}{2}$$ = -3
number of terms (n) = 6
We know that,
Sn = $$\frac {a(1 - r^n)}{1 - r}$$
Sn = $$\frac {2(1 - (-3)^6)}{1 - (-3)}$$
Sn = $$\frac {2(1 -729)}{1 + 3}$$
Sn = $$\frac {2× (-729))}{4}$$
∴Sn= -364.5Ans
Here,
first term (a) = 3
second term (b) = 6
common ratio (r) = $$\frac ba$$ =$$\frac 63$$ = 2
last term (l) = 1535
sum of the series (Sn) = ?
Using formula,
Sn = $$\frac{lr - a}{r - 1}$$ = $$\frac {1535 × 2 - 3}{2 - 1}$$ = $$\frac {3070 - 3}{1}$$ = 3067
∴ Sum (Sn) = 3067 Ans
0%
• ### If f = {(1,2) , (2,3) , (4,5)} , find f (^-1) and the domain and range of f (^-1) .
f (^-1) = {( 2,1) , (1,2) , (3,4)}
Domain of f (^-1) = {(5 ,3, 5)}
Range of f (^-1) = {1,3,4}
f (^-1) = {( 1,1) , (3,3) , (5,4)}
Domain of f (^-1) = {(2 ,3, 3)}
Range of f (^-1) = {1,2,3}
f (^-1) = {( 2,1) , (3,2) , (5,4)}
Domain of f (^-1) = {(2 ,3, 5)}
Range of f (^-1) = {1,2,4}
f (^-1) = {( 2,3) , (1,2) , (5,4)}
Domain of f (^-1) = {5,3,2)}
Range of f (^-1) = {4,2,1}
• ### If f = {(1,4) , (2,4) , (3,5)} and g = {(4,2) , (5,2) }, then show the function gof in arrow diagram and find it ordered pair form.
{(3,2) , (2,2) , (0,1)}
{(3,2) , (2,2) , (0,2)}
{(1,2) , (2,2) , (3,2)}
{(2,2) , (3,2) , (4,2)}
• ### If f = {(b,q) , (a,p) , (c,r)} and gof = {(a,1) , (c,3) , (b,2)} , find the function g in terms of ordered pairs .
{(1,1) , (q,2) , (q,3)}
{(p,q) , (1,2) , (r,3)}
{(p,1) , (q,q) , (r,r)}
{(p,1) , (q,2) , (r,3)}
• ### If f is an identity function and g= {(1,3) , (2,4) , (3,5)} then write the composite function gof in ordered pair form .
{(1) , (2) ,(3,5)}
{(1,2) , (2,4) ,(4,5)}
{(1,3) , (2,4) ,(3,5)}
{(1,2) , (2,4) ,(4,5)}
• ### If gof = {(2,9) , (3,13) , (4,17)} and f (x) = 2x+1 , find the function 'g' in order pair form.
g = {(5,7) , (6,13) , (16,17)}
g = {(5,9) , (7,13) , (9,17)}
g = {(5,8) , (7,14) , (0,17)}
g = {(5,5) , (7,3) , (2,17)}
• ### If f = {(1,5) , (2,4) , (3,6)} and fog = {(5,6) , (4,5) , (6,4)} then find g (^-1) .
g (^-1) = {(1,2) ,(1,4) (2,6)}
g (^-1) = {(2,5) ,(4,4) (4,6)}
g (^-1) = {(5,3) ,(4,4) (,6,2)}
g (^-1) = {(3,5) ,(1,4) (2,6)}
• ### If f = {(2,3) (4,5) , (5,6)} and g = {(4,3) , (2,5) , (5,8)} then find gof (^-1) in ordered pair form .
{(3,5) , (5,3) , (6,8)}
{(3,5) , (5,3) , (5,3)}
{(5,5) , (3,3) , (8,8)}
{(1,5) , (2,5) , (,3,8)}
x2 , -2
2x , -4
4x, -2
3x, -3
80
8
4
2
5,5
-5
4,5
-4,-5
2x - 3
x-2
23 - x
3x - 2
• ### If f (x) = 5 - (frac{6}{x}) then what is the value of x at f (x) = -f (^-1) (-1) ?
-1
-5
1
(frac{-1}{-5})
2
35
30
12
1
-2
16
2
• ### If f (x) = 2x + 3 and fog (x) = 2x + 5 , find (gof) (^-1) .
(frac{4-x}{2})
(frac{x-4}{2})
(frac{x-2}{2})
(frac{x-4}{4})
• ## You scored /15
Forum Time Replies Report
##### Asmi
Find the value of k if (k 2), (4k-6), (3k-6) are first term of an AP.
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Basic Arithmetic : Addition and Subtraction with Money
Example Questions
← Previous 1
Example Question #1 : Addition And Subtraction With Money
Jane bought a pencil for , a skirt for , a coat for , and a headband for . How much did Jane pay in total for all the items she bought?
Explanation:
Add up all the individual items.
Example Question #1 : Addition And Subtraction With Money
Billy earned for shoveling his driveway. He wants to buy a new toy and a candy bar, and then add the rest to his piggy bank. The new toy costs and the candy bar costs How much money will Billy have left over to add to his piggy bank?
Explanation:
.
After buying the candy bar he then had left for his piggy bank,
.
Example Question #1 : Addition And Subtraction With Money
John recently bought the following school supplies: a calculator for , a pencil bag for , a mechanical pencil for , and a set of pens for
How much did John spend in total on school supplies?
Explanation:
In order to solve this problem, you must add all of the costs of school supplies together. This can be done through the following equation:
Therefore, the total amount that John spent on school supplies is
Example Question #1 : Money And Time
Jenny has a new job at the mall. Her first paycheck was . She bought a pair of shoes for and a new dress for on the weekend. How much money from her first paycheck does Jenny have left on Monday?
Explanation:
You must subtract the cost of the two purchases from the overall paycheck amount.
Begin by subtracting the shoe purchase from the total paycheck amount:
Then, subtract the cost of the dress:
is the amount of money Jenny has left from her paycheck on Monday.
Example Question #1 : Addition And Subtraction With Money
Approximately how much will John need to pay if a pair of shoes cost $42.64, a pair of pants costs$37.99, and a shirt costs $21.34. Possible Answers:$120
$110$80
$100$90
$100 Explanation: The key word, "approximately," indicates that we must round each of these values. Let's round to the nearest tens place because each of our answer choices is rounded to the tens place. Shoes cost$42.64.
$42.64 rounded (down) to the nearest tens place is$40.
Pants cost $37.99.$37.99 rounded (up) to the nearest tens place is $40. A shirt costs$21.34.
$21.34 rounded (down) to the nearest tens place is$20.
Let's add up our estimated prices.
$40 +$40 + $20 =$100.
Example Question #1 : Addition And Subtraction With Money
Approximately how much will Sophie need to buy a shirt at $12.99, a loaf of bread at$3.99, and a television at $43.76? Possible Answers:$100
$80$61
$60 Correct answer:$61
Explanation:
Since we're looking for an approximate value, we need to round our numbers up. Adding the three items gives a total of $60.74, so at least$61 is needed - thus $61 is the correct answer. Example Question #13 : Money And Time Laura is going bowling with her friends this weekend and she is using some of the money that she has saved up from allowance and her birthday. She will have to pay a flat fee of$8 to rent her shoes and $10.50 per game she bowls. If she has planned to bowl 4 games and she is only taking$50 with her to the bowling alley will she have enough money? If yes, how much money will she have when she leaves the bowling alley?
No
Yes,
Yes,
Yes, None.
Yes, None.
Explanation:
The flat fee for the shoes will only be paid once since the shoes will be used for all four games. The cost per game will be paid each time one game finishes and a new one is started.
To calculate her cost add the fee for the shoes and the fee for each game played. That would be $8 (shoes) +$10.50 (game 1) + $10.50 (game 2) +$10.50 (game 3) + $10.50 (game 4) =$50 (total cost).
Laura had \$50 with her so there was enough money and nothing left over when she left the bowling alley.
Example Question #1 : Addition And Subtraction With Money
Danielle buys four items costing , , , and . What is the estimated total of Danielle's items?
Between and
Between and
Between and
Between and
Between and
Explanation:
To solve this problem, one should round either to the tenths place, or to the nearest dollar. can be rounded to , to , to , and to . When one adds up the estimated costs, it yields about .
Example Question #151 : Basic Arithmetic
Sam buys four items, costing , , and . What is the estimated cost of Sam's items?
Between and dollars
Between and dollars
Between and dollars
Between and dollars
Between and dollars
Explanation:
In this problem, the costs of the items can be rounded to either the nearest dollar, or the nearest ten cents.
and can be rounded to and , respectively.
can be rounded to , and can be rounded to .
When the new, estimated costs of the items are added, the sum is , which is between and dollars.
Example Question #152 : Basic Arithmetic
Approximately how much does he pay in total?
More than
Between and
Between and
Less than
Between and
Between and
Explanation:
First, add up all of the values:
Then, decide which of the answer choices approximates this value. Our answer is between and .
Another way to solve this problem is to round the four values to their nearest whole numbers and then sum those rounded values:
← Previous 1
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# Forming and Solving Algebraic Expressions: Class 7 Math
## Algebra
- In algebra, letters are used to represent unknowns and are used together with numbers.
In algebra, always remember the following:
n means 1 x n
2p means 2 x p
3y2 means 3 x y2 which is same as 3 x y x y
4a2d means 4 x a2 x d
2(2a + 5b) = 2 x 2a + 2 x 5b = 4a + 10b
### Simplifying Algebraic Expressions
##### Example 1
Work out: 3(2a + 4b) - 2(a + 2b)
= (3 x 2a + 3 x 4b) - (2 x a + 2 x 2b)
= 6a + 12b - 2a + 4b
Putting like terms together
= 6a - 2a + 12b + 4b
= 4a + 16b
### Forming and Simplifying Algebraic Expressions
##### Example 2
Albert, Wilson and Charles shared some apples. Albert got 5 more apples than Charles who got twice as much as Wilson. How many apples did they get all together?
From the information above, we gather that:
Wilson has a apples. Hence:
Charles has (2 x a = 2a) apples
Albert = (2a + 5) apples
Hence
= a + 2a + 2a + 5
= 5a + 5
### Solving Algebraic Expressions
When solving a quadratic equation, you find the value of the unknown in the equation.
##### Example 3
Find the value of y in 3(y + 2) = 24
We start by opening the brackets:
3(y + 2) is the same as (3 x y) + (3 x 2)
3y + 6 = 24
3y = 24 - 6
3y = 18
Divide both sides by 3:
3y / 3 = 18 / 3
y = 6
### Forming and Solving Algebraic Expressions
Now that you have learnt how to solve algebraic expressions. Let's try solving real-life mathematical problems.
In this section, you will come across words such as more than, greater than, less than, younger than, shorter than, twice as much and similar comparative words.
##### Example 4
Mwilu is 6 years older than Kyalo. If Kyalo is 12 years old, how old is Mwilu.
Let Mwilu's age be b:
Then b = 6 years + Kyalo's age.
b = 6 + 12
b = 18
Mwilu is 18 years age
### Inequalities
Let us say an unknown number p is between 8 and 14 but is neither 8 or 14. Can you tell which number p is likely to be?
p can be 9, 10, 11, 12, 13
From this, we can say that p is greater than 8 and less than 14.
This can be mathematically expressed as p > 8 and p < 14. This are called inequalities.
#### Simplifying Inequalities
##### Example 5
Simplify the following inequality: 3p - 2 > 19
3p - 2 > 19
Collecting like terms together:
3p > 19 + 2
3p > 21
Divide both sides by 3:
3p / 3 > 21 / 3
21 / 3 = 7
p > 7
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How To Do Substitution??In math class we are doing these substitution questions and I don't understand how to do them at all. They are all like this: The sum of two numbers is 8. Three times the...
How To Do Substitution??
In math class we are doing these substitution questions and I don't understand how to do them at all. They are all like this:
The sum of two numbers is 8. Three times the first plus four times the second is 29. Find the numbers.
If you could explain how to do it in steps that would really help! Thanks!! :)
Posted on
Since these are two pretty simple equations, you could also do the following:
For (1) a + b = 8 and (2) 3a + 4b = 29
solve (1) for either variable (in this case they both have a coefficient of one so both are just as easy). So subtract b from each side and get a = 8 - b.
Now plug in 8 - b for a in (2). You will get 3 ( 8 - b ) + 4b = 29. Now simplify and solve for b.
3 ( 8 - b ) + 4b = 24 - 3b + 4b = 24 + b = 29 => subtract 24 from both sides => b = 5
Now you can substitue b=5 in (1) and get a + 5 = 8 => a =3.
This approach is really best when you are given two simple equations without exponents and a lot of dividing. There are so many ways to do substitution that as long as you are doing legitimate math "moves" and being very careful in simplifying, you will get the same answers. Practice really is the best way to learn it. The link below should help a good deal.
Posted on
well you use the key words that the expression says to help you for example
the word sum means addition or the word of means to multiply like that. I put some other words down below to help you a little more
Posted on
The sum of two numbers is 8. Three times the first plus four times the second is 29. Find the numbers.
First we must parse the sentence and write it algebraically.
The sum of two numbers is 8: a + b = 8
Three times the first (3a) plus four times the second (+ 4b) is 29 (=29) so:
3a + 4b = 29.
a + b = 8
Given our two equations, we can solve for a and b. In general, you need as many unique equations as you have variables.
We know a in terms of b: a = 8 - b
Plug this into the second equation:
3(8-b) + 4b = 29 Solve for b.
We know b in terms of a: b = 8 - a
Plug into second equation:
3a + 4(8-a) = 29 Solve for a.
Posted on
Substitution and elimation are the techniques employed to used while solving two (or more) simultaneous equations with two (or more) unknwons.Once we detrmined the value of a variable, we substitute the value in any one of the equations to detrmine the other unknown.
We assume two unknown numbers x and y.There two conditions given which is obeyed by these numbers:
By first condition the sum of the numbers, x+y =8 .......... (1).
By 2nd condition, 3times first or 3x and 4times the 2nd or 4y add up 3x+4y and this is 29. So the the second codition is:
3x+4y = 29...............(2).
The two given conditone are the reation between two numbers x and y in the form of above 2equations in 2 variables.
Substitution is a technique, by which we reduce the two equations in two variables to one equation in one variable.
We can use from the first equation x+y = 8, x = 8-y in the 2nd equation and replace x in 2nd equation by 8-y, as 3(8-y)+4y = 29 and solve for y. You get now 24-3y+4y =29. Or 24+y = 29. Or y =5.
Using this already detrmined value of y=5 in any of the two equations, x+y=8 or 3x+4y = 29, we get x+5= 8 . Or x = 8-5 =3. Here also, when we determined one of the value of the two variables, we use the technique of substitution of that velue of the varible to find the value of the other variable.
Posted on
In the type of question there are two variables to be found out - the two numbers. To find value of two numbers using algebraic equation we need a pair of equation. These two equation can be formed by the two conditions given in the example. That is:
1. The sum of two numbers is 8.
2. Three times the first plus four times the second is 29.
Let us assume that the two numbers are x and y.
Then, using the first given condition we form the equation:
x + y = 8 .... (1)
And using the second given condition we form the equation:
3x + 4y = 29 .... (2)
We can sole these equation for x an y as follows.
Multiplying equation (1) by 3 we get:
3x + 3 y = 24 ... (3)
subtracting equation 3 from equation 2 we get:
3x - 3x + 4y - 3y = 29 -24
Simplifying the above equation we get:
y = 5.
Substituting this value of y in equation (1), that is using the number 5 instead of y, we get:
x + 5 = 8
Therefore x = 8 - 5 = 3
Therefore the two numbers are 3 and 5.
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# Tips And Tricks And Shortcuts For Clocks And Calendars
## Clocks and Calendar Tips and Tricks and Shortcuts
Here we have some of the tips and tricks and shortcuts on problems based on clocks and calendar. The tricks and shortcuts mentioned below will help you solve problems based on them easily, and efficiently.
## Type 1: Tips and Tricks for Clocks
### Question 1.
What is the angle between the minute hand the hour hand when the time is 4:30.
Options:
A. 350
B. 100
C. 200
D. None of the above
#### Solution:
Tip: It is easy to calculate the angle between the minute and the hour hand by using a simple formula,
Angle = (X*30)-((Y*11)/2)
Multiplying hours with minutes, we get = 4 x 30 = 120
Applying the formula, we get (Yx11)/2
= 30 x 11/ 2
= 165
When we subtract the two values, we get,
= 165 – 130
= 350
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## Type 2: Tips ad Tricks for Clocks
### Question 2.
Calculate the time between 6 and 7 o’clock when the hands of a clock are in the same straight line but are not together?
Options:
A. 65.45 min past 6
B. 60 minutes past 6
C. 50 minutes past 6
D. Cannot be determined
#### Solution:
Tip: You can use a short formula to calculate the time when the angle is given
Angle = (minutes) -30 (hours)
Using the above formula, we get
180 = (minutes) -30 (hours)
180 = (minutes) – 30 (6)
180 + 180 = minutes
Minutes = 2 x 360/ 11
= 65.45
## Type 1: Calendars Tips and Tricks
### Question 1.
Find the day for a given date ,What was the day on 26th May 2006?
Options:
A. Monday
B. Friday
C. Wednesday
D. Saturday
#### Solution:
26 May 2006 = (2005 years + time period from 1/1/2006 to 26/5/2006)
To calculate number of odd days till the year 2000, we require
Number of odd days in 1600 years = 0
Number of odd days in 400 years = 0
In the next step, for calculating odd days in the next five years,
5 years = (4 ordinary years + 1 leap year)
= 4 +2
= 6 odd days
Now, we have to calculate the number of odd days from 1st January 2006 to 26th May 2006.
January (31 days) + February (28 days because 2006 is not a leap year) + March (31 days) + April (30 days) + May 26 days = 146 days
Total number of odd days in 146 days = (146/7) = 20 weeks + 1 odd day
Total number of odd days in the entire period = 0 (1600 years) + 0 (400 years) + 5 (5 years) + 0 (time from 1/1/2006 to 2/2/2006) = 5 odd days
According to the table, on 26th May 2006, the day was Friday.
>
DaysSundayMondayTuesdayWednesdayThursdayFridaySaturday
Number of odd days0123456
## Type 2: Calendar Tips and Shortcuts
### Question 2.
Find out a day when some other day is given ,It was Friday on 7th December 2007. What was the day on 7th December 2006?
Options:
A. Tuesday
B. Monday
C. Friday
D. Thursday
#### Solution:
2006 was not a leap year
Hence, the number of odd days is 1
Now 7th December 2007 will be 1 day beyond the day on 7th December 2006 due to one odd day.
Since, 7th December 2007 was a Friday, hence, 7th December 2006 was a Thursday.
## Type 3: Tips and Tricks for Calendar
### Question 3.
Which year after 2005 will have the same calendar as of 2005?
Options:
A. 2011
B. 2022
C. 2015
D. 2054
#### Solution:
The given year is 2005 which is not a leap year
We add 11 years to the given year and get (2005 + 11) = 2016 which is a leap year
Also add 6 years to the given year (2005 + 6) = 2011
Hence, the calendar for 2005 will be same as for the year 2011.
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## Is the product of two integers always positive?
Answer: The product of two integers can either be positive, negative, or zero. To carry out the multiplication of two integers: Multiply their signs and get the resultant sign.
## Why is the product of two integers an integer?
y+x=x+y=x−(−y′) is the difference of two naturals and therefore an integer. xy=x(−y′)=−(xy′) is the negative of the product of two naturals so is an integer. x−y=x+y′ the sum of two naturals and thus an integer.
Is the product of two negative integers is positive?
Answer: Yes the product of two negative integer is always positive.
### Why is the product of 2 negative numbers always positive?
The fact that the product of two negatives is a positive is therefore related to the fact that the inverse of the inverse of a positive number is that positive number back again.
### What is the product of two integers?
Step by Step Explanation: We know that integers are closed with respect to multiplication. This means that on multiplying two integers we get the result as an integer. Thus, the product of two integers is always an integer.
What is the product of 2 negative integers?
positive integer
The product of two negative integers is a positive integer. Example: (-2)×(-2) = +4.
## Why is the product of two numbers positive?
When you multiply two negative numbers or two positive numbers then the product is always positive. 3 times 4 equals 12. Since there is one positive and one negative number, the product is negative 12. Now we have two negative numbers, so the result is positive.
## What is the product of two negative integers?
Are integers positive and negative numbers?
Whole numbers, figures that do not have fractions or decimals, are also called integers. They can have one of two values: positive or negative. Positive integers have values greater than zero. Negative integers have values less than zero.
### What is the product of a positive and negative numbers?
Rule 1: The product of a negative integer and a positive integer is a negative integer. Rule 2: The product of two positive integers or two negative integers is a positive integer. That means if you multiply two OF the same sign numbers, the product is always positive.
### What do you call the product of a positive integer?
factorial, in mathematics, the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point.
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# TIME AND WORK PROBLEMS TYPE 6 SHORTCUT TRICK TO SOLVE IN 10 SECONDS
Here is Time and work problems shortcut type 6.
This time, we are posting Time and work problems type #6 as a part of the series. As said earlier, there are approximately more than 20 types of time and work problems from which there are some questions that can be asked in SSC, IBPS or other competitive exams.
The time and work problem shortcut trick given below is type 6. You can click on SITEMAP given on top of site and on sitemap page under quantitative aptitude section, to learn all those 5 types of time and work questions and All OTHER TRICKS.
See TIME AND WORK PROBLEMS TYPE 5 Here.
So here is a fast math shortcut trick to solving time and work problems type #6.
Suppose you are given a question like the one given below:
Q: A can do a piece of work in 15 days but if B helps him, he can finish the work in 10 days. Determine in how many days, B can finish that work working alone?
Shortcut method for solving this question is given below:
Given
A=15 days
And
A+B=10 days
Step 1.
Take LCM of 15 and 10 by shortcut of LCM described here. The LCM will be 30 which is total work.
Step 2. Divide this LCM i.e. 30 by 15 and 10 respectively. The values comes out are A’s and A+B’s one day work capacity. Write it as
A= 15 (30/15=2)
So A=15 (2)
And
A+B= 10 (30/10=3)
So A+B=10 (3)
The values 2 and 3 are A and A+B’s one day work.
Step 3. Now A=2 and A+B=3 (equating to values given in brackets)
We get B=1
So B will finish that work in 30/1=30 days Ans. (because B=1)
If it is asked that in how many days, will A finish that work, then
30/2=15 (because A=2)
When solving in paper, don’t write as long as I do in this post as I was doing it just to explain the trick completely. If you practice these questions with this trick then you can easily solve them in 10 seconds.
Stay in touch for more time and work problems and other types of tricks.
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## What you need to know
Things to remember:
– There are 1000ml in 1L.
We need to remember that:
There are 1000 millilitres in 1 litre
Writing these out can be time consuming and take up more space, so we have a shorthand way of writing them:
Litres — L
Millilitres — ml
What is 7L in ml?
We know that 1L is the same as 1000ml, and that 7L is just 7 lots of 1L, so we need 7 lots of 1000.
$$7\times1000=7000$$
7L = 7000ml
So really, we just need to multiply the litres by 1000 to turn them into millilitres.
We can draw a little diagram to help us remember.
Looking at this, how do you think we might turn millilitres into litres? What is the opposite of multiplication? Division! So, we can go backwards by dividing by 1000.
What is 3700ml in L?
To go from millilitres to litres we divide by 1000.
$$3700\div1000=3.7$$
3700ml = 3.7L
## Example Questions
#### Question 1: What is 4.5L in millilitres?
$$4.5\times1000=4500$$
4.5L = 4500ml
#### Question 2: What is 1300 millilitres in L?
$$1300\div1000=1.3$$
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# 2009 AMC 12B Problems/Problem 16
## Problem
Trapezoid $ABCD$ has $AD||BC$, $BD = 1$, $\angle DBA = 23^{\circ}$, and $\angle BDC = 46^{\circ}$. The ratio $BC: AD$ is $9: 5$. What is $CD$?
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$
## Solutions
### Solution 1
Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$. Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$. Because $\overline {AD} \parallel \overline {BC}$, it follows that the triangles $BCE$ and $ADE$ are similar. Therefore $$\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,$$ so $CD = \boxed{\frac 45}.$
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# Factors of 441—with division and prime factorization
## Factors of 441
Factors of 441 are the integers that can be equally divided by 441.
The factors with a negative sign of 441 are negative factors. Do you know that 441 is the sum of the cubes of the first six natural numbers, which is also the square of 21? We will explore the factors of 441 as well as its prime factors, as well as its factors in pairs, in this chapter.
Factors of 441: 1, 3, 7, 9, 21, 49, 63, 147 and 441.
Negative Factors of 441: -1, -3, -7, -9, -21, -49, -63, -147 and -441.
Prime Factorization of 441: 3 × 3 × 7 × 7
## What are Factors of 441?
Factors of 441 are all the numbers that multiply to give 441 as their product.
## What are Factors of -441?
When we multiply two numbers, and we get -441 as a product, that numbers are the factors of -441.
Here is the explanation:
-1 × 441 = -441
1 × -441 = -441
-3 × 147 = -441
3 × -147 = -441
-7 × 63 = -441
7 × -63 = -441
-9 × 49 = -441
9 × -49 = -441
-21 × 21 = -441
21 × -21 = -441
-49 × 9 = -441
49 × -9 = -441
-63 × 7 = -441
63 × -7 = -441
-147 × 3 = -441
147 × -3 = -441
-441 × 1 = -441
441 × -1 = -441
We can get -441 by having (-1, 441), (-3, 147), (-7, 63), (-9, 49), (-21, 21), (-49, 9), (-63, 7), (-147, 3), and (-441, 1) as a pair of factors. Similarly, we can also get -441 by having(1, -441), (3, -147), (7, -63), (9, -49), (21, -21), (49, -9), (63, -7), (147, -3), and (441, -1) as a pair of factors.
## How to Calculate the Factors of 441?
Here are the steps to calculating 35 factors:
Step 1:
Determine the number to be multiplied. The factoring number is 441.
Step 2:
Calculate the product of two numbers.
Step 3:
441 = 3 × 147, 441 = 7 × 63, 441 = 9 × 49 and 441= 21 × 21.
Therefore, the factors of 441 are 1, 3, 7, 9, 21, 49, 63, 147, and 441. Using illustrations and interactive examples, explore the factors of 441:
### More Factors:
• Factors of 37: The factors of 37 are 1 and 37.
• Factors of 15: The factors of 15 are 1, 3, 5, and 15.
• Factors of 45: The factors of 45 are 1, 3, 5, 9, 15, and 45.
• Factors of 40: The factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40
• Factors of 25: The factors of 25 are 1, 5 and 25
## Factors of 441 by Prime Factorization
You can find the prime factorization factors of 35 by following these steps:
Step 1:
Determine the pair of factors which, when multiplied, give 441
Factoring 441 can be done by multiplying 21 and 21.
Step 2:
Because 21 isn’t a prime number, 3×7 is factorable.
Since 3 and 7 are both prime numbers, they cannot be factored further.
Step 3: T
herefore, 441 can be written as 441 = 21 × 21 = 3 × 7 × 3 × 7
Thus, 441 = 32 × 72 is the prime factorization of 441.
## Factors of 441 in Pairs
The pair factors of the number 441 are two whole numbers that multiply each other. There could be positive or negative pair factors, but not fractions or decimals.
• Positive factors of 441:(1, 441), (3, 147), (7, 63), (9, 49), (21, 21), (49, 9), (63, 7), (147, 3), (441, 1)
• Negative factors of 441: (- 1, -441), (-3, -147 ), (-7, -63), (-9, -49), (-21, -21), (-49, -9), (-63, -7), (-147, -3), (-441, -1)
### Tips and Tricks:
A number with a factor of 1 is the smallest. Therefore, it is a factor of 441.
Because all the digits in 441 add up to 9, 3 is a factor of 441 as well. Hence, 3 is one of its factors.
21 is a factor of 441 since it is divisible by 3 and 7.
### Challenging Questions
• John has \$441. He has to distribute it among his 7 friends. If 5 of his friends get an equal amount of money and the rest 2 get the half of the amount which 5 others got. How much money will be left with John?
• David has to travel from Town A to Town B. The distance between both towns is 441 miles. He drives at an average speed of 49 mph. He takes 10 pitstops of 20 minutes each. How long will he take to reach Town B?
## Solved Examples
1. Example 1: Clarke’s school has 441 students. If there are 21 classrooms in the entire school. Help him calculate how many students can sit in each classroom?
Solution:
Clarke’s school has 441 students. There are 21 classrooms in the entire school.
To calculate the number of students in each classroom Clarke has to divide 441 by 21, (441 ÷ 21 = 21 ). He found that 21 students can sit in each classroom.
2. Example 2: Williamson went to a cinema hall having 441 chairs. If there were 9 rows in the hall, help Williamson calculate how many seats are there in each row?
Solution:
There were 441 chairs in the cinema hall. Williamson saw there were 9 rows in the hall.
To calculate the number of seats in each row he will divide 441 by 9 as given below. (441 ÷ 9 = 49 ). There will be 49 seats in each row.
## FAQs
### What are the Factors of 441?
Factors of 441 = 1, 3, 7, 9, 21, 49, 63, 147, and 441.
### Is 441 a Perfect Square?
By finding the prime factorization of 441, we can express it as 21 × 21. It is therefore a perfect square.
### What is the Prime Factorization of 441 using Exponents?
Prime factorization of 441 by exponents is given by 441 = 32 × 72
### What are the Common Factors of 441 and 567?
Factors of 441 = 1, 3, 7, 9, 21, 49, 63, 147, and 441.
Factors of 567 = 1, 3, 7, 9, 21, 27, 63, 81, 189, and 567.
Thus, the common factors of 441 and 567 are 1, 3, 7, 9, 21, and 63.
### What are the Prime Factors of 441?
There are two distinct prime factors in 441: 3 and 7.
### What is the Product of the Prime Factors of 441?
The number 441 is equal to 3 × 3 × 7 × 7. 3 and 7 are distinct prime factors. Hence, the product is 21.
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Need Help?
# Kagome - The development of a pattern
## Introduction
We use the same language to describe uniform tilings and polyhedra: Platonic if all the regular polygons are the same, Archimedean if there's a mixture, and a tiling - like the kagome pattern, 3.6.3.6 - can be thought of as an infinite polyhedron. What controls the size of the polyhedron is the angular defect, d, the difference from 360°, at each vertex. They total 720°, so, if there are v vertices, vd = 720°.
## From 3.6.3.6 to 3.5.3.5
We start with 3.6.3.6 with each '3' in blue and each '6' divided into 6 yellow '3s'. If we remove one yellow triangle from each hexagon and the necessary blue triangles in a patch of the right size, we have the net for a solid whose angular defect is 60°, and number of vertices therefore 720°/60° = 12:
This is the icosahedron. It has a feature we shall find throughout: a blue triangle at the centre of each face. What were straight lines in the tiling are polygons in the solid. If we inscribe the solid in a sphere, such polygons are circumscribed by great circles. We shall call these polygons 'equatorial' sections of the solid. We show one such polygon, a 10-gon, in green:
There is one for each pair of opposite vertices, so 12/2 = 6 in all.
The yellow triangles form pentagonal pyramids with their apices at the vertices of the icosahedron. If we cut these off, we can replace them with pentagons and we have the solid 3.5.3.5:
There are 20 triangles - one for each face of the icosahedron, and 12 pentagons - one for each vertex of the icosahedron. We can work out the number of vertices in two ways.
There are 5 for each of the 12 pentagons, but two pentagons share a vertex, total 12 x 5/2 = 30.
There are 3 for each of the 20 triangles, but two triangles share a vertex, total 20 x 3/2 = 30.
The angle defect is therefore 720°/30 = 24°.
## From 3.5.3.5 to 3.4.3.4
We go back to our tiling and remove a second yellow triangle from each hexagon, and the necessary blue triangles to form the net of a solid which has an angle defect of 120°, and therefore 720°/120° = 6 vertices. The result is an octahedron:
The equatorial sections are now hexagons. There is one for each pair of opposite faces, 8/2 = 4 in all. We replace the yellow pyramids with squares and we have the solid 3.4.3.4:
There are 8 triangles and 6 squares. Calculating as before, we have v = 8 x 3/2 or 6 x 4/2 = 12, and an angle defect at each vertex of 720°/12 = 60°.
## From 3.5.3.5 to 3.4.3.4
Back we go and remove yet a third yellow triangle. The angular defect is now 180° and the solid therefore has 720°/180° = 4 vertices. It is a tetrahedron:
The equatorial section is now a square. There is one for each pair of opposite edges, therefore 6/2 = 3 in all. The yellow pyramids are tetrahedra. Replacing each with a triangle, we have 3.3.3.3, the octahedron:
We know the angle defect for this solid, 120°.
## Conclusion
In this exercise we have taken a sequence of numerical strings, where the numbers represent the polygons which surround each vertex in a uniform tiling or solid, and seen what it means in terms of the geometry.
Paul Stephenson, Magic Mathworks Travelling Circus
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# Before Beginning
## Four Fours
Make numbers using four fours
• $$0=4 + 4 - 4 -4$$
• $$1=\frac{44}{44}$$
• $$2-=\frac 4 4 + \frac 4 4$$
• $$3 = \frac{4 \times 4 -4}{4} = \frac{4+4+4}{4}$$
## An Analytic Problem
Arrange 10 soldiers in 5 rows with 4 in each row.
Solution
# Chapter 4.1
Chapter name: Multiplication of Algebraic Expressions
#### Topics
• Multiplication Law
• Exponential Law
• Multiplication of Signs
• Multiplication of Monomial and Polynomial
• Exercises
## Laws of Multiplication
• Cumulative law: $$a \times b = b \times a \space , e.g., 2 \times 3 = 3 \times 2$$
• Associative law: $$(a \times b) \times c = a \times (b \times c), e.g., (2 \times 3) \times 4 = 2 \times (3 \times 4)$$
• Exponential law: $$a \times a \times a = a^3, a^2 \times a^4 = a^6$$
• $$a^m \times a^n = a^{m+n}, (a^m)^n = a^{mn}$$
• Distributive law: 2(a+b) = 2a + 2b
• Caution: $$2^3 \ne 2 \times 3 = 6$$, but $$2^3 = 2 \times 2 \times 2$$
## Sign Laws
• $$(+1) \times (+1) = +1$$
• $$(+1) \times (-1) = -1$$
• $$(-1) \times (+1) = -1$$
• $$(-1) \times (-1) = +1$$
• $$+ve \rightarrow friend$$
• $$-ve \rightarrow foe$$
• $$(+1) \times (+1) \rightarrow$$ friend of friend
• $$(+1) \times (-1) \rightarrow$$ friend of foe
## Problems
Multiply
1. $$5a^2x^2$$ with $$3ax^5y$$
2. $$2x+3y$$ with $$5xy$$
3. $$x^2+2xy+y^2$$ with $$x+y$$
4. $$x^2+xy+y^2$$ with $$x^2-xy+y^2$$
5. $$y^2-y+1$$ with $$1+y+y^2$$
6. $$2^{-3}x^5y^3$$ with $$2^3 x^{-4}y^{-1}$$
1. $$15a^3x^7y$$
1. $$xy^2$$
## Chapter 4.2: Division
Divide the first expression by the second one
1. $$30a^4x^3, -6a^2x$$
2. $$36x^4y^3+9x^5y^2, 9xy$$
3. $$6x^2+x-2, 2x-1$$
4. $$6y^2+3x^2-11xy, 3x-2y$$
5. $$x^4y^4-1, x^2y^2+1$$
6. $$a^5+11a-12, a^2-2a+3$$
1. $$-5a^2x^2$$
1. $$4x^3y^2+x^4y$$
1. $$3x+2$$
1. $$x-3y$$
1. $$x^2y^2-1$$
1. $$a^3+2a^2+a-4$$
## Div
Questions
1. $$30a^4x^3, -6a^2x$$
2.$$36x^4y^3+9x^5y^2, 9xy$$
# 4.3: Simplification
## Rule of PEMDAS/BEDMAS/BODMAS
Notation varies by countries
• $$P/B \rightarrow$$ ()
• $$E/O \rightarrow$$ Exponent/Order $$\rightarrow 2^3$$
• MD $$\rightarrow$$ Multiplication/Division
• AS $$\rightarrow$$ Addition/Subtraction
• BO $$\ne$$ Bracket OFF! O is order/exponent
• $$3 \div \frac 1 2=?$$ $$3 \div 1\div 2=?$$
• $$6 \times 3 \div3=?$$
• $$(2+3)^2 + 6 \times 3 \div3 + 4-3=?$$
## Simplify
Questions
• $$1. 6-2\{5-(8-3)+(5+2)\}$$
• $$2. 7-2[-6+3\{-5+2(4-3)\}]$$
• $$3. 3x+(4y-z)-{a-b-(2c-4a)-5a}$$
• $$4. [8b-3\{2a-3(2b+5)-5(b-3)\}]-3b$$
## Simplify (cntd.)
$$A=x^2-xy+y^2$$
$$B = x^2+xy+y^2$$
$$C = x^4+x^2y^2+y^4$$
1. Find $$A \times B$$
1. Determine $$BC \div B^2-C$$
# Formulae
MEMORIZE FEEL
• $$(a+b)^2 = a^2+2ab+b^2$$
• $$(a-b)^2 = a^2-2ab+b^2$$
• $$a^2-b^2=(a+b)(a-b)$$
# Corollary
• $$a^2+b^2= ?$$
• $$a^2+b^2= ?$$ (in another form)
• $$(a+b)^2 = ?$$ (in terms of) $$(a-b)^2$$
• $$(a-b)^2 = ?$$ (in terms of) $$(a+b)^2$$
• $$(a+b)^2+(a-b)^2 = ?$$
• $$(a+b)^2+(a-b)^2 = ?$$
• $$ab=?$$
# Geometric Interpretation
$$(a+b)^2 = a^2+2ab+b^2$$
# Relationship between (a+b)2 and (a-b)2
$\begin{eqnarray} (a+b)^2 &=& a^2+2ab+b^2 \nonumber \\ &=& a^2-2ab+b^2 + 2ab + 2ab \nonumber \\ &=& (a-b)^2 +4ab \nonumber \\ \end{eqnarray}$
• Find (a-b)2 similarly
• (2x+y)
• (5m-3p)
• ax+b+2
# Simplify
Questions
• $$1.\space (x+y)^2+2(x+y)(x-y)+(x-y)^2$$
• $$2. \space (2a+1)^2-4a(2a+1)+4a^2$$
• $$3. \space (5a+3b)^2+2(5a+3b)(4a-3b)+(4a-3b)^2$$
• $$4. \space (8x+y)^2-(16x+2y)(5x+y)+(5x+y)^2$$
# Find Values Using Formula
Questions
• $$1.\space 25x^2+36y^2-60xy$$ if $$x=-4, y=-5$$
• $$2. \space if \space (a-b)=7$$ and $$ab=3, (a+b)^2=?$$
• $$3. \space if \space x+\frac 1 x = 5$$ find $$(x^2-\frac 1 {x^2})^2$$
• $$4. \space m+\frac 1 m = 2, m^4+\frac 1 {m^4}=?$$
• $$5. \space p^2-3p+1=0, p^2+\frac 1 {p^2}=?$$
# Formula
• $$(a+b)(a-b)=a^2-b^2$$
• $$(x+a)(x+b) = x^2+x(a+b)+ab$$
# Multiply Using Formula
Questions
• $$1. \space (10+xy)(10-xy)$$
• $$2. \space (3x+2y+1)(3x-2y+1)$$
• $$3. \space (x^2-x+1(x^2+x+1)$$
• $$4. \space (y+4)(y+7)$$
# Chapter 5.3 (Factorization)
## Factorization Concept
• What is a factor? What’s another name?
• $$12 = 3 \times 4 = 3 \times 2 \times 2$$
• $$a^2+ab = a(a+b)$$
• $$a^2-b^2 = (a+b)(a-b)$$
• $$a^2+5a+6=?$$
## Resolve into Factors
Questions
• $$1. \space 2x-6x^2$$
• $$2. \space 25(a+2b)^2-36(2a-5b)^2$$
• $$3. \space 2bd-a^2-c^2+b^2+d^2+2ac$$
• $$4. \space x^2+11x+30$$
• $$5. \space 2x^2+11x+12$$
• $$6. \space 6x^2+17x+5$$
# Chapter 5.2: LCM-HCF
• Factor
• Dividend
• Quotient
• Divisor
## LCM Example
Multiples
$$12 \to 12, 24, 36, 48, 60, 72, 84, 96$$
$$16 \to 16, 32, 48, 64, 80, 96$$
• Common multiples: 48, 96
• Lowest Common multiple: 48
• $$8x^2yz^2, 10x^3yz^3$$
## HCF Example
Factors
$$12 \to 2, 3, 4, 6, 12$$
$$16 \to 2, 4, 8, 16$$
• Common factors: 2, 4
• Highest common factor (HCF): 4
## Find LCM
Questions
• $$1. 4x^2y^3z, 6xy^3z^2, 8x^3yz^3$$
• $$2. a-2, a^2-4, a^2-a-2$$
• $$3. x^3-3x^2-10x, x^3+6x^2+8x, x^4-5x^3-14x^2$$
• $$4. xy-y, x^3y-xy, x^2-2x+1$$
• $$5. x^2-8x+15, x^2-25, x^2+2x-15$$
• $$6. a^2-7a+12, a^2+a-20, a^2+2a-15$$
## Find HCF
Questions
• $$1. 4x^2y^3z, 6xy^3z^2, 8x^3yz^3$$
• $$2. a^2+ab, a^2-b^2$$
• $$3. x^3-3x^2-10x, x^3+6x^2+8x, x^4-5x^3-14x^2$$
• $$4. xy-y, x^3y-xy, x^2-2x+1$$
• $$5. x^2-8x+15, x^2-25, x^2+2x-15$$
• $$6. a^2-7a+12, a^2+a-20, a^2+2a-15$$
## Can HCF be > LCM
Consider 0, 9
Factors
• $$0 \times 1 = 0$$
• $$0 \times 2 = 0$$
• $$0 \times 3 = 0$$
$$\cdots$$
$$\cdots$$
Factors of 0: 0, 1, 2, 3, …, 9, 10,…
Factors of 9: 1, 3, 9
Multiples
We get multiples by multiplying the numbers by $$1, 2, 3, \cdots$$
• $$0 \times 1 = 0$$
• $$0 \times 2 = 0$$
• $$0 \times 3 = 0$$
Multiple of 0: 0
Multiple of 9: 0, 9, 18, 27, …(0 is a multiple of any number)
• HCF = 9
• LCM = 0
# Chapter 5.2: Algebraic Fractions
## Reduce to Lowest Form
Questions
• $$1.\frac{a^2b}{a^3b}$$
• $$2. \frac{x^2+x}{xy+y}$$
• $$3. \frac{x^2+2x-15}{x^2+9x+20}$$
## Express with Common Denominator
Questions
• $$1. \frac{a}{bc}, \frac{a}{ac}$$
• $$2. \frac{a}{a-b}, \frac{b}{a+b}$$
• $$3. \frac{3}{a^2-4}, \frac{2}{a(a+2)}$$
• $$4. \frac{2}{x^2-x-2}, \frac{3}{x^2+x-6}$$
Questions
• $$1. \frac{x}{2a} + \frac{y}{3b}$$
• $$2. \frac{3}{x^2-4x-5} + \frac{4}{x+1}$$
• $$3. \frac{3}{x+3}-\frac{2}{x+2}$$
• $$4. \frac{1}{x^2-1}+\frac 1 {x^4-1}+\frac 4 {x^8-1}$$
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# NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions InText Questions
These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions InText Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions InText Questions
NCERT Intext Question Page No. 204
Question 1.
Observe the following tables and find if x and y are directly proportional.
Answer:
i. e. each ratio is the same
.’. x and y are directly proportional
i. e. all the ratios are not the same x
and y are not directly proportional.
i.e. all the ratios are not the same
x and y are not directly proportional.
Question 2.
Principal = ₹ 1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) change is in direct proportion with time period.
Answer:
Case of Simple Interest
[P = ₹ 1000, r = 8% p.a.]
In each case the ratio is the same.
The simple interest changes in direct proportion with time period.
Case of Compound Interest [P = ₹ 1000, r = 8% p.a.]
$$\frac{\mathrm{CI}}{\mathrm{T}}$$ is not the same in each case.
∴ The compound interest does not change in direct proportion with time period.
NCERT Intext Question Page No. 211
Question 1.
Observe the following tables and find which pair of variables (here x and y) are in the inverse proportion.
Answer:
(i) x1y1 = 50 x 5 = 250
x2y2 = 40 x 6 = 240
x1y1 ≠ x2y2
Hence, x and y are not in inverse proportion.
(ii) x1y1 = 100 x 60 = 6000
x2y2 = 200 x 30 = 6000
x3y3 = 300 x 20 = 6000
x4y4 = 400 x 15 = 6000
x1y1 = x2y2 = x3y3 = x4y4
Hence x and y are in inverse proportion.
(iii) x1y1 = 90 x 10 = 900
x2y2 = 60 x 15 = 900
x3y3 = 45 x 20 = 900
x4y4 = 30 x 25 = 750
x1y1 = x2y2 = x3y3 ≠ x4y4
.’. x and y are not in inverse proportion.
Note: (1) When two quantities x and y are in direct proportion (or vary directly), they are also written is x ∝ y.
(2) When two quantities x and y are in inverse proportion (or very inversely), they are also written as x ∝ $$\frac{1}{y}$$.
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# 1.1 Factors, products, and exponents
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
## Overview
• Factors
• Exponential Notation
## Factors
Let’s begin our review of arithmetic by recalling the meaning of multiplication for whole numbers (the counting numbers and zero).
## Multiplication
Multiplication is a description of repeated addition.
$7+7+7+7$
the number 7 is repeated as an addend* 4 times. Therefore, we say we have four times seven and describe it by writing
$4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
The raised dot between the numbers 4 and 7 indicates multiplication. The dot directs us to multiply the two numbers that it separates. In algebra, the dot is preferred over the symbol $×$ to denote multiplication because the letter $x$ is often used to represent a number. Thus,
$4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7=7+7+7+7$
## Factors and products
In a multiplication, the numbers being multiplied are called factors. The result of a multiplication is called the product. For example, in the multiplication
$4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7=28$
the numbers 4 and 7 are factors, and the number 28 is the product. We say that 4 and 7 are factors of 28. (They are not the only factors of 28. Can you think of others?)
Now we know that
$\begin{array}{ccc}\left(\text{factor}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(\text{factor}\right)& =& \text{product}\end{array}$
This indicates that a first number is a factor of a second number if the first number divides into the second number with no remainder. For example, since
$4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7=28$
both 4 and 7 are factors of 28 since both 4 and 7 divide into 28 with no remainder.
## Exponential notation
Quite often, a particular number will be repeated as a factor in a multiplication. For example, in the multiplication
$7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
the number 7 is repeated as a factor 4 times. We describe this by writing ${7}^{4}.$ Thus,
$7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7={7}^{4}$
The repeated factor is the lower number (the base), and the number recording how many times the factor is repeated is the higher number (the superscript). The superscript number is called an exponent.
## Exponent
An exponent is a number that records how many times the number to which it is attached occurs as a factor in a multiplication.
## Sample set a
For Examples 1, 2, and 3, express each product using exponents.
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3.$ Since 3 occurs as a factor 6 times,
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3={3}^{6}$
$8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8.$ Since 8 occurs as a factor 2 times,
$8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8={8}^{2}$
$5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9.$ Since 5 occurs as a factor 3 times, we have ${5}^{3}.$ Since 9 occurs as a factor 2 times, we have ${9}^{2}.$ We should see the following replacements.
$\underset{{5}^{3}}{\underbrace{5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5}}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\underset{{9}^{2}}{\underbrace{9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9}}$
Then we have
$5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9={5}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{9}^{2}$
Expand ${3}^{5}.$ The base is 3 so it is the repeated factor. The exponent is 5 and it records the number of times the base 3 is repeated. Thus, 3 is to be repeated as a factor 5 times.
${3}^{5}=3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3$
Expand ${6}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{10}^{4}.$ The notation ${6}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{10}^{4}$ records the following two facts: 6 is to be repeated as a factor 2 times and 10 is to be repeated as a factor 4 times. Thus,
${6}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{10}^{4}=6\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10$
## Exercises
For the following problems, express each product using exponents.
$8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8$
${8}^{3}$
$12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12$
$5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
${5}^{7}$
$1\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1$
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4$
${3}^{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{4}^{2}$
$8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15$
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9$
${2}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{9}^{8}$
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}10$
Suppose that the letters $x$ and $y$ are each used to represent numbers. Use exponents to express the following product.
$x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y$
${x}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{y}^{2}$
Suppose that the letters $x$ and $y$ are each used to represent numbers. Use exponents to express the following product.
$x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y$
For the following problems, expand each product (do not compute the actual value).
${3}^{4}$
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3$
${4}^{3}$
${2}^{5}$
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2$
${9}^{6}$
${5}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{6}^{2}$
$5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6$
${2}^{7}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{4}$
${x}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{y}^{4}$
$x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y$
${x}^{6}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{y}^{2}$
For the following problems, specify all the whole number factors of each number. For example, the complete set of whole number factors of 6 is 1, 2, 3, 6.
20
$1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}20$
14
12
$1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}6,\text{\hspace{0.17em}}12$
30
21
$1,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}7,\text{\hspace{0.17em}}21$
45
11
$1,\text{\hspace{0.17em}}11$
17
19
$1,\text{\hspace{0.17em}}19$
2
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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Miscellaneous
Chapter 7 Class 11 Permutations and Combinations
Serial order wise
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Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Misc 5 (Method 1) How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated? A number is divisible by 10 if 0 is at the units place Thus, We need to form 6 digit number whose unit place is 0 So, We need to fill up 5 places with the remaining digits 1, 3, 5, 7, & 9 120 is divisible by 10 as last digit is 0 Hence, n = Number of digits = 5 & r = Number of places to fill = 5 Number of 6 digit numbers = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 5 × 4 × 3 × 2 × 1 = 120 Misc 5 (Method 2) How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated? A number is divisible by 10 if 0 is at the units place Thus, We need to form 6 digit number whose unit place is 0 So, We need to fill up 5 places with the remaining digits 1, 3, 5, 7, & 9 120 is divisible by 10 as last digit is 0 Thus, Number of 6 digit numbers = 1 × 5 × 4 × 3 × 2 × 1 = 120
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# Predicted Trend Value of a Time Series
A time series shows data on a variable’s outcome in different periods—for example, a time series showing EURUSD exchange rates between a given time interval.
Time series models explain the past and predict the future of a time series using a trend, which is a long-term pattern in a specific direction.
## Types of Trends
### Linear Trend Models
In a linear trend, the dependent variable changes at a constant amount with time. Consider a time series with a linear trend. The value of the time series at time $$t$$ can be expressed as:
$$\text{y}_{\text{t}}=\text{b}_{0}+\text{b}_{1}\text{t}+\epsilon_{\text{t}}$$
Where:
$$\text{y}_{\text{t}}$$ = Value of the time series at time $$t$$.
$$\text{b}_{0}$$ = Intercept term.
$$\text{b}_{1}$$ = Trend coefficient.
$$\text{t}$$ = Time, the independent variable. $$t=1, 2, …, \text{T}$$.
$$\epsilon_{\text{t}}$$ = Random error term.
The following shows a linear trend of the EURUSD between 05 January 2021 to 06 February 2021.
#### Predicted Value of a Linear Trend
The predicted value of a time series with a linear trend in period five is expressed as:
$$\widehat{y_{5}}=\widehat{b_{0}}+\widehat{b_{1}}(5)$$
#### Out of Sample Prediction
The predicted value of a time series for period $$T+1$$ is given by:
$$\widehat{\text{y}}_{T+1}=\widehat{b}_{0}+\widehat{b_{1}}(T+1)$$
#### Example: Predicted Value of a Linear trend
Given that $$b_{0}=2$$ and $$b_{1}=1.6$$, the predicted value of y after four periods is closest to:
$$y_{t}=b_{0}+b_{1}t+\epsilon_{t}$$
$$y_{4}=2+1.6\times4=8.4$$
## Log-Linear Trend Models
A log-linear trend analysis relies on an independent variable that changes exponentially over time or grows at a constant rate. It often works better with financial time series.
A time series with exponential growth can be modeled as:
$$y_{t}=e^{b_{0}+b_{1}(t)}$$
Where:
$$y_{t}$$ = The value of the dependent variable at time $$t$$.
$$b_{0}$$ = Intercept term.
$$b_{1}$$ = Constant growth rate.
$$t$$ = time = $$1,2,3,…,T$$.
Taking the natural log on both sides of the above equation generates a log-linear model expressed as:
$$\text{lny}_{\text{t}}=\text{ln ln}(e^{b_{0}+b_{1}t})$$
$$\text{ln ln}\ \text{y}_{\text{t}}=\ \text{b}_{0}+\text{b}_{1}\text{t}$$
The following figure shows different forms of log-linear trends.
#### Example: Predicted Value of a Log-Linear Trend
If $$b_{0}=3$$ and $$b_{1}=1.5$$, the predicted value of y after four periods is:
$$\text{ln y}_{t}=b_{0}+b_{1}t+\epsilon_{t}$$
$$\text{ln y}_{4}= 3+1.5\times4$$
$$\text{y}_{4}=e^{9}=8103.08$$
## Question
Consider a log-linear trend model with an intercept term of 0.5 and a slope coefficient of 0.10. The predicted value of the trend after six periods is closest to:
A. 1.1.
B. 1.6.
C. 3.0.
### Solution
$$y_{t}=e^{b_{0}+b_{1}(t)}$$
$$y_{t}=e^{0.5+(0.10\times6)}=3.0$$
LOS 5 (a) Calculate and evaluate the predicted trend value for a time series, modeled as either a linear trend or a log-linear trend, given the estimated trend coefficients.
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2021-03-24
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In these lessons, we will learn some models for decimals
• how to write decimals in expanded form
• how to compare and order decimals
• how to estimate and round decimals.
### Model Decimals
1. Using decimals extends the place value system to represent parts of a whole.
The use of a decimal point is a symbol that separates the tenths from the ones, or the ‘part from the whole’. For example, in the number 4.2, the decimal point separates the 4 ones and the 2 tenths.
2. The base ten place value system is built on symmetry around the ones place and the decimal.
3. Decimals can represent parts of a whole, as well as mixed numbers.
4. Decimals can be interpreted and read in more than one way. For example, 4.3 may be renamed 43 tenths.
5. Decimals can be renamed as other decimals or fractions. For example, 800/1000 or 0.800 can be renamed as 80/100 or 0.80. It can also be renamed as 8/10 or 0.8
Take note that
0.5 or 0.50 or 0.500 are all equal to 1/2
0.25 or 0.250 are both equal 1/4
0.75 and 0.750 are equal to 3/4
We can use a thousands grid to model decimals.
The following shaded grid can be used to represent the number 3.146
We can also use base ten blocks to model decimals.
The following can be used to represent the number 3.231
Meter sticks can also be used to represent decimals. Measuring to the nearest millimeter is one thousandths of a meter. Centimeters are hundredths of a meter and decimeters are tenths of a meter.
Model Decimals
Model decimals on grids and on number lines.
### Decimals in Expanded Form
For Example:
45.23 = 40 + 5 + 0.2 + 0.03
or
45.23 = (4 × 10) + (5 × 1) + (2 × 0.1) + (3 × 0.01).
50.302 = 50 + 0.3 + 0.002
or
50.302 = (5 × 10) + (3 × 0.1) + (2 × 0.001).
Decimals - Expanded Form
Reading/writing decimals in numeric, word, and expanded form.
Example:
1. Write in expanded form:
a. 9.45
b. 0.21
c. 8.878
d. 0.3911
### Compare Decimals
We can compare decimals using place value. For example:
0.02 < 0.2 because 0 tenths is less than 2 tenths
0.021 > 0.01 because 2 hundredths is greater that 1 hundredths
We can compare decimals using a benchmark number. For example:
0.021 < 0.2 because 0.021 is less than 0.1 and 0.2 is greater than 0.1
0.8 > 0.423 because 0.8 is more than half and 0.423 is less than half.
We can compare using equivalent decimals with same number of digits. For example:
0.34 > 0.3 because 0.34 > 0.30 (34 hundredths > 30 hundredths)
8.302 < 8.32 because 8.302 < 8.320 (302 thousandths < 320 thousandths)
Ordering & Comparing Decimals
### Estimate and Round Decimals
We can round decimals to simpler decimals.
For Example: 2.9286
can be rounded to 2.929 (nearest thousandths)
can be rounded to 2.93 (nearest hundredths)
can be rounded to 2.9 (nearest tenths)
can be rounded to 3 (nearest ones)
Estimating and Rounding with Decimals
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.
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# Ellipse
What is an ellipse? You might describe it as a squashed circle, which would be correct. But, in mathematics, an ellipse has a much more specific definition, which is “a closed curve where the sum of the distances from the two focus points inside the curve to every point on the curve is constant.” An ellipse is not to be confused with an oval, which does not have a lovely mathematical definition like the ellipse does.
Let’s take a look at an ellipse so we can see those two focus points, which together we call foci:
We can illustrate the mathematical definition of an ellipse by showing a point on the ellipse along with connecting lines to the foci:
If we measure these two line segments and then add them together, that will be the total length from any point on this ellipse to the two focus points. Depending on where the point is on the ellipse, the blue line will sometimes be shorter and the red line longer, but they’ll always add to the same amount.
Now let’s talk more about those focus points. We find them on the major axis of the ellipse, which is always the longer of the two axes:
Depending on the ellipse, the foci get closer to or farther from the center! Look at this ellipse that’s not quite as squashed:
Notice how the focus points are closer to the center of the ellipse in this one. This is due to something we call eccentricity, which is expressed as a value between 0 and 1. We can calculate the eccentricity by creating a ratio. We measure the distance from either focus point to the center and place that over the distance from that focus point to either end of the minor axis. When the eccentricity is 0, we have a very special kind of ellipse that we call a circle. When that happens, it means that the two focus points have converged into one point at the center of the ellipse.
At the opposite end of the spectrum is a totally flattened ellipse and those two measures will be the same, resulting in an eccentricity of 1. It also means our “ellipse” will look like a line segment; it’s totally flattened with the foci as the endpoints of the line segment.
Here is the standard form of an equation of an ellipse:
$$\frac{\left ( x-h \right )^{2}}{a^{2}}+\frac{\left ( y-k \right )^{2}}{b^{2}}=1$$
We have a lot of letters to define here. The center of the ellipse is represented by $$h$$ and $$k$$, which form an ordered pair. The coefficient $$a$$ is the length from the center of the ellipse to either end of the ellipse parallel to the $$x$$-axis, and $$b$$ is the length from the center to either end of the ellipse parallel to the $$y$$-axis.
We can see that, in this case, $$b$$ is equal to half the length of the major axis, which in this case is the height of the ellipse, and $$a$$ is half the minor axis, which is the width. So if we want to create an ellipse with a height of 14 units and a width of 10 units, $$b$$ would be equal to half of 14 which is 7, and $$a$$ would be equal to half of 10 which is 5.
If we wanted that ellipse to be located with a center at $$(-3,2)$$, our equation would look like this:
$$\frac{\left ( x+3 \right )^{2}}{25}+\frac{\left ( y-2 \right )^{2}}{49}=1$$
We substituted the $$x-$$ and $$y-$$ values of our center $$(-3,2)$$ for $$(h,k)$$. Notice that since $$h$$ is negative, it changes the sign inside the left binomial to a plus sign. This is because subtracting -3 is the same thing as adding 3. Our denominators are just the squared versions of our $$a$$ and $$b$$. $$a$$ is 5, which squares to 25, and $$b$$ is 7, which squares to 49.
Okay, to wrap things up, let’s go through a couple of practice problems. Feel free to pause the video after each question to work it out on your own.
1. What is the equation of an ellipse that has a center at $$(6,4)$$, has a height of 12, and a width of 8?
We start with the standard form of an equation for an ellipse.
$$\frac{\left ( x-6 \right )^{2}}{a^{2}}+\frac{\left ( y-4 \right )^{2}}{b^{2}}=1$$
Next, we substitute the $$x-$$ and $$y-$$values of the center for $$h$$ and $$k$$ in our general equation.
$$\frac{\left ( x-6 \right )^{2}}{a^{2}}+\frac{\left ( y-4 \right )2}{b^{2}}=1$$
Finally, we take half of the width, which is the minor axis, to find our value for $$a$$ (half of 8 is 4). We then square it after substituting it into the general equation for our first denominator of 16. We do the same thing for $$b$$. Take half of the height (half of 12 is 6) and substitute it in for $$b$$, where it is squared to become 36. Done!
$$\frac{\left ( x-6 \right )^{2}}{16}+\frac{\left ( y-4 \right )^{2}}{36}=1$$
2. What is the equation of this ellipse:
This one is a little tougher because we need to find the length of both axes, as well as find the center of the ellipse. We start with the standard form for an equation of an ellipse:
$$\frac{\left ( x-h \right )^{2}}{a^{2}}+\frac{\left ( y-k \right )^{2}}{b^{2}}=1$$
Next, we need to determine how tall and wide our ellipse is. By looking at the diagram, we can see that it is 8 units tall and 6 units wide. We also need to find the center, which we can do by drawing the major and minor axes. Here, we find the center at $$(-1,1)$$. Now we can substitute the $$x-$$ and $$y-$$values of the center for $$h$$ and $$k$$ in our general equation.
$$\frac{\left ( x-(-1) \right )^{2}}{a^{2}}+\frac{\left ( y-1 \right )^{2}}{b^{2}}=1$$
When we substitute the $$h$$, we end up with $$(x – (-1))^{2}$$ , which can be rewritten as $$(x+1)^{2}$$ because subtracting a negative is the same thing as adding. Finally, we take half of the width, which is the minor axis, to find our value of $$a$$ (half of 6 is 3). We then square it after substituting it into the standard form equation for our first denominator of 9. We do the same thing for $$b$$. Take half of the height (half of 8, which equals 4) and substitute it in for $$b$$, where it is squared to become 16.
$$\frac{\left ( x+1 \right )^{2}}{9}+\frac{\left ( y-1 \right )^{2}}{16}=1$$
That’s all for this review of ellipses. Thanks for watching, and happy studying!
## Ellipse Practice Questions
Question #1:
What is the equation of an ellipse that has a center at $$(3,1)$$, has a height of 8 units, and has a width of 4 units?
$$\frac{{(x-3)}^2\ }{16}+\frac{{(y-1)}^2}{64}=1$$
$$\frac{{(x-3)}^2\ }{4}+\frac{{(y-1)}^2}{16}=1$$
$$\frac{{(x+3)}^2\ }{4}+\frac{{(y+1)}^2}{16}=1$$
$$\frac{{(x-3)}^2\ }{64}+\frac{{(y-1)}^2}{16}=1$$
$$\frac{{(x-h)}^2\ }{a^2}+\frac{{(y-k)}^2}{b^2}=1$$
Next, substitute the $$x$$– and $$y$$-values of the center point for $$h$$ and $$k$$ in our general equation:
$$\frac{{(x-3)}^2\ }{a^2}+\frac{{(y-1)}^2}{b^2}=1$$
The value of $$a$$ is half the width. Since half of 4 is 2, substitute 2 for $$a$$. Square this value after substituting it into the general equation to get a denominator of 4 in the first fraction.
$$\frac{{(x-3)}^2\ }{4}+\frac{{(y-1)}^2}{b^2}=1$$
The value of $$b$$ is half the height. Since half of 8 is 4, substitute 4 for $$b$$. Square this value after substituting it into the general equation to get a denominator of 16 in the second fraction. Therefore, the equation of this ellipse is:
$$\frac{{(x-3)}^2\ }{4}+\frac{{(y-1)}^2}{16}=1$$
Question #2:
What is the equation of this ellipse?
$$\frac{x^2\ }{4}+\frac{y^2}{10}=1$$
$$\frac{x^2\ }{16}+\frac{y^2}{100}=1$$
$$\frac{x^2\ }{100}+\frac{y^2}{16}=1$$
$$\frac{x^2\ }{10}+\frac{y^2}{4}=1$$
$$\frac{{(x-h)}^2\ }{a^2}+\frac{{(y-k)}^2}{b^2}=1$$
We need to find the center, which is plotted on the coordinate plane. This center point is located at the origin, $$(0,0)$$. Substitute the $$x$$– and $$y$$-values of the center point for $$h$$ and $$k$$ in our general equation. Since $$h=0$$, write the first numerator as $$x^2$$. Since $$k=0$$ as well, write the second numerator as $$y^2$$.
$$\frac{x^2\ }{a^2}+\frac{y^2}{b^2}=1$$
Next, determine how tall and wide the ellipse is. By looking at the graph, we can see that it is 20 units tall and 8 units wide. Take half of the width to find the value for $$a$$. Since half of 8 is 4, substitute 4 for $$a$$. Square this value after substituting it into the general equation to get a denominator of 16 in the first fraction.
$$\frac{x^2\ }{16}+\frac{y^2}{b^2}=1$$
Take half of the height to find the value for $$b$$. Since half of 20 is 10, substitute 10 for $$b$$. Square this value after substituting it into the general equation to get a denominator of 100 in the second fraction. Therefore, the equation of this ellipse is:
$$\frac{x^2\ }{16}+\frac{y^2}{100}=1$$
Question #3:
What is the equation of an ellipse that has a center at (1,-5), has a height of 6 units, and has a width of 18 units?
$$\frac{{(x-1)}^2\ }{81}+\frac{{(y+5)}^2}{9}=1$$
$$\frac{{(x-1)}^2\ }{9}+\frac{{(y+5)}^2}{3}=1$$
$$\frac{{(x+1)}^2\ }{81}+\frac{{(y-5)}^2}{9}=1$$
$$\frac{{(x+1)}^2\ }{3}+\frac{{(y-5)}^2}{9}=1$$
$$\frac{{(x-h)}^2\ }{a^2}+\frac{{(y-k)}^2}{b^2}=1$$
Next, substitute the $$x$$– and $$y$$-values of the center point for $$h$$ and $$k$$ in our general equation. Since subtracting –5 is the same thing as adding 5, write the second numerator as $${(y+5)}^2$$.
$$\frac{{(x-1)}^2\ }{a^2}+\frac{{(y+5)}^2}{b^2}=1$$
Then, take half of the width to find the value for $$a$$. Since half of 18 is 9, substitute 9 for $$a$$. Square this value after substituting it into the general equation to get a denominator of 81 in the first fraction.
$$\frac{{(x-1)}^2\ }{81}+\frac{{(y+5)}^2}{b^2}=1$$
Take half the height to find the value for $$b$$. Since half of 6 is 3, substitute 3 for $$b$$. Square this value after substituting it into the general equation to get a denominator of 9 in the second fraction. Therefore, the equation of this ellipse is:
$$\frac{{(x-1)}^2\ }{81}+\frac{{(y+5)}^2}{9}=1$$
Question #4:
Ralph is installing a window above his front door. A scaled version of the window is graphed on the coordinate plane as the ellipse shown below. What is the equation of this ellipse?
$$\frac{{(x-1)}^2\ }{1}+\frac{{(y+5)}^2}{16}=1$$
$$\frac{{(x-5)}^2\ }{16}+\frac{{(y-1)}^2}{1}=1$$
$$\frac{{(x+5)}^2\ }{1}+\frac{{(y-1)}^2}{4}=1$$
$$\frac{{(x+5)}^2\ }{4}+\frac{{(y-1)}^2}{1}=1$$
$$\frac{{(x-h)}^2\ }{a^2}+\frac{{(y-k)}^2}{b^2}=1$$
We need to find the center, which is plotted on the coordinate plane. The coordinates of this center point are $$(-5,1)$$. Substitute the $$x$$– and $$y$$-values of the center point for $$h$$ and $$k$$ in our general equation. Since subtracting –5 is the same thing as adding 5, write the first numerator as $${(x+5)}^2$$. Write the second numerator as $$\left(y-1\right)^2$$.
$$\frac{{(x+5)}^2\ }{a^2}+\frac{{(y-1)}^2}{b^2}=1$$
Next, determine the height and width of the ellipse. By looking at the graph, we can see that it is 2 units tall and 4 units wide. Take half of the width to find the value of $$a$$. Since half of 4 is 2, substitute 2 for $$a$$. Square this value after substituting it into the general equation to get a denominator of 4 in the first fraction.
$$\frac{{(x+5)}^2\ }{4}+\frac{{(y-1)}^2}{b^2}=1$$
Take half the height to find the value for $$b$$. Since half of 2 is 1, substitute 1 for $$b$$. Square this value after substituting it into the general equation to get a denominator of 1 in the second fraction. Therefore, the equation of the ellipse for Ralph’s window is:
$$\frac{{(x+5)}^2\ }{4}+\frac{{(y-1)}^2}{1}=1$$
Question #5:
Charlotte is installing a swimming pool in her backyard. A scaled version of the pool can be graphed on the coordinate plane as an ellipse. The graph of Charlotte’s swimming pool has its center at $$(0,3)$$, a width of 14 units, and a height of 10 units. What is the equation of this ellipse?
$$\frac{{(x-3)}^2\ }{7}+\frac{y^2}{5}=1$$
$$\frac{{(x-3)}^2\ }{49}+\frac{y^2}{25}=1$$
$$\frac{x^2\ }{7}+\frac{{(y-3)}^2}{5}=1$$
$$\frac{x^2\ }{49}+\frac{{(y-3)}^2}{25}=1$$
$$\frac{{(x-h)}^2\ }{a^2}+\frac{{(y-k)}^2}{b^2}=1$$
The coordinates of the center point are $$(0,3)$$. Substitute the $$x$$– and $$y$$-values of the center point for $$h$$ and $$k$$ in our general equation. Since $$h=0$$, write the first numerator as $$x^2$$. Since $$k=3$$, write the second numerator as $$\left(y-3\right)^2$$.
$$\frac{x^2\ }{a^2}+\frac{{(y-3)}^2}{b^2}=1$$
Take half of the width to find the value for $$a$$. Since half of 14 is 7, substitute 7 for $$a$$. Square this value after substituting it into the general equation to get a denominator of 49 in the first fraction.
$$\frac{x^2\ }{49}+\frac{{(y-3)}^2}{b^2}=1$$
Finally, take half the height to find the value for $$b$$. Since half of 10 is 5, substitute 5 for $$b$$. Square this value after substituting it into the general equation to get a denominator of 25 in the second fraction. Therefore, the equation of the ellipse for Charlotte’s swimming pool is:
$$\frac{x^2\ }{49}+\frac{{(y-3)}^2}{25}=1$$
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FutureStarr
4 Out of 14 As a Percentage OR
## 4 Out of 14 As a Percentage
A quarter of focus groups—14 out of a total of 46—expressed support for a white colonization of Mars.
### Percentage
Before we get started in the fraction to percentage conversion, let's go over some very quick fraction basics. Remember that a numerator is the number above the fraction line, and the denominator is the number below the fraction line. We'll use this later in the tutorial.
I've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below). (Source: visualfractions.com)
### Use
The concept of percent increase is basically the amount of increase from the original number to the final number in terms of 100 parts of the original. An increase of 5 percent would indicate that, if you split the original value into 100 parts, that value has increased by an additional 5 parts. So if the original value increased by 14 percent, the value would increase by 14 for every 100 units, 28 by every 200 units and so on. To make this even more clear, we will get into an example using the percent increase formula in the next section.
Although we have just covered how to calculate percent increase and percent decrease, sometimes we just are interested in the change in percent, regardless if it is an increase or a decrease. If that is the case, you can use the percent change calculator or the percentage difference calculator. A situation in which this may be useful would be an opinion poll to see if the percentage of people who favor a particular political candidate differs from 50 percent. (Source: www.omnicalculator.com)
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1. Chapter 3 Class 11 Trigonometric Functions
2. Serial order wise
Transcript
Example 23 Solve sin 2x โ sin 4x + sin 6x = 0. sin 2x โ sin 4x + sin 6x = 0 (sin 6x + sin 2x) โ sin 4x = 0 2 sin ((6๐ฅ + 2๐ฅ)/2) cos ((6๐ฅ โ 2๐ฅ)/2) โ sin 4x = 0 2 sin (8๐ฅ/2) cos (4๐ฅ/2) โ sin 4x = 0 2 sin 4x cos (2x) โ sin 4x = 0 sin 4x (2 cos (2x) โ 1) = 0 Hence sin 4x = 0 or 2cos 2x โ 1 = 0 sin 4x = 0 or 2cos 2x = 1 sin 4x = 0 or cos 2x = 1/2 We need to find general solution both separately General solution for sin 4x = 0 Let sin x = sin y โ sin 4x = sin 4y Given sin 4x = 0 From (1) and (2) sin 4y = 0 sin 4y = sin (0) 4y = 0 โ y = 0 General solution for sin 4x = sin 4y is 4x = nฯ ยฑ (-1)n 4y where n โ Z Put y = 0 4x = nฯ ยฑ (-1)n 0 4x = nฯ x = ๐๐/4 where n โ Z General solution for cos 2x = ๐/๐ Let cos x = cos y โ cos 2x = cos 2y Given cos 2x = 1/2 From (3) and (4) cos 2y = 1/2 cos (2y) = cos (๐/3) โ 2y = ๐/3 General solution for cos 2x = cos 2y is 2x = 2nฯ ยฑ 2y where n โ Z putting 2y = ๐/3 2x = nฯ ยฑ ๐/3 x = 1/2 (2nฯ ยฑ ๐/3) x =2๐๐/2 ยฑ 1/2 ร ๐/3 x = nฯ ยฑ ๐/6 where n โ Z Hence General Solution is For sin4x = 0, x = ๐๐/4 and for cos 2x = 1/2 , x = nฯ ยฑ ๐/6 where n โ Z
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Home > Number prompts >
### Rectangle ratios 1 inquiry
In the prompt we are told that the ratio of the length to the width of the large rectangle equals the ratio of length to width of the smaller rectangles. Moreover, the length of the bigger rectangle is twice the width of a smaller rectangle (a = 2c) and the width of the bigger rectangle equals the length of a smaller one. These constraints are satisfied by the square root of two - that is, both ratios equal √2:1 in their simplest forms. Once students have understood the meaning of the prompt during the question and observation phase of the inquiry, they might explore by using a kind of iterative process. To start the process, the teacher might encourage the class to speculate about the lengths and widths. For example, if a = 8 and b = 4, then a:b = 8:4 = 2:1 and b:c = 4:2. As a = 8 and c = 2, this does not satisfy the constraint that a = 2c. Let's try a = 8, b = 6 and c = 4, which gives a:b = 8:6 = 4:3 = 1:0.75 and b:c = 6:4 = 3:2 = 1:0.67 (accurate to 2 decimal places). Or we could could compare the ratios using a multiplier and common part, which gives a:b = 16:12 and b:c = 18:12. This is closer, but we need to reduce b in the first ratio, say to b = 5.5. Now a:b = 8:5.5 and b:c = 5.5:4, which simplify to a:b = 32:22 and b:c = 30.25:22. Let's change b to 5.4 or 5.6 and so on. Ultimately, the teacher uses algebraic reasoning, co-constructing as many of the steps as possible, to show that the solution involves an irrational number. In so doing, students encounter important algebraic manipulations. In the first two steps of the approach that follows, a ratio is converted into a fraction and then the fractions are 'cross multiplied'. Both manipulations will need an explanation before students develop the confidence to reproduce the procedure in extending this line of inquiry (see below).Students can be convinced of the fact that 2:√2 = √2:1 by dividing both terms in the first ratio by √2. Extending the line of inquiryOnce the class has gone through the first phase of the inquiry, students could be expected to work more independently to show that the ratios in the following two cases are a:b:c = 3:√3:1 and a:b:c = 4:√4:1 respectively. Topics covered during the inquiry:Simplifying ratiosConverting ratios to the form 1:nComparing ratios using multipliersConverting between fractions and ratiosAlgebraic manipulationPercentage errorIrrational numbers.Resources Prompt sheetPowerPointNotice: Please be aware that, to the knowledge of the authors, this prompt has not been used in a classroom. If you do use it with a class, we would very much appreciate hearing about the students' questions and observations and how the inquiry developed. Please contact Inquiry Maths here. From abstract to concreteThe inquiry could become more concrete if the teacher decides to introduce A-sized paper. The ratio of the length and width of each size is the same as for the rectangles in the prompt.Inquiry 1Students might divide the length by the width to verify the ratio for all sizes from A0 to A8. As they do so, they could record the degree of accuracy (in decimal places) compared to √2, which is 1.4142 (rounded to four decimal places). For example, the calculation for A0 is 1189 ÷ 841 = 1.413793, which is accurate to 2 decimal places. This line of inquiry ends with a discussion about why the integer dimensions of A-sized paper can never give the exact value of √2.Inquiry 2Alternatively, students could work out the size of an A0 sheet given the dimensions of, for example, an A8 sheet. The A8 sheet is 74 mm by 52 mm, so the A7 sheet is (74 x 1.4142) mm by 74 mm. One question that arises in this pathway of the inquiry relates to the percentage error caused by rounding √2 in the calculations. How close are the dimensions of the students' A0 sheet to the ones given in the diagram below?
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### Statistics Knowledge Portal
A free online introduction to statistics
# Correlation Coefficient
## What is the correlation coefficient?
The correlation coefficient is the specific measure that quantifies the strength of the linear relationship between two variables in a correlation analysis. The coefficient is what we symbolize with the r in a correlation report.
## How is the correlation coefficient used?
For two variables, the formula compares the distance of each datapoint from the variable mean and uses this to tell us how closely the relationship between the variables can be fit to an imaginary line drawn through the data. This is what we mean when we say that correlations look at linear relationships.
## What are some limitations to consider?
Correlation only looks at the two variables at hand and won’t give insight into relationships beyond the bivariate data. This test won’t detect (and therefore will be skewed by) outliers in the data and can’t properly detect curvilinear relationships.
### Correlation coefficient variants
In this section, we’re focusing on the Pearson product-moment correlation. This is one of the most common types of correlation measures used in practice, but there are others. One closely related variant is the Spearman correlation, which is similar in usage but applicable to ranked data.
## What do the values of the correlation coefficient mean?
The correlation coefficient r is a unit-free value between -1 and 1. Statistical significance is indicated with a p-value. Therefore, correlations are typically written with two key numbers: r = and p = .
• The closer r is to zero, the weaker the linear relationship.
• Positive r values indicate a positive correlation, where the values of both variables tend to increase together.
• Negative r values indicate a negative correlation, where the values of one variable tend to increase when the values of the other variable decrease.
• The values 1 and -1 both represent "perfect" correlations, positive and negative respectively. Two perfectly correlated variables change together at a fixed rate. We say they have a linear relationship; when plotted on a scatterplot, all data points can be connected with a straight line.
• The p-value helps us determine whether or not we can meaningfully conclude that the population correlation coefficient is different from zero, based on what we observe from the sample.
What is a p-value?
A p-value is a measure of probability used for hypothesis testing. The goal of hypothesis testing is to determine whether there is enough evidence to support a certain hypothesis about your data. Actually, we formulate two hypotheses: the null hypothesis and the alternative hypothesis. In the case of correlation analysis, the null hypothesis is typically that the observed relationship between the variables is the result of pure chance (i.e. the correlation coefficient is really zero — there is no linear relationship). The alternative hypothesis is that the correlation we’ve measured is legitimately present in our data (i.e. the correlation coefficient is different from zero).
The p-value is the probability of observing a non-zero correlation coefficient in our sample data when in fact the null hypothesis is true. A low p-value would lead you to reject the null hypothesis. A typical threshold for rejection of the null hypothesis is a p-value of 0.05. That is, if you have a p-value less than 0.05, you would reject the null hypothesis in favor of the alternative hypothesis—that the correlation coefficient is different from zero.
## How do we actually calculate the correlation coefficient?
The sample correlation coefficient can be represented with a formula:
$$r=\frac{\sum\left[\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)\right]}{\sqrt{\mathrm{\Sigma}\left(x_i-\overline{x}\right)^2\ \ast\ \mathrm{\Sigma}(y_i\ -\overline{y})^2}}$$
View Annotated Formula
Let’s step through how to calculate the correlation coefficient using an example with a small set of simple numbers, so that it’s easy to follow the operations.
Let’s imagine that we’re interested in whether we can expect there to be more ice cream sales in our city on hotter days. Ice cream shops start to open in the spring; perhaps people buy more ice cream on days when it’s hot outside. On the other hand, perhaps people simply buy ice cream at a steady rate because they like it so much.
We start to answer this question by gathering data on average daily ice cream sales and the highest daily temperature. Ice Cream Sales and Temperature are therefore the two variables which we’ll use to calculate the correlation coefficient. Sometimes data like these are called bivariate data, because each observation (or point in time at which we’ve measured both sales and temperature) has two pieces of information that we can use to describe it. In other words, we’re asking whether Ice Cream Sales and Temperature seem to move together.
As before, a useful way to take a first look is with a scatterplot:
We can also look at these data in a table, which is handy for helping us follow the coefficient calculation for each datapoint. When talking about bivariate data, it’s typical to call one variable X and the other Y (these also help us orient ourselves on a visual plane, such as the axes of a plot). Let’s call Ice Cream Sales X, and Temperature Y.
Notice that each datapoint is paired. Remember, we are really looking at individual points in time, and each time has a value for both sales and temperature.
Ice Cream Sales (X)Temperature °F (Y)
370
675
980
## 1. Start by finding the sample means
Now that we’re oriented to our data, we can start with two important subcalculations from the formula above: the sample mean, and the difference between each datapoint and this mean (in these steps, you can also see the initial building blocks of standard deviation).
The sample means are represented with the symbols and , sometimes called “x bar” and “y bar.” The means for Ice Cream Sales () and Temperature () are easily calculated as follows:
$$\overline{x} =\ [3\ +\ 6\ +\ 9] ÷ 3 = 6$$
$$\overline{y} =\ [70\ +\ 75\ +\ 80] ÷ 3 = 75$$
## 2. Calculate the distance of each datapoint from its mean
With the mean in hand for each of our two variables, the next step is to subtract the mean of Ice Cream Sales (6) from each of our Sales data points (xi in the formula), and the mean of Temperature (75) from each of our Temperature data points (yi in the formula). Note that this operation sometimes results in a negative number or zero!
Ice Cream (X)Temperature °F (Y)$x_i-\overline{x}$$y_i-\overline{y} 3$$70$$3 - 6 = -3$$70 - 75 = -5$
$6$$75$$6 - 6 = 0$$75 - 75 = 0 9$$80$$9 - 6 = 3$$80 - 75 = 5$
## 3. Complete the top of the coefficient equation
This piece of the equation is called the Sum of Products. A product is a number you get after multiplying, so this formula is just what it sounds like: the sum of numbers you multiply.
$$\sum[(x_i-\overline{x})(y_i-\overline{y})]$$
We take the paired values from each row in the last two columns in the table above, multiply them (remember that multiplying two negative numbers makes a positive!), and sum those results:
$$[(-3)(-5)] + [(0)(0)] + [(3)(5)] = 30$$
### INSIGHT:
How does the Sum of Products relate to the scatterplot?
The Sum of Products calculation and the location of the data points in our scatterplot are intrinsically related.
Notice that the Sum of Products is positive for our data. When the Sum of Products (the numerator of our correlation coefficient equation) is positive, the correlation coefficient r will be positive, since the denominator—a square root—will always be positive. We know that a positive correlation means that increases in one variable are associated with increases in the other (like our Ice Cream Sales and Temperature example), and on a scatterplot, the data points angle upwards from left to right. But how does the Sum of Products capture this?
• The only way we will get a positive value for the Sum of Products is if the products we are summing tend to be positive.
• The only way to get a positive value for each of the products is if both values are negative or both values are positive.
• The only way to get a pair of two negative numbers is if both values are below their means (on the bottom left side of the scatter plot), and the only way to get a pair of two positive numbers is if both values are above their means (on the top right side of the scatter plot).
So, the Sum of Products tells us whether data tend to appear in the bottom left and top right of the scatter plot (a positive correlation), or alternatively, if the data tend to appear in the top left and bottom right of the scatter plot (a negative correlation).
## 4. Complete the bottom of the coefficient equation
The denominator of our correlation coefficient equation looks like this:
$$\sqrt{\mathrm{\Sigma}{(x_i\ -\ \overline{x})}^2\ \ast\ \mathrm{\Sigma}(y_i\ -\overline{y})^2}$$
Let's tackle the expressions in this equation separately and drop in the numbers from our Ice Cream Sales example:
$$\mathrm{\Sigma}{(x_i\ -\ \overline{x})}^2=-3^2+0^2+3^2=9+0+9=18$$
$$\mathrm{\Sigma}{(y_i\ -\ \overline{y})}^2=-5^2+0^2+5^2=25+0+25=50$$
When we multiply the result of the two expressions together, we get:
$$18\times50\ =\ 900$$
This brings the bottom of the equation to:
$$\sqrt{900}=30$$
## 5. Finish the calculation, and compare our result with the scatterplot
Here's our full correlation coefficient equation once again:
$$r=\frac{\sum\left[\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)\right]}{\sqrt{\mathrm{\Sigma}\left(x_i-\overline{x}\right)^2\ \ast\ \mathrm{\Sigma}(y_i\ -\overline{y})^2}}$$
Let's pull in the numbers for the numerator and denominator that we calculated above:
$$r=\frac{30}{30}=1$$
A perfect correlation between ice cream sales and hot summer days! Of course, finding a perfect correlation is so unlikely in the real world that had we been working with real data, we’d assume we had done something wrong to obtain such a result.
But this result from the simplified data in our example should make intuitive sense based on simply looking at the data points. Let's look again at our scatterplot:
Now imagine drawing a line through that scatterplot. Would it look like a perfect linear fit?
## A picture can be worth 1,000 correlation coefficients!
Scatterplots, and other data visualizations, are useful tools throughout the whole statistical process, not just before we perform our hypothesis tests.
In fact, it’s important to remember that relying exclusively on the correlation coefficient can be misleading—particularly in situations involving curvilinear relationships or extreme outliers. In the scatterplots below, we are reminded that a correlation coefficient of zero or near zero does not necessarily mean that there is no relationship between the variables; it simply means that there is no linear relationship.
Similarly, looking at a scatterplot can provide insights on how outliers—unusual observations in our data—can skew the correlation coefficient. Let’s look at an example with one extreme outlier. The correlation coefficient indicates that there is a relatively strong positive relationship between X and Y. But when the outlier is removed, the correlation coefficient is near zero.
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A radical equation or irrational equation, is any one that has an unknown quantity under the radical sign (Figure I). To solve such equations, the roots containing the unknowns must be isolated in one member of the equation and then raise both members to a power equal to the root index. Sometimes, when doing the above, “wrong” solutions may appear, called strange roots. Therefore, always check the initial equation to detect and discard invalid solutions.
The equation »√x = a» is said to be a radical equation or one irrational equation. In these, the unknowns they are under a radical sign, that is, in the part subradical. To solve an irrational equation it is necessary remove the radical sign, which is achieved by raising the two members of the equation to a power equal to the root index:
Steps to solve an equation with radicals or irrational
To solve this type of equations the following steps are carried out:
1. Group the terms of the variable that are under the radical sign in one member of the equation and the rest of the terms in the other.
2. Raise both limbs to a power equal to the index of the root involved.
3. If the equation obtained in step 2 does not contain radicals, it must be solved normally. If, on the other hand, it has one or more radicals, steps 1 and 2 are repeated until obtaining an equation for radicals.
4. Substitute the values obtained in the previous step into the initial equation and evaluate if they comply with the latter.
Sometimes, solutions, called strange roots, may appear that do not meet the original equation. Therefore, each of the solutions in the original equation must be verified and those that do not comply with it must be discarded.
Example 1: Solve 3√ (4x + 3) = 3
We raise both members to the cube:
[3√ (4x + 3)]3 = 3= 4x + 3 = 27
We subtract 3 from both members of the equation:
4x = 27 - 3
We divide by 4 both members of the equation:
x = 24/4
Possible solution:
x = 6
Substituting x = 6 into the original equation to evaluate whether it is a strange root or not, we find that 3√ (4x + 3) = 3, is correct. So:
Solution: x = 6
Example 2: Solve 3√ [2 + 5√ (x + 5)] = 2
We raise both members to the cube:
{ 3√ [2 + 5√ (x + 5)]}3 = 23
2 + 5√ (x + 5) = 8
We subtract 2 from both members of the equation:
√ (x + 5) = 8 - 2
Let us observe that the unknown is under a radical sign:
√ (x + 5) = 6
Therefore, we square both members:
[√ (x + 5)]2 = 62
x + 5 = 36
We subtract 5 from both members of the equation:
x = 36 - 1
Possible solution:
x = 31
Substituting in the original equation to evaluate if it is a strange root or not, we check that 3√ {2 + 5√ [(31) + 5]} = 2, is correct. So:
Solution: x = 31
Example 3: Solve √ [2 + √ (x - 5)] = √ (13 - x)
We square both members:
{√ [2 + √ (x - 5)]}2 = {√ (13 - x)}2
2 + √ (x - 5) = 13 - x
We subtract 2 from both members of the equation:
√ (x - 5) = 13 - x - 2
Let us observe that the unknown is under a radical sign:
√ (x - 5) = 11 - x
Therefore, we square both members:
[√ (x - 5)]2 = (11 - x)2
We remove the radical squared; remember that (a - b)= to- 2ab + b:
x - 5 = 121 - 22x + x2
x2 - 23x + 126 = 0
Solving the equation, we obtain two possible solutions:
x = [- b ± √ (b2 - 4ac)] / 2a
With a = 1, b = -23, c = 126:
x = {23 ± √ [(- 23)2 - 4 (1) (126)]} / 2a
x1 = [23 + √ (529 - 504)] / 2 = (23 + √25) / 2 = (23 + 5) / 2 = 28/2 = 14
x2 = [23 - √ (529 - 504)] / 2 = (23 - √25) / 2 = (23 - 5) / 2 = 18/2 = 9
By substituting x1 = 14 in the original equation to evaluate whether it is a strange root or not, we check that √ [2 + √ (14 - 5)] ≠ √ (13 - 14), is not true. So:
x1 = 14 is a strange root
Now by substituting x2 = 9 in the original equation to evaluate if it is a strange root or not, we check that √ [2 + √ (9 - 5)] = √ (9 - 14), is correct. So:
Solution: x = 31
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Introduction
The quadratic formula is a formula in getting the roots of the quadratic equation $ax^2 + bx + c = 0$, where $a$, $b$ and $c$ are real numbers and $a \neq 0$. The quadratic formula is a generalization of completing the square, and it is usually used as a calculation strategy if a quadratic equation is not factorable.
In the graph of the quadratic function $f(x) = ax^2 + bx + c$, the values of $x$ in $ax^2 + bx + c = 0$, are the coordinates of $x$ where the curve pass through.
But how did mathematicians come up with the quadratic formula? How were they able to invent such complicated formula? In this post, we discuss the derivation of the quadratic formula.
Theorem
If $ax^2 + bx + c = 0$ where $a \neq 0$, then
$x = \displaystyle \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$
Proof
Given: $ax^2 + bx + c = 0$ and $a \neq 0$.
Divide both sides by $a$.
$\displaystyle x^2 + \frac{b}{a}x + \frac{c}{a} = 0$.
Add $\displaystyle \frac{-c}{a}$ to each side.
$\displaystyle x^2 + \frac{b}{a}x = \frac{-c}{a}$.
Complete the square by getting half of $\frac{b}{a}$, squaring it and then adding the result to both sides.
$x^2 + \displaystyle \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}$.
Write the left side as a binomial squared.
$\left ( x + \displaystyle \frac{b}{2a} \right )^2 = \displaystyle \frac{b^2}{4a^2} - \frac{c}{a}$
Simplify the right hand side of the equation.
$\left ( x + \displaystyle \frac{b}{2a} \right )^2 = \displaystyle \frac{b^2-4ac}{4a^2}$
Take the square root of both sides.
$x + \displaystyle \frac{b}{2a} = \frac{\pm \sqrt{b^2 - 4ac}}{2a}$.
Add $\frac{-b}{2a}$ to both sides.
$x = \displaystyle\frac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \blacksquare$.
***
Image via Wikipedia
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For many people, Mathematics has been an extremely challenging subject. However, these useful Math hacks will guide you! Ready to solve some equations? These math hacks will blow your mind!
These math hacks are something you weren’t taught back in school and I assure you that it’s something you wished you should have known earlier. Life in math class would have been so much at ease.
Maybe maths could have been a favorite subject if these tips and tricks were handy.
1. All 1’s to a Numeric Palindrome:
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321
2. Through 9 to 9 Through 1:
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
3. Sequential Input to All 1’s:
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 + 10 = 1111111111
4. Multiply 63 X 11
63 x 11 = 693
Step 1 = 63 = 6 () 3
Step 2 = 6+3 = 9
Step 3= 6 (9) 3
5. Multiply 5 digit number with 11
13432 x 11 = 147752
Step 1 = 13432 = 1()2
Step 2 = 1+3 = 4
Step 3 = 3+4 = 7
Step 4 = 4+3 = 7
Step 5 = 3+2 = 5
Step 6 = 1(4775)2
6. Multiples of 9
Then list the numbers 0 to 9 down the paper and next to that then write the numbers 0 to 9 up the paper. It means you just have to write 0 to 9 in both ascending and descending order in parallel lines. You’ll see in the photo above that this will give you the answer for each multiplying factor x 9!
That’s the order of 9 tables! Look for yourself!
0 9
1 8
2 7
3 6
4 5
5 4
6 3
7 2
8 1
9 0
7. Easy way to find 97 square
97 x 97 = 9409
First digit 97 = 100 -3
Second digit 97 – 3 = 94
Now 3 x 3 = 9
Result = 9409
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8. Easy fraction
Step 1 = 1/7 + 1/6
Step 2 = (7 + 6) / (7 x 6)
Step 3 = 13/42
9 Turn complex multiplication into simple
56 x 54 = 3024
Step 1 = 5 x (5 + 1) = 30
Step 2 = 6 x 4 = 24
Result = 3024
11. Multiply by 15
34 x 15
Step1: Place one zero after the number = 340
Step 2: Find half of the number = 170
Step 3: Add both numbers = 340+170 = 510
12. Subtracting from 1000
1,000 – 648
Step 1 = 9 – 6 = 3
Step 2 = 9 – 4 = 5
Step 3= 10 – 8 = 2
Result = 352
13. Simple multiplication by 4
The trick here is to multiply your number by 2, and then multiply it by 2 again.
58 x 4 = (58 x 2) + (58 x 2) = 116 +116 = 232
14. Quick square calculation
This trick will help you find the square root of a two-digit number that ends with a five. Multiply the first digit by itself and add 1 to it. Then write 25. That’s it!
For example – 352
3 x (3 + 1) & 25
3 x 4 & 25
12 & 25
1225
15. Sequential Inputs of 9
9 x 9 = 81
99 x 99 = 9801
999 x 999 = 998001
9999 x 9999 = 99980001
99999 x 99999 = 9999800001
999999 x 999999 = 999998000001
9999999 x 9999999 = 99999980000001
99999999 x 99999999 = 9999999800000001
999999999 x 999999999 = 999999998000000001
16. Sequential Inputs of 6
6 x 7 = 42
66 x 67 = 4422
666 x 667 = 444222
6666 x 6667 = 44442222
66666 x 66667 = 4444422222
666666 x 666667 = 444444222222
6666666 x 6666667 = 44444442222222
66666666 x 66666667 = 4444444422222222
666666666 x 666666667 = 444444444222222222
17. Sequential 8’s with 9
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
37 x 3 =111
37 x 6 =222
37 x 9 =333
37 x12 = 444
37 x 15 = 555
37 x 18 = 666
37 x 21 = 777
37 x 24 = 888
37 x 27 = 999
19. Quick multiplication 5
5887 x 5 = (5887/2) & 5 or 0
5887 x 5 = 2943.5 (fraction) & 5
5887 x 5 = 29,435
Divide a given number by 2. If the result is a whole number, add 0 at the end. If the result is not a whole number, ignore the decimal and just add 5 at the end.
20. Easy math trick to calculate tips
If you need to leave 15% of your order as tips at the restaurant, then here is a simple way to do it. First, find 10% of your sum (divide your sum by 10), and then add another half of the result. That’s your tip.
Say 15% of \$25
(10% of \$25) + (10% of \$25)/2
Results = \$2.50 + \$1.25 = \$3.75
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21. Division by 5
Dividing large numbers by five is, in fact, pretty simple. All you need to do is multiply your number by 2 and then move the decimal point.
197 /5 = ?
Step 1 = 197 x 2 = 394
Step 2 = 39.4
Results = 39.4
22. How to remember pi value
Pi is 3.1415926 (remembered with the mnemonic “May I have a large container of coffee“). Note that the number of letters in each word is equal to the corresponding digit of pi.
If you found this brilliant Math Hacks then let us know in comments. Also share it with your friends and help them too.
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Responses
1. Ashwani Kumar Kapoor
Excellent, would like to receive more of these.Tnx.
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# The minimum value of f(x,y)=x^2+13y^2-6xy-4y-2 is?
Jul 23, 2018
$f \left(x , y\right) = {x}^{2} + 13 {y}^{2} - 6 x y - 4 y - 2$
$\implies f \left(x , y\right) = {x}^{2} - 2 \cdot x \cdot \left(3 y\right) + {\left(3 y\right)}^{2} + {\left(2 y\right)}^{2} - 2 \cdot \left(2 y\right) \cdot 1 + {1}^{2} - 3$
$\implies f \left(x , y\right) = {\left(x - 3 y\right)}^{2} + {\left(2 y - 1\right)}^{2} - 3$
Minimum value of each squared expression must be zero.
So ${\left[f \left(x , y\right)\right]}_{\text{min}} = - 3$
Jul 23, 2018
There is a relative minimum at $\left(\frac{3}{2} , \frac{1}{2}\right)$ and $f \left(\frac{3}{2} , \frac{1}{2}\right) = - 3$
#### Explanation:
I think that we must calculate the partial derivatives.
Here,
$f \left(x , y\right) = {x}^{2} + 13 {y}^{2} - 6 x y - 4 y - 2$
The first partial derivatives are
$\frac{\partial f}{\partial x} = 2 x - 6 y$
$\frac{\partial f}{\partial y} = 26 y - 6 x - 4$
The critical points are
$\left\{\begin{matrix}2 x - 6 y = 0 \\ 26 y - 6 x - 4 = 0\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}3 y = x \\ 26 y - 6 \cdot 3 y - 4 = 0\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}3 y = x \\ 8 y = 4\end{matrix}\right.$
$\iff$, $\left\{\begin{matrix}x = \frac{3}{2} \\ y = \frac{1}{2}\end{matrix}\right.$
The second partial derivatives are
$\frac{{\partial}^{2} f}{\partial {x}^{2}} = 2$
$\frac{{\partial}^{2} f}{\partial {y}^{2}} = 26$
$\frac{{\partial}^{2} f}{\partial x \partial y} = - 6$
$\frac{{\partial}^{2} f}{\partial y \partial x} = - 6$
The determinant of the Hessian matrix is
$D \left(x , y\right) = | \left(\frac{{\partial}^{2} f}{\partial {x}^{2}} , \frac{{\partial}^{2} f}{\partial x \partial y}\right) , \left(\frac{{\partial}^{2} f}{\partial {y}^{2}} , \frac{{\partial}^{2} f}{\partial y \partial x}\right) |$
$= | \left(2 , - 6\right) , \left(- 6 , 26\right) |$
$= 52 - 36$
$= 16 > 0$
As $D \left(x , y\right) > 0$
and
$\frac{{\partial}^{2} f}{\partial {x}^{2}} = 2 > 0$
There is a relative minimum at $\left(\frac{3}{2} , \frac{1}{2}\right)$
And
$f \left(\frac{3}{2} , \frac{1}{2}\right) = {1.5}^{2} + 13 \cdot {0.5}^{2} - 6 \cdot 1.5 \cdot 0.5 - 4 \cdot 0.5 - 2 = - 3$
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Suppose S and T are mutually exclusive events P(S)=20.
This question aims to find the P (S) or P (T) of two mutually exclusive events S and T if the probability of P (S) is given.
Two events are called mutually exclusive if they do not occur at the same time or simultaneously. For Example, when we toss a coin, there are two possibilities whether the head will be displayed or the tail will be displayed on its return. It means both head and tail cannot occur at the same time. It is a mutually exclusive event and the probability of these events occurring at the same time becomes zero. There is another name for mutually exclusive events and that is the disjoint event.
The representation of mutually exclusive events is given as:
$P (A \cap B) = 0$
The disjoint events have a rule of addition that is only true only one event is occurring at a time and the sum of this event is the probability of occurrence. Assume two events $A$ or $B$ are occurring then their probability is given by:
$P (A Or B) = P (A) + P (B)$
$P (A \cup B) = P (A) + P (B)$
When two events $A$ and $B$ are not mutually exclusive events then the formula changes to
$P (A \cup B) = P (A) + P (B) – P (A \cap B)$
If we consider that $A$ and $B$ are mutually exclusive events which means the probability of their occurrence at the same time becomes zero. It can be shown as:
$P (A \cap B) = 0$
The addition rule of probability is as follows:
$P (A \cup B) = P (A) + P (B) – P (A \cap B)$
This rule in terms of S and T can be written as:
$P (S \cup T) = P (S) + P (T) – P (S \cap T)$
Consider the probability of event T is $P (T) = 10$.
By putting values:
$P (S \cup T) = 20 + 10 – P (S \cap T)$
$P (S \cup T) = 30 – P (S \cap T)$
According to the definition of mutually exclusive events:
$P (S \cap T) = 0$
$P (S \cup T) = 30 – 0$
$P (S \cup T) = 30$
Numerical Solution
The probability of occurrence of mutually exclusive events is $P (S \cup T) = 30$
Example
Consider two mutually exclusive events M and N having P (M) = 23 and P (N) = 20. Find their P (M) or P (N).
$P (M \cup N) = 23 + 20 – P (M \cap N)$
$P (M \cup N) = 43 – P (M \cap N)$
According to the definition of mutually exclusive events:
$P (M \cap N) = 0$
$P (M \cup N) = 43 – 0$
$P (M \cup N) = 43$
Image/Mathematical drawings are created in Geogebra.
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# 11.6 Solving systems with gaussian elimination
Page 1 / 13
In this section, you will:
• Write the augmented matrix of a system of equations.
• Write the system of equations from an augmented matrix.
• Perform row operations on a matrix.
• Solve a system of linear equations using matrices.
Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries.
We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables . In this section, we will revisit this technique for solving systems, this time using matrices.
## Writing the augmented matrix of a system of equations
A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix .
For example, consider the following $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ system of equations.
$\begin{array}{l}3x+4y=7\\ 4x-2y=5\end{array}$
We can write this system as an augmented matrix:
We can also write a matrix containing just the coefficients. This is called the coefficient matrix .
$\left[\begin{array}{cc}3& 4\\ 4& -2\end{array}\right]$
A three-by-three system of equations such as
has a coefficient matrix
$\left[\begin{array}{rrr}\hfill 3& \hfill -1& \hfill -1\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill -3\end{array}\right]$
and is represented by the augmented matrix
Notice that the matrix is written so that the variables line up in their own columns: x -terms go in the first column, y -terms in the second column, and z -terms in the third column. It is very important that each equation is written in standard form $\text{\hspace{0.17em}}ax+by+cz=d\text{\hspace{0.17em}}$ so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.
Given a system of equations, write an augmented matrix.
1. Write the coefficients of the x -terms as the numbers down the first column.
2. Write the coefficients of the y -terms as the numbers down the second column.
3. If there are z -terms, write the coefficients as the numbers down the third column.
4. Draw a vertical line and write the constants to the right of the line.
## Writing the augmented matrix for a system of equations
Write the augmented matrix for the given system of equations.
The augmented matrix displays the coefficients of the variables, and an additional column for the constants.
Write the augmented matrix of the given system of equations.
$\begin{array}{l}4x-3y=11\\ 3x+2y=4\end{array}$
$\left[\begin{array}{cc}4& -3\\ 3& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}|\begin{array}{c}11\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}4\end{array}\right]$
## Writing a system of equations from an augmented matrix
We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form.
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
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Math and Arithmetic
Education
# How do you simplify 3 2x 4y 7x 3 10x?
012
###### 2015-02-22 11:29:38
Unfortunately, the browser used by Answers.com for posting questions is incapable of accepting mathematical symbols. This means that we cannot see the mathematically critical parts of the question. We are, therefore unable to determine what exactly the question is about and so cannot give a proper answer to your question.
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## Related Questions
7x - 4y + x = 8x - 4y Depending on what your teacher is looking for, I might even go a step further to simplify it to 4(2x-y).
3x - 2y + 7x + 4y = 3x + 7x - 2y +4y = (3+7)x + (-2+4)y = 10x + 2y
10x - 3 - 2x = 4 7x - 3 = 4 7x = 7 x = 1
According to the distributive property, 7x(4y + 4) - 8(4y + 4) = (7x - 8)(4y + 4). Is this question true or false?>
it means to put together the terms that have the same variable. so if you had the equation: 2x+4y+5x it would simplify to 7x+4y because you would combine to the two x-variables.
7x + 13 = 2x + 3 First, subtract 3 from both sides.7x + 13 - 3 = 2x + 3 - 3Simplify both sides.7x + 10 = 2xNow, subtract 2x from both sides.7x - 2x + 10 = 2x - 2xSimplify again.5x + 10 = 0Finally, subtract 10 from both sides, simplify, and divide both sides by 5.5x + 10 - 10 = 0 - 105x = -10x = -2So the answer is x = -2.
(14x^3 - 10x^2) = 2x^2(7x - 5) 63x - 45 = 9(7x - 5) 14x^3 - 10x^2 + 63x - 45 = (2x^2 + 9)(7x - 5)
10x² - 7x - 12 = (5x + 4)(2x - 3)
18 + 7x - 12 + 5x = ?18 - 12+ 7x - 5x = ?6 + 2x = ?
To function form. - 7X - 4Y = 16 - 4Y = 7X + 16 Y = - 7/4X - 4 ------------------------ The slope is, - 7/4 --------
10x + 5 which is also equal to 5(2x + 1)
simplify it, 7x-12=x+2+2x-3x 7x=2+12+3x-3x 7x=14 x=2
7x+4y = 39 2x+4y = 14 Subtract the equations in order to eliminate y: 5x = 25 Divide both sides by 5 to find the value of x: x = 5 Substitute the value of x into the original equations to find the value of y: Therefore: x = 5 and y = 1
the simplest form of 5x + 3y +2x - 6y is 7x-3y
2x2 + 7x - 15 = 02x2 + 10x - 3x - 15 = 0(2x2 - 3x) + (10x - 15) = 0x(2x - 3) + 5(2x - 3) = 0(2x - 3)(x + 5) = 0(2x - 3) = 0 or (x + 5) = 0x = 3/2 or x = -5
12 - 3x = 24 - 10x so 7x = 12 making x = 12/7 or one and five-sevenths.
The greatest common factor in the two terms is [ 4y ].The factored form of the expression is4y (y + 7x)
"Combine like terms" means you have to join 2 or more similar terms into one term. Doing this can simplify an equation or expression. For example, 10x+3+7x 10x and 7x are like terms. Therefore, the coefficients 10 and 7 can be added together. So: 10x+3+7x=17x+3
7x-4020+3x= 7x+-4020+3x= 7x+3x+-4020= 10x+-4020= 10x-4020
Answers to pick: -2x - y = 4 2x + y = -2 -8x + 4y = -16 -4x + 2y = -8 -x - 2y = 6 -7x + 4y = 16 2x + 3y = 6 3x + 2y = 4
###### Math and ArithmeticFactoring and MultiplesAlgebraSchool Subjects
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## LCM and HCF
#### LCM and HCF
1. Three men step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. The minimum distance each should cover so that all can cover the distance in complete steps is
1. We find LCM of 63, 70 and 77 cm.
Required distance = LCM of 63, 70 and 77 cm.
##### Correct Option: C
We find LCM of 63, 70 and 77 cm.
∴ LCM = 7 × 9 × 10 × 11 = 6930
Required distance = LCM of 63, 70 and 77 cm. = 6930
1. Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case.
1. As we know that When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.
Required number = (LCM of 15, 20, 36 and 48) + 3
∴ LCM ( k ) = 2 × 2 × 3 × 5 × 3 × 4 = 720
##### Correct Option: D
As we know that When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.
Required number = (LCM of 15, 20, 36 and 48) + 3
∴ LCM ( k ) = 2 × 2 × 3 × 5 × 3 × 4 = 720
Here , r = 3
∴ Required number = k + r = 720 + 3 = 723
1. The greatest 4-digit number exactly divisible by 10, 15, 20 is
1. We know that Greatest n digit number which when divided by three numbers p,q,r leaves no remainder will be Required Number = (n – digit greatest number) – R , R is the remainder obtained on dividing greatest n digit number by L.C.M of p.q,r.
LCM of 10, 15 and 20 = 60
Largest 4-digit number = 9999
##### Correct Option: B
We know that Greatest n digit number which when divided by three numbers p,q,r leaves no remainder will be Required Number = (n – digit greatest number) – R , R is the remainder obtained on dividing greatest n digit number by L.C.M of p.q,r.
LCM of 10, 15 and 20 = 60
Largest 4-digit number = 9999
∴ Required number = 9999 – remainder = 9999 – 39 = 9960
1. The smallest perfect square divisible by each of 6, 12 and 18 is
1. The LCM of 6, 12 and 18 = 36 = 62
##### Correct Option: D
The LCM of 6, 12 and 18 = 36 = 62 = 36
Hence , required answer is 36.
1. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is
1. We find LCM of 20, 28, 32 and 35
∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120
∴ Required number = 5834 – LCM of 20, 28, 32 and 35
##### Correct Option: B
We find LCM of 20, 28, 32 and 35
∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120
∴ Required number = 5834 – LCM of 20, 28, 32 and 35
∴ Required number = 5834 – 1120 = 4714
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# 4.7 - Perimeter and Area
The student will
• solve practical problems that involve determining perimeter and area in U.S. Customary and metric units.
### BIG IDEAS
• Doctors measure our height to see how tall we have gotten.
• Linear measurement is key when finding area, perimeter, or volume. If we want to buy a rug, paint a wall, or frame a picture, we have to measure in order to know how much to buy.
• Scientist use linear measurements when collecting data on experiments.
• Measurements also help students to understand the concepts of fractions. When we are using measurement tools such as a ruler we are practicing how to use fractions on a number line.
### UNDERSTANDING THE STANDARD
• Perimeter is the path or distance around any plane figure.
• To determine the perimeter of any polygon, determine the sum of the lengths of the sides.
• Area is the surface included within a plane figure. Area is measured by the number of square units needed to cover a surface or plane figure.
• Students should have opportunities to investigate and discover, using manipulatives, the formulas for the area of a square and the area of a rectangle.
• Area of a square = side length × side length
• Area of rectangle = length × width
• Perimeter and area should always be labeled with the appropriate unit of measure.
### ESSENTIALS
The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to
• Determine the perimeter of a polygon with no more than eight sides, when the lengths of the sides are given, with diagrams.
• Determine the perimeter and area of a rectangle when given the measure of two adjacent sides, with and without diagrams.
• Determine the perimeter and area of a square when the measure of one side is given, with and without diagrams.
• Solve practical problems that involve determining perimeter and area in U.S. Customary and metric units.
### KEY VOCABULARY
Updated: Aug 22, 2018
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# Download Lesson 1: Factors and Multiples of Whole Numbers 48 48
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```Date: ______________
Lesson 1: Factors and Multiples of Whole Numbers
Multiples of 12:
Factors of 12:
Prime number:
________________________________________________________________
Composite number:
Prime factors:
Prime factorization:
How to Determine the Prime Factor of a Number:
Factor tree method
Hockey stick method
(split composite factors)
(repeated division by prime factors)
48
48
Try these:
a) 108
Math 10FP
b) 252
Marsh
Greatest Common Factor (GCF) - ________________________________________________
Example 1: Determine the GCF of 24 and 42.
Method 1 – list all the factors of each number (“rainbow”).
24
42
So the greatest common factor is ______.
Method 2 – Write the prime factorizations of each number.
24
42
24 =
1) Highlight the factors that appear in both factorizations.
42 =
2) Multiply them together: GCF = ___________
Example 2: Determine the GCF of 27, 126 and 144.
27
27 = ________________=________
126 = ________________=_________
144 = ________________=_______
126
144
Take common factor to lowest exponent.
GCF = __________ = _____
Math 10FP
Marsh
Least Common Multiple (LCM) - ________________________________________________
________________________________________________
Example 3: Determine the LCM of 28, 42 and 63.
Method 1 – List multiples of each number until the same multiple appears on all lists
28:
42:
63:
So the LCM is __________.
Method 2 – Write the prime factorizations of each number
28
42
63
28 =
The greatest power of 2 in any list is _________________.
42 =
The greatest power of 3 in any list is_________________. 63 =
The greatest power of 7 in any list is _________________.
1) Highlight the greatest power of each prime factor in any list.
2) Multiply them together.
LCM = _____________ = ______________ = ______
page 140 #3-5ace,8-11ac,13,17
Math 10FP
Marsh
```
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In a cyclic quadrilateral ABCD, where ∠ABC = 69° and ∠ACB = 31°, we can find ∠BDC using the properties of cyclic quadrilaterals.
In a cyclic quadrilateral, the opposite angles are supplementary, meaning they add up to 180°. Since ∠ACB = 31°, this angle is part of triangle ABC which is inscribed in the circle. The angle ∠ACB is subtended by the same arc as angle ∠ADB (the angle at the opposite side of the circle). Therefore, ∠ADB = 31°.
Since ∠ADB and ∠BDC are opposite angles in cyclic quadrilateral ABCD, ∠BDC = 180° – ∠ADB = 180° – 31° = 149°.
Let’s discuss in detail
## Cyclic Quadrilaterals and Angle Relationships
In the study of circle geometry, cyclic quadrilaterals offer fascinating insights into angle relationships. A cyclic quadrilateral is a four-sided figure with all its vertices lying on the circumference of a circle. In such a quadrilateral, certain angle properties hold true, which are crucial for solving problems like the one at hand. We are given a cyclic quadrilateral ABCD with angles ∠ABC = 69° and ∠ACB = 31°, and our task is to find the measure of ∠BDC.
### Understanding the Properties of Cyclic Quadrilaterals
One of the key properties of cyclic quadrilaterals is that the sum of each pair of opposite angles is 180°. This supplementary relationship is a direct consequence of the quadrilateral’s vertices lying on a circle. This property will be instrumental in determining the unknown angle ∠BDC in quadrilateral ABCD.
#### Analyzing Triangle ABC within the Cyclic Quadrilateral
In the cyclic quadrilateral ABCD, triangle ABC is a critical component. The given angles, ∠ABC = 69° and ∠ACB = 31°, belong to this triangle. Since triangle ABC is inscribed in the circle, the angle ∠ACB is subtended by arc AD. This subtended angle relationship is vital for understanding the angles formed at other points on the circle.
The angle ∠ACB (31°) in triangle ABC subtends the same arc as angle ∠ADB in the cyclic quadrilateral. Since these angles are subtended by the same arc, they are equal. Therefore, ∠ADB, which is the angle at the opposite side of the circle from ∠ACB, is also 31°. This equality is a direct application of the angle subtended by the same arc theorem in circle geometry.
##### Calculating Angle ∠BDC
With ∠ADB determined as 31°, we can now calculate ∠BDC. Since ∠ADB and ∠BDC are opposite angles in the cyclic quadrilateral ABCD, and opposite angles in a cyclic quadrilateral are supplementary, ∠BDC equals 180° – ∠ADB. Therefore, ∠BDC = 180° – 31° = 149°. This calculation is based on the supplementary nature of opposite angles in a cyclic quadrilateral.
###### Geometric Principles in Cyclic Quadrilaterals
In conclusion, by applying the properties of cyclic quadrilaterals and the theorem of angles subtended by the same arc, we find that ∠BDC in the cyclic quadrilateral ABCD is 149°. This exercise demonstrates the elegance and interconnectedness of geometric principles, particularly in the context of cyclic quadrilaterals. Understanding these relationships allows for a deeper appreciation of geometry’s role in elucidating spatial relationships and solving complex problems.
Discuss this question in detail or visit to Class 9 Maths Chapter 9 for all questions.
Questions of 9th Maths Exercise 9.3 in Detail
In Figure, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. In Figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. In Figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. In Figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that ∠ACP = ∠QCD. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. Prove that a cyclic parallelogram is a rectangle.
Last Edited: January 6, 2024
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## Simplifying Algebraic Fractions involving Quadratic Terms: Common Factors
As an experienced mathematics tutor, I have found that simplifying algebraic fractions involving quadratic terms can be a challenging task for many students. However, with a solid understanding of common factors and the right approach, anyone can master this essential skill. In this article, we will explore the process of simplifying algebraic fractions with quadratic terms, focusing on identifying and utilising common factors.
## Understanding Algebraic Fractions
Before we dive into the simplification process, let’s briefly review what algebraic fractions are. An algebraic fraction is a fraction in which the numerator and/or denominator contain algebraic expressions, such as polynomials. For example, $\displaystyle \frac{2x^2 + 6x + 4}{x + 1}$ is an algebraic fraction with a quadratic numerator and a linear denominator.
## Simplifying Algebraic Fractions with Common Factors
The key to simplifying algebraic fractions lies in identifying common factors in the numerator and denominator. By dividing both the numerator and denominator by their common factors, we can often simplify the fraction to its lowest terms.
### Step 1: Factorise the numerator and denominator
To begin, we must factorise the numerator and denominator of the algebraic fraction. This involves breaking down the quadratic expressions into their simplest factors.
For quadratic expressions, we can use various factorisation techniques such as:
• Trial and error
• Grouping
• Difference of squares
• Sum or difference of cubes
Let’s consider the example mentioned earlier:
$\displaystyle \frac{2x^2 – 8}{3x^2 + 15x + 18}$
To factorise the numerator, we can use the difference of squares formula:
$\displaystyle 2x^2 – 8 = 2(x^2 – 4) = 2(x + 2)(x – 2)$
For the denominator, we can use the following steps:
$\displaystyle 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 2)(x + 3)$
### Step 2: Identify common factors
Once we have factorised the numerator and denominator, the next step is to identify any common factors between them. These common factors can be numbers, variables, or even more complex expressions.
In our example, we can see that there is a common factor of 2 between the numerator and denominator:
$\displaystyle \frac{2(x + 2)(x – 2)}{3(x + 2)(x + 3)}$
### Step 3: Divide out common factors
To simplify the algebraic fraction, we divide both the numerator and denominator by their common factors.
In our example, we can divide out the common factor of 2:
$\displaystyle \frac{2(x + 2)(x – 2)}{3(x + 2)(x + 3)} = \frac{2}{3} \times \frac{(x + 2)(x – 2)}{(x + 2)(x + 3)} = \frac{2}{3} \times \frac{x – 2}{x + 3}$
Thus, the simplified fraction is $\displaystyle \frac{2(x – 2)}{3(x + 3)}$.
## More Examples
Let’s work through a few more examples to reinforce the concept of simplifying algebraic fractions with common factors.
### Example 1
$\displaystyle \frac{3x^2 + 6x}{9x}$
First, let’s factorise the numerator and denominator:
• Numerator: $\displaystyle 3x^2 + 6x = 3x(x + 2)$
• Denominator: $\displaystyle 9x = 3x \times 3$
Now, we can identify the common factor $3x$ and divide it out:
$\displaystyle \frac{3x(x + 2)}{3x \times 3} = \frac{x + 2}{3}$
### Example 2
$\displaystyle \frac{2x^2 – 18}{3x^2 + 21x + 36}$
Let’s factorise the numerator and denominator:
• Numerator: $\displaystyle 2x^2 – 18 = 2(x^2 – 9) = 2(x + 3)(x – 3)$
• Denominator: $\displaystyle 3x^2 + 21x + 36 = 3(x^2 + 7x + 12) = 3(x + 3)(x + 4)$
The common factor between the numerator and denominator is $(x + 3)$. Dividing it out, we get:
$\displaystyle \frac{2(x + 3)(x – 3)}{3(x + 3)(x + 4)} = \frac{2}{3} \times \frac{(x + 3)(x – 3)}{(x + 3)(x + 4)} = \frac{2}{3} \times \frac{x – 3}{x + 4}$
### Example 3
$\displaystyle \frac{4x^2 – 1}{2x – 1}$
Factorising the numerator and denominator:
• Numerator: $\displaystyle 4x^2 – 1 = (2x + 1)(2x – 1)$
• Denominator: $\displaystyle 2x – 1$
The common factor is $(2x – 1)$, so we divide it out:
$\displaystyle \frac{(2x + 1)(2x – 1)}{2x – 1} = 2x + 1$
## Practice Problems
To solidify your understanding of simplifying algebraic fractions with common factors, try solving these practice problems:
1. $\displaystyle \frac{x^2 – 25}{x – 5}$
2. $\displaystyle \frac{6x^2 + 18x}{12x}$
3. $\displaystyle \frac{x^2 + 7x + 12}{x^2 + 4x + 3}$
4. $\displaystyle \frac{9x^2 – 16}{3x + 4}$
Solutions:
1. $x + 5$
2. $\displaystyle \frac{x + 3}{2}$
3. $\displaystyle \frac{x + 4}{x + 1}$
4. $3x-4$
## Combining Common Factors with Other Techniques
In some cases, you may need to combine the common factor technique with other factorisation methods to simplify algebraic fractions fully. For example, you might encounter a fraction where the numerator and denominator share a common factor, but the remaining terms require factorisation using the difference of squares or sum/difference of cubes formulas.
Let’s consider an example:
$\displaystyle \frac{2x^3 – 16x}{x^2 – 9}$
In this case, both the numerator and denominator share a common factor of $x$. Let’s factorise them:
• Numerator: $\displaystyle 2x^3 – 16x = 2x(x^2 – 8)$
• Denominator: $\displaystyle x^2 – 9 = (x + 3)(x – 3)$
Now, we can divide out the common factor $x$:
$\displaystyle \frac{2x(x^2 – 8)}{x(x + 3)(x – 3)} = \frac{2(x^2 – 8)}{(x + 3)(x – 3)}$
To simplify further, we need to factorise $x^2 – 8$ using the difference of squares formula:
$\displaystyle x^2 – 8 = x^2 – (\sqrt{8})^2 = (x + 2\sqrt{2})(x – 2\sqrt{2})$
Substituting this back into our fraction:
$\displaystyle \frac{2(x + 2\sqrt{2})(x – 2\sqrt{2})}{(x + 3)(x – 3)}$
The fraction cannot be simplified further, so this is our final answer.
## Conclusion
Simplifying algebraic fractions with quadratic terms may seem intimidating at first, but by following the steps outlined in this article, you can break down the process into manageable parts. Remember to factorise the numerator and denominator, identify common factors, and divide them out to simplify the fraction. With practice and perseverance, you’ll soon be able to confidently tackle even the most complex algebraic fractions.
• When factorising quadratic expressions, always look for common factors first. This can save you time and simplify the process.
• If you’re having trouble factorising a quadratic expression, try using the quadratic formula: $\displaystyle x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic expression in standard form $(ax^2 + bx + c)$.
• Remember that if the numerator and denominator have no common factors, the fraction is already in its simplest form.
• When simplifying algebraic fractions, be careful not to divide by zero. If you end up with a denominator that equals zero for a specific value of the variable, note that the fraction is undefined for that value.
By keeping these tips in mind and practising regularly, you’ll soon be able to simplify algebraic fractions with quadratic terms like a pro!
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## Induction Made Simple: The Ultimate Guide
“Induction Made Simple: The Ultimate Guide” is your gateway to mastering the art of mathematical induction, demystifying a powerful tool in mathematics. This ultimate guide…
Simplifying Algebraic Fractions involving Quadratic and Linear Terms: Difference of Squares As an experienced mathematics tutor, I have noticed that many students struggle with simplifying…
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# CLASS-4CONCEPT OF FRACTION
CONCEPT OF FRACTION
If anything is divided into two or more pieces, then each piece is called the fraction of the whole thing.
In the above picture we can observe that, the rectangular diagram has been divided into 9 small rectangular blocks and among them, 5 blocks have been colored.
Colored Blocks
So, proportionate/fraction of colored block is = -----------------
Total blocks
= 5/9
It is happen only through 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 5/9
And , similarly proportionate/fraction of color less block is = 1/9 + 1/9 + 1/9 + 1/9 = 4/9.
In the above picture big blocks are divided into 12 blocks, so the value of every block is 1/12. There are three different colors, so we can say that there are three colors of blocks with the same value.
imagine the value of big block 1
Each small block’s value defines = ----------------------------
Big block divided into no. of pieces
= 1/12
Total no. of Orange blocks = 1/12 + 1/12 + 1/12 + 1/ 12 = 4/12
Total no. of Blue blocks = 1/12 + 1/12 + 1/12 = 3/12
Total no. of Red blocks = 1/12 + 1/12 + 1/12 + 1/12 + 1/12
= 5/12.
So, Big blocks = Tot. Orange blocks + Tot. Blue Blocks + Tot. Red Blocks
= 4/12 + 3/12 + 5/12
= (4+3+5)/12
= 12/12 = 1
So, it can be proved that 24 small different color blocks by addition makes big blocks or small blocks are the part of the said big blocks.
In the above picture big blocks are divided into 24 blocks, so the value of every block is 1/24. There are three different colors, so we can say that there are three colors of blocks with the same value.
Imagine value of big block 1
Each small block’s value defines = -----------------------------
Big block divided into no. of pieces
= 1 / 24
Total no. of Purple blocks = 1/24 + 1/24 + 1/24 + 1/ 24 + 1/24
= 5/24
Total no. of Blue blocks = 1/24 + 1/24 + 1/24 + 1/24 + 1/24
= 5/24
Total no. of Red blocks = 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24
= 8/24
Total no. of Green blocks = 1/24 + 1/24 + 1/24 + 1/24 + 1/24 + 1/24
= 6/24.
So, Big blocks = Tot. Purple blocks + Tot. Blue Blocks + Tot. Red Blocks + Tot. Green blocks
= 5/24 + 5/24 + 8/24 + 6/24
= (5+5+8+6) / 24
= 24/24 = 1
So, it can be proved that 24 small different color blocks by addition makes big blocks or small blocks are the part of the said big blocks.
Now we will learn, how to add & subtract fraction value
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# Algebra II : Basic Single-Variable Algebra
## Example Questions
### Example Question #1 : Other Mathematical Relationships
Sarah notices her map has a scale of . She measures between Beaver Falls and Chipmonk Cove. How far apart are the cities?
Explanation:
is the same as
So to find out the distance between the cities
### Example Question #2 : Direct Proportionality
If an object is hung on a spring, the elongation of the spring varies directly as the mass of the object. A 20 kg object increases the length of a spring by exactly 7.2 cm. To the nearest tenth of a centimeter, by how much does a 32 kg object increase the length of the same spring?
Explanation:
Let be the mass of the weight and the elongation of the spring. Then for some constant of variation
We can find by setting from the first situation:
so
In the second situation, we set and solve for :
which rounds to 11.5 centimeters.
### Example Question #1 : Other Mathematical Relationships
Sunshine paint is made by mixing three parts yellow paint and one part red paint. How many gallons of yellow paint should be mixed with two quarts of red paint?
(1 gallon = 4 quarts)
Explanation:
First set up the proportion:
x =
Then convert this to gallons:
### Example Question #1 : Other Mathematical Relationships
Sally currently has 192 books. Three months ago, she had 160 books. By what percentage did her book collection increase over the past three months?
Explanation:
To find the percentage increase, divide the number of new books by the original amount of books:
### Example Question #1 : Proportionalities
Find for the proportion .
Explanation:
To find x we need to find the direct proportion. In order to do this we need to cross multiply and divide.
From here we mulitply 100 and 1 together. This gets us 100 and now we divide 100 by 4 which results in
### Example Question #1 : Basic Single Variable Algebra
On a map of the United States, Mark notices a scale of . If the distance between New York City and Los Angeles in real life is , how far would the two cities be on Mark's map?
Explanation:
If the real distance between the two cities is , and = , then we can set up the proportional equation:
### Example Question #1 : Basic Single Variable Algebra
If and , find and .
Explanation:
We cannot solve the first equation until we know at least one of the variables, so let's solve the second equation first to solve for . We therefore get:
With our , we can now find x using the first equation:
We therefore get the correct answer of and .
### Example Question #8 : Direct Proportionality
If an object is hung on a spring, the elongation of the spring varies directly with the mass of the object. A 33 kilogram object increases the length of a spring by exactly 6.6 centimeters. To the nearest tenth of a kilogram, how much mass must an object posess to increase the length of that same spring by exactly 10 centimeters?
Explanation:
Let be the mass of the weight and the elongation of the spring, respectively. Then for some constant of variation
.
We can find by setting :
Therefore .
Set and solve for :
kilograms
### Example Question #9 : Direct Proportionality
If is directly proportional to and when at , what is the value of the constant of proportionality?
Explanation:
The general formula for direct proportionality is
where is the proportionality constant. To find the value of this , we plug in and
Solve for by dividing both sides by 12
So .
### Example Question #10 : Direct Proportionality
The amount of money you earn is directly proportional to the nunber of hours you worked. On the first day, you earned $32 by working 4 hours. On the second day, how many hours do you need to work to earn$48.
Explanation:
The general formula for direct proportionality is
where is how much money you earned, is the proportionality constant, and is the number of hours worked.
Before we can figure out how many hours you need to work to earn $48, we need to find the value of . It is given that you earned$32 by working 4 hours. Plug these values into the formula
Solve for by dividing both sides by 4.
So . We can use this to find out the hours you need to work to earn $48. With , we have Plug in$48.
Divide both sides by 8
So you will need to work 6 hours to earn \$48.
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0
# What is one and two thirds add four and five sixths?
Updated: 4/28/2022
Wiki User
9y ago
Before you can add fractions, you first need to convert the fractions so they have common denominators. Two thirds is equal to four sixths, so now we can add the fractions that have common denominators. Four sixths and five sixths equals eleven sixths, or one and five sixths left over. Now, add your whole numbers...one, four and the one gained when we converted the fraction equals six and five sixths.
Wiki User
9y ago
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Q: What is one and two thirds add four and five sixths?
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Related questions
### What is the answer to two thirds plus one sixth?
five sixths you change 2/3 to 4/6 then add
### What does two thirds plus four sixths equal?
When you simplify four sixths (4/6), you get two thirds (2/3), making this problem easier. Since both fractions now have the common denominators of 3, you just add the numerators together to get four thirds (4/3), or one and one third (1 1/3).
### What is 1 third plus 1 half?
to get this answer you need to turn them both into sixths and add them 1 thied is 2 sixths and 1 half is 3 sixths add them and you get5 sixths,which leaves 1 sixth to be the answer.The answer is 1 sixth.
### What is One half plus one sixths?
Four sixths. ^^ You see, 6 is the common denominator, so, you work with one half. Now, you have to make 6 the denominator, right? So, you multiply 2 by 3 and 1 also by three, and then you have to add three sixths and one sixths and you get four sixths.
1 and 1/6 or 7/6
### What formula generates this sequence four sevenths five eighths two thirds seven tenths?
Each time, you add one to the numerator and one to the denominator.
### Is five sixths a full dispenser?
No, five sixths is not a full dispenser. A full dispenser would have all six equal parts filled.
2 1/6
1 and 1/6
### How do you simplify one third plus one third?
if you add one third and one third you get two sixths then if you divide it by 2 each you will get it back to one thirds
### How do you add three fractions without the same denominators?
When you want to add fractions that have different denominators, you have to convert them into fractions that do have the same denominators. So for example, how much is a half plus a third? Both of these can be converted into sixths, so one half equals three sixths, and one third equals two sixths. Three sixths plus two sixths equals five sixths. That's the answer. If you have three fractions with different denominators, the procedure is the same. Convert all three into fractions with the same denominators, then add them.
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# Basics: Vectors, the Other Dimensional Number
There's another way of working with number-like things that have multiple dimensions in math, which is very different from the complex number family: vectors. Vectors are much more intuitive to most people than the the complex numbers, which are built using the problematic number i.
A vector is a simple thing: it's a number with a direction. A car can be going 20mph north - 20mph north is a vector quantity. A 1 kilogram object experiences a force of 9.8 newtons straight down - 9.8n down is a vector quantity.
To be precise about the definition, a vector is a quantity qualitatively different parts: a magnitude (aka, size) and a direction.
When we're talking about vectors, we also generally add the term scalar. A scalar is what we were calling a number before we got to vectors - it's a quantity without a direction. 20mph is a scalar quantity called speed; 20mph northwest is a vector quantity called velocity.
Vectors are fascinating things, which can be used to describe all sorts of physical phenomena. In physics, you simple can't get away from them - they're everywhere. Velocity, acceleration, force, momentum - they're all vectors.
Normally, we draw vectors as arrows, where the length of the arrow corresponds to the magnitude of the vector, and the direction of the arrow is (obviously) the direction of the vector.
We generally represent vectors in one of two ways. One of them is by length + angle - for example, for a vector in a plane, we might use 10 @ 60 degrees, where the angle is measured relative to the X axis. The other representation is based on components. Take that vector 10@60 degrees, and put its starting point on the origin of a 2d graph. Draw a horizontal line from its tip to the y axis - that's it's y component, written Ay; draw a vertical line from its tip to the x axis - that's it's X component, written Ax. The result is a pair [5.00,8.66]. In general, an n-dimensional vector can be written as n scalars between brackets, so a three dimnesional vector could be written [x,y,z], and so on. If we aren't using the components, we describe the magnitude of a vector A by |A|.
And of course, you can do algebra with vectors. After all, given a new thing
to play with, what do mathematicians do? We invent algebras for it!
Vector algebra is built mainly on addition and two kinds of multiplication.
Vector addition is really amazingly easy. To compute the sum A+B of two vectors A and B, draw them on a graph, with the tip of A touching the tail of B. Then draw an arrow from the tail of A to the tip of B. That arrow is the sum.
Multiplication is a bit weird. There are two different kinds of multiplication of vectors. One, called dot product or scalar product multiplies two vectors, and results in a scalar. The other, the cross product is non-commutative, and only works in at least three dimensions.
Given two vectors A and B, their dot product A⋅B is easiest to define in terms of
their components.If they're n dimensional vectors, then the dot product is:
Σi=1...n Ai*Bi
So if A were a vector [A1, A2, A3], and B were
a vector [B1, B2, B3], then A⋅B=A1*B1+A2*B2+A3*B3. For example, let's look at two vectors: A=[2,4,1,5], and B=[6,2,3,-4]. A⋅B=2*6+4*2+1*3+5*-4 = 12+8+3+-20 = 3.
The cross product is more interesting. Given two vectors A and B, their cross product A×B is a vector C where C's magnitude is the area of the parallelogram
formed by putting the tail of B to the tip of A, and then adding parallels for both A and B. C's direction is perpendicular to both A and B. There is one little trick - there are two possible directions for the cross product A×B; which one to use is a matter of convention. The common convention that I was taught is called the right hand rule: take the pair of vectors, and put their tails together, so that they form a V. Then take your right hand, with the fingers curled from A towards B, the direction that your thumb is pointing is the direction of the cross-product.
In three dimension, there's a simple formula in terms of components. If A=[Ax, Ay, Az], and B=[Bx, By, Bz], then A×B=C where:
• Cx = Ay*Bz - Az*By
• Cy = Az*Bx - Ax*Bz
• Cz = Ax*By - Ay*Bx
So, for example, if A=[3,4,5] and B=[9,7,6] then A×B=
[4*6-7*5, 5*9-3*6, 3*7-4*9] = [24-35,45-18,21-36] = [-11,27,-15].
Now, as I mentioned, in vector algebra cross-product isn't commutative. But it's not a total loss like quaternions; vector multiplication is anti-commutative: A×B = -(B×A).
Cross product isn't associative either - A×(B×C)≠(A×B)×C. But we do at least have the very interestingly odd property:
A×(B×C) + B×(C×A) + C×(A×B) = 0
Where 0 is the 0 vector - that is, the vector with length 0. (Originally, one of the parens was mispositioned in the above equation; thanks for commenter "rory" for the catch.)
Finally - we can multiply a scalar times a vector - multiplying a scalar s by a vector A = [x,y,z] gives [sx,sy,sz].
Tags
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the cross product ... only works in at least three dimensions.
The cross product works in only three dimensions. There's a lot more to be said about exactly why, but that's much more than a comment will bear.
But...but...you didn't mention vector spaces, and all the other exciting things about vectors! What's interesting to me is that so many other mathematical concepts can be related to vectors (in R^n) in the sense of isomorphism. Maybe you can do another post discussing vectors in a more general setting?
Hi, by the way. I've recently discovered your blog and love it. It's wonderful to see someone so passionate about mathematical concepts, and so interested in teaching it.
Q: What do you get when you cross a chicken with a turkey?
A: (chicken)*(turkey)*sin(θ)
Q: What do you get when you cross a mountain climber with a mosquito?
A: Nothing -- you can't cross a scaler with a vector.
...
Sorry...
xander
I am so slow! I was just about to ask you if vectors can also be considered multi-dimensional numbers? Since vectors are a more complete description of something we see (velocity tells us more about it than speed right?) are numbers among other things also descriptions of things we see?
The cross product works in only three dimensions.
I thought that there were cross products also in 7 dimensions.
I would hazard a guess (and I may vaguely recall having read it somewhere) that this has something to do with the 1, 2, 4 and 8 dimensions of the previous multidimensional topic.
By Ãrjan Johansen (not verified) on 11 Feb 2007 #permalink
Just to be ultra-anal, this is the Gibbs vector calculus. It's not the only way to understand vectors, just the most common. (This is "basics", after all.) You can also understand vectors using linear algebra or geometric algebra.
Geometric algebra, incidentally, answers the question that bugged me since high school: Why is it when you (cross) multiply a vector (directed length) with a vector (directed length), you get a vector (directed length; shouldn't that be an area)?
By Pseudonym (not verified) on 11 Feb 2007 #permalink
The cross product works in only three dimensions. There's a lot more to be said about exactly why, but that's much more than a comment will bear.
I'm rather fond of considering the cross product a special case of the wedge product. It does let you define an analogue to the cross-product in higher dimensions, though I suppose it's a bit unfortunate to have resulting objects that don't live in the same realm as the ones you started with.
Pseudonym
Why is it when you (cross) multiply a vector (directed length) with a vector (directed
length), you get a vector (directed length; shouldn't that be an area)?
You do: the area of a parallelogram is the norm of the cross product of the two spanning vectors.
Why cross products only work in dimension 3? 1 +2 =3.
In three dimensions, 1-forms are dual to 2-forms. A constant 1-form is a vector. A constant 2-form is a cross (wedge) product. Thus there is a canonical way to associate a vector with a cross product. In dim 5 for instance, 4-forms are dual to 1-forms, so there is a "cross product" which takes four vectors and returns a vector.
Oh and by the way, the best vector spaces are over the complex numbers, not over the reals. (Scalars are complex)
By Anonymous (not verified) on 11 Feb 2007 #permalink
We can also define subtraction A-B as the difference vector between the tip of B to the tip of A, when the tails are put on the same point.
Given two vectors A and B, their cross product AÃB is a vector C
More precisely a pseudovector, because it doesn't behave as a proper vector under rotations. The difference is important in physics. ( http://en.wikipedia.org/wiki/Cross_product , http://en.wikipedia.org/wiki/Pseudo-vector )
A 1 kilogram object experiences a force of 9.8 newtons straight down - 9.8n down is a vector quantity.
Now I'm going to nitpick considering that this is a math blog. On the other hand we discussed ambigousness of symbols in the last basic post (i vs j for imaginary unit). And here we have a rare and basic case of an unambigious symbolisation in form of SI units - which can be important if we don't want Mars probes to crash unnecessarily. (Nudge, nudge.) Besides, it is one of my most beloved pet peeves.
An SI measurement consists of a value and a unit, with space between. All units are spelled lowercase, i.e. "meter", "newton". Units derived from persons have uppercase symbols, i.e "m", "N".
The value contains the plural and the unit should reject the article (makes software formatting simpler, btw). The above now becomes "a force of 9.8 newton straight down - 9.8 N down".
Also, prefixes are often misused. All prefixes larger than 1 are uppercase, all smaller are lowercase. There is one exception, due to a base unit exception. For historical reasons, the unit for mass is kilogram, not gram. Since this unit must be spelled "kilogram (kg)", the "kilo (k)" prefix is lowercase.
So it is "1 kN" or "2 GN", or "3 mN" or "4 uN". Really anal, but to see '5 GG' and '6kn' could be enough to make some grown men cry. Uh, but not me, btw. :-)
I know that none of the commenters needed this "basics" post, but assuming you're going to use it in teaching others math, here's a handy trick I use when teaching dot product (to high school students, but would work for college students too):
Start with the first row of A and transpose it (you can describe it as rotating it 90 degrees or quarter-turn clockwise if you haven't done transposes). Now, "overlay" this column with the first column of B. I think this helps them visualize what to do -- when numbers are written side-by-side (like "3x"), they know that means multiply; and when they have a vertical list of numbers, they've only seen that in adding a bunch of numbers by hand. Ta-da: that's the first entry in the answer. Repeat.
It also reinforces the property that the number of columns of A must "feed into" the number of rows of B. If you had an m-by-n matrix multiplied by another m-by-n matrix (m =/= n), they'd see that they couldn't "overlay" the transposed row of A with the column of B: it would be too short or too long. I have yet to find a better way to get beginning students to internalize this idea.
The cross product works in only three dimensions.
One generalization of a cross-product works in any dimension. It's an (n-1)-ary operator with the property that:
x1&sdot cross(x2,&hellip ,xn) = det(x1,&hellip ,xn)
With this version, the only reason 3 is special is because 3 = 2+1, and people prefer binary operators.
By Antendren (not verified) on 12 Feb 2007 #permalink
Marc did say AT LEAST three dimensions for the cross roduct.
"AÃ(BÃC) + (BÃC)ÃA + CÃ(AÃB) = 0"
That looks wrong. Let BÃC=D, then you have AÃD + DÃA + CÃ(AÃB) = 0. But AÃD + DÃA = 0, so that would require that CÃ(AÃB) = 0 for all vectors A,B,C.
Did you mean AÃ(BÃC) + BÃ(CÃA) + CÃ(AÃB) = 0 (i.e., rearrange the brackets in the second expression)?
It's a while since I did anything with cross products, but I think that's right.
Rory:
Good catch, the parens are wrong there - the parens should be in the same position in all three parts of the cross-product. I'll fix it ASAP.
I am so slow! I was just about to ask you if vectors can also be considered multi-dimensional numbers?
Actually, the way I learned to represent vector components was not as ordered pairs but as multi-diminsional numbers. ie (Vx,Vy) = Vî + Vĵ where î and ĵ are called 'unit vectors' (vectors of magnitude 1 and direction parallel to the x and y axis respectively). As for the second part of you post, I'm not exactly sure what your asking but one possible answer is that numbers only describe things we see, or anything infact, when they have units. 1 is an abstract concept but 1 meter describes something. It says "for whatever a 'meter' is, we've got 1 of them." Units act as a kind of interface between the quantitative and qualitative aspects of a mathematical description. Infact units of measurement might actually be an interesting topic for a Basics post, MCC.
Also, MCC,
AÃ(BÃC) + (BÃC)ÃA + CÃ(AÃB) = 0
is that right? I mean if BxC = D and AxD = -(DxA) doesn't that mean AxD + DxA + Cx(AxB) = Cx(AxB)? maybe you meant Ax(BxC) + Bx(CxA) + Cx(AxB)= 0?
You seem to have left an italics tag open when you made your correction. Also it looks like I am whispering. Cool.
Incidentally, here's how to make sure that you never mess up the Jacobi identity.
A derivation D on a product BÃC is an operator such that:
D(BÃC) = D(B)ÃC + BÃD(C)
You may recognise this as the product rule from high-school calculus.
The Jacobi identity says, that the operator AÃ is a derivation:
AÃ(BÃC) = (AÃB)ÃC + BÃ(AÃC)
Apply a little reordering and there you have it.
By Pseudonym (not verified) on 12 Feb 2007 #permalink
Infact units of measurement might actually be an interesting topic for a Basics post, MCC.
I certainly agree, but perhaps it is more appropriate for a physics blog. And Mark has been very productive already.
So I have proposed the same on a physics scienceblog, Chad Orzel's IIRC. We'll see if he or anyone else picks up the torch here.
here's how to make sure that you never mess up the Jacobi identity.
Another way is to go back to the permutation definition, and observe that A,B,C simply rotates through the expression.
Your way is more elegant and certain, since it explains more and places the parenthesis correctly.
Cool. But what about all the statistical uses? Including vectors with many more dimensions than 2 or 3. Could easily be thousands of dimensions, or, given data mining, millions.....
you have a very minor typo I think.
"you simple can't get away from them"
should be:
"you simply can't get away from them"
Oh look, a post where my pseudonym makes sense! Also...
The common convention that I was taught is called the right hand rule
ObXKCD (and really, on this blog, XKCD really is obligatory): Right-Hand Rule
By Johnny Vector (not verified) on 13 Feb 2007 #permalink
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# How do you derive the 4 kinematic equations?
## What is deriving an equation?
Derive means to obtain the result from specified or given sources. For example, you might have other formulas that have those variables in it, and you’re supposed to use those formulas, and manipulate them algebraically, to get the final result in your link.
## What are derivations in physics?
Derivation means the action of obtaining something from a source or origin. Through derivation, we find a logical connection between a natural phenomenon and a mathematical description of that phenomenon. In general, this points to an important conclusion about nature itself.
## What are the 3 equations of motion?
The three equations are, v = u + at. v² = u² + 2as. s = ut + ½at²
## What are the 5 equations of motion?
The equations are as follows: v=u+at,s=(u+v2)t,v2=u2+2as,s=ut+12at2,s=vt−12at2.
## Are there 4 or 5 kinematic equations?
There are four basic kinematics equations: v = v 0 + a t. Δ x = ( v + v 0 2 ) t. Δ x = v 0 t + 1 2 a t 2. v 2 = v o 2 + 2 a Δ x.
## Why do we need to derive the formula?
Without these equations, there would be no knowledge about the workings of DNA, medicine would not exist in the form that we have today. We would not have cars, planes, or electricity to power our homes. Life would be drastically different. In terms of evolution, this all happened in an instant.
## How do you derive an equation from a graph?
To find the equation of a graphed line, find the y-intercept and the slope in order to write the equation in y-intercept (y=mx+b) form. Slope is the change in y over the change in x.
## What is derivative in basic calculus?
derivative, in mathematics, the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations.
## How do you derive the second equation of motion?
1. s = u + v 2 × t.
2. s = ( u + ( u + a t ) ) 2 × t.
3. s = 2 u + a t 2 × t.
4. s = ( 2 u 2 + a t 2 ) × t.
## How do you find the equation of motion?
Newton’s second law, which states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration a of its centre of mass, F = ma, is the basic equation of motion in classical mechanics.
## What is the easiest way to memorize derivations?
1. Do problems related to that formula(use it)
2. Understand the formulae (physical meaning-what does it mean)
3. Prepare a list containg formule to avoid ambiguity among your formulae.
## How can I memorize physics fast?
1. Mastering the basics is necessary: One of the easiest ways of learning physics is to master the basic theories.
2. Simplification is a Key Rule:
3. Create Flashcards:
5. A good tutor makes a BIG difference!
6. Use diagrammatic representations to learn the concepts:
## How can I memorize physics?
1. Understand the concept and meaning of each formula.
2. Apply and practice.
3. Make notes and paste them up.
4. Use memory techniques.
5. Sleep on it.
6. Take care of your body.
## How do you derive the three equations of motion graphically?
The following are the three equation of motion: First Equation of Motion: v = u + at. Second Equation of Motion: s = ut + 1/2(at2) Third Equation of Motion: v2 = u2 – 2as.
## How do you derive the third equation of motion graphically?
Draw a line parallel to x-axis DA from point, D from where object starts moving. Draw another line BA from point B parallel to Y-axis which meets at E at y-axis. Third Equation of Motion: The Distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDOE.
## What formula is V u at?
Given equation is v = u + a t . Simply it is first newton’s equation of motion.
## Does MCAT give physics equations?
No, you do not need to memorize all equations for the MCAT. For MCAT Physics, there are many equations that you should memorize in order to maximize your score. Occasionally, the AAMC will provide physics formulas during the exam itself, but never basic formulas.
## What does T in physics mean?
In the field of physics, the letter ‘T’ is often used to denote the period of oscillation (commonly reffered to as simply ‘period’). The period of oscillation can be defined as the duration of time between one wave and the next one passing the same spot. Its units are usually expressed in seconds.
## What does u stand for in physics?
The letter u is used in physics to denote the sign of potential energy, as well as the initial velocity and the object distance in ray optics.
## Is velocity first or second derivative?
If position is given by a function p(x), then the velocity is the first derivative of that function, and the acceleration is the second derivative. By using differential equations with either velocity or acceleration, it is possible to find position and velocity functions from a known acceleration.
## What is the derivative of acceleration?
Acceleration is the derivative of velocity with respect to time: a(t)=ddt(v(t))=d2dt2(x(t)). Momentum (usually denoted p) is mass times velocity, and force (F) is mass times acceleration, so the derivative of momentum is dpdt=ddt(mv)=mdvdt=ma=F.
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# Finding the Radius of Convergence through Ratio Test
• JustinDaniels
In summary: At x=2, the series converges since ##f(u+2) = f(u) + u^{4/3}##.At x=inf, the series diverges since ##f(u+2) = -u^{4/3}##.
## Homework Statement
Let f(x)= (1+x)4/3 - In this question we are studying the Taylor series for f(x) about x=2.
This assignment begins by having us find the first 6 terms in this Taylor series. For time, I will omit them; however, let's note that as we continuously take the derivative of this function, we are applying the chain rule over and over. The derivative of the 'inside function' is 1, so in essence, we are really just applying the power rule again and again.
(b) Let ck = f(k)(2)/k! be the coefficient of (x-2)k in the Taylor series. It is not so easy to write a closed-form expression for ck. However, find a recursive formula to express ck+1 in terms of ck.
(c) Use your recursive formula, together with the ratio test, to determine which values for x would make the Taylor series Σk=0ck(x-2)k converge or diverge absolutely.
## The Attempt at a Solution
(b) I note that the coefficient of each function is being changed by the power in which the previous coefficient is raised to--which always contains a (3-1. Furthermore, we must handle the factorial. As a result, I observe the following formula for the general term:
(4-3k)(1/(k+1))(3-1)ck = ck+1
(c) Applying the ratio test: lim k→∞ |ck+1|/|ck|
I note that ck+1 contains a ck in it. As a result they cancel out. Leaving us with:
lim k→∞ (4-3k)(1/(k+1))(3-1)
Here is where I get stuck.. I believe we can pull the (1/3) out from the limit, as it is a constant. The other two parts of the limit approach -∞ and 0 respectively. Thus, we would apply L'Hopital's rule? This gives us
1/3 lim k→∞ [-3(-(k+1)-2)]
From here, do we pull out the -3, and evaluate the limit? If so, our limit approaches 0 again?
Any help would be appreciated. Here is the next question, to give you an idea of what I need.
(d) You should have observed a range of x values that extends for the same distance on either side of x=2. This distance is called the Radius of Convergence. Try to find some property of the function that explains why the radius of convergence is what it is.
Note: I believe the for 0<x<2 it converges.
At 2 it is inconclusive.
2<x<inf it diverges.
I heard this from a classmate. Any help is appreciated!
Thanks,
Calculus 2 Student.
I didn't get the same relationship between ##c_k## and ##c_{k+1}## you did. It looks like you're missing a factor of 1/3. You also need a factor of ##x-2## because the terms of the series also contain powers of ##x-2##. Finally, you misapplied L'Hopital's rule, but you don't really need to use it.
The limit of the expression you got is more straightforwardly evaluated as follows:
$$\lim_{k\to\infty} \frac{-3k+4}{3k+3} = \lim_{k\to\infty} \frac{-3+\frac 4k}{3+\frac 3k} = \frac{-3}{3} = -1.$$ Just divide the top and bottom by ##k##. Of course, as I noted, your expression is incorrect, so you need to fix that first.
vela said:
I didn't get the same relationship between ##c_k## and ##c_{k+1}## you did. It looks like you're missing a factor of 1/3. You also need a factor of ##x-2## because the terms of the series also contain powers of ##x-2##. Finally, you misapplied L'Hopital's rule, but you don't really need to use it.
The limit of the expression you got is more straightforwardly evaluated as follows:
$$\lim_{k\to\infty} \frac{-3k+4}{3k+3} = \lim_{k\to\infty} \frac{-3+\frac 4k}{3+\frac 3k} = \frac{-3}{3} = -1.$$ Just divide the top and bottom by ##k##. Of course, as I noted, your expression is incorrect, so you need to fix that first.
So I need an additional factor of (1/3)? I have the formula containing a 3-1, which is 1/3. Also, ##c_k## is the coefficient of the kth term in the series (i.e. the number before the ##x-2##), therefore I do not think it needs a factor of ##x-2##.
Note, the coefficients of the series are:
f(x) = 1
f'(x) = 4/3 --as a result of the power rule being applied the first time.
f''(x) = 4/9
f'''(x) = -8/27.
Notice, the denominator changes by a factor of (1/3) each time, and the numerator by a factor of the numerator of the previous terms power (4-3k). Finally, we have to handle the factorial each time, and as a result I observe the recursive function to be
##(1/3)(1/(k+1))(4-3k)c_k=c_{k+1}##
I hope that makes sense...
Thank you again. I appreciate your time,
jd
Your series can't be right. Since ##f(x) = (1+x)^{4/3}##, you have ##f(2) = 3^{4/3}##. For a series expanded about ##x=2##, that value is the constant term. Remember that f and its derivatives have to be evaluated at ##x=2## since that's the point you're expanding about.
It might help you to make the substitution ##u=x-2##, which gives ##f(x) = f(u+2) = (3+u)^{4/3}## and expand about ##u=0##.
I'm quite confused.. Here is what I get for the first six terms in the Taylor series:
Term 0: ##(1+(2))^{4/3}##
Term 1: ##4/3 (1+2)^{1/3}(x-2)##
Term 2: ##\frac{(4/9) (1+2)^{-2/3}}{2!} (x-2)^2##
Term 3: ##\frac{-(8/27) (1+2)^{-5/3}}{3!} (x-2)^3##
Term 4: ##\frac{(40/81) (1+2)^{-8/3}}{4!} (x-2)^4##
Term 5: ##\frac{-(320/243) (1+2)^{-11/3}}{5!} (x-2)^5##
Term 6: ##\frac{(3520/72)9 (1+2)^{-14/3}}{6!} (x-2)^6##
I need to find a formula that gives me the coefficient of some term, in respect to its previous term.
The value of the constant term is approximately: 4.326748
The value of the coefficient of the (x-1) term is approximately: 1.922999
The value of the coefficient of the ##(x+1)^2## term is approximately: .1068333
Applying the formula I suggested, we observe:
##k=1##
##c_k = 1.922999##
##c_k(4-3k)(1/3)(1/2)=.3204998333##, which does not equal my suggested ##c_{k+1}##. As you noted, I am off by a factor of (1/3). This suggests our formula should be:
##c_{k+1}=(4-3k)(1/9)(\frac{1}{k+1}){c_k}##
Let us apply this to the ##(x+1)^2## term and ##(x+1)^3## term's coefficients.
The value of the coefficient of the ##(x+1)^2## term is approximately: .1068333
The value of the coefficient of the ##(x+1)^3## term is approximately: -.00791357788
Applying the new formula, we observe:
##k=2##
##c_k = .1068333##
##c_k(4-(3*2))(1/9)(\frac{1}{2+1})=-.00791357788##, which was the suggested ##c_{k+1}##.
Given this, we use the recursive formula, together with the ratio test, to determine which values for x would make the Taylor series ##\sum_{k=0}^\infty c_k (x-2)^k## converge or diverge absolutely. Applying the ratio test, we will note the ##c_k##'s cancel out.
##\lim_{k\rightarrow +\infty}## ##{(4-3k)(1/9)}## ##\frac{1}{k+1}## which can be simplified as:
##\lim_{k\rightarrow +\infty}## ##\frac{-3k+4}{9k+9}## I am assuming we would again divide by ##k## and then evaluate the limit. In doing so:
##\lim_{k\rightarrow +\infty}## ##\frac{-3+\frac{4}{k}}{9+\frac{9}{k}} = \frac{-3}{9} = \frac{-1}{3}##
I am now confused as to how this helps me know which values this series converges or diverges absolutely for.. It appears as though the series' convergence is not dependent on ##x## and since ##\frac{-1}{3}## is smaller than one, this series converges for all values of x. However, the next question on this assignment suggests that the series only converges for some values on x on either side of 2.
edit: I guess in applying the Ratio Test, we would be taking the absolute value of the limit. As a result, we would observe (1/3) and not a negative (1/3)... this doesn't change the result much however..
Thanks again!
jd
Last edited:
Okay, so I incorrectly defined ##a_n## and ##a_{n+1}##; as a result, I misapplied the Ratio Test.
##a_{n+1}## should be ##\frac{1}{9} \frac{1}{k+1} (4-3k) c_{k} (x-2)^{k+1}##
In reapplying the Ratio Test, I observe the following:
##lim_{k \rightarrow \infty} |\frac {\frac{1}{9} \frac{1}{k+1} (4-3k)c_k(x-2)^{k+1}}{c_k(x-2)^k}|##
We can pull the constant multiple of (1/9) out of the limit and cancel out the ##c_k##'s. Furthermore, ##\frac{(x-2)^{k+1}}{(x-2)^k}## simplifies to ##(x-2)##.
##\frac{1}{9} lim_{k \rightarrow \infty} |\frac{-3k+4}{k+1} (x-2)|##
I will omit much of the algebra; however, we will again divide by k and the resulting ##4/k## and ##1/k## will go to zero. I observe the following:
##\frac{1}{9} lim_{k \rightarrow \infty} |\frac{-3}{1} (x-2)|##
Here is where I am stuck again.. Is the following correct?:
Do we pull the ##-3## out of our limit as a negative or positive 3, due to the absolute value? Given we do, I observe the following:
##\frac{3}{9} lim_{k \rightarrow \infty} |(x-2)|##
The 3/9 simplifies to 1/3. Applying the Ratio Test, the series should converge for x values that make the limit less than one. As a result of the absolute value the following bounds satisfy this requirement.
##(1,3)##--a distance of 2. Do we call this the Radius of Convergence, and have I done everything correct up to this point?
I really appreciate your help. I would be completely lost otherwise.
Thanks again,
jd
Okay, spoke to my professor. The limit of the ratio test evaluates out to ##\frac{1}{3} |x-2|##, which I essentially had in the previous post. We note that this value is not dependent on ##k##, so that when k goes to infinity the limit stays as ##\frac{1}{3} |x-2|##. The range of x values that satisfies the ratio being less than 1 is:
##-1<x<5##
We note that when the x values becomes either -1 or 5 the ratio evaluates to 1--telling us nothing about the convergence on the sequence.
I believe we have solved the problem.
Thank you again for your help.
jd
JustinDaniels said:
Okay, spoke to my professor. The limit of the ratio test evaluates out to ##\frac{1}{3} |x-2|##, which I essentially had in the previous post. We note that this value is not dependent on ##k##, so that when k goes to infinity the limit stays as ##\frac{1}{3} |x-2|##.
Note that the limit can't depend on ##k##. You found the limit as ##k\to\infty.##
We note that when the x values becomes either -1 or 5 the ratio evaluates to 1--telling us nothing about the convergence on the sequence.
The ratio test tells you nothing, so you have to go back to the original series, plug in the each value and see if the resulting series converge.
I observed the range of values to be ##-1 < x < 5##. Thus, the radius of convergence is 3. The next question in this assignment asks us to find a property of the function that explains why the radius is what it is. I see no relationship between the two however.
Consider what I suggested in post #4.
## 1. What is the Ratio Test method used for?
The Ratio Test method is used for determining the convergence or divergence of an infinite series. It specifically helps in finding the radius of convergence for power series.
## 2. How do you apply the Ratio Test to find the radius of convergence?
To apply the Ratio Test, you need to take the limit of the absolute value of the ratio of consecutive terms in the series. If this limit is less than 1, then the series converges. The radius of convergence is then equal to the reciprocal of this limit.
## 3. Can the Ratio Test be used for all series?
No, the Ratio Test can only be used for series with positive terms. It may also fail to give a definite answer for some series, in which case other convergence tests must be used.
## 4. What is the significance of the radius of convergence?
The radius of convergence tells us the range of values for which the power series converges. If the value of x is within the radius of convergence, then the series will converge. If it is outside the radius, the series will diverge.
## 5. Is there a shortcut method for finding the radius of convergence?
Yes, there are a few shortcut methods such as the Root Test and the Integral Test that can be used to find the radius of convergence. However, these methods may not always give a definite answer and the Ratio Test is still considered the most reliable method for finding the radius of convergence.
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# Uniform Acceleration
## Define and calculate uniform acceleration using formulas
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Uniform Acceleration
Credit: Justin De La Ornellas
Source: http://www.flickr.com/photos/ornellas/8869268939/
After jumping, this cliff diver undergoes effective free fall. Cliff diving is a sport in which athletes twist and flip on their way down. Even with the air resistance, these divers will accelerate the whole way down.
### Uniform Acceleration
Acceleration that does not change in time is called uniform or constant acceleration. The velocity at the beginning of the time interval is called initial velocity, vi\begin{align*}v_i\end{align*}, and the velocity at the end of the time interval is called final velocity, vf\begin{align*}v_f\end{align*}. In a velocity versus time graph for uniform acceleration, the slope of the line is the acceleration. The equation that describes the curve is vf=vi+at\begin{align*}v_f = v_i + at\end{align*}.
Example 1
If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s2 for 2.5 s, what is the final velocity?
vf=vi+at=4.0 m/s+(4.0 m/s2)(2.5 s)=4.0 m/s+10. m/s=14 m/s\begin{align*}v_f = v_i + at = 4.0 \ \text{m/s} + (4.0 \ \text{m/s}^2)(2.5 \ \text{s}) = 4.0 \ \text{m/s} + 10. \ \text{m/s} = 14 \ \text{m/s}\end{align*}
Example 2
If a cart slows from 22.0 m/s with an acceleration of -2.0 m/s2, how long does it require to get to 4 m/s?
t=vfvia=18 m/s2.0 m/s2=9.0 s\begin{align*}t=\frac{v_f-v_i}{a}=\frac{-18 \ \text{m/s}}{-2.0 \ \text{m/s}^2}=9.0 \ \text{s}\end{align*}
### Summary
• Acceleration that does not change in time is uniform, or constant, acceleration.
• The equation relating initial velocity, final velocity, time, and acceleration is vf=vi+at\begin{align*}v_f = v_i + at\end{align*}.
### Review
1. If an object has zero acceleration, does that mean it has zero velocity? Give an example.
2. If an object has zero velocity, does that mean it has zero acceleration? Give an example.
3. If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?
4. The friction of the water on a boat produces an acceleration of -10. m/s2. If the boat is traveling at 30. m/s and the motor is shut off, how long it take the boat to slow down to 5.0 m/s?
### Notes/Highlights Having trouble? Report an issue.
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# In a bundle of 20 sticks, there are 4 sticks each of length 1 m 50cm, 10 sticks each of length 2 m, and each of the rest of length 1 m. what is the average length of the sticks in the bundle?a) ${\rm{1}}{\rm{.2 m}}$b) ${\rm{1}}{\rm{.5 m}}$c) ${\rm{1}}{\rm{.6 m}}$d) ${\rm{1}}{\rm{.8 m}}$
Last updated date: 13th Jun 2024
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Hint: In this question we have to find the value of the average length of the sticks in the bundle. For that, we are going to solve using the average formula. To use the average formula we will initially find the length of the given number of sticks by multiplying it with the length and summing them and then divide the sum with the total number of sticks. This is explained in a complete step by step solution.
Formulas used:
${\text{Average number = }}\dfrac{{{\text{Sum of observations}}}}{{{\text{Total number of observations}}}}$
It is given that total number of sticks = $20$
Here also it is given that there are $4$ sticks each of length ${\rm{1 m 50}}\;{\rm{cm}}$.
That is the length of four sticks is given by $4 \times 1.5$
Also, it is given that there are $10$ sticks each of length ${\rm{2 m}}$.
That is the length of ten sticks is $10 \times 2$.
We should find the remaining number of sticks since it is given that the remaining sticks are of length ${\rm{1 m}}$.
${\text{Remaining number of sticks = total number of sticks - the number of sticks we have}}$
${\text{Remaining sticks = 20 - (10 + 4) = 6}}$
Also we have there are 6 sticks each of length ${\rm{1 m}}$
That is the length of six sticks $6 \times 1$.
As we know that ${\text{average length = }}\dfrac{{{\text{sum of all lengths}}}}{{{\text{total number of sticks}}}}$
${\text{Average length}} = \dfrac{{4 \times 1.5 + 10 \times 2 + 6 \times 1}}{{20}}$
Let us now multiply the terms in the numerator we get,
$\Rightarrow \dfrac{{6 + 20 + 6}}{{20}}$
Now by adding the terms in the numerator we get the following fraction,
$\Rightarrow \dfrac{{32}}{{20}}$
By dividing the elements in the numerator and denominator we get,
${\text{Average length = 1}}{\rm{.6}}$
$\therefore$ The average length of the given 20 sticks is $1.6m$.
Note:
Here we can initially multiply the length of the sticks and then add it, that is we can multiply $4 \times 1.5$, $10 \times 2$, and $6 \times 1$ then add them together in the numerator. Also, we should be careful in finding the remaining number of sticks which are of length ${\rm{1 m}}$.
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# STANDARD FORM OF A RATIONAL NUMBER
Observe the following rational numbers.
⅔, ⅘, ⁻³⁄₇, ¼, ⁻⁶⁄₁₁, ⁻⁴⁷⁄₅₁
(i) Both numerator and denominator of all the rational numbers above are integers.
(i) The denominators of the rational numbers are all positive integers
(ii) 1 is the only common factor between the numerator and the denominator of each of them.
(iii) The negative sign occurs only in the numerator.
Such rational numbers are said to be in standard form.
A rational number is said to be in standard form, if its denominator is a positive integer and both the numerator and denominator have no common factor other than 1.
If a rational number is not in the standard form, then it can be simplified to arrive at the standard form.
Note :
In a rational number, if there is a negative sign in denominator, it can be moved to the numerator.
For example,
ᵃ⁄₋b = ⁻ᵃ⁄b
Examples 1-6 : Reduce the given rational number to its standard form.
Example 1 :
⁶⁄₈
Solution :
In the rational number 6/8, both 6 and 8 are even numbers, so they re evenly divisible by 2.
⁶⁄₈ = ⁽⁶ ÷ ²⁾⁄₍₈ ÷ ₂₎
¾
The standard form of ⁶⁄₈ is ¾.
Example 2 :
³⁄₄.₅
Solution :
In the rational number ³⁄₄.₅, the denominator is not an integer, it is a decimal number. To make the denominator 4.5 as an integer, multiply both numerator and denominator by 10. Since there is one digit after the decimal point in 4.5, we have to multiply by 10.
³⁄₄.₅ = ⁽³ ˣ ¹⁰⁾⁄₍₄.₅ ₓ ₁₀₎
= ³⁰⁄₄₅
= ⁽³⁰ ÷ ⁵⁾⁄₍₄₅ ÷ ₅₎
= ⁶⁄₉
= ⁽⁶ ÷ ³⁾⁄₍₉ ÷ ₃₎
=
The standard form of ³⁄₄.₅ is .
Example 3 :
⁴⁸⁄₋₈₄
Solution :
Method 1 :
In the rational number ⁴⁸⁄₋₈₄, since both the numerator and denominator are two digit numbers, we can do successive division by smaller numbers to make the process easier.
⁴⁸⁄₋₈₄ = ⁻⁴⁸⁄₈₄
= ⁻⁽⁴⁸ ÷ ²⁾⁄₍₈₄ ÷ ₂₎
= ⁻²⁴⁄₄₂
= ⁻⁽²⁴ ÷ ²⁾⁄₍₄₂ ÷ ₂₎
= ⁻¹²⁄₂₁
= ⁻⁽¹² ÷ ³⁾⁄₍₂₁ ÷ ₃₎
⁻⁴⁄₇
Method 2 :
⁴⁸⁄₋₈₄ = ⁻⁴⁸⁄₈₄
The highest common factor of 48 and 84 is 12. Thus, we can get its standard form by dividing it by 12.
= ⁻⁽⁴⁸ ÷ ¹²⁾⁄₍₈₄ ÷ ₁₂₎
⁻⁴⁄₇
The standard form of ⁴⁸⁄₋₈₄ is ⁻⁴⁄₇.
Example 4 :
⁻¹⁸⁄₋₄₂
Solution :
In the rational number -18/(-42), since both the numerator and denominator are negative even numbers, we can start dividing both -18 and -42 by -2.
⁻¹⁸⁄₋₄₂ = ⁽⁻¹⁸ ÷ ⁻²⁾⁄₍₋₄₂ ÷ ₋₂₎
= ⁹⁄₂₁
= ⁽⁹ ÷ ³⁾⁄₍₂₁ ÷ ₃₎
= ³⁄₇
The standard form of ⁻¹⁸⁄₋₄₂ is ³⁄₇.
Example 5 :
.⁴⁄₀.₆
Solution :
In the rational number 0.4/0.6, both the numerator and denominator are not integers, they are decimal numbers. To make both the numerator and denominator as integers, multiply both numerator and denominator by 10. Since there is one digit after the decimal point in both the numerator and denominator, we have to multiply by 10.
.⁴⁄₀.₆ = . ˣ ¹⁰⁾⁄₍₀.₆ ₓ ₁₀₎
= ⁴⁄₆
= ⁽⁴ ÷ ²⁾⁄₍₆ ÷ ₂₎
=
The standard form of .⁴⁄₀.₆ is .
Example 6 :
.⁰³⁄₀.₉
Solution :
In the rational number 0.03/0.9, both the numerator and denominator are not integers, they are decimal numbers. Comparing the numerator 0.03 and denominator 0.9, there are more number of digits (two digits) after the decimal point in the numerator 0.03. Since there are two digits after the decimal point in numerator 0.03, we have to multiply both numerator and denominator by 100 to get rid of the decimal points in both.
.⁰³⁄₀.₉ = .⁰³ ˣ ¹⁰⁰⁾⁄₍₀.₉ ₓ ₁₀₀₎
= ³⁄₉₀
= ⁽³ ÷ ³⁾⁄₍₉₀ ÷ ₃₎
= ¹⁄₃₀
The standard form of .⁰³⁄₀.₉ is ¹⁄₃₀.
Examples 7-10 : Write the given decimal number as a rational number in standard form.
Example 7 :
0.2
Solution :
Since there is one digit after the decimal point in 0.2, it can be written as a fraction with the denominator 10.
0.2 = ²⁄₁₀
= ⁽² ÷ ²⁾⁄₍₁₀ ÷ ₂₎
=
Example 8 :
0.05
Solution :
Since there are two digits after the decimal point in 0.05, it can be written as a fraction with the denominator 100.
0.05 = ⁵⁄₁₀₀
= ⁽⁵ ÷ ⁵⁾⁄₍₁₀₀ ÷ ₅₎
= ¹⁄₂₀
Example 9 :
0.001
Solution :
Since there are three digits after the decimal point in 0.001, it can be written as a fraction with the denominator 1000.
0.001 = ¹⁄₁₀₀₀
Example 10 :
0.502
Solution :
Since there are three digits after the decimal point in 0.502, it can be written as a fraction with the denominator 1000.
0.502 = ⁵⁰²⁄₁₀₀₀
= ⁽⁵⁰² ÷ ²⁾⁄₍₁₀₀₀ ÷ ₂₎
= ²⁵¹⁄₅₀₀
Example 11 :
In the standard form of a rational number, what is the common factor of numerator and denominator?
(A) 0
(B) 1
(C) -2
(D) 2
Solution :
In the standard form of any rational number, the common factor of numerator and denominator is always 1.
Example 12 :
The rational number 2/3 is in standard form, because
(A) both numerator and denominator are integers.
(B) the denominator is a positive integer
(C) the common factor of numerator and denominator is always 1
(D) all the above
Solution :
Kindly mail your feedback to v4formath@gmail.com
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# Rules of Exponents - Problem 2
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Using the rules of exponents to simplify an expression with negative exponents. So the main thing you have to remember when dealing negative exponents is it basically takes anything that is in the numerator, puts in the denominator, anything that's in the denominator and moves it to the numerator.
So 5 to the -2. What that tells me is that this term which is in the numerator right now this is 5 to the -2 over 1 is just going to have to move down to the denominator leaving me with 1 over 5² which is the same thing as 1 over 25.
Similarly when we have a negative exponent in the denominator it just moves it to the denominator, so this ends up being 4² which is going to be 16 and what a negative does on a fraction is just basically flip it over. There's a number of different ways you could deal with this problem. You could distribute this -3 in, we have a fraction to a power, this -3 has to go both the numerator and the denominator or you could just say okay this is 2/3 to the -1 to the third.
Basically remember when you have a power to a power you have to multiply so -1 to the third is the same thing as -3. What this -1 does is flip over my fraction so what I have here then is, 3/2 to the third, this 3/2 will get distributed into both leaving us with 3 to the third, 27 over 2 to the third which is 8.
Like I said you could also distribute this in so you'd end up with 2 to the -3, the negative is going to bring it down to the denominator leaving us with 2 to the third. We'd also have 3 to the -3 in the denominator and they would bring it up to the top leaving us with the 3 to the third in the top or 27.
A number of different ways of dealing with this, what I tend to do is if I see a negative sign on a fraction or a statement I just know that everything is going to have to get flipped to the opposite spot, but if you want to distribute it in first that's perfectly fine too.
|
# Sequences
If we assign each $$n\in\mathbb{N}$$ a real number $$a_n\in\mathbb{R}$$, we get a sequence $$(a_n)$$ with elements
$(a_n) = a_1, a_2, a_3, \dots a_n$
If a sequence starts at index $$k$$, we call the sequence
$(a_n)_{n\geq k} = a_k, a_{k+1}, a_{k+2}, \dots a_n = a_n - a_{k-1}$
If $$n < \infty$$, we call the sequence finite and otherwise infinite.
To define a sequence, we can state elements in explicit form
$(a_n) = 1,2,3,4,\dots$
or in closed form
$(a_n) = n^2 = 1, 4, 9, 16, \dots$
Additionally, we can define a sequence as a recurrence relation, where each term depends on the previous term:
$$a_1=0, a_2=a_1+1, a_3=a_2+2, a_4=a_3+3, \dots$$
A sequence is called a first-order recurrence if each element depends on only one previous term. A second-order recurrence for example is the famous Fibonacci-Sequence:
$(a_n) = a_{n-1} + a_{n-2}\text{ with } a_0=a_1 = 1$
## Arithmetic Sequences
If the difference between two elements $$d$$ is constant, such that
$a_{n+1} - a_{n} = d$
we call the sequence
$(a_{n+1}) = a_{n} + d$
an arthmetic sequence or arithmetic progression with $$a_1$$ being the first term.
Arithmetic sequences can also be defined in a closed form:
$(a_n) = a_1 + (n - 1)\cdot d$
If $$d=0$$ the sequence is constant. If $$d>0$$, the sequence is monotonically increasing and if $$d<0$$, the sequence is monotonically decreasing.
The name Aritmetic Sequence arises because each term other than the first is the arithmetic mean of its two neighbouring terms:
$a_n = \frac{a_{n-1} + a_{n+1}}{2}$
## Geometric Sequences
If the quotient $$q$$ of two consecutive elements is constant, such that
$\frac{a_{n+1}}{a_n} = q$
we call the sequence
$(a_{n+1}) = a_n\cdot q$
a geometric squence with $$a_1$$ being the first term.
Geometric squences can also be defined in a closed form:
$(a_n) = a_1\cdot q^{n-1}$
Geometric series have the following behavior:
• If $$0<q<1$$, the sequence is monotonically decreasing and converges to zero as $$q^n\to 0$$ as $$n\to\infty$$.
• If $$-1<q<0$$, the sequence is alternating in sign and converges to zero as $$q^n\to 0$$ as $$n\to\infty$$.
• If $$q>1$$, the sequence is monotonically increasing as $$q^n\to\infty$$ as $$n\to\infty$$.
• If $$q<-1$$, the sequence is unbound and alternates in sign.
• If $$q=0$$ the sequence is constant $$0$$.
• If $$q=1$$ the sequence is constant $$a_1$$.
• If $$q=-1$$ the sequence is alternating between $$a_1$$ and $$-a_1$$.
The name Geometric Sequence arises because each term other than the first is the geometric mean of its two neighbouring terms:
$|a_n| = \sqrt{a_{n-1} \cdot a_{n+1}}$
## Linear Recurrence Sequences
In practice often a combination of arithmetic and geometric sequences aries. If we have a constant $$d$$ and a constant $$q$$ we call the sequence
$(a_{n+1}) = a_n \cdot q + d$
a linear recurrence sequence with $$a_1$$ beng the first term.
Linear recurrence sequences can be expressed in a closed form as
$\begin{array}{rl} x_2 &= a_1 \cdot q + d\\ x_3 &= (a_1 \cdot q + d) \cdot q + d = a_1 \cdot q^2 + d\cdot q + d\\ x_4 &= (a_1 \cdot q^2 + d\cdot q + d) \cdot q + d = a_1 \cdot q^3 + d\cdot q^2 + d\cdot q + d\\ & \vdots\\ x_n &= a_1\cdot q^{n-1} + d\cdot(q^{n-2} + \dots + r + 1)\\ &= a_1\cdot q^{n-1} + d\cdot\left(\frac{r^{n-1}-1}{r-1}\right)\\ &= \left(a_1 + \frac{d}{q-1}\right)\cdot q^{n-1} - \frac{d}{q-1} \,\,\forall q\neq 1 \end{array}$
Note that arithmetic sequences are linear recurrence sequence with $$q=1$$ and geomtric sequenecs are linear recurrence sequences with $$d=0$$.
### Inferring Parameters for Linear Recurrence Sequences
Let’s say we get the sequence $$a_1, a_2, a_3, \dots$$, how can we infer the parameter $$q$$ and $$d$$ of this sequence?
We know that
$\begin{array}{rl} a_2 &= a_1 \cdot q + d\\ a_3 &= a_2 \cdot q + d \end{array}$
Subtracting both equations from another gives
$a_3 - a_2 = (a_2 - a_1)\cdot q$
From which follows that
$q = \frac{a_3 - a_2}{a_2 - a_1}$
Taking the first equation again gives
$d = a_2 - a_1 \cdot q$
## Monotonicity of Sequences
An important criterion to distinguish sequences is the monotonicity of sequences. If the elements get larger with increasing index, the sequence is called monotonically increasing, if the elements get smaller with increasing index, the sequence is called monotonically decreasing. If the sequence is changing the sign, we call it an alternating sequence.
For a sequnce $$(a_n)$$ we say
$\begin{array}{l} a_{n+1}\geq a_n \,\forall n\Leftrightarrow (a_n)\text{ is monotonically increasing}\\ a_{n+1}\leq a_n \,\forall n\Leftrightarrow (a_n)\text{ is monotonically decreasing}\\ a_{n+1}= a_n \,\forall n\Leftrightarrow (a_n)\text{ is constant increasing}\\ \end{array}$
If the relations hold for $$>$$ and $$<$$ instead of $$\geq$$ and $$\leq$$, we call the sequences strictly monotonically increasing and decreasing respectively.
## Boundedness of Sequences
A sequence $$(a_n)$$ is called bounded if there exist two real numbers $$s$$ and $$S$$ such that all elements of the sequence are between lower bound $$s$$ and upper bound $$S$$:
$s\leq a_n\leq S\,\forall n$
## Limits of Sequences
A sequence $$(a_n)$$ has a limit $$a$$ if an infinite neighborhood of $$a$$ results in a finite set of elements. In this case we say
$\lim\limits_{n\to\infty} a_n=a$
If a series has a limit, we call it convergent, otherwise divergent.
We can say the following about sequences:
• A limit is always distinct.
• $$\pm\infty$$ is no valid limit
• Every monotonical and bounded sequence is convergent.
• Every convergent sequence is bounded.
Sequences that converge to zero are called null sequence.
We can say that a sequence $$(a_n)$$ converges to $$a$$ if the sequence $$(a_n-a)$$ is a null sequence.
### Cauchy convergence criterion
Augustin-Louis Cauchy formulated a more general criterion for the case that the limit $$a$$ is not known beforehand. It says that a sequence $$(a_n)$$ is convergent iff for every arbitrary small $$\epsilon>0$$ exists a $$n_0\in\mathbb{N}$$ such that all all sequence elements $$a_i, a_j$$ follow $$|a_i-a_j|<\epsilon$$, begining with element $$a_{n_0}$$. More formally this means
$\forall\epsilon>0\exists n_0\in\mathbb{N}\forall i,j\geq n_0 : |a_i-a_j|<\epsilon$
### Important sequence limits
$\begin{array}{rl} \lim\limits_{n\to\infty}\frac{a}{n} &= 0\forall a\in\mathbb{R}\\ \lim\limits_{n\to\infty}a^n &= 0\,\forall |a|<1\\ \lim\limits_{n\to\infty}\sqrt[n]{a}&=1\,\forall a\in\mathbb{R}^+\\ \lim\limits_{n\to\infty}\sqrt[n]{n} &= 1\\ \lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n &= e\\ \lim\limits_{n\to\infty}\left(1-\frac{1}{n}\right)^n &= \frac{1}{e}\\ \lim\limits_{n\to\infty}\left(1+\frac{k}{n}\right)^n &= e^k\\ \end{array}$
Where $$e=2.71828\dots$$ is the irrational number called Euler’s number.
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## Engage NY Eureka Math 5th Grade Module 2 Lesson 6 Answer Key
### Eureka Math Grade 5 Module 2 Lesson 6 Pattern Sheet Answer Key
Solve
5 x 100 = 500, 500 – 5 = 495, 5 x 99 = 495, 3 x 100 = 300, 300 – 3 = 297, 3 x 99 = 297, 2 x 100 = 200, 200 – 2 = 198, 2 x 99 = 198, 6 x 100 = 600, 600 – 6 = 594, 6 x 99 = 594, 4 x 100 = 400, 4 x 99 = 396, 7 x 100 = 700, 7 x 99 = 693, 9 x 100 = 900, 9 x 99 = 891, 8 x 100 = 800, 8 x 99 = 792, 5 x 100 = 500, 50 x 100 = 5000, 5000 – 50 = 4950, 50 x 99 = 4950, 80 x 100 = 8000, 80 x 99 = 7920, 60 x 100 = 6000, 60 x 99 = 5940, 11 x 100 = 1100, 1100 – 11 = 1089, 11 x 99 = 1089, 21 x 100 = 2100, 2100 – 21 = 2079, 21 x 99 = 2079, 31 x 100 = 3100, 31 x 99 = 3069, 71 x 100 = 7100, 71 x 99 = 7029, 42 x 100 = 4200, 42 x 99 = 4158, 53 x 99 = 5247, 64 x 99 = 6336, 75 x 99 = 7425, 97 x 99 = 9603.
Explanation:
In the above-given question,
given that,
5 x 100 = 500, 500 – 5 = 495, 5 x 99 = 495, 3 x 100 = 300, 300 – 3 = 297, 3 x 99 = 297, 2 x 100 = 200, 200 – 2 = 198, 2 x 99 = 198, 6 x 100 = 600, 600 – 6 = 594, 6 x 99 = 594, 4 x 100 = 400, 4 x 99 = 396, 7 x 100 = 700, 7 x 99 = 693, 9 x 100 = 900, 9 x 99 = 891, 8 x 100 = 800, 8 x 99 = 792, 5 x 100 = 500, 50 x 100 = 5000, 5000 – 50 = 4950, 50 x 99 = 4950, 80 x 100 = 8000, 80 x 99 = 7920, 60 x 100 = 6000, 60 x 99 = 5940, 11 x 100 = 1100, 1100 – 11 = 1089, 11 x 99 = 1089, 21 x 100 = 2100, 2100 – 21 = 2079, 21 x 99 = 2079, 31 x 100 = 3100, 31 x 99 = 3069, 71 x 100 = 7100, 71 x 99 = 7029, 42 x 100 = 4200, 42 x 99 = 4158, 53 x 99 = 5247, 64 x 99 = 6336, 75 x 99 = 7425, 97 x 99 = 9603.
Question 1.
5 x 100 = 500.
5 x 100 = 500.
Explanation:
In the above-given question,
given that,
5 x 100.
500.
Question 2.
500 – 5 = 495.
500 – 5 = 495.
Explanation:
In the above-given question,
given that,
500 – 5.
495.
Question 3.
5 x 99 = 495.
5 x 99 = 495.
Explanation:
In the above-given question,
given that,
5 x 99.
495.
Question 4.
3 x 100 = 300.
3 x 100 = 300.
Explanation:
In the above-given question,
given that,
3 x 100.
300.
Question 5.
300 – 3 = 297.
300 – 3 = 297.
Explanation:
In the above-given question,
given that,
300 – 3.
297.
Question 6.
3 x 99 = 297.
3 x 99 = 297.
Explanation:
In the above-given question,
given that,
3 x 99.
297.
Question 7.
2 x 100 = 200.
2 x 100 = 200.
Explanation:
In the above-given question,
given that,
2 x 100.
200.
Question 8.
200 – 2 = 198.
200 – 2 = 198.
Explanation:
In the above-given question,
given that,
200 – 2.
198.
Question 9.
2 x 99 = 198.
2 x 99 = 198.
Explanation:
In the above-given question,
given that,
2 x 99.
198.
Question 10.
6 x 100 = 600
6 x 100 = 600.
Explanation:
In the above-given question,
given that,
6 x 100.
600.
Question 11.
600 – 6 = 594.
600 – 6 = 594.
Explanation:
In the above-given question,
given that,
600 – 6.
594.
Question 12.
6 x 99 = 594.
6 x 99 = 594.
Explanation:
In the above-given question,
given that,
6 x 99.
594.
Question 13.
4 x 100 = 400.
4 x 100 = 400.
Explanation:
In the above-given question,
given that,
4 x 100.
400.
Question 14.
4 x 99 = 396.
4 x 99 = 396.
Explanation:
In the above-given question,
given that,
4 x 99.
396.
Question 15.
7 x 100 = 700.
7 x 100 = 700.
Explanation:
In the above-given question,
given that,
7 x 100.
700.
Question 16.
7 x 99 = 693.
5 x 100 = 500.
Explanation:
In the above-given question,
given that,
5 x 100.
500.
Question 17.
9 x 100 = 900
9 x 100 = 900.
Explanation:
In the above-given question,
given that,
9 x 100.
900.
Question 18.
9 x 99 = 891.
9 x 99 = 891.
Explanation:
In the above-given question,
given that,
9 x 99.
891.
Question 19.
8 x 100 = 800
8 x 100 = 800.
Explanation:
In the above-given question,
given that,
8 x 100.
800.
Question 20.
8 x 99 = 792.
8 x 99 = 792.
Explanation:
In the above-given question,
given that,
8 x 99.
792.
Question 21.
5 x 100 = 500.
5 x 100 = 500.
Explanation:
In the above-given question,
given that,
5 x 100.
500.
Question 22.
50 x 100 = 5000
50 x 100 = 5000.
Explanation:
In the above-given question,
given that,
50 x 100.
5000.
Question 23.
5000 – 50 = 4950.
5000 – 50 = 4950.
Explanation:
In the above-given question,
given that,
5000 – 50.
4950.
Question 24.
50 x 99 = 4950.
50 x 99 = 4950.
Explanation:
In the above-given question,
given that,
50 x 99.
4950.
Question 25.
80 x 100 = 8000
80 x 100 = 8000.
Explanation:
In the above-given question,
given that,
80 x 100.
8000.
Question 26.
80 x 99 = 7920.
80 x 99 = 7920.
Explanation:
In the above-given question,
given that,
80 x 99.
7920.
Question 27.
60 x 100 = 6000
60 x 100 = 6000.
Explanation:
In the above-given question,
given that,
60 x 100.
6000.
Question 28.
60 x 99 = 5940.
60 x 99 = 5940.
Explanation:
In the above-given question,
given that,
60 x 99.
5940.
Question 29.
11 x 100 = 1100
5 x 100 = 500.
Explanation:
In the above-given question,
given that,
5 x 100.
500.
Question 30.
1100- 11 = 1089
1100 – 11 = 1089.
Explanation:
In the above-given question,
given that,
1100 – 11.
1089.
Question 31.
11 x 99 = 1089.
11 x 99 = 1089.
Explanation:
In the above-given question,
given that,
11 x 99.
1089.
Question 32.
21 x100 = 2100.
21 x 100 = 2100.
Explanation:
In the above-given question,
given that,
21 x 100.
2100.
Question 33.
2100-21 = 2079
2100 – 21 = 2079.
Explanation:
In the above-given question,
given that,
2100 – 21.
2079.
Question 34.
21 x 99 = 2079.
21 x 99 = 2079.
Explanation:
In the above-given question,
given that,
21 x 99.
2079.
Question 35.
31 x100 = 3100.
31 x 100 = 3100.
Explanation:
In the above-given question,
given that,
31 x 100.
3100.
Question 36.
31 x 99 = 3069.
31 x 99 = 3069.
Explanation:
In the above-given question,
given that,
31 x 99.
3069.
Question 37.
71 x100 = 7100.
71 x 100 = 7100.
Explanation:
In the above-given question,
given that,
71 x 100.
7100.
Question 38.
71 x 99 = 7029.
71 x 99 = 7029.
Explanation:
In the above-given question,
given that,
71 x 99.
7029.
Question 39.
42 x 100 = 4200
42 x 100 = 4200.
Explanation:
In the above-given question,
given that,
42 x 100.
4200.
Question 40.
42 x 99 = 4158.
42 x 99 = 4158.
Explanation:
In the above-given question,
given that,
42 x 99.
4158.
Question 41.
53 x 99 = 5247.
53 x 99 = 5247.
Explanation:
In the above-given question,
given that,
53 x 99.
5247.
Question 42.
64 x 99 = 6336.
64 x 99 = 6336.
Explanation:
In the above-given question,
given that,
64 x 99.
6336.
Question 43.
75 x 99 = 7425.
75 x 99 = 7425.
Explanation:
In the above-given question,
given that,
75 x 99.
7425.
Question 44.
97 x 99 = 9603.
97 x 99 = 9603.
Explanation:
In the above-given question,
given that,
97 x 99.
9603.
### Eureka Math Grade 5 Module 2 Lesson 5 Problem Set Answer Key
Question 1.
Draw an area model. Then, solve using the standard algorithm. Use arrows to match the partial products from your area model to the partial products in the algorithm.
a. 48×35
48 x 35 = 1680.
Explanation:
In the above-given question,
given that,
48 x 35.
1680.
b. 648 x 35
648 x 35 = 22680.
Explanation:
In the above-given question,
given that,
648 x 35.
22680.
Question 2.
Solve using the standard algorithm.
a. 758 x 92
758 x 92 = 69736.
Explanation:
In the above-given question,
given that,
758 x 92.
69736.
b. 958 x 94
958 x 94 = 90052.
Explanation:
In the above-given question,
given that,
958 x 94.
90052.
c. 476 x 65
476 x 65 = 30940.
Explanation:
In the above-given question,
given that,
476 x 65.
30940.
d. 547 x 64
547 x 64 = 35008.
Explanation:
In the above-given question,
given that,
547 x 64.
35008.
Question 3.
Carpet costs $16 a square foot. A rectangular floor is 16 feet long by 14 feet wide. How much would it cost to carpet the floor? Answer: The cost to carpet the floor = Explanation: In the above-given question, given that, Carpet costs$16 a square foot.
AÂ rectangular floor is 16 feet long by 14 feet wide.
16 x 14 = 224.
224 x 16 = $3584. Question 4. General admission to The American Museum of Natural History is$19.
a. If a group of 125 students visits the museum, how much will the group’s tickets cost?
The group’s ticket cost = $2375. Explanation: In the above-given question, given that, If a group of 125 students visits the museum. 125 x$ 19.
$2375. b. If the group also purchases IMAX movie tickets for an additional$4 per student, what is the new total cost of all the tickets? Write an expression that shows how you calculated the new price.
The new total cost of all the tickets = $9500. Explanation: In the above-given question, given that, If the group also purchases IMAX movie tickets for an additional$4 per student.
2375 x 4.
9500.
### Eureka Math Grade 5 Module 2 Lesson 6 Exit Ticket Answer Key
Draw an area model. Then, solve using the standard algorithm. Use arrows to match the partial products from your area model to the partial products in the algorithm.
a. 78 x 42
78 x 42 = 3276.
Explanation:
In the above-given question,
given that,
78 x 42.
3276.
b. 783 x 42
783 x 42 = 32886.
Explanation:
In the above-given question,
given that,
783 x 42.
32886.
### Eureka Math Grade 5 Module 2 Lesson 6 Homework Answer Key
Question 1.
Draw an area model. Then, solve using the standard algorithm. Use arrows to match the partial products from your area model to the partial products in the algorithm.
a. 27 x 36
27 x 36 = 972.
Explanation:
In the above-given question,
given that,
27 x 36.
972.
b. 527 x 36
527 x 36 = 18972.
Explanation:
In the above-given question,
given that,
527 x 36.
18972.
Question 2.
Solve using the standard algorithm.
a. 649 x 53
649 x 53 = 34397.
Explanation:
In the above-given question,
given that,
649 x 53.
34397.
b. 496 x 53
496 x 53 = 26288.
Explanation:
In the above-given question,
given that,
496 x 53.
26288.
c. 758 x 46
758 x 46 = 34868.
Explanation:
In the above-given question,
given that,
758 x 46.
34868.
d. 529 x 48
529 x 48 = 25392.
Explanation:
In the above-given question,
given that,
529 x 48.
25392.
Question 3.
Each of the 25 students in Mr. McDonald’s class sold 16 raffle tickets. If each ticket costs $15, how much money did Mr. McDonald’s students raise? Answer: The money did Mr. McDonald’s students raise =$375.
Explanation:
In the above-given question,
given that,
Each of the 25 students in Mr. McDonald’s class sold 16 raffle tickets.
if each ticket costs $15. 25 x 15 = 375. Question 4. Jayson buys a car and pays by installments. Each installment is$567 per month. After 48 months, Jayson owes $1,250. What was the total price of the vehicle? Answer: The total price of the vehicle = 12,358. Explanation: In the above-given question, given that, Jayson buys a car and pays by installments. Each installment is$567 per month.
After 48 months Jayson owes \$1250.
567 x 12 = 6804.
6804 + 6804 = 13608.
13608 – 1250 = 12,358.
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Type a math problem
Solve for x
Steps for Completing the Square
All equations of the form can be solved using the quadratic formula: . The quadratic formula gives two solutions, one when is addition and one when it is subtraction.
Subtract from both sides of the equation.
Subtracting from itself leaves .
This equation is in standard form: . Substitute for , for , and for in the quadratic formula, .
Square .
Multiply times .
The opposite of is .
Now solve the equation when is plus. Add to .
Now solve the equation when is minus. Subtract from .
The equation is now solved.
Solve for y
Steps for Solving Linear Equation
Subtract from both sides.
The equation is in standard form.
Divide both sides by .
Dividing by undoes the multiplication by .
Solve for x (complex solution)
Graph
x^{2}-5x+3y=20
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x^{2}-5x+3y-20=20-20
Subtract 20 from both sides of the equation.
x^{2}-5x+3y-20=0
Subtracting 20 from itself leaves 0.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(3y-20\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 3y-20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\left(3y-20\right)}}{2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25+80-12y}}{2}
Multiply -4 times 3y-20.
x=\frac{-\left(-5\right)±\sqrt{105-12y}}{2}
x=\frac{5±\sqrt{105-12y}}{2}
The opposite of -5 is 5.
x=\frac{\sqrt{105-12y}+5}{2}
Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is plus. Add 5 to \sqrt{105-12y}.
x=\frac{-\sqrt{105-12y}+5}{2}
Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is minus. Subtract \sqrt{105-12y} from 5.
x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}
The equation is now solved.
x^{2}-5x+3y=20
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
x^{2}-5x+3y-3y=20-3y
Subtract 3y from both sides of the equation.
x^{2}-5x=20-3y
Subtracting 3y from itself leaves 0.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=20-3y+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}=-2.5. Then add the square of -\frac{5}{2}=-2.5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-5x+\frac{25}{4}=20-3y+\frac{25}{4}
Square -\frac{5}{2}=-2.5 by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=\frac{105}{4}-3y
\left(x-\frac{5}{2}\right)^{2}=\frac{105}{4}-3y
Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{105}{4}-3y}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{\sqrt{105-12y}}{2} x-\frac{5}{2}=-\frac{\sqrt{105-12y}}{2}
Simplify.
x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}
Add \frac{5}{2}=2.5 to both sides of the equation.
-5x+3y=20-x^{2}
Subtract x^{2} from both sides.
3y=20-x^{2}+5x
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# How to Introduce Fractions to Your Child
Oct 22, 2020 | Red Deer
“What is ½ plus ½ ?” Ask that to a child, in higher elementary or middle school, and too often they’d answer “2/4”. Even some high school students would answer “2/5” when asked “what is ½ plus 1/3?”
When a whole is divided into a number of equal parts, they are called fractions. When a person encounters the word “fractions” the images “part of a whole” and “equal parts” must come to mind.
This concept must be second nature and deeply embedded in the child’s consciousness, and the earlier in life the better.
Half: Two Parts The Same
Start by teaching children the basic principles of half. Once the idea of half is learned thoroughly, its attributes can be easily applied to all other fractions.
Teaching Two Parts The Same
Children often call any division into two pieces “half” whether the pieces are the same (equal) or not. Correct the child and say that the two parts must be the same to have halves.
Next, ask ”what is half of 4?” then half of 6, 8. Then ask “what is half of 3?” Most likely they will answer “there is no such thing as half of 3”. Ask the child to imagine 3 cookies and to divide them into “two parts the same”. They will discover that one cookie must be broken into two pieces.
Now you have introduced the child to a very important mathematical concept: All wholes can be broken down into parts.
Next ask “what is half of 5? … of 7?”
In a couple of weeks the child will have mastered the skill.
Next, ask a reverse question: “half of what number is 3?” and the next reverse question is “half of what number is two and a half?”
If your child has trouble with these reverse questions, have them draw pictures of the parts, both separate and combined. Or break up and recombine cookies until they see the concept.
Bring on the Reinforcements
• “What is half of 10?”
• “Half of what number is 4?”
• “I have 8 books. If you take half, how many will you have?”
• “Half of the cookies are missing and we have 6 left. How many did we start with?”
Movin’ On
Draw five sandwiches in a row. Then draw with a line dividing them all in half and ask:
• “If you have 5 whole sandwiches, how many half sandwiches can you make?”
• “If you have 8 half sandwiches, how many whole sandwiches can you make?”
• “If you have 3 half sandwiches, how many sandwiches do you have?"
• “If you have 5 sandwiches and eat one and a half of them, how many do you have left?
• “If you have two and a half sandwiches and I have one and a half sandwiches, how many do we have altogether?”
After mastering the above questions, comes “what is half of a half?”; when explaining this, remind them that half means two parts the same.
Next, try a more complex reverse question like:
“You have a box of cookies. You eat half of them after lunch and half of what’s left after dinner. After dinner, you have two cookies left. How many cookies did you start with?”
If you do these things on a daily basis, one day it will “just happen” that your child will be able to intelligently talk about half. They already understand the basic concepts of fractions and ready to move on to the next level.
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7.7 Segments of chords secants and tangents Theorem : If two chords, seg.AB and seg.CD intersect inside or outside a circle at P then l (seg. PA) ´ l (seg. PB) = l (seg. PC) ´ l (seg. PD) Figure 7.21 (a) Figure 7.21 (b) In figure 7.21 (a) P is in the interior of the circle. Join AC and BD and consider DAPC and DBDP. m Ð APC = m Ð BPD (vertical angles). m Ð CAP = m Ð BDP (angles inscribed in the same arc). \ DAPC ~ D BPD ( A A test ) \ ( corresponding sides of similar triangles). l (seg. PA) ´ l (seg. PB) = l (seg. PC) ´ l (seg. PD). Now consider figure 7.21 (b). P is in the exterior of the circle. Join A to C and B to D. Consider D PAC and D PBD m Ð APC = m Ð BPD (same angle). m Ð CAP = m Ð PDB (exterior angle property of a cyclic quadrilateral). \ D PAC ~ D PDB ( A A test ) \ (corresponding sides of similar triangles). \ l (seg. PA) ´ l (seg. PB) = l (seg. PC) ´ l (seg. PD). Consider a secant PAB to a circle, (figure 7.22) intersecting the circle at A and B and line PT is a tangent then l (seg. PA) ´ l (seg. PB) = l (seg. PT)2. Figure 7.22 P is a point in the exterior of the circle. A secant passes through P and intersects the circle at points A & B. Tangent through P touches the circle in point T. To prove that l (seg. PA) ´ l (seg. PB) = l (seg. PT)2 Consider D PTA and D PTB. m Ð TPA = Ð TPB ( same angle) According to Tangent Secant theorem, m Ð ATP = m (arc AT) = m Ð PTB ( inscribed angle ) \ D PTA ~ D PTB ( A A test ) \ (corresponding sides of similar triangles). \ l (seg. PA) ´ l (seg. PB) = l (seg. PT)2. Theorem: The lengths of two tangent segments from an external point to a circle are equal. As shown in figure 7.23 seg. QR and seg. QS are two tangents on a circle with P as its center. Figure 7.23 To prove that l (seg.QR) = l (seg.QS) join P to Q and R to S. m Ð PRQ = m Ð PSQ = 900. The radius and the tangent form a right angle at the point of tangency, \ D PRQ and D PSQ are right triangles such that seg. PR @ seg PS (radii of the same circle). seg. PQ @ seg. PQ (same side). \ D PRQ @ D PSQ (H.S) \ seg.QR @ seg.QS (corresponding sides of congruent triangles are congruent). \ l (seg.QR) = l (seg.QS). Index 7.1 Introduction 7.2 Lines of circle 7.3 Arcs 7.4 Inscribed angels 7.5 Some properties od tangents, secants and chords 7.6 Chords and their arcs 7.7 Segments of chords secants and tangents 7.8 Lengths of arcs and area of sectors Chapter 8
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# Solving Linear Equations
## Techniques for Solving Linear Equations
Writing down every step when solving an equation is not always necessary. Solving an equation is often part of a larger problem, and anything that we can do to make the process more efficient will make solving the entire problem fastar and easier. For example, we can combine some steps.
Combining Steps Writing Every Step 4x - 5 = 23 4x - 5 = 23 4x = 28 Add 5 to each side. 4x - 5 + 5 = 23 + 5 x = 7 Divide each side by 4. 4x = 28 x = 7
The same steps are used in each of the solutions. However, when 5 is added to each side in the solution on the left, only the result is written. When each side is divided by 4, only the result is written.
The equation -x = -5 says that the additive inverse of x is -5. Since the additive inverse of 5 is -5, we conclude that x is 5. So instead of multiplying each sideof -x = -5 by -1, we solve the equation as follows:
-x = -5 x = 5 Additive inverse property
Sometimes it is simpler to isolate x on the right-hand side of the equation:
3x + 1 = 4x - 5 6 = x Subtract 3x from each side and add 5 to each side.
You can rewrite 6 = x as x = 6 or leave it as is. Either way, 6 is the solution.
For some equations with fractions it is more efficient to multiply by a multiplicative inverse instead of multiplying by the LCD:
Multiply each side by the reciprocal of x
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# Sequences
## Add or multiply the same number to get the next term in a sequence.
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Arithmetic and Geometric Sequences
A sequence is a list of numbers with a common pattern. The common pattern in an arithmetic sequence is that the same number is added or subtracted to each number to produce the next number. The common pattern in a geometric sequence is that the same number is multiplied or divided to each number to produce the next number.
Are all sequences arithmetic or geometric?
### Sequences
A sequence is just a list of numbers separated by commas. A sequence can be finite or infinite. If the sequence is infinite, the first few terms are followed by an ellipsis \begin{align*}(\ldots)\end{align*} indicating that the pattern continues forever.
An infinite sequence: \begin{align*} 1, 2, 3, 4, 5, \ldots\end{align*}
A finite sequence: \begin{align*}2, 4, 6, 8\end{align*}
In general, you describe a sequence with subscripts that are used to index the terms. The \begin{align*}k^{th}\end{align*} term in the sequence is \begin{align*}a_k\end{align*}.
\begin{align*}a_1, a_2, a_3, a_4,\ldots, a_k, \ldots\end{align*}
#### Arithmetic Sequences
Arithmetic sequences are defined by an initial value \begin{align*}a_1\end{align*} and a common difference \begin{align*}d\end{align*}.
\begin{align*}a_1 &=a_1 \\ a_2 &=a_1+d \\ a_3 &=a_1+2d \\ a_4 &=a_1+3d \\ & \ \ \vdots \\ a_n &=a_1+(n-1)d\end{align*}
#### Geometric Sequences
Geometric sequences are defined by an initial value \begin{align*}a_1\end{align*} and a common ratio \begin{align*}r\end{align*}.
\begin{align*}a_1 &=a_1\\ a_2 &=a_1 \cdot r\\ a_3 &=a_1\cdot r^2\\ a_4 &=a_1 \cdot r^3\\ & \ \ \vdots \\ a_n &=a_1 \cdot r^{n-1}\end{align*}
When trying to determine what kind of sequence it is, first test for a common difference and then test for a common ratio. If the sequence does not have a common difference or ratio, it is neither an arithmetic or geometric sequence.
\begin{align*}0.135, 0.189, 0.243, 0.297, \ldots\end{align*} is an arithmetic sequence because the common difference is 0.054.
\begin{align*}\frac{2}{9}, \frac{1}{6}, \frac{1}{8}, \ldots\end{align*} is a geometric sequence because the common ratio is \begin{align*}\frac{3}{4}\end{align*}.
\begin{align*}0.54, 1.08, 3.24, \ldots\end{align*} is not arithmetic because the differences between consecutive terms are 0.54 and 2.16 which are not common. The sequence is not geometric because the ratios between consecutive terms are 2 and 3 which are not common.
### Examples
#### Example 1
Earlier, you were asked if all sequences are arithmetic or geometric. The sequence above shows that not all sequences are arithmetic or geometric. Two famous sequences that are neither arithmetic nor geometric are the Fibonacci sequence and the sequence of prime numbers.
Fibonacci Sequence: \begin{align*} 1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots\end{align*}
Prime Numbers: \begin{align*}2, 3, 5, 7, 11, 13, 17, 19, 23, \ldots\end{align*}
#### Example 2
For the following sequence, determine the common ratio or difference and the next three terms.
\begin{align*}\frac{2}{3}, \frac{4}{9}, \frac{6}{27}, \frac{8}{81}, \frac{10}{243}, \ldots\end{align*}
This sequence is neither arithmetic nor geometric. The differences between the first few terms are \begin{align*}-\frac{2}{9}, -\frac{2}{9}, -\frac{10}{81}, -\frac{14}{243}\end{align*}. While there was a common difference at first, this difference did not hold through the sequence. Always check the sequence in multiple places to make sure that the common difference holds up throughout.
The sequence is also not geometric because the ratios between the first few terms are \begin{align*}\frac{2}{3}, \frac{1}{2}, \frac{4}{9}\end{align*}. These ratios are not common.
Even though you cannot get a common ratio or a common difference, it is still possible to produce the next three terms in the sequence by noticing the numerator is an arithmetic sequence with starting term of 2 and a common difference of 2. The denominators are a geometric sequence with an initial term of 3 and a common ratio of 3. The next three terms are:
\begin{align*}\frac{12}{3^6}, \frac{14}{3^7}, \frac{16}{3^8}\end{align*}
#### Example 3
What is the tenth term in the following sequence?
\begin{align*}-12, 6, -3, \frac{3}{2}, \ldots\end{align*}
The sequence is geometric and the common ratio is \begin{align*}-\frac{1}{2}\end{align*}. The equation is:
\begin{align*} a_n=-12 \cdot \left(-\frac{1}{2}\right)^{n-1}\end{align*}
The tenth term is:
\begin{align*}-12 \cdot \left(-\frac{1}{2}\right)^9=\frac{3}{128}\end{align*}
#### Example 4
What is the tenth term in the following sequence?
\begin{align*}-1, \frac{2}{3}, \frac{7}{3}, 4, \frac{17}{3}, \ldots\end{align*}
The pattern might not be immediately recognizable, but try ignoring the \begin{align*}\frac{1}{3}\end{align*} in each number to see the pattern a different way.
\begin{align*}-3, 2, 7, 12, 17, \ldots\end{align*}
You should see the common difference of 5. This means the common difference from the original sequence is \begin{align*}\frac{5}{3}\end{align*}. The equation is \begin{align*}a_n=-1+(n-1)\left(\frac{5}{3}\right)\end{align*}. The \begin{align*}10^{th}\end{align*} term is:
\begin{align*}-1+9 \cdot \left(\frac{5}{3}\right)=-1+3 \cdot 5=-1+15=14\end{align*}
#### Example 5
Find an equation that defines the \begin{align*}a_k\end{align*} term for the following sequence.
\begin{align*}0, 3, 8, 15, 24, 35, \ldots\end{align*}
The sequence is not arithmetic nor geometric. It will help to find the pattern by examining the common differences and then the common differences of the common differences. This numerical process is connected to ideas in calculus.
0, 3, 8, 15, 24, 35
3, 5, 7, 9, 11
2, 2, 2, 2
Notice when you examine the common difference of the common differences the pattern becomes increasingly clear. Since it took two layers to find a constant function, this pattern is quadratic and very similar to the perfect squares.
\begin{align*}1, 4, 9, 16, 25, 36, \ldots\end{align*}
The \begin{align*}a_k\end{align*} term can be described as \begin{align*}a_k=k^2-1\end{align*}
### Review
Use the sequence \begin{align*}1, 5, 9, 13, \ldots\end{align*} for questions 1-3.
1. Find the next three terms in the sequence.
2. Find an equation that defines the \begin{align*}a_k\end{align*} term of the sequence.
3. Find the \begin{align*}150^{th}\end{align*} term of the sequence.
Use the sequence \begin{align*}12, 4, \frac{4}{3}, \frac{4}{9}, \ldots\end{align*} for questions 4-6.
4. Find the next three terms in the sequence.
5. Find an equation that defines the \begin{align*}a_k\end{align*} term of the sequence.
6. Find the \begin{align*}17^{th}\end{align*} term of the sequence.
Use the sequence \begin{align*}10, -2, \frac{2}{5}, -\frac{2}{25}, \ldots\end{align*} for questions 7-9.
7. Find the next three terms in the sequence.
8. Find an equation that defines the \begin{align*}a_k\end{align*} term of the sequence.
9. Find the \begin{align*}12^{th}\end{align*} term of the sequence.
Use the sequence \begin{align*}\frac{7}{2}, \frac{9}{2}, \frac{11}{2}, \frac{13}{2}, \ldots\end{align*} for questions 10-12.
10. Find the next three terms in the sequence.
11. Find an equation that defines the \begin{align*}a_k\end{align*} term of the sequence.
12. Find the \begin{align*}314^{th}\end{align*} term of the sequence.
13. Find an equation that defines the \begin{align*}a_k \end{align*} term for the sequence \begin{align*}4, 11, 30, 67, \ldots\end{align*}
14. Explain the connections between arithmetic sequences and linear functions.
15. Explain the connections between geometric sequences and exponential functions.
To see the Review answers, open this PDF file and look for section 12.2.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
arithmetic sequence An arithmetic sequence has a common difference between each two consecutive terms. Arithmetic sequences are also known are arithmetic progressions.
common difference Every arithmetic sequence has a common or constant difference between consecutive terms. For example: In the sequence 5, 8, 11, 14..., the common difference is "3".
common ratio Every geometric sequence has a common ratio, or a constant ratio between consecutive terms. For example in the sequence 2, 6, 18, 54..., the common ratio is 3.
Explicit Explicit formulas define each term in a sequence directly, allowing one to calculate any term in the sequence without knowing the value of the previous terms.
Explicit formula Explicit formulas define each term in a sequence directly, allowing one to calculate any term in the sequence without knowing the value of the previous terms.
geometric sequence A geometric sequence is a sequence with a constant ratio between successive terms. Geometric sequences are also known as geometric progressions.
sequence A sequence is an ordered list of numbers or objects.
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# Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations)
Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) help students get a clear idea of operations and methods to be followed in solving problems effortlessly. The solutions contain explanations of basic concepts in an understandable language. Shortcut tricks and important formulas are covered in a stepwise manner for better conceptual knowledge among students. Selina Solutions Concise Maths Class 7 Chapter 11 Fundamental Concepts (Including Fundamental Operations) PDF can be downloaded from the links provided below.
Chapter 11 explains the elementary treatment and operations on various terms, as per the latest syllabus of the ICSE board. The main aim of preparing chapter-wise solutions is to provide a better hold on the subject among students.
## Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations) Download PDF
### Exercises of Selina Solutions Concise Maths Class 7 Chapter 11 – Fundamental Concepts (Including Fundamental Operations)
Exercise 11A Solutions
Exercise 11B Solutions
Exercise 11C Solutions
Exercise 11D Solutions
Exercise 11E Solutions
Exercise 11F Solutions
## Access Selina Solutions Concise Maths Class 7 Chapter 11: Fundamental Concepts (Including Fundamental Operations)
#### Exercise 11A page: 121
1. Separate constant terms and variable terms from the following:
8, x, 6xy, 6 + x, – 5xy2, 15az2, 32z/ xy, y2/ 3x
Solution:
The constant term is 8.
The variable terms are x, 6xy, 6 + x, – 5xy2, 15az2, 32z/ xy, y2/ 3x.
2. For each expression, given below, state whether it is a monomial, binomial or trinomial:
(i) 2x ÷ 15
(ii) ax + 9
(iii) 3x2 × 5x
(iv) 5 + 2x – 3b
(v) 2y – 7z/3 ÷ x
(vi) 3p × q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a3 -3ab2 × c
Solution:
(i) 2x ÷ 15 = 2x/15
It has one term and hence it is a monomial.
(ii) ax + 9
It has two terms and hence it is a binomial.
(iii) 3x2 × 5x = 15x3
It has one term and hence it is a monomial.
(iv) 5 + 2x – 3b
It has three terms and hence it is a trinomial.
(v) 2y – 7z/3 ÷ x = 2y – 7z/3x
It has two terms and hence it is a binomial.
(vi) 3p × q ÷ z = 3pq/z
It has one term and hence it is a monomial.
(vii) 12z ÷ 5x + 4 = 12z/5x + 4
It has two terms and hence it is a binomial.
(viii) 12 – 5z – 4 = 8 – 5z
It has two terms and hence it is a binomial.
(ix) a3 -3ab2 × c = a3 – 3ab2c
It has two terms and hence it is a binomial.
3. Write the coefficient of:
(i) xy in -3axy
(ii) z2 in p2yz2
(iii) mn in – mn
(iv) 15 in -15p2
Solution:
(i) xy in -3axy
The coefficient of xy in -3axy = -3a
(ii) z2 in p2yz2
The coefficient of z2 in p2yz2 = p2y
(iii) mn in – mn
The coefficient of mn in – mn = -1
(iv) 15 in -15p2
The coefficient of 15 in -15p2 = -p2
4. For each of the following monomials, write its degree:
(i) 7y
(ii) –x2y
(iii) xy2z
(iv) -9y2z3
(v) 3m3n4
(vi) -2p2q3r4
Solution:
(i) The degree of 7y is 1.
(ii) The degree of –x2y = 2 + 1 = 3
(iii) The degree of xy2z = 1 + 2 + 1 = 4
(iv) The degree of -9y2z3 = 2 + 3 = 5
(v) The degree of 3m3n4 = 3 + 4 = 7
(vi) The degree of -2p2q3r4 = 2 + 3 + 4 = 9
5. Write the degree of each of the following polynomials:
(i) 3y3 – x2y2 + 4x
(ii) p3q2 – 6p2q5 + p4q4
(iii) -8mn6 + 5m3n
(iv) 7 – 3x2y + y2
(v) 3x – 15
(vi) 2y2z + 9yz3
Solution:
(i) The degree of 3y3 – x2y2 + 4x is 4
x2y2 is the term which has the highest degree.
(ii) The degree of p3q2 – 6p2q5 + p4q4 is 8
p4q4 is the term which has the highest degree.
(iii) The degree of -8mn6 + 5m3n is 7
-8mn6 is the term which has the highest degree.
(iv) The degree of 7 – 3x2y + y2 is 3
– 3x2y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1
3x is the term which has the highest degree.
(vi) The degree of 2y2z + 9yz3 is 4
9yz3 is the term which has the highest degree.
6. Group the like terms together:
(i) 9x2, xy, -3x2, x2 and -2xy
(ii) ab, -a2b, -3ab, 5a2b and -8a2b.
(iii) 7p, 8pq, -5pq, -2p and 3p
Solution:
(i) 9x2, xy, -3x2, x2 and -2xy
9x2, -3x2 and x2 are like terms
xy and -2xy are like terms.
(ii) ab, -a2b, -3ab, 5a2b and -8a2b
-a2b, 5a2b and -8a2b are like terms
ab and – 3ab are like terms.
(iii) 7p, 8pq, -5pq, -2p and 3p
7p, -2p and 3p are like terms
8pq and -5pq are like terms.
7. Write the numerical coefficient of each of the following:
(i) y
(ii) – y
(iii) 2x2y
(iv) -8xy3
(v) 3py2
(vi) -9a2b3
Solution:
(i) The numerical coefficient of y is 1.
(ii) The numerical coefficient of – y is – 1.
(iii) The numerical coefficient of 2x2y is 2.
(iv) The numerical coefficient of -8xy3 is -8.
(v) The numerical coefficient of 3py2 is 3.
(vi) The numerical coefficient of -9a2b3 is -9.
8. In -5x3y2z4; write the coefficient of:
(i) z2
(ii) y2
(iii) yz2
(iv) x3y
(v) –xy2
(vi) -5xy2z
Also, write the degree of the given algebraic expression.
Solution:
(i) The coefficient of z2 is -5x3y2z2.
(ii) The coefficient of y2 is -5x3z4.
(iii) The coefficient of yz2 is -5x3yz2.
(iv) The coefficient of x3y is -5yz4.
(v) The coefficient of –xy2 is 5x2z4.
(vi) The coefficient of -5xy2z is x2z3.
So the degree of the given algebraic expression = 3 + 2 + 4 = 9.
#### Exercise 11B page: 125
1. Fill in the blanks:
(i) 8x + 5x = …….
(ii) 8x – 5x = ……
(iii) 6xy2 + 9xy2 = …….
(iv) 6xy2 – 9xy2 = …….
(v) The sum of 8a, 6a and 5b = …….
(vi) The addition of 5, 7xy, 6 and 3xy = ……..
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ……..
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = ………
(x) 16x2 – 9x2 = ……. and 25xy2 – 17xy2 = ………
Solution:
(i) 8x + 5x = 13x
(ii) 8x – 5x = 3x
(iii) 6xy2 + 9xy2 = 15xy2
(iv) 6xy2 – 9xy2 = -3xy2
(v) The sum of 8a, 6a and 5b = 14a + 5b
It can be written as
8a + 6a + 5b = 14a + 5b
(vi) The addition of 5, 7xy, 6 and 3xy = 11 + 10xy
It can be written as
5 + 7xy + 6 + 3xy = 11 + 10xy
(vii) 4a + 3b – 7a + 4b = 7b – 3a
It can be written as
4a + 3b – 7a + 4b = (4 – 7)a + (3 + 4)b
= -3a + 7b
(viii) – 15x + 13x + 8 = 8 – 2x
It can be written as
-15x + 13x + 8 = (-15 + 13) x + 8 = -2x + 8
(ix) 6x2y + 13xy2 – 4x2y + 2xy2 = 2x2y + 15xy2
It can be written as
6x2y + 13xy2 – 4x2y + 2xy2 = (6 – 4) x2y + (13 + 2) xy2
= 2x2y + 15xy2
(x) 16x2 – 9x2 = 7x2 and 25xy2 – 17xy2 = 8xy2
(i) -9x, 3x and 4x
(ii) 23y2, 8y2 and – 12y2
(iii) 18pq, -15pq and 3pq
Solution:
(i) -9x, 3x and 4x
It can be written as
= -9x + 3x + 4x
So we get
= -9x + 7x
= -2x
(ii) 23y2, 8y2 and – 12y2
It can be written as
= 23y2 + 8y2 – 12y2
So we get
= 31y2 – 12y2
= 19y2
(iii) 18pq, -15pq and 3pq
It can be written as
= 18pq – 15pq + 3pq
So we get
= 3pq + 3pq
= 6pq
3. Simplify:
(i) 3m + 12m – 5m
(ii) 7n2 – 9n2 + 3n2
(iii) 25zy – 8zy – 6zy
(iv) -5ax2 + 7ax2 – 12ax2
(v) – 16am + 4mx + 4am – 15mx + 5am
Solution:
(i) 3m + 12m – 5m
It can be written as
= 15m – 5m
So we get
= 10m
(ii) 7n2 – 9n2 + 3n2
It can be written as
= (7 + 3) n2 – 9n2
So we get
= 10n2 – 9n2
= n2
(iii) 25zy – 8zy – 6zy
It can be written as
= 25zy – 14zy
So we get
= 11zy
(iv) -5ax2 + 7ax2 – 12ax2
It can be written as
= (-5 – 12) ax2 + 7ax2
So we get
= -17ax2 + 7ax2
= -10ax2
(v) – 16am + 4mx + 4am – 15mx + 5am
It can be written as
= (-16 + 4 + 5) am + (4 – 15) mx
So we get
= – 7am – 11mx
(i) a + b and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii) -3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x
Solution:
(i) a + b and 2a + 3b
It can be written as
= a + b + 2a + 3b
So we get
= a + 2a + b + 3b
= 3a + 4b
(ii) 2x + y and 3x – 4y
It can be written as
= 2x + y + 3x – 4y
So we get
= 2x + 3x + y – 4y
= 5x – 3y
(iii) -3a + 2b and 3a + b
It can be written as
= -3a + 2b + 3a + b
So we get
= -3a + 3a + 2b + b
= 3b
(iv) 4 + x, 5 – 2x and 6x
It can be written as
= 4 + x + 5 – 2x + 6x
So we get
= x – 2x + 6x + 4 + 5
= 5x + 9
5. Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z – 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b – c and a + b + c
(iii) 4x2 + 8xy – 2y2 and 8xy – 5y2 + x2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
(v) 5x2 – 2xy + 3y2, -2x2 + 5xy + 9y2 and 3x2 – xy – 4y2
Solution:
(i) 3x + 8y + 7z, 6y + 4z – 2x and 3y – 4x + 6z
It can be written as
= 3x + 8y + 7z + 6y + 4z – 2x + 3y – 4x + 6z
By further calculation
= 3x – 2x – 4x + 8y + 6y + 3y + 7z + 4z + 6z
So we get
= 3x – 6x + 17y + 17z
= -3x + 17y + 17z
(ii) 3a + 5b + 2c, 2a + 3b – c and a + b + c
It can be written as
= 3a + 5b + 2c + 2a + 3b – c + a + b + c
By further calculation
= 3a + 2a + a + 5b + 3b + b + 2c – c + c
So we get
= 6a + 9b + 3c – c
= 6a + 9b + 2c
(iii) 4x2 + 8xy – 2y2 and 8xy – 5y2 + x2
It can be written as
= 4x2 + 8xy – 2y2 + 8xy – 5y2 + x2
By further calculation
= 4x2 + x2 + 8xy + 8xy – 2y2 – 5y2
So we get
= 5x2 + 16xy – 7y2
(iv) 9x2 – 6x + 7, 5 – 4x and 6 – 3x2
It can be written as
= 9x2 – 6x + 7 + 5 – 4x + 6 – 3x2
By further calculation
= 9x2 – 3x2 – 6x – 4x + 7 + 5 + 6
So we get
= 6x2 – 10x + 18
(v) 5x2 – 2xy + 3y2, -2x2 + 5xy + 9y2 and 3x2 – xy – 4y2
It can be written as
= 5x2 – 2xy + 3y2 – 2x2 + 5xy + 9y2 + 3x2 – xy – 4y2
By further calculation
= 5x2 – 2x2 + 3x2 – 2xy + 5xy – xy + 3y2 + 9y2 – 4y2
So we get
= 6x2 + 2xy + 8y2
6. Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) -4x2 and + 7x
(iv) +4a and -7b
(v) x3, 3x2y and 2y2
(vi) 11 and –by
Solution:
(i) x and 3y
The sum of x and 3y is x + 3y.
(ii) -2a and +5
The sum of -2a and + 5 is -2a + 5.
(iii) -4x2 and + 7x
The sum of -4x2 and + 7x is -4x2 + 7x.
(iv) +4a and -7b
The sum of +4a and -7b is + 4a – 7b.
(v) x3, 3x2y and 2y2
The sum of x3, 3x2y and 2y2 is x3 + 3x2y + 2y2.
(vi) 11 and –by
The sum of 11 and -by is 11 – by.
7. The sides of a triangle are 2x + 3y, x + 5y and 7x -2y. Find its perimeter.
Solution:
It is given that
Sides of a triangle are 2x + 3y, x + 5y and 7x -2y
We know that
Perimeter = Sum of all three sides of a triangle
Substituting the values
= 2x + 3y + x + 5y + 7x – 2y
By further calculation
= 2x + x + 7x + 3y + 5y – 2y
So we get
= 10x + 8y – 2y
= 10x + 6y
8. The two adjacent sides of a rectangle are 6a + 9b and 8a – 4b. Find its perimeter.
Solution:
It is given that
Sides of a rectangle are 6a + 9b and 8a – 4b
So length = 6a + 9b and breadth = 8a – 4b
We know that
Perimeter = 2 (length + breadth)
Substituting the values
= 2 (6a + 9b + 8a – 4b)
By further calculation
= 2 (14a + 5b)
So we get
= 28a + 10b
9. Subtract the second expression from the first:
(i) 2a + b, a + b
(ii) -2b + 2c, b + 3c
(iii) 5a + b, -6b + 2a
(iv) a3 – 1 + a, 3a – 2a2
(v) p + 2, 1
Solution:
(i) 2a + b, a + b
It can be written as
= (2a + b) – (a + b)
So we get
= 2a + b – a – b
= 2a – a + b – b
= a
(ii) -2b + 2c, b + 3c
It can be written as
= (-2b + 2c) – (b + 3c)
So we get
= -2b + 2c – b – 3c
= -2b – b + 2c – 3c
= -3b – c
(iii) 5a + b, -6b + 2a
It can be written as
= (5a + b) – (-6b + 2a)
So we get
= 5a + b + 6b – 2a
= 5a – 2a + b + 6b
= 3a + 7b
(iv) a3 – 1 + a, 3a – 2a2
It can be written as
= (a3 – 1 + a) – (3a – 2a2)
So we get
= a3 – 1 + a – 3a + 2a2
= a3 + 2a2 + a – 3a – 1
= a3 + 2a2 – 2a – 1
(v) p + 2, 1
It can be written as
= p + 2 – 1
So we get
= p + 1
10. Subtract:
(i) 4x from 8 – x
(ii) -8c from c + 3d
(iii) – 5a – 2b from b + 6c
(iv) 4p + p2 from 3p2 – 8p
(v) 5a – 3b + 2c from 4a – b – 2c
Solution:
(i) 4x from 8 – x
It can be written as
= (8 – x) – 4x
By further calculation
= 8 – x – 4x
= 8 – 5x
(ii) -8c from c + 3d
It can be written as
= (c + 3d) – (-8c)
By further calculation
= c + 3d + 8c
= 9c + 3d
(iii) – 5a – 2b from b + 6c
It can be written as
= (b + 6c) – (-5a – 2b)
By further calculation
= b + 6c + 5a + 2b
= 5a + 3b + 6c
(iv) 4p + p2 from 3p2 – 8p
It can be written as
= (3p2 – 8p) – (4p + p2)
By further calculation
= 3p2 – 8p – 4p – p2
= 2p2 – 12p
(v) 5a – 3b + 2c from 4a – b – 2c
It can be written as
= (4a – b – 2c) – (5a – 3b + 2c)
By further calculation
= 4a – b – 2c – 5a + 3b – 2c
= -a + 2b – 4c
11. Subtract -5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.
Solution:
We know that
Sum of 4a2 + 3 – 8a and 9a – 7 can be written as
= 4a2 + 3 – 8a + 9a – 7
By further calculation
= 4a2 + a – 4
Here
(4a2 + a – 4) – (-5a2 – 3a + 1) = 4a2 + a – 4 + 5a2 + 3a – 1
By further calculation
= 4a2 + 5a2 + a + 3a – 4 – 1
So we get
= 9a2 + 4a – 5
12. By how much does 8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x – 3?
Solution:
We know that
8x3 – 6x2 + 9x – 10 exceed 4x3 + 2x2 + 7x – 3
It can be written as
= (8x3 – 6x2 + 9x – 10) – (4x3 + 2x2 + 7x – 3)
By further calculation
= 8x3 – 6x2 + 9x – 10 – 4x3 – 2x2 -7x + 3
So we get
= 8x3 – 4x3 – 6x2 – 2x2 + 9x – 7x – 10 + 3
= 4x3 – 8x2 + 2x – 7
13. What must be added to 2a3 + 5a – a2 – 6 to get a2 – a – a3 + 1?
Solution:
The answer can be obtained by subtracting 2a3 + 5a – a2 – 6 from a2 – a – a3 + 1
= (-a3 + a2 – a + 1) – (2a3 + 5a – a2 – 6)
It can be written as
= -a3 + a2 – a + 1 – 2a3 – 5a + a2 + 6
By further calculation
= – a3 – 2a3 + a2 + a2 – a – 5a + 1 + 6
= -3a3 + 2a2 – 6a + 7
14. What must be subtracted from a2 + b2 + 2ab to get – 4ab + 2b2?
Solution:
The answer can be obtained by subtracting – 4ab + 2b2 from a2 + b2 + 2ab
= a2 + b2 + 2ab – (– 4ab + 2b2)
It can be written as
= a2 + b2 + 2ab + 4ab – 2b2
By further calculation
= a2 + b2 – 2b2 + 2ab + 4ab
= a2 – b2 + 6ab
15. Find the excess of 4m2 + 4n2 + 4p2 over m2 + 3n2 – 5p2.
Solution:
The answer can be obtained by subtracting m2 + 3n2 – 5p2 from 4m2 + 4n2 + 4p2
= (4m2 + 4n2 + 4p2) – (m2 + 3n2 – 5p2)
It can be written as
= 4m2 + 4n2 + 4p2 – m2 – 3n2 + 5p2
By further calculation
= 4m2 – m2 + 4n2 – 3n2 + 4p2 + 5p2
= 3m2 + n2 + 9p2
#### Exercise 11C page: 129
1. Multiply:
(i) 3x, 5x2y and 2y
(ii) 5, 3a and 2ab2
(iii) 5x + 2y and 3xy
(iv) 6a – 5b and – 2a
(v) 4a + 5b and 4a – 5b
Solution:
(i) 3x, 5x2y and 2y
Product = 3x × 5x2y × 2y
We can write it as
= 3 × 5 × 2 × x × x2 × y × y
So we get
= 30x3y2
(ii) 5, 3a and 2ab2
Product = 5 × 3a × 2ab2
We can write it as
= 5 × 3 × 2 × a × ab2
So we get
= 30a2b2
(iii) 5x + 2y and 3xy
Product = 3xy (5x + 2y)
We can write it as
= 3xy × 5x + 3xy × 2y
So we get
= 15x2y + 6xy2
(iv) 6a – 5b and – 2a
Product = – 2a (6a – 5b)
We can write it as
= -2a × 6a + 2a × 5b
So we get
= -12a2 + 10ab
(v) 4a + 5b and 4a – 5b
Product = (4a + 5b) (4a – 5b)
So we get
= 16a2 – 25b2
2. Copy and complete the following multiplications:
Solution:
3. Evaluate:
(i) (c + 5) (c – 3)
(ii) (3c – 5d) (4c – 6d)
(iii) (1/2a + 1/2b) (1/2a – 1/2b)
(iv) (a2 + 2ab + b2) (a + b)
(v) (3x – 1) (4x3 – 2x2 + 6x – 3)
Solution:
(i) (c + 5) (c – 3)
It can be written as
= c (c – 3) + 5 (c – 3)
By further calculation
= c2 – 3c + 5c – 15
= c2 + 2c – 15
(ii) (3c – 5d) (4c – 6d)
It can be written as
= 3c (4c – 6d) – 5d (4c – 6d)
By further calculation
= 12c2 – 18cd – 20cd + 30d2
= 12c2 – 38cd + 30d2
(iii) (1/2a + 1/2b) (1/2a – 1/2b)
It can be written as
= 1/2a (1/2a – 1/2b) + 1/2b (1/2a – 1/2b)
By further calculation
= 1/4a2 – 1/4ab + 1/4ab – 1/4b2
= 1/4a2 – 1/4b2
(iv) (a2 + 2ab + b2) (a + b)
It can be written as
= a (a2 + 2ab + b2) + b (a2 + 2ab + b2)
By further calculation
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + b3 + 3a2b + 3ab2
(v) (3x – 1) (4x3 – 2x2 + 6x – 3)
It can be written as
= 3x (4x3 – 2x2 + 6x – 3) – 1 (4x3 – 2x2 + 6x – 3)
By further calculation
= 12x4 – 6x3 + 18x2 – 9x – 4x3 + 2x2 – 6x + 3
= 12x4 – 6x3 – 4x3 + 18x2 + 2x2 – 9x – 6x + 3
So we get
= 12x4 – 10x3 + 20x2 – 15x + 3
4. Evaluate:
(i) (a + b) (a – b).
(ii) (a2 + b2) (a + b) (a – b), using the result of (i).
(iii) (a4 + b4) (a2 + b2) (a + b) (a – b), using the result of (ii).
Solution:
(i) (a + b) (a – b).
It can be written as
= a (a – b) + b (a – b)
By further calculation
= a2 – ab + ab – b2
= a2 – b2
(ii) (a2 + b2) (a + b) (a – b)
Substituting the result of (i)
= (a2 + b2) (a2 – b2)
It can be written as
= a2 (a2 – b2) + b2 (a2 – b2)
So we get
= a4 – a2b2 + a2b2 – b4
= a4 – b4
(iii) (a4 + b4) (a2 + b2) (a + b) (a – b)
Substituting the result of (ii)
= (a4 + b4) (a4 – b4)
It can be written as
= a4 (a4 – b4) + b4 (a4 – b4)
By further calculation
= a8 – a4b4 + a4b4 – b8
= a8 – b8
5. Evaluate:
(i) (3x – 2y) (4x + 3y)
(ii) (3x – 2y) (4x + 3y) (8x – 5y)
(iii) (a + 5) (3a – 2) (5a + 1)
(iv) (a + 1) (a2 – a + 1) and (a – 1) (a2 + a + 1); and then: (a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
(v) (5m – 2n) (5m + 2n) (25m2 + 4n2)
Solution:
(i) (3x – 2y) (4x + 3y)
It can be written as
= 3x (4x + 3y) – 2y (4x + 3y)
By further calculation
= 12x2 + 9xy – 8xy – 6y2
So we get
= 12x2 + xy – 6y2
(ii) (3x – 2y) (4x + 3y) (8x – 5y)
Substituting result of (i)
= (12x2 + xy – 6y2) (8x – 5y)
It can be written as
= 8x (12x2 + xy – 6y2) – 5y (12x2 + xy – 6y2)
By further calculation
= 96x3 + 8x2y – 48xy2 – 60x2y – 5xy2 + 30y3
So we get
= 96x3 + 8x2y – 60x2y – 48xy2 – 5xy2 + 30y3
= 96x3 – 52 x2y – 53xy2 + 30y3
(iii) (a + 5) (3a – 2) (5a + 1)
It can be written as
= a (3a – 2) + 5 (3a – 2) (5a + 1)
By further calculation
= (3a2 – 2a + 15a – 10) (5a + 1)
So we get
= (3a2 + 13a – 10) (5a + 1)
We can write it as
= 5a (3a2 + 13a – 10) + 1 (3a2 + 13a – 10)
By further calculation
= 15a3 + 65a2 – 50a + 3a2 + 13a – 10
= 15a3 + 68a2 – 37a – 10
(iv) (a + 1) (a2 – a + 1) and (a – 1) (a2 + a + 1); and then: (a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
Consider
(a + 1) (a2 – a + 1)
It can be written as
= a (a2 – a + 1) + 1 (a2 – a + 1)
By further calculation
= a3 – a2 + a + a2 – a + 1
So we get
= a3 + 1
(a – 1) (a2 + a + 1)
It can be written as
= a (a2 + a + 1) – 1 (a2 + a + 1)
By further calculation
= a3 + a2 + a – a2 – a – 1
So we get
= a3 – 1
Here
(a + 1) (a2 – a + 1) + (a – 1) (a2 + a + 1)
= a3 + 1 + a3 – 1
= 2a3
(v) (5m – 2n) (5m + 2n) (25m2 + 4n2)
It can be written as
= [5m (5m + 2n) – 2n (5m + 2n)] (25m2 + 4n2)
By further calculation
= (25m2 + 10mn – 10mn – 4n2) (25m2 + 4n2)
So we get
= (25m2 – 4n2) (25m2 + 4n2)
We can write it as
= 25m2 (25m2 + 4n2) – 4n2 (25m2 + 4n2)
By multiplying the terms
= 625m4 + 100m2n2 – 100m2n2 – 16n4
= 625m4 – 16n4
6. Multiply:
(i) mn4, m3n and 5m2n3
(ii) 2mnpq, 4mnpq and 5mnpq
(iii) pq – pm and p2m
(iv) x3 – 3y3 and 4x2y2
(v) a3 – 4ab and 2a2b
Solution:
(i) mn4, m3n and 5m2n3
It can be written as
= 5m2n3 × mn4 × m3n
By further calculation
= 5m(2 + 1 + 3) n(3 + 4 + 1)
= 5m6n8
(ii) 2mnpq, 4mnpq and 5mnpq
It can be written as
= 5mnpq × 2mnpq × 4mnpq
By further calculation
= 5 × 2 × 4 m(1 + 1 + 1) n(1 + 1 + 1) p(1 + 1 + 1) q(1 + 1 + 1)
= 40m3n3p3q3
(iii) pq – pm and p2m
It can be written as
= p2m × (pq – pm)
So we get
= p3qm – p3m2
(iv) x3 – 3y3 and 4x2y2
It can be written as
= 4x2y2 × (x3 – 3y3)
By further calculation
= 4x5y2 – 12x2y5
(v) a3 – 4ab and 2a2b
It can be written as
= 2a2b × (a3 – 4ab)
By further calculation
= 2a5b – 8a3b2
7. Multiply:
(i) (2x + 3y) (2x + 3y)
(ii) (2x – 3y) (2x + 3y)
(iii) (2x + 3y) (2x – 3y)
(iv) (2x – 3y) (2x – 3y)
(v) (-2x + 3y) (2x – 3y)
Solution:
(i) (2x + 3y) (2x + 3y)
It can be written as
= 2x (2x + 3y) + 3y (2x + 3y)
By further calculation
= 4x2 + 6xy + 6xy + 9y2
= 4x2 + 12xy + 9y2
(ii) (2x – 3y) (2x + 3y)
It can be written as
= 2x (2x + 3y) – 3y (2x + 3y)
By further calculation
= 4x2 + 6xy – 6xy – 9y2
= 4x2 – 9y2
(iii) (2x + 3y) (2x – 3y)
It can be written as
= 2x (2x – 3y) + 3y (2x – 3y)
By further calculation
= 4x2 – 6xy + 6xy – 9y2
= 4x2 – 9y2
(iv) (2x – 3y) (2x – 3y)
It can be written as
= 2x (2x – 3y) – 3y (2x – 3y)
By further calculation
= 4x2 – 6xy – 6xy + 9y2
= 4x2 – 12xy + 9y2
(v) (-2x + 3y) (2x – 3y)
It can be written as
= -2x (2x – 3y) + 3y (2x – 3y)
By further calculation
= – 4x2 + 6xy + 6xy – 9y2
= – 4x2 + 12xy – 9y2
#### Exercise 11D page: 132
1. Divide:
(i) – 16ab2c by 6abc
(ii) 25x2y by – 5y2
(iii) 8x + 24 by 4
(iv) 4a2 – a by – a
(v) 8m – 16 by – 8
Solution:
(i) – 16ab2c by 6abc
We can write it as
= – 16ab2c/ 6abc
= -8/3 b
(ii) 25x2y by – 5y2
We can write it as
= 25x2y/ -5y2
= -5 x2/y
(iii) 8x + 24 by 4
We can write it as
= (8x + 24)/4
Separating the terms
= 8x/4 + 24/4
= 2x + 6
(iv) 4a2 – a by – a
We can write it as
= (4a2 – a)/ -a
Separating the terms
= 4a2/-a – a/-a
= – 4a + 1
(v) 8m – 16 by – 8
We can write it as
= (8m – 16)/ -8
Separating the terms
= 8m/-8 – 16/-8
= – m + 2
2. Divide:
(i) n2 – 2n + 1 by n – 1
(ii) m2 – 2mn + n2 by m – n
(iii) 4a2 + 4a + 1 by 2a + 1
(iv) p2 + 4p + 4 by p + 2
(v) x2 + 4xy + 4y2 by x + 2y
Solution:
(i) n2 – 2n + 1 by n – 1
n2 – 2n + 1 by n – 1 = n – 1
(ii) m2 – 2mn + n2 by m – n
m2 – 2mn + n2 by m – n = m – n
(iii) 4a2 + 4a + 1 by 2a + 1
4a2 + 4a + 1 by 2a + 1 = 2a + 1
(iv) p2 + 4p + 4 by p + 2
p2 + 4p + 4 by p + 2 = p + 2
(v) x2 + 4xy + 4y2 by x + 2y
x2 + 4xy + 4y2 by x + 2y = x + 2y
3. The area of a rectangle is 6x2 – 4xy – 10y2 square unit and its length is 2x + 2y unit. Find its breadth.
Solution:
It is given that
Area of a rectangle = 6x2 – 4xy – 10y2 square unit
Length = 2x + 2y unit
We know that
So we get
= (6x2 – 4xy – 10y2)/ (2x + 2y)
= 3x – 5y units
4. The area of a rectangular field is 25x2 + 20xy + 3y2 square unit. If its length is 5x + 3y unit, find its breadth. Hence, find its perimeter.
Solution:
It is given that
Area of a rectangular field = 25x2 + 20xy + 3y2 square unit
Length = 5x +3y unit
We know that
So we get
= (25x2 + 20xy + 3y2)/ (5x + 3y)
= 5x + y units
Now the perimeter of the rectangular field = 2 (length + breadth)
Substituting the values
= 2 (5x + 3y + 5x + y)
So we get
= 2 (10x + 4y)
= 20x + 8y
5. Divide:
(i) 2m3n5 by – mn
(ii) 5x2 – 3x by x
(iii) 10x3y – 9xy2 – 4x2y2 by xy
(iv) 3y3 – 9ay2 – 6ab2y by – 3y
(v) x5 – 15x4 – 10x2 by – 5x2
Solution:
(i) 2m3n5 by – mn
It can be written as
= 2m3n5/ -mn
= -2m2n4
(ii) 5x2 – 3x by x
It can be written as
= (5x2 – 3x)/ x
Separating the terms
= 5x2/x – 3x/x
= 5x – 3
(iii) 10x3y – 9xy2 – 4x2y2 by xy
It can be written as
= (10x3y – 9xy2 – 4x2y2)/ xy
Separating the terms
= 10x3y/xy – 9xy2/xy – 4x2y2/xy
= 10x2 – 9y – 4xy
(iv) 3y3 – 9ay2 – 6ab2y by – 3y
It can be written as
= (3y3 – 9ay2 – 6ab2y)/ -3y
Separating the terms
= 3y3/-3y – 9ay2/-3y – 6ab2y/ -3y
= -y2 + 3ay + 2ab2
(v) x5 – 15x4 – 10x2 by – 5x2
It can be written as
= (x5 – 15x4 – 10x2)/ -5x2
Separating the terms
= x5/-5x2 – 15x4/-5x2 – 10x2/-5x2
= -1/5x3 + 3x2 + 2
#### Exercise 11E page: 133
Simplify:
1. x/2 + x/4
Solution:
x/2 + x/4
Taking LCM
= (2x + x)/4
= 3x/4
2. a/10 + 2a/5
Solution:
a/10 + 2a/5
Taking LCM
= (a + 4a)/ 10
= 5a/10
= a/2
3. y/4 + 3y/5
Solution:
y/4 + 3y/5
Taking LCM
= (5y + 12y)/ 20
= 17y/20
4. x/2 – x/8
Solution:
x/2 – x/8
Taking LCM
= (4x – x)/ 8
= 3x/8
5. 3y/4 – y/5
Solution:
3y/4 – y/5
Taking LCM
= (15y – 4y)/ 20
= 11y/20
6. 2p/3 – 3p/5
Solution:
2p/3 – 3p/5
Taking LCM
= (10p – 9p)/ 15
= p/15
7. k/2 + k/3 + 2k/5
Solution:
k/2 + k/3 + 2k/5
Here the LCM of 2, 3 and 5 is 30
= (15k + 10k + 12k)/ 30
= 37k/30
8. 2x/5 + 3x/4 – 3x/5
Solution:
2x/5 + 3x/4 – 3x/5
Here the LCM of 5 and 4 is 20
= (8x + 15x – 12x)/ 20
= 11x/20
9. 4a/7 – 2a/3 + a/7
Solution:
4a/7 – 2a/3 + a/7
Here the LCM of 3 and 7 is 21
= (12a – 14a + 3a)/ 21
= a/21
10. 2b/5 – 7b/15 + 13b/3
Solution:
2b/5 – 7b/15 + 13b/3
Here the LCM of 3, 5 and 15 is 15
= (6b – 7b + 65b)/ 15
= 64b/15
11. 6k/7 – (8k/9 – k/3)
Solution:
6k/7 – (8k/9 – k/3)
Here the LCM of 7, 9 and 3 is 63
= [54k – (56k – 21k)]/ 63
By further calculation
= (54k – 35k)/ 63
= 19k/63
12. 3a/8 + 4a/5 – (a/2 + 2a/5)
Solution:
3a/8 + 4a/5 – (a/2 + 2a/5)
Here the LCM of 8, 5 and 2 is 40
= [15a + 32a – (20a + 16a)]/ 40
By further calculation
= (47a – 36a)/ 40
= 11a/40
13. x + x/2 + x/3
Solution:
x + x/2 + x/3
Taking LCM
= (6x + 3x + 2x)/ 6
= 11x/6
14. y/5 + y – 19y/15
Solution:
y/5 + y – 19y/15
Here the LCM of 5 and 15 is 15
= (3y + 15y – 19y)/ 15
So we get
= – y/15
15. x/5 + (x + 1)/2
Solution:
x/5 + (x + 1)/2
Here the LCM of 5 and 2 is 10
= (2x + 5x + 5)/ 10
= (7x + 5)/ 10
#### Exercise 11F page: 136
Enclose the given terms in brackets as required:
1. x – y – z = x – (…………)
2. x2 – xy2 – 2xy – y2 = x2 – (………….)
3. 4a – 9 + 2b – 6 = 4a – (………..)
4. x2 – y2 + z2 + 3x – 2y = x2 – (…………)
5. -2a2 + 4ab – 6a2b2 + 8ab2 = -2a (…………….)
Solution:
1. x – y – z = x – (y + z)
2. x2 – xy2 – 2xy – y2 = x2 – (xy2 + 2xy + y2)
3. 4a – 9 + 2b – 6 = 4a – (9 – 2b + 6)
4. x2 – y2 + z2 + 3x – 2y = x2 – (y2 – z2 – 3x + 2y)
5. -2a2 + 4ab – 6a2b2 + 8ab2 = -2a (a – 2b + 3ab2 – 4b2)
Simplify:
6. 2x – (x + 2y – z)
Solution:
2x – (x + 2y – z)
We can write it as
= 2x – x – 2y + z
So we get
= x – 2y + z
7. p + q – (p – q) + (2p – 3q)
Solution:
p + q – (p – q) + (2p – 3q)
We can write it as
= p + q – p + q + 2p – 3q
So we get
= 2p – q
8. 9x – (-4x + 5)
Solution:
9x – (- 4x + 5)
We can write it as
= 9x + 4x – 5
So we get
= 13x – 5
9. 6a – (- 5a – 8b) + (3a + b)
Solution:
6a – (- 5a – 8b) + (3a + b)
We can write it as
= 6a + 5a + 8b + 3a + b
So we get
= 6a + 5a + 3a + 8b + b
= 14a + 9b
10. (p – 2q) – (3q – r)
Solution:
(p – 2q) – (3q – r)
We can write it as
= p – 2q – 3q + r
So we get
= p – 5q + r
11. 9a (2b – 3a + 7c)
Solution:
9a (2b – 3a + 7c)
We can write it as
= 18ab – 27a2 + 63ca
12. – 5m (- 2m + 3n – 7p)
Solution:
– 5m (- 2m + 3n – 7p)
We can write it as
= 10m2 – 15mn + 35mp
13. – 2x (x + y) + x2
Solution:
– 2x (x + y) + x2
We can write it as
= -2x2 – 2xy + x2
So we get
= – x2 – 2xy
14. b (2b – 1/b) – 2b (b – 1/b)
Solution:
b (2b – 1/b) – 2b (b – 1/b)
We can write it as
= 2b2 – 1 – 2b2 + 2
So we get
= 1
15. 8 (2a + 3b – c) – 10 (a + 2b + 3c)
Solution:
8 (2a + 3b – c) – 10 (a + 2b + 3c)
We can write it as
= 16a + 24b – 8c – 10a – 20b – 30c
So we get
= 6a + 4b – 38c
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# Coin Toss Problem | AMC 10A, 2017| Problem No 18
Try this beautiful Problem on Probability based on Coin toss from AMC 10 A, 2017. You may use sequential hints to solve the problem.
## Coin Toss - AMC-10A, 2017- Problem 18
Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $q-p ?$
,
• $1$
• $2$
• $3$
• $4$
• $5$
combinatorics
Coin toss
Probability
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
AMC-10A, 2017 Problem-18
#### Check the answer here, but try the problem first
$4$
## Try with Hints
#### First Hint
Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins.
Now can you finish the problem?
#### Second Hint
Let $P$ be the probability Amelia wins. Note that $P=$ chance she wins on her first turn $+$ chance she gets to her second turn $\cdot \frac{1}{3}+$ chance she gets to her third turn $\cdot \frac{1}{3} \ldots$ This can be represented by an infinite geometric series,
Therefore the value of $P$ will be $P=\frac{\frac{1}{3}}{1-\frac{2}{3} \cdot \frac{3}{5}}=\frac{\frac{1}{3}}{1-\frac{2}{5}}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3} \cdot \frac{5}{3}=\frac{5}{9}$ which is of the form $\frac{p}{q}$
Now Can you finish the Problem?
#### Third Hint
Therefore $q-p=9-5=4$
## Subscribe to Cheenta at Youtube
Try this beautiful Problem on Probability based on Coin toss from AMC 10 A, 2017. You may use sequential hints to solve the problem.
## Coin Toss - AMC-10A, 2017- Problem 18
Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $q-p ?$
,
• $1$
• $2$
• $3$
• $4$
• $5$
combinatorics
Coin toss
Probability
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
AMC-10A, 2017 Problem-18
#### Check the answer here, but try the problem first
$4$
## Try with Hints
#### First Hint
Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins.
Now can you finish the problem?
#### Second Hint
Let $P$ be the probability Amelia wins. Note that $P=$ chance she wins on her first turn $+$ chance she gets to her second turn $\cdot \frac{1}{3}+$ chance she gets to her third turn $\cdot \frac{1}{3} \ldots$ This can be represented by an infinite geometric series,
Therefore the value of $P$ will be $P=\frac{\frac{1}{3}}{1-\frac{2}{3} \cdot \frac{3}{5}}=\frac{\frac{1}{3}}{1-\frac{2}{5}}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3} \cdot \frac{5}{3}=\frac{5}{9}$ which is of the form $\frac{p}{q}$
Now Can you finish the Problem?
#### Third Hint
Therefore $q-p=9-5=4$
## Subscribe to Cheenta at Youtube
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# Parametric Form of Hyperbola
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Last updated date: 23rd Jul 2024
Total views: 183.9k
Views today: 3.83k
## Parametric Equation of Hyperbola: Preface
Hyperbola is a subdivision of conic sections in the field of Mathematics. When the surface of a cone intersects a plane, curves are formed, and these curves are known as conic sections. There are three categories of conic sections: the eclipse, the hyperbola, and the parabola.
We use conic sections to study 3D geometry which has a vast number of applications in various fields of engineering. Also, when a spacecraft uses the gravitational slingshot technique, the path followed by the craft is a hyperbola. In this article, we will get to know about the parametric form of the hyperbola. We will also see some interesting facts about the hyperbola and also answer some of the questions.
## Parametric Coordinates of Hyperbola
In this topic, we will find how to calculate the parametric equation of the hyperbola.
To understand this in a simple way, let’s take the help of the following diagram.
Parametric Coordinates of Hyperbola
The circle inscribed in between the hyperbola and on the transverse axis is called an auxiliary circle. If the equation of the hyperbola shown in the figure is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ then the equation of the auxiliary circle inscribed between the auxiliary circle will be $x^{2}+y^{2}=a^{2}$.
Let M$(x,y)$ be any point on the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$.
Now, from M make a line MQ to the transverse axis such that it is perpendicular to it and then select a point on the circle $x^{2}+y^{2}=a^{2}$ such that the angle OCQ is 90 degrees.
Now, join (C and Q) and (O and C), The length of OC is a. Now let the length ∠SOC,
Here the ∠SOC is called the eccentric angle of the point M on the hyperbola.
Now from the right triangle OCQ, we get
$\dfrac{OC}{OQ}=cos\Theta$
$\dfrac{a}{OQ}=\dfrac{1}{sec\Theta }$ , (The radius of the circle is a=OC)
$OQ=asec\Theta$
Therefore, the abscissa of the M(x,y) on the hyperbola is asecθ.
As the point M lies on the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, so it will satisfy the equation .
So, $\dfrac{(asec\Theta )^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ (as x = sec)
$\Rightarrow \dfrac{y^{2}}{b^{2}}=sec^{2}\Theta -1$
$\Rightarrow \dfrac{y^{2}}{b^{2}}=tan^{2}\Theta$
$\Rightarrow y^{2}=b^{2}tan^{2}\Theta$
$\Rightarrow y=btan\theta$
Hence, the coordinate of the point M is (asecθ,btanθ) and for all the values of θ This point lies on the hyperbola and hence the polar coordinates of a hyperbola is represented by $(asec\Theta ,btan\Theta )$.
## Parametric Form of Hyperbola
If we want to write a parametric form of the hyperbola, we can write it as
Case 1: if the hyperbola is horizontal then:
$f(t)=(x(t),y(t))$
$x(t)=asec\Theta$
$y(t)=btan\Theta$
Case 2: if the hyperbola is vertical then:
$f(t)=(x(t),y(t))$
$x(t)=atan\Theta$
$y(t)=asec\Theta$
## Interesting Facts
• When an object, let's say a jet, moves faster than the speed of sound it creates a conical form of a wave in space. When that wave intersects the ground, the curve we get from that intersection is a hyperbola.
• The cooling towers are generally made of hyperbolic shape to achieve 2 things; first, the least amount of material used to make it and second, the structure should be strong enough to withstand strong winds.
## Solved Examples
Q1. Find the parametric coordinates of the point $(3\sqrt{2},2)$ on the hyperbola $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$.
Ans The equation of the hyperbola is $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$.
If we compare it with general equation of hyperbola
$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ we get
$a^{2}=9\Rightarrow a=3$
$b^{2}=4\Rightarrow b=2$
So the parametric coordinates of the hyperbola will be $(3sec\Theta ,2tan\Theta )$.
Now, to calculate the parametric coordinates,
$3sec\Theta =3\sqrt{2}\Rightarrow sec\Theta =\sqrt{2}\Rightarrow \Theta =45^{\circ}$
$2tan\Theta =2\Rightarrow tan\Theta =1\Rightarrow \Theta =45^{\circ}$
So the parametric coordinates will be $(3sec45^{\circ},2tan45^{\circ})$.
## Practice Questions
Question Write parametric for each of the hyperbolas below:
1. $\dfrac{(x-3)^{2}}{256}-\dfrac{(y+3)^{2}}{625}=1$
Ans:
$x=3+16sect$
$y=-5+25tant$
2. $\dfrac{(y+7)^{2}}{324}-\dfrac{(x-8)^{2}}{169}=1$
Ans:
$x=8+13sect$
$y=-7+18tant$
## Summary
The article summarises the concept of parametric coordinates of hyperbola and its form.
Now we know that if the equation a hyperbola is
$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ then the parametric coordinates will be
$(x(t)=asec\Theta,y(t)=btan\Theta)$ where $\theta$ is the parameter.
Competitive Exams after 12th Science
## FAQs on Parametric Form of Hyperbola
1. What is the formula of the equation of parametric coordinates of a rectangular hyperbola?
If the equation of the hyperbola is $xy=c^2$then the parametric coordinates of the hyperbola will be $x(t)=c(t)$, $y(t)=\dfrac{c}{t}$.
2. What is the second parameter other than (asec,btan) that can be used in hyperbola?
The second parameter that can be used against $(x(t)=asec\Theta,y(t)=btan\Theta)$ is $(x(t)=acosec\Theta,y(t)=bcot\Theta)$ , they can be used if the equation of hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ .
3. What is a rectangular hyperbola?
When the length of the major axis and minor axis of a hyperbola are equal, the hyperbola is known as a rectangular hyperbola.
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# How do you evaluate 710,000\times \frac { 15} { 16} ?
Dec 17, 2016
$= 665 , 625$
#### Explanation:
There are some facts in maths are useful to know.
When working with fractions and decimals, powers of 2, 5 and 10 are easy to relate to each other.
Note the following:
$2 \times 5 = 10$
$4 \times 25 = 100 \text{ "rarr" } {2}^{2} \times {5}^{2} = {10}^{2}$
$8 \times 125 = 1000 \text{ "rarr " } {2}^{3} \times {5}^{3} = {10}^{3}$
$16 \times 625 = 10 , 000 \text{ "rarr" } {2}^{4} \times {5}^{4} = {10}^{4}$
$710 , 000 \times \frac{15}{16} = \frac{71 \times 10 , 000}{1} \times \frac{15}{16}$
$= \frac{71 \times {\cancel{10 , 000}}^{625}}{1} \times \frac{15}{16}$
$= \textcolor{red}{71} \times 625 \times \textcolor{red}{3} \times 5$
$= \textcolor{red}{213} \times 3125$
$= \left(\textcolor{red}{200 + 10 + 3}\right) \times 3125 \text{ } \leftarrow$ distributive law
$= 625 , 000 + 31 , 250 + 9 , 375$
$= 665 , 625$
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### Home > PC > Chapter 5 > Lesson 5.2.3 > Problem5-74
5-74.
Use the graph of $y = x^{2} − 4$ to sketch the graph of $f ( x ) = \frac { 1 } { x ^ { 2 } - 4 }$ without using a graphing calculator. State the equations of the asymptotes in the graph of $f\left(x\right)$
Graph the $y =$ function and identify the zeros ($x$-intercepts).
For the reciprocal function, $\frac{1}{0}$ is undefined, so these are the locations of the vertical asymptotes.
Sketch the vertical asymptotes.
Upward parabola, vertex at (0, comma negative 4), x intercepts at negative 2 & 2, dashed vertical segments at x = negative 2 and x = 2.
Mark the places where $y = 1$ or $−1$. These points do not change because $\frac{1}{1} = 1$ and $\frac{1}{−1} = −1$.
Added approximate highlighted points on parabola, (negative 2.2, comma 1), (2.2, comma 1), (negative 1.7, comma negative 1), & (1.7, comma negative 1).
The reciprocal at the point $\left(0, −4\right)$ is $\left(0, −\frac{1}{4}\right)$. Mark it.
Draw a smooth curve from it to where $y = −1$ is marked and to where $y = 1$ is marked.
Added, downward curve between highlighted points below x axis, going through the point (0, comma 1 fourth).
When the $y$-values of the original function are small, $\frac{1}{\left(± \text{ small number}\right)}$ is a $±$ large number. Show this on your graph.
Added, 4 almost vertical curves, from each of the highlighted points above x axis, to infinity, and from each of the highlighted points below the x axis, to negative infinity.
When the $y$-values of the original function are large, $\frac{1}{\text{large number}}$ is a small number. Show this on your graph.
Added, upward curves from approximate point, (negative 2.2, comma 1), going left above x axis, from approximate point (2.2, comma 1), going right above x axis.
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Volume with Unit Cubes MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
Volume is the amount of three-dimensional space an object occupies, in cubic units, within a container.
In this section, we are going to be investigating how many small cubes (called unit cubes) can be placed inside of a right rectangular prism. A connection will be made to the volume of the right rectangular prism.
Volume and Unit Cubes: (in Right Rectangular Prisms)
Volume is measured in cubic units. A "unit cube" with side lengths of one unit each, will have a volume of 1 cubic unit. If we can determine how many of these cubes will perfectly "fit" inside of a container, we will know the volume of the container, in cubic units. So, the volume of an object can be represented by the number of unit cubes that can be placed within the object. Volume = 1 cubic unit
Unit cubes are used when investigating the volume of right rectangular prisms (boxes).
To determine the volume, we can ask, "How many unit cubes can fit inside this box?"
The dimensions of this box are 2 units by 4 units by 2 units. The volume of this right rectangular prism ("box") is the total of the number of unit cubes it holds. By counting, we can see that there are are a total of 16 unit cubes within this solid. The volume = 16 cubic units Check: V = l x w x h = 2 x 4 x 2 = 16
For some figures, the unit cubes, with a volume of 1, fit "nicely" into the object,
while other objects may require fractional unit cubes.
The dimensions of this box are 3½ units by 4 units by 2 units. Because of the fractional side, 3½, the large box cannot be filled "exactly" with the unit cubes that have a volume of one. You cannot use half of a unit cube to finish filling the large box. To deal with this problem, unit cubes with a with a side of ½ unit and a volume of cubic units are needed. The number of fractional cubes needed to fill the box is: 7 cubes x 8 cubes x 4 cubes = 224 cubes BUT ... this is NOT the volume of the box .With this "new" fractional unit cube, the number of cubes used to fill the box, is NOT the volume of the box, as it was in the previous example. Each cube does not represent 1 cubic unit. Each cube now only represents cubic units. We will need to multiply the number of cubes used by to get the actual volume of the big box ( 224 x = 28 ). There are 28 cubic units in the big box. Check: V = l x w x h = 3½ x 4 x 2 = 28
Volume as Area times Depth:
While counting the number of unit cubes within the rectangular prism to determine volume,
a pattern can be seen that will help to determine the count of the cubes quickly.
Pattern: Find the number of cubes seen in one face and multiply times the number of rows of that face, going to the back of the figure. Volume = (cubes in face) • (rows going back) = 8 • 2 = 16 square units
Pattern: A pattern similar to that shown above can be expressed using the term "area". The number of cube faces seen in the face of the figure, comprise the area of the face of the figure. The new pattern is expressed as:
Volume = (area of face) • (depth of face)
= (4 • 2) • 2 = 16 square units
This pattern will be a popular strategy to determine the volume of many solids.
• Flip the "box" over so it sits on its base. • The area of the base is its length (l) times its width (w). • Consider stacking copies of the area of the base (cross sections) on top of one another until they fill the solid. • This pattern utilizes what is called a "limiting argument" with the area being a 2-dimensional cross section with no thickness. • The thickness, however, can be theoretically considered very, very, very small so as not to affect calculations. • These cross sections are stacked to the height (h) of the figure, creating the formula V = B • h, with B the area of the base and h = height. The area of the Base is l • w, call it B. The volume of a "box" is V = l • w • h Substitute B: V = B • h:
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
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## Describing Triangles Using Sides
### Equilateral triangles
These are the most ‘perfect’ looking triangles – an equilateral triangle has three sides which are all the same length. This means that the three angles inside the triangle are the same. Equilateral triangles can also be called equiangular triangles. ‘Equi’ sort of means ‘equal’, and ‘angular’ obviously refers to angles, so the word means that all the angles are equal.
In an equilateral triangle, we know that all the interior angles are the same. This means that we can work out what each angle is by:
If you look at the diagram you should see 3 small straight lines drawn through the middles of each side. These short straight lines indicate that the side they’re drawn through is the same length as any other sides with a short line drawn through it.
### Isosceles triangles
OK, first of all, the most important thing to do with isosceles triangles is to learn how to spell that word – isosceles. I find it easiest to remember as:
isos + celes
Now, once you’ve learnt how to spell this word, you need to know what an isosceles triangle is. Well, it’s quite simple. Isosceles triangles have exactly two sides the same length. So an equilateral triangle is not an isosceles triangle for instance, because it has three equal sides, not two.
Because two of the sides are equal in length, this means that all isosceles triangles will also have two equal angles. Looking at the diagram, you should see two angles marked out, with short straight lines drawn through them. Like for sides, these short straight lines are used to show that angles are the same size. The two equal angles in an isosceles triangle are always in the corners of the triangle where the two equal sides do not meet.
### Scalene triangles
Scalene triangles are ones where none of the sides are the same length. This also means that none of the angles are the same. I like to think of them as ‘messy’ triangles because nothing matches. Here’s one example of a scalene triangle:
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 7.5: Solving Two-Step Equations
Difficulty Level: At Grade Created by: CK-12
## Introduction
A Trip to Fenway Park
Marc is going to get his wish. After much ado, his Grandpa managed to pull a few strings and get tickets for the Red Sox game. Marc can hardly wait for Saturday’s game.
“I hope that it is as good as I think it will be,” Marc said.
“Well for $535.00 I hope we don’t get rained out.” “Five hundred thirty-five dollars!! Wow, that is a lot of money!” Marc exclaimed. “Yes, but that does include a$15.00 service fee. I was able to secure four tickets and that is all that matters. Don’t worry Marc; having your dream come true is worth the money.”
“Thanks Grandpa,” Marc said, smiling.
Afterwards, Marc began to wonder what the price of each ticket really was. He knew that they got field box seats and that they had to be pretty expensive given the final bill. How could he figure out the ticket price?
This is where a two step equation will help Marc. This lesson will teach you all about writing and solving two-step equations. When finished, you will know how to figure out the ticket price for Marc.
What You Will Learn
In this lesson you will learn how to complete the following:
• Write two-step single variable equations from verbal models.
• Solve two-step single variable equations.
• Identify properties used to solve multi-step equations.
• Model and solve real-world problems using two-step equations.
Teaching Time
I. Write Two-Step Single Variable Equations from Verbal Models
In the last two lessons you worked on solving one-step single variable equations. Here is an example:
Example
3x=15\begin{align*}3x=15\end{align*}
Now if you think about this equation, you only had to perform one operation to solve it. We used an inverse or opposite operation to solve it. Because this is a multiplication equation, we used division to solve for the unknown variable. We could say that we performed one step to solve this equation. The one step was to divide. We call an equation where one operation is needed to solve for the unknown variable a one-step equation.
What happens when there is more than one operation needed to solve an equation?
When this happens, we have two-step equation. A two-step equation has two operations in it. Let’s look at a few examples of two-step equations.
3x+5x32=20=5\begin{align*}3x+5 &= 20\\ \frac{x}{3}-2 &= 5\end{align*}
If you look at both of these equations, you will see that there are two operations present in each. The first equation has multiplication and addition. The second equation has division and subtraction. You will see other variations of two-step equations too, but this gives you an idea.
Before we begin solving equations, let’s look at how we can write a two-step equation from a verbal model.
First, let’s think about some of the words that mean addition, subtraction, multiplication and division. Identifying these key words is going to assist us when writing equations from verbal models. Remember that a verbal model uses words.
Sum
Altogether
In all
Plus
and
Difference
Less than
More than
Take away
subtract
Product
Times
Groups of
Quotient
Split Up
Divided
Take a few minutes to write down these key words in your notebook.
Now let’s look at an example.
Example
Six times a number, plus five, is forty-one.
First, identify any key words that identify operations.
Six times a number, plus five, is forty-one.
Next, begin to translate the words into an equation.
Six = 6
Times ="."\begin{align*}= "."\end{align*}
A number = x
Plus = +
Five = 5
Is means =
Forty-one is 41
Next, we put it altogether.
6.x+5=41\begin{align*}6.x+5=41\end{align*}
Example
Four less than two times a number is equal to eight.
First, identify any key words that identify operations.
Four less than two times a number is eight.
Now we can translate each part.
Four becomes 4
Less than means subtraction
Two becomes 2
Times ="."\begin{align*}= "."\end{align*}
A number = x
Is equal to means =
Eight = 8
Notice that the one tricky part is in the words “less than” because it is less than two times a number, the two times a number needs to come first in the equation. Then we can subtract.
Put it altogether.
2.x4=8\begin{align*}2.x-4=8\end{align*}
Example
A number divided by two, plus six, is equal to fourteen.
First, look for key words that identify operations.
A number divided by two, plus six, is equal to fourteen.
Next, translate each word into an equation.
A number means variable (x)
Divided means ÷\begin{align*}\div\end{align*}
By two means 2 is our divisor
Six means 6
Is means =
Fourteen means 14
Put it all together.
x2+6=14\begin{align*}\frac{x}{2}+6=14\end{align*}
7K. Lesson Exercises
Write an equation for each.
1. The product of five and a number, plus three, is twenty-three.
2. Six times a number, minus four, is thirty-two.
3. A number, y\begin{align*}y\end{align*}, divided by 3, plus seven, is ten.
Take a few minutes to check your work with a partner. Do your equations represent the verbal model?
II. Solve Two-Step Single Variable Equations
Now that you know how to write a two-step single variable equation it is time to learn how to solve them. When we solve the equation, we will be looking for the value of the variable in the equation.
Sometimes, you will be able to solve a two-step equation using mental math. Particularly when the numbers are small and are all positive integers, you might be able to solve them in your head. This is the first method that we are going to look at.
Example
3x+3=9\begin{align*}3x+3=9\end{align*}
Here the numbers are small. You can probably look at this one and say to yourself, “What number times three plus three is nine?” The logical answer is 2.
You can check your work by substituting 2 in for x\begin{align*}x\end{align*} to see if both sides of the equation are the same. If they are, then we can say that the equation balances and your work is accurate.
3(2)+36+39=9=9=9\begin{align*}3(2)+3 &= 9\\ 6+3 &= 9\\ 9 &= 9\end{align*}
The answer checks out. The value of x\begin{align*}x\end{align*} is 2.
That is a great question! This is where our second strategy comes in.
When you have a two-step equation that you can’t figure out in your head, you can use the strategy isolate the variable.
Example
2+3n=11\begin{align*}2+3n=11\end{align*}.
Notice that there are two terms on the left side of the equation, 2 and 3n\begin{align*}3n\end{align*}. Our first step should be to use inverse operations to get the term that includes a variable, 3n\begin{align*}3n\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 2 is added to 3n\begin{align*}3n\end{align*}. So we can use the inverse of addition, which is subtraction, and subtract 2 from both sides of the equation. We need to subtract 2 from both sides of the equation because of the Subtraction Property of Equality. You probably remember that property states that in order to keep the values on both sides of the equation equal, whatever we subtract from one side of the equation must also be subtracted from the other side.
Let's see what happens when we subtract 2 from both sides of the equation.
2+3n22+3n0+3n3n=11=112=9=9\begin{align*}2 + 3n &= 11\\ 2 - 2+3n &= 11-2\\ 0+3n &= 9\\ 3n &= 9\end{align*}
Now, the term that includes a variable, 3n\begin{align*}3n\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the n\begin{align*}n\end{align*} by itself. Since 3n\begin{align*}3n\end{align*} means 3×n\begin{align*}3 \times n\end{align*}, we can use the inverse of multiplication––division. We can divide both sides of the equation by 3. We need to divide both sides of the equation by 3 because of the Division Property of Equality. That property states that if we divide one side of the equation by a number, we must divide the other side of the equation by the same number in order to keep the values on both sides of the equation equal.
Let's see what happens when we divide both sides of the equation by 3.
3n3n31nn=9=93=3=3\begin{align*}3n &= 9\\ \frac{3n}{3} &= \frac{9}{3}\\ 1n &= 3\\ n &= 3\end{align*}
The value of n\begin{align*}n\end{align*} is 3.
Example
2x5=11\begin{align*}2x-5=11\end{align*}
First, notice that there are two terms on the left side of the equal sign. To solve this equation, we want to isolate the variable and get x\begin{align*}x\end{align*} alone. Let’s start with the term that is not connected to the x\begin{align*}x\end{align*}, "-5".
We use the Addition Property of Equality to cancel out the -5. We do this by using the inverse operation and adding 5 to both sides of the equation. Remember because of the properties of equality what we do to one side of the equals must match on the other side.
2x5+52x+02x=11+5=16=16\begin{align*}2x-5+5 &= 11+5\\ 2x+0 &= 16\\ 2x &= 16\end{align*}
Now we have a problem that is a multiplication problem. We can use the Division Property of Equality to solve for x\begin{align*}x\end{align*} by dividing both sides of the equation by 2. Think about this and it makes perfect sense. Since we want to figure out what number times two is sixteen we can divide sixteen by two to figure it out.
2x2x=162=8\begin{align*}\frac{2x}{2} &= \frac{16}{2}\\ x &= 8\end{align*}
The value of x\begin{align*}x\end{align*} is 8.
Example
x58=17\begin{align*}\frac{x}{5}-8=17\end{align*}.
Notice that there are two terms on the left side of the equation, \begin{align*}\frac{x}{5}\end{align*} and 8. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{x}{5}\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 8 is subtracted from \begin{align*}\frac{x}{5}\end{align*}. So, we can use the inverse of subtraction, which is addition, and add 8 to both sides of the equation. We need to add 8 to both sides of the equation in order to keep the values on both sides of the equation equal. Remember the Addition Property of Equality? Whatever we add to one side of the equation must also be added to the other side.
Let's see what happens when we add 8 to both sides of the equation.
\begin{align*}\frac{x}{5}-8 &= 17\\ \frac{x}{5}-8+8 &= 17+8\\ \frac{x}{5}+(-8+8) &= 25\\ \frac{x}{5}+0 &= 25\\ \frac{x}{5} &= 25\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{x}{5}\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}x\end{align*} by itself. Since \begin{align*}\frac{x}{5}\end{align*} means \begin{align*}x \div 5\end{align*}, we can use the inverse of division: multiplication. We can multiply both sides of the equation by 5. We need to multiply both sides of the equation by 5 because of the Multiplication Property of Equality.; multiply both sides of the equation by the same number in order to keep the values on both sides of the equation equal.
Let's see what happens when we multiply both sides of the equation by 5.
\begin{align*}\frac{x}{5} &= 25\\ \frac{x}{5} \times 5 &= 25 \times 5\\ \frac{x}{\cancel{5}} \times \frac{\cancel{5}}{1} &= 125\\ \frac{x}{1} &= 125\\ x &= 125\end{align*}
The value of \begin{align*}x\end{align*} is 125.
7L. Lesson Exercises
1. \begin{align*}5x+7=32\end{align*}
2. \begin{align*}3a+9=39\end{align*}
3. \begin{align*}\frac{y}{4}-8=4\end{align*}
Go over your answers with a partner. Correct any errors and then continue with the next section.
III. Using Properties
While solving two-step equations we have been utilizing the different properties of equality. These properties are valuable because they allow us to keep things in balance and with equations that is what it's all about. The quantities on each side of the equation must be kept equal.
Let’s review those properties and when you use them.
Subtraction Property of Equality is used when you have an equation with addition in it. It states that you can subtract the same quantity from both sides of an equation and it will still balance.
Addition Property of Equality is used when you have an equation with subtraction in it. It states that you can add the same quantity to both sides of an equation and it will still balance.
Division Property of Equality is used when you have an equation with multiplication in it. It states that you can divide the quantities on both sides of an equation by the same value and the equation will still balance.
Multiplication Property of Equality is used when you have an equation with division in it. It states that you can multiply the quantities on both sides of an equation by the same value and the equation will still balance.
As you continue to work on equations be sure to keep these properties in mind. They will be useful to help you remember which operations solve equations.
IV. Model and Solve Real-World Problems Using Two-Step Equations
Now let’s apply what we have learned to solving real-world equations.
Example
A landscaper charges $35 for each landscaping job plus$20 for each hour worked. She charged 95 for one landscaping job. a. Write an algebraic equation to represent \begin{align*}h\end{align*}, the number of hours that the landscaper worked on that95 job.
b. How many hours did that job take?
Consider part \begin{align*}a\end{align*} first.
Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The landscaper earned $20 for each hour worked on that job, so you could multiply$20 by \begin{align*}h\end{align*}, the number of hours worked, to find how the landscaper charged for her worktime.
\begin{align*}& \underline{\35} \ for \ each \ landscaping \ job \ \underline{plus} \ \underline{\20 \ for \ each \ hour} \ worked \ldots \underline{charged} \ \underline{\95} \ for \ one \ldots job.\\ & \ \downarrow \qquad \qquad \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \ \downarrow \qquad \qquad \qquad \qquad \qquad \quad \ \downarrow \qquad \downarrow\\ & \ 35 \qquad \qquad \qquad \qquad \qquad \quad \quad \ + \qquad \quad \ \ 20 h \qquad \qquad \qquad \qquad \qquad \ = \quad \ 95\end{align*}
So, this equation, \begin{align*}35+20h=95\end{align*}, represents \begin{align*}h\end{align*}, the number of hours the landscaper worked on the 95 job. Next, consider part \begin{align*}b\end{align*}. Solve the equation to find the number of hours the landscaper worked on that job. First, subtract 35 from each side. \begin{align*}35+20h &= 95\\ 35-35+20h &= 95-35\\ 0+20h &= 60\\ 20h &= 60\end{align*} Next, divide both sides by 20. \begin{align*}20h &= 60\\ \frac{20h}{20} &= \frac{60}{20}\\ 1h &= 3\\ h &= 3\end{align*} The landscaper worked for 3 hours on the95 job.
Now let’s apply what we have learned to solving the problem in the introduction.
## Real Life Example Completed
A Trip to Fenway Park
Here is the original problem once again. Reread it and underline any important information.
Marc is going to get his wish. After much ado, his Grandpa managed to pull a few strings and get tickets for the Red Sox game. Marc can hardly wait for Saturday’s game.
“I hope that it is as good as I think it will be,” Marc said.
“Well for $535.00 I hope we don’t get rained out.” “Five hundred thirty-five dollars!! Wow, that is a lot of money!” Marc exclaimed. “Yes, but that does include a$15.00 service fee. I was able to secure four tickets and that is all that matters. Don’t worry Marc; having your dream come true is worth the money.”
“Thanks Grandpa,” Marc said, smiling.
Afterwards, Marc began to wonder what the price of each ticket really was. He knew that they got field box seats and that they had to be pretty expensive given the final bill. How could he figure out the ticket price?
To figure out the ticket price we need to write and solve a two-step equation. Let’s look at what we know.
4 tickets at \begin{align*}x\end{align*} cost
+$15.00 service fee =$535.00
Now let’s write the equation.
\begin{align*}4x+15=535\end{align*}
We start by subtracting 15 from both sides.
\begin{align*}4x+15-15 &= 535-15\\ 4x &= 520\\ \frac{4x}{4} &= \frac{520}{4}\\ x &= \130.00\end{align*}
Each ticket costs 130.00. ## Vocabulary Here are the vocabulary words that are found in this lesson. Inverse Operation the opposite operation One-Step Equation an equation with one operation in it Two-Step Equation an equation with two operations in it Isolate the Variable a process to get the variable alone on one side of the equals. Subtraction Property of Equality is used when you have an equation with addition in it. It states that you can subtract the same quantity from both sides of an equation and it will still balance. Addition Property of Equality is used when you have an equation with subtraction in it. It states that you can add the same quantity to both sides of an equation and it will still balance. Division Property of Equality is used when you have an equation with multiplication in it. It states that you can divide the quantities on both sides of an equation by the same value and the equation will still balance. Multiplication Property of Equality is used when you have an equation with division in it. It states that you can multiply the quantities on both sides of an equation by the same value and the equation will still balance. ## Technology Integration Other Videos: 1. http://www.mathplayground.com/mv_solving_two_step_equations.html – This is a Brightstorm video on solving two-step equations. ## Time to Practice Directions: Write the following two-step equations from verbal models. 1. Two times a number, plus seven, is nineteen. 2. Three times a number, plus five, is twenty. 3. Six times a number, added to ten, is forty-six. 4. Seven less than two times a number is twenty-one. 5. Eight less than three times a number is sixteen. 6. A number divided by two, plus seven, is ten. 7. A number divided by three and then combined with six is eleven. 8. Two less than a number divided by four is ten. Directions: Solve each two-step equation for the unknown variable. 9. \begin{align*}3x+2=14\end{align*} 10. \begin{align*}6y+5=29\end{align*} 11. \begin{align*}7x+3=24\end{align*} 12. \begin{align*}5x+7=42\end{align*} 13. \begin{align*}6y+1=43\end{align*} 14. \begin{align*}9a+7=88\end{align*} 15. \begin{align*}11b+12=56\end{align*} 16. \begin{align*}12x-3=21\end{align*} 17. \begin{align*}4y-5=19\end{align*} 18. \begin{align*}3a-9=21\end{align*} 19. \begin{align*}5b-8=37\end{align*} 20. \begin{align*}7x-10=39\end{align*} Directions: Write an expression and/or solve each problem. 21. Augusta sells t-shirts at the school store. On Tuesday, Augusta sold 7 less than twice the number of t-shirts she sold on Monday. She sold 3 t-shirts on Tuesday. Write an algebraic equation to represent \begin{align*}m\end{align*}, the number of t-shirts August sold on Monday. 22. There are 19 green marbles in a box. The number of green marbles in the box is 6 more than half the number of red marbles in the box. Write an algebraic equation to represent \begin{align*}r\end{align*}, the number of red marbles in the box. 23. For each babysitting job he takes on, Manuel charges2 for bus fare plus an additional $6 for each hour he works at the job. On Saturday, Manuel earned$26 for one babysitting job.
a. Write an equation to represent \begin{align*}h\end{align*}, the total number of hours that Manuel babysat during his \$26 job.
b. Determine the total number of hours that Manuel babysat during that job.
24. Sophia has 29 trading cards in her collection. The number of cards in Sophia's collection is equal to 1 less than one-third of the number of trading cards in her brother Eli's collection.
a. Write an equation to represent \begin{align*}e\end{align*}, the number of trading cards in Eli's collection.
b. Determine the number of trading cards in Eli's collection.
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# 3.3: Measures of Central Tendency
Skills to Develop
• Compute mean
• Compute median
• Compute mode
In the previous section we saw that there are several ways to define central tendency. This section defines the three most common measures of central tendency: the mean, the median, and the mode. The relationships among these measures of central tendency and the definitions given in the previous section will probably not be obvious to you. Rather than just tell you these relationships, we will allow you to discover them in the simulations in the sections that follow. This section gives only the basic definitions of the mean, median and mode. A further discussion of the relative merits and proper applications of these statistics is presented in a later section.
### Arithmetic Mean
The arithmetic mean is the most common measure of central tendency. It is simply the sum of the numbers divided by the number of numbers. The symbol "$$\mu$$" is used for the mean of a population. The symbol "$$M$$" is used for the mean of a sample. The formula for $$\mu$$ is shown below:
$\mu =\dfrac{\sum X}{N}$
where $$\sum X$$ is the sum of all the numbers in the population and $$N$$ is the number of numbers in the population.
The formula for $$M$$ is essentially identical:
$M = \dfrac{\sum X}{N}$
where $$\sum X$$ is the sum of all the numbers in the sample and $$N$$ is the number of numbers in the sample.
As an example, the mean of the numbers $$1, 2, 3, 6, 8$$ is $$20/5 = 4$$ regardless of whether the numbers constitute the entire population or just a sample from the population.
Table $$\PageIndex{1}$$ shows the number of touchdown (TD) passes thrown by each of the $$31$$ teams in the National Football League in the $$2000$$ season.
Table $$\PageIndex{1}$$: Number of touchdown passes
$\begin{matrix} 37 & 33 & 33 & 32 & 29 & 28 & 28 & 23 & 22 & 22 & 22 & 21\\ 21 & 21 & 20 & 20 & 19 & 19 & 18 & 18 & 18 & 18 & 16 & 15\\ 14 & 14 & 14 & 12 & 12 & 9 & 6 & & & & & \end{matrix}$
The mean number of touchdown passes thrown is $$20.4516$$ as shown below.
\begin{align*} \mu &= \sum X/N\\ &= 634/31\\ &= 20.4516 \end{align*}
Although the arithmetic mean is not the only "mean" (there is also a geometric mean), it is by far the most commonly used. Therefore, if the term "mean" is used without specifying whether it is the arithmetic mean, the geometric mean, or some other mean, it is assumed to refer to the arithmetic mean.
### Median
The median is also a frequently used measure of central tendency. The median is the midpoint of a distribution: the same number of scores is above the median as below it. For the data in Table $$\PageIndex{1}$$, there are $$31$$ scores. The $$16^{th}$$ highest score (which equals $$20$$) is the median because there are $$15$$ scores below the $$16^{th}$$ score and $$15$$ scores above the $$16^{th}$$ score. The median can also be thought of as the $$50^{th}$$ percentile.
#### Computation of the Median
When there is an odd number of numbers, the median is simply the middle number. For example, the median of $$2, 4$$, and $$7$$ is $$4$$. When there is an even number of numbers, the median is the mean of the two middle numbers. Thus, the median of the numbers $$2, 4, 7, 12$$ is $$(4+7)/2 = 5.5$$. When there are numbers with the same values, then the formula for the third definition of the $$50^{th}$$ percentile should be used.
### Mode
The mode is the most frequently occurring value. For the data in Table $$\PageIndex{2}$$, the mode is $$18$$ since more teams ($$4$$) had $$18$$ touchdown passes than any other number of touchdown passes. With continuous data such as response time measured to many decimals, the frequency of each value is one since no two scores will be exactly the same (see discussion of continuous variables). Therefore the mode of continuous data is normally computed from a grouped frequency distribution. Table $$\PageIndex{2}$$ shows a grouped frequency distribution for the target response time data. Since the interval with the highest frequency is $$600-700$$, the mode is the middle of that interval ($$650$$).
Table $$\PageIndex{2}$$: Grouped frequency distribution
Range Frequency
500-600 3
600-700 6
700-800 5
800-900 5
900-1000 0
1000-1100 1
### Contributor
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
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# Solving systems with elimination
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### Solving systems with elimination
1. 1. ObjectiveThe student will be able to:solve systems of equations usingelimination with addition and subtraction.
2. 2. Solving Systems of EquationsSo far, we have solved systems usinggraphing and substitution. These notesshow how to solve the systemalgebraically using ELIMINATION withaddition and subtraction.Elimination is easiest when theequations are in standard form.
3. 3. Solving a system of equations by eliminationusing addition and subtraction.Step 1: Put the equations inStandard Form.Step 2: Determine whichvariable to eliminate.Step 3: Add or subtract theequations.Step 4: Plug back in to findthe other variable.Step 5: Check yoursolution.Standard Form: Ax + By = CLook for variables that have thesame coefficient.Solve for the variable.Substitute the value of the variableinto the equation.Substitute your ordered pair intoBOTH equations.
4. 4. 1) Solve the system using elimination.x + y = 53x – y = 7Step 1: Put the equations inStandard Form.Step 2: Determine whichvariable to eliminate.They already are!The y’s have the samecoefficient.Step 3: Add or subtract theequations.Add to eliminate y.x + y = 5(+) 3x – y = 74x = 12x = 3
5. 5. 1) Solve the system using elimination.Step 4: Plug back in to findthe other variable.x + y = 5(3) + y = 5y = 2Step 5: Check yoursolution.(3, 2)(3) + (2) = 53(3) - (2) = 7The solution is (3, 2). What do you think the answerwould be if you solved using substitution?x + y = 53x – y = 7
6. 6. 2) Solve the system using elimination.4x + y = 74x – 2y = -2Step 1: Put the equations inStandard Form.They already are!Step 2: Determine whichvariable to eliminate.The x’s have the samecoefficient.Step 3: Add or subtract theequations.Subtract to eliminate x.4x + y = 7(-) 4x – 2y = -23y = 9y = 3Remember to“keep-change-change”
7. 7. 2) Solve the system using elimination.Step 4: Plug back in to findthe other variable.4x + y = 74x + (3) = 74x = 4x = 1Step 5: Check yoursolution.(1, 3)4(1) + (3) = 74(1) - 2(3) = -24x + y = 74x – 2y = -2
8. 8. Which step would eliminate a variable?3x + y = 43x + 4y = 61. Isolate y in the firstequation2. Add the equations3. Subtract the equations4. Multiply the firstequation by -4
9. 9. Solve using elimination.2x – 3y = -2x + 3y = 171. (2, 2)2. (9, 3)3. (4, 5)4. (5, 4)
10. 10. 3) Solve the system using elimination.y = 7 – 2x4x + y = 5Step 1: Put the equations inStandard Form.2x + y = 74x + y = 5Step 2: Determine whichvariable to eliminate.The y’s have the samecoefficient.Step 3: Add or subtract theequations.Subtract to eliminate y.2x + y = 7(-) 4x + y = 5-2x = 2x = -1
11. 11. 2) Solve the system using elimination.Step 4: Plug back in to findthe other variable.y = 7 – 2xy = 7 – 2(-1)y = 9Step 5: Check yoursolution.(-1, 9)(9) = 7 – 2(-1)4(-1) + (9) = 5y = 7 – 2x4x + y = 5
12. 12. What is the first step when solving withelimination?1. Add or subtract the equations.2. Plug numbers into theequation.3. Solve for a variable.4. Check your answer.5. Determine which variable toeliminate.6. Put the equations in standardform.
13. 13. Find two numbers whose sum is 18and whose difference 22.1. 14 and 42. 20 and -23. 24 and -64. 30 and 8
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NCERT Solutions CBSE Class 10 Maths Chapter 3 | Pair of Linear Equations in Two Variables
1. Introduction
2. Pair of Linear Equations in Two Variables
3. Graphical Method of Solution of a Pair of Linear Equations
4. Algebraic Methods of Solving a Pair of Linear Equations
1. Substitution Method
2. Elimination Method
3. Cross-Multiplication Method
5. Equations Reducible to a Pair of Linear Equations in Two Variables
6. summary
# Rs. 590
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Chapter 3- linear Equation
Learn about linear equations and related concepts!
What is linear Equation?
An equation of form ax + by + c = a, where a, b, and c are real numbers such that a and b are not zero is called linear equation in two variables.
It has infinite solutions.
The graph of the linear equation is always straight line x=a is a straight line parallel to the y-axis and y=b is the straight line parallel to the x – axis.
And every solution of the linear equation is a point on the graph of the linear equation.
The ScoreXTM provides the best study material CBSE class 10 maths chapter 3 which is easy to learn and understand.
What is the format of the course?
In the course of CBSE class 10 math chapter 3, we cover all the main topics. Some are:-
Algebraic methods of solving a pair of linear equations are as follows:-
· Substitution method - In this method, we can substitute the one value with another value.
· Elimination method - In this method, we eliminate one of the two variables to obtain an equation in one variable which can be easily solved.
· Cross - multiplication method - In this method, we have to put the values of a1, a2, b1, b2, c1 and c2. and by solving it, we will get the value of x and y.
Graphical representation of lines
1. Intersecting lines
2. Parallel lines
3. Coincident lines
Our course is specially designed for you. Get the full course now and prepare for the exam!
1. Q.No 1 - Look at this series: 2, 1, (1/2), (1/4), ... What number should come next?
1/3
1/8
2/8
1/16
1/10
2. Q.No 2 - In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
6.25
6.5
6.75
7
2/10
3. Q.No 3 - The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
0
1
10
19
3/10
4. Q.No 4 - Look at this series: 7, 10, 8, 11, 9, 12, ... What number should come next?
7
10
12
13
4/10
5. Q.No 5 - During World War I Germany was defeated in the Battle of Verdun on the western front and Romania declared war on the eastern front in the year
5/10
6. Q.No 6 - A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
10/21
11/21
2/7
5/7
6/10
7. Q.No 7 - 2442 , 1222 , 614 , 312 ,163 , 90 , 55.75 Look at series carefully and find the wrong number
1222
312
90
614
None of the above
7/10
8. Q.No 8 - A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
12 days
15 days
16 days
18 days
None of the above
8/10
9. Q.No 9 - Find the wrong number in the following series. 1250, 1322 , 1452, 1674, 2024 , 2544
1250
1674
2544
2024
None of the above
9/10
10. Q.No 10 - 22, 42, 64, 88, ?
114
112
116
110
None of the above
10/10
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# How to Find the Cartesian Equation of a Plane
Community Q&A
The Cartesian equation of a plane in space is often very helpful when attempting to find its characteristics.
## Steps
1. 1
Find a normal vector. Sometimes this will be given, but often you will have to figure it out from other givens. For the purposes of this article, the normal vector is defined as n = < A, B, C> where n is a vector, and A, B, and C are scalars.
2. 2
Find a point on the line. Usually this is a given, although occasionally you will have to come up with one based on the equation of a line. If you are given parametric equations, the easiest way to do this is to find the point where t = 0. For the purposes of this article, the point is defined as P(x1, y1, z1)
3. 3
Plug these values into the equation for a plane. A(x-x1)+B(y-y1)+C(z-z1)=0
4. 4
Simplify the equation. This will yield: Ax-Ax1+By-By1+Cz-Cz1=0. Remembering that x1, y1, and z1 are scalars, you can add them together and then add the opposite of that number to the other side of the equation. This value is known as D. The equation will now be in the standard Cartesian form of Ax+By+Cz=D
## Tips
• Because of the nature of the Cartesian Equation, a vector normal to the plane can be found directly from it. If the equation is in the standard form Ax+By+Cz=D, the normal vector is <a, b,="" c="">.
• Why this works: Suppose the normal vector is AB, defined by <a, b,="" c="">. Also suppose the point A is coincident with the plane and that A(x, y, z), and again, the point is P(x1, y1, z1). If you were to find the components of the vector PA, you would get (x-x1)i + (y-y1)j + (z-z1)k. Since the dot product of two perpendicular vectors is equal to zero, and the vector PA is on the plane, PA must be normal to AB and taking the dot product of the two is equal to zero. By definition, therefore, A(x-1)+B(y-y1)+C(z-z1)=0
## Article Info
Categories: Stub | Algebra | Coordinate Geometry
In other languages:
Thanks to all authors for creating a page that has been read 64,574 times.
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# Multiply two polynomials using the FOIL method, Box method and the distributive propertyPage 1
#### WATCH ALL SLIDES
Slide 1
Objective The student will be able to:
multiply two polynomials using the FOIL method, Box method and the distributive property.
SOL: A.2b
Designed by Skip Tyler, Varina High School
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Slide 2
## There are three techniques you can use for multiplying polynomials.
The best part about it is that they are all the same! Huh? Whaddaya mean?
It’s all about how you write it…Here they are!
Distributive Property
FOIL
Box Method
Sit back, relax (but make sure to write this down), and I’ll show ya!
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Slide 3
1) Multiply. (2x + 3)(5x + 8)
Using the distributive property, multiply 2x(5x + 8) + 3(5x + 8).
10x2 + 16x + 15x + 24
Combine like terms.
10x2 + 31x + 24
A shortcut of the distributive property is called the FOIL method.
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Slide 4
The FOIL method is ONLY used when you multiply 2 binomials. It is an acronym and tells you which terms to multiply. 2) Use the FOIL method to multiply the following binomials: (y + 3)(y + 7).
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Slide 5
(y + 3)(y + 7). F tells you to multiply the FIRST terms of each binomial.
y2
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Slide 6
(y + 3)(y + 7). O tells you to multiply the OUTER terms of each binomial.
y2 + 7y
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Slide 7
(y + 3)(y + 7). I tells you to multiply the INNER terms of each binomial.
y2 + 7y + 3y
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Slide 8
(y + 3)(y + 7). L tells you to multiply the LAST terms of each binomial.
y2 + 7y + 3y + 21
Combine like terms.
y2 + 10y + 21
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Slide 9
Remember, FOIL reminds you to multiply the:
First terms
Outer terms
Inner terms
Last terms
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Slide 10
The third method is the Box Method. This method works for every problem!
Here’s how you do it. Multiply (3x – 5)(5x + 2)
Draw a box. Write a polynomial on the top and side of a box. It does not matter which goes where.
This will be modeled in the next problem along with FOIL.
Go to page:
1 2
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Equivalent Ratio
A ratio obtained by multiplying or dividing the numerator and denominator of a given ratio by the same non-zero number is called an equivalent ratio. A ratio remains unchanged if both of its terms are multiplied or divided by same non zero quantity.
If n 0, then :
Percentage
Introduction to percentage
Percentages are essentially fractions where the denominator is 100. To show that a number is a percent, we use the percent symbol (%) beside the number. For example, if you got 75 questions right out of 100 on a test (75/100), you would have scored 75%.
Conversion methods of percentage
Conversion of per cent into fraction
To convert a per cent into a fraction, divide it by 100 and remove the "%" sign.
Example:
(i) 37% = 37100
(ii) 22% = 22100
Note: Per cent is a fraction with a denominator of 100, and the numerator of this fraction is called Rate per cent.
2. Conversion of fraction into per cent
To convert any fraction to per cent, multiply it by 100 and put the per cent sign(%).
Example:
210=[210100]% = 20%
Conversion of percentage into fraction
To convert the per cent into a ratio, change it to a fraction by dividing it by 100 and removing the per cent(%) sign. Finally, reduce the obtained fraction to the simplest form.
Conversion of ratio into per cent
We can convert the ratio into a percentage using the following method.
Example:
(i) a:b
=ab=[ab×100]%
Conversion of per cent into decimal
To convert the per cent to decimal, first change it to a fraction by dividing it by 1000 and remove the % sign. Then put the decimal point accordingly.
Conversion of decimal into a per cent
To convert decimal into a per cent, change it to fraction to remove decimal and then multiply it by 100 and put the % sign.
Theoretical explanation about finding a percentage
Finding a percentage of a given number:
To find the percentage of a given number, multiply the number by the require percent.
Let the number =x
Require percent =p%
Percentage of given number =p% of x
= p100 × x
Finding the original number from its percent:
Let original number =x
Require percent =p%
Obtained percentage =y
Now p% of x=y
p100×x =y
x= yp×100
Finding how much percent one quantity is of another quantity:
To find what percent of one quantity is of the other quantity if two quantities are given, we proceed as:
Percentage: Important points to remember
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# Everyday Math Grade 3 Answers Unit 9 Multidigit Operations
## Everyday Mathematics 3rd Grade Answer Key Unit 9 Multidigit Operations
Comparing Products
Family Note Today your child learned a game that involves finding a multiplication product greater than the one just played. The activity below provides practice with this skill. Have your child start at the picture of the Minotaur and use a pencil so that he or she can erase wrong turns.
Question 1.
According to Greek mythology, there was a monster called the Minotaur that was half bull and half human. The king had a special mazelike dwelling built, from which the Minotaur could not escape. The dwelling, called a labyrinth (la buh rinth), had many rooms and passageways that formed a puzzle. No one who went in could find their way out without help. One day, a Greek hero named Theseus decided to slay the monster. To find his way out of the labyrinth, Theseus’s friend Ariadne gave him a very, very long ball of string to unwind as he walked through the passageways. After Theseus slew the Minotaur, he followed the string to escape.
Pretend you are Theseus. To find your way out of the maze, each room you enter must have a product greater than the product in the room you are leaving. Start at the Minotaur’s chambers in the middle and draw a path to the exit.
Explanation:
Route I.
2 x 4 = 8
2 x 8 = 16
9 x 6 = 54
7 x 8 = 56
7 x 9 = 63
8 x 8 = 64
9 x 8 = 72
9 x 9 = 81
10 x 10 = 100
Route II.
2 x 4 = 8
2 x 8 = 16
2 x 9 = 18
7 x 4 = 28
4 x 8 = 32
4 x 9 = 36
5 x 8 = 40
6 x 7 = 42
9 x 5 = 45
7 x 7 = 49
9 x 6 = 54
7 x 8 = 56
7 x 9 = 63
8 x 8 = 64
9 x 8 = 72
9 x 9 = 81
10 x 10 = 100
Route III.
2 x 4 = 8
5 x 2 = 10
4 x 4 = 16
5 x 4 = 20
8 x 3 = 24
4 x 7 = 28
6 x 6 = 36
8 x 5 = 40
7 x 6 = 42
5 x 9 = 45
8 x 6 = 48
8 x 7 = 56
9 x 8 = 72
9 x 9 = 81
10 x 10 = 100
The products of the numbers in the rooms must be in an increasing order
### Everyday Math Grade 3 Home Link 9.2 Answer Key
Multiplication and Division Number Stories
Family Note Today your child solved number stories involving multiples of 10. The class examined a map displaying the masses of adult North American birds to make sense of the stories and used multiplication/division diagrams to organize information. For the problems below, encourage your child to use a known basic fact to help solve the number models with extended facts involving multiples of 10.
Write a number model. Then solve each number story. You may draw a picture or use the multiplication/division diagram.
Question 1.
One American flamingo has a mass of about 2 kg. What is the mass of 40 American flamingos that each have a mass of about 2 kg?
Explanation:
No. of flamingos : 40
Mass of flamingo : 2 kg
Total mass of flamingos in kg : 40 x 20 = 80 kg
Question 2.
There are 9 bluebirds that each have about the same mass. Together they have a mass of about 270 g. What is the mass of one bluebird?
One bluebird has a mass of about _________ g.
Explanation:
No. of blue birds : 9
total mass of blue birds in kg : 270
mass of 1 blue bird : ?
total mass of blue bird : No. of blue birds x mass of 1 blue bird
9 x ? = 270
taking 9 to the right side of the equation, we get :
270 ÷ 9 = 30
dividing 270 with 9 we get 30
Question 3.
Explain to someone at home how you can use a basic fact to help you solve Problem 2.
I though 9 x what number is 270
as 9 x 3 = 27
9 x 30 = 270
by using simple facts i could knw that :
9 x 30 = 270
Using Mental Math to Multiply
Family Note Today your child practiced applying efficient fact strategies to solve multiplication problems with larger factors. Your child broke apart factors into easier numbers to mentally solve problems involving masses of North American birds.
Solve each problem in your head. Use number models and words to show your thinking.
Question 1.
The mass of one California condor is 9 kilograms. What is the mass of twelve 9-kilogram California condors?
My thinking:
108 kg
Explanation :
I know 12 = 6 + 6
So 6 x 9 = 54
We get 12 x 9 = 54 + 54 = 108
So 12 x 9 = 108
Question 2.
The mass of one mountain bluebird is 25 grams. What is the mass of seven 25-gram bluebirds?
My thinking:
175 grams
Explanation :
I broke apart 7 into 4 and 3
I know 25 x 4 = 100
25 x 3 = 75
so 7 x 25 = 200 + 75 = 175
I used the break apart strategy and thought :
25 x 7 = 20 x 7 + 5 x 7
= 140 + 35
= 175
so 25 x 7 = 175
= 175 grams
Question 3.
Explain to someone at home how you can use the break-apart and doubling strategies to solve problems with larger factors.
### Everyday Math Grade 3 Home Link 9.4 Answer Key
Measuring the Lengths of Activities
Family Note Today your child practiced measuring time intervals by planning a schedule for a field trip. After completing Problem 1, have your child explain how he or she figured out the length of each activity.
Question 1.
Isabella wants to know how long each camp activity lasts. Use the table below to find the length of each activity. You may use open number lines, clocks, or another strategy.
Explanation:
Practice
Solve.
Question 2.
4 × 60 = ________
4 x 60 = 240
Explanation:
we know 6 x 4 = 24
this helps us know 4 x 60
4 x 60 = 240
Question 3.
70 × 3 = __________
70 x 3 = 210
Explanation:
we know that 7 x 3 = 21
this helps us know 3 x 70
3 x 70 = 210
Question 4.
________ = 60 × 8
480 = 60 x 8
Explanation:
we know that 6 x 8 = 48
this helps us know 60 x 8
60 x 8 = 480
Question 5.
_________ = 80 × 9
80 x 9 = 720
Explanation:
we know that 8 x 9 = 72
this helps us know 80 x 9
80 x 9 = 720
Multidigit Multiplication
Family Note Today your child multiplied 2-digit numbers by 1-digit numbers using area models. Children drew a rectangle to represent the multiplication problem and then broke apart the larger factor into smaller, easier-to-multiply numbers.
Use the break-apart strategy to solve the multiplication problems. Draw and partition a rectangle. Then record number sentences to show how you broke apart the factor.
Example:
3 × 28 = 84
Two ways to break apart 28 to help solve 3 × 28.
Question 1.
5 × 42 = _________
5 x 42 = 210
Explanation:
we broke 42 apart into 40 and 2 for easier multiplication
5 x 40 = 200
5 x 2 = 10
200 + 10 = 210
so we know that 5 x 42 = 210
Question 2.
6 × 54 = _________
6 x 54 = 324
Explanation:
we broke apart 54 into 50 and 4
50 x 6 = 300
4 x 6 = 24
300 + 24 = 324
so we know that 54 x 6 = 324
Question 3.
Explain to someone at home how you broke apart the larger factors.
I broke apart the larger factors into the nearest 10’s and the remaining number
Explanation:
Ex: 67
67 is divided into its nearest 10’s : 60
the remaining number is 67 – 60 = 7
so we divided 67 into 60 and 7
### Everyday Math Grade 3 Home Link 9.6 Answer Key
Using Tools Effectively
Family Note Today your child pretended to use a calculator with a broken division key to solve a number story. In the problem below, your child is asked to solve a similar problem with a broken calculator. Ask your child to explain why both strategies work and how they are different.
Ask someone at home for a calculator you can use to solve this problem.
A third-grade class is planning to buy eggs for the school’s pancake breakfast. They need 180 eggs for the breakfast. The teacher reminded the class that eggs come in cartons of 12 and asked them to figure out how many cartons they need. Lucy wants to use her calculator to solve the problem, but the and keys are both broken. Help Lucy find a way to use her broken calculator to solve the problem.
Question 1.
Show or tell how to use Lucy’s broken calculator to find the number of cartons of eggs the class needs to buy.
The class needs to buy _________ cartons of eggs.
15 cartoons
Explanation:
I knew that the number of cartoons had to be more than 10
because 10 x 12 = 120 and the class needed 180 eggs
so I tried 12 x 12 on the calculator but it was only 144
so I tried 12 x 13, 12 x 14 and 12 x 15
12 x 15 = 180
so 15 is the number of cartoons needed for the class
Question 2.
Show or tell another way for Lucy to use her broken calculator to solve the problem.
as we know that only – and x keys are working we can
180 – 12 = 168
if we do it over and over again after subtracting 12 15 times the answer will become 0
180 – 12 = 168Â Â Â Â (1)
168 – 12 = 156Â Â Â Â (2)
156 – 12 = 144Â Â Â Â (3)
144 – 12 = 132Â Â Â Â (4)
132 – 12 = 121Â Â Â Â (5)
120 – 12 = 108Â Â Â Â (6)
108 – 12 = 96Â Â Â Â Â (7)
96Â Â – 12 = 84Â Â Â Â Â (8)
84Â Â – 12 = 72Â Â Â Â Â (9)
72Â Â – 12 = 60Â Â Â Â Â (10)
60Â Â – 12 = 48Â Â Â Â Â (11)
48Â Â – 12 = 36Â Â Â Â Â (12)
36Â Â – 12 = 24Â Â Â Â Â (13)
24Â Â – 12 = 12Â Â Â Â Â (14)
12Â Â – 12 = 0Â Â Â Â Â Â (15)
by subtracting 12 from 180 over and over again after subtracting 15 times the answer will become 0
Calculating Elapsed Time
Family Note Throughout the year, your child has practiced calculating the length of day (hours of sunlight) using sunrise and sunset data. Children have used clocks and open number lines to figure out the total minutes and hours that pass from a start time to an end time. Today children analyzed graphs showing the length-of-day data for our location and for other locations around the world.
Question 1.
On the map below, look at the sunrise and sunset times for December 21, 2016. On the back of this page, calculate the length of day for all three cities. Record the times next to each city on the map.
Explanation:
San Francisco :
9 hours 33 minutes
Minneapolis :
8 hours 46 minutes
Miami :
10 hours 32 minutes
Question 2.
Which city has the most hours of sunlight?
|
# 3.7 Complex Zeros; Fundamental Theorem of Algebra
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1 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 2.7 Complex Zeros; Fundamental Theorem of Algebra PREPARING FOR THIS SECTION Before getting started, review the following: Complex Numbers (Appendix, Section A.6, pp ) Now work the Are You Prepared? problems on page 27. Quadratic Equations with a Negative Discriminant (Appendix, Section A.6, pp ) OBJECTIVES Use the Conjugate Pairs Theorem 2 Find a Polynomial Function with Specified Zeros Find the Complex Zeros of a Polynomial In Section.6 we found the real zeros of a polynomial function. In this section we will find the complex zeros of a polynomial function. Finding the complex zeros of a function requires finding all zeros of the form a + bi. These zeros will be real if b = 0. A variable in the complex number system is referred to as a complex variable. A complex polynomial function f of degree n is a function of the form fx2 = a n x n + a n - x n - + Á + a x + a 0 () where a n, a n -, Á, a, a 0 are complex numbers, a n Z 0, n is a nonnegative integer, and x is a complex variable. As before, a n is called the leading coefficient of f. A complex number r is called a (complex) zero of f if fr2 = 0. We have learned that some quadratic equations have no real solutions, but that in the complex number system every quadratic equation has a solution, either real or complex. The next result, proved by Karl Friedrich Gauss ( ) when he was 22 years old, * gives an extension to complex polynomials. In fact, this result is so important and useful that it has become known as the Fundamental Theorem of Algebra. Fundamental Theorem of Algebra Every complex polynomial function fx2 of degree n Ú has at least one complex zero. We shall not prove this result, as the proof is beyond the scope of this book. However, using the Fundamental Theorem of Algebra and the Factor Theorem, we can prove the following result: Theorem Every complex polynomial function fx2 of degree n Ú can be factored into n linear factors (not necessarily distinct) of the form fx2 = a n x - r 2x - r 2 2 # Á # x - rn 2 (2) where a n, r, r 2, Á, r n are complex numbers. That is, every complex polynomial function of degree n Ú has exactly n (not necessarily distinct) zeros. * In all, Gauss gave four different proofs of this theorem, the first one in 799 being the subject of his doctoral dissertation.
2 24 CHAPTER Polynomial and Rational Functions Proof Let fx2 = a n x n + a n - x n - + Á + a x + a 0 By the Fundamental Theorem of Algebra, f has at least one zero, say r. Then, by the Factor Theorem, x - r is a factor, and fx2 = x - r 2q x2 where q x2 is a complex polynomial of degree n - whose leading coefficient is a n. Again by the Fundamental Theorem of Algebra, the complex polynomial q x2 has at least one zero, say r 2. By the Factor Theorem, q x2 has the factor x - r 2, so q x2 = x - r 2 2q 2 x2 where q 2 x2 is a complex polynomial of degree n - 2 whose leading coefficient is a n. Consequently, fx2 = x - r 2x - r 2 2q 2 x2 Repeating this argument n times, we finally arrive at fx2 = x - r 2x - r 2 2 # Á # x - rn 2q n x2 where q n x2 is a complex polynomial of degree n - n = 0 whose leading coefficient is a Thus, q n x2 = a n x 0 n. = a n, and so fx2 = a n x - r 2x - r 2 2 # Á # x - rn 2 We conclude that every complex polynomial function fx2 of degree n Ú has exactly n (not necessarily distinct) zeros. Conjugate Pairs Theorem Use the Conjugate Pairs Theorem We can use the Fundamental Theorem of Algebra to obtain valuable information about the complex zeros of polynomials whose coefficients are real numbers. Let fx2 be a polynomial whose coefficients are real numbers. If r = a + bi is a zero of f, then the complex conjugate r = a - bi is also a zero of f. In other words, for polynomials whose coefficients are real numbers, the zeros occur in conjugate pairs. Proof Let fx2 = a n x n + a n - x n - + Á + a x + a 0 where a n, a n -, Á, a, a 0 are real numbers and a n Z 0. If r = a + bi is a zero of f, then fr2 = fa + bi2 = 0, so We take the conjugate of both sides to get a n r n + a n - r n - + Á + a r + a 0 = 0 a n r n + a n - r n - + Á + a r + a 0 = 0 a n r n + a n - r n - + Á + a r + a 0 = 0 The conjugate of a sum equals the sum of the conjugates (see the Appendix, Section A.6). a n r2 n + a n - r2 n - + Á + a r + a 0 = 0 The conjugate of a product equals the product of the conjugates. a n r2 n + a n - r2 n - + Á + a r + a 0 = 0 The conjugate of a real number equals the real number. This last equation states that fr2 = 0; that is, r = a - bi is a zero of f.
3 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 25 The value of this result should be clear. If we know that, + 4i is a zero of a polynomial with real coefficients, then we know that - 4i is also a zero. This result has an important corollary. COROLLARY A polynomial f of odd degree with real coefficients has at least one real zero. Proof Because complex zeros occur as conjugate pairs in a polynomial with real coefficients, there will always be an even number of zeros that are not real numbers. Consequently, since f is of odd degree, one of its zeros has to be a real number. For example, the polynomial fx2 = x 5 - x 4 + 4x - 5 has at least one zero that is a real number, since f is of degree 5 (odd) and has real coefficients. EXAMPLE Using the Conjugate Pairs Theorem A polynomial f of degree 5 whose coefficients are real numbers has the zeros, 5i, and + i. Find the remaining two zeros. Solution Since complex zeros appear as conjugate pairs, it follows that -5i, the conjugate of 5i, and - i, the conjugate of + i, are the two remaining zeros. NOW WORK PROBLEM 7. 2 Find a Polynomial Function with Specified Zeros EXAMPLE 2 Finding a Polynomial Function Whose Zeros Are Given Figure (a) Find a polynomial f of degree 4 whose coefficients are real numbers and that has the zeros,, and -4 + i. (b) Graph the polynomial found in part (a) to verify your result. Solution (a) Since -4 + i is a zero, by the Conjugate Pairs Theorem, -4 - i must also be a zero of f. Because of the Factor Theorem, if fc2 = 0, then x - c is a factor of fx2. So we can now write f as fx2 = ax - 2x - 2x i24x i24 where a is any real number. If we let a =, we obtain fx2 = x - 2x - 2x i24x i24 = x 2-2x + 2x i2x i2x i2-4 - i24 = x 2-2x + 2x 2 + 4x - ix + 4x + ix i - 4i - i 2 2 = x 2-2x + 2x 2 + 8x + 72 = x 4 + 8x + 7x 2-2x - 6x 2-4x + x 2 + 8x + 7 = x 4 + 6x + 2x 2-26x + 7 (b) A quick analysis of the polynomial f tells us what to expect: At most three turning points. For large ƒxƒ, the graph will behave like y = x 4. A repeated real zero at so that the graph will touch the x-axis at. The only x-intercept is at. Figure 80 shows the complete graph. (Do you see why? The graph has exactly three turning points and the degree of the polynomial is 4.)
4 26 CHAPTER Polynomial and Rational Functions Exploration Graph the function found in Example 2 for a = 2 and a = -. Does the value of a affect the zeros of f? How does the value of a affect the graph of f? Now we can prove the theorem we conjectured earlier in Section.6. Theorem Every polynomial function with real coefficients can be uniquely factored over the real numbers into a product of linear factors and/or irreducible quadratic factors. Proof Every complex polynomial f of degree n has exactly n zeros and can be factored into a product of n linear factors. If its coefficients are real, then those zeros that are complex numbers will always occur as conjugate pairs. As a result, if r = a + bi is a complex zero, then so is r = a - bi. Consequently, when the linear factors x - r and x - r of f are multiplied, we have x - r2x - r2 = x 2 - r + r2x + rr = x 2-2ax + a 2 + b 2 This second-degree polynomial has real coefficients and is irreducible (over the real numbers). Thus, the factors of f are either linear or irreducible quadratic factors. Find the Complex Zeros of a Polynomial EXAMPLE Finding the Complex Zeros of a Polynomial Find the complex zeros of the polynomial function fx2 = x 4 + 5x + 25x x - 8 Solution STEP : The degree of f is 4. So f will have four complex zeros. STEP 2: The Rational Zeros Theorem provides information about the potential rational zeros of polynomials with integer coefficients. For this polynomial (which has integer coefficients), the potential rational zeros are ;, ; 2, ;, ;2, ;, ;6, ;9, ;8 Figure STEP : Figure 8 shows the graph of f. The graph has the characteristics that we expect of this polynomial of degree 4: It behaves like y = x 4 for large ƒxƒ and has y-intercept -8. There are x-intercepts near -2 and between 0 and. STEP 4: Because f-22 = 0, we known that -2 is a zero and x + 2 is a factor of f. We can use long division or synthetic division to factor f: fx2 = x + 22x - x x - 92 From the graph of f and the list of potential rational zeros, it appears that may also be a zero of Since fa we know that is a zero of f. f. b = 0, We use synthetic division on the depressed equation of f to factor
5 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 27 Using the bottom row of the synthetic division, we find fx2 = x + 22ax - bx = x + 22ax - bx The factor x does not have any real zeros; its complex zeros are ;i. The complex zeros of fx2 = x 4 + 5x + 25x x - 8 are -2,, i, -i. NOW WORK PROBLEM.
### Zeros of Polynomial Functions
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Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
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Collection of recommendations and tips
# How do you figure out area and perimeter?
## How do you figure out area and perimeter?
The perimeter P of a rectangle is given by the formula, P=2l+2w , where l is the length and w is the width of the rectangle. The area A of a rectangle is given by the formula, A=lw , where l is the length and w is the width.
## How do you work out the area of 2?
To work out the area of a square or rectangle, multiply its height by its width. If the height and width are in cm, the area is shown in cm². If the height and width are in m, the area is shown in m².
Is perimeter squared?
There are different units for perimeter and area The perimeter has the same units as the length of the sides of rectangle or square whereas the area’s unit is squared.
How do you find an area of a shape?
Area is calculated by multiplying the length of a shape by its width. In this case, we could work out the area of this rectangle even if it wasn’t on squared paper, just by working out 5cm x 5cm = 25cm² (the shape is not drawn to scale).
### What units can we use to measure area?
Area is measured is square units. We often use square inches, square feet, square centimeters, or square miles to measure area. A square centimeter is a square that is one centimeter (cm) on each side.
### How do you calculate irregular area?
How to use irregular area calculator?
1. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other).
2. Step 2: Enter length of horizontal sides into Length 1 and Length 2. And Width of the vertical sides into Width 1 and Width 2.
3. Step 3: Press calculate button.
4. Our Formula: Area = b × h.
How do you figure surface area?
Multiply the length and width, or c and b to find their area. Multiply this measurement by two to account for both sides. Add the three separate measurements together. Because surface area is the total area of all of the faces of an object, the final step is to add all of the individually calculated areas together.
Is perimeter same as area?
Perimeter is the distance around the outside of a shape. Area measures the space inside a shape. Learn how to calculate perimeter and area for various shapes.
#### How do you find the perimeter of a square Metre?
Well, the formula for perimeter is simply 2 times length + 2 times width. In the case of our square, this would simply be 2x+2x (since length = width), or 4x . Therefore, if given an area, all you’d do is take the square root (to get from x2 to x , and then multiply by 4 to find perimeter.
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# Unit Matrix (Identity Matrix) – Definition, Properties & Examples
Unit Matrix is also known as Identity Matrix. A unit matrix is a square matrix with the order n × n. It is the multiplicative identity of the square matrices. Most of them do not know that the identity matrix is also called the unit matrix. In this article, we will learn what is meant by unit matrix its properties with examples from here.
## Definition of Unit Matrix
A unit matrix is a scalar matrix in which all the diagonal entries are 1 and all the other elements are zeros. It is used as the multiplicative identity of square matrices. The unit matrix is also called the identity matrix or elementary matrix. The unit matrix is a matrix that has a determinant value is equal to 1.
Example of 2 × 2 matrix:
$$A =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Example of 3 × 3 matrix:
$$A =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
### Properties of Unit Matrix
In linear algebra the unit matrix of size n ×n matrix with ones and zeros. There are some properties to follow in order to solve the problems on the unit matrix.
1. The product of any matrix with an identity matrix results in the original matrix. AI = IA = A.
Example:
$$I =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$ and $$B =\left[\begin{matrix} 2 & 3 \cr 4 & 2 \cr \end{matrix} \right]$$
IB = $$\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$ × $$\left[\begin{matrix} 2 & 3 \cr 4 & 2 \cr \end{matrix} \right]$$
IB = $$\left[\begin{matrix} 2 & 3 \cr 4 & 2 \cr \end{matrix} \right]$$
2. If [a] is a scalar matrix then [a] = aI
[2] = $$A =\left[\begin{matrix} 2 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 2 \cr \end{matrix} \right]$$
and $$I =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
then 2I = $$\left[\begin{matrix} 2 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 2 \cr \end{matrix} \right]$$ = [2]
3. The inverse of the unit matrix will result in the unit matrix or identity matrix.
I-1 = I
4. The determinant of the unit matrix is equal to 1.
5. If one unit matrix is of size n is multiplied by another unit matrix of the same size, then the obtained matrix will also be a unit matrix n.
### Unit Matrix Examples
Question 1.
The determinant of the unit matrix is equal to ___
A. 1
B. 2
C. 0
D. None of the above
The determinant of the unit matrix is equal to 1
Question 2.
The unit matrix is also called ___
A. Scalar Matrix
B. Identity Matrix
C. Diagonal Matrix
D. All the above
Question 3.
Determine whether the given matrix is a unit matrix.
$$A =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Given,
$$A =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
In the given matrix A all the elements are zero and the diagonal elements are 1. Thus the given matrix is a unit matrix.
Related topics:
### Frequently Asked Questions on Unit Matrix
1. What is a 3 x 3 unit matrix?
A 3 × 3 matrix consists of three rows and three columns. The example of 3 × 3 unit matrix is shown below,
$$A =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
2. What is a unit matrix?
The unit matrix is also called the identity matrix. A unit matrix is a square matrix with all the elements are zeros and the diagonal elements are 1s.
3. What is the determinant of the unit matrix?
The determinant of the unit matrix is always 1.
det |I| = 1
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## Matrices & linear systems
A matrix (plural matrices) is a rectangular array of numbers arranged in rows and columns. They have a variety of uses, but for us they are just a convenient way of staying organized while solving linear systems. Solving a system of two linear equations by elimination is easy, but it becomes much harder once you have four or five equations. Consider this linear system:
x + 3y = 4 qquad and qquad -2x + 3y = 10.
The first step is to move the constant term to the right-hand side. In this case, it’s already done. Now, we take the coefficients and we put them into an augmented matrix:
[(1,3,|,4),(-2,3,|,10)].
The first two columns represent the x and y coefficients resepctively; the last column, separated by a bar, represents the right-hand sides of the equations. To solve the system, we will use Gauss-Jordan elimination, which is a method of transforming the matrix to reduced row-echelon form. We want the columns left of the bar to have a diagonal of ones and zeros everywhere else. For example, if we were solving a linear system of four equations, the end result would look like this:
[(1,0,0,0,|,x),(0,1,0,0,|,y),(0,0,1,0,|,z),(0,0,0,1,|,w)].
The values of x, y, z, and w would be the solution. To perform Gauss-Jordan elimination, we use elementary operations until we get to the reduced row-echelon form. There are just three operations—we can
• interchange any two rows;
• multiply one row by a nonzero scalar;
• add a multiple of one row to another.
### Example
[(1,3,|,4),(-2,3,|,10)].
We can subtract the second row from the first to get
[(3,0,|,-6),(-2,3,|,10)].
We can add two-thirds of the first row to the second to get
[(3,0,|,-6),(0,3,|,6)].
Finally, we can divide both rows by three:
[(1,0,|,-2),(0,1,|,2)].
The values x=-2 and y=2 are indeed the solution to this system.
|
Geometry Proofs worksheets
Geometry plays an important role in mathematics. Geometry is broadly classified into analytical and practical geometry. In proof, the condition should be solved analytically to show the given condition should be always true, the condition must be true under any circumstances. There are some predefined geometry theorems to show the given condition as a true statement. Let us discuss some geometry proofsworksheets.
Theorem 1 (geometry proofs worksheets):
Angles at a point add to 360o
Proof
In the above figure, the angles make up a full turn, and the sum of all the angles is 360o , so
a + b + c + d + e = 360o
This will fit for any number of angles. We are here proved by using five angles
Theorem 2 (geometry proofs worksheets):
The angle on a straight line is 180o.
Proof:
In the above figure, a and b are angles at a given point both are on the same vertex. So that a + b = 360o, by Theorem 1.
But a = b, so a + a = 360 degrees or 2a = 360 degrees, which gives
a = 180o.
Theorem 3 (geometry proofs worksheets):
The sum of angles on a straight line is 180
Proof:
In the above figure, both a and b are angles on a straight line and on a single vertex, so a+ b = 180o by theorem 2
Theorem 4 (geometry proofs worksheets):
Vertically opposite angles are equal
Proof:
In the above figure, angles a and b make up a straight line.
So, a + b = 180o, by theorem 2.
Angles a and c makes a straight line.
a + c = 180o, by theorem 2
so a + b = a + c
Subtracting a from both sides we get, b = c.
The perpendicular opposite angles b and c are equal.
Similarly,
a + b = 180o (angles on a straight line)
d + b = 180o (angles on a straight line)
so, a + b = d + b
Subtracting b from both sides we get, a = d.
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# FIND EQUATION OF PERPENDICULAR BISECTOR WITH TWO POINTS
A perpendicular bisector of a line segment is a line segment perpendicular to and passing through the midpoint of the line.
To find equation of perpendicular bisector, we follow the steps given below.
Step 1 :
Find the midpoint of the line segment for which we have to find the perpendicular bisector.
Step 2 :
Find the slope of the line segment.
Step 3 :
Find the slope of the perpendicular line using the formula -1/m. Here m is slope of the given line.
Step 4 :
To find equation of the perpendicular line, we use the formula given below.
(y - y1) = -1/m (x - x1)
Write an equation for the perpendicular bisector of the line segment joining
Problem 1 :
A(-3, 4) and B(5, 6)
Solution :
Step 1 :
Midpoint of AB :
Step 2 :
Slope of AB :
Step 3 :
Slope of perpendicular line = -1/m
= -1/(1/4)
= -4
Step 4 :
Equation of the perpendicular bisector of AB
y - y1 = m(x - x1)
y - 5 = -4(x - 1)
y = -4x + 4 + 5
y = -4x + 9
So, equation of perpendicular bisector is y = -4x + 9.
Problem 2 :
A(3, 8) and B(7, 14)
Solution :
Step 1 :
Midpoint of AB :
Step 2 :
Slope of AB :
Step 3 :
Slope of perpendicular line = -1/m
= -1/(3/2)
= -2/3
Step 4 :
Equation of the perpendicular bisector of AB
y - y1 = m(x - x1)
y - 11 = -2/3(x - 5)
3(y - 11) = -2(x - 5)
3y - 33 = -2x + 10
2x + 3y = 10 + 33
2x + 3y = 43
So, equation of perpendicular bisector is 2x + 3y = 43.
Problem 3 :
A(-5, 6) and B(1, 8)
Solution :
Step 1 :
Midpoint of AB :
Step 2 :
Slope of AB :
Step 3 :
Slope of perpendicular line = -1/m
= -1/(1/3)
= -3
Step 4 :
Equation of the perpendicular bisector of AB
y - y1 = m(x - x1)
y - 7 = -3(x + 2)
y - 7 = -3x - 6
3x + y = 7 - 6
3x + y = 1
y = -3x + 1
So, equation of perpendicular bisector is y = -3x + 1.
Problem 4 :
A(-3, -6) and B(-1, 2)
Solution :
Step 1 :
Midpoint of AB :
Step 2 :
Slope of AB :
Step 3 :
Slope of perpendicular line = -1/m
= -1/4
Step 4 :
Equation of the perpendicular bisector of AB
y - y1 = m(x - x1)
y + 2 = -1/4(x + 2)
4(y + 2) = -1(x + 2)
4y + 8 = -x - 2
x + 4y = -8 - 2
x + 4y = -10
So, equation of perpendicular bisector is x + 4y = -10.
## Recent Articles
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2. ### Solving Two Step Inequality Word Problems
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Solving Two Step Inequality Word Problems
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# Difference between revisions of "2021 AIME II Problems/Problem 11"
## Problem
A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$, and a different number in $S$ was divisible by $7$. The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$. Find the sum of all possible values of the greatest element of $S$.
## Solution 1
Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(""+string(i+0.5)+"",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(""+string(i+42.5)+"",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(""+string(i+84.5)+"",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\begin{array}{c||c|c|l} & & & \\ [-2.5ex] \textbf{Max Value} & \boldsymbol{S} & \textbf{Valid?} & \hspace{16.25mm}\textbf{Reasoning/Conclusion} \\ [0.5ex] \hline & & & \\ [-2ex] 49 & \{46,47,48,49\} & & \text{The student who gets } 46 \text{ will reply yes.} \\ 50 & \{47,48,49,50\} & \checkmark & \text{Another possibility is } S=\{89,90,91,92\}. \\ 51 & \{48,49,50,51\} & & \text{The student who gets } 51 \text{ will reply yes.} \\ 56 & \{53,54,55,56\} & & \text{The student who gets } 53 \text{ will reply yes.} \\ 57 & \{54,55,56,57\} & & \text{The student who gets } 57 \text{ will reply yes.} \\ 63 & \{60,61,62,63\} & & \text{The students who get } 60,61,62 \text{ will reply yes.} \\ 66 & \{63,64,65,66\} & & \text{The students who get } 64,65,66 \text{ will reply yes.} \\ 72 & \{69,70,71,72\} & & \text{The student who gets } 69 \text{ will reply yes.} \\ 73 & \{70,71,72,73\} & & \text{The student who gets } 73 \text{ will reply yes.} \\ 78 & \{75,76,77,78\} & & \text{The student who gets } 75 \text{ will reply yes.} \\ 79 & \{76,77,78,79\} & \checkmark & \text{Another possibility is } S=\{34,35,36,37\}. \\ 80 & \{77,78,79,80\} & & \text{The student who gets } 80 \text{ will reply yes.} \end{array}$$ Finally, all possibilities for $S$ are $\{34,35,36,37\}, \{47,48,49,50\}, \{76,77,78,79\},$ and $\{89,90,91,92\},$ from which the answer is $37+50+79+92=\boxed{258}.$
Remarks
1. Alternatively, we can reconstruct the second table in this solution as follows, where Y and N denote the replies of "yes" and "no", respectively. Notice that this table has some kind of symmetry! $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real j=5.5; j<41.5; j+=6) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,yellow); } for (real j=6.5; j<41.5; j+=7) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,green); } fill((4,1)--(8,1)--(8,2)--(4,2)--cycle,mediumgray); fill((33,1)--(37,1)--(37,2)--(33,2)--cycle,mediumgray); fill((42,4)--(41,4)--(41,3)--cycle,yellow); fill((42,4)--(42,3)--(41,3)--cycle,green); for (real i=0.5; i<=41.5; ++i) { label(""+string(i+42.5)+"",(i,3.5),fontsize(9pt)); } draw(circle((6.5,3.5),0.45)); draw(circle((7.5,3.5),0.45)); draw(circle((8.5,3.5),0.45)); draw(circle((13.5,3.5),0.45)); draw(circle((14.5,3.5),0.45)); draw(circle((20.5,3.5),0.45)); draw(circle((23.5,3.5),0.45)); draw(circle((29.5,3.5),0.45)); draw(circle((30.5,3.5),0.45)); draw(circle((35.5,3.5),0.45)); draw(circle((36.5,3.5),0.45)); draw(circle((37.5,3.5),0.45)); label("Y",(3.5,2.5),blue); label("N",(4.5,2.5),blue); label("N",(5.5,2.5),blue); label("N",(6.5,2.5),blue); label("N",(4.5,1.5),blue); label("N",(5.5,1.5),blue); label("N",(6.5,1.5),blue); label("N",(7.5,1.5),blue); label("N",(5.5,0.5),blue); label("N",(6.5,0.5),blue); label("N",(7.5,0.5),blue); label("Y",(8.5,0.5),blue); label("Y",(10.5,2.5),blue); label("N",(11.5,2.5),blue); label("N",(12.5,2.5),blue); label("N",(13.5,2.5),blue); label("N",(11.5,1.5),blue); label("N",(12.5,1.5),blue); label("N",(13.5,1.5),blue); label("Y",(14.5,1.5),blue); label("Y",(17.5,2.5),blue); label("Y",(18.5,2.5),blue); label("Y",(19.5,2.5),blue); label("N",(20.5,2.5),blue); label("N",(20.5,1.5),blue); label("Y",(21.5,1.5),blue); label("Y",(22.5,1.5),blue); label("Y",(23.5,1.5),blue); label("Y",(26.5,2.5),blue); label("N",(27.5,2.5),blue); label("N",(28.5,2.5),blue); label("N",(29.5,2.5),blue); label("N",(27.5,1.5),blue); label("N",(28.5,1.5),blue); label("N",(29.5,1.5),blue); label("Y",(30.5,1.5),blue); label("Y",(32.5,2.5),blue); label("N",(33.5,2.5),blue); label("N",(34.5,2.5),blue); label("N",(35.5,2.5),blue); label("N",(33.5,1.5),blue); label("N",(34.5,1.5),blue); label("N",(35.5,1.5),blue); label("N",(36.5,1.5),blue); label("N",(34.5,0.5),blue); label("N",(35.5,0.5),blue); label("N",(36.5,0.5),blue); label("Y",(37.5,0.5),blue); for (real i=0; i<=42; ++i) { draw((i,4)--(i,3)); } draw((0,4)--(42,4)); draw((0,3)--(42,3)); [/asy]$
2. As a confirmation, we can verify that each student will be able to deduce what $S$ is upon hearing the four replies of "no" in unison. For example, if $S=\{47,48,49,50\},$ then all students will know that no one gets $46$ or $51,$ otherwise that student will reply yes (as discussed). Therefore, all students will conclude that $S$ has only one possibility.
~MRENTHUSIASM
## Solution 2
We know right away that $42\not\in S$ and $84\not\in S$ as stated above. But where to go from here?
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\{11,12,13,14\}$ and $\{12,13,14,15\}$. Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets.
Now we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\{11,12,13,14\}$ and $\{12,13,14,15\}$. The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S. Take for example $\{33,34,35,36\}$, $\{34,35,36,37\}$, and $\{35,36,37,38\}$. 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1. Since 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$; $\{34,35,36,37\}$, $\{47,48,49,50\}$, $\{76,77,78,79\}$ and $\{89,90,91,92.\}$. Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=\boxed{258}$
~KingRavi
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# Find the integral of $\cos 2x$.
Last updated date: 20th Jun 2024
Total views: 414.3k
Views today: 6.14k
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414.3k+ views
Hint: Integration is a way of adding slices to find the whole.
Integration can be used to find areas, volumes, central points and many useful things.
But it is easiest to start with finding the area under the curve of a function like this
$y = f(x)?$
The symbol for ‘integral’ is $'\int '$.
For example
$\int {2x\,dx}$
After the integral symbol we put the function we want to find the integral of and then finish with dx to mean the slice go in the x direction (and approach zero in width)
$\int {2x\,dx = {x^2}} + c$
Where ‘c’ is the ‘constant integration.’
Therefore,
To integrate cos2x, also written as $\sqrt {\cos 2x\,dx}$ and cos2x we usually use a ’u’ substitution to build a new integration in term of u
$\begin{gathered} Let\,u = 2x \\ \dfrac{{du}}{{dx}} = 2 \\ \end{gathered}$
Then $\begin{gathered} du = 2 \\ dx \\ dx = \dfrac{1}{2}du \\ \end{gathered}$
We arrange to get and expression for ‘dx’ in terms of u
Now,
$\int {\cos 2x\,dx = \int {\cos u\dfrac{1}{2}} du}$
We get this by replacing 2x with u and replacing ‘dx’ will ${Y_2}$du
$\int {\cos u\dfrac{1}{2}du} = \dfrac{1}{2}\int {\cos u\,du}$
We move the $\dfrac{1}{2}$ outside of the integral as it is simply a multiplier.
$\int {\operatorname{Cos} u = \operatorname{Sin} u}$
Now we have a simple integration with $\operatorname{Sin} \,u$
$\dfrac{1}{2}\int {\cos \,u\,du = \dfrac{1}{2}\operatorname{Sin} \,u}$
We remember the $\dfrac{1}{2}$ outside the integral sign and reintroduce it here
$= \dfrac{1}{2}\operatorname{Sin} 2x + C$
Hence, this is the final answer where C is the integration constant.
Note: Finding an integral is the reverse of finding a derivative.
In the differential we are given a function and we have to find the derivative or difference of the function, but in the integral, we are to find a function whose differential is given.
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## Algebra: A Combined Approach (4th Edition)
Published by Pearson
# Chapter 5 - Section 5.7 - Dividing Polynomials - Exercise Set: 1
#### Answer
Given $\frac{(12x^{4} + 3x^{2})}{x}$, the answer, simplified, is $12x^{3} +3x$
#### Work Step by Step
Given $\frac{(12x^{4} + 3x^{2})}{x}$, the answer, simplified, is $12x^{3} +3x$ • We can separate the polynomial (top) by its terms to divide it by the monomial (bottom). So, if we separate the polynomial (the quotient, which is also the numerator of the equation), it looks like this: Term #1: $12x^{4}$ Term #2: $3x^{2}$ Adding its divisor, our equation would then look like this: $\frac{12x^{4}}{x}$ $+$ $\frac{3x^{2}}{x}$ • Now we need to simplify each fraction, consisting of the one term of the polynomial and the monomial. And we can further simplify each term into its coefficients and exponents to make it a little easier. So, our equation will then look like this: $(\frac{12}{1})$$(\frac{x^{4}}{x}) + (\frac{3}{1})$$(\frac{x^{2}}{x})$ • Now, let's take the 1st term that we have set up, and divide the coefficients and the exponents. (Remember that when we divide exponents, we are really just subtracting the denominator's exponent from the numerator's exponent.) For $(\frac{12}{1})$$(\frac{x^{4}}{x}), (\frac{12}{1}) =12 and (\frac{x^{4}}{x}) is x^{4-1} =x^{3}, which, combined, can be simplified to 12x^{3} • Now, let's take the 2nd term that we have set up, and divide the coefficients and the exponents. (Remember that when we divide exponents, we are really just subtracting the denominator's exponent from the numerator's exponent.) For (\frac{3}{1})$$(\frac{x^{2}}{x})$, $(\frac{3}{1})$ $=3$ and $(\frac{x^{2}}{x})$ is $x^{2-1}$ $=x^{1}$, which $=x$ Combined, it may be simplified to: $3x$ • Now that we have simplified each polynomial's terms when each is divided by the monomial, our equation of $(\frac{12}{1})$$(\frac{x^{4}}{x}) + (\frac{3}{1})$$(\frac{x^{2}}{x})$ is then further simplified to the final answer of: $12x^{3} +3x$
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## how do you multiply fractions?
I would like to know how to multiply / divide, and add / subtract fractions please
How to add or subtract fractions:
1. Look at the denominators. Are they the same or are they different?
2. If they are different, then that means each fraction is divided into different numbers of equal parts -- so different that you won't know how many equal parts you will end up with when you add or subtract. That's why the denominators have to be the same.
3. To make the denominators the same, you must determine the least common denominator of the two fractions.
4. Then you divide it by the denominator of one fraction to determine by which number to multiply both the numerator and the denominator of that fraction and thereby change it without affecting the value. For example, when you multiply 1/2 by 2/2 (i.e. 1/2 * 2/2), you will get 2/4. However, you will not have a different value because you will have multiplied 1/2 by a fraction that's equivalent to 1 when the numerator is the same as the denominator.
5. Repeat step 4 with the other fraction.
6. Now that the denominators of both fractions are the same, you finally know how many equal parts you will end up with when you add or subtract. So you can perform either one of those operations with the new numerators of both fractions, but you keep the denominator since that denotes the number of equal parts.
How to multiply fractions:
1. You multiply both the numerators and the denominators. If you don't understand how this works, I will explain in further steps with an example.
2. Here's what fraction multiplication looks like: (2/3)*(4/5).
3. 2 is the numerator of the first fraction, and 3 is the denominator of that fraction. The denominator means into how many equal parts the whole is divided, and the numerator means how many of those equal parts are visible. Therefore, 2 out of all 3 equal parts are visible.
4. Now, you can divide the whole into further equal parts by multiplying both the numerator and the denominator by any number you want. That is to say, you can divide each of 3 equal parts into 5 equal parts so that the final number of equal parts of the entire whole will be 3*5=15 to make the final denominator of the fraction product.
5. Now, since only 2 equal parts out of 3 were visible, and each of those 3 equal parts was divided into 5, each of those 2 visible equal parts had to be divided into 5 so that we can take a certain number out of each of the 2 groups of 5 equal parts, and that is 4 according to the numerator of the second fraction. This is how you multiply the numerators of the fractions, so 4*2=8 to make the final numerator of the fraction product.
6. So now the fraction product is 8/15, and in conclusion, you can take a fraction of a fraction.
Another way to explain fraction multiplication is this:
You can multiply both the numerator and denominator of the first fraction by the denominator of the second fraction (e.g. you can multiply both the 2 and 3 in 2/3 by 5, the denominator of the other fraction and get 10/15). Then you can multiply both the numerator and denominator of the other fraction by the numerator of the first fraction (e.g. you can multiply both the 4 and 5 in 4/5 by 2, the numerator of the first fraction and get 8/10). Now, when you multiply (2/3)*(4/5), you can change it to (10/15)*(8/10) without affecting the fractions' values. Now that you have 10/15, you can take 8 out of those 10/15, and your final answer will be 8/15.
How to divide fractions:
1. You keep the first fraction but multiply by the reciprocal of the other fraction. If you don't understand how this works, I will explain with an example in further steps:
2. Here's an example: (1/2) / (3/4).
3. This can mean what fraction of 3/4 is 1/2?
4. It would be a lot easier to see if the divisor were 1 because any number divided by 1 is that number.
5. Let's imagine the fraction divisor, 3/4, as 1 by converting it to that number. First, you take the reciprocal by swapping positions of both the numerator and denominator. So the reciprocal of 3/4 is 4/3. Then you multiply the fraction by its reciprocal to get a product of 1.
6. Now that you have multiplied the fraction divisor by its reciprocal to get 1, you use the latter by which to multiply the fraction dividend as well because all fractions must be multiplied by anything equivalent to 1. That's how you end up multiplying the fraction dividend by the reciprocal of the fraction divisor.
Fractions:
1.You need the least common denominator (commonly referred to as LCD). This means that you want the smallest possible number in the denominator, in which both of the original denominators will divide into.
2. Set your equation up so that you see the LCD on the bottom and simply add the two numerators.
Ex: 2/3 + 8/12 --> 8/12 + 8/12= 16/12 or 1 1/3 (unless the directions say not to, always simplify your answer)
Subtraction:
1. Again, find your least common denominator. Really, any common denominator is acceptable but the LCD helps when you simplify the answer later.
2. Write you equation horizontally and not vertically( like a normal subtraction problem), with the common denominator on bottom and the converted numerators on top.
3. Perform the subtraction accross the top and place the common denominator on bottom.
4. Simplify if possible.
Ex: 4/5 - 1/3 --> 12/15 - 5/15 = 7/15
Multiplication:
1. Don't waste your time with common denominators. Just multiply the numerators together and place the result in the numerator position on the other side of the = sign.
2. Multiply your denominators together and place the result in the denominator position on the other side of the = sign.
3. Simplify if possible.
Ex: 5/12 x 2/3 = 10/36 or 5/18
Division:
1. Always express your work. This will help you keep track of what you did and show that you know what you are talking about.
2. Write you equation horizontally.
3. Take one fraction and invert it. (ex. 4/5 --> 5/4)
4. Then multiply your equations using the same method as above.
Ex: (4/5)/(1/3) = x --> (5/4) * (1/3) = 5/12
That fraction division in that example above was not right. You actually invert only the 2nd fraction and multiply it by the first.
Well if you are adding or subtracting fractions, you want to first get common denomenators (or bottom number of fraction) before you can do so.
Example: 5/4 + 7/8.
We know that in order to make 4 in the first denomenator equal to the 8 in the other denomenator, we have to multiply it by 2. However, the only way to do so without changing the value of the fraction is to do it in a form that equals one. For instance: 2/2 equals 1. So in essence, if I multiply 5/4 by 2/2 I am actually just multiplying it by 1. I am not changing the value of the fraction, I am just making it to where I can add it to 7/8.
So, if I multiply 5/4 by 2/2, what I am doing is saying (2*5)/(2*4). This equals 10/8.
If I know add 10/8 to 7/8 that is saying (10+7)/8. This equals 17/8.
Since we also covered multiplication in the previous example I will skip to division:
When you divide a fraction by another fraction, you are multiplying by its reciprical.
Ex:
(4/5)/(3/2) is the same as saying (4/5)*(2/3). Which would be (4*2)/(5*3) = 8/15
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# HOW TO EXPRESS PRODUCTS OF TRIG FUNCTIONS AS SUM OR DIFFERENCE
## About "How to Express Products of Trig Functions as Sum or Difference"
How to Express Products of Trig Functions as Sum or Difference :
Here we are going to see some example problems to show expressing the product of trig functions as sum or difference.
sin(A + B) = sin A cos B + cos A sin B -------(1)
sin(A − B) = sin A cos B − cos A sin B -------(2)
cos(A + B) = cos A cos B − sin A sin B ------(3)
cos(A − B) = cos A cos B + sin A sin B -------(4)
(1) + (2)
sin (A + B) + sin (A - B) = 2 sin A cos B
(1) - (2)
sin (A + B) - sin (A - B) = 2 cos A sin B
(3) + (4)
cos (A + B) + cos (A - B) = 2 cos A cos B
(3) - (4)
cos (A + B) - cos (A - B) = -2 sin A sin B
## Expressing Products of Trig Functions as Sum or Difference
Question 1 :
Express each of the following as a sum or difference
(i) sin 35°cos 28°
Solution :
= sin 35°cos 28°
Multiply and divide the given trigonometric ratio by 2.
= (2/2) sin 35°cos 28°
= (1/2) (2 sin 35°cos 28°)
It exactly matches the formula 2 sin A cos B
2 sin A cos B = sin (A + B) + sin (A - B)
= (1/2) [sin (35°+28°) + sin (35°-28°)]
= (1/2) [sin 63°sin 7°]
(ii) sin 4x cos 2x
Solution :
= sin 4x cos 2x
Multiply and divide the given trigonometric ratio by 2.
= (2/2) sin 4x cos 2x
= (1/2) (2 sin 4x cos 2x)
It exactly matches the formula 2 sin A cos B
2 sin A cos B = sin (A + B) + sin (A - B)
= (1/2) [sin (4x+2x) + sin (4x-2x)]
= (1/2) [sin 6x sin 2x]
(iii) 2 sin 10θ cos 2θ
Solution :
= 2 sin 10θ cos 2θ
It exactly matches the formula 2 sin A cos B
2 sin A cos B = sin (A + B) + sin (A - B)
= (1/2) [sin (10θ+2θ) + sin (10θ+2θ)]
= (1/2) [sin 12θ sin 8θ]
(iv) cos 5θ cos 2θ
Solution :
= cos 5θ cos 2θ
Multiply and divide the given trigonometric ratio by 2.
= (2/2) cos 5θ cos 2θ
= (1/2) (2 cos 5θ cos 2θ)
It exactly matches the formula 2 cos A cos B
2 cos A cos B = cos (A + B) + cos (A - B)
= (1/2) [cos (5θ + ) + cos (5θ - )]
= (1/2) [cos 7θ + cos 3θ]
(v) sin 5θ sin 4θ.
Solution :
= sin 5θ sin 4θ
Multiply and divide the given trigonometric ratio by 2.
= (-2/-2) sin 5θ sin 4θ
= (-1/2) (-2 sin 5θ sin 4θ)
It exactly matches the formula -2 sin 5θ sin 4θ
-2 sin A sin B = cos (A + B) - cos (A - B)
= (-1/2) [cos (5θ + 4θ) - cos (5θ - 4θ)]
= (-1/2) [cos 9θ - cos θ]
= (1/2)[cos θ - cos 9θ]
After having gone through the stuff given above, we hope that the students would have understood, "How to Express Products of Trig Functions as Sum or Difference"
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# Little Gauss and the Dividing Time Puzzler
I gave this problem to the Precalc Honors class, right before their lunch (the block is split into two parts). A couple of them came back with the solution, but most had chosen to eat and socialize (how dare they!). An epiphany hit me when I saw how a student had started to solve the problem. He added 1 + 2 + 3 + 4 + 5 + … + 11 + 12 to find out if the sum is divisible by 3 or 4.
My brain found the link to one of my very favorite math tricks / math stories: Little Gauss finding the sum of all the integers from 1 to 100.
So I put the following up on the board:
`S = 1 + 2 + 3 + 4 + 5 + ... + 98 + 99 + 100. Find S.`
Naturally a couple of the kids take the calculators out, so I go over and swipe the calculator out of their hands. I go through the whole story: An elementary teacher needed some quiet time to get stuff done, so the teacher asks the class to add all the integers between 1 and 100. Little Carl Gauss walks up in under a minute with the answer. Teacher is losing patience with this know-it-all, and asks him to add all the integers between 1 and 1000. Gauss strolls up in under a minute with the answer again. (Yes I dramatize the story. Sue me.)
## So how’d he do it?
“You guys agree that I can write the numbers backwards and the sum will be the same right?”
```S = 1 + 2 + 3 + 4 + 5 + ... + 98 + 99 + 100 S = 100 + 98 + 97 + 96 + 95 + ... + 3 + 2 + 1```
What about if I add from top to bottom, what do I get?
`2S = 101 + 101 + 101 + 101 + 101 ... + 101 + 101`
How many 101’s is that?
`2S = 101 x 100`
`S = 5,050`
Go on to ask them what is the sum of all integers between 1 to 1,000 and find the following sum: 2 + 5 + 8 + 11 + … + 98 + 101.
So back to the puzzler. Find the sum of the clock digits:
`1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 `(then reverse them)
`12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1`
How many pairs (after you divide by the 2)?
6 pairs.
So how many divisions of the clock by the two lines?
4 divisions wouldn’t work, so 3 divisions.
What are the divisions? Lets keep the pairs together so the sums are equal in each section.
Neat:
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# What is the rule for dividing by 100?
## What is the rule for dividing by 100?
To divide by 100, move each digit two place value columns to the right. If the number ends in two ‘0’ digits in the tens and units columns, dividing by 100 has the same effect as removing these digits. To divide by 100, move all digits two place value columns to the right.
## How do you divide decimals step by step?
Dividing Decimals
1. Step 1: Estimate the answer by rounding .
2. Step 2: If the divisor is not a whole number, then move the decimal place n places to the right to make it a whole number.
3. Step 3: Divide as usual.
4. Step 4: Put the decimal point in the quotient directly above where the decimal point now is in the dividend.
How do you divide dividends?
Divide the dividend by the whole-number divisor to find the quotient.
1. Multiply the divisor by a power of 10 to make it a whole number.
2. Multiply the dividend by the same power of 10. Place the decimal point in the quotient.
3. Divide the dividend by the whole-number divisor to find the quotient.
How do you solve 100 divided by 5?
100 divided by 5 is 20.
### How do you work out 18 divided by 100?
Using a calculator, if you typed in 18 divided by 100, you’d get 0.18. You could also express 18/100 as a mixed fraction: 0 18/100.
### When dividing numbers by 100 mentally what do we do with the decimal point of the dividend?
If you divide any whole number by 10, copy the dividend and make it have one decimal digit. If you divide any whole number by 100, copy the dividend and make it have two decimal digits.
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# Question Video: Finding the Distance Covered by a Body on a Rough Inclined Plane until It Comes to Rest Mathematics
A body of mass 74 kilograms was projected at 8.5 m/s along the line of greatest slope up a plane inclined at 30° to the horizontal. Given that the resistance of the plane to its motion was 7.4 N, find the distance the body traveled until it came to rest. Take 𝑔 = 9.8 m/s².
04:46
### Video Transcript
A body of mass 74 kilograms was projected at 8.5 meters per second along the line of greatest slope up a plane inclined at 30 degrees to the horizontal. Given that the resistance of the plane to its motion was 7.4 newtons, find the distance the body traveled until it came to rest. Take 𝑔 to equal 9.8 meters per second squared.
Alright, so let’s say that this is our slope with its inclination angle of 30 degrees. And on this slope, we have a body being projected along the slope line at 8.5 meters per second. As the body moves up the slope, it encounters a frictional force we can call 𝐹 with a given magnitude of 7.4 newtons. We’re told further that the body has a mass we’ll call 𝑚 of 74 kilograms. And knowing that the body eventually comes to a rest, we want to calculate the distance along the slope, we’ll call it 𝑑, that it takes to do this.
As we get started, let’s clear a bit of space on screen and recall that Newton’s second law of motion tells us that the net force acting on a body is equal to its mass times its acceleration. Considering an up-close view of our body as it moves up the incline, let’s draw a free-body diagram showing all the forces acting on it. We know that the body experiences a weight force, its mass times the acceleration due to gravity, a normal or reaction force perpendicular to the surface of the plane. And because of the body is moving uphill, it experiences a frictional force downhill. These are the forces that act on our body as it rises up the incline.
And let’s say that, directionally, the positive 𝑥-direction points down the incline and the positive 𝑦-direction points perpendicularly away from it. Applying Newton’s second law to this scenario, we can say that the sum of forces in the 𝑥-direction equals our body’s mass times its acceleration in this direction. These forces include the frictional force 𝐹, as well as a component of the weight force that acts in the 𝑥-direction.
To solve for this component, we need to realize that this angle in our right triangle here is identical to the 30-degree angle of inclination of our plane. Because that’s true, the component of the weight force we’re interested in is 𝑚 times 𝑔 times the sin of 30 degrees. The sin of 30 degrees is one-half. So, when we write out all the forces acting on our body in the 𝑥-direction, we have 𝐹 plus 𝑚 times 𝑔 divided by two. The second law tells us that this sum equals our body’s mass times its acceleration in-in this case, the 𝑥-direction. Dividing both sides of this equation by 𝑚, we find that factor cancelling on the right. And we find that 𝑎 sub 𝑥 is equal to 𝐹 plus 𝑚 times 𝑔 over two all divided by the mass 𝑚.
Since we’re given the force 𝐹, the mass 𝑚, and we also know that the constant 𝑔 is equal to 9.8 meters per second squared, we can substitute those values into this equation to solve for 𝑎 sub 𝑥. When we enter this expression on our calculator, we find a result of exactly five with units of meters per second squared. This means that as our body moves up the incline, it decelerates at 5 meters per second squared. And this will help us solve for the distance 𝑑 it takes to come to a rest.
Because our body experiences a constant acceleration, that means its motion can be described by what are called the equations of motion. These are sometimes also called the kinematic equations or SUVAT equations. And in our case, we’ll look at a specific equation of motion that tells us that the final velocity of a body squared equals its original velocity squared plus two times its acceleration multiplied by its displacement. Rearranging this equation to solve for 𝑑, it’s equal to 𝑣 sub 𝐹 squared minus 𝑣 sub zero squared all over two 𝑎.
And as we consider our scenario with respect to these variables, we know that 𝑣 sub 𝐹 is zero because our body ends up at rest. We’re told the value of what’s called here 𝑣 sub zero. That’s 8.5 meters per second. And we’ve just recently solved for the acceleration 𝑎. If we plug in these values and leave out units for now, we get this expression. And notice that it’s a negative value. The reason for this is that, technically, we’re solving for displacement, which, according to our sign convention, where positive values are down the incline is actually truly a negative value.
What we actually want to solve for, though, is the distance that our body travels before it comes to rest. To calculate that in this case, we just need to take the absolute value of this fraction. Entering this expression on our calculator, we get a result of 7.225. The units for this distance are meters. So, we can say that this body moves 7.225 meters up this incline before it comes to a stop.
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## How to Find the Distance Between Any Two Places
written by: vishalseafarerโขedited by: Lamar Stonecypherโขupdated: 7/31/2011
Finding the distance between your house to your friend's house in the next block is always easy, but the approach that you use to calculate the distance between two places holds true only to a certain extent. So what's the right approach?
• slide 1 of 6
Today, we have GPS, internet, and other gadgets and tools that will help us calculate distances between one point and another. But imagine the situation nearly two thousand years back. How would you have done it?
• slide 2 of 6
Eratosthenes, a greek mathematician, did just that. He came up with a way and found the circumference of the Earth. In other words, the distance between the point where he started (let us call it point A), all the way around the world, back to the point (point A) where he started. He found this using basic mathematics, but with a very interesting approach. Hunderds of years later, it was found that Eratosthenes was very close to the real value that was found using modern tools and equipment.
• slide 3 of 6
### The Approach Used to Calculate the Distance Between Two Points on Earth
All of us would have learned in our school days that the shortest distance between two points is a straight line. Hence, if you have to find the distance between two points, you will draw a straight line and measure the distance of the straight line between those points. However, this rule applies only for planar surfaces.
Let's take two reference points- point A and point B- separated by a large distance, for easier understanding of the concept.
What will you do if you have to find the distance between two points on Earth? In other words, what approach will you use to calculate the distance between two cities?
Will you stand in a place (point A) and measure the distance to point B by walking in a straight line? You will get the wrong values if you are going to use that approach.That is because the shortest distance on a sphere will appear as a curve on the surface of the sphere and not a straight line. Here is a visual guide on the shortest distance between two points on a sphere.
The straight line approach (in mathematical terms it is called Euclidean Geometry) is much different from the spherical geometry. Given the fact that the Earth is a huge sphere, the approach of using a straight line between two points to calculate the shortest distance will hold true only to a certain distance (ten or twenty kilometers), above which you will be having errors that will affect your further calculations.
• slide 4 of 6
### How to Calculate the Distance Between Two Points on Earth
The great circle distance or the orthodromic distance is the shortest distance between two points on a sphere (or the surface of Earth). In order to use this method, you need to have the co-ordinates of point A and point B. (The great circle method is chosen over other methods because it provides a greater accuracy.)
First, convert the latitude and longitude values from decimal degrees to radians. For this divide the values of longitude and latitude of both the points by 180/pi. The value of pi is 22/7. The value of 180/pi is approximately 57.29577951. If you want to calculate the distance between two places in miles, use the value 3,963, which is the radius of Earth. If you want to calculate the distance between two places in kilometers, use the value 6,378.8, which is the radius of Earth.
Find the value of the latitude in radians:
Value of Latitude in Radians, lat = Latitude / (180/pi) OR
Value of Latitude in Radians, lat = Latitude / 57.29577951
Find the value of longitude in radians:
Value of Longitude in Radians, long = Longitude / (180/pi) OR
Value of Longitude in Radians, long = Longitude / 57.29577951
Get the co-ordinates of point A in terms of latitude and longitude. Use the above conversion method to convert the values of latitude and longitude in radians. I will call it as lat1 and long1. Do the same for the co-ordinates of Point B and get lat2 and long2.
Now, to get the distance between point A and point B use the following formula:
Distance, d = 3963.0 * arccos[(sin(lat1) * sin(lat2)) + cos(lat1) * cos(lat2) * cos(long2 - long1)]
The obtained distance, d, is in miles. If you want your value to to be in units of kilometers, multiple d by 1.609344.
d in kilometers = 1.609344 * d in miles
Thus you can have the shortest distance between two places on Earth using the great circle distance approach.
• slide 5 of 6
### Tools Used to Find the Distance Between Two Places
Here are a few very reliable tools available on the internet that will help you to easily find the distance between two places.
2) Bing Maps.
There are many other tools available on the internet. Feel free to explore them.
• slide 6 of 6
### References
1) Wikipedia - Eratosthenes
2) Meridian World Data - How to Calculate the Distance Between Two Points on Earth
3) Wikipedia - Great Circle Distance
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# The Multiplication Principle of Equations
Imagine again our fulcrum and lever from the previous section:
Instead of adding a third shape, lets add another square to the left side of the fulcrum. What happens?
Similarly to adding the circle, it will fall to the left. But if we add a triangle to the right, since the triangle is equal to the square, we reach equilibrium again:
In this situation, we have multiplied each side by two. This brings about a similar, equally important property of equations:
The Multiplication Principle of Equations
The Multiplication Principle of Equations states that if you multiply both sides of an equation by the same value, you maintain the equation. Mathematically, this is:
If
Then
This property is particularly useful in equations such as . We saw that in the additive property, we could also subtract numbers from each side. In the multiplication property, we may also divide numbers from each side. For this section, we will primarily focus on the fact that we can divide each side by the same number. We will revisit this property again in a later section for multiplication specific application.
Solve the equation
Let’s divide each side by 3:
Solve
In many applications, we will need to use the Addition and Multiplication principles together:
Solve:
First use the addition property to isolate the variable term:
Then use the multiplication principle:
And we might also need to distribute and simplify as well:
Solve:
Solve:
Solve:
Solve:
Here, we can divide each side by the fraction , but recall that dividing by a fraction is the same as multiplying by its reciprocal. Thus, we can multiply both sides by the fraction :
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Solving Simultaneous Linear Equations by Matrix Method
Find the value of $\displaystyle m$ by matrix method for which the simultaneous equations $\displaystyle 3x + my = 5$ and $\displaystyle (m + 2) x + 5y = m$ have (i) an infinite number of solutions (ii) no solution.
Solution
$\displaystyle \left. \begin{array}{l}3x+my=5\\(m+2)x+5y=m\end{array} \right\}\ \ ---------(1)$
$\displaystyle \text{Transforming into matrix form},$
$\displaystyle \left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right)------(2)$
$\displaystyle \text{Let }A=\left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right),X=\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ \text{and }B=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right).$
$\displaystyle \text{Then we have, }AX=B.$
$\displaystyle \text{If }\det A\ne 0,\text{then }{{A}^{{-1}}}\ \text{exists}\text{.}$
$\displaystyle {{A}^{{1}}}AX={{A}^{{-1}}}B$
$\displaystyle IX={{A}^{{-1}}}B$
$\displaystyle X={{A}^{{-1}}}B$ $\displaystyle \text{which is a unique solution for }X.$
$\displaystyle \begin{array}{l}\text{It is given that }X\text{ has}\\\text{(i) an infinite number of solutions and}\\\text{(ii) no solution}\text{.}\end{array}$
$\displaystyle \text{It means that the system has no unique solution}\text{.}$
$\displaystyle \text{Thus, }\det A=0.$
$\displaystyle \therefore 15-2m-{{m}^{2}}=0$
$\displaystyle \therefore (5+m)(3-m)=0$
$\displaystyle \therefore m=-5\ \text{or }m=3$
$\displaystyle \text{When }m=-5,\ \text{System (1) becomes}$
$\displaystyle 3x-5y=5\text{ and }-3x+5y=-5$
$\displaystyle \begin{array}{l}\therefore \text{The two equations represents the same }\\\text{straight line and there will be infinite }\\\text{number of solutions}\text{.}\end{array}$
$\displaystyle \text{When }m=3,\ \text{System (1) becomes}$
$\displaystyle 3x+3y=5\text{ and }5x+5y=-3$
$\displaystyle \begin{array}{l}\therefore \text{The two lines are parallel and }\\\text{there will be no solution}\text{.}\\\text{ }\end{array}$
|
# Number Systems
TYPES OF NUMBER SYSTEMS
In the world of electronics math is very important. In order to know how to calculate voltage, current, power, resistance, and much more you must be able to use different types of math.
In these tutorials we will cover the basics that will help to introduce you to numbering systems that will help you in your future electronics projects.
Until now, you have probably used only one system, the decimal system. You may also be familiar with the Roman numeral system, even though you seldom use it.
THE DECIMAL NUMBER SYSTEM
In this section you will be studying modern systems. You should realize that these systems have certain things in common. These common terms will be defined using the decimal system as our base. Each term will be related to each system as that system is introduced.
Each of the number systems you will study is built around the following components: the UNIT, NUMBER, and BASE (RADIX).
Unit and Number
The terms unit and number when used with the decimal system are almost self-explanatory. By definition the unit is a single object; that is, an apple, a dollar, a day. A number is a symbol representing a unit or a quantity. The figures 0, 1, 2, and 3 through 9 are the symbols used in the decimal system. Thesesymbols are called Arabic numerals or figures. Other symbols may be used for different systems.
For example, the symbols used with the Roman numeral system are letters - V is the symbol for 5, X for 10, M for 1,000, and so forth. We will use Arabic numerals and letters in the number system discussions in this section.
The base, or radix, of a number system tells you the number of symbols used in that system. The base of any system is always expressed in decimal numbers. The base, or radix, of the decimal system is 10. This means there are 10 symbols - 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 - used in the system.
A system using three symbols - 0, 1, and 2 - would be base 3; four symbols would be base 4; and so forth. Remember to count the zero or the symbol used for zero when determining the number of symbols used in a system.
The base of a number system is indicated by a subscript (decimal number) following the value of the number. The following are examples of numerical values in different bases with the subscript to indicate the base:
7592 base 10
214 base 5
123 base 4
656 base 7
You should notice the highest value symbol used in a number system is always one less than the base of the system. In base 10 the largest value symbol possible is 9; in base 5 it is 4; in base 3 it is 2.
Positional Notation and Zero
You must observe two principles when counting or writing quantities or numerical values. They are the POSITIONAL NOTATION and the ZERO principles.
Positional notation is a system where the value of a number is defined not only by the symbol but by the symbol’s position. Let’s examine the decimal (base 10) value of 427.5. You may know from experience that this value is four hundred twenty-seven and one-half. Now examine the position of each number:
If 427.5 is the quantity you wish to express, then each number must be in the position shown. If you exchange the positions of the 2 and the 7, then you change the value.
You can see that the power of the base is multiplied by the number in that position to determine the value for that position.The following graph illustrates the progression of powers of 10:
All numbers to the left of the decimal point are whole numbers, and all numbers to the right of the decimal point are fractional numbers. A whole number is a symbol that represents one, or more, complete objects, such as one apple or \$5. A fractional number is a symbol that represents a portion of an object,such as half of an apple (.5 apples) or a quarter of a dollar (\$0.25). A mixed number represents one, or more, complete objects, and some portion of an object, such as one and a half apples (1.5 apples).
When you use any base other than the decimal system, the division between whole numbers and fractional numbers is referred to as the RADIX POINT. The decimal point is actually the radix point of the decimal system, but the term radix point is normally not used with the base 10 number system.
Just as important as positional notation is the use of the zero. The placement of the zero in a number can have quite an effect on the value being represented. Sometimes a position in a number does not have a value between 1 and 9. Consider how this would affect your next paycheck.
If you were expecting a check for \$605.47, you wouldn’t want it to be \$65.47. Leaving out the zero in this case means a difference of \$540.00. In the number 605.47, the zero indicates that there are no tens. If you place this value on a bar graph, you will see that there are no multiples of 101.
Most Significant Digit and Least Significant Digit (MSD and LSD)
Other important factors of number systems that you should recognize are the MOST SIGNIFICANT DIGIT (MSD) and the LEAST SIGNIFICANT DIGIT (LSD).
The MSD in a number is the digit that has the greatest effect on that number.
The LSD in a number is the digit that has the least effect on that number.
Look at the following examples:
You can easily see that a change in the MSD will increase or decrease the value of the number the greatest amount. Changes in the LSD will have the smallest effect on the value.
The nonzero digit of a number that is the farthest LEFT is the MSD, and the nonzero digit farthest RIGHT is the LSD, as in the following example:
In a whole number the LSD will always be the digit immediately to the left of the radix point.
|
# Find the general solution of the differential equation? y dy/dx-2e^x=0
Apr 15, 2018
$y = \pm 2 \sqrt{{e}^{x} + A}$
#### Explanation:
We wish to solve the differential equation $y \cdot y ' - 2 {e}^{x} = 0$.
This is a separable, first-order ordinary differential equation. As such, it can be solved using techniques suitable for separable 1st-order ODE's.
The most straightforward technique is to get our equation in the form $f \left(y\right) \mathrm{dy} = g \left(x\right) \mathrm{dx}$. We can then integrate both sides, ridding ourselves of the $y '$ term.
We get our function into this form as such:
$y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {e}^{x} = 0$
$y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {e}^{x}$
$\left(y\right) \mathrm{dy} = \left(2 {e}^{x}\right) \mathrm{dx}$
See that our left-hand side is a function of just $y$ and our right-hand side a function of just $x$. Integrate both sides.
$\int \left(y\right) \mathrm{dy} = \int \left(2 {e}^{x}\right) \mathrm{dx}$
$\frac{1}{2} {y}^{2} = 2 {e}^{x} + C$
${y}^{2} = 4 {e}^{x} + 2 C$
$y = \pm \sqrt{4 {e}^{x} + 2 C} = \pm 2 \sqrt{{e}^{x} + \frac{C}{2}}$
Since $C$ is an arbitrary constant, let $A = \frac{C}{2}$. Then our final answer is $y = \pm 2 \sqrt{{e}^{x} + A}$.
|
# Section 1.5 - Exploring Angle Pairs
Section 1.5 Exploring Angle Pairs 1.5 Exploring Angle Pairs Objective: Students will be able to: identify special angle pairs and use their relationships to find angle measures
1.5 Exploring Angle Pairs Vocabulary: Key Vocabulary adjacent angles vertical angles complementary angles supplementary angles linear pair angle bisector 1.5 Exploring Angle Pairs
Special angle pairs can help you identify geometric relationships. We use them to find angle measures. 1.5 Exploring Angle Pairs Problem 1: Use the diagram at the right. Is the statement true? Explain a.
b.
a.
1800 c.
Problem 2: What can you conclude from the information in the diagram? 1.5 Exploring Angle Pairs Problem 2 Solution: < 1 and < 2 are congruent. This means they have the same measurement in degrees. < 1 + < 2 are congruent to < 4.
This means they have the same measurement in degrees. They are also opposites which means they are vertical angles. < 3 and < 5 are vertical and congruent. 1.5 Exploring Angle Pairs Problem 2b: Can you make each conclusion from the information in the diagram? Explain.
a. b. c. d. Segment TW is congruent to Segment WV Segment PW is congruent to Segment WQ
Problem 2b Solutions: Can you make each conclusion from the information in the diagram? Explain. a. Segment TW is congruent to Segment WV Since they have a red mark , YES they are congruent. b. Segment PW is congruent to Segment WQ We cannot conclude this. No marks are given and we do not know the length. c.
d. Segment TV bisects Segment PQ No. Segment PQ is bisecting segment TV. 1.5 Exploring Angle Pairs A linear pair is a pair of adjacent angles whose noncommon sides are opposite rays. The angles of a linear pair form a straight angle. Remember a straight line = 1800 So supplementary angles = 1800 1.5 Exploring Angle Pairs
Problem 3:
1. Make a sketch (drawing) 2. Label 2x + 24 3. Use algebra to solve. (2x + 24) + (4x + 36) = 180 6x + 60 = 180 6x = 120 x = 20 K
L 4x + 36 P J 4. Plug in the variable and find the measure of the angle. < KPL = (2(20) + 24) = 40 + 24 = 64 < JPL = (4(20) + 36) = 80 + 36 = 116 1.5 Exploring Angle Pairs
An angle bisector is a ray that divides an angle into two congruent (same degree) angles. Its endpoint is a the angle vertex. Within the ray, a segment with the same endpoint is also an angle bisector. The ray or segment bisects the angle. In the diagram, Ray AY is the angle bisector of
Ray AC bisects
58 D 3. Use algebra to solve. Since AC bisects (cuts in half) ADB, we know that
A 1.5 Exploring Angle Pairs Lesson Check 1.5 Exploring Angle Pairs Lesson Check 1.5 Exploring Angle Pairs
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# Texas Go Math Grade 1 Lesson 4.2 Answer Key Use Model Adding To
Refer to our Texas Go Math Grade 1 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 1 Lesson 4.2 Answer Key Model Adding To.
## Texas Go Math Grade 1 Lesson 4.2 Answer Key Model Adding To
Essential Question
How can you model odding to, or joining to, a group?
Explanation:
With the help of blocks we can add
Joined the cubes for addition
Explore
Use to show adding to, or joining.
Draw to show your work.
For The Teacher
• Read the following problem. Have children use connecting cubes to model the problem, and draw to show their work. There are 6 children on the playground. 2 more children join them. How many children are on the playground?
Explanation:
6 + 2 = 8
8 children are on the playground
Math Talk
Mathematical Processes
Explain how you use cubes to find your answer.
Explanation:
Joined the cubes for addition
Model and Draw
Explanation:
5 + 2 = 7
The sum of 5 and 2 is 7
Share and Show
Use to show adding to, or joining. Draw the . Write the sum.
Question 1.
3 cats and 1 more cat
3 + 1 = ___
Explanation:
The sum of 3 and 1 is 4
3 + 1 = 4
Question 2.
2 birds and 3 more birds
2 + 3 = __
Explanation:
The sum of 2 and 3 is 5
2 + 3 = 5
Question 3.
4 bugs and 4 more bugs
4 + 4 = ___
Explanation:
The sum of 4 and 4 is 8
4 + 4 = 8
Question 4.
4 fish and 2 more fish
4 + 2 = ___
Explanation:
The sum of 4 and 2 is 6
4 + 2 = 6
Problem Solving
Use to show adding to, or joining. Draw the Write the sum.
Question 5.
5 bees and 4 more bees
5 + 4 = ___
Explanation:
The sum of 5 and 4 is 9
5 + 4 = 9
Question 6.
4 dogs and 3 more dogs
4 + 3 = ___
Explanation:
The sum of 4 and 3 is 7
4 + 3 = 7
Write or draw to solve.
Question 7.
Molly puts 3 grapes in the bowl. Then she puts 5 more grapes in the bowl. How many grapes are in the bowl?
____ grapes
Explanation:
3 +5 = 8
8 grapes are in the bowl
Question 8.
H.O.T. Multi-Step Julia has 4 books on the table. She puts 1 more book on the table. Then she puts 2 more BSra books on the table. How many books are on the table?
____ books
Explanation:
4 + 1 + 2 = 7
7 books on the table
Question 9.
Chase sees 4 frogs. 2 frogs are green. 2 frogs are brown. Then he sees 1 more brown frog. How many brown frogs does he see?
___ brown frogs
Answer: 3 brown frogs
Explanation:
2 + 1 = 3
3 brown frogs that he see
Choose the correct answer.
Question 10.
There are 6 zebras eating grass. Then 3 more zebras join them. How many zebras are eating grass?
(A) 8
(B) 9
(C) 10
Explanation:
6 + 3 = 9
9 zebras are eating grass
Question 11.
Use Tools Which number sentence does the picture show?
(A) 3 + 1 = 4
(B) 3 + 5 = 8
(C) 5 + 5 = 10
Explanation:
3 + 5 = 8
is the sentence that the picture show
Question 12.
Annie picks 5 flowers. Then she picks 2 more. How many flowers does Annie have?
(A) 7
(B) 9
(C) 8
Explanation:
5 + 2 = 7
7 flowers that Annie have
Question 13.
Texas Test Prep Use to show adding to, or joining. What is the sum of 4 and 3?
(A) 1
(B) 7
(C) 6
Explanation:
4 + 3 = 7
the sum of 4 and 3 is 7
Take Home Activity
• Put 3 pennies in one group and 2 pennies in another group. Have your child write an addition sentence to tell about the pennies. Repeat for other combinations of pennies with sums up to 10.
### Texas Go Math Grade 1 Lesson 4.2 Homework and Practice Answer Key
Use to show adding to. Draw the . Write the sum.
Question 1.
1 monkey and 5 more monkeys
1 + 5 = ___
Explanation:
The sum of 1 and 5 is 6
1 + 5 = 6
Question 2.
3 camels and 6 more camels
3 + 6 = ___
Explanation:
The sum of 3 and 6 is 9
3 + 6 = 9
Problem Solving
Write or draw to solve.
Question 3.
Kevin throws the ball to his dog 4 times. Then he throws the ball 4 more times. How many times does he throw the ball?
___ times
Explanation:
4 + 4 = 8
8 times that he throw the ball
Question 4.
Multi-Step Emily buys 5 shirts. 2 are pink. 3 are blue. Then she buys 2 more shirts. They are red. How many shirts does she buy?
___ shirts
Explanation:
5 + 2 = 7 shirts she buy
Lesson Check
Texas Test Prep
Choose the correct answer.
Question 5.
There are 8 children reading. Then I more child joins them. How many children are reading?
(A) 7
(B) 10
(C) 9
Explanation:
8 + 1 = 9
9 children are reading
Question 6.
Which number sentence does the picture show?
(A) 4 + 3 = 7
(B) 4 + 2 = 6
(C) 4 + 5 = 9
Explanation:
4 + 2 = 6
Is the number sentence that the picture show
Question 7.
What is the sum of 3 and 6?
(A) 3
(B) 9
(C) 8
|
# Decimals
Decimals are used to denote non-integer, ‘partial’ quantities. Decimals are directly related to fractions and percentages as shown below.
$\frac{1}{2}=0.5=1/2=50\%$.
When working with decimals using a calculator, you can enter and treat them just like integers; the same rules of addition, subtraction, multiplication, and division all apply. Since calculators are permitted on the COMPASS Test, it is not necessary for you to practice multiplying and dividing decimal numbers by hand.
To convert a fraction into a decimal using your calculator just preform the division implied by the fraction. For example:
$\frac{5}{8}=5/8=0.625$
One time-saving trick to know about decimals involves multiplying or dividing by powers of ten. If you multiply any number by 10, just move the decimal place once to the right. Examples:
$123\cdot10=1,230 \qquad 315.87\cdot10=3,158.7$
If you divide by 10, move the decimal place once to the left:
$123/10=12.3 \qquad 315.87/10=31.587$
If you multiply by 100, you are multiplying by 10 twice since $10\cdot10=100$, so you move the decimal place two spots to the right:
$315.87\cdot100=31,587$.
This rule applies when multiplying and dividing by any powers of ten (10,100,1000, 10000,…).
Because of the exclusive relationship 10 has to our base 10 numbering system, the decimal places are each referred to by their relationship to the powers of ten. For the number: 31.587, 5 is in the tenths place, 8 is in the hundredths place, and 7 is in the thousandths place.
If you were asked to round 31.587 to the nearest hundredth you would round it to 31.59. You essentially chop off everything past the hundredths place, and adjust the number in the hundredths place up 1 if the number in the thousandths place is greater than or equal to 5.
Practice Questions
Use a calculator as needed. Round to the nearest hundredth.
Convert the following fractions to decimal numbers:
1.) $\frac{5}{7}$
2.) $\frac{7}{5}$
3.) $3 \frac{3}{5}$
4.) $\frac{13}{4}$
Solve:
5.) $12.41\cdot3.5$
6.) $211.7/19.4$
7.) Paul has 12.75 cups of rice. His wife Susan brings home $2\frac{3}{4}$ cups from the store. They then eat 5.8 cups of rice during the week. How much rice do they have left?
8.) Mick buys 4 movie tickets for $7.25 each. He then buys 2 sodas for$3.50 each, 1 popcorn for $2.65, and 3 boxes of candy for$1.89 each. How much money has Mick spent?
|
# Anyone can explain to me what's the difference between "limit", "limsup" and "liminf" of a function? It would be helpful to explain with concrete example.
##### 1 Answer
Feb 15, 2017
I'll try to give an example below.
#### Explanation:
Example1 :
$f \left(x\right) = \sin \left(\frac{1}{x}\right)$ as $x \rightarrow 0$
Every deleted $\epsilon$ ball around $0$ has supremum $1$, so
${\lim}_{x \rightarrow 0} \text{sup} f \left(x\right) = 1$
Every deleted $\epsilon$ ball around $0$ has infimum $- 1$, so
${\lim}_{x \rightarrow 0} \text{inf} f \left(x\right) = - 1$
As we know ${\lim}_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$ does not exist.
Example 2:
$g \left(x\right) = x \sin \left(\frac{1}{x}\right)$ as $x \rightarrow 0$
Every deleted $\epsilon$ ball around $0$ has supremum $\epsilon$, so
${\lim}_{x \rightarrow 0} \text{sup} f \left(x\right) = {\lim}_{\epsilon \rightarrow 0} \epsilon = 0$
Every deleted $\epsilon$ ball around $0$ has infimum $- \epsilon$, so
${\lim}_{x \rightarrow 0} \text{inf} f \left(x\right) = {\lim}_{\epsilon \rightarrow 0} - \epsilon = 0$
We know that ${\lim}_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0$, for two reasons.
We can use the squeeze theorem on the left and right to get the result.
If we have lim sup = lim inf, then that value is also the limit.
|
## About "How to find the missing coordinates of a triangle"
Here we are going to see how to find the missing coordinates of a triangle.
We may follow the steps given below to get the missing coordinate.
Step 1 :
Take the given points as (x1, y1) (x2, y2) and (x3, y3).
Step 2 :
Use the formula for area of triangle and apply the above values.
Step 3 :
Equate them to the given area, and solve for unknown.
Example 1 :
Find the value of "k" for which the given points are collinear.
(k, -1) ( 2, 1) and (4, 5)
Solution :
If the given points are collinear then the area of triangle is zero
(1/2) [(k + 10 – 4) – (-2 + 4 + 5k)] = 0
Multiply by 2 on both sides,
(k + 6) – (2 + 5k) = 0
(k + 6 – 2 - 5k) = 0
-4 k + 4 = 0
-4k = -4
k = (-4)/(-4)
k = 1
Hence the value of k is 1
Example 2 :
Find the value of "k" for which the given points are collinear.
(2, -5) (3, -4) and (9, k)
Solution :
If the given points are collinear then the area of triangle is zero.
(1/2) [(-8 + 3k – 45) – (-15 - 36 + 2k)] = 0
Multiply by 2 on both sides,
(3k - 53) – (-51+ 2k) = 0 x 2
(3k - 53 + 51- 2k) = 0
k - 2 = 0
k = 2
Hence, the value of k is 2.
Example 3 :
Find the value of "k" for which the given points are collinear.
(k, k) (2, 3) and (4, -1)
Solution :
If the given points are collinear then the area of triangle is zero.
(1/2) [(3k - 2 + 4k) – (2k + 12 - k)] = 0
Multiply by 2 on both sides
[(7k - 2) – (k+12)] = 0 x 2
(7k - 2 – k - 12) = 0
6 k - 14 = 0
6k = 14
k = 14/6
k = 7/3
Hence, the value of k is 7/3
Example 4 :
Vertices of the triangle taken in order and its area is 17 square units, find the value of a.
(0, 0) (4, a) and (6, 4)
Solution :
If the given points are collinear then the area of triangle is zero.
Area of triangle = 17 sq.units
17 = (1/2) [ (0 + 16 + 0) – (0 + 6 a + 0)]
(1/2)(16 – 6 a) = 17
(1/2) x 2 (8 - 3a) = 17
8 – 3 a = 17
-3a = 17 – 8
-3a = 9
a = 9/(-3)
a = -3
Hence, the value of a is -3.
After having gone through the stuff given above, we hope that the students would have understood "How to find the missing coordinates of a triangle"
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# Angle relationships between intersecting lines and parallel lines example
We are given that the angle EDG is equal to α.
Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles EDG and CDE are supplementary and EDG+ CDE=180°.
The angle CDE=180°-α.
Two parallel lines combined with two intersecting lines form 2 similar triangles. We are given that the lines HI and DE are parallel, therefore the triangles ABC and CDE are similar and the angles CDE and ABC are equal and their value is 180°-α (we found that CDE=180°-α).
Note that instead of using similar triangles we can use the statement that two parallel lines combined with two intersecting lines form 2 equal alternate angles. We are given that the lines HI and DE are parallel, therefore the angles CDE and ABC are equal (these angles are alternate) and their value is 180°-α (we found that CDE=180°-α).
We are given that the angle FAH is equal to 40°, therefore the angle BAC is also equal to 40° (the angles are vertical, therefore they are equal).
We are given that AC=BC, therefore the angle BAC=ABC.
We found that the angle BAC=ABC, the angle BAC=40° and the angle ABC=CDE=180°-α. Therefore, the angle ABC=CDE= BAC=FAH=180°-α=40°
We can solve the equation 180°-α=40°, getting α=180°-40°=140°
|
## High School Math Lesson Plan: Solving Multi-Step Equations
written by: Ginean Royal • edited by: Carly Stockwell • updated: 2/24/2014
This lesson plan will demonstrate to students how to solve multi-step equations.
• slide 1 of 2
Prior Knowledge
Before this lesson your students should have previously learned how to write equations. This involves:
(a) translating sentences into equations,
(b) translating equations into sentences,
(c) solving one-step equations using addition and subtraction and
(d) solving simple equations using multiplication or division. Now students will learn how to solve multi-step equations.
Common Core State Standards
A.REI.1: Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
A.REI.3: Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Mathematical Practice(s)
Look for and express regularity in repeated reasoning.
Learning Target(s)
• Applying order of operations and inverse operations to solve equations
• Constructing an argument to justify my solution process
Essential Question(s)
What is one problem-solving strategy used to solve multi-step equations?
Answer: Working backwards or using the order of operations in reverse.
Vocabulary
Inverse operation, isolate, variable, constant, reciprocal, coefficient
• slide 2 of 2
### LESSON
Notes:
• Instruct that minus and negative are equivalent
• GOAL: To isolated the specified variable
• Instruct that what is done on one side of the equation (= sign), must be done on the other side of the = sign.
Utilize chilimath.com for the lesson today. There are two pages to view. Use all of the examples or as many as you choose.
* Reminder for students: that –x is the same as -1x & x is the same as 1x.
Guided Practice
3-6 practice problems. You can do 1or 2 problems with the students at the board (Smart Board, Elmo, etc.) and then put them in small groups of no more than 3 to do the rest. These problems can be pulled from any textbook or other resource.
Independent Practice
Approximately 5 problems to be done alone.
Closure/Review
Ask 1-3 questions relating to today’s lesson to be answered by the class as a whole. This will give you a general idea of the class’ understanding of today’s topic.
Exit Ticket
This is to be done the last 3-5 minutes of class and given to you (by hand or in a designated area of your room) as they leave class. Possible question(s):
Solve this problem: 5k + 3 = 23
|
#### You may also like
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
### Making Rectangles, Making Squares
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
This gives a short summary of the properties and theorems of cyclic quadrilaterals and links to some practical examples to be found elsewhere on the site.
# Property Chart
### Why do this problem?
This game provides an interesting context in which to consider the properties of quadrilaterals (or triangles), and has a particular focus on the combinations of properties that are possible.
### Possible approach
Quadrilaterals Game could be used either at the beginning or at the end of this problem.
Use the instruction on the problem page to set up and play the game for about half a lesson, then move the group on to the questions at the end of the problem.
### Key questions
• Which shapes are most useful in this game?
• Which property cards are 'good' and 'bad' and why?
• Tell me two cards where there is no shape that works for both.
### Possible support
The game could be played as a whole class - shuffle and arrange the property cards on the board so that everyone has the same question. Groups of 3 or 4 then work together filling in the grid and checking each others work. A correct shape or gap will earn 10 points, but each incorrect shape or gap will lose 10 points. After a set time, all the grids are displayed, and students try to find errors in the other groups' work, in order to establish the scores and the winners. They may be ready to try the problem as stated after this!
Another way in to the problem could be to produce some partly completed grids and ask students to finish them, or produce some completed grids with a few deliberate errors for students to find and correct.
### Possible extension
If only the quadrilaterals are visible on the board can you identify the property cards in each position? In what other ways can you adapt/invert/develop this game to make new and possibly harder challenges?
Suggest students have a go at Shapely Pairs
|
Tweaking a proof of the infinitude of primes
This blog presents two proofs of the infinitude of primes. The first one is the standard proof that is due to Euclid (around 300 BC). The second one is due to Christian Goldbach (from a letter to Leonhard Euler in 1730). This post attempts to generalize the second one. The goal is to capture the essence of the proof and in the process gains a better perspective. Reading the second proof is not a prerequisite. The reader is welcome to compare the argument here with the previous proof to appreciate the economy that is in the tweaked proof.
____________________________________________________________________________
The Tweak
The existence of the following sequence will imply that there are infinitely many prime numbers.
There is a sequence of positive odd integers $x_0,x_1,x_3, \cdots$ such that
• $x_0 < x_1 < x_3 < \cdots$,
• any two elements of the sequence are relatively prime, i.e. for $i \ne j$, the terms $x_i$ and $x_j$ have no common divisor other than 1.
If such a sequence exists, we can infer an infinite sequence of prime numbers. Note that each term $x_j$ is a product of distinct primes. Let $p_j$ be the smallest prime factor of $x_j$. The prime numbers $p_j$ must be distinct since the elements in the sequence $x_0,x_1,x_3, \cdots$ are pairwise relatively prime.
How do we know if such a sequence exists? It can be defined recursively. Let $x_0 be any odd positive integers that are relatively prime. For example, $x_0=3$ and $x_1=5$. Another example: $x_0=21$ and $x_1=55$. Then for each $n>1$, let $x_n=x_0 \times x_1 \times \cdots \times x_{n-1}+2$. Clearly the sequence is increasing. To see that any two terms of the sequence are relatively prime, let $j and let $d$ be a common divisor of $x_j$ and $x_n$. Since $d$ divides $x_j$, it is clear that $d$ divides $x_0 \times x_1 \times \cdots \times x_{n-1}$. This means that $d$ divides $x_n-x_0 \times x_1 \times \cdots \times x_{n-1}=2$. With $d$ being odd and $d$ dividing 2, it must be that $d=1$.
The proof in the previous post essentially uses the above argument with the first two terms fixed at $x_0=3$ and $x_1=5$. With these two initial choices, it can be shown that the sequence $x_0,x_1,x_3, \cdots$ consists of the Fermat numbers. It is nice to know that the resulting sequence consists of the Fermat numbers. However, for proving that primes are infinite, the fact about Fermat numbers is immaterial. For example, if the initial two terms are $x_0=21$ and $x_1=55$, the same argument will still lead to an infinite sequence of prime numbers. Using Fermat numbers is one way to derive the sequence in question. Sometimes a broader picture picture can be obscured by being too specific in the argument.
____________________________________________________________________________
An Equivalent Statement
The above argument for the infinitude of primes points to the fact that the existence of an infinite sequence of prime numbers is reduced to the existence of a sequence as described in the above argument. We have the following theorem.
Theorem
The following two statements are equivalent.
• There exists a sequence of positive odd integers $x_0,x_1,x_3, \cdots$ such that $x_0 < x_1 < x_3 < \cdots$ and such that any two terms of the sequence are relatively prime.
• There are infinitely many prime numbers.
It is clear that if there are infinitely many primes, the sequence $x_0,x_1,x_3, \cdots$ can simply be made the primes. So one direction is obvious. The other direction is proved by the above argument.
The theorem essentially says that the “infinity” of primes is identical to the “infinity” of numbers that are pairwise relatively prime. Of course, the purpose of discussing the latter is to give an argument that there are infinitely many primes.
____________________________________________________________________________
$\copyright \ 2016 \text{ by Dan Ma}$
|
These activity sheets will help your beginning students learn how to compare different mathematical expressions.
#### Algebra is fun, not for everyone, though! Algebraic expressions have integers, constants, and variables, followed by their methods of operations. To compare two algebraic expressions, you require a basic understanding of algebra, i.e., its addition, multiplication, etc. Here is how we compare to algebraic expressions: Solving equations simultaneously - These expressions have variables, and to compare these expressions, you need to find the values of these variables. For example, you solve one expression for 'x' and then put its value in the second expression. Step I However, for comparison, we solve the first expression in terms of 'x' or whichever variable the equation has. Then repeating the same, we solve the second expression in terms of 'y' and name these expressions as iii or iv Step II - We equate the values we have derived. Step III - In the last step, we find out the values of linear equations.
When you compare algebraic expressions whether they are super simple or complex the first thing you need to do is to treat each expression as separate and then evaluate each expression. You will want to have each value in the simplest form possible. From that moment further, you are free to just compare each value. These worksheets explain how to compare different mathematical expressions, and rank them as greater than, less than, or equal to. Students will also practice their rounding skills.
# Print Comparing Algebraic Expressions Worksheets
## Comparing Algebraic Expressions Lesson
This worksheet explains how to compare expressions. A sample problem is solved.
## Lesson and Practice
Students will review and practice comparing algebraic expressions. A sample problem is solved and two practice problems are provided.
## Worksheet
Students will compare expressions using math symbols. Ten problems are provided.
## Practice Sheet
Students will estimate the total number using rounded numbers. Example: There are 52 people riding on a train. 30 more people are waiting to get on at the next stop. Estimate the total number of people that will be riding the train altogether, using rounded numbers.
## Drills Worksheet
Students will write the number that is described. Eight problems are provided.
## Algebraic Show Off
Students will round off numbers to the nearest whole ten. Twenty-five problems are provided.
## Warm Up
Students will order the numbers from smallest to greatest. Three problems are provided.
## Math Symbols Lesson
This worksheet explains the concept of using math symbols to compare numbers. A sample problem is solved.
## Math Symbols Lesson and Practice
Students will use math symbols to compare two pictures. A sample problem is solved and two practice problems are provided.
## Math Symbols Worksheet
Students will use math symbols to compare pairs of numbers. Ten problems are provided.
## Practice
Students will COMPARE numbers using math symbols. Ten problems are provided.
## Drill
Students will practice using math symbols to compare numbers. Eight problems are provided.
## Warm Up
Write these numbers in order from the smallest to the greatest. Three problems are provided.
|
0
# How can you find out how many solutions an equation has?
Updated: 4/28/2022
Wiki User
14y ago
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.
Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:
sin(x) = 0.5
has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.
Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:
sin(x) = 0.5
has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.
Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:
sin(x) = 0.5
has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.
Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:
sin(x) = 0.5
has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
Wiki User
14y ago
Wiki User
14y ago
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.
Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:
sin(x) = 0.5
has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
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Q: How can you find out how many solutions an equation has?
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Related questions
### How can someone find how many solutions there are in an equation?
you can find it by counting how many numbers they are in the equation
### If an equation is an identity how many solutions does it have?
An identity equation has infinite solutions.
### How many solutions do a equation have?
It will depend on the equation.
### If the discriminant of a quadratic equation is -4 how many solutions does the equation have?
If the discriminant of a quadratic equation is less then 0 then it will have no real solutions.
### How many solutions for x does the following equation have?
It has the following solutions.
### How many solutions will the equation 25 equals 2x squared 6x have?
The quadratic equation will have two solutions.
Infinitely many
2
### How many solutions are there in the equation 2x equals 10?
There is one solution. To find it, divide both sides of the equation by 2. This leaves you with x=5, where 5 is your solution.
### What is the quadriatic formula used for?
To find the solutions of x in a quadratic equation.
x, y ,z
### How many solutions does 240 have as a discriminant?
240 is not an equation and so the concept of solutions is meaningless.
|
Precalculus Functions: Domain, Range, Composite, and Inverse
Key Terms
• Domain
• Range
• Composition of functions (composite function)
• Mapping
• One-to-one (function)
• Horizontal line test
• Vertical line test
• Inverse (of a function)
Objectives
• Determine the domain and range of various functions
• Define the composition of two functions
• Identify functions that are one-to-one by applying the horizontal line test
• Calculate the inverse of a one-to-one function
•
Domain and Range of a Function
We will deal with real-valued functions of real variables--that is, the variables and functions will only have values in the set of real numbers. Furthermore, by just looking at a few examples, we can see that for a given function, sometimes the function or the variable (or both) is limited in the interval of values it can take. In other words, sometimes the function would otherwise not be real-valued for a given real value of the variable, or the function may simply only take on a limited range of values regardless of the value of the independent variable. Consider the case of f(x) = x2. Here, f is defined for any real value x, but f is always greater than or equal to zero--characteristics we can extrapolate by looking at the graph of the function.
We call the set of values of the independent variable for which the function is defined (typically meaning real-valued) the domain of the function. Concomitantly, we call the set of values of the function itself (corresponding to the entire domain) the range of the function. In the case of f(x) = x2, the domain is the entire set of real numbers (often expressed as ), since for any real value x, the function f is also real-valued. We can also express this domain as the interval (–∞,∞). The range of f is (as we can see from the graph above) all real values greater than or equal to 0, which we can also express as the interval [0, ∞). To reiterate, the domain of a function f(x) is the set of all values of x for which the function is also real-valued. The range of f(x) is the set of all values of f corresponding to the domain of f.
Practice Problem: Find the domain and range of each function below.
a. b. c.
Solution: To find the domain, determine which values for the independent variable will yield a real value for the function. For the range, one option is to graph the function over a representative portion of the domain--alternatively, you can determine the range by inspection.
a. The domain is the interval (–∞, 1), since the denominator must be non-zero and the expression under the radical must be greater than or equal to zero. The graph of the function is shown below to illustrate the range.
Note that the function approaches (but never reaches) zero as c approaches –∞, and it approaches ∞ as c approaches 1. Thus, the range is (0, ∞).
b. Following an approach similar to that of part a, the domain of h(r) is (–∞, ∞) (or ) and the range is also (–∞, ∞) (or ).
c. Because y must be non-negative (greater than or equal to zero), the domain is [0, ∞). The range is also [0, ∞).
Compositions of Functions
A useful tool is a composition of functions (or composite function), which we can describe in one sense as a "function of a function." Consider two functions f(x) and g(x). We define the composite function as follows:
For instance, consider f(x) = 3x and g(x) = -x + 4. The composite function is then
Interested in learning more? Why not take an online Precalculus course?
Practice Problem: Given the functions and , find the following composite functions.
a. b.
Solution: Follow the pattern described above. Plug the expression for one function into the variable for the other according to the required order.
a. b.
One-to-One Functions
A function can be described as a mapping of values in the domain to values in the range, as the diagram below illustrates for a function f.
Each value in the domain would be "connected" by an arrow (representing the function) to exactly one value in the range. (Recall that this is a necessary condition for an algebraic relation to be a function: for each value of the independent variable-i.e., the domain-the dependent variable can have only one corresponding value.) But the reverse is not always true: for some functions, certain values in the range may correspond to more than one value in the domain. When each value in the domain corresponds to only one value in the range and each value in the range corresponds to only one value in the domain, the function is called one-to-one. The diagrams below illustrate these differences.
Determining if a function is one-to-one simply by looking at the algebraic expression may be difficult. A simpler way to determine if it is one-to-one is the horizontal line test. Draw a graph of the function and then check if any horizontal line intersects the function more than once. If every horizontal line intersects the function a maximum of one time, then that function is one-to-one. (Note that you can use the nearly identical vertical line test to determine if an algebraic relation is a function: if each vertical line intersects the relation no more than one time, then that relation is a function.)
Inverse of a Function
A function "maps" values in the domain to values in the range by taking a value from the domain as input and outputting the value in the range. But what about a mapping in the opposite direction (from the range to the domain)? This "reverse" mapping is called the inverse of a function. For a function f(x), the inverse is often written as f -1(x). For a function to have an inverse, it must be one-to-one (otherwise, the inverse would not meet the criterion of a function).
To calculate the inverse, first ensure that the function is one-to-one. Next, perform the following procedure on the original function (which, for simplicity, we'll call f(x)):
1. Substitute f -1(x) for all instances of x.
2. Substitute x for f(x).
3. Solve for f -1(x).
Consider, for instance, f(x) = 5x + 2. This function is one-to-one, and we can find the inverse as follows.
Specifically, for each value x, note the relationship shown in the diagram below.
Algebraically, this relationship means that if we plug f(x) into f -1(x), we should get x as the result! (Alternatively, if we plug f -1(x) into f(x), we should also get x.) That is,
Let's try it for the example above:
You can also show that the other composite function yields a result of x as well.
Practice Problem: Determine if the function is one-to-one, and if so, calculate its inverse.
Solution: First, we must determine if k(z) is one-to-one. Since this is not easily done just by inspection, let's plot a graph of the function.
As it turns out, the function passes the horizontal line test (and the vertical line test), so it is one-to-one. Let's calculate the inverse:
Check the result:
The inverse checks out.
|
Floating point/Lesson Three
Absolute and Relative Error
Before delving further into a discussion of storing a number, how can we measure how wrong a number is can be an important issue. If the correct answer is 100, is 100.9 might not be a bad approximation. What if the answer is 1, and we store the number as 1.9?
Denote the correct answer as x, and the stored value as x0. Then, the absolute error is simply | x - x0 |. Thus, in the previous example, our absolute error was 0.9 (in both cases).
On the other hand, it may be important to see what the error is in relation to how large the number is. An error of 0.9 is usually less important when the number is 100 instead of 1. The relative error is defined as | x - x0 |/| x |.
In our two examples, the relative errors are | 100.9 - 100 |/| 100 | = 0.009 and | 1 - 1.9 | / | 1 | = 0.9.
Binary representation in a Computer
In a computer, numbers are stored in normalized floating-point representation. The numbers are converted to binary, and are represented as ±x × 2n, where x is a number with only decimal digits and a 1 as the first digit before the decimal, and n is an exponent of 2.
The number, stored as the portion after the 1 (1.f), is called the normalized mantissa. The exponent is simply called the exponent.
For instance, 55.5 would be 55.5 base 10 = 110111.1 base 2 = 1.101111 base 2 × 25 = 1.101111 × (2)101.
Limited Storage
The computer has a limited amount of storage space for a number. Let's say that this amount is three digits (0.xyz), with a 1 digit exponent, and the exponent can be either positive or negative. Thus, we have the following numbers (table from Cheney & Kincaid):
```0.000 × 2° = 0 0.000 × 2¹ = 0 0.000 × 2-¹ = 0
0.001 × 2° = 1/8 0.001 × 2¹ = 1/4 0.001 × 2-¹ = 1/16
0.010 × 2° = 1/4 0.010 × 2¹ = 1/2 0.010 × 2-¹ = 1/8
0.011 × 2° = 3/8 0.011 × 2¹ = 3/4 0.011 × 2-¹ = 3/16
0.100 × 2° = 1/2 0.100 × 2¹ = 1 0.100 × 2-¹ = 1/4
0.101 × 2° = 5/8 0.101 × 2¹ = 5/4 0.101 × 2-¹ = 5/16
0.110 × 2° = 3/4 0.110 × 2¹ = 3/2 0.110 × 2-¹ = 3/8
0.111 × 2° = 7/8 0.111 × 2¹ = 7/4 0.111 × 2-¹ = 7/16
```
We have the following numbers:
```0 1/16 1/8 3/16 1/4 5/16 3/8 7/16 1/2 5/8 3/4 7/8 1 5/4, 7/4
X X X X X X X X X X X X X ... X...X(off screen)
```
As you can tell, the numbers are not equally spaced.
You may ask, why are we including 0.000, 0.001, 0.010, and 0.011? That is a good question, as computers assume that the first number is a '1'. Computers do this so that they can save space. This means that, in actuality, the numbers in our machine are the following (we also note that computers normally store as 1.f, but that's not the exercise now):
```0.100 × 2° = 1/2 0.100 × 2¹ = 1 0.100 × 2-¹ = 1/4
0.101 × 2° = 5/8 0.101 × 2¹ = 5/4 0.101 × 2-¹ = 5/16
0.110 × 2° = 3/4 0.110 × 2¹ = 3/2 0.110 × 2-¹ = 3/8
0.111 × 2° = 7/8 0.111 × 2¹ = 7/4 0.111 × 2-¹ = 7/16
```
Now, the number line looks like this:
```0 1/4 5/16 3/8 7/16 1/2 5/8 3/4 7/8 1 5/4, 7/4
X X X X X X X X X X ... X...X (off screen)
```
This phenomenon, where computers miss really small numbers, is known in computer science as the hole at zero. When a number is in the hole and rounded down to zero, it is considered part of underflow.
Rounding
If the computer can only process a finite amount of numbers, it has to choose to round answers to another number. It can choose to round up always, round down always, or round to the nearest computer number. We will assume that the computer rounds to the nearest number. This is done in practice as well.
Rounding down to the nearest integer always is known as chopping.
Computer Epsilon
A special number in computers is computer epsilon. This number is defined as the smallest number such that:
1 + ε ≠ 1.
In our special computing system developed in the previous section, the space between 1 and 5/4 is 1/4. So the computer epsilon is 1/8, because anything above 1 + 1/8 rounds to 5/4, and anything below 1 + 1/8 is rounded to 1.
In lesson four, we will outline a proof that states that the roundoff error when digits are rounded up always or rounded down always is ε, but when the rounding is done to the nearest machine number, the error is ε/2.
Overflow
Another important number is overflow. When the computer cannot process a number because it is too large, it runs to overflow. An overflow error is extremely important, because it stops most computer problems. If a number is underflowed, it is considered not as serious, and the computer will simply round to zero. This makes sense because underflow is bounded by 1/2 the distance between 0 and the next computer number; whereas in overflow, there is no bound on how large the relative error is.
Matlab Example
Matlab has the code for the computer epsilon, the real max, or the last machine number before overflow, and the real min, the first machine number after zero:
```
>> eps
ans =
2.2204e-016
>> realmax
ans =
1.7977e+308
>> realmin
ans =
2.2251e-308
```
Homework
1. Assume a system of 2 binary digits with an exponent of 0 or ±1. Draw a number line indicating the available numbers to the computer, if a zero is allowed as the first number. Then, remove the appropriate numbers. How large is the hole at zero? What is computer epsilon?
2. In the three digit number system we developed in this lesson (with the hole at zero), perform the following operations:
a) 1/12 + 1/3 (assume that 1/12 is stored in the system BEFORE it is operated on)
b) 1/2 + 1/2
c) 1/8 + 1/3 (assume that the number is added BEFORE it is stored)
d) 1 + 9
Source
Cheney Ward and David Kincaid. Numerical Methods and Computing. Belmont, CA: Thomson, 2004.
|
# What is the second derivative of secx?
Recall that ##sec(x)=1/{cos(x)}##. You can use the formula which states that
##d/{dx} 1/{f(x)} = -frac{f’}{f^2}##.
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This formula can easily be obtained by using the usual formula
##d/{dx} {a(x)}/{b(x)}= frac{a’ b – a b’}{b^2}##, where ##a(x)equiv 1##, and ##b(x)=f(x)##.
Since ##d/{dx} cos(x)=-sin(x)##, we have that
##d/{dx} 1/{cos(x)} = -frac{ -sin(x)}{cos^2(x)} = frac{sin(x)}{cos^2(x)}##
For the second derivative of ##sec(x)##, let’s derive one more time the first derivative: again, by the rule for the derivation of rational functions, we have
##d/{dx} -frac{sin(x)}{cos^2(x)} = frac{sin'(x)cos^2(x) – sin(x)(cos^2(x))’}{cos^4(x)}##
Since ##sin'(x)=cos(x)## and ##(cos^2(x))’=-2cos(x)sin(x)##, we have
##frac{cos^3(x)+2sin^2(x)cos(x)}{cos^4(x)}##
Simplifying ##cos(x)##, we get
##frac{cos^2(x)+2sin^2(x)}{cos^3(x)}##
Writing ##cos^2(x)## as ##1-sin^2(x)##, we have
##frac{1-sin^2(x)+2sin^2(x)}{cos^3(x)}=frac{1+sin^2(x)}{cos^3(x)}##
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