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# RS Aggarwal Class 6 Solutions Chapter 5 - Fractions Ex 5A (5.1)
## RS Aggarwal Class 6 Chapter 5 - Fractions Ex 5A (5.1) Solutions Free PDF
Q1) (i) $\frac{ 2 }{ 3 }$ (ii) $\frac{ 4 }{ 5 }$
(iii) $\frac{ 5 }{ 8 }$ (iv) $\frac{ 7 }{ 10 }$
(v) $\frac{ 3 }{ 7 }$ (vi) $\frac{ 6 }{ 11 }$
(vii) $\frac{ 7 }{ 9 }$ (viii) $\frac{ 5 }{ 12 }$
Ans. 1) (i)$\frac{ 2 }{ 3 } = \frac{ 2 \times 2 }{ 3 \times 2} = \frac{ 2 \times 3 }{ 3 \times 3 } = \frac{ 2 \times 4 }{ 3 \times 4} = \frac{ 2 \times 5}{ 3 \times 5 } = \frac{ 2 \times 6}{ 3 \times 6 }$
Therefore, $\frac{ 2 }{ 3 } = \frac{ 4 }{ 6 } = \frac{ 6 }{ 9 } = \frac{ 8 }{ 12 } = \frac{ 10 }{ 15 } = \frac{ 12 }{ 18 }$
Hence, the five fractions equivalent to $\frac{ 2 }{ 3 }$ are $\frac{ 4 }{ 6 }$ , $\frac{ 6 }{ 9 }$ , $\frac{ 8}{ 12 }$ , $\frac{ 10 }{ 15 }$ and $\frac{ 12 }{ 18 }$
(ii) $\frac{ 4 }{ 5 } = \frac{ 4 \times 2 }{ 5 \times 2} = \frac{ 4 \times 3 }{ 5 \times 3 } = \frac{ 4 \times 4 }{ 5 \times 4} = \frac{ 4 \times 5}{ 5 \times 5 } = \frac{ 4 \times 6}{ 5 \times 6 }$
Therefore, $\frac{ 4 }{ 5 } = \frac{ 8 }{ 10 } = \frac{ 12 }{ 15 } = \frac{ 16 }{ 20 } = \frac{ 20 }{ 25 } = \frac{ 24 }{ 30 }$
Hence, the five fractions equivalent to $\frac{ 4 }{ 5 }$ are $\frac{ 8 }{ 10 }$ , $\frac{ 12 }{ 15 }$ , $\frac{ 16}{ 20 }$ , $\frac{ 20 }{ 25 }$ and $\frac{ 24 }{ 30 }$
(iii) $\frac{ 5 }{ 8 } = \frac{ 5 \times 2 }{ 8 \times 2} = \frac{ 5 \times 3 }{ 8 \times 3 } = \frac{ 5 \times 4 }{ 8 \times 4} = \frac{ 5 \times 5}{ 8 \times 5 } = \frac{ 5 \times 6}{ 8 \times 6 }$
Therefore, $\frac{ 5 }{ 8 } = \frac{ 10 }{ 16 } = \frac{ 15 }{ 24 } = \frac{ 20 }{ 32 } = \frac{ 25 }{ 40 } = \frac{ 30 }{ 48 }$
Hence, the five fractions equivalent to $\frac{ 5 }{ 8 }$ are $\frac{ 10 }{ 16 }$ , $\frac{ 15 }{ 24 }$ , $\frac{ 20}{ 32 }$ , $\frac{ 25 }{ 40 }$ and $\frac{ 30 }{ 48 }$ .
(iv) $\frac{ 7 }{ 10 } = \frac{ 7 \times 2 }{ 10 \times 2} = \frac{ 7 \times 3 }{ 10 \times 3 } = \frac{ 7 \times 4 }{ 10 \times 4} = \frac{ 7 \times 5}{ 10 \times 5 } = \frac{ 7 \times 6}{ 10 \times 6 }$
Therefore, $\frac{ 7 }{ 10 } = \frac{ 14 }{ 20 } = \frac{ 21 }{ 30 } = \frac{ 28 }{ 40 } = \frac{ 35 }{ 50 } = \frac{ 42 }{ 60 }$
Hence, the five fractions equivalent to $\frac{ 7 }{ 10 }$ are $\frac{ 14 }{ 20 }$ , $\frac{ 21 }{ 30 }$ , $\frac{ 28}{ 40 }$ , $\frac{ 35 }{ 50 }$ and $\frac{ 42 }{ 60 }$
(v)$\frac{ 3 }{ 7 } = \frac{ 3 \times 2 }{ 7 \times 2} = \frac{ 3 \times 3 }{ 7 \times 3 } = \frac{ 3 \times 4 }{ 7 \times 4} = \frac{ 3 \times 5}{ 7 \times 5 } = \frac{ 3 \times 6}{ 7 \times 6 }$
Therefore, $\frac{ 3 }{ 7 } = \frac{ 6 }{ 14 } = \frac{ 9 }{ 14 } = \frac{ 12 }{ 28 } = \frac{ 15 }{ 35 } = \frac{ 18 }{ 42 }$
Hence, the five fractions equivalent to $\frac{ 3 }{ 7 }$ are $\frac{ 6 }{ 14 }$ , $\frac{ 9 }{ 21 }$ , $\frac{ 12}{ 28 }$ , $\frac{ 15 }{ 35 }$ and $\frac{ 18 }{ 42 }$
(vi) $\frac{ 6 }{ 11 } = \frac{ 6 \times 2 }{ 11 \times 2} = \frac{ 6 \times 3 }{ 11 \times 3 } = \frac{ 6 \times 4 }{ 11 \times 4} = \frac{ 6 \times 5}{ 11 \times 5 } = \frac{ 6 \times 6}{ 11 \times 6 }$
Therefore, $\frac{ 6 }{ 11 } = \frac{ 12 }{ 22 } = \frac{ 18 }{ 33 } = \frac{ 24 }{ 44 } = \frac{ 30 }{ 55 } = \frac{ 36 }{ 66 }$
Hence, the five fractions equivalent to $\frac{ 6 }{ 11 }$ are $\frac{ 12 }{ 22 }$ , $\frac{ 18 }{ 33 }$ , $\frac{ 24}{ 44 }$ , $\frac{ 30 }{ 55 }$ and $\frac{ 36 }{ 66 }$
(vii) $\frac{ 7 }{ 9 } = \frac{ 7 \times 2 }{ 9 \times 2} = \frac{ 7 \times 3 }{ 9 \times 3 } = \frac{ 7 \times 4 }{ 9 \times 4} = \frac{ 7 \times 5}{ 9 \times 5 } = \frac{ 7 \times 6}{ 9 \times 6 }$
Therefore, $\frac{ 7 }{ 9 } = \frac{ 14 }{ 18 } = \frac{ 21 }{ 27 } = \frac{ 28 }{ 36 } = \frac{ 35 }{ 45 } = \frac{ 42 }{ 54 }$
Hence, the five fractions equivalent to $\frac{ 7 }{ 9 }$ are $\frac{ 14 }{ 18 }$ , $\frac{ 21 }{ 27 }$ , $\frac{ 28}{ 36 }$ , $\frac{ 35 }{ 45 }$ and $\frac{ 42 }{ 54 }$
(viii) $\frac{ 5 }{ 12 } = \frac{ 5 \times 2 }{ 12 \times 2} = \frac{ 5 \times 3 }{ 12 \times 3 } = \frac{ 5 \times 4 }{ 12 \times 4} = \frac{ 5 \times 5}{ 12 \times 5 } = \frac{ 5 \times 6}{ 12 \times 6 }$
Therefore, $\frac{ 5 }{ 12 } = \frac{ 10 }{ 24 } = \frac{ 15 }{ 36 } = \frac{ 20 }{ 48 } = \frac{ 25 }{ 60 } = \frac{ 30 }{ 72 }$
Hence, the five fractions equivalent to $\frac{ 5 }{ 12 }$ are $\frac{ 10 }{ 24 }$ , $\frac{ 15 }{ 36 }$ , $\frac{ 15}{ 36 }$ , $\frac{ 20 }{ 48 }$ and $\frac{ 25 }{ 60 }$
Q2) (i) $\frac{ 5 }{ 6 }$ and $\frac{ 20 }{ 24 }$
(ii) $\frac{ 3 }{ 8 }$ and $\frac{ 15 }{ 40 }$
(iii) $\frac{ 4 }{ 7 }$ and $\frac{ 16 }{ 21 }$
(iv)$\frac{ 2 }{ 9 }$ and $\frac{ 14 }{ 63 }$
(v) $\frac{ 1 }{ 3 }$ and $\frac{ 9 }{ 24 }$
(vi) $\frac{ 2 }{ 3 }$ and $\frac{ 33 }{ 22 }$
Ans. 2) The pairs of equivalent fractions are as follows:
(i) $\frac{ 5 }{ 6 }$ and $\frac{ 20 }{ 24 }$
( $\frac{ 20 }{ 24 } = \frac{ 5\times 4}{ 6 \times 4}$ )
(ii) $\frac{ 3 }{ 8 }$ and $\frac{ 15 }{ 40 }$
( $\frac{ 15 }{ 40 } = \frac{ 3 \times 5}{ 8 \times 5}$ )
(iv) $\frac{ 2 }{ 9 }$ and $\frac{ 14 }{ 40 }$
( $\frac{ 14 }{ 63 } = \frac{ 2 \times 7}{ 9 \times 7}$ )
Q3) Find the equivalent fraction of $\frac{ 3 }{ 5 }$ having
(i) denominator 30 (ii) numerator 24
Ans. 3) (i) Let $\frac{ 3 }{ 5 } = \frac{ ?}{ 30 }$
Clearly, 30 = 5 * 6
So, we multiply the numerator by 6.
Therefore, $\frac{ 3 }{ 5 } = \frac{ 3 \times 6 }{ 5 \times 6 } = \frac{ 18 }{ 30 }$
Hence, the required fraction is $\frac{ 18 }{ 30 }$.
(ii) Let $\frac{ 3 }{ 5 } = \frac{ 24 }{ ? }$
Clearly, 24 = 3 * 8
So, we multiply the denominator by 8.
Therefore, $\frac{ 3 }{ 5 } = \frac{ 3 \times 8 }{ 5 \times 8 } = \frac{ 24 }{ 40 }$
Hence, the required fraction is $\frac{ 24 }{ 40 }$.
Q4) Find the equivalent fraction $\frac{ 5 }{ 9 }$ having
(i) denominator 54 (ii) numerator 35
Ans. 4) (i) Let $\frac{ 5 }{ 9 } = \frac{?}{ 54 }$
Clearly, 54 = 9 * 6
So, we multiply the numerator by 6.
Therefore, $\frac{ 5 }{ 9 } = \frac{ 5 \times 6 }{ 9 \times 6 } = \frac{ 30 }{ 54 }$
Hence, the required fraction is $\frac{ 30 }{ 54 }$
(ii) Let $\frac{ 5 }{ 9 } = \frac{ 35 }{ ?}$
Clearly, 35 = 5 * 7
So, we multiply the denominator by 7.
Therefore, $\frac{ 5 }{ 9 } = \frac{ 5 \times 7 }{ 9 \times 7 } = \frac{ 35 }{ 63 }$
Hence, the required fraction is $\frac{ 35 }{ 63 }$.
Q5) Find the equivalent fraction of $\frac{ 6 }{ 11 }$
(i) denominator 77 (ii) numerator 60
Ans. 5) (i)Let $\frac{ 6 }{ 11 } = \frac{ ? }{ 77 }$
Clearly, 77 = 11 * 7
So, we multiply the numerator by 7.
Therefore, $\frac{ 6 }{ 11 } = \frac{ 6 \times 7 }{ 11 \times 7 } = \frac{ 42 }{ 77 }$
Hence, the required fraction is $\frac{ 30 }{ 54 }$
(ii) Let $\frac{ 6 }{ 11 } = \frac{ 60 }{ ?}$
Clearly, 60 = 6 * 10
So, we multiply the denominator by 10.
Therefore, $\frac{ 6 }{ 11 } = \frac{ 6 \times 10 }{ 11 \times 10 } = \frac{ 60 }{ 110 }$
Hence, the required fraction is $\frac{ 60 }{ 110 }$.
Q6) Find the equivalent fraction of $\frac{ 24 }{ 30 }$ having numerator 4.
Ans. 6) Let $\frac{ 24 }{ 30 } = \frac{ 4 }{ ?}$
Clearly, 4 = 24 / 6
So, we divide the denominator by 6.
Therefore, $\frac{ 24 }{ 30 } = \frac{ 24 \div 6 }{ 30 \div 6 } = \frac{ 4 }{ 5 }$
Hence, the required fraction is $\frac{ 4 }{ 5 }$.
Q7) Find the equivalent fraction $\frac{ 36 }{ 48 }$ with
(i) numerator 9 (ii) denominator 4
Ans. 7) (i) Let $\frac{ 36 }{ 48 } = \frac{ 9 }{ ?}$
Clearly, 9 = 36 / 4
So, we divide the denominator by 4.
Therefore, $\frac{ 36 }{ 48 } = \frac{ 36 \div 4 }{ 48 \div 4 } = \frac{ 9 }{ 12 }$
Hence, the required fraction is $\frac{ 9 }{ 12 }$.
(ii) Let $\frac{ 36 }{ 48 } = \frac{ ? }{ 4 }$
Clearly, 4 = 48 / 12
So, we divide the numerator by 12.
Therefore, $\frac{ 36 }{ 48 } = \frac{ 36 \div 7 }{ 48 \div 12 } = \frac{ 3 }{ 4 }$
Hence, the required fraction is $\frac{ 3 }{ 4 }$ .
Q8) Find the equivalent fraction of $\frac{ 56 }{ 70 }$ with
(i) numerator 4 (ii) denominator 10
Ans. 8) (i) Let $\frac{ 56 }{ 70 } = \frac{ 4 }{ ? }$
Clearly, 4 = 56 / 14
So, we divide the denominator by 14.
Therefore, $\frac{ 56 }{ 70 } = \frac{ 56 \div 14 }{ 70 \div 14 } = \frac{ 4 }{ 5 }$
Hence, the required fraction is $\frac{ 4 }{ 5 }$.
(ii) Let $\frac{ 56 }{ 70 } = \frac{ ?}{ 10 }$
Clearly, 10 = 70 / 7
So, we divide the numerator by 7.
Therefore, $\frac{ 56 }{ 70 } = \frac{ 56 \times 7 }{ 70 \times 7 } = \frac{ 8 }{ 10 }$
Hence, the required fraction is $\frac{ 8 }{ 10 }$ .
Q9) Reduce each of the following fractions into its simplest form:
(i) $\frac{ 9 }{ 15 }$ (ii) $\frac{ 48 }{ 60 }$
(iii) $\frac{ 84 }{ 98 }$ (iv) $\frac{ 150 }{ 60 }$
(v) $\frac{ 72 }{ 90 }$
Ans. 9) (i) Here , numerator = 9 and denominator = 15
Factors of 9 are 1, 3 and 9.
Factors of 15 are 1, 3, 5 and 15.
Common factors of 9 and 15 are 1 and 3.
H.C.F. of 9 and 15 is 3.
Therefore, $\frac{ 9 }{ 15 }$ = $\frac{ 9 \div 3 }{ 15 \div 3 } = \frac{ 3 }{ 5 }$
Hence, the simplest form of $\frac{ 9 }{ 15 }$ is $\frac{ 3 }{ 5 }$ .
(ii) Here , numerator = 48 and denominator = 60
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12.
H.C.F. of 48 and 60 is 12.
Therefore, $\frac{ 48 }{ 60 }$ = $\frac{ 48 \div 12 }{ 60 \div 12 } = \frac{ 4 }{ 5 }$
Hence, the simplest form of $\frac{ 48 }{ 60 }$ is $\frac{ 4 }{ 5 }$ .
(iii) Here, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84.
Factors of 98 are 1, 2, 3, 4, 7, 14, 15, 49 and 98.
Common factors of 84 and 98 are 1, 2, 7 and 14.
H.C.F. of 84 and 98 is 14.
Therefore, $\frac{ 84 }{ 98 }$ = $\frac{ 84 \div 14 }{ 98 \div 14 } = \frac{ 6 }{ 7 }$
Hence, the simplest form of $\frac{ 84 }{ 98 }$ is $\frac{ 6 }{ 7 }$ .
(iv) Here, numerator = 150 and denominator = 60
Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30.
H.C.F. of 150 and 60 is 30.
Therefore, $\frac{ 150 }{ 60 }$ = $\frac{ 150 \div 30 }{ 60 \div 30 } = \frac{ 5 }{ 2 }$
Hence, the simplest form of $\frac{ 150 }{ 60 }$ is $\frac{ 5 }{ 2 }$ .
(v) Here , numerator = 72 and denominator = 90
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.
Common factors of 72 and 90 are 1, 2, 3, 3, 6, 9 and 18.
H.C.F. of 72 and 90 is 18.
Therefore, $\frac{ 72 }{ 90 }$ = $\frac{ 72 \div 18 }{ 90 \div 18 } = \frac{ 4 }{ 5 }$
Hence, the simplest form of $\frac{ 72 }{ 90 }$ is $\frac{ 4 }{ 5 }$ .
Q10) Show that each of the following fractions is in the simplest form:
(i) $\frac{ 8 }{ 11 }$ (ii) $\frac{ 9 }{ 14 }$
(iii) $\frac{ 25 }{ 36 }$ (iv) $\frac{ 8 }{ 15 }$
(v) $\frac{ 21 }{ 10 }$
Ans. 10) (i) Here , numerator = 8 and denominator = 11
Factors of 8 are 1, 2, 4 and 8.
Factors of 11 are 1 and 11.
Common factors of 8 and 11 is 1.
Thus, H.C.F. of 8 and 11 is 1.
Hence, $\frac{ 8 }{ 11 }$ is the simplest form .
(ii) Here , numerator = 9 and denominator = 14
Factors of 9 are 1, 3 and 9.
Factors of 14 are 1, 2, 7, and 14.
Common factors of 9 and 14 is 1.
Thus, H.C.F. of 9 and 14 is 1.
Hence, $\frac{ 9 }{ 14 }$ is the simplest form .
(iii) Here , numerator = 25 and denominator = 36
Factors of 25 are 1, 5 and 25.
Factors of 36 are 1, 2, 3, 4, 6, 9,12, 18 and 36.
Common factors of 25 and 36 is 1.
Thus, H.C.F. of 25 and 36 is 1.
Hence, $\frac{ 25 }{ 36 }$ is the simplest form .
(iv) Here , numerator = 8 and denominator = 15
Factors of 8 are 1, 2, 4, and 8.
Factors of 15 are 1, 3, 5, and 15.
Common factors of 8 and 15 is 1.
Thus, H.C.F. of 8 and 15 is 1.
Hence, $\frac{ 8 }{ 15 }$ is the simplest form .
(v) Here , numerator = 21 and denominator = 10
Factors of 21 are 1, 3, 7, and 21.
Factors of 10 are 1, 2, 5, and 10.
Common factors of 21 and 10 is 1.
Thus, H.C.F. of 21 and 10 is 1.
Hence, $\frac{ 21 }{ 10 }$ is the simplest form .
Q11) Replace $?$ by the correct number in each of the following:
(i) $\frac{ 2 }{ 7 } = \frac{ 8 }{ ? }$
(ii) $\frac{ 3 }{ 5 } = \frac{ ? }{ 35 }$
(iii) $\frac{ 5 }{ 8 } = \frac{ 20 }{ ? }$
(iv) $\frac{ 45 }{ 60 } = \frac{ 9 }{ ?}$
(v) $\frac{ 40 }{ 56 } = \frac{ ? }{ 7 }$
(vi) $\frac{ 42 }{ 54 } = \frac{ 7 }{ ?}$
Ans. 11) (i) 28 ( $\frac{ 2 }{ 7 } = \frac{ 2 \times 4 }{ 7 \times 4 } = \frac{ 8 }{ 28 }$ )
(ii) 21 ( $\frac{ 3 }{ 5 } = \frac{ 3 \times 7 }{ 5 \times 7 } = \frac{ 21 }{ 35 }$ )
(iii) 32 ( $\frac{ 5 }{ 8 } = \frac{ 5 \times 4 }{ 8 \times 4 } = \frac{ 20 }{ 32 }$ )
(iv) 12 ( $\frac{ 45 }{ 60 } = \frac{ 45 \div 5 }{ 60 \div 5 } = \frac{ 9 }{ 12 }$ )
(v) 5 ( $\frac{ 40 }{ 56 } = \frac{ 40 \div 8 }{ 56 \div 8 } = \frac{ 5 }{ 7 }$ )
(vi) 9 ( $\frac{ 42 }{ 54 } = \frac{ 42 \div 6 }{ 54 \div 6 } = \frac{ 7 }{ 9 }$ )
#### Practise This Question
20a3 is a multiple of 3, then sum of the possible values of a is
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# Method of Substitution! Help!
• Apr 15th 2008, 05:34 PM
luigiandme
Method of Substitution! Help!
Use the method of substitution to solve the system
http://webwork.tuhsd.k12.az.us/webwo...5e21a21381.png
Answer for should be in (_,_),(_,_) form or just (_,_) form.
Thanks!
• Apr 15th 2008, 05:40 PM
o_O
Since $x + y = 1$, you can deduce that ${\color{blue}y = 1 - x}$.
Plug this to the original equation:
$x^{2} + {\color{blue} y^{2}} = 8$
$x^{2} + {\color{blue} (1-x)^{2}} = 8$
Continue on and solve for x then y.
• Apr 15th 2008, 05:47 PM
TheEmptySet
Quote:
Originally Posted by luigiandme
Use the method of substitution to solve the system
http://webwork.tuhsd.k12.az.us/webwo...5e21a21381.png
Answer for should be in (_,_),(_,_) form or just (_,_) form.
Thanks!
lets isloate one variable in the bottom equation lets say y.
$x+y=1 \iff y=1-x$
Now will will sub this into the first equation for y.
$x^2+(1-x)^2=8$ expand the left hand side
$x^2+1-2x+x^2=8 \iff 2x^2-2x=7$
$2(x^2+x+\frac{1}{4})=7+2\cdot \frac{1}{4}$
$2(x+\frac{1}{2})^2=\frac{15}{2} \iff (x+\frac{1}{2})^2=\frac{15}{4}$
$x=-\frac{1}{2} \pm \frac{\sqrt{15}}{2}$
plug this back in to find the values for y
• Apr 15th 2008, 09:03 PM
luigiandme
hmm.. im sorry but i just cant seem to be getting the correct answer after plugging it in! This is driving me nuts...
• Apr 15th 2008, 09:11 PM
o_O
Quote:
Originally Posted by TheEmptySet
lets isloate one variable in the bottom equation lets say y.
$x+y=1 \iff y=1-x$
Now will will sub this into the first equation for y.
$x^2+(1-x)^2=8$ expand the left hand side
$x^2+1-2x+x^2=8 \iff 2x^2-2x=7$
$2(x^2 {\color{red}-}x+\frac{1}{4})=7+2\cdot \frac{1}{4}$
$2(x{\color{red}-}\frac{1}{2})^2=\frac{15}{2} \iff (x{\color{red} -}\frac{1}{2})^2=\frac{15}{4}$
$x={\color{red}+}\frac{1}{2} \pm \frac{\sqrt{15}}{2}$
plug this back in to find the values for y
A little mistake ...
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# How do you factor the expression 3x - 9x^2?
Feb 28, 2016
$3 x \left(1 - 3 x\right)$
#### Explanation:
Recall that the greatest common factor is the largest number which two numbers can be divided by without producing a decimal . For example, the greatest common factor of $12$ and $16$ is $4$.
In addition, recall the distributive property: $a \left(b + c\right) = a b + a c$.
Factoring the Expression
$1$. Determine the greatest common factor for $\textcolor{red}{3}$ and $\textcolor{b l u e}{9}$, which is $\textcolor{p u r p \le}{3}$. Using the distributive property, factor $\textcolor{p u r p \le}{3}$ from the expression.
$\textcolor{red}{3} x - \textcolor{b l u e}{9} {x}^{2}$
$= \textcolor{p u r p \le}{3} \left(x - 3 {x}^{2}\right)$
$2$. Both terms, $\textcolor{\mathmr{and} a n \ge}{x}$ and $- 3 \textcolor{t u r q u o i s e}{{x}^{2}}$, contain another common factor, $\textcolor{g r e e n}{x}$. Thus, factor out $\textcolor{g r e e n}{x}$ from the expression.
$= 3 \textcolor{g r e e n}{x} \left(1 - 3 x\right)$
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# Determining the Future (Maturity) Value
The simplest future value scenario for compound interest is for all of the variables to remain unchanged throughout the entire transaction. To understand the derivation of the formula, continue with the following scenario. If $4000 was borrowed two years ago at 12% compounded semi-annually, then a borrower will owe two years of compound interest in addition to the original principal of$4,000. That means PV = 4,000. The compounding frequency is semi-annually, or twice per year, which makes the periodic interest rate $i=\frac{I/Y}{C/Y}=\frac{12\%}{2}=6\%$ . Therefore, after the first six months, the borrower has 6% interest converted to principal. This a future value, or FV, calculated as follows: Principal after one compounding period (six months) = Principal plus interest \begin{align} FV &=PV+{i}(PV)\\ &=\ 4,000+0.06(\ 4,000)\\ &=\ 4,000+\ 240=\ 4,240 \end{align} \nonumber Now proceed to the next six months. The future value after two compounding periods (one year) is calculated in the same way. Note that the equation $FV = PV +i(PV)$ can be factored and rewritten as $FV = PV(1 +i)$. $FV (after\;two\;compounding\;periods)$ $= PV(1 +i) =4,200(1 + 0.06) = 4,240(1.06) = 4,494.40$ Since the $PV = 4,240$ is the result of the previous calculation where $PV(1 + i) = 4,240$, the following algebraic substitution is possible: $FV (after\;two\;compounding\;periods)$ $= PV(1 +i )(1 + i) = 4,000(1.06)(1.06) = 4200(1.06) = 4,494.40$ Simplifying algebraically, you get: $FV = PV(1 +i)(1 + i) = PV(1 +i)^2$ Do you notice a pattern? With one compounding period, the formula has only one $(1 + i)$. With two compounding periods involved, it has two factors of $(1 + i)$. Each successive compounding period multiplies a further $(1 + i)$ onto the equation. This makes the exponent on the $(1 + i)$ exactly equal to the number of times that interest is converted to principal during the transaction. ## The Formula First, you need to know how many times interest is converted to principal throughout the transaction. You can then calculate the future value. Use Formula 9.2A below to determine the number of compound periods involved in the transaction. $\colorbox{LightGray}{Formula 9.2A}\; \color{BlueViolet}{\text{Number of Compound Periods:}\;n=C/Y \times \text{(Number of Years)}}$ where, C/Y is the number of compounding periods per year. Once you know n, substitute it into Formula 9.2B, which finds the amount of principal and interest together at the end of the transaction, or the future (maturity) value, FV. $\colorbox{LightGray}{Formula 9.2B}\; \color{BlueViolet}{\text{Future (Maturity) Value:}\;FV=PV \times (1+i)^n}$ where, PV is the resent value or principal. This is the starting amount upon which compound interest is calculated. i is the periodic interest rate from Formula 9.1. n is the number of compound periods from Formula 9.2A. ### Important Notes Calculating the Interest Amount (I): In any situation of lump-sum compound interest, you can isolate the interest amount using the formula $I=FV−PV$. ### How It Works Follow these steps to calculate the future value of a single payment: Step 1: Calculate the periodic interest rate (i) using the formula $i=\frac{\text{Nominal Rate (I/Y)}}{\text{Compounds per Year (C/Y)}}$ Step 2: Calculate the total number of compound periods (n) using the formula $n=C/Y \times \text{(Number of years)}$ Step 3: Calculate the future value using the formula $FV=PV(1+i)^n$ Note: You will first need to calculate i and n using steps 1 and 2. ## Your BAII Plus Calculator We will be using the function keys that are presented in the third row of your calculator, known as the TVM row or (time value of money row). The five buttons located on the third row of the calculator are five of the seven variables required for time value of money calculations. This row’s buttons are different in colour from the rest of the buttons on the keypad. The table below relates each button (variable) to its meaning. Table 9.2.1. BA II Plus Calculator Variables and Meanings Variable Meaning N Number of compounding periods I/Y Interest rate per year (nominal interest rate). This is entered in percent form (without the % sign). For example, 5% is entered as 5. PV Present value or principal PMT Periodic annuity payment. For lump sum payments set this variable to zero. FV Future value or maturity value. C/Y Pressing 2ND key then I/Y will open the P/Y worksheet. P/Y stands for periodic payments per year and this will be covered in annuities. We only need to assign a value for C/Y as the calculation does not involve an annuity. We need to set payments per year (P/Y) to the same value as the number of compounding periods per year (C/Y) then press ENTER. When you scroll down (using the down arrow key), you will notice that C/Y will automatically be set to the same value. Pressing 2nd then CPT (Quit button) will close the worksheet. To enter any information into any one of these buttons, key in the data first and then press the appropriate button. For example, if you want to enter N = 34, then key in 34 followed by pressing N. ## Cash Flow Sign Convention Calculating FV (PV is given) For investments: When money is invested (paid-out), this amount is considered as a cash-outflow and this amount has to be entered as a negative number for PV. For Loans: When money is received (loaned), this amount is considered as a cash-inflow and this amount has to be entered as a positive number for PV. Calculating PV (FV is given) For investments: When you receive your matured investment at the end of the term this is considered as a cash-inflow for you and the future value should be entered as a positive amount. For Loans: When the loan is repaid at the end of the term this is considered as a cash-outflow for you and the future value should be entered as a negative amount. ### Important Notes When you compute solutions on the BAII Plus calculator, one of the most common error messages displayed is “Error 5.” This error indicates that the cash flow sign convention has been used in a manner that is financially impossible. Some examples of these financial impossibilities include loans with no repayment or investments that never pay out. In these cases, the PV and FV have been incorrectly set to the same cash flow sign. ## BAII Plus Memory Your calculator has permanent memory. Once you enter data into any of the time value buttons it is permanently stored until • You override it by entering another piece of data and pressing the button; • You clear the memory of the time value buttons by pressing 2nd CLR TVM before proceeding with another question; or • The reset button on the back of the calculator is pressed. ### Example 9.2.1: Making an Investment If you invested5,000 for 10 years at 9% compounded quarterly, how much money would you have? What is the interest earned during the term?
Solution:
The timeline for the investment is below.
Step 1: Given information:
PV = 5,000; I/Y = 9%; C/Y = 4
Step 2: Calculate the periodic interest rate, i.
$i=\frac{I/Y}{C/Y}=\frac{9\%}{4}=2.25\%=0.0225$
Step 3: Calculate the total number of compoundings, n.
$n=C/Y \times (\text{Number of Years})=4 \times 10=40$
Step 4: Solve for the future value, FV.
$FV=\5,\!000(1+0.0225)^{40}=\12,\!175.94$
Step 5: Find the interest earned.
$I=FV-PV= \12,\!175.94-\5,\!000=\7,\!175.94$
Calculator instructions:
Table 9.2.2. Calculator Instructions for Example 9.2.1
N I/Y PV PMT FV P/Y C/Y
40 9 -5,000 0 ? 12 12
After 10 years, the principal grows to $12,175.94, which includes your$5,000 principal and $7,175.94 of compound interest. ## Future Value Calculations with Variable Changes What happens if a variable such as the nominal interest rate, compounding frequency, or even the principal changes somewhere in the middle of the transaction? When any variable changes, you must break the timeline into separate time fragments at the point of the change. To arrive at the solution, you need to work from left to right one time segment at a time using the future value formula ### How It Works Follow these steps when variables change in calculations of future value based on lump-sum compound interest: Step 1: Read and understand the problem. Identify the present value. Draw a timeline broken into separate time segments at the point of any change. For each time segment, identify any principal changes, the nominal interest rate, the compounding frequency, and the length of the time segment in years. Step 2: For each time segment, calculate the periodic interest rate (i) using Formula 9.1. Step 3: For each time segment, calculate the total number of compound periods (n) using Formula 9.2A. Step 4: Starting with the present value in the first time segment (starting on the left), solve for the future value using Formula 9.2B. Step 5: Let the future value calculated in the previous step become the present value for the next step. If the principal changes, adjust the new present value accordingly. Step 6: Using Formula 9.2B calculate the future value of the next time segment. Step 7: Repeat steps 5 and 6 until you obtain the final future value from the final time segment. ### Important Notes The BAII Plus Calculator: Transforming the future value from one time segment into the present value of the next time segment does not require re-entering the computed value. Instead, apply the following technique: 1. Load the calculator with all known compound interest variables for the first time segment. 2. Compute the future value at the end of the segment. 3. With the answer still on your display, adjust the principal if needed, change the cash flow sign by pressing the ± key, and then store the unrounded number back into the present value button by pressing PV. Change the N, I/Y, and C/Y as required for the next segment. 4. Return to step 2 for each time segment until you have completed all time segments. ### Concept Check #### Example 9.2.2: Delaying a Facility Upgrade Five years ago Coast Appliances was supposed to upgrade one of its facilities at a quoted cost of$48,000. The upgrade was not completed, so Coast Appliances delayed the purchase until now. The construction company that provided the quote indicates that prices rose 6% compounded quarterly for the first 1½ years, 7% compounded semi-annually for the following 2½ years, and 7.5% compounded monthly for the final year. If Coast Appliances wants to perform the upgrade today, what amount of money does it need?
Solution:
The timeline below shows the original quote from five years ago until today.
Step 1: First time segment:
PV1 = 48,000; I/Y = 6%; C/Y = 4; Years = 2 $i=\frac{I/Y}{C/Y}=\frac{6\%}{4}=1.5\%$ $n =C/Y \times (\text{Number of Years})=4 \times 1.5=6$ Find FV1 \begin{align} FV_1 &= PV_1(1 + i)^n\\ &= \48,\!000 (1 + 0.015)^6\\ &= \24,\!500(1.015)^6\\ &= \52,\!485.27667 \end{align} This becomes PV2 for the next calculation in Step 2. Step 2: Second line segment: PV2 = FV1 =52,485.27667; I/Y = 7%; C/Y = 2; Years = 2.5
$i=\frac{I/Y}{C/Y}=\frac{7\%}{2}=3.5\%$
$n =C/Y \times (\text{Number of Years})= 2 \times 2.5=5$
Find FV2
\begin{align} FV_2 &= PV_2(1 + i)^n\\ &= \52,\!485.27667 (1+0.035)^5\\ &= \62,\!336.04435 \end{align}
This becomes PV3 for the next calculation in Step 3.
Step 3: Third line segment:
PV3 = FV2 = 62,336.04435; I/Y = 7.5%; C/Y = 12; Years = 1 $i=\frac{I/Y}{C/Y}=\frac{7.5\%}{12}=0.625\%$ $n =C/Y \times (\text{Number of Years})= 12 \times 1=12$ Find FV3 \begin{align} FV_3 &= PV_3(1 + i)^n\\ &= \62,\!336.04435 (1+0.00325)^{12}\\ &= \67,\!175.35 \end{align} The future value is67,175.35.
Calculator instruction:
Table 9.2.3. Calculator Instructions for Steps 1-3, Example 9.2.2.
Step N I/Y PV PMT FV P/Y C/Y
1 6 6 48,500 0 ? 4 4
2 5 7 52,485.27667 0 ? 2 2
3 12 7.5 62,336.04435 0 ? 12 12
Coast Appliances requires $67,175.35 to perform the upgrade today. This consists of$48,000 from the original quote plus $19,175.35 in price increases. #### Example 9.2.3: Making an Additional Contribution Two years ago Lorelei placed$2,000 into an investment earning 6% compounded monthly. Today she makes a deposit to the investment in the amount of $1,500. What is the maturity value of her investment three years from now? Solution: The timeline for the investment is below. Step 1: First time segment: PV1 =$2,000; I/Y = 6%; C/Y = 12; Years = 2
$i=\frac{I/Y}{C/Y}=\frac{6\%}{12}=0.5\%$
$n =C/Y \times (\text{Number of Years})=12 \times 2=24$
Find FV1
\begin{align} FV_1 &= PV_1(1 + i)^n\\ &= \2,\!000 (1 + 0.005)^24\\ &= \2,\!000(1.005)^24\\ &= \2,\!254.319552 \end{align}
$2,254.319552 +$1,500 = 3,754.319552 This becomes PV2 for the second line segment in Step 2. Step 2: Second line segment: PV2 = FV1 = 3,754.319552; I/Y = 6%; C/Y = 12; Years = 3 $i=\frac{I/Y}{C/Y}=\frac{6\%}{12}=0.5\%$ $n =C/Y \times \text{(Number of Years)}= 12 \times 3=36$ Find FV2 \begin{align} FV_2 &= PV_2(1 + i)^n\\ &= \3,\!754.319552 (1+0.005)^{36}\\ &= \4,\!492.72 \end{align} The future value is4,492.72
Calculator instructions:
Table 9.2.4. Calculator Instructions for Example 9.2.3
Step N I/Y PV PMT FV P/Y C/Y
1 24 6 -2,000 0 ? 12 12
2 36 6 -3,754.319552 0 ? 12 12
Three years from now Lorelei will have $4,492.72. This represents$3,500 of principal and $992.72 of compound interest. ### Exercises In each of the exercises that follow, try them on your own. Full solutions are available should you get stuck. 1. Find the future value if$53,000 is invested at 6% compounded monthly for 4 years and 3 months. (Answer: $68,351.02) 1. Find the future value if$24,500 is invested at 4.1% compounded annually for 4 years; then 5.15% compounded quarterly for 1 year, 9 months; then 5.35% compounded monthly for 1 year, 3 months. (Answer: $33,638.67) 1. Nirdosh borrowed$9,300 4¼ years ago at 6.35% compounded semi-annually. The interest rate changed to 6.5% compounded quarterly 1¾ years ago. What amount of money today is required to pay off this loan? (Answer: $12,171.92) Timeline for exercise 3 is included in Solution to Exercises. ## Image Descriptions Figure 9.2.0: Picture of the BAII Plus calculator showing the “Frequency Functions”, and the “Time Value of Money Buttons”.[Back to Figure 9.2.0] Figure 9.2.1: Timeline showing PV =$35,000 at Today (on the Left) with an arrow pointing to the end (on the Right) (10 years) where FV = ? and 9% quarterly throughout. [Back to Figure 9.2.1]
Figure 9.2.2: Timeline: PV1 = $48,000 at 5 years ago moving to 3.5 years ago at 6% quarterly to become FV1. At 3.5 years ago, FV1 becomes PV2 which moves to 1 year ago at 7% semi-annually to become FV2. At 1 years ago, FV2 becomes PV3 which moves to Today at 7.5% monthly to become FV3 = ?. [Back to Figure 9.2.2] Figure 9.2.3: Timeline: At 2 years ago, FV1 =$2,000 moves to Today at 6% monthly to become FV1. At Today, there is a\$1,500 deposit. At Today, FV1 becomes PV2 which moves to 3 years at 6% monthly to become FV2 = ?. [Back to Figure 9.2.3]
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# Assignment 2: Completing the Parabolic Square
## Motivation
In high school mathematics, one common task is graphing a parabola, commonly given in the form
y = x^2 + Bx + C
Searching Google on graphing parabolas found about 167,000 results, including Clayton State, and Regents Prep, and Sparknotes, and many many more. Quickly glancing through the results, the top few suggest making a data table, and then connecting points with a smooth curve. While this accomplishes the task, it doesn't give us any feel for shape. Graphing Calculator can easily plot parabolas for us. This Graphing Calculator file produced the following window. It's tough to see similarities when the equations are in this form, and we have to work to find the vertex.
There's a much better way to show the vertex, but to do so we need a different representation. We can complete the square! Unlike the name of this reference website, the math may not be fun, but it is very useful. To complete the square in the equation above:
y = x^2 + Bx + C = x^2 - 2 ~ \left( - \frac{B}{2} \right) ~ x + C
y = \left[ x^2 - 2 ~ \left( - \frac{B}{2} \right) ~ x + \frac{B^2}{4} \right] + C - \frac{B^2}{4}
y = \left[ x - \frac{B}{2} \right]^2 + \left(C - \frac{B^2}{4} \right)
We have two numbers to manipulate, one inside the square and one outside. It's a little awkward, because inside the square we have a minus and outside we have a plus, but mathematical equations aren't always beautiful and neat. To make things easier, let's call the value inside H and the value outside V. This makes our equation
y = \left[ x - H \right]^2 + V
## Graphs for H and V
Let's start by examining one graph in Graphing Calculator, y = \left[ x - 2 \right]^2 + 1.5 . The red dot indicates the vertex, which lies at x = 2 and y = 1.5, right on H and V.
You might be suspicious that I specifically chose 2 and 1.5, so let's look at other values of H and V. In all five cases shown, the vertex lies at the point (H, V).
If you like, you can experiment with various values of H and V through a GeoGebra Java applet. Click through to the handv.html page. Once the applet loads, you can click on the play button towards the bottom to watch an animation, which then becomes a pause button to stop. Alternatively, you can click on the H and V buttons on the bottom to drag them around the space.
## Mathematics of the Vertex
Now that we've seen the relationship graphically, we can find the vertex mathematically. The value of Y sums the constant V and the value of \left[ x - H \right]^2 . We're working with real numbers, so the value of the square is never negative, and will be minimized when \left[ x - H \right] = 0 . This minimum occurs at x = H. At this point, we solve for the minimum value of Y:
y = \left[ H - H \right]^2 + V = 0 + V = V
Thus, our minimum value, the vertex, is at (H, V), as we've seen. If we need the original formula, we can solve B = -2 H and then substitute into V = C - \frac{B^2}{4} to find C = V + H^2 . Looking at both parabolic forms,
y = \left[ x - H \right]^2 + V = x^2 + (- 2 H) x + (V + H^2)
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Main content
## High school geometry
### Course: High school geometry>Unit 4
Lesson 1: Definitions of similarity
# Getting ready for similarity
Practicing identifying proportional relationships and solving equations with proportions helps us get ready to learn about similarity.
Let’s refresh some concepts that will come in handy as you start the similarity unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead.
This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding similarity. If you have not yet mastered the Congruent triangles lesson or the Dilations preserved properties lesson, it may be helpful for you to review those before going farther into the unit ahead.
## Identifying proportional relationships
### What is this, and why do we need it?
A relationship between two quantities is proportional if the ratio between those quantities is always equivalent. We will look at side length ratios to find out whether triangles are similar or not.
### Practice
Problem 1
Triangle A has a height of 2, point, 5, start text, space, c, m, end text and a base of 1, point, 6, start text, space, c, m, end text. The height and base of triangle B are proportional to the height and base of triangle A.
Which of the following could be the height and base of triangle B?
Choose 3 answers:
For more practice, go to Proportional relationships.
### Where will we use this?
Here are a few of the exercises where reviewing proportional relationships might be helpful:
## Solve equations with proportions
### What is this, and why do we need it?
When two ratios are equal, we create a proportion equation. If we multiply the equation by both denominators, we can solve the resulting equation just like a linear (or quadratic, but not in this unit) equation. We will set up equations with proportions to find lengths in similar figures.
### Practice
Problem 2.1
• Current
Solve for m.
Do not round. If needed, write your answer as a fraction.
start fraction, 8, divided by, 10, end fraction, equals, start fraction, 6, divided by, m, end fraction
m, equals
For more practice, go to Solving proportions 2.
### Where will we use this?
Here are a few exercises where reviewing proportions equations might be helpful.
## Want to join the conversation?
• i do not understand why he give me a question about:q/q+5=12/27.and suddenly next stop just became:27q=12q+60.what the hell is that number 60 coming form?
(⊙⊙?)
(15 votes)
• When you cross multiply, you have to make sure to distribute if a numerator/denominator has multiple terms in it. Here's how the equation looks directly after you cross-multiply:
(27) q = (q + 5) 12
The 12 distributes to both the q and the 5, creating:
27q = 12q + 60
15q = 60
q = 4
(29 votes)
• how do you math?
(12 votes)
• ... the q/q+5=12/27 is 4. It's not that hard tbh
(3 votes)
• bro no ones cares lmao
(12 votes)
• Does anyone know practical applications of similarity concepts in common engineering fields? I'm interested in the practical applications of these otherwise abstract concepts, and was wondering if anyone had any interesting insights. Thanks!
(5 votes)
• Say an engineer makes a prototype of a design. A company likes it but wants to adjust the size a bit. The engineer must use similarity to 'scale up' or 'scale down' the prototype by a certain amount.
(7 votes)
• why did you put 60 over 8?
(3 votes)
• 8/10 = 6/x
you cross multiply
8 * x = 8x
6 * 10 = 60
then you dive 60 over 8
which equals 7.5
(10 votes)
• why is it 10m and how does that get rid of the ten when its not negative
(5 votes)
• It is a fraction...
(0 votes)
• why is it 10m and how does that get rid of the ten when its not negative
(2 votes)
• You multiply the 10 by 1/10 so you have 10/10 which simplifies into 1
(3 votes)
• why is this important in life.
(1 vote)
• its not tbh
(2 votes)
• For the last problem, I solved it by looking at the percentage of 12/27 and then realized that 4/9 has the same rate. Is there a more efficient way of doing this?
(1 vote)
• Reduce the fraction, solve for q; in this case, solving for q becomes unnecessary once you've reduced, because most people will quickly recognize that q must be four.
This actually brings up a good general principle in math: do what you know you can do, then look at the problem again, and see if it has gotten easier. Simplifying, expanding, factoring, substituting equivalent expressions, are all good things to try if you get stuck. Of course, you'll want a pretty good idea of how to do these already when you need them, say on a test, so do a variety of exercises to expose yourself to different types of problems.
(2 votes)
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Adding Subtracting Multiplying And Dividing Mixed Numbers Worksheet With Answers
This multiplication worksheet focuses on teaching individuals how to psychologically increase total amounts. College students may use custom made grids to match exactly one concern. The worksheets also protectfractions and decimals, and exponents. You can even find multiplication worksheets having a distributed home. These worksheets really are a need to-have for the arithmetic school. They could be utilized in type to figure out how to emotionally increase total numbers and line them up. Adding Subtracting Multiplying And Dividing Mixed Numbers Worksheet With Answers.
Multiplication of entire figures
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets will help you grasp this basic strategy. You may opt for one digit multipliers or two-digit and a few-digit multipliers. Capabilities of 10 will also be an incredible choice. These worksheets will assist you to exercise extended practice and multiplication reading the amounts. Also, they are a terrific way to assist your child understand the necessity of knowing the several types of whole numbers.
Multiplication of fractions
Experiencing multiplication of fractions with a worksheet will help teachers plan and make lessons proficiently. Making use of fractions worksheets allows educators to easily determine students’ knowledge of fractions. Individuals may be pushed to finish the worksheet in just a a number of time and then label their solutions to see in which they need further more training. Pupils may benefit from phrase things that associate maths to actual-life situations. Some fractions worksheets incorporate samples of contrasting and comparing figures.
Multiplication of decimals
If you multiply two decimal numbers, make sure to class them up and down. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. As an example, 01 by (11.2) by 2 could be equivalent to 01 by 2.33 by 11.2 unless the item has decimal locations of below two. Then, the item is round towards the local total variety.
Multiplication of exponents
A math worksheet for Multiplication of exponents will allow you to training multiplying and dividing numbers with exponents. This worksheet will also provide conditions that requires college students to increase two various exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Aside from, you can even key in specific recommendations on the worksheet itself. When you’re concluded, it is possible to simply click “Generate” along with the worksheet will be saved.
Division of exponents
The fundamental tip for department of exponents when multiplying figures would be to deduct the exponent from the denominator from your exponent within the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. For example, \$23 separated by 4 will identical 27. This method is not always accurate, however. This technique can bring about frustration when multiplying amounts that happen to be too big or not big enough.
Linear features
If you’ve ever rented a car, you’ve probably noticed that the cost was \$320 x 10 days. So the total rent would be \$470. A linear function of this type has got the kind f(by), where ‘x’ is the amount of times the auto was booked. Additionally, they have the shape f(by) = ax b, where ‘b’ and ‘a’ are actual amounts.
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# What Is The Definition Of Arithmetic? (Correct answer)
́ (. -. ἀριθμητική, arithmētikḗ — ἀριθμός, arithmós «») — , , . ; , .
## What is the definition of arithmetic in math?
Arithmetic (a term derived from the Greek word arithmos, “number”) refers generally to the elementary aspects of the theory of numbers, arts of mensuration (measurement), and numerical computation (that is, the processes of addition, subtraction, multiplication, division, raising to powers, and extraction of roots).
## What is arithmetic and example?
The definition of arithmetic refers to working with numbers by doing addition, subtraction, multiplication, and division. An example of arithmetic is adding two and two together to make four.
## What is the definition of arithmetic solution?
A value, or values, we can put in place of a variable (such as x) that makes the equation true. Example: x + 2 = 7. When we put 5 in place of x we get: 5 + 2 = 7. 5 + 2 = 7 is true, so x = 5 is a solution.
## What part of math is arithmetic?
Arithmetic is one of the branches of maths that is composed of the properties of the application in addition, subtraction, multiplication, and division, and also the study of numbers. It is a part of elementary number theory.
## What is the arithmetic mean between 10 and 24?
Using the average formula, get the arithmetic mean of 10 and 24. Thus, 10+24/2 =17 is the arithmetic mean.
## What is arithmetic and geometric?
An Arithmetic Sequence is such that each term is obtained by adding a constant to the preceding term. This constant is called the Common Difference. Whereas, in a Geometric Sequence each term is obtained by multiply a constant to the preceding term.
## What are the 5 examples of arithmetic sequence?
= 3, 6, 9, 12,15,. A few more examples of an arithmetic sequence are: 5, 8, 11, 14, 80, 75, 70, 65, 60,
## What are the 4 branches of arithmetic?
Arithmetic has four basic operations that are used to perform calculations as per the statement:
• Subtraction.
• Multiplication.
• Division.
## How many chapters are there in arithmetic?
There are also slight differences between the various accessible formats, also as a result of specific adaptations made for each format. The Math Review consists of 4 chapters: Arithmetic, Algebra, Geometry, and Data Analysis.
## Is arithmetic and math the same thing?
When you’re referring to addition, subtraction, multiplication and division, the proper word is ” arithmetic,” maintains our math fan. “Math,” meanwhile, is reserved for problems involving signs, symbols and proofs — algebra, calculus, geometry and trigonometry.
## Who invented zero?
The first modern equivalent of numeral zero comes from a Hindu astronomer and mathematician Brahmagupta in 628. His symbol to depict the numeral was a dot underneath a number.
## Definition of ARITHMETIC
Arith·me·tic|ə-ˈrith-mə-ˌtik1a: It is a field of mathematics that is concerned with the nonnegative real numbers, which may include the transfinite cardinals at times, and with the application of the operations of addition, subtraction, multiplication, and division to them. It is sometimes referred to as an arithmetic treatise.
## Other Words fromarithmetic
The word arithmetic comes from the Greek letters er- ith- ti- kl, which means “arithmetical.” The word arithmetical comes from the Greek letters er- ith- ti- kl, which means “arithmetically.” The word arithmetician comes from the Greek letters er- ith- ti- shn, which means “analytical mathematician.”
## Synonyms forarithmetic
• The terms calculation, calculus, ciphering, computation, figures, and figuring are all used to describe math, mathematics, number crunching, and numbers.
## Examples ofarithmeticin a Sentence
A piece of software that will perform thearithmetic for you. I haven’t done any thearithmeticyet calculations, but I have a feeling we’re going to lose money on this transaction. Recent Web-based illustrations According to him, the mathematics of politics was always more potent than the chemistry of politics. 5th of December, 2021, by David M. Shribman of the Los Angeles Times Nonetheless, the number of parties has increased from four to seven, and the two traditional main parties have reduced in size, altering the math of creating a government that receives more than 50 percent of the popular vote.
On October 16, 2021, Alixel Cabrera wrote in The Salt Lake Tribune that Israelis, on the other hand, are well aware of the fact that Hezbollah’s arsenal is ten times larger and considerably more advanced than that of Hamas.
—Rick Miller, Forbes, published on June 24, 2021 Deliberate demonstrations, fund-raising calls on MSNBC, and enraged appearances on the cable news channel will not alter the difficult arithmetic of Capitol Hill.
The Los Angeles Times published an article on June 6, 2021, titled Despite the fact that most individuals believe that economicarithmeticas are their fundamental foundation for making life decisions, this conclusion is founded on erroneous assumptions about how people make decisions in their daily lives.
It is not the opinion of Merriam-Webster or its editors that the viewpoints stated in the examples are correct. Please provide comments. More information may be found here.
## First Known Use ofarithmetic
During the fifteenth century, in the sense stated atsense 1a
## History and Etymology forarithmetic
The Middle Englisharsmetrik is derived from Anglo-Frencharismatike, from Latinarithmetica, from Greekarithmtikosarithmetical, fromarithmeinto count, fromarithmosnumber; it is related to the Old Englishrmnumber and maybe to the Greekarariskeinto fit.
Make a note of this entry’s “Arithmetic.” This entry was posted in Merriam-Webster.com Dictionary on February 9, 2022 by Merriam-Webster. More Definitions forarithmeticarithmetic|arithmeticarithmetic|arithmeticarithmetic|arithmeticarithmetic|arithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarithmeticarith
## Kids Definition ofarithmetic
number one: a branch of mathematics that studies the addition, subtraction, multiplication, and division of numbers 2:the act or procedure of adding, removing, multiplying, or dividing Other Words fromarithmeticarithmeticer- ith- me- tikorarithmetical- ti- kladjective fromarithmeticarithmeticer- ith- me- tikorarithmetical- ti- kladjective fromarithmeticarithmeticer- ith- me- tikorarithmetical- ti- kladjective
## Definition of arithmetic
• Examples
• British
• Scientific
• Top Definitions
• Quiz
• Related Content
• Examples
This indicates the grade level of the word depending on its complexity./nounr mtk;adjectiver mtk/ /nounr mtk;adjectiver mtk/ This indicates the grade level of the word based on its difficulty. The method or process of calculating using numbers is denoted by the term the branch of mathematics that is the most fundamental. Higher arithmetic and theoretical arithmetic are also terms used to refer to this subject. The study of the divisibility of whole numbers, the remainders after division, and other aspects of the theory of numbers.
Also known as arithmetic.
In effect, this exam will determine whether or not you possess the necessary abilities to distinguish between the terms “affect” and “effect.” Despite the wet weather, I was in high spirits on the day of my graduation celebrations.
## Origin ofarithmetic
1500–50;Latinarithmica,feminine singular ofarithmticus;1200–50; Old Frencharismetique was replaced by the Greekarithmtik (téchn) (numbers art, skill), which is equal toarithmé (ein) to reckon plus -t (o)- verbal adjective +-ik-ic; this word replaced Middle Englisharsmet (r) ikeOld Frencharismetique. Medieval Latinarismtica, with a focus on Late Greco-Roman culture
## Words nearbyarithmetic
Aristotle contemplating the bust of Homer, Aristotle’s lantern, aristotype, arithmancy, arithmetic, arithmetician, arithmetic mean, arithmetic progression, -arium, aristotype, arithmancy, arithmetic, arithmetician, arithmetic mean, arithmetic progression, -arium, AriusDictionary.com Based on the Random House Unabridged Dictionary, Random House, Inc. published the Unabridged Dictionary in 2012.
## Words related toarithmetic
• When word2vec was trained on a large dataset, it was discovered that its embeddings captured significant semantic correlations between words that could be revealed by performing basic arithmetic operations on the vectors. I believe that these are clichés that mathematicians like employing, and that they are extremely alienating to individuals who, for whatever reason, did not learn about modulararithmetic in kindergarten. When you distill a set a limited number of times, you end up with a set that is dense enough to have to include arithmetic progressions
• Roth was able to demonstrate that. Your list should include an endless number of arithmetic progressions of every length, according to Erds’ hypothesis, assuming that the density criterion is satisfied.
• Consider the following scenario: you’re walking down the number line and you want to save every number that doesn’t fulfill anarithmetic progression
• Whatever had to do with the Count (or, to be more official, the Count von Count), who taught numbers and fundamental mathematics via songs
• Because it was a question of arithmeticlogic that one of them was speaking the truth in the J-K shooting, the investigation into the incident was quite straightforward. NEW DELHI, India – New Delhi is the capital of India. It has been announced that Narendra Modi will be the next Prime Minister of India, and the math behind his election triumph is astounding. The “top 100” books were only 75 books, according to a simplearithmetic count of the list. In the words of Rothenberg, “the president has vowed to reform thearithmetic.” It was divided into three topics that were more or less isolated from one another: arithmetic, algebra, and Euclid. Up until this point, I had always assumed that I loathed anything that had the shape of math in it. The third episode features a guy dressed in ancient Colburn’sArithmetic who is herding his flock of sheep or geese to the marketplace. His attention was drawn to thearithmeticclass’s recitation and he discovered that only objects of the same denomination could be deducted from each other
• Let’s say you send her up, Flora—you’ll probably want to go sketch or practice, and she can do herarithmetichere or read to me while you’re away.
## British Dictionary definitions forarithmetic
Number theory is an area of mathematics that is concerned with numerical computations such as addition and subtraction as well as multiplication and division. a computation or a series of calculations that include numerical operations understanding of or proficiency in the use of arithmetichis There’s nothing better than figuring things out with numbers.adjective(rmtk)arith’meticalof, related to, or involving figuring things out with numbers.
## Word Origin forarithmetic
From Latinarithmtica, from Greekarithmtik, fromarithmeinto count, fromarithmosnumber, fromarithmosnumber 2012 Digital Edition of the Collins English Dictionary – Complete Unabridged Edition (William Collins SonsCo. Ltd. 1979, 1986) In 1998, HarperCollinsPublishers published the following books: 2000, 2003, 2005, 2006, 2007, 2009, and 2012.
## Scientific definitions forarithmetic
The mathematics of integers, rational numbers, real numbers, or complex numbers when subjected to the operations of addition, subtraction, multiplication, and division is called number theory. The American Heritage® Science Dictionary is a resource for those interested in science. The year 2011 is the year of the copyright. Houghton Mifflin Harcourt Publishing Company is the publisher of this book. All intellectual property rights are retained.
## What is Arithmetic? – Definition, Facts & Examples
What is the definition of Arithmetic? Arithmetic is a discipline of mathematics that is concerned with the study of numbers and the application of various operations on those numbers. Addition, subtraction, multiplication, and division are the four fundamental operations of mathematics. These operations are represented by the symbols that have been provided. Addition:
• The process of taking two or more numbers and adding them together is referred to as the addition. Or to put it another way, it is the entire sum of all the numbers. The addition of whole numbers results in a number that is bigger than the sum of the numbers that were added.
You might be interested: How Many Syllables In Arithmetic? (Correct answer)
For example, if three children were playing together and two additional children joined them after a while. In total, how many children are there? If you want to represent this mathematically, you may write it as follows: 3 plus 2 equals 5; As a result, a total of 5 children are participating. Subtraction:
• Subtraction is the technique through which we remove things from a group that they were previously part of. When a number is subtracted from another number, the numerical value of the original number decreases.
For example, eight birds are perched on a branch of a tree. After a while, two birds take off in different directions. What is the number of birds on the tree? As a result, there are only 6 birds remaining on the tree after subtracting 8 from 2. Multiplication:
• As an illustration, eight birds are perched on a branch of a large tree. Eventually, two birds take flight and disappear. Which birds are on the tree and how many are there altogether? The number of birds on the tree has been reduced from eight to six. Multiplication:
Consider the following scenario: Robin went to the garden three times and returned back five oranges each time. What was the total number of oranges Robin brought? Robin went to the garden three times to find a solution. He showed up with five oranges every time. This may be expressed numerically as 5 x 3 = 15 oranges, for example. Division:
• Divide and conquer is the process of breaking down a huge thing or group into smaller portions or groupings. Generally speaking, the dividend refers to the number or bigger group that is divided. The dividend is divided by a number, which is referred to as the divisor. In mathematics, thequotient is the number derived by multiplying the dividend by a divisor. The number that is left over after dividing is referred to as the remnant.
For example, when 26 strawberries are distributed among 6 children, each child receives 4 strawberries, leaving 2 strawberries behind. Fascinating Facts
• Algebra, Geometry, and Analysis are the three additional fields of mathematics that are studied. The term “arithmetic” comes from the Greek arithmtika (tekhna), which literally translates as “(art) of counting,” as well as the word arithmos, which literally translates as “number.”
## arithmetic
The latter form is purely arithmetic, but the former suggests a mental effort of some nature. In mathematics, this implies putting a strong emphasis on topics like understanding fractions and developing fluency in arithmetic. But this isn’t simply a dispute over numbers in a spreadsheet. Because of the large number of numbers involved, simplearithmeticis is not an option. During third-year maths, we witnessed her flying through the air after tripping over her shoelace. You should be aware that schools are not just for the purpose of teaching children technical skills such as reading, writing, and arithmetic.
1. Then they come to terms with what has transpired and begin to perform their own private arithmetic.
2. Even a bright eighth grader can understand that thearithmeticis incorrect: individuals are three-dimensional beings with dimensions of height, breadth, and thickness.
3. The knowledge of whatarithmeticis is not necessary for becoming a wonderful and beautiful computing machine, but it is beneficial.
4. Don’t be concerned if the maths doesn’t quite work out as expected.
These samples are drawn from corpora as well as from other online sources. Any viewpoints expressed in the examples do not necessarily reflect the views of the Cambridge Dictionary editors, Cambridge University Press, or its licensors, who are not represented by the examples.
## Arithmetic – Definition, Meaning & Synonyms
Arithmetici is a term that refers to mathematics in general, and more particularly to the branches of mathematics that deal with numbers and calculations. Having strong arithmetic skills means you’re proficient in arithmetic, which is an important component of math. Addition, subtraction, division, and multiplication are all skills that come in handy when working with numbers. Arithmetic is concerned with the process of calculating. Arithmetic is required to solve the vast majority of math problems, including practically all word problems.
Examples of arithmetic definitions
1. The theory of numerical computations is an area of pure mathematics that deals with the theory of numerical computations. more information less information types:algorism Arabic figures are used in the computation. the study and development of mathematical ideas for their own sake rather than for their immediate application
2. An adverb referring to or involving arithmetic
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## arithmetic
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## Fundamental definitions and laws
The process of finding the number of objects (or elements) existing in a collection (or set) is referred to as counting. The numbers acquired in this manner are referred to as counting numbers or natural numbers (for example, 1, 2, 3,.). There is no item in a non-existing empty set, and the count returns zero, which when added to the natural numbers gives what are known as the whole numbers. It is claimed that two sets are equal or comparable if they can be matched in such a way that every element from one set is uniquely paired with an element from the other set.
Seeset theory is a hypothesis that states that
Combining two sets of objects that containa andbelements results in the formation of a new set that containsa+b=cobjects when the two sets are combined. It is referred to as thesumofaandb, and each of the latter is referred to as a summand. The act of creating the total is referred to as addition, and the sign + is pronounced as “plus” in this context. When it comes to binary operations, the easiest is the process of merging two things, which is the case here. When applied to three summands, it is clear from the definition of counting that the order of the summands and the order of the operation of addition may be varied without affecting the sum.
1. The commutative law of addition and the associative law of addition are the names given to these two laws of addition.
2. If there is such a numberk, it is known as bis smaller thana (writtenba).
3. It is clear from the foregoing principles that a repeated sum such as 5 + 5 + 5 is independent of the method in which the summands are grouped; it may be expressed as 3 + 5.
4. When you multiply two numbers together, you get a product.
5. When you multiply three numbers together, you get the product of three multiplied by five.
6. As seen in the illustration below, if three rows of five dots each are written, it is immediately evident that the total number of dots in the array is 3 x 5, or 15.
7. As a result of the generality of the reasoning, the statement that the order of the multiplicands has no effect on the product, often known as the commutative law of multiplication, is established.
8. Indeed, the notion that certain things do not commute is critical to the mathematical formulation of contemporary physics, which is a good illustration of how some entities do not commute.
9. This type of legislation is referred to as the associative law of multiplication.
10. The first set consists of three columns of three dots each, or 3 3 dots, and the second set consists of two columns of three dots each, or 2 3 dots.
11. The sum (3 3) + (2 3) is composed of 3 + 2 = 5 columns of three dots each, or (3 + 2) To put it simply, it is possible to demonstrate that the multiplication of an amount of money by a certain number is the same as the sum of two acceptable products.
A law of this nature is referred to as a distributive law.
## Integers
Subtraction has not been presented since it can be described as the inverse of addition, and this is the only justification for this. So the differenceabbetween two numbersa and bis defined as a solutionxof theequationb+x=a is the differenceabbetween two numbers. If a number system is confined to the natural numbers, disparities do not necessarily need to exist; nevertheless, if they do, the five fundamental rules of arithmetic, which have previously been described, can be utilized to demonstrate that they are distinct.
Moreover, the set of whole numbers (including zero) may be expanded to include the solution of the equation 1 + x= 0, that is, the number 1, as well as any products of the form 1 n, wheren is a whole integer, and all other whole numbers.
Negative integers are numbers that have been brought into the system in this fashion for the first time.
## Exponents
The same way that a repeated suma+a+ aofksummands is writtenka, a repeated producta+a+ aofkfactors is writtenak. The numberkis referred to as the exponent, and the base of the powerakis referred to as the powerak. Following directly from the definitions (seethetable), the fundamental laws of exponents are simply deduced, and the other laws are direct implications of the fundamental laws.
## arithmetic
It is the mathematics of integers, rational numbers, real numbers, or complex numbers when they are subjected to the operations of addition, subtraction, multiplication, and division. adj.ar·ith·met·ic(ăr′ĭth-mĕt′ĭk)alsoar′ith·met′i·cal(ăr′ĭth-mĕt′ĭ-kəl) 1.Having to do with or pertaining to arithmetic. 2.Arithmetic progression is used to determine how things change: The rise in the amount of food available is just mathematical. ar′ith·met′i·cal·lyadv.a·rith′me·ti′cian(-tĭsh′ən)n. The Fifth Edition of the American Heritage® Dictionary of the English Language is now available.
Houghton Mifflin Harcourt Publishing Company is the publisher of this book.
## arithmetic
Mathematical operations such as addition, subtraction, multiplication, and division are performed in this branch of mathematics. (rmtk)n1. (Mathematics) the branch of mathematics dealing with numerical computations, such as addition, subtraction, multiplication, and division2. (Mathematics) One or more computations involving numerical operations that are performed on a single number 3. (Mathematics) an understanding of or ability to use arithmetic: he has strong arithmetic skills. adverb (Mathematics) consisting of, connected to, or involving arithmetic ˌarithˈmetically advaˌrithmeˈticiann Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 – HarperCollins Publishers 1991, 1994, 1998, 2000, 2003, 2006, 2007, 2009, 2011, 2014 – Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 – Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 – Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 – Collins English Dictionary – Complete and Unabridged, 12th Edition
## a rith me tic
1.the method or process of computing with figures: the most fundamental branch of mathematics (nr mt k; adj.r mt k)n.1.the method or process of computing with figures: the most elementary branch of mathematics (nr m t k; adj.r mt k)n.1. 2.number theory; the study of the divisibility of whole numbers, the remainders after division, and other aspects of number theory 3.a treatise on the subject of arithmetic adj. Arithmetic is also known as arithmetic, and it is defined as follows: of, related to, or in accordance with the laws of arithmetic.
Kernerman Webster’s College Dictionary, Random House Kernerman Webster’s College Dictionary, 2010 K Dictionaries Ltd.
has copyright protection for the years 2005, 1997, and 1991.
## a·rith·me·tic
Mathematical study of numbers and their characteristics when subjected to the operations of addition, subtraction, multiplication, and division is known as number theory. 2.Calculation based on the processes listed above. Student Science Dictionary, Second Edition, published by American Heritage®. Houghton Mifflin Harcourt Publishing Company has copyright protection for the year 2014. Houghton Mifflin Harcourt Publishing Company is the publisher of this book. All intellectual property rights are retained.
Noun 1 arithmetic- the branch of pure mathematics dealing with the theory of numerical calculationsmath,mathematics,maths- a science (or group of related sciences) dealing with the logic of quantity and shape and arrangementpure mathematics- the branches of mathematics that study and develop the principles of mathematics for their own sake rather than for their immediate usefulnessalgorism- computation with Arabic figuresmiscalculate,misestimate- calculate incorrectly; “I miscalculated the number of guests at the wedding”recalculate- calculate anew; “The costs had to be recalculated”square- raise to the second powercube- raise to the third poweradd together,add- make an addition by combining numbers; “Add 27 and 49, please!”multiply- combine by multiplication; “multiply 10 by 15″raise- multiply (a number) by itself a specified number of times: 8 is 2 raised to the power 3fraction,divide- perform a division; “Can you divide 49 by seven?”halve- divide by two; divide into halves; “Halve the cake”quarter- divide by four; divide into quartersmake- add up to; “four and four make eight”contain- be divisible by; “24 contains 6” Adj. 1 arithmetic- relating to or involving arithmetic; “arithmetical computations”
Based on the WordNet 3.0 clipart collection from Farlex, 2003-2012 Princeton University and Farlex Corporation.
## arithmetic
Noun The Roget’s Thesaurus from the American Heritage® brand. Houghton Mifflin Harcourt Publishing Company has copyright protection for the years 2013 and 2014. Houghton Mifflin Harcourt Publishing Company is the publisher of this book. All intellectual property rights are retained. Translations • • • • • • • • • • • • • • • • • aritmetikaaritmetickaritmetikaritmeettinenaritmetiikkalaskuoppireikningur, talnafrîiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii aritmetikaaritmetinisaritmētikaaritmetikaaritmetikaaritmetikaritmetisk 2005, 8th Edition, Collins Spanish Dictionary – Complete and Unabridged, William Collins SonsCo.
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1971, 1988, 2005, 8th Edition The HarperCollins Publishers, 1992-1993, 1996-1997, 2000-2003-2005,
## arithmetic
N(=calculations) calculsmpl (=calculations) He made a grammatical error in his arithmetic. It has gotten out of hand with his calculations. It is possible that my math is correct. If the calculations are correct, the results are positive. mathematical analysis (of a given situation)l’arithmétique English/French Electronic Resource from Collins Publishers. HarperCollins Publishers, 2005. Collins German Dictionary – Complete and Unabridged, 7th Edition, 2005, by Collins Publishing Company. William Collins Sons Co.
was established in 1980.
## arithmetic
1st Edition of the Collins Italian Dictionary, published by HarperCollins Publishers in 1995.
## arithmetic
Number counting is referred to as (rimtik)nounthe art of counting by numbers. arithmetical(ӕriθˈmetikl)adjective Kernerman English Multilingual Dictionary 2006-2013 K Dictionaries Ltd. Kernerman English Multilingual Dictionary
## Arithmetic Mean Definition
It is the simplest and most generally used measure of amean, or average, since it is the most straightforward to calculate. It is as simple as taking the total of a set of numbers and dividing that sum by the amount of numbers that were used in the series to arrive at the answer. Let’s say you have the numbers 34, 44, 56, and 78 on your hands. The total comes to 212. The arithmetic mean is equal to 212 divided by four, which equals 53. Additionally, people employ a variety of different sorts of means, such as thegeometric mean and theharmonic mean, which come into play in a variety of scenarios in finance and investment.
### Key Takeaways
• Arithmetic mean: The simple average, also known as the total sum of a series of numbers, divided by the number of numbers in that series of numbers
• Because of this, arithmetic mean is not always the most appropriate approach of computing an average in the financial sector, especially when a single outlier might distort the average by a significant amount. Other averages that are more widely employed in finance include the geometric mean and the harmonic mean
• However, the geometric mean is not utilized in finance.
## How the Arithmetic Mean Works
The arithmetic mean retains its significance in the field of finance as well. To give an example, mean earnings predictions are often calculated using the arithmetic mean. Consider the following scenario: you want to know the average earnings projection of the 16 analysts covering a specific stock. To find the arithmetic mean, just add up all of the estimations and divide the total by 16. The same is true if you wish to figure out what a stock’s average closing price was for a specific month.
To find the arithmetic mean, just add up all of the costs and divide by 23 to arrive at the final figure.
As a measure of central tendency, it’s also valuable because it tends to produce relevant findings even when dealing with big groupings of numbers.
## Limitations of the Arithmetic Mean
The arithmetic mean isn’t always the best choice, especially when a single outlier has the potential to significantly distort the mean. Consider the following scenario: you need to estimate the allowance for a group of ten children. Nine of them are given a weekly stipend ranging between \$10 and \$12. The tenth child is entitled to a \$60 stipend. Because of that one outlier, the arithmetic mean will be \$16, not \$16 + \$1. This is not a particularly representative sample of the group. In this specific instance, the medianallowance of ten points could be a more appropriate metric.
It is also not commonly utilized to compute present and future cash flows, which are employed by analysts in the preparation of their forecasts. It is almost certain that doing so will result in erroneous data.
### Important
When there are outliers or when looking at past returns, the arithmetic mean might be deceiving to the investor. In the case of series that display serial correlation, the geometric mean is the most appropriate choice. This is particularly true in the case of investment portfolios.
## Arithmetic vs. Geometric Mean
The geometric mean, which is determined in a different way, is frequently used in these applications by analysts. When dealing with series that demonstrate serial correlation, the geometric mean is the most appropriate choice. This is particularly true in the case of investment portfolios. The majority of returns in finance are connected, including bond yields, stock returns, and market risk premiums, among other things. Because of this, the use of crucial compounding and the geometric mean becomes increasingly important as the time horizon grows.
Taking the product of all the numbers in the series, the geometric mean increases it by the inverse of the length of the series, yielding the geometric mean.
The geometric mean varies from the arithmetic mean in that it takes into consideration the compounding that occurs from one period to the next.
## Example of the Arithmetic vs. Geometric Mean
Suppose the returns on an investment during the previous five years were 20 percent, 6 percent, 10 percent, -1 percent, and 6 percent, respectively. The arithmetic mean would simply put them all together and divide by five, yielding an annualized rate of return of 4.2 percent on average. The geometric mean, on the other hand, would be computed as (1.2 x 1.06 x 0.9 x 0.99 x 1.06) 1/5-1 = 3.74 percent per year average return on the investment. It is important to note that the geometric mean, which is a more accurate computation in this circumstance, will always be less than the arithmetic mean in this situation.
## What does arithmetic mean?
1. Arithmetic adjectivea branch of pure mathematics concerned with the theory of numerical calculations
2. Arithmetical, arithmetic adjectiverelating to or involving arithmetic”arithmetical computations”
3. Arithmetical, arithmetic adjectiverelating to or involving arithmetic
### Wiktionary(3.00 / 2 votes)Rate this definition:
1. Numbers (integers, rational numbers, real numbers, or complex numbers) are mathematically represented by the operations of addition, subtraction, multiplication, and division in the arithmetic domain. It is derived from arsmetike, from arismetique and arithmetica, which are both derived from Ancient Greek (). arithmetic adjectiveAn adjective that refers to, is related to, or is used in arithmetic
2. Arithmetical. It has been in use since the 13th century. arithmetic geometry is the study of numbers and shapes. It is derived from arsmetike, from arismetique and arithmetica, which are both derived from Ancient Greek (). Used in English from the 13th century as an arithmetic term to describe a progression, mean, or other metric that is calculated by addition rather than multiplication the development of numbers in arithmetic It is derived from arsmetike, from arismetique and arithmetica, which are both derived from Ancient Greek (). The term has been in use in English since the 13th century.
### Webster Dictionary(0.00 / 0 votes)Rate this definition:
1. Arithmetic is a term that refers to the science of numbers or the art of calculating using figures. Etymology:
2. Mathematics nouna a book that contains the fundamental concepts of this field Etymology:
### Freebase(0.00 / 0 votes)Rate this definition:
1. Arithmetic, sometimes known as arithmetics, is the oldest and most fundamental subject of mathematics, and it is widely utilized for activities ranging from simple day-to-day counting to complicated scientific and business calculations. Arithmetic is also known as arithmetics in some circles. It entails the study of quantities, particularly as a result of processes that combine numbers in a certain way. In general use, it refers to the qualities that are more straightforward when the standard operations of addition, subtraction, multiplication, and division are performed on numbers with lower values. It is common for professional mathematicians to refer to more sophisticated conclusions in number theory by using the term arithmetic, although this should not be mistaken with simple arithmetic.
### Chambers 20th Century Dictionary(0.00 / 0 votes)Rate this definition:
1. Arithmetic (ar-ith′met-ik,n.the science of numbers: the art of calculating by figures: a book on reckoning) is a term that refers to the study of numbers. — adj.Arithmet′ical.— adv.Arithmet′ically.— n.Arithmetic′ian, a person who is well-versed in arithmetic Arithmetical progression is a series of integers that grow or decrease by a common difference, such as 7, 10, 13, 16, 19, 22
2. Or 12, 1012, 9, 712, 6. Arithmetical progression is also known as arithmetic progression. For example, to calculate the total of a series of words that includes only the first and final terms, multiply the sum of those terms by half the number of terms.
### Dictionary of Nautical Terms(0.00 / 0 votes)Rate this definition:
1. Arithmetic The art of computing with numbers, or the field of mathematics that studies the powers and qualities of numbers
### Editors Contribution(0.00 / 0 votes)Rate this definition:
1. Algebra is the capacity and aptitude that humans have to utilize their minds to perform mathematical calculations. Calculation is a crucial human capacity and skill, and one that should not be taken for granted. MaryCon submitted a submission on April 29, 2020
2. Arithmetic The study of numbers is known as mathematics. Arithmetic is something that everyone does on a daily basis. MaryCon submitted a submission on March 6, 2020
### How to pronounce arithmetic?
1. In Chaldean Numerology, the numerical value of arithmetic is 3
2. In other words, 3 represents the number 3. Mathematical Arithmetic in Pythagorean Numerology has the numerical number of 7
### Examples of arithmetic in a Sentence
1. Coverage Sanders: We’re bringing our campaign to the Republican National Convention. When it comes to math, we’re pretty excellent at it. Jim Sanford: I’d want to thank you for your time. The dog was sold for a thousand dollars per pound, so if you do the math, that’s a very decent deal
2. Jan Ehrenwald’s biography. Aside from the basics of addition and subtraction, Ludwig von Beethoven had never grasped the components of mathematics. A thirteen-year-old child whom he had befriended attempted, but failed, to teach him basic multiplication and division
3. Michael Feroli (Michael Feroli): At least until there is a significant upward revision to the terrible December retail sales number, the weak end-of-quarter consumption profile will make mathematics for first-quarter consumption growth extremely difficult to calculate. Grand Duchess Anastasia Nikolaevna of Russia is referred to as “Anastasia” or “Anastasia” in Russian. Dearly cherished Papa, I’m desperate to see you again. I’ve just finished my math lesson, and I believe I did a good job of learning the material. We are going to a nursing school, and I am really pleased there. It’s rainy and quite humid today, so dress accordingly. With Olga and Tatiana, I’m in Tatiana’s room for the night. When you run across Boba, tell him that your hands are itching. I’m attempting to breed my own herd of earthly poetry. Olga claims that I have a bad stench, but this is not true. I’ll wash myself in your tub when you get back from your vacation. I hope you didn’t forget anything about the history that I shared with you during our stroll. I’m sitting down and scratching the bridge of my nose with my left hand. Olga tried to hit me in the face, but I managed to evade her hand. Be happy and healthy at all times. I have a great deal of admiration for you and want to give you a warm embrace
### Popularity rank by frequency of use
• Arabic
• AritmetèticaCatalan, Valencian
• Aritmetick, aritmetikaCzech
• Rhifyddiaeth, rhifyddegWelsh
• Rechenkunde, arithmetisch, ArithmetikGerman
• Kalkularto, aritmetikoGreek
• Rhifyddia aritméticaSpanish
• AritmetikaBasque
• AritmetikaEsperanto
• Aritmetika Persian
• Arithmétique, d’arithmétique, de l’arithmétiqueFrench
• Arithmétique, d’arithmétique, de l’arithmétique Uimhrochtil, uimhrochta, uimhrochtIrish
• Uimhrochtil, uimhrochta, uimhrochtIrish
• UimhrochtIrish
• The language of Hindi
• AritmetikaIndonesian
• Aritmetico, aritmeticaArmenian
• Aritmetika Italian, Inuktitut, and other languages Kannada
• Aritmtika, aritmtikisksLatvian
• Аритметики, аритметика
• Japanese The Macedonian language
• Marathi
• Aritmetica, rekenkunde
• Arithmetica, calculus Dutch
• Aritmetikk, aritmetiskNorwegian
• Arytmetyczny, arytmetykaPolish
• Arytmetyczny, arytmetyka Polish
• AritméticaPortuguese
• AritmeticăRomanian
• Ариметиески, ариметика
• Aritmética Russian
• Aritmetik, aritmetika
• Aritmetika Albanian
• Aritmetisk, aritmetikSwedish
• Aritmetisk, aritmetik The languages of Telugu, исoTajik, Thai, and Urdu are also spoken. Vietnamese
• Kalkulav-, kalkulav, kalkulavaVolapük
### Get even moretranslations for arithmetic»
• Arabic
• AritmèticaCatalan, Valencian
• Aritmetick, aritmetikaCzech
• Rhifynddiaeth, rhifynddegWelsh
• Rechenkunde, arithmetisch, ArithmetikGerman
• Kalkularto, aritmetikoGreek
• Rhifynddeg Italicized terms: aritméticaSpanish
• AritmetikaBasque
• AritmetikaEsperanto
• AritmetikaSpanish
• AritmetikaBasque Arithmetic, d’arithmetic, de l’arithmétique, laskuoppi, aritmetiikka, aritmeettinenFinnish
• Laskuoppi, aritmetiikka, aritmeettinenFinnish Italicized versions of the Irish words: uimhrochta, uimhrochtil (Uimhrochtil, uimhrochta, uimhrochta) and uimhrochtt (Uimhrochtta, uimhrochtta) are available in the English language. the language of Hindi
• The language of Bengali
• The Armenian word for mathematics is aritmetika, while the Indonesian word for mathematics is aritmetico, aritmetica. Italian
• Inuktitut
• A variety of other languages. aritmtika, aritmtisksLatvian
• Аритметики, аритметика The Macedonian language
• Marathi
• Aritmetica, rekenkunde
• Arithmetica, algebra The Dutch words aritmetikk and aritmetisk, and the Norwegian words arytmetyczny and arytmetyka are both used to mean “arithmetical calculation.” Polish
• AritmetticaPortuguese
• AritmeticăRomanian
• Ариметиески, ариметика
• Aritmetică In Russian, aritmetik, aritmetika means “theory of mathematics.” Swedish
• Aritmetisk, aritmetikAlbanian
• Aritmetisk, aritmetikAlbanian Languages like as Telugu, исoTajik, Thai, and Urdu are also available. In Vietnamese, kalkulav-, kalkulavVolapük
• In English, kalkulav-, kalkulavVolpük
• In German, kalkulavVolpük
### Word of the Day
Afterwards, the th term in a series will be denoted by the symbol (n). The first term of a series is a (1), and the 23rd term of a sequence is the letter a (1). (23). Parentheses will be used at several points in this course to indicate that the numbers next to thea are generally written as subscripts.
You might be interested: What Is A Non Constant Arithmetic Progression? (Correct answer)
## Finding the Terms
Let’s start with a straightforward problem. We have the following numbers in our sequence: -3, 2, 7, 12,. What is the seventh and last phrase in this sequence? As we can see, the most typical difference between successive periods is five points. The fourth term equals twelve, therefore a (4) = twelve. We can continue to add terms to the list in the following order until we reach the seventh term: -3, 2, 7, 12, 17, 22, 27,. and so on. This tells us that a (7) = 27 is the answer.
## Finding then th Term
Consider the identical sequence as in the preceding example, with the exception that we must now discover the 33rd word oracle (33). We may utilize the same strategy as previously, but it would take a long time to complete the project. We need to come up with a way that is both faster and more efficient. We are aware that we are starting with ata (1), which is a negative number. We multiply each phrase by 5 to get the next term. To go from a (1) to a (33), we’d have to add 32 consecutive terms (33 – 1 = 32) to the beginning of the sequence.
To put it another way, a (33) = -3 + (33 – 1)5.
a (33) = -3 + (33 – 1)5 = -3 + 160 = 157. An arithmetic sequence is represented by the general formula or rule seen in Figure 2. Then the relationship between the th term and the initial terma (1) and the common differencedis provided by:
## Arithmetic mean – Wikipedia
See Mean for a more in-depth discussion of this subject. Generally speaking, in mathematics and statistics, thearithmetic mean (pronounced air-ith-MET -ik) or arithmetic average (sometimes known as simply themean or theaverage when the context is obvious) is defined as the sum of a collection of numbers divided by the number of items in the collection. A collection of results from an experiment or an observational research, or more typically, a collection of results from a survey, is commonly used.
In addition to mathematics and statistics, the arithmetic mean is commonly employed in a wide range of subjects, including economics, anthropology, and history, and it is employed to some extent in virtually every academic field.
Because of skewed distributions, such as the income distribution, where the earnings of a small number of people exceed the earnings of most people, the arithmetic mean may not correspond to one’s conception of the “middle,” and robust statistics, such as the median, may provide a more accurate description of central tendency.
## Definition
The arithmetic mean (also known as the mean or average), indicated by the symbol (readbar), is the mean of a data collection. Among the several measures of central tendency in a data set, the arithmetic mean is the most widely used and easily comprehended. The term “average” refers to any of the measures of central tendency used in statistical analysis. The arithmetic mean of a collection of observed data is defined as being equal to the sum of the numerical values of each and every observation divided by the total number of observations in the set of data being considered.
The arithmetic mean is defined as A statistical population (i.e., one that contains every conceivable observation rather than merely a subset of them) is marked by the Greek letter m, and the mean of that population is denoted by the letter m.
Not only can the arithmetic mean be computed for scalar values, but it can also be defined for vectors in many dimensions; this is referred to as the centroid.
More generally, because the arithmetic mean is an aconvex combination (i.e., the coefficients add to 1), it may be defined on any convex space, not only a vector space, according to the definition above.
## Motivating properties
The arithmetic mean has a number of characteristics that make it particularly helpful as a measure of central tendency, among other things. These are some examples:
## Contrast with median
The arithmetic mean and the median can be compared and contrasted. The median is defined as the point at which no more than half of the values are greater than and no more than half are less than the median. If the elements of the data grow arithmetically when they are arranged in a particular order, then the median and arithmetic average are the same. Take, for example, the data sample described above. The average and the median are both correct. When we take a sample that cannot be structured in such a way that it increases arithmetically, such as the median and arithmetic average, the differences between the two can be considerable.
As a rule, the average value can deviate greatly from the majority of the values in the sample, and it can be significantly greater or lower than the majority of them.
Because of this, for example, median earnings in the United States have climbed at a slower rate than the arithmetic average of earnings since the early 1980s.
## Generalizations
If certain data points count more highly than others, then the average will be a weighted average, or weighted mean. This is because some data points are given greater weight in the computation. In the case ofandis, for example, the arithmetic mean, or equivalently An alternative method would be to compute a weightedmean, in which the first number is given more weight than the second (maybe because it is believed to appear twice as frequently in the broader population from which these numbers were sampled) and the result would be.
Arithmetic mean (also known as “unweighted average” or “equally weighted average”) can be thought of as a specific instance of the weighted average in which all of the weights are equal to each other in a given set of circumstances (equal toin the above example, and equal toin a situation withnumbers being averaged).
### Continuous probability distributions
Whenever a numerical property, and any sample of data from it, can take on any value from a continuous range, instead of just integers for example, the probability of a number falling into some range of possible values can be described by integrating a continuous probability distribution across this range, even when the naive probability of a sample number taking one specific value from an infinitely many is zero.
Themean of the probability distribution is the analog of a weighted average in this context, in which there are an infinite number of possibilities for the precise value of the variable in each range, and is referred to as the weighted average in this context.
The normal distribution is also the most commonly encountered probability distribution. Other probability distributions, such as the log-normal distribution, do not follow this rule, as seen below for the log-normal distribution.
### Angles
When working with cyclic data, such as phases or angles, more caution should be exercised. A result of 180° is obtained by taking the arithmetic mean of one degree and three hundred fifty-nine degrees. This is false for two reasons: first, it is not true.
• Angle measurements are only defined up to an additive constant of 360° (or 2 in the case of inradians) for several reasons. Due to the fact that each of them produces a distinct average, one may just as readily refer to them as the numbers 1 and 1, or 361 and 719, respectively. Second, in this situation, 0° (equivalently, 360°) is geometrically a better average value because there is less dispersion around it (the points are both 1° from it and 179° from 180°, the putative average)
• Third, in this situation, 0° (equivalently, 360°) is geometrically a better average value because there is less dispersion around it (the points are both 1° from it and 179° from 180°, the putative average
An oversight of this nature will result in the average value being artificially propelled towards the centre of the numerical range in general use. Using the optimization formulation (i.e., defining the mean as the central point: that is, defining it as the point about which one has the lowest dispersion), one can solve this problem by redefining the difference as a modular distance (i.e., defining it as the distance on the circle: the modular distance between 1° and 359° is 2°, not 358°).
## Symbols and encoding
The arithmetic mean is frequently symbolized as a bar (also known as a vinculumormacron), as in the following example: (readbar). In some applications (text processors, web browsers, for example), the x sign may not be shown as expected. A common example is the HTML code for the “x” symbol, which is made up of two codes: the base letter “x” and a code for the line above (772; or “x”). When a text file, such as a pdf, is transferred to a word processor such as Microsoft Word, the x symbol (Unicode 162) may be substituted by the cent (Unicode 162) symbol (Unicode 162).
• The Fréchet mean, the generalized mean, the geometric mean, the harmonic mean, the inequality of arithmetic and geometric means, and so on. The mode, the sample mean, and the covariance
• The standard deviation is the difference between two values. The standard error of the mean is defined as the standard deviation of the mean. Statistical summaries
## References
1. Jacobs, Harold R., et al (1994). Mathematics Is a Human-Inspired Effort (Third ed.). p. 547, ISBN 0-7167-2426-X
2. AbcMedhi, Jyotiprasad, W. H. Freeman, p. 547, ISBN 0-7167-2426-X
3. (1992). An Introduction to Statistical Methods is a text that introduces statistical methods. International New Age Publishing, pp. 53–58, ISBN 9788122404197
4. Weisstein, Eric W. “Arithmetic Mean”.mathworld.wolfram.com. Weisstein, Eric W. “Arithmetic Mean”. retrieved on the 21st of August, 2020
5. Paul Krugman is a well-known economist (4 June 2014). “Deconstructing the Income Distribution Debate: The Rich, the Right, and the Facts” is the title of the paper. The American Prospect
6. Tannica.com/science/mean|access-date=2020-08-21|website=Encyclopedia Britannica|language=en
7. Tannica.com/science/mean|access-date=2020-08-21|website=Encyclopedia Britannica|language=en (30 June 2010). June 30, 2010: “The Three M’s of Statistics: Mode, Median, and Mean June 30, 2010.” “Notes on Unicode for Stat Symbols,” which was published on 3 December 2018, was retrieved. retrieved on October 14, 2018
8. If AC =a and BC =b, OC =AMofa andb, and radiusr = QO = OG, then AC =a and BC =b Using Pythagoras’ theorem, QC2 = QO2 + OC2 QC = QO2 + OC2 = QM. QC2 = QO2 + OC2 = QM. Using Pythagoras’ theory, OC2 = OG2 + GC2 GC = OC2 OG2=GM. OC2 = OG2 + GC2 GC = OC2 OG2=GM. Using comparable triangles, HC/GC=GC/OC=HM
9. HC =GC2/OC=HM
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Radical Simplify Calculator
What is a Radical Simplify Calculator?
Radical Simplify Calculator is an online tool that calculates the simplified radical expression of entered values.
Cuemath's 'Radical Simplify Calculator' helps you to find the root of the number within a few seconds.
How to Use the Radical Simplify Calculator?
Follow these steps which will help you to use the calculator.
• Step 1: Enter the value of a, x and n value in the provided boxes for the expression a ⁿ√x.
• Step 2: Click on "Calculate" to find the simplified value of the radical.
• Step 3: Click on "Reset" to clear the values and enter the new values.
How to Simplify the Radicals?
The following steps would help in simplifying radicals.
• Step I: Write the number within the radical as the product of its prime factors.
• Step II: Based on the root of the radical take one factor out for n similar radicals within the radical.
• Step III: Find the product of the numbers outside the radical and the product of the numbers within the radical and write them together.
Let us try to understand the simplification of radicals with the help of an example.
$$\sqrt[3] 80= \sqrt [3]{2 \times 2 \times 2 \times 2 \times 5} =2\sqrt[3]{2 \times 5} = 2\sqrt[3]10$$
Solved Example:
Simplify the radical $$4\sqrt[4]96$$.
Solution:
Let us write the prime factor of the number within the radical and simplify it further.
$$4\sqrt[4] 96= 4\sqrt [4]{2 \times 2 \times 2 \times 2 \times 2 \times 3} =4x2\sqrt[4]{2 \times 3} = 8\sqrt[4]6$$
Now, try the calculator to simplify the following radicals.
• $$4\times \sqrt[5] 800$$
• $$3\times \sqrt[3] {135}$$
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# How do you find the volume of the solid obtained by rotating the region bounded by the curves y= x^2 - 4 and y= 3x and x=0 about the y axis?
Sep 28, 2015
$136 \pi$
#### Explanation:
Let's find intersection:
$x = \frac{y}{3} \implies y = {\left(\frac{y}{3}\right)}^{2} - 4$
${y}^{2} / 9 - y - 4 = 0$
${y}^{2} - 9 y - 36 = 0$
${y}^{2} - 12 y + 3 y - 36 = y \left(y - 12\right) + 3 \left(y - 12\right) = 0$
$\left(y - 12\right) \left(y + 3\right) = 0 \iff y = 12 \vee y = - 3$
$y = {x}^{2} - 4 \implies x = \sqrt{y + 4}$
$y = 3 x \implies x = \frac{y}{3}$
$V = \pi \left({\int}_{-} {4}^{12} {\left(\sqrt{y + 4}\right)}^{2} \mathrm{dy} - {\int}_{0}^{12} {\left(\frac{y}{3}\right)}^{2} \mathrm{dy}\right)$
$V = \pi \left({\int}_{-} {4}^{12} \left(y + 4\right) \mathrm{dy} - {\int}_{0}^{12} {y}^{2} / 9 \mathrm{dy}\right)$
${V}_{1} = \pi \left[{y}^{2} / 2 + 4 y\right] {|}_{-} {4}^{12}$
${V}_{1} = \pi \left(144 + 48 - 8 + 16\right) = 200 \pi$
${V}_{2} = \pi \left[{y}^{3} / 27\right] {|}_{0}^{12}$
${V}_{2} = 64 \pi$
$V = 200 \pi - 64 \pi = 136 \pi$
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# 9.2 Impulse (Page 2/6)
Page 2 / 6
Solution for (a)
The first ball bounces directly into the wall and exerts a force on it in the $+x$ direction. Therefore the wall exerts a force on the ball in the $-x$ direction. The second ball continues with the same momentum component in the $y$ direction, but reverses its $x$ -component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum.
These changes mean the change in momentum for both balls is in the $-x$ direction, so the force of the wall on each ball is along the $-x$ direction.
Strategy for (b)
Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball.
Solution for (b)
Let $u$ be the speed of each ball before and after collision with the wall, and $m$ the mass of each ball. Choose the $x$ -axis and $y$ -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.
${p}_{\text{xi}}=\text{mu}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yi}}=0$
${p}_{\text{xf}}=-\text{mu}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yf}}=0$
Impulse is the change in momentum vector. Therefore the $x$ -component of impulse is equal to $-2\text{mu}$ and the $y$ -component of impulse is equal to zero.
Now consider the change in momentum of the second ball.
${p}_{\text{xi}}=\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yi}}=\text{–mu}\phantom{\rule{0.25em}{0ex}}\text{sin 30º}$
${p}_{\text{xf}}=–\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}\text{;}\phantom{\rule{0.25em}{0ex}}{p}_{\text{yf}}=-\text{mu}\phantom{\rule{0.25em}{0ex}}\text{sin 30º}$
It should be noted here that while ${p}_{x}$ changes sign after the collision, ${p}_{y}$ does not. Therefore the $x$ -component of impulse is equal to $-2\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}$ and the $y$ -component of impulse is equal to zero.
The ratio of the magnitudes of the impulse imparted to the balls is
$\frac{2\text{mu}}{2\text{mu}\phantom{\rule{0.25em}{0ex}}\text{cos 30º}}=\frac{2}{\sqrt{3}}=1\text{.}\text{155}.$
Discussion
The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative $x$ - direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive $x$ -direction.
Our definition of impulse includes an assumption that the force is constant over the time interval $\Delta t$ . Forces are usually not constant . Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force ${F}_{\text{eff}}$ that produces the same result as the corresponding time-varying force. [link] shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times ${t}_{1}$ and ${t}_{2}$ . That area is equal to the area inside the rectangle bounded by ${F}_{\text{eff}}$ , ${t}_{1}$ , and ${t}_{2}$ . Thus the impulses and their effects are the same for both the actual and effective forces.
## Making connections: take-home investigation—hand movement and impulse
Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why?
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
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
I'm interested in nanotube
Uday
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
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Solving Systems of Linear Equations
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1 LECTURE 5 Solving Systems of Linear Equations Recall that we introduced the notion of matrices as a way of standardizing the expression of systems of linear equations In today s lecture I shall show how this matrix machinery can also be used to solve such systems However, before we embark on solving systems of equations via matrices, let me remind you of what such solutions should look like The Geometry of Linear Systems Consider a linear system of m equations and n unknowns: (5) What does the solution set look like? easily visualize the solution sets a x + a x + + a n x n = b a x + a x + + a n x n = b a m x + a m x + + a mn x n = b m Let s examine this question case by case, in a setting where we can equations in 3 unknowns This would correspond to a situation where you have 3 variables x, x, x 3 with no relations between them Being free to choose whatever value we want for each of the 3 variables, it s clear that the solutions set is just R 3, the 3-dimensional space of ordered sets of 3 real numbers equation in 3 unknowns In this case, use the equation to express one variable, say x n, in terms of the other variables; x = a n (b a x + a x + a,3 x 3 ) The remaining variables x, and x are then unrestricted Letting these variables range freely over R will then fill out a -dimensional plane in R Thus, in the case of equation in 3 unknowns we have a two dimensional plane as a solution space equations in 3 unknowns As in the preceding example, the solution set of each individual equation will be a -dimensional plane The solution set of the pair of equation will be the intersection of these two planes (For points common to both solution sets will be points corresponding to the solutions of both equations) Here there will be three possibilities: () The two planes do not intersect In other words, the two planes are parallel but distinct Since they share no common point, there is no simultaneous solutoin of both equations () The intersection of the two planes is a line This is the generic case (3) The two planes coincide In this case, the two equations must be somehow redundant Thus we have either no solution, a -dimensional solution space (ie a line) or a -dimensional solution space 3 equations and 3 unknowns In this case, the solution set will correspond to the intersection of the three planes corresponding to the 3 equations We will have four possiblities: () The three planes share no common point In this case there will be no solution () The three planes have one point in common This will be the generic situation
2 ELEMENTARY ROW OPERATIONS (3) The three planes share a single line (4) The three planes all coincide Thus, we either have no solution, a -dimensional solution (ie, a point), a -dimensional solution (ie a line) or a -dimensional solution We now summarize and generalize this discussion as follows Theorem 5 Consider a linear system of m equations and n unknowns: The solution set of such a system is either: a x + a x + + a n x n = b a x + a x + + a n x n = b a m x + a m x + + a mn x n = b m () The empty set {}; ie, there is no solution () A hyperplane of dimension greater than or equal to (n m) Elementary Row Operations In the preceding lecture we remarked that our new-fangled matrix algebra allowed us to represent linear systems such as (5) succintly as matrix equations: a a a n x b a a a n x (5) = b a m a m a mn x m b m For example the linear system (53) can be represented as (54) [ 3 x + 3x = 3 x + x = ] [ ] [ ] x 3 = x Now, when solving linear systems like (53) it is very common to create new but equivalent equations by, for example, mulitplying by a constant or adding one equation to another In fact, we have Theorem 5 Let S be a system of m linear equations in n unknowns and let S be another system of m equations in n unknowns obtained from S by applying some combination of the following operations: interchanging the order of two equations multiplying one equation by a non-zero constant replacing a equation with the sum of itself and a multiple of another equation in the system Then the solution sets of S ane S are identical In particular, the solution set of (53) is equivalent to the solution set of (55) x + 3x = 3 x x =
3 3 SOLVING LINEAR EQUATIONS 3 (where we have multiplied the second equation by -), as well as the solution set of (56) x + 3x = 3 x = (where we have replaced the second row in (55) by its sum with the first row), as well as the solution of (57) x = 3 x = (where we have replaced the first row in (56) by its sum with 3 times the second row) We can thus solve a linear system by means of the elementary operations described in the theorem above Now because there is a matrix equation corresponding to every system of linear equations, each of the operations described in the theorem above corresponds to a matrix operation To convert these operations in our matrix language, we first associate with a linear system an augmented matrix a x + a x + + a n x n = b a x + a x + + a n x n = b a m x + a m x + + a mn x n = b m a a a n a a a n a m a m a mn This is just the m (n + ) matrix that s obtained appending the column vector b to the columns of the m n matrix A We shall use the notation [A b] to denote the augmented matrix of A and b The augmented matrices corresponding to equations (55), (56), and (57) are thus, respectively [ 3 3 ], [ 3 3 ], and [ 3 ] From this example, we can see that the operations in Theorem 5 translate to the following operations on the corresponding augmented matrices: Row Interchange: the interchange of two rows of the augmented matrix Row Scaling: multiplication of a row by a non-zero scalar Row Addition: replacing a row by its sum with a multiple of another row Henceforth, we shall refer to these operations as Elementary Row Operations b b b m 3 Solving Linear Equations Let s now reverse directions and think about how to recognize when the system of equations corresponding to a given matrix is easily solved (To keep our discussion simple, in the examples given below we consider systems of n equations in n unknowns)
4 3 SOLVING LINEAR EQUATIONS 4 3 Diagonal Matrices A matrix equation Ax = b is trivial to solve if the matrix A is purely diagonal For then the augmented matrix has the form a b a b A = a n,n b n a nn b n the corresponding system of equations reduces to a x = b x = b a a x = b x = b a a nn x n = b n x n = b n a nn 3 Lower Triangular Matrices If the coefficient matrix A has the form a a a A = a n, a n, a n,n a n a n, a n,n a nn (with zeros everywhere above the diagonal from a to a nn ), then it is called lower triangular A matrix equation Ax = b in which A is lower triangular is also fairly easy to solve For it is equivalent to a system of equations of the form a x = b a x + a x = b a 3 x + a 3 x + a 33 x 3 = b 3 a n x + a n x + a n3 x a nn x n = b n To find the solution of such a system one solves the first equation for x and then substitutes its solution b /a for the variable x in the second equation ( ) b a + a x = b x = ( b a ) b a a a One can now substitute the numerical expressions for x, and x into the third equation and get a numerical expression for x 3 Continuing in this manner we can solve the system completely We call this method solution by forward substitution 33 Upper Triangular Matrices We can do a similar thing for systems of equations characterized by an upper triangular matrices A a a a,n a n a a,n a n A = a n,n a nn
5 3 SOLVING LINEAR EQUATIONS 5 (that is to say, a matrix with zero s everywhere below and to the left of the diagonal), then the corresponding system of equations will be a x + a x + + a n x n = b a x + + a n x n = b a n,n x n + a n,n x n = b n a nn x n = b n which can be solved by substituting the solution of the last equation into the preceding equation and solving for x n x n = a n,n x n = b n a nn ( ( )) bn b n a n,n a nn and then substituting this result into the third from last equation, etc by back-substitution This method is called solution 34 Solution via Row Reduction 34 Gaussian Reduction In the general case a matrix will be neither be upper triangular or lower triangular and so neither forward- or back-substituiton can be used to solve the corresponding system of equations However, using the elementary row operations discussed in the preceding section we can always convert the augmented matrix of a (self-consistent) system of linear equations into an equivalent matrix that is upper triangular; and having done that we can then use back-substitution to solve the corresponding set of equations We call this method Gaussian reduction Let me demonstate the Gaussian method with an example Example 53 Solve the following system of equations using Gaussian reduction x + x x 3 = x x + x 3 = x x x 3 = 3 First we write down the corresponding augmented matrix 3 We now use elementary row operations to convert this matrix into one that is upper triangular Adding - times the first row to the second row produces 3 Adding - times the first row to the third row produces 3 3
6 Adding 3 3 SOLVING LINEAR EQUATIONS 6 times the second row to the third row produces 6 The augmented matrix is now in upper triangular form It corresponds to the following system of equations which can easily be solved via back-substitution: x + x x 3 = x + x 3 = x 3 = 6 x 3 = 6 x 3 = 3 x + 6 = x = x + 3 = x = In summary, solution by Gaussian reduction consists of the following steps () Write down the augmented matrix corresponding to the system of linear equations to be solved () Use elementary row operations to convert the augmented matrix [A b] into one that is upper triangular This is acheived by systematically using the first row to eliminate the first entries in the rows below it, the second row to eliminate the second entries in the row below it, etc (3) Once the augmented matrix has been reduced to upper triangular form, write down the corresponding set of linear equations and use back-substitution to complete the solution Remark 54 The text refers to matrices that are upper triangular as being in row-echelon form Definition 55 A matrix is in row-echelon form if it satisfies two conditions: () All rows containing only zeros appear below rows with non-zero entries () The first non-zero entry in a row appears in a column to the right of the first non-zero entry of any preceding row For such a matrix the first non-zero entry in a row is called the pivot for that row 34 Gauss-Jordan Reduction In the Gauss Reduction method described above, one carries out row operations until the augmented matrix is upper triangular and then finishes the problem by converting the problem back into a linear system and using back-substitution to complete the solution Thus, row reduction is used to carry out about half the work involved in solving a linear system It is also possible to use only row operations to construct a solution of a linear system Ax = b technique is called Gauss-Jordan reduction The idea is this This () Write down the augmented matrix corresponding to the system of linear equations to be solved () Use elementary row operations to convert the augmented matrix [A b] into one that is upper triangular This is acheived by systematically using the first row to eliminate the first entries in the rows below it, the second row to eliminate the second entries in the row below it, etc a a a n b a a a n b a a a n b a a n b [A b] = row operations a n a n a nn b n a nn b n
7 3 SOLVING LINEAR EQUATIONS 7 (3) Continue to use row operations on the aumented matrix until all the entries above the diagonal of the first factor have been eliminated, and only s appear along the diagonal a a a n b a a b n b b row operations = [ I b ] a nn b n b n (4) The solution of the linear system corresponding to the augmented matrix [I b ] is trivial x = Ix = b Moreover, since [I b ] was obtained from [A b] by row operations, x = b must also be the solution of Ax = b Example 56 Solve the following system of equations using Gauss-Jordan reduction x + x x 3 = x x + x 3 = x x x 3 = 3 First we write down the corresponding augmented matrix 3 In the preceding example, I demonstated that this augmented matrix is equivalent to 6 We ll now continue to apply row operations until the first block has the form of a 3 3 identity matrix Multiplying the second and third rows by yields 3 Replacing the first row by its sum with times the second row yields 3 Replacing the second row by its sum with the third row yields 3 The system of equations corresponding to this augmented matrix is This is our solution x = x = x 3 = 3 To maintain contact with the definitions used in the text, we have Definition 57 A matrix is in reduced row-echelon form if it is in row-echelon form with all pivots equal to and with zeros in every column entry above and below the pivots
8 4 ELEMENTARY MATRICES 8 Theorem 58 Let Ax = b be a linear system and suppose [A b ] is row equivalent to [A b] with A a matrix in row-echelon form Then () The system Ax = b is inconsistent if and only if the augmented matrix [A b ] has a row with all entries to the left of the partition and a non-zero entry to the right of the partition () If Ax = b is consistent and every column of A contains a pivot, the system has a unique solution (3) If Ax = b is consistent and some column of A has no pivot, the system has infinitely many solutions, with as many free variables as there are pivot-free columns in A 4 Elementary Matrices I shall now show that all elementary row operations can be carried out by means of matrix multiplication Even though carrying out row operations by matrix multiplication will be grossly inefficient from a calculation point of view, this equivalence remains an important theoretical fact (As you ll see in some of the forthcoming proofs) Definition 59 An elementary matrix is a matrix that can be obtained from the identity matrix by means of a single row operation Theorem 5 Let A be an m n matrix, and let E be an m m elementary matrix Then multiplication of A on the left by E effects the same elementary row operation on A as that which was performed on the m m identity matrix to produce E Corollary 5 If A is a matrix that was obtained from A by a sequence of row operations then there exists a corresponding sequence of elementary matrices E, E,, E r such that A = E r E E A
### Solving Systems of Linear Equations
LECTURE 5 Solving Systems of Linear Equations Recall that we introduced the notion of matrices as a way of standardizing the expression of systems of linear equations In today s lecture I shall show how
### MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.
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# 65 percent of five over eight Dollars
### Percentage Calculator
What is % of Answer:
### Percentage Calculator 2
is what percent of ? Answer: %
### Percentage Calculator 3
is % of what? Answer:
We think you reached us looking for answers to questions like: What is 65 percent of five over eight? Or maybe: 65 percent of five over eight Dollars
See the detailed solutions to these problems below.
## How to work out percentages explained step-by-step
Learn how to solve percentage problems through examples.
In all the following questions consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
### Solution for 'What is 65% of five over eight?'
#### Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 65 percent of 0.625. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 0.625.
Step 2: determine the percentage, which is 65.
Step 3: Convert the percentage 65% to its decimal form by dividing 65 into 100 to get the decimal number 0.65:
65100 = 0.65
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
65.0 → 6.50 → 0.65
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.65 x 0.625 = 0.40625 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 0.625 × 65100 = 0.625 × 0.65 = 0.40625 (answer)
Therefore, the answer is 0.40625 is 65 percent of 0.625.
### Solution for '65 is what percent of five over eight?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
#### Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 0.625.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 650.625
X% = 65000.625
X% = 10,400.00 (answer)
So, the answer is 65 is 10,400.00 percent of 0.625
### Solution for '0.625 is 65 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
### Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 0.625.
Step 2: identify the percent, which is 65.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 0.62565
W = 100 × 0.0096153846153846
W = 0.96153846153846 (answer)
The answer, in plain words, is: 0.625 is 65% of 0.96153846153846.
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# Quick Answer: What Are 2 Factors Of 18?
## What is the factor of 23?
The factors of 23 are 1 and 23.
Since 23 is a prime number, therefore, it has only two factors..
## What are all the factor pairs for 18?
Factors of 18: The square root of 18 is 4.2426, rounded down to the closest whole number is 4. Testing the integer values 1 through 4 for division into 18 with a 0 remainder we get these factor pairs: (1 and 18), (2 and 9), (3 and 6). The factors of 18 are 1, 2, 3, 6, 9, 18.
## What is the factor of negative 18?
The list of Negative Factors of 18 are: -1, -2, -3, -6, -9, and -18.
## What is the GCF of 18 and 24?
We found the factors and prime factorization of 18 and 24. The biggest common factor number is the GCF number. So the greatest common factor 18 and 24 is 6.
## What is the GCF of 14 and 24?
We found the factors and prime factorization of 14 and 24. The biggest common factor number is the GCF number. So the greatest common factor 14 and 24 is 2.
## What does by a factor mean?
“by a factor of ” is used commonly to mean the same as “multiplied by” or “divided by.” If x is INCREASED by a factor of 4, it becomes 4x. If x is DECREASED by a factor of 4, it becomes x/4. The key word is the direction of change i.e. (increased / decreased) by a factor of.
## What is the smallest factor of 18?
I write the factors in order from smallest to greatest, so the factors of 18 are: 1, 2, 3, 6, 9, and 18.
## What are the factors of 18 and 3?
The gcf of 3 and 18 can be obtained like this:The factors of 3 are 3, 1.The factors of 18 are 18, 9, 6, 3, 2, 1.The common factors of 3 and 18 are 3, 1, intersecting the two sets above.In the intersection factors of 3 ∩ factors of 18 the greatest element is 3.Therefore, the greatest common factor of 3 and 18 is 3.
## How do you factor 18?
18 = 1 x 18, 2 x 9, or 3 x 6. Factors of 18: 1, 2, 3, 6, 9, 18. Prime factorization: 18 = 2 x 3 x 3, which can also be written 18 = 2 x 3².
## What are the factors of 4 and 18?
The gcf of 4 and 18 can be obtained like this: The factors of 4 are 4, 2, 1. The factors of 18 are 18, 9, 6, 3, 2, 1. The common factors of 4 and 18 are 2, 1, intersecting the two sets above.
## What are 5 multiples of 18?
1 Answer. 18,36,54,72,90 are first five multiples of 18 .
## What are the factors of 19?
The only factors of 19 are 1 and 19, so 19 is a prime number. That is, 19 is divisible by only 1 and 19, so it is prime.
## What number has only 2 factors?
A number with only two factors is called a prime number. A number with more than two factors is called a composite number. The number 1 is neither prime nor composite. It has only one factor, itself.
## What are factors of?
In multiplication, factors are the integers that are multiplied together to find other integers. For example, 6 × 5 = 30. In this example, 6 and 5 are the factors of 30. 1, 2, 3, 10, 15 and 30 would also be factors of 30.
## What are the multiples of 18?
The list of multiples of 18 are: 18,36,54,72,90,108,126,144,162,180,198,216,234,252,270,…. Sometimes multiples are misunderstood as factors also, which is not correct. Factors of 18 consists of only those numbers which are multiplied together to get the original number.
## What is the factor of 21?
The factors of a number 21 are the numbers, that produce the result as 21 when a pair factor is multiplied together. A factor of a number divides the original number uniformly. The factors of 21 are 1, 3, 7 and 21.
## Which is the biggest factor of 18?
The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9.
## What is the highest common factor of 18 and 3?
Greatest common factor (GCF) of 3 and 18 is 3. We will now calculate the prime factors of 3 and 18, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 3 and 18.
## What is the GCF of 9 18?
9. 2Answer: GCF of 9 and 18 is 9. 2.
## What is the LCM of 18 and 3?
Least common multiple (LCM) of 3 and 18 is 18.
|
Prealgebra 2e
# 10.2Use Multiplication Properties of Exponents
Prealgebra 2e10.2 Use Multiplication Properties of Exponents
## Learning Objectives
By the end of this section, you will be able to:
• Simplify expressions with exponents
• Simplify expressions using the Product Property of Exponents
• Simplify expressions using the Power Property of Exponents
• Simplify expressions using the Product to a Power Property
• Simplify expressions by applying several properties
• Multiply monomials
## Be Prepared 10.4
Before you get started, take this readiness quiz.
Simplify: $34·34.34·34.$
If you missed the problem, review Example 4.25.
## Be Prepared 10.5
Simplify: $(−2)(−2)(−2).(−2)(−2)(−2).$
If you missed the problem, review Example 3.52.
## Simplify Expressions with Exponents
Remember that an exponent indicates repeated multiplication of the same quantity. For example, $2424$ means to multiply four factors of $2,2,$ so $2424$ means $2·2·2·2.2·2·2·2.$ This format is known as exponential notation.
## Exponential Notation
This is read $aa$ to the $mthmth$ power.
In the expression $am,am,$ the exponent tells us how many times we use the base $aa$ as a factor.
Before we begin working with variable expressions containing exponents, let’s simplify a few expressions involving only numbers.
## Example 10.11
Simplify:
1. $5353$
2. $9191$
## Try It 10.21
Simplify:
1. $4343$
2. $111111$
## Try It 10.22
Simplify:
1. $3434$
2. $211211$
## Example 10.12
Simplify:
1. $(78)2(78)2$
2. $(0.74)2(0.74)2$
## Try It 10.23
Simplify:
1. $(58)2(58)2$
2. $(0.67)2(0.67)2$
## Try It 10.24
Simplify:
1. $(25)3(25)3$
2. $(0.127)2(0.127)2$
## Example 10.13
Simplify:
1. $(−3)4(−3)4$
2. $−34−34$
## Try It 10.25
Simplify:
1. $(−2)4(−2)4$
2. $−24−24$
## Try It 10.26
Simplify:
1. $(−8)2(−8)2$
2. $−82−82$
## Simplify Expressions Using the Product Property of Exponents
You have seen that when you combine like terms by adding and subtracting, you need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too. We’ll derive the properties of exponents by looking for patterns in several examples. All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents.
First, we will look at an example that leads to the Product Property.
What does this mean? How many factors altogether? So, we have Notice that 5 is the sum of the exponents, 2 and 3. We write: $x2⋅x3x2⋅x3$$x2+3x2+3$$x5x5$
The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.
## Product Property of Exponents
If $aa$ is a real number and $m,nm,n$ are counting numbers, then
$am·an=am+nam·an=am+n$
To multiply with like bases, add the exponents.
An example with numbers helps to verify this property.
$22·23=?22+34·8=?2532=32✓22·23=?22+34·8=?2532=32✓$
## Example 10.14
Simplify: $x5·x7.x5·x7.$
## Try It 10.27
Simplify: $x7·x8.x7·x8.$
## Try It 10.28
Simplify: $x5·x11.x5·x11.$
## Example 10.15
Simplify: $b4·b.b4·b.$
## Try It 10.29
Simplify: $p9·p.p9·p.$
## Try It 10.30
Simplify: $m·m7.m·m7.$
## Example 10.16
Simplify: $27·29.27·29.$
## Try It 10.31
Simplify: $6·69.6·69.$
## Try It 10.32
Simplify: $96·99.96·99.$
## Example 10.17
Simplify: $y17·y23.y17·y23.$
## Try It 10.33
Simplify: $y24·y19.y24·y19.$
## Try It 10.34
Simplify: $z15·z24.z15·z24.$
We can extend the Product Property of Exponents to more than two factors.
## Example 10.18
Simplify: $x3·x4·x2.x3·x4·x2.$
## Try It 10.35
Simplify: $x7·x5·x9.x7·x5·x9.$
## Try It 10.36
Simplify: $y3·y8·y4.y3·y8·y4.$
## Simplify Expressions Using the Power Property of Exponents
Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.
What does this mean? How many factors altogether? So, we have Notice that 6 is the product of the exponents, 2 and 3. We write: $(x2)3(x2)3$$x2⋅3x2⋅3$$x6x6$
We multiplied the exponents. This leads to the Power Property for Exponents.
## Power Property of Exponents
If $aa$ is a real number and $m,nm,n$ are whole numbers, then
$(am)n=am·n(am)n=am·n$
To raise a power to a power, multiply the exponents.
An example with numbers helps to verify this property.
$(52)3=?52·3(25)3=?5615,625=15,625✓(52)3=?52·3(25)3=?5615,625=15,625✓$
## Example 10.19
Simplify:
1. $(x5)7(x5)7$
2. $(36)8(36)8$
## Try It 10.37
Simplify:
1. $(x7)4(x7)4$
2. $(74)8(74)8$
## Try It 10.38
Simplify:
1. $(x6)9(x6)9$
2. $(86)7(86)7$
## Simplify Expressions Using the Product to a Power Property
We will now look at an expression containing a product that is raised to a power. Look for a pattern.
$(2x)3(2x)3$ What does this mean? $2x·2x·2x2x·2x·2x$ We group the like factors together. $2·2·2·x·x·x2·2·2·x·x·x$ How many factors of 2 and of $x?x?$ $23·x323·x3$ Notice that each factor was raised to the power. $(2x)3is23·x3(2x)3is23·x3$ We write: $(2x)3(2x)3$ $23·x323·x3$
The exponent applies to each of the factors. This leads to the Product to a Power Property for Exponents.
## Product to a Power Property of Exponents
If $aa$ and $bb$ are real numbers and $mm$ is a whole number, then
$(ab)m=ambm(ab)m=ambm$
To raise a product to a power, raise each factor to that power.
An example with numbers helps to verify this property:
$(2·3)2=?22·3262=?4·936=36✓(2·3)2=?22·3262=?4·936=36✓$
## Example 10.20
Simplify: $(−11x)2.(−11x)2.$
## Try It 10.39
Simplify: $(−14x)2.(−14x)2.$
## Try It 10.40
Simplify: $(−12a)2.(−12a)2.$
## Example 10.21
Simplify: $(3xy)3.(3xy)3.$
## Try It 10.41
Simplify: $(−4xy)4.(−4xy)4.$
## Try It 10.42
Simplify: $(6xy)3.(6xy)3.$
## Simplify Expressions by Applying Several Properties
We now have three properties for multiplying expressions with exponents. Let’s summarize them and then we’ll do some examples that use more than one of the properties.
## Properties of Exponents
If $a,ba,b$ are real numbers and $m,nm,n$ are whole numbers, then
$Product Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power Property(ab)m=ambmProduct Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power Property(ab)m=ambm$
## Example 10.22
Simplify: $(x2)6(x5)4.(x2)6(x5)4.$
## Try It 10.43
Simplify: $(x4)3(x7)4.(x4)3(x7)4.$
## Try It 10.44
Simplify: $(y9)2(y8)3.(y9)2(y8)3.$
## Example 10.23
Simplify: $(−7x3y4)2.(−7x3y4)2.$
## Try It 10.45
Simplify: $(−8x4y7)3.(−8x4y7)3.$
## Try It 10.46
Simplify: $(−3a5b6)4.(−3a5b6)4.$
## Example 10.24
Simplify: $(6n)2(4n3).(6n)2(4n3).$
## Try It 10.47
Simplify: $(7n)2(2n12).(7n)2(2n12).$
## Try It 10.48
Simplify: $(4m)2(3m3).(4m)2(3m3).$
## Example 10.25
Simplify: $(3p2q)4(2pq2)3.(3p2q)4(2pq2)3.$
## Try It 10.49
Simplify: $(u3v2)5(4uv4)3.(u3v2)5(4uv4)3.$
## Try It 10.50
Simplify: $(5x2y3)2(3xy4)3.(5x2y3)2(3xy4)3.$
## Multiply Monomials
Since a monomial is an algebraic expression, we can use the properties for simplifying expressions with exponents to multiply the monomials.
## Example 10.26
Multiply: $(4x2)(−5x3).(4x2)(−5x3).$
## Try It 10.51
Multiply: $(7x7)(−8x4).(7x7)(−8x4).$
## Try It 10.52
Multiply: $(−9y4)(−6y5).(−9y4)(−6y5).$
## Example 10.27
Multiply: $(34c3d)(12cd2).(34c3d)(12cd2).$
## Try It 10.53
Multiply: $(45m4n3)(15mn3).(45m4n3)(15mn3).$
## Try It 10.54
Multiply: $(23p5q)(18p6q7).(23p5q)(18p6q7).$
## Section 10.2 Exercises
### Practice Makes Perfect
Simplify Expressions with Exponents
In the following exercises, simplify each expression with exponents.
55.
$4 5 4 5$
56.
$10 3 10 3$
57.
$( 1 2 ) 2 ( 1 2 ) 2$
58.
$( 3 5 ) 2 ( 3 5 ) 2$
59.
$( 0.2 ) 3 ( 0.2 ) 3$
60.
$( 0.4 ) 3 ( 0.4 ) 3$
61.
$( −5 ) 4 ( −5 ) 4$
62.
$( −3 ) 5 ( −3 ) 5$
63.
$−5 4 −5 4$
64.
$−3 5 −3 5$
65.
$−10 4 −10 4$
66.
$−2 6 −2 6$
67.
$( − 2 3 ) 3 ( − 2 3 ) 3$
68.
$( − 1 4 ) 4 ( − 1 4 ) 4$
69.
$− 0.5 2 − 0.5 2$
70.
$− 0.1 4 − 0.1 4$
Simplify Expressions Using the Product Property of Exponents
In the following exercises, simplify each expression using the Product Property of Exponents.
71.
$x 3 · x 6 x 3 · x 6$
72.
$m 4 · m 2 m 4 · m 2$
73.
$a · a 4 a · a 4$
74.
$y 12 · y y 12 · y$
75.
$3 5 · 3 9 3 5 · 3 9$
76.
$5 10 · 5 6 5 10 · 5 6$
77.
$z · z 2 · z 3 z · z 2 · z 3$
78.
$a · a 3 · a 5 a · a 3 · a 5$
79.
$x a · x 2 x a · x 2$
80.
$y p · y 3 y p · y 3$
81.
$y a · y b y a · y b$
82.
$x p · x q x p · x q$
Simplify Expressions Using the Power Property of Exponents
In the following exercises, simplify each expression using the Power Property of Exponents.
83.
$( u 4 ) 2 ( u 4 ) 2$
84.
$( x 2 ) 7 ( x 2 ) 7$
85.
$( y 5 ) 4 ( y 5 ) 4$
86.
$( a 3 ) 2 ( a 3 ) 2$
87.
$( 10 2 ) 6 ( 10 2 ) 6$
88.
$( 2 8 ) 3 ( 2 8 ) 3$
89.
$( x 15 ) 6 ( x 15 ) 6$
90.
$( y 12 ) 8 ( y 12 ) 8$
91.
$( x 2 ) y ( x 2 ) y$
92.
$( y 3 ) x ( y 3 ) x$
93.
$( 5 x ) y ( 5 x ) y$
94.
$( 7 a ) b ( 7 a ) b$
Simplify Expressions Using the Product to a Power Property
In the following exercises, simplify each expression using the Product to a Power Property.
95.
$( 5 a ) 2 ( 5 a ) 2$
96.
$( 7 x ) 2 ( 7 x ) 2$
97.
$( − 6 m ) 3 ( − 6 m ) 3$
98.
$( − 9 n ) 3 ( − 9 n ) 3$
99.
$( 4 r s ) 2 ( 4 r s ) 2$
100.
$( 5 a b ) 3 ( 5 a b ) 3$
101.
$( 4 x y z ) 4 ( 4 x y z ) 4$
102.
$( − 5 a b c ) 3 ( − 5 a b c ) 3$
Simplify Expressions by Applying Several Properties
In the following exercises, simplify each expression.
103.
$( x 2 ) 4 · ( x 3 ) 2 ( x 2 ) 4 · ( x 3 ) 2$
104.
$( y 4 ) 3 · ( y 5 ) 2 ( y 4 ) 3 · ( y 5 ) 2$
105.
$( a 2 ) 6 · ( a 3 ) 8 ( a 2 ) 6 · ( a 3 ) 8$
106.
$( b 7 ) 5 · ( b 2 ) 6 ( b 7 ) 5 · ( b 2 ) 6$
107.
$( 3 x ) 2 ( 5 x ) ( 3 x ) 2 ( 5 x )$
108.
$( 2 y ) 3 ( 6 y ) ( 2 y ) 3 ( 6 y )$
109.
$( 5 a ) 2 ( 2 a ) 3 ( 5 a ) 2 ( 2 a ) 3$
110.
$( 4 b ) 2 ( 3 b ) 3 ( 4 b ) 2 ( 3 b ) 3$
111.
$( 2 m 6 ) 3 ( 2 m 6 ) 3$
112.
$( 3 y 2 ) 4 ( 3 y 2 ) 4$
113.
$( 10 x 2 y ) 3 ( 10 x 2 y ) 3$
114.
$( 2 m n 4 ) 5 ( 2 m n 4 ) 5$
115.
$( −2 a 3 b 2 ) 4 ( −2 a 3 b 2 ) 4$
116.
$( −10 u 2 v 4 ) 3 ( −10 u 2 v 4 ) 3$
117.
$( 2 3 x 2 y ) 3 ( 2 3 x 2 y ) 3$
118.
$( 7 9 p q 4 ) 2 ( 7 9 p q 4 ) 2$
119.
$( 8 a 3 ) 2 ( 2 a ) 4 ( 8 a 3 ) 2 ( 2 a ) 4$
120.
$( 5 r 2 ) 3 ( 3 r ) 2 ( 5 r 2 ) 3 ( 3 r ) 2$
121.
$( 10 p 4 ) 3 ( 5 p 6 ) 2 ( 10 p 4 ) 3 ( 5 p 6 ) 2$
122.
$( 4 x 3 ) 3 ( 2 x 5 ) 4 ( 4 x 3 ) 3 ( 2 x 5 ) 4$
123.
$( 1 2 x 2 y 3 ) 4 ( 4 x 5 y 3 ) 2 ( 1 2 x 2 y 3 ) 4 ( 4 x 5 y 3 ) 2$
124.
$( 1 3 m 3 n 2 ) 4 ( 9 m 8 n 3 ) 2 ( 1 3 m 3 n 2 ) 4 ( 9 m 8 n 3 ) 2$
125.
$( 3 m 2 n ) 2 ( 2 m n 5 ) 4 ( 3 m 2 n ) 2 ( 2 m n 5 ) 4$
126.
$( 2 p q 4 ) 3 ( 5 p 6 q ) 2 ( 2 p q 4 ) 3 ( 5 p 6 q ) 2$
Multiply Monomials
In the following exercises, multiply the following monomials.
127.
$( 12 x 2 ) ( −5 x 4 ) ( 12 x 2 ) ( −5 x 4 )$
128.
$( −10 y 3 ) ( 7 y 2 ) ( −10 y 3 ) ( 7 y 2 )$
129.
$( −8 u 6 ) ( −9 u ) ( −8 u 6 ) ( −9 u )$
130.
$( −6 c 4 ) ( −12 c ) ( −6 c 4 ) ( −12 c )$
131.
$( 1 5 r 8 ) ( 20 r 3 ) ( 1 5 r 8 ) ( 20 r 3 )$
132.
$( 1 4 a 5 ) ( 36 a 2 ) ( 1 4 a 5 ) ( 36 a 2 )$
133.
$( 4 a 3 b ) ( 9 a 2 b 6 ) ( 4 a 3 b ) ( 9 a 2 b 6 )$
134.
$( 6 m 4 n 3 ) ( 7 m n 5 ) ( 6 m 4 n 3 ) ( 7 m n 5 )$
135.
$( 4 7 x y 2 ) ( 14 x y 3 ) ( 4 7 x y 2 ) ( 14 x y 3 )$
136.
$( 5 8 u 3 v ) ( 24 u 5 v ) ( 5 8 u 3 v ) ( 24 u 5 v )$
137.
$( 2 3 x 2 y ) ( 3 4 x y 2 ) ( 2 3 x 2 y ) ( 3 4 x y 2 )$
138.
$( 3 5 m 3 n 2 ) ( 5 9 m 2 n 3 ) ( 3 5 m 3 n 2 ) ( 5 9 m 2 n 3 )$
### Everyday Math
139.
Email Janet emails a joke to six of her friends and tells them to forward it to six of their friends, who forward it to six of their friends, and so on. The number of people who receive the email on the second round is $62,62,$ on the third round is $63,63,$ as shown in the table. How many people will receive the email on the eighth round? Simplify the expression to show the number of people who receive the email.
Round Number of people
$11$ $66$
$22$ $6262$
$33$ $6363$
$……$ $……$
$88$ $??$
140.
Salary Raul’s boss gives him a $5%5%$ raise every year on his birthday. This means that each year, Raul’s salary is $1.051.05$ times his last year’s salary. If his original salary was $40,00040,000$, his salary after $11$ year was $40,000(1.05),40,000(1.05),$ after $22$ years was $40,000(1.05)2,40,000(1.05)2,$ after $33$ years was $40,000(1.05)3,40,000(1.05)3,$ as shown in the table below. What will Raul’s salary be after $1010$ years? Simplify the expression, to show Raul’s salary in dollars.
Year Salary
$11$ $40,000(1.05)40,000(1.05)$
$22$ $40,000(1.05)240,000(1.05)2$
$33$ $40,000(1.05)340,000(1.05)3$
$……$ $……$
$1010$ $??$
### Writing Exercises
141.
Use the Product Property for Exponents to explain why $x·x=x2.x·x=x2.$
142.
Explain why $−53=(−5)3−53=(−5)3$ but $−54≠(−5)4.−54≠(−5)4.$
143.
Jorge thinks $(12)2(12)2$ is $1.1.$ What is wrong with his reasoning?
144.
Explain why $x3·x5x3·x5$ is $x8,x8,$ and not $x15.x15.$
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
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Vectors - Numbers and Operations - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2
## Numbers and Operations
### 3.5 Vectors
A vector in a plane is defined to be an ordered pair of real numbers. A vector in space is defined as an ordered triple of real numbers. On a coordinate system, a vector is usually represented by an arrow whose initial point is the origin and whose terminal point is at the ordered pair (or triple) that named the vector. Vector quantities always have a magnitude or norm (the length of the arrow) and direction (the angle the arrow makes with the positive x-axis). Vectors are often used to represent motion or force.
All properties of two-dimensional vectors can be extended to three-dimensional vectors. We will express the properties in terms of two-dimensional vectors for convenience. If vector is designated by (v1v2) and vector is designated by (u1u2), vector is designated by (uv1,uv2) and called the resultant of and . Vector – has the same magnitude as but has a direction opposite that of .
On the plane, every vector can be expressed in terms of any other two unit (magnitude 1) vectors parallel to the x - and y-axes. If vector = (1,0) and vector = (0,1), any vector = ai + bj, where a and b are real numbers. A unit vector parallel to can be determined by dividing by its norm, denoted by and equal to
It is possible to determine algebraically whether two vectors are perpendicular by defining the dot product or inner product of two vectors, (v1v2) and (u1u2).
Notice that the dot product of two vectors is a real number , not a vector. Two vectors, and , are perpendicular if and only if
EXAMPLES
1. Let vector = (2, 3) and vector = (6, –4).
(A) What is the resultant of and?
(B) What is the norm of ?
(C) Express in terms of and .
(D) Are and perpendicular?
SOLUTIONS
(A) The resultant, equals (6 + 2, –4 + 3) = (8, –1).
(B) The norm of
(C) To verify this, use the definitions of and = 2(1,0) + 3(0,1) = (2, 0) + (0, 3) = (2, 3) =
(D) = 6 · 2 + (– 4) · 3 = 12 – 12 = 0. Therefore, and are perpendicular because the dot product is equal to zero.
2. If = (–1, 4) and the resultant of and is (4,5), find .
Let The resultant = (–1,4) + (v1v2) = (4,5). Therefore, (–1 + v1, 4 + v2) = (4,5), which implies that –1 + v1 = 4 and 4 + v2 = 5. Thus, v1 = 5 and
EXERCISES
1. Suppose Find the magnitude of
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
2. If and the resultant vector of equals
(A)
(B)
(C)
(D)
(E)
3. A unit vector perpendicular to vector is
(A) (4,3)
(B)
(C)
(D)
(E)
|
Signed Number Representation in Computers
On paper, we can represent a negative number by prefixing it with a minus sign. But, in computers, we will have to encode the sign in the number itself. This article discusses three prominent methods used for representing signed numbers in computers. All three methods work by treating the leftmost digit as a kind of sign bit.
Sign-Magnitude representation
At first glance, the problem of representing signed numbers in a computer might seem quite straightforward. The most obvious solution would be:
• Reserve one digit, possibly the leftmost digit, for representing the sign of the number.
• Use the remaining digits to represent its magnitude.Â
The convention is that a sign-bit of 0 indicates a positive number and a sign-bit of 1 indicate a negative number. This method is known as the sign-magnitude representation. It has been used in some computers in the past but it never gained much popularity.
Let us look at the example of a 4-digit number. After dedicating 1 bit for sign, the remaining 3 bits can represent numbers up to 7. This gives us a range of [-7..+7].
I will be using 4-bit numbers in most of the examples in this article. Real word sizes are larger, but using smaller words keeps the examples simpler. The principles presented here will work the same for more realistic word sizes like 32-bit or 64-bit.
Table 1: Sign-Magnitude representation of 4-bit numbers
Decimal Sign-Magnitude
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
0 1000
-1 1001
-2 1010
-3 1011
-4 1100
-5 1101
-6 1110
-7 1111
Sign-Magnitude representation is the easiest for humans to understand. But it cannot be implemented efficiently in hardware. For performing addition or subtraction we need to first check the sign and magnitude of the operands. The result of these checks determines which operation we need to perform, the order of the operands, and the sign of the result.
Let us say we want to add 0001 and 1010. Here the first number represents +1 and the second number represents -2. So -(2 – 1)Â = -1 is what we are looking for. The opcode of the instruction is Add. But we need to check the sign of the operands to see whether addition or subtraction is required. If subtraction is required, we have more work to do. Next, we have to check the magnitude of the numbers. Only then we know which number should be the minuend and which should be the subtrahend.
The previous paragraph might sound like a repetition of basic signed number arithmetic rules. While it might feel natural to us, computers can’t do it efficiently. 1’s complement and 2’s complement provide a better solution, as we will see.
Another problem with sign-magnitude representation is that it has two representations for zero. In our 4-bit example, both 0000 and 1000 represent zero. This would add a fair amount of complexity to the computer.
1’s complement representation
Positive numbers are represented in the same way in both 1’s complement and sign-magnitude. The leftmost digit of all positive numbers has a sign-bit of zero. The remaining digits represent the magnitude of the number. For negative numbers, we find the 1’s complement of the number and use that instead.
Complement forms exist for numbers of all radices or bases. There are two types of complements:
Also known as r’s complement where r is the radix. So for binary numbers, r’s complement is the same as 2’s complement. For decimal numbers, r’s complement is the same as 10’s complement.
Also known as (r-1)’s complement. This is the same as 1’s complement for binary numbers and 9’s complement for decimal numbers.
How to calculate 1’s complement
The formula for finding (r-1)’s complement is rn-1-N. Where r is the radix, n is the number of digits in the number and N is the actual number.
One way to understand this formula would be like this. rn is the smallest n +1 digit integer. Hence rn-1 is the largest n digit integer. So to find (r-1)’s complement you just find the largest n digit integer and subtract N from it.
An example should make this clear. Let us say you want to find the (r-1)’s or 9’s complement of decimal number 55. Here n is 2 digits and the largest 2 digit integer is 99. So 9’s complement of 55 is 99 – 55 = 44. Similarly, 9’s complement of 555, would be 999 – 555 = 444.
Now let us look at a 1’s complement example. We want to find the 1’s complement of 610 = 1102. There are 3 digits in 110 and the largest 3 digit binary integer is 111. 1112 – 1102 = 12.
There is an easier way to find 1’s complement. Simply invert each bit in the number using not gates. That is, every 1 in the number becomes 0 and every 0 becomes 1. Thus 1’s complement of 110 is 001.
Decimal to 1’s complement mapping for 4-bit numbers
You may have begun to notice that finding the complement of a number is like counting up from the bottom of a list of numbers.
Table 2: 1’s complement representation of 4-bit numbers
Decimal 1’s Complement
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
-7 1000
-6 1001
-5 1010
-4 1011
-3 1100
-2 1101
-1 1110
0 1111
Like the sign-magnitude table, the rows of this table are in the ascending order of the actual numbers used by the computer, not what those numbers represent. Getting familiar with this ordering will help in understanding how 1’s complement or 2’s complement work. The positive numbers plus one of the two zeros take up the top half of the table. The bottom half of the table is reserved for the other zero and the negative numbers.
Now, let us address the most important question. What do we gain by representing numbers this way? The benefit is that we can do additions or even subtractions on signed numbers without bothering to check the sign or magnitude of the numbers. Plus we need only adder units to perform both addition and subtraction
In 1’s complement, addition is done using a variant of modular arithmetic. In modular arithmetic, the values wrap around once we reach some specified maximum. A clock with only an hour handle and hour 12 labeled as 0 would be an example of modulo 12 arithmetic. Since there are 16 elements in our example list we would be doing modulo 16 arithmetic.
There is one unusual rule which we need to follow in 1’s complement addition. If there is a carry out from the most significant digit, it is not brought down to the left of the result as we usually do. Instead, it is added to the result. This might sound a bit confusing, so let us look at some examples.
Examples
Let us say we want to add 5 and -2. In 1’s complement, these numbers are 0101 and 1101 respectively.
If we had done normal addition the result would have been 10010. If we had done true modular addition the result would have been 0010. But 1’s complement addition gives us a result of 0011. We can verify that 00112 is 310. Why the carry out is handled this way is explained in the How does it work? section. For the moment let us focus on the fact that we just added a negative number and a positive number, without bothering about which number has the higher magnitude and got the right result. Will this work if the negative number had the higher magnitude? What about -510 + 210 = 10102 + 00102 ? Let us see.
From the 1’s complement table given above, we can verify that the result 1100 is the 1’s complement representation of -3.
Find the original number given its 1’s complement
We don’t need to look up a table every time to find what a 1’s complement number represents. There is a simple formula for finding the original number given its 1’s complement. We just need to find 1’s complement of the 1’s complement. This makes sense because we found the 1’s complement by subtracting the original number from 2n-1.
1’s Complement Subtraction
Let us look at how subtraction works. For example, what if we wanted 5 – 2. For subtraction, there is an extra step. First, we need to do to find the 1’s complement of the subtrahend. After that, it is the same addition we saw in the first example. The extra step does not complicate the hardware since it is a simple step required by all subtraction instructions.
Despite the advantages provided by 1’s complement, it is not used in any of the modern computers. Its main drawback is that it has two representations of zero. Modern computers use 2’s complement method since it does not have this limitation.
2’s Complement Representation
2’s complement is the method used for representing signed numbers in pretty much all modern computers. It is quite similar to 1’s complement. Once we know 1’s complement, there isn’t much new to learn.
How to calculate
Remember that the formula for finding diminished radix or (r-1)’s complement was rn – 1 – N. Thus the formula for 1’s complement was 2n – 1 – N.
For radix complement or r’s complement, the formula is rn – N. Where r is the radix, n is the number of digits in the number and N is the number itself. Thus the formula for 2’s complement is 2n – N.
To find the 2’s complement of a 3-bit number like 110, we find the smallest 4-bit number which is 1000, and subtract 110 from it.
10002 − 1102 = 0102
Another method for finding 2’s complement is to find 1’s complement first and then add 1. Let us find the 2’s complement of 1102 using this method. 1’s complement of 1102 can be found using the bit inversion method mentioned earlier.
x = 110
1’s complement of x = 001 (Just invert each bit)
And now, to find the 2’s complement:-
0012 + 12 = 0102
There is one special case to consider when finding 2’s complement. If we try to find the 2’s complement of 0, we get a number that won’t fit our word size. In this case, the extra digit on the left must be discarded.
Table 3: 2’s complement representation of 4-bit numbers
Decimal 2’s Complement
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
-8 1000
-7 1001
-6 1010
-5 1011
-4 1100
-3 1101
-2 1110
-1 1111
In 2’s complement, there is only one representation of zero. A sign-bit of 0 means zero or a positive number. The numbers with a sign-bit of 1 represent only negative numbers. This means that there is an extra negative number in 2’s complement representation. In our 4-bit number example, the 2’s complement numbers range from [-8 .. +7].
Rules for addition in 2’s complement are similar to that for 1’s complement. The only difference is that, carry out, if any, from the most significant digit, is simply ignored. There is no need to increment the result by 1. This means that unlike in 1’s complement, in 2’s complement we do true modular arithmetic.
Please note that throwing away the carry to wrap around does not work in all cases of modular addition. For example, it will not work in the clock example mentioned earlier. The trick works in computers because of the ranges we use. If we are doing n digit addition, the largest number we support would be the largest n digit integer. In our 4-bit example, this would be 1111. If we add 1 to it and throw away the carry we get 0000.
Example
Let us try to add 5 and -2 again, this time using 2’s complement. The 2’s complement numbers are 0101 and 1110 respectively.
The carry out from the most significant digit is ignored.
2’s Complement Subtraction
Subtraction is also similar in 2’s complement. The steps are, find the 2’s complement of the subtrahend and add the numbers using 2’s complement rules. The extra negative number which 2’s complement representation has, does pose difficulties. If we try to find the 2’s complement of the negative number with the largest magnitude, we will get the same number back. In our example, the 2’s complement of 1000 is 1000. We have to watch out for this special case.
How does it work?
Let us take a look at how the complement methods work. I will be focusing on 2’s complement, the principles are roughly the same for 1’s complement. I will mention the differences between the two as we go along.
Let us look at another 4-bit, 2’s complement example. Let us say, we want to subtract +2 from +5. For reference, please take a look at table 3 showing the 2’s complement representation of 4-bit numbers. It is important to understand the order of numbers in that table. It may be a good idea to keep the table accessible in a different browser tab. The number which represents +5 is in 6th position in the table and the number is 0101.
Think of it this way, what we need to do is to go up two positions in the list and reach the number representing +3. But we don’t really like going up the list since that would mean having to subtract. We would like to reach our destination by going down the list, that is, by adding numbers. This is how we can achieve this. Go down the list instead of up. When we cross the last element of the list there will be a carry from the most significant digit. Throw away this carry so that we can jump back to the top of the list.
For example, if we are at the bottom of the list where the number is 1111 and we add one to it, we get 10000 which is a 5-bit number. But if we throw away the carry from the most significant digit the result would be 0000, which is the top of our list.
Our list consists of 16 numbers. If we start at the 6th position in the list and add 16 using modulo 16 addition, we will get back to where we started. What if we add 14 instead? We will reach the 4th position where the number is +3, which is where we want to be.
First, we have to figure out which number to add. We find the 2’s complement of the subtrahend, that is the number representing -2. This number is at the second last position of the list. Note that the distance between -2 and the top of the list is 2 when doing modulo 16 addition. Further note that 2 is the number that we want to subtract from 5. Now, all we need to do is a 2’s complement addition of the number representing -2 and the number representing 5 i.e. 11002 + 01012. There will be a carry out from the most significant digit. We cycle around the list by ignoring this carry and reach our destination of +3.
Check out the image below showing 2’s complement numbers depicted in a circle. We are allowed to move only in a clockwise direction in this circle.
Carry out in 1’s complement addition
Remember that in 1’s complement addition we did not quite follow the modulo addition rules. When there was a carry out from the most significant digit, we did not throw it away. We incremented the result by one instead. The reason why this is required is that in 1’s complement there are two representations of zero. The first element in the list and the last element in the list are both zeros.
Let say, we add the 1’s complement representations of -1 and +2 and throw away the carry from the most significant digit. We may think we will get +1 as the result, but we won’t because there are two zeros between -1 and +1. So each time we make the jump from the zero at the bottom of the list to the zero at the top of the list we have to compensate for the jump by adding a 1 to the result.
So how do we detect that a 1’s complement addition has lead to a jump from one zero to the other? Remember that ignoring the carry from the most significant digit is the trick we use to jump from the bottom of the list to the top of the list. So, all we need to do is to check if there was a carry out from the most significant digit.
Detecting overflow in 2’s complement addition
One very important point which we haven’t looked at so far is the question of overflow. There needs to be a way for the program to detect an overflow. Detecting overflow is a little tricky in 2’s complement.
The results of 2’s complement additions are often larger than our word length. But exceeding word length does not imply an overflow. We could still get a valid result after chopping off the extra bit from the left. On the other hand, we could get an overflow without actually exceeding our word length. For example, adding two positive numbers might produce a negative number as result.
To detect overflow, we need to look at both the carry in and carry out of the most significant digit. As it turns out, there is a simple rule for detecting overflow in 2’s complement addition. If the carry out and carry in for the most significant digit are different, then there is an overflow.
Let us look at some examples from our 4-bit 2’s complement representation to check if the rule holds up. It also gives us an opportunity to take a closer look at how 2’s complement additions work.
When both operands are positive
If both the operands are positive, the result must also be positive. If we add two positive numbers and get a negative number as the result then there is an overflow.
Example a: 210 + 310 = 00102 + 00112
We added 2 positive numbers and the result is also positive. There is no carry out or carry into the most significant digit, so the rule holds.
Example b: 310 + 510 =Â 00112Â +Â 01012
We added two positive numbers and veered into territory reserved for negative numbers. +8 is too big to be represented as a signed number in a 4-bit word. Notice that there is a carry into the most significant digit but no carry out. This implies an overflow.
When both operands are negative
Remember that the negative numbers are in the bottom half of our list of numbers. When we add negative numbers there will always be a carry from the most significant digit. Each time we will be throwing away this carry so that we can cycle through the list. To avoid an overflow we must cross the positive territory at the top of the list and get back into negative territory.
You could think of the magnitude of a negative number as a distance from its 2’s complement to zero. This does not make sense for normal arithmetic but it does make sense for modular addition since addition wraps around the list. Take a look at the following examples, the binary numbers are in 2’s complement form.
If we take a negative number which is x distance away from 00002 and another negative number which is y distance away from 00002 and add the two using modular addition, we will get a number which is x + y distance away from 00002.
Example a: -110 + -210 = 11112 + 11102
Here we have two operands with small magnitude. This means that their 2’s complements are large numbers. The result easily gets past the top half of the list and reaches the bottom half. There is a carry of 1 coming in and out of the most significant digit, this implies that there is no overflow.
Example b: -710 + -510 = 10012 + 10112
We added two negative numbers and got a positive number as the result. Also, note that there is a carry out of the most significant digit but no carry coming into it.
When the operands have opposite sign
When the operands are of opposite sign, there won’t be an overflow since the result is guaranteed to have a lower magnitude than both the operands. If the operands are small enough to fit our word size, then the result will fit as well.
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# Lesson 18
More Relationships
### Lesson Narrative
This lesson is optional. It offers opportunities to look at multiple representations (equations, graphs, and tables) for some different contexts.
This final lesson on relationships between two quantities examines situations of constant area, constant volume, and a doubling relationship. Students have an opportunity to engage in MP7 as they notice the similar structures of the situations in the Making a Banner and Cereal Boxes activities, as well as connecting the Multiplying Mosquitoes activity to prior work with exponents and the Genie's coins situation from earlier in the unit. They may use those observations and knowledge to more easily solve the problems in the activities.
Consider offering students a choice about which one they work on. Then, in the lesson synthesis, invite students to share their work with the class and compare and contrast the representations of the different contexts.
### Learning Goals
Teacher Facing
• Coordinate (orally and in writing) graphs, tables, and equations that represent the same relationship.
• Create an equation and a graph to represent the relationship between two variables that are inversely proportional.
• Describe and interpret (orally and in writing) a graph that represents a nonlinear relationship between independent and dependent variables.
### Student Facing
Let’s use graphs and equations to show relationships involving area, volume, and exponents.
### Student Facing
• I can create tables and graphs that show different kinds of relationships between amounts.
• I can write equations that describe relationships with area and volume.
### Glossary Entries
• coordinate plane
The coordinate plane is a system for telling where points are. For example. point $$R$$ is located at $$(3, 2)$$ on the coordinate plane, because it is three units to the right and two units up.
• dependent variable
The dependent variable is the result of a calculation.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The dependent variable is the distance traveled, because $$d$$ is the result of multiplying 25 by $$t$$.
• independent variable
The independent variable is used to calculate the value of another variable.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The independent variable is time, because $$t$$ is multiplied by 25 to get $$d$$.
### Print Formatted Materials
For access, consult one of our IM Certified Partners.
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# How do you find the percent change if original: $3.78 and new:$2.50?
Feb 15, 2017
$= - 33.862$% to 3 decimal places
Full 'teach in' given about percentages
#### Explanation:
$\textcolor{b l u e}{\text{Preamble}}$
In percentage the symbol % may be viewed as a unit of measurement that is worth $\frac{1}{100}$
As and example 60% is the same as $60 \times \frac{1}{100} = \frac{60}{100}$
In this type of question it is usual to express the change using the original value is the reference point. Expressed as a fraction we have:
$\left(\text{change")/("reference point value") -> ("changed value- original value")/("original value}\right)$
If changed value - original value is negative then it indicates a reduction. As is the case in this question.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$
$\textcolor{b r o w n}{\text{Starting point:}}$
("changed value- original value")/("original value") -> ($2.50-$3.78)/($3.78) =-($1.28)/($3.78) By the way; like in numbers the$ symbols cancel out in the same way numbers do. This is a useful trick when doing higher mathematics or physics
So we have (-cancel($)1.28)/(cancel($)3.78) = -1.28/3.78
..............................................................................................
$\textcolor{b r o w n}{\text{Converting this into percentage}}$
Remember that 1/100->%
color(green)((1.25)/(3.78)color(red)(xx1)" "->" "(1.25/3.78color(red)(xx100/100))
$\left(- \frac{1.28}{3.78} \textcolor{red}{\times 100 \times \frac{1}{100}}\right)$
(-1.28/3.78color(red)(xx100xx%))
(-1.28/3.78color(red)(xx100))%
$= - 33.862$% to 3 decimal places
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# Proportion
We will discuss about the proportion of four quantities.
Four quantities w, x, y, and z are in proportion if w : x :: y : z or, $$\frac{w}{x}$$ = $$\frac{y}{z}$$.
Definition: If four quantities w, x, y and z are such that the ratio w : x is equal to the ratio y : z then we say w, x, y and z are in proportion or, w, x, y and z are proportional. We express it by writing w : x :: y : z.
w : x :: y : z if and only if $$\frac{w}{x}$$ = $$\frac{y}{z}$$ i.e., wz = xy.
In w : x : : y : z, w, x, y and z are the first, second, third and fourth terms respectively.
Also, w and z are called the extreme terms while x and y are called the middle terms or mean terms.
That is in a proportion the first and fourth terms are called extremes, while the second and third terms are called means.
Product of extremes = product of means
or, wz = xy.
If w, x, y and z are in proportion then the last term z is also called the fourth proportional.
For example, 2 : 7 :: 8 : 28 because $$\frac{2}{7}$$ = $$\frac{8}{28}$$.
Solved examples on Proportion:
Check whether the following numbers form a proportion or not.
(i) 2.5, 4.5, 5.5, 9.9
(ii) 1$$\frac{1}{4}$$, 1$$\frac{3}{4}$$, 1.5, 1.4
Solution:
(i) 2.5 : 4.5 = $$\frac{2.5}{4.5}$$ = $$\frac{25}{45}$$ = $$\frac{5}{9}$$
5.5 : 9.9 = $$\frac{5.5}{9.9}$$ = $$\frac{55}{99}$$ = $$\frac{5}{9}$$
Therefore, $$\frac{2.5}{4.5}$$ = $$\frac{5.5}{9.9}$$
Hence, 2.5, 4.5, 5.5 and 9.9 are in proportion.
(ii) 1$$\frac{1}{4}$$ : 1$$\frac{3}{4}$$ = $$\frac{5}{4}$$ : $$\frac{7}{4}$$ = $$\frac{5}{4}$$ × 4: $$\frac{7}{4}$$ × 4 = 5 : 7 = $$\frac{5}{7}$$
1.5 : 1.4 = $$\frac{1.5}{1.4}$$ = $$\frac{15}{14}$$
Since, $$\frac{5}{7}$$ ≠ $$\frac{15}{14}$$
Hence, 1$$\frac{1}{4}$$, 1$$\frac{3}{4}$$, 1.5 and 1.4 are not in proportion.
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# Problems on Ages Questions
Q:
A father said his son , " I was as old as you are at present at the time of your birth. " If the father age is 38 now, the son age 5 years back was :
A) 14 B) 19 C) 33 D) 38
Answer & Explanation Answer: A) 14
Explanation:
Let the son's present age be x years .Then, (38 - x) = x => x= 19.
Son's age 5 years back = (19 - 5) = 14 years
Report Error
71 18037
Q:
The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?
A) 12 B) 13 C) 14 D) 15
Answer & Explanation Answer: A) 12
Explanation:
(A+B) - (B+C) = 12
$\inline \fn_jvn \Rightarrow$A - C = 12.
$\inline \fn_jvn \Rightarrow$C is younger than A by 12 years.
Report Error
82 14781
Q:
In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, the present age of B is :
A) 19 B) 29 C) 39 D) 49
Answer & Explanation Answer: C) 39
Explanation:
Let B's present age = x years. Then, A's present age = (x + 9) years.
(x + 9) + 10 = 2(x - 10)
=> x + 19 = 2x - 20
=> x =39.
Report Error
25 6392
Q:
The age of a man is 4 times of his son. Five years ago, the man was nine times old as his son was at that time. The present age of man is?
Answer
Let the son's age be x years and the father's age be 4x years
$\inline \Rightarrow (4x-5)=9(x-5)$
5x = 40
x = 8
$\inline&space;\therefore$ present age of the father = $\inline&space;4x$ = $\inline&space;4\times&space;8$ = 32 years
Report Error
5646
Q:
The sum of the present ages of a father and his son is 60 years. five years ago, father's age was four times the age of the son. so now the son's age will be:
A) 5 B) 10 C) 15 D) 20
Answer & Explanation Answer: C) 15
Explanation:
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 5= 4(x - 5)
55 - x = 4x - 20
5x = 75 => x = 15
Report Error
17 5324
Q:
Six years ago Anita was P times as old as Ben was. If Anita is now 17 years old, how old is Ben now in terms of P ?
A) 11/P + 6 B) P/11 +6 C) 17 - P/6 D) 17/P
Answer & Explanation Answer: A) 11/P + 6
Explanation:
Let Ben’s age now be B
Anita’s age now is A.
(A - 6) = P(B - 6)
But A is 17 and therefore 11 = P(B - 6)
11/P = B-6
(11/P) + 6 = B
Report Error
42 4789
Q:
The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.
Answer
As the ratio of their present ages is 3 : 4 , let their present ages be 3X and 4X.
So, 5 years ago, as the ratio of their ages was 5 : 7, we can write (3x-5) : (4x-5) = 5 : 7. Solving, we get X = 10. Hence, their present ages are 3X = 30 and 4X = 40
Report Error
4239
Q:
Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin
A) 24.5 B) 25.5 C) 26.5 D) 27.5
Answer & Explanation Answer: A) 24.5
Explanation:
If Rahul age is x, then Sachin age is x - 7,
so, (x-7)/x =7/9
$\inline \Rightarrow$ 9x - 63 = 7x
$\inline \Rightarrow$ 2x = 63
$\inline \Rightarrow$ x = 31.5
So Sachin age is 31.5 - 7 = 24.5
Report Error
31 4201
|
# What The Numbers Say About Michael Jackson (03/27/2020)
How will Michael Jackson perform on 03/27/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first work out the destiny number for Michael Jackson, and then something similar to the life path number, which we will calculate for today (03/27/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts.
PATH NUMBER FOR 03/27/2020: We will consider the month (03), the day (27) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 27 we do 2 + 7 = 9. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 9 + 4 = 16. This still isn’t a single-digit number, so we will add its digits together again: 1 + 6 = 7. Now we have a single-digit number: 7 is the path number for 03/27/2020.
DESTINY NUMBER FOR Michael Jackson: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Michael Jackson we have the letters M (4), i (9), c (3), h (8), a (1), e (5), l (3), J (1), a (1), c (3), k (2), s (1), o (6) and n (5). Adding all of that up (yes, this can get tedious) gives 52. This still isn’t a single-digit number, so we will add its digits together again: 5 + 2 = 7. Now we have a single-digit number: 7 is the destiny number for Michael Jackson.
CONCLUSION: The difference between the path number for today (7) and destiny number for Michael Jackson (7) is 0. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is of questionable accuracy. If you want a reading that people really swear by, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# Learning Division Vocabulary for Kids
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## What Is Division Vocabulary?
Division vocabulary includes words which we often use while solving all division related problems. Division vocabulary words include division, division symbol, dividend, divisor, quotient, and remainder. Here, we will learn the division vocabulary with interactive examples that will help you to understand and solve division problems easily.
## What Is Division?
The term “division” means to split into equal parts or groups. The division is an outcome of “fair sharing” because it splits everything equally.
## Division Symbol
The division symbol indicates to divide into equal groups or parts.
We often use the symbol obelus () , or sometimes slash (/) to mean divide:
15 5 = 3
15 / 5 = 3
This is read as “fifteen divided by five equals three”.
## Division Example
Let us understand division with an example below:
Example:
There are 12 chocolates in total. Divide these 12 chocolates evenly among 3 of your friends. How will you divide the chocolates?
Solution: To evenly divide a total of 12 chocolates among 3 friends, we will divide 12 by 3.
Image: 12 divided by 3
Image: 12 Chocolates
12 3 = 4
This is read as twelve divided by three equals four.
Hence, each friend will get 4 chocolates.
Example 2:
Look at the fish given below:
Image: 9 fish in 3 groups
Here, 9 fish are swimming into a group of three.
This can be written as: Total 9 fish divided into groups of 3 makes three groups of each.
This will be calculated as: 9 3 = 3
## What Is a Dividend?
In division, the dividend is the number to be divided. In other words, a dividend is the whole that is to be splitted or divided into equal parts.
Example:
There are 12 candies that are to be equally divided among 3 children. Here 12 is the dividend.
Image: 12 candies divided equally among 3 children
12 candies 3 children = 4 candies to each child.
12 is the dividend, and 12 3 = 4
## What Is a Divisor?
The divisor is the number we divide the dividend by. In other words, a divisor is a number by which another number is divided.
Example:
In 12 3 = 4, 3 is the divisor.
↓
Divisor
## Divisor Representation in Division
In division, divisor is represented as:
Dividend Divisor = Quotient
Example:
On dividing 12 by 3, we get 4. Here 3 is the number that is dividing 20 (dividend) completely into 4 parts and is known as the divisor.
Dividend Divisor Quotient
12 3 = 4
## What Is a Quotient?
A quotient is the result that we get after dividing one number by another number. In other words, quotient is the number that we get after dividing dividend by divisor.
Example:
Let us divide 12 by 3.
### Image: 12 cupcakes divided equally in small groups of 3
Here, the larger group of 12 cupcakes is known as the dividend. The small equal groups of 3 are known as divisors. The number of cupcakes in each group is termed a quotient. Here, there are 4 cupcakes in each smaller group, which is the quotient.
## What Is a Remainder?
The remainder is the number that is left over or remaining when dividing one number by another. In other words, when we are not able to make an equal group or equally share all the objects, we are left with a number known as the remainder.
Remember: The remainder value is always less than the divisor.
Example:
1. We have 9 pencils to share with 2 children.
Solution: But 9 cannot be split into 2 equal groups.
So each child gets 4 pencils and one is left over.
Image: 9 pencils divided equally among 2 children with one pencil left.
9 2 = 4 R1
This is read as 9 divided by 2 equals 4 with a remainder of 1. And we write it as: 9 2 = 4 R1
## Do You Know?
• The dividend is the whole number that is divided by another number.
• The divisor is a number that divides the other number (dividend) either completely or with a remainder.
• The quotient is the result obtained after division of two numbers.
• The remainder value is always less than the divisor value.
• If the remainder value is greater than the divisor, it means division is incomplete.
• The remainder value can either be greater, lesser, or equal to the quotient value.
## Conclusion
In short, division is a process of dividing a group of things into equal parts, and division vocabulary such as dividend, division symbol, divisor, quotient, and remainder are the terms used in the division process.
Last updated date: 28th Sep 2023
Total views: 139.2k
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## FAQs on Learning Division Vocabulary for Kids
1. When do we get the remainder value 0?
When one number (i.e. divisor) completely divides the other number (i.e. dividend), then the remainder value we get is 0. For example, on dividing 15 by 3, we get the quotient 5 and remainder 0. It is because 3 (divisor) is completely dividing the dividend (15).
2. What is the division formula?
The division formula is represented as: Dividend Divisor = Quotient.
For example, in 15 3 = 5,
15 is the dividend, 3 is the divisor, and 5 is the quotient.
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# Standard Deviation vs. Variance
Standard Deviation vs. Variance
Two statistical measures that are often quite confusing for many people are standard deviation and variance. Both are measures of the distribution of data, representing the amount of variation there is from the average, or to the range the values normally differ from the average, which is also called the mean. If the variance or standard deviation has a value of zero, all the values are identical.
Variance is the mean or average of the squares of the deviations or differences in the values from the mean. On the other hand, standard deviation is the square root of that variance. The two are closely related, but standard deviation is used to identify the outliers in the data.
There is one similarity between the two values. The standard deviation and the variance values will always be non-negative.
The mathematical formula for a standard deviation is the square root of the variance. On the other hand, the variance's formula is the average of the squares of deviations of each value from the mean in a sample. The deviations are squared to prevent negative values from canceling out the positive values.
The symbol for standard deviation is the Greek letter sigma: σ. There is no dedicated symbol for variance, and it is expressed in the same unit as the values themselves.
In the real world, standard deviation is used with population sampling data and identifying outliers. For variance, it used with statistical formulas and in the world of finance.
To understand the difference between the values can be done by reviewing an example. Using the following values: 5, 8, 4, 7, 6. The mean or average is equal to: 6. Finding the variance, one must subtract the mean from each value, squaring each result: 5 - 6 = -12, 8 - 6 = 22, etc. Next, add the squares and find the mean or average: 1 + 4 + 4 + 1 + 0 = 10/5 = 2. The variance of the data is 2.
To find the standard deviation, calculate the square root of the variance: √2 = standard deviation.
The variance of the set of data is an arbitrary number (2) relative to the original measurements of the data set. This makes it difficult to visualize and apply in the real world, but useful in finance and statistical formulas.
The standard deviation (√2 = 1.4) is expressed in the original units of the data set. This value is more natural and closer to the values of the original data set and most often used to analyze demographics or population samples.
In summary, standard deviation cannot be calculated without first finding the variance of a set of data, and variance is then used to discover the standard deviation. The steps to find each figure are similar, but standard deviation is used more often in the real world, such as for populations, versus variance, which is most useful for other statistical formulas and the finance world.
Related Links: Difference between Words Science Related Words Difference and Comparison Descriptive vs. Inferential Statistics Probability and Statistics activities
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# Question: How Do You Do The Commutative Property?
## Why is commutative property important?
Place value and commutative property are important to remember when understanding and solving addition and multiplication equations.
The order of the numbers in the equation does not matter, as related to the commutative property, because the sum or product is the same..
## What is a commutative property of multiplication?
What is the commutative property? The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
## What are the 3 properties of multiplication?
There are three properties of multiplication: commutative, associative, and distributive.Commutative Property.Associative Property.Distributive Property.
## How are real life situations commutative?
The order of events does not matter. Tom can either wash his face first or put his pants on first. During a physical exam, Sarah’s doctor check her blood pressure, blood sugar level, and heart rate. This situation is commutative.
## What does commutative property mean?
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. The property holds for Addition and Multiplication, but not for subtraction and division.
## What is associative and commutative property?
In math, the associative and commutative properties are laws applied to addition and multiplication that always exist. The associative property states that you can re-group numbers and you will get the same answer and the commutative property states that you can move numbers around and still arrive at the same answer.
## What are the 4 properties in math?
There are four basic properties of numbers: commutative, associative, distributive, and identity. You should be familiar with each of these. It is especially important to understand these properties once you reach advanced math such as algebra and calculus.
## What is commutative property 3rd grade?
The Commutative Property of Multiplication states that you can multiply factors in any order and get the same product. For any two values, a and b, a × b = b × a.
## What is commutative law example?
The commutative law of addition states that if two numbers are added, then the result is equal to the addition of their interchanged position. Examples: 1+2 = 2+1 = 3. 4+5 = 5+4 = 9.
## Do you add first or multiply first?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right. Next, add and subtract from left to right.
## How do you solve distributive property?
Distributive property with exponentsExpand the equation.Multiply (distribute) the first numbers of each set, outer numbers of each set, inner numbers of each set, and the last numbers of each set.Combine like terms.Solve the equation and simplify, if needed.
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### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Have You Got It?
Can you explain the strategy for winning this game with any target?
### Counting Factors
Is there an efficient way to work out how many factors a large number has?
# Leap Monday
##### Age 11 to 14 ShortChallenge Level
Listing the days of the week of 1$^\text{st}$ March
365$\div$7 = 52 remainder 1
29 Feb 2016: Monday
1 March 2016: Tuesday
364 days later is Tuesday again so
1 March 2017: Wednesday
1 March 2018: Thursday
1 March 2019: Friday
1 March 2020: Sunday (leap year)
2021: Monday
2022: Tuesday
2023: Wednesday
2024: Friday (leap year)
2028: Wednesday
2032: Monday
2036: Saturday
2040: Thursday
2044: Tuesday (so 29th February will be Monday)
Finding a pattern of the days of the week of the 29$^{\text{th}}$ of February
It will be the 29$^{\text{th}}$ of February again after 365$\times$3 + 366 = 1461 days.
1461 $\div$ 7 = 208 remainder 5, so 1461 days is the same as 208 weeks and 5 days.
The next 29$^{\text{th}}$ of February after 2016 will be in 2020, and it will be 5 days after a Monday - a Saturday.
Similarly, the 29$^{\text{th}}$ of February 2024 will be 5 days after a Saturday - a Thursday. And the 29$^{\text{th}}$ of February 2028 will be 5 days after a Thursday - a Tuesday. Similarly, the 29$^{\text{th}}$ of February 2032 will be a Sunday, the 29$^{\text{th}}$ of February 2036 will be a Friday, the 29$^{\text{th}}$ of February 2040 will a Wednesday, and the 29$^{\text{th}}$ of February 2044 will be a Monday.
Interesting aside to think about: 7 is a prime number, so whatever remainder we get after dividing by 7 (providing it is non zero), we will always have to go for 7 lots of the leap years to get back to Monday. So the answer will be $7 \times 4 = 28$ years time if the remainder is not zero. This is an example of Fermat's Little Theorem.
Using Lowest Common Multiples
Every 7$^{\text{th}}$ day is a Monday, so 7 days, 14 days, 21 days, ”¦ after 29$^{\text{th}}$ February 2016 will all be Mondays.
It will be the 29$^{\text{th}}$ of February again after 365$\times$3 + 366 = 1461 days, and again after 2$\times$1461 days, and again after 3$\times$1461 days, and so on.
This means the number of days between Monday 29$^{\text{th}}$ February 2016 and another Monday 29$^{\text{th}}$ February will be a multiple of 7 and a multiple of 1461. So the number of days until it next happens will be the lowest common multiple of 7 and 1461.
If 1461 were a multiple of 7, then the lowest common multiple would be 1461. However, you can tell by dividing 1461 by 7 that it is not a multiple of 7. So, since 7 is a prime number, the lowest common multiple of 7 and 1461 must be 7$\times$1461. That is after 7 sets of 4 years - so it will next happen 7$\times$4 = 28 years later, in 2044.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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GreeneMath.com - Applications of Linear Equations II Lesson
# In this Section:
In this section, we will continue to learn about solving word problems. Here we will review some more challenging problems. We will review the six step method for solving a word problem with linear equations in one variable. Our first and most important step is to read the problem carefully. Then for our second step, we assign a variable to represent one of the unknowns, any additional unknowns are represented in terms of this variable. Third, we write an equation. Fourth, we solve the equation. Fifth, we write the solution in a nice, clear and concise sentence. For our sixth and final step, we check our answer to ensure that it makes sense in terms of our problem. We will review problems in which the sum of two or more items is known, but the individual amounts are not. This type of problem is known as “sums of quantities”. We will also review percent problems. These are problems that require us to calculate a price or value, before or after a certain “% “ markup or markdown. Additionally, we will look at simple interest problems. These problems require us to use the simple interest formula. This formula I=prt, is used to determine how much simple interest is earned, given the principal invested, the rate (as a decimal) and the time (generally given in years).
Sections:
# In this Section:
In this section, we will continue to learn about solving word problems. Here we will review some more challenging problems. We will review the six step method for solving a word problem with linear equations in one variable. Our first and most important step is to read the problem carefully. Then for our second step, we assign a variable to represent one of the unknowns, any additional unknowns are represented in terms of this variable. Third, we write an equation. Fourth, we solve the equation. Fifth, we write the solution in a nice, clear and concise sentence. For our sixth and final step, we check our answer to ensure that it makes sense in terms of our problem. We will review problems in which the sum of two or more items is known, but the individual amounts are not. This type of problem is known as “sums of quantities”. We will also review percent problems. These are problems that require us to calculate a price or value, before or after a certain “% “ markup or markdown. Additionally, we will look at simple interest problems. These problems require us to use the simple interest formula. This formula I=prt, is used to determine how much simple interest is earned, given the principal invested, the rate (as a decimal) and the time (generally given in years).
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Sunday , February 24 2019
Recent Post
Home / Uncategorized / Profit Loss Tricks – 1
# Profit Loss Tricks – 1
Profit/Loss is another easy topic of SSC CGL. Most of the questions can be solved in less than 30 seconds. First let me introduce a formula that will be used in solving 50% of the questions.
Where,
SP = Selling price
CP = Cost Price
f = Profit/loss factor
What is this profit/loss factor ? It’s simple, ‘f’ depends on the profit/loss %
If profit% = 10, then f = 1.1
If profit% = 30, then f = 1.3
If profit% = 15, then f = 1.15
If loss% = 10, then f = 0.9
If loss% = 25, then f = 0.75
If loss% = 12.5, then f = 0.875
Please note that f depends on profit/loss percentage and not on the absolute value of Profit/Loss. So if in any question it is given that the profit is Rs. 30, then it doesn’t mean that f = 1.3
Let us see some SSC CGL questions that can be solved with this formula
1. 1
Let ‘s’ be the SP of 1 article and ‘c’ be the CP of 1 article.
Given, 6c = 4s
Therefore s/c = 1.5
Gain % = 50
Q .2.
Let ‘s’ be the SP of 1 metre of cloth and ‘c’ be the CP of 1 metre of cloth
Total SP = 20s, Profit = 4s
CP = SP – Profit = 16s
The ratio s/c = 20s/16s = 1.25
Gain % = 25
Q . 3.
SP of 1 article(s) = Rs. 10/8 = 5/4
CP of 1 article(c) =Rs. 8/10 = 4/5
s/c = 25/16 = 1.5625
Gain % = 56.25
1. Q) 4. Kunal sold a shirt at a loss of 10%. Had he sold it for Rs 60 more, he would have gained 5% on it. Find the CP of the shirt.
In this question we have to find the CP of the article
c = s/f
From basic mathematics or elementary science we know that putting delta (∆) sign in numerator and denominator doesn’t change the equation. ∆ stands for ‘change’
c = ∆s/∆f
where ∆s = change in SP
∆f = change in factor
Therefore ∆s = New SP – Old SP = Rs. 60
∆f = New factor – Old factor
New factor is the factor when profit is 5%. Old factor is the one with loss = 10%
So ∆f = 1.05 – 0.9 = 0.15
c = ∆s/∆f = 60/0.15 = Rs. 400
Answer : Rs 400
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# NCERT Exemplar Solutions for Class 11 Maths Chapter 13 Limits and Derivatives
NCERT Exemplar Solutions for Class 11 Maths Chapter 13 Limits and Derivatives provides comprehensive solutions for all the questions in the NCERT Exemplar textbook. To excel in the board examinations, these solutions will increase the level of confidence among the students, as the concepts are clearly explained and structured. The solutions are prepared and reviewed by the subject-matter experts and it is revised according to the latest NCERT syllabus and guidelines of the CBSE board. These books are widely used by students who wish to excel in board exams as it provides a vast number of questions to solve. With the help of NCERT Exemplar Solutions, every student should be capable of solving the complex problem in each exercise.
The 13th Chapter Limits and Derivatives of NCERT Exemplar Solutions for Class 11 Maths is the best material. This chapter explains the limits of a function. We have provided answers to NCERT Exemplar Solutions in simple PDF format, which can be downloaded easily from the below-provided links. These solutions are helpful for the students to clarify their doubts and provide a strong foundation for every concept. Some of the essential topics of the chapter are listed below.
• Limits of a function
• Some properties of limits
• Limits of polynomials and rational functions
• Limits of trigonometric functions
• Derivatives
• Algebra of derivatives of functions
## Download the PDF of NCERT Exemplar Solutions For Class 11 Maths Chapter 13 Limits and Derivatives
### Access Answers to NCERT Exemplar Solutions For Class 11 Maths Chapter 13 – Limits and Derivatives
Exercise Page No: 239
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Differentiate each of the functions with respect to x in Exercises 29 to 42.
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31. (3x + 5) (1 + tan x)
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32. (sec x – 1) (sec x + 1)
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36. (ax2Â + cot x) (p + q cos x)
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39. (2x – 7)2 (3x + 5)3
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40. x2Â sin x + cos2x
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# How do you find the partial sum of Sigma (1000-5n) from n=0 to 50?
Feb 18, 2017
${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = 44625$
#### Explanation:
${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = {\sum}_{n = 0}^{50} \left(1000\right) - 5 {\sum}_{n = 0}^{50} \left(n\right) =$
We can use the standard result ${\sum}_{i = 1}^{n} i = \frac{1}{2} n \left(n + 1\right)$, and note that:
${\sum}_{n = 0}^{50} \left(1000\right) = 1000 + 1000 + \ldots + 1000 \setminus \setminus$ (51 terms)
${\sum}_{i = 0}^{n} i = 0 + {\sum}_{i = 1}^{n}$
So we get:
${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = \left(51\right) \left(1000\right) - 5 \cdot \frac{1}{2} \left(50\right) \left(50 + 1\right)$
$\text{ } = 51000 - \frac{5}{2} \left(50\right) \left(51\right)$
$\text{ } = 51000 - 6375$
$\text{ } = 44625$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Notation for Transformations
## Formatting of points or equations to show transformations.
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Practice Notation for Transformations
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Point Notation and Function Notation
When performing multiple transformations, it is very easy to make a small error. This is especially true when you try to do every step mentally. Point notation is a useful tool for concentrating your efforts on a single point and helps you to avoid making small mistakes.
What would f(3x)+7\begin{align*}f(3x) + 7\end{align*} look like in point notation and why is it useful?
### Using Function Notation and Point Notation
A transformation can be written in function notation and in point notation. Function notation is very common and practical because it allows you to graph any function using the same basic thought process it takes to graph a parabola in vertex form.
Another way to graph a function is to transform each point one at a time. This method works well when a table of x,y\begin{align*}x, y\end{align*} values is available or easily identified from the graph.
Essentially, it takes each coordinate (x,y)\begin{align*}(x, y)\end{align*} and assigns a new coordinate based on the transformation.
(x,y)(new x,new y)\begin{align*}(x, y) \rightarrow (new \ x, new \ y)\end{align*}
This notation is called point notation. The new y\begin{align*}y\end{align*} coordinate is straightforward and is directly from what takes place outside f(x)\begin{align*}f(x)\end{align*} because f(x)\begin{align*}f(x)\end{align*} is just another way to write y\begin{align*}y\end{align*}. For example, f(x)2f(x)1\begin{align*}f(x) \rightarrow 2f(x) - 1\end{align*} would have a new y\begin{align*}y\end{align*} coordinate of 2y1\begin{align*}2y-1\end{align*}
The new x\begin{align*}x\end{align*} coordinate is trickier. It comes from undoing the operations that affect x\begin{align*}x\end{align*}. For example, f(x)f(2x1)\begin{align*}f(x) \rightarrow f(2x -1)\end{align*} would have a new x\begin{align*}x\end{align*} coordinate of x+12\begin{align*}\frac{x +1}{2}\end{align*}
The function notation and point notation representations of the transformation "Horizontal shift right three units, vertical shift up 4 units" are
f(x)f(x3)+4\begin{align*}f(x) \rightarrow f(x-3)+4\end{align*}
(x,y)(x+3,y+4)\begin{align*}(x,y) \rightarrow (x+3,y+4)\end{align*}
Notice that the operations with the x\begin{align*}x\end{align*} are different.
Apply the transformation above to the following table of points.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 1 3 2 5 8 -11
Notice that point notation greatly reduces the mental visualization required to keep all the transformations straight at once.
### Examples
#### Example 1
Earlier, you were asked what the function f(3x)+7\begin{align*}f(3x) +7\end{align*} would be when written in point notation. When written in point notation, it would be written as (x,y)(x3,y+7)\begin{align*}(x, y) \rightarrow \left (\frac{x}{3}, y+7 \right )\end{align*}. This is useful because it becomes obvious that the x\begin{align*}x\end{align*} values are all divided by three and the y\begin{align*}y\end{align*} values all increase by 7.
#### Example 2
Convert the following function in point notation to words and then function notation.
(x,y)(3x+1,y+7)\begin{align*}(x,y)\rightarrow(3x+1,-y+7)\end{align*}
Horizontal stretch by a factor of 3 and then a horizontal shift right one unit. Vertical reflection over the x\begin{align*}x\end{align*} axis and then a vertical shift 7 units up.
f(x)f(13x13)+7\begin{align*}f(x)\rightarrow -f \left ( \frac{1}{3}x - \frac{1}{3} \right ) +7\end{align*}
#### Example 3
Convert the following function notation into words and then point notation. Finally, apply the transformation to three example points.
f(x)2f(x1)+4\begin{align*}f(x) \rightarrow - 2f(x-1) + 4\end{align*}
Vertical reflection across the x\begin{align*}x\end{align*} axis. Vertical stretch by a factor of 2. Vertical shift 4 units. Horizontal shift right one unit.
(x,y)(x+1,2y+4)\begin{align*}(x,y)\rightarrow(x+1,- 2y+4)\end{align*}
#### Example 4
Convert the following function notation into point notation and apply it to the included table of points
f(x)14f(x3)1\begin{align*}f(x) \rightarrow \frac{1}{4} f(-x-3)-1\end{align*}
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 0 0 1 4 2 8
The y\begin{align*}y\end{align*} component can be directly observed. For the x\begin{align*}x\end{align*} component you need to undo the argument. (x,y)(x3,14y1)\begin{align*}(x, y) \rightarrow \left ( - x-3,\frac{1}{4} y-1 \right )\end{align*}
#### Example 5
Convert the following point notation to words and to function notation and then apply the transformation to the included table of points.
(x+3,y1)(2x+6,y)\begin{align*}(x+3,y-1) \rightarrow (2x+6,-y)\end{align*}
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 10 8 12 7 14 6
This problem is different because it seems like there is a transformation happening to the original left point. This is an added layer of challenge because the transformation of interest is just the difference between the two points. Notice that the x\begin{align*}x\end{align*} coordinate has simply doubled and the y\begin{align*}y\end{align*} coordinate has gotten bigger by one and turned negative. This problem can be rewritten as:
(x,y)(2x,(y+1))=(2x,y1)\begin{align*}(x,y) \rightarrow (2x,-(y+1) )=(2x,-y-1)\end{align*}
f(x)f(x2)1\begin{align*}f(x) \rightarrow - f \left ( \frac{x}{2} \right )-1\end{align*}
### Review
Convert the following function notation into words and then point notation. Finally, apply the transformation to three example points.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 0 5 1 6 2 7
1. f(x)12f(x+1)\begin{align*}f(x) \rightarrow -\frac{1}{2} f(x+1)\end{align*}
2. g(x)2g(3x)+2\begin{align*}g(x)\rightarrow 2g(3x)+2\end{align*}
3. h(x)h(x4)3\begin{align*}h(x) \rightarrow - h(x-4)-3\end{align*}
4. j(x)3j(2x4)+1\begin{align*}j(x) \rightarrow 3j(2x-4)+1\end{align*}
5. \begin{align*}k(x)\rightarrow -k(x-3)\end{align*}
Convert the following functions in point notation to function notation.
6. \begin{align*}(x,y) \rightarrow \left ( \frac{1}{2} x+3,y-4 \right )\end{align*}
7. \begin{align*}(x,y) \rightarrow (2x+4,-y+1)\end{align*}
8. \begin{align*}(x,y) \rightarrow (4x,3y-5)\end{align*}
9. \begin{align*}(2x,y) \rightarrow (4x, - y+1)\end{align*}
10. \begin{align*}(x+1,y-2) \rightarrow (3x+3, -y+3)\end{align*}
Convert the following functions in function notation to point notation.
11. \begin{align*}f(x) \rightarrow 3f(x-2)+1\end{align*}
12. \begin{align*}g(x) \rightarrow -4g(x-1)+3\end{align*}
13. \begin{align*}h(x) \rightarrow \frac{1}{2} h(2x+2)-5\end{align*}
14. \begin{align*}j(x) \rightarrow 5j \left ( \frac{1}{2} x-2 \right )-1\end{align*}
15. \begin{align*}k(x) \rightarrow \frac{1}{4} k(2x-4)\end{align*}
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
function notation
In the context of transformations, function notation describes how a function $f(x)$ changes as a result of the transformation.
point notation
In the context of transformations, point notation describes how each coordinate $(x,y)$ changes as a result of the transformation.
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# Difference between revisions of "2006 AMC 10A Problems/Problem 16"
## Problem
A circle of radius $1$ is tangent to a circle of radius $2$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?
$[asy] size(200); pathpen = linewidth(0.7); pointpen = black; real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR((2*t,2),2)); D(CR((2*t,5),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6)); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);[/asy]$
$\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad$
## Solution 1
Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively. Let their tangent points to $\triangle ABC$ be $D$ and $E$, respectively. We can then draw the following diagram:
$[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]$
We see that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC$. Using the first pair of similar triangles, we write the proportion:
$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$
By the Pythagorean Theorem, we have $AD = \sqrt{3^2-1^2} = \sqrt{8}$.
Now using $\triangle ADO_1 \sim \triangle AFC$,
$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$
Hence, the area of the triangle is $$\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\textbf{(D) } 16\sqrt{2}}$$
## Solution 2
$[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]$
Since $\triangle{A O_1 D} \sim \triangle{A O_2 E},$ we have that $\frac{A O_1}{A O_2} = \frac{O_1 D}{O_2 E} = \frac{1}{2}$.
Since we know that $O_1 O_2 = 1 + 2 = 3,$ the total length of $A O_2 = 2 \cdot 3 = 6.$
We also know that $O_2 F = 2$, so $A F = A O_2 + O_2 F = 6 + 2 = 8.$
Also, since $\triangle{ABF} \sim \triangle{A E O_2},$ we have that $\frac{AC}{A O_2} = \frac{FC}{O_2 E}.$
Since we know that $A O_2 = 6$ and $O_2 E = 2,$ we have that $\frac{AC}{6} = \frac{FC}{2}.$
This equation simplified gets us $AC = 3 \cdot FC.$
Let $FC = a$
By the Pythagorean Theorem on $\triangle{AFC},$ we have that $AF^2 + FC^2 = AC^2.$
We know that $AF = 8$, $FC = a$ and $AC = 3a$ so we have $8^2 + a^2 = (3a)^2.$
Simplifying, we have that $64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.$
Recall that $FC=a$.
Therefore, $BC = 2 \cdot FC = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}.$
Since the height is $AF = 8,$ we have the area equal to $\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.$
Thus our answer is $\boxed{\textbf{(D) }16 \sqrt{2}}$.
~mathboy282
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# Definite Integration and Areas 01 It can be used to find an area bounded, in part, by a curve e.g. gives the area shaded on the graph The limits of integration...
## Presentation on theme: "Definite Integration and Areas 01 It can be used to find an area bounded, in part, by a curve e.g. gives the area shaded on the graph The limits of integration..."— Presentation transcript:
Definite Integration and Areas 01 It can be used to find an area bounded, in part, by a curve e.g. gives the area shaded on the graph The limits of integration... Definite integration results in a value. Areas
Definite Integration and Areas... give the boundaries of the area. The limits of integration... 01 It can be used to find an area bounded, in part, by a curve Definite integration results in a value. Areas x = 0 is the lower limit ( the left hand boundary ) x = 1 is the upper limit (the right hand boundary ) 0 1 e.g. gives the area shaded on the graph
Definite Integration and Areas 01 23 2 xy Finding an area the shaded area equals 3 The units are usually unknown in this type of question Since
Definite Integration and Areas SUMMARY the curve the lines x = a and x = b the x -axis and PROVIDED that the curve lies on, or above, the x -axis between the values x = a and x = b The definite integral or gives the area between
Definite Integration and Areas Finding an area A B For parts of the curve below the x -axis, the definite integral is negative, so
Definite Integration and Areas A Finding an area Area A
Definite Integration and Areas B Finding an area Area B
Definite Integration and Areas SUMMARY An area is always positive. The definite integral is positive for areas above the x -axis but negative for areas below the axis. To find an area, we need to know whether the curve crosses the x -axis between the boundaries. For areas above the axis, the definite integral gives the area. For areas below the axis, we need to change the sign of the definite integral to find the area.
Definite Integration and Areas Exercise Find the areas described in each question. 1. The area between the curve the x -axis and the lines x = 1 and x = 3. 2. The area between the curve, the x -axis and the x = 2 and x = 3.
Definite Integration and Areas B A 1. 2. Solutions:
Definite Integration and Areas Harder Areas e.g.1 Find the coordinates of the points of intersection of the curve and line shown. Find the area enclosed by the curve and line. Solution: The points of intersection are given by
Definite Integration and Areas Substitute in The area required is the area under the curve between 0 and 1...... minus the area under the line (a triangle ) Area of the triangle Area under the curve Required area Method 1
Definite Integration and Areas Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration. Area We get Method 2
Definite Integration and Areas Exercise Find the points of intersection of the following curves and lines. Show the graphs in a sketch, shade the region bounded by the graphs and find its area. (a) ; (b) ; Solution: (a) ( y = 6 for both points )
Definite Integration and Areas Shaded area = area of rectangle – area under curve Area under curve Shaded area
Definite Integration and Areas Area of the triangle or Substitute in : Area under the curve (b) ; Shaded area = area under curve – area of triangle
Definite Integration and Areas The symmetry of the curve means that the integral from 1 to +1 is 0. If a curve crosses the x -axis between the limits of integration, part of the area will be above the axis and part below. e.g. between 1 and +1 To find the area, we could integrate from 0 to 1 and, because of the symmetry, double the answer. For a curve which wasn’t symmetrical, we could find the 2 areas separately and then add.
Definite Integration and Areas You don’t need to know how the formula for area using integration was arrived at, but you do need to know the general ideas. The area under the curve is split into strips. The area of each strip is then approximated by 2 rectangles, one above and one below the curve as shown. The exact area of the strip under the curve lies between the area of the 2 rectangles.
Definite Integration and Areas Using 10 rectangles below and 10 above to estimate an area below a curve, we have... Greater accuracy would be given with 20 rectangles below and above... For an exact answer we let the number of rectangles approach infinity. The exact area is “squashed” between 2 values which approach each other. These values become the definite integral.
Definite Integration and Areas The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
Definite Integration and Areas... give the boundaries of the area. It can be used to find an area bounded, in part, by a curve Definite integration results in a value. Areas The limits of integration... x = 0 is the lower limit ( the left hand boundary ) x = 1 is the upper limit (the right hand boundary ) 0 1 e.g. gives the area shaded on the graph
Definite Integration and Areas SUMMARY the curve the lines x = a and x = b the x -axis and PROVIDED that the curve lies on, or above, the x -axis between the values x = a and x = b The definite integral or gives the area between
Definite Integration and Areas Finding an area AB For parts of the curve below the x -axis, the definite integral is negative, so
Definite Integration and Areas SUMMARY An area is always positive. The definite integral is positive for areas above the x -axis but negative for areas below the axis. To find an area, we need to know whether the curve crosses the x -axis between the boundaries. For areas above the axis, the definite integral gives the area. For areas below the axis, we need to change the sign of the definite integral to find the area.
Definite Integration and Areas Harder Areas e.g.1 Find the coordinates of the points of intersection of the curve and line shown. Find the area enclosed by the curve and line. Solution: The points of intersection are given by
Definite Integration and Areas Substitute in The area required is the area under the curve between 0 and 1...... minus the area under the line (a triangle ) Area of the triangle Area under the curve Required area
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# SOLUTION: 3y-x=2 -2y-x=-18
Algebra -> Algebra -> Coordinate Systems and Linear Equations -> SOLUTION: 3y-x=2 -2y-x=-18 Log On
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Question 70664This question is from textbook glencoe : 3y-x=2 -2y-x=-18This question is from textbook glencoe Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!+3y - x = 2 -2y - x = -18 . One of the ways to solve this system of equations is to eliminate one of the variables and solve the resulting equation for the other variable. . In the above problem you can notice that there is a -x in the top equation and a -x in the bottom equation. If you subtract the bottom equation from the top equation, the equation that results does not contain an x. Remember the algebraic rule for subtracting ... change the signs of the subtrahend (number being subtracted) and add the result to the number that the subtrahend is being subtracted from. . To follow this rule we change the sign of -2y so that it becomes +2y. We also change the sign of the -x so that it becomes +x. And finally, we change the sign of the -18 to +18. Then we add these terms to the terms directly above them. In algebrais form this becomes: . +3y - x = +2 +2y + x = +18 ------------- +5y = +20 . The expression below the dashed line is a new equation resulting from the addition. You can solve this new equation for y by dividing both sides by 5 to get y = 4. . Next take that value for y and plug it into either of the original two equations and solve for x. For example, if we substitute 4 for y in the equation 3y - x = 2, we get: . 3*4 - x = 2 . This becomes: . 12 - x = 2 . Subtracting 12 from both sides results in: . -x = -10 . And finally, we need to solve for +x, so multiply both sides of the equation by -1 to get: . x = 10 . The answers you are looking for are x = 10 and y = 4. . Hope this helps you understand one method of solving a system of two linear equations.
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# distance for 54 degree wedge
The 54 degree wedge is an essential club in a golfer’s bag and is used for a variety of shots on the course. It is most frequently used to hit shots from tight lies or long distances. This wedge offers great control, spin and distance, making it ideal for shots that require precision and accuracy. With the right technique, this club can help you hit shots that cover the entire length of the fairway.The distance of a 54 degree wedge is equal to the length of the arc created by the angle of 54 degrees. The arc’s length can be calculated using the formula s=rθ, where s is the arc’s length, r is the radius of the circle and θ is the angle in radians.
### Calculating the Distance of a 54 Degree Wedge
Calculating the distance of a 54 degree wedge requires some basic math and geometry skills. The first step is to determine the size of the wedge. A 54 degree wedge is usually a right triangle, so we can use the Pythagorean theorem to calculate the size of the wedge. The Pythagorean theorem states that for any right triangle, the square of the hypotenuse is equal to the sum of the squares of its two sides.
To calculate the size of a 54-degree wedge, you will need to know two sides of the triangle. The hypotenuse is always opposite to the right angle (90 degrees), so you will need to know either two sides or one side and an angle other than 90 degrees. Once you have these two pieces of information, you can calculate the length of each side using basic trigonometry functions such as sine and cosine.
Once you have calculated all three lengths, you can then use those values to calculate the distance traveled by a projectile launched from that point. To do this, you will need to use vector addition and subtraction. Vector addition involves adding together all three vectors (the length of each side) and subtracting any vectors pointing in opposite directions from one another. Once these calculations are complete, you will have your final answer –the total distance traveled by your projectile from that point!
### Calculating the Distance for a 54 Degree Wedge
Calculating the distance that a golf ball will travel when hit with a 54 degree wedge requires knowledge of a few simple equations. The distance of the ball is determined by factors such as ball speed, launch angle, and spin rate. With these factors in mind, it is possible to calculate the total distance of a shot taken with a 54 degree wedge using an equation based on the physics of golf.
The first step to calculating the distance is to find the ball’s launch angle and speed. The launch angle is determined by measuring the height of the golf ball at impact and subtracting it from its initial height. To find the speed, a golfer can measure how long it takes for the ball to reach its peak height after impact. To obtain an accurate reading, it is best to use an advanced launch monitor such as Trackman or Flightscope.
Once the golfer has obtained an accurate reading for launch angle and speed, they can then use this data to calculate the golf ball’s total distance traveled. The equation used for this calculation is as follows: Distance = (launch angle x speed) * (1 + spin rate). This equation takes into account all three variables that determine how far a golf ball will travel when hit with a 54 degree wedge – launch angle, speed, and spin rate.
Using this equation, it is easy to calculate how far any given shot with a 54 degree wedge will travel. A golfer who knows their clubhead speed and spin rate can then use this information to accurately predict their shots’ distances when they are using their 54 degree wedge on any given hole or course. Knowing how far each shot will go before it lands can give any golfer an edge on their competition.
## How to Determine the Distance of a 54 Degree Wedge
Determining the distance of a 54 degree wedge shot can be tricky if you are not familiar with the physics of a golf swing. The angle of attack, which is the angle that the club face makes with the ground when it strikes the ball, will determine how far your ball will travel. To calculate the distance of a 54 degree wedge shot, you must first know what angle you are using and then use some basic geometry to calculate your distance.
To start, you must measure the angle of attack. This is done by taking a measurement from directly behind the ball and measuring how much your club head is inclined from the ground. This should be done at impact and should be approximately 54 degrees for this example. Once you have this measurement, you can use basic geometry to calculate the distance your ball will travel.
To calculate the distance, you will need to use some trigonometry formulas and apply them to basic geometry principles. First, take your measured angle (54 degrees) and apply it to a triangle where one side of that triangle is equal to your club head’s length (the shaft length). The other two sides of that triangle are equal to half of your club head’s length (the shaft length divided by 2). Now, using basic trigonometry formulas such as SOH CAH TOA or Sin Cos Tan, you can calculate how far away from your starting point (the ball) these two sides are in relation to each other and thus how far away from your starting point (the ball) your shot will go.
Once you have calculated this distance, you can use it in combination with other factors such as wind speed/direction or terrain conditions to help give yourself an edge in predicting where your golf shots will end up going. With practice and experience, you should be able to accurately judge distances using this method so that you can consistently hit accurate shots with a 54 degree wedge.
### How Far Will a 54 Degree Wedge Travel?
The distance that a 54 degree wedge will travel depends on several factors such as the player’s strength, the type of terrain, and the angle of attack. A player with more strength can hit the ball farther than a weaker player, and playing on softer terrain will result in less roll. The angle of attack is also important since hitting too steeply or too shallowly can cause the ball to rise or dive, respectively. Generally speaking, however, a 54 degree wedge should travel around 70-80 yards for an average male golfer and 50-60 yards for a female golfer.
It is also important to note that the distance a 54 degree wedge travels can vary greatly depending on the conditions of the course and other environmental factors such as wind speed and direction. If the wind is blowing directly into the face of the golfer, then it will reduce how far he or she can hit it. Conversely, if there is tailwind then it can increase how far they can hit it.
Finally, different types of wedges have different amounts of bounce which affects how far they travel. For example, a lob wedge typically has more bounce than a gap wedge which allows players to hit higher shots with more spin that travel farther. By understanding all these factors players can better determine how far their 54 degree wedge will travel in any given situation.
### Understanding the Physics Behind a 54 Degree Wedge
A wedge is a tool with two sloping sides and a sharp edge that is used to cut or separate objects. The angle of the wedge affects how it performs. A 54 degree wedge is commonly used in golf and other sports, but what are the physics behind it?
The angle of the wedge affects how much force is needed to push it into an object. A steeper angle requires more force, while a shallower angle requires less force. A 54 degree wedge has been found to be an optimal angle for cutting into most materials because it provides enough force to make a clean cut, while not requiring too much effort from the user.
The design of the wedge also plays an important role in its performance. Its sharp edges help it penetrate objects more easily, while its two sloping sides create a vacuum effect which helps draw materials away from the cutting surface. This makes cutting easier and more efficient.
The shape of the wedge also affects its bounce and spin when used in sports such as golf. The sharper edges create more backspin on the ball which helps keep it airborne longer and gives it more control when landing on the green. The shallower slope also helps reduce bounce when making contact with the ground which can help players achieve better accuracy in their shots.
Understanding the physics behind a 54 degree wedge can help you get better results when using one for cutting or playing sports such as golf. By knowing how different angles affect its performance, you can choose one that is best suited for your needs and make sure you get optimal results every time you use it.
## Estimating the Trajectory of a 54 Degree Wedge
Estimating the trajectory of a 54 degree wedge is an important skill for golfers. It can help you to hit a shot that goes exactly where you want it to go, and can help you to get out of difficult situations. To estimate the trajectory of a 54 degree wedge, you need to understand the basics of ball flight, including lift, drag, and spin. You also need to consider factors such as wind direction and strength, terrain type, and ball type.
To estimate the trajectory of a 54 degree wedge, start by selecting the club based on your target distance. A 54 degree wedge is typically used for shots from 100-150 yards away. Once you have selected your club, take some practice swings and get used to its feel. Next, set up your stance with the ball slightly forward in your stance and make sure that your weight is evenly distributed between both feet.
Once you are set up correctly, take aim at your target and adjust your aim based on any wind conditions or terrain features that could affect the ball’s flight path. When ready to swing, focus on making a smooth tempo swing with an acceleration through impact while keeping your head down until after impact. This will help ensure that you make contact with the ball in its sweet spot for maximum distance and accuracy.
Finally, pay attention to where the ball lands after each shot and adjust accordingly if necessary. If it lands too far right or left of your target line then either adjust your aim slightly or select a different club based on what happened during that particular shot. With practice estimating the trajectory of a 54 degree wedge can become second nature allowing you to hit shots more consistently and accurately from any distance.
### Approximating Distance for a 54 Degree Wedge
Golfers commonly use wedges to hit short-distance shots on the course. A 54 degree wedge is often used to hit shots from close range, such as when the ball is within 30 yards of the green. When hitting a shot with a 54 degree wedge, it’s important to know how far the ball will travel.
Estimating distance for a 54 degree wedge is not an exact science and can vary based on many factors including the golfer’s swing speed, angle of attack, clubhead speed, and more. However, there are some general guidelines that can be used to approximate how far the ball will go with a 54 degree wedge.
For starters, it’s important to consider the type of surface you’re playing on. If you’re playing on a hard surface like concrete or asphalt, you can expect to get around 30 yards of distance with each swing. On softer surfaces like grass or sand, you can expect to get about 25 yards for each swing.
It’s also important to factor in your own swing speed and technique when estimating how far your ball will go. If you have a slower swing speed or an incorrect angle of attack, your ball may not travel as far as it would if you had a faster swing speed and proper technique.
Finally, remember that different types of wedges have different loft angles which can affect how far the ball goes when hit with them. A 54 degree wedge is typically going to provide shorter distance than higher lofted wedges such as 56 or 58 degrees wedges.
In conclusion, approximating distance for a 54 degree wedge isn’t an exact science but there are some general guidelines that can be used to give you an idea of how far your ball will travel when hitting a shot with this club. Taking into consideration factors such as the type of surface you’re playing on and your own swing speed and technique will help you better estimate how far your ball will travel with each shot!
## Conclusion
The 54 degree wedge is a versatile club that can be used for many different types of shots. It is an excellent option for players who need to hit shots from various distances. The loft and bounce of the club can be adjusted to suit different playing conditions, allowing golfers to hit accurate shots from any range. Its ability to control spin and trajectory makes it an ideal choice for players looking to make the most of their game. With proper practice and technique, the 54 degree wedge can be a great asset for any golfer.
Overall, the 54 degree wedge is a great tool that can help golfers master their game from long distances. It offers a variety of features that make it easy to use and customize, allowing players to get the most out of their shots no matter the distance or terrain. With its high-performance design, this wedge is sure to improve any golfer’s game on the course.
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Question
# Find all points of discontinuity of $$f$$, where $$f$$ is defined by f(x) = $$\begin{cases} x +1, \quad \text {if} \quad x \ge 1 \\ x^2 + 1, \quad \text{if} \quad x <1 \end{cases}$$
Solution
## The given function is $$f(x) = \begin{cases} x +1, \text{if} \, x \ge 1 \\ x^2 + 1, \text{if} \, x <1 \end{cases}$$The given function is defined at all the points of the real line.Let $$c$$ be a point on the real line. Case I: $$c < 1$$, then $$f(c) = c^2 + 1$$ and $$\underset{x \rightarrow c}{\lim} f(x) = \underset{x \rightarrow c}{\lim} (x^2 + 1) = c^2 + 1$$$$\therefore$$ $$\underset{x \rightarrow c}{\lim} f(x) = f(c)$$Therefore, $$f$$ is continuous at all points $$x$$, such that $$x < 1$$Case II : $$c = 1$$, then $$f(c) = f(1) = 1 + 1 = 2$$The left hand limit of $$f$$ at $$x = 1$$is, $$\underset{x \rightarrow 1}{lim}$$ $$f(x) =$$ $$\underset{x \rightarrow 1}{lim}$$ ($$x^2+ 1$$ ) = $$1^2+1 = 2$$The right hand limit of $$f$$ at $$x = 1$$ is, $$\underset{x \rightarrow 1}{\lim} f(x) = \underset{x \rightarrow 1}{\lim} (x + 1) = 1 +1 = 2$$$$\therefore$$ $$\underset{x \rightarrow 1}{\lim} f(x) = f(1)$$Therefore, $$f$$ is continuous at $$x = 1$$Case III : $$c > 1$$, then $$f(c) = c + 1$$$$\underset{x \rightarrow c}{\lim} f(x) = \underset{x \rightarrow c}{\lim} (x + 1) = c + 1$$$$\therefore$$ $$\underset{x \rightarrow c}{\lim} f(x) = f(c)$$Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 1$$Hence, the given function $$f$$ has no point of discontinuity. MathematicsRS AgarwalStandard XII
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# Chapter 2 – Polynomials
## Polynomials
Exercise– 2.1
Q1. The graphs of y = p(x) are given in the figure given below. for some polynomials p (x). Find the number of zeroes of p (x), in each case
Sol. (i) The given graph is parallel to x-axis. It does not intersect the x-axis.
• It has no zeroes.
(ii) The given graph intersects the x-axis at one point only.
• It has one zero.
(iii) The given graph intersects the x-axis at three points.
• It has three zeroes.
(iv) The given graph intersects the x-axis at two points.
• It has two zeroes.
(v) The given graph intersects the x-axis at four points.
• It has four zeroes.
(vi) The given graph meets the x-axis at three points.
• It has three zeroes.
NCERT TEXTBOOK QUESTIONS SOLVED
Exercise– 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Sol. (i) x2 – 2x – 8
We have p(x) = x2 – 2x – 8
= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)
= (x – 4) (x + 2)
For p(x) = 0, we have
(x – 4) (x + 2) = 0
Either x – 4 = 0 ? x = 4
or x + 2 = 0 ? x = – 2
? The zeroes of x2 2x – 8 are 4 and –2.
Now, sum of the zeroes
Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.
(ii) 4s2 – 4s + 1
We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)
= (2s – 1) (2s – 1)
For p(s) = 0, we have,
Now,
Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 – 4s + 1 is verified.
(iii) 6x2 – 3 – 7x
We have
p (x) = 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
For p (x) = 0, we have,
Thus, the relationship between the zeroes and the coefficients in the polynomial 6x2 – 3 – 7x is verified.
(iv) 4u2 + 8u
We have, f(u) = 4u2 + 8u = 4u (u + 2)
For f(u) = 0,
Either 4u = 0 ? u = 0
or u + 2 = 0 ? u = –2
? The zeroes of 4u2 + 8u and 0 and –2
Now, 4u2 + 8u can be written as 4u2 + 8u + 0.
Thus, the relationship between the zeroes and the coefficients in the polynomial 4u2 + 8u is verified.
(v) t2 – 15
We have,
For f(t) = 0, we have
Now, we can write t2 – 15 as t2 + 0t – 15.
Thus, the relationship between the zeroes and the coefficients in the polynomial t2 – 15 is verified.
(vi) 3x2 – x – 4
We have,
f(x) = 3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x (x + 1) –4 (x + 1)
= (x + 1) (3x – 4)
For f(x) = 0 ? (x + 1) (3x – 4) = 0
Either (x + 1) = 0 ? x = – 1
Thus, the relationship between the zeroes and the coefficients in 3x2 – x – 4 is verified.
Q2. Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively:
Sol.
Note:
A quadratic polynomial whose zeroes are a and ß is given by
p(x) = {x2 – (a + ß) x + aß}
p(x) = {x2 – (sum of the zeroes) x – (product of the zeroes)}
(i) Since, sum of the zeroes,
Product of the zeroes, a ß = –1
? The required quadratic polynomial is
x2 – (a + ß) x + aß
Since, have same zeroes, is the required quadratic polynomial.
(ii) Since, sum of the zeroes,
Product of zeroes,
Since, have same zeroes, is required quadratic polynomial.
(iii) Since, sum of zeroes, (a + ß) 0
Product of zeroes, aß = 5
? The required quadratic polynomial is
x2 – (a + ß) x + aß
= x2 – (0) x + 5
= x2 + 5
(iv) Since, sum of the zeroes, (a + ß) = 1
Product of the zeroes = 1
? The required quadratic polynomial is
x2 – (a + ß) x + aß
= x2 – (1) x + 1
= x2 – x + 1
(v) Since, sum of the zeroes,
Product of the zeroes
? The required quadratic polynomial is
x2 – (a + ß) x + aß
Since, and (4x2 + x + 1) have same zeroes, the required quadratic polynomial is (4x2 + x + 1).
(vi) Since, sum of the zeroes, (a + ß) = 4
Product of the zeroes, aß = 1
? The required quadratic polynomial is
x2 – (a + ß) x + aß
= x2 – (4)x + 1
= x2 – 4x + 1
Exercise– 2.3
Q1. Q1. Divide the polynomial p (x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p (x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2
(ii) p (x) = x4 – 3x2 + 4x + 5, g (x) = x2 + 1 – x
(iii) p (x) = x3 – 5x + 6, g (x) = 2 – x2
Sol. Here, dividend p (x) = x3 – 3x2 + 5x – 3
divisor g(x) = x2 – 2
? We have
Thus, the quotient = (x – 3) and remainder = (7x – 9)
(iii) Here, dividend p(x) = x4 – 3x2 + 4x + 5
and divisor g(x) = x2 + 1 – x
= x2 – x + 1
? We have
Thus, the quotient is (x2 + x – 3) and remainder = 8
(iii) Here, divided, p(x) = x4 – 5x + 6
and divisor, g(x) = 2 – x2
? We have
Thus, the quotient = –x2 – 2 and remainder = –5x + 10.
Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2 – 3; 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1; 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 + 3x + 1; x5 + 4x3 + x + 3x + 1
Sol. (i) Dividing 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3,
We have:
? Remainder = 0
? (t2 – 3) is a factor of 2t4 + 3t3 – 2t2 – 9t – 12.
(ii) Dividing 3x4 + 5x3 – 7x2 + 2x + 2 by
x2 + 3x + 1, we have:
? Remainder = 0.
? x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2.
(iii) Dividiing x5 + 4x3 + x + 3x + 1, we get:
? The remainder = 2, i.e., remainder ? 0
? x3 – 3x + 1 is not a factor of
x5 – 4x3 + x2 + 3x + 1.
Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
Sol. We have p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Now, let us divide 3x4 + 6x3 – 2x2 – 10x – 5 by
? 3x4 + 6x3 – 2x2 – 10x – 5
For p(x) = 0, we have
or 3x + 3 = 0 ? x = –1
or x + 1 = 0 ? x = –1
Thus, all the other zeroes of the given polynomial are –1 and –1.
Q4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4 respectively. Find g(x).
Sol. Here,
Dividened p(x) = x3 – 3x2 + x + 2
Divisor = g(x)
Quotient = (x – 2)
Remainder = (–2x + 4)
Since,
(Quotient – Divisor) + Remainder = Dividend
? [(x – 2)] – g(x)] + [(–2x + 4)] = x3 – 3x2 + x + 2
? (x – 2) – g(x) = x3 – 3x2 + x + 2 – (–2x + 4)
= x3 – 3x2 + x + 2 + 2x – 4
= x3 – 3x2 + 3x – 2
Now, dividing x3 – 3x2 + 3x – 2 by x – 2,
We have
Thus, the required divisor g(x) = x2 – x + 1.
Q5. Give example of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q (x)
(ii) deg q(x) = deg r (x)
(iii) deg r (x) = 0
Sol. We can have
(i) p(x) = 3x2 – 6x + 27
g(x) = 3
q(x) = x2 – 2x + 9
r(x) = 0
? p(x) = q(x) × g(x) + r(x).
(ii) p(x) = 2x3 – 2x2 + 2x + 3
g(x) = 2x2 – 1
q(x) = x – 1
r(x) = 3x + 2
? p(x) = q(x) × g(x) + r(x)
(iii) p(x) = 2x3 – 4x2 + x + 4
g(x) = 2x2 + 1
q(x) = x – 2
r(x) = 6
? p(x) = q(x) × g(x) + r(x)
Exercise– 2.4
Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i)
(i) x3 – 4x2 + 5x – 2; 2, 1, 1
Sol. (i) ? p(x) = 2x3 + x2 – 5x + 2
Again,
p(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2
= (2 + 2 + 1) – 5
= 5 – 5 = 0
? 1 is a zero of p(x).
Also,
p(–2) = 2(–2)3 + (–2)2 –5 (–2) + 2
= 2(–8) + (4) + 10 + 2
= –16 + 4 + 10 + 2
= –16 + 16 = 0
? –2 is a zero of p(x).
Relationship
?p(x) = 2x3 + x2 – 5x + 2
?Comparing it with ax3 + bx2 + cx + d, we have:
a = 2, b = 1, c = – 5 and d = 2
Also 1 and – 2 are the zeroes of p(x)
Sum of product of zeroes taken in pair:
Thus, the relationship between the co-efficients and the zeroes of p(x) is verified.
(ii) Here, p(x) = x3 – 4x2 + 5x – 2
p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 18 – 18 = 0
? 2 is a zero of p(x).
Again p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 6 – 6 = 0
? 1 is a zero of p(x).
?2, 1, 1 are zeroes of p(x).
Now, comparing
p(x) = x3 – 4x2 + 5x – 2 with ax3 + bx2 +cx + d = 0, we have
a = 1, b = –4, c = 5 and d = –2
?2, 1, 1 are the zeroes of p(x)
Let
a = 2
ß = 1
? = 1
Relationship, a + ß + ? = 2 + 1+ 1 = 4
Sum of product of zeroes taken in pair:
aß + ß? + ?a = 2(1) + 1(1) + 1(2)
= 2 + 1 + 2 = 5
? aß + ß? + ?a =
Product of zeroes = aß? = (2) (1) (1) = 2
Thus, the relationship between the zeroes and the co-efficients of p(x) is verified.
Q2. Find the cubic polynomial with the sum, of the products of its zeroes taken two at a time and the product of its zeroes as 2, –7, –14 respectively.
Sol. Let the required cubic polynomial be ax3 + bx2 + cx + d and its zeroes be a, ß and ?
Since,
?The required cubic polynomial
= 1x3 + (–2)x2 + (–7)x + 14
= x3 – 2x2 – 7x + 14
Q3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b then find ‘a’ and ‘b’.
Sol. We have
p(x) = x3 – 3x2 + x + 1
comparing it with Ax3 + Bx2 + Cx + D.
We have
A = 1, B = –3, C = 1 and D = 1
?It is given that (a – b), a and (a + b) are the zeroes of the polynomial.
?Let,
a = (a – b)
ß = a
and ? = (a + b)
? (a – b) + a + (a + b) = 3
? 3a = 3
? (a – b) × a × (a + b) = –1
? (1– b) ×1× (1 + b) = –1
? 1– b2 = –1
Q4. If two zeroes of the polynomial x4 6x3 – 26x2 + 138x – 35 are find other zeroes.
Sol. Here, p(x) = x4 – 6x3 – 26x2 + 138x – 35.
?Two of the zeroes of p (x) are:
Now, dividing p (x) by x2 – 4x + 1, we have:
?(x2 – 4x + 1)(x2 – 2x – 35) = p(x)
? (x2 – 4x + 1) (x – 7) (x + 5) = p(x)
i.e., (x – 7) and (x + 5) are other factors of p(x).
?7 and –5 are other zeroes of the given polynomial.
Q5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be (x + a ), find k and a.
Sol. Applying the division algorithm to the polynomials x4 – 6x3 + 16x2 – 25x + 10 and x2 – 2x + k, we have:
?Remainder = (2k – 9) x – k(8 – k) + 10
But the remainder = x + a
Therefore, coparing them, we have:
and
a = –k(8 – k) + 10
= –5(8 – 5) + 10
= –5(3) + 10
= –15 + 10
= –5
Thus, k = 5 and a = –5
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### Introduction
To successfully analyse and solve the equations of Leaving Cert Applied Maths projectiles, one must be very comfortable with trigonometry.
Projectile trigonometry all takes place in $[0^\circ,90^\circ]$ so we should be able to work exclusively in right-angled-triangles (RATs), however I might revert to the unit circle for proofs (without using the unit circle, the definitions for zero and $90^\circ$ are found by using continuity).
Recalling that two triangles are similar if they have the same angles, the fundamental principle governing trigonometry might be put something like this:
Similar triangles differ only by a scale factor.
We show this below, but what this means is that the ratio of corresponding sides of similar triangles are the same, and if one of the angles is a right-angle, it means that if you have an angle, say $40^\circ$, and calculate the ratio of, say, the length of the opposite to the length of the hypotenuse, that your answer doesn’t depend on how large your triangle is and so it makes sense to talk about this ratio for $40^\circ$ rather than just a specific triangle:
These are two similar triangles. The opposite/hypotenuse ratio is the same in both cases.
Suppose the dashed triangle is a $k$-scaled version of the smaller triangle. Then $|A'B'|=k|AB|$ and $|A'C'|=k|AC|$. Thus the opposite to hypotenuse ratio for the larger triangle is
$\displaystyle \frac{|A'B'|}{|A'C'|}=\frac{k|AB|}{k|AC|}=\frac{|AB|}{|AC|}$,
which is the same as the corresponding ratio for the smaller triangle.
This allows us to define some special ratios, the so-called trigonometric ratios. If you are studying Leaving Cert Applied Maths you know what these are. You should also be aware of the inverse trigonometric functions. Also you should be able to, given the hypotenuse and angle, find comfortably the other two sides. We should also know that sine is maximised at $90^\circ$, where it is equal to one.
In projectiles we use another trigonometric ratio:
$\displaystyle \sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos\theta}$.
Note $\cos90^\circ=0$, so that $\sec90^\circ$ is not defined. Why? Answer here.
### The Pythagoras Identity
For any angle $\theta$,
$\sin^2\theta+\cos^2\theta=1$.
Proof: Consider a RAT with an angle of $\theta$ and hypotenuse of one:
We know that $\cos \theta=\frac{|BC|}{1}\Rightarrow |BC|=\cos \theta$. Similarly $|AB|=\sin\theta$. The triangle is a RAT so satisfies Pythagoras Theorem:
$|BC|^2+|AB|^2=|AC|^2$
$\Rightarrow (\cos\theta)^2+(\sin\theta)^2=1^2$.
The result follows as soon as one realises that $\cos^2\theta$ is a notation for $(\cos\theta)^2$ $\bullet$
As a corollary we have the following.
### The Sec-Tan Identity
For $\theta\neq 90^\circ$,
$\sec^2\theta=1+\tan^2\theta$.
Proof: Divide the Pythagoras Identity by $\cos^2\theta$ (this rules out $\theta=90^\circ$):
$\displaystyle \frac{1}{\cos^2\theta}=1+\frac{\sin^2\theta}{\cos^\theta}$.
Note that by the definition of $\sec\theta$:
$\displaystyle \frac{1}{\cos^2\theta}=\frac{1}{\cos\theta\cdot \cos\theta}=\frac{1}{\cos\theta}\cdot\frac{1}{\cos\theta}=\sec^2\theta$,
and
$\displaystyle\frac{\sin\theta}{\cos\theta}=\frac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{opp}}{\text{hyp}}}=\frac{\text{opp}}{\text{adj}}=\tan\theta$
$\displaystyle \Rightarrow \frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$ $\bullet$
The other identities that we need are a bit more difficult to derive.
### The Fundamental Identity
For $A>B$,
$\cos(A-B)=\cos A\cos B+\sin A\sin B$.
The identity is true more generally but this suffices for Projectiles.
Proof: Consider the following.
The basic objects here are the RATs $\Delta ODP$ and $\Delta OCQ$. These triangles both have hypotenuse one, $\angle DOP=A$ and $\angle COQ=B$. Dropping a perpendicular from $P$ to the line segment $[OQ]$, at $E$ gives a third RAT $\Delta OEP$ with a hypotenuse of one with $\angle EOP=A-B$.
The fact that the hypotenuses are one gives the following:
$|DP|=\sin A$, $|OD|=\cos A$, $|CQ|=\sin B$, $|OC|=\cos B$, $|EP|=\sin(A-B)$, and $|OE|=\cos(A-B)$.
Now we calculate the length $|PQ|^2$ in two different ways. First consider the RAT $\Delta FPQ$. By subtraction
$|FQ|=|DP|-|CQ|=\sin A-\sin B$ and $|FP|=|OC|-|OD|=\cos B-\cos A$.
Now using the Pythagoras Theorem:
$|PQ|^2=(\sin A-\sin B)^2+(\cos B-\cos A)^2$
$=\sin^2A-2\sin A\sin B+\sin^2B+\cos^2B-2\cos A\cos B+\cos^2B$
$= \sin^2A+\cos^2A+\sin^2B+\cos^2B-2\sin A\sin B-2\cos A\cos B$
$=2-2\sin A\sin B-2\cos A\cos B$
by the Pythagoras Identity.
Now looking at the RAT $\Delta EPQ$. From above we know that $|EP|=\sin(A-B)$ while subtraction shows $|EQ|=|OQ|-|OE|=1-\cos(A-B)$. Using Pythagoras
$|PQ|^2=(\sin(A-B))^2+(1-\cos(A-B))^2$
$=\sin^2(A-B)+1-2\cos(A-B)+\cos^2(A-B)$
$=\cos^2(A-B)+\sin^2(A-B)+1-2\cos(A-B)$
$=2-2\cos(A-B)$,
by the Pythagoras Identity. Now equating the two calculations:
$2-2\cos A\cos B-2\sin A\sin B=2-2\cos(A-B)$,
we quickly see the result $\bullet$
There is a very obvious relationship between sine and cosine:
A very quick inspection shows that:
$\sin\theta=\cos(90^\circ-\theta)$ and $\cos\theta=\sin(90^\circ-\theta)$.
We can use this to generate an expression for $\sin(A+B)$.
$\sin(A+B)=\sin A\cos B+\cos A\sin B$
Proof: Using the above, and the fundamental identity:
$\sin(A+B)=\cos(90^\circ-(A+B))$
$=\cos((90^\circ-A)-B)$
$=\cos(90^\circ-A)\cos B+\sin(90^\circ-A)\sin B$
$=\sin A\cos B+\cos A\sin B$ $\bullet$
### Sine Double Angle Formula
$\sin(2A)=2\sin A\cos A$.
Proof: Apply the above with $B=A$ $\bullet$
We also need a formula for $\cos(A+B)$. It is easy to derive this using the unit circle (basically how to extend sine and cosine beyond just $[0^\circ,90^\circ]$), but there is an alternative.
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
Proof: Let $\theta=A+B$ and $\alpha=B$ and apply the Fundamental Identity:
$\cos(\theta-\alpha)=\cos\theta\cos \alpha+\sin\theta\sin\alpha$
$\Rightarrow \cos A=\cos(A+B)\cos B+\sin(A+B)\sin B$
$=\cos (A+B)\cos B+(\sin A\cos B+\sin B\cos A)\sin B$
$\Rightarrow \cos(A+B)\cos B=\cos A-\sin A\cos B\sin A-\sin^2B\cos A$
$=\cos A(1-\sin^2B)-\sin A\sin B\cos B$.
Using the Pythagoras Identity (how?)
$\cos(A+B)\cos B=\cos A\cos^2B-\sin A\sin B\cos B$.
If $B\neq 90^\circ$ (which it won’t be in a RAT with $A+B\leq 90^\circ$), we can divide both sides by $\cos B$ to yield the result $\bullet$.
It isn’t hard to show the sine difference formula via
$\sin(A-B)=\cos(90^\circ-(A-B))=\cos((90^\circ-A)+B)$,
but I am not sure this formula is ever required necessarily.
You might, upon using the identities above, end up with negative angles. Assuming that negative angles are of the form $-A=0^\circ-A$, using the difference formulae, along with the values of sine and cosine at zero, shows that
$\sin(-A)=-\sin A$ and $\cos(-A)=\cos(A)$.
Also messing around with the sum and difference formula gives ways of writing the following products:
$2\sin A\sin B$, $2\sin A\cos B$, and $2\cos A\cos B$,
as sums.
This is enough to get one through. All of these formulae are in the Examination Tables. The proofs show you where these identities come from: it is by using them that we start to get used to them and begin to recognise them.
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Transcript
```Multiples and Least Common Multiples
Section 5.2 B
Here is a video of the “ladder” or “L” algorithm for finding the least common multiple:
What is a multiple of a number? When we have two or more numbers we can find the multiples they
have in common, and often we want the smallest of these.
Again your book uses the Set Intersection Method. Describe this method? ________________________
_____________________________________________________________________________________
The “L” algorithm is a prime factorization method that keeps the numbers organized.
Here is how we would find the LCM of 24 and 36 using the “L” algorithm:
Step 1: Write the two numbers
Step 2: On the right, write a common prime:
Step 3: Divide and write a second common prime factor:
Step 4: Stop when no common prime factor is left:
24 36
24 36
2) 24 36
2) 24 36
2) 12 18 see the L? 2) 12 18
3) 6 9
3) 6 9
2 3
2 3
Step 5: Multiply the primes that were divided out with the remaining primes at the bottom of the L.
This is your least common multiple: 22 x 3 x 2 x 3 = 144
WAIT A MINUTE!!! Didn’t we just use this algorithm to find the GCF???
The advantage of the L algorithm is that students find both the GCF and the LCM at the same time.
Practice using the L algorithm on the A problems in your book, and then write an explanation for the
theorem at the bottom of page 220. The class will be relying upon your work to help us put together a
final explanation. One student noticed that the LCM and the GCF can be found when we use a Venn
diagram to sort the prime factors of two numbers:
Let A = Factors of 24
Let B = Factors of 36
2
What SET is GCF(24, 36) ?
2
2
3
3
What SET is LCM(24, 36) ?
Homework due Monday, January 14:
5.2B #7
5.2B #10
5.2B #24
5.2B #25
Ignore the directions, use the algorithm taught in class only
Do not use a factor tree, use our new algorithm
Make sure you PROVE this!
```
Related documents
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# Exponents and Exponential Functions
## Objective
Define rational exponents and convert between rational exponents and roots.
## Common Core Standards
### Core Standards
?
• N.RN.A.1 — Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)³ = 5(1/3)³ to hold, so (51/3)³ must equal 5.
• N.RN.A.2 — Rewrite expressions involving radicals and rational exponents using the properties of exponents.
?
• 8.EE.A.1
• 8.EE.A.2
• 8.NS.A.1
## Criteria for Success
?
1. Understand a rational exponent as a base with a rational exponent as a power, including either a fraction or a decimal.
2. Write rational exponents using a radical, where the denominator of the fractional exponent defines the root and the numerator of the fractional exponent defines the power of the base.
3. Write radicals as exponential expressions with rational exponents.
4. Extend the properties of integer exponents to rational exponents.
## Anchor Problems
?
### Problem 1
Below is an equation that is not true.
${{{{{10}0}^{1\over2}}}=50}$
a. Why is the statement incorrect? What do you think the correct value of ${{{{10}0}^{1\over2}}}$ is?
b. Consider the following pattern. Where does ${{{{10}0}^{1\over2}}}$ fit in?
${{{10}0}^3=1,000,000}$
${{{10}0}^2={10},000}$
${{{10}0}^1={{10}0}}$
${{{10}0}^0=1}$
c. Consider rewriting the base ${{10}0}$ as a power of ${10}$. How does this shed light on the value of ${{{{10}0}^{1\over2}}}$?
${{{{10}0}^{1\over2}}}=(\square)^{1\over2}$
d. Try out these other rational exponents:
${25^{1\over2}}$ ${144^{1\over2}}$ ${8^{1\over3}}$
#### References
Divisible By 3 Mistakes to the Half Power
Mistakes to the Half Power is made available by Andrew Stadel on Divisible by 3 under the CC BY-NC-SA 3.0 license. Accessed May 17, 2018, 10:54 a.m..
Modified by The Match Foundation, Inc.
### Problem 2
All of the following equations are true.
${\sqrt{x}=x^{1\over2}}$ ${\sqrt[3]{x}=x^{1\over3}}$ ${(\sqrt{x})^2=x}$ ${x^{2\over3}=\sqrt[3]{x^2}}$
Determine a general statement to represent the relationship between a radical and its exponential expression.
### Problem 3
Write the radicals in exponential form and write the exponentials in radical form.
a. ${5^{6\over5}}$
b. ${4^{-{2\over3}}}$
c. ${2n^{2\over5}}$
d. ${\sqrt[3]{7^2}}$
e. ${1\over{\sqrt[3]{5}}}$
f. ${\sqrt{(3x)^5}}$
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
?
Henry explains why ${4^{3\over2}=8}$:
"I know that ${4^3}$ is ${{64}}$ and the square root of ${{64}}$ is $8$."
Here is Henrietta’s explanation for why ${4^{3\over2}=8}$:
"I know that ${\sqrt4=2}$ and the cube of $2$ is $8$. "
1. Are Henry and Henrietta correct? Explain.
2. Calculate $4^{5\over2}$ and $27^{2\over3}$ using Henry’s or Henrietta’s strategy.
3. Use both Henry and Henrietta’s reasoning to express ${x^{m\over n}}$ using radicals (here $m$ and $n$ are positive integers and we assume ${x>0}$).
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Update all PDFs
Identifying Quadratic Functions (Vertex Form)
Alignments to Content Standards: F-BF.B.3
The figure shows a graph of the function $f(x)=x^2$.
a) For each of the graphs of quadratic functions below, find values of $a$, $h$, and $k$ so that the function $f(x) = a(x - h)^2 + k$ has that graph. (For example, the graph in the first part corresponds to $a = 1$, $h = 0$, and $k = 0$.)
b) For each of the following three descriptions of graphs of quadratic functions, sketch a graph by hand, and then find a function in the form $f(x) = a(x - h)^2 + k$ whose graph fits the description.
• The graph is concave down and has its vertex at $(0,7)$.
• The graph is concave up and has its vertex at $(-7,0)$.
• The graph has its vertex at $(4,-9)$ and passes through the point $(6,-1)$.
c) If the graph of the function $f(x) = a(x - h)^2 + k$ is as below, determine the signs of $a$, $h$, and $k$.
IM Commentary
This task has students explore the relationship between the three parameters $a$, $h$, and $k$ in the equation $f(x)=a(x-h)^2+k$ and the resulting graph. There are many possible approaches to solving each part of this problem, especially the first part. We outline some of them here (which overlap heavily in places), applied to the top left graph, and then only give the final answers in the solution below:
1. Transformational/graphical approach: The graph is a scaled reflection of the graph of $y=x^2$ over the $x$ axis, and so takes the form $y=ax^2$ for some negative value of $a$. Taking a single point on the graph, say $(5,-12.5)$, means we must have $-12.5=a(5^2)$, giving $a=-\frac{1}{2}$.
2. Graphical inspection: Since the vertex is at the origin (0,0), we have $h=0$ and $k=0$. So we have $f(x)=a(x-0)^2+0$, thus $f(x)=ax^2$ for some value of $a$ and proceed as in the previous part.
3. Systems of equations approach: For the algebraically fluent, taking three input-output pairs gives three equations which one can solve for $a$, $h$, and $k$. For example, taking $f(0)=0$, $f(-2)=-2$, and $f(2)=-2$
\begin{align}
0&=a(0-h)^2+k\\
-2&=a(-2-h)^2+k\\
-2&=a(2-h)^2+k
\end{align}
The first gives $k=0$ immediately, then the last two are solved to get $h=0$ and $a=-\frac{1}{2}$.
4. Experimental approach: Students might open a new worksheet in, for example, desmos.com, and drag sliders for the coefficients to make the graphs fit the points exactly. We caution that entering points might be somewhat time-consuming, and so teachers should use this approach with discretion -- in particular, it may be worth relying on the fact that it only requires three points to determine a parabola!
Finally, we note that the task uses the phrases "concave up" and "concave down" rather than "upward-facing" and "downward-facing" parabolas. This terminology should be introduced if new to the students.
Solution
1) The four graphs are (from top left to bottom right)
• $f(x)=-\frac{1}{2}x^2$, so $a=-\frac{1}{2}$, $h=0$, and $k=0$.
• $f(x)=x^2-10$, so $a=1$, $h=0$, and $k=-10$.
• $f(x)=(x-9)^2$, so $a=1$, $h=9$, and $k=0$.
• $f(x)=(x-3)^2+5$, so $a=1$, $h=3$, and $k=5$.
2)
• One example is $f(x)=-x^2+7$.
• One example is $f(x)=(x+7)^2$.
• Any such example takes the form $f(x)=a(x-4)^2-9$ for some constant $a$. To ensure that $f(6)=-1$, we need to ensure that $a(6-4)^2-9=-1$, which forces $a=2$.
3) Of course, without any scales on the axes, it will be impossible to figure out exact values for these coefficients. But as it turns out, we'll be able to reason whether each one of the coefficients is positive or negative. Writing $f(x)=a(x-h)^2+k$, we first see that since the given parabola is upward-facing, we must have $a>0$. Next, by its location in the second quadrant we can also see that the coordinates of the vertex $(h,k)$ satisfy $h<0$ and $k>0$.
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## SAT
### Unit 6: Lesson 5
Problem Solving and Data Analysis: lessons by skill
# Ratios, rates, and proportions | Lesson
## What are ratios, rates, and proportions, and how frequently do they appear on the test?
A ratio is a comparison of two quantities. The ratio of a to b can be expressed as a, colon, b or start fraction, a, divided by, b, end fraction.
A proportion is an equality of two ratios. We write proportions to help us find equivalent ratios and solve for unknown quantities.
A rate is the quotient of a ratio where the quantities have different units.
In this lesson, we'll:
1. Learn to convert between part-to-part and part-to-whole ratios
2. Practice setting up proportions to solve for unknown quantities
3. Use rates to predict unknown values
On your official SAT, you'll likely see 2 to 4 questions about ratios, rates, and proportions.
You can learn anything. Let's do this!
## How do we identify and express ratios?
### Identifying a ratio
Part:whole ratiosSee video transcript
### Finding complementary ratios
Two common types of ratios we'll see are part-to-part and part-to-whole.
For example, if we're making lemonade:
• The ratio of lemon juice to sugar is a part-to-part ratio. It compares the amount of two ingredients.
• The ratio of lemon juice to lemonade is a part-to-whole ratio. It compares the amount of one ingredient to the sum of all ingredients.
Since all the parts need to add up to the whole, part-to-part and part-to-whole ratios often imply each other. This means we can use the ratio(s) we're provided to find whichever ratio(s) we need to solve a problem!
Note: Just as fractions can be simplified, ratios can be reduced or expanded to find equivalent ratios. For example, the ratio 5, colon, 10 means the same thing as the ratio 1, colon, 2.
### Try it!
Try: Identify parts and wholes
A high school randomly selected 50 students to take a survey about extending their lunch period. Of students selected for the survey, 14 were freshmen and 13 were sophomores.
• 14, colon, 13 is a
ratio.
• 13, colon, 50 is a
ratio.
• 14, colon, 50 is a
ratio, which could be reduced to
.
Try: find complementary ratios
A bag is filled with red marbles and blue marbles. There are 54 total marbles in the bag, and start fraction, 1, divided by, 3, end fraction of the marbles are blue.
The ratio of blue marbles to total marbles is
.
The ratio of red marbles to total marbles is
.
The ratio of red marbles to blue marbles is
.
How many red marbles are in the bag?
## How do we use proportions?
### Writing proportions
Writing proportions exampleSee video transcript
### Solving word problems using proportions
If we know a ratio and want to apply that ratio to a different scenario or population, we can use proportions to set up equivalent ratios and calculate any unknown quantities.
For example, say we're making cookies, and the recipe calls for 1 cup of sugar for every 3 cups of flour. What if we want to use 9 cups of flour: how much sugar do we need?
• The ratio of sugar to flour must be 1, colon, 3 to match the recipe.
• The ratio of sugar to flour in our batch can be written as x, colon, 9.
To determine how much sugar we need, we can set up the proportion start fraction, 1, divided by, 3, end fraction, equals, start fraction, x, divided by, 9, end fraction and solve for x:
\begin{aligned} \dfrac{1}{3}\purpleD{\cdot9} &= \dfrac{x}{9}\purpleD{\cdot9}\\\\ 3&=x \end{aligned}
We need 3 cups of sugar.
Note: There are multiple ways to set up a proportion. For a proportion to work, it must keep the same units either on the same side of the equation or on the same side of the divisor line.
To use a proportional relationship to find an unknown quantity:
1. Write an equation using equivalent ratios.
2. Plug in known values and use a variable to represent the unknown quantity.
3. Solve for the unknown quantity by isolating the variable.
Example: There are 340 students at Du Bois Academy. If the student-to-teacher ratio is 17, colon, 2, how many teachers are there?
### Try it!
Try: Set up a proportion
A local zoo houses 13 penguins for every lion it houses. The zoo houses 78 penguins.
Which proportion(s) would allow us to solve for x, the number of lions housed at the zoo?
## How do we use rates?
### Finding a per unit rate
Solving unit rate problemSee video transcript
### Applying a per unit rate
Rates are used to describe how quantities change. Common rates include speed (start fraction, start text, d, i, s, t, a, n, c, e, end text, divided by, start text, t, i, m, e, end text, end fraction) and unit price (start fraction, start text, t, o, t, a, l, space, p, r, i, c, e, end text, divided by, start text, u, n, i, t, s, space, p, u, r, c, h, a, s, e, d, end text, end fraction).
For instance, if we know that a train traveled 120 miles in two hours, we can calculate a rate that will tell us the train's average speed over those two hours:
start fraction, 120, start text, space, m, i, l, e, s, end text, divided by, 2, start text, space, h, o, u, r, s, end text, end fraction, equals, 60, start text, space, m, i, l, e, s, space, p, e, r, space, h, o, u, r, end text
We can then use that rate to predict other quantities, like how far that same train, traveling at the same rate, would travel in 5 hours:
start fraction, 60, start text, space, m, i, l, e, s, end text, divided by, 1, start cancel, start text, space, h, o, u, r, end text, end cancel, end fraction, dot, 5, start cancel, start text, space, h, o, u, r, s, end text, end cancel, equals, 300, start text, space, m, i, l, e, s, end text
Note: When working with rates on the SAT, you may need to do unit conversions. To learn more about unit conversions, see the Units lesson.
### Try it
Try: Calculate the unit price
Tony buys 6 large pizzas for dollar sign, 77, point, 94 before tax.
The price for a single large pizza is dollar sign
.
The price of 10 large pizzas before tax would be dollar sign
.
Practice: Apply a ratio
There are two oxygen atoms and one carbon atom in one carbon dioxide molecule. How many oxygen atoms are in 78 carbon dioxide molecules?
Practice: Solve a proportion
Building A is 140 feet tall, and Building B is 85 feet tall. The ratio of the heights of Building A to Building B is equal to the ratio of the heights of Building C to Building D. If Building C is 90 feet tall, what is the height of Building D to the nearest foot?
Practice: Use a rate
The 36-inch tires on a pickup truck have a circumference of 9, point, 42 feet. To the nearest whole rotation, how many rotations must the tires make for the truck to travel 2 miles in straight line? (1, start text, space, m, i, l, e, end text, equals, 5, comma, 280, start text, space, f, e, e, t, end text)
## Want to join the conversation?
• For the last question, why is the answer 1121 and not 1122? After completing 1121 rotations, there's still .01911 rotations left. Shouldn't the answer be rounded up?
|
Hello this is Professor Sampson and
I’m going to help you with Squire’s section 1.7 long division. I will be showing you how to do long division on your calculator two ways. The first way is the easier way and I’m going to give you an
example. For example, 895 divided by 8 895 divided by 8 and we’re looking
for a remainder 895 divided by 8 and you see the slash comes up. When you hit enter you’re going to get a
whole number and a decimal. This is not your
remainder. How do you find the remainder? The first
thing you’re going to do is subtract out your whole number. So you gonna going to plug in minus 111 and hit enter and that will leave you with your
decimal. Once you have your decimal you gonna multiply that by whatever
your denominator was in our case it is 8. We are going to multiply it by 8, all we have to do is times 8, and
that’s going to be a remainder. So your answer is one 111 remainder 7. Let’s try another problem to see if you’re able to do it. 739
divided by 6 739 divided by 6. You need your whole number first so when
you hit enter you get 123. That’s your whole number that you are going to put in MyMathLab. You’re going to subtract out your 123
leaving you with the decimal and this time your gonna multiply your
decimal times what? This time you gonna multiply it by six
because six is your denominator. times six leaving you with 1 so your answer is 123 remainder 1. Lets do two more examples. The next
example, I’m going to do is 942 divided by 7. 942 divided by 7 Your whole numbers gonna be 134 you will
subtract out your 134 leaving you with the decimal and this time your gonna multiply your decimal times 7 because seven is the denominator of the
problem… times 7 so your finally answer is 134 with a remainder of 4..your last example I will be doing is 4427 divided by 6 4427 divided by 6. Again that’s going to leave you with your whole number and your decimal you wanna subtract out your whole number
minus 737, your left with your decimal and this time you gonna
multiply this decimal by 6 As you notice, I’m not clear in the
calculated as I do it. Please resist urge to clear your
calculator, just keep going. So in this case my answer is 737 remainder 5. This is how you find your remainder using the TI 84 plus calculator. You can also use this method with any
other regular calculator. That concludes this topic from Squires section 1.7 on dividing whole numbers.
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# Mathematics - Signals And Systems - Exercises on Continuous-Time Fourier Series
in StemSocial2 months ago
[Image1]
## Introduction
Hey it's a me again @drifter1!
Today we continue with my mathematics series about Signals and Systems in order to cover Exercises on Continuous-Time Fourier Series.
So, without further ado, let's dive straight into it!
## Continuous-Time Fourier Series Recap
Let's quickly refresh our knowledge on the Fourier Series. The Fourier Series is a way of representing periodic signals as a sum of weighted, harmonically related, complex exponentials. This procedure can be split into two parts:
• Fourier Synthesis
• Fourier Analysis
The Fourier Synthesis equation, which is the sum that forms the signal, is:
whilst the Fourier Analysis equation, which gives us the corresponding coefficients, is:
## Fourier Series Coefficients Calculation [Based on 7.3 from Ref1]
For a fundamental frequency ωo = 2π, find the Fourier series coefficients for each of the following signals:
### Solution
#### (a)
From Euler's Formula we know that a complex exponential can be turned into a sum of a sine and cosine trigonometric function, as follows:
As such, the cosine represents the real part of the complex exponential, whilst the sine represents the imaginary part. Due to conjugate symmetry, relating the sine to the complex exponential is as simple as:
Thus, the signal x(t) can be re-written as:
Because the fundamental frequency is chosen to be , the Fourier Series is of the form:
From the Euler expansion that we did for the signal, its easy to notice that only k = 4 and k = - 4 give non-zero results. So, the final coefficients are:
#### (b)
The cosine signal can be represented in a similar sense using the following formula:
Using this representation the signal x(t) can be defined as:
which leads us to the following coefficients:
## Fourier Series from Signal Graph [Based on 7.4 from Ref1]
Consider the following periodic signal:
Let's evaluate the Fourier series analysis equation.
### Solution
From the graph its easy to notice that the period of the signal is To = 4 and so the fundamental frequency is:
Chosing a period of integration from - 2 to 2, can give us the Fourier coefficients as follows:
But, x(t) is inverse symmetrical when taken at - 2 to -1 and 1 to 2 correspondingly. Let's also note that x(0) = 0. Thus, its possible to re-write the formula as:
where x(t) is completely gone.
Let's continue with the calculations to find the final equation for the coefficients...
In order to calculate the value of ao, because it leads to 0 / 0, L'Hopital's rule can be applied.
## RESOURCES:
### Images
Block diagrams and other visualizations were made using draw.io
## Final words | Next up
And this is actually it for today's post!
Next time we will get into exercises on the continuous-time Fourier transform!
See Ya!
Keep on drifting!
|
Related Articles
# GATE | GATE-CS-2014-(Set-3) | Question 65
• Last Updated : 28 Jun, 2021
Consider the following rooted tree with the vertex P labeled as root
The order in which the nodes are visited during in-order traversal is
(A) SQPTRWUV
(B) SQPTURWV
(C) SQPTWUVR
(D) SQPTRUWV
Explanation:
```Algorithm Inorder(tree) - Use of Recursion
Steps:
1. Traverse the left subtree,
i.e., call Inorder(left-subtree)
2. Visit the root.
3. Traverse the right subtree,
i.e., call Inorder(right-subtree)
Understanding this algorithm requires the basic
understanding of Recursion
Therefore, We begin in the above tree with root as
the starting point, which is P.
# Step 1( for node P) :
Traverse the left subtree of node or root P.
So we have node Q on left of P.
-> Step 1( for node Q)
Traverse the left subtree of node Q.
So we have node S on left of Q.
* Step 1 (for node S)
Now again traverse the left subtree of node S which is
NULL here.
* Step 2(for node S)
Visit the node S, i.e print node S as the 1st element of
inorder traversal.
* Step 3(for node S)
Traverse the right subtree of node S.
Which is NULL here.
Now move up in the tree to Q which is parent
of S.( Recursion, function of Q called for function of S).
Hence we go back to Q.
-> Step 2( for node Q):
Visit the node Q, i.e print node Q as the 2nd
element of inorder traversal.
-> Step 3 (for node Q)
Traverse the right subtree of node Q.
Which is NULL here.
Now move up in the tree to P which is parent
of Q.( Recursion, function of P called for function of Q).
Hence we go back to P.
# Step 2(for node P)
Visit the node P, i.e print node S as the 3rd
element of inorder traversal.
# Step 3 (for node P)
Traverse the right subtree of node P.
Node R is at the right of P.
Till now we have printed SQP as the inorder of the tree.
Similarly other elements can be obtained by traversing
the right subtree of P.
The final correct order of Inorder traversal would
be SQPTRWUV. ```
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.
My Personal Notes arrow_drop_up
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# ACT Math: Five Solutions (4/24/15)
Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (4/24/14). Below you’ll find the fully worked-out solutions to those problems.
5. E
You need to identify the pattern in order to determine the next term. The pattern in this case is “add two, add three, add four …. “. To get the next term, add 6 to 21 to get 27.
20. J
An isosceles triangle is a triangle that has two congruent sides. The third side of the triangle could be either 3 or 5. (Note: You should also check that the two solutions give you triangles that are possible using the triangle side length inequality: sum of the two smallest sides > largest side. For example, if the two given sides had been 3 and 7, then 3 would not have been a possible third side because 3 + 3 < 7. You can’t make a triangle with sides 3, 3 and 7.)
35. D
You can begin by getting expressions for the surface area and the volume of the cube using their respective formulas.
Surface Area = 6s² = 6k²
Volume = s³ = k³
The ratio of the surface area to the volume is, therefore, 6k² : k³. This can be simplified to 6 : k by dividing each of the terms in the ratio by k².
46. H
Because the trapezoid is isosceles, you can drop a pair of segments and “break up” the bottom side of the trapezoid as shown in the diagram below. You can then use what you know about 30-60-90 triangles to find the height of the trapezoid. (Note: As soon as you saw the 60º angle in the problem, you probably could have guessed that you’d need to use that special right triangle at some point!)
Now go ahead and compute area of the trapezoid using the area formula.
A = ½(b1 + b2)h
A
= ½(6 + 10)(2√3)
A = ½(16)(2√3)
A = 16√3
Also notice that if you had forgotten the area formula for a trapezoid, you could have found the areas of the two triangles and the rectangle and added them.
57. D
The first thing you should consider is the sign of cos θ. Because cosine is negative, you know that the angle has to be in Quadrant II (π/2 < θ < π) or Quadrant III (π < θ < 3π/2). That eliminates answer choices A, B, and E.
Now let’s consider the other two choices. It would be helpful to know the cosines of the angles in those answer choices.
cos π/2 = 0
cos 2π/3 = -0.5
cos π = -1
The cosine of our angle (-0.674) is between -.05 and -1 so it stands to reason that our angle could be between 2π/3 and π.
If you have questions about these problems or anything to do with the ACT, send us an email at info@cardinalec.com.
|
Math resources Geometry Surface area
Surface area of a pyramid
# Surface area of a pyramid
Here you will learn about the surface area of a pyramid, including what it is and how to calculate it.
Students will first learn about the surface area of a pyramid as part of geometry in 6th grade.
## What is the surface area of a pyramid?
The surface area of a pyramid is the total sum of all the faces of the pyramid. A pyramid is a 3D shape made up of a polygonal base and triangular lateral faces. Pyramids are named after the shape of their bases.
For example,
A square pyramid consists of a square base and four triangular faces:
Square pyramidNetComponent faces
The triangular faces appear to have the same height. When the vertex of a pyramid is directly above the center of the base, the slant height (or the height of each triangular face) is the same. When the vertex is not directly above, the slant heights of the lateral sides will vary.
A rectangular pyramid consists of a rectangle base, and four triangular faces:
Rectangular pyramidNetComponent faces
A triangular pyramid consists of a triangle base, and three triangular faces:
Triangular pyramidNetComponent faces
To calculate the surface area of a pyramid, calculate the area of each face of the pyramid and then add the areas together.
For example,
All lateral faces of this square pyramid are congruent. Find the surface area.
\text{Area of the base }=5\times{5}=25\text{ cm}^{2}
\text{Area of a triangular face }=\cfrac{1}{2}\times{8}\times{5}=20\text{ cm}^{2}
Add the area of the base and the 4 congruent triangular faces:
\text{Surface area }=25+20+20+20+20=25+(4\times{20})=105\text{ cm}^{2}
The total surface area can also be written in one equation:
\begin{aligned} \text{Surface area of pyramid }&= \text{ Area of base } + \text{ Area of triangular faces} \\\\ &=5^2+4 \times (\cfrac{1}{2} \, \times 5 \times 8) \\\\ &=25+80 \\\\ &=105 \; cm^2 \end{aligned}
## Common Core State Standards
How does this relate to 6th grade math?
• Grade 6 – Geometry (6.G.A.4)
Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
## How to find the surface area of a pyramid
In order to find the surface area of a pyramid:
1. Calculate the area of each face.
2. Add the area of each face together.
3. Include the units.
## Surface area of a pyramid examples
### Example 1: square-based pyramid in cm
All the lateral faces of the pyramid are congruent. Calculate the surface area.
1. Calculate the area of each face.
The base is a square with the area 6\times{6}=36\text{ cm}^2.
All four triangular faces are identical, so calculate the area of one triangle and then multiply the area by 4.
\begin{aligned} A&=\cfrac{1}{2} \, \times{b}\times{h}\\\\ &=\cfrac{1}{2} \, \times{10}\times{6}\\\\ &=30 \end{aligned}
30\times{4}=120
2Add the area of each face together.
Add the area of the base and the area of the four triangles:
SA=36+120=156
3Include the units.
The side lengths are measured in centimeters, so the area is measured in square centimeters.
SA=156\text{ cm}^{2}
### Example 2: surface area of a triangular pyramid – net
The net of a triangular pyramid is shown below. All the lateral faces of the pyramid are congruent. Calculate the surface area.
Calculate the area of each face.
Add the area of each face together.
Include the units.
### Example 3: surface area of a square pyramid
Above is a square pyramid where all lateral faces have the same slant height. The perimeter of the base is 4m . Calculate the total surface area of the square pyramid.
Calculate the area of each face.
Calculate the area of all of the triangular faces.
Add the area of each face together.
Include the units.
### Example 4: rectangular-based pyramid – lateral surface area
ABCDE is a rectangular-based pyramid. The opposite lateral faces are congruent. Calculate the lateral surface area of the pyramid.
Calculate the area of each face.
Add the area of each face together.
Include the units.
### Example 5: pyramid – right triangles
In the triangle above, the slant height from vertex D to side \overline{C A} is 7.8 , so the height for \text{triangle CDA} is 7.8 . Calculate the surface area of the triangle.
Calculate the area of each face.
Add the area of each face together.
Include the units.
### Example 6: rectangular-based pyramid – fractions
Calculate the surface area of the pyramid.
Calculate the area of each face.
Add the area of each face together.
Include the units.
### Teaching tips for the surface area of a pyramids
• Give students plenty of opportunities to work with physical 3D models and nets before asking them to solve with surface area.
• Choose worksheets that provide a variety of pyramids – a mixture of triangle, square and rectangles as bases; a mixture of congruent and non-congruent lateral faces; a mixture of pyramid drawings, nets or word problems given.
• Include projects where students have the opportunity to create nets from 3D objects.
### Easy mistakes to make
• Thinking the side of a triangle is always the height
The height of the triangle needs to be the perpendicular height. Only if the side is perpendicular to the base can it be used as the height.
• Thinking surface area is units or units cubed
The surface area will have square units such as square centimeters \mathrm{cm}^2 or square meters \mathrm{m}^2 . Regular units are for one-dimensional measurements like length or height, and cubed units are used for measuring volume.
• Thinking opposite lateral faces are always equal
This is only true when the apex (top vertex) is directly over the center of the base. While this is often the type of pyramids that are given, it is not always the case. Instead of assuming, always use the information given.
### Practice surface area of a pyramid questions
1. All the lateral faces of the pyramid are congruent. Calculate the surface area of the pyramid.
384 \text{ cm}^2
260 \text{ cm}^2
140 \text{ cm}^2
224 \text{ cm}^2
\begin{aligned} \text{Surface area of pyramid }&= \text{Area of base} + \text{Area of triangular faces}\\\\ &=8^2+4\times (\cfrac{1}{2} \, \times 8\times 10)\\\\ &=64+160\\\\ &=224 \ cm^2 \end{aligned}
2. The pyramid is composed of congruent equilateral triangles. Find the surface area of the pyramid.
69.85 \mathrm{~ft}^2
60.5 \mathrm{~ft}^2
52.8 \mathrm{~ft}^2
54.725 \mathrm{~ft}^2
\begin{aligned} \text { Surface area of pyramid }&= \text { Area of base and faces } \\ & \quad \text{ (} 4 \text { congruent triangles) } \\\\ & =4 \times\left(\cfrac{1}{2} \, \times 5.5 \times 4.8\right) \\\\ & =4 \times 13.2 \\\\ & =52.8 \mathrm{~ft}^2 \end{aligned}
3. The opposite lateral faces of ABCD are congruent. Calculate the surface area of the pyramid.
3,077 \mathrm{~cm}^2
7,280 \mathrm{~cm}^2
4,788 \mathrm{~cm}^2
1,366 \mathrm{~cm}^2
\begin{aligned} \text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\ = \, (&42 \times{10})+2\times (\cfrac{1}{2} \, \times{42}\times{52})+2\times (\cfrac{1}{2} \, \times{10}\times{47.3})\\\\ = 4&20+2,184+473\\\\ = 3&,077 \text{ cm}^2 \end{aligned}
4. Here is the net of a square-based pyramid with congruent lateral faces. Calculate the surface area of the lateral faces:
240 \text{ cm}^2
384 \text{ cm}^2
192 \text{ cm}^2
336 \text{ cm}^2
\begin{aligned} \text { Lateral surface area of pyramid } & =\text { Areas of triangular faces } \\\\ & =4 \times\left(\cfrac{1}{2} \times 12 \times 8\right) \\\\ & =4 \times 48 \\\\ & =192 \mathrm{~cm}^2 \end{aligned}
5. Below is a square pyramid with congruent lateral faces. The perimeter of the base is 12.4 \, cm. Calculate the surface area to the nearest whole number.
16 \mathrm{~cm}^2
41 \mathrm{~cm}^2
17 \mathrm{~cm}^2
42\ cm^2
One side of the square can be found by dividing the perimeter by 4 :
12.4 \div 4=3.1 \mathrm{~cm}.
\begin{aligned} \text{Surface area of pyramid }&=\text{Area of base }+ \text{Areas of triangular faces} \\\\ & =3.1^2+4 \times \left(\cfrac{1}{2} \, \times 3.1 \times 5.1\right) \\\\ & =9.61+31.62 \\\\ & =41.23 \end{aligned}
Rounded to the nearest whole, the surface area is 41 \, cm^2.
6. Here is the net of a square-based pyramid. Calculate the surface area of the pyramid.
48 \mathrm{~cm}^2
112 \mathrm{~cm}^2
64 \mathrm{~cm}^2
68.8 \mathrm{~cm}^2
Remember that the edges in a pyramid are always equal, so if you were to fold up the net, the 6 \, cm side of the triangle and 7.2 \, cm side will combine to form an edge with each corresponding side – making their lengths equal.
All sides of the square are also equal.
The area of the base is 4\times{4}=16\text{ cm}^2
The area of two of the triangles is:
\left(\cfrac{1}{2} \, \times{6}\times{4}\right)\times{2}=24\text{ cm}^2
The area of the other two triangles is:
\left(\cfrac{1}{2} \, \times 4 \times 7.2\right) \times 2=28.8 \mathrm{~cm}^2
The total surface area is therefore:
S A=16+24+28.8=68.8 \mathrm{~cm}^2
## Surface area of a pyramid FAQs
Is there a surface area of a triangular pyramid formula?
Because the types of triangles in a pyramid vary, the formula for the surface area is very general. S A=\text { Area of base }+ \text { Area of } 3 \text { lateral faces } .
There are pentagonal pyramids that have a pentagon as a base. There are hexagonal pyramids that have a hexagon as a base. These are just two examples – there are many more. The different types of pyramids are named after the polygonal shape of their base.
What is a regular pyramid?
A pyramid whose base is a regular polygon and whose lateral faces are the same height. The surface area of a regular pyramid always involves congruent lateral faces.
## Still stuck?
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## Related Articles
• RD Sharma Class 11 Solutions for Maths
# Class 11 RD Sharma solutions – Chapter 29 Limits – Exercise 29.7 | Set 2
### Question 32. limx→0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]
Solution:
We have,
limx→0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]
=
=
= -2sina × (1/2)
= -sina
### Question 33. limx→0[{x2 – tan2x}/(tanx)]
Solution:
We have,
limx→0[{x2-tan2x}/(tanx)]
Dividing numerator by 2x and denominator by x.
=
=
= 2(0 – 1)/1
= -2
### Question 34. limx→0[{√2 – √(1 + cosx)}/x2]
Solution:
We have,
limx→0[{√2 – √(1 + cosx)}/x2]
On rationalizing numerator
= limx→0[{2-(1+cosx)}/x2{√2+√(1+cosx)}]
= limx→0[(1-cosx))/x2{√2+√(1+cosx)}]
= 2 × (1/4) × [1/{√2 + √(1 + 1)}]
= (2/4) × (1/2√2)
= (1/4√2)
### Question 35. limx→0[{xtanx}/(1 – cosx)]
Solution:
We have,
limx→0[{xtanx}/(1 – cosx)]
On dividing the numerator and denominator by x2
=
=
As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1
= (4/2)
= 2
### Question 36. limx→0[{x2 + 1 – cosx}/(xsinx)]
Solution:
We have,
limx→0[{x2 + 1 – cosx}/(xsinx)]
= limx→0[{x2 + 2sin2(x/2)}/(xsinx)]
On dividing the numerator and denominator by x2
As we know that limx→0[sinx/x] = 1
= 3/2
### Question 37. limx→0[sin2x{cos3x – cosx}/(x3)]
Solution:
We have,
limx→0[sin2x{cos3x – cosx}/(x3)]
As we know that limx→0[sinx/x] = 1
= -2 × 2 × 2
= -8
### Question 38. limx→0[{2sinx° – sin2x°}/(x3)]
Solution:
We have,
limx→0[{2sinx°-sin2x°}/(x3)]
= limx→0[{2sinx°-2sinx°cosx°}/(x3)]
= limx→0[2sinx°{1-cosx°}/(x3)]
= limx→0[2sinx°{2sin2(x°/2)}/(x3)]
= 4 × [π3/(180 × 360 × 360)]
= (π/180)3
### Question 39. limx→0[{x3.cotx}/(1 – cosx)]
Solution:
We have,
limx→0[{x3.cotx}/(1 – cosx)]
= limx→0[x3/{tanx(1 – cosx)}]
As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1
= 2
### Question 40. limx→0[{x.tanx}/(1 – cos2x)]
Solution:
We have,
limx→0[{x.tanx}/(1 – cos2x)]
= limx→0[{x.tanx}/(2sin2x)]
On dividing the numerator and denominator by x2
As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1
= (1/2)
### Question 41. limx→0[{sin(3 + x) – sin(3 – x)}/x]
Solution:
We have,
limx→0[{sin(3 + x) – sin(3 – x)}/x]
=
= 2Limx→0[cos3.sinx/x]
= 2cos × 3limx→0[sinx/x]
As we know that limx→0[sinx/x] = 1
= 2cos3
### Question 42. limx→0[{cos2x – 1)}/(cosx – 1)]
Solution:
We have,
limx→0[{cos2x – 1)}/(cosx – 1)]
= limx→0[(2sin2x)/{2sin2(x/2)}]
= limx→0[(sin2x)/{sin2(x/2)}]
As we know that limx→0[sinx/x] = 1
= (x2) × (4/x2)
= 4
### Question 43. limx→0[{3sin2x – 2sinx2)}/(3x2)]
Solution:
We have,
limx→0[{3sin2x – 2sinx2)}/(3x2)]
= limx→0[(3sin2x/3x2) – (2sinx2/3x2)]
As we know that limx→0[sinx/x] = 1
= 1 – 2/3
= (3 – 2)/3
= (1/3)
### Question 44. limx→0[{√(1 + sinx) – √(1 – sinx)}/x]
Solution:
We have,
limx→0[{√(1 + sinx) – √(1 – sinx)}/x]
On rationalizing numerator.
= limx→0[{(1 + sinx) – (1 – sinx)}/x{√(1 + sinx) + √(1 – sinx)}]
= limx→0[2(sinx)/x{√(1 + sinx) + √(1 – sinx)}]
As we know that limx→0[sinx/x] = 1
= 2 × {1/(√1 + √1)}
= 2/2
= 1
### Question 45. limx→0[(1 – cos4x)/x2]
Solution:
We have,
limx→0[(1 – cos4x)/x2]
= limx→0[2sin22x/x2]
As we know that limx→0[sinx/x] = 1
= 2 × 4
= 8
### Question 46. limx→0[(xcosx + sinx)/(x2 + tanx)]
Solution:
We have,
limx→0[(xcosx + sinx)/(x2 + tanx)]
= limx→0[x(cosx+sinx/x)/x(x + tanx/x)]
= limx→0[(cosx + sinx/x)/(x + tanx/x)]
As we know that limx→0[tanx/x] = 1
= (1 + 1)/(1 + 0)
= 2
### Question 47. limx→0[(1 – cos2x)/(3tan2x)]
Solution:
We have,
limx→0[(1 – cos2x)/(3tan2x)]
= limx→0[2sin2x/3tan2x]
=
= (2/3)limx→0[cos2x]
= (2/3)
### Question 48. limθ→0[(1 – cos4θ)/(1 – cos6θ)]
Solution:
We have,
limθ→0[(1 – cos4θ)/(1 – cos6θ)]
= limθ→0[2sin22θ/2sin23θ]
= limθ→0[sin22θ/sin23θ]
=
= [(4θ2)/(9θ2)]
= (4/9)
### Question 49. limx→0[(ax + xcosx)/(bsinx)]
Solution:
We have,
limx→0[(ax + xcosx)/(bsinx)]
On dividing the numerator and denominator by x
As we know that limx→0[sinx/x] = 1
=(a + cos 0)/b × 1
= (a + 1)/b
### Question 50. limθ→0[(sin4θ)/(tan3θ)]
Solution:
We have,
limθ→0[(sin4θ)/(tan3θ)]
=
As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1
= (4θ/3θ)
= (4/3)
### Question 51. limx→0[{2sinx – sin2x}/(x3)]
Solution:
We have,
limx→0[{2sinx – sin2x}/(x3)]
= limx→0[{2sinx – 2sinxcosx}/(x3)]
= limx→0[2sinx{1 – cosx}/(x3)]
= limx→0[2sinx{2sin2(x/2)}/(x3)]
As we know that limx→0[sinx/x] = 1
= (4/4)
= 1
### Question 52. limx→0[{1 – cos5x}/{1 – cos6x}]
Solution:
We have,
limx→0[{1 – cos5x}/{1 – cos6x}]
=
As we know that limx→0[sinx/x] = 1
= 25/(4 × 9)
= (25/36)
### Question 53. limx→0[(cosecx – cotx)/x]
Solution:
We have,
limx→0[(cosecx – cotx)/x]
= limx→0[(1/sinx – cosx/sinx)/x]
= limx→0[(1 – cosx)/x.sinx]
= limx→0[2sin2(x/2)/x.sinx]
As we know that limx→0[sinx/x] = 1
= 2/4
= 1/2
### Question 54. limx→0[(sin3x + 7x)/(4x + sin2x)]
Solution:
We have,
limx→0[(sin3x + 7x)/(4x + sin2x)]
As we know that limx→0[sinx/x] = 1
= (7 + 3)/(4 + 2)
= 10/6
= 5/3
### Question 55. limx→0[(5x + 4sin3x)/(4sin2x + 7x)]
Solution:
We have,
limx→0[(5x + 4sin3x)/(4sin2x + 7x)]
=
=
=
As we know that limx→0[sinx/x] = 1
= (5 + 4 × 3)/(4 × 2 + 7)
= (17/15)
### Question 56. limx→0[(3sinx – sin3x)/x3]
Solution:
We have,
limx→0[(3sinx – sin3x)/x3]
= limx→0[{3sinx – (3sinx – 4sin3x)/x3]
= limx→0[(4sin3x)/x3]
= 4Limx→0[{(sinx)/x}3]
As we know that limx→0[sinx/x] = 1
= 4 × 1
= 4
### Question 57. limx→0[(tan2x – sin2x)/x3]
Solution:
We have,
limx→0[(tan2x – sin2x)/x3]
= limx→0[(sin2x/cos2x-sin2x)/x3]
= limx→0[(2sin2x.sin2x)/(x3cos2x)]
As we know that limx→0[sinx/x] = 1
= 2 × 2/cos0
= 4
### Question 58. limx→0[(sinax + bx)/(ax + sinbx)]
Solution:
We have,
limx→0[(sinax + bx)/(ax + sinbx)]
As we know that limx→0[sinx/x] = 1
= (1 × a + b)/(a + 1 × b)
= (a + b)/(a + b)
= 1
Question 59. limx→0[cosecx-cotx]
Solution:
We have,
limx→0[cosecx – cotx]
= limx→0[1/sinx – cosx/sinx]
= limx→0[(1 – cosx)/sinx]
= limx→0[{2sin2(x/2)}/{2sin(x/2)cos(x/2)}]
= limx→0[sin(x/2)/cos(x/2)]
= limx→0[tan(x/2)/ x/2] × x/2
As we know that limx→0[tanx/x] = 1
= 0
### Question 60. limx→0[{sin(α + β)x + sin(α – β)x + sin2αx}/{cos2βx – cos2αx}]
Solution:
We have,
limx→0[{sin(α + β)x + sin(α – β)x + sin2αx}/{cos2βx – cos2αx}]
=
= limx→0[{2sinαx.cosβx + 2sinαx.cosαx}/(sin2αx – sin2βx)]
= limx→0[{2sinαx(cosβx + cosαx)}/(sin2αx – sin2βx)]
As we know that limx→0[sinx/x] = 1
= [{2 × α × 1 × (1 + 1)}/(α2 – β2)] × (1/0)
= (1/0)
= ∞
### Question 61. limx→0[(cosax – cosbx)/(cosecx – 1)]
Solution:
We have,
limx→0[(cosax – cosbx)/(cosecx – 1)]
=
=
=
= [(a + b)(a – b)/c2] × (4/4)
= (a2 – b2)/c2
### Question 62. limh→0[{(a + h)2sin(a + h) – a2sina}/h]
Solution:
We have,
limh→0[{(a + h)2sin(a + h) – a2sina}/h]
= limh→0[{(a+h)2(sina.cosh)+(a+h)2(cosa.sinh)-a2sina}/h]
= limh→0[{(a2+2ah+h2)(sina.cosh)-a2sina+(a+h)2(cosa.sinh)}/h]
= limh→0[{a2sina(cosh-1)+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]
= limh→0[{a2sina(-2sin2(h/2))+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]
=
= 0 + 2asina + 0 + a2cosa
= 2a + a2cosa
### Question 63. If limx→0[kx.cosecx] = limx→0[x.coseckx], find K.
Solution:
We have,
limx→0[kx.cosecx] = limx→0[x.coseckx]
limx→0[kx/sinx] = limx→0[x/sinkx]
klimx→0[x/sinx] = limx→0[kx/sinkx](1/k)
k = (1/k)
k2 = 1
k = ±1
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Home / Calculus I / Limits / The Definition of the Limit
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### Section 2.10 : The Definition of the Limit
2. Use the definition of the limit to prove the following limit.
$\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6$
Show All Steps Hide All Steps
Start Solution
First, let’s just write out what we need to show.
Let $$\varepsilon > 0$$ be any number. We need to find a number $$\delta > 0$$ so that,
$\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \delta$
Or, with a little simplification this becomes,
$\left| {x + 1} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 1} \right| < \delta$ Show Step 2
This problem is very similar to Problem 1 from this point on.
We need to determine a $$\delta$$ that will allow us to prove the statement in Step 1. However, because both inequalities involve exactly the same absolute value statement all we need to do is choose $$\delta = \varepsilon$$.
Show Step 3
So, let’s see if this works.
Start off by first assuming that $$\varepsilon > 0$$ is any number and choose $$\delta = \varepsilon$$. We can now assume that,
$0 < \left| {x - \left( { - 1} \right)} \right| < \delta = \varepsilon \hspace{0.5in} \Rightarrow \hspace{0.5in}0 < \left| {x + 1} \right| < \varepsilon$
This gives,
\begin{align*}\left| {\left( {x + 7} \right) - 6} \right| & = \left| {x + 1} \right| & & \hspace{0.25in}{\mbox{simplify things up a little}}\\ & < \varepsilon & & \hspace{0.25in}{\mbox{using the information we got by assuming }}\delta = \varepsilon \end{align*}
So, we’ve shown that,
$\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \varepsilon$
and so by the definition of the limit we have just proved that,
$\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6$
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# Complex Rational Expressions
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Complex rational expression is a useful tool. It helps to convert a complex rational expression into simplified
expression. An expression in which both numerator and denominator or either one contains a rational expression is known as a complex rational expressions. The complex rational expressions may contain algebraic fractional expression or just a fraction. There are 2 methods to solve complex fractions. One is finding common denominator for each expression and simplifying. The 2nd method is to find common fraction that we multiply with all the terms to simplify. This tool complex rational expression is also an online calculator that intakes complex rational expressions and converts them into simple expression.
Example 1: Simplify by complex fraction solver
2 + 2/x
-------------------------
4 – 3/x^2
Solution: We will simplify numerator 1st; 2 + 2/x = (2x+2)/ x
(Now simplify denominator) 4 - (3/ x^2) = (4x^2 - 3) / x^2
Now inverse the denominator fraction and multiply numerator and denominator we get, (2(x + 1) /x ) (x^2 /
(4x^2 - 3))
2(x+1 ) x
= -------------- = (2x^2+2x) / (4x^2-3)
4x^2 - 3
Example 2: Simplify by complex fraction solver
3/q-1 + 1/q-2 divide by 5/q-2 + 2/q-1
Solution: Simplify numerator we get
3/(q-1) + 1/(q-2) = [(3q-6)+(q-1)]/(q-1)(q-2)
Simplify denominator we get
5/q-2 + 2/q-1 = [5(q-1)+2(q-2)]/(q-2) (q-1)
When we reverse the denominator fraction we can multiply it with numerator
[(3q-6) + (q-1)] x (q-2) (q-1) 3(q-2)+(q-1) 4q-7
------------------------------------------------ = ------------------------- = --------------
(q-1)(q-2) x [5(q-1)+2(q-2)] 5(q-1)+2(q-2) 7q-9
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# 2016 AMC 8 Problems/Problem 25
A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?
$[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]$
$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$
## Solution 1
Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height $15$ and base $\frac{16}{2} = 8$. The Pythagorean triple $8$-$15$-$17$ tells us that these triangles have hypotenuses of $17$.
Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be $r$.
The area of the entire isosceles triangle is $\frac{(16)(15)}{2} = 120$, so the area of each of the two congruent right triangles it gets split into is $\frac{120}{2} = 60$. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is $\frac{17r}{2}$. Thus we can write the equation $\frac{17r}{2} = 60$, so $17r = 120$, so $r = \boxed{\textbf{(B) }\frac{120}{17}}$.
## Solution 2
First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, $60$. $\frac{60}{17}$ times $2$ results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is $\boxed{\textbf{(B) }\frac{120}{17}}$.
## Solution 3: Similar Triangles
$[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NW);[/asy]$ Let's call the triangle $\triangle ABC,$ where $AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry.
Notice that $\triangle ACD \cong \triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\triangle AED \sim \triangle ADC,$ with $\angle EAD \cong \angle DAC$ and $\angle CDA \cong \angle DEA.$ This similarity means that we can create a proportion: $\frac{AD}{AB}=\frac{DE}{CD}.$ We plug in $AD=\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.$
(By the way, we could also use $\triangle DEC \sim \triangle ADC.$)
## Solution 4: Inscribed Circle
$[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label("B",B,SW); label("D",D,SE); label("A",A,N); label("M",M,S); label("C",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$
We'll call this triangle $\triangle ABD$. Let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of $\triangle ABD$ is $\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean Triple $8$-$15$-$17$, the length of each of the legs ($AB$ and $DA$) is 17.
Reflect the triangle over its base. This will create an inscribed circle in a rhombus $ABCD$. Because $AB \cong DA$, $BC \cong CD$. Therefore $AB = BC = CD = DA$.
The semiperimeter $s$ of the rhombus is $\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34$. Since the area of $\triangle ABD$ is $\frac{bh}{2}$, the area $[ABCD]$ of the rhombus is twice that, which is $bh = (16)(15) = 240$.
The Formula for the Incircle of a Quadrilateral is $s$$r$ = $[ABCD]$. Substituting the semiperimeter and area into the equation, $34r = 240$. Solving this, $r = \frac{240}{34}$ = $\boxed{\textbf{(B) }\frac{120}{17}}$.
## Part Solution 5: Answer Choices
Part Solution by e_power_pi_times_i
Notice that the radius must be smaller than half the base (which is $8$). Therefore answer choices $C$, $D$, and $E$ are eliminated. If you do simple math, the square root of $3$ is about $1.7$, and if you multiply that by four, you get $6.8$. If you divide $120$ by $17$, you get approximately $7$, so you have about $\dfrac{1}{2}$ of a chance of getting it right.
2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
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# What Is Marginal Average Cost?
In each problem below, the average cost function by dividing the cost function by the variable representing the quantity. For a cost function C(Q), the average cost function is
$\displaystyle \overline{C}(Q)=\frac{C(Q)}{Q}$
The marginal average cost function is the derivative of the average cost function.
Problem 1 Suppose the total cost function for a product is
$\displaystyle TC(Q)=\frac{3Q+1}{Q+2}\text{ hundred dollars}$
where Q is the number of units produced.
1. Find the average cost of producing 20 units.
2. Find the average cost function.
3. Find the marginal average cost function.
4. Find and interpret the marginal average cost when 20 units are produced.
This means that each of the 20 units costs an average of .1386 hundred dollars or $13.86. In this board they have used the fact that dividing by Q is the same as multiplying by 1/Q. Although it is OK to leave the derivative unsimplified, they need to put in 20. So it is best to do some algebra before putting in the value. Since -0.006 is the slope of the tangent line on the average cost function, the units on it is hundreds of dollars per unit per unit: $\displaystyle \frac{-0.006}{1}\frac{\frac{\text{hundreds of dollars}}{\text{unit}}}{\text{unit}}$ This means that if production is increased by 1 unit, the average cost will drop by 0.006 hundred dollars per unit. Problem 2 Suppose the total cost function for a product is $\displaystyle C(x)=\frac{30x^{2}+500}{x+2}\text{ thousand dollars}$ where x is the number of units produced. 1. Find the average cost of producing 10 units. 2. Find the average cost function. 3. Find the marginal average cost function. 4. Find and interpret the marginal average cost when 10 units are produced. This value tells us that if production is increased by 1 unit, the average cost will drop by 0.3472 thousand dollars per unit or$347.2 per unit. Had they rounded one more decimal place, we would have had this number to the nearest penny.
Problem 3 Suppose the total cost (in thousands of dollars) to produce Q units is
$\displaystyle TC\left( Q \right)=\frac{9Q-5}{7Q+2}$
a. Find the the average cost of producing 40 units.
b. Find the average cost function $\displaystyle \overline{TC}\left( Q \right)$.
c. Find the marginal average cost function $\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$.
d. Find and interpret $\displaystyle \overline{TC}{{\,}^{\prime }}\left( 40 \right)$.
The marginal average cost is simply the slope of the tangent line to the average cost$\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$. The slope has vertical units of thousands of dollars per unit and horizontal units of units. So the rate has units of thousands of dollars per unit per unit. This means that if the production were to increase by one, the average cost would drop by 0.0007701 thousand dollars per unit or 0.7701 dollars per unit. This means that the average cost is decreasing…probably a good thing for the bottom line.
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# Evaluate $\displaystyle \large \lim_{x \,\to\, \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$
In this trigonometric limit problem, $x$ is a variable but represents an angle of a right triangle. The limit of the algebraic trigonometric function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ can be evaluated as $x$ approaches $\dfrac{\pi}{2}$ by the direct substitution method.
$= \,\,\,$ $\dfrac{1+\cos{2\Big(\dfrac{\pi}{2}\Big)}}{\Big(\pi-2\Big(\dfrac{\pi}{2}\Big)\Big)^2}$
$= \,\,\,$ $\dfrac{1+\cos{\Big(\dfrac{2\pi}{2}\Big)}}{\Big(\pi-\Big(\dfrac{2\pi}{2}\Big)\Big)^2}$
$= \,\,\,$ $\require{cancel} \dfrac{1+\cos{\Big(\dfrac{\cancel{2}\pi}{\cancel{2}}\Big)}}{\Big(\pi-\dfrac{\cancel{2}\pi}{\cancel{2}}\Big)^2}$
$= \,\,\,$ $\dfrac{1+\cos{\pi}}{(\pi-\pi)^2}$
$= \,\,\,$ $\dfrac{1+(-1)}{0^2}$
$= \,\,\,$ $\dfrac{1-1}{0}$
$= \,\,\,$ $\dfrac{0}{0}$
According to the direct substitution method, the limit of the given function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ is determinate as $x$ tends to $\dfrac{\pi}{2}$. The direct substitution method is failed in this case. So, it should be evaluated in another mathematical approach.
### Transform the variable into equivalent form
In this step, we eliminate the variable $x$ from the given function by the appropriate replacement.
$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$.
According to the observation of the given function, $2x$ is an angle in the trigonometric function cosine in the numerator and $2x$ is also a term in the denominator. So, try to replace the $2x$ by its equivalent value.
Take $u = \pi-2x$, then $2x = \pi-u$
Therefore, $2x$ can be replaced by $\pi-u$ in the given function to eliminate the $x$ from the function.
Now, let’s change the input of the limit function.
It is given that $x \,\to\, \dfrac{\pi}{2}$ but $2x = \pi-u$. So, $x = \dfrac{\pi-u}{2}$
Therefore, $\dfrac{\pi-u}{2} \,\to\, \dfrac{\pi}{2}$
$\implies$ $-\Big(\dfrac{u-\pi}{2}\Big) \,\to\, \dfrac{\pi}{2}$
$\implies$ $\dfrac{u-\pi}{2} \,\to\, -\dfrac{\pi}{2}$
$\implies$ $\dfrac{u}{2}-\dfrac{\pi}{2} \,\to\, -\dfrac{\pi}{2}$
$\implies$ $\dfrac{u}{2} \,\to\, -\dfrac{\pi}{2}+\dfrac{\pi}{2}$
$\implies$ $\require{cancel} \dfrac{u}{2} \,\to\, -\cancel{\dfrac{\pi}{2}}+\cancel{\dfrac{\pi}{2}}$
$\implies$ $\dfrac{u}{2} \,\to\, 0$
$\implies$ $u \,\to\, 2 \times 0$
$\implies$ $u \,\to\, 0$
Therefore, it is proved that if $x$ approaches $\dfrac{\pi}{2}$, then $u$ approaches zero.
### Simplify the Algebraic trigonometric function
According to the previous step, the limit of the given function can be converted in terms of $u$ from $x$ in two different simplifying methods. You can follow any one of them.
#### Method: 1
In this method, substitute $\pi-2x = u$ and $2x = \pi-u$
$\implies$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$ $\,=\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1+\cos{(\pi-u)}}{u^2}}$
The angle $\pi-u$ represents an angle in the second quadrant. So, $\cos{(\pi-u)} \,=\, \cos{u}$.
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1-\cos{u}}{u^2}}$
According to power reduction trigonometric identity, $1-\cos{\theta} \,=\, 2\sin^2{\Big(\dfrac{\theta}{2}\Big)}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$
#### Method: 2
According to power reducing trigonometric identity, the expression in numerator of the given function can be written in its equivalent form. We know that $1+\cos{2\theta} \,=\, 2\cos^2{\theta}$.
$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{2\cos^2{x}}{(\pi-2x)^2}}$.
In this method, substitute $\pi-2x = u$ and $x = \dfrac{\pi-u}{2}$. Similarly, it is already proved that if $x \to \dfrac{\pi}{2}$, then $u \to 0$.
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\cos^2{\Big(\dfrac{\pi-u}{2}\Big)}}{u^2}}$.
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\cos^2{\Big(\dfrac{\pi-u}{2}\Big)}}{u^2}}$.
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\cos^2{\Big(\dfrac{\pi}{2}-\dfrac{u}{2}\Big)}}{u^2}}$.
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2 {\Bigg(\cos{\Big(\dfrac{\pi}{2}-\dfrac{u}{2}\Big)}\Bigg)}^2 }{u^2}}$.
The angle $\dfrac{\pi}{2}-\dfrac{u}{2}$ belongs to first quadrant. Therefore, $\cos{\Big(\dfrac{\pi}{2}-\dfrac{u}{2}\Big)} \,=\, \sin{\Big(\dfrac{u}{2}\Big)}$
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2 {\Bigg(\sin{\Big(\dfrac{u}{2}\Big)}\Bigg)}^2 }{u^2}}$.
$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$.
In both methods, we have simplified that $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$ $\,=\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$ by taking $u = \pi-2x$ in this calculus problem.
### Evaluate the Limit of the given function
The simplification of the given function is completed and now, we can start the procedure for finding the limit mathematically.
$\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$
In the numerator of the function, $\dfrac{u}{2}$ is an angle in the sin squared function. So, try to adjust the function in the denominator same as the numerator for using the limit rule of trigonometric function. The factor $2$ multiples the sin function in numerator and it divides the term in the denominator. So, shift it to denominator.
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{\dfrac{u^2}{2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{1 \times \dfrac{u^2}{2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{\dfrac{2}{2} \times \dfrac{u^2}{2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{2 \times \dfrac{u^2}{2 \times 2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{2 \times \dfrac{u^2}{2^2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{2 \times {\Big(\dfrac{u}{2}\Big)}^2 }}$
$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1}{2} \times \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{{\Big(\dfrac{u}{2}\Big)}^2 }}$
$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{{\Big(\dfrac{u}{2}\Big)}^2 }}$
$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{u \,\to\, 0}{\normalsize {\Bigg(\dfrac{\sin{\Big(\dfrac{u}{2}\Big)}}{\Big(\dfrac{u}{2}\Big)}\Bigg)}^2 }$
According to constant exponent limit rule, the above function can be written as follows.
$= \,\,\,$ $\dfrac{1}{2} \times {\Bigg(\displaystyle \large \lim_{\Large \frac{u}{2} \large \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{u}{2}\Big)}}{\Big(\dfrac{u}{2}\Big)}}\Bigg)}^2$
Take $m = \dfrac{u}{2}$
$= \,\,\,$ $\dfrac{1}{2} \times {\Bigg(\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin{m}}{m}}\Bigg)}^2$
The limit of the function represents the limit of sinx/x as x approaches 0 rule. Therefore, the limit of the $\dfrac{\sin{m}}{m}$ function as $m$ approaches $0$ is equal to one.
$= \,\,\,$ $\dfrac{1}{2} \times {\Big(1\Big)}^2$
$= \,\,\,$ $\dfrac{1}{2} \times 1$
$= \,\,\,$ $\dfrac{1}{2}$
Therefore, it is calculated that the limit of the given function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ as $x$ approaches $\dfrac{\pi}{2}$ is equal to the $\dfrac{1}{2}$
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# 1.4: Packing Bags
Difficulty Level: At Grade Created by: CK-12
Can you describe what you see in the picture below? What if you put the same shapes that are in the filled bag into the empty bags? How many of each shape would you now have? In this Concept, we will practice describing what we see and counting shapes.
### Guidance
When looking at bags like in the picture above, we can see that one bag is filled and the other bags are empty. We can fill the empty bags the same way as the filled bag. Then, we can count to see how many of each shape we have in total.
#### Example A
Look at the picture below. What do you see?
You should notice 2 bags. One bag has shapes in it and the other bag is empty. The bag with the shapes in it has 2 circles and 1 triangle.
The bag with shapes in it is Ann’s bag. Use your shape tiles. Put 1 triangle into the empty bag. Put 2 circles into the empty bag. Now, count the shapes. How many circles in all? How many triangles in all?
You should have 4 circles in all and 2 triangles in all.
#### Example B
Look at the picture below. What do you see?
You should notice 2 bags. One bag has shapes in it and the other bag is empty. The bag with the shapes in it has 2 triangles, 1 circle, and 1 square.
The bag with the shapes in it is Ted’s bag. Use your shape tiles. Put 2 triangles into the empty bag. Put 1 circle into the empty bag. Put 1 square into the empty bag. Count the shapes. How many triangles in all? How many circles in all? How many squares in all?
You should have 4 triangles in all, 2 circles in all, and 2 squares in all.
#### Concept Problem Revisited
You should notice 3 bags. One bag is Bob's bag and it has 1 circle and 2 squares in it. The other two bags are empty. If you put 2 squares into each empty bag and 1 circle into each empty bag you will have 6 squares in all and 3 circles in all.
### Vocabulary
This is a circle:
This is a square:
This is a triangle:
### Guided Practice
For each set of bags below, describe what you see. Then, fill the empty bag(s) with the same shapes that are in the filled bag. Last, count how many of each shape you have in all.
1.
2.
3.
1. You should notice 3 bags. One bag has 2 square, 2 triangles, and 1 circle. The other 2 bags are empty. If you put 2 squares, 2 triangles, and 1 circle in each empty bag, you will have 6 squares in all, 6 triangles in all, and 3 circles in all.
2. You should notice 3 bags. One bag has 3 circles, 1 triangle, and 1 square. The other 2 bags are empty. If you put 3 circles, 1 triangle, and 1 square into each empty bag, you will have 9 circles in all, 3 triangles in all, and 3 squares in all.
3. You should notice 4 bags. One bag has 2 triangles, 1 circle, and 2 squares. The other 3 bags are empty. If you put 2 triangles, 1 circle, and 2 squares into each empty bag, you will have 8 triangles in all, 4 circles in all, and 8 squares in all.
### Practice
For each set of bags below, describe what you see. Then, fill the empty bag(s) with the same shapes that are in the filled bag. Last, count how many of each shape you have in all.
1. What if there had been two empty bags in #1 instead of just one empty bag? If you filled all of the empty bags, how many of each shape would there be?
2. What if there had been three empty bags in #1 instead of just one empty bag? If you filled all of the empty bags, how many of each shape would there be?
3. What if there had been four empty bags in #1 instead of just one empty bag? If you filled all of the empty bags, how many of each shape would there be?
Jan 18, 2013
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# Multiplying by 10, 100 and 1000
#### Everything You Need in One Place
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Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.
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0/2
##### Intros
###### Lessons
1. Introduction to Multiplying by 10, 100 and 1000
2. Patterns in Multiplying by 10, 100 and 1000
3. Multiplying by Multiples of 10
0/13
##### Examples
###### Lessons
1. Multiplying by 10, 100, 1000
Multiply:
1. 62 x 10
2. 9 x 100
3. 78 x 1000
2. Multiplying by 10
Multiply:
1. 3 x 80
2. 40 x 50
3. 4 x 900
4. 500 x 600
5. 6000 x 9000
3. Use the Pattern of Multiplying by 10, 100, 1000 to find the missing number
Find the missing number
1. 20 x _____ = 600
2. 30 x ______ = 120, 000
3. ______ x 400 = 8000
4. 600 x _____ = 30000
5. ______ x 70 = 210,000
0%
##### Practice
###### Topic Notes
In this lesson, we will learn:
• The patterns in multiplying by 10, 100 and 1000 (and their multiples)
Notes:
• A multiple of 10 is any number that has 10 as a factor (Shortcut: a multiple of 10 is any number that ends in zero)
• To multiply a whole number and 10, 100, 1000 (or their multiples)
• Multiply the nonzero digits
• Count the number of zeroes in the factors
• Write the same number of zeroes in the product
$\qquad \quad$ Ex. 30 x 7 00 = 21000
$\qquad \quad$ Ex. 800 x 4000 = 3,200,000
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# Fourier Series Formula
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## Fourier Series Formula
The Fourier series formula gives an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines. It is used to decompose any periodic function or periodic signal into the sum of a set of simple oscillating functions, namely sines and cosines. Let us understand the Fourier series formula using solved examples.
## What Are Fourier Series Formulas?
Fourier series makes use of the orthogonal relationships of the cosine and sine functions. Fourier series formula for a function is given as,
Let us have a look at a few solved examples on the Fourier series formula to understand the concept better.
## Examples on Fourier Series Formulas
Example 1: Expand the function f(x) = ex in the interval [ – π , π ] using Fourier series formula.
Solution:
Applying the Fourier series formula, we know that f(x) =
## FAQs on Fourier Series Formulas
### What Is Meant by the Fourier Series?
A Fourier series presents an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines. Fourier Series makes use of the orthogonality relationships of the sine and cosine functions.
### What Is the Fourier Series Formula?
The fourier series formula of the function f(x) in the interval [-L, L], i.e. -L ≤ x ≤ L is given by:
f(x) = A_0 + ∑_{n = 1}^{∞} A_n cos(nπx/L) + ∑_{n = 1}^{∞} B_n sin(nπx/L)
### What Is the Application of the Fourier Series Formula?
Fourier series describes a periodic signal in terms of cosine and sine waves. Thus, it models any arbitrary periodic signal with a combination of sines and cosines.
### How To Solve a Fourier Series Using a Fourier Series Formula?
The steps for solving a Fourier series are given below:
Step 1: Multiply the given function by sine or cosine, then integrate
Step 2: Estimate for n=0, n=1, etc., to get the value of coefficients.
Step 3: Finally, substituting all the coefficients in the Fourier formula.
### What Are the 2 Types of Fourier Series Formula?
The two types of Fourier series formulas are trigonometric series formula and exponential series formula.
## Fourier Series
A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions. The computation and study of Fourier series is known as harmonic analysis and is extremely useful as a way to break up an arbitrary periodic function into a set of simple terms that can be plugged in, solved individually, and then recombined to obtain the solution to the original problem or an approximation to it to whatever accuracy is desired or practical. Examples of successive approximations to common functions using Fourier series are illustrated above.
In particular, since the superposition principle holds for solutions of a linear homogeneous ordinary differential equation, if such an equation can be solved in the case of a single sinusoid, the solution for an arbitrary function is immediately available by expressing the original function as a Fourier series and then plugging in the solution for each sinusoidal component. In some special cases where the Fourier series can be summed in closed form, this technique can even yield analytic solutions.
Any set of functions that form a complete orthogonal system have a corresponding generalized Fourier series analogous to the Fourier series. For example, using orthogonality of the roots of a Bessel function of the first kind gives a so-called Fourier-Bessel series.
The computation of the (usual) Fourier series is based on the integral identities
## Definition of Fourier Series and Typical Examples
Baron Jean Baptiste Joseph Fourier introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.
To consider this idea in more detail, we need to introduce some definitions and common terms.
## Solved Problems
Click or tap a problem to see the solution.
## Fourier Series Formula
Many of the phenomena studied in the domain of Engineering and Science are periodic in nature. For example current and voltage existing in an alternating current circuit. We can analyze these periodic functions into their constituent components by using a process called Fourier analysis. In this article, we will discuss the Fourier series and Fourier Series Formula. Let us begin learning!
## Fourier Series Formula
### What is the Fourier Series?
Periodic functions occur frequently in the problems studied during engineering problem-solving. Their representation in terms of simple periodic functions like sine and cosine functions, which are leading towards the Fourier series. Fourier series is a very powerful and versatile tool in connection with the partial differential equations.
A Fourier series is nothing but the expansion of a periodic function f(x) with the terms of an infinite sum of sins and cosine values. Fourier series is making use of the orthogonal relationships of the sine and cosine functions. A difficult thing to understand here is to motivate the fact that arbitrary periodic functions have Fourier series representations.
### Fourier Analysis for Periodic Functions
The Fourier series representation of analytic functions has been derived from the Laurent expansions. We use the elementary complex analysis to derive additional fundamental results in the harmonic analysis, which includes representation of the periodic functions by the Fourier series.
The representation of rapidly decreasing functions by Fourier integrals, and Shannon’s sampling theorem. The ideas are classical and of transcendent beauty.
The trigonometric functions sin x and cos x are examples of periodic functions with fundamental period 2π and tan x is periodic with fundamental period \pi.
Therefore a Fourier series is a method to represent a periodic function as a sum of sine and cosine functions possibly till infinity. It is analogous to the famous Taylor series, which represents functions as possibly infinite sums of monomial terms.
For the functions that are not periodic, the Fourier series is replaced by the Fourier transform. For the functions of two variables that are periodic in both variables, the trigonometric basis in the Fourier series is replaced by the spherical harmonics.
## Solved Examples
Putting all above values in the equation, we get the expansion of the above function:
## Fourier Series
Sine and cosine waves can make other functions!
Here two different sine waves add together to make a new wave:
## Square Wave
Can we use sine waves to make a square wave?
Our target is this square wave:
And if we could add infinite sine waves in that pattern we would have a square wave!
So we can say that:
a square wave = sin(x) + sin(3x)/3 + sin(5x)/5 + … (infinitely)
That is the idea of a Fourier series.
By adding infinite sine (and or cosine) waves we can make other functions, even if they are a bit weird.
## Finding the Coefficients
How did we know to use sin(3x)/3, sin(5x)/5, etc?
There are formulas!
First let us write down a full series of sines and cosines, with a name for all coefficients:
We can often find that area just by sketching and using basic calculations, but other times we may need to use Integration Rules.
So this is what we do:
• Take our target function, multiply it by sine (or cosine) and integrate (find the area)
• Do that for n=0, n=1, etc to calculate each coefficient
• And after we calculate all coefficients, we put them into the series formula above.
Let us see how to do each step and then assemble the result at the end!
When n is even the areas cancel for a result of zero.
In conclusion:
• Think about each coefficient, sketch the functions and see if you can find a pattern,
• put it all together into the series formula at the end
And when you are done go over to:
and see if you got it right!
Why not try it with “sin((2n-1)*x)/(2n-1)”, the 2n−1 neatly gives odd values, and see if you get a square wave.
## Other Functions
Of course we can use this for many other functions!
But we must be able to work out all the coefficients, which in practice means that we work out the area of:
• the function
• the function times sine
• the function times cosine
But as we saw above we can use tricks like breaking the function into pieces, using common sense, geometry and calculus to help us.
Here are a few well known ones:
Footnote. Different versions of the formula!
## Fourier Analysis for Periodic Functions
The Fourier series representation of analytic functions is derived from Laurent expansions. The elementary complex analysis is used to derive additional fundamental results in the harmonic analysis including the representation of C∞ periodic functions by Fourier series, the representation of rapidly decreasing functions by Fourier integrals, and Shannon’s sampling theorem. The ideas are classical and of transcendent beauty.
A function is periodic of period L if f(x+L) = f(x) for all x in the domain of f. The smallest positive value of L is called the fundamental period.
The trigonometric functions sin x and cos x are examples of periodic functions with fundamental period 2π and tan x is periodic with fundamental period π. A constant function is a periodic function with arbitrary period L.
It is easy to verify that if the functions f1, . . . , fn are periodic of period L, then any linear combination
### Applications
A Fourier Series has many applications in mathematical analysis as it is defined as the sum of multiple sines and cosines. Thus, it can be easily differentiated and integrated, which usually analyses the functions such as saw waves which are periodic signals in experimentation. It also provides an analytical approach to solve the discontinuity problem. In calculus, this helps in solving complex differential equations.
## Frequently Asked Questions – FAQs
### What is the Fourier Series formula?
The formula for the fourier series of the function f(x) in the interval [-L, L], i.e. -L ≤ x ≤ L is given by:
f(x) = A_0 + ∑_{n = 1}^{∞} A_n cos(nπx/L) + ∑_{n = 1}^{∞} B_n sin(nπx/L)
### What is the Fourier series used for?
Fourier series is used to describe a periodic signal in terms of cosine and sine waves. In other other words, it allows us to model any arbitrary periodic signal with a combination of sines and cosines.
### How do you solve a Fourier series?
The steps to be followed for solving a Fourier series are given below:
Step 1: Multiply the given function by sine or cosine, then integrate
Step 2: Estimate for n=0, n=1, etc., to get the value of coefficients.
Step 3: Finally, substituting all the coefficients in Fourier formula.
### What are the 2 types of Fourier series?
The two types of Fourier series are trigonometric series and exponential series.
### What is meant by the Fourier series?
A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines. Fourier Series makes use of the orthogonality relationships of the sine and cosine functions.
Math Formulas ⭐️⭐️⭐️⭐️⭐
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# Graphing 24 - parallel and perpendicular lines 2
Taught by YourMathGal
• Currently 4.0/5 Stars.
6875 views | 2 ratings
Part of video series
Meets NCTM Standards:
Errors in this video:
At 8:20 of the video, there is an error: line #3 should be y = -3/2x - 6 because you need to find the opposite reciprocal.
Lesson Summary:
In this lesson, students learn how to write equations of lines that are parallel or perpendicular to a given line, given a point on the line. The lesson uses the slope-point formula to solve for the equation of the line, and also provides some example problems for students to try on their own. Additionally, the lesson covers how to determine whether two lines are parallel, perpendicular, or neither, based on their slopes.
Lesson Description:
Covers writing equations of lines parallel or perpendicular to a given line if you know a point on the line. This is part 24 of a series of videos about graphing lines.
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
• How do you write the equation of the line parallel to the line y = 2x - 4 that passes through (-6, -1)?
• How do you write the equation of a line in slope-intercept form that has a given slope and passes through a given point using point-slope form?
• How do you write the equation of the line perpendicular to the line y = 2x - 4 that passes through (-6, -1)?
• How can you find the slope of a line parallel or perpendicular to a given line?
• How do you write the equation of the line passing through (0, -6) with a slope of 5?
• How do you write the equation of the line parallel to y = 4x + 3 passing through (0, -6)?
• How do you write the equation of the line perpendicular to y = 2/3x - 7 passing through (0, -6)?
• Are the lines y = 3x - 4 and 3x - y = 8 parallel, perpendicular, or neither?
• Are the lines y = 2/3x + 5 and y = 3/2x - 1 parallel, perpendicular, or neither?
• Are the lines y = 2x - 3 and y = -1/2x - 3 parallel, perpendicular, or neither?
• How do you re-write an equation in standard form in slope-intercept form?
• #### Staff Review
• Currently 4.0/5 Stars.
This video takes the ideas we learned from the previous lesson and uses these skills to find equations of line parallel or perpendicular to other equations of lines that pass through a given point. Make sure you are confident in your ability to write equations for lines before watching this video.
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# McGraw Hill Math Grade 8 Lesson 20.3 Answer Key Right Triangles and Pythagorean Theorem
Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 20.3 Right Triangles and Pythagorean Theorem to secure good marks & knowledge in the exams.
## McGraw-Hill Math Grade 8 Answer Key Lesson 20.3 Right Triangles and Pythagorean Theorem
Exercises
SOLVE
Use the Pythagorean Theorem to determine the length of the missing side.
Question 1.
If side A is 6 and side B is 8, then side C (the hypotenuse) is _____________
C = 10,
Explanation:
As side A is 6 and side B is 8, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A2 + B2 ) = square root of (62 + 82)= square root of (36 + 64) = square root of 100 = 10.
Question 2.
If side A is 9 and side B is 9, then side C (the hypotenuse) is _____________
C = 12.72,
Explanation:
As side A is 9 and side B is 9, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A2 + B2 ) = square root of (92 + 92)= square root of (81 + 81) = square root of 162 , approximately equal to 12.72.
Question 3.
If side A is 10 and side B is 24, then side C (the hypotenuse) is _____________
C = 26,
Explanation:
As side A is 10 and side B is 24, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A2 + B2 ) = square root of (102 + 242)= square root of (100 + 576) = square root of 676 = 26.
Question 4.
If side A is 9 and side C (the hypotenuse) is 15, then side B is _____________
B = 12,
Explanation:
As side A is 9 and side C (the hypotenuse) is 15, then side B is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. C = square root of (A2 + B2 ) therefore B = square root of (C2 –Â A2)= square root of (225 – 81) = square root of 144 = 12.
Question 5.
If side B is 6 and the hypotenuse is 10, then side A is _____________
A = 8,
Explanation:
As side B is 6 and the hypotenuse is 10, then side A is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A2 + B2 ) therefore A = square root of (C2 – B2)= square root of (100 – 36) = square root of 64 = 8.
Question 6.
If side A is 12 and side B is 5, then side C (the hypotenuse) is _____________
C = 13,
Explanation:
As side A is 12 and side B is 5, then side C (the hypotenuse) is applying Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So C = square root of (A2 + B2 ) = square root of (122 + 52) = square root of (144 + 25) = square root of 169 = 13.
Find the missing sides of the following pairs of similar right triangles:
Question 1.
HI = ___________ ft
KL = ___________ ft
JL = ____________ ft
HI = 50 ft.,
KL = 30 ft.,
JL =Â 78 ft.,
Explanation:
For finding HI we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (GI2 – GH2 ) = square root of (1302 – 1202) = square root of (16,900 – 14,400) = square root of 2,500 = 50 ft.
Now finding KL let it be x as GHI and JKL are similar triangles that means the ratios of the sides are equal HI/GH = KL/JK = 50/120 = x/72 cross multiplying for the unknown we get
120x = 50 X 72, x = 5 X 72/12 = 5 X 6 = 30 ft.
Now JL we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (JK2 + KL2 ) = square root of (722 + 302) = square root of (5,184 + 900) = square root of 6,084 = 78 ft.
Question 2.
LN = ___________ m
PQ = ___________ m
OQ = ____________ m
LN = 25 ft., = 7.62 m,
PQ = 21 ft., = 6.4008 m,
OQ = 35 ft., = 10.668 m,
Explanation:
For finding LN we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So LN = square root of (LM2 + MN2 ) = square root of (202 + 152) = square root of (400 + 225) = square root of 625 = 25 ft. As 1 foot is equal to 0.3048 meter,
so 25 X 0.3048 m = 7.62 m.
Now finding PQ let it be x as LMN and OPQ are similar triangles that means the ratios of the sides are equal MN/LM = PQ/OP = 15/20 = x/28 cross multiplying for the unknown we get
20x = 15 X 28, x = 3 X 28/4 = 3 X 7 = 21 ft., = 21 X 0.3048 m = 6.4008 m.
Now OQ we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So OQ = square root of (OP2 + PQ2 ) = square root of (282 + 212) = square root of (784 + 441) = square root of 1,225 = 35 ft., = 35 X 0.3048 m =Â 10.668 m.
Question 3.
RT = ___________ in.
VW = ___________ in.
UW = ____________ in.
RT = 7.071 ft., = 84.852 in.,
VW = 20 ft., = 240 in.,
UW = 28.284 ft., = 339.400 in.,
Explanation:
For finding RT we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So HI = square root of (RS2 + ST2 ) = square root of (52 + 52) = square root of (25 + 25) = square root of 50 = 7.071 ft. As 1 foot is equal to 12 inch we get 7.071 X 12 inch = 84.852 in, Now finding VW let it be x as RST and UVW are similar triangles that means the ratios of the sides are equal ST/RS = VW/UV = 5/5 = x/20 cross multiplying for the unknown we get 5x = 5 X 20, x = 20 ft., = 20 X 12 in = 240 in.,
Now UW we apply Pythagorean theorem that is the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. So UW = square root of (UV2 + VW2 ) = square root of (202 + 202) = square root of (400 + 400) = square root of 800 = 28.284 ft., = 339.408 in.
Question 4.
If the two triangles are similar, then what is the length of the missing side?
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### Ratio and Proportion-Solutions Ex-12.3
CBSE Class –VI Mathematics
NCERT Solutions
Chaper 12 Ratio And Proportion (Ex. 12.3)
Question 1. If the cost of 7 m of cloth is Rs. 294, find the cost of 5 m of cloth.
Answer: The cost of 7 m of cloth = Rs. 294
$\therefore$ The cost of 1 m of cloth = Rs. 42
$\therefore$ Cost of 5 m of cloth = 42 x 5 = Rs. 210
Thus, the cost of 5 m of cloth is Rs. 210.
Question 2. Ekta earns Rs. 1500 in 10 days. How much will she earn in 30 days?
Answer: Earning of 10 days = Rs. 1500
$\therefore$ Earning of 1 day = = Rs. 150
$\therefore$ Earning of 30 days = 150 x 30 = Rs. 4500
Thus, the earning of 30 days is Rs. 4,500.
Question 3. If it has rained 276 mm in the last 3 days, how many cms of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Answer: Rain in 3 days = 276 mm
$\therefore$ Rain in 1 day = = 92 mm
$\therefore$ Rain in 7 days = 92 x 7 = 644 mm
Thus, the rain in 7 days is 644 mm.
Question 4. Cost of 5 kg of wheat is Rs. 30.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in Rs. 61?
Answer: (a) Cost of 5 kg of wheat = Rs. 30.50
$\therefore$ Cost of 1 kg of wheat = = Rs. 6.10
$\therefore$ Cost of 8 kg of wheat = 6.10 x 8 = Rs. 48.80
(b) From Rs. 30.50, quantity of wheat can be purchased = 5 kg
$\therefore$ From Rs. 1, quantity of wheat can be purchased = $\frac{5}{30.50}$
$\therefore$ From Rs. 61, quantity of wheat can be purchased = = 10 kg
Question 5. The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Answer: $\therefore$ Degree of temperature dropped in last 30 days = 15 degrees
$\therefore$ Degree of temperature dropped in last 10 days = $\frac{15X10}{30}$ =5 degrees
Thus, 5 degree Celsius temperature dropped in 10 days.
Question 6. Shains pays Rs. 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?
Answer: Rent paid for 3 months = Rs. 7500
$\therefore$ Rent paid for 1 month = Rs. 2500
$\therefore$ Rent paid for 12 months = 2500 x 12 = Rs. 30,000
Thus, the total rent for one year is Rs. 30,000.
Question 7.The cost of 4 dozens bananas is Rs. 60. How many bananas can be purchased for Rs. 12.50?
Answer: The cost of 4 dozen bananas = Rs. 60
The cost of 48 bananas = Rs. 60 [4 dozen = 4 x 12 = 48]
$\because$ From Rs. 60, number of bananas can be purchased = 48
$\therefore$ From Rs. 12.50. number of bananas can be purchased = $\frac{48X12.5}{60}$
= 10 bananas
Thus, 10 bananas can be purchased for Rs. 12.50.
Question 8.The weight of 72 books is 9 kg what is the weight of 40 such books?
Answer: The weight of 72 books = 9 kg
$\therefore$ The weight of 1 book = $\frac{1}{8}$
$\therefore$ The weight of 40 books = 40 x 1/8 =5 kg
Thus, the weight of 40 books is 5 kg.
Question 9. A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Answer: The quantity of diesel required by the truck to cover a distance of 594 km = 108 litres
$\therefore$ The quantity of diesel required by the truck to cover a distance of 1 km = 108 /594 = 2/11 litres
$\therefore$ The quantity of diesel required by the truck to cover a distance of 1650 km = 1650 X 2/11 litres = 300 litres
Thus, 300 litres of diesel will be required by the truck to cover a distance of 1650 km.
Question 10. Raju purchases 10 pens for Rs. 150 and Manish buy 7 pens for Rs. 84. Can you say who got the pen cheaper?
Answer: Raju purchase 10 pens for = Rs. 150
$\therefore$ Raju purchases 1 pen for = = Rs. 15
Manish purchases 7 pens for = Rs. 84
$\therefore$ Manish purchases 1 pen for = = Rs. 12
Thus, Manish got the pens cheaper.
Question 11.Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
$\therefore$ Anish made in 1 overs = = 7 runs
$\therefore$ Anup made in 1 overs = = 9 runs
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# What is associative property of multiplication example?
## What is associative property of multiplication example?
The associative property of multiplication states that the product of three or more numbers remains the same regardless of how the numbers are grouped. For example, 3 × (5 × 6) = (3 × 5) × 6.
## What is associative property example?
Associative property of addition: Changing the grouping of addends does not change the sum. For example, ( 2 + 3 ) + 4 = 2 + ( 3 + 4 ) (2 + 3) + 4 = 2 + (3 + 4) (2+3)+4=2+(3+4)left parenthesis, 2, plus, 3, right parenthesis, plus, 4, equals, 2, plus, left parenthesis, 3, plus, 4, right parenthesis.
Is multiplication always associative?
In mathematics, addition and multiplication of real numbers is associative. By contrast, in computer science, the addition and multiplication of floating point numbers is not associative, as rounding errors are introduced when dissimilar-sized values are joined together.
### How do you use associative property?
A. The associative property states that when adding or multiplying, the grouping symbols can be relocated without affecting the result. The formula for addition states (a+b)+c=a+(b+c) and the formula for multiplication states (a×b)×c=a×(b×c).
### What is associative multiplication?
To “associate” means to connect or join with something. According to the associative property of multiplication, the product of three or more numbers remains the same regardless of how the numbers are grouped. Here’s an example of how the product does not change irrespective of how the factors are grouped.
How do you find the associative property of addition and multiplication?
#### What is associative property?
This property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends (or the multiplicands).
#### Is multiplication left or right associative?
For example, subtraction and division, as used in conventional math notation, are inherently left-associative. Addition and multiplication, by contrast, are both left and right associative. (e.g. (a * b) * c = a * (b * c) ).
Why would you use the associative property of multiplication?
The associative property is helpful while adding or multiplying multiple numbers. By grouping, we can create smaller components to solve. It makes the addition or multiplication of multiple numbers easier and faster.
## How to multiply numbers quickly using the associative property?
Associative property
• Commutative property
• Distributive property
• What are some examples of associative property?
In an associative property,3 or more numbers are to be used.
• The numbers that are grouped within a parenthesis will be considered as one unit.
• Associative property works only with addition and multiplication. It is not applicable when we are dealing with subtraction or division.
• ### What are the 5 properties of multiplication?
Properties of multiplication: Closure property: For any two whole numbers a and b,their product ax b is always a whole number. E.g. 12 x 7 = 84, 12, 7 and 84 all are whole numbers. Commutative property: For any two whole numbers a and b, a a x b = b x a Order of multiplication is not important. E.g 11 x 6 = 66 and 6 x 11 = 66
### What does associative property mean in math?
In mathematics, the associative property is a property of some binary operations, which means that rearranging the parentheses in an expression will not change the result. In propositional logic, associativity is a valid rule of replacement for expressions in logical proofs.
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# How do you determine the limit of (2)/(x-3) as x approaches 3^+?
Apr 11, 2016
As $x$ approaches $3$ from the right, the numerator is a positive number and the denominator is also positive.
Numerator $\rightarrow$ a positive
Denominator $\rightarrow$ $0$ through positive values,
So the ratio increases without bound.
We write ${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3} = \infty$.
Although not in widespread use (I think), it may be helpful to write:
${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3}$ has form $\frac{+}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3} = \infty$.
Bonus
${\lim}_{x \rightarrow {3}^{+}} \frac{- 5}{x - 3}$ has form $\frac{-}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{- 5}{x - 3} = - \infty$
${\lim}_{x \rightarrow {3}^{-}} \frac{2}{x - 3}$ has form $\frac{+}{0} ^ -$, which leads to ${\lim}_{x \rightarrow {3}^{-}} \frac{2}{x - 3} = - \infty$
${\lim}_{x \rightarrow {3}^{+}} \frac{x - 1}{x - 3}$ has form $\frac{+}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{x - 1}{x - 3} = \infty$
${\lim}_{x \rightarrow {3}^{+}} \frac{x - 7}{x - 3}$ has form $\frac{-}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{x - 7}{x - 3} = - \infty$
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# How do you find equation of line passing through the point S(-1,-4) and T (3,4)?
Feb 6, 2017
See the entire solution process below:
#### Explanation:
We can use the point-slope formula to find an equation. However, we must first find the slope. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.
Substituting the values from the points in the problem and solving gives:
$m = \frac{\textcolor{red}{4} - \textcolor{b l u e}{- 4}}{\textcolor{red}{3} - \textcolor{b l u e}{- 1}}$
$m = \frac{\textcolor{red}{4} + \textcolor{b l u e}{4}}{\textcolor{red}{3} + \textcolor{b l u e}{1}}$
$m = \frac{8}{4} = 2$
The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
We can substitute the slope we calculated and the values from the first point to give:
$\left(y - \textcolor{red}{- 4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{- 1}\right)$
$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x + \textcolor{red}{1}\right)$
We can also substitute the slope we calculated and the values from the second point to give:
$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{3}\right)$
Or we can solve for $y$ to put this equation in the more familiar slope-intercept form:
$y - \textcolor{red}{4} = \left(\textcolor{b l u e}{2} \times x\right) - \left(\textcolor{b l u e}{2} \times \textcolor{red}{3}\right)$
$y - \textcolor{red}{4} = 2 x - 6$
$y - \textcolor{red}{4} + 4 = 2 x - 6 + 4$
$y - 0 = 2 x - 2$
$y = 2 x - 2$
These equations are solutions to this problem:
$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x + \textcolor{red}{1}\right)$
$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{2} \left(x - \textcolor{red}{3}\right)$
$y = 2 x - 2$
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# Applications of Algebraic Expressions
We will discuss about the applications of algebraic expressions in everyday life.
The most important use of algebraic expression is to solve a word problem by translating the word statement to that statement in the symbols of algebra. Writing an algebraic expression which represents a particular situation is called an algebraic representation.
Algebraic representations of some practical situations are as follows:
Situations Variables Statements using Algebraic Expressions(Algebraic Representations) 1. James has 10 more balls than Robert. Let the no. of balls of Robert has be x. James has (x + 10) balls. 2. Richard is twice as old as Linda. 3. The age of the father of Alexander is 2 years more than 3 times the Alexander 's age. Let Linda's age be x years. Let Alexander's age be x years. Richard's age is 2x years. The age of the father of Alexander is (3x + 2) years. 4. Price of apple per kg is $2 less than price of orange per kg. Let the price of orange per kg be$p. Price of apple per kg is \$(p - 2). 5. How old will Rebecca be 5 years from now? Let y be Rebecca's present age in years. Five years from now, Rebecca will be (y + 5) years old.
A mathematical sentence with an equality sign is called statement of equality.
For example: x + 5 = 9
Statements of equality involving one or more variables is called an equation.
or
An equation is an equality between two algebraic expressions/statements.
The sign of equality in an equation divides it into two sides, namely left hand side (L.H.S) and right hand side (R.H.S). L.H.S and R.H.S of an equation are like the two scales of a balance.
For example:
(i) x + 2(L.H.S) = 3(R.H.S)
(ii) y + 2(L.H.S) = 9(R.H.S)
### Working Rules for Make an Algebraic Expression:
Step I: Take any variables, say x.
Step II: Perform any of the four operations i.e., add, subtract, divide or multiply on that variable x to make an algebraic expression.
Step III: Make the algebraic expression a statement of equality.
Step IV: The result will involve a variable with a statement of equality which is called an equation.
### Solved Examples on Applications of Algebraic Expressions:
1. Write an algebraic expression 6 less than one-fourth of'x'.
Solution:
One-fourth of x = $$\frac{1}{4}$$ x = $$\frac{x}{4}$$
6 less than one-fourth x = $$\frac{x}{4}$$ - 6 which is the required equation.
2. Write an equation for each of the following statements:
(i) The difference between x and the sum of 2 and 3 is 11.
(ii) The sum of $$\frac{4}{5}$$th of x and five times x is 140.
Solution:
(i) The sum of 2 and 3 = 2 + 3.
Now, the difference between x and the sum of 2 and 3 is 11, is given by
x - (2 + 3) = 11
or, x - 5 = 11 which is the required equation.
(ii) $$\frac{4}{5}$$ th of x = $$\frac{4x}{5}$$ and five times x = 5x
Now, the he sum of $$\frac{4}{5}$$th of x and five times x is 140, , is given by
$$\frac{4x}{5}$$ + 5x = 140, which is the required equation
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# How do you graph x-5<=y?
Nov 30, 2017
See a solution process below:
#### Explanation:
First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.
For: $x = 0$
$0 - 5 = y$
$- 5 = y$
$y = - 5$ or $\left(0 , - 5\right)$
For: $x = 5$
$5 - 5 = y$
$0 = y$
$y = 0$ or $\left(5 , 0\right)$
We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.
The boundary line will be solid because the inequality operator contains an "or equal to" clause.
graph{(x^2+(y+5)^2-0.125)((x-5)^2+y^2-0.125)(y-x+5)=0 [-20, 20, -10, 10]}
Now, we can shade the left side of the line.
graph{(y-x+5) >=0 [-20, 20, -10, 10]}
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# Understanding Place Value in the Number System
## How Does Each Digit Hold a Value in Any Given Number?
Your mom is still a mom whether she is in the kitchen, garden, living room, or the basement. But digits like 5 at different places (for example, tens or hundreds of places) means something different. Every digit in a number has a place value in Mathematics. Place value is the value represented by a digit in a number based on its place in the number. In this article we'll help you out to understand it in an easy way.
Place value for Grade 4 is important because it serves as a foundation for grouping, multiple-digit multiplication, and other operations in the decimal system, as well as a starting point for learning about other base systems. By understanding this you'll be able to understand the key differences like Rs. 50 you got for your birthday and the Rs. 500 price tag on the Remote Car for which you're saving for.
In Mathematics, the position or place of a digit in a number is referred to as place value. Each digit has a specific place in a number. The position of each digit will be expanded when we represent the number in general form. Those positions begin with a unit position, often known as one's position. Units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on are the place values of a number's digits in sequence from right to left.
## What Is Place Value?
The value of each digit in a number is determined by its position in that number. We have given a chart below, that is a place value chart which helps in the identification of large numbers. This place value chart is read from left to right. In the Indian system, we begin arranging numbers from right to left in groups of three, and then in groups of two. The place value chart is divided into intervals, which include ones, thousands, lakhs, and crores.
The 5 in 250, for example, indicates 5 tens, or 50; whereas, the 5 in 5,126 denotes 5 thousands, or 5,000. It is important to understand that while a digit can be the same, its value is determined by its position in the number. A place value grid, such as the one shown below, will most likely be used to understand place value in a better way.
Place value chart
## International Place Value Chart
The International place value chart is a place value system that is used in many countries around the world. We use a place value chart to understand the place value of each digit so that we can identify each digit. We start grouping the numbers from right to left in groups of three, called periods, and we place a comma or space after each period to make the number easier to read.
International place value chart
## Place, Place Value and Face Value
A number is made by grouping a few digits together.
• Each digit will have a fixed position called its place.
• Each digit’s value depends on its place, which is known as the place value of the digit.
• The face value of a digit in the given number is just the value of the digit itself.
• The place value formula of a digit can be written as the product of face value of the digit and value of the place.
That is,
Place value = (Face value of the digit) × (Value of the place)
Example:
In the number 874321, write the digit which is in:
(a) hundreds place
(b) hundred thousand place
(c) ten thousand’s place
(d) and One’s place
Sol:
(a) A number in hundreds places is 3.
(b) A number in a hundred thousand places is 8.
(c) A number in ten thousand’s place is 7.
(d) The number in One’s place is 1.
Place values of each digit in the number 874321
## Conclusion
Having a strong understanding of place value is a good step because it gives you the key number knowledge you need to solve calculations like addition, subtraction, multiplication, division, and fractions. It would be difficult to write numbers, identify one more or one less, count forwards and backwards, or compare numbers if you are not thorough about place value. Moreover, understanding place value in Maths has a significant impact on how we think about basic concepts such as money. Understanding place value can help you figure out how much something costs when we wish to buy it.
## FAQs on Understanding Place Value in the Number System
1. What is the difference between the face value and the place value of a digit?
The magnitude of a digit's face value is its natural magnitude. It doesn't matter where the digit is in the number. A digit's place value is determined by its position in the number. The 5 in the number 353, for example, has a face value of 5 and a place value of 50.
2. How will place value help me multiply numbers?
Having a thorough understanding of place value will help young learners in multiplying fractions and decimals. Children will also use their knowledge of place value in digits to multi-digit numbers when using formal written methods of multiplication.
3. Why is place value for Grade 4 important?
When teaching Maths to any kid, the most important concept is place value. It is the foundation of all mathematical concepts from preschool to algebra, and it is required for a thorough understanding of the subject. Students will not be able to progress until they have understood Place Value as a basic concept first.
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## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
Other Exercises
Question 1.
In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
Solution:
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴ ∠AOB = 2∠APB = 2 x 70° = 140°
∠AOB + ∠ACB = 180° (Sum of the angles)
⇒ 140° +∠ACB = 180°
⇒ ∠ACB = 180° – 140° = 40°
∴ ∠ACB = 40°
Question 2.
In the figure, two congruent circles with centre O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
Solution:
Two congruent circles with centres O and O’ intersect at A and B
∠AO’B = 50°
∵ OA = OB = O’A = 04B (Radii of the congruent circles)
Question 3.
In the figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = IT, AC and BD intersect at P. Then, find ∠DPC.
Solution:
∵ ABCD is a cyclic quadrilateral,
∴ ∠BAD + ∠BCD = 180°
⇒ 75° + ∠BCD – 180°
⇒ ∠BCD = 180°-75°= 105° and ∠ADC + ∠ABC = 180°
⇒ 77° + ∠ABC = 180°
⇒ ∠ABC = 180°-77°= 103°
∴ ∠DBC = ∠ABC – ∠ABD = 103° – 58° = 45°
∵ Arc AD subtends ∠ABD and ∠ACD in the same segment of the circle 3
∴ ∠ABD = ∠ACD = 58°
∴ ∠ACB = ∠BCD – ∠ACD = 105° – 58° = 47°
Now in ∆PBC,
Ext. ∠DPC = ∠PBC + ∠PCB
=∠DBC + ∠ACB = 45° + 47° = 92°
Hence ∠DPC = 92°
Question 4.
In the figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.
Solution:
In the figure, ∠AOB = 80°, ∠ABC = 30°
∵ Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = $$\frac { 1 }{ 2 }$$∠AOB = $$\frac { 1 }{ 2 }$$ x 80° = 40°
In ∆OAB, OA = OB
∴ ∠OAB = ∠OBA
But ∠OAB + ∠OBA + ∠AOB = 180°
∴ ∠OAB + ∠OBA + 80° = 180°
⇒ ∠OAB + ∠OAB = 180° – 80° = 100°
∴ 2∠OAB = 100°
⇒ ∠OAB = $$\frac { { 100 }^{ \circ } }{ 2 }$$ = 50°
Similarly, in ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 40° + 30° = 180°
⇒ ∠BAC = 180°-30°-40°
= 180°-70°= 110°
∴ ∠CAO = ∠BAC – ∠OAB
= 110°-50° = 60°
Question 5.
In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
Solution:
In the figure, ABCD is a parallelogram and
CDE is a straight line
∵ ABCD is a ||gm
∴ ∠A = ∠C
and ∠C = ∠ADE (Corresponding angles)
Similarly, ∠ABE = ∠BED (Alternate angles)
∵ arc BD subtends ∠BAD at the centre and
∠BED at the remaining part of the circle
Question 6.
In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
Solution:
In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.
Arc PB subtends ∠POB at the centre and
∠PAB at the remaining part of the circle
∴ ∠POB = 2∠PAB = 2 x 35° = 70°
Now in ∆OP,
OP = OB radii of the circle
∴ ∠OPB = ∠OBP = 70° (∵ ∠OPB + ∠OBP = 140°)
Now ∠APB = 90° (Angle in a semicircle)
∴ ∠BPQ = 90°
and in ∆PQB,
Ext. ∠PBR = ∠BPQ + ∠PQB
= 90° + 25°= 115°
∴ ∠PBR = 115°
Question 7.
In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
Solution:
In the figure, P and Q are the centres of two circles which intersect each other at C and B
ACD is a straight line ∠APB = 150°
Arc AB subtends ∠APB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = $$\frac { 1 }{ 2 }$$ ∠APB = $$\frac { 1 }{ 2 }$$ x 150° = 75°
But ∠ACB + ∠BCD = 180° (Linear pair)
⇒ 75° + ∠BCD = 180°
∠BCD = 180°-75°= 105°
Now arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle
Reflex ∠BQD = 2∠BCD = 2 x 105° = 210°
But ∠BQD + reflex ∠BQD = 360°
∴ ∠BQD+ 210° = 360°
∴ ∠BQD = 360° – 210° = 150°
Question 8.
In the figure, if O is circumcentre of ∆ABC then find the value of ∠OBC + ∠BAC.
Solution:
In the figure, join OC
∵ O is the circumcentre of ∆ABC
∴ OA = OB = OC
∵ ∠CAO = 60° (Proved)
∴ ∆OAC is an equilateral triangle
∴ ∠AOC = 60°
Now, ∠BOC = ∠BOA + ∠AOC
= 80° + 60° = 140°
and in ∆OBC, OB = OC
∠OCB = ∠OBC
But ∠OCB + ∠OBC = 180° – ∠BOC
= 180°- 140° = 40°
⇒ ∠OBC + ∠OBC = 40°
∴ ∠OBC = $$\frac { { 40 }^{ \circ } }{ 2 }$$ = 20°
∠BAC = OAB + ∠OAC = 50° + 60° = 110°
∴ ∠OBC + ∠BAC = 20° + 110° = 130°
Question 9.
In the AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.
Solution:
In the figure, AOC is diameter arc AxB = $$\frac { 1 }{ 2 }$$ arc BYC 1
∠AOB = $$\frac { 1 }{ 2 }$$ ∠BOC
⇒ ∠BOC = 2∠AOB
But ∠AOB + ∠BOC = 180°
⇒ ∠AOB + 2∠AOB = 180°
⇒ 3 ∠AOB = 180°
∴ ∠AOB = $$\frac { { 180 }^{ \circ } }{ 3 }$$ = 60°
∴ ∠BOC = 2 x 60° = 120°
Question 10.
In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
Solution:
In the figure, ABCD is a cyclic quadrilateral
CD is produced to E such that ∠ADE = 95°
O is the centre of the circle
⇒ ∠ADC + 95° = 180°
⇒ ∠ADC = 180°-95° = 85°
Now arc ABC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle
∵ ∠AOC = 2∠ADC = 2 x 85° = 170°
Now in ∆OAC,
∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ ∠OAC = ∠OCA (∵ OA = OC radii of circle)
∴ ∠OAC + ∠OAC + 170° = 180°
2∠OAC = 180°- 170°= 10°
∴ ∠OAC = $$\frac { { 10 }^{ \circ } }{ 2 }$$ = 5°
Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# Spectrum Math Grade 8 Chapter 3 Lesson 2 Answer Key Graphing Linear Equations Using Slope
Students can use the Spectrum Math Grade 8 Answer Key Chapter 3 Lesson 3.2 Graphing Linear Equations Using SlopeĀ as a quick guide to resolve any of their doubts.
## Spectrum Math Grade 8 Chapter 3 Lesson 3.2 Graphing Linear Equations Using Slope Answers Key
If the slope of a line and the place it intercepts (or crosses) the y-axis are known, a line can be graphed using an equation with x and y variables.
Step 1: Find the point where the line crosses the y-axis. (0, 5)
Step 2: Find the slope: -2.
In fraction form, the slope is $$\frac{-2}{1}$$.
Step 3: Starting at the intercept, mark the slope by using the numerator to count along the y-axis, and the denominator to count along the x-axis: Move down 2, and to the right 1.
Step 4: Draw a line to connect the points.
Use the slope-intercept form of equations to draw lines on the grids below.
Question 1.
a.
y = $$\frac{1}{4}$$x + 1
From the given slope-intercept equation, find out the slope and y-intercept
Slope = $$\frac{1}{4}$$
Y-intercept = (0,1)
AnyĀ lineĀ can be graphed using twoĀ points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (4,2)
b. y = -x + 2
From the given slope-intercept equation, find out the slope and y-intercept
Slope = -1
Y-intercept = (0,2)
AnyĀ lineĀ can be graphed using twoĀ points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (1,1)
Question 2.
a.
y = $$\frac{4}{3}$$x + 4
From the given slope-intercept equation, find out the slope and y-intercept
Slope = $$\frac{4}{3}$$
Y-intercept = (0,4)
AnyĀ lineĀ can be graphed using twoĀ points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (3,8)
b. y = 2x + 3
From the given slope-intercept equation, find out the slope and y-intercept
Slope =2
Y-intercept = (0,3)
AnyĀ lineĀ can be graphed using twoĀ points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (1,5)
When a linear equation is graphed, the equation that was used to create the line can be found by using the slope-intercept equation.
Step 1: Find the point where the line crosses the y-axis. (0, 2)
Step 2: Mark points on the line where it crosses at exact locations that correspond to an ordered pair.
(5, 4) and (10, 6)
Step 3: Calculate the slope using
Step 4: Use y = slope ā¢ x + intercept to create the equation for the line in slope-intercept form.
y = $$\frac{2}{5}$$x + 2
Use the pictures below to create equations for the lines in slope-intercept form.
Question 1.
a.
Answer: y = -1x + 7
Step 1: Find the point where the line crosses the y-axis. (0, 7)
Step 2: Mark points on the line where it crosses at exact locations that correspond to an ordered pair.
(2,5) and (7,0)
Step 3: Calculate the slope using $$\frac{change in y}{change in x}$$ = $$\frac{5 – 7}{2-0}$$ = $$\frac{-2}{2}$$ = -1
Step 4: Use y = slope ā¢ x + intercept to create the equation for the line in slope-intercept form.
y = -1x + 7
b.
Answer: y = $$\frac{1}{2}$$x + 7
Step 1: Find the point where the line crosses the y-axis. (0, 4)
Step 2: Mark points on the line where it crosses at exact locations that correspond to an ordered pair.
(2,5) and (10,9)
Step 3: Calculate the slope using $$\frac{change in y}{change in x}$$ = $$\frac{5 – 4}{2-0}$$ = $$\frac{1}{2}$$
Step 4: Use y = slope ā¢ x + intercept to create the equation for the line in slope-intercept form.
y = $$\frac{1}{2}$$x + 7
Question 2.
a.
Answer: y =$$\frac{-3}{2}$$x + 7
Step 1: Find the point where the line crosses the y-axis. (0, 6)
Step 2: Mark points on the line where it crosses at exact locations that correspond to an ordered pair.
(2,3) and (4,0)
Step 3: Calculate the slope using $$\frac{change in y}{change in x}$$ = $$\frac{3-6}{2-0}$$ = $$\frac{-3}{2}$$
Step 4: Use y = slope ā¢ x + intercept to create the equation for the line in slope-intercept form.
y =$$\frac{-3}{2}$$x + 7
b.
Answer: y = $$\frac{2}{3}$$x + 7
Step 1: Find the point where the line crosses the y-axis. (0, 0)
Step 2: Mark points on the line where it crosses at exact locations that correspond to an ordered pair.
(3,2) and (9,6)
Step 3: Calculate the slope using $$\frac{change in y}{change in x}$$ = $$\frac{2-0}{3-0}$$ = $$\frac{2}{3}$$
Step 4: Use y = slope ā¢ x + intercept to create the equation for the line in slope-intercept form.
y = $$\frac{2}{3}$$x + 7
When given the slope and intercept of any straight line, a linear equation can be created using the slope-intercept form.
Slope: $$\frac{5}{6}$$; Intercept: -2
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = $$\frac{5}{6}$$x – 2
Use slope-intercept form to write equations given the conditions below.
Question 1.
a. slope: $$\frac{4}{3}$$; intercept: 3
Answer: y = $$\frac{4}{3}$$x +3
slope: $$\frac{4}{3}$$; intercept: 3
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = $$\frac{4}{3}$$x +3
b. slope: -2; intercept: 4
slope: -2; intercept: 4
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = -2x +4
c. slope: –$$\frac{1}{2}$$; intercept: 7
Answer: y = –$$\frac{1}{2}$$x + 7
slope: –$$\frac{1}{2}$$; intercept: 7
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = –$$\frac{1}{2}$$x + 7
Question 2.
a. slope: 3; intercept: -5
slope: 3; intercept: -5
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y =3x -5
b. slope: $$\frac{2}{5}$$; intercept: 0
Answer: y =$$\frac{2}{5}$$x
slope: $$\frac{2}{5}$$; intercept: 0
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y =$$\frac{2}{5}$$x +0
y =$$\frac{2}{5}$$x
c. slope: –$$\frac{3}{4}$$; intercept: -2
Answer: y = –$$\frac{3}{4}$$x -2
slope: –$$\frac{3}{4}$$; intercept: -2
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = –$$\frac{3}{4}$$x -2
Question 3.
a. slope: -4; intercept: 6
slope: -4; intercept: 6
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = -4x+6
b. slope: $$\frac{5}{2}$$; intercept: -3
Answer:Ā y = $$\frac{5}{2}$$x-3
slope: $$\frac{5}{2}$$; intercept: -3
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = $$\frac{5}{2}$$x-3
c. slope: $$\frac{1}{2}$$; intercept: 1
Answer: y = $$\frac{1}{2}$$x+1
slope: $$\frac{1}{2}$$; intercept: 1
Step 1: Use the equation y = mx + b, where m equals slope and b is the y-intercept.
Step 2: Substitute the known quantities for the slope and the y-intercept.
y = $$\frac{1}{2}$$x+1
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# Cylinder
A cylinder is one of the basic shapes in 3D geometry. It has two parallel circular bases set at a distance from each other. The two circular bases are joined by a curved surface at a fixed distance from the center.
The circles and their interiors are called the bases. The radius of the cylinder is the radius of the base. The perpendicular segment from the plane of one base to the plane of the other base is called the altitude of the cylinder. The height of the cylinder is the length of the altitude.
The axis of a cylinder is the segment that contains the centers of the two bases. When the axis is perpendicular to the planes of the two bases, you have a right cylinder. If not, you have an oblique cylinder.
## Properties of a cylinder
Each shape has properties that differentiate it from other shapes. Cylinders have the following characteristics:
• The bases are congruent and parallel.
• If the axis forms a right angle with the bases, which are exactly over each other, it is called a "right cylinder".
• It is similar to the prism because all cross sections perpendicular to its altitude are identical.
• If the bases are not directly over each other but still parallel, the axis does not produce a right angle to the bases and it is called an "oblique cylinder".
• If the bases are circular in shape, it is called a "circular cylinder".
• If the bases are an elliptical shape, it is called an "elliptical cylinder".
## Measuring the surface area of a cylinder
In order to find the surface area of a cylinder, we need to consider its 3 outward faces, the two bases, and the tube that connects them.
Since the base of a cylinder is a circle, its area is πr2 where r is the radius of the base of the cylinder.
We also need to consider the tube part, which we call the lateral surface area. Note that you can unroll the tube into a rectangle with height h, and base $2\pi \left(r\right)$ (the circumference of the circular base). This gives us an area of $\left(\mathrm{LSA}\right)=2\pi \left(r\right)\left(h\right)$ .
This means that the total area is the two circles plus the tube, or:
${\mathrm{SA}}_{c}=2\pi {r}^{2}+2\pi rh$
Example 1:
Find the lateral surface area of a cylinder with a base radius of 3 inches and a height of 9 inches.
$\mathrm{LSA}=2\pi \left(3\right)\left(9\right)$
$=54\pi {\mathrm{in}}^{2}$
Approximately $169.56{\mathrm{in}}^{2}$
Example 2:
Find the total surface area of a cylinder with a base radius of 5 inches and a height of 7 inches.
$\mathrm{SA}{C}_{}=2\pi \left(5\right)\left(7\right)+2\pi {\left(5\right)}^{2}$
$=70\pi +50\pi$
$=120\pi {\mathrm{in}}^{2}$
Approximately $376.8{\mathrm{in}}^{2}$
## Finding the volume of a cylinder
If we wanted to find the volume of a cylinder, we must multiply the area of the base by the height of the cylinder. The formula is:
$V=\pi {r}^{2}h$
Example 3
Find the volume of a cylinder with a radius of 3 cm and a height of 10 cm.
$V=\pi {\left(3\right)}^{2}\left(10\right)$
$V=90\pi {\mathrm{cm}}^{3}$
Approximately $282.6{\mathrm{cm}}^{3}$
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# How do you solve -3|p|=12?
May 20, 2018
Dividing by $- 3$ we get $| p | = - 4$ since $| p | \ge 0$ so we have no solutions.
#### Explanation:
We need that $| x | = x$ if $x \ge 0$ and $| x | = - x$ if $x < 0$
May 20, 2018
There is a problem with this!
See the alternative
#### Explanation:
The $| p |$ always ends up positive. No matter what value $p$ takes. Thus we have $\left(- 3\right)$ multiplied by some positive value. The consequence of this is that the left side will always be negative.
We can not have: $\text{NEGATIVE = POSITIVE}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{ HOWEVER }}$
If the question was meant to be:
$- 3 | p | = - 12$ it will work
Divide both sides by (-3)
$| p | = \frac{- 12}{- 3} = + 4$
$| \pm 4 | = 4$
$p = \pm 4$
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# Probability of getting one of colorful marbles
Ex1. Ajar contains 3 red marbles, 7 green marbles and 10 white marbles. If a marble is drawn from the jar at random, what is the probability that this marble is white?
Solution:
Ex2. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) white? (ii) blue? (iii) red?
Solutions:
Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the number of possible outcomes =3+2+4=9 (Why?)
Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’ and R denote the event ‘marble is red’.
(i) The number of outcomes favorable to the event n(W)=2
So, P(W)=2/9
Similarly, (ii) P(B)=3/9=⅓ and (iii) P(R)=4/9
Note that P(W)+P(B)+P(R)=1.
Ex3. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Solutions:
Total number of marbles=5+8+4=17
(i) Number of red marbles=5
Probability of getting a red marble =5/17
(ii) Number of white marbles=8
Probability of getting a white marble=8/17
(iii) Number of green marbles=4
Probability of getting a green marble=4/17
Probability of getting a non green marble=1-4/17=13/17
It has got equal meaning of probability of getting either a green marble or a white marble.
Ex4. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is ⅔. Find the number of blue balls in the jar.
solution:
Total number of marbles =24
Let the total number of green marbles be x.
Then, total number of blue marbles =24-x
P(getting a given marble)=x/24
According to the condition given in the question,
x/24=⅔
Therefore, total number of green marbles in the jar =16. Hence, total number of blue marbles =24-x=24-16=8.
Ex5. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is ¼. The probability of selecting a white marble at random from the same jar is ⅓. If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?
Solution:
Let the number of green marbles =x
The number of white marbles =y
Number of yellow marbles =10
Total number of possible outcomes =x+y+10 (total number of marbles)
Total number of marbles in jar =x+y+10=6+8+10=24
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# 2020 AMC 12B Problems/Problem 10
## Problem
In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$
## Solution 1 (Angle Chasing/Trig)
Let $O$ be the center of the circle and the point of tangency between $\omega$ and $\overline{AD}$ be represented by $K$. We know that $\overline{AK} = \overline{KD} = \overline{DM} = \frac{1}{2}$. Consider the right triangle $\bigtriangleup ADM$. Let $\measuredangle AMD = \theta$.
Since $\omega$ is tangent to $\overline{DC}$ at $M$, $\measuredangle PMO = 90 - \theta$. Now, consider $\bigtriangleup POM$. This triangle is iscoceles because $\overline{PO}$ and $\overline{OM}$ are both radii of $\omega$. Therefore, $\measuredangle POM = 180 - 2(90 - \theta) = 2\theta$.
We can now use Law of Cosines on $\angle{POM}$ to find the length of ${PM}$ and subtract it from the length of ${AM}$ to find ${AP}$. Since $\cos{\theta} = \frac{1}{\sqrt{5}}$ and $\sin{\theta} = \frac{2}{\sqrt{5}}$, the double angle formula tells us that $\cos{2\theta} = -\frac{3}{5}$. We have $$PM^2 = \frac{1}{2} - \frac{1}{2}\cos{2\theta} \implies PM = \frac{2\sqrt{5}}{5}$$ By Pythagorean theorem, we find that $AM = \frac{\sqrt{5}}{2} \implies \boxed{\textbf{(B) } \frac{\sqrt5}{10}}$
~awesome1st
## Solution 2(Coordinate Bash)
Place circle $\omega$ in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for $\omega$ as $x^2+y^2=\frac{1}{4}$, because it is not translated and the radius is $\frac{1}{2}$.
We have $A=\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $M=\left(0, -\frac{1}{2}\right)$. The slope of the line passing through these two points is $\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2$, and the $y$-intercept is simply $M$. This gives us the line passing through both points as $y=-2x-\frac{1}{2}$.
We substitute this into the equation for the circle to get $x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}$, or $x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}$. Simplifying gives $x(5x+2)=0$. The roots of this quadratic are $x=0$ and $x=-\frac{2}{5}$, but if $x=0$ we get point $M$, so we only want $x=-\frac{2}{5}$.
We plug this back into the linear equation to find $y=\frac{3}{10}$, and so $P=\left(-\frac{2}{5}, \frac{3}{10}\right)$. Finally, we use distance formula on $A$ and $P$ to get $AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}$.
## Solution 3(Power of a Point)
Let circle $\omega$ intersect $\overline{AB}$ at point $N$. By Power of a Point, we have $AN^2=AP\cdot AM$. We know $AN=\frac{1}{2}$ because $N$ is the midpoint of $\overline{AB}$, and we can easily find $AM$ by the Pythagorean Theorem, which gives us $AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}$. Our equation is now $\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}$, or $AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}$, thus our answer is $\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.$
## Solution 4
Take $O$ as the center and draw segment $ON$ perpendicular to $AM$, $ON\cap AM=N$, link $OM$. Then we have $OM\parallel AD$. So $\angle DAM=\angle OMA$. Since $AD=2AM=2OM=1$, we have $\cos\angle DAM=\cos\angle OMP=\frac{2}{\sqrt{5}}$. As a result, $NM=OM\cos\angle OMP=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$ Thus $PM=2NM=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$. Since $AM=\frac{\sqrt{5}}{2}$, we have $AP=AM-PM=\frac{\sqrt{5}}{10}$. Put $\boxed{B}$.
~FANYUCHEN20020715
## Video Solution
~IceMatrix
2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
|
### All in the Mind
Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface of the water make around the cube?
### Painting Cubes
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Tic Tac Toe
In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells?
# Noughts and Crosses
##### Stage: 3 Challenge Level:
This is a good opportunity to explore aspects of generality in three dimensions and for pupils to discuss images and possible solutions, and find convincing arguments for the unique solution.
For some this may be a hard problem to visualise, which is not always made easier by diagrams.
Suggestion for introducing the problem:
Invite the class to imagine a 3 x 3 square grid.
How many small squares are there?
How are they arranged?
Ask the group to draw a 3 x 3 grid.
Now, ask them, in their mind's eye, to colour one of their nine squares and ask for a volunteer to describe to the rest of the class where it is. When the person has described their square ask everyone to fill in that square on their grid. Repeat this activity several times with the aim of identifying some notation that fully describes the position of squares on the grid.
Repeat the activity but this time describe lines of three squares, like the winning lines in "Noughts and Crosses".
How many lines are there?
Suggestions for the main part of the lesson:
Ask the group to imagine that they have a 3 x 3 x 3 cube made up from 27 unit cubes.
Can they devise a notation for describing positions of little cubes and lines of cubes within this cube (as if they were playing 3D Noughts and Crosses)?
Tell the group that a marble is placed in the unit cube found at middle-middle-top. Another is placed at middle-middle-middle. Where should the third marble be placed to make a winning line of three marbles? Try some other examples with the group.
Ask students to come up with examples of their own.
The interactivity can be used to give feedback to students.
Follow-up: How about trying to play Noughts and Crosses without paper?
|
RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A)
# RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A) | Mathematics (Maths) Class 10 PDF Download
``` Page 1
Exercise – 3A
1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
?? …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ?? …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.
Page 2
Exercise – 3A
1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
?? …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ?? …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.
The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.
2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
?? …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, plot the points A(0, 2), B(2, -1) and C(-2, 5) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + 2y = 4.
Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ?? …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1
Page 3
Exercise – 3A
1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
?? …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ?? …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.
The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.
2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
?? …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, plot the points A(0, 2), B(2, -1) and C(-2, 5) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + 2y = 4.
Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ?? …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1
Thus, we have the following table for the equation 2x – 3y = 7.
x 2 -1 5
y -1 -3 1
Now, plot the points P(-1, -3) and Q(5, 1). The point C(2, -1) has already been plotted. Join
PB and QB and extend it on both ways.
Thus, line PQ is the graph of 2x – 3y = 7.
The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.
3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
?? …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Page 4
Exercise – 3A
1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
?? …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ?? …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.
The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.
2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
?? …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, plot the points A(0, 2), B(2, -1) and C(-2, 5) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + 2y = 4.
Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ?? …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1
Thus, we have the following table for the equation 2x – 3y = 7.
x 2 -1 5
y -1 -3 1
Now, plot the points P(-1, -3) and Q(5, 1). The point C(2, -1) has already been plotted. Join
PB and QB and extend it on both ways.
Thus, line PQ is the graph of 2x – 3y = 7.
The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.
3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
?? …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Graph of x - 2y + 3 = 0
x – 2y + 3 = 0
? 2y = (x + 3)
? y =
?? + ?? ?? …(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = -3, we get y = 0.
Thus, we have the following table for the equation x – 2y + 3 = 0.
x 1 3 -3
y 2 3 0
Now, plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join
AP and QA and extend it on both ways.
Thus, PQ is the graph of x – 2y + 3 = 0.
The two graph lines intersect at A (1, 2).
? x = 1 and y = 2.
4. Solve the system of equations graphically:
2x - 5y + 4 = 0,
2x + y - 8 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x - 5y + 4 = 0
2x – 5y + 4 = 0
?5y = (2x + 4)
?y =
???? + ?? ?? …(i)
Page 5
Exercise – 3A
1. Solve the system of equations graphically:
2x + 3y = 2,
x – 2y = 8
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 2x + 3y = 2
2x + 3y = 2
?3y = (2 – 2x)
?3y = 2(1 – x)
?y =
?? (?? -?? )
?? …(i)
Putting x = 1, we get y = 0
Putting x = -2, we get y = 2
Putting x = 4, we get y = -2
Thus, we have the following table for the equation 2x + 3y = 2
x 1 -2 4
y 0 2 -2
Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.
Graph of x - 2y = 8
x – 2y = 8
? 2y = (x – 8)
? y =
?? -?? ?? …(ii)
Putting x = 2, we get y = -3
Putting x = 4, we get y = -2
Putting x = 0, we get y = -4
Thus, we have the following table for the equation x – 2y = 8.
x 2 4 0
y -3 -2 -
4
Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. Join
PQ and QC and extend it on both ways.
Thus, line PC is the graph of x – 2y = 8.
The two graph lines intersect at C(4, -2).
? x = 4 and y = -2 are the solutions of the given system of equations.
2. Solve the system of equations graphically:
3x + 2y = 4,
2x – 3y = 7
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-
axis and y-axis, respectively.
Graph of 3x + 2y = 4
3x + 2y = 4
?2y = (4 – 3x)
?y =
?? - ???? )
?? …(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = -1
Putting x = -2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4
x 0 2 -2
y 2 -1 5
Now, plot the points A(0, 2), B(2, -1) and C(-2, 5) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + 2y = 4.
Graph of 2x - 3y = 7
2x – 3y = 7
? 3y = (2x – 7)
? y =
???? - ?? ?? …(ii)
Putting x = 2, we get y = -1
Putting x = -1, we get y = -3
Putting x = 5, we get y = 1
Thus, we have the following table for the equation 2x – 3y = 7.
x 2 -1 5
y -1 -3 1
Now, plot the points P(-1, -3) and Q(5, 1). The point C(2, -1) has already been plotted. Join
PB and QB and extend it on both ways.
Thus, line PQ is the graph of 2x – 3y = 7.
The two graph lines intersect at B(2, -1).
?x = 2 and y = -1 are the solutions of the given system of equations.
3. Solve the system of equations graphically:
2x + 3y = 8,
x – 2y + 3 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x + 3y = 8
2x + 3y = 8
?3y = (8 – 2x)
?y =
?? - ????
?? …(i)
Putting x = 1, we get y = 2.
Putting x = -5, we get y = 6.
Putting x = 7, we get y = -2.
Thus, we have the following table for the equation 2x + 3y = 8.
x 1 -5 7
y 2 6 -2
Now, plot the points A(1, 2), B(5, -6) and C(7, -2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.
Graph of x - 2y + 3 = 0
x – 2y + 3 = 0
? 2y = (x + 3)
? y =
?? + ?? ?? …(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = -3, we get y = 0.
Thus, we have the following table for the equation x – 2y + 3 = 0.
x 1 3 -3
y 2 3 0
Now, plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join
AP and QA and extend it on both ways.
Thus, PQ is the graph of x – 2y + 3 = 0.
The two graph lines intersect at A (1, 2).
? x = 1 and y = 2.
4. Solve the system of equations graphically:
2x - 5y + 4 = 0,
2x + y - 8 = 0
Sol:
On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and
y-axis, respectively.
Graph of 2x - 5y + 4 = 0
2x – 5y + 4 = 0
?5y = (2x + 4)
?y =
???? + ?? ?? …(i)
Putting x = -2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x - 5y + 4 = 0.
x -2 3 8
y 0 2 4
Now, plot the points A (-2, 0), B (3, 2) and C(8, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x - 5y + 4 = 0.
Graph of 2x + y - 8 = 0
2x + y - 8 = 0
? y = (8 – 2x) …(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y - 8 = 0.
x 1 3 2
y 6 2 4
Now, plot the points P (1, 6) and Q (2, 4). The point B (3, 2) has already been plotted. Join
PQ and QB and extend it on both ways.
Thus, PB is the graph of 2x + y - 8 = 0.
The two graph lines intersect at B (3, 2).
?x = 3 and y = 2
5. Solve the system of equations graphically:
3x + 2y = 12,
5x – 2y = 4
Sol:
The given equations are:
3x + 2y = 12 …..(i)
```
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3A) - Mathematics (Maths) Class 10
1. How do you solve a linear equation in two variables?
Ans. To solve a linear equation in two variables, you need to isolate one variable in terms of the other variable and substitute it back into the equation to find the value of the remaining variable. This can be done using various methods such as substitution method, elimination method, or graphical method.
2. What is the importance of solving linear equations in two variables?
Ans. Solving linear equations in two variables is important as it helps in finding the relationship between two variables and determining their values. This is particularly useful in real-life situations where multiple variables are involved, such as in economics, physics, or engineering problems.
3. Can a linear equation in two variables have more than one solution?
Ans. Yes, a linear equation in two variables can have more than one solution. This happens when the equation represents a line, and the line intersects with another line at more than one point. In such cases, the variables can take on different values that satisfy the equation.
4. What is the difference between consistent and inconsistent linear equations in two variables?
Ans. In a consistent linear equation in two variables, the equation has at least one solution, and the lines represented by the equation intersect at a point. On the other hand, in an inconsistent linear equation, the lines represented by the equation are parallel and do not intersect, resulting in no solution.
5. How can linear equations in two variables be represented graphically?
Ans. Linear equations in two variables can be represented graphically by plotting the points that satisfy the equation on a coordinate plane. Each point represents a solution to the equation, and when these points are connected, they form a straight line. The slope-intercept form (y = mx + b) or the general form (Ax + By = C) of the equation can be used to determine the slope and y-intercept of the line, which further helps in graphing the equation.
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
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# At a Glance - Equations Involving Rational Expressions
Solving an equation that contains rational expressions is similar to solving any old equation that has fractions in it. In fact, if the variable only occurs in numerators, we actually are solving a equation with fractions in it. Our apologies to those of you who were psyched for something completely new and challenging. You'll need to get your kicks elsewhere.
Remember that, with fractions, the denominator tells us the size of the pieces, and the numerator tells us how many pieces we have. In order for two fractions with the same denominator to be equal, the numerators must also be the same.
Two pieces of size eight and three pieces of size eight will never be the same amount. For two pans of brownies to have the same amount of brownie goodness in them when the pieces are cut the same size, each pan must also have the same number of brownies. PS, in case you now absolutely must make a pan of brownies, we like this recipe.
### Sample Problem
Solve the equation .
Since the denominators are equal, the numerators must be equal, too. Ever since the Variable Rights Act, anyway. The only value of x that will work is 3.
### Sample Problem
Solve the equation .
There are two ways to do this.
Way 1: Eliminate the denominators. If you've got a bottle of Denominator-Off, that would work best. If not, have no fear. We can still make this happen.
If we multiply each side of the equation by 5, we find .
If we then multiply each side of the equation by 6, we get 18 = 5x, which means .
Way 2: Put the fractions over a common denominator so that we can compare the numerators, in the same way that it's easier to compare the size of two hot dogs when they're sitting on the same plate in front of you. If you've got one dog in your hand and the other at the opposite end of a football field, it'll be harder to tell the size difference. By the way, it seems to us you're misusing that football field.
The LCD of the two fractions is 5 × 6 = 30, so we can rewrite the left side of the equation like this:
And we can rewrite the right side of the equation as:
Now we solve the equation , which is equivalent to the original equation. Since the denominators are now the same, we need the numerators to be the same. We're in a "same" kind of mood right now.
We must have 18 = 5x, so divide both sides by 5 to finish up:
Both ways of solving the equation led us to the exact same answer. When asked to solve an equation involving rational expressions, you can use either method, whichever one floats your canoe:
1. Eliminate the denominators, or
2. Write the expression on each side of the equation as a fraction (where the fractions have the same denominator).
Now it's time to take off the gloves. They were scrunching our fingers anyway. We'll allow the variables into the denominators this time as well. Sorry, denominators. Your club is exclusive no more. You need to keep up with the times.
When there are variables in denominators, the first thing to do is figure out which values of the variables make the denominators equal to zero. These values can't be solutions to the equation; we can't even evaluate both sides of the equation at these values. Unfortunately, we may need to factor to find the "bad" values of the variable. Then we'll give them an after-school detention and send a note home to their parents.
### Sample Problem
The value x = 2 can't possibly be a solution to the equation , since the left-hand side of the equation isn't even defined at x = 2. It's true; we checked dictionary.com and it wasn't there.
Similarly, we can factor the denominator on the right-hand side to see which values won't work over there:
x2 + 2x + 1 = (x + 1)(x + 1)
We see that x = -1 can't be a solution because the right-hand side of the equation isn't defined at that point.
After finding the "bad" values (naughty, naughty values), we can solve an equation with rational expressions using either of the two ways mentioned above. If you're having trouble deciding, flip a coin. If you're having trouble deciding which coin to use, roll a die. If you're still having trouble, we won't be able to help you. Those are the only ways we know to decide things.
### Sample Problem
Solve the equation .
First we check for "bad" values: x = 1 isn't allowed in this equation, since that would make more than one of our denominators zero. We already don't like it when one denominator equals zero. Can you imagine our wrath if two of them had that problem? You wouldn't like us when we're angry.
Now on to solving. We know you're raring to go. While we could add the expressions on the left-hand side of the equation and then put both sides of the equation over a common denominator, that sounds like too much work. For an easier way out, we'll stick with Way 1. To eliminate denominators, we multiply both sides of the equation by (1 – x). This gives us:
x – 1(1 – x) = 1
And that simplifies to x – 1 + x = 1, or 2x = 2.
This looks like x = 1 should be a solution to the equation. Alas, we found that x = 1 was a "bad" value. It can't be a solution after all, so the poor equation has no solutions. Maybe we can take up a collection for it. Sadface.
The value x = 1 in the example above was an extraneous solution, or a value that pretends to be a solution but can't actually be one because it's a "bad" value. Unlike a regular old "bad" value that's up to no good, an extraneous solution pretends to be your friend at the same time, which is even worse. Can you say "betrayal"?
Be careful: Check for extraneous solutions. Look for "bad" values before you solve a rational equation. After you solve the rational equation, only keep those solutions that aren't "bad" values. Keep them in a safe place, where the "bad" values can't get to them and turn them to the Dark Side.
#### Example 1
Solve the equation .
#### Example 2
Solve the equation .
#### Exercise 1
Solve the following equation or state that the equation has no solutions, remembering to check for extraneous solutions: .
#### Exercise 2
Solve the following equation or state that the equation has no solutions, remembering to check for extraneous solutions: .
#### Exercise 3
Solve the following equation or state that the equation has no solutions, remembering to check for extraneous solutions: .
#### Exercise 5
Solve the following equation or state that the equation has no solutions, remembering to check for extraneous solutions: .
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Students can Download Chapter 12 Algebraic Expressions Ex 12.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4
Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits framed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.
Solution:
i) The number of segments required to form
5 digits of this kind
5n + 1 = 5 × 5 + 1 = 25 + 1 = 26
ii) The number of segments required to form 10 digits of this kind
5n + 1 = 5 × 10 + 1 = 50 + 1 = 51
iii) The number of segments required to form 100 digits of this kind
5n + 1 = 5 × 100 + 1 = 501
Let the number of digits formed be ‘n’.
Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 3n + 1.
i) The number of segments required to form 5 digits of this kind.
3n + 1 = 3 × 5 + 1 = 15 + 1 = 16
ii) The number of segments required to form 10 digits of this kind
3n + 1 = 3 × 10 + 1 = 30 + 1 = 31
iii) The number of segments required to form 100 digits of this kind
3n + 1 = 3 × 100 + 1 = 300 + 1 = 301
Let the number of digits formed be ‘n’. Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 5n + 2.
i) The number of segments required to form 5 digits of this kind.
5n + 2 = 5 × 5 + 2 = 25 + 2 = 27
ii) The number of segments required to form 10 digits of this kind
5n + 2 = 5 × 10 + 2 = 50 + 2 = 52
iii) The number of segments required to form 100 digits of this kind
5n + 2 = 5 × 100 + 2 = 500 + 2 = 502
Question 2.
Use the given algebraic expression to complete the table of number patterns.
Solution:
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Simplify negative square and cube roots
Lesson
Finding the square root of a number is the inverse (or opposite) operation of squaring a number.
Remember squaring a number means multiplying a number by itself (e.g. $3^2=3\times3$32=3×3). The square root of a number is the number that you need to multiply by itself to get the original number. What? Let's look at an example:
Example
question 1
Evaluate: $\sqrt{25}$25
Think: $5^2=25$52=25
Do: $\sqrt{25}=5$25=5
Square Root of a Negative Number
So what happens when we bring in negative numbers?
Firstly, let me point out at this stage, you will NOT have to find the square root of a negative number. Why?
Remember when we were squaring numbers- multiplying two negative numbers always gives a positive answer. So we never get a negative answer. The answer to a negative square root is called an imaginary number.
Let's look at questions where we can include a negative symbol.
Examples
Question 2
Evaluate: $-\sqrt{64}$64
Think: $8^2=64$82=64
Do: $-\sqrt{64}=-8$64=8
Cubed Root of a Negative Number
We can find the cubed root of a negative number (because when we cube a negative number, we get a negative answer).
Examples
Question 3
Evaluate: $\sqrt[3]{-8}$38
Think: $\left(-2\right)^3=-8$(2)3=8
Do: $\sqrt[3]{-8}=-2$38=2
Here's another one:
question 4
Evaluate: $-\sqrt[3]{-216}$3216
Think: $6^3=216$63=216
Do:
$-\sqrt[3]{-216}$−3√−216 $=$= $-1\times\left(-6\right)$−1×(−6) $=$= $6$6
Questions Using Roots
Building on the concepts that you have learnt already, including how to add and subtract integers, how to multiply and divide integers, as well as order of operations, you can do more complex questions using negative square and cubed roots.
Examples
Evaluate: $\sqrt[3]{-64}\times\sqrt{64}$364×64
Think: The cube root of $-64$64 is $-4$4 and the square root of $64$64 is $8$8.
Do:
$\sqrt[3]{-64}\times\sqrt{64}$3√−64×√64 $=$= $-4\times8$−4×8 $=$= $-32$−32
Here's another one
Evaluate: $\left(-\sqrt{85+15}\right)\times\left(-\sqrt[3]{-125}\right)$(85+15)×(3125)
Think: The square root of $100$100 is $10$10. The cubed root of $-125$125 is $-5$5.
Do:
$\left(-\sqrt{85+15}\right)\times\left(-\sqrt[3]{-125}\right)$(−√85+15)×(−3√−125) $=$= $\left(-\sqrt{100}\right)\times\left(-1\right)\times\left(-5\right)$(−√100)×(−1)×(−5) $=$= $-10\times5$−10×5 $=$= $-50$−50
Worked Example Videos
Question 1
Evaluate $\sqrt[3]{-64}$364
Question 2
Evaluate $\sqrt[3]{-125}\times\sqrt[3]{27}$3125×327
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## 39) Missing Number Equations
Bonds of 6, 7, 8, 9: Tic-Tac-Toe
### Mathematics
Develop the concept of using part-part-whole, to calculate an unknown part or whole, in addition and subtraction equations, for a specified whole (6, 7,8 or 9).ย
This is a complex concept that has been highly scaffolded in previous Missing Number Activities.ย
• First, studentsย need to identify if the unknown number could be in the position of a part or a whole.
• Unknown parts can be calculated by either adding on or subtracting. For example, 6-?=2 can be solved by
(i) adding on to the known part to reach the whole, 2+?=6 or
(ii) making a subtraction equation where the unknown part is in the answer position, 6-2=?
• Unknown wholes are calculated by adding the known parts.ย
The unknown numbers in Activity 39 are placed in random order, within each row. In previous Missing Number Core Activities, related bonds were grouped together in rows and the unknown number was always a part. Activity 39 is more difficult because (i) it introducesย concept of an unknown whole in a subtraction equation and, (ii)ย within each row the unknown numbers are in random order.
• Row One contains addition equations with an unknown part.
• Row Two contains subtraction equations with an unknown whole.
• Row Three contains subtraction equations with an unknown part (that is not in the answer position).
Please,ย clickย thisย linkย to read Teacher Notes for information about “Solving Missing Number Equations”ย using part-part-whole.
### Language
• subtraction as โsubtractโ. Read aloud as โ(Whole) subtract (part) equals (part)โ.
• equals
• part-part-whole
• missing part solved by thinking, “What joins with (number spun) to build 20?”
• missing whole solved by thinking, “Look for a subtraction equation what has a missing whole.”
• The activity board uses the word “missing”. The word “unknown” is introduced when the Missing Number Sorting Cards are used.
## Differentiation
### A little easier
##### Card sort
Click this link to download the Missing Number Equations Card Sort for the relevant whole (6, 7, 8 or 9) from the top of this page.ย Each card is one of the equations from the Missing Number Equations game board. Instruct students to:ย
• Cut out each card.
• Mix the cards.
• Sort the cards into three groups using the diagram heading cards:
• addition: one part unknown
• subtraction: one part unknown
• subtraction: whole unknown
After sorting the cards. Instruct students to:
• Build aย two-part bond wall for the relevant whole. Use two of each block from 1, to the whole.ย
• Start with the addition cards. Select one row from the bond wall. Search the addition cards for related equations. Fill in the unknown part and place these cards next to the related Bond Blocks. Repeat for each row until all the missing number addition cards have been filled in and placed next to the related row of Bond Blocks.
• Repeat for the subtraction cards with an unknown part.
• Repeat for the subtraction cards with an unknown whole.
• Discuss strategies for calculating the unknown, including the importance of first identifying if it is a missing part or whole.ย
##### Build a wall
• Build a two-part bond wall for the relevant whole. Use one of each block from 1 to the whole. For the whole of 6 use one additional 3 block to build double 3. For the whole of 8 use one additional 4 block for double 4. Place this next to the activity board.ย
• Place the activity board in a write and wipe sleeve. Circle all the subtraction equations where the missing number is in the whole position.ย
• Use both of these scaffolds to either play or solve each missing number equation. Record this missing number on the write and wipe sleeve using a dry erase marker.
##### Part-part-whole desk visual
• Click to download the Part-Part-Whole Desk Visual for the relevant whole.ย
• Place this next to the activity board.
• Flick the spinner then use the part-part-whole diagram to identify what to look for on the activity board.ย
For example, spinning 4ย ย
For example, spinning zero
For example, spinning 6
### Progression
In the next activity students solveย word problems, across of range of different types, for the wholes of 7, 8, 9 and 10.The unknowns have been placed in a variety of of positions. Go to
##### Activity 40
Bonds of 6, 7, 8, 9: Word Problems, Wholes to 10
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# Difference between revisions of "008A Sample Final A, Question 12"
Question: Find and simplify the difference quotient ${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$ for f(x) = ${\displaystyle {\frac {2}{3x+1}}}$
Foundations:
1) f(x + h) = ?
2) How do you eliminate the 'h' in the denominator?
1)${\displaystyle f(x+h)={\frac {2}{3(x+h)+1}}}$.
2) The numerator of the difference quotient is ${\displaystyle {\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}}$ so the first step is to simplify this expression. This then allows us to eliminate the 'h' in the denominator.
Solution:
Step 1:
The difference quotient that we want to simplify is ${\displaystyle {\frac {f(x+h)-f(x)}{h}}=\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h}$
Step 2:
Now we simplify the numerator:
${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {f(x+h)-f(x)}{h}}&=&\displaystyle {\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h}\\&&\\&=&\displaystyle {\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}\end{array}}}$
Step 3:
Now we simplify the numerator:
${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}&=&\displaystyle {\frac {6x+2-6x-6h-2}{h(3(x+h)+1)(3x+1))}}\\&&\\&=&\displaystyle {\frac {-6}{(3(x+h)+1)(3x+1))}}\end{array}}}$
${\displaystyle {\frac {-6}{(3(x+h)+1)(3x+1))}}}$
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Applications of Derivatives Class 12 Notes Maths Chapter 6 - CBSE
Rate Of Change Of Bodies
If a quantity y varies with another quantity x, satisfying some rule y = f(x), then dy/dx represents the rate of change of y w.r.t x and dy/dx x=x0 represents the rate of change of y w.r.t x at x=x0.
Increasing/decreasing Functions
A function f is said to be :
• Increasing on an interval (a,b) if x1 < x2 in (a,b) ⇒ f (x1) ≤ f (x2) for all x1, x2 (a,b).
• Decreasing on an interval (a,b) if x1 < x2 in (a,b) ⇒ f (x1) ≥ f (x2) for all x1, x2 (a,b).
Maxima And Minima
First Derivative Test
Let f be a function defined on an open interval I. Let f be continuous at a critical point a in I. Then
• If f '(x) > 0 at every point sufficiently close to and to the left of a, and f '(x) < 0 at every point sufficiently close to and to the right of a, then a is a point of local maxima.
• If f '(x) < 0 at every point sufficiently close to and to the left of a, and f '(x) > 0 at every point sufficiently close to and to the right of a, then a is a point of local minima.
• If f '(x) does not change sign as x increases through a, then a is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflexion.
Second Derivative Test
Let f be a function defined on an interval I and a I. Let f be twice differential at c. Then
• x = a is a point of local maxima if f '(a) = 0 and f ''(a) < 0 The values f(a) is local maximum value of f.
• x = a is a point of local minima if f '(a) = 0 and f ''(a) > 0 In this case, f(a) is local minimum value of f.
• The test fails if f '(a) = 0 and f ''(a) = 0 In this case, we go back to the first derivative test and find whether a is a point of maxima, minima or a point of inflexion.
Working rule for finding absolute maxima and/or absolute minima
• Step: 1 Find all critical points of f in the interval, i.e., find points x where either f '(x) = 0 or f is not differentiable.
• Step: 2 Take the end points of the interval.
• Step: 3 At all these points (listed in Step 1 and 2), calculate the values of f.
• Step: 4 Identify the maximum and minimum values if f out of the values calculated in step 3. This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f.
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## The Inscribed Angle Theorem and Its Applications
An inscribed angle is an angle formed by two chords of a circle with the vertex on its circumference. In the first circle in Figure 1, segments AB and AC are chords of a circle and the vertex A is on its circumference. Hence, angle A is an inscribed angle. In the second circle in Figure 1, angle Q is also an inscribed angle.
Angles formed by the two radii of a circle are called central angles. The vertex of a central angle is on the center of the circle. In the Figure 1 below, angles BOC and SOT are central angles.
The arc determined by the endpoints of two chords or two radii, and is opposite of the angle, is called the intercepted arc. In the first circle in Figure 1, arc BC, denoted by the red curve, is the intercepted arc of angle BAC (or angle BOC). Similarly, in the second circle, arc ST denoted by the green curve, is the intercepted arc of the central angle SOT. Saying the other way around, angle BAC is the subtended angle of arc BC and and angle SOT is the subtended angle of arc ST.
The Inscribed Angle Theorem
In this article, we are going to discuss the relationship between an inscribed angle and a central angle (I have created a GeoGebra applet about it) having the same intercepted arc. This is shown in the first circle in Figure 1. Angle BAC and angle BOC have the same intercepted arc BC. Therefore, if a central and an inscribed angle have the same intercepted arc, the measure of that central angle is twice that of the measure of the inscribed angle.
Figure 1 – Inscribed angles, central angles and their intercepted arcs.
One strategy to solve a problem in geometry is to sometimes draw lines from one point to another. These lines drawn are called auxiliary lines. » Read more
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Lessons
1. Basics
2. Deductive Reasoning
3 - Parallel lines
4 - Congruent Triangles
6 - Inequalities
7 - Similar Polygon
8 - Rt. Triangles
9 - Circles
10 - Constructions
11 - Areas of 2D objects
12 - Areas and Volumes
13 - Coordinates
Medians, Altitudes, and Perpendicular Bisectors
Objective:
• Understanding about the properties of medians, altitudes, and perpendicular ( |_) bisectors.
Lesson 4-3 Medians, Altitudes, and Perpendicular Bisectors
Median: A segment from a vertex to the midpoint of the opposite side
Altitude: The perpendicular segment from a vertex to the line that contains the opposite side.
In an acute triangle all of the altitudes are all inside the triangle.
In a right triangle, two of the altitudes are the legs of the triangle and the third is inside the triangle.
In an obtuse triangle, two of the altitudes are outside of the triangle.
Perpendicular Bisector: a line (or ray or segment) that is perpendicular to the segment at its midpoint.
Theorem 4-5 If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment. Segment AB is Segment BC by Theorem 4-5 Theorem 4-6 If a point is equidistant from the endpoints of a segment, then the point lies on the perpendicular bisector of the segment. Theorem 4-7 If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle. Theorem 4-8 If a point is equidistant from the sides of an angle, then the point lies on the bisector of the angle.
Quickie Math Copyright (c) 2000 Team C006354
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# The Tigers won twice as many football games as they lost. They played 96 games. How many games did the win?
Dec 7, 2016
They Tigers won 64 games.
#### Explanation:
Let's call the games the Tigers won $w$ and the games they lost $l$.
With the information provided in the question we can write two equations we can solve using substitution:
Because we know they played 96 games we know we can add the wins and losses to equal 96:
$w + l = 96$
And, because we know they won twice as many games as they lost we can write:
$w = 2 l$
Because the second equation is already in terms of $w$ we can substitute $2 l$ for $w$ in the first equation and solve for $l$:
$2 l + l = 96$
$3 l = 96$
$\frac{3 l}{3} = \frac{96}{3}$
$l = 32$
We can now substitute $32$ for $l$ in the first equation and calculate $w$.
$w = 2 \times 32$
$w = 64$
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Chapter 10.5, Problem 25E
### Calculus (MindTap Course List)
11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Chapter
Section
### Calculus (MindTap Course List)
11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem
# Finding the Area of a Polar Region In Exercises 19-26, use a graphing utility to graph the polar equation. Find the area of the given region analytically.Between the loops of r = 3 − 6 sin θ
To determine
To calculate: The value of the area between the loops of polar equation r=36sinθ and draw the area by the use of graphing utility.
Explanation
Given:
The polar equation is r=3−6sinθ.
Formula used:
The area of the polar equation is given by;
A=12∫αβ[f(θ)]2dθ
Where, α and β are limits of the integration.
Calculation:
Consider the polar equation r=3−6sinθ.
Now, use the following steps in the TI-83 calculator to obtain the graph:
Step 1: Press ON button to open the calculator.
Step 2: Press MODE button and then scroll down to press pol and press ENTER button.
Step 3: Now, press the button Y= and enter the provided equation.
Step 4: Press WINDOW button and then set the window as follows:
Xmin=−9,Xmax=9,Ymin=−10 and Ymax=2
Step 5: Press ENTER to get the graph.
The graph obtained is:
Since, it can be seen that there is symmetry in above graph.
So, the area inside the outer loop is twice the area obtained by integration of the polar equation r=3−6sinθ from r=0 to r=9.
At r=0, the value of θ is;
0=3−6sinθ3=6sinθ12=sinθ
This gives,
θ=π6 and θ=5π6
And, at r=9, the value of θ is;
9=3−6sinθ6=−6sinθ−1=sinθ
This gives;
θ=3π2
So, to get the area inside the outer loop, twice the area obtained by integration of the polar equation r=3−6sinθ from θ=5π6 to θ=3π2.
The area inside the loop is,
Aouter=2[12∫5π/63π/2[f(θ)]2dθ]=∫5π/63π/2[3−6sinθ]2dθ=∫5π/63π/2[9+36sin2θ−36sinθ]dθ
Use the identity 1−cos2θ=2sin2θ;
Aouter=∫5π/63π/2[9+18(1−cos2θ)−36sinθ]dθ=∫5π/63π/2[27−18cos2θ−36sinθ]dθ=[27θ−9sin2θ−36cosθ]5π/63π/2
Further solve and get,
Aouter=[27θ−9sin2θ−36cosθ
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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th
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Question Video: Solving Absolute Value Inequalities Algebraically Mathematics • 10th Grade
Find algebraically the solution set of the inequality 1/|4𝑥 − 9| > 17.
08:04
Video Transcript
Find algebraically the solution set of the inequality one over the absolute value of four 𝑥 minus nine is greater than 17.
First, let’s look closely at the inequality. Here we notice that we have an 𝑥 in the denominator. And that means there is one place where 𝑥 will not exist. We cannot have a zero in the denominator. And that means four 𝑥 minus nine cannot equal zero.
We want to first find out what 𝑥 would be to make this statement zero. So we solve for 𝑥. We add nine to both sides of the equation. On the left, it cancels out, leaving us with four 𝑥 cannot equal nine. We still need to isolate 𝑥, and we do that by dividing by four on both sides. Four divided by four is one, leaving us with 𝑥 cannot equal nine-fourths. There is no solution at 𝑥 equals nine-fourths.
But now we need to find out where there is a solution. And to do that, we need to think about what we know about absolute value. When we solve an inequality with absolute value, we deal with two cases. The positive case: one over four 𝑥 minus nine is greater than 17. And the negative case: one over negative four 𝑥 minus nine is greater than 17.
Let’s start with our positive case. Our goal is to isolate 𝑥. And to do that, we’ll need to get it out of the denominator. We multiply both sides of our inequality by four 𝑥 minus nine over one. On the left side of our equation, these two things cancel out, leaving us with one is greater than 17 times four 𝑥 minus nine.
To get rid of our parentheses, we’ll distribute our 17. 17 times four 𝑥 equals 68𝑥. 17 times negative nine equals negative 153. Bring down our inequality symbol. We now have one greater than 68𝑥 minus 153.
Continuing as we try to isolate 𝑥, we add 153 to the right side and the left side. Negative 153 plus positive 153 cancels each other out. They’re equal to zero. One plus 153 is 154. 154 is greater than 68𝑥. We can isolate 𝑥 by dividing the right side by 68. To keep everything equal, we have to also divide the left by 68. 68 divided by 68 equals one. 154 over 68 is greater than 𝑥.
I noticed that both the numerator and the denominator here are even, which means we can simplify this by dividing the numerator and the denominator by two. 154 divided by two equals 77. 68 divided by two equals 34. 77 over 34 is greater than 𝑥. Or in the more common format, 𝑥 is less than 77 over 34. And this is one-half of the solution set.
We now need to look at the negative case. The first thing we wanna do is distribute this negative to both the four 𝑥 and the negative nine. If I do that, I end up with negative four 𝑥 plus nine is greater than 17.
Again, we have an 𝑥 in the denominator, and we want to get it out of the denominator. We can do that by multiplying its reciprocal on the left and the right. On the left side, negative four 𝑥 plus nine over one cancels out one over negative four 𝑥 plus nine. We’re left with just one. One is greater than 17 times negative four 𝑥 plus nine. We distribute our 17. 17 times negative four 𝑥 equals negative 68𝑥. 17 times nine equals positive 153.
Our inequality now says one is greater than negative 68𝑥 plus 153. We subtract 153 from both sides. Positive 153 minus 153 equals zero. One minus 153 equals negative 152, which is greater than negative 68𝑥.
In our last step to isolate 𝑥, we divide the right side by negative 68 and we divide the left side by negative 68. And here is our key moment. When we divide by negatives and we have inequalities, we must flip the sign. We divided both sides by a negative number. Therefore, we have to flip the inequality symbol.
On the left, we have negative 152 over negative 68. A negative over a negative would give us a positive number. Negative 68 over negative 68 cancels out, leaving us with 𝑥. 152 and 68 are both divisible by four. If we divide 152 by four, we get 38. If we divide 68 by four, we get 17. 38 over 17 is less than 𝑥. Again, we might want to write this in its more common format. 𝑥 is greater than 38 over 17.
Now how do we take these three pieces of information and put them together? Looking at the inequality symbol, we see that we’re dealing with less than and greater than, but not equal to. And that means the brackets we’ll use will face outward. 𝑥 must be greater than 38 over 17 and less than 77 over 34.
What we’ve said so far, if you imagine on a number line, we’re saying that 𝑥 cannot be equal to 38 over 17, but everything between 38 over 17 and 77 over 34. However, we need to exclude nine over four. There’s a hole in this domain at nine-fourths. To exclude nine-fourths, we can say minus nine-fourths. 𝑥 falls between thirty-eight seventeenths and seventy-seven thirty-fourths, with the exception of nine-fourths.
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#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 14 Maths Textbook Solution.
ANSWER: Equation of tangent, $x+y-2=0$
Equation of normal , $y-x=0$
HINTS:
Differentiating the given curve with respect to x and find its slope first.
GIVEN:
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=2 \text { at }(1,1)$
SOLUTION:
\begin{aligned} &\frac{2}{3} x^{\frac{-1}{3}}+\frac{2}{3} y^{\frac{-1}{3}} \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=-\frac{x^{\frac{-1}{3}}}{y^{\frac{-1}{3}}}=\frac{-y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \end{aligned}
Slope of tangent,
$m=\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-1}{1}=-1$
Equation of tangent is
\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-1=-1(x-1) \\ &\Rightarrow y-1=-x+1 \\ &\therefore x+y-2=0 \end{aligned}
Equation of Normal is ,
\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-1=1(x-1) \\ &\Rightarrow y-1=x-1 \\ &\therefore y-x=0 \end{aligned}
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# What is integral 1/(x-1)(x+2)?
sciencesolve | Certified Educator
You need to evaluate the indefinite intregral, such that:
`int (dx)/((x-1)(x+2))`
You need to use the partial fraction expansion to split the given integral into two simpler integrals, such that:
`1/((x-1)(x+2)) = a/(x-1) + b/(x+2)`
`1 = a(x+2) + b(x-1) `
`1 = ax + 2a + bx - b`
`1 = x(a+b) + 2a - b`
Equating the coefficients of like powers yields:
`{(a + b = 0),(2a - b = 1):} => {(a = -b),(-2b - b = 1):}`
`{(a = -b),(-3b = 1):} => {(a = -b),(b = -1/3):} => {(a = 1/3),(b = -1/3):}`
`1/((x-1)(x+2)) = (1/3)(1/(x-1) - 1/(x+2))`
Integrating both sides yields:
`int 1/((x-1)(x+2)) dx = (1/3)int(1/(x-1) - 1/(x+2))dx`
Using the property of linearity of intgeral, yields:
`int 1/((x-1)(x+2)) dx = (1/3)int(1/(x-1)dx) - (1/3)(int 1/(x+2))dx`
`int 1/((x-1)(x+2)) dx = (1/3)(ln|x-1| - ln|x+2|) + c`
`int 1/((x-1)(x+2)) dx = (1/3)(ln|(x-1)/(x+2)|) + c`
Hence, evaluating the indefinite integral, using partial fraction expansion, yields `int 1/((x-1)(x+2)) dx = (1/3)(ln|(x-1)/(x+2)|) + c.`
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Lesson Objectives
• Learn how to find the partial fraction decomposition with quadratic factors
## How to Find the Partial Fraction Decomposition with Quadratic Factors
In this lesson, we will learn how to find the partial fraction decomposition with quadratic factors. At this point in our course, one should be fully comfortable with the process of combining two or more rational expressions into a single rational expression. Here, we will consider how to reverse this process and turn a single rational expression into the sum of two or more rational expressions. Let's look at this process with an example.
Example #1: Find the partial fraction decomposition. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. If you do not have a proper fraction, you must do long division and then apply the following steps to the remainder.
Step 2) Factor the denominator completely. $$x^4 - 3x^2 - 10=(x^2 + 2)(x^2 - 5)$$ Step 3) For each distinct linear or quadratic factor, we set up a fraction with the factor as the denominator and a variable as the numerator. It is typical to see capital letters as the variables:{A, B, C, D, E,...}$$\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ Step 4) Set this equal to the original rational expression. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression. $$5x^2 - 4=A(x^2 - 5) + B(x^2 + 2)$$ Match up the two sides. To do this, we will simplify the right side and change it into the same form as the left side. $$5x^2 - 4=Ax^2 - 5A + Bx^2 + 2B$$ $$5x^2 - 4=(A + B)x^2 + (2B - 5A)$$ These two expressions can only be equal if: $$A + B=5$$ $$2B - 5A=-4$$ We can easily solve this with substitution. $$A=5 - B$$ Plug into the second equation: $$2B - 5(5 - B)=-4$$ $$2B - 25 + 5B=-4$$ $$7B=21$$ $$B=3$$ Plug into the first equation: $$A + 3=5$$ $$A=2$$ We just replace A with 2 and B with 3 and we are done: $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{2}{x^2 + 2}+ \frac{3}{x^2 - 5}$$
#### Skills Check:
Example #1
Find the partial fraction decomposition.
Please choose the best answer. $$\frac{-7x^2 + 1}{x^4 - x^2 - 6}$$
A
$$\frac{2}{x^2 - 3}+ \frac{4}{x^2 + 2}$$
B
$$\frac{2}{x^2 - 3}- \frac{3}{x^2 + 2}$$
C
$$-\frac{4}{x^2 - 3}- \frac{3}{x^2 + 2}$$
D
$$-\frac{6}{x^2 - 3}- \frac{3}{x^2 + 2}$$
E
$$-\frac{6}{x^2 - 3}+ \frac{1}{x^2 + 2}$$
Example #2
Find the partial fraction decomposition. $$\frac{3x^2 + x - 14}{x^3 - 5x + x^2 - 5}$$
A
$$-\frac{2}{x + 1}- \frac{2}{x^2 - 5}$$
B
$$-\frac{6}{x + 1}+ \frac{3}{x^2 - 5}$$
C
$$\frac{3}{x + 1}+ \frac{1}{x^2 - 5}$$
D
$$\frac{4}{x + 1}- \frac{7}{x^2 - 5}$$
E
$$\frac{3}{x - 1}+ \frac{2}{x^2 - 3}$$
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# Simple interest calculator
Simple interest is calculated only on the initial amount (principal) that you invested.
Example: Suppose you give \$100 to a bank which pays you 5% simple interest at the end of every year. After one year you will have \$105, and after two years you will have \$110. This means that you will not earn an interest on your interest. Your interest payments will be$5 per year no matter how many years the initial sum of money stays in a bank account.
This calculator can be used to solve various types of simple interest problems. The calculator will print easy to understand step-by-step explanation.
problem
You deposit $12000 into a bank account paying 1.5% simple interest per month. You left the money in for 210 days. Find the interest earned and the amount at the end of those 210 days? solution The interest is$1242.734 and the amount is $13242.734. explanation STEP 1: Convert interest rate of 1.5% per month into rate per year. $$\text{rate per year = rate per month} \cdot 12 = 1.5 \% \cdot 12 = 18 \%$$ STEP 2: Convert 210 days into years. $$\text{ 210 days } = \frac{ 210 }{365} \text{ years} = 0.57534 \text{ years}$$ STEP 3: Find an interest by using the formula$ I = P \cdot i \cdot t $, where I is interest, P is total principal, i is rate of interest per year, and t is total time in years. In this examplee P =$12000, i = 18% and t = 0.57534 years, so
\begin{aligned} I &= P \cdot i \cdot t \\ I &= 12000 \cdot 0.18 \cdot 0.57534 \\ I &= 1242.734 \end{aligned}
STEP 4: Find an amount by using the formula $A = P + I$.
Since P = $12000 and I =$1242.734 we have
\begin{aligned} A &= P + I \\ A &= 12000 + 1242.734 \\ A &= 13242.734 \end{aligned}
## Report an Error !
Script name : simple-interest-calculator
Form values: 0 , 12000 , 1.5 , 3 , 210 , 3 , g , , ,
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Simple Interest Calculator
How much one will get after investment time.
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examples
example 1:ex 1:
You deposit $\$ 1000$into a bank account paying$7\,\%$simple interest per year. You left the money in for$3 \text{years}$. Find the interest earned and the amount at the end of those$3 \text{years}$? example 2:ex 2: You deposit$\$12000$ into a bank account paying $1.5\%$ simple interest $\text{per month}$. You left the money in for $210 \text{days}$. Find the interest earned and the amount.
example 3:ex 3:
You deposit some money into a bank account paying $4\%$ simple interest $\text{per year}$. You received $\$72$in interest after$3 \, \text{years}$. How much the deposit (principal) was? example 4:ex 4: You deposit some money into a bank account paying$2\%$simple interest$\text{per 6 months}$. You received$\$15$ in interest after $\text{9 month}$. How much the principal was?
example 5:ex 5:
You deposit $\$1000$into a bank account and received$\$50$ simple interest after $\text{3 months}$. What had been the interest rate?
example 6:ex 6:
You deposit $\$350$into a bank account paying$1.2\%$simple interest$\text{per month}$. If you received$\$9$ as interest, find the time for which the money stayed in the bank.
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A line segment has endpoints at (7 , 4) and (2 , 5). If the line segment is rotated about the origin by (pi)/2 , translated horizontally by -3, and reflected about the y-axis, what will the line segment's new endpoints be?
Nov 23, 2016
$\left(7 , 7\right)$ and $\left(8 , 2\right)$.
Explanation:
Moving a line segment is equivalent to moving its endpoints.
When a point $\left({x}_{0} , {y}_{0}\right)$ is rotated about the origin by $\frac{\pi}{2}$, the new point is always $\left({x}_{1} , {y}_{1}\right) = \left(\text{-} {y}_{0} , {x}_{0}\right)$. Think of it like this: if you're walking in the woods and holding a map so that "forward" is east, that's the same situation. The map's "right" (east) is your "forward", the map's "forward" (north) is your "left", etc.
So in a 1/4 turn counterclockwise, old right $\left({x}_{0}\right)$ becomes new up $\left({y}_{1}\right)$, and old up $\left({y}_{0}\right)$ becomes new left $\left(\text{-} {x}_{1}\right)$. This is the same as $\left({x}_{1} , {y}_{1}\right) = \left(\text{-} {y}_{0} , {x}_{0}\right)$.
So after rotating our points $\frac{\pi}{2}$ about the origin, the new points are:
$\left(\text{-} 4 , 7\right)$ and $\left(\text{-} 5 , 2\right)$.
Horizontal translations only affect your $x$-value, because they are a left-right ($x$-axis) shift, and not an up-down ($y$-axis) shift.
After translating both points horizontally by -$3$, our new points are:
$\left(\text{-} 7 , 7\right)$ and $\left(\text{-} 8 , 2\right)$.
Finally, reflecting a point about the $y$-axis simply means changing the sign of its $x$-coordinate. This reflection is a left-to-right flip, so the up-down $\left(y\right)$ location will not change. (If you flip a map over so that north and south stay "up" and "down", the map's "east" becomes left, and its "west" becomes right.)
After reflection about the $y$-axis, our final points will be:
$\left(7 , 7\right)$ and $\left(8 , 2\right)$.
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# Differential Equation
## What is Differential Equation?
A differential equation is a mathematical equation that involves one or more functions and their derivatives. The rate of change of a function at a point is defined by its derivatives. It's mostly used in fields like physics, engineering, and biology. The analysis of solutions that satisfy the equations and the properties of the solutions is the primary goal of differential equations. Using explicit formulas is one of the simplest ways to solve the differential equation. In this article, let us discuss the differential equation meaning, types, methods to solve the differential equation, order and degree of the differential equation, differential equations formulas and few solved problems.
### Differential Equation Definition
A differential equation has one or more terms as well as the derivatives of one variable (the dependent variable) in relation to another variable (i.e., independent variable)
$\frac{dy}{dx}$ = f(x)
Here "x" is the independent variable, and "y" is the dependent variable.
For example, $\frac{dy}{dx}$ = 5x
Partially derivatives and ordinary derivatives are also present in a differential equation. The differential equation defines a relationship between a quantity that is continuously varying with respect to a change in another quantity, and the derivative represents a rate of change.
### What is a Differential Equation?
An equation involving an unknown function y=f(x) and one or more of its derivatives is known as a differential equation. To put it another way, it's an equation that involves derivatives of one or more dependent variables with respect to one or more independent variables. Differential equations are important for describing nature mathematically, and they are at the heart of many physical theories.
### Order of Differential Equation
The order of differential equation is the order of the equation's highest order derivative present in the equation. Here are some examples of differential equations in various orders.
$\frac{d^{3}x}{dx^{3}}$ + 3x$\frac{dy}{dx}$ = e$^{y}$
The order of the highest derivative in this equation is 3, indicating that it is a third-order differential equation.
Example - ($\frac{d^{2}y}{dx^{2}}$)$^{4}$ + $\frac{dy}{dx}$ = 3
This equation represents a second-order differential equation.
This way we can have higher-order differential equations i.e n$^{th}$ order differential equations.
### First Order Differential Equation
As you can see in the first example, the differential equation is a First Order Differential Equation with a degree of 1. In the form of derivatives, all linear equations are in the first order. It only has the first derivative, dy/dx, where x and y are the two variables, and is written as:
dy/dx = f(x, y) = y’
For example $\frac{dy}{dx}$ + (x$^{2}$ + 5)y = $\frac{x}{5}$
This also represents a First-order Differential Equation.
### Second-Order Differential Equation
The equation which includes second-order derivatives is the second-order differential equation. It is represented as;
$\frac{d}{dx}$($\frac{dy}{dx}$) = $\frac{d^{2}y}{dx^{2}}$ = f’’(x) = y’’
For example $\frac{d^{2}y}{dx^{2}}$ + (x$^{3}$ + 3x)y = 9
In this case, the highest derivative's order is 2. As a result, the equation is a second-order differential equation.
### Types of Differential Equations
Differential equations are classified into many categories. They are:
• Ordinary Differential Equations
• Partial Differential Equations
• Linear Differential Equations
• Non-linear Differential Equations
• Homogeneous Differential Equations
• Non-homogeneous Differential Equations
### Differential Equations Solutions
The solution to the differential equation can be found using one of two methods.
• Separation of variables
• Integrating factor
The variable is isolated when the differential equation can be written in the form dy/dx = f(y)g(x), where f is the function of y only and g is the function of x only. Rewrite the problem as 1/f(y)dy= g(x)dx and then integrate on both sides using an initial condition.
When the differential equation is of the form dy/dx + p(x)y = q(x), where p and q are both functions of x only, the integrating factor technique is used.
y'+ P(x)y = Q is a first-order differential equation (x). P and Q are functions of x and the first derivative of y, respectively. An equation containing partial or ordinary derivatives of an unknown function is referred to as a higher-order differential equation. It is possible to represent it in any order.
### Degree of Differential Equation
The power of the highest order derivative, where the original equation is defined as a polynomial equation in derivatives such as y',y”, y”', and so on, is the Degree of Differential Equation.
Assume that $\frac{d^{2}y}{dx^{2}}$ + 2($\frac{dy}{dx}$) + y = 0 is a differential equation, in which case the degree of this equation is 1. Here are some more examples:
dy/dx + 1 = 0, degree is 1
(y”’)3 + 3y” + 6y’ – 12 = 0, in this equation, the degree is 3.
### Ordinary Differential Equation
The function and its derivatives are involved in an ordinary differential equation. Just one independent variable and one or more of its derivatives with respect to the variable are used.
Ordinary differential equations have an order that is defined as the order of the highest derivative in the equation. The n$^{th}$ order ODE's general form is as follows:
F(x, y, y’,…., yn ) = 0
2$\frac{d^{2}y}{dx^{2}}$ + ($\frac{dy}{dx}$)$^{3}$ = 0 is an ordinary differential equation.
### Linear Differential Equations
A differential equation of the form: $\frac{dy}{dx}$ + My = N
The first-order linear differential equation, where M and N are constants or functions of x only, The following is an example of first-order linear differential equations: $\frac{dy}{dx}$ + y = sinx
### Linear Differential Equations Real World Example
Find this basic example to better understand differential equations. Have you ever wondered why a hot cup of coffee cools down when held at room temperature? A hot body's cooling is proportional to the temperature difference between its temperature T and the temperature T$_{0}$ of its surroundings, according to Newton. In terms of mathematics, this sentence can be written as:
dT/dt ∝ (T – T$_{0}$)…………(1)
A linear differential equation takes this shape.
With the addition of a proportionality constant k, the above equation becomes:
dT/dt = k(T – T$_{0}$) …………(2)
T is the body temperature, and t is the time in this equation.
T$_{0}$ is the temperature of the surrounding,
The rate of cooling of the body is dT/dt.
### Applications of Differential Equations:
1) Differential equations are used to explain the growth and decay of various exponential functions.
2) They can also be used to explain how a return on investment changes over time.
3) They're used in medical science to model cancer growth and disease spread across the body.
4) It may also be used to explain the movement of electricity.
5) They assist economists in determining the most effective investment strategies.
6) These equations may also be used to explain the motion of waves or a pendulum.
### Differential Equations Examples
1. Form the Differential Equation y=mx, Where m is an Arbitrary Constant, to Describe the Family of Curves.
Sol: Here we will eliminate constant by differentiating y,
Given y=mx
First, calculate the value of $\frac{du}{dx}$
$\frac{dy}{dx}$ = $\frac{d(mx)}{dx}$
⇒ $\frac{dy}{dx}$ = m
Or m = $\frac{dy}{dx}$
Now, the given equation is y=mx
Put the value of m in the above equation,
y = $\frac{dy}{dx}$(x)
⇒ x($\frac{dy}{dx}$) - y = 0
Hence differential equation is x($\frac{dy}{dx}$) - y = 0
2. Determine the Order and Degree of the Differential Equation 2x$^{2}$$\frac{d^{2}y}{dx^{2}}$ - 3$\frac{dy}{dx}$ + y = 0 is
Sol: Given equation is 2x$^{2}$$\frac{d^{2}y}{dx^{2}}$ - 3$\frac{dy}{dx}$ + y = 0
We can write it as 2x$^{2}$y’’ - 3y’ + y = 0
The highest order of the derivative is 2 and the degree of the highest derivative is 1 as shown in the equation.
Hence order is 2 and degree 1 of the differential equation 2x$^{2}$$\frac{d^{2}y}{dx^{2}}$ - 3$\frac{dy}{dx}$ + y = 0
3. The Integrating Factor of the Differential Equation x$\frac{dy}{dx}$ - y = 2x$^{2}$ is
Sol: Given equation x$\frac{dy}{dx}$ - y = 2x$^{2}$
Divide both sides by x,
⇒ $\frac{dy}{dx}$ - $\frac{y}{x}$ = 2x
Now compare the above equation with the standard equation of the form,
$\frac{dy}{dx}$ + Py = Q
Here P = - $\frac{1}{x}$ and Q = 2x
So Integrating Factor (IF) = e$^{\int pdx}$
IF = e$^{\int -\frac{1}{x}dx}$
IF = e$^{-logx}$
IF = e$^{log x^{-1}}$
IF = x$^{-1}$
IF = $\frac{1}{x}$
Hence integrating factor = $\frac{1}{x}$
### Conclusion:
From above discussion we can say a differential equation is defined as an equation containing the derivative or derivatives of the dependent variable with respect to the independent variable. We have discussed differential equations formulas and different methods to solve differential equations by using basic differential equations formulas.
Last updated date: 06th Jun 2023
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## FAQs on Differential Equation
1. What is the Differential Equation?
Ans: A differential equation is a mathematical equation that has one or more derivatives of a function. The function's derivative is given by dy/dx. A solution to a differential equation is a function y=f(x) that contains the differential equation when f and its derivatives are substituted into the equation.
2. What is the Use of a Differential Equation?
Ans: The differential equation's main aim is to calculate the function over its entire domain. It's a term for the exponential growth or decay of a system over time. It has the power to foresee what will happen in the world around us. It is commonly used in a variety of areas, including physics, chemistry, biology, and economics.
3. What is the General Solution of Differential Equations?
Ans: In the case of an ODE, the general solution contains all possible solutions and usually includes arbitrary constants or arbitrary functions (in the case of a PDE.) A specific solution does not contain any arbitrary constants or functions.
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# Let’s extend our knowledge of trigonometric functions…
## Presentation on theme: "Let’s extend our knowledge of trigonometric functions…"— Presentation transcript:
Let’s extend our knowledge of trigonometric functions…
…in Section 4.3a
Trigonometric Functions of Any Angle
(not just positive or acute angles) A more “dynamic” view of angles: Initial Side – beginning position of a ray (part of the angle) Vertex – the endpoint of this ray Terminal Side – final position of a ray, after being rotated about the vertex Measure of an Angle – a number that describes the amount of rotation from the initial side to the terminal side of the angle Positive angles are generated by counterclockwise rotations and negative angles are generated by clockwise rotations.
Trigonometric Functions of Any Angle
(not just positive or acute angles) terminal side terminal side initial side initial side A positive angle in standard position A negative angle in standard position
Trigonometric Functions of Any Angle
(not just positive or acute angles) Coterminal Angles – angles that have the same initial side and the same terminal side, but have different measures. Positive and negative coterminal angles Two positive coterminal angles
Guided Practice Find and draw a positive angle and a negative angle that are coterminal with the given angle. Note: there are infinitely many possible solutions… Here are two: (a)
Guided Practice Find and draw a positive angle and a negative angle that are coterminal with the given angle. (b) The diagram? (c) The diagram? I believe that the diagram should always come first!
Definition: Trigonometric Functions of any Angle
Let be any angle in standard position and let P(x, y) be any point on the terminal side of the angle (except the origin). Let r denote the distance from P to the origin ( ). Then P(x,y) y r x
Guided Practice Let be the acute angle in standard position whose terminal side contains the point (5, 3). Find the six trig functions of . The diagram? P(5,3) 3 5
Guided Practice Let be any angle in standard position whose terminal side contains the point (–5, 3). Find the six trig functions of . The diagram? P(–5,3) 3 –5
Guided Practice Find the six trigonometric functions of .
First, let’s draw a diagram, complete with a reference triangle: P(1,–1)
Whiteboard Practice Find each of the following without a calculator by using ratios in a reference triangle: (a) (b) (c)
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Chapter 2: Approaches to Problem Solving Lecture notes Math 1030 Section B
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1 Section B.1: Standardized Unit Systems: U.S. and Metric Different system of standardized units There are different systems of standardized units: the international metric system, called SI (from the French Système International d Unités) and the English system, called the U.S. customary system (USCS). The U.S. Customary System The U.S. Customary System Units of length are inches, feet, yards, and miles. Units of mass are ounces and pounds. For volumes we use quarts, gallons, and barrels. Ex.1 The Kentucky Derby. In the Kentucky Derby, horses race a distance of 10 furlongs. How many miles is the race? (Hint: 1 furlong = 1/8 miles.) 1
2 Ex.2 20,000 Leagues Under the Sea. In the Jules Verne s novel 20,000 Leagues Under the Sea, does the title refer to an ocean depth? How do you know? (Hint: 1 league = 3 miles.) The International Metric System The International Metric System The basic units of length, mass, time and volume in the metric system are meter for length (m); kilogram for mass (kg); second for time (s); liter for volume (L). In the international metric system we use powers of 10 to find all the other units, and we write the new units adding a prefix which indicates multiplication by a power of 10. 2
3 Ex.3 (1) Convert 2759 centimeters to metes. (2) How many nanoseconds are in a microsecond? Metric-USCS Conversions This is a table of conversions: USCS to Metric Metric to USCS 1 in. = 2.54 cm 1 cm = in. 1 ft = m 1 m = 3.28 ft 1 yd = m 1 m = yd 1 mi = km 1 km = mi 1 lb = kg 1 kg = lb 1 gal = L 1 L = gal Ex.4 Gas Price per Liter. At a gas station in Mexico, the price of gasoline is 8 pesos per liter. What is the price in dollars per gallons? We know that 1 peso = \$
4 Ex.5 Square Kilometers to Square Miles. How many square kilometers are in a square mile? Temperature Units: Fahrenheit, Celsius and Kelvin Temperature Units There are three temperature scales commonly used today: Fahrenheit, Celsius and Kelvin. The Fahrenheit scale is used in the United States and it is defined so that water freezes at 32 F and boils at 212 F. Internationally, temperature is measured on the Celsius scale, which places the freezing point of water at 0 C and the boiling point at 100 C. The Kelvin scale is used in science and is the same as the Celsius scale except for its zero point: a temperature of 0 K is the coldest possible temperature, known as absolute zero and it correspond to C or F. The conversions are: from Celsius to Fahrenheit: F = 1.8C + 32 from Fahrenheit to Celsius: C = F from Celsius to Kelvin: K = C from Kelvin to Celsius: C = K
5 Ex.6 Human Body Temperature. Average human body temperature is 98.6 F. What is it in Celsius and Kelvin? Ex.7 The local weather report says that tomorrow the temperature will be 59, but does not specify whether it is in Celsius or Fahrenheit. Can you tell which it is? Why? 5
6 Section B.2: Units of Energy and Power Definition of energy Energy is what makes matter move or heat up. The international metric unit of energy is the joule. Calories For example, we need energy from food to walk or run. The most familiar energy unit is the food Calorie used to measure the energy our body can draw from food: 1 Calorie = 4184 joules. Definition of power Power is the rate at which energy is used. The international metric unit of energy is the watt: 1 watt = 1 joule. s Ex.8 You are riding an exercise bicycle at a fitness center. The readout states that you are using 500 Calories per hour. Are you generating enough power to light a 100-watt bulb? 6
7 Definition of kilowatt-hour A kilowatt-hour is a unit of energy: 1 kilowatt-hour = 1000 joule 1 s 1 hr = 1000 joule 1 s 3,600 s = 3,600,000 joule. Ex.9 Your utility company charges 8 cents per kilowatt-hour of electricity. How much does it cost to keep a 100-watt light bulb on for a week? 7
8 Section B.3: Units of Density and Concentration Definition of density Density describes compactness or crowding. There are three different kinds of densities: ( g Material density is given in units of mass per unit volume, such as grams per cubic centimeter cm ). 3 For example, the density of water is 1cm g. Objects with densities less than 1 g 3 cm float in the water, 3 while objects with densities more than 1 g cm 3 sink. Population density is given by the number of people per unit area. For example, if 500 people live in a square region that is 2 miles on a side, the population density of the area is 500 people 4mi 2 = 125 people mi 2. Information density is used to describe how much memory can be stored by digital media. For example, each square inch surface of a DVD hold about 100 megabytes of information, so we say that a DVD has an information density of 100 MB. in 2 Definition of concentration Concentration describes the amount of one substance mixed with another. There are many types of concentration, for example: The concentration of an air pollutant is often measured by the numbers of molecules of the pollutant per million molecules of air. For example, if there are 20 molecules of carbon monoxide in each 1 million molecules of air, we state the carbon monoxide concentration as 20 parts per million (ppm). Blood alcohol content (BAC) describes the concentration of alcohol in a person s body. It is usually measured in units of grams of alcohol per 100 milliliter of blood. For example, in most states a person is considered legally intoxicated if his/her blood alcohol content is at or above 0.08 gram of alcohol per 100 milliliters of blood ( 0.08 g 100 ml ). Ex.10 Manhattan Island has a population of about 1.5 million people living in an area of about 57 square kilometers. What is its population density? If there were no high-rise apartments, how much space would be available per person? 8
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## 3rd Grade Number and Operations
### Links on each of the pages below were verified in December, 2014
To work on third grade number and operations standards, click on the numbers below to visit pages with plenty of internet resources for each of the learning standards on the right.
Checks for Understanding are at the top of this page. Scroll down to find internet resources related to the State Performance Indicators (SPIs).
At the very bottom of this page is a link to Number and Operations resources for grades 3-5 from the National Library of Virtual Manipulatives.
### Checks for Understanding (Formative/Summative Assessment)
Whole Numbers - Represent whole numbers up to 10,000 using various models (such as base-ten blocks, number lines, place-value charts) and in standard form, written form, and expanded form. Symbols - Understand and use the symbols =, < and > to signify order and comparison. Parentheses - Use parentheses to indicate grouping. Mental Computations - Use a variety of methods to perform mental computations and compare the efficiency of those methods. Simple Estimates - Use highest order value (such as tens or hundreds digit) to make simple estimates. Story Problems - Solve a variety of addition and subtraction story problems including those with irrelevant information. Multiplications - Represent multiplication using various representations such as equal-size groups, arrays, area models, and equal jumps on number lines. Division - Represent division using various representations such as successive subtraction, the number of equal jumps, partitioning, and sharing. Contexts for Facts - Describe contexts for multiplication and division facts. Unit Fractions - Understand that symbols such as 1/2, 1/3, and 1/4 represent numbers called unit fractions. Identify Fractions - Identify fractions as parts of whole units, as parts of sets, as locations on number lines, and as division of two whole numbers. Compare Fractions - Compare fractions using drawings, concrete objects, and benchmark fractions. Sum of Parts - Understand that when a whole is divided into equal parts to create unit fractions, the sum of all the parts adds up to one.
### State Performance Indicators
Read and Write Numbers - Read and write numbers up to 10,000 in numerals and up to 1,000 in words. Place Value - Identify the place value of numbers in the ten-thousands, thousands, hundreds, tens, and ones positions. Expanded Numbers - Convert between expanded and standard form with whole numbers to 10,000. Order Numbers - Compare and order numbers up to 10,000 using the words less than, greater than, and equal to, and the symbols <, >, =. Multiplication and Division Representations - Identify various representations of multiplication and division. Multiplication Facts - Recall basic multiplication facts through 10 times10 and the related division facts. Multiplication Problems - Compute multiplication problems that involve multiples of ten using basic number facts. Inverse Relationship - Solve problems that involve the inverse relationship between multiplication and division. Contextual Problems - Solve contextual problems involving the addition (with and without regrouping) and subtraction (with and without regrouping) of two- and three digit whole numbers. Equivalent Fractions - Identify equivalent fractions given by various representations. Fractions Interpretations - Recognize and use different interpretations of fractions. Name Fractions - Name fractions in various contexts that are less than, equal to, or greater than one. Order Fractions - Recognize, compare, and order fractions (benchmark fractions, common numerators, or common denominators). Add and Subtract Fractions - Add and subtract fractions with like denominators.
### National Library of Virtual Manipulatives
Number & Operations - for Grades 3 through 5.
Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles.
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Explain why you get this result. Constructing Orthocenter of a Triangle - Steps. How do you construct the orthocenter of an obtuse triangle? You can find where two altitudes of a triangle intersect using these four steps: Find the equations of two line segments forming sides of … Draw arcs on the opposite sides AB and AC. Drawing (Constructing) the Orthocenter Let's build the orthocenter of the ABC triangle in the next app. For a GSP script that constructs the orthocenter of any triangle, click here. Here the 'line' is o… This is done because, this being an obtuse triangle, the altitude will be outside the triangle, where it intersects the extended side PQ.After that, we draw the perpendicular from the opposite vertex to the line. Time-saving video on how to define and construct the incenter of a triangle. Then, go to CONSTRUCT on the toolbar and select Perpendicular Line from the list. It is the reflection of the orthocenter of the triangle about the circumcenter. To construct the orthocenter of an obtuse triangle, call it triangle ABC, we use the following steps: Step 1: construct an altitude of the obtuse... To find the orthocenter, you need to find where these two altitudes intersect. How to construct the orthocenter of a triangle with compass and straightedge or ruler. We No other point has this quality. Each line you constructed above contains an altitude of the triangle. Step 3: Use to construct the line through C and perpendicular to AB. Application, Who Step 4: Use to place a point where the altitudes intersect. Set them equal and solve for x: Now plug the x value into one of the altitude formulas and solve for y: Therefore, the altitudes cross at (–8, –6). Univ. Incenters, like centroids, are always inside their triangles.The above figure shows two triangles with their incenters and inscribed circles, or incircles (circles drawn i… Step 2: Use to construct the line through B and perpendicular to AC. Concept explanation. Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. Construct triangle ABC whose sides are AB = 6 cm, BC = 4 cm and AC = 5.5 cm and locate its orthocenter. The circumcenter, centroid, and orthocenter are also important points of a triangle. Construct the orthocenter of triangle HBC. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side. Video explanation and sample problem on how to construct the orthocenter in an obtuse triangle. This location gives the incenter an interesting property: The incenter is equally far away from the triangle’s three sides. more. Remember, the altitude of a triangle is a perpendicular segment from the vertex of the triangle to the opposite side. How to Protect Your Health from Covid-19? If you're seeing this message, it means we're having trouble loading external resources on our website. 5.4 Orthocenter Compass Construction / obtuse triangle – How do you make a Circumcenter on geogebra? Repeat step 2 using first point A and segment CB; then using point C and segment AB. Orthocenter. (–2, –2) The orthocenter of a triangle is the …, Inquiries around A Euclidean construction The slope of the line AD is the perpendicular slope of BC. Draw intersecting arcs from B and D, at F. Join CF. For a more, see orthocenter of a triangle.The orthocenter is the point where all three altitudes of the triangle intersect. To construct the orthocenter of a triangle, there is no particular formula but we have to get the coordinates of the vertices of the triangle. of WisconsinJ.D. To unlock all 5,300 videos, Copyright © 2018-2020 All rights reserved. Consider a triangle with circumcenter and centroid . How To Construct The Orthocenter. Definition of "supporting line: The supporting line of a certain segment is the line Using this to show that the altitudes of a triangle are concurrent (at the orthocenter). Now consider the triangle HBC. GSP then constructs a line perpendicular to point B and segment AC. how to find the orthocenter with coordinates, http://www.mathopenref.com/printorthocenter.html#:~:text=1%20Set%20the%20compasses%27%20width%20to%20the%20length,half%20the%20distance%20to%20P.%20More%20items...%20, https://www.mathopenref.com/constorthocenter.html, https://www.onlinemath4all.com/construction-of-orthocenter-of-a-triangle.html, https://mathopenref.com/printorthocenter.html, https://www.brightstorm.com/math/geometry/constructions/constructing-the-orthocenter/, http://jwilson.coe.uga.edu/EMAT6680/Evans/Assignment%208/HowToConstructOrthocenter.htm, https://brilliant.org/wiki/triangles-orthocenter/, https://byjus.com/orthocenter-calculator/, https://www.youtube.com/watch?v=oXojD8Uwp9g, https://www.onlinemathlearning.com/geometry-constructions-4.html, https://www.onlinemathlearning.com/triangle-orthocenter.html, https://study.com/academy/lesson/orthocenter-in-geometry-definition-properties.html, http://jwilson.coe.uga.edu/EMAT6680Fa09/DeGeorge/Assign4TdeG/Orthocenters.html, https://study.com/academy/answer/how-to-construct-the-orthocenter-of-an-obtuse-triangle.html, https://www.brightstorm.com/math/geometry/constructions/constructing-the-orthocenter/all, https://www.dummies.com/education/math/geometry/orthocenter-coordinates-in-a-triangle-practice-geometry-questions/, Read A Euclidean construction. The circumcenter of a triangle is the center of a circle which circumscribes the triangle.. This page shows how to construct (draw) the circumcenter of a triangle with compass and straightedge or ruler. The orthocenter is one of the four most common centers of a triangle. © 2021 Brightstorm, Inc. All Rights Reserved. Let be the midpoint of . Draw nABC with obtuse /C and construct its orthocenter O. Th en fi nd the orthocenters of nABO, nACO, and nBCO. Ruler. Will your conjecture be true for any - the answers to estudyassistant.com To draw the perpendicular or the altitude, use vertex C as the center and radius equal to the side BC. Construct the altitude from the obtuse vertex just as you normally would do. more ››. Now you can see the intersection point of the three constructed lines which is the orthocenter. This will help convince you that all three altitudes do in fact intersect at a single point. To construct the orthocenter for a triangle geometrically, we have to do the following: Find the perpendicular from any two vertices to the opposite sides. For example, for the given triangle below, we can construct the orthocenter (labeled as …. The orthocenter is the point of concurrency of the altitudes in a triangle. A point of concurrency is the intersection of 3 or more lines, rays, segments or planes. Step 1 : The first thing we have to do is find the slope of the side BC, using the slope formula, which is, m = y2-y1/x2-x1 2. 5.4 Orthocenter Compass Construction / obtuse triangleThis is a compass construction of the three altitudes of an arbitrary obtuse triangle. How to construct the orthocenter of a triangle? Let's learn these one by one. 1. So not only is this the orthocenter in the centroid, it is also the circumcenter of this triangle right over here. An example of constructing an incenter using a compass and straightedge included. The line segment needs to intersect point C and form a right angle (90 degrees) with the "suporting line" of the side AB. The construction starts by extending the chosen side of the triangle in both directions. How To Download Roblox On Nintendo Switch. A point of concurrency is the intersection of 3 or more lines, rays, segments or planes. Problem 1. (You may need to extend the altitude lines so they intersect if the orthocenter is outside the triangle) Optional Step 11. For obtuse triangles, the orthocenter falls on the exterior of the triangle. Previous Post: How To Find Ka From Ph? The orthocenter is the point of concurrency of the altitudes in a triangle. The orthocenter is the point where all three altitudes of the triangle intersect. Repeat steps 7,8,9 on the third side of the triangle. In this video I show you how to do just that. The orthocenter is located by constructing three altitudes in a triangle. Recall the orthocenter of a triangle is the common intersection of the three lines containing the altitudes. Theorthocenter is just one point of concurrency in a triangle. It follows that is parallel to and is therefore perpendicular to ; i.e., it is the altitude fro… Now, from the point, A and slope of the line AD, write the stra… Then the triangles , are similar by side-angle-side similarity. If you're behind a web filter, ... And I wanted to show that you can always construct that. Then follow the below-given steps; 1. Here $$\text{OA = OB = OC}$$, these are the radii of the circle. The others are the incenter, the circumcenter and the centroid. Orthocenter-1)Construct the orthocenter of the given triangle ABC.Set the compass width to the length of a side of the triangle. 3. It is also the center of the circumcircle, the circle that passes through all three vertices of the triangle. Reasoning, Diagonals, Angles and Parallel Lines, Univ. The orthocenter of a triangle is the intersection of the three altitudes of a triangle. See Orthocenter of a triangle. Answer: 3 question a. 2)From B draw an arc across AC creating point F. 3)From C draw an arc across BA creating point P. 4)Set the compass width to more than half the distance BP. Depending on your construction method, you may need to extend one of the triangle sides to construct … The orthocenter of a triangle is the point where all three of its altitudes intersect. Next construct the orthocenter, H, of triangle ABC. Are, Learn What did you discover? 5)From B and P, draw arcs that intersect at point Q. The orthocenter is just one point of concurrency in a triangle. The point where the two altitudes intersect is the orthocenter of the triangle. 2. Improve your math knowledge with free questions in "Construct the centroid or orthocenter of a triangle" and thousands of other math skills. Note: The orthocenter's existence is a trivial consequence of the trigonometric version Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful. start your free trial. Let be the point such that is between and and . How to Fix Blue Screen of Death Error in Windows 10? Now, let us see how to construct the orthocenter of a triangle. You find a triangle’s incenter at the intersection of the triangle’s three angle bisectors. The others are the incenter, the circumcenter and the centroid. Grades, College https://www.brightstorm.com/.../constructions/constructing-the-orthocenter How to construct the circumcenter of a triangle in Geogebra – Post navigation. It is located at the point where the triangle's three altitudes intersect called a point of concurrency . This page shows how to construct the orthocenter of a triangle with compass and straightedge or ruler. Circumscribed and Inscribed Circles and Polygons, Constructing a Perpendicular at a Point on a Line. Finding it on a graph requires calculating the slopes of the triangle sides. First, construct any triangle ABC. of Wisconsin Law school, Brian was a geometry teacher through the Teach for America program and started the geometry program at his school. The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. One of the four main points of concurrencyof a triangle is the orthocenter.The orthocenter is where thethree altitudes intersect.If we look at three different types of triangles,if I look at an acute triangleand I drew in one of the altitudes orif I dropped an altitude as somemight say, if I drew in another altitude,then this point right here willbe the orthocenter.I could also draw in the third altitude,but I know that since this is a pointof concurrency the three altitudes mustintersect there so I only haveto draw two.If we look at a right triangle, if I wereto draw in an altitude from that vertex,well, that just happens to be thisleg of this right triangle.If I drew in the altitude of this triangle,then I would see -- excuse me, thisside, then this leg wouldbe its altitude.And if we drew in this last one from our90-degree angle, we see that the pointwhere they are concurrent is rightat the vertex of that right angle.So in a right triangle your orthocenterwill be at the vertex of the rightangle.And, last, if we look another an obtusetriangle, we remember in order to findthe altitude of this side we have to extendthat side drop down an altitudewhich is outside of our triangle to find-- and I'm just going to extendthis -- to find the ortho -- to findthe altitude from this vertex, I'mgoing to draw a perpendicularsegment through the vertex.So it looks like it's going to intersectright over there, and for this thirdside I would have to extend it untilwe could find our 90-degree angle.Okay.So in an obtuse triangle your orthocenterwill be outside of your triangle.So expect that on a quiz. But with that out of the way, we've kind of marked up everything that we can assume, given that this is an orthocenter and a center-- although there are other things, other properties of especially centroids that we know. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line. Circumcenter. How to Save Living Expenses for College Students. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side. Step 1: Use to construct the line through A and perpendicular to BC. This is identical to the constructionA perpendicular to a line through an external point. 3. How to construct the orthocenter of acute, right and obtuse triangles. altitudes intersect in a single point, called the orthocenter of the triangle, usually denoted by H. The orthocenter lies inside the triangle if and only if the mathematician Gaston Albert Gohierre de Longchamps. Suppose we have a triangle ABC and we need to find the orthocenter of it. The orthocenter of a triangle is the point of intersection of any two of three altitudes of a triangle (the third altitude must intersect at the same spot). The orthocenter is located inside an acute triangle, on a right triangle, and outside an obtuse triangle. To construct orthocenter of a triangle, we must need the following instruments. b. Step 5: Use to add a label to this point where … Compass. 4. Get Better In construct the altitude, Use vertex C as the center of a triangle side! Called a point where the altitudes right triangle, click here free in! Over here, –2 ) how to construct orthocenter circumcenter of a side of the triangle ’ s three sides triangle and perpendicular. Post: how to find the orthocenter falls on the toolbar and select perpendicular line from list. Draw ) the circumcenter of a triangle is the common intersection of 3 or lines. Go to construct the orthocenter ( labeled as …, let us see how construct. So they intersect if the orthocenter of the triangle by side-angle-side similarity outside. Altitudes do in fact intersect at point Q above contains an altitude of the triangle nACO, and outside obtuse! Behind a web filter,... and I wanted to show that you can always that... Segment AC to show that you can how to construct orthocenter construct that, rays, segments or planes orthocenter falls the... F. Join CF on a right triangle, click here,... I..., BC = 4 cm and locate its orthocenter along that line go to construct the orthocenter is the,... Altitude of a triangle ’ s how to construct orthocenter at the point where all three altitudes of the altitudes intersect line... Application, Who we are, Learn more over here similar triangles you! To BC or orthocenter of a triangle using point C and segment ;. Over here your free trial line through B and perpendicular to a line which passes through vertex... Who we are, Learn more orthocenter falls on the third side of the line a. A gsp script that constructs the orthocenter of a triangle is the (! See how to find Ka from Ph is also the center of the sides intersect to place point... Geogebra – Post navigation, click here and I wanted to show you. Perpendicular at a single point ABC along that line incenter using a compass and straightedge ruler. Whose sides are AB = 6 cm, BC = 4 cm and AC = 5.5 cm and AC 5.5! I show you how to construct the orthocenter of any triangle, we must need following. Was a geometry teacher through the Teach for America program and started geometry! Opposite sides AB and AC triangle ABC whose sides are AB = 6 cm BC... To define and construct the altitude, Use vertex C as the center of a triangle is point. The orhtocenter of ABC along that line the obtuse vertex just as you normally would do straightedge or ruler an. Do in fact intersect at a point of concurrency here the 'line ' is o…:! / ABC and the centroid example, for the given triangle ABC.Set compass... Triangle intersect intersect if the orthocenter of the altitudes intersect of Death Error in 10! Steps 7,8,9 on the opposite side to do just that make a circumcenter on?!, rays, segments or planes here \ ( \text { OA = OB OC... Triangle sides orthocenters of nABO, nACO, and nBCO to Fix Screen. And select perpendicular line from the vertex of the line through a vertex of the triangle = OC \. You that all three altitudes of a triangle you how to construct orthocenter a triangle ’ three. A point of concurrency in a triangle this will help convince you all... To a line which passes through a and perpendicular to a line perpendicular point! Recall the orthocenter of a triangle this page shows how to construct the centroid a circumcenter on?. Which circumscribes the triangle gives the incenter is equally far away from the list 1! Which circumscribes the triangle ’ s three altitudes do in fact intersect at a single point one of. Construct Euler line between the othocenter of PQR / ABC and we need to extend the altitude so... You how to construct the circumcenter of a circle which circumscribes the triangle.! And sample problem on how to construct the orthocenter of a triangle math skills points between the othocenter of and... Who we are, Learn more step 11 page shows how to construct orthocenter the... Center and radius equal to the length of a triangle ’ s incenter at intersection... Graph requires calculating the slopes of the triangle in the centroid the constructionA perpendicular to B. Then constructs a line perpendicular to AC between and and repeat step 2 using first point a and AB. Behind a web filter,... and I wanted to show that you can see intersection. Loading external resources on our website, at F. Join CF draw the perpendicular or altitude! On our website exterior of the three lines containing the altitudes intersect called a point where the... Us see how to construct on the opposite side, Who we,! Let us see how to construct ( draw ) the orthocenter, H, of triangle and. Compass and straightedge included orthocenter of a triangle is the intersection of the triangle to opposite! Interesting property: the incenter of a triangle in geogebra – Post navigation, we construct... Problem on how to do just that, right and obtuse triangles falls on the toolbar and perpendicular! Is identical to the opposite side intersection of the triangle repeat steps 7,8,9 on the exterior of the.. Incenter of a triangle ABC and segment AC draw how to construct orthocenter perpendicular or the altitude so! 5,300 videos, start your free trial external resources on our website where all three its... The list Constructing ) the circumcenter and the centroid at his school, these are incenter. Start your free trial teacher through the Teach for America program and the... Triangle ’ s three angle bisectors triangle ) Optional step 11 an external point & apos ; s altitudes... Are, Learn more how do how to construct orthocenter construct the orthocenter of an obtuse triangle incenter an interesting:. Incenter of a triangle or the altitude lines so they intersect if the orthocenter an. construct the orthocenter falls on the exterior of the triangle example, for the given triangle below we! Triangle intersect the given triangle below, we must need the following instruments be. Circle which circumscribes the triangle sides a perpendicular at a single point and straightedge or ruler altitudes.... 'Re behind a web filter,... and I wanted to show that can. Third side of the triangle and is perpendicular to a line perpendicular to a line passes... Recall the orthocenter in the next app three vertices of the line AD is the point of in... That intersect at a single point AB = 6 cm, BC = 4 cm and locate orthocenter... 2: Use to place a point of concurrency sides are AB = 6 cm BC... Is this the orthocenter of the triangle sides altitudes intersect called a point concurrency... The common intersection of the sides intersect if you 're seeing this,... Above contains an altitude is a line perpendicular to the opposite side nACO and! The side BC the two orthocenter / circumcenter of a triangle, on graph... Use vertex C as the center of the triangle and is perpendicular to AC fact intersect at a point the... Behind a web filter,... and I wanted to show that can..., start your free trial perpendicular bisectors of the triangle \ ), these the! You 're seeing this message, it is the point where the perpendicular or the lines! Of a triangle ABC and we need to extend the altitude, Use vertex C the. This location gives the incenter an interesting property: the incenter of a triangle with compass and or... Abc whose sides are AB = 6 cm, BC = 4 cm and locate its orthocenter O. Th fi. Construct ( draw ) the circumcenter construction starts by extending the chosen side of the ABC triangle geogebra... Normally would do the chosen side of the triangle that you can see the intersection of the.... Make a circumcenter on geogebra locate its orthocenter O. Th en fi nd the orthocenters of nABO, how to construct orthocenter! A compass and straightedge or ruler are the radii of the sides intersect videos, start your free trial Teach! The following instruments Constructing a perpendicular at a single point, these the... 4: Use to place a point where all three of its altitudes intersect is the orthocenter falls on exterior... Triangle sides place a point of the altitudes in a triangle ’ three...: the incenter of a triangle for the given triangle below, we must the... Acute, right and obtuse triangles, are similar by side-angle-side similarity compass! Th en fi nd the orthocenters of nABO, nACO, and nBCO calculating the slopes of sides... Our website and there are corresponding points between the two altitudes intersect a! ( Constructing ) the orthocenter of a triangle '' and thousands of other math skills Post... Let be the point of concurrency of the triangle ’ s three angle bisectors altitudes a! Fact intersect at a single point Who we are, Learn more passes through a vertex of circumcircle... Having trouble loading external resources on our website right triangle, click here of 3 more... Where the triangle circumcenter and the centroid, creating more similar triangles draw nABC with obtuse and... A point of concurrency in a triangle is a perpendicular at a single point chosen side of the triangle Optional... Three altitudes of the triangle calculating the slopes of the ABC triangle in both.!
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# Properties of Multiplication of Integers
The properties of multiplication of integers are discussed with examples. All properties of multiplication of whole numbers also hold for integers.
The multiplication of integers possesses the following properties:
### Property 1 (Closure property):
The product of two integers is always an integer.
That is, for any two integers m and n, m x n is an integer.
For example:
(i) 4 × 3 = 12, which is an integer.
(ii) 8 × (-5) = -40, which is an integer.
(iii) (-7) × (-5) = 35, which is an integer.
### Property 2 (Commutativity property):
For any two integer’s m and n, we have
m × n = n × m
That is, multiplication of integers is commutative.
For example:
(i) 7 × (-3) = -(7 × 3) = -21 and (-3) × 7 = -(3 × 7) = -21
Therefore, 7 × (-3) = (-3) × 7
(ii) (-5) × (-8) = 5 × 8 = 40 and (-8) × (-5) = 8 × 5 = 40
Therefore, (-5) × (-8) = (-8) × (-5).
### Property 3 (Associativity property):
The multiplication of integers is associative, i.e., for any three integers a, b, c, we have
a × ( b × c) = (a × b) × c
For example:
(i) (-3) × {4 × (-5)} = (-3) × (-20) = 3 × 20 = 60
and, {(-3) × 4} × (-5) = (-12) × (-5) = 12 × 5 = 60
Therefore, (- 3) × {4 × (-5)} = {(-3) × 4} × (-5)
(ii) (-2) × {(-3) × (-5)} = (-2) × 15 = -(2 × 15)= -30
and, {(-2) × (-3)} × (-5) = 6 × (-5) = -(6 × 5) = -30
Therefore, (- 2) × {(-3) × (-5)} = {-2) × (-3)} × (-5)
### Property 4 (Distributivity of multiplication over addition property):
The multiplication of integers is distributive over their addition. That is, for any three integers a, b, c, we have
(i) a × (b + c) =a × b + a × c
(ii) (b + c) × a = b × a + c × a
For example:
(i) (-3) × {(-5) + 2} = (-3) × (-3) = 3 × 3 = 9
and, (-3) × (-5) + (-3) × 2 = (3 × 5 ) -( 3 × 2 ) = 15 - 6 = 9
Therefore, (-3) × {(-5) + 2 } = ( -3) × (-5) + (-3) × 2.
(ii) (-4) × {(-2) + (-3)) = (-4) × (-5) = 4 × 5 = 20
and, (-4) × (-2) + (-4) × (-3) = (4 × 2) + (4 × 3) = 8 + 12 = 20
Therefore, (-4) × {-2) + (-3)} = (-4) × (-2) + (-4) × (-3).
Note: A direct consequence of the distributivity of multiplication over addition is
a × (b - c) =a × b - a × c
### Property 5 (Existence of multiplicative identity property):
For every integer a, we have
a × 1 = a = 1 × a
The integer 1 is called the multiplicative identity for integers.
### Property 6 (Existence of multiplicative identity property):
For any integer, we have
a × 0 = 0 = 0 × a
For example:
(i) m × 0 = 0
(ii) 0 × y = 0
### Property 7:
For any integer a, we have
a × (-1) = -a = (-1) × a
Note: (i) We know that -a is additive inverse or opposite of a. Thus, to find the opposite of inverse or negative of an integer, we multiply the integer by -1.
(ii) Since multiplication of integers is associative. Therefore, for any three integers a, b, c, we have
(a × b) × c = a × (b × c)
In what follows, we will write a × b × c for the equal products (a × b) × c and a × (b × c).
(iii) Since multiplication of integers is both commutative and associative. Therefore, in a product of three or more integers even if we rearrange the integers the product will not change.
(iv) When the number of negative integers in a product is odd, the product is negative.
(v) When the number of negative integers in a product is even, the product is positive.
### Property 8
If x, y, z are integers, such that x > y, then
(i) x × z > y × z, if z is positive
(ii) x × z < y × z , if z is negative.
These are the properties of multiplication of integers needed to follow while solving the multiplication of integers.
Numbers - Integers
Integers
Multiplication of Integers
Properties of Multiplication of Integers
Examples on Multiplication of Integers
Division of Integers
Absolute Value of an Integer
Comparison of Integers
Properties of Division of Integers
Examples on Division of Integers
Fundamental Operation
Examples on Fundamental Operations
Uses of Brackets
Removal of Brackets
Examples on Simplification
Numbers - Worksheets
Worksheet on Multiplication of Integers
Worksheet on Division of Integers
Worksheet on Fundamental Operation
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# NCERT Solutions For Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4
Here, Below you all know about NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 Question Answer. I know many of you confuse about finding this Chapter 8 Linear Equations in Two Variables Ex 8.4 Of Class 9 NCERT Solutions. So, Read the full post below and get your solutions.
## NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4
NCERT TEXTBOOK EXERCISES
Question 1. Give the geometric representations of y = 3 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
y=3 …(i)
(i) The representation of the solution on the number line is shown in the figure below, where y = 3 is treated as an equation in one variable.
(ii) We know that, y = 3 can be written as
0 . x + y = 3
which is a linear equation in the variables x and y. This is represented by a line. Now, all the values of x are permissible because 0 . x is always 0.
However, y must satisfy the equation y = 3.
Note that, the graph AB is a line parallel to the x-axis and at a distance of 3 units to the upper side of it.
Question 2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
2x + 9=0
⇒ x = –9/2 …. ( i)
(i) The representation of the solution on the number line is shown in the figure below, where x = –9/2 is treated as an equation in one variable.
(ii) We know that, 2x + 9= 0 can be written as
2x + 0y + 9 = 0
which is a linear equation in two variables x and y.
This is represented by a line.
Now, all the values of y are permissible because 0 .
y is always 0.
However, x must satisfy the equation 2x + 9 = 0.
Note that, the graph AB is a line parallel to the y-axis and at a distance of – 9/2 = – 4.5 to the left of it.
## NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Exercise 8.4 PDF
For NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4, you may click on the link below and get your NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Exercise pdf file.
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# Mathematical Proof/Methods of Proof/Direct Proof
A constructive proof is the most basic kind of proof there is. It is a proof that starts with a hypothesis, and a person uses a series of logical steps and a list of axioms, to arrive at a conclusion.
## Parts of a Theorem
A theorem is a proven statement that was constructed using previously proven statements, such as theorems, or constructed using axioms. Some theorems are very complicated and involved, so we will discuss their different parts.
### The Hypothesis
The hypothesis is the "if" statement of a theorem. In a way, it is similar to an axiom because it is assumed to be true in order to prove a theorem. We will consider a simple example.
Theorem 2.1.1. If A and B are sets such that ${\displaystyle A\subseteq B}$ and ${\displaystyle B\subseteq A}$ , then A=B.
In this theorem, the hypothesis is everything before the word "then." This is a very simple proof. We need to prove that for every x, ${\displaystyle x\in A\Leftrightarrow x\in B}$ . For the purpose of analyzing proofs, we will define ${\displaystyle P(x)=x\in A}$ and ${\displaystyle Q(x)=x\in B}$ . The most common way to prove an "if and only if" statement is to prove necessity and sufficiency separately
• So we start by showing that ${\displaystyle P(x)\Rightarrow Q(x).}$ Therefore, we assume that P(x) is true. That is, ${\displaystyle x\in A.}$ Since we assumed, by hypothesis, that ${\displaystyle A\subseteq B,}$ , we know that ${\displaystyle x\in B}$ , which means that Q(x) is true.
• Now we show that ${\displaystyle Q(x)\Rightarrow P(x)}$ , so we assume that Q(x) is true. This means that ${\displaystyle x\in B}$ . Since we know that ${\displaystyle B\subseteq A,}$ we know that ${\displaystyle x\in A,}$ so P(x) is true.
By these two conclusions, we see that ${\displaystyle P(x)\Leftrightarrow Q(x).}$
Now, by axiom 3, A=B, since ${\displaystyle x\in A\Leftrightarrow x\in B.}$ This concludes the proof. This is a very trivial proof, but its point was to show how to use a hypothesis or set of hypotheses in order to reach the desired conclusion. This method here is the most common in proving that two sets are equal. You prove that each set is a subset of the other.
### The Conclusion
The part of the theorem after the word "then" is called the conclusion. The proof of a theorem is merely the logical connection between the hypothesis and the conclusion. Once you've seen and proved a few theorems, a conclusion is almost predictable. For example, what conclusion would you naturally draw from the following two statements?
1. All Americans are people.
2. All people live on Earth.
These two statements are the hypotheses. To word this as a theorem, we would have "If all Americans are people and all people live on Earth, then all Americans live on Earth." This statement is what most people would call completely obvious and requires no proof. However, to show how this concept is applied in mathematics, we will abstract this theorem and prove it.
Theorem 2.1.2. If ${\displaystyle A\subset B}$ and ${\displaystyle B\subset C}$ , then ${\displaystyle A\subset C}$ .
To see how this relates to our problem, let A be the set of all Americans, B the set of all people, and C the set of all things that live on Earth.
To show that ${\displaystyle A\subset C}$ , we need to show that ${\displaystyle \forall x\in A,x\in C.}$ So we suppose ${\displaystyle x\in A.}$ By hypothesis, ${\displaystyle A\subset B,}$ so ${\displaystyle x\in B.}$ Also by hypothesis, ${\displaystyle B\subset C}$ , so ${\displaystyle x\in C.}$ Since this was true for any arbitrary ${\displaystyle x\in A,}$ we have shown that ${\displaystyle A\subset C.}$
### The Definition
While a definition is not usually part of a theorem, they are commonly introduced immediately before a theorem, in order to help define the symbols you use or to help prove it.
Definition 2.1.3. If a set A has only finitely many elements, then the order of A, denoted by |A|, is the number of elements in A.
This definition gives meaning to the following theorem.
Theorem 2.1.4. If A and B are finite sets such that A = B, then |A|=|B|.
Here we take advantage of the fact that A is a finite set. Let n be the integer such that |A| = n. Then index the elements of A so that ${\displaystyle A=\{x_{1},x_{2},\ldots ,x_{n}\}.}$ Now ${\displaystyle \forall i=1,2,\ldots ,n}$ , we have ${\displaystyle x_{i}\in B}$ . So we see that B has at least n elements, that is ${\displaystyle |B|\geq n.}$ Also, every element of B is in A (by hypothesis), so it follows that there are no more elements in B than there are in A, so ${\displaystyle |B|\leq n}$ , thus |B| = n = |A|, which concludes the proof.
### The Given
Sometimes, the first part of a theorem lists what is required in order for a theorem to be proven. Hence, this list is described in the theorem as what is given. This helps readers understand exactly what is the hypothesis and what is the conclusion in a theorem statement. For example, Theorem 2.1.4 can be rewritten so that it lists what is required beforehand.
Theorem 2.1.4. Given finite sets A and B, if A = B, then |A|=|B|.
This statement clearly highlights what the hypothesis is and what the conclusion is, in this example.
## Classifying Theorems
There are different terms that mathematicians like to use in stating mathematical results. Theorem is probably the most common and well-known, especially to non-mathematicians. There are, however, a few other related terms used in mathematics. They are all theorems, but have more specific uses.
### Lemma
A lemma is a "small theorem." When a result is less profound, more trivial, or boring, it can be called a lemma. A lemma is also used to make the proof of a theorem shorter. That is, if a chunk of a proof can be pulled off and proved separately, then it is called a lemma and the proof of the theorem will say something to the effect of "as proved in the lemma."
For example, the following lemma will help to make the proof of Theorem 2.1.4 more concise.
Lemma 2.1.5. If A and B are finite sets and ${\displaystyle A\subset B}$ then ${\displaystyle |A|\leq |B|}$ .
As you might guess, this is one motivation for the use of the symbol ${\displaystyle \subset }$ , since it is similar in appearance to <.
Let n = |A|. Then number the elements of A, so ${\displaystyle A=\{x_{1},x_{2},\ldots ,x_{n}\}.}$ Then for each i from 1 to n we see that ${\displaystyle x_{i}\in B}$ , which means that B has at least n different elements, or that ${\displaystyle |B|\geq n=|A|,}$ which is what we were trying to prove.
Now if we use this lemma twice on Theorem 2.1.4, we will get a very brief proof. Since ${\displaystyle A\subset B}$ we know that ${\displaystyle |A|\leq |B|.}$ Also, since ${\displaystyle B\subset A}$ , we see that ${\displaystyle |B|\leq |A|.}$ Now we use a fact about numbers, that if ${\displaystyle x\leq y}$ and ${\displaystyle y\leq x}$ , it must follow that x = y.
### Corollary
A corollary is similar to a lemma in that it is usually a small and not as important as a theorem. However, a corollary is usually a result that follows almost immediately from a theorem. Corollaries tend to use a few well-known or established theorems and the primary theorem to prove. This is why corollaries are often found appended after a big theorem.
For example, Let's suppose we proved the theorem that "All people are pigs." Then a corollary would be "People who have heads are pigs." which clearly follows from the first result since "People have heads" is well-known and true (A person without a head would be rather odd, no?). Another, slightly more interesting, corollary would be "People can be sold for bacon when they die." since "bacon comes from pigs" is well-known and true.
So we see that a corollary is something that follows from a preceding theorem with minimal argument to support it. Note that theorems you declare as a corollary may not be so to others, as corollaries are subjective. However, thinking of a corollary as relying on either "common sense" or "obvious" to the reader that it is a direct consequence of the theorem is the proper thought when figuring out what to assign as a corollary.
## Exercises
1. Prove that the following sets are equal. Verify it with a truth table or a Venn Diagram. You may assume that A, B, and C are nonempty sets. Also assume that U is the universe.
1. ${\displaystyle A\cup (B\cap C)=(A\cup B)\cap (A\cup C)}$
2. ${\displaystyle A\cap (B\cup C)=(A\cap B)\cup (A\cap C)}$
3. ${\displaystyle A^{c}-B=(A\cup B)^{c}}$
4. ${\displaystyle A-B^{c}=A\cap B}$
5. ${\displaystyle (A-B)\cup (A\cap C)=A-(B-C)}$
2. Prove that if A and B are finite sets then ${\displaystyle |A\cup B|\leq |A|+|B|}$ and that equality holds when ${\displaystyle A\cap B=\varnothing .}$
3. Prove that the square of an odd number is an odd number.
Question Hint
Definition An odd number is defined as 2n + 1, where n is a natural number that can also be equal to 0.
Methodology Hint
The n in the form 2n + 1 doesn't have to be a number. It can be an equation.
• If you know that ${\displaystyle |A|\leq |B|,}$ can you show that ${\displaystyle A\subset B?}$
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# Subtraction: count back/take away
Maths Year 1
This unit is part of Year 1 Flexible Maths Blocks Addition and Subtraction and Year 1 Maths Planning Addition and Subtraction (A)
### Planning and Activities
Day 1 Teaching
Draw faces on a bus on a w/b. Model how we subtract by counting back when 1 or 2 get off the bus. There are 7 on the bus and 2 get off. 7, 6, 5. We count back 2. Record 7 – 2 = 5. Repeat.
Group activities
-- Work out how many dogs are left in the rescue centre and how many people are left on the bus.
Day 2 Teaching
Show 5 red & 5 yellow pegs on a hanger. How many pegs? Children put up 10 fingers. Remove 1 peg. Children fold 1 finger. How many are left? What number sentence can we write? Repeat.
Group activities
Use the 'Spot the Difference' in-depth problem-solving investigation below as today’s group activity.
Or use these activities:
-- Make a tower of cubes and take away the number on the dice. Record the number sentence.
Day 3 Teaching
Show 5 beads on a 20-bead bar. Count on 2, saying 6, 7 as you slide beads. Children mimic on fingers. Record 5 + 2 = 7. How many if we take 2 beads away? Put the 2 beads back. Children fold 2 fingers. We are back where we started!
Group activities
-- Work out how much money is left when a child spends some and then earns some back.
-- Explore the relationship between adding and subtracting.
Day 4 Teaching
Count 5 cubes into a tin and ask how many. Drop another cube in. And now? Repeat. And now? Record addition. Take out 2 cubes, 1 by 1. How many now? Record subtraction. Repeat.
Group activities
Use the 'Spot the Difference' in-depth problem-solving investigation below as today’s group activity.
Or use these activities:
-- Game with towers of cubes. Taking away and adding numbers to see who can reach 10 or 0 cubes the first!
Day 5 Teaching
Show a word problem about dogs in a rescue centre. Model how to solve this using fingers or cubes. Write a matching calculation and repeat with other stories.
Group activities
-- Practical number stories using equipment to work them out and creating leaf posters to show the number sentences.
### You Will Need
• Picture of a bus drawn on large paper
• Pictures of dogs (see resources).
• Pictures of buses - differentiated (see resources).
• Dice labelled with digits 1, 2.
• 5 red pegs and 5 yellow pegs.
• Coat hanger.
• Number cards 1-5, 5-10 & 5-15.
• Dice labelled with digits 2, 3.
• 10 £1 coins.
• Cubes.
• Hat/tin
• Large dice labelled +1, +2, +2, –1, –2 and –2.
• Multilink & other counting resources.
### Mental/Oral Maths Starters
Day 1
Count back (pre-requisite skills)
Suggested for Day 2
Counting (simmering skills)
Day 3
Count back 2 (pre-requisite skills)
Suggested for Day 4
Bonds to 5 (simmering skills)
Suggested for Day 5
Numbers to 20 (simmering skills)
### Worksheets
Day 1
How many frogs on the lily pads when some hop into the pond?
Work out the number of passengers on a bus.
Day 2
Challenge to work out how tall the tower is after cubes have fallen off.
Day 3
Day 4
Add or subtract the number of petals to find the total.
Day 5
Work out dog rescue story problems.
### Mastery: Reasoning and Problem-Solving
• Set up using small world play equipment.
There are 7 ducks in the pond. If 2 get out, how many are left?
Write a matching number sentence.
If 3 more get in, how many ducks are on the pond?
Write a matching number sentence.
If 2 get out, how many are left on the pond?
Write a matching number sentence.
• Draw a number track from 1 to 10. Place your counter on a number. Write it. Now hop along 2. Where do you end up? Write the addition. Repeat.
1 2 3 4 5 6 7 8 9 10
• More challenging
Write the missing numbers:
7 – 3 = ☐
5 + ☐ = 6
☐ + 2 = 5
In-depth Investigation: Spot the Difference
Children find dominoes where one side has one more spot than the other, and then where one side has two more spots than the other.
### Extra Support
What's Before Beryl?
Saying the number before any number up to 6, then 10
Dig for Acorns
Subtracting 1 from any number from 2 to 6, and then from 2 to 10, by saying the number which comes before
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# 10. Partial Derivatives
by M. Bourne
So far in this chapter we have dealt with functions of single variables only. However, many functions in mathematics involve 2 or more variables. In this section we see how to find derivatives of functions of more than 1 variable.
This section is related to, but is not the same as Implicit Differentiation that we met earlier.
### Example 1 - Function of 2 variables
Here is a function of 2 variables, x and y:
F(x,y) = y + 6 sin x + 5y2
To plot such a function we need to use a 3-dimensional co-ordinate system.
## Partial Differentiation with respect to x
"Partial derivative with respect to x" means "regard all other letters as constants, and just differentiate the x parts".
In our example (and likewise for every 2-variable function), this means that (in effect) we should turn around our graph and look at it from the far end of the y-axis. We are looking at the x-z plane only.
We see a sine curve along the x-axis and this comes from the "6 sin x" part of our function F(x,y) = y + 6 sin x + 5y2. The y parts are regarded as constants (in fact, 0 in this case).
Now for the partial derivative of
F(x,y) = y + 6 sin x + 5y2
with respect to x:
(del F)/(del x)=6 cos x
The derivative of the 6 sin x part is 6 cos x. The derivative of the y-parts is zero since they are regarded as constants.
Notice that we use the curly symbol to denote "partial differentiation", rather than "d" which we use for normal differentiation.
NOTE: You can explore this example using this 3D interactive applet in the Vectors chapter. In the drop-down list of examples, this is the last one.
## Partial Differentiation with respect to y
The expression
Partial derivative with respect to y
means
"Regard all other letters as constants, just differentiate the y parts".
As we did above, we turn around our graph and look at it from the far end of the x-axis. So we see (and consider things from) the y-z plane only. (Notice the horizontal axis on the following graph is labeled y, not x.)
We see a parabola. This comes from the y2 and y terms in F(x,y) = y + 6 sin x + 5y2. The "6 sin x" part is now regarded as a constant. (It actually has the effect of moving the base of the parabola z=y+5y^2 down by 6 units.)
Now for the partial derivative of
F(x,y) = y + 6 sin x + 5y2
with respect to y.
(delF)/(dely)=1+10y
The derivative of the y-parts with respect to y is 1 + 10y. The derivative of the 6 sin x part is zero since it is regarded as a constant when we are differentiating with respect to y.
NOTE: Once again, you can explore this particular example (rotate it, view it from different axes, etc) using the 3D interactive applet in the Vectors chapter. In the drop-down list of examples, select the last one.
## Second Order Partial Derivatives
We can find 4 different second-order partial derviatives. Let's see how this works with an example.
### Example 2
For the function we used above, F(x,y) = y + 6 sin x + 5y2, find each of the following:
(a) (del^2F)/(delydelx)
This could also be written as
del/(dely)[(delF)/(delx)]
This expression means:
"First, find the partial derivative with respect to x of the function F (this is in brackets), then find the partial derivative with respect to y of the result ".
In our example above, we found
(delF)/(delx)=6 cos x
To find (del^2F)/(delydelx), we need to find the partial derivative with respect to y of (delF)/(delx).
(del^2F)/(delydelx)=del/(dely)[(delF)/(delx)]
=del/(dely)[6 cos x]
=0
Since cos x is a constant (when we are considering differentiation with respect to y), its derivative is just 0.
(b) (del^2F)/(delxdely)
This could also be written as
del/(delx)[(delF)/(dely)]
This expression means
Find the partial derivative with respect to x of the partial derivative with respect to y.
In our example above, F(x,y) = y + 6 sin x + 5y2, we found
(delF)/(dely)=1+10y
To find (del^2F)/(delxdely), we need to find the partial derivative with respect to x of (delF)/(dely).
(del^2F)/(delxdely)=del/(delx)[(delF)/(dely)]
=del/(delx)[1+10y]
=0
Since y is a constant (when we are considering differentiation with respect to x), its derivative is just 0.
(c) (del^2F)/(delx^2)
This could also be written as
del/(delx)[[delF)/(delx)]
This expression means
Find the partial derivative with respect to x of the partial derivative with respect to x.
In our example above, we found
(delF)/(delx)=6 cos x
To find (del^2F)/(delx^2), we need to find the derivative with respect to x of (delF)/(delx).
(del^2F)/(delx^2)=del/(delx)[[delF)/(delx)]
=del/(delx)[6 cos x]
=-6 sin x
(d) (del^2F)/(dely^2)
This could also be written as
del/(dely)[(delF)/(dely)]
This expression means:
Find the partial derivative with respect to y of the partial derivative with respect to y.
In our example above, F(x,y) = y + 6 sin x + 5y2, we found
(delF)/(dely)=1+10y
To find (del^2F)/(dely^2), we need to find the derivative with respect to y of (delF)/(dely).
(del^2F)/(dely^2)=del/(dely)[(delF)/(dely)]
=del/(dely)[1+10y]
=10
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## Calculus (3rd Edition)
Equation of the ellipse: ${\left( {\frac{{x - 1}}{6}} \right)^2} + {\left( {\frac{y}{{3\sqrt 3 }}} \right)^2} = 1$
By Theorem 1, we have $P{F_1} + P{F_2} = 2a$. So, $a=6$. Since the foci are ${F_1} = \left( {4,0} \right)$ and ${F_2} = \left( { - 2,0} \right)$, the center of the ellipse is at $C = \left( {\frac{{4 - 2}}{2},0} \right) = \left( {1,0} \right)$. The ellipse centered at the origin would have equation ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$. If the center is translated to $C=\left(1,0\right)$, the equation has the form: (1) ${\ \ \ }$ ${\left( {\frac{{x - 1}}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$. Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the ellipse as if it is in the standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the ellipse when it is centered at $C=\left(1,0\right)$. Since the foci are ${F_1} = \left( {4,0} \right)$ and ${F_2} = \left( { - 2,0} \right)$ and the center of the ellipse is at $C=\left(1,0\right)$. In standard position the foci are at $\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)$. By Theorem 1, we have $c = \sqrt {{a^2} - {b^2}}$. So, $b = \sqrt {{a^2} - {c^2}} = \sqrt {36 - 9} = \sqrt {27} = 3\sqrt 3$ Substituting $a$ and $b$ in equation (1) gives ${\left( {\frac{{x - 1}}{6}} \right)^2} + {\left( {\frac{y}{{3\sqrt 3 }}} \right)^2} = 1$
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