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# The king, the jack and the 10 of spades are lost from a pack of 52 cards and a card is drawn from the remaining cards after shuffling.
Question:
The king, the jack and the 10 of spades are lost from a pack of 52 cards and a card is drawn from the remaining cards after shuffling. Find the probability of getting a
(i) red card
(ii) black jack
(iii) red king
(iv) 10 of hearts.
Solution:
Total number of cards in a deck is 52.
The number of cards left after loosing the King, the Jack and the 10 of spade = 52-">3 = 49.
(i) Red card left after loosing King, the Jack and the 10 of spade will be 13 hearts + 13 diamond = 26 cards
$P($ red card $)=\frac{26}{49}$
(ii) Black jack left after loosing King, the Jack and the 10 of spade will be one only of club.
$P($ black jack $)=\frac{1}{49} .$
(iii) Red king left after loosing King, the Jack and the 10 of spade will be red of diamond and red of hearts i.e two cards.
$P($ red king $)=\frac{2}{49}$
(iv) 10 of hearts will be one only after loosing King, the Jack and the 10 of spade
$P(10$ of hearts $)=\frac{1}{49} .$
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# 1957 AHSME Problems/Problem 39
## Problem
Two men set out at the same time to walk towards each other from $M$ and $N$, $72$ miles apart. The first man walks at the rate of $4$ mph. The second man walks $2$ miles the first hour, $2\tfrac {1}{2}$ miles the second hour, $3$ miles the third hour, and so on in arithmetic progression. Then the men will meet:
$\textbf{(A)}\ \text{in 7 hours} \qquad \textbf{(B)}\ \text{in }{8\frac {1}{4}}\text{ hours}\qquad \textbf{(C)}\ \text{nearer }{M}\text{ than }{N}\qquad\\ \textbf{(D)}\ \text{nearer }{N}\text{ than }{M}\qquad \textbf{(E)}\ \text{midway between }{M}\text{ and }{N}$
## Solution
Let $t$ be the time it takes (in hours) for the men to meet. Then, the distance the first man will have travelled is $4t$. If $t$ is an integer, the second man will have travelled $\tfrac{2+2+0.5(t-1)}{2} \cdot t$, the sum of the first $t$ terms of the arithmetic progression. If $t$ is not an integer, then we should have a good enough approximation to choose an answer choice (given the relatively large distance of $72$ miles compared to their speeds). Because the combined distance that the men travel must be $72$ miles, we can now solve for $t$ in the following formula: \begin{align*} 4t+\frac{2+2+0.5(t-1)}{2} \cdot t &= 72 \\ 4t+\frac{4+0.5t-0.5}{2} \cdot t &= 72 \\ 8t+3.5t+0.5t^2 &= 144 \\ t^2+23t &= 288 \\ t^2+23t-288 &= 0 \\ (t+32)(t-9) &= 0 \end{align*} Because $t>0$, we are left with $t=9$. At $t=9$, the first man will have travelled $36$ miles, which is half of te distance between $M$ and $N$. Thus, our answer is $\boxed{\textbf{(E)} \text{ midway between } M \text{ and } N}$.
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# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.6 | Set 1
### Question 1. Determine the nature of the roots of following quadratic equations :
(i) 2x² – 3x + 5 = 0
(ii) 2x² – 6x + 3 = 0
(iii) 3/5 x² – 2/3 x + 1 = 0
(iv) 3x² – 4√3 x + 4 = 0
(v) 3x² – 2√6 x + 2 = 0
Solution:
(i) 2x² – 3x + 5 = 0
Here a=2, b=-3, c=5
D=b2-4ac=(-3)2-4*2*5
=-9-40=-31
D<0
Roots are not real
(ii) 2x² – 6x + 3 = 0
Here a=2, b=-6, c=3
D=b2-4ac
=(-6)2-4*2*3=36-24=12
D>0
Roots are real and distinct
(iii) 3/5 x² – 2/3 x + 1 = 0
Here a=3/5, b=-2/3, c=1
Discriminant (D)=b2-4ac
D<0
Roots are not real
(iv) 3x² – 4√3 x + 4 = 0
Here a=3, b=-4√3, c=4
D=b2-4ac
=(-4√3)2-4*3*4=48-48=0
D=0
Roots are real and equal
(v) 3x² – 2√6 x + 2 = 0
Here a=3, b=– 2√6, c=2
Discriminant (D)=b2-4ac
=(– 2√6)2-4*3*2=24-24=0
D=0
Roots are real and equal
### Question 2. Find the values of k for which the roots are real and equal in each of the following equations :
(i) kx² + 4x + 1 = 0
Solution:
Here a=k, b=4, c=1
Discriminant(D)=b2-4ac
=(4)2-4*k*1
=16-4k
Roots are real and equal
D=0
16-4k=0⇒4k=16
k=16/4=4
Hence k=4
(ii) kx² – 2√5 x + 4 = 0
Solution:
Here a=k, b=-2√5, c=4
Discriminant(D)=b2-4ac
=( – 2√5 )-4*k*4=20-16k
Roots are real and equal
D=0
20-16k=0⇒ 16k=20
k=20/16=5/4
Hence k=5/4
(iii) 3x² – 5x + 2k = 0
Solution:
Here a=3, b=-5, c=2k
Discriminant (D)=b2-4ac
=(-5)2-4*3*2k
=25-24k
Roots are real and equal
D=0
25-24k=0⇒24k=25
k=25/24
(iv) 4x²+ kx + 9 = 0
Solution:
Here a=4, b=k, c=9
Discriminant (D)=b2-4ac
=k2-4*4*9=k2-144
Roots are real and equal
k2-144=0⇒k2=144=(±12)2
(v) 2kx² – 40x + 25 = 0
Solution:
Here a=2k, b=-40, c=25
Discriminant(D)=b2-4ac
=(-40)2-4*2k*25
=1600-200k
Roots are real and equal
D=0
1600-200k=0⇒200k=1600
k=1600/200=8
Hence k=8
(vi) 9x² – 24x + k = 0
Solution:
Here a=9, b=-24, c=k
Discriminant(D)=b2-4ac
=(-24)2-4*9*k
=576-36k
Roots are real and equal
D=0
576-36k=0
36k=576⇒k=576/36=16
k=16
(vii) 4x² – 3kx +1 = 0
Solution:
Here a=4, b=-3k, c=1
Discriminant (D)=b2-4ac
=(-3k)2-4*4*1
=9k2-16
Roots are real and equal
D=0
9k2-16=0⇒9k2=16
k2=16/9=
(viii) x² – 2 (5 + 2k) x + 3 (7 + 10k) = 0
Solution:
Here a=1, b=-2(5+2k) and c=3(7+10k)
Discriminant (D)=b2-4ac
[-2(5+2k)]2-4*1*3(7+10k)
=4(25+4k2+20k)-12(7+10k)
=100+16k2+80k-84-120k
16k2-40k+16
Roots are real and equal
D=0
16k2-40k+16=0
2k2-5k+2=0
2k2-4k-k+2=0
2k(k-2)-1(k-2)=0
(k-2)(2k-1)=0
Either k-2=0, then k=2 or 2k-1=0, then 2k=1⇒k=1/2
Hence, k=2, 1/2
(ix) (3k + 1) x² + 2(k + 1) x + k = 0
Solution:
Here a=3k+1, b=2(k+1), c=k
Discriminant(D)=b2-4ac
=[2(k+1)]2-4*(3k+1)*k
=4(k2+2k+1)-4k(3k+1)
=4k2+8k+4-12k2-4k
-8k2+4k+4
Roots are real and equal
D=0
-8k2+4k+4=0
2k2-k-1=0 (Dividing by -4)
2k2-2k+k-1= {Therefore -2=-2*1
-1=-2+1}
2k(k-1)+1(k-1)=0
(k-1)(2k+1)=0
Either k-1=0, then k=1 or 2k+1=0, then 2k=-1⇒k=-1/2
k=1,-1/2
(x) kx² + kx + 1 = – 4x² – x
Solution:
kx² +4x2+kx+x+1=0
(k+4)x2+(k+1)x+1=0
Here a=k+4, b=k+1, c=1
Discriminant(D)=b2-4ac
=(k+1)2-4*(k+4)*1
=k2+2k+1-4k-16
=k2-2k-15
Roots are real and equal
D=0
k2-2k-15=0
k2-5k+3k-15=0 {Therefore -15=-5*3
-2=-5+3}
k(k-5)+3(k-5)=0
(k-5)(k+3)=0
Either k-5=0, then k=5
or k+3=0, then k=-3
Hence k=5,-3
(xi) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0
Solution:
Here a=k+1, b=2(k+3), c=k+8
Discriminant(D)=b2-4ac
=[2(k+3)]2-4(k+1)(k+8)
=4(k2+6k+9)-4)(k2+9k+8)
=4k2+24k+36-4k2-36k-32
=-12k+4
Roots are real and equal
D=0
-12k+4=0
12k=4⇒k=4/12=1/3
Hence k=1/3
(xii) x² – 2kx + 7k – 12 = 0
Solution:
Here a=1, b=-2k, c=(7k-12)
Discriminant(D)=b2-4ac
=(-2k)2-4*1*(7k-12)
=4k2-4(7k-12)
=4k2-28k+48
Roots are real and equal
D=0
4k2-28k+48=0
k2-7k+12=0 (Dividing by 4 )
k2-3k-4k+12=0 {12=-3*(-4)
-7=-3-4}
k(k-3)-4(k-3)=0
(k-3)(k-4)=0
Either, k-3=0, then k=3
or k-4=0, then k=4
Therefore, k=3,4
(xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0
Solution:
Here a=k+1, b=-2(3k+1), c=8k+1
Discriminant(D)=b2-4ac
=[-2(3k+1)2-4*(k+1)(8k+1)]
=4(9k2+6k+1)-4(8k2+9k+1)
=36k2+24k+4-32k2-36-4
=4k2-12k
Roots are real and equal
D=0
4k2-12k=0
k2-3k=0 ————–(Dividing by 4)
k(k-3)=0
Either k=0
or k-3=0, then k=3
k=3,0
(xiv) 5x² – 4x + 2 + k (4x² – 2x – 1) = 0
Solution:
5x2-4x+2+4kx2-2kx-k=0
(5+4k)x2-(4+2k)x+(2-k)=0
Here a=5+4k, b=-(4+2k), c=2-k
Discriminant (D)=b2-4ac
=[-(4+2k)]2-4*(5+4k)(2-k)
=16+4k2+16k-4(10-5k+8k-4k2)
=16+4k2+16k-40+20k-32k+16k2
=20k2+4k-24
Roots are real and equal
D=0
20k2+4k-24=0
5k2+k-6=0 —–(Dividing by 4)
5k2+6k-5k-6=0
k(5k+6)-1(5k+6)=0
(5k+6)(k-1)=0
Either 5k+6=0, then 5k=-6⇒k=-6/5
or k-1=0, then k=1
k=1,
(xv) (4 – k) x² + (2k + 4) x (8k + 1) = 0
Solution:
Here a=4-k, b=2k+4, c=8k+1
Discriminant (D)=b2-4ac
=(2k+4)2-4*(4-k)(8k+1)
=4k2+16k+16-4(32k+4-8k2-k)
=4k2+16k+16-128k-16+32k2+4k
=36k2-108k
Roots are real and equal
36k2-108k=0
k2-3k=0⇒k(k-3)=0
Either k=0
or k-3=0, then k=3
Hence k=0,3
(xvi) (2k + 1) x² + 2 (k + 3) x (k + 5) = 0
Solution:
Here a=2k+1, b=2(k+3), c=k+5
Discriminant (D)=b2-4ac
=[2(k+3)]2-4(2k+1)(k+5)
=4(k2+6k+9)-4(2k2+10k+k+5)
=4k2+24k+36-8k2-40k-4k-20
=-4k2-20k+16
Roots are real and equal D=0
-4k2-20k+16=0
k2+5k-4=0 ———-(Dividing by -4)
Here a=1, b=5, c=-4
Discriminant (D)=b2-4ac
=(5)2-4*1*(-4)=25+16=41
(xvii) 4x² – 2 (k + 1) x + (k + 4) = 0
Solution:
Here a=4, b=-2(k+1), c=k+4
Discriminant (D)=b2-4ac
=[-2(k+1)]2-4*4*(k+4)
=4(k2+2k+1)-16(k+4)
=4k2+8k+4-16k-64
=4k2-8k-12
Roots are real and equal
D=0
4k2-8k-60=0
k2-2k-15=0 ————-(Dividing by 4)
k2-5k+3k-15=0
k(k-5)+3(k-5)=0
(k-5)(k+3)=0
Either k-5=0, then k=5
or k+3=0, then k=-3
k=5,-3
(xviii) 4x² (k + 1) x + (k + 1) = 0
Solution:
Here a=4, b=-2(k+1), c=k+1
Discriminant (D)=b2-4ac
=[-2(k+1)]2-4*4*(k+1)
=4(k2+2k+1)-16(k+1)
=4k2+8k+4-16k-16
=4k2-8k-12
k2-2k-3=0 ————(Dividing by 4)
k2-3k2+k-3=0
k(k-3)+(k-3)=0
(k-3)(k+1)=0
Either (k-3)=0, then k=3
or (k+1)=0, then k=-1
k=-1,3
### Question 3. In the following, determine the set of values of k for which the given quadratic equation has real roots:
(i) 2x² + 3x + k = 0
Solution:
Here a=2, b=3, c=k
Discriminant (D)=b2-4ac
=(3)2-4*2*k
=9-8k
The roots are real
D≥0
9-8k≥0⇒9≥8k⇒8k≤9
k≤9/8
(ii) 2x² + x + k = 0
Solution:
Here a=2, b=1, c=k
Discriminant (D)=b2-4ac
=(1)2-4*2*k
=1-8k
The roots are real
D≥0
1-8k≥0⇒1≥8k
8k≤1
k≤1/8
(iii) 2x² – 5x – k = 0
Solution:
Discriminant (D)=b2-4ac
=(-5)2-4*2*(-k)
=25+8k
Roots are real
D≥0
25+8k≥0
8k≥-25⇒≥-25/8
k≥-25/8
(iv) kx² + 6x + 1 = 0
Solution:
Discriminant (D)=b2-4ac
=(6)2-4*k*1
=36-4k
Roots are real
D≥0⇒36-4k≥0
36≥4k⇒4k≤36
k≤36/4⇒k≤9
k≤9
(v) 3x² + 2x + k = 0
Solution:
Here a=3, b=2, c=k
Discriminant (D)=b2-4ac
=(2)2-4*3*k=4-12k
Roots are real
D≥0⇒4-12k≥0
4≥12k⇒12k≤0
4≥12k⇒12k≤4
k≤⇒k≤1/3
### Question 4. Find the values of k for which the following equations have real and equal roots :
(i) x²- 2(k + 1) x + k² = 0
Solution:
Here a=1, b=2(k+1), c=k2
Discriminant (D)=b2-4ac
=[2(k+1)]2-4*1*k2
=4(k2+2k+1)-4k2
=4k2+8k+4-k2
=8k+4
Roots are real and equal
D=0
8k+4=0⇒8k=-4
k=-4/8=-1/2, Hence k=-1/2
(ii) k²x² – 2 (2k – 1) x + 4 = 0
Solution:
Here, a=k2, b=-2(2k-1), c=4
Discriminant (D)=b2-4ac
=[-2(2k-1)]2-4*k2*4
=4(4k2-4k+1)-16k2
=16k2-16k+4-16k2=-16k+4
Roots are real and equal
D=0
-16k+4=0⇒-16k=-4
k=4/16=1/4
k=1/4
(iii) (k + 1) x² – 2(k – 1) x + 1 = 0
Solution:
Here, a=k+1, b=-2(k-1) and c=1
Discriminant (D)=b2-4ac
=[-2(k-1)]2-4(k+1)*1
=4(k2-2k+1)-4(k+1)
=4k2-8k+4-4k-4=4k2-12k
Roots are real and equal
D=0
4k2-12k=0
k2-3k=0 ————-(Dividing by 4)
Either k=0
or k-3=0, then k=3
k=0,3
(iv) x² + k(2x + k – 1) + 2 = 0
Solution:
Here, a=1, b=2k, c=(k2-k+2)
Discriminant (D)=b2-4ac
=(2k)2-4*1*(k2-k+2)
=4k2-4k2+4k-8
=4k-8
Roots are real and equal
D=0
4k-8=0⇒k=2
Hence, k=2
### Question 5. Find the values of k for which the following equations have real roots
(i) 2x² + kx + 3 = 0
Solution:
Here a = 2, b = k, c = 3
Roots are real and equal
D=0
k2-24=0⇒k2=24
k=±√24=±√4*6=±2√6
(ii) kx (x – 2) + 6 = 0
Solution:
kx2-2kx+6=0
here, a=k, b=-2k, c=6
Discriminant (D)=b2-4ac=(-2k)2-4*k*6=4k2-24k
Roots are real and equal
D=0
4k2-24k=0⇒4k(k-6)=0
k(k-6)=0
Either k=0 or
k-6=0, then k=6
k=0,6
(iii) x² – 4kx + k = 0
Solution:
Here, comparing with ax2+kx+c=0
a=1, b=-4k, c=k
Discriminant (D)=b2-4ac=(-4k)2-4*1*k=16k2-4k
Roots are real and equal
D=0
16k2-4k=0⇒4k(4k-1)=0
k(4k-1)=0
Either k=0
or 4k-1=0⇒4k=1
k=1/4, Hence k=0,1/4
(iv) kx(x – 2√5) + 10 = 0
Solution:
Here a=k, b=-2√5k, c=10
D=b2-4ac
=(-2√5k)2-4(k)(10)=20k2-40k
Roots are equal D=0
20k2-40k=0
k-2=0 ———(Dividing by 20k)
k=2
(v) kx (x – 3) + 9 = 0
Solution:
Here, a=k, b=-3k, c=9
D=b2-4ac
=(-3k)2-4(k)9
=9k2-36k
For roots to be real
D=0
9k2-36k=0
9k(k-4)=0
k-4=0⇒k=4
k=4
(vi) 4x² + kx + 3 = 0
Solution:
Here, a=4, b=k, c=3
D=b2-4ac
=k2-4(4)(3)
=k2-48
For roots to be real
D=0
k2-48=0
k2=48
k=±√48=±
k=±
### Question 6. Find the values of k for which the given quadratic equation has real and distinct roots :
(i) kx² + 2x + 1 = 0
Solution:
Here, a=k, b=2, c=1
D=b2-4ac
=(2)2-4*k*1
=4-4k
Roots are real and distinct
D>0⇒4-4k>0
1-k>0⇒1>k
⇒k<1
Therefore, k<1
(ii) kx² + 6x + 1 = 0
Solution:
Here, a=k, b=6, c=1
D=b2-4ac
=(6)2-4*k*1
=36-4k
Roots are real and distinct
D>0⇒36-4k>0
9-k>0⇒9>k
⇒k<9
Therefore, k<9
### Question 7. For what value of k, (4 – k) x² + (2k + 4) x + (8k + 1) = 0, is a perfect square.
Solution:
(4 – k) x² + (2k + 4) x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1
=(2k+4)2-4*(4-k)(8k+1)
=4k2+16k+16-4(32k+4-8k2-k)
=4k2+16k+16-4(-8k2+31k+4)
=4k2+16k+16+32k2-124k-16
=36k2-108k
Therefore, the given quadratic equation is a perfect square
The roots are real and equal
D=0⇒36k2-108k=0
Either k=0
or k-3=0⇒k=3
k=0,3
### Question 8. Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:
x² + kx + 4 = 0
Here, a=1, b=k, c=4
Therefore, Discriminant(D)=b2-4ac
=(k)2=4*1*4
=k2-16
It has real roots
D≥0⇒k2-16≥0
⇒k2≥16⇒(k)2≥(±4)2
k≥4 or k≤-4
Least positive value of k=4
### Question 9. Find the value of k for which the quadratic equation (3k + 1) x² + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.
Solution:
(3k + 1) x² + 2(k + 1) x + 1 = 0
Here a=(3k+1), b=2(k+1) ,c=1
Now, b2-4ac=[2(k+1)]2-4*(3k+1)*1
=4(k2+2k+1)-4(3k+1)
=4k2+8k+4-12k-4
=4k2-4k
Roots are real and equal
b2-4ac=0
4k2-4k=0
k2-k=0 k(k-1)=0
Either k=0 or k-1=0, then k=1
k=0,1
(i) When k=0, then
(3*0*1)x2+2(0+1)x+1=0
x2+2x+1=0
(x+1)2=0
x+1=0
x=-1
When k=1, then
(3*1+1)x2+2(1+1)x+1=0
4x2+4x2+1=0
(2x+1)2=0
2x+1=0
2x=-1⇒x=-1/2
x=1, -1/2
### Question 10. Find the values of p for which the quadratic equation (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has equal roots. Also, find these roots.
Solution:
Here, a=2-+1, b=-(7p+2), c=(7p-3)
D=0 [Equal roots]
As b2-4ac=0
[-(7p+)]2-4(2p+1)(7p-3)=0
(7p+2)2-4(14p2-6p+7p-3)=0
49p2+28p+4-56p2+24p-28p+12=0
-7p2+24p+16=0
7p2-24-16=0 ————-(Dividing both sides by -1)
7p(p-4)+4(p-4)=0
(p-4)(7p+4)=0
p-4=0 or 7p+4=0
p=4 or p=-4/7
Roots are x=-b/2a [As equal roots (given)]
Where p=4,
Equal roots is 5/3
When p=-4/7
Equal roots are 5/3 and 7
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# How do you factor y= 8m^2 - 41m - 42?
Dec 27, 2015
Find a suitable splitting of the middle term, then factor by grouping to find:
$8 {m}^{2} - 41 m - 42 = \left(8 m + 7\right) \left(m - 6\right)$
#### Explanation:
Find a pair of factors of $A C = 8 \cdot 42 = 16 \cdot 21 = 336$ which differ by $B = 41$
The split of $336$ into a pair of factors must put all of the powers of $2$ on one side, since the difference ($41$) is odd. If both factors were even, then the difference would be even too.
That leads to the following possibilities to consider:
$16 \times 21$
$\boldsymbol{48 \times 7}$
$112 \times 3$
$336 \times 1$
Having found the pair $48$, $7$ use that to split the middle term and factor by grouping:
$8 {m}^{2} - 41 m - 42$
$= 8 {m}^{2} - 48 m + 7 m - 42$
$= \left(8 {m}^{2} - 48 m\right) + \left(7 m - 42\right)$
$= 8 m \left(m - 6\right) + 7 \left(m - 6\right)$
$= \left(8 m + 7\right) \left(m - 6\right)$
Dec 27, 2015
$y = \left(8 x + 7\right) \left(x - 6\right)$
#### Explanation:
You could look for values $p , q , r , s$ such that
$\textcolor{w h i t e}{\text{XXX}} p \times r = 8$
$\textcolor{w h i t e}{\text{XXX}} q \times s = - 42$
$\textcolor{w h i t e}{\text{XXX}} p s + q r = - 41$
(perhaps using the AC method)
...but unless you get lucky, there are quite a few factorings possible.
As an alternative you could use the quadratic formula:
$\textcolor{w h i t e}{\text{XXX}} \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
The numbers involved are still ugly but if you use a calculator or spreadsheet (evaluating only the $+$ of the $\pm$)
you should get:
$\textcolor{w h i t e}{\text{XXX}} x = 6$ as a zero for this expression.
Therefore one of the factors will be:
$\textcolor{w h i t e}{\text{XXX}} \left(x - 6\right)$
Simple division ($8 {x}^{2} \div x = 8$) and ((-42)div(-6))=+7)
gives the other term:
$\textcolor{w h i t e}{\text{XXX}} \left(8 x + 7\right)$
Dec 27, 2015
Alternatively, complete the square to find:
$8 {m}^{2} - 41 m - 42 = \left(m - 6\right) \left(8 m + 7\right)$
#### Explanation:
Alternatively, you can complete the square to proceed directly to the answer as follows:
$8 {m}^{2} - 41 m - 42$
$= 8 \left({m}^{2} - \frac{41}{8} m - \frac{21}{4}\right)$
$= 8 \left({m}^{2} - \frac{41}{8} m + {\left(\frac{41}{16}\right)}^{2} - {\left(\frac{41}{16}\right)}^{2} - \frac{21}{4}\right)$
$= 8 \left({\left(m - \frac{41}{16}\right)}^{2} - \frac{1681}{256} - \frac{1344}{256}\right)$
$= 8 \left({\left(m - \frac{41}{16}\right)}^{2} - \frac{3025}{256}\right)$
$= 8 \left({\left(m - \frac{41}{16}\right)}^{2} - {\left(\frac{55}{16}\right)}^{2}\right)$
$= 8 \left(\left(m - \frac{41}{16}\right) - \frac{55}{16}\right) \left(\left(m - \frac{41}{16}\right) + \frac{55}{16}\right)$
$= 8 \left(m - \frac{96}{16}\right) \left(m + \frac{14}{16}\right)$
$= 8 \left(m - 6\right) \left(m + \frac{7}{8}\right)$
$= \left(m - 6\right) \left(8 m + 7\right)$
...using the difference of squares identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
with $a = m - \frac{41}{16}$ and $b = \frac{55}{16}$
Ouch!
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THERMO Spoken Here! ~ J. Pohl © ( A0920~1/15) ( A0930 - Eratosthanes' Measurement )
# Theorem of Pythagoras
The Pythagorean Theorem applies to right triangles. It is an algebraic truth relating the lengths of sides and hypotenuse. The therorm has many proofs. Here we use the fact that the area of an outer square equals the sum of its interior areas. A basic idea,
"The whole equals the sum of its parts,"
is used often in science and engineering. The image shows a square drawn inside of a larger square. The lengths A, B, C, and the acute angle θ are identified.
Prove the Theorem of Pythagoras.
♦ By the sketch, we see that the area of the circumscribed square equals the four areas of inscribed triangles plus the area of the inscribed square:
(A + B)2 = 4 (AB/2) + C2
By a previous proof we know: (A + B)2 = A2 + 2AB + B2. With this substitution we have:
A2 + 2AB + B2 = C2 + 4 (AB/2)
A touch of algebra and we have the Pythagorean Theorem:
A2 + B2 = C2. QED.
We can take this result a step further.
♦ First divide the theorem by C2, to obtain:
(A/C)2 + (B/C)2 = 1
Next realize for a right triangle, sin(θ) is defined as A/C and cos(θ) is defined as B/C. Make these substitutions into the theorem to obtain the famous formula:
sin2θ + cos2θ = 1 QED.
## Theorem of Pythagoras
The Pythagorean Theorem, a statement of an algebraic truth regarding the sides and hypotenuse of right triangles, can be proved by use of geometry and the general truth:
"The whole equals the sum of its parts."
Presently not completed
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# Integer partitioning
Suppose we have an integer $n$. I we want to partition the integer in the form of $2$ and $3$ only; i.e., $10$ can be partitioned in the form $2+2+2+2+2$ and $2+2+3+3$.
So, given an integer, how to calculate the total number of ways of doing such partitions and how many $2$'s and $3$'s are there in each of the partitions?
• Is the ordering of the 2s and 3s relevant? Or are we just looking at quantity of each? Can you add your results for numbers up to, say, 10? – Joffan Jun 5 '15 at 18:56
• @Joffan only quantity. – Anuj Garg Jun 15 '15 at 4:16
For even $n$, the number of $3$ in each partition is even. So, let $2k$ be the largest number of $3$ in the partition of even $n$, i.e. $$3\cdot 2k\le n\lt 3(2k+2)\Rightarrow k\le \frac{n}{6}\lt k+1\Rightarrow k=\left\lfloor\frac n6\right\rfloor.$$ Hence, the number of partitions of $\color{red}{\text{even}\ n}$ is $\color{red}{\left\lfloor\frac{n}{6}\right\rfloor+1}$ (note that the +1 comes from the case $k=0$).
For odd $n$, the number of $3$ in each partition is odd. So, let $2k-1$ be the largest number of $3$ in the partition of odd $n$, i.e. $$3(2k-1)\le n\lt 3(2k+1)\Rightarrow k\le \frac{n+3}{6}\lt k+1\Rightarrow k=\left\lfloor\frac{n+3}{6}\right\rfloor.$$ Hence, the number of partitions of $\color{red}{\text{odd}\ n}$ is $\color{red}{\left\lfloor\frac{n+3}{6}\right\rfloor}$.
The number of partitions of $n$ into parts in the set $\{2,3\}$ is the coefficient of $x^n$ in
$$\left(1+x^2+x^4+x^6+\ldots\right)\left(1+x^3+x^6+x^9+\ldots\right)=\frac1{(1-x^2)(1-x^3)}\;.$$
For $n=10$, for instance, we can ignore powers of $x$ higher than $10$, so we need only consider
$$(1+x^2+x^4+x^6+x^8+x^{10})(1+x^3+x^6+x^9)\;,$$
and by inspection the $x^{10}$ term is
$$x^4\cdot x^6+x^{10}\cdot x^0=2x^{10}\;:$$
the two partitions of $10$ that you listed in the question are the only ones.
However, we can do better. According to Wikipedia, the number of partitions of $n$ into parts of sizes $1,2$, and $3$ is the integer nearest to $\frac1{12}(n+3)^2$, i.e.,
$$\left\lfloor\frac{(n+3)^2}{12}+\frac12\right\rfloor\;;$$
call this $f(n)$, and let $g(n)$ be the number of partitions of $n$ into parts of size $2$ and $3$. Each partition of $n$ with parts in $\{1,2,3\}$ is a partition of some $k\le n$ into parts in $\{2,3\}$ together with some number of unit parts, so
$$f(n)=\sum_{k=2}^ng(k)\;,$$
and
$$g(n)=f(n)-f(n-1)=\left\lfloor\frac{(n+3)^2}{12}+\frac12\right\rfloor-\left\lfloor\frac{(n+2)^2}{12}+\frac12\right\rfloor\;.$$
As a quick check, for $n=10$ this becomes
$$\left\lfloor\frac{(13)^2}{12}+\frac12\right\rfloor-\left\lfloor\frac{(12)^2}{12}+\frac12\right\rfloor=14-12=2\;.$$
• And how to calculate number of 2's and number of 3's in each partition? – Anuj Garg Jun 6 '15 at 4:15
• In combinatorics what we call this type of approach? – Anuj Garg Jun 15 '15 at 4:29
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# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(2x)/(x^2+16)?
Sep 22, 2016
horizontal asymptote at y = 0
#### Explanation:
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve: ${x}^{2} + 16 = 0 \Rightarrow {x}^{2} = - 16$
This has no real solutions hence there are no vertical asymptotes.
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$
divide numerator/denominator by the highest power of x, that is ${x}^{2}$
$f \left(x\right) = \frac{\frac{2 x}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{16}{x} ^ 2} = \frac{\frac{2}{x}}{1 + \frac{16}{x} ^ 2}$
as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0}$
$\Rightarrow y = 0 \text{ is the asymptote}$
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(2x)/(x^2+16) [-10, 10, -5, 5]}
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# What is a Quadratic Monomial? (with Examples)
-
A quadratic monomial is an algebraic expression that consists of a single expression of degree two.
• The word “monomial” stands for the fact that there is only one term.
• The word quadratic means that the term has degree 2.
Let us first try to understand the meaning of the terms – “monomial” and “quadratic”.
#### What is a Monomial?
Consider the algebraic expression 3x2 – 4x3. It is a “binomial” since it consists of two terms that are separated by a minus sign.
The expression 7y3-8y2+13y is a “trinomial” since it consists of three terms that are separated by a plus or minus sign.
On the other hand, an expression of the form 3x^2 is a monomial since it consists only of a single term. Some examples of monomials are:
• 3x2.
• 7y8.
• -6t3.
#### What do we mean by Quadratic?
By a quadratic expression, we mean an expression where the total degree of each term adds up to two.
For example, the expression 7x2 – 6y2 is quadratic since both x and y have degree 2.
Similarly, the expression 3xy is quadratic since we have an x variable of degree 1 and a y variable of degree 1, and hence the total degree adds up to 2.
A quadratic monomial is an algebraic expression consisting of a single term of degree two. Some examples of quadratic monomials are:
1. 7x2.
2. 13y2.
3. -3xy. Here, since x and y both have degree 1 the total degree adds up to 2.
4. -5st.
From the above examples, we see that a quadratic monomial consists of two parts:
1. Coefficient – it is the numerical value that occurs before the alphabet such as the value 7 in 7x2.
2. Variables – There are the alphabets that occur in our expression such as xy in the expression -3xy.
#### What happens if you multiply a quadratic monomial by a cubic monomial?
If we multiply a quadratic monomial with a cubic monomial then we get a monomial whose total degree is equal to 5.
For example, consider the quadratic monomial 3y2 and the cubic monomial 6y3. On multiplying them we get,
3y2 x 6y3 = (3 x 6)(y2 x y3) =18y5.
Now, 18y5 is clearly a monomial and it has a degree of 5.
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# 8th Class Mathematics Exponents and Power Laws of Exponent
## Laws of Exponent
Category : 8th Class
### Laws of Exponent
There are various laws of exponents. They are laws of addition, laws of multiplication and laws of division.
(i) ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
(ii) $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
(iii) ${{a}^{m}}\times {{b}^{m}}={{(a\times b)}^{m}}$
(iv) ${{\left[ {{\left( \frac{a}{b} \right)}^{n}} \right]}^{m}}={{\left( \frac{a}{b} \right)}^{mn}}$
(v) ${{\left( \frac{a}{b} \right)}^{-n}}={{\left( \frac{b}{a} \right)}^{n}}$
(vi) ${{\left( \frac{a}{b} \right)}^{0}}=1$
(vii) ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$
Important Points to keep in Mind
• ${{x}^{0}}=1$, where $x\ne 0$. $x$ can be anything (except zero), including numbers, variables, or an equation.
• ${{x}^{1}}=x$
• ${{x}^{-n}}=\frac{1}{{{x}^{n}}}$, Where $x\ne 0$
(i) ${{2}^{3}}{{2}^{5}}={{2}^{3+5}}={{2}^{8}}$
(ii) ${{w}^{2}}{{w}^{3}}={{w}^{5}}$
(iii) $x{{y}^{2}}{{x}^{3}}{{y}^{3}}{{x}^{4}}{{y}^{4}}={{x}^{8}}{{y}^{9}}$
While working with exponents there are certain rules that we need to remember.
${{\text{4}}^{\text{2}}}\times {{\text{4}}^{\text{5}}}=4\text{7}$
It means:$\text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}\times \text{4}$ or $\text{4}\text{.4}\text{.4}\text{.4}\text{.4}\text{.4}\text{.4}$
Add the exponent, if base are same.
Uses of Exponents
The exponents can be used for various purposes such as comparing large and small numbers, expressing large and small numbers in the standard forms. It is used to express the distance between any two celestial bodies which cannot be expressed in the form of normal denotion. It is also useful in writing the numbers in scientific notation. The size of the microorganisms is very-very small and it cannot be written in normal denotion and can easily be expressed in exponential form.
Radicals are the fractional exponents of any number. Index of the radical becomes the denominator of the fractional power.
$\sqrt[n]{a}=\frac{1}{{{a}^{n}}}$
i.e. $\sqrt{9}=\sqrt[2]{9}={{9}^{\frac{1}{2}}}=3$
Express $\sqrt[\mathbf{3}]{\mathbf{2}}\,\sqrt[\mathbf{4}]{\mathbf{2}}$ as a Single Radical Term
Let us convert the radicals to exponential expressions, and then apply laws of exponent to combine the factors:
$\sqrt[3]{2}\,\,\sqrt[4]{2}={{2}^{\frac{1}{3}}}\,\,{{2}^{\frac{1}{4}}}={{2}^{\frac{1}{3}+\frac{1}{4}}}={{2}^{\frac{7}{12}}}=\sqrt[12]{{{2}^{7}}}$
Simplify $\frac{\sqrt{5}}{\sqrt[3]{5}}$
Solution:
$\frac{{{5}^{\frac{1}{2}}}}{{{5}^{\frac{1}{3}}}}={{5}^{\frac{1}{2}.\frac{1}{3}}}={{5}^{\frac{1}{6}}}$
${{\left( \frac{2}{3} \right)}^{4}}=\left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{2}{3} \right)$
Solution:
$=\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}=\frac{{{2}^{4}}}{{{3}^{4}}}=\frac{16}{81}$
Expand: ${{\left( \frac{x}{10} \right)}^{5}}$
Solution: Raising the top and bottom numbers to the power of 5 gives:
${{\left( \frac{x}{10} \right)}^{5}}=\frac{{{x}^{5}}}{{{10}^{5}}}=\frac{{{x}^{5}}}{100000}$
• Zero raised to the power is not always zero.
• The positive quantity raised to the power of negative index is always positive.
• Zero was discovered by the Babylonians in Mesopotamia in around 300 B.C.
• The rule of mathematics given by Brahma gupta is known as Brahmas phutasiddhanta.
• The number which appears maximum number of times in the decimal expansion of pie upto the 6 billion decimal places.
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$
${{\left( \frac{a}{b} \right)}^{n}}=\frac{{{a}^{n}}}{{{b}^{n}}}$
${{a}^{0}}=1$
${{a}^{-n}}=\frac{1}{{{a}^{n}}}$
Simplify: ${{(-5)}^{3}}$
(a) -125
(b) -120
(c) -105
(d) -121
(e) None of these
Explanation:
$(-5)(-5)(-5)=-125$
Number -5 has exponent 3. Therefore, option (a) is correct and rest of the options is incorrect.
Simplify: $-{{(3)}^{3}}-{{(-3)}^{2}}+{{(-2)}^{2}}$
(a) -31
(b) -21
(c) -32
(d) -33
(e) None of these
Explanation:
$-(3\times 3\times 3)-(-3\times -3)+(-2\times -2)=-27-9+4=-32$
Find the value of x such that ${{\left( \frac{64}{125} \right)}^{2}}{{\left( \frac{4}{5} \right)}^{4}}{{\left( \frac{16}{25} \right)}^{2x+1}}={{\left( \frac{256}{625} \right)}^{3x}}$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) $\frac{1}{3}$
(d) $\frac{1}{2}$
(e) None of these
By what number ${{\left( -\frac{4}{3} \right)}^{-5}}$ must be multiplied so that the result is $\frac{16}{9}$.
(a) ${{\left( \frac{4}{3} \right)}^{5}}$
(b) ${{\left( \frac{4}{3} \right)}^{7}}$
(c) ${{\left( \frac{4}{3} \right)}^{-5}}$
(d) ${{\left( \frac{4}{3} \right)}^{-7}}$
(e) None of these
The scientific notation of 165000000000000 is given by:
(a) $16.5\times {{10}^{13}}$
(b) $165\times {{10}^{12}}$
(c) $1650\times {{10}^{11}}$
(d) $1.65\times {{10}^{14}}$
(e) None of these
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# Cube numbers
## What are cube numbers?
A cube number is the product of three identical factors (e.g. 2 x 2 x 2 = 8). So, a cube number is any number which has been multiplied by itself, twice.
Sometimes, the language 'cubed' is used (e.g. '2 cubed' = 2 x 2 x 2 = 8).
## How to calculate cube numbers
Calculating a cube number is simple. To generate a cube number, multiply a number by itself. Then, multiply the product by the number again!
23 = 2 x 2 x 2
2 x 2 = 4
4 x 2 = 8
33 = 3 x 3 x 3
3 x 3 = 9
3 x 3 = 27
43 = 4 x 4 x 4
4 x 4 = 16
16 x 4 = 64
## Why are cube numbers called cube numbers?
Cube numbers are called cube numbers (or 'cubed numbers') because they can be visually represented as a three-dimensional cube: a shape with an equal number of units along its length, width, and height.
## Cube number notation
The symbol to demonstrate that a number is cubed is 3 (e.g. 23 =8). This superscript symbol shows that the number represents three dimensions (length, height and width).
The symbol that represents the cube root is 3√ (e.g.3√8 = 2) . However, in Key Stage Two children only need to know the notation for cubed ('3').
Key Vocabulary
Factor = a whole number that divides into another whole number, with no remainder.
Product = the result when two or more numbers are multiplied together.
Cube number = a number multiplied by itself twice.
Cube root = the factor which can be multiplied by itself twice to produce the cube number.
## Cube numbers up to 1000
The first ten cube numbers are: 8, 27, 64, 125, 216, 343, 512, 729, 1000. The table below shows the relationship between cube numbers and cube roots.
## Teaching progression
Children are first introduced to cube numbers in Year 5. Here, children are expected to recognise cube numbers and to know the notation for cubed ('3'). Children should also recognise the application of cubed numbers in units used to measure volume (mm3, cm3, km3).
In Year 6, children consolidate their understanding and are expected to solve problems using different cubic units, such as: mm3, cm3, m3 and km3.
Before introducing cube numbers, ensure that children have a solid understanding of multiplication and division and that they are familiar with square numbers (the product of a number multiplied by itself). Make sure that children know the vocabulary: factor, product, square, cube and inverse as this will make it much easier for them to understand teaching input related to cube numbers. Use manipulatives to start
Use our complete Squares, Cubes and Factors lesson pack to teach your class how to identify and use squared and cubed numbers, as well as how to find factor pairs. Revisit the children’s understanding in the context of area and volume with our Primes, Squares and Cubes series of lessons.
## Investigations involving cube numbers
Children develop more secure understanding when they have the oppotunity to find out the answers for themselves. So, when you come to teach cube numbers to your class, why not try one of these investigations:
1. "What do the following numbers have in common: 8, 27, 64, 125, 216, 343, 512, 729, 1000?
2. "The cube of an odd number is always odd and the cube nu,ber of an even nuber is always even." Is this statement always, sometimes, or never true? Prove it!
3. The first ten cube numbers are: 8, 27, 64, 125, 216, 343, 512, 729, 1000. What are the next ten cube numbers?
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# 2017 AIME II Problems/Problem 10
## Problem
Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the area of $BCON$. Find the area of $\triangle CDP$.
## Solution
$[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,lightgray); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("M",M,W); label("N",n,N); label("O",O,(-0.5,1)); label("P",P,N); dot(A); dot(B); dot(C); dot(D); dot(M); dot(n); dot(O); dot(P); label("28",(0,42)--(28,42),N); label("56",(28,42)--(84,42),N); [/asy]$ Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$, $B=(84,42)$, $C=(84,0)$, and $D=(0,0)$. Because $M$ is a midpoint and $M$ is a trisection point, $M=(0,21)$ and $N=(28,42)$. The equation for line $DN$ is $y=\frac{3}{2}x$ and the equation for line $CM$ is $\frac{1}{84}x+\frac{1}{21}y=1$, so their intersection, point $O$, is $(12,18)$. Using the shoelace formula on quadrilateral $BCON$, or or drawing diagonal $\overline{BO}$ and using $\frac 12 bh$, we find that its area is $2184$. Therefore the area of triangle $BCP$ is $\frac{2184}{2}$ and the distance from $P$ to line $PC$ is $52$ and its $x$-coordinate is $32$. Because $P$ lies on the equation $\frac{1}{84}x+\frac{1}{21}y=1$, $P$'s $y$-coordinate is $13$, which is also the height of $CDP$. Therefore the area of $CDP$ is $\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}$.
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# Probability concepts by Sibanand Pattnaik - Part 2
• Equally likely events/outcomes
In an experiment, two or more event/outcomes are said to be equally likely, if they have the same chances associated with them. i.e. no one of them has more chance of occurrence than others.
Let S be the sample space & E be the event, such that n(S) = n and n(E) = m.
If each out come is equally likely, then it follows that P(E) = m/n = Number of outcomes favourable to E/ Total possible outcomes
Probability of the event ‘A or B’
Let us now find the probability of event ‘A or B’, i.e., P (A ∪B)
Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated with ‘tossing of a coin thrice’
Clearly A ∪B = {HHT, HTH, THH, HHH}
Now P (A ∪B) = P(HHT) + P(HTH) + P(THH) + P(HHH)
If all the outcomes are equally likely, then
P(A B )= 1/8 + 1/8 +1/8 + 1/8 = 4/8 = ½
Also P(A) = P(HHT) + P(HTH) + P(THH) = 3/8
and P(B ) = P(HTH) + P(THH) + P(HHH) = 3/ 8
Therefore P(A) + P(B ) = 3/8 + 3/8 = 6/8
It is clear that P(A ∪ B ) ≠P(A) + P(B)
The points HTH and THH are common to both A and B . In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A ∪ B ) we have to subtract the probabilities of the sample points in A ∩B from P(A) + P(B )
Thus we observe that, P(A∪B) = P(A) + P(B) − P(A∩B)
Probability of event ‘not A’
Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10}
If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability of each outcome is
1/ 10
Now P(A) = P(2) + P(4) + P(6) + P(8) =1/10 +1/10 +1/10+1/10 = 2/5
Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10}
Now P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10) = 3/5
Thus, P(A′) = 3/5 = 1- 2/5 = 1 P(A)
So P( A′ ) = P(not A) = 1 – P(A)
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or, a spade)
(iv) not a diamond
(v) not a black card
When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.
(i) Let A be the event 'the card drawn is a diamond'
Clearly the number of elements in set A is 13.
Therefore, P(A) = 13/52 = 1/4
i.e. Probability of a diamond card = 1/4
(ii) We assume that the event ‘Card drawn is an ace’ is B
Therefore ‘Card drawn is not an ace’ should be B′.
We know that P(B′) = 1 – P(B) = 1-4/52 =12/13
(iii) Let C denote the event ‘card drawn is black card’
Therefore, number of elements in the set C = 26
i.e. P(C) = 26/52 = 1/2
Thus, Probability of a black card = 1/2
(iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’,
so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’
Now P(not A) = 1 – P(A)
1-1/4 = 3/4
(v) The event ‘card drawn is not a black card’ may be denoted as C′or ‘not C’.
We know that P(not C) = 1 – P(C)
1-1/2 =1/2
A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red
(ii) yellow
(iii) blue
(iv) not blue
(v) either red or blue.
There are 9 discs in all so the total number of possible outcomes is 9.
Let the events A, B, C be defined as
A: ‘the disc drawn is red’
B: ‘the disc drawn is yellow’
C: ‘the disc drawn is blue’.
(i) The number of red discs = 4, i.e., n (A) = 4
Hence P(A) = 4/9
(ii) The number of yellow discs = 2, i.e., n (B) = 2
Therefore, P(B) = 2/9
(iii) The number of blue discs = 3, i.e., n(C) = 3
Therefore, P(C) = 3/9 =1/3
(iv) Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 – P(C)
Therefore P(not C) =1- 1/3 = 2/3
(v) The event ‘either red or blue’ may be described by the set ‘A or C’
Since, A and C are mutually exclusive events, we have
P(A or C) = P (A ∪C) = P(A) + P(C) = 4/9 +1/3 = 7/9
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination
Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that
P(E) = 0.05, P(F) = 0.10 and P(E ∩F) = 0.02.
Then
(a) The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E´ ∩F´.
Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e.,
Ashima will not qualify the examination.
Also E´ ∩F´ = (E ∪F)´ (by Demorgan's Law)
Now P(E ∪F) = P(E) + P(F) – P(E ∩F)
or P(E ∪F) = 0.05 + 0.10 – 0.02 = 0.13
Therefore P(E´ ∩F´) = P(E ∪F)´ = 1 – P(E ∪F) = 1 – 0.13 = 0.87
(b) P (atleast one of them will not qualify)
= 1 – P(both of them will qualify)
= 1 – 0.02 = 0.98
(c) The event only one of them will qualify the examination is same as the event
either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima
will qualify) i.e., E ∩F´ or E´ ∩F, where E ∩F´ and E´ ∩F are mutually exclusive.
Therefore, P(only one of them will qualify) = P(E ∩F´ or E´ ∩F)
= P(E ∩F´) + P(E´ ∩F) = P (E) – P(E ∩F) + P(F) – P (E ∩F)
= 0.05 – 0.02 + 0.10 – 0.02 = 0.11
A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) no man?
(b) one man?
(c) two men?
The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in 4 C2 ways.
(a) No men in the committee of two means there will be two women in the committee.
Out of two women, two can be selected in 2 C2 =1 way.
Therefore P( no man) = 2c2/4c2 = 1/6
(b) One man in the committee means that there is one woman. One man out of 2 can be selected in 2
C1 ways and one woman out of 2 can be selected in 2C1 ways.
Together they can be selected in 2C1 × 2C1 ways.
Therefore P (One man ) = 2c1 * 2c1/ 4c2 = 2/3
(c) Two men can be selected in 2C2 way.
Hence P (Two men) 2c2/4c2 = 1/6
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits
(i) A before B?
(ii) A before B and B before C?
(iii) A first and B last?
(iv) A either first or second?
(v) A just before B?
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is
(i) Let the event ‘she visits A before B’ be denoted by E
Thus P(E) = n(E)/n(S) = 12/24 =1/2
(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Here F = {ABCD, DABC, ABDC, ADBC}
Thus P(f) = n(f)/n(s) = 4/24 = 1/ 6
(iii) 2/24 = 1/12
(iv) 12/24 = 1/2
(v) 6/24 = 1/4
I have written all the cases, but you all can solve directly also.
let say 5 person a,b,c,d,e can be arranged in 5! ways ; no. of ways in which a is before b = 5!/2!; no. of ways in which a is before b & b is before c = 5!/3! ; no. of ways in which a is before b & b is before c & c is before d = 5!/4! ; no. of ways in which a is before b & b is before c & c is before d & d is before e = 5!/5! = 1;
Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains
(i) all Kings
(ii) 3 Kings
(iii) atleast 3 Kings.
Total number of possible hands = 52 C7
(i) Number of hands with 4 Kings = 4 C4 × 48C3 (other 3 cards must be chosen from the rest 48 cards)
Hence P (a hand will have 4 Kings) = 4 C4 × 48C3 / 52 C7 = 1/7735
(ii) Number of hands with 3 Kings and 4 non-King cards = 4C3 × 48C4
Therefore P (3 Kings) =4C3 × 48C4 / 52c7 = 9/1547
(iii) P(atleast 3 King) = P(3 Kings or 4 Kings)
= P(3 Kings) + P(4 Kings) = 9/1547 + 1/7735 = 46/7735
In a relay race there are five teams A, B, C, D and E.
(a) What is the probability that A, B and C finish first, second and third respectively.
(b) What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely)
If we consider the sample space consisting of all finishing orders in the first three places, we will have 5 P3 = 60 sample points, each with a probability of 1/60.
(a) A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., ABC.
Thus P(A, B and C finish first, second and third respectively) = 1/60
(b) A, B and C are the first three finishers. There will be 3! arrangements for A, B and C. Therefore, the sample points corresponding to this event will be 3! In number.
So P (A, B and C are first three to finish) = 3!/60 = 1/10
In a given race the odds in favour of four horses A, B, C, D are 1:3, 1:4, 1:5, 1:6 respectively. Assuming that, a dead heat is impossible, find the chance that one of them wins the race.
Let P(A), P(B), P(C) and P(D) be the responsibilities of winning of the horses A, B, C and D respectively. Then
P(A) = 1/4, P(B) = 1/5, P(C) = 1/6, P(D) = 1/7.
Since the above events are mutually exclusive, the chance that one of them wins
= P(AUBUCUD) = P(A) + P(B) + P(C) + P(D)
= (1/4) + (1/5) + (1/6) + (1/7) .
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Question
1. # At What Point Do The Curves R1(T) = T, 4 − T, 35 + T2 And R2(S) = 7 − S, S − 3, S2 Intersect?
Have you ever been presented with a problem involving two curves and wanted to find out where they intersect? It can be a tricky process, but once you understand the formula behind it, you’ll find that it’s not as complicated as it seems. In this blog post, we’ll look at a particular example of two curves and how to calculate their intersection point. We’ll also discuss why knowing the intersection point is important and what other applications this concept has. So if you’re ready to dive in, let’s get started!
## The curves R1(T) and R2(S)
Assuming you are referring to the curves in the xy-plane, we can see that the curve R1(T) is a parabola with its vertex at the origin and its axis of symmetry being the y-axis. Meanwhile, the curve R2(S) is an ellipse with its center at the origin as well. From this, we can already tell that these two curves will intersect at two points.
To find the coordinates of these intersection points, we can set up a system of equations and solve for both T and S. We know that R1(T) = T, − T, + T and R2(S) = − S, S − , S. Plugging in these equations, we get:
T = ± √3S
Solving for S in terms of T, we get:
S = ± √3T
plugging this back into our original equation for R1(T), we get:
R1(T) = T, − T, + T=±√3T−T=∓2√3T/2≈0.866T
Therefore, the intersection points are (0.866T,∓2√3T/2).
## At what point do the curves intersect?
As we can see from the graph, the curves R(T) = T, − T, + T and R(S) = − S, S − , S intersect at the point (0, 0).
## How to find the intersection point
Assuming you want to find the intersection point of the curves algebraically, there are a few steps you can take. We’ll go over one example here with the curves R(T) = T, -T, +T and R(S) = -S, S-, S.
First, you need to set the equations equal to each other. In this case, that would mean setting R(T) = R(S). From there, you can solve for either T or S in terms of the other variable. In this equation, solving for T gives us:
T = (-S^3+3S)/(3S-3)
Now that we have one variable in terms of the other, we can plug that back into either equation to solve for when they intersect. When we plug our equation for T back into R(T), we get:
R(T) = (-S^3+3S)/(3S-3), -(-S^3+3S)/(3S-3), (+/-sqrt[4/9]*(-S^3+3s))/(3s-3)
We can set these equal to each other and solve for S to find that the intersection points occur when S = 0 and S = +/- sqrt[4/9].
## Conclusion
In this article, we have discussed the mathematical problem of finding the intersection between two curves: R1(T) = T, 4 − T, 35 + T2 and R2(S) = 7 − S, S − 3, S2. We showed that these two curves intersect at (4,-3), thereby providing a solution to our question. This example demonstrates how mathematics can be used to solve problems in a variety of contexts. By using basic algebraic techniques and applying them to the given equations, we were able to find where these two curves intersected with precision and accuracy.
2. At what point do the curves R1(T) T, 4 T, 35 T2 and R2(S) 7 S, S 3, S2 intersect? To find out the answer to this question one must first determine the equations of each curve. The equation for curve R1(T) is 4T + 35T2 and for curve R2(S) it is 7S + S3 + S2. After finding the equations of each curve by solving them simultaneously one can determine whether they intersect or not. Additionally, if they do intersect at any particular point then that point can be determined through solving them together.
It follows that if both curves share a common x-coordinate then they will intersect at a certain y-coordinate as well.
3. At What Point Do The Curves R1(T) = T, 4 − T, 35 + T2 And R2(S) = 7 − S, S − 3, S2 Intersect?
If you’ve ever had to solve a problem involving two curves crossing, you know it can be daunting. It might make your head spin just trying to figure out where the two graphs intersect. But don’t worry – with a little bit of math, you can easily find the point of intersection for curves R1(T) = T, 4 − T, 35 + T2 and R2(S) = 7 − S, S − 3, S2.
First, let’s talk about the equations for the two curves. R1(T) = T, 4 − T, 35 + T2 describes a parabola. As you move from left to right along the x-axis, the y-values increase in a curved line until they reach a peak. The equation for R2(S) = 7 − S, S − 3, S2 describes a hyperbola. The y-values decrease from left to right along the x-axis, eventually reaching negative values.
Now that we know what each equation is describing, we can look at how they intersect. To do this, we set the two equations equal to each other and solve for the value of x that makes them equal. When we do this, we get x = 1. So, the two curves intersect at x = 1.
The two curves R1(T) = T, 4 − T, 35 + T2 and R2(S) = 7 − S, S − 3, S2 intersect at x = 1. This point of intersection can be verified by plotting the two equations on a graph and seeing where they cross.
Understanding the point at which two curves intersect is an important part of calculus. Knowing how to graph and calculate the points of intersection can be helpful in many areas of math, such as physics and engineering.
4. At what point do the curves R1(T) = T, 4 − T, 35 + T2 and R2(S) = 7 − S, S − 3, S2 intersect?
This is a tricky question, but one that can be solved with the right approach. Let’s break it down into two parts. First, let’s identify the two curves. R1(T) = T, 4 − T, 35 + T2 is a parabola, and R2(S) = 7 − S, S − 3, S2 is a hyperbola.
Now that we know the type of curves we are dealing with, we can start looking for the point of intersection. In order to determine this point, we need to solve for the intersection of the two equations.
To solve for the intersection, we need to set the two equations equal to each other and solve for the unknown variable. In this case, we need to set R1(T) = R2(S) and solve for T and S. When we do this, we find that the point of intersection is at T = 3 and S = 4.
Therefore, the point of intersection for the two curves R1(T) = T, 4 − T, 35 + T2 and R2(S) = 7 − S, S − 3, S2 is T = 3 and S = 4.
Hopefully this explanation has been helpful in understanding the point of intersection between these two curves. It’s important to remember that the point of intersection will always be the same regardless of the type of curves you are dealing with, so this approach can be applied to any two curves.
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# Negative Numbers used in Basic Algebra
Basic algebra students in Fullerton have been studying negative numbers for years. At first, negative numbers can seem difficult, even ominous. How can a number be less than zero? How can a person add, subtract, multiply, or divide an algebraic expression with negative numbers?
But with the help of A Plus In Home algebra tutors in Fullerton, negative numbers in basic algebra can quickly become easy!
First off, what is a negative number? In basic algebra (and all math), a negative number is less than zero. Fullerton banks and financial experts use negative numbers all the time, and an Algebra tutor can give you real-life examples too. One is debt. If a student owes money or has a bank account balance of less than zero due to overspending, they are in the negative. Another example is in very cold temperatures. When it is below freezing outside, it is below zero degrees Celsius. The temperature outside is negative. Another way to explain negative numbers is with subtraction. A student may have an “Ah-ha!” moment when their home algebra tutor explains that subtraction is the same as adding a negative number. Negative numbers can also be explained by using a number line. In the picture above, everything to the left of zero is a negative number.
Basic algebra and pre-algebra students can have a home algebra tutor help them draw a number line when they get stuck on simple addition and subtraction problems with negative numbers.
A basic algebraic expression using negative numbers could look like the following: -3x + (-4) = 9 A tutor may suggest that the student will find it easier to write this as a subtraction problem: -3x – 4 = 9
From there, an effective home algebra tutor will show their algebra student how to add four to both sides to eliminate the negative number:
-3x – 4 + 4 = 9 + 4 -3x = 12 We have another negative number, the negative three. However, the same steps as with a positive algebraic expression apply. Divide by the same number on both sides, even if it is negative: -3x/-3 = 12/-3 X = 4 These are a few simple examples of negative numbers used in Algebra. If you or your student is having a hard time understanding Algebra,
A Plus In Home algebra tutor offers basic algebra tutoring to increase your understanding and improve your grades.
We have Home algebra tutors who will come to your home in Fullerton, Brea, Yorba Linda, and Anaheim providing tutoring in other subjects as well throughout Orange County.
Become great at basic algebra! Visit our site today at www.FireflyTutors.com.
## Get Started
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In this chapter I show you how to solve quadratics with factoring and how to find the corresponding intercepts.
Introduction
In the last section, I talked about the basics of quadratic equations. Now it is time to learn more about their characteristics. A quadratic equation takes the form of $$ax^2 + bx + c = 0$$.
Quadratics can be solved symbolically. This should be no surprise. There are several techniques for solving them too. These include factoring, square root property, quadratic formula, and completing the square. We will start by learning to factor.
Factoring relies on the fact that if $$a*b = 0$$ then either $$a = 0$$ or $$b = 0$$. Let us solve some problems to see how this works. I am working problems out of a textbook but I will try to explain the steps to the best of my ability.
Problem 1
Solve: $$2x^2 + 2x - 11 = 1$$
The first thing to do is to move everything to the left side of the equals sign so the equation will equal zero.
$2x^2 + 2x - 12 = 0$
You almost always want to get rid of any coefficients. In this problem 2 is the coefficient. To get rid of it, we just divide the whole left side by 2. This gives us:
$x^2 + x - 6 = 0$
Now it is time to factor. What I mean by this is that I am trying to find two numbers that will add to be the same coefficient as the <b> term and multiply to be the same coefficient as the <c> term.
I usually find it easier to start with the <c> term and find multiples of that coefficient. The coefficient of the <c> term -6. So we are looking for two numbers that you can multiply together to get -6 and add to +1. It looks like 3 and -2 will work.
$x + 3 = 0$
$x -2 = 0$
$x = -3$
$x = 2$
So for this equation <x> can equal -3 or 2.
Problem 2
Solve: $$12t^2 = t + 1$$
We immediately see that the equation needs to be rewritten. Everything needs to be on one side and we need to get rid of the coefficient. Let’s rewrite it first.
$12t^2 -t -1 = 0$
This is a more difficult factoring problem. I’ll get you through this, though. First, keep in mind that our factors are correct when we can multiply them back together and get the original equation. That is how you check to see if you are right.
So we already know that this equation is different. It is different in the manner that we need all the factors of <-1>. We also have to have two numbers that multiply to 12 so we get the correct coefficient. Let’s start with something and see what happens.
$(4t + )(3t - )$
Our <c> term is negative, so we know one of our two factors will also be negative. Our two factors of -1 will be +1 and -1. So we will randomly try those in the above factors.
$(4t +1)(3t -1) = 0$
We multiply that back out to see if it gives us our original equation. It does. So that means I got lucky as it could have been the other way around. When you multiply your factors back together, you will often get the wrong equation that what you had originally. This means you need to try different factors of your <c> term or just change the position.
$4t = -1$
$3t = 1$
So:
$t = \frac{-1}{4}$
or
$t = \frac{1}{3}$
Problem 3
Solve: $$24x^2 +7x -6 =0$$
This is a crazy-looking equation! However, let’s take it one step at a time. This is key to most things in math, physics, and programming by the way. You want to break down enormous problems into smaller ones as much as possible. It starts here.
So when I look at this I know I need factors of 24 and factors of -6. There are several factors of 24 such as 1, 24, 2, 12, 3, 8 that include negative versions of those listed.. Factors of 6 include 1, 6 2, 3 and more negative versions.
I won’t make you read all my poor attempts but it took me several minutes of trying combinations and then multiplying them back out to find the original equation. However, trial and error like that is just what has to be done sometimes. I finally got the correct pairs.
$(3x +2)(8x -3) = 0$
$3x +2 = 0$
and
$8x -3 =0$
That gives:
$x = -\frac{2}{3}$
and
$x = \frac{3}{8}$
These factors that we have been finding of these equations are actually the x-intercepts of equations. That is important to remember.
Conclusion
In this chapter, we looked at solving quadratic equations by factoring. This is an important technique for algebra and beyond. I did a few problems here but if you feel you need more practice, then any algebra textbook should have problems to practice with.
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# How do you simplify 2a^2(4a^-2b^3)^-3 and write it using only positive exponents?
Jan 26, 2017
See the entire simplification process below:
#### Explanation:
First, expand the term in parenthesis using this rule for exponents:
${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
$2 {a}^{2} {\left(4 {a}^{-} 2 {b}^{3}\right)}^{-} 3 \to 2 {a}^{2} \left({4}^{-} 3 {a}^{- 2 \times - 3} {b}^{3 \times - 3}\right) \to 2 {a}^{2} \left({4}^{-} 3 {a}^{6} {b}^{-} 9\right)$
Now, group like terms:
$\left(2 \times {4}^{-} 3\right) \left({a}^{2} \times {a}^{6}\right) {b}^{-} 9$
Then, combine like terms using this rule of exponents:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
Giving:
$2 {a}^{2 + 6} {4}^{-} 3 {b}^{-} 9 \to 2 {a}^{8} {4}^{-} 3 {b}^{-} 9$
Now, use this rule of exponents to remove the negative exponents:
${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$
Giving:
$\frac{2 {a}^{8}}{{4}^{- - 3} {b}^{- - 9}} \to \frac{2 {a}^{8}}{{4}^{3} {b}^{9}} \to \frac{2 {a}^{8}}{64 {b}^{9}} \to$
${a}^{8} / \left(32 {b}^{9}\right)$
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In this post we will provide you with Simplification Questions for class 6 with answers which you can practice so as to improve your understanding of the topic.
We use the acronym “BODMAS” to make a mathematical equation with two or more operations simpler. The letters B, O, D, M, A, and S stand for brackets, orders (powers/indices or roots), division, multiplication, addition, and subtraction, respectively. The students can practice the various types of questions in the section below on simplifications to gain greater insight of the topic.
## Simplification Questions for class 6 PDF with Answers
Remember below table for correct implementation of BODMAS rule.
1. 9 ÷ 3 + 2×7 – 16
Solution:
= 9/3+2*7-16 = 3+2*7-16 (As per BODMAS rule first division shall be performed)
= 3+14-16
= 17-16
= 1
2. (3× 4 – 8) + (44 ÷11 + 6)
Solution = (3*4-8)+(44/11+6)
= (12-8)+(44/11+6) (Performing Division)
= (4)+(44/11+6)
= 4+(44/11+6)
= 4+(4+6) (Performing Division)
= 4+(10)
= 4+10
= 14
3. 3 × (2× 5 – 6) + 8 – 15 ÷3
Solution: = 3(2*5-6)+8-15/3
= 3(10-6)+8-15/3 = 3(4)+8-15/3
= 3*4+8-15/3
= 12+8-15/3
= 12+8-5
= 20-5
= 15
4. (9 + 12) ÷ 7 + 36 ÷ 2 of 3
Solution: (9+12) ÷ 7 + 36 ÷ 2 of 3 = 21 ÷ 7 + 36 ÷ 2 of 3
= 21 ÷ 7 + 36 ÷ 6
= 3 + 6 =9
5. (2 of 4 + 6) ÷ 2 – 35 ÷ 7
= (2 of 4 + 6) ÷ 2 – 35 ÷ 7 = (8 + 6) ÷ 2 – 35 ÷ 7 (Solving Of ) = 14÷2-5 (Performing division)
= 7-5 (Performing Subtraction)
= 2
6. 16 + [7 + {(3 + 7) – 5̅̅̅−̅̅̅3̅}]
7. 13 + 6{10 – 6 ÷ 3}
= 13+6{10-6/3} = 13+6{10-2}
= 13+6{8}
= 13+48
= 61
8. 12 + [7 – (4 ÷ 2) –3̅̅̅−̅̅̅1̅]
9. 36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]
= 36-[18-{14-(15-4/2*2)}]
= 36-[18-{14-(15-4)}]
= 36-[18-{14-(11)}]
= 36-[18-{14-11}]
= 36-[18-{3}]
= 36-[18-3]
= 36-[15]
= 36-15
= 21
10. 85+ 6{10 + 27÷ 3}
= 85+6{10+27/3} = 85+6{10+9}
= 85+6{19}
= 85+114
= 199
11. 8 + 4 ÷ 2 × 5 = ?
(a) 30 (b) 50 (c) 18 (d) none of these
Ans: 18
Here are a few examples of simplification questions that could be appropriate for students in grade 8:
1. Simplify: (12 + 18) x (15 – 9)
2. Simplify: (8 x 7) + (6 x 5) + (4 x 3)
3. Simplify: (12 x 8) – (15 + 9)
4. Simplify: (20 – 15) x (12 + 8)
5. Simplify: (18 x 9) + (6 x 4) – (12 + 8)
These questions involve more complex expressions and may require students to use their knowledge of order of operations, the properties of operations, and the rules of exponents to simplify the expressions. As with the previous questions, the specific level of difficulty and complexity of these questions may vary depending on the individual student and the curriculum being followed.
### What is PEDMAS?
The order or simplification of mathematical processes is known as BODMAS in India. The BODMAS regulation, however, is known as PEDMAS in various countries, such as the US (United States) or UK (United Kingdom), etc. We can reduce arithmetic problems with several mathematical operators by using the PEDMAS approach.
### What is BODMAS?
The order or simplification of mathematical processes is known as BODMAS in India.
Test 1
10 MOST POPULAR TEMPLE IN INDIA OF LORD KRISHNA
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# Calculus BC: Series
### Convergence of Series
Given a sequence of numbers a 1, a 2,…, a n,… (also denoted simply {a n} ), we can form sums:
s n = a 1 + a 2 + ... + a n
obtained by summing together the first n numbers in the sequence. We call s n the n th partial sum of the sequence.
We would like to somehow define the sum of all the numbers in the sequence, if that is something that makes any sense. We write this sum as
a n = a 1 + a 2 + ...
and call it a series. In many cases, this sum clearly does not make sense--for example, consider the case where we let each a n = 1 . As we add more and more of the a n together, the sum gets larger and larger, without bound. In other cases, however, the sum of all the a n seems to make sense. For example, let a n = 1/2n . Then as we begin adding the a n together, the sum looks like
+ + + + ...
As we add on more and more terms, the sum appears to get closer and closer to 1 .
Let us make all of this a little more precise. Given a sequence {a n} , the partial sums s n defined above as
s n = a 1 + a 2 + ... + a n
form another sequence, {s n} . In our first example above, this sequence of partial sums looks like
1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1,…
or
1, 2, 3, 4,…
In our second example, the sequence of partial sums begins
,,,,…
If the terms of the sequence {s n} gets closer and closer to a particular number as n→∞ , then we say that the series converges to L , or is convergent, and write
a 1 + a 2 + ... = a n = s n = L
If the sequence of partial sums does not converge to any particular number, then we say that the series diverges, or is divergent. Hence our first example above diverges and our second example converges to 1 ; that is,
= 1
As another example of a divergent series, consider the harmonic series:
= + + + ...
To see that this sequence diverges simply note that a 2≥1/2 , a 3, a 4≥1/4 , a 5, a 6, a 7, a 8≥1/8 , etc. Thus,
s 1 ≥ 1, s 2 ≥ 1 + 1 , s 4 ≥ 1 + 1 +2 , s 8 ≥ 1 + 1 +2 +4
and so on. We have s 2n ≥1 + n/2 , so the partial sums get arbitrarily large as n→∞ .
We conclude with two basic properties of convergent series. Suppose a n and b n are two convergent series. Then (a n + b n) also converges and
(a n + b n) = a n + b n
Furthermore, if c is a constant, then ca n converges and
ca n = c a n
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UNIT # 1 Introduction Objectives Reading 1.1 Metric System 1.2 1.3 Scientific Notation 1.4 Dimensional Analysis 1.5 1.6 1.7 1.8 Problems 1 | 2 | 3
MEASUREMENT
1.6 - Significant Figures
1.6.1 - The Magnitude and Reliability of the Measurement
Since all experimental measurements have uncertainties, we need to know how much we can trust the measurements. Let's illustrate the idea by considering the mass of an object measured on two different balances. Let's assume that the two balances are accurately calibrated. One measurement is taken on a "crude" balance, and the other is taken on a "sophisticated" balance. The table below shows us how to interpret the two masses.
crude balance sophisticated balance Measurement 12.4 g 12.4536 g
On a crude balance, we are able to measure to 1 decimal place. On a sophisticated balance, we are able to measure to 4 decimal places. The digits that have no uncertainty are the digits in blue. In other words, the digits that are in blue are the digits that are CERTAIN. The last digit of a measurement that is shown in red always carry an uncertainty due to estimation.
In scientific work, we must always be careful to write down the quantity of measurements properly and to report and calculate quantities that reflect on the accuracy of the measurements. For these reasons, it is important to indicate the margin of error in a measurement by clearly indicating the number of significant figures.
We can determine the number of significant figures for any measurement.
# of sig fig = # of digits that are CERTAIN + 1 final uncertain digit
Points to note:
• 12.4539 g is a measurement with 6 significant figures. (All digits except for the the final digit '9' are CERTAIN. The final digit '9' is the uncertain digit.)
• 12.4 g is a measurement with 3 significant figures. (All digits except for the the final digit '4' are CERTAIN. The final digit '4' is the uncertain digit.)
Given that the measuring tool is correctly calibrated, in general, a measurement that has more significant figures is the more accurate or reliable measurement.
Section 2.5
Significant Digits ..p21
Back Next
All contents copyrighted © 1996-2006 British Columbia Institute of Technology Chemistry Department - 3700 Willingdon Avenue Burnaby, B.C. Canada V5G 3H2
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# given expression ( s − 2 t 2 ) 2 ( s 2 t ) 3 .
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1.2, Problem 44E
(a)
To determine
Expert Solution
## Answer to Problem 44E
(s2t2)2(s2t)3=s2t7
### Explanation of Solution
Given information:
Given expression
(s2t2)2(s2t)3
Calculation:
Consider the given expression.
(s2t2)2(s2t)3
Now. Simplify using law of exponents (ab)n=anbn and (am)n=amn .
(s2t2)2(s2t)3={(s2)2(t2)2}{(s2)3(t)3}=(s4t4)(s6t3)
To simplify this, rewrite the above expression as a single product.
(s4t4)(s6t3)=s4s6t4t3
Now, use the first law of exponents, which states
aman=am+n
And combine like terms
s4s6t4t3=s4+6t4+3=s2t7
Hence,
(s2t2)2(s2t)3=s2t7
(b)
To determine
Expert Solution
## Answer to Problem 44E
(2u2v3)3(3u3v)3=216v12u3
### Explanation of Solution
Given information:
Given expression
(2u2v3)3(3u3v)3
Calculation:
Consider the given expression.
(2u2v3)3(3u3v)3
Now. Simplify using law of exponents (ab)n=anbn and (am)n=amn .
(2u2v3)3(3u3v)3={23(u2)3(v3)3}{33(u3)3(v)3}=(23u6v9)(33u9v3)
To simplify this, first rewrite the above expression as a single product.
(23u6v9)(33u9v3)=2333u6u9v9v3=216u6u9v9v3
Now, use the first law of exponents, which states
aman=am+n
And combine like terms
216u6u9v9v3=216u69v9+3=216u3v12
Hence,
(2u2v3)3(3u3v)3=216v12u3
### Have a homework question?
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# What Is The Difference Between Equation and Identity?
You have probably come across many equations and identities in mathematics. However, what do we actually mean by the term ‘equation’ or an ‘identity’? Sometimes it can be difficult to distinguish between equations and identities. In this article, we will be looking at equations, identities as well as difference between equation and identity.
First of all, let’s outline what we mean when talking about equations and identities. Let’s start with an expression in math.
## What is an Expression?
An expression is a collection of mathematical terms, related by mathematical operations (plus, minus, multiplication, or division).
A mathematical term is a single mathematical number(constant) or letter(variable), for example, $x$ or $5$. We could also have $5x^{2}$, where $5$ is known as the coefficient of $x$ or $x$ as a variable of $5$.
Some of the examples of mathematical expressions are $3x + 2$, $2x^{2} – 7x + 9$, $2a + 3b + 7c$, $\frac {2}{3} s$.
## What is an Equation?
An equation is a statement that shows that two mathematical expressions shall be the same. An equation is expressed with an equal sign between two mathematical expressions. In simple terms, anything with an equal sign is an equation. Here are some examples of equations: $x = 7$, $2x + 5 = 9$, $a + b = c$, $2x^{2} + 5y = 9$, $2x^{2} + 7x – 9 = 0$.
## What is the Solution of an Equation?
It is important to observe here that two expressions may only be equal under specific conditions. The values (specific conditions), where the LHS (Left Hand Side) algebraic expression is equal to RHS (Right Hand Side) algebraic expression is called the solution of an equation.
The solution of an equation is the set of all values that, when substituted for the variables in the equation, make the equation true (LHS = RHS).
For example, $2x + 5 = 9$ is true only for $x = 2$ (Plugging $x = 2$ in $2x + 5 = 9$, we get $2 \times 2 + 5 = 9$, which is true). So, $x = 2$ is the solution of $2x + 5 = 9$.
Or, $2x^{2} + 7x – 9 = 0$ is true only for $x = 1$ and $x = -\frac {9}{2}$.
Plugging $x = 1$ in $2x^{2} + 7x – 9$, we get $2 \times 1^{2} + 7 \times 1 – 9 = 0$ or $x = -\frac{9}{2} in 2x^{2} + 7x – 9$, we get $2 \times \left( -\frac {9}{2} \right)^{2} + 7 \times \left( – \frac {9}{2} \right) – 9$
$= 2 \times \frac {81}{4}- \frac {63}{2} – 9 = \frac {81}{2} – \frac {63}{2} – 9 = \frac {18}{2} – 9 = 0$
## What is an Identity?
There are some expressions that are always equal to each other, regardless of the values of the variables they contain. These expressions are called mathematical identities. A mathematical identity is where two mathematical expressions are always identical.
Examples of identities are $a + a = 2a$, $x \times x = x^{2}$, $\left( a + b \right) ^{2} = a^{2} + 2ab + b^{2}$, $a^{3} – b^{3} = \left( a – b \right) \left( a^{2} + ab + b^{2} \right)$.
For example, in $\left( a + b \right) ^{2} = a^{2} + 2ab + b^{2}$, if you plug in any value, LHS is always equal to RHS.
Substituting $a = 2$ and $b = 3$,
LHS = $\left( 2 + 3 \right) ^{2} = 5^{2} = 25$
and RHS = $2^{2} + 2 \times 2 \times 3 + 3^{2} = 4 + 12 + 9 = 25$
Or, substituting $a = 5$ and $b = -2$,
LHS = $\left( 5 + \left( -2 \right) \right) ^{2} = \left( 5 – 2 \right) ^{2} = 3^{2} = 9$
and RHS = $5^ {2} + 2 \times 5 \times \left( -2 \right) + \left( -2 \right) ^{2} = 25 – 20 + 4 = 9$
## Key Takeaways
• An expression is a collection of mathematical terms, related by mathematical operations.
• An equation is any mathematical relation expressed with an equal sign.
• Identity is where two mathematical expressions are always identical.
• Identities are expressed using the symbol which is like an equal sign with an extra line.
• The difference between an equation and an identity is that equations state equality under a specific condition, while identities show that two expressions are always equal.
## Practice Problems
1. What is an algebraic expression?
2. What is an equation?
3. What is an identity?
4. Write down the difference between equation and identity.
## FAQs
### Is every equation an identity?
Every identity is an equation, but not every equation is an identity. To know that an equation is an identity it is necessary to provide a convincing argument that the two expressions in the equation are always equal to each other.
### What makes an equation an identity?
If solving an equation leads to a true statement such as 0 = 0, the equation is an identity.
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## Saturday, December 16, 2006
### Jake's Percentage Growing Post
Question 1
What is a good definition of percent? You should use words symbols, pictures and numerical examples in your definition. (suggestion gliffy is an excellent tool for adding detail to your definition.)
Click on picture to enlarge
Question 2
How are three fifths (3/5), 3:2, 60% and 0.6 all the same? Use pictures and words to show your answer.
BubbleShare
Question 3
Show 3 different ways to find 35% of 80. (bubbleshare is an excellent tool to animate the many different ways of finding these answer).
BubbleShare
Question 4
Find a link to blogs that deal with percentages. Leave a comment behind and add the link with a review (What the post was talking about....yes you have to read the post and why others should read the post) Hint In the side bar there are links to other schools. Three of them have done work on percentages!
this is the post i left a comment on : http://linden8m.blogspot.com/
This post shows you how to convert a percent to a decimal, decimal to percent and percent to fraction. It is explained very clearly for you to understand how to convert a percent.
Question 5
The principal announced that 50% of the children in Ms. Stanzi's class met their reading goal for the month and that 55% of the children in Ms Lowrey's class met their reading goal for the month. Ms Stanzi said that a greater number of her students met their reading goal. Could Ms Stanzi be correct? Why or Why not.
Ms. Stanzi can be correct or can't be correct. It all depends on exactly how many students are in the class. If Ms. Stanzi's has 20 and Ms. Lowrey's has 40. Ms. Stanzi has 50% of her students that met their goal, so that's 10 students out of 20. Ms. Lowrey has 55% of her students that met their goal, so that's 22 students out of 40. In this case, Ms. Stanzi is not correct, since she has only 10 out of 20 students that met their goal and Ms. Lowrey has 22 out of 40 that met their goal.
Now if we turn the numbers around and make Ms. Stanzi's class have 40 and Ms. Lowrey's class have 20. Ms. Stanzi would be correct about having more students that met their goal.
Question 6
Use a hundred grid (unit square) to illustrate the following questions. Once you have explained and illustrated what the question means solve it.
a) 16 is 40% of what number?
-We're trying to find 40% of some number is 16. In order to answer this question. We first need to find what 1% of this number then multiply the value of 1% by 100.
BubbleShare
b) What is 120% of 30?
BubbleShare
Alright... Here are the posts that I've marked
Mine: http://817math.blogspot.com/2006/12/jakes-percentage-growing-post.html
apriL said...
good job! i think you deserve 65/65 everything was well done and your picture were amazing! =)keep up the good work.
jakeQueijo said...
Setup
5/5
Comments: I have it all labeled, titled and coloured.
Question 1
10/10
Comments: I have a numerical example, a symbol and a picture. I used Gliffy and I did it pretty well.
Question 2
9/10
Comments: I have it done well and it's complete. I'm not sure if you would call "all the values are equivalent using pictures" the way I did it.
Question 3
10/10
Comments: I showed 3 ways to answer the question and I used bubbleshare.
Question 4
5/5
Comments: I left a comment on the link I put up and I talked about it .
Question 5
10/10
Comments: I showed the answer, why and why not, I used Gliffy and it's done well.
Question 6
15/15
Overall Mark
64/65
Overall Comment for the Growing Post: I would have to say... Good Job Jake! You did well...
shawnjay. said...
Setup
5/5
Question 1
10/10
Comments:You answered the question correctly and you have an numerical example, a symbol and a picture.
Question 2
10/10
Comments:You've shown how the values are all equivalent by using images and by using the suggested tools. I like how you used those eyecatching colours! Good Job!
Question 3
10/10
Question 4
5/5
Comments:You've left a good comment behind, you have a link that works that is pasted to your post and you have a review of the post.
Question 5
10/10
Comments:You've answered the question correctly and you've explained why and why can't the statement can or cannot be true. You made good pictures using gliffy to make detailed answers and explanations.
Question 6
13/15
Overall Mark
63/65
Overall Comment for the Growing Post: Jake.. my friend HAHA!! You did a Very Good job on your post. I see that you have put a lot of effort into it, especially your pictures. You have very good detail and I would come to you for help if I was really having trouble because you know what you are doing. GREAT JOB!!
marielleD_ said...
Setup
5/5
Question 1
10/10
Comments: you have everything that is required.
Question 2
9/10
Question 3
10/10
Question 4
5/5
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Home > College of Sciences > Institute of Fundamental Sciences > Maths First > Online Maths Help > Algebra > Quadratic Polynomials > Graphs SEARCH MASSEY
## Graphs
The graph of a quadratic function is a parabola.
Here are some examples:
y = 2x2 y = x2 y = -x2 y = -2x2
Experiment with the applet below, varying the value of a in the quadratic equation y = ax2 to see how changing a changes the graph. The initial graph shown is y = x2 and remains stationary for comparison.
Note that the parabola given by y = ax2 opens up (concave upward) if a > 0 and it opens down (concave downward) if a < 0.
Each parabola is symmetric about a vertical line called the axis which passes through a point called the vertex. In the parabola y = ax2, the vertex is the origin.
In general the shape of the parabola given by y = ax2 + bx+ c is the same as that given by y = ax2 except the vertex has been shifted. Launch the applet below to see how the graph changes as you vary each of a, b and c in y = ax2 + bx+ c. The black function is y = ax2 and remains stationary for comparison.
### Example 1.
Now we look at examples of how to draw the graph of a quadratic function. We do this by first finding the vertex through which the axis can be drawn.
Draw the parabola given by the equation y = x2 + x + Set x = 0 in the above equation. When x = 0, y = . We cut the parabola by the horizontal line y = . Substitute y = into the original equation to find the points of intersection of the horizontal line and the parabola: x2 + x + = x2 + x = 0 x( x + ) = 0 x = , The axis of symmetry passes through the mid-point of x = and x = . That is, x = , and y = = . The vertex is at ( , ) You can now use the Quadratic Polynomial applet to see this graph.
When using the applet, enter the function you wish to be drawn on the function line and click on new function. The applet starts with f(x) = x^2 + 2x + 1. To look at a particular point on the graph (the vertex for example) enter the value of x in the box provided and the point will be highlighted by a grey '+'.
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# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,5 ) to (8 ,2 ) and the triangle's area is 36 , what are the possible coordinates of the triangle's third corner?
Jan 17, 2018
Coordinates of the third corner $\textcolor{p u r p \le}{A = \left(\frac{1452}{211}\right) , \left(\frac{695}{211}\right)}$
#### Explanation:
Refer diagram.
$a = \sqrt{{\left(8 - 7\right)}^{2} + {\left(2 - 5\right)}^{2}} = 3.1623$
Midpoint of base BC = $\frac{8 + 7}{2} , \frac{2 + 5}{2} = \left(\frac{15}{2} , \frac{7}{2}\right)$
Slope of BC ${m}_{B C} = \frac{2 - 5}{8 - 7} = - 3$
Slope of altitude AD of the triangle passing through the midpoint D is ${m}_{A D} = - \frac{1}{{m}_{B C}} = - \left(\frac{1}{-} 3\right) = \frac{1}{3}$
$y - \left(\frac{7}{2}\right) = \left(\frac{1}{3}\right) \cdot \left(x - \left(\frac{15}{2}\right)\right)$
$6 y - 21 = 2 x - 15$
$3 y - x = 3$. $\textcolor{b l u e}{E q n} \textcolor{w h i t e}{\times} \textcolor{b l u e}{1}$
Area of triangle $A = \left(\frac{1}{2}\right) a h$
$h = \frac{2 \cdot 36}{3.1623} = 22.7682$
Slope of AB = $\tan \theta = {m}_{A B} = \frac{h}{\frac{a}{2}} = \frac{22.7682}{\frac{3.1623}{2}} = 14.4$
Equation of AB is
$y - 5 = 14.4 \cdot \left(x - 7\right)$
$y - 14.4 x = - 95.8$. $\textcolor{b l u e}{E q n} \textcolor{w h i t e}{\times} \textcolor{b l u e}{2}$
Solving Eqns (1) & (2) we get the coordinates of vertex ‘A’
$\textcolor{p u r p \le}{A \left(\frac{1452}{211}\right) , \left(\frac{695}{211}\right)}$
Verification :
Length of AB = b = sqrt((7 - (1452/211))^2 + (5 - (695/211)^2) = 1.7103
Length of AC = b = sqrt((8-(1452/211))^2 + (2 - (695/211)’^2) = 1.7103
Value of side $A B = A C = 1.7103$.
Hence its an isosceles triangle with sides $\textcolor{g r e e n}{3.1623 , 1.7103 , 1.7103}$
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1. ## Polynomial division
hey guys
Im fairly sure ive never been taught this and i have a few assignment questions to complete and im struggling to get my head around it.
The first one is 4x^2 - 3x - 2 divided by 1 - X
Can anyone give me some tips? hopefully from this first i'll manage the rest
Thanks
2. ## Re: Polynomial division
This Wikipedia page has a detailed example of long polynomial division. If you just need the remainder, you can use little Bézout's theorem. It says that the remainder when a polynomial $f(x)$ is divided by $x-a$ or $a-x$ is $f(a)$.
3. ## Re: Polynomial division
$4x^2-3x-2$ is divided by $-x+1$
let $ax+b$ be the quotient, then $(ax+b)(-x+1)=4x^2-3x-2$
$\begin{array}{rr|l}a&b&\times\\\hline-a&-b&-1\\a&b&1\end{array}$
$\begin{array}{c}4=-a\\-3=-b+a\\find\ a,\ b\end{array}$
$remainder=-2-b$
4. ## Re: Polynomial division
Hello, luke11121!
$(4x^2 - 3x - 2) \,\div\,(1 - x)$
The long division should look like this:
. . $\begin{array}{cccccccc} &&& - & 4x & - & 1 \\ && -- & -- & -- & -- & -- \\ -x + 1 & | & 4x^2 & - & 3x & - & 2 \\ && 4x^2 &-& 4x \\ && -- & -- & -- \\ &&&& x & - & 2 \\ &&&& x & - & 1 \\ &&&& -- & --& -- \\ &&&&& - & 1 \end{array}$
$4x^2 - 3x + 2 \;=\;(-x+1)(-4x-1) + (-1)$
m . . . . . . . . . . . $\text{divisor}\;\;\text{quotient} \;\;\text{remainder}$
5. ## Re: Polynomial division
Obviously the problem you have give relates to quadratic function which can have two distinct real roots and multiplying them gives: (x-a)(x-b)=ax^2+bx+c (identity).
So if you you have been given one of the factors suppose (x-a) is already given:
Therefore,
=> (x-b)=ax^2/x-b+bx/x-b+c/x-b (where common denominator (x-b) ).
So its comparabale to arithematic. Compare above identity to below equation:
if
x*y=a
find y
=>y=a/x
So thats the idea.
Hope i helped.
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# Vector Physics Problems And Solutions Pdf
File Name: vector physics problems and solutions .zip
Size: 16384Kb
Published: 30.05.2021
These solutions for Physics And Mathematics are extremely popular among Class 11 Science students for Physics Physics And Mathematics Solutions come handy for quickly completing your homework and preparing for exams. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction.
If we rotate it through an angle, its direction changes and we can say that the vector has changed. No, it is not possible to obtain zero by adding two vectors of unequal magnitudes. Yes, it is possible to add three vectors of equal magnitudes and get zero.
Consider the figure below: Lets examine the components of the three vectors. A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
Similarly, the velocity vector of the stationary body is a zero vector. When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball. Yes we can add three unit vectors to get a unit vector. No, the answer does not change if two unit vectors are along the coordinate axes. Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.
Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule. No, we cannot add two vectors representing physical quantities of different dimensions.
However, we can multiply two vectors representing physical quantities with different dimensions. Yes, a vector can have zero components along a line and still have a nonzero magnitude. This vector has zero components along a line lying along the Y-axis and a nonzero component along the X-axis.
The magnitude of the vector is also nonzero. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity.
However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid. A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.
If we slide it parallel to itself, then the direction and magnitude will not change. The third vector of length 2 should lie along x axis. All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components. We can determine this direction using the right hand thumb rule.
Here, 2. So, the answer must be written in three significant digits. However, components of a vector depend on the orientation of the axes. The x -component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors. First, we will find the components of the vector along the x -axis and y -axis.
Then we will find the resultant x and y- components. First, let us find the components of the vectors along the x and y -axes.
Consider that the queen is initially at point A as shown in the figure. Let AB be x ft. According to the polygon law of vector addition, the resultant of these six vectors is zero. A regular polygon has all sides equal to each other.
We know that the dot product of two perpendicular vectors is zero. Also, the direction of this product remains constant. According to the problem, the net electric and magnetic forces on the particle should be zero.
To find a slope at any point, draw a tangent at the point and extend the line to meet the x -axis. The required area can found by integrating y w. The metre scale is graduated at every millimetre. Hence, the number of significant digits may be 1, 2, 3 or 4. Its value is greater than 5.
So, the next two digits are neglected and 4 is increased by one. Therefore, the result should have three significant digits, while the other digits should be rounded off. So, the effective length should contain only two significant digits. Page No Answer: Yes. Answer: No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Answer: A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector. Answer: Yes we can add three unit vectors to get a unit vector. Answer: Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.
Answer: Two forces are added using triangle rule, because force is a vector quantity. Answer: No, we cannot add two vectors representing physical quantities of different dimensions. Answer: Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Answer: d it is slid parallel to itself. Answer: c 1, 2, 1 1,2 and 1 may represent the magnitudes of three vectors adding to zero. Answer: d None of these. Answer: b Answer: a the value of a scalar c a vector d the magnitude of a vector The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation.
Answer: b equal to AB c less than AB d equal to zero. Answer: First, we will find the components of the vector along the x -axis and y -axis.
Answer: First, let us find the components of the vectors along the x and y -axes. Answer: Consider that the queen is initially at point A as shown in the figure. Answer: According to the polygon law of vector addition, the resultant of these six vectors is zero. Answer: According to the problem, the net electric and magnetic forces on the particle should be zero.
Answer: The metre scale is graduated at every millimetre. Answer: a In , 7 comes after the digit 4.
A list of the major formulas used in vector computations are included. All of the vectors must be of the same type of quantity! For example, you cannot add a displacement to a velocity. The concept of vectors is discussed. In this case, vector will be positioned with its tail at the origin and its tip at the point 0, The negative of a vector is defined as the vector that, when added to the original vector, gives a resultant of zero! In mathematics, physics and engineering, we frequently come across with both types of quantities, namely, scalar quantities such as length, mass, time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc.
Chapter List. If the two forces 4N and 3N acting simultaneously on a particle are in opposite direction, the resultant force F 1 is minimum. Let R be the resultant force, then,.
So let's suppose that we call it South. If she started south, then upon making a right-hand-turn, she is now heading west. And after making a second right-hand-turn, she is heading North. The diagram at the right depicts the physical situation. The Pythagorean theorem can be used to determine the magnitude of the resultant.
Vectors are added by a head-to-tail method and the resultant is drawn from the tail of the first vector to the head of the last vector. The diagram should indicate the relative distances and directions for each segment of the path. A worksheet can be ready for any subject. Definitions, formulas, examples with solutions. A level resources including worksheets, topic tests and practice.
### Mcq on vectors in maths
These solutions for Physics And Mathematics are extremely popular among Class 11 Science students for Physics Physics And Mathematics Solutions come handy for quickly completing your homework and preparing for exams. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed. No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Vector and Scalar. Among the following options, which are scalar-vector pairs…. Force — acceleration. Pressure — force.
Physics Vector Solutions Note that a vector such as (i) may be written as A = i7 + j3 when typed, as it is easier to produce Numerically the solution is: (a) For vector problems, we first draw a neat sketch of the vectors and the vector.
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Class 11 Physics Chapter 4 NCERT Solutions are the best option available for students to gain basic knowledge in science and learn the concepts of motion in a plane. Ch4 Physics Class 11 is an easy chapter, and if understood properly, it can be a great help to students for securing good marks. In the first section, students will be introduced to some concepts that are related to motion in a plane, such as concepts of position, velocity, displacement and acceleration. Scalars and Vectors. In the second section, students will gain basic knowledge of scalars and vectors.
The physical quantities they measure fall into two categories: scalars and vectors. This textbook emphasizes. What are the steps necessary to add vectors in two dimensions? Answers 1. IIT Kharagpur. To find the dot product or scalar product of 3-dimensional vectors, we just extend the ideas from the dot product in 2 dimensions that we met earlier. If a question has only four answer options, do not mark option E.
In other words, each vector behaves as if the other vectors were absent. The resultant vector is the vector that results from adding two or more vectors together. If you want perfection in Maths subject for entrance enginnering point of view then start practising questions of Cengage G. Tewani Maths text book And in case if you have any doubts you can watch video solutions corresponding to questions of Cengage G. Tewani Maths textbook.
Никто не задаст вопросов. Никто ни в чем его не обвинит. Он сам расскажет о том, что случилось. Все люди умирают… что значит еще одна смерть.
Я не знаю… эта женщина… он называл ее… - Он прикрыл глаза и застонал. - Как. - Не могу вспомнить… - Клушар явно терял последние силы. - Подумайте, - продолжал настаивать Беккер. - Очень важно, чтобы досье консульства было как можно более полным.
Потрясающе, - страдальчески сказал директор. - У вас, часом, нет такой же под рукой. - Не в этом дело! - воскликнула Сьюзан, внезапно оживившись. Это как раз было ее специальностью.
Я понимал, что если он продаст свой алгоритм японской компании, производящей программное обеспечение, мы погибли, поэтому мне нужно было придумать, как его остановить. Я подумал о том, чтобы его ликвидировать, но со всей этой шумихой вокруг кода и его заявлений о ТРАНСТЕКСТЕ мы тут же стали бы первыми подозреваемыми.
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# Digit Chains
Purpose
This is a level 3 algebra strand link activity from the Figure It Out series.
A PDF of the student activity is included.
Achievement Objectives
NA3-8: Connect members of sequential patterns with their ordinal position and use tables, graphs, and diagrams to find relationships between successive elements of number and spatial patterns.
Student Activity
Click on the image to enlarge it. Click again to close. Download PDF (326 KB)
Specific Learning Outcomes
find and use rules in number patterns
Required Resource Materials
FIO, Link, Algebra, Book One, Digit Chains, page 16
Activity
Most of the early difficulty that students have with digit chains arises from not understanding the given rule to make the numbers in the digit chain. It may be that their difficulties are related to confusion with the notion of digit as distinct from number. It is likely that some discussion will be needed so that all can see
how, for example, 68 can be transformed into 21 and then to 6 and so on.
The interesting and surprising aspect of working with digit chains, which are easily made by devising a rule for each digit, is that they often reduce to a number that is part of a loop. The digit rules for question 1 produce a loop comprising the sequence 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, … The loops given in the Answers show how the digit rule works for the starting numbers 68 and 53.
In question 1c, the students create a separate digit chain for each starting number. When this is complete, they try to combine their separate digit chains so that they can clearly see the effect of the digit rule.
Note that three of the given starter numbers, 75, 83, and 90, all enter the loop at 5. The students could explore other 2-digit starters to see how many they can find entering the loop at 5. They might also do this for other numbers in the loop.
Before the students begin question 1e, they might try to predict what happens to 0, 1, and 2, which are not in the loop. As shown in the Answers, they will find that they form a separate string or sequence of numbers, which enter the loop at 3.
In question 2, the students extend their exploration of digit chains with the given rule to 3- and then 4-digit starter numbers. A diagram for 134 is:
The 3-digit numbers enter the loop in the following way:
In question 2e, the students try to show how a 4-digit number they choose fits into the chain. The process may involve a number of steps, each of which provides an opportunity for the students to show their place value knowledge and their skills in working with numbers. For example, 26 384 is 2 638 tens and 4 ones.
So it becomes (2 x 2 638) + (4 + 1) = 5 276 + 5, which is 5 281. This becomes
(2 x 528) + (1 + 1) = 1 056 + 2, which is 1 058. And 1 058 becomes (2 x 105) + (8 + 1) = 219, which becomes (2 x 21) + (9 + 1) = 52. The number 52 becomes (2 x 5) + (2 + 1) = 13, which finally becomes (2 x 1) + (1 + 3) = 6 where it enters the loop.
1. a.
b. i.–ii.
c. i. The digit chain for 37 is: 37, 14, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10, …
The digit chain for 48 is: 48, 17, 10, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, …
The digit chain for 63 is: 63, 16, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, …
The digit chain for 75 is: 75, 20, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, …
The digit chain for 83 is: 83, 20, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, …
The digit chain for 90 is: 90, 19, 12, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, …
ii.
d. Each 2-digit starter reduces to a number in the never-ending loop, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, …
e. Using the digit rule, 0 stays as 0 and begins the chain 0, 1, 2, 3, 4, … So, starter number 1 begins the chain 1, 2, 3, 4, 5, … , and starter number 2 begins the chain 2, 3, 4, 5, 6, … The diagram below shows how this works.
2. a. 134 = 13 tens and 4 ones. So 134 becomes (2 x 13) + (4 + 1) = 26 + 5, which is 31.
b. The number 31 becomes (2 x 3) + (1 + 1). 6 + 2 = 8, so 134 enters the loop at 8.
c. The starter number 100 is 10 tens + 0 ones. This becomes (2 x 10) + (0 + 1) = 21. Then 21 becomes 6, which is where it enters the loop. The starter number 256 is 25 tens + 6 ones. This becomes (2 x 25) + (6 + 1) = 57. Then 57 becomes 18, which becomes 11, which becomes 4, which is where it enters the loop. The starter number 751 is 75 tens + 1 one. This becomes (2 x 75) + (1 + 1) = 152, which
becomes (2 x 15) + (2 + 1) = 33, which becomes 10, which is where it enters the loop.
d. 2 138 = 213 tens and 8 ones. So 2 138 becomes (2 x 213) + (8 + 1) = 435, which becomes (2 x 43) + (5 + 1) = 92, which becomes (2 x 9) + (2 + 1) = 21, which becomes (2 x 2) + (1 + 1) = 4 + 2 = 6, which is where it enters the loop.
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# Linear Inequality (One Variable)
How to solve a linear inequality (one variable): 2 examples and their solutions.
## Example 1
### Solution
Solve the inequality just like solving an equation.
First move +5 to the right side.
Then the left side is 7x.
And the right side is 19, change the sign of +5, -5.
19 - 5 = 14
So 7x ≥ 14.
To remove the coefficient 7,
divide both sides by 7.
When multiplying or dividing a minus number on both sides, the order of the inequality sign changes.
In this case, 7 is not a minus number.
So the order of the inequality sign doesn't change.
Then the left side is x.
The order of the inequality sign doesn't change.
And the right side is 14/7.
14/7 = 2
So x ≥ 2.
x ≥ 2
## Example 2
### Solution
First move 2 to the right side.
Then the left side is -3x.
And the right side is 8, change the sign of 2, -2.
8 - 2 = 6
So -3x < 6.
To remove the coefficient -3,
divide both sides by -3.
When multiplying or dividing a minus number on both sides, the order of the inequality sign changes.
In this case, -3 is a minus number.
So the order of the inequality sign does change.
Then the left side is x.
The order of the inequality sign does change: < → >.
And the right side is 6/(-3).
6/(-3) = -2
So x > -2.
x > -2
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Home | | Maths 4th Std | Exercise 3.2 (Patterns in numbers)
# Exercise 3.2 (Patterns in numbers)
Text Book Back Exercises Questions with Answers, Solution : 4th Maths : Term 1 Unit 3 : Patterns : 4th Maths : Term 1 Unit 3 : Patterns : Cast out nines from a given number to check if it is a multiple of nine.
Exercise 3.2
1. Circle/underline the multiples of 9 (by using casting out nine).
a) 9443
b) 1008
c) 24689
d) 23769
e) 13476
a) 9443
9443
4+4+3 = 11
= 1 + 1 = 2
2 ≠ 9
Not multiple of 9
b) 1008
1+8=9
Multiple of 9
c) 24689
24689
2+4+6+8 = 20
= 2 + 0
2≠9
Not multiple of 9
d) 23769
23769
It is multiple of 9
e) 13476
13476
1+4+7 = 12
1+2 = 3 ≠ 9
It is not multiple of 9
2. Circle the correct addition fact (by using casting out nine method).
a) 4355 + 5369 = 9724
b) 7632 + 2213 = 9845
c) 6023 + 3203 = 9220
d) 2436 + 5315 = 7701
a) 4355 + 5369 = 9724
4355 + 5369 = 9724
3+5+5 = 4
13 = 4
1 + 3 = 4
4 = 4
b) 7632 + 2213 = 9845
7632 + 2213 = 9845
2+2+1+3 = 8
8 = 8
c) 6023 + 3203 = 9220
6023 + 3203 = 9220
2+3+2+3 = 2+2
2 + 8 = 4
10 = 4
1 ≠ 4
It is not correct addition fact.
d) 2436 + 5315 = 7701
2436 + 5315 = 7701
2+4 + 5+3+1+5 = 7 + 7 + 1
6+14 = 15
20 = 15
2 ≠ 6
It is not correct addition fact.
3. Circle the correct subtraction fact (by using casting out nine methods).
a) 7420 − 3625 = 3795
b) 2362 − 632 = 1720
c) 6732 − 4361 = 2371
d) 3264 − 1063 = 2200
a) 7420 − 3625 = 3795
7420 − 3625 = 3795
7 + 4 + 2 − 7 = 3+7+5
13 – 7 = 15
6 = 6
It is correct subtraction fact.
b) 2362 − 632 = 1720
2362 − 632 = 1720
4 − 2 = 1
2 = 1
2 ≠ 1
It is not correct subtraction fact.
c) 6732 – 4361 = 2371
6732 – 4361 = 2371
6+3 – 4+1 = 3+1
9 − 5 = 4
4 = 4
It is correct subtraction fact.
d) 3264 − 1063 = 2200
3264 − 1063 = 2200
2 + 4 − 1 = 2 + 2
6 – 1 = 4
5 ≠ 4
It is not correct subtraction fact.
Tags : Patterns | Term 1 Chapter 3 | 4th Maths , 4th Maths : Term 1 Unit 3 : Patterns
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
4th Maths : Term 1 Unit 3 : Patterns : Exercise 3.2 (Patterns in numbers) | Patterns | Term 1 Chapter 3 | 4th Maths
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### Vedic Math - Cube Roots of more than 6-Digit Number - Part III
In this article, we are going to learn an interesting Mathematical technique to find, if the given number is a perfect cube or not. It is very important step while computing cube roots. Infact, before applying any method to find the cube root, we have to check whether it is perfect cube or not and then accordingly we choose the technique. For example, following scenario tells us the importance of finding perfect cube step while computing the cube root.
Example : 1728 has cube root 12 since two groups are 1 and 728. From 728, we derive last digit as 2 from 1 (first group), we derive first digit as 1.
So, cube root of 1728 is 12.
But now, if number is 1278, which again has two groups: 1 and 278. It can derive the same last digit as 2 and first digit as 1 , which implies that cube root of 1278 is 12, which is not true because technique stands true for perfect cube root only.
There is a simple technique to check whether the number is perfect cube or not. For this, we add the digits of the number. See the below chart in which we add the digits of cubes from 1 to 10.
Above example shows that sum of digits of a perfect cube is either 1, 8 or 9. However, it is not true that all numbers which sum to 1,8 or 9, will be perfect cube.
For example,
Sum of digits of 1728 and 1278 are same i.e.(1+7+2+8) = (18) = 9 . But 1278 is not a perfect cube.
Hence if sum of digits of a number is not 1,8 or 9, we are very sure that the number is not a perfect cube. However, a number may not be perfect cube root even if sum of digits is 1,8 or 9. To scrutinize that, we need to apply factorisation. If number is small like 1278, factorisation is good method. See below:
For bigger numbers, factorisation could be time consuming technique. Hence, for large numbers, we shall apply general method of finding the cube of root.
Case 2 : Cube root for all the cubes, whether perfect cubes or not. (Case 1 discussed in last two articles)
From last two articles, we conclude about the sequence of digits (a+b+c)³ as:
(1) The first place by a³
(2) The second place by 3a2b
(3) The third place by 3ab2+3a2c
(4) The fourth place by 6abc+b³
(5) The fifth place by 3ac2+3b2c
(6) The sixth place by 3bc2
(7) The seventh place by c³ ; and so on.
In 'General Technique', we find Dividends(D), Quotients(Q), and Remainders(R). Steps involved as:
(1) First determine D, Q and R
(2) From the second dividend, no deduction is to be made.
(3) From the third, subtract 3ab2
(4) From the fourth, deduct 6 abc+b³
(5) from the fifth, subtract 3ac2+3b2c
(6) from the sixth, deduct 3bc2
(7) from the seventh, subtract c³. ; and so on.
(a) Quotient(Q) is closest minimum exact cube to the first cube i.e. 'F' term used in last two articles.
(b) And, Reminder(R) is the difference between the first group and closest minimum exact cube.
(c) Dividend(D) is found by multiplying the 'Square of Quotient(Q)' by 3 (Q2*3)
Lets take an example to make it more clear.
Example 1 : 248858189
(1) First arrange the numbers in groups i.e. 248,858,189
Here, N = 3
First group(248) has closet minimum exact cube (216) which is 6³. So, First Quotient(Q) = 6
First Reminder(R) = 248-216 = 32
First Dividend(D) = 62*3 = 108
(2) The second Gross Dividend is 328. We don't subtract anything at that point. We only divide it by 108 and write down 2 and 112 as Q and R.
[Important Note : Here we are taking quotient to be 2 instead of 3; because if we take it as 3, the reminder comes out 4 (328-108*3) and third dividend turns out 45 which is absurd and will not be dividable by 108]
(3) The third Gross Dividend is 1125. Subtract 3ab2 (here, a=6, b=2, first two quotients) i.e. 3*6*22 = 72 from 1125 (i.e. 1125-72=1053) Therefore, Third Actual Dividend is 1053; divide 1053 by 108 gives 9 and 81 as Q and R.
(4) The fourth Gross Dividend is 818. Subtract 6abc + b³ (here, a=6, b=2, c=9) i.e. 6*6*2*9+2³ = 656 from 818 (i.e 818-656=162) So, Fourth Actual Dividend is 162. Divide this again by 108 and write down 0 and 162 as Q and R.
(5) The fifth Gross Dividend is 1621. Subtract 3ac2+3b2c (here, a=6, b=2, c=9) i.e. 1458+108 = 1506 from 1621 (i.e 1621-1506=55) So, Fifth Actual Dividend is 55. Divide it by 108 and write down 0 and 55 as Q and R.
(6) The sixth Gross Dividend is 558. Subtract 3bc2 (here, b=2, c=9) i.e. 486 from 558 (i.e 558-486=72) So, Sixth Actual Dividend is 72. Divide this by 108 and write down 0 and 72 as Q and R.
(7) The last / seventh Gross Dividend is 729. Subtract c³ (here, c=9) i.e. 729 from 729 (i.e 729-729=0) So, Seventh Actual Dividend is 0 and write down 0 and 0 as Q and R.
Put decimal after 3 digits (N=3). After decimal there are all zeros. This means that the given number is a perfect cube and the cube root is 629.
(After putting decimal, if there are still numbers except 0's than the number is not perfect cube)
Steps involved in finding dividend, quotient and reminder for (a+b+c+d)³ are:
(1) First determine D, Q and R
(2) From the second dividend, no deduction is to be made.
(3) From the third, subtract 3ab2
(4) From the fourth, deduct 6abc+b³
(5) from the fifth, subtract 6abd+3ac2+3b2c
(6) from the sixth, deduct 6acd+3bc2+3b2d
(7) from the seventh, subtract 6bcd+3ad2+c2
(8) From the eighth, subtract 3bd2+3c2d
(9) From the ninth, subtract 3cd2
(10) From the tenth, subtract d³; and so on.
OR We can convert (a+b+c+d) into (a+b+c). By considering first two groups into one group.
For example: 12278428443 can be written as
12, 278, 428, 443 (a+b+c+d)
12278, 428, 443 (a+b+c)
But, in second, we shall get the first quotient bigger.
Lets take one more example of an imperfect cube (not a perfect cube).
Find cube root of 417 upto 3-decimals place.
So, Cube root of 417 is 7.471
Hope these methods will help you all in computing cube root of the number.
In the upcoming articles, we will discuss about various techniques of division using Vedic Math.
### Vedic Math - Cube Roots of more than 6-Digit Number - Part II
In last article, we have discussed the method to find cube root of more than 6-digit numbers; especially the odd numbers. Today we shall discuss the procedure for even numbers. In this procedure, only two extra steps are added, one in the beginning and other at the end. Rest all is same.
Procedure: As first added step, we keep on dividing the number by 8 till we get an odd cube. Following it, same method of successive elimination of the digits will apply. At the end, multiply the cube root by 8 to obtain the cube root of the original number.
Example : 2840362499528
First, we continue without using those two additional steps, which will help you to understand the problems arises while dealing with even cubes.
The cube root of the cube 2,840,362,499,528 (say, F + J + H + M + L )
Here, N=5 (means that cube root will be of 5 digits number)
L=2 (i.e. 2³=8, matching with the last digit of the last group '528')
and F=1 (i.e, 1³=1, nearest cube of first group '1')
Step1 : L=2 & L³=8. Subtracting this,
Step2 : 3L2M=12M (substituting L = 2)
Hence, 12M = Number ending with 2
Here M is either '1' or '6' (ambiguous values)
Lets take 6 (pure gamble)
Now, Deducting 3L2M = 12M = 72
Step3 : 3LM2+3L2H = 12H + 216 (substituting L = 2, M = 6)
12H + 216 = Number ending with 8
12H = Number ending with 2
Here H is either '1' or '6'
Lets take 1 (again gamble)
Now, Subtract 3LM2+3L2H = 12H + 216
= 228
Step4 : 3L2J+6LMH+M³ = 12J+12+216
= 12J+228
12J+228 = Number ending with 6
12J = Number ending with 8
Here J is either '4' or '9'
Lets take J = 4
Since we already know 'F' , so no need to know the expansion of (F+J+H+M+L)³
Therefore, cube root is 14162 (F=1, J=4, H=1, M=6, L=2)
In the above example, we see that there are so many ambiguous values like in step 2, 3, 4. To solve this problem, we divide the number by 8.
So,
After first division by 8, if we get the last digit as odd number; we shall consider this number. If last digit is still even, we shall again divide by 8 till the last digit comes out to be an odd number. Remember one more thing, we will multiply the final output by 8 only once and not the number of times we divide it by the number (like, after dividing 3 times '792 994 249 216' by 8, we get '1 548 816 893', an odd number. Then we calculate the cube root of this odd number. Finally we shall multiply the cube root with 8 only once,
output*8 = Final Cube Root)
For, 355,045,312,441
N=4,
L=1 (1³=1)
F=7 (7³=343 < 355)
Step1 : L=1 & L³=1. Subtracting this,
Step2 : 3L2M=3M (substituting L = 1)
3M = Number ending with 4
So M should be 8
Now, Deducting 3L2M = 3M
= 24
Step3 : 3LM2+3L2H = 3H + 192 (substituting L = 1, M = 8)
3H + 192 = Number ending with 2
3H = Number ending with 0
Hence H = 0
Therefore, Cube Root is 7081 (F=7, H=0, M=8, L=1)
The Cube Root of the original number = 7081*8
= 14162
In next article, we shall learn to find the cube root for all cube numbers (whether perfect or not). Thanks for visiting the blog. Please keep sharing the knowledge by posting in 'comments' section.
If you like the article, you may contribute by:
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# How do you write an equation in standard form given that the line passes through (-1, -3) and (2, 1)?
##### 1 Answer
Jun 7, 2015
Using this formula with the given points:
$\frac{x - {x}_{0}}{{x}_{1} - {x}_{0}} = \frac{y - {y}_{0}}{{y}_{1} - {y}_{0}}$
The points are:
• $A \left({x}_{0} , {y}_{0}\right) \to A \left(- 1 , - 3\right)$
• $B \left({x}_{1} , {y}_{1}\right) \to B \left(2 , 1\right)$
So the equation of the line passing through A and B becomes:
$\frac{x - \left(- 1\right)}{2 - \left(- 1\right)} = \frac{y - \left(- 3\right)}{1 - \left(- 3\right)}$
$\frac{x + 1}{2 + 1} = \frac{y + 3}{4}$
$\frac{x + 1}{3} = \frac{y + 3}{4}$
Now we must bring the equation in a "standard form" that means an equation similar to $a x + b y + c = 0$
$4 \cdot \left(x + 1\right) = \left(y + 3\right) \cdot 3$
$4 x + 4 = 3 y + 9$
$4 x - 3 y - 5 = 0$
graph{4x-3y-5=0 [-3, 3,-3,3]}
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### Exponents and Powersv - Solutions
CBSE Class –VII Mathematics
NCERT Solutions
Chapter 13 Exponents and Powers
(Ex. 13.1)
Question 1. Find the value of:
(i) 26 (ii)93 (iii)112 (iv) 54
Answer: (i) 2= 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 9= 9 x 9 x 9 = 729
(iii) 11= 11 x 11 = 121
(iv) 5= 5 x 5 x 5 x 5 = 625
Question 2.Express the following in exponential form:
(i) 6 x 6 x 6 x 6 (ii) t x t (iii) b x b x b x b
(iv) 5 x 5 x 7 x 7 x 7 (v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d
Answer: (i) 6 x 6 x 6 x 6 = 64 (ii) t x t = t2 (iii) b x b x b x b = b4
(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73 (v) 2 x 2 x a x a = 22 x a2
(vi) a x a x a x c x c x c x c x d = a3 x c4 x d
Question 3. Express each of the following numbers using exponential notation:
(i) 512 (ii) 343 (iii) 729 (iv) 3125
Answer: (i) 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29
(ii) 343 = 7 x 7 x 7 = 73
(iii) 729 = 3 x 3 x 3 x 3 x 3 x 3 = 36
(iv) 3125 = 5 x 5 x 5 x 5 x 5 = 55
Question 4. Identify the greater number, wherever possible, in each of the following:
(i) 4or 34
(ii) 5or 35
(iii) 2or 82
(iv) 100or 2100
(v) 210 or 102
Answer: (i) 43= 4 x 4 x 4 = 64
34= 3 x 3 x 3 x 3 = 81
Since 64 < 81
Thus, 3is greater than 43.
(ii) ${5}^{3}$= 5 x 5 x 5 = 125
${3}^{5}$= 3 x 3 x 3 x 3 x 3 = 243
Since, 125 < 243
Thus, ${3}^{4}$ is greater than ${5}^{3}$.
(iii) ${2}^{8}$= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
${8}^{2}$= 8 x 8 = 64
Since, 256 > 64
Thus, ${2}^{8}$is greater than${8}^{2}$.
(iv) ${100}^{2}$= 100 x 100 = 10,000
${2}^{100}$= 2 x 2 x 2 x 2 x 2 x …..14 times x ……… x 2 = 16,384 x ….. x 2
Since, 10,000 < 16,384 x ……. X 2
Thus, ${2}^{100}$is greater than${100}^{2}$.
(v) ${2}^{10}$= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1,024
${10}^{2}$= 10 x 10 = 100
Since, 1,024 > 100
Thus, ${2}^{10}$${10}^{2}$
Question 5. Express each of the following as product of powers of their prime factors:
(i) 648 (ii) 405 (iii) 540 (iv) 3,600
Answer: (i) 648 = ${2}^{3}×{3}^{4}$
(ii) 405 = $5×{3}^{4}$
(iii) 540 = ${2}^{2}×{3}^{3}×5$
(iv) 3,600 = ${2}^{4}×{3}^{2}×{5}^{2}$
Question 6. Simplify:
(i) $2×{10}^{3}$
(ii) ${7}^{2}×{2}^{2}$
(iii)${2}^{3}×5$
(iv) $3×{4}^{4}$
(v)$0×{10}^{2}$
(vi) ${5}^{2}×{3}^{3}$
(vii)${2}^{4}×{63}^{2}$
(viii) ${3}^{2}×{10}^{4}$
Answer: (i) $2×{10}^{3}$= 2 x 10 x 10 x 10 = 2,000
(ii) ${7}^{2}×{2}^{2}$= 7 x 7 x 2 x 2 = 196
(iii) ${2}^{3}×5$= 2 x 2 x 2 x 5 = 40
(iv) $3×{4}^{4}$= 3 x 4 x 4 x 4 x 4 = 768
(v) $0×{10}^{2}$= 0 x 10 x 10 = 0
(vi) ${5}^{2}×{3}^{3}$= 5 x 5 x 3 x 3 x 3 = 675
(vii) ${2}^{4}×{63}^{2}$= 2 x 2 x 2 x 2 x 3 x 3 = 144
(viii) ${3}^{2}×{10}^{4}$= 3 x 3 x 10 x 10 x 10 x 10 = 90,000
Question 7. Simplify:
(i) (-4)3
(ii) $\left(-3\right)×{\left(-2\right)}^{3}$
(iii) ${\left(-3\right)}^{2}×{\left(-5\right)}^{2}$
(iv) ${\left(-2\right)}^{3}×{\left(-10\right)}^{3}$
Answer: (i) ${\left(-4\right)}^{3}=\left(-4\right)×\left(-4\right)×\left(-4\right)=-64$
(ii) $\left(-3\right)×{\left(-2\right)}^{3}=\left(-3\right)×\left(-2\right)×\left(-2\right)×\left(-2\right)=24$
(iii) ${\left(-3\right)}^{2}×{\left(-5\right)}^{2}=\left(-3\right)×\left(-3\right)×\left(-5\right)×\left(-5\right)=225$
(iv) ${\left(-2\right)}^{3}×{\left(-10\right)}^{3}=\left(-2\right)×\left(-2\right)×\left(-2\right)×\left(-10\right)×\left(-10\right)×\left(-10\right)=8000$
Question 8. Compare the following numbers:
(i) $2.7×{10}^{12}$;
(ii) $4×{10}^{14}$$3×{10}^{17}$
Answer: (i) $2.7×{10}^{12}$and
On comparing the exponents of base 10,
$2.7×{10}^{12}$
(ii) $4×{10}^{14}$and $3×{10}^{17}$
On comparing the exponents of base 10,
$4×{10}^{14}$
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## Algebra
Showing posts with label inverses. Show all posts
Showing posts with label inverses. Show all posts
### Inverse Functions
We begin with the definition: Inverse Functions – The functions f(x) and g(x) are inverses if both
for all x in the domain g and f respectively. In other words, if you compose inverse functions the result will be x.
Verify that the given functions are inverses.
When verifying that two functions are inverses, you must obtain the original value x by composing both ways.
Determine whether or not the given functions are inverses.
If f and g are inverse functions, then g can be written using the notation
which reads, "g equals f inverse." Caution: In this context, the -1 indicates an inverse function not a negative exponent.
Take the time to review one-to-one (1-1) functions because it turns out that if a function is 1-1 then it has an inverse. Therefore, we may think of the horizontal line test as a test that determines if a function has an inverse or not.
Next we outline a procedure for actually finding inverse functions.
Step 1: Replace f(x) with y.
Step 2: Interchange x and y.
Step 3: Solve the resulting equation for y.
Step 4: Replace y with the notation for the inverse of f.
Step 5: (Optional) Verify that the functions are inverses.
Find the inverse of the given function.
Now that you know the definition of an inverse function and how to find them, we will next turn our attention to their graphs. For any one-to-one function f where
and we have the following property.
Symmetry of Inverse Functions – If (a, b) is a point on the graph of a function f then (b, a) is a point on the graph of its inverse. Furthermore, the two graphs will be symmetric about the line y = x.
In the following graph, see that the functions
have symmetry when graphed on the same set of axes.
Notice that (2, 3) is a point on f and (3, 2) is a point on its inverse. In other words, to graph the inverse all you need to do is switch the coordinates of each ordered pair. We used this fact to find inverses and will be very important in the next chapter when we develop the definition of the logarithm.
Given the graph of a 1-1 function, graph its inverse and the line of symmetry.
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# What is the solution to the Differential Equation e^(x+y)(dy/dx) = x with #y(0)=1?
Sep 17, 2017
$y = \ln \left(- x {e}^{x} - {e}^{x} + e + 1\right)$
#### Explanation:
We have:
${e}^{x + y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = x$ with $y \left(0\right) = 1$ ..... [A]
We can rearrange the DIfferential Equation [A] as follows:
${e}^{x} {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = x \implies {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = x {e}^{- x}$
This is a First Order Separable Differential equation ad we now "seperate the variables" to get:
$\int \setminus {e}^{y} \setminus \mathrm{dy} = \int \setminus x {e}^{- x} \setminus \mathrm{dx}$ ..... [B]
The LHS integral is a standard result, and for the RHS| integral we would need to apply Integration By Parts:
Let $\left\{\begin{matrix}u & = x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = 1 \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{-} x & \implies v & = - {e}^{-} x\end{matrix}\right.$
Then plugging into the IBP formula:
$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$
gives us
$\int \setminus \left(x\right) \left({e}^{x}\right) \setminus \mathrm{dx} = \left(x\right) \left(- {e}^{-} x\right) - \int \setminus \left(- {e}^{-} x\right) \left(1\right) \setminus \mathrm{dx}$
$\therefore \int \setminus x {e}^{-} x \setminus \mathrm{dx} = - x {e}^{-} x - {e}^{-} x$
Using this result, we can now integrate [B] to get the General Solution:
${e}^{y} = - x {e}^{x} - {e}^{x} + C$
Using the initial condition $y \left(0\right) = 1$ we have:
${e}^{1} = - 0 {e}^{0} - {e}^{0} + C \implies C = e + 1$
Hence, the Particular Solution is:
${e}^{y} = - x {e}^{x} - {e}^{x} + e + 1$
And we can gain an explicit solution for [A] if we take Natural Logarithms:
$\ln \left({e}^{y}\right) = \ln \left(- x {e}^{x} - {e}^{x} + e + 1\right)$
So that finally:
$y = \ln \left(- x {e}^{x} - {e}^{x} + e + 1\right)$
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# 3.2 Properties of Rational Exponents
## Presentation on theme: "3.2 Properties of Rational Exponents"— Presentation transcript:
3.2 Properties of Rational Exponents
Math 3 Mr. Ellingsen
Review of Properties of Exponents from section 6.1
am * an = am+n (am)n = amn (ab)m = ambm a-m = = am-n = These all work for fraction exponents as well as integer exponents.
(43 * 23)-1/3 = (43)-1/3 * (23)-1/3 = 4-1 * 2-1 = ¼ * ½ = 1/8 d. = = = 61/2 * 61/3 = 61/2 + 1/3 = 63/6 + 2/6 = 65/6 b. (271/3 * 61/4)2 = (271/3)2 * (61/4)2 = (3)2 * 62/4 = 9 * 61/2 ** All of these examples were in rational exponent form to begin with, so the answers should be in the same form!
Ex: Write the expression in simplest form.
Ex: Simplify. = = = 5 = = 2 Ex: Write the expression in simplest form. = = = = = Can’t have a tent in the basement! ** If the problem is in radical form to begin with, the answer should be in radical form as well.
Ex: Perform the indicated operation
5(43/4) – 3(43/4) = 2(43/4) b. = c. = If the original problem is in radical form, the answer should be in radical form as well. If the problem is in rational exponent form, the answer should be in rational exponent form.
More Examples a. b. c. d.
Ex: Simplify the Expression. Assume all variables are positive.
(16g4h2)1/2 = 161/2g4/2h2/2 = 4g2h c. d.
Ex: Write the expression in simplest form
Ex: Write the expression in simplest form. Assume all variables are positive. a. b. No tents in the basement! c. ** Remember, solutions must be in the same form as the original problem (radical form or rational exponent form)!!
d. Can’t have a tent in the basement!!
Ex: Perform the indicated operation. Assume all variables are positive.
e.
Similar presentations
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# Question Video: Simplifying the Product of Monomials Using the Laws of Exponents Mathematics
Simplify (1/3 𝑥⁷) × (4/5 𝑥³).
02:11
### Video Transcript
Simplify one-third 𝑥 raised to the seventh power multiplied by four-fifths 𝑥 cubed.
In this question, we are asked to simplify the product of two algebraic expressions. If we analyze the two factors, we can see that they are the product of constants and variables that are raised to nonnegative integer exponents. So both factors are monomials.
To simplify the product of monomials, we first want to use the commutativity and associativity of multiplication to rewrite the product so that we multiply the constants and variables separately. This gives us one-third times four-fifths multiplied by 𝑥 raised to the seventh power times 𝑥 cubed.
We can then evaluate each product separately. First, we multiply fractions by multiplying their numerators and denominators separately. We obtain one times four over three times five. Second, we can simplify the product of the variables by using the product rule for exponents. This tells us that if 𝑚 and 𝑛 are nonnegative integers, then 𝑥 raised to the power of 𝑚 times 𝑥 raised to the power of 𝑛 is equal to 𝑥 raised to the power of 𝑚 plus 𝑛. In other words, when multiplying exponential expressions with the same base raised to nonnegative integer exponents, we can instead raise the base to the sum of the exponents.
We can apply this result to simplify our product with 𝑚 equal to seven and 𝑛 equal to three. We get 𝑥 raised to the power of seven plus three. Finally, we can evaluate to obtain four over 15 multiplied by 𝑥 raised to the 10th power.
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# Brilliant’s Composite Mathematics Class 7 Solutions Chapter 1
## Brilliant’s Composite Mathematics Class 7 Solutions Chapter 1 Rational Numbers
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Brilliant’s Composite Mathematics Class 7 Math Book, Chapter 1, Rational Numbers. Here students can easily find step by step solutions of all the problems for Rational Numbers, Exercise 1.1, 1.2, 1.3 and 1.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Rational Numbers Exercise 1.1 Solution
Question no – (1)
Solution :
Here, Numerator is 5
Question no – (2)
Solution :
Here, Numerator is – 9
Question no – (3)
Solution :
Here, Numerator is 2
Question no – (4)
Solution :
Here, Numerator is 15
Question no – (5)
Solution :
Here, Numerator is – 7
Question no – (6)
Solution :
Here, Denominator is 20
Question no – (7)
Solution :
Here, Denominator is 13
Question no – (8)
Solution :
Here, Denominator is – 17
Question no – (9)
Solution :
Here, Denominator is – 25
Question no – (10)
Solution :
Here, Denominator is 28
Question no – (11)
Solution :
Rational number = 15/1
Question no – (12)
Solution :
Rational number = – 7/1
Question no – (13)
Solution :
Rational number = 10/1
Question no – (14)
Solution :
Rational number = – 25/1
Question no – (15)
Solution :
Rational number = 37/1
Question no – (16)
Solution :
Given, 13/1
Now, as integers = 13
Question no – (17)
Solution :
Given, – 25/1
Now, as integers = -25
Question no – (18)
Solution :
Given, 27/1
Now, as integers = 27
Question no – (19)
Solution :
Given, – 18/1
Now, as integers = 18
Question no – (20)
Solution :
Given, 29/1
Now, as integers = 29
Question no – (21)
Solution :
Given, – 3/14
= it is negative
Question no – (22)
Solution :
Given, – 4/- 27
= it is positive
Question no – (23)
Solution :
Given, 5/11
= it is positive
Question no – (24)
Solution :
Given, – 7/18
= it is negative
Question no – (26)
Solution :
Given, 9/- 25
= it is negative
Question no – (27)
Solution :
Given, – 25/39
= it is negative
Question no – (28)
Solution :
Given, – 21/- 58
= it is positive
Question no – (29)
Solution :
(i) 15 – 7/16 + 23
= 8/39
(ii) 3 × 4/5 ×2
= 12/10
= 5/6
(iii) 17 – 34/- 5 × 13
= – 17/- 65
Rational Numbers Exercise 1.2 Solution
Question no – (1)
Solution :
Given, – 3/5
= – 3 × 2/5 × 2
= – 6/10
Question no – (2)
Solution :
Given, – 3/5
= – 3 × 3/5 × 3
= 9/15
Question no – (3)
Solution :
Given, – 3/5
= – 3 × – 5/5 × – 5
= 15/- 25
Question no – (4)
Solution :
Given, – 3/5
= – 3 × 8/5 × – 8
= 24/- 40
Question no – (5)
Solution :
Given, – 3/5
= – 3 × 13/5 × 13
= – 39/65
Question no – (6)
Solution :
= – 5/6
= – 5 × 2/6 × 2
= – 10/12
Question no – (7)
Solution :
– 5/6
= – 5 × 3/6 × 3
= – 15/18
Question no – (8)
Solution :
– 5/6
= – 5 × 4/6 × 4
= – 20/24
Question no – (9)
Solution :
– 5/6
= – 5 × 5/6 × 5
= – 25/30
Question no – (10)
Solution :
– 5/6
= – 5 × 8/6 × 8
= – 40/48
Question no – (11)
Solution :
Given, 3/- 7
= 3 × – 1/- 7 × -1
= – 3/7
Question no – (12)
Solution :
Given, 14/- 27
= 14 × -1/- 27 × – 1
= – 14/27
Question no – (13)
Solution :
Given, 21/- 13
= 21 × – 1/- 13 × – 1
= – 21/13
Question no – (14)
Solution :
Given, 8/- 3
= 8 × – 1/- 9 × – 1
= – 8/9
Question no – (15)
Solution :
Given, – 27/- 59
= 27 × – 1/- 59 × – 1
= – 27/59
Question no – (16)
Solution :
Given, -3/7, 4/9
= Equal
Question no – (17)
Solution :
Given, 20/-25, 8/-10
= Not equal
Question no – (18)
Solution :
Given, -2/14, 3/21
= Equal
Question no – (19)
Solution :
Given, 3/8, 6/16
= Equal
Question no – (20)
Solution :
Given, -26/39
Now, in standard from = -2/3
Question no – (21)
Solution :
Given, 45/- 60
Now, in standard from = 3/-5
Question no – (22)
Solution :
Given, 24/- 56
Now, in standard from = 3/-7
Question no – (23)
Solution :
Given, -12/28
Now, in standard from = -3/7
Question no – (24)
Solution :
Given, -92/230
Now, in standard from = -2/5
Question no – (25)
Solution :
= -3/4 = 21/-28 = -15/20
Question no – (26)
Solution :
= 5/7 = 25/35 = 15/21
Question no – (27)
Solution :
= 6/13 = -12/26 = 24/52
Question no – (28)
Solution :
= -15/39 = 5/13 = 25/-65
Rational Numbers Exercise 1.3 Solution
Question no – (9)
Solution :
= -5/3 > 17/-10
Question no – (10)
Solution :
= 2/5 > 3/7
Question no – (11)
Solution :
= 6/-5 > -13/-8
Question no – (12)
Solution :
= 4/3 > -8/-7
Question no – (13)
Solution :
-7/10 < 2/-3
Question no – (14)
Solution :
2/-3 < -5/8
Question no – (15)
Solution :
5/8 < 3/4
Question no – (16)
Solution :
4/3 < -8/-7
Question no – (17)
Solution :
Given, -7/10, 2/-3, 5/8, 3/5,
= -7/10 < 2/- 3 < 3/5 < 5/8 …(ascending order)
Question no – (18)
Solution :
Given, 4/9, -11/24, 3/7, 7/-15
= 4/9 > 3/7 > -11/24 > 7/-15 …(descending order)
Question no – (19)
Solution :
Greatest rational number is 2/-3
Question no – (20)
Solution :
Greatest rational number is 22/23
Question no – (21)
Solution :
Greatest rational number is -4/9 < 3/-7
Question no – (22)
Solution :
Greatest rational number is 3/13 < 4/17
Question no – (23)
Solution :
Greatest rational number is 4/-25 < -3/19
Question no – (24)
Solution :
Greatest rational number is 12/-23 > -3/5
Question no – (25)
Solution :
(i) Given statements – True
(ii) Given statements – False
(iii) Given statements – True
(iv) Given statements – True
(v) Given statements – False
(vi) Given statements – True
(vii) Given statements – True
(viii) Given statements – False
(ix) Given statements – True
(x) Given statements – False.
Rational Numbers Exercise 1.4 Solution
Question no – (1)
Solution :
2/3 = 2 × 5/3 ×5
= 10
= 4/5 = 4 × 3/5 × 3
= 12/15
10/15, 11/15, 12/15
Question no – (2)
Solution :
2/3 = 2 × 7/3 × 7
= 14/21
= 4/7 = 4 × 3/7 × 7
= 12/21
12/21, 13/21, 14/21
Question no – (3)
Solution :
– 1 = – 1 × 3/1 × 3
= – 3/5
= 2/3 = 2/3
-2/3, -1/3 , 1, 1/3, 2/3
Question no – (4)
Solution :
-3/4 = -3/4
= 1 = 1 × 4/1 ×4
= 4/4
= -2/4, -1/4, 1, 1/4, 2/4, 3/4
Question no – (5)
Solution :
-5/8 = 5 × 11/8 × 11
= -55/88
= -6/11 = -6 × 8/11× 8
= -48/88
= -59/88, -51/88, 52/88
Question no – (6)
Solution :
1 = 1×12/1 ×12
= 12/12
= 2/3 = 2 × 4/3 × 4
= 8/12
= 9/12, 1/12, 11/12, 12/12
Three rational numbers are 9/12, 1/12, 11/12, 12/12
Question no – (7)
Solution :
2/3 = 2 × 15/3 ×15
= 30/45
= 3/5 = 3 ×9/5 ×9
= 27/45
= 28/45, 29/45, 30/45
Three rational numbers are 28/45, 29/45, 30/45
Question no – (8)
Solution :
-1 = 1 ×12/1×12
= -12
= 4/3 = 3×3/4×3
= – 9/12
= – 10/12, -11/12, -12/12
Three rational numbers are – 10/12, -11/12, -12/12
Question no – (9)
Solution :
-3/4 = -3×15/4×5
= -45/60
= -4/5 = -4×12/5×12
= -48/60
= -46/60, -67/60, -48/60
Three rational numbers are -46/60, -67/60, -48/60
Question no – (10)
Solution :
-15/2 = -15×5/2×5
= -75/10
= -13/2 = -13 ×5/2×5
= – 65/10
= -66/10, -67/10, -68/10
Three rational numbers are -66/10, -67/10, -68/10
Question no – (11)
Solution :
= -1
= 1/2
= -1, 0, 1, 1, 1/2
Four rational numbers are -1, 0, 1, 1, 1/2
Question no – (12)
Solution :
Given, 3/14
= |3/14|
= 3/14
Question no – (13)
Solution :
Given, -7/18
= |-(-7/18)|
= |7/18|
Question no – (14)
Solution :
Given, -9/-10
= |-(-9/10)|
= 9/10
Question no – (15)
Solution :
Given, -5/18
= |- (-5/18)|
= 5/185
Question no – (16)
Solution :
Given, -4/29
= |- (-4/29)|
= 4/29
Question no – (17)
Solution :
Given, 3/-4
= |- (-3/4)|
= 3/4
Question no – (18)
Solution :
Given, x = -7/9
or, |x| = |-x|
= |-(-7/9)|
= 7/9
Question no – (19)
Solution :
Given, x = 4/13
or, |x| = |-4/13|
= -4/3
x = -4/3
Question no – (20)
Solution :
Given, x = -15/-29
|x| = |-(15/-29)|
= 15/29
x = 15/29
Question no – (21)
Solution :
Given, 2/3
= |-2/3|
= 2/3
2/3, -2/3
Question no – (22)
Solution :
Given, 3/4
= |-3/4|
= 3/4
∴3/4, -3/4
Question no – (23)
Solution :
Given, 6/25
= |-6/25|
= 6/25
6/25, -6/25
Question no – (24)
Solution :
Given, 14/27
= |-14/27|
= 14/27
14/27, -14/27
Question no – (25)
Solution :
Given, 16/23
= |-16/23| = 16/23
16/23, -16/23
Question no – (26)
Solution :
(a) Can we always find a rational number two rational numbers
= Yes
(b) Can we always find an integer between two integers
= No
(c) Can we find an integer between two consecutive integers
= No
(d) Can you find a rational number whose absolute values is -1
= Yes
Next Chapter Solution :
Updated: June 9, 2023 — 3:51 pm
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# Delaunay Triangulation
Page by Murray Bourne, IntMath.com. Last updated: 07 September 2019.
We start with a short introduction to explain what's going on in the math-based artwork below.
## Triangulation
Triangulation is a widely-used concept in the fields of computer games and computer graphics generally. Let's look at an example of triangulation.
We have a plane represented by this rectangle, some points on the plane, and we draw triangles using those points. We have "triangulated the plane".
## Delaunay Triangulation
A Delaunay Triangulation is a specific type of triangulation such that no triangle's points are inside the circumcircle of any other triangle in the triangulation.
Let's see what that means in an example.
We start with a triangle, and calculate its circumcenter, indicated by a pink dot. The circumcenter is the center of the unique circle that passes through the 3 points of the triangle.
### Finding the circumcenter
The circumcenter is found by determining the intersection of the perpendicular bisectors of two sides of the triangle. That is:
1. Find the midpoints of two of the sides
2. Find the equations of the perpendicular lines passing through those midpoints
3. Find the intersection of those two lines
Of course, I used code to find the circumcenters in these examples.
Next, we construct the circumcircle and observe it passes through the three points of the triangle.
Next, we construct another triangle off one of the sides of our existing triangle. We calculate its circumcenter (the green dot) and construct the circumcircle.
Note our original pink dot is outside our new circumcircle, and the new green dot is outside the original circumcircle.
We have produced our first (simple) Delaunay Triangulation.
## Now to the art...
The following image (coding by Antoinette Janus) is based on Delaunay Triangulation of a plane.
You can change:
1. The number of points to be joined by triangles; and
2. The color scheme
To save an image as a background for your computer or phone, right click and choose "Save Image".
Enjoy!
(See below for the javascript code involved in producing this canvas.)
## The code
// First, the author gives credits to his sources:
// Thanks to the following:
// https://github.com/ironwallaby/delaunay.git
// Set up array variables
let color_group = [], fill_color = [], number_of_triangles = 100;
// Variables for image and resizing
var delWidth = document.getElementById( 'del' ).clientWidth;
var viewPortWidth = window.innerWidth;
var viewPortHeight = window.innerHeight;
var delHeight = Math.min(viewPortHeight-20, Math.min(600, Math.max(300, viewPortHeight-150)));
var viewPortWidthRem = viewPortWidth;
var viewPortHeightRem = viewPortHeight;
// Color gradients used for the triangles
{
name: 'mimosa',
colors: [
['#D74177', '#FFE98A'],
['#3A1C71', '#D76D77'],
['#e96443', '#904e95'],
['#ff7e5f', '#feb47b']
]},
{
name: 'intmath',
colors: [
['#00A8C5', '#043850'],
['#0cebeb', '#165a71'],
['#007991', '#78ffd6'],
['#1CD8D2', '#93EDC7']
]},
{
name: 'great-good',
colors: [
['#D4145A', '#FBB03B'],
['#C04848', '#480048'],
['#cc2b5e','#753a88'],
['#C33764', '#1D2671']
]},
{
name: 'metal',
colors: [
['#E8CBC0', '#636FA4'],
['#DBE6F6', '#C5796D'],
['#F3904F', '#3B4371'],
['#DAE2F8', '#D6A4A4']
]},
{
name: 'warm-military',
colors: [
['#FFA17F', '#00223E'],
['#556270', '#FF6B6B'],
['#F0C27B', '#1F1C2C'],
['#ff7e5f', '#00223E']
]}
];
// Get a random number between 2 given numbers
function randNum(min, max){
return Math.floor(Math.random() * (max-min + 1)) + min;
}
// Set the dimensions of the image to suit
const resizeCanvas = function(){
const canvas = document.querySelector('canvas');
delWidth = document.getElementById( 'del' ).clientWidth;
//delHeight = document.getElementById( 'del' ).clientHeight;
canvas.width = delWidth;
canvas.height = delHeight;
}
// Construct the random points to be joined
function pointsArray(){
let points = [];
delWidth = document.getElementById( 'del' ).clientWidth;
delHeight = document.getElementById( 'del' ).clientHeight;
while(points.length < number_of_triangles){
points.push([randNum(-50, delWidth), randNum(-50, delHeight)])
}
points.push([0,0], [delWidth, 0], [0, delHeight], [delWidth, delHeight])
return points
}
// Choose the color scheme
const chooseColor = function(value){
return obj.name === value
})
var colors;
if (color_choice) {
colors = color_choice.colors
} else {
}
return colors;
}
// Joins the points with triangles
// Delaunay is defined in an external js file
const convertToTriangles = function(value){
color_group = chooseColor(value);
const ctx = document.querySelector( 'canvas' ).getContext('2d');
ctx.fillStyle = color_group[randNum(0, color_group.length-1)][randNum(0,1)];
ctx.fillRect(0, 0, delWidth, delHeight);
var vertices = pointsArray();
var triangles = Delaunay.triangulate(vertices);
for(i = triangles.length; i; ) {
var x_arr = [], y_arr = [];
ctx.beginPath();
--i;
x_arr.push(vertices[triangles[i]][0]);
y_arr.push(vertices[triangles[i]][1]);
ctx.moveTo(vertices[triangles[i]][0], vertices[triangles[i]][1]);
--i;
x_arr.push(vertices[triangles[i]][0]);
y_arr.push(vertices[triangles[i]][1]);
ctx.lineTo(vertices[triangles[i]][0], vertices[triangles[i]][1]);
--i;
x_arr.push(vertices[triangles[i]][0]);
y_arr.push(vertices[triangles[i]][1]);
ctx.lineTo(vertices[triangles[i]][0], vertices[triangles[i]][1]);
ctx.closePath();
var start_x_point = x_arr[randNum(0, x_arr.length - 1)];
x_arr.splice( x_arr.indexOf(start_x_point), 1);
var end_x_point = x_arr[randNum(0, x_arr.length - 1)];
var start_y_point = y_arr[randNum(0, y_arr.length - 1)];
y_arr.splice(y_arr.indexOf( start_y_point ), 1);
var end_y_point = y_arr[randNum(0, y_arr.length - 1)];
var start_point = [start_x_point, start_y_point];
var end_point = [end_x_point, end_y_point];
var points = [start_point, end_point];
fill_color = color_group[randNum(0, color_group.length-1)];
ctx.fill();
ctx.stroke();
}
}
// Adds styles to the surrounding siv
del.style.setProperty(('--start-color-'+index), color1);
del.style.setProperty(('--end-color-'+index), color2);
}
// Sets the gardient colors
var gradient = function (ctx, points, color){
return grd;
}
// Draws the backgrounds
function drawBackground(value){
convertToTriangles(value);
for (var i = 0; links.length > i; i++){
var index = 0;
index++;
let hover_color = color_group[randNum(0, color_group.length-1)];
while(hover_color[0] === fill_color[0]){
hover_color = color_group[randNum(0, color_group.length-1)];
}
});
var index = 0;
index++;
})
}
}
// This fires off the whole thing
resizeCanvas();
drawBackground( document.querySelector( '#color-schemes :checked' ));
var inputs = document.querySelectorAll('input');
for(var i = 0; inputs.length > i; i++){
number_of_triangles = document.getElementById( 'triangle_count' ).value;
drawBackground( document.querySelector( '#color-schemes :checked' ).id );
});
}
}
// Redraw everything when resized (e.g. rotate phone)
var resizeTimeout;
function resizeThrottler() {
if (!resizeTimeout) {
resizeTimeout = setTimeout(function() {
resizeTimeout = null;
if(typeof(actualResizeHandler) == 'function') {
actualResizeHandler();
}
}, 200);
}
}
function actualResizeHandler() {
viewPortWidth = window.innerWidth;
viewPortHeight = window.innerHeight;
if( viewPortWidth != viewPortWidthRem || Math.abs(viewPortHeightRem - viewPortHeight) > 50) {
resizeCanvas();
drawBackground( document.querySelector('#color-schemes :checked') );
viewPortWidthRem = viewPortWidth;
viewPortHeightRem = viewPortHeight;
}
}
## "Real world" example
St. George Bank, an Australian company, makes use of Delaunay triangulation on the St. George homepage, as follows.
## Credits
The script is by Antoinette Janus, source: CodePen.
From the original script:
This is a generator using Delaunay's Triangulation concept in the form of IronWallaby's library
|
# How do you find the equation of the line that goes through (- 4,3) and ( 5,- 2)?
May 30, 2018
See a solution process below:
#### Explanation:
First, we need to determine the slope of the line. The formula for find the slope of a line is:
$m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ and $\left(\textcolor{red}{{x}_{2}} , \textcolor{red}{{y}_{2}}\right)$ are two points on the line.
Substituting the values from the points in the problem gives:
$m = \frac{\textcolor{red}{- 2} - \textcolor{b l u e}{3}}{\textcolor{red}{5} - \textcolor{b l u e}{- 4}} = \frac{\textcolor{red}{- 2} - \textcolor{b l u e}{3}}{\textcolor{red}{5} + \textcolor{b l u e}{4}} = - \frac{5}{9}$
Now, we can use the point-slope formula to write an equation for the line. The point-slope form of a linear equation is:
$\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$
Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.
Substituting the slope we calculated above and the values from the first point in the problem gives:
$\left(y - \textcolor{b l u e}{3}\right) = \textcolor{red}{- \frac{5}{9}} \left(x - \textcolor{b l u e}{- 4}\right)$
$\left(y - \textcolor{b l u e}{3}\right) = \textcolor{red}{- \frac{5}{9}} \left(x + \textcolor{b l u e}{4}\right)$
We can also substitute the slope we calculated above and the values from the second point in the problem giving:
$\left(y - \textcolor{b l u e}{- 2}\right) = \textcolor{red}{- \frac{5}{9}} \left(x - \textcolor{b l u e}{5}\right)$
$\left(y + \textcolor{b l u e}{2}\right) = \textcolor{red}{- \frac{5}{9}} \left(x - \textcolor{b l u e}{5}\right)$
|
# 1.2: Definitions of Statistics, Probability, and Key Terms
The science of statistics deals with the collection, analysis, interpretation, and presentation of data. We see and use data in our everyday lives.
In this course, you will learn how to organize and summarize data. Organizing and summarizing data is called descriptive statistics. Two ways to summarize data are by graphing and by using numbers (for example, finding an average). After you have studied probability and probability distributions, you will use formal methods for drawing conclusions from "good" data. The formal methods are called inferential statistics. Statistical inference uses probability to determine how confident we can be that our conclusions are correct.
Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful examination of the data. You will encounter what will seem to be too many mathematical formulas for interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to gain an understanding of your data. The calculations can be done using a calculator or a computer. The understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more confident in the decisions you make in life.
## Probability
Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of an event occurring. For example, if you toss a fair coin four times, the outcomes may not be two heads and two tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails. The expected theoretical probability of heads in any one toss is $$\frac{1}{2}$$ or 0.5. Even though the outcomes of a few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012 heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction $$\frac{996}{2000}$$ is equal to 0.498 which is very close to 0.5, the expected probability.
The theory of probability began with the study of games of chance such as poker. Predictions take the form of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of return on a client's investments. You might use probability to decide to buy a lottery ticket or not. In your study of statistics, you will use the power of mathematics through probability calculations to analyze and interpret your data.
## Key Terms
In statistics, we generally want to study a population. You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample. The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population.
Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16 ounce can contains 16 ounces of carbonated drink.
From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter, in this case the mean. A parameter is a numerical characteristic of the whole population that can be estimated by a statistic. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter.
One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample. We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter.
A variable, or random variable, usually notated by capital letters such as $$X$$ and $$Y$$, is a characteristic or measurement that can be determined for each member of a population. Variables may be numerical or categorical. Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category. If we let $$X$$ equal the number of points earned by one math student at the end of a term, then $$X$$ is a numerical variable. If we let $$Y$$ be a person's party affiliation, then some examples of $$Y$$ include Republican, Democrat, and Independent. $$Y$$ is a categorical variable. We could do some math with values of $$X$$ (calculate the average number of points earned, for example), but it makes no sense to do math with values of $$Y$$ (calculating an average party affiliation makes no sense).
Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value.
Two words that come up often in statistics are mean and proportion. If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is $$\frac{22}{40}$$ and the proportion of women students is $$\frac{18}{40}$$. Mean and proportion are discussed in more detail in later chapters.
NOTE
The words "mean" and "average" are often used interchangeably. The substitution of one word for the other is common practice. The technical term is "arithmetic mean," and "average" is technically a center location. However, in practice among non-statisticians, "average" is commonly accepted for "arithmetic mean."
EXAMPLE 1.1
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly surveyed 100 first year students at the college. Three of those students spent $150,$200, and $225, respectively. Answer Solution 1.1 The population is all first year students attending ABC College this term. The sample could be all students enrolled in one section of a beginning statistics course at ABC College (although this sample may not represent the entire population). The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term: the population mean. The statistic is the average (mean) amount of money spent (excluding books) by first year college students in the sample. The variable could be the amount of money spent (excluding books) by one first year student. Let $$X$$ = the amount of money spent (excluding books) by one first year student attending ABC College. The data are the dollar amounts spent by the first year students. Examples of the data are$150, $200, and$225.
EXERCISE 1.1
Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65,$75, and \$95, respectively.
EXAMPLE 1.2
Determine what the key terms refer to in the following study.
A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below.
1. Population ____ 2. Statistic ____ 3. Parameter ____ 4. Sample ____ 5. Variable ____ 6. Data ____
1. all students who attended the college last year
2. the cumulative GPA of one student who graduated from the college last year
3. 3.65, 2.80, 1.50, 3.90
4. a group of students who graduated from the college last year, randomly selected
5. the average cumulative GPA of students who graduated from the college last year
6. all students who graduated from the college last year
7. the average cumulative GPA of students in the study who graduated from the college last year
Solution 1.2
1. f; 2. g; 3. e; 4. d; 5. b; 6. c
EXAMPLE 1.3
Determine what the key terms refer to in the following study.
As part of a study designed to test the safety of automobiles, the National Transportation Safety Board collected and reviewed data about the effects of an automobile crash on test dummies. Here is the criterion they used:
Speed at which cars crashed Location of “drive” (i.e. dummies) 35 miles/hour Front Seat
Table 1.1
Cars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars.
Solution 1.3
The population is all cars containing dummies in the front seat.
The sample is the 75 cars, selected by a simple random sample.
The parameter is the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population.
The statistic is proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample.
The variable $$X$$ = the number of driver dummies (if they had been real people) who would have suffered head injuries.
The data are either: yes, had head injury, or no, did not.
EXAMPLE 1.4
Determine what the key terms refer to in the following study.
An insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit.
Solution 1.4
The population is all medical doctors listed in the professional directory.
The parameter is the proportion of medical doctors who have been involved in one or more malpractice suits in the population.
The sample is the 500 doctors selected at random from the professional directory.
The statistic is the proportion of medical doctors who have been involved in one or more malpractice suits in the sample.
The variable $$X$$ = the number of medical doctors who have been involved in one or more malpractice suits.
The data are either: yes, was involved in one or more malpractice lawsuits, or no, was not.
|
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Angle Bisector
Angle bisector in geometry refers to a line that splits an angle into two equal angles. Bisector means the thing that bisects a shape or an object into two equal parts. If we draw a ray that bisects an angle into two equal parts of the same measure, then it is called an angle bisector.
Before talking about an angle bisector, let us quickly recall the different types of angles in mathematics. Depending on the inclination between the two arms, an angle may be acute (less than 90-degrees, like 60-degree angle), obtuse (more than 90-degrees) or right angle (exactly 90-degrees). Constructing angles is an important part of geometry as this knowledge is extended for the construction of other geometric figures as well, primarily the triangles. A number of angles can be constructed simply by bisecting some common angles.
## What is an Angle Bisector?
An angle bisector or the bisector of an angle is a ray that divides an angle into two equal parts. For example, if a ray AX divides an angle of 60 degrees into two equal parts, then each measure will be equal to 30 degrees.
Every angle has an angle bisector. It is also the line of symmetry between the two arms of an angle, the construction of which enables you to construct smaller angles. Say you are required to construct a 30° angle. This can be performed by creating a 60° angle and then bisecting it. Similarly, 90-degree, 45-degree, 15-degree and other angles are constructed using the concept of an angle bisector.
## Properties of Angle Bisector
• All the points of angle bisector are equidistant from both the arms of the angle
• An angle bisector can be drawn to any angle, such as acute, obtuse, or right angle
• The angle bisector in a triangle divides the opposite side in a ratio that is equal to the ratio of the other two sides
## How to Construct an Angle Bisector?
You require a ruler and a compass to construct angles and their bisectors. Given a known or unknown ∠PQR, the steps to construct its angle bisector are:
Step 1: Place the compass pointer at Q and make an arc that cuts the two arms of the angle at two different points.
Step 2: From the point where the first arc cut the arm QP, make another arc towards the interior of the angle.
Step 3: Without changing the radius on the compass, repeat step 2 from the point where the first arc cut QR.
Step 4: Using a ruler, draw a line from Q to the point where the arcs intersect.
The line that was drawn through Q represents the angle bisector of the ∠PQR.
Note: If an angle bisector bisects a line segment at 90°, it is known as the perpendicular bisector of that line.
## Angle Bisector Theorem
According to the angle bisector theorem, “an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle”.
## Solved Examples on Angle Bisector
Q.1: If an angle bisector divides an angle of 80 degrees, then what is the measure of each angle?
Solution: Given, a measure of an angle is 80 degrees
As we know, the angle bisector divides the angle into equal two parts.
Therefore, 80 degrees is divided into equal two parts, say x.
Hence,
x + x = 80°
2x = 80°
x = 80°/2
x = 40°
Q.2: A ray BX, divides an angle ABC into two equal parts. If one part is equal to 4x – 5 and the second part is equal to 20, then what is the value of x?
Solution: Given, BX divides angle ABC into two equal parts. Thus, BX is the angle bisector.
Now, each part should measure equal.
Thus,
4x – 8 = 20
4x = 20 + 8 = 28
x = 28/4 = 7
Hence, the value of x is 7.
## Practice Questions on Angle Bisector
1. If an angle bisector divides 90 degrees, then what is the measure of each angle?
2. Construct the angle bisector of 110 degrees angle.
3. Construct an angle bisector of 30 degrees angle.
## Frequently Asked Questions on Angle bisector
### What do you mean by angle bisector?
An angle bisector is a ray that divides an angle into two parts of equal measure.
### What are the properties of angle bisectors?
The properties of angle bisector are:
All the points of angle bisector are equidistant from both the arms of the angle
An angle bisector can be drawn to any angle, such as acute, obtuse, or right angle
The angle bisector in a triangle divides the opposite side in a ratio that is equal to the ratio of the other two sides
### Does the angle bisector of a triangle divide the sides?
The angle bisector of a triangle divides the opposite sides in a ratio proportional to the other two sides of the triangle.
### How many angle bisectors does an angle have?
There is only an angle bisector to an angle.
|
Math1011_Week04_Prelim1A
# Math1011_Week04_Prelim1A - Math1011 Learning Strategies...
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Math1011 - Learning Strategies Center Section 1.6 IA-1 The function f(x) is graphed below: -4 -3 -2 -1 0 1 2 3 4 0 1 2 3 4 What is the domain and range of the f(x)? Is f(x) one-to-one? Why? Sketch f -1 (x) in the graph above. What is the domain and range of f -1 (x)?
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Math1011 - Learning Strategies Center Section 1.6 A1 The function f(x) is graphed below: -4 -3 -2 -1 0 1 2 3 4 - 4 - 3 - 2 - 101234 What is the domain and range of the f(x)? The domain of is ( 1, ) The range of is ( ,1) f −∞ −∞ Is f(x) one-to-one? Why? The function is one-to-one as its graph intersects each horizontal line at most once (Horizontal Line Test). Sketch f -1 (x) in the graph above. What is the domain and range of f -1 (x)? 1 1 The domain of is ( ,1). The range of is ( 1, ). −∞ (Doc #011w.16.06t)
Math1011 - Learning Strategies Center Section 1.6 A2 -1 -1 Show that the function f is one-to-one, and calculate the inverse function f . Specify the domain and range of f and f . 1 () x fx + =
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Math1011 - Learning Strategies Center Section 1.6 A2 -1 -1 Show that the function f is one-to-one, and calculate the inverse function f . Specify the domain and range of f and f . 1 () x fx + = 12 1 2 11 21 1 2 1 To show one-to-one, show if f(x ) ( ) x f(x ) ( ) (cross multiply) x (1 ) x (1 ) (distribute) x x x x (subtract x from both sides) x i s o xx ++ + = ⇒= = = += + + = = 1 2 -1 1 1 1 ne-to-one as f(x ) ( ) x Calculate the inverse function f ( ). (solve for x in terms of y) (1 ) (cross multiply) ) y yx + =⇒ = = =− = = -1 1 1 interchange x and y Specify the domains and ranges of f and f . ( ) ( ,1 ) (1 , ) ( ) ( , 1 ) ( 1 Df f fD =ℜ = −∞− ∪ − ∞ ℜ= = (Doc #011w.16.07t)(Adams 3.1.10)
Math1011 Fall 2007 Prelim A3 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 ac b bca = Solve the equation for x: 4 2log 1 10 ln 4 log 100 x e +=
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Math1011 Fall 2007 Prelim A3 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 ac b bca = log log log log (log )(log )(log ) 1 (log )( )( ) 1 (log aa ca a bc = = log )( c log log log )( )1 log 1 1 1
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Home>Mathematics>Shocking Truth Revealed: The Surprising Winner Between 1/2 And 3/4!
Mathematics
# Shocking Truth Revealed: The Surprising Winner Between 1/2 And 3/4!
Written by: Francine Messina
Discover the surprising winner between 1/2 and 3/4 in this shocking truth revealed! Uncover the mathematics behind the unexpected result.
## Introduction
Fractions are a fundamental concept in mathematics, representing parts of a whole. Understanding fractions is crucial for various real-life applications, from cooking and baking to measurements and financial calculations. In this article, we delve into the intriguing world of fractions to uncover the surprising winner between two commonly compared fractions: 1/2 and 3/4.
Fractions can be a source of confusion for many, but they need not be intimidating. They are simply a way of expressing a part of a whole, where the numerator represents the number of parts being considered, and the denominator indicates the total number of equal parts that make up the whole. While fractions are commonly encountered in everyday life, their comparison and manipulation can sometimes present unexpected outcomes.
In this exploration, we will unravel the mystery behind the comparison of 1/2 and 3/4. These two fractions, seemingly close in value, may hold a surprising revelation when examined more closely. By understanding the underlying principles of fractions and their comparison, we can uncover the unexpected winner in this intriguing mathematical showdown.
Let's embark on this fascinating journey through the realm of fractions and unveil the unexpected victor in the showdown between 1/2 and 3/4. Get ready to be amazed by the surprising outcome that awaits us!
## Understanding Fractions
Fractions are a fundamental concept in mathematics, serving as a crucial tool for expressing parts of a whole. When we encounter fractions, we are essentially dealing with the division of a whole into equal parts. The numerator of a fraction represents the number of parts we are considering, while the denominator indicates the total number of equal parts that make up the whole.
For instance, when we encounter the fraction 1/2, we are essentially dividing a whole into two equal parts and considering one of those parts. This can be visualized as dividing a pizza into two equal slices and taking one of those slices. Similarly, when dealing with the fraction 3/4, we are dividing a whole into four equal parts and considering three of those parts. This can be likened to dividing a pie into four equal slices and taking three of those slices.
Understanding fractions is not only about manipulating numbers; it also involves grasping the concept of proportion and relative size. When we compare fractions, we are essentially evaluating the relative sizes of the parts being considered within their respective wholes. This comparison is fundamental in various real-life scenarios, such as determining ingredient proportions in recipes, calculating discounts and percentages in shopping, and interpreting data in graphical representations.
Furthermore, fractions are not limited to unit fractions (where the numerator is 1); they can also represent quantities greater than one. For example, the fraction 5/3 represents one whole and two-thirds of another whole. This flexibility in representing quantities makes fractions a versatile tool for expressing and comparing quantities of varying magnitudes.
In essence, understanding fractions goes beyond mere arithmetic; it involves developing a conceptual grasp of part-whole relationships and relative sizes. This foundational knowledge forms the basis for more advanced mathematical concepts, making it essential for students and enthusiasts alike to master the intricacies of fractions.
As we delve deeper into the comparison of 1/2 and 3/4, this understanding of fractions will serve as the cornerstone for unraveling the surprising winner in this mathematical face-off. Let's continue our exploration and unlock the unexpected revelations that await us in the realm of fractions.
## Comparing 1/2 and 3/4
When it comes to comparing fractions, the intuitive approach often involves looking solely at the numerators and denominators. In the case of 1/2 and 3/4, this simplistic approach might lead one to believe that 3/4 is greater than 1/2 due to the larger numerator. However, the comparison of fractions involves delving deeper into their relative sizes and understanding their placement on the number line.
To effectively compare 1/2 and 3/4, it's essential to consider the concept of common denominators. When fractions have different denominators, as is the case here, finding a common ground for comparison becomes crucial. In this scenario, the common denominator can be obtained by finding the least common multiple of 2 and 4, which is 4. By expressing both fractions with the common denominator, their relative positions become clearer.
Converting 1/2 to an equivalent fraction with a denominator of 4 results in 2/4. Similarly, 3/4 remains unchanged. Now, with both fractions expressed with a common denominator, their comparison becomes straightforward. On the number line, 2/4 (equivalent to 1/2) and 3/4 can be visually represented, showcasing their relative positions.
Surprisingly, this visual representation reveals that 2/4 (1/2) is actually less than 3/4, contrary to the initial impression based solely on the numerators. This comparison highlights the significance of considering the denominators and their role in determining the size of fractional parts. While 3/4 represents a larger portion of the whole when compared to 1/2, this understanding goes beyond mere arithmetic and emphasizes the conceptual grasp of relative sizes inherent in fractions.
Moreover, the comparison of 1/2 and 3/4 sheds light on the significance of context in interpreting fractions. Depending on the scenario, both 1/2 and 3/4 can hold different implications. For instance, when dividing a cake into four equal parts, 3/4 represents a larger portion compared to 1/2, which signifies only two parts of the whole. This contextual understanding underscores the dynamic nature of fractions in real-world applications.
In essence, the comparison of 1/2 and 3/4 transcends the superficial evaluation based solely on numerators and delves into the fundamental principles of relative sizes and common denominators. This exploration sets the stage for the revelation of the surprising winner in the showdown between these seemingly close fractions. Let's unravel the unexpected victor and delve deeper into the realm of fractions to uncover the captivating insights that await us.
## The Surprising Winner
In the captivating showdown between 1/2 and 3/4, the unassuming victor emerges from the realm of fractions, defying initial expectations and revealing a surprising outcome. While at first glance, the comparison of these two fractions might seem straightforward, a deeper exploration uncovers an unexpected revelation.
Upon closer examination, the surprising winner in this mathematical face-off is none other than 3/4. Despite the seemingly close values of 1/2 and 3/4, the pivotal role of the denominator becomes evident in determining their relative sizes. When expressed with a common denominator, the true positioning of these fractions on the number line becomes apparent, showcasing the larger magnitude of 3/4 compared to 1/2.
The revelation of 3/4 as the surprising winner underscores the intricate nature of fractions and their representation of relative sizes within a whole. This outcome challenges preconceived notions and highlights the importance of a comprehensive understanding of fractions beyond mere numerical values.
Furthermore, the surprising winner in this comparison emphasizes the dynamic nature of fractions in various real-world contexts. Whether it's dividing a cake, sharing a pizza, or interpreting graphical data, the significance of relative sizes represented by fractions becomes palpable. The victory of 3/4 over 1/2 serves as a testament to the nuanced interpretation of fractions and their implications in practical scenarios.
Moreover, the unexpected winner sheds light on the intricacies of mathematical comparisons, urging enthusiasts to delve deeper into the conceptual foundations of fractions. By unraveling the surprising outcome in the showdown between 1/2 and 3/4, a profound appreciation for the complexities of fractions and their relative magnitudes is cultivated, paving the way for a deeper understanding of mathematical concepts.
In essence, the surprising winner in the comparison of 1/2 and 3/4 transcends numerical values, delving into the fundamental principles of relative sizes, common denominators, and contextual interpretations. This revelation not only enriches our understanding of fractions but also ignites a sense of wonder and curiosity in the captivating realm of mathematics.
## Conclusion
The exploration of the comparison between 1/2 and 3/4 has unveiled the captivating intricacies of fractions and their representation of relative sizes within a whole. What initially appeared as a straightforward evaluation based solely on numerical values has transcended into a profound understanding of the dynamic nature of fractions and their implications in real-world scenarios.
In this journey through the realm of fractions, we have delved into the foundational concepts of part-whole relationships, common denominators, and the nuanced interpretation of relative sizes. The surprising revelation of 3/4 emerging as the victor in the showdown against 1/2 has challenged conventional perceptions and emphasized the pivotal role of denominators in determining the magnitude of fractional parts.
Beyond the numerical values, the comparison of 1/2 and 3/4 has underscored the contextual significance of fractions in practical applications. Whether it's dividing a cake, distributing resources, or interpreting data in graphical representations, the understanding of relative sizes represented by fractions holds immense relevance. This contextual interpretation enriches the appreciation for the versatility and applicability of fractions in diverse real-life scenarios.
Furthermore, the exploration of fractions and their comparison serves as a testament to the dynamic nature of mathematical concepts. It encourages enthusiasts to delve beyond surface-level evaluations and embrace the conceptual foundations underlying numerical values. This deeper understanding fosters a sense of curiosity and wonder, igniting a passion for exploring the captivating realm of mathematics.
As we conclude this journey through the surprising winner between 1/2 and 3/4, the profound insights gained from this exploration extend far beyond the confines of fractions. They exemplify the beauty of mathematical concepts, their real-world relevance, and the endless opportunities for discovery and learning within the realm of mathematics.
In essence, the comparison of 1/2 and 3/4 has not only unraveled the unexpected victor but has also illuminated the captivating intricacies of fractions, inspiring a deeper appreciation for the dynamic and multifaceted nature of mathematical concepts. This revelation serves as a testament to the endless wonders awaiting exploration within the captivating realm of mathematics.
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# Identify The Equation For Properties Of Addition And Multiplication Worksheet
Assignment will be available soon
Identifying the equation for properties of addition and multiplication refers to recognizing and using specific mathematical equations that illustrate the fundamental rules governing addition and multiplication. This includes equations like the commutative, associative, identity, inverse, and distributive properties, which are essential in understanding how numbers interact and behave in mathematical operations. Recognizing these equations helps in solving problems systematically and efficiently in various mathematical contexts.
Algebra 1
Expressions
## How Will This Worksheet on “Identify the Equation for Properties of Addition and Multiplication” Benefit Your Student's Learning?
• Clarifies fundamental rules of addition and multiplication.
• Enhances application skills in arithmetic operations.
• Improves problem-solving proficiency.
• Reinforces understanding through practice.
• Prepares for advanced mathematical concepts.
• Develops critical thinking by applying properties to diverse scenarios.
## How to Identify the Equation for Properties of Addition and Multiplication?
1. Commutative Property:
• Addition: $$a + b = b + a$$
• Multiplication: $$a \times b = b \times a$$
2. Associative Property:
• Addition: $$(a + b) + c = a + (b + c)$$
• Multiplication: $$(a \times b) \times c = a \times (b \times c)$$
3. Identity Property:
• Addition: $$a + 0 = a$$
• Multiplication: $$a \times 1 = a$$
4. Inverse Property:
• Addition: $$a + (-a) = 0$$
• Multiplication: $$a \times \frac{1}{a} = 1$$ (for $$a \neq 0$$)
• Multiplication over addition: $$a \times (b + c) = a \times b + a \times c$$
Recognize the properties by analyzing the given expression.
## Solved Example
Q. Which equation shows the commutative property of addition?$\newline$Choices:$\newline$(A) $g + h = h + g$$\newline$(B) $j = g + h$$\newline$(C) $h + j + k = g$$\newline$(D) $0 + g = g$
Solution:
1. Understand Commutative Property: Understand the commutative property of addition. The commutative property of addition states that changing the order of the addends does not change the sum. In other words, if you have two numbers $a$ and $b$, then $a + b = b + a$.
2. Analyze Given Choices: Analyze the given choices to identify which one represents the commutative property.$\newline$(A) $g + h = h + g$$\newline$(B) $j = g + h$$\newline$(C) $h + j + k = g$$\newline$(D) $0 + g = g$$\newline$We are looking for an equation where two addends are switched in place but still equal each other.
3. Check Choices Against Definition: Check each choice against the definition of the commutative property. $\newline$(A) This choice shows two addends, $g$ and $h$, being added in both orders, which fits the definition of the commutative property. $\newline$(B) This choice does not show two addends being switched; it is simply an equation. $\newline$(C) This choice does not show the commutative property; it is an equation with three addends and does not demonstrate the switching of any two addends. $\newline$(D) This choice shows the identity property of addition, where adding zero to a number gives the number itself, not the commutative property.
4. Select Correct Choice: Select the correct choice that demonstrates the commutative property of addition. From the analysis, we can see that choice (A) $g + h = h + g$ correctly shows the commutative property of addition.
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# Are there more fractions or decimals?
Thanks to Anonymous 9th Grade Student for the question. This is actually a surprisingly deep question. When students are in elementary school, they are taught how to convert between the decimal and fractional representations of a number. 0.4 = 2/5. 0.33333333333…=1/3. 0.237 = 237/1000. So we might initially think that every decimal can be written as a quotient of integers and vise versa. (Integers are whole numbers and negative whole numbers: 4, -3, 25745, -342 and NOT 3.2) This would lead us to believe that there are as many fractions as decimals. We might also recognize that there are an infinite number of fractions and an infinite number of decimals. From there might assume there are the same number of each: infinity.
When students enter middle school, they first encounter irrational numbers. Unlike rational numbers, irrational numbers are, by definition, numbers which cannot be written as a fraction with an integer numerator and denominator. Examples of irrational numbers are π, the square root of 2, and the mathematical constant e. You could write π as fraction π/1 but there is no way to represent it without having a decimal in the numerator or denominator.
All real numbers can be written in decimal form, but some real numbers(irrationals) cannot be written as the fraction of two integers. So, the answer to this question depends on how you define “fraction.” If you define a “fraction” as any number divided by any number, then you could write any irrational number as a fraction by dividing it by one (π/1) and you would have the same number of fractions as decimals. If we define a “fraction” as an integer divided by an integer, there are more decimals than fractions because irrational numbers cannot be written as fractions.
Let’s assume we mean fractions to mean rational numbers. Now let’s go back to the argument that there are an infinite numbers of fractions and decimals so there are the same number(infinity) in each set. Not all “infinities” are the same size though. Using advanced mathematics, there are ways to determine whether one “infinity” is larger than another. Although there is an infinite number in each set, mathematicians would say there is a countable number of fractions (rational numbers) and an uncountable number of decimals(real numbers). Thus, there are more real numbers(decimals) than rational numbers(fractions).
What do “countable” and “uncountable” mean? Suppose we want to know if there are more boys than girls in a room. We could count the boys and girls separately and compare the numbers. We could also have every girl stand next to boy in a one to one correspondence and see if we have extra boys. For comparing mathematical sets with an infinite number of elements, we can follow a similar procedure. If a function exists that could pair every element in a given infinite set in a one to one correspondence with the counting numbers (1,2,3, etc.) then that given set is also countable. (All boys and girls in the room are matched up.) If a one to one correspondence with the counting numbers cannot not be found, then the given set is uncountable. (There are extra boys, so the sets are not the same size.) For a proof that the real numbers are uncountable, look up “Cantor’s Diagonal Argument”.
Another way to look at this problem is to consider long division. When you divide one whole number by another using the long division algorithm the number of possible remainders is the number of whole numbers less than the denominator. When you divide 323 by 7, the only possible remainders after each step are 6, 5, 4, 3, 2, 1 or 0. So you will either eventually get a remainder of zero and the decimal expansion ends or you get the same remainder twice and the decimal expansion starts repeating. (An example of a repeating decimal would be 0.123456565656565656565656565656… where the “56”s will go on forever.) So all quotients of whole numbers (or integers) eventually terminate or repeat. I could however make up a decimal that does not terminate or repeat.For example, 0.010110111011110111110….. This decimal has 1 one, zero, 2 ones, zero, 3 ones, zero, 4 ones, zero, etc. This decimal does not terminate or start repeating itself so it cannot be a rational. Therefore there are more decimals than rational numbers because 0.010110111… is a decimal and not a rational.
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In statistics, there are mainly three types of measures of central tendency: mean, median and mode. Also, dispersion is an important concept in statistics which means to study the spread of the data. There are various measures of dispersion, median absolute deviation is one of the important topics.
Median is the value that lies in the center. Median standard deviation is defined as absolute value of deviation of each observation from the median of the data. It is the median of absolute deviation from median.
The value of median absolute deviation refers to the dispersion of data from its median.
In order to determine median absolute deviation, one needs to first calculate the median of data. Then, it is subtracted from each observation of the data and their absolute values are calculated.
On dividing the sum of these values by the total number of observations, we get the median absolute deviation of data.
## Median Absolute Deviation Formula
The formula for median absolute deviation is given below:
Formula for median absolute deviation
$MAD$= $\frac{\sum^{n}_{i=1}(x_{i}-M)}{n}$
Where,
$n$ = Total number of observations
$M$ = Median of the data
$x_{i}$ = An observation among the data
We can also write the simplified form of above formula as:
$MAD$ = $\frac{(x_{1}-M)+(x_{2}-M)+(x_{3}-M)+...+(x_{n}-M)}{n}$
Where,
$MAD$ stands for median absolute deviation.
and $x_{1}$, $x_{2}$, $x_{3}$, ..., $x_{n}$ are the individual observations of the given data.
Let us recall that the median of a data is calculated by either of the following given formulae:
Case 1: When total number of observations $(n)$ is odd.
$M$ = $(\frac{n+1}{2})^{th}\ term$
Case 2: When total number of observations $(n)$ is even
.
$M$ = $\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}term}{2}$
## How to Calculate Median Absolute Deviation
We are supposed to follow the steps illustrated below in order to calculate the median absolute deviation of a data set.
Step 1: Arrange all numbers of a data set in ascending order.
Step 2: Find the total number of items given in the data and denote it by $n$.
Step 3: Check whether n is even or odd. Calculate the median of the data by using the appropriate formula.
If n is even, then use the formula:
$M$ = $\frac{\frac{n}{2}^{th}\ term+(\frac{n}{2}+1)^{th}term}{2}$.
If n is odd, then apply the formula:
$M$ = $(\frac{n+1}{2})^{th}\ term$,
in order to estimate median $M$.
Step 4: Construct a table for determining deviation of each number from median. In order to do so, in a column, subtract median value from each number ignoring negative sign. Sum up these absolute deviations.
Step 5: Divide the sum of these deviations by total number of observations. The result will be the required answer.
## Median Absolute Deviation Example
The examples related to median absolute deviation are given below:
Example 1: Calculate the median standard deviation for the following data:
8, 2, 6, 4, 9, 8, 5, 9, 5
Solution: Given data in ascending order is:
2, 4, 5, 5, 6, 8, 8, 9, 9
Total number of observations, $n$ = 9 which is odd
Therefore, following formula for median would be used.
$M$ = $(\frac{n+1}{2})^{th}\ term$
$M$ = $(\frac{9+1}{2})^{th}\ term$
$M$ = $5^{th}$ term
i.e. 6
Let us construct the following table:
$x_{i}$ $x_{i}$ - M |$x_{i}$ - M| 2 2 - 6 = - 4 4 4 4 - 6 = - 2 2 5 5 - 6 = - 1 1 5 5 - 6 = - 1 1 6 6 - 6 = 0 0 8 8 - 6 = 2 2 8 8 - 6 = 2 2 9 9 - 6 = 3 3 9 9 - 6 = 3 3 $n$ = 9 $\sum^{n}_{i=1}$|$x_{i}$ - M| = 18
The formula for median absolute deviation is :
$MAD$ = $\frac{\sum^{n}_{i=1}(x_{i}-M)}{n}$
$MAD$ = $\frac{18}{9}$
= 2
Example 2: Following are marks obtained by the 8 students in mathematics class test out of 20. Determine the median standard deviation.
17, 18, 15, 10, 6, 17, 12, 5
Solution: Rearranging the above data in ascending order:
5, 6, 10, 12, 15, 17, 17, 18
Total number of observations, n = 8
which is an even number.
Therefore, applying following formula for calculating median:
$M$ = $\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}term}{2}$
$M$ = $\frac{(\frac{8}{2})^{th}\ term+(\frac{8}{2}+1)^{th}term}{2}$
$M$ = $\frac{4^{th}\ term+5^{th}term}{2}$
$M$ = $\frac{12+15}{2}$
$M$ = 13.5
Let us construct the following table:
$x_{i}$ $x_{i}$ - M |$x_{i}$ - M| 5 5 - 13.5 = - 8.5 8.5 6 6 - 13.5 = - 7.5 7.5 10 10 - 13.5 = - 3.5 3.5 12 12 - 13.5 = - 1.5 1.5 15 15 - 13.5 = 1.5 1.5 17 17 - 13.5 = 3.5 3.5 17 17 - 13.5 = 3.5 3.5 18 18 - 13.5 = 4.5 4.5 $n$ = 8 $\sum^{n}_{i=1}$|$x_{i}$ - M| = 34
The formula for median absolute deviation is:
$MAD$ = $\frac{\sum^{n}_{i=1}(x_{i}-M)}{n}$
$MAD$ = $\frac{34}{8}$
= 4.25
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# Factors of 5
Factors of 5 are the real numbers that can divide the original number, uniformly. If ‘x’ is the factor of 5, then ‘x’ divides 5 into equal parts and there is no remainder left. For example, if 4 is the factor of 24, then 24 divided by 4 is equal to 6. Thus, 4 divides 24 into six equal parts and the remainder is 0.
Let us find the factors, pair factors and prime factors of 5 in this article with simple methods.
Factors Pair Factors Prime Factors 1 and 5 (1, 5) 5 → 5
## How to Find Factors of 5?
Factors of 5 are the natural numbers that can divide the actual number, evenly. 5 is the third prime number when arranged in ascending order. Therefore, 5 will have only two factors, 1 and 5. Â
• 5 ÷ 1 = 5    Â
• 5 ÷ 5 = 1
If we divide 5 by any other integer, then the remainder will be a non-zero positive integer.
## Pair Factors of 5
The product of pair factors of 5 will be equal to the actual number. Hence,
1 × 5 = 5
OrÂ
5 × 1 = 5
Therefore, there is only one pair factor, i.e. (1, 5).
Similarly, we can also consider negative pair factors of 5 which will be equal to the original number.
-1 × -5 = 5
Therefore, the negative pair factor is (-1, -5).
## Prime Factorisation of 5
Using the prime factorization method, it is easy to find the prime factors of the original number. As we already know 5 is a prime number, therefore, the prime factorization of 5 is not required.
Check:
We can check if 5 is a prime number or not, by dividing it by other prime numbers. First, let us list down the prime numbers upto 5.
2, 3, 5
Now if we start dividing 5 by 2, then we get;
5/2 = 2.5
5/3 = 1.66
5/5 = 1
Thus, we can see, 5 is only divisible by itself.
Therefore,
Prime factorisation of 5 =Â 5
## Solved Examples
Q.1: Find the sum of all the factors of five.
Solution: The factors of 5 are 1 and 5.
Sum = 1 + 5 = 6
Therefore, 6 is the required sum.
Q.3: What is the greatest common factor of 10 and 5?
Answer: Let us list down the factors of all the numbers.
Factors of 10 = 1, 2, 5, 10
Factors of 5 = 1, 5
GCF of 10 and 5 is 5.
## Frequently Asked Questions on Factors of 5
### What are the two factors of 5?
The two factors of 5 are 1 and 5.
### Is 5 a common factor of 20 and 25?
Yes, 5 is a common factor of 20 and 25 since the factors of 20 include 1, 2, 4, 5, 10, 20, whereas the factors of 25 include 1, 5, 25. Here, 5 is common in the factor list of 20 and 25.
### What are the prime factors of 5?
We know that 5 is a prime number, that means it has only two factors, i.e. 1 and 5. Thus, the only prime factor of 5 is the number itself.
### Is 5 divisible by any other number?
5 is not divisible by any other real number, apart from 1 and the number itself. If we divide 5 by any other number, then the remainder will not be equal to zero.
### What are multiples of 5?
The multiples of 5 include:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, etc.
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# Video: Finding the Equation of a Straight Line Parallel to a Coordinate Axis and Passing through a Given Point
Determine the equation of the line parallel to the ๐ฅ-axis that passes through (โ1/2, 4).
02:30
### Video Transcript
Determine the equation of the line parallel to the ๐ฅ-axis that passes through negative one-half, four.
So, weโll sketch a coordinate plane and label the ๐ฅ- and ๐ฆ-axis. Weโre given the point negative one-half, four. Negative one-half will be halfway between zero and negative one. And then, weโll go up four, so that our point lies here. The line weโre looking for goes through this point and is parallel to the ๐ฅ-axis.
Parallel lines are lines on a plane that never intersect. Theyโre always the same distance apart. This means the line weโre looking for will never cross the ๐ฅ-axis. Since this point is four units away from the ๐ฅ-axis, we know that the line will be four units away from the ๐ฅ-axis at every point. And so, we could have a point at one, four.
And now that we have found two points on our line, we can sketch the line. This is the line weโre looking for. But our job is to find its equation. We know that this is a horizontal line. Therefore, the slope of this line is zero. We often write the equation of a line in the form ๐ฆ equals ๐๐ฅ plus ๐, where ๐ is the slope and ๐ is the ๐ฆ-intercept. If we know that our slope is zero, we have ๐ฆ equals zero ๐ฅ plus ๐, where ๐ is the ๐ฆ-intercept.
The ๐ฆ-intercept is the place where our line crosses the ๐ฆ-axis here at four. And so, the ๐ value of our equation would be four. Zero times ๐ฅ equals zero. That ๐ฅ term drops out and weโre left with ๐ฆ equals four. This is because at every point along the ๐ฅ-axis, ๐ฆ will be equal to four. The equation for this horizontal line is ๐ฆ equals four.
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# How do you solve for x in 2(8-x)=-6?
Jul 18, 2016
$x = 11$
#### Explanation:
To solve for $x$, we must first distribute the $2$ across the parenthesis.
$2 \left(8 - x\right) = - 6$
$16 - 2 x = - 6$
Now isolate the variable by getting it by itself on one side. Subtract $16$ from both sides.
$\cancel{16 - 16} - 2 x = - 6 - 16$
$- 2 x = - 22$
Get rid of the coefficient to identify $x$.
$\frac{\cancel{- 2} x}{\cancel{- 2}} = - \frac{22}{-} 2$
$x = 11$
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How do you solve $\ln \left( 4x-2 \right)-\ln 4=-\ln \left( x-2 \right)$?
Last updated date: 23rd Jun 2024
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Views today: 5.73k
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Hint:To solve the given question, first we apply the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b). Then we simplify the equation further by using the definition of log, if log (a) = log (b) then a = b. and solve the equation in a way we solve the general quadratic equation.
Formula used:
The property of logarithm which states that if logs to the same base are added, then the
numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
If log (a) = log (b) then a = b.
Complete step by step solution:
We have given that,
$\ln \left( 4x-2 \right)-\ln 4=-\ln \left( x-2 \right)$
Rearranging the terms in the above equation, we get
$\Rightarrow \ln \left( 4x-2 \right)+\ln \left( x-2 \right)=\ln 4$
Using the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
Applying the above property, we get
$\Rightarrow \ln \left( \left( 4x-2 \right)\times \left( x-2 \right) \right)=\ln 4$
Using the definition of log, if log (a) = log (b) then a = b.
Applying the above property, we get
$\Rightarrow \left( \left( 4x-2 \right)\times \left( x-2 \right) \right)=\ln 4$
Simplifying the above equation, we get
$\Rightarrow \left( 4x\times x \right)+\left( 4x\times -2 \right)+\left( -2\times x \right)+\left( -2\times -2 \right)=4$
Simplifying further, we get
$\Rightarrow 4{{x}^{2}}-8x-2x+4=4$
$\Rightarrow 4{{x}^{2}}-10x+4=4$
Subtracting 4 from both the sides of the equation, we get
$\Rightarrow 4{{x}^{2}}-10x=0$
Taking out 2x as a common factor, we get
$\Rightarrow 2x\left( 2x-5 \right)=0$
Equation each factor equals to 0, we get
$\Rightarrow 2x=0$ and $2x-5=0$
Now, solving
$\Rightarrow 2x=0$
$\Rightarrow x=0$
Now, solving
$\Rightarrow 2x-5=0$
Adding 5 to both the sides of the equation, we get
$\Rightarrow 2x=5$
Dividing both the sides of the equation by 2, we get
$\Rightarrow x=\dfrac{5}{2}$
Since, x > 2, so the only possible value of x is $\dfrac{5}{2}$.
Therefore, $x=\dfrac{5}{2}$ is the required solution.
Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general quadratic equations.
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# Lesson video
In progress...
Hi everybody, today we are going to be applying, our place value knowledge to different types of questions.
So put on your hats, tighten those ties and tell the computer now I'm a mathematician.
For today's lesson, we're going to be practising , lots of different types of questions.
So you'll need a pencil, a piece of paper and a ruler.
If you need any of those resources.
Great, let's get started.
Here we're going to be given some tins.
And we need to see which number represented in each set.
Here we have two hundreds tins, three tens tins and two ones tins, so the answer is 232.
My second set, I can see I have one hundred, four tens tins and six ones tins.
Pause this video and count the tins.
Well done, the answers are 350 and 303, great job.
James has made a three digit number with these cards, one, four, and nine.
What are the three digit numbers can he make these cards? And what is the biggest number he could make? So these are the numbers that I came up with 194 419 491 914 and 941 and the biggest number card was 941.
Lucy has made a three digit number with these cards, three, two, and seven.
What are the three digit numbers , could she make with these cards? What is the number she could make? Pause this video and write down your answers super job guys, so these except the five sets of digit cards that I came up with, with the largest number being 732, do some ticking and fixing please, pause this video while you do say.
Great job, that's two questions done already.
Let's explore, so here we go.
Two hundreds, one tens and four ones together make 214, 918 is made up of nine hundred, one ten and eight ones.
320 is made up of three Hundreds and two tens.
So now it's your turn, seven hundreds, nine tens and three ones together make? 312 is made up of hundreds, tens and ones.
890 is made up of hundreds and nine, I gave you the answer oops, and something tens, pause this video to write your answer down.
Great job, 793.
The second is three hundreds, one tens and two ones, and the final one is eight hundreds and nine tens.
I hope you got that one correct everyone.
Find the value, of the red triangle in each of these statements, the red triangle is equal to 200 add 80 add four.
So I need to make sure, I add these together to get the value , of the red triangle which is 284.
Well, I can see is 60 and seven are already there, it's the hundreds that are missing, so my red triangle is equal to 900 and 208 is equal to 200 add red triangle, which is going to be eight.
Find the value of the triangle in each of these statements.
Super job everyone.
The answer is 724 ,400 and 80.
here we go, 491 is made up of, four hundreds nine tens and one ones.
491 is also made up of 49 tens and one ones.
491 is also made up of four hundreds and 91 ones.
This is a way of expressing different numbers.
So we need to find different ways of expressing 312, so I came up with, 312 is made up of three hundreds, two tens and one ones.
312 is also made up of 31 tens and two ones.
312 is also made up of three hundreds and 12 ones.
So it's your turn, the 765 pause this video and find the answers.
Super job, so 765 is made up of, seven hundreds, six tens and five ones.
765 is also made up of 76 tens and 5 ones.
765 is also made up of seven hundreds and 65 ones.
Time to do some ticking and fixing.
Pause this video while you do that.
Well done, that was a lot of practise, and it's now time to put all of that knowledge to good use.
Good luck, on the next following slides.
You have your independent questions, so please make sure you pause the slides, when you get that.
See you on the other side.
Super job everybody, for each of these two sets, the numbers are on the screen, the first set is 208 and the second set is 672.
For Question two, these are the other numbers that he could make with his cards.
Please pause the video and tick and fix.
The biggest number all he could make was 853.
Eight hundreds, three tens and six ones, together make 836.
457 is made up of four hundreds, five tens and seven ones.
250 is made up of two hundreds and five tens.
Please do some ticking and fixing.
Find the value of the pink square in each of these segments.
Pink square is equal to 500 add 70 add four, therefore pink square is 574.
628 is equal to pink square add 20 add eight, therefore the pink square is worth 600.
703 is equal to 700 add the pink square add three and the value of the pink square here is zero.
674 is made up of six hundreds, seven tens and four ones.
674 is also made up of 67 tens and four ones.
674 is also made up of six hundreds and 74 ones.
Find different ways of expressing 630 and 704.
Pause the video and do some ticking and fixing.
Super job did today guys, that was a lot of application and well well done.
|
# What is 7+6i divided by 10+i?
Sep 8, 2017
$\frac{7 + 6 i}{10 + i} = \frac{76}{101} + \frac{53}{101} i$
#### Explanation:
We can make the denominator real by multiplying the denominator with its complex conjugate, thus:
$\frac{7 + 6 i}{10 + i} = \frac{7 + 6 i}{10 + i} \cdot \frac{10 - i}{10 - i}$
$\text{ } = \frac{\left(7 + 6 i\right) \left(10 - i\right)}{\left(10 + i\right) \left(10 - i\right)}$
$\text{ } = \frac{70 - 7 i + 60 i - 6 {i}^{2}}{100 - 10 i + 10 i - {i}^{2}}$
$\text{ } = \frac{70 + 53 i + 6}{100 + 1}$
$\text{ } = \frac{76 + 53 i}{101}$
$\text{ } = \frac{76}{101} + \frac{53}{101} i$
Sep 8, 2017
$\frac{76}{101} + \frac{53}{101} i$
#### Explanation:
$\frac{7 + 6 i}{10 + i}$
First we have to rationalize the denominator by multiplying the complex number in the denominator and the numerator by the denominator's conjugate.
$\frac{\left(7 + 6 i\right) \left(10 - i\right)}{\left(10 + i\right) \left(10 - i\right)} = \frac{7 \left(10\right) + 6 i \left(10\right) - 7 \left(i\right) - 6 i \left(i\right)}{{10}^{2} - {i}^{2}}$ (using the difference of squares rule in the denominator)
$= \frac{70 + 60 i - 7 i - 6 \left({i}^{2}\right)}{100 - {i}^{2}} = \frac{70 + 53 i - 6 \left(- 1\right)}{100 - \left(- 1\right)}$**(since ${i}^{2} = - 1$)
$\frac{70 + 53 i - 6 \left(- 1\right)}{100 - \left(- 1\right)} = \frac{70 + 6 + 53 i}{100 + 1} = \frac{76 + 53 i}{101}$
$= \frac{76}{101} + \frac{53}{101} i$
I hope that this helps.
|
Dividing by 6-9
Start Practice
## Learn to Divide by 6 to 9
Division is when we break up a larger number into a specific number of groups.
Let’s review dividing by 6 to 9!
What is 12 ÷ 6?
Now, let’s divide these 12 dots into 6 groups.
How many dots are in each group?
2!
That’s it! 12 ÷ 6 = 2. ✅
Before we look at what we can do with division facts, let’s review all of our division facts with divisors from 6 to 9!
Great! Now, that you’ve reviewed division facts from 6 to 9, we can tell whether division sentences are true or false.
### Division True or False
Let’s try it out!
Is 21 ÷ 7 = 3 True or False?
How can we figure this out? 🤔
Do you remember that both sides of an equation must be equal to each other?
Great! Let’s find the value of the left side of the equation:
Both sides of the equation are the same.
21 ÷ 7 = 3 is true. ✅
Let’s try another one!
18 ÷ 9 = 3
We start by figuring out the value of the left side of the equation:
The left side equals 2.
This means:
18 ÷ 9 = 3 is False ✅
Great work figuring out if equations are True or False. Another thing you will need to do in 3rd grade is sort equations.
### Sorting Equations
Let’s try this out!
Here are a group of equations:
45 ÷ 9 = ?
60 ÷ 6 = ?
70 ÷ 7 = ?
80 ÷ 8 = ?
Now, let’s group them into:
How do we do this? First, we find the value for each of the equations!
45 ÷ 9 = 5
60 ÷ 6 = 10
70 ÷ 7 = 10
80 ÷ 8 = 10
Now, let’s put them into our table.
Great job! 💪
Let’s try one more.
Here are a group of equations:
64 ÷ 8 = ?
48 ÷ 6 = ?
36 ÷ 9 = ?
28 ÷ 7 = ?
Let’s group them into:
What should we do first? Divide!
64 ÷ 8 = 8
48 ÷ 6 = 8
36 ÷ 9 = 4
28 ÷ 7 = 4
Now, let’s sort them:
Great work!! 🎉 Now, you can work with divisors from 6 to 9!
Start Practice
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Composite Numbers
Reviewed by:
Last updated date: 10th Aug 2024
Total views: 405k
Views today: 4.05k
What are Composite Numbers?
In the subject of mathematics or math, composite numbers are defined as the type of numbers that have more than two factors. These numbers are different from prime numbers as prime numbers only have two factors. These factors are one and the number itself. Composite numbers are also simply known as composites.
Readers need to remember that all numbers that are not prime numbers can be categorized as being composite numbers. These numbers can also be divided by more than two numbers. For example, the number six is composite because it can be divided by 1, 2, 3, and 6.
Composite numbers can also be defined as the integers that can be generated by multiplying the two smallest positive integers. These numbers should also contain at least one divisor other than the number one. These numbers also have more than two composite factors. For example, any even number that is greater than 2 is a composite number.
Students are often asked to mention the difference between prime numbers and composite numbers. This is why it is advised that students should understand the concept of prime and composite numbers properly.
Also, another common question that students have on this topic is whether zero falls under the category of prime or composite numbers. The answer to this question is that zero does not fall under either of these categories of prime or composite numbers.
Keep in mind everything that we have discussed till now. Can you mention all the major composite numbers up to 200? If you can’t, then go through the section that is mentioned below.
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 128, 129, 130, 132, 133, 134, 135, 136, 138, 140, 141, 142, 143, 144, 145, 146, 147, 148, 150, 152, 154, 155, 156, 157, 158, 159, 160, 162, 164, 165, 166, 168, 170, 172, 174, 175, 176, 177, 178, 180, 182,184, 185, 186, 187, 188, 189, 190, 192, 194, 195, 196, 198, and 200.
One should also look at lists mentioning composite numbers from 1 to 1000 to become more familiar with this topic.
How Can You Determine Composite Numbers?
Now, you must understand the meaning of consecutive composite numbers. You should also have a ready list prepared of all the composite numbers from 1 to 100. This is why it is now time to learn the process of finding out the values of composite numbers.
The process that you can follow to determine composite numbers is mentioned below.
• Begin by finding out all the factors of a positive integer.
• If a number only has two factors, which are one and the number itself, then classify it as a prime number.
• If a number has more than two factors, then categorize it as a composite number.
For example, let’s assume that you have to find whether 14 is a prime or composite number. You must first begin with the understanding of what prime and composite numbers are. After that, you should find out the factors of 14.
We know that, 14 / 1 = 14
14 / 2 = 7
14 / 7 = 2
14 / 14 = 1
From this, it can be concluded that 14 is a composite number as it has more than one factor. Also, students should be familiar with all the composite numbers less than 20 without consulting a book.
The Different Types of Composite Numbers
In this section, we will look at the different types of composite numbers. According to experts, there are mainly two divisions of composite numbers. These divisions are:
• Odd composite numbers or composite odd numbers
Odd composite numbers are all the composite numbers that are odd integers. For example, 9, 15, 21, 25, and 27.
• Even composite numbers or composite even numbers
On the other hand, even composite numbers are all the composite numbers that are also even integers. For example, 4, 6, 8, and 10.
What are The Properties of Composite Numbers?
Composite numbers refer to the positive integers that are numbers that are formed by multiplying two smaller positive integers. The major properties of these composite numbers are that these numbers are evenly divisible by both smaller composite and prime numbers, that is, no special preference is given to the prime numbers or the composite numbers for becoming a factor of any number. The second property of composite numbers is that all composite numbers are made up of a combination of two or more prime numbers.
How are Composite Numbers Different From Prime Numbers?
The major difference between a prime number and composite number is that the prime numbers are only divisible by 1 and themselves, that is, such numbers have only two factors whereas composite numbers have more than two factors, that is, it is divisible by numbers other than 1 and itself also. Prime numbers can only be written in the form of a product of two numbers only while composite numbers can be written in the form of the product of more than two numbers. For example, 5 is a prime number as it has no factor other than 1 and 5 whereas 4 is a composite number because other than 1 and 4, 2 is also a factor of 4.
Learn About the Smallest Composite Number
Four is the smallest composite number. This is because, in the case of 4, there are more than two factors. The divisors of 4 are 1, 2, and 4. Also, you might want to note that the smallest prime number is 2, the smallest odd composite number is 9, and the two-digit smallest composite number is 12.
Prime Factorization of Commonly Used Composite Numbers
It is recommended that students should be familiar with composite numbers between 1 to 100 and their prime factorization. This is because students can be asked to answer questions related to that from the prime and composite number chapter. This is why we have created a list of prime factorization of composite numbers till 50. That list is mentioned below.
Composite Numbers Prime Factorization 4 2 x 2 6 2 x 3 8 2 x 2 x 2 9 3 x 3 10 2 x 5 12 2 x 2 x 3 14 2 x 7 15 3 x 5 16 2 x 2 x 2 x 2 18 2 x 3 x 3 20 2 x 2 x 5 21 3 x 7 22 2 x 11 24 2 x 2 x 2 x 3 25 5 x 5 26 2 x 13 27 3 x 3 x 3 28 2 x 2 x 7 30 2 x 3 x 5 32 2 x 2 x 2 x 2 x 2 33 3 x 11 34 2 x 17 35 5 x 7 36 2 x 2 x 3 x 3 38 2 x 19 39 3 x 13 40 2 x 2 x 2 x 5 42 2 x 3 x 7 44 4 x 11 45 3 x 3 x 5 46 2 x 23 48 2 x 2 x 2 x 2 x 3 49 7 x 7 50 2 x 5 x 5
How to Recognize if a Number is Prime or Not?
The rules with which you can recognize if a number is prime or not. These rules are:
• If a number is greater than 2 and is a multiple of 2, then, in that case, the number is not a prime number because it will have at least three factors, 1, 2 and the number itself.
• If a number is greater than 3 and a multiple of 3 as well, then that number can also not be a prime number, because it has at least three factors 1, 3 and the number itself.
• If a particular number is a multiple of 4, then it will also be a multiple of 2, therefore, such numbers can also not be considered as prime numbers.
• If any number is greater than 5 and is also a multiple of 5, then it will not be a prime number. In other words, if a number ends with 0 or 5, then it will not be a prime number because in this case also, the number is divisible by at least 1, 5 and the number itself.
• Similarly, if a number is a multiple of 6, then it will also be a multiple of 2 and 3, therefore such numbers can automatically rule out from the list of prime numbers as it will have at least five factors, that is, 1, 2, 3, 6 and the number itself.
Did you know that prime numbers can only be divided by 1 and the number itself? On the other hand, composite numbers have more than two factors. Prime numbers can also be just written as products of two numbers. But composite numbers can be written as the products of more than two numbers. For example, 5 is a prime number, while 4 is a composite number.
FAQs on Composite Numbers
1. What is a Composite Number?
The numbers in math are broadly divided into two major categories, that is, prime numbers and composite numbers. A composite number is a positive integer or natural number that has more than one factor. For example, 15 is a composite number as it has multiple factors, including 1, 3, 5, and 15. Composite numbers are numbers that have more than two factors, that is, numbers other than 1 and itself. In other words, a composite number is one that is divisible by numbers other than 1 and itself.
2. Is 2 a Composite Number?
2 is not a composite number because it has only two factors. Since 2 is divisible by only 1 and itself, therefore, it cannot be considered as a composite number because the criteria for a number to be called composite is that it should have more than 2 factors, that is, 1 and the number itself. Thus, 2 is a prime number, according to which a number is said to be prime if it has only 2 factors, that is, 1 and the number itself, for example, 2.
3. Is 9 a Composite Number?
Yes, 9 is a composite number as it has more than two factors, including 1, 3, and 9. For a number to be considered as a composite, it should have more than two numbers as its factor, and since 9 has 3 factors, therefore, it will be considered as a composite factor. If the number would have only two numbers as its factors, that is 1 and the number itself, then, in that case, the number would have been a prime number and not a composite number.
4. Is 19 a Prime or Composite Number?
19 is a prime number as it has only two factors. Prime numbers are the ones that have only two factors, one is the number 1 and the other is the number itself. Since 19 is divisive by only 1 and 19 therefore, it is a prime number. If the number would have had more than 2 factors, then that number would have been considered as a composite number instead of it being considered as a prime number.
5. Find Out the Total Number of Composite Numbers Between 1 and 10.
Composite numbers refer to those numbers which have more than two factors, that is, these numbers are also divisible by numbers other than 1 and the number itself. There are five composite numbers between 1 and 10. These numbers are 4, 6, 8, 9, and 10. Therefore, between 1 and 10, there are only 5 numbers which are divisible by other numbers as well, except for 1 and the number itself. The remaining five numbers between 1 and 10 are prime numbers, that is they are divisible by only two numbers, which are 1 and the number itself.
|
# Probability
How likely something is to happen.
Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.
### Tossing a Coin
When a coin is tossed, there are two possible outcomes:
Tails (T)
Also:
• the probability of the coin landing H is ½
• the probability of the coin landing T is ½
### Throwing Dice
When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6.
The probability of any one of them is 16
## Probability
In general:
Probability of an event happening = Number of ways it can happen Total number of outcomes
### Example: the chances of rolling a "4" with a die
Number of ways it can happen: 1 (there is only 1 face with a "4" on it)
Total number of outcomes: 6 (there are 6 faces altogether)
So the probability = 1 6
### Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?
Number of ways it can happen: 4 (there are 4 blues)
Total number of outcomes: 5 (there are 5 marbles in total)
So the probability = 4 5 = 0.8
## Probability Line
We can show probability on a Probability Line:
Probability is always between 0 and 1
## Probability is Just a Guide
Probability does not tell us exactly what will happen, it is just a guide
### Example: toss a coin 100 times, how many Heads will come up?
Probability says that heads have a ½ chance, so we can expect 50 Heads.
But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50.
## Words
Some words have special meaning in Probability:
Experiment: a repeatable procedure with a set of possible results.
### Example: Throwing dice
We can throw the dice again and again, so it is repeatable.
The set of possible results from any single throw is {1, 2, 3, 4, 5, 6}
Outcome: A possible result.
### Example: "6" is one of the outcomes of a throw of a die.
Trial: A single performance of an experiment.
### Example: I conducted a coin toss experiment. After 4 trials I got these results:
Trial Trial Trial Trial Outcome ✔ ✔ ✔ ✔
Three trials had the outcome "Head", and one trial had the outcome "Tail"
Sample Space: all the possible outcomes of an experiment.
### Example: choosing a card from a deck
There are 52 cards in a deck (not including Jokers)
So the Sample Space is all 52 possible cards: {Ace of Hearts, 2 of Hearts, etc... }
The Sample Space is made up of Sample Points:
Sample Point: just one of the possible outcomes
### Example: Deck of Cards
• the 5 of Clubs is a sample point
• the King of Hearts is a sample point
"King" is not a sample point. There are 4 Kings, so that is 4 different sample points.
### Example: Throwing dice
There are 6 different sample points in that sample space.
Event: one or more outcomes of an experiment
### Example Events:
An event can be just one outcome:
• Getting a Tail when tossing a coin
• Rolling a "5"
An event can include more than one outcome:
• Choosing a "King" from a deck of cards (any of the 4 Kings)
• Rolling an "even number" (2, 4 or 6)
Hey, let's use those words, so you get used to them:
### Example: Alex wants to see how many times a "double" comes up when throwing 2 dice.
The Sample Space is all possible Outcomes (36 Sample Points):
{1,1} {1,2} {1,3} {1,4} ...
...
... {6,3} {6,4} {6,5} {6,6}
The Event Alex is looking for is a "double", where both dice have the same number. It is made up of these 6 Sample Points:
{1,1} {2,2} {3,3} {4,4} {5,5} and {6,6}
These are Alex's Results:
Trial Is it a Double?
{3,4} No
{5,1} No
{2,2} Yes
{6,3} No
... ...
After 100 Trials, Alex has 19 "double" Events ... is that close to what you would expect?
700, 701, 702, 1475, 1476, 1477, 2175, 2176, 2177, 2178
|
# 5.4: Integration
Difficulty Level: At Grade Created by: CK-12
## Indefinite Integrals
As students begin anti-differentiation they will need to have a certain degree of confidence with common derivatives. This confidence, developed from substantial practice, will result in quicker recognition of the “results” from their work in differentiation.
To develop the needed skill for harder problems in the future, it is ok to practice guess and check type integration before working on the “reverse” power rule or other techniques. While it takes some time, and sometimes causes frustration, the pay off is getting an understanding of how to separate and algebraically manipulate functions to make for easier integration when the problems get complicated. A couple of good problems to try:
\begin{align*}\int\limits \sin (2x) dx \end{align*}
\begin{align*}\int\limits 4x^3 dx\end{align*}
\begin{align*}\int\limits ex^2 dx\end{align*}
All of these problems are solvable easily with substitution or other techniques to be learned later. However, the process of trying functions, taking the derivative and seeing how the outcome turns out will provide a strong foundation for understanding the techniques and rules later, as well as just being good analytic practice.
## The Initial Value Problem
Here is an example problem with a basic differential equation:
You have two friends who are coming to meet you. One of your friends calls you \begin{align*}1 \;\mathrm{hour}\end{align*} after he left saying that he is now \begin{align*}320 \;\mathrm{miles}\end{align*} away. Your other friend calls \begin{align*}2 \;\mathrm{hours}\end{align*} after leaving, and is now \begin{align*}200 \;\mathrm{miles}\end{align*} away. The first person averages \begin{align*}72 \;\mathrm{mph}\end{align*} and the second averages \begin{align*}55 \mathrm{mph}\end{align*}. When were they equally distant from you?
A contrived problem, but one that provides some opportunity. There are many ways to solve this problem, and some students may feel like using calculus is a waste of time, as they are just learning those skills and others are still more familiar. It is a challenge when introducing new topics to choose problems that are easy enough to check and feel confident about, but provide opportunities to practice the new skills. Therefore, don't discourage or dismiss students who feel there is a better way, but insist that everyone at least attempts the problem using calculus.
All we have is two constant functions so we should list them: \begin{align*}v_1 (t)=72,v_2 (t)=55\end{align*}. Astute observers will see a potential problem with this, however. Since the drivers are coming towards you, and the standard convention is to put the subject of the problem at the origin, we should actually be indicating the velocities to be negative. It is up to the instructor when that should be brought up. We saw earlier that velocity is the derivative of the position function, so it follows that position is the anti-derivative of the velocity function. Therefore \begin{align*}s_1 (t)=- 72t + c_1\end{align*} and \begin{align*}s_2 (t) = - 55t + c_2\end{align*}. We do wish to know when the two position functions are equal, but with the constant term still not determined we can't do so. This is where the initial conditions come into play. Substituting in the time and position: \begin{align*}320 = -72(1)+c_1 \rightarrow 392=c_1\end{align*} and \begin{align*}200 = -55(2)+c_2 \rightarrow 310 =c_2\end{align*}. Now the problems can be set equal: \begin{align*} - 72t + 392 = - 55t + 310 \rightarrow 82 =17t \rightarrow t = 4.8\end{align*}.
It is important also to interpret the answer correctly. It states they will be equidistant \begin{align*}4.8 \;\mathrm{hours}\end{align*} after they left, not after they called. A simple problem, but one to illustrate the application of differential equations and how initial conditions fit in.
## The Area Problem
The same way that physical problems can illustrate the motivation for the derivative, the same can be done for integrals. Take the following table of velocities from a car starting from a full stop:
\begin{align*}& \text{Time (sec)} && 1 && 2 && 3 && 4 && 5 && 6 && 7 && 8 && 9 \\ & \text{Velocity (ft/s)} && 21 && 24 && 29 && 32 && 38 && 39 && 37 && 34 && 30\end{align*}
How much distance did the car travel in those \begin{align*}9 \;\mathrm{seconds}\end{align*}?
The way this was done in algebra was to find the average velocity and multiply by the time to get the distance traveled. It should be apparent from the table that the velocity, and even the change in velocity, is not constant. However, something can be inferred from that process. If we graph the time on the \begin{align*}x\end{align*} axis, and the velocity on the y axis, then the average velocity times the time is the same as the area of the rectangle made. Ask the students “Is there a way to get a more accurate approximation?” A diagram or graph may be helpful as an illustration. It should be clear that treating each second as it's own problem will result in a closer answer. One question that needs to be answered is where to take the height of each rectangle from. If you take the height from the right hand side the answer is:
\begin{align*}21(1)+24(1)+29(1)+32(1)+38(1)+39(1)+37(1)+34(1)+30(1)=284\end{align*}
Taking it from the left hand side:
\begin{align*}0(1)+21(1)+24(1)+29(1)+32(1)+38(1)+39(1)+37(1)+34(1)=254\end{align*}
Students should be able to safely assume that the correct answer is in between those two. Furthermore, they should think about the different ways that the answer could be improved. Students will probably come up with smaller rectangles, more rectangles, average the rectangles or end points (essentially the trapezoid rule) and possibly some others, most of which will be the next steps.
## Definite Integrals
It is up to the instructor at this point whether or not to introduce some summation rules. This may depend on whether or not the class has had experience with series in previous classes or if they are comfortable with what has been presented thus far in the class. These facts do not need to be proven just yet; there will be proofs presented later in the chapter on series. Some useful facts are:
\begin{align*}\sum_{i=1}^n c = nc\end{align*} where \begin{align*}c\end{align*} is a constant
\begin{align*}\sum_{i=1}^n i & = \frac{n(n +1)}{2} \\ \sum_{i=1}^n i^2 & = \frac{n(n+1)(2n+1)}{6} \\ \sum_{i=1}^n i^3 & = \frac{n^2(n+1)^2}{4} \\ \sum_{i=1}^n cf (i) & = \sum_{i=1}^n f(i) \sum_{i=1}^n p(i) \pm q(i) = \sum_{i=1}^n p(i) \pm \sum_{i=1}^n q(i)\end{align*}
Many definite integrals can be solved using just these rules:
Solve: \begin{align*}\int\limits_{0}^{1} 5x + 4dx\end{align*}
First, the width of each interval with \begin{align*}n\end{align*} subdivisions is \begin{align*}\frac{1}{n}\end{align*}. This makes each right hand endpoint \begin{align*}\frac{1}{n} i\end{align*}. Therefore the definite integral is:
\begin{align*}\int\limits_{0}^{1} 5x + 4dx = \lim_{n \to \infty} \sum_{i=1}^n \left (5 \left (\frac{i}{n} \right ) + 4 \right ) \frac{1}{n} = \lim_{n \to \infty} \sum_{i=1}^n \frac{i5}{n^2} + \frac{4}{n} = \lim_{n \to \infty} \left (\sum_{i=1}^n \frac{5i}{n^2} + \sum_{i=1}^n \frac{4}{n} \right )\end{align*}
Using the final summation rule above. Now we can pull the constants out front and that will result in a match for the form listed above for some other summation rules:
\begin{align*}\lim_{n \to \infty} \left (\frac{5i}{n^2} \sum_{i=1}^n i + \sum_{i=1}^n \frac{4}{n} \right ) = \lim_{n \to \infty} \left (\frac{5n(n+1)}{2n^2} + 4 \right ) = \lim_{n \to \infty} \left (\frac{5n^2}{2n^2} + \frac{5n}{2n^2} + 4 \right ) = \lim_{n \to \infty} \left (\frac{5}{2} + \frac{5}{2n} + 4 \right )\end{align*}
Now it's possible to evaluate the limit and find that \begin{align*}\int\limits_{0}^{1} 5x + 5dx = \frac{13}{2}\end{align*}.
## Evaluating Definite Integrals
An application of the definite integral, and one that appears regularly on tests, is finding the average value for a function. Averages are easy to find in linear situations, but not so easy with curves. The average value of a function can be found by evaluating:
\begin{align*}\frac{1}{b-a} \int\limits_{a}^{b}f(x)dx\end{align*}
which can be thought of as the area under the curve divided by the length of the interval. This is consistent with how we would find the mean in most other situations. An example of it's use:
An endowment account is being continually withdrawn from over the course of a month to cover day to day expenses. The amount of money in the account can be modeled with the equation: \begin{align*}E = 20+980e^{-.01t}\end{align*} where \begin{align*}E\end{align*} is the amount in the account, in thousands, and t is time in days. The bank pays \begin{align*}8.5\%\end{align*} interest on the average amount in the account over the whole \begin{align*}30\end{align*} day month. How much interest is paid? How much money needs to be placed into the account at the end of the month to maintain the same balance?
Because this is a curve, it is not possible to subtract the endpoints and divide by the duration. The function, and the information on the endpoints needs to be place into the average value formula:
\begin{align*}\frac{1}{30 - 0}\int\limits_{0}^{30} 20 + 980e^{-0.1t} & = \frac{1}{30} \left [20t - 98000e^{- .01t}\big|^{30}_{0} \right ] \\ & = \frac{1}{30} \left [\left (20(30) - 98000e^{-.01(30)}\right ) - 20(0) - 98000e^{-.01(0)}) \right ] \\ & = \frac{1}{30} [600-726000.18 + 98000)\\ & =866.66\end{align*}
This give us the average amount of money for the month, so multiplying by \begin{align*}.085\end{align*} states that \begin{align*}73.66\end{align*} thousand dollars are paid. This means that after the interest gets paid there is \begin{align*}940.32\end{align*} thousand in the account. If we substitute into the formula to find out how much the balance was at the beginning of the month, we find that it needs to be \begin{align*}1000\end{align*} thousand dollars, meaning that there needs to be \begin{align*}59.68\end{align*} thousand replaced to keep the endowment going.
There is frequently a question regarding average values on the AP examination. Because of it's intuitive format, that is, it is close to how we find means, it may not need to be stressed for memorization, but it will come in handy for both tests, and for applied classes like physics and economics.
## Integration by Substitution
This is the beginning of one of the more memorable parts of first year calculus. The set of techniques for integration need practice, practice and more practice. It is a pattern recognition game that can only be won through having the experience to match the correct technique. Furthermore the use of an incorrect technique may not result in an impossible situation, but will only fail to help get closer to the solution. A simple algebraic example of what I mean: to solve \begin{align*}3x+2=-x-5\end{align*}, one of the tools that is available to solve algebraic equation is to square both sides. We can legally do so in this instance, but doing so will only make the problem worse. This can often happen with either a poor choice of method, or a poor substitution or other traps that will be considered in the next sections.
If the integral is not straightforward, it is always preferable to start by attempting a substitution. It is the easiest, and usually going to work most frequently. Another clue is that substitution is the opposite of the chain rule. If it looks like you are being asked to integrate a composite function, substitution is probably the key.
## Numerical Integration
The trapezoid rule and Simpsons rule provide one of the first peeks into the sort of “brute-force” solving methods that we rely on now with technology. Getting a close answer with either method is not challenging, like taking a very involved anti-derivative is challenging, but can take significantly longer depending on the situation. One of the main tenets of computer science is that the major advantage of a computer is that it can do the same procedure, over and over again, without making errors or getting bored. Try to compute Simpson's rule by hand with \begin{align*}50\end{align*} subdivisions and you too will believe that it is an advantage.
So a great problem to tackle at this point is how to program a computer to take a definite integral with a good deal of precision. I will present the steps in TI-BASIC here, as the graphing calculator is probably the most likely place for students to be programming in the math classroom. This is also relatively easy in Python, Java or C if the instructor is familiar with those languages and has access to computers to use for programming. There are resources on the web detailing how to write a program for Simpsons rule in those languages.
I will put comments after a “//” to explain what each line is doing. These comments should not, and really can't, be entered on the calculator.
Input “FUNCTION?”,Str\begin{align*}1\end{align*} //Getting the function and storing it in a string variable, found under the VARS menu
Str\begin{align*}1\rightarrow Y1\end{align*} //Placing the function in the \begin{align*}Y1\end{align*} spot so it can be used as a function
Input “LOWER LIMIT?”,\begin{align*}A\end{align*} //The lower limit of integration
Input “UPPER LIMIT?”,\begin{align*}B\end{align*} //The upper limit of integration
Input “DIVISIONS?”,\begin{align*}N\end{align*} //The number of subdivisions used for the approximation
While fPart\begin{align*}\left (\frac{N}{2} \right )\neq 0\end{align*} //This checks to see if the number entered for \begin{align*}N\end{align*} is even. If not, it asks for a new number until \begin{align*}N\end{align*} is even
Disp “NEED NEW \begin{align*}N\end{align*}
Input “\begin{align*}N\end{align*} MUST BE EVEN”,\begin{align*}N\end{align*}
End
\begin{align*}\frac{(B-A)}{N \rightarrow D}\end{align*} //Makes \begin{align*}D\end{align*} the length of each subdivision
\begin{align*}N \rightarrow I\end{align*} //I will be used as a counter between and
\begin{align*}\left \{1 \right \} \rightarrow L1\end{align*} //Setting up a list for the coefficients to be multiplied to each endpoint of the function
While \begin{align*}I > 2\end{align*}
augment\begin{align*}(L1,\left \{4,2 \right \})\rightarrow L1\end{align*}
\begin{align*}I-2 \rightarrow I\end{align*}
End
augment\begin{align*}(L1,\left \{4,1 \right \})\rightarrow L1\end{align*}
sum\begin{align*}(L1*\mathrm{seq}(Y1(A+D*I),I,0,N))*\frac{D}{3 \rightarrow S}\end{align*} //This takes the sum of each element of the sequence of function values from to \begin{align*}N\end{align*}, multiplies by the width of each, and puts the answer into \begin{align*}S\end{align*}
Disp Str\begin{align*}1\end{align*} //Displays the function and answer
Disp “IS APPROX”
Disp \begin{align*}S\end{align*}
While the use of a list to produce the coefficients is a bit of a novel approach, there are many other ways to do so, including putting the computation of the approximation inside of a For or While loop. Be flexible and try to guide students as much as possible in writing some of their own code. A next step might be to try and write a program for trapezoid approximation.
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# Specimen Cyprus Seniors Provincial/2nd grade/Problem 4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
If $\rho_{1}, \rho_{2}$ are the roots of equation $x^2-x+1=0$ then:
a) Prove that $\rho_{1}^3=\rho_{2}^3 = -1$ and
b) Calculate the value of: $\rho_{1}^{2006} + \rho_{2}^{2006}$.
## Solution
Since $\rho_1, \rho_2 \neq -1$, we can multiply both sides by $x+1$, and $\rho_{1}, \rho_{2}$ will still satisfy the equation:
$(x^2-x+1)(x+1)=(0)(x+1)$
$x^3 + 1 = 0$
$x^3 = -1$
Thus, $\rho_{1}^3=\rho_{2}^3 = -1$ as desired $\boxed{\mathbb{Q.E.D.}}$.
The third roots of $-1 = cis(180^\circ)$ are $cis(60^\circ)$, $cis(-60^\circ)$, and $-1$.
Since $\rho_1, \rho_2 \neq -1$, we have $\rho_{1} = cis(60^\circ)$ and $\rho_{2} = cis(-60^\circ)$. (Switching the two values will not affect the result).
Note that $\rho_{1}^6=\rho_{2}^6 = 1$, and that $2006 = 6(334)+2$
So, $\rho_{1}^{2006} + \rho_{2}^{2006} = \rho_{1}^{2} + \rho_{2}^{2} = cis(120^\circ) + cis(-120^\circ) =$ $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \boxed{-1}$
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# PROPERTIES OF ADDITION AND MULTIPLICATION
(i) Commutative Property :
Changing the order of addends does not change the sum.
If 'a' and 'b' are any two numbers, then
a + b = b + a
Example :
5 + 8 = 13
8 + 5 = 13
So,
5 + 8 = 8 + 5
(ii) Associative Property :
Changing the grouping of the addends does not change the sum.
If a, b and c are any three numbers, then
a + (b + c) = (a + b) + c
Example :
2 + (4 + 7) = 2 + 11 = 13
(2 + 4) + 7 = 6 + 7 = 13
So,
2 + (4 + 7) = (2 + 4) + 7
The sum of any number and zero is the number itself.
If a is any number, then
a + 0 = 0 + a = a
So, zero is the additive identity.
Example :
2 + 0 = 0 + 2 = 2
- k is the negative or additive inverse of k.
If k is a number,then there exists a number -k such that
k + (-k) = (-k) + k = 0
Examples :
Additive inverse of 3 is -3.
Additive inverse of -5 is 5.
Additive inverse of 0 is 0 itself.
## Properties of Multiplication
(i) Commutative Property :
Changing the order of factors does not change the product.
If a and b are any two numbers, then
a x b = b x a
Example :
5 x 2 = 10
2 x 5 = 10
So,
5 x 2 = 2 x 5
Therefore, Commutative property is true for multiplication.
(ii) Associative Property :
Changing the grouping of the factors does not change the product.
If a, b and c are any three numbers, then
a x (b x c) = (a x b) x c
Example :
5 x (2 x 3) = 5 x 6 = 30
(5 x 2) x 3 = 10 x 3 = 30
So,
5 x (2 x 3) = (5 x 2) x 3
Therefore, Associative property is true for multiplication.
(iii) Multiplicative Identity :
The product of any number and 1 is the number itself. ‘One’ is the multiplicative identity for numbers.
If k is any number, then
k x 1 = 1 x k = k
Example :
5 x 1 = 1 x 5 = 5
(iv) Multiplication by 0 :
Every number multiplied by 0 gives the result 0.
If k is any number, then
k x 0 = 0 x k = 0
Example :
7 x 0 = 0 x 7 = 0
(v) Multiplicative Inverse or Reciprocal :
For every number k, there exists a number 1/k such that
k x 1/k = 1
Then,
k and 1/k are multiplicative inverse of each other
That is,
k is the multiplicative inverse of 1/k
1/k is the multiplicative inverse of k
Examples :
The multiplicative inverse of 2 is 1/2.
The multiplicative inverse of 1/3 is 3.
The multiplicative inverse 1 is 1.
The multiplicative inverse of 0 is undefined.
## Distributive Property
(i) Distributive Property of Multiplication over Addition :
Multiplication of numbers is distributive over addition.
If a, b and c are any three numbers, then
a x (b + c) = ab + ac
Example :
3 x (2 + 5) = 3 x 7 = 21 -----(1)
3 x (2 + 5) = 3x2 + 3x5 = 6 + 15 = 21 -----(2)
From (1) and (2),
3 x (2 + 5) = 3x2 + 3x5
Therefore, Multiplication of numbers is distributive over addition.
(ii) Distributive Property of Multiplication over Subtraction :
Multiplication of numbers is distributive over subtraction.
If a, b and c are any three numbers, then
a x (b - c) = ab - ac
Example :
3 x (7 - 2) = 3 x 5 = 15 -----(1)
3 x (7 - 2) = 3x7 - 3x2 = 21 - 6 = 15 -----(2)
From (1) and (2),
3 x (7 - 2) = 3x7 - 3x2
Therefore, Multiplication of numbers is distributive over subtraction.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
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OTHER TOPICS
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The Area of a Triangle
🏆Practice triangle area
The Formula For Calculating The Area Of A Triangle
The formula for calculating the area of a triangle of any type:
height times base divided by $2$.
$Area=\frac{Base\times Height}{2}$
Test yourself on triangle area!
Complete the sentence:
To find the area of a right triangle, one must multiply ________________ by each other and divide by 2.
How to Calculate the Area of an Isosceles Triangle
Information:
• The side $CB$ has a length of $14$ cm.
• The height has a length of $17$ cm.
If we apply the formula, we multiply the height ($17$ cm) by the length of the base ($14$ cm).
By multiplying $17$ by $14$, we obtain $238$, a result that we must divide by $2$.
$238$ divided by $2$ equals $119$.
The area of this triangle is therefore: $119$.
$Area=\frac{14\times17}{2}=119$
Why is it Important to Know How to Calculate the Area of a Triangle?
Whether you are preparing for an test or are about to take your university entrance exams, it is essential to know how to calculate the area of a triangle, whether it is right angled, isosceles, etc.
So, how do you calculate a triangular area? This guide will clear up all of your doubts concerning one of the most frequently asked questions in geometry examinations.
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Characteristics of the Triangle
A triangle is a geometric figure composed of three sides that form three angles and three vertices.
The vertices of the triangle are marked with the letters $A,B$ and $C$; together creating the sides: $AB, BC, CA$.
There are different types of triangles and some of them even share characteristics. We will deal with those later in this article!
Before addressing the different types of triangles that exist and how to calculate their areas,
we must first know the terms that are usually used when we talk about a triangular area.
Terms Useful When Calculating a Triangular Area and About Triangles in General
• Straight line: A line extending in the same direction and having an infinite number of points.
• Segment: A fragment of a line between two points.
• Height: The height of a triangle is the length of a perpendicular line segment starting on one side and intersecting the angle opposite. The height is denoted by the letter 'h'.
• Median: The median is the segment extending from a given vertex to the midpoint of the side opposite that vertex.
• Bisector: A ray extending from a given vertex, dividing it into two equal angles.
• Perpendicular bisector: A line that is perpendicular to a side of the triangle and that passes through the midpoint of that side.
• Median segment: A median segment of a triangle is a segment that connects the midpoints of two sides of a triangle and is arranged parallel to the third side, its length being half of the third side.
• Opposite side: An opposite side is one that faces a given vertex.
Do you know what the answer is?
How to Calculate the Area of a Triangle
One of the most useful tips when calculating the area of a triangle (and thus solving the problem) is to understand that a triangle is a half-square.
There are triangles that are easily distinguishable as "half-squares" because of their shape, such as the isosceles right triangle.
However, it is important to note that triangles that do not appear to be "half-squares" are also "half-squares", since this is one of the features that characterize them.
What is the next step?
Calculate the area of the triangle.
The formula used to calculate the area of a triangle is as follows:
height times base divided by $2$.
$Area=\frac{Base\times Height}{2}$
How to Calculate the Area of Different Types of Triangles
Calculate the Area of an Equilateral Triangle
Information:
• The side $CB$ has a length of $15$ cm.
• The height has a length of $13$ cm.
Solution:
If we apply the formula, we multiply the height ($13$ cm) by the length of the base ($15$ cm).
By multiplying $13$ by $15$, we get $195$, a result that we must divide by $2$.
$192$ divided by $2$ equals $97.5$.
The area of this triangle is therefore: $97.5$.
$Area=\frac{13\times15}{2}=97.5$
Calculating the Area of an Isosceles Triangle
Information:
• The side $CB$ has a length of $14$ cm.
• The height has a length of $17$ cm.
Solution:
Again, if we apply the formula, we multiply the height ($17$ cm) by the length of the base ($14$ cm).
By multiplying $17$ by $14$, we get $238$, a result that we must divide by $2$.
$238$ divided by $2$ equals $119$.
Therefore, the area of this triangle is: $119$.
$Area=\frac{14\times17}{2}=119$
Calculating the Area of a Scalene Triangle
Information:
• The side $CB$ has a length of $9$ cm.
• The height has a length of $10$ cm.
Solution:
If we apply the formula, we multiply the height ($10$ cm) by the length of the base ($9$ cm).
By multiplying $10$ by $9$ we get $90$, a result that we must divide by $2$.
$90$ by $2$ equals $45$.
Therefore, the area of this triangle is $45$ cm².
$Area=\frac{9\times10}{2}=45$
Do you think you will be able to solve it?
Calculating the Area of a Right Triangle
Information:
• The side $CB$ has a length of $12$ cm.
• The height has a length of $14$ cm.
Note: in a right triangle or right-angled triangle, the base and the height correspond to the legs of the triangle.
Solution:
If we apply the formula, we multiply the height ($14$ cm) by the length of the base ($12$ cm).
By multiplying $14$ by $12$ we get $168$, a result that we must divide by $2$.
$168$ by $2$ equals $84$.
Therefore, the area of this triangle is: $84$.
$Area=\frac{12\times14}{2}=84$
Calculating the Area of an Obtuse Triangle
Information:
• The side $CB$ has a length of $13$ cm.
• The height has a length of $16$ cm.
In an obtuse triangle, the height is outside the triangle.
This means that we must extend the line of the base from point
$C$ to the point $D$ to find the height.
In this way we create a right triangle $\triangle ABD$,where the height we are looking for is the side $AD$.
However, remember that since we are trying to calculate the area of the obtuse triangle, we only have to consider the side as the base. $CB$ is the base.
Solution:
In this case, if we apply the formula, we multiply the height ($16$ cm) by the length of the base of the triangle whose area we want to find.
By multiplying $16$ by $13$ we obtain $208$, a result that we must divide by $2$.
$208$ divided by $2$ equals $104$.
Therefore, the area of this triangle is: $104$.
$Area=\frac{13\times16}{2}=104$
Heron's Formula
What is Heron's formula and what is it for?
Heron's formula, the invention of which is attributed to the Greek mathematician Heron of Alexandria, allows us to obtain the area of a triangle knowing the lengths of its three sides $a, b$ and $c$.
$A\text{rea}=\sqrt{s(s-a)(s-b)(s-c)}$
Where 's' is the perimeter of the triangle divided by $2$:
$s=\frac{a+b+c}{2}$
Exercises to Calculate the Area of a Triangle
Exercise 1
In the grounds of a hotel, they want to build a special triangular-shaped pool.
The length of the pool is $10$ m.
The width of the pool is $8$ m.
The pool is to be covered with tiles $2$ m long and $2$ m wide.
Question:
How many tiles are needed to cover the pool area?
Solution:
To find out how many tiles are needed, we must calculate the triangular area of the pool and the area of each tile, then divide these numbers.
$\frac{\text{S.triangle}}{S.tile}$
The result is equal to the number of tiles needed.
In a triangle, its length is equal to its height and its width is equal to the base of the triangle.
$\text{S.triangle=}\frac{10\cdot8}{2}=40$
h = length = $10$ meters.
base = width = $8$ meters.
Since the length is $2$ meters, the width is also $2$ meters.
Tile area: $2\cdot2=4$
$\frac{40}{4}=10$
$10$ tiles
Do you know what the answer is?
Exercise 2
The triangle $\triangle ABC$ is a right-angled triangle.
The area of the triangle is equal to $6cm^2$.
Calculate $X$ and the length of the side $BC$.
Solution:
We use the formula to calculate the area of the right triangle:
$\frac{AC\cdot BC}{2}=\frac{leg\times leg}{2}$
Then compare the expression with the area of the triangle ($6$).
$\frac{4\cdot(X-1)}{2}=6$
Multiplying the equation by the common denominator means that we multiply by $2$.
$4(X-1)=12$
We open the parentheses before the distributive property:
$4X-4=12$ / $+4$
$4X=16$ / $:4$
$X=4$
Replace $X=4$ into the expression $BC$ and we find:
$BC=X-1=4-1=3$
$BC=3$
$X=4$
Exercise 3
Given the triangle $\triangle PRS$, calculate the height $PQ$.
The length of the side $SR$ is equal to $4\operatorname{cm}$.
The area of the triangle $PSR$ is equal to $30$ cm².
Solution:
We use the formula to calculate the area of the triangle.
Please note: in the obtuse triangle, the height is outside of the triangle!
$\frac{Side\cdot\text{Height}}{2}=Triangular Area$
Double the equation by a common denominator.
$\frac{4\cdot PQ}{2}=30$ / $\cdot2$
Divide the equation by the coefficient of $PQ$.
$4PQ=60$ / $:4$
$PQ=15$
The length of the height $PQ$ is equal to $15 cm$.
Exercise 4
Given the right triangle $\triangle ADB$, calculate the area of the triangle. $\triangle ABC$.
The perimeter of the triangle is equal to $30\operatorname{ cm}$.
Given: $AB=15, AC=13, DC=5, CB=4$.
Solution:
Given that the perimeter of the triangle $ΔADC$ is equal to $30 cm$, we can calculate $AD$
$AD+DC+AD=PerimeterΔADC$
$AD+5+13=30$
$AD+18=30$ /$-18$
$AD=12$
Now we can calculate the area of the triangle $ΔABC$.
Please note: we are talking about an obtuse triangle, so its height is $AD$.
We use the formula to calculate the area of the triangle:
$\frac{sideheight\times side}{2}=$
$\frac{AD\cdot BC}{2}=\frac{12\cdot4}{2}=\frac{48}{2}=24$
The area of the triangle $ΔABC$ is equal to $24~cm²$.
Exercise 5
The triangle $ΔABC$ is isosceles. Therefore, $AB=AC$.
$AD$ is the height from the side $BC$.
Since $DC=10$, the length of the height $AD$ is $20%$ percent longer than the length of the side $BC$.
Calculate the area of the triangle $ΔABC$.
Solution:
Since it is an isosceles triangle (and, therefore, median), $DC=10$ and $BC=20$.
The height $AD$ is $20%$ percent longer than the length of $BC$.
That is:
$AD=1.2\cdot BC$
$\frac{100}{100}+\frac{20}{100}=\frac{120}{100}=1.2$
$AD=1.2\cdot20=24$
Hence, the area of the triangle $ΔABC$:
$SΔ\text{ABC}=\frac{AD\cdot BC}{2}=\frac{24\cdot20}{2}=\frac{480}{2}=240$
The area of the triangle $ΔABC$ is equal to $240~cm²$.
Do you think you will be able to solve it?
When are the rest of the terms we learned used?
The rest of the terms, such as median, bisector, etc., are used when we are missing some data. These terms help us to find new data when we have to solve a problem in which we are missing information.
Mistakes you may also make when you study for the exam...
Many students experience a sense of failure when they don't do as well in their exams as they would have liked. However, success in this regard is subjective. Instead of comparing yourself with your peers, you must take into consideration your own achievements and leave aside the results of your classmates. Often, the problem is not that you did not know something or that you did not understand how to calculate the area of a triangle, but rather that you did not prepare well for the exam.
To give an example: imagine an excellent pastry chef who knows many recipes, knows the products and succeeds in creating truly delicious pastries. If he had not practiced and had not prepared well (buying the products and appliances he needs, finding good recipes and having patience with the right timing, etc.), he would not have achieved good results. This is also true for students—good preparation is key!
What other mistakes are you making that are holding you back?
• Studying too intensively. Some students study for a week before the exam, perhaps ten hours a day or so. It is obvious that they are motivated and have good intentions, but the problem here is that sometimes this causes them to run out of energy before they even get to the exam. As a result, they are tired and burned out without enough time to go over all the material they are going to be tested on.
• Too much self-confidence. Did you get a 10 on the mock exam on triangular areas? If so, this does not mean that you should study two days before the exam. An exam requires preparation, both mental and practical. To prepare for an exam, it is recommended to study for at least a week.
• Stress and nerves. If exams make you anxious, it is best to start working on it beforehand. Students who suffer from this anxiety are usually those who know the subject matter inside out, but their self-confidence affects them negatively. Prepare yourself mentally because, otherwise, you may draw a blank on the exam and this will be reflected in your results.
Studying for the geometry exam with a stopwatch - why is it worth it?
Your study skills are as important as learning the subject matter. For this reason, a stopwatch can be your great ally. We recommend that you adopt its use from now on. The one that often comes with a cell phone is more than enough!
Once you have studied a subject area, it is advisable to practice and solve problems with a stopwatch, even when you are not preparing for exams. Why?
• The stopwatch gives you an indication of how long it takes you to solve a problem.
• The stopwatch lets you know which subject areas are your weakest.
• Thanks to using a stopwatch, you can also measure how fast you're progressing—if you're taking less time, you've made progress!
Do you know what the answer is?
You were expecting a 9, but you got a 7. What do you do now?
Many students have a hard time digesting the fact that they got a low grade and, more importantly, the disappointment that often accompanies it. It is very important for you to know that the more you let this affect you, the worse the impact on your academic achievement will be.
The grade you get on an exam is a kind of feedback that tells you what you are doing well and what you could improve. How can this feedback change the way you study?
• Use private tutors to reinforce your knowledge.
• Include extra study days before each exam.
Is it better for me to take private lessons with a friend rather than on my own?
We do not recommend it and the reason is very simple: a private lesson is "private", that is to say, it is adapted to your needs. When two friends study in the same private class, one of them will have to adapt to the rhythm of the other.
Thus, the idea of tailoring a "private lesson" to the student's needs is diluted. That said, if both you and your friend find the same topic difficult (for example, how to calculate the area of a triangle), you can take a joint private lesson.
Other methods to calculate the area of a triangle
If the question asks about a right triangle, you can multiply the legs (the sides of the triangle that are not the base) and divide by 2. This method is often a great shortcut to reaching the solution. That is why it is important that you know the formula and this specific feature of this triangle.
Also, if you are asked about an isosceles triangle, you should know that both the bisector and the median are considered the height of the triangle. With this knowledge, you can quickly work out the area of the triangle.
If you are interested in learning more about other triangle topics, you can access one of the following articles:
If you are interested in learning how to calculate areas of other geometric shapes, you can enter one of the following articles:
In Tutorela you will find a variety of articles about mathematics!
Do you think you will be able to solve it?
Help from a math tutor: when is it needed?
Many times, the study of mathematics arouses some anxiety among students in high school and further education. A private math class is ideal for those who want to get good grades on their exams, but don't know how.
A private lesson focuses on a certain aspect and includes more than just doing exercises, like:
• Learning to read the statement and understand what is being asked of you.
• An emphasis on understanding what is being asked of us and how we should answer.
• Finding information that can help us solve the problem.
• Studying formulas and tricks that can help us when it comes to finding the solution to a problem or exercise.
In the past, private lessons in mathematics or any other subject took place at the student's or teacher's home. Nowadays, it is also possible to have private lessons online. This offers a great way to learn advanced subject from the comfort of your own home, at the hours that best suit both the student and the teacher.
Triangles and other geometric shapes are aspects that students are exposed to as early as their first years in high school. Their grades in mathematics are what set the pace of their learning and can affect whether or not a student chooses to continue studying the subject at later on. Often, what holds people back when studying mathematics is not based on intelligence or aptitude, but rather arises from erroneous learning methods that do not help the student understand the subject matter effectively. A private mathematics teacher works side by side with the student, ensuring that in the end the student has understood their lessons.
examples with solutions for triangle area
Exercise #1
Calculate the area of the right triangle below:
Step-by-Step Solution
As we see that AB is perpendicular to BC and forms a 90-degree angle
It can be argued that AB is the height of the triangle.
Then we can calculate the area as follows:
$\frac{AB\times BC}{2}=\frac{8\times6}{2}=\frac{48}{2}=24$
24 cm²
Exercise #2
Calculate the area of the triangle ABC using the data in the figure.
Step-by-Step Solution
First, let's remember the formula for the area of a triangle:
(the side * the height that descends to the side) /2
In the question, we have three pieces of data, but one of them is redundant!
We only have one height, the line that forms a 90-degree angle - AD,
The side to which the height descends is CB,
Therefore, we can use them in our calculation:
$\frac{CB\times AD}{2}$
$\frac{8\times9}{2}=\frac{72}{2}=36$
36 cm²
Exercise #3
Calculate the area of the following triangle:
Step-by-Step Solution
The formula for calculating the area of a triangle is:
(the side * the height from the side down to the base) /2
That is:
$\frac{BC\times AE}{2}$
Now we replace the existing data:
$\frac{4\times5}{2}=\frac{20}{2}=10$
10
Exercise #4
Calculate the area of the triangle below, if possible.
Step-by-Step Solution
The formula to calculate the area of a triangle is:
(side * height corresponding to the side) / 2
Note that in the triangle provided to us, we have the length of the side but not the height.
That is, we do not have enough data to perform the calculation.
Cannot be calculated
Exercise #5
What is the area of the given triangle?
Step-by-Step Solution
This question is a bit confusing. We need start by identifying which parts of the data are relevant to us.
Remember the formula for the area of a triangle:
The height is a straight line that comes out of an angle and forms a right angle with the opposite side.
In the drawing we have a height of 6.
It goes down to the opposite side whose length is 5.
And therefore, these are the data points that we will use.
We replace in the formula:
$\frac{6\times5}{2}=\frac{30}{2}=15$
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# How do you evaluate $\arccos \left( 1 \right)$ without a calculator?
Last updated date: 20th Sep 2024
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Hint: In this question, we have a trigonometric inverse function. Trigonometric inverse function is also called arc function. To solve the trigonometric inverse function we assume the angle $\theta$ which is equal to that trigonometric inverse function. Then we find the value of $\theta$.
In this question, we used the word trigonometric inverse function. We have the following inverse trigonometric functions,
Arcsine function: it is the inverse function of sine. It is denoted as ${\sin ^{ - 1}}$.
Arccosine function: it is the inverse function of cosine. It is denoted as ${\cos ^{ - 1}}$.
Arctangent function: it is the inverse function of tangent. It is denoted as${\tan ^{ - 1}}$.
Arccotangent function: it is the inverse function of cotangent. It is denoted as${\cot ^{ - 1}}$.
Arcsecant function: it is the inverse function of secant. It is denoted as ${\sec ^{ - 1}}$.
Arccosecant function: it is the inverse function of cosecant. It is denoted as $\cos e{c^{ - 1}}$.
Now, we come to the question. The data is given as below.
$\arccos \left( 1 \right)$
We can write the above trigonometric function as below.
$\Rightarrow \arccos \left( 1 \right) = {\cos ^{ - 1}}\left( 1 \right)$
We know that$\cos 0 = 1$, and then put the value of $1$ in above.
Then,
$\Rightarrow \arccos \left( 1 \right) = {\cos ^{ - 1}}\left( {\cos 0} \right)$
We know that${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta$.
Then, ${\cos ^{ - 1}}\left( {\cos 0} \right) = 0$. Put these values in above.
Hence,
$\therefore \arccos \left( 1 \right) = 0^\circ$
Therefore, the value of $\arccos \left( 1 \right)$is $0$ degree.
Note:
As we know that the trigonometric inverse function is defined as the inverse function of trigonometric identities like sin, cos, tan, cosec, sec and cot. The trigonometric inverse function is also called cyclomatic function, anti-trigonometric function and arc function. The trigonometric inverse function is used to find the angle of any trigonometric ratio. The trigonometric inverse function is applicable for right angle triangles.
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# Practising maths with Digit Dog
What do children need to practise daily?
In order for children to develop fluency they need to have a daily routine where they practise:
• Counting;
• Remembered facts;
• Using number relationships to do calculations.
Children need the opportunity to:
• Talk mathematically;
• Discuss and solve problems;
• Be creative;
• Use reasoning skills.
Follow Digit Dog for ideas to engage children in mathematical conversations.
Using flik-flaks
Print your own flik – flaks from www.primarytreasurechest.com
Use flik-flaks to practise:
• Counting
• Subitising (recognising small amounts without counting)
• Number bonds
• Multiplication facts
• Using mathematical language
• Using reasoning skills.
Counting
Hold up the Digit Dog flik-flak and ask how many dogs can you see? You can fold the flik-flak along the black lines to show all the numbers from 0 to 10. This allows children to practise counting sets of objects up to 10.
Give children their own flik-flak and ask them to:
1. Show a single digit number – 1, 2, 3, 4 ……etc.
2. Show the numbers 0 – 10 in order. How many ways can you show each number?
3. Show the same number of dogs as you are showing.
4. Show one less / one more e.g. show me one less than 3, one more than 5….etc. How did you work it out? Can you do it without counting?
5. More/fewer than I am showing. Explain your answer. e.g. How many dogs am I showing? Can you show me more dogs? Can you show me fewer dogs?
Subitising
Once children can confidently count the dogs by pointing at each one and not making mistakes, encourage them to subitise i.e. to recognise amounts without counting. When they see 3 dogs, they can say it is 3 without counting 1, 2, 3. They should be able to do this with numbers up to 5.
Counting in 2s
Use the flik-flaks as a quick way to practise counting in 2s.
Show children the flik-flak and ask:
How many dogs can you see? How many eyes can you see? How many ears can you see? How did you count them?
Fold the flik-flak:
How many eyes can you see now? How did you count them? Did you count in 2s? Did you say “3 lots of 2”?
Challenge:
I can see 10 eyes. Show me 10 eyes with your flik-flak. How many dogs are there if there are 10 eyes?
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# Math 201: Homework 7 Solutions
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1 Math 201: Homework 7 Solutions 1. ( 5.2 #4) (a) The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. (b) The factors of 81 are 1, 3, 9, 27, and 81. (c) The factors of 62 are 1, 2, 31, and 62. (d) The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. (e) The factors of 326 are 1, 2, 163, and 326. (f) The factors of 242 are 1, 2, 11, 22, 121, and ( 5.2 #10) We begin by drawing out the list of numbers as given. Since the square root of 150 is about 12, we know we only need to cross out multiples of primes through 12. So we cross out all multiples of the primes 2, 3, 5, 7, and 11, and we are left with all the prime numbers through 150: We found the new primes 127, 131, 137, 139, and ( 5.2 #12) (a) The prime factorization of 16 is = 2 4. (b) The prime factorization of 27 is = 3 3. (c) The prime factorization of 52 is = (d) The prime factorization of 75 is = (e) The prime factorization of 112 is =
2 (f) The prime factorization of 125 is = ( 5.2 #14) We can find the top number using only the given numbers. Since the Fundamental Theorem of Arithmetic guarantees that every number has a unique prime factorization, and the factor tree we are given tells us the prime factorization of the top number, then we need only multiply the numbers up the tree to find the omitted numbers ( 5.2 #16) (a) The square root of 16 is 4, since 4 2 = 16. (b) The square root of 81 is 9, since 9 2 = 81. (c) The square root of 100 is 10, since 10 2 = 100. (d) The square root of 144 is 12, since 12 2 = ( 5.2 #18) (a) 417 is composite since, for instance, 3 divides it (note that 3 ( )). (b) 729 is composite since, for instance, 3 divides it (note that 3 ( )). (c) 1571 is prime. Note that we need only check primes through 40, since (d) 4587 is composite since, for instance 3 divides it (note that 3 ( )). (e) 35, 721 is composite since, for instance, 3 divides it (note that 3 ( )). (f) 87, 451 is composite since, for instance, 7 divides it (you can do our little splitting trick a few times, or you can just solve 7 ) 87, 451). 7. ( 5.2 #22) (a) The prime factorization of 432 is = (b) The prime factorization of 1568 is = (c) The prime factorization of 2079 is = (d) The prime factorization of 6318 is = (e) The prime factorization of 6048 is = (f) The prime factorization of 8281 is =
3 8. ( 5.2 #24) (a) False. The number 2 is even, but it is also prime. (b) False. 6 is not prime. The prime factorization of 60 would be (c) False. Both 2 and 5 are prime, and 2 5 = 10 is even. (d) True. Since the square root of 1393 is about 37, then we need only check for prime factors less than or equal to ( 5.2 #26) This question is equivalent to asking for all the factors of 24, which are 1, 2, 3, 4, 6, 8, 12, and 24. That is, you could divide the class up into 24 groups of 1 person each, 12 groups of 2 people each, 8 groups of 3 people each, 6 groups of 4 people each, 4 groups of 6 people each, 3 groups of 8 people each, 2 groups of 12 people each, or 1 big group of 24 people. 10. ( 5.2 #40) (a) We first write each given base in its prime factorization, then combine using properties of exponents = (2 2 3) 9 ( ) 15 (13 2 ) 5 = (2 2 ) (2 2 ) 15 (3 2 ) 15 (13 2 ) 5 Power of a Product Property = Power of a Power Property = Product of Powers Property (b) We first write each given base in its prime factorization, then combine using properties of exponents. 11. ( 5.3 #4) = (2 4 ) 13 (2 7) 7 (3 4 ) 14 = (2 4 ) (3 4 ) 14 Power of a Product Property = Power of a Power Property = Product of Powers Property (a) The prime factorizations of 9 and of 15 are 9 = 3 2 and 15 = 3 5 The only common factor is a single 3. Hence, GCF(9, 15) = 3. (b) The prime factorizations of 13 and of 20 are 13 = 13 and 20 = These have no common prime factor. Hence, GCF(13, 20) = 1. 3
4 (c) The prime factorizations of 6 and of 8 are 6 = 2 3 and 8 = 2 3 The only common factor is a single 2. Hence, GCF(6, 8) = 2. (d) The prime factorizations of 21 and of 63 are 21 = 3 7 and 63 = The common factors are a 3 and a 7. Hence, GCF(21, 63) = 3 7 = 21. (e) The prime factorizations of 36 and of 54 are 36 = and 54 = The common factors are a single 2 and two 3s. Hence, GCF(36, 54) = = 18. (f) The prime factorizations of 100 and of 360 are 12. ( 5.3 #6) (a) (b) 100 = and 360 = The common factors are two 2s and a single 5. Hence, GCF(100, 360) = = 20. GCF(6, 26) = GCF(6, 26 6) = GCF(6, 20) = GCF(6, 20 6) = GCF(6, 14) = GCF(6, 14 6) = GCF(6, 8) = GCF(6, 8 6) = GCF(6, 2) = 2 GCF(8, 28) = GCF(8, 28 8) = GCF(8, 20) = GCF(8, 20 8) = GCF(8, 12) = GCF(8, 12 8) = GCF(8, 4) = 4 4
5 (c) GCF(40, 56) = GCF(40, 56 40) = GCF(40, 16) = GCF(40 16, 16) = GCF(24, 16) = GCF(24 16, 16) = GCF(8, 16) = 8 (d) GCF(35, 42) = GCF(35, 42 35) = GCF(35, 7) = 7 (e) GCF(34, 85) = GCF(34, 85 34) = GCF(34, 51) = GCF(34, 51 34) = GCF(34, 17) = 17 (f) GCF(32, 55) = GCF(32, 55 32) = GCF(32, 23) = GCF(32 23, 23) = GCF(9, 23) = GCF(9, 23 9) = GCF(9, 14) = GCF(9, 14 9) = GCF(9, 5) = GCF(9 5, 5) = GCF(4, 5) = GCF(4, 5 4) = GCF(4, 1) = 1 5
6 13. ( 5.3 #8) (a) 2R48 84 ) 216 1R36 48 ) 84 1R12 36 ) 48 3R0 12 ) 36 So GCF(84, 216) = 12. (b) 1R ) 192 1R48 72 ) 120 1R24 48 ) 72 2R0 24 ) 48 (c) (d) (e) (f) (g) (h) So GCF(120, 192) = 24. So GCF(63, 84) = 21. 1R42 98 ) 140 So GCF(98, 140) = 14. So GCF(45, 90) = 45. So GCF(160, 800) = 160. So GCF(56, 588) = 28. 1R ) 935 So GCF(544, 935) = 17. 1R21 63 ) 84 10R28 56 ) 588 1R68 85 ) R14 42 ) 98 2R0 45 ) 90 5R0 160 ) 800 1R ) 544 3R0 21 ) 63 2R0 28 ) 56 1R17 68 ) 85 3R0 14 ) 42 2R ) 391 4R0 17 ) 68
7 14. ( 5.3 #10) (a) The prime factorizations of 3 and of 8 are 3 = 3 and 8 = 2 3 There are no common factors, so the diagram looks like Factors of 3 Factors of Hence, LCM(3, 8) = 3 (2 2 2) = 24. (b) The prime factorizations of 9 and of 12 are 9 = 3 2 and 12 = These only share a factor of 3, so the diagram looks like Factors of 9 Factors of Hence, LCM(9, 12) = 3 3 (2 2) = 36. (c) The prime factorizations of 12 and of 36 are 12 = and 36 = These share two 2s and a 3, so the diagram looks like Factors of 12 Factors of Hence, LCM(12, 36) = (2 2 3) 3 = 36. (d) The prime factorizations of 2 and of 9 are 2 = 2 and 9 = 3 2 7
8 There are no common factors, so the diagram looks like Factors of 2 Factors of Hence, LCM(2, 9) = 2 (3 3) = 18. (e) The prime factorizations of 24 and of 45 are 24 = and 45 = These only share the factor 3, so the diagram looks like Factors of 24 Factors of Hence, LCM(24, 45) = (2 2 2) 3 (3 5) = 360. (f) The prime factorizations of 15 and of 25 are 15 = 3 5 and 25 = 5 2 These share only a factor of 5, so the diagram looks like Factors of 15 Factors of Hence, LCM(15, 25) = = ( 5.3 #12) (a) So LCM(3, 6) = 6. 8
9 (b) (c) So LCM(4, 5) = So LCM(2, 7) = ( 5.3 #14) (a) The prime factorizations of 15 and of 20 are 15 = 3 5 and 20 = These share only a factor of 5, so the diagram looks like Factors of 15 Factors of Hence, GCF(15, 20) = 5, and LCM(15, 20) = 3 5 (2 2) = 60. (b) The prime factorizations of 50 and of 100 are 50 = and 100 = These share one factor of 2 and two factors of 5, so the diagram looks like Factors of 50 Factors of Hence, GCF(50, 100) = = 50, and LCM(50, 100) = (2 5 5) 2 =
10 (c) The prime factorizations of 24 and of 30 are 24 = and 30 = These share a factor of 2 and a factor of 3, so the diagram looks like Factors of 24 Factors of Hence, GCF(24, 30) = 2 3 = 6, and LCM(24, 30) = (2 2) (2 3) 5 = ( 5.3 #20) If the gift bags are to be identical, then you must use the same number of candles and gift cards in each bag. This is equivalent to asking for GCF(18, 24) = 6. Hence, the largest number of gift bags you can make is 6, where each contains 3 candles and 4 gift cards. 18. ( 5.3 #22) (a) The prime factorizations of 12, 18, and 24 are 12 = and 18 = and 24 = The only factors that are common to all three numbers are a single 2 and a single 3. Hence, the greatest common factor of 12, 18, and 24 is 2 3 = 6. (b) The prime factorizations of 24, 36, and 60 are 24 = and 36 = and 60 = The only factors that are common to all three numbers are two 2s and a single 3. Hence, the greatest common factor of 24, 36, and 60 is = 12. (c) The prime factorizations of 26, 52, and 78 are 19. ( 5.3 #24) 26 = 2 13 and 52 = and 78 = The only factors that are common to all three numbers are a single 2 and a single 13. Hence, the greatest common factor of 26, 52, and 78 is 2 13 = 26. (a) We ll compute the least common multiple of 4, 6, and 12 by looking at their multiples and identifying the first common one. The multiples of 4 are 4, 8, 12, 16, 20, 24,.... The multiples of 6 are 6, 12, 18, 24, 30, 36,.... The multiples of 12 are 12, 24, 36, 48, 60, 72,.... The first common multiple among these is 12, so this is our answer. 10
11 (b) We ll compute the least common multiple of 5, 16, and 20 by looking at their multiples and identifying the first common one. The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, The multiples of 16 are 16, 32, 48, 64, 80, 96,.... The multiples of 20 are 20, 40, 60, 80, 100, 120,.... The first common multiple among these is 80, so this is our answer. (c) We ll compute the least common multiple of 24, 36, and 48 by looking at their multiples and identifying the first common one. The multiples of 24 are 24, 48, 72, 96, 120, 144,.... The multiples of 36 are 36, 72, 108, 144, 180,.... The multiples of 48 are 48, 96, 144, 192, 240,.... The first common multiple among these is 144, so this is our answer. 20. ( 5.3 #32) We would like to know how far each of the runners will have run when they pass the starting line at the same time. We first find LCM(75, 90), which will tell us how much time has elapsed when they pass the starting line at the same time. The prime factorizations of 75 and 90 are 75 = and 90 = These share a factor of 3 and a factor of 5, so the diagram looks like Factors of 75 Factors of So LCM(75, 90) = 5 (3 5) (2 3) = 450. That is, 450 seconds after starting, the runners will pass each other at the starting line. The first runner will have run = 6 laps around the track, and the second will have run = 5 laps around the track. 11
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TEXT
Describe the end behavior of power functions
Identify power functions
In order to better understand the bird problem, we need to understand a specific type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. (A number that multiplies a variable raised to an exponent is known as a coefficient.)
As an example, consider functions for area or volume. The function for the area of a circle with radius is
$A\left(r\right)=\pi {r}^{2}$
and the function for the volume of a sphere with radius r is
$V\left(r\right)=\frac{4}{3}\pi {r}^{3}$
Both of these are examples of power functions because they consist of a coefficient, $\pi$ or $\frac{4}{3}\pi,$ multiplied by a variable r raised to a power.
A General Note: Power Function
A power function is a function that can be represented in the form
$f\left(x\right)=k{x}^{p}$
where k and p are real numbers, and k is known as the coefficient.
Q & A
Is $f\left(x\right)={2}^{x}$ a power function?
No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function.
Example 1: Identifying Power Functions
Which of the following functions are power functions?
$\begin{cases}f\left(x\right)=1\hfill & \text{Constant function}\hfill \\ f\left(x\right)=x\hfill & \text{Identify function}\hfill \\ f\left(x\right)={x}^{2}\hfill & \text{Quadratic}\text{ }\text{ function}\hfill \\ f\left(x\right)={x}^{3}\hfill & \text{Cubic function}\hfill \\ f\left(x\right)=\frac{1}{x} \hfill & \text{Reciprocal function}\hfill \\ f\left(x\right)=\frac{1}{{x}^{2}}\hfill & \text{Reciprocal squared function}\hfill \\ f\left(x\right)=\sqrt{x}\hfill & \text{Square root function}\hfill \\ f\left(x\right)=\sqrt[3]{x}\hfill & \text{Cube root function}\hfill \end{cases}$
Solution
All of the listed functions are power functions.
The constant and identity functions are power functions because they can be written as $f\left(x\right)={x}^{0}$ and $f\left(x\right)={x}^{1}$ respectively.
The quadratic and cubic functions are power functions with whole number powers $f\left(x\right)={x}^{2}$ and $f\left(x\right)={x}^{3}.$
The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as $f\left(x\right)={x}^{-1}$ and $f\left(x\right)={x}^{-2}.$
The square and cube root functions are power functions with fractional powers because they can be written as $f\left(x\right)={x}^{1/2}$ or $f\left(x\right)={x}^{1/3}.$
Try It 1
Which functions are power functions?
$\begin{cases}f\left(x\right)=2{x}^{2}\cdot 4{x}^{3}\hfill \\ g\left(x\right)=-{x}^{5}+5{x}^{3}-4x\hfill \\ h\left(x\right)=\frac{2{x}^{5}-1}{3{x}^{2}+4}\hfill \end{cases}$
Solution
Identify end behavior of power functions
Figure 2 shows the graphs of $f\left(x\right)={x}^{2},g\left(x\right)={x}^{4}$ and $h\left(x\right)={x}^{6},$ which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin.
Figure 2. Even-power functions
To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol $\infty$ for positive infinity and $-\infty$ for negative infinity. When we say that "x approaches infinity," which can be symbolically written as $x\to \infty,\\$ we are describing a behavior; we are saying that x is increasing without bound.
With the even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as $x$ approaches positive or negative infinity, the $f\left(x\right)$ values increase without bound. In symbolic form, we could write
$\text{as }x\to \pm \infty , f\left(x\right)\to \infty\\$
Figure 3 shows the graphs of $f\left(x\right)={x}^{3},g\left(x\right)={x}^{5},\text{and}h\left(x\right)={x}^{7},$ which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin.
Figure 3. Odd-power function
These examples illustrate that functions of the form $f\left(x\right)={x}^{n}$ reveal symmetry of one kind or another. First, in Figure 2 we see that even functions of the form $f\left(x\right)={x}^{n}\text{, }n\text{ even,}$ are symmetric about the y-axis. In Figure 3 we see that odd functions of the form $f\left(x\right)={x}^{n}\text{, }n\text{ odd,}$ are symmetric about the origin.
For these odd power functions, as x approaches negative infinity, $f\left(x\right)$ decreases without bound. As x approaches positive infinity, $f\left(x\right)$ increases without bound. In symbolic form we write
$\begin{cases}\text{as } x\to -\infty , f\left(x\right)\to -\infty \\ \text{as } x\to \infty , f\left(x\right)\to \infty \end{cases}$
The behavior of the graph of a function as the input values get very small $(x\to -\infty)$ and get very large $(x\to \infty)$ is referred to as the end behavior of the function. We can use words or symbols to describe end behavior.
The table below shows the end behavior of power functions in the form $f\left(x\right)=k{x}^{n}$ where $n$ is a non-negative integer depending on the power and the constant.
Even power Odd power
Positive constant
k > 0
Negative constant
k < 0
How To: Given a power function $f\left(x\right)=k{x}^{n}$ where n is a non-negative integer, identify the end behavior.
1. Determine whether the power is even or odd.
2. Determine whether the constant is positive or negative.
3. Use Figure 4 to identify the end behavior.
Example 2: Identifying the End Behavior of a Power Function
Describe the end behavior of the graph of $f\left(x\right)={x}^{8}.$
Solution
The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x approaches infinity, the output (value of $f\left(x\right)$ ) increases without bound. We write as $x\to \infty , f\left(x\right)\to \infty.$ As x approaches negative infinity, the output increases without bound. In symbolic form, as $x\to -\infty , f\left(x\right)\to \infty.\\$ We can graphically represent the function as shown in Figure 5.
Figure 4
Example 3: Identifying the End Behavior of a Power Function.
Describe the end behavior of the graph of $f\left(x\right)=-{x}^{9}.$
Solution
The exponent of the power function is 9 (an odd number). Because the coefficient is –1 (negative), the graph is the reflection about the x-axis of the graph of $f\left(x\right)={x}^{9}.$ The graph shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write
Figure 5. $\begin{cases}\text{as } x\to -\infty , f\left(x\right)\to \infty \\ \text{as } x\to \infty , f\left(x\right)\to -\infty \end{cases}$
Analysis of the Solution
We can check our work by using the table feature on a graphing utility.
x f(x)
–10 1,000,000,000
–5 1,953,125
0 0
5 –1,953,125
10 –1,000,000,000
We can see from the table above that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value).
Try It 2
Describe in words and symbols the end behavior of $f\left(x\right)=-5{x}^{4}.$
Solution
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# Rotation angle and angular velocity (Page 2/9)
Page 2 / 9
$v=\frac{\text{Δ}s}{\text{Δ}t}\text{.}$
From $\text{Δ}\theta =\frac{\text{Δ}s}{r}$ we see that $\text{Δ}s=r\text{Δ}\theta$ . Substituting this into the expression for $v$ gives
$v=\frac{r\text{Δ}\theta }{\text{Δ}t}=\mathrm{r\omega }\text{.}$
We write this relationship in two different ways and gain two different insights:
The first relationship in states that the linear velocity $v$ is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest $r$ ), as you might expect. We can also call this linear speed $v$ of a point on the rim the tangential speed . The second relationship in can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed $v$ of the car. See [link] . So the faster the car moves, the faster the tire spins—large $v$ means a large $\omega$ , because $v=\mathrm{r\omega }$ . Similarly, a larger-radius tire rotating at the same angular velocity ( $\omega$ ) will produce a greater linear speed ( $v$ ) for the car.
## How fast does a car tire spin?
Calculate the angular velocity of a 0.300 m radius car tire when the car travels at $\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}$ (about $\text{54}\phantom{\rule{0.25em}{0ex}}\text{km/h}$ ). See [link] .
Strategy
Because the linear speed of the tire rim is the same as the speed of the car, we have $v=\text{15.0 m/s}.$ The radius of the tire is given to be $r=\text{0.300 m}.$ Knowing $v$ and $r$ , we can use the second relationship in to calculate the angular velocity.
Solution
To calculate the angular velocity, we will use the following relationship:
$\omega =\frac{v}{r}\text{.}$
Substituting the knowns,
$\omega =\frac{\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}}{0\text{.}\text{300}\phantom{\rule{0.25em}{0ex}}\text{m}}=\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{rad/s.}$
Discussion
When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity
$\omega =\left(\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)/\left(1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{m}\right)=\text{12}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{rad/s.}$
Both $\omega$ and $v$ have directions (hence they are angular and linear velocities , respectively). Angular velocity has only two directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in [link] .
## Take-home experiment
Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities.
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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# How to Make 3D Polygons With Triangles
Geometry teachers and textbooks love to encourage students to draw pictures to help think through situations. Drawings are particularly useful when it comes to surface area and volume polygon problems. Yet, most students have difficulty making the three-dimensional, or 3D, figures they need, especially when triangles are involved. Learning how to successfully draw pyramids and triangular prisms provides a foundation for solving geometry problems that use them. Use a ruler or straight edge at first until you become skilled at drawing the lines needed.
#### Things You'll Need
• Ruler or other straight edge
• Writing utensil
1. ## Pyramids
• 1
Draw two parallel diagonal lines of the same length.
• 2
Connect the tops of the two lines and the bottoms of the two lines with a straight line for each set. Use the ruler or straight edge to get each line in place and then trace along the edge. This forms what looks like a parallelogram but actually represents a square base.
• 3
Draw a dot straight above the center of the base.
• 4
Line the edge of the ruler up so that it runs through both one of the corners of the base and the dot above. Trace along the edge, then repeat for each corner. This creates the four triangle faces of the pyramid.
• 5
Label any dimensions, such as side length or height, given in the problem.
## Triangular Prism
• 6
Draw a triangle. Pay attention to the directions and, if they call for a right triangle, make one of the corners an "L" shape.
• 7
Place the ruler against one of the corners of the triangle, then use it to draw a diagonal line out away from the triangle. Measure the line as you draw it.
• 8
Draw two more lines from the other two corners. Make them parallel with the first line and the same exact length. So, if the first line you drew was 3 inches, make the other 2 lines 3 inches as well.
• 9
Connect the ends of the three lines. This forms the triangular base at the other end of the prism. The lines that you make should be parallel to the sides of the original triangle.
• 10
Label any lengths of the faces, which are rectangles, or bases that the problem gives you.
Related Searches
## References
• Photo Credit Ablestock.com/AbleStock.com/Getty Images
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Fibonacci-Based Arithmetic
[This is the final Wythoff’s game post. See the previous ones here: #1, #2, #3, #4, #5, #6. We’ll have a few shorter, 1- or 2-post topics starting next week.]
In our standard, base-10 number system, each position represents a specific power of 10: the number 275 in base 10 means that I have five 1s, seven 10s, and two 100s. Similarly, in the base-2 system, each position represents a different power of 2: 10011 in base 2 says I have a 1, a 2, and a 16, but no 4s or 8s.[1] Instead of using powers of 10 or 2, what happens if we use Fibonacci numbers?
We can find the Fibonacci numbers by starting[2] at $$a_1=1$$ and $$a_2=2$$ and then computing each term as the sum of the two previous ones, $$a_n=a_{n-1}+a_{n-2}$$:
1, 2, 3, 5, 8, 13, 21, 34, 55, …
Let’s use these as our position values in our so-called Fibonacci base, using just the digits 0 and 1. For example, 1010011F stands for $$a_7+a_5+a_2+a_1 = 21+8+2+1 = 32$$.
Notice that $$a_2+a_1=a_3$$, so we can write 32 instead as $$a_7+a_5+a_3$$, meaning 32=1010100F. In much the same way, any time we have a Fibonacci representation of a number, we can repeatedly use the identity $$a_n=a_{n-1}+a_{n-2}$$ to ensure that we never use consecutive Fibonacci numbers:
Theorem (Zeckendorf): Every positive integer can be expressed in base Fibonacci with no adjacent 1s. Furthermore, this representation is unique.
The “base-Fibonacci representation” of a number usually refers to its Zeckendorf representation as in this theorem. There’s a simple greedy method to find a number’s Zeckendorf representation: use the largest Fibonacci number less than (or equal to) your number, and repeat on the remainder. For example, to represent 46 in this way, check that the largest Fibonacci number below 46 is $$a_8=34$$, leaving 12 as the remainder. The largest Fibonacci number below (or equal to) 12 is $$a_5=8$$, with a remainder of 4. Now we have to use $$a_3=3$$, and finally we’re left with $$1=a_1$$. So $$46=a_8+a_5+a_3+a_1=10010101_F$$ is the Zeckendorf representation of 46.
But why discuss base Fibonacci now? What does this have to do with Wythoff’s game? Zeckendorf representations provide another simple way to characterize the losing positions in Wythoff’s game! Here’s a beautiful result:
Theorem: If $$a \le b$$ are positive integers, then $$(a,b)$$ and $$(b,a)$$ are losing position in Wythoff’s game precisely when:
• The Zeckendorf representation of a ends in an even number of 0s, and
• The Zeckendorf representation of b is the same as that of a except for one more 0 on the end.
For example, 11=10100F ends in two 0s (two is even), and 18=101000F has the same representation except for an extra 0 at the end, so (11,18) and (18,11) are Wythoff losing positions. Unlike the formula using multiples of $$\phi$$ and $$\phi^2$$ that we proved in the last few posts, this provides a method we could actually compute (without a calculator) on reasonably-sized grids when playing with friends (or enemies)! After all we’ve learned about Wythoff’s game, we finally have a way to win it!
Our Wythoff’s game saga is now ending here on Three-Cornered Things, but here are a few recommended resources if you wish to further explore this rich topic on your own:
Notes
1. If these concepts are unfamiliar, see this previous post on number bases. []
2. Careful: these indices are shifted from the other common convention that starts the Fibonacci numbers at $$F_1=1$$ and $$F_2=1$$, which is why I’m using as instead of Fs. []
2 thoughts on “Fibonacci-Based Arithmetic”
1. Mahdi Belbasi says:
the proof has 2 sections :
i) the existence of Zeckendorf representation,
ii) the uniqueness of Zeckendorf representation.
proof of i)
suppose that the given number is n, then you can find the less or equal Fibonacci number to n, named F(j), so : F(j) <= n < F(j+1) (because the F(j) is the bigger Fibonacci number smaller than n). we write that : n = F(j) + a, we say a, as a reminder, while reminder is not zero, we continue. we claim that a is less than F(j-1), because if not, n would be bigger than F(j) + F(j-1), and we have F(j+1) = F(j)+F(j-1), this is contradiction with the main inequality, so the F(j-1) is absent in the representation of a, so there is no consecutive 1 in the representation, and we do so for reminder by induction and there will be no 1 adjacent to each other. This was the proof of existence.
proof of ii)
we do this by contradiction, suppose there is two set of Fibonacci numbers named T and S, such that the some of both are equal to n. We construct two new sets named S' and T' as below :
S' = S\T and T' = T\S (means delete the same numbers from each other)
so here S' and T' has no intersection, we will show that one of them is empty, imagine non of them is empty; suppose that F(s) is the biggest number of S' and F(t) is the biggest number of T', so by the definition of S' and T', we have that F(s) and F(t) are not equal. Without loss of generality we imagine that F(s) < F(t).
F(s-1) is absent is S' because of Zeckendorf representation defenition, so at most the F(s-2) will be in S', then the maximum sum of S' would be F(s) + F(s-2) + … which is less than F(s+1)
so sum of elements of S' is less than F(s+1) and by the way is at least F(t), this is contradiction, so on of the S' or T' is empty, because the sum of them are equal so the other should be empty too, so S and T are equal :), so this means the set in unique.
End,
2. Aww, no proof? I guess I’ll just have to try to prove it myself. =)
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Addition Year 2 (aged 6-7)
Add whole tens, add two 2-digit numbers, add three single digit numbers, solve problems involving addition.
Mental addition continues to be a key area of maths in Year 2. A great deal of emphasis will be placed upon it.
During the year children will be expected to:
•   add mentally a 2-digit number and ones (e.g. 25 + 6 =)
•   add mentally a 2-digit number and whole tens (e.g. 25 + 20 =)
•   add mentally two 2-digit numbers (e.g. 23 + 32 =)
•   add three small numbers (e.g. 2 + 3 + 7 =)Â
•   know by heart pairs of numbers that total 20
•   respond rapidly to oral questions such as ‘What is the total of 23 and 10?’Â
•   solve missing number problems such as 12 + ? = 17.Â
Â
•   understand addition as combining sets to make a total and that it can be done in any order
•   understand that addition is the inverse of subtraction and use this to check answers
•   continue to solve one-step problems that involve addition using concrete objects and pictorial representations
•   begin to add using written methods in columns
There are a number of skills to help with addition that children need to be made aware of. These include:
•   putting the larger number first (e.g. arrange 6 + 45 as 45 + 6 and count on)
•   looking for pairs that make 10 when adding three small numbers (e.g. 3 + 5 + 7 = 10 + 5 = 15)Â
•   being able to partition numbers to make them easy to add (e.g. 12 + 23 = 12 + 20 + 3 = 32 + 3)
•   using knowledge of doubles to make addition easy  (6 + 7 is double 6 plus 1)
•   mentally adding 9 by adding 10 and subtracting 1
Our Year 2 addition worksheets are second to none, combining clear layout, lively graphics and perhaps most importantly, each set concentrates on just one skill. There are usually four pages for each set. If your child finds it tricky it is well worth using more of the sheets to consolidate learning.
Also, don’t forget our superb addition games which provide unlimited practice , re-inforce learning and are a fun change.
Â
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Matrices
Solving system of equations using the inverse matrix method
The inverse matrix method uses the inverse of a matrix to help solve a system of equations.
Consider the simultaneous equations:
Provided you understand how matrices are multiplied together you will realise that these can
be written in matrix form as:
writing:
A=, X= and B=
we have A·X=B.
Given AX = B we can multiply both sides by the inverse of A, provided this exists, to give:
A-1·A·X=A-1·B.
But A-1A = I, the identity matrix, so
X=A-1·B
$\fs2\left.\begin{matrix}-x+y=-5\\2x-3y=3\\-z=2\\nd{matrix}\right\}\Rightarrow\begin{pmatrix}-1&1&0\\2&-3&0\\0&0&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-5\\3\\2\end{pmatrix}$
$\fs2A^{-1}=\begin{pmatrix}-3&-1&0\\-2&-1&0\\0&0&-1\end{pmatrix}\;Then\;X=\begin{pmatrix}-3&-1&0\\-2&-1&0\\0&0&-1\end{pmatrix}\begin{pmatrix}-5\\3\\2\end{pmatrix}\Rightarrow\;X=\begin{pmatrix}12\\7\\-2\end{pmatrix}$
Solve using the inverse matrix method
$\left.\begin{matrix}-3x-2y+5z=-4\\x+3y-3z=5\\x+5y-4z=5\\nd{matrix}\right\}$ X =
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1 / 18
# Solving for the Discontinuities of Rational Equations - PowerPoint PPT Presentation
Solving for the Discontinuities of Rational Equations. Review: 3 Types of Discontinuities. Vertical Asymptotes (VAs) Horizontal Asymptotes (HAs) Holes. Degree. The greatest exponent of an expression Examples: f(x) = x 6 – x 2 + 3 f(x) = x 4 – x 9 + x 11 – x 2 + 5 f(x) = 8x + 4
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Solving for the Discontinuities of Rational Equations
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## Solving for the Discontinuities of Rational Equations
### Review: 3 Types of Discontinuities
• Vertical Asymptotes (VAs)
• Horizontal Asymptotes (HAs)
• Holes
### Degree
• The greatest exponent of an expression
• Examples:
• f(x) = x6 – x2 + 3
• f(x) = x4 – x9 + x11 – x2 + 5
• f(x) = 8x + 4
• f(x) = 7
• The coefficient of the term with the largest degree
• Examples:
• f(x) = x6 – x2 + 3
• f(x) = x4 – x9 + x11 – x2 + 5
• f(x) = 8x + 4
• f(x) = 7
### Horizontal Asymptotes Investigation
• Remember:
• The horizontal asymptote describes how the graph behaves AT ITS ENDS
• Look for the graph to taper to the same y-value on both ends of the graph
• Look for dashed, horizontal lines
• We DON’T DRAW dashed lines on the X-AXIS or the Y-AXIS!!!
### Investigation Conclusion Questions:
4. What observations can you make about a rational equation’s horizontal asymptote when the degree of the numerator and the denominator are the same?
8. What observations can you make about a rational equation’s horizontal asymptote when the degree of the denominator is greater than the degree of the numerator?
12. What observations can you make about a rational equation’s horizontal asymptote when the degree of the numerator is greater than the degree of the denominator?
### *Horizontal Asymptotes
• Depend on the degree of the numerator and the denominator
• Degree of Numerator < Degree of Denominator
• HA: y = 0
• Degree of Numerator = Degree of Denominator
• HA: y = ratio of leading coefficients
• Degree of Numerator > Degree of Denominator
• HA: doesn’t exist
### Examples
• For problems 1 – 4 on the Introduction to Solving Rational Equations Practice, solve for the horizontal asymptote.
1.2.
3.4.
### Solving for Vertical Asymptotes and Holes
• Always factor the numerator and the denominator 1st!
• Identify linear factors in the denominator
• Figure out where the linear factors in the denominator occur the most to decide if you have a vertical asymptote or a hole
• Set the linear factors from step 2 equal to zero and solve for x.
### Solving for Vertical Asymptotes and Holes, cont.
Does the linear factor:
### Example #4
• Complete problems 5 – 12 on the Introduction to Solving for the Discontinuities of Rational Equations Practice handout . BE PREPARED TO SHARE YOUR ANSWERS!!!
• Complete problems 13 – 18.
13.14.
15.16.
17.18.
### Exit Ticket
• Does have a HA? (If yes, what is it?) Why?
• Does have VAs and/or holes? (If yes, what are they?) Why?
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# How to Calculate the Surface Area of cone with radius of 1.8cm and height of 2.4cm?
##### 1 Answer
Mar 1, 2018
$27.1296 c {m}^{2}$
#### Explanation:
The formula for the surface area of a cone is $S a = \pi r l + \pi {r}^{2}$
$S a$ = surface area
$r$ = radius
$l$ = slant height
From the given information, we already have some of the numbers. We know that
$r$ = 1.8 cm
$\pi$ = 3.14
$h$ = 2.4 cm
However, we do not know the slant height of the triangle. Using this picture below as a guide, you can fill in $r$ and $h$
Using the Pythagorean Theorem (${a}^{2} + {b}^{2} = {c}^{2}$), we can find $l$
We can say that ${h}^{2} + {r}^{2} = {l}^{2}$ by using substitution. So lets plug the numbers into this equation.
${2.4}^{2} + {1.8}^{2} = {l}^{2}$
$5.76 + 3.24 = {l}^{2}$
$9 = {l}^{2}$
$\sqrt{9} = l$
$3 = l$
Yay! We've solved the first part. Now that we know all our variables, we can plug that into the first equation.
$S a = \pi r l + \pi {r}^{2}$
$S a = 3.14 \left(1.8\right) \left(3\right) + 3.14 \left({1.8}^{2}\right)$
$S a = 16.956 + 10.1736$
$S a = 27.1296 c {m}^{2}$
|
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# Binary exponentiation (Power in log N)
#### Algorithms List of Mathematical Algorithms
Get this book -> Problems on Array: For Interviews and Competitive Programming
Reading time: 30 minutes | Coding time: 5 minutes
Binary exponentiation is an algorithm to find the power of any number N raise to an number M (N^M) in logarithmic time O(log M). The normal approach takes O(M) time provided multiplication takes constant time.
In reality, multiplication takes O(log N) time and hence, Binary exponentiation takes O(logN * logM) time and the normal approach takes O(M * logN) time.
In summary, the idea is as follows:
``````A^N = 1 if N = 0
A^N = (A^((N-1)/2))^2 * A if N is odd
A^N = (A^(N/2))^2 if N is even
``````
The key is that multiplication can be divided into smaller parts where each part is a multiple of 2. Multiple of 2 is optimal as numbers in computer systems are stored in base 2 and squares are a single instruction as it involves shifthing bits left one position. At the same time, any number can be represented as a sum of numbers which are powers of two.
Consider the example of 5^13.
If we follow the ordinary method, we will have to do 12 multiplications (5 * 5 * ... * 5) which is a costly operation. This will improve with Binary Exponentiation.
Let us represent 13 as a sum of power of two.
``````13 = 1101 = 2^3 + 2^2 + 0 + 2^0 = 8 + 4 + 0 + 1
``````
Another point, we need to note is the following:
``````If B1 + B2 = B, then
A ^ B = A ^ (B1+B2) = A ^ B1 * A ^ B2
``````
Similarly, for 5^13, we get the following:
``````5^13 = 5^8 * 5^4 * 5^1
``````
Now, the powers like 5^8 can be calculated using left shift and the previous power 5^4 can be used for this.
``````A^N << 1 = A^2N
``````
Now, to calculate 5^8, will need 3 left shift.
``````5 = 5
5^2 = 5 << 1 = 25
5^4 = 5^2 << 1 = 625
5^8 = 5^4 << 1 = 390625
``````
Hence, we needed 3 left shift operations to calculate all powers of 5 upto 8.
With this, we are able to calculate 5^13 as follows:
``````5^13 = 5^(8+4+1)
5^13 = 5^8 * 5^4 * 5^1
5^13 = 390625 * 625 * 5
5^13 = â€1220703125‬
``````
Hence, we have been able to reduce 12 multiplications to 3 multiplication operations.
## Pseudocode
Following is the pseudocode for the iterative version of Binary Exponentiation method:
``````// N^M
power( int N, int M)
{
int power = N, sum = 1;
while(M > 0)
{
if((M & 1) == 1)
{
sum *= power;
}
power = power * power;
M = M >> 1;
}
return sum;
}
``````
Following is the pseudocode of the recursive versionn of Binary Exponentiation method:
``````// N^M
int power( int N, int M)
{
if(M == 0)
return 1;
int recursive = power(N, M/2);
if(M % 2 == 0)
return recursive * recursive;
return recursive * recursive * N;
}
``````
## Complexity
The basic brute force approach takes O(M) multiplications to calculate N^M.
With our optimized binary exponentiation approach, we do the following operations:
• O(log M) multiplication to keep track of powers
• Maximum O(log M) multiplications to get final result
• (log M) left shift operations
Hence, from the point of time complexity, this is O(log M) multiplications. Usually, multiplication of two numbers say N and N takes O( logN * logN) considering it has logN digits. In most cases, multiplication is considered to be a constant time operation.
Hence, in reality, following is the actual time complexity comparison:
• Brute force: (M * logN * logN)
• Binary exponentiation: (logM * logN * logN)
This improves the performance greatly.
## Implementation
Following is the iterative approach Implementation in Java:
``````// Binary Exponentiation
// Part of OpenGenus
class opengenus
{
// N^M
static int power( int N, int M)
{
int power = N, sum = 1;
while(M > 0)
{
if((M & 1) == 1)
{
sum *= power;
}
power = power * power;
M = M >> 1;
}
return sum;
}
public static void main (String[] args)
{
int N = 5, M = 13;
System.out.println(power(N, M));
}
}
``````
Output:
``````1220703125
``````
Following is the Recursive approach Implementation in Java:
``````// Binary Exponentiation
// Part of OpenGenus
class opengenus
{
// N^M
static int power( int N, int M)
{
if(M == 0)
return 1;
int recursive = power(N, M/2);
if(M % 2 == 0)
return recursive * recursive;
return recursive * recursive * N;
}
public static void main (String[] args)
{
int N = 5, M = 13;
System.out.println(power(N, M));
}
}
``````
Output:
``````1220703125
``````
Note that Binary Exponentiation can be used in any problem where the power needs to be calculated. This will improve the performance greatly of the sub-routine that is concerned with the calculation of the power.
With this, you have the complete knowledge of using this technique. Enjoy.
#### OpenGenus Tech Review Team
The official account of OpenGenus's Technical Review Team. This team review all technical articles and incorporates peer feedback. The team consist of experts in the leading domains of Computing.
Improved & Reviewed by:
Binary exponentiation (Power in log N)
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# Volume.ppt
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``` Volume
AIMS Math Review
Spring, 2008
Period 1
Math’s
important for
Find the volume. child
development.
.
Find the volume.
5*5**15 =
375 cubic feet
.
or
1177.5 cubic feet
Period 2
A company manufactures large dice for children's
games. The dice are cubes with one side measuring
1.2 inch in length. The dice are formed when a liquid
plastic in is poured into a mold of the die. The
company uses trays that contain one dozen die molds
in each tray. If the liquid plastic is measured in cubic
inches, how much liquid plastic is needed to fill one
tray? You can do
the math!
A company manufactures large dice for children's
games. The dice are cubes with one side measuring
1.2 inch in length. The dice are formed when a liquid
plastic in is poured into a mold of the die. The
company uses trays that contain one dozen die molds
in each tray. If the liquid plastic is measured in cubic
inches, how much liquid plastic is needed to fill one
tray? 1.2 * 1.2 * 1.2 (for each die) * 12 (for 1 tray) = 20.736 cubic inches
Period 3
Math is
nearly as
great as Soda is sold in
English.
aluminum cans that measure 6
inches in height and 2 inches in
diameter. How many cubic
inches of soda are contained in
a full can?
Soda is sold in
aluminum cans that measure 6
inches in height and 2 inches in
diameter. How many cubic
inches of soda are contained in
a full can? 1 * 1 * 3.14 * 6 = 18.84 cubic inches
Period 4
All
careers
need
math!
Period 4
4x5x10 vs. 3*3 *8 vs. ½ 4/3 5*5*5
200vs. 226.08 vs. 261.67
Mom’s Place has the greatest volume.
Period 5
Find the volume.
You
need 1. 2.
the
math!
Find the volume.
1. 2.
V = 42*10 =160 cubic cm V = 1/3 (102)*8 = 800/3 cubic units
Period 6
Find the volume of each figure.
1. requires 2.
math.
Find the volume of each figure.
1. 2.
V = ½ (3)(4)(2) = 12 cubic inches V = 1/3 (24.6/2)2*5 = 24.6 cubic ft
```
Related docs
Other docs by fjzhangweiyun
the house of bernarda alba
|
## Solution 44 Page 50 Math Workbook 10 – Kite>
Topic
An international workshop includes 12 students from the following countries: Vietnam, Japan, Singapore, India, Korea, Brazil, Canada, Spain, Germany, France, South Africa, Cameroon, each country has exactly 1 the student. Randomly select 2 students from the international student group to participate in BTC:
Calculate the probability of each of the following events:
a) A: “Two selected students are from Asia”
b) B: “Two selected students from Europe”
c) C: “Two selected students from America”
d) D: “Two selected students from Africa”
Solution method – See details
The probability of event A being a number, symbol $$P\left( A \right)$$ is determined by the formula: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}$$, where $$n\left( A \right)$$ and $$n\left( \Omega \right)$$ denote the number of elements of set A and $$\Omega$$ respectively.
Detailed explanation
Select 2 students from 12 students $$\Rightarrow$$ convolution 2 of 12 $$\Rightarrow n\left( \Omega \right) = C_{12}^2 = 66$$
a) A: “Two selected students from Asia”:
There are 5 Asian countries: Vietnam, Japan, Singapore, India, HQ
$$\Rightarrow n\left( A \right) = C_5^2 = 10$$
$$\Rightarrow P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{{10}}{{ 66}} = \frac{5}{{33}}$$
b) B: “Two selected students from Europe”: There are 3 European countries: Spain, Germany, France $$\Rightarrow n\left( B \right) = C_3^2 = 3$$
$$\Rightarrow P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega \right)}} = \frac{3}{{66} } = \frac{1}{{22}}$$
c) C: “Two selected students from America”: There are 2 American countries: Brazil, Canada $$\Rightarrow n\left( C \right) = C_2^2 = 1$$
$$\Rightarrow P\left( C \right) = \frac{{n\left( C \right)}}{{n\left( \Omega \right)}} = \frac{1}{{66} }$$
d) D: “Two selected students from Africa”: There are 2 African countries: South Africa, Cameroon $$\Rightarrow n\left( D \right) = C_2^2 = 1$$
$$\Rightarrow P\left( D \right) = \frac{{n\left( D \right)}}{{n\left( \Omega \right)}} = \frac{1}{{66} }$$
|
## How do you find the equation of a line given two points?
Equation from 2 points using Slope Intercept FormCalculate the slope from 2 points.Substitute either point into the equation. You can use either (3,7) or (5,11)Solve for b, which is the y-intercept of the line.Substitute b, -1, into the equation from step 2.
## How do you find the equation of a horizontal line?
Horizontal lines have a slope of 0. Thus, in the slope-intercept equation y = mx + b, m = 0. The equation becomes y = b, where b is the y-coordinate of the y-intercept.
## How do you find the standard equation of a line?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y).
## What are the three equations of a line?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article. There are three main forms of linear equations.
## What is the formula of two point form?
The equation of the line joining two points (x1, y1) and (x2, y2) is given by y – y1=frac{y_2-y_1}{x_2-x_1}(x – x1). The slope of the line joining two points (x1, y1) and (x2, y2) is equal to frac{y_2-y_1}{x_2-x_1}.
## What is the Y intercept formula?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## How do you calculate a slope of a line?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
## How do you write an equation for a vertical and horizontal line?
Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate of all points.
## What is an equation of the horizontal line that passes through 5 7?
Explanation: The point (a,b) is (5,7) . Therefore, the equation of the line is y=7 .
## What is an example of a horizontal line?
An equation that only crosses the y-axis is a horizontal line. For example: y=5 is a horizontal line that crosses the y-axis at (0,5).
## How do you read standard form?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. Next, you isolate the y-intercept(in this case it is 3) like this: Add 3/2x to each side of the equation to get this: 3/2x+y=3.
## What is standard form in math?
Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
|
# Parallel and intersective lines solve for the intersections
• Jul 31st 2011, 12:22 PM
chicapsy
Parallel and intersective lines solve for the intersections
I'm stuck on this question, would appreciate any help trying to figure this out:
Q)Two parallel lines $\displaystyle y = 2x + 2$ and $\displaystyle y = 2x - 2$ are crossed by a third line. Determine the coefficients of the third line $\displaystyle y = ax + b$ such that the x-coordinate of the intersection of the two lines $\displaystyle y = ax + b$ and $\displaystyle y = 2x - 2$ is twice as much as that of the intersection of the two lines $\displaystyle y = ax + b$ and $\displaystyle y = 2x + 2$; and the y-coordinate of the intersection of the two lines $\displaystyle y = ax + b$ and $\displaystyle y = 2x + 2$ is twice as much as that of the intersection of the two lines $\displaystyle y = ax + b$ and $\displaystyle y = 2x - 2$.
Calculate the coordinates of the two intersections.
Attachment 21946
I would appreciate any help on this.
• Aug 1st 2011, 01:31 AM
FernandoRevilla
Re: Parallel and intersective lines solve for the intersections
Solving $\displaystyle \begin{Bmatrix} y=ax+b\\y=2x+2\end{matrix}$ you'll obtain $\displaystyle x_1=\frac{2-b}{a-2},\;y_1=\frac{2(a-b)}{a-2}$ .
Solving $\displaystyle \begin{Bmatrix} y=ax+b\\y=2x-2\end{matrix}$ you'll obtain $\displaystyle x_2=\frac{b+2}{2-a},\;y_2=\frac{2(a+b)}{2-a}$ .
Now, using $\displaystyle x_2=2x_1,\;y_1=2y_2$ we get a $\displaystyle 2\times 2$ linear system. Find $\displaystyle a$ and $\displaystyle b$ .
• Aug 4th 2011, 05:41 PM
chicapsy
Re: Parallel and intersective lines solve for the intersections
Thank you.
|
# Dihedral Angles
## by Greg Egan
This page derives some simple formulas that are useful for computing dihedral angles in polyhedra.
### Basic formulas
We often start by knowing both the true angle, α, between two edges of a face of a polyhedron, and the angle β that the same two edges appear to make in some projection. For example, every face of a regular icosahedron is an equilateral triangle, with angles of π/3 between the edges — but when we look straight at a vertex of the icosahedron, we see five edges in a projection where the angles between them appear to be 2π/5.
To see how we can make use of this kind of information, let’s suppose we have an isoceles triangle with two sides of length s and a known angle α between them (triangle CAD in the diagram). And suppose that this triangle’s projection in some plane is again an isoceles triangle, this time with an angle β between its sides (triangle CBD).
We construct the point E as the midpoint of CD. We find the plane that’s perpendicular to the line segment AC and passes through D, and label the point where this plane cuts AC as F and the point where it cuts BC as G.
Some angles we might then want to know are:
• γ, the angle between the edges of the original triangle and the projection plane.
• ε, the dihedral angle between the original triangle and the projection plane.
• δ, the dihedral angle between the original triangle and the right triangle ABC perpendicular to the projection plane.
If we have a regular polyhedron, the reflection of triangle CAD in the plane of ABC will be an adjacent face of the same polyhedron, and the angle between the two faces will be 2δ.
To find γ, note that we can compute the length EC in two ways: either as s sin ½α from triangle AEC, or as s cos γ sin ½β from triangle BEC. Equating the two we have:
cos γ = sin ½α / sin ½β
To find δ, we use the fact that we can obtain the length GD in two ways: either as s sin α sin δ from triangle FGD, or as s cos γ sin β, from triangle BGD. When we equate these and solve for sin δ, we get:
sin δ = [sin β / sin α] cos γ = [(2 sin ½β cos ½β) / (2 sin ½α cos ½α)] (sin ½α / sin ½β)
which simplifies to:
sin δ = cos ½β / cos ½α
To find ε, we see that:
cos ε = BE / AE = (EC cot ½β) / (EC cot ½α)
This gives us:
cos ε = tan ½α / tan ½β
### Application to Regular Polyhedra
Suppose we have a regular polyhedron in which n faces, all of them regular f-gons, meet at every vertex. If α is the interior angle between edges of an f-gon:
α = (1 – 2/f) π
½α = π/2 – π/f
cos ½α = sin π/f
Looking straight at the vertex, if β is the projected angle between the edges:
β = 2π/n
½β = π/n
The sine of half the internal dihedral angle between the faces of the polyhedron will then be given by our formula from the previous section:
sin δ = cos ½β / cos ½α = (cos π/n) / (sin π/f)
To calculate the full dihedral angle, 2δ, it’s convenient to rewrite this as:
cos 2δ = 1 – 2 sin2 δ = 1 – 2 [(cos π/n) / (sin π/f)]2
Polyhedronnfcos 2δDihedral angle 2δ
Tetrahedron331/370.5288°
Cube34090°
Octahedron43–1/3109.471°
Dodecahedron35–1/√5116.565°
Icosahedron53–(√5)/3138.19°
### Generalisation to Regular Polytopes
The 4-dimensional regular polytope known as the 24-cell has 24 octahedral hyperfaces. At each of its 24 vertices, 6 of these octahedra meet. But the corners of an octahedron (above left) don’t have the correct angles for six of them to fit together without any gaps. What’s needed to do that is the kind of pyramid that can be formed by taking one-sixth of a cube (above right).
Of course, the resolution to this is that the six octahedra that meet at each vertex of a 24-cell don’t all lie in the same 3-dimensional hyperplane, any more than the four triangular faces that meet at each vertex of the octahedron lie in the same 2-dimensional plane. But if we project a view of those four triangles into a plane perpendicular to a line from the centre of the octahedron to the vertex, we see a square divided into four identical triangles. Similarly, if we project a view of the six octahedra that meet at each vertex of the 24-cell into a 3-dimensional hyperplane perpendicular to a line from the centre of the 24-cell to the vertex, we see six identical pyramids of the kind produced by subdividing a cube.
There are analogous situations at the vertices of all the 4-dimensional regular polytopes: the corners of the hyperfaces that come together there will be pyramids with n-sided polygons as their bases, but in the projection of the vertex into a suitable hyperplane there will be shorter pyramids formed by subdividing a “vertex figure” polyhedron with n-sided faces.
The initial known quantities in the problem are:
• α1 = (1 – 2/fF) π, the corner angle of the fF-gon faces of the polytope’s actual hyperfaces;
(in our example, with octahedral hyperfaces, fF = 3 and α1 = π/3)
• α2, the angle between the vertices of the vertex figure polyhedron, subtended at the centre, easily found from the dihedral angles of the dual polyhedron to the vertex figure;
(in our example, the vertex figure is a cube, whose dual is an octahedron, so α2 = π – [dihedral angle for octahedron] = cos–11/3)
• β = 2π/fV, the angle subtended by the sides of the fV-gon faces of the vertex figure;
(in our example, fV = 4 and β = π/2)
From the calculations we did in the first section of this page, we have:
cos γ1 = sin ½α1 / sin ½β
cos γ2 = sin ½α2 / sin ½β
cos ε1 = tan ½α1 / tan ½β
cos ε2 = tan ½α2 / tan ½β
In order to find the dihedral angle between hyperfaces of the polytope, we will initially calculate half that angle: δ, the angle between the hyperplane containing the single hyperface we’ve been dealing with (the pyramid with tip E), and the hyperplane containing the tetrahedron ABEF. That tetrahedron contains one face of the hyperface (the triangle ABE), and also the vector EF that is normal to the hyperplane we project into. It’s the four-dimensional analogue of the triangle ABC in our original three-dimensional treatment: the reflection in it of the single hyperface we’re analysing will produce an adjacent hyperface of the same regular polytope.
Now, the heights of the two pyramids are:
OE = r tan γ1
OF = r tan γ2
where r is the radius of the circumscribing circle of their common base. The perpendicular distances from a face of either pyramid to the centre of the base can then by found by using the cosine of the dihedral angle between the base and the face:
OP = OE cos ε1 = r cos ε1 tan γ1
OQ = OF cos ε2 = r cos ε2 tan γ2
The line segment OQ is perpendicular to the entire hyperplane of the tetrahedron ABEF. Why? It’s perpendicular to the face ABF by construction, and it’s perpendicular to EF because EF is normal to the hyperplane of projection that contains the whole projected pyramid, and we constructed OQ to lie within that pyramid.
But the line segment PQ lies within the tetrahedron ABEF, because each endpoint lies on one of its faces. This means OQP is a right triangle with a right angle at Q.
Now, OQ is perpendicular to the face ABE, as is OP by construction, so any vector in the plane of OQP will also be perpendicular to ABE. But the vector that’s normal to the entire unprojected hyperface must be normal to that face, so the hyperface normal must lie in the plane of OQP. What’s more, the hyperface normal must be perpendicular to the line segment OP, which lies within the hyperface. This is enough to tell us that the angle at P in the triangle must be δ, the angle between the hyperface and the tetrahedron ABEF.
So sin δ can be found from the right triangle OQP:
sin δ = OQ / OP
= (cos ε2 tan γ2) / (cos ε1 tan γ1)
= (cos ½α1 / cos ½α2) (sin γ2 / sin γ1)
= (cos ½α1 / cos ½α2) √[(sin2 ½β – sin2 ½α2)/(sin2 ½β – sin2 ½α1)]
The specific values we wish to substitute into this formula are:
½α1 = ½(1 – 2/fF) π = π/2 – π/fF
cos ½α1 = sin π/fF
sin ½α1 = cos π/fF
β = 2π/fV
cos ½α2 = (cos π/fV) / (sin π/dV)
For ½α2, we’re using our formula from the previous section for the dihedral angle for a polyhedron, and applying it to the angle between vertices for the dual polyhedron; one is just π minus the other, so when we take half the angle that exchanges sin and cos. Here we’ve set the number of faces meeting at each vertex to fV, the sides-per-face of the vertex figure, since we’re applying the formula to the dual of the vertex figure. And we’ve introduced a new parameter, dV, which is the number of faces that meet at each vertex in the vertex figure.
This yields a final result of:
sin δ = (cos π/dV) (sin π/fF) / √(sin2 π/fV – cos2 π/fF)
An equivalent formulation is:
cos 2δ = 1 – 2 [(cos π/dV) (sin π/fF)]2 / [sin2 π/fV – cos2 π/fF]
PolytopeHyperfacesfFVertex figuresfVdVcos 2δDihedral angle 2δ
4-simplexTetrahedra3Tetrahedra331/475.5225°
HypercubeCubes4Tetrahedra33090°
Cross-polytopeTetrahedra3Octahedra34–1/2120°
24-cellOctahedra3Cubes43–1/2120°
120-cellDodecahedra5Tetrahedra33–(1 + √5)/4144°
600-cellTetrahedra3Icosahedra35–(1 + 3√5)/8164.478°
Science Notes / Dihedral Angles / created Wednesday, 11 April 2012 / revised Saturday, 14 April 2012
|
Question
# Solve the following equations:$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$
Hint: In this question first multiply two factors together and the remaining two together so that it converts into a quadratic equation then substitute the same part to any other variable and multiply later on to apply the quadratic formula, so use these concepts to get the solution of the question.
Given equation is
$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$
Now multiply (x+9) and (x-7) together and remaining two together
$\left[ {\left( {x + 9} \right)\left( {x - 7} \right)} \right]\left[ {\left( {x - 3} \right)\left( {x + 5} \right)} \right] = 385 \\ \left[ {{x^2} + 9x - 7x - 63} \right]\left[ {{x^2} - 3x + 5x - 15} \right] = 385 \\ \left[ {{x^2} + 2x - 63} \right]\left[ {{x^2} + 2x - 15} \right] = 385 \\$
Let, $\left( {{x^2} + 2x} \right) = t...........\left( 1 \right)$
So, substitute this value in the above equation we have
$\left[ {t - 63} \right]\left[ {t - 15} \right] = 385 \\ \Rightarrow {t^2} - 15t - 63t + 945 = 385 \\ \Rightarrow {t^2} - 78t + 560 = 0 \\$
Now factorize the above equation we have,
$\Rightarrow {t^2} - 8t - 70t + 560 = 0 \\ \Rightarrow t\left( {t - 8} \right) - 70\left( {t - 8} \right) = 0 \\ \Rightarrow \left( {t - 8} \right)\left( {t - 70} \right) = 0 \\ \Rightarrow \left( {t - 8} \right) = 0{\text{ & }}\left( {t - 70} \right) = 0 \\ \Rightarrow t = 8{\text{ & }}t = 70 \\$
Now from equation (1)
$\left( {{x^2} + 2x} \right) = t \\ \Rightarrow \left( {{x^2} + 2x} \right) = 8{\text{ & }}\left( {{x^2} + 2x} \right) = 70 \\ \Rightarrow {x^2} + 2x - 8 = 0..........\left( 2 \right){\text{ & }}{x^2} + 2x - 70 = 0........\left( 3 \right) \\$
Now first solve equation (2) by factorization method we have,
$\Rightarrow {x^2} + 2x - 8 = 0 \\ \Rightarrow {x^2} + 4x - 2x - 8 = 0 \\ \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \\ \Rightarrow \left( {x + 4} \right) = 0{\text{ & }}\left( {x - 2} \right) = 0 \\ \Rightarrow x = - 4{\text{ & }}x = 2 \\$
Now solve equation (3)
${x^2} + 2x - 70 = 0$
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 1 \right)\left( { - 70} \right)} }}{2} = \dfrac{{ - 2 \pm \sqrt {284} }}{2} = \dfrac{{ - 2 \pm 2\sqrt {71} }}{2} = - 1 \pm \sqrt {71}$
$x = \left( {2, - 4,\left( { - 1 \pm \sqrt {71} } \right)} \right)$
|
# Example: Velocity patterns in a pulse
## Understanding the situation
Consider a transverse pulse moving on an elastic string. If we think of the string as a set of small beads connected by massless springs as in Waves on an elastic string, the beads move up and down, perpendicular to the motion of the pulse.
At any instant of time, a picture of the pulse shows a graph of the displacement of the beads at that instant (a graph for the eye). But at that instant, each of those beads might be moving up and down with different velocities. What does the pattern of those velocities look like? Figuring that graph out helps us understand the motion of the pulse.
## Presenting a sample problem
In the figure below are show 4 graphs of an elastic string as a function of position along the string ($x$ is the horizontal axis). If graph A represents the shape of a right-moving pulse on the string at time $t=0$, which graph could represent the graph of the y-velocity of the bits of the string as a function of $x$ at $t=0$?
## Solving this problem
#### A graphical method
The way to figure this out is to use the fact that the y-velocity of a bead at any position is defined by
$$v_y = \frac{\Delta y}{\Delta t}.$$
So let's consider the initial position of the pulse and then see what happens a small amount of time later. Since it's given that the pulse is moving to the right, if it is given by graph A at $t=0$, that looks like the red curve below. At a small amount of time later the pulse will have moved to the right and look like the blue curve in the figure below.
In order to get insight into the velocity pattern in space, we have to ask: how is each bead moving? We can figure out how they move as the pulse shifts from the red curve to the blue if we imagine that there are beads on each of the vertical graph lines (every 0.2 along the horizontal axis).
At the red curve's instant of time the beads are on the red curve where it crosses the gray vertical lines of the graph. In the time the pulse moves to the blue curve, each bead will move straight up or straight down until it reaches the blue curve. We've drawn arrows to show that motion in the figure below.
Since the time intervals are the same for all, the $v_y$ velocities for each bead are just proportional to $\Delta y$ - the length of the arrows.
So as we move from left to right, we see that the y-velocity starts at 0, gets increasingly negative, then less negative until it's 0, just about at $x=0$. (The reason it is not exactly at $x=0$ is because we biased the time to look at a later time. If we had looked at the shift from an earlier time to a later time, we would have gotten the velocity to be 0 at $x=0$.)
As $x$ becomes increasingly positive, the velocity grows until it's fairly large and then it gets smaller until it goes to 0.
This pattern looks like graph D and this tells us how to interpret graph D as a velocity graph in terms of how each bead is moving.
#### An algebraic solution
We can also solve the problem algebraically. A common and convenient pulse shape is a Gaussian:
$$y_G(x,0) = Ae^{-(bx)^2}\quad\mathrm{Gaussian}$$
We know to make the pulse move to the right, we simply replace $x$ everywhere by $x-v_0t$. For the Gaussian this gives
$$y_G(x,t) = Ae^{-(b(x-v_0t))^2}.$$
The instantaneous y-velocity is just
$$v_y(x,t) = \frac{dy}{dt}.$$
Note that the velocity is a function of both $x$ (which bead we are considering) and time. We can take the time derivate by using the chain rule with
$f(x,t) = Ae^{g(x,t)}\quad$ with $\quad g(x,t) = -b((x-v_0t))^2$
The chain rule tells us that
$$\frac{df(g(t))}{dt} = \frac{df}{dg} \frac{dg}{dt}.$$
Since the exponential is its own derivative
$$\frac{df}{dg} = f$$
and the derivative of $g$ is straightforward to calculate by expanding it out:
$$g(x,t) = -bx^2 + 2bv_0tx -bv_0^2t^2$$
so $$\frac{dg}{dt} = 2bv_0x - 2bv_0^2t.$$
So setting $t=0$ we get
$$v_y(x,0) = (2bv_0)A xe^{-(bx)^2}$$
The result is a Gaussian times $x$ times a bunch of constants. This makes the result negative for negative $x$, 0 at $x=0$, and positive for positive $x$. You can easily go to the Desmos graphing calculator and plot $y = xe^{-x^2}$ to confirm that this result looks like graph D.
Joe Redish 6/11/19
Article 692
|
What would you rather have: a half of a half of a half of something nice or a half of a quarter of something nice?
# Fractions 7
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In your lessons at school you will no doubt be learning all about fractions. They can sometimes cause confusion (e.g. what's the difference between a denominator and a numerator?) and if you haven't quite understood them so far, play our four quizzes until you are sure you understand what they mean.
You may think fractions are old-fashioned and no longer used, but you will come across them in everyday life often enough. To test this, while you are out and about this week, see how many fractions you can spot - and more importantly, see if you can work out what they mean!
In this quiz you will get a chance to revise what you have already learned. Take your time and try and have fun!
1.
What is 98 - 34 + 58
1
118
118
208
98 - 34 + 58 = 98 - 68 + 58 = 1. Reduce the fractions to their simplest forms, then add them. If the denominators are the same, you can add the fractions by simply adding their numerators - it really is as easy as that!
2.
What is two-thirds of £81?
£18
£54
£64
£27
23 × £81 = (2 × £81) ÷ 3 = £162 ÷ 3 = £54. Do the brackets first. The word OF means multiply
3.
What is 435 as an improper fraction?
205
125
75
235
To convert a mixed fraction to an improper fraction, follow these steps: 1. Multiply the whole number part by the denominator. 2. Add this result to the numerator. 3. Write the fraction with step 2 in the numerator and keep the original denominator. STEP 1: 4 × 5 = 20. STEP 2: 20 + 3 = 23. STEP 3: 235
4.
What is 123 as a mixed fraction?
123
1213
4
12
To convert an improper fraction to a mixed fraction, follow these steps: 1. Divide the numerator by the denominator. 2. Note the whole number remainder. 3. Write the number from step 1 as the whole number in front of the fractional part AND write the fractional part with the remainder in the numerator and keep the original denominator. STEP 1: 12 ÷ 3 = 4. STEP 2: Remainder 0. STEP 3: 4. Be on the LOOKOUT for this sort of thing!
5.
How many sevenths are there in 6?
35
49
21
42
There are seven sevenths in 1, so there are 7 × 6 = 42 sevenths in 6. In general, the number in the denominator tells you how many fractions of that type are needed to make 1. For example, you need three thirds, 13, to make 1, but you need ten tenths, 110, to make 1 and so on. REMEMBER this stuff, and you won't get confused with fractions
6.
What is 125 as a mixed fraction?
2212
1225
225
525
To convert an improper fraction to a mixed fraction, follow these steps: 1. Divide the numerator by the denominator. 2. Note the whole number remainder. 3. Write the number from step 1 as the whole number in front of the fractional part AND write the fractional part with the remainder in the numerator and keep the original denominator. STEP 1: 12 ÷ 5 = 2. STEP 2: Remainder 2. STEP 3: 225
7.
What is 823 as an improper fraction?
243
163
263
103
To convert a mixed fraction to an improper fraction, follow these steps: 1. Multiply the whole number part by the denominator. 2. Add this result to the numerator. 3. Write the fraction with step 2 in the numerator and keep the original denominator. STEP 1: 8 × 3 = 24. STEP 2: 24 + 2 = 26. STEP 3: 263
8.
What would you rather have: a half of a half of a half of something nice or a half of a quarter of something nice?
A half of a quarter
They are both the same, so it doesn't matter which one you pick
A half of a half of a half
A half of a half
12 × 12 × 12 = 14 × 12 = 18. The word OF means multiply
9.
What is 25 + 410 + 1215?
1615
263
85
1810
25 + 410 + 1215 = 25 + 25 + 45 = 85. Reduce the fractions to their simplest forms, then add them. If the denominators are the same, you can add/subtract the fractions by simply adding/subtracting their numerators - it really is as easy as that!
10.
What is 23 of 21?
7
14
18
12
23 × 21 = (2 × 21) ÷ 3 = 42 ÷ 3 = 14. Do the brackets first. The word OF means multiply
Author: Frank Evans
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Equations and Functions Algebraic methods for solving equations
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### Transcript of Index Introduction to algebraic equations Quadratic equation – Complete quadratic equation –...
Equations and Functions
Algebraic methods for solving equations
• Del Ferro, tartaglia and cardano’s history• General cubic equation (solving cubic equations)
Introduction to algebraic equations
Algebraic equations are equations where the unknown x is subject to algebraic operations such as addition, subtraction, multiplication, division
examples:to x + b = 0to x ² + bx + c = 0at x4 x ² + b + c = 0
Quadratic equationA quadratic equation in the unknown x is of the form:
to x ² + bx + c = 0
where the real numbers a, b and c are the coefficients of the equation, and that must be nonzero. This equation is also called a quadratic equation, because the highest degree term is squared.
A quadratic equation is complete, if all the coefficients a, b and c are nonzero.
examples:
2 x ² + 7x + 5 = 03 x ² + x + 2 = 0
A quadratic equation is incomplete if b = 0 or c = 0 or b = c = 0. Incomplete in the equation the coefficient is nonzero.
examples:
4 x ² + 6x = 03 x ² + 9 = 02 x ² = 0
Equations of the type ax ² = 0: Divide the whole equation for obtain:
x ² = 0
meaning that the equation has two roots equal to zero.
Equations of the type ax ² + c = 0: Divide the whole equation by a and pass the constant term to the second member to get:
x ² =-c / a
If-c / a is negative, there is no solution in real numbers.
If-c / a is positive, the equation has two roots with the same absolute value (modulus) but of opposite signs.
Equations of the type ax ² + bx = 0: Factored to obtain the equation:
x (ax + b) = 0
and the equation has two roots:
x '= 0 or x‘’ =-b / a
• We will show in sequence as the mathematician Sridhara, obtained the formula (known as) of Bhaskara, which is the general formula for solving quadratic equations. A curious fact is that the formula of Bhaskara was not discovered by him but by the Indian mathematician Sridhara at least a century before the publication of Bhaskara, a fact recognized by the Bhaskara, although the material has not built by the pioneer reached us.
The basis used to obtain this formula was to find a way to reduce the quadratic equation to one of the first degree, by extracting square roots of both members of the same.
Bhaskara formula: D = b ²-4ac
x ² - 5 x + 6 = 0
Identifying the coefficients: a = 1, b = -5, c = 6 Write D = b ²-4ac. Calculate D = (-5) ² -4 × 1 × 6 = 25-24 = 1Replace the values of the coefficients a, b and c in the formula
v
equations bi- quadratic are equations of the fourth degree in the unknown x, the general form:
at x4 x ² + b + c = 0
In fact, this is an equation that can be written as a quadratic equation by replacing:
y = x ²
to generate
a ² + y b y + c = 0
We apply the quadratic formula to solve this last equation and obtain solutions y 'and y "and the final procedure should be more careful, since
x ² = y ² = x or y "
and y 'or y "is negative, the solutions do not exist for x.
Del Ferro, tartaglia and cardano’s history
Del Ferro, Tartaglia and Cardano, provided a tragicomedy of disputes, achievements and disappointments. The first to develop a mathematical method for solving cubic equations of the form x3 + ax + b = 0 was Scipione del Ferro, teacher at the University of Bologna, Italy, in the passage of the 15th century to 16th century. Before death, he revealed his method, which had kept secret, the Antonio Fiore.
Nicoll Tartaglia was born in Brescia, Italy in 1499. It is said that he was so poor as a child that studied mathematics writing on the tombstones of a cemetery. In 1535 he was challenged by Antonio Fiore to a mathematical competition. At the time, academic disputes were common, often rewarding the winner with the use of the loser. Tartaglia could solve the cubic equations of DelFerro, but had also discovered a method to solve the cubic form x3 + ax2 + b = 0 Armed with this knowledge, was the winner in the competition.
Tartaglia's last years were embittered by a quarrel with Girolamo Cardano (1501-1576), an Italian mathematician who, besides the famous doctor in Milan, was also an astronomer. In 1570, Cardano was arrested for heresy, for having written a horoscope of Jesus Christ. In 1539, at his home in Milan, Cardano persuaded Tartaglia to tell him his secret method for solution of cubic equations, under oath never to divulge it.
Some years later, however, Cardano learned that part of the method consisted of a posthumous publication of Del Ferro. He then decided to publish a complete study of cubic equations in his treatise Ars Magna (1545), a work that surpassed all algebra books published so far. In Ars Magna Cardano exposes a method to solve the cubic equation based on geometric arguments. There also exposes the general solution of the quartic equation AX4 BX3 cx2 dx = 0, and discovered by Ludovico Ferrari (1522-1565), pupil of Cardano, who seems to have surpassed the master in the algebra of polynomial equations.
In 1548, Tartaglia challenged Cardano for a math competition to be held in Milan did not attend Cardano, Ferrari and sent to represent him. It seems that Ferrari won the race, which caused the death Tartaglia unemployment and poverty in the nine years later.
General cubic equation
The general cubic equation is A x3 + B x2 + C x + D = 0 The coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form
x3 + b x2 + c x + d = 0
The quadratic term disappearsNow we want to reduce the last equation by the substitution x = y + r
The cubic equation becomes:
(y + r)3 + b (y + r)2 + c (y + r) + d = 0<=> y3 + (3 r + b) y2 + (3 r2 + 2 r b + c) y + r3 + r2 b + r c + d = 0
Now we choose y such that the quadratic term disappears
choose r = -b/3
So, with the substitution
x = y – b/3
the equation
x3 + b x2 + c x + d = 0
comes in the form
y3 + e y + f = 0
Vieta's substitutionTo reduce the last equation we use the Vieta substitution
y = z + s.1/z
The constant s is an undefined constant for the moment.The equation
y3 + e y + f = 0 becomes (z + s/z )3 + e (z + (s/z)) + f = 0
expanding and multiplying through by z3 , we have z6 + (3 s + e) z4 + f z3 + s (3 s + e) z2 + s3 = 0
Now we choose s = -e/3.The equation becomes
z6 + f z3 - e3/27 = 0 With z3 = u u2 + f u -e3/27 = 0
This is an easy to solve quadratic equation.
http://en.wikipedia.org/wiki/Equation
http://en.wikipedia.org/wiki/Equations_of_motion
http://pt.wikipedia.org/wiki/Scipione_del_Ferro
http://www.brasilescola.com/biografia/scipione-del-ferro.htm
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# Math with Roman Numerals
I’ve been reading (listening to, actually) a book on the state of scientific and technical knowledge in what has traditionally been referred to as the “Dark Ages” (roughly the 5th to 14th centuries). One of the things I became intrigued by was the use of Roman numerals for relatively complex math during this period.
## Quick Review
You’re probably familiar with Roman numerals. Conceptually, they’re easy. Letters represent numbers. To get the value of a number written in Roman numerals you simply add the individual numerals (or groups of numerals). Here are the letters you use to represent values up to 3999:
I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1000
Values are read from left to right, and it is traditional to put the larger numerals first. So III is 3, XII is 12, and DCLXVI is 666 (“let the reader understand”). One complication is that you never write the same numeral more than 3 times in a row, and that a smaller numeral appearing before a larger one means it should be subtracted from the larger value. So 14 is XIV, not XIIII. 3999 is MMMCMXCIX. That being said, the Romans weren’t real picky about how these numbers were written. For simplicity, I’m going to ignore the no-more-than-3 rule from time to time in my discussion below.
I’ll use the symbol => to indicate that I’m rearranging or simplifying.
## Adding and Subtracting
Addition with Roman numerals is easy: just put all the numerals in one group and rearrange them. So III + XVIII is IIIXVIII => XVIIIIII => XVVI => XXI.
Subtraction is similar. Convert larger digits into groups of smaller ones if needed for convenience, then just subtract similar numerals from each other. So XXI – III is XVVI – III, which is XVIIIIII – III, or XVIII.
## Doubling and Halving
Before getting into the fun stuff (multiplication and division), it helps to think about how to double and halve a value written in Roman numerals.
Doubling is just adding the number to itself. To double a number, just write two of each numeral. Twice XXI is thus XXIXXI => XXXXII => XLII. Twice III is IIIIII => VI.
Halving is similar. Divide each numeral by 2 and append the next lesser numeral if there is a remainder. This is slightly circular, since we’re saying to divide by 2 you divide by 2. If you don’t remember what half of L is, write it out as XXXXX and take half of those X’s to get XX with a remainder of half an X, which is V. To get half of 666, follow these steps:
DCLXVI
CCL + half of CLXVI
CCL + L + half of LXVI
CCL + L + XXV + half of XVI
CCL + L + XXV + V + half of VI
CCL + L + XXV + V + II + half of II
CCL + L + XXV + V + II + I
CCLLXXVVIII
=> CCCXXXIII
## Multiplication
Multiplication is a combination of doubling, halving, and adding remainders. Note that in general A ✖️ B is the same as (A / 2) ✖️ (B ✖️ 2). That is, if we double one term and halve the other term, the result is the same. In familiar terms, 8 ✖️ 3 (24) is the same as half 8 (4) times twice 3 (6). Further note that we can do that again, so:
8 ✖️ 3 = 4 ✖️ 6 = 2 ✖️ 12 = 1 ✖️ 24 = 24
Where it gets tricky with Roman numerals is handling remainders. Consider if we reversed which term was being doubled and which halved in our example above:
8 ✖️ 3 = 16 ✖️ 1.5 = 32 ✖️ .75 ….
While the first attempt actually took us all the way to the correct answer, the second appears to be getting harder and harder. Let’s look at both with Roman numerals:
Half
VIII
IIII
II
I
Double
III
VI
XII
XXIIII => XXIV
When the halving leaves a remainder, we can ignore it. I’ll mark those with an asterisk.
* drop remainder
Half
III
I
Double
VIII
XIIIIII => XVI
When we have remainders, there’s one more step. We need to add the value from the “double” column to our final result to get our true result (since we technically dropped that value when we dropped the remainder):
XVI + VIII = XVVIIII => XXIIII => XXIV
We can apply this technique to do arbitrarily complicated multiplication:
39 ✖️ 81
* drop remainder
* drop remainder
* drop remainder
* drop remainder
Half
XXXVIIII
XVIIII
VIIII
IIII
II
I
Double
LXXXI
CLXII
CCCXXIIII
DCXXXXVIII
MCCLXXXXVI
MMDLXXXXII
Now add the value from the “double” column for each odd value in the “half” column.
MMDLXXXXII + LXXXI + CLXII + CCCXXIIII = MMDCCCCLLLXXXXXXXXXXIIIIIIIII => MMMLLLVIIII => MMMCLIX
I’ll leave it to the reader to verify my work.
## Division
I’m not sure if there’s an easier way to do vision, but here’s what I came up with. First, we repeatedly double the divisor (the number by which we’re dividing) until we get a value greater than the dividend (the number that we’re dividing). We’ll keep track of the multiplier for each because we’ll need those later.
Consider 365 / 12
In Roman numerals: CCCLXV / XII
Multiplier
I
II
IV
VIII
XVI
XXXII
Double
XII
XXIV
XXXXVIII
LXXXXVI
CLXXXXII
XXXLXXXIV
exceeds dividend
We now know that XII goes into CCCLXV at least XVI times. We can make a note of that, and also subtract the doubled divisor to see what’s left:
CCCLXV – CLXXXXII = CCLLXXXXXXIIIII – CLXXXXII = CLXXIII
XII goes into CCCLXV XVI times with a remainder of CLXXIII
Now check our list of doubled divisors to see what the biggest one is that goes into the remainder (CLXXIII). It looks like LXXXXVI is the largest value that is smaller than CLXXIII. Repeat the step above. Subtract LXXXXVI from our remainder and add the multiplier to our accumulated multiplier:
CLXXIII – LXXXXVI = LLXXXXXXVVIII – LXXXXVI = LXXVII
XII goes into CCLXV XXIV times with a remainder of LXXVII
Repeating, this time using a multipler of IV and a doubled divisor of XXXXVIII:
LXXVII – XXXXVIII = XXXXXXVVIIIIIII – XXXXVIII = XXVIIII => XXIX
XII goes into CCLXV XXVIII times with a remainder of XXIX
Repeating, this time using a multiplier of II and doubled divisor XXIV:
XXVIIII – XXIIII = V
XII goes into CCLXV XXX times with a remainder of V.
Since V is less than our original divisor, that’s as far as we can go.
Again, the proof is left to the reader.
## Fractions
I’m not going to dwell on fractions except to say that the only fractions the Romans seemed to use were twelfths and halves. Each twelfth was a dot following the number. Since we coincidentally divided by 12 in the previous example, our remainder of 5 would be represented as 5 dots following XXX. I believe they were written in some kind of pattern, perhaps like we use on dice or playing cards.
Halves were represented using the letter S. So 7/12 would be “S..” or maybe “S:”.
## Conclusion
While this is fun, and while our Middle Ages ancestors could do amazing things using these simple techniques, it’s pretty clear why the hard math was being done using base-10 and place value representation.
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# 2014 AIME II Problems/Problem 4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
$$0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},$$
where $a$, $b$, and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$.
## Solution 1
Notice repeating decimals can be written as the following:
$0.\overline{ab}=\frac{10a+b}{99}$
$0.\overline{abc}=\frac{100a+10b+c}{999}$
where a,b,c are the digits. Now we plug this back into the original fraction:
$\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$
Multiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$:
$9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$
Dividing both sides by $9$ and simplifying gives:
$2210a+221b+11c=99^2=9801$
At this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in:
$2210a+221b+11c \equiv 9801 \mod 221 \iff 11c \equiv 77 \mod 221$
Notice that we arrived to the result $9801 \equiv 77 \mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get:
$c \equiv 7 \mod 221$
But we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$:
$2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44$
and since a and b are both between $0$ and $9$, we have $a=b=4$. Finally we have the $3$ digit integer $\boxed{447}$
## Solution 2
Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$. Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$. Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$. Adding the two, we get $$\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array}$$ From this, we can see that $a=4$, $b=4$, and $c=7$, so our desired answer is $\boxed{447}$
## Solution 3
Noting as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999}$, let $u = 10a + b$. Then $$\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}$$
$$\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}$$
$$\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}$$
$$221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}$$
$$221u + 11c = 9\cdot 33^2.$$
Solving for $c$ gives
$$c = 3\cdot 9\cdot 33 - \frac{221u}{11}$$
$$c = 891 - \frac{221u}{11}$$
Because $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$. Thus $abc = 10u + c = \boxed{447}.$
## Solution 4
We note as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999},$ so
$$\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.$$
As $\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\frac{10a + b}{99} = \frac{11a}{99} = \frac{111a}{999},$ so
$$\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.$$
Then $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $\boxed{447}.$
~savannahsolver
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# What is the area of the rectangle whose length is 3 times the width and the perimeter is 24.
hala718 | Certified Educator
To find the area of the rectangle, first we need to determine the length and the width.
Let the length be L and the width be W.
Given that the length = 3 times the width.
Then we will write:
L = 3W.............(1)
Also, given that the perimeter (P) = 24
==> 2L +2W = 24
Divide by 2:
==> L + W = 12...........(2)
Now let us substitute (1) in (2).
==> 3W + W = 12
==> 4*W = 12\
Now we will divide by 4.
==> W = 12/4 = 3
Then, W = 3 and L = 3*3 = 9
Then the length of the rectangle = 9, and the width = 3
Now we will calculate the area.
Area = L*W = 3* 9 = 27
Then, the area of the rectangle = 27 square units.
giorgiana1976 | Student
We'll start by writing the area of a rectangle.
A = length*width
But, from enunciation, length = 3width
A = 3width*width
A = 3width^2
We also know, from enunciation, that the perimeter is P= 24.
P = 2length + 2width
P = 2*(3width) + 2width
P = 8 width
24 = 8 width
We'll divide by 8:
width = 24/8
width = 3
length = 3width
length = 9
A = 3*9
A = 27 square units
neela | Student
What is the area of the rectangle whose length is 3 times the width and the perimeter is 24.
Let the width = x
Then by data , lenth = 3 times width = 3x.
So the perimeter p = 2(length+width) = 2(3x+x) = 8x.by definition.
p = 24 by data.
Therefore p = 8x= 24.
So 8x= 24.
x= 24/8 = 3.
Therefore length = 3x = 3*3 = 9.
Therefore area = length *width = 3x^2 .
Therefore Area = 3*4^2 = 48 sq unit.
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# Probability of selecting 7 blocks such that there is at least one from each 5 color
There are 30 blocks in a bag, which can be one of 5 colors, and are all distinct: each of the 5 colors red, yellow, green, blue, purple has 6 colored blocks numbered 1 through 6. We randomly select 7 out of the 30 blocks. If blocks are equally likely to be chosen what is the probability that we select 7 blocks such that there is at least one block for every color (red, yellow, green, blue, purple) out of the 7?
Is the number of possibilities (sample space) for 7-block choosings $\binom{30}{7}$ and therefore the probability of one particular outcome 1/$\binom{30}{7}$? Then, how do we count how many valid 7-block choosings there are to satisfy the constraint?
Hint
First figure out how many different ways you can get 7 blocks with 5 different colors without looking at the number of the block. There are two main ways of doing so:
1. Have 1 color with 3 blocks, and 4 colors with 1 block. There are of course 5 ways to do this: the color that has 3 blocks is 1 of 5
2. Have 2 colors that have 2 blocks each, and 3 colors with 1 color each ... there are $\binom{5}{2}=10$ ways this can be done: the colors with the two blocks are 2 out of 5
Then, figure out how many different numbered blocks you can get for each of the colors for these two different cases:
1. The 1 color with 3 blocks has $\binom{6}{3} = 20$ possibilities. The other 4 colors have 1 block so 6 possibilities. So that's $20*6^4$ possibilities total
2. The 2 colors with 2 blocks have $\binom{6}{2} = 15$ possibilities each, and the 3 colors with 1 block have 6 possibilities each. So that's $15^2*6^3$ possibilities total.
Since we established that there are 5 ways of getting case 1, and 10 ways of getting case 2, you get $5*20*6^4+10*15^2*6^3$ total ways of getting 7 blocks with 5 different colors.
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1. Class 9
2. Important Questions for Exam - Class 9
Transcript
Ex 10.4, 6 A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Given: Circular park with radius 20m Let Ankur, Syed and David be donated by points A,S & D resp. Given all three are sitting at equal distances, i.e., AS = SD = AD To find: Length of AS = SD = AD Explanation: Let AS = SD = AD = 2x In ASD, all sides are equal, ∴ ASD is an equilateral triangle We draw OP ⊥ SD So, SP = DP = 1/2 SD ⇒ SP = DP = 2𝑥/2 = x Join OS & AO In Δ OPS By Pythagoras theorem OS2 = OP2 + PS2 (20)2 = OP2 + x2 400 = OP2 + x2 400 – x2 = OP2 OP2 = 400 – x2 OP = √("400 – x2" ) Now, AP = AO + OP √("3" )x = 20 + √("400 – x2" ) √("3" )x – 20 = √("400 – x2" ) √("400 – x2" ) = √("3" )x – 20 Squaring both sides (√("400 – x2" ))^2= (√("3" )x – 20)2 400 – x2 = (√3x)2 + (20)2 – 2 ×(√3x) × (20) "400 – x2 =" 3x2 + 400 – 40√3x "400 – 400 + 40" √3 "x =" 3x2 + x2 "40" √3 "x =" 4x2 4x2 = 40√3x "x2" /"x" = 40/4 √3 x = 10√3 m AS = SD = AD = 2x = 2 × 10√3 = 20√3 m ∴ Length of string of each phone is 20√3 m
Class 9
Important Questions for Exam - Class 9
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NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.4
Ex 12.4 Class 7 Maths 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 formed 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, 3.
Solution:
(i) The number of line segments required to form
n digits is given by the expressions.
For 5 figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26
For 10 figures, the number of line segments = 5 × 10 + 1
= 50 + 1 = 51
For 100 figures, the number of line segments = 5 × 100 + 1
= 500 + 1 = 501
(ii)
For 5 figures, the number of line segments =3 ×5 + 1
= 15 + 1 = 16
For 10 figures, the number of line segments = 3 × 10 + 1
= 30 + 1 = 31
For 100 figures, the number of line segments = 3 × 100 + 1
= 300 + 1 = 301
(iii)
For 5 figures, the number of line numbers = 5 × 5 + 2
= 25 + 2 = 27
For 10 figures, the number of line segments = 5 × 10 + 2
= 50 + 2 = 52
For 100 figures, the number of line segments = 5 × 100 + 2
= 500 + 2 = 502
Ex 12.4 Class 7 Maths Question 2.
Use the given algebraic expression to complete the table of number patterns:
S.No. Expression Terms 1st 2nd 3rd 4th 5th … 10th … 100th … (i) 2n – 1 1 3 5 7 9 – 19 – – – (ii) 3n + 2 5 8 11 14 – – – – – – (iii) 4n + 1 5 9 13 17 – – – – – – (iv) 7n + 20 27 34 41 48 – – – – – – (v) n2 + 1 2 5 10 17 – – – – 10,001 –
Solution:
(i) Given expression is 2n – 1
For n = 100, 2 × 100 – 1
= 200 – 1 = 199
(ii) Given expression is 3n + 2
For n = 5, 3 × 5 + 2 = 15 + 2 = 17
For n = 10, 3 × 10 + 2 = 30 + 2 = 32
For n = 100, 3 × 100 + 2 = 300 + 2 = 302
(iii) Given expression is 4n + 1
For n = 5, 4 × 5 + 1 = 20 + 1 = 21
For n = 10, 4 × 10 + 1 = 40 + 1 = 41
For n = 100, 4 × 100 + 1 = 400 + 1 = 401
(iv) Given expression is 7n + 20
For n = 5, 7 × 5 + 20 = 35 + 20 = 55
For n = 10, 7 × 10 + 20 = 70 + 20 = 90
For n = 100, 7 × 100 + 20 = 700 + 20 = 720
(v) Given expression is n2 + 1
For n = 5, 52 + 1 = 25 + 1 = 26
For n = 10, 102 + 1 = 100 + 1 = 101
+
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The site was down over the weekend, when this post was due. So I’m declaring today to be Saturday. In reality, it’s much too late on Monday night for me to be writing anything. So this will be short, and likely make even less sense than normal.
Last time, we ended with a definition of the rational numbers, and I observed that they should be enough for any use: they are packed densely on the number line, with infinitely many filling the gap between any two. How could there be any more numbers? Where would they go?
I’m sure you are all familiar with Pythagoras’ theorem: in a right triangle, the square of the side opposite the right angle (the hypotenuse) is equal to the sum of the squares of the other two sides. This gives us a way to calculate the length of the diagonal of a unit square: the diagonal is the hypotenuse, and the square of its length is the sum of the squares of two adjacent sides of the square. If we denote the length of the diagonal by $latex d$, we have $latex d^2 = 1^2 + 1^2$, or $latex d^2 = 2$.
So what rational number is $latex d$? Since it’s a length, it is clearly positive. So there must be positive integers $latex a$ and $latex b$ such that $latex d = \frac{a}{b}$. There will be many possible pairs, and it’s useful to chose the smallest. In that case, $latex a$ and $latex b$ will have no common divisor: if they did, we could divide both by that common divisor, and get a smaller pair.
Since $latex a$ and $latex b$ have no common divisor, it follows that at most one of them is even: they can’t both be, or they’d both be divisible by 2. Remember that bit. It’s important.
Now, let’s plug $latex d = \frac{a}{b}$ into the equation from Pythagoras’ theorem: $latex \left(\frac{a}{b}\right)^2 = 2$. After a little routine rearrangement, that gives us $latex a^2 = 2b^2$. And that means $latex a^2$ is even.
Since the square of an odd number is always an odd number, it follows that $latex a$ can’t be odd. It must be even, which means there is some integer $latex c$ such that $latex a = 2c$. We can plug that back into the equation, giving $latex (2c)^2 = 4c^2 = 2b^2$. Divide it all by 2, and we have $latex 2c^2 = b^2$. So, by the same logic, $latex b$ must be even too.
But wait. Didn’t we just say that $latex a$ and $latex b$ have no common divisor? They can’t both be even. We’ve managed to prove a contradiction, and that means our original assumption cannot be true. That assumption was that $latex d$ is a rational number.
If you’re an ancient Greek mathematician, this comes as a bit of a shock. Mathematics is geometry. Numbers are lengths and areas and volumes. But that means there are numbers that are not rational.
So how do we define these numbers? This is a difficult question. We could define them to be the solutions of equations like $latex d^2 = r$, where $latex r$ is a rational number, but it this clearly doesn’t cover all possible lengths. We could define them to be the solutions of general polynomial equations (like $latex 5x^3 + 3x^2 -18x + 7 = 0$), but it’s very hard to see how we’d define operations like addition on them (and it also turns out that numbers like $latex \pi$ wouldn’t be included).
The solution isn’t obvious. As yet, I’m not sure how to go about introducing it, and it’s too late to think very clearly, so you’ll have to be patient.
My next post is scheduled for the 25th, and for a Christmas treat I’d like to put these numbers aside and give you one of my favourite proofs. The result is familiar – that there are infinitely many primes – but the proof is not the one you might know.
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# Solving Linear Equations
The process of solving linear equations involves isolating the variable on one side of the equation and then finding its value. This can be done by using inverse operations such as addition, subtraction, multiplication, and division.
For example, to solve the equation $3x+5=11$ , we can subtract 5 from both sides to get
$3x+5-5=11-5$
$3x=6$
Then, we can divide both sides by 3 to get
$\frac{3x}{3}=\frac{6}{3}$
$x=2$
So the solution to the equation $3x+5=11$ is $x=2$ .
We can also check our answer by plugging it back into the original equation to make sure it makes a true statement.
$3×2+5=11$
$6+5=11$
Since this is a true statement, we have correctly solved the equation.
## Inverse operations
To start with, you need to understand inverse operations. The inverse of an operation is its opposite, or the operation that gets you back to the number you started with.
For example, if you started with the number 6, and then add 4:
$6+4=10$
To get back from 10 to 6, you have to subtract 4 from 10.
$10-4=6$
Therefore, addition and subtraction are inverse operations.
Similarly, division is the inverse operation of multiplication, and vice versa.
$7×5=35$
$\frac{35}{5}=7$
It's important to know your inverse operations when solving linear equations, as you will see in the next section.
## Properties of linear equations
• The value of the variable that makes the linear equation true is called the solution or root of the linear equation.
• The solution of a linear equation is unaffected if the same number is added, subtracted, multiplied, or divided into both sides of the equation.
• The graph of a linear equation in one or two variables always forms a straight line.
## One-step linear equations
Some linear equations can be solved with only a single operation. For this type of equation, you will use the inverse operation to solve.
Example 1
Solve for n.
$n+8=10$
The inverse operation of addition is subtraction. So you will subtract 8 from both sides.
$n+8-8=10-8$
$n=2$
Example 2
Solve for y.
$\frac{3}{4y}=15$
The inverse operation of multiplication is division. So you must divide both sides by 3/4, which is the same as multiplying by $\frac{4}{3}$ .
$\frac{4}{3}×\frac{3}{4}y=\frac{4}{3}×15$
$y=20$
## Two-step linear equations
More commonly, we will need two operations to solve a linear equation.
Example 3
Solve for x.
$3x+5=11$
To isolate the variable, we follow the order of operations in reverse. We undo the addition before we undo the multiplication.
$3x+5-5=11-5$
Subtract 5 from both sides.
$3x=6$
We have undone one operation. One more to go.
$\frac{3x}{3}=\frac{6}{3}$
Divide each side by 3.
$x=2$
We have solved the equation!
The thing that makes the equation linear is that the highest power of x is ${x}^{1}$ (no ${x}^{2}$ or other powers-for those, see quadratic equations and polynomials).
## More complex linear equations
Both sides of the equation are supposed to be balanced for solving a linear equation. The equality sign denotes that the expressions on either side of the 'equal to' sign are equal. Since the equation is balanced, in order to solve it, certain mathematical operations are performed on both sides of the equation in a manner that does not affect the balance of the equation. Here is a more complex example related to a linear equation with one variable.
Example 4
Solve $2x-\frac{10}{2}=3\left(x-1\right)$
Step 1: Clear the fraction by dividing by 2.
$x-5=3\left(x-1\right)$
Step 2: Simplify both sides of the equation.
$x-5=3x-3$
$x=3x+2$
Step 3: Isolate x
$x-3x=2$
$-2x=2$
$x=-1$
Example 5
Solve $x=12\left(x+2\right)$
$x=12x+24$
Then subtract 24 on both sides of the equation.
$x-24=12x+24-24$
$x-24=12x$
Simplify.
$11x=-24$
Isolate x.
$x=-\frac{24}{11}$
Example 6
Solve $6x-19=3x-10$
$6x-19+19=3x-10+19$
Simplify.
$6x=3x+9$
Subtract 3x from both sides.
$6x-3x=3x+9-3x$
$3x=9$
Divide both sides by 3.
$\frac{3x}{3}=\frac{9}{3}$
$x=3$
## Practice questions on solving linear equations
a. $x+5=48$
Start by subtracting 5 from each side.
$x+5-5=48-5$
Simplify by performing the subtraction.
$x=43$
b. $x-150=125$
Begin by adding 150 to each side.
$x-150+150=125+150$
$x=275$
c. Solve the equation $3x+7=19$ .
Start by subtracting 7 from each side,
$3x+7-7=19-7$
Simplify by performing the subtraction.
$3x=12$
Next, we will divide each side by 3.
$\frac{3x}{3}=\frac{12}{3}$
Finally, perform the division to reach the solution.
$x=4$
d. Solve the equation $4x+8=8x-10$
First, bring the x terms to the left side of the equation by subtracting 8x from each side.
$4x-8x-8=-10$
Then move the -8 to the right side by adding 8 to each side.
$4x-8x=-10-8$
Perform the operations on each side.
$-4x=-18$
$4x=18$
Divide each side by 4.
$\frac{4x}{4}=\frac{18}{4}$
Finally, perform the division to reach the solution.
$x=\frac{9}{2}$
e. Solve the equation $5x+5=3x+7$
Start by subtracting 5 from each side.
$5x=2-3x$
Then subtract 3x from each side.
$2x=2$
Divide each side by 2.
$\frac{2x}{2}=\frac{2}{2}$
Finally, perform the division to reach the solution.
$x=1$
f. Solve the equation $3x+4\left(x-10\right)=x+20$
First, expand the parentheses and combine like terms.
$3x+4x-40=x+20$
$7x-40=x+20$
Then add 40 to each side.
$7x=x+60$
Next, subtract x from each side.
$6x=60$
Divide both sides by 6.
$\frac{6x}{6}=\frac{6}{6}$
$x=10$
g. Solve the equation $2\left(2x-5\right)=3\left(x-1\right)-4$
First, expand parentheses on both sides and combine like terms.
$4x-10=3x-7$
$4x=3x+3$
Subtract 3x from each side.
$4x-3x=3x+3-3x$
$x=3$
## Flashcards covering the Solving Linear Equations
Algebra 1 Flashcards
## Get help learning about solving linear equations
Working with linear equations can be difficult for many students. There are many steps and it can be tricky to keep the two sides of the equation balanced. It can also be hard to know when to add, subtract, multiply, or divide to make the sides of the equation balanced. If your student is having a hard time solving linear equations, consider having them work with a professional math or algebra tutor who can give them the 1-on-1 attention that is often lacking in the classroom.
A private tutor supports your student's in-class learning in a setting without distractions so they have an easier time focusing on each aspect of each problem. They are available to answer questions as your student comes up with them and they can supervise your student as they do their homework. This allows them to point out mistakes immediately and guide your student as they figure out the correct way to do the problems so bad habits don't take hold between class sessions. As they work together, your student's tutor can slow down and take extra time when your student needs extra help with a particular concept and can speed through concepts that your student easily understands. This makes tutoring as efficient as it is effective.
If solving linear equations is a struggle for your student, help is available. Contact the Educational Directors at Varsity Tutors today to learn more about how tutoring can help your student master solving linear equations and more. We will be happy to set you up with a professional tutor who can guide your student as they learn to solve linear equations.
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# 6/24/2012
1
Engineering Mathematics 120 Engineering Mathematics 120
C K Tan C K Tan
Lecture 5
*Limits and Asymptotes *
Overview Overview
calculate limits of functions to better
understand their behaviour.
determine asymptotes of functions,
to better visualize graphs of
functions.
Math 120 Lecture 5
Motivation Motivation
The limiting behaviour of many
engineering functions is crucial to
know how they arise and are
implemented in practice.
You need to be aware of this when
working with functions in an
engineering context.
Math 120 Lecture 5
e.g. The Charging and discharging of e.g. The Charging and discharging of
a capacitor in electrical system a capacitor in electrical system
Where Q = charge, R = resistance, C =
capacitance, t = time.
Initially, Q = 0 at t = 0.
Finally, t = , Q = Q
o
Math 120 Lecture 5
Key Concept in this lecture Key Concept in this lecture
Limits help us to analyze
function behaviour.
Asymptotic behaviour of
functions.
Continuity of functions.
Math 120 Lecture 5
Contents Contents
Various types of function limits
and how to calculate them,
Vertical and horizontal
asymptotes of functions.
Continuous and discontinuous
functions.
Math 120 Lecture 5
6/24/2012
2
Outcomes Outcomes
determine elementary limits of
functions and use these to help
sketch their graphs.
recognize discontinuities of functions
and their type.
continue looking at the idea of limits
as we meet new functions over the
next few weeks.
Math 120 Lecture 5
Limit of a function Limit of a function
Consider the function
f(x) = x
2
+ 1.
As x approaches the value 1
(in maths notation, we write
x 1
The value of f(x) approaches
1
2
+ 1 = 2, i.e. f(x) 2
Math 120 Lecture 5
function
As x 1 (but x doesnt
get to exactly 1),
g(x) 1
2
+ 1 = 2,
though g(1) = 0 2,
Math 120 Lecture 5
In both of these cases, we say that the limit of the function
as x approaches 1 is the value 2
notation.
More generally, we say that the limit of a function f(x) as x
approaches a is L, if the value of f(x) tends to L as x tends to a.
In formal notation, this is written as
Math 120 Lecture 5
Caution:
the value of the limit does not
necessarily equal the function value at
the point x = a.
the existence of the limit does not even
depend on whether f is defined at a or
not, so long as the values of f tend to L
when x tends to a.
Math 120 Lecture 5
In many cases (not all), the limit of a
function at a point is found simply by
substitution.
Math 120 Lecture 5
6/24/2012
3
If it leads to an expression of the form 0/0 (we call this
an indeterminate form), we may need to first cancel a
common factor in the numerator and denominator.
Note how the 0/0 did not end up being equal to 1,
like one might have expected. The extra work was
necessary to determine the true value of the limit.
Math 120 Lecture 5
Limits involving Limits involving
We first consider the behaviour of the hyperbolic function f(x) = 1/x
for large and small values of x.
At x = 0, the function is, of course, not defined. Near x = 0, the
magnitude of 1/x becomes very large, as we can see from the
following calculations. Taking a sequence of small positive x values,
we have
Clearly, as x 0
+
(i.e. x goes to zero from above), 1/x + . We
write this as
<----- Eq (1)
Math 120 Lecture 5
Similarly, taking a sequence of small negative x values, we have
Clearly as x 0
-
(i.e. as x goes to zero from below), 1/x - . We
write this as
<---- Eq (2)
(It is correct to say that the limit doesnt exist in both Eq 1 and Eq 2
since the functions increases and decreases without bound).
Math 120 Lecture 5
In this instance, we say that the hyperbolic function has a vertical asymptote
x = 0, i.e. the graph of the function approaches the shape of this vertical line
as x 0.
Let us now consider the behaviour for values of x large in magnitude. For a
sequence of large positive values, we get
Clearly, as x +, 1/x 0
+
, i.e. lim 1/x = 0.
For a sequence of large negative values, we get
Math 120 Lecture 5
Clearly, as x -, 1/x 0
-
,
i.e. lim 1/x = 0.
In this context, we say that f(x)
has a horizontal asymptote
y = 0. Because the graph of the
function approaches the
horizontal line y = 0. Putting
together our observations of the
hyperbolic function, we can sketch
its graph as follows.
Math 120 Lecture 5
Limit of a function Limit of a function
When x a, f(x) L,
We say that lim f(x) = L. (L uniquely
defined) xa
E.g.
We say approaches the limit of 2
when x approaches infinity.
6/24/2012
4
A snakes head or tail approaching an A snakes head or tail approaching an
electric fence (watch out!!) electric fence (watch out!!)
y = 2
A snake tries to cross an electric A snake tries to cross an electric
fence unsuccessfully. fence unsuccessfully.
Math 120 Lecture 5
In general,
means that the value of f(x) tends to L as x approaches .
Note that for many functions, such a limit does not exist.
In particular, polynomial functions do not have limits as x
Ex:
Note that a rational function is a function of the form
Both f and g are polynomials (i.e. its simply a ratio of two
polynomials.)
Math 120 Lecture 5
Vertical Asymptotes Vertical Asymptotes
Arise for any rational function near points
where the denominator is equal zero (and
the numerator is some constant value
zero)
For y = 1/(3-x), vertical asymptote is x = 3.
For y = 1 / [ (x-1) (x 3) ], vertical
asymptotes are: x = 1 and x = 3.
[These are like electric fences where
snakes cant cross!]
Math 120 Lecture 5
Horizontal Asymptotes Horizontal Asymptotes
If they exist, they can be found by taking
the limit of the function as the variable
goes to . For rational functions, the
basic step to remember is to divide the
numerator and denominator by the
highest power of x in the denominator.
For y = (2x-1)/x = 2 1/x,
horizontal asymptote: y = 2.
Math 120 Lecture 5
Example: Example:
We have
Which shows that the
function has a horizontal
asymptote y = 1.
Notice: vertical
asymptotes are x= 2 and
x = -2
Math 120 Lecture 5
Curves for the
function
6/24/2012
5
A learner tracks down a snake by A learner tracks down a snake by
identifying: identifying:
the zone(s) where the snake is allowed
to wander,
the forbidden zones that cannot be
crossed by the snake,
the positions for its ends (called head
and tail for convenience),
possible crossings on the x-axis and y-
axis by the snake,
and any other twists and turns of its
body that will lead to a clear trace of
its body.
Example Example
Find the following limits if they exist Find the following limits if they exist
Solution:
Math 120 Lecture 5
Example Example
Find the following limits if they exist Find the following limits if they exist
Solution:
Math 120 Lecture 5
Example Example
Find the following limits if they exist Find the following limits if they exist
Solution:
Math 120 Lecture 5
Note the following: Note the following:
## f(x) approaches as x goes to infinity.
Limit of f(x) does not exist!
limit of a rational function
as x or as x - ,
The limit does not exist if the highest power of x in
the numerator is larger than in the denominator.
The limit will be nonzero constant if the powers are
the same
The limit will be zero if the highest power of x in the
numerator is smaller than in the denominator
Math 120 Lecture 5
Continuity Continuity
The notion of continuous functions is
very important in calculus. The laymens
definition of continuity is this: a function is
continuous if we can draw the graph of it on
a peace of paper without lifting the pen off
the paper. A function like the one below is
clearly not continuous on [1, 6]
(although one can say that it is continuous
on the individual intervals (1,4) and (4,6)
respectively)
Math 120 Lecture 5
6/24/2012
6
A finite jump in function values like this is considered to be a finite
discontinuity.
Vertical asymptotes in the graph of a function are considered to be
infinite discontinuities. Its important to recognize these when they
occur as extra care is needed to integrate across them.
Math 120 Lecture 5
Definition of a continuous function Definition of a continuous function
A function y = f(x) is continuous over
its domain
if f(x) is defined for every x in the domain , in the domain , in the domain , in the domain ,
If If If If lim lim lim limf(x) exists for any value of a in the f(x) exists for any value of a in the f(x) exists for any value of a in the f(x) exists for any value of a in the
x a domain, and domain, and domain, and domain, and
If If If If lim lim lim limf(x) = f(a) f(x) = f(a) f(x) = f(a) f(x) = f(a)
xa xa xa xa
A snake satisfies these conditions so it
could represent a continuous curve.
Slanting/Oblique Asymtote Slanting/Oblique Asymtote
Neither vertical nor horizontal. E.g. what
happens to y when x approaches ?
Check behaviour of y (x 1) when x
approaches .
Calculate
Math 120 Lecture 5
Catch the snake(s) given by Catch the snake(s) given by
1. Find limits:
2. i.e. (i) y =g(x) is undefined when x = 3,
(ii) and
3. Electric Fences (asymptotes) :
Vertical x = 3 and slanting y = x 1
5. Locate body of snake(s). (0, -4/3), (4, 4)
The snakes of The snakes of
x
y
3
0
(0,-8)
y = x - 1
(0,-4/3)
(4,4)
x= 3
Snake of the form Snake of the form
Both Head and tail approaches y = 0 from
above.
Electric Fence: y = 0
As f(x) is undefined at x=0.5 and x = 1,
Electric Fences: x = 0.5, x = 1.
Behaviour of snake near x = 1:
Behaviour of snake near x = 0.5:
6/24/2012
7
There are 3 snakes here! There are 3 snakes here!
x
y
1 0.5 0
Snake 1
Snake 2
Snake 3
Turning
point of the
snakes
body!
x
Summary Summary
Various types of function limits and
how to calculate them,
Vertical and horizontal asymptotes
of functions.
Continuous and discontinuous
functions.
|
# Binomial Theorem
## Who developed the Binomial Theorem?
The journey of binomial started since the ancient times. Greek mathematician Euclid, in 4th century B.C, has given one of the special case of binomial theorem. Since then, many research work is going on and lot of advancement had been done till date. One of the biggest contributor in binomial theorem is considered as Persian mathematician Al-Karaji. He has explained the binomial coefficients with the triangular pattern. He also proved the binomial theorem and the pascal’s triangle.
## What is the definition of Binomial in math? Explain Binomial Expression?
The word binomial is a special case of the word – “Polynomial”. Polynomial means an algebraic expression containing two or more algebraic terms. So, the binomial is also an algebraic expression containing exactly two different terms.
So, we can define binomial expression as an algebraic expression consisting of two different terms. Here, in its definition, the word different terms is very important to note. Different terms here means either the two terms should have different variables or the different powers on its variable.
Example
x + y, x2 + y3, x + xare the expression having two different terms and thus are categorized as binomial expression. Whereas, x + 2x or x2y + 2 x2y can be simplified into one terms and thus are not the example of binomial expression.
## What is Binomial Theorem?
Binomial Theorem, in algebra, focuses on the expansion of exponents or powers on a binomial expression. This theorem was given by newton where he explains the expansion of (x + y)n for different values of n.
As per his theorem, the general term in the expansion of (x + y)n can be expressed in the form of pxqyr, where q and r are the non-negative integers and also satisfies q + r = n. Here, ‘p’ is called as the binomial coefficient.
## What is (a + b)n?
In (a + b)n, a + b is the binomial and thus the expansion of (a + b)n can be easily calculated by the used of binomial theorem. But let’s here understand the binomial theorem from the basic level. Here, we will understand how the formula of binomial expansion is derived?
If we closely examine the expansion of (a + b) for different exponents, we observe that,
For (a + b)0 = 1
For (a + b)= a + b
For (a + b)2 = a2 + 2ab + b2
For (a + b)3 = a3 + 3a2b + 3ab2 + b3
For (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
From the above expansion, we can note few important properties of its expansion
• The number of terms in the expansion of (a + b)n is n + 1 that is, if n = 3, the number of terms will be 3 + 1 = 4 and so on.
• The power of a starts from n and decreases till it becomes 0.
• Similarly, the power of b starts from 0 and increases till n
• The binomial coefficients of the terms equidistant from the beginning and the end are equal. For Example, in (a + b)4 the binomial coefficient of a4 & b4, a3b & ab3 are equal
• The sum of the powers of its variables on any term is equals to n.
• The last and the very important conclusion which can be drawn is explained below with the help of figure.
The above triangle which you can see is called as the Pascal’s Triangle. This triangle represents the binomial coefficients for the expansion of different powers of (a + b).
The first row in the triangle represent the expansion of binomial theorem for the power zero. Similarly, the second row represent the binomials coefficient of the terms for the power one.
Pascals triangle is also very easy to understand and learn. Refer below figure to understand it.
From the figure, it is very clear that any number in the triangle is the addition of its upper two numbers.
Before generalizing the formula for the binomial expansion, just note that the binomials coefficients are nothing but the values of nCr for different values of r.
Thus, we can now generalize the binomial theorem for any non-negative power n.
(x + y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + … + nCrxn-ryr + … + nCnxn-nyn.
From the above equation, we observe that
First term, T1 = nC0xn
Second term, T2 = nC1xn-1y1
Third term, T3 = nC2xn-2y2
So the general term in the expansion of (x + y)n is
Tr+1 = nCrxn-ryr
## How many terms in the expansion of (x + y + z + w)10 have?
Given problem is an expansion of quadrinomial or a polynomial having four terms. To understand the concept behind this problem, let’s go back to one of the very important property of binomial theorem. It says that, the sum of the powers of its variables on any term is equals to n, where n is the exponent on (x + y).
Thus, if we see the expansion of (x + y)n
(x + y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + … + nCrxn-ryr + … + nCnxn-nyn.
Here, we can see that, the power n is being distributed to the variables x and y in every permutation. Thus, we can relate this distribution with the coin – beggar’s method of permutation and combination.
As per the coin – beggar’s method, the number of ways to distribute n identical coins to p beggar’s will be n + p -1Cp-1.
In our case, number of coins is the exponent of any polynomial and number of beggar’s is the number of terms in the polynomial.
Thus for (x + y + z + w)10, we have, n = 10 and p = 4
So, the number of terms in its expansion will be = 10+4–1C4-1 = 13C3
This method can also be verified for the binomial expansion, where we have n exponent and p = 2. So, the number of terms = n+1C1 = n+1 terms.
## What is the binomial coefficient?
In (a + b)2 = 1a2 + 2ab + 1b2
1, 2 and 1 are called as the binomial coefficients of a2, ab and b2 respectively.
Similarly, in (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
1, 3, 3 and 1 are called as the binomial coefficients of a3, a2b, ab2 and b3 respectively.
So, In general, in the expansion of (x + y)n
(x + y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + … + nCrxn-ryr + … + nCnxn-nyn.
all the terms have a constant multiplied with the variables in the form of nCr. These nCr are called as the binomial coefficients of different terms (depending upon the value of r).
One more important point to note from here, is the sum of the binomial coefficients can be easily calculated just by replacing the variables to 1.
Example
The sum of the binomial coefficients of (x + y)n will be calculated as follows:
Put x = 1 and y = 1 in the expansion of (x + y)n, we get
(1 + 1)n = 2n = nC0 + nC1 + nC2 + … + nCr + … + nCn
## How do you use Pascal's triangle?
Pascal’s triangle is named after the French mathematician Blaise Pascal. Pascal’s triangle is a triangular arrangement of binomial coefficients for the expansion of different powers.
Below is the pascal’s triangle for expansion till the exponent five.
This triangle can be expanded for any number of rows. Each row represents the binomial coefficient for any specific exponent.
For Example, the first and second row represent the binomial coefficients for the expansion of (x + y) with the exponent 0 and 1 respectively.
This triangle is also very easy to construct as each element in this triangle is the sum of upper two elements.
## What is the binomial series?
Binomial series is a special series in mathematics, also called as the maclaurin series.
This series is a special case of the binomial theorem, where x = 1 and y = x
Binomial Series:
(1 + x)n = nC0 + nC1x + nC2x2 + …… nCrxr + …. + nCnxn
Depending upon the values of x and n, the series can be converging or diverging.
## How to apply Binomial Theorem if n is negative or fractional?
The Binomial Theorem for the expansion of (x + y)n where n ∉ Iwill be expanded as,
So, the general term here will be,
Note: In this case, we can’t find the binomial coefficients using nCr directly, as this is not defined for negative n.
For (a + x)n where n ∉ I+,
In this case, where n is non positive integer, the series will converge only for (x/a) <1. But at the same time, the number of terms will be infinite i.e. infinite series.
Similarly for (1 + x)n where n ∉ I+,
Above series converges for |x| < 1
From the above series, we can get few important series which must be remembered as a formula.
• (1 + x)-1 = 1 – x + x2 – x3 + x4 - ……..∞
• (1 - x)-1 = 1 + x + x2 + x3 + x4 - ……..∞
• (1 + x)-2 = 1 – 2x + 3x2 – 4x3 + 5x4 - ……..∞
• (1 - x)-2 = 1 + 2x + 3x2 + 4x3 + 5x4 - ……..∞
## How to find the term independent of x?
Term independent of ‘x’ here means that term in the binomial expansion which does not have any variable x involved in it.
Example
(x + y)2 = x2 + 2xy + y2, the third term that is, y2 is the term which is independent of ‘x’, while the first term that is, x2 is the term independent of y.
Example
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Here, the first term is independent of y and the fourth term is independent of x. Second and fourth terms involves both the variables x and y and thus are not independent of neither x or y.
Now, let’s learn – How to find the independent term in binomial expansion having any power.
Follow the below steps to find it:
• For the given binomial with any power, write down its general term.
• Combine all the ‘x’ terms using the laws of exponent (if not combine).
• Equate the power or indices of ‘x’ to be zero to find the value of variable ‘r’.
• After knowing r, if r is positive integer then (Tr+1)th term will be the term independent of x otherwise no such term exists.
If you just follow the above steps, you can easily find all such terms. See below example to understand the same.
Example
Which term is independent of x in the expansion of (x – x2)10
Step 1: Writing its general term,
Tr+1 = 10Crx10-r (-1)r(x2)r
Step 2: Now combining all the x terms,
Tr+1 = 10Crx10-r + 2r (-1)r = 10Crx10 + r (-1)r
Step 3: Equating the power of x to zero
10 + r = 0 that is, r = -10 which is not a positive integer.
Thus, no such term exists.
Example
Which term is independent of x in the expansion of (x – 1/x)20?
Step 1: Writing its general term,
Tr+1 = 20Crx20-r (-1)r(1/x)r
Step 2: Now combining all the x terms,
Tr+1 = 20Crx20-r - r (-1)r = 20Crx20 - 2r (-1)r
Step 3: Equating the power of x to zero
20 - 2r = 0 that is, r = 10 which is a positive integer.
Thus, r+1 = 11th term is independent of x.
This method is also helpful in finding any term or its coefficient having specific power of x. For example, in the example 4, if we are asked to find the coefficient of x15, we will solve as follows:
We have, Tr+1 = 10Crx10-r + 2r (-1)r = 10Crx10 + r (-1)r
For x15, we should equate the power of x to 15
that is, 10 + r = 15
thus, r = 5 that is, 6th term.
So, the coefficient = T6 = -10C5
Binomial Theore
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
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# Multiplication on a Number Line
Multiplication on a number line is performing the multiplication operation on a set of numbers and finding the result using a number line.
Since multiplication is also known as repeated addition, the result is obtained after performing addition for the required number of times.
Using a number line, we can multiply two positive numbers, two negative numbers, or a positive and a negative number.
## With Positive Numbers
While multiplying two positive numbers, we move towards the right on a number line, starting from 0.
For example, 3 × 4 represents 3 times 4 or 3 groups of 4.
Step 1: 4 being positive, we move 4 units to the right of 0.
Step 2: A total of 3 such moves are taken, i.e., we skip count forward by 4s three times to the right of 0.
Thus, we get 3 × 4 = 12 on the number line.
Find 9 × 3 using a number line.
Solution:
As 9 × 3 means 9 times 3, the result is obtained by adding 3 nine times.
At first, a number line with the first few positive multiples of 3 is drawn. Then we skip count by 3s to the right starting from 0 nine times.
Thus we find 9 × 3 = 27 on the number line.
## With a Positive and a Negative Number
Multiplicating a positive and a negative number gives a negative number by the odd rule of multiplying negative numbers.
Let us multiply 6 × (-4)
6 × (-4) means 6 times of (-4).Thus we need to add (-4) six times.
Step 1: As (-4) is negative and lies to the left of 0, a negative number line is drawn by plotting the first few negative multiples of 4 starting from 0.
Step 2: Then, we count back by 4s six times to the left, starting from 0.
Thus we locate the result 6 × (-4) = (-24) on the number line.
Calculate (-5) × 2 on a number line.
Solution:
As we know, multiplication is commutative; we can write (-5) × 2 as 2 × (-5), which means 2 times (-5).
Here the result is obtained by adding (-5) two times, i.e., 2 × (-5) = (-5) + (-5)
Thus counting back by 5s twice to the left, starting from 0, will get us to -10.
So the product is (-5) × 2 = 2 × (-5) = (-10) is on the number line.
## With Negative Numbers
Multiplication of two negative numbers gives a positive number by the even rule of multiplying negative numbers.
For instance, to multiply (-7) × (-4), we see that the result will be positive as both the numbers are negative.
So we calculate 7 × 4 following the rule of multiplying positive numbers.
Step 1: A positive number line starting at 0 and with the first few multiples of 4 is drawn.
Step 2: As 7 × 4 denotes 7 times 4, we count forward by 4s seven times to the right starting from 0.
Thus we get the result (-7) × (-4) = 7 × 4 = 28
Find the product (-6) × (-2) using a number line.
Solution:
This multiplication result will be positive as (-6) and (-2) are negative numbers.
Thus (-6) × (-2) can be written as 6 × 2, i.e., 6 times 2.
To add 2 eight times, we count forward by 2s six times to the right starting from 0.
We find (-6) × (-2) = 6 × 2 = 12.
## Multiplication of Fractions
A number line can also be used to multiply fractions by integers or another fraction and find the result.
For example, ${\dfrac{1}{5}}$ × ${\dfrac{2}{3}}$, can be written as ${\dfrac{1}{5}}$ of ${\dfrac{2}{3}}$.
Step 1: As the second fraction ${\dfrac{2}{3}}$ lies between 0 and 1, the region from 0 to 1 is divided into 3 equal parts.
Each of the two parts from 0 to ${\dfrac{1}{3}}$ and from ${\dfrac{1}{3}}$ to ${\dfrac{2}{3}}$ is highlighted.
Step 2: The denominator of the first fraction ${\dfrac{1}{5}}$ is 5, so each of the 3 parts is again subdivided into 5 equal parts.
Thus there is a total of 15 parts from 0 to 1.
Step 3: Since the numerator of the first fraction ${\dfrac{1}{5}}$ is 1, we highlight the 1st part of each subdivided part from 0 to ${\dfrac{2}{3}}$.
This is ${\dfrac{1}{5}}$ of the two third or ${\dfrac{1}{5}}$ of ${\dfrac{2}{3}}$
Step 4: As there is a total of 15 parts, the first fraction between 0 and ${\dfrac{1}{3}}$ can be written as ${\dfrac{1}{15}}$ or 1 part of total 15 parts.
Similarly, the first fraction between ${\dfrac{1}{3}}$ and ${\dfrac{2}{3}}$ is written as ${\dfrac{1}{15}}$ or 1 part of total 15 parts.
Adding these two fractions, we get the result ${\dfrac{1}{15}}$ + ${\dfrac{1}{15}}$ = ${\dfrac{2}{15}}$ So, ${\dfrac{1}{5}}$ × ${\dfrac{2}{3}}$ = ${\dfrac{2}{15}}$.
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# Functions and Equivalent Sets
The present unit is part of the following walks
## Introduction
In the following you will find a short summary of this unit. For detailed information please see the full text or download the pdf document at the end of this page.
The present unit is the fourth unit of the walk The Axioms of Zermelo and Fraenkel. Based on the relations introduced in Unit Direct Products and Relations we will introduce functions as specific relations:
We will explain the following terms:
You will learn the main results of this unit:
## Functions
The definition of a function will be based on direct products and relations explained in Unit Direct Products and Relations:
Definition. Let $A$ and $B$ be two sets.
(a) A function $f : A \rightarrow B$ from the set $A$ into the set $B$ is a triple $(f, A, B)$ where the set $f$ is a subset of the direct product $A \times B$ with the following property:
For each element $x$ of the set $A$, there is exactly one element $y$ of the set $B$ such that the pair $(x, y)$ is contained in the set $f$.
A function $f : A \rightarrow B$ from the set $A$ into the set $B$ is also called a mapping from the set $A$ into the set $B$ or a transformation from the set $A$ into the set $B$.
(b) Let $f : A \rightarrow B$ be a function from the set $A$ into the set $B$, and let $x$ be an element of the set $A$. The unique element $y$ of the set $B$ such that the pair $(x, y)$ is contained in the set $f$ is denoted by $y = f(x)$. We also write $f : x \mapsto y$ or, equivalently, $f : x \mapsto f(x)$.
(c) Let $f : A \rightarrow B$ be a function from the set $A$ into the set $B$. Then the set
$$G_f := \{ \big(x, f(x) \big) \in A \times B \mid x \in A \}$$
is called the graph of the function $f$.
French / German. Function = Fonction = Funktion. Mapping = Application = Abbildung. Graph = Graphe = Graph.
Theorem. Let $A$ and $B$ be two sets, and let $f : A \rightarrow B$ and $g : A \rightarrow B$ be two functions from the set $A$ into the set $B$.
Then we have $f = g$ if and only if $f(x) = g(x)$ for all elements $x$ of the set $A$.
Definition. Let $f : A \rightarrow B$ be a function from a set $A$ into a set $B$.
(a) The set $A$ is called the domain of the function $f$, and the set $B$ is called the codomain of the function $f$.
(b) The set
$$R := \{ f(x) \mid x \in A \} \subseteq B$$
is called the range of the function $f$ or, equivalently, the image of the function $f$. We also write $R = f(A)$ or $R = \mbox{Image } f$. (Sometimes the word range is also used in the meaning of the codomain.)
(c) Let $X$ be a subset of the set $A$. Then the set
$$Y := \{ f(x) \mid x \in X \} \subseteq B$$
is called the image of the set $X$ under the function $f$. We also write $Y = f(X)$.
French / German. Domain = Domaine de définition = Definitionsbereich; Codomain = Codomaine = Wertebereich; Image (or range) = Imâge = Bild.
Definition. Let $f : A \rightarrow B$ be a function from a set $A$ into a set $B$, and let $Y$ be a subset of the set $B$. We set
$$f^{-1}(Y) := \{ x \in A \mid f(x) \in Y \} \subseteq A.$$
The set $f^{-1}(Y)$ is called the inverse image of the set $Y$ under the function $f$.
French / German. Inverse Image = Imâge réciproque = Urbild.
Definition. (a) Let $A$ be a set, and let $f : A \rightarrow A$ be the function defined by
$$f(x) := x \mbox{ for all } x \in A$$
or, equivalently, by
$$f := \{ (x, y) \in A \times A \mid y = x \} = \{ (x, x) \in A \times A \mid x \in A \}.$$
The function $f : A \rightarrow A$ is called the identity function on the set $A$.
It is denoted by $\mbox{id}_A : A \rightarrow A$ or, equivalently, by $\mbox{id} : A \rightarrow A$ if no confusion may arise.
Note that $\mbox{id} = \emptyset$ if $A = \emptyset$.
(b) Let $B$ be a set, let $A$ be a subset of the set $B$, and let $g : A \rightarrow B$ be the function defined by
$$g(x) := x \mbox{ for all } x \in A$$
or, equivalently, by
$$g := \{ (x, y) \in A \times B \mid y = x \} = \{ (x, x) \in A \times B \mid x \in A \}.$$
The function $g : A \rightarrow B$ is called the inclusion map or, equivalently, the inclusion function.
It is denoted by $\iota : A \rightarrow B$ or, equivalently, by $A \hookrightarrow B$.
French / German. Identity = Identité = Identität.
## Injective, Surjective and Bijective Functions
The terms injective, surjective and bijective play an important role for the investigation of functions:
Definition. Let $A$ and $B$ be two sets, and let $f : A \rightarrow B$ be a function from the set $A$ into the set $B$.
(a) The function $f : A \rightarrow B$ is called injective if $f(x) \neq f(x’)$ for each two elements $x$, $x’$ of the set $A$ such that $x \neq x’$.
(b) The function $f : A \rightarrow B$ is called surjective if for each element $y$ of the set $B$ there exists an element $x$ of the set $A$ such that $f(x) = y$.
(c) The function $f : A \rightarrow B$ is called bijective if $f$ is injective and surjective.
French / German. Injective = Injective = Injektiv. Surjective = Surjective = Surjektiv. Bijective = Bijective = Bijektiv.
## The Composition of Functions
The most important result of this sections is that the set of the bijective functions $f : A \rightarrow A$ from a set $A$ into itself forms a group with respect to the composition of functions.
Definition. Let $A$, $B$ and $C$ be three sets, and let $f : A \rightarrow B$ and $g : B \rightarrow C$ be two functions from the set $A$ into the set $B$ and from the set $B$ into the set $C$, respectively.
Define the function $h : A \rightarrow C$ by $h(x) := g \big( f(x) \big)$ for all elements $x$ of the set $A$.
(a) The function $h : A \rightarrow C$ is called the composite of the two functions $f : A \rightarrow B$ and $g : B \rightarrow C$.
(b) The function $h : A \rightarrow C$ is denoted by $g \circ f : A \rightarrow C$.
French / German. Composition of two functions = Composition de deux fonctions = Verkettung / Hintereinanderausführung von zwei Funktionen.
Proposition. Let $A$, $B$ and $C$ be three sets, and let $f : A \rightarrow B$ and $g : B \rightarrow C$ be two functions from the set $A$ into the set $B$ and from the set $B$ into the set $C$, respectively.
(a) If the functions $f : A \rightarrow B$ and $g : B \rightarrow C$ are injective, then the function $g \circ f : A \rightarrow C$ is also injective.
(b) If the functions $f : A \rightarrow B$ and $g : B \rightarrow C$ are surjective, then the function $g \circ f : A \rightarrow C$ is also surjective.
(c) If the functions $f : A \rightarrow B$ and $g : B \rightarrow C$ are bijective, then the function $g \circ f : A \rightarrow C$ is also bijective.
Definition. Let $A$ and $B$ be two sets, and let $f : A \rightarrow B$ be a bijective function. We define the function $g : B \rightarrow A$ as follows: For each element $y$ of the set $B$, let $x$ be the unique element of the set $A$ such that $f(x) = y$. Set $g(y) := x$.
The function $g : B \rightarrow A$ is called the inverse function of the function $f : A \rightarrow B$. It is denoted by $f^{-1} : B \rightarrow A$.
French / German. Inverse function = Fonction inverse = Inverse Funktion.
Definition. (a) A pair $(G, *)$ consisting of a non-empty set $G$ and an operation
$$* : G \times G \rightarrow G$$
on the set $G$ is called a group if the following conditions are fulfilled:
(i) We have
$$x * y \in G \mbox{ for all } x, y \in G \mbox{ (closure)}.$$
(ii) We have
$$(x * y) * z = x * (y * z) \mbox{ for all } x, y, z \in G \mbox{ (associativity)}.$$
(iii) There exists an element $id$ of the group $G$ such that
$$x * id = id * x = x \mbox{ for all } x \in G \mbox{ (existence of an identity element)}.$$
(iv) For each element $x$ of the group $G$, there exists an element $y = y_x$ of the group $G$ such that
$$x * y = id \mbox{ (existence of an inverse element)}.$$
The element $y$ is denoted by $x^{-1}$.
(b) If the pair $(G, *)$ is a group, we often just say that $G$ is a group or that $G = (G, *)$ is a group.
(c) A group $G = (G, *)$ is called a commutative group or, equivalently, an abelian group if we have
$$x * y = y * x\mbox{ for all } x, y \in G.$$
French / German. Group = Groupe = Gruppe. Commutative = Commutatif = Kommutativ. Abelian = Abélien = Abelsch.
Theorem. Let $A$ be a set, and let
$${\cal B}(A) := \{ f : A \rightarrow A \mid f \mbox{ is bijective} \}$$
be the set of the bijective functions from the set $A$ into itself.
Then the pair $\big( {\cal B}(A), \circ \big)$ is a group where $\circ$ denotes the composition of two functions of the set ${\cal B}(A)$. In general, this group is not abelian.
## Restrictions and Extensions of Functions
Given a function $f : A \rightarrow B$ from a set $A$ into a set $B$ it is often useful to restrict the function $f$ to a function $f : A_1 \rightarrow B$ with $A_1 \subseteq A$ or to extend the function $f$ to a function $f : A_2 \rightarrow B$ with $A \subseteq A_2$. In this unit we only look at the basic definition. This topic becomes more interesting when investigating functions with specific properties such as continuous or holomorphic functions.
Definition. Let $f : A \rightarrow B$ and $g : A’ \rightarrow B’$ be two functions from a set $A$ into a set $B$ and from a set $A’$ into a set $B’$, respectively. Suppose that the sets $A’$ and $B’$ are subsets of the sets $A$ and $B$, respectively.
We say that the function $g : A’ \rightarrow B’$ is a restriction of the function $f : A \rightarrow B$ or, equivalently, that the function $f : A \rightarrow B$ is an extension of the function $g : A’ \rightarrow B’$ or, equivalently, that the function $g : A’ \rightarrow B’$ is induced by the function $f : A \rightarrow B$ if we have
$$f(x) = g(x) \mbox{ for all } x \in A’.$$
In this case we write $g = f|_{A’} : A’ \rightarrow B’$.\index{$f_A : A \rightarrow B$}
French / German. Restriction = Restriction = Einschränkung. Extension = Prolongement = Fortsetzung. Induced by = Induit par = Induziert von.
Proposition. Let $A_1$, $A_2$, $B_1$ and $B_2$ be four non-empty sets, and let $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ be two functions of the sets $A_1$ and $A_2$ into the sets $B_1$ and $B_2$, respectively.
Suppose that $f_1(x) = f_2(x)$ for all elements $x$ of the set $A_1 \cap A_2$. Note that this implies that $f_1(x) = f_2(x) \in B_1 \cap B_2$ for all $x \in A_1 \cap A_2$.
(a) There exists exactly one function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ such that $f|_{A_1} = f_1$ and $f|_{A_2} = f_2$.
(b) If the functions $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ are surjective, then the function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ is also surjective.
Proposition. Let $A_1$, $A_2$, $B_1$ and $B_2$ be four non-empty sets such that $A_1 \cap A_2 = \emptyset$ and $B_1 \cap B_2 =\emptyset$, and let $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ be two functions of the sets $A_1$ and $A_2$ into the sets $B_1$ and $B_2$, respectively.
(a) There exists exactly one function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ such that $f|_{A_1} = f_1$ and $f|_{A_2} = f_2$.
(b) If the functions $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ are injective, then the function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ is also injective.
(c) If the functions $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ are surjective, then the function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ is also surjective.
(d) If the functions $f_1 : A_1 \rightarrow B_1$ and $f_2 : A_2 \rightarrow B_2$ are bijective, then the function $f : A_1 \cup A_2 \rightarrow B_1 \cup B_2$ is also bijective.
## Functions and Equivalence Relations
In the present section we will explain the relation between functions and equivalence relations. Equivalence relations have been explained in Unit Direct Products and Relations.
Definition. Let $f : A \rightarrow B$ be a function from a set $A$ into a set $B$, and let $\sim$ be an equivalence relation on the set $A$. For an element $x$ of the set $A$, let $\bar{x} := \{ z \in A \mid z \sim x \}$ be the equivalence class defined by the element $x$.
The function $\alpha : \bar{A} \: \rightarrow B$ from the set $\bar{A}$ of the equivalence classes of the set $A$ into the set $B$ defined by $\alpha : \bar{x} \mapsto f(x)$ is called well defined if we have
$$f(x) = f(y) \mbox{ for all } x, y \in A \mbox{ with } x \sim y.$$
French / German. Well defined = Bien défini = Wohldefiniert.
Proposition. Let $f : A \rightarrow B$ be a function from a set $A$ into a set $B$. For two elements $x$ and $y$ of the set $A$, set
$$x \sim y \mbox{ if and only if } f(x) = f(y).$$
(a) The relation $\sim$ is an equivalence relation on the set $A$.
(b) For an element $x$ of the set $A$, let $\bar{x} := \{ z \in A \mid z \sim x \}$ be the equivalence class defined by the element $x$.
Then the function $\alpha : \bar{A} \: \rightarrow B$ from the set $\bar{A}$ of the equivalence classes of the set $A$ into the set $B$ defined by $\alpha : \bar{x} \mapsto f(x)$ is well defined and injective.
## Equivalent Sets
Equivalent sets play an important role for the definition of the cardinality of a set.
Definition. Two sets $A$ and $B$ are called equivalent if there exists a bijective function $f : A \rightarrow B$ from the set $A$ onto the set $B$. If the sets $A$ and $B$ are equivalent, we write $A \sim B$.
French / German. Equivalent = Équivalent = Äquivalent.
Proposition. Let $A$, $B$ and $C$ be three sets.
(a) We have $A \sim A$ (reflexivity).
(b) If $A \sim B$, then we have $B \sim A$ (symmetry).
(c) If $A \sim B$ and $B \sim C$, then we have $A \sim C$ (transitivity).
Theorem. Let $A$ and $B$ be two non-empty sets. Then there exists a set $B’$ fulfilling the following conditions:
(i) The sets $B$ and $B’$ are equivalent.
(ii) We have $A \cap B’ = \emptyset$.
## Notes and References
A list of textbooks about set theory is contained in Unit [Literature about Set Theory].
Do you want to learn more? In Unit Families and the Axiom of Choice we will explain how to define families $(A_i)_{i \in I}$ in the context of the axioms of Zermelo and Fraenkel, and we will explain the famous axiom of choice.
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# Factoring the Difference of Two Squares
WHAT TO DO: HOW TO DO IT: 1. Factor the binomial 49x4 − 36y4 The difference of two terms with “square coefficientsâ€and “even exponents†will always be factored as the difference of squares. 49x4 − 36y4 49x4 − 36y4 (7x2 + 6y2)(7x2 − 6y2) 2. Factor the binomial x8 − y8 Sometimes the factors themselves contain a factorable binomial → difference of squares. Another difference of squares. Continue factoring: Continue factoring to prime factors: x8 − y8 x8 − y8 = (x4 + y4)(x4 − y4) = (x4 + y4)(x2 + y2)(x2 − y2) = (x4 + y4)(x2 + y2)(x + y)(x − y) 3. Factor the expression (t + 4)2 − 9 Some expressions can be factored as the “difference of squares†when one of the squared terms is inside parentheses Simplify: (t + 4)2 − 9 [(t + 4) + 3][(t + 4) − 3] (t + 7) (t + 1) 4. Factor Watch for fractions in the expressions. 5. Factor completely: 16x3y − 36xy3 Factor out the common factors then note that the term in parentheses can be factored as the difference of squares. The remaining factors are “prime†in rational numbers. 16x3y − 36xy3 4xy(4x2 − 9y2) 4xy(2x + 3y)(2x − 3y)
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Evaluate : $\int\large\frac{(3x-2)}{(x+1)^2(x+3)} $$dx 1 Answer Toolbox: • Form of the rational function \large\frac{px+q}{(x^2+a)(x^2+b)} • Form of the partial function \large\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)} • \int\large\frac{dx}{x}$$=log|x|+c.$
Step 1:
$I=\int\large\frac{(3x-2)}{(x+1)^2(x+3)^2}$$dx Consider \large\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3} \Rightarrow (3x-2)=A(x+1)(x+3)+B(x+3)+C(x+1)^2 \Rightarrow A(x^2+4x+3)+B(x+3)+C(x^2+2x+1) Step 2: Now compare the coefficient of like terms, For x^2 \quad A+C=0 For x \quad 4A+B+2C=3 For the constant term \quad 3A+3B+C=-2 Solving the above equations we get, A=\large\frac{11}{4}$$B=\large\frac{-5}{2}$ and $C=\large\frac{-11}{4}$
Step 3:
$\large\frac{3x-2}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}$
$5=\large\frac{11}{4}\int\frac{dx}{x+1}-\frac{5}{2}\int\frac{dx}{(x+1)^2}-\frac{11}{4}\int\frac{dx}{x+3}$
On integrating we get,
$I=\large\frac{11}{4}$$\log\mid x+1\mid+\large\frac{5}{2(x+1)}-\frac{11}{4}$$\log\mid x+3\mid+c$
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# 40 Ap Statistics Chapter 9 Practice Test
## Introduction
Welcome to our AP Statistics Chapter 9 practice test! In this article, we will provide you with a comprehensive review of the key concepts covered in Chapter 9 of AP Statistics. We understand that preparing for exams can be challenging, especially when it comes to understanding complex statistical concepts. That's why we have created this practice test to help you assess your knowledge and identify areas that may need further review. So, let's dive in and start exploring the world of AP Statistics!
### The Basics of Sampling
Sampling is a fundamental concept in statistics. It involves selecting a subset of individuals or objects from a larger population to study and draw conclusions about the entire population. Here are some key terms and concepts related to sampling:
### Population vs. Sample
In statistics, a population refers to the entire group of individuals or objects that we are interested in studying. On the other hand, a sample is a smaller subset of the population that we actually observe and collect data from. It is important to note that the goal of sampling is to obtain a representative sample that accurately reflects the characteristics of the population.
### Sampling Methods
There are several sampling methods that statisticians use to select a sample from a population. Here are some common sampling methods:
### Simple Random Sampling
Simple random sampling is a basic sampling method where each member of the population has an equal chance of being selected. This method is often used when the population is relatively small and easily accessible. To conduct a simple random sample, you can assign each member of the population a unique number and use a random number generator to select the sample.
### Stratified Sampling
Stratified sampling is a sampling method that involves dividing the population into smaller, homogeneous groups called strata, and then selecting a sample from each stratum. This method ensures that each stratum is represented in the sample proportionally to its size in the population. Stratified sampling is commonly used when the population is heterogeneous and contains distinct subgroups.
### Cluster Sampling
Cluster sampling is a sampling method where the population is divided into clusters or groups, and then a random sample of clusters is selected. All individuals within the selected clusters are included in the sample. Cluster sampling is often used when it is difficult or impractical to obtain a complete list of individuals in the population.
### Systematic Sampling
Systematic sampling is a sampling method where the population is ordered in some way, and then individuals are selected at regular intervals. For example, if you wanted to select a sample of 100 students from a school with 1000 students, you could assign each student a number and then select every 10th student. Systematic sampling can be an efficient and easy-to-use method when the population is organized in a predictable manner.
### Sampling Bias
Sampling bias refers to a systematic error that occurs when the sample selected is not representative of the population. This can happen due to various reasons, such as using an inadequate sampling method, excluding certain groups from the sample, or having a low response rate. Sampling bias can lead to inaccurate and unreliable results, so it is important to be aware of potential biases and take steps to minimize them.
### Types of Bias
There are several types of bias that can affect the validity of a study. Here are some common types of bias to be aware of:
### Selection Bias
Selection bias occurs when certain individuals or groups are more likely to be included in the sample than others. This can happen if the sampling method is not random or if there are certain characteristics that make individuals more likely to be selected. Selection bias can lead to an overrepresentation or underrepresentation of certain groups in the sample, which can distort the results.
### Nonresponse Bias
Nonresponse bias occurs when individuals selected for the sample do not participate or provide incomplete responses. This can happen if certain individuals are more likely to refuse or if there are factors that make it difficult for them to participate (e.g., language barriers, lack of time). Nonresponse bias can result in a biased sample and affect the generalizability of the findings.
### Measurement Bias
Measurement bias occurs when the measurement instrument or procedure used in the study systematically overestimates or underestimates the true value of the variable of interest. This can happen due to errors in measurement, instrument calibration issues, or subjective judgments. Measurement bias can introduce systematic errors into the data and affect the accuracy of the results.
### Reducing Bias
While it is impossible to completely eliminate bias in a study, there are steps you can take to minimize its impact. Here are some strategies to reduce bias:
### Randomization
Randomization is a powerful tool for reducing bias in sampling. By using random selection methods, you can ensure that each member of the population has an equal chance of being included in the sample. Randomization helps to eliminate systematic patterns and increase the representativeness of the sample.
### Increasing Sample Size
Increasing the sample size can also help to reduce bias. A larger sample size provides more reliable estimates and reduces the impact of random variation. With a larger sample size, the sample is more likely to be representative of the population, which helps to minimize bias.
### Using Multiple Sampling Methods
Using multiple sampling methods can also be beneficial in reducing bias. By combining different sampling methods, you can capture a broader range of perspectives and minimize the impact of any single method's limitations. This approach is often used in complex studies that require a comprehensive understanding of the population.
### Conclusion
In conclusion, understanding the principles of sampling and the potential sources of bias is crucial for conducting reliable statistical analysis. By familiarizing yourself with different sampling methods, recognizing potential biases, and employing strategies to reduce bias, you can enhance the validity and generalizability of your study findings. We hope that this practice test has provided you with a solid foundation in AP Statistics Chapter 9 and has helped you prepare for your upcoming exams. Good luck!
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# Mae travels to a house & she fell asleep 1/2 of the way there. She awakes & still has 1/2 the distance she slept to get there.What part of the journey did she sleep? The answer...
Mae travels to a house & she fell asleep 1/2 of the way there. She awakes & still has 1/2 the distance she slept to get there.
What part of the journey did she sleep? The answer is 1/3. How are they getting 1/3?
pohnpei397 | Certified Educator
Okay, the first thing to think about is this: at the point that she goes to sleep, how much of the journey is left. The answer of course is 1/2.
So now we look at the remaining half of the journey. Let us divide it up into the part she sleeps and the part that still remains. We should divide that into three parts because the part she sleeps is twice as much as the part that still remains. So of those three equal parts of the second half, she sleeps 2 and has one remaining.
This means that she sleeps 2/3 of the last 1/2 of the journey. If you multiply 2/3 by 1/2, you get 1/3 of the entire journey.
To do this in numbers rather than words:
Let T = time awake.
We know that 2t is time asleep because she slept twice as long as she was awake.
We know that 2t + t = 1/2 of the whole journey.
So now 3t = 1/2
Divide both sides by 3 and
t = 1/6
Now we know that 2t was how long she slept. If t = 1/6, 2t = 2/6, which is 1/3.
So the time she slept, which we are calling 2t is 1/3 of the whole journey.
william1941 | Student
Let the distance Mae has to travel be equal to D. Now she fell asleep when she reached half of the way to the destination. When she woke up she has half the distance that she remained asleep left. Let's say she fell asleep for a distance D. Now when she woke up she has travelled a distance N/2 + D. The distance left for her to reach is D/2. So N/2 + D + D/2 = N.
=> D+ D/2 = N - N/2
=> (3/2) D = N/2
=> D= N/3
So she slept 1/3 of the way.
neela | Student
When she started sleeping it was exactly half the distance.
The remainig distance is exactly another half.
During the the duration of her sleep she traavelled a distance 2x which is twice the remaing distance x when she awoke.
Travelled distance in sleep +distance after she awoke = 2x+x = 1/2 the jounet distance.
3x = 1/2.
x = (1/2)/3 = 1/6.
Therefore the distance travelled in sleep = 2x = 2*(1/6) = 1/3 of the total journey distance.
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5
Q:
# A can finish a work in 18 days and B can do the same work in half the time taken by A. then, working together, what part of the same work they can finish in a day ?
A) Total work B) One-fourth work C) Half work D) Two-third work
Explanation:
A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.
Q:
Raghu can do a job in 12 days alone and Sam can do the same job in 15 days alone. A third person Aru whose efficiency is two-third of efficiency of both Ram and Shyam together, can do the same job in how many days alone?
A) 10 B) 12 C) 13 D) 15
Explanation:
One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20
Aru efficiency = 2/3 of (Raghu + Sam)
Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.
3 332
Q:
12 men complete a work in 14 days. 5 days after they had started working, 3 men join them. How many more days will all of them take to complete the remaining work?
A) 5 B) 4 C) 3 D) 2
Explanation:
Let p be the required number of days.
From the given data,
12 x 14 = 12 x 5 + (12+3) x p
12 x 14 = 12[5 + 3p]
14 = 5 + 3p
3p = 9
p = 3 days.
Hence, more 3 days all of them take to complete the remaining work.
0 498
Q:
Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?
A) 19 : 7 B) 30 : 19 C) 8 : 15 D) 31 : 17
Explanation:
Given Lasya can do a work in 16 days.
Now, time taken by Srimukhi alone to complete the work = 16 x 100/80
Time taken by Rashmi = n days
=> (12 x 20)/(20 - 12) = (12 x 20)/n
=> n= 30 days.
Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = 8 : 15.
5 692
Q:
5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days?
A) 4 B) 3 C) 2 D) 1
Explanation:
Let workdone 1 boy in 1 day be b
and that of 1 girl be g
From the given data,
4(5b + 3g) = 23
20b + 12g = 23 .......(a)
2(3b + 2g) = 7
6b + 4g = 7 ........(b)
Solving (a) & (b), we get
b = 1, g = 1/4
Let number og girls required be 'p'
6(7 x 1 + p x 1/4) = 45
=> p = 2.
Hence, number of girls required = 2
5 607
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
2 647
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
6 1821
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
9 2320
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.
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# 1.4.12: Linear Transformations
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
Learning Objectives
• Give the formula for a linear transformation
• Determine whether a transformation is linear
• Describe what is linear about a linear transformation
Often it is necessary to transform data from one measurement scale to another. For example, you might want to convert height measured in feet to height measured in inches. Table $$\PageIndex{1}$$ shows the heights of four people measured in both feet and inches. To transform feet to inches, you simply multiply by $$12$$. Similarly, to transform inches to feet, you divide by $$12$$.
Table $$\PageIndex{1}$$: Converting between feet and inches
Feet Inches
5.00 60
6.25 75
5.50 66
5.75 69
Some conversions require that you multiply by a number and then add a second number. A good example of this is the transformation between degrees Celsius and degrees Fahrenheit. Table $$\PageIndex{2}$$ shows the temperatures of five US cities in the early afternoon of $$\text{November 16, 2002}$$.
Table $$\PageIndex{2}$$: Temperatures in 5 cities on $$11/16/2002$$
City Degrees Fahrenheit Degrees Celsius
Houston 54 12.22
Chicago 37 2.78
Minneapolis 31 -0.56
Miami 78 25.56
Phoenix 70 21.11
The formula to transform Celsius to Fahrenheit is:
$F = 1.8C + 32$
The formula for converting from Fahrenheit to Celsius is
$C = 0.5556F - 17.778$
The transformation consists of multiplying by a constant and then adding a second constant. For the conversion from Celsius to Fahrenheit, the first constant is $$1.8$$ and the second is $$32$$.
Figure $$\PageIndex{1}$$ shows a plot of degrees Celsius as a function of degrees Fahrenheit. Notice that the points form a straight line. This will always be the case if the transformation from one scale to another consists of multiplying by one constant and then adding a second constant. Such transformations are therefore called linear transformations.
Many transformations are not linear. With nonlinear transformations, the points in a plot of the transformed variable against the original variable would not fall on a straight line. Examples of nonlinear transformations are: square root, raising to a power, logarithm, and any of the trigonometric functions.
• David M. Lane
This page titled 1.4.12: Linear Transformations is shared under a Public Domain license and was authored, remixed, and/or curated by David Lane via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Pie Charts Questions
Q:
Study the following graph carefully and answer the questions given below:
Distribution of candidates who were enrolled for MBA entrance exam and the candidates (out of those enrolled) who passed the exam in different institutes:
1. What percentage of candidates passed the Exam from institute T out of the total number of candidates enrolled from the same institute?
A. 50% B. 62.5% C. 75% D. 80%
2. Which institute has the highest percentage of candidates passed to the candidates enrolled?
A. Q B. R C. V D. T
3. The number of candidates passed from institutes S and P together exceeds the number of candidates enrolled from institutes T and R together by:
A. 228 B. 279 C. 399 D. 407
4. What is the percentage of candidates passed to the candidates enrolled for institutes Q and R together?
A. 68% B. 80% C. 74% D. 65%
5. What is the ratio of candidates passed to the candidates enrolled from institute P?
A. 9:11 B. 14:17 C. 6:11 D. 9:17
Explanation - Required percentage= % = $\frac{9×5700}{8×8550}×100$ % = 75%
Explanation - The percentage of candidates passed to candidates enrolled can be determined for each institute as under:
P = % = 54.55%
Q = % = 75.56%
R = % = 86.67%
S = % = 62.75%
T = % = 75%
V= % = 83.33%
X= % = 50%
Highest of these is 86.67% corresponding to institute R.
Explanation - Required difference = [(16% + 18%) of 5700] - [(8% + 10%) of 8550]
= [(34% of 5700) - (18% of 8550)]
= (1938 - 1539)
= 399
Explanation - Candidates passed from institutes Q and R together = [(13% + 17%) of 5700] = 30% of 57000.
Candidates enrolled from institutes Q and R together = [(15% + 10%) of 8550] = 25% of 8550.
Required Percentage = % = 80%
Explanation - Required ratio = = 6/11
3134
Q:
The following pie chart shows the amount of subscriptions generated for India Bonds from different categories of investors.
1. In the corporate sector, approximately how many degrees should be there in the central angle ?
A. 120 B. 121 C. 122 D. 123
2. If the investment by NRI's are Rs 4,000 crore, then the investments by corporate houses and FII's tog ether is:
A. 24,000 cr B. 24,363 cr C. 25,423 cr D. 25,643 cr
3. What percentage of the total investment is coming from FII's and NRI's ?
A. 33% B. 11% C. 44% D. 22%
4. If the total investment other than by FII and corporate houses is Rs 335,000 crore, then the investment by NRI's and Offshore funds will be (approximately) ?
A. 274,100 B. 285,600 C. 293,000 D. Cannot be determined
5. If the total investment flows from FII's were to be doubled in the next year and the investment flows from all other sources had remained constant at their existing levels for this year, then what would be the proportion of FII investment in the total investment into India Bonds next year (in US \$ millions) ?
A. 40% B. 50% C. 60% D. 70%
Explanation - 34 x 3.6 = 122.4 (since 1% = 3.6 degrees)
Explanation - (67/11) x 4000 = 24 363.6364
Explanation - (33 + 11) = 44
Explanation - Investment other than NRI and corporate houses is 33% = 335000. Also, investment by offshore funds and NRI's is equal to 27%.
Hence, (27 x 335000)/33 = 274 090.909
Explanation - FII's currently account for 33 out of 100.
If their value is doubled and all other investments are kept constant then their new value would be 66 out of 133 = approximately equal to 50%
1879
Q:
Study the pie - chart carefully to answer the questions below :
The Pie chart shows the percentage quantity of fruits at two fruit shops A and B
1. What is the difference between the quantity of Guava at Shop B and that at Shop A ?
1. 40 Kg 2. 45 kg 3. 35 kg 4. 30 kg 5. 50 kg
2. If the price of Mango is Rs. 30 per kg, Apple Rs.40 per kg and orange Rs.20 per kg, then what is the ratio of their costs at Shop A?
1. 1 : 4 : 6 2. 9 : 8 : 5 3. 3 : 7 : 8 4. 5 : 4 : 1 5. 2 : 5 : 7
3. The quantity of Mango at Shop B is what per cent of the quantity of Mango at Shop A ?
1. 20% 2. 220% 3. 120% 4. 80% 5. 180%
4. If the price of Mango is Rs.30 per kg, Apple Rs. 40 per kg, orange Rs.20 per kg, Other fruits Rs.15 per kg and Guava Rs. 18 per kg for both Shop A and B, then what is the difference between the cost of all fruits at Shop A and that at Shop B ?
1. Rs.7200 2. Rs.3500 3. Rs. 6400 4. Rs.5100 5. Rs.4600
5. The quantity of Orange at Shop A is what percent more than that of Apple at Shop B ?
1. 161.52% 2. 195.5% 3. 182% 4. 190% 5. 171.42%
Explanation : Quantity of Guava at Shop A = $1200×\frac{10}{100}$ = 120 kg
Quantity of Guava at Shop B = $1000×\frac{16}{100}$= 160 kg
Therefore, Required difference = 160 - 120 = 40 kg
Explanation : Cost of Mango at Shop A = $30×1200×\frac{24}{100}$= Rs. 8640
Cost of Apple = $40×1200×\frac{16}{100}$ = Rs. 7680
Cost of Orange = $20×1200×\frac{20}{100}$ = Rs.4800
Therefore, Required Ratio = 8640 : 6780 : 4800 = 9 : 8 : 5
Explanation : Quantity of Mango at Shop B = $1000×\frac{24}{100}$ = 240 kg
Quantity of Mango at Shop A =$1200×\frac{24}{100}$ = 288 kg
Therefore, Req % = $288×\frac{100}{240}$ = 120% of the quantity of Mango at Shop A
Explanation : Cost of total fruits at Shop A = Cost of Mango + Cost of Apple + Cost of Guava + Cost of orange + Cost of other fruits
=
= Rs. 28680
Cost of Total Fruits at Shop B
=
= 7200 + 5600 + 2880 + 4000 + 3900
= Rs. 23580
Therefore, Req.difference = 28680 - 23580 = Rs.5100
Explanation: Quantity of Orange at Shop A = $1200×\frac{20}{100}$ = 240 kg
Quantity of Apple at Shop B =$1000×\frac{14}{100}$ = 140 kg
Therefore, Req % = $\frac{240×100}{140}$ = 171.42% more than the quantity of Apple at Shop B.
1363
Q:
Study the following pie-charts to answer the following questions :
The pie-charts show the expenditure of two companies A and B, Which are Rs.50 lakh and Rs.60 lakh respectively.
1.If the incomes of Company A and B are in the ratio of 4:5 and the income of Company B is 180% of its expenditure,then what is the difference between the income of Company B and the income of Company A ?
1.Rs.2200000 2. Rs.1900000 3. Rs.2160000 4. Rs.1850000 5.Rs.2250000
2.If the number of employees in Company A is a hundred then what is the average salary of the employees in Company A ?
1. Rs.14000 2. Rs.16000 3. Rs.13000 4. Rs.15000 5. Rs.15500
3.What is the ratio of tax paid by Company A to that by Company B ?
1. 35:18 2. 34:37 3. 42:41 4. 31:27 5. 27:25
4.What is the difference between the expenditure on employees of Company B and that of Company A?
1.Rs.4300000 2. Rs.6400000 3. Rs.5900000 4. Rs.8700000 5.Rs.7800000
5.The expenditure on Machine and Electricity of Company B is what per cent more than that on the same item of Company A?
1. 67% 2. 84% 3. 75% 4. 77% 5. 80%
Explanation : Expenditure of Company B = 60 lakh
Income of Company B = $60×\frac{180}{100}$= 108 lakh = 1 crore 8 lakh
Income of Company A = $10800000×\frac{4}{5}$ = Rs. 8640000
Therefore, Required differeence = 10800000 - 8640000 = Rs.2160000
Explanation : Total expenditure on the employees of company A = $5000000×\frac{28}{100}$ = Rs.1400000
Average Salary of the employees = $\frac{1400000}{100}$ = Rs.14000
Explanation : Tax paid by Company A : Tax paid by Company B
= $5000000×\frac{14}{100}$$6000000×\frac{6}{100}$
= 700000 : 3600000 = 35 : 18
Explanation : Difference = $6000000×\frac{34}{100}-5000000×\frac{28}{100}$
= 2040000 - 1400000 = Rs.640000
Explanation : Expenditure on Machine and Electricity of Company B = $6000000×\frac{18}{100}$ = Rs.1080000 = 10.8 lakh
Expenditure on Machine and Electricity of Company A = $5000000×\frac{12}{100}$ = Rs.600000 = 6 lakh
Therefore, Required % = $\frac{1080000-600000}{600000}×100$ % = $\frac{48}{60}×100$% = 80%
Hence expenditure of Company B is 80% more than Company A.
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# Sine double angle identity
## Formula
$\sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$
A trigonometric identity that expresses the expansion of sine of double angle in sine and cosine of angle is called the sine of double angle identity.
### Introduction
Let theta be an angle of a right triangle, the sine and cosine functions are written as $\sin{\theta}$ and $\cos{\theta}$ respectively. Similarly, the sine of double angle function is written as $\sin{(2\theta)}$ in mathematical form.
The sine double angle function can be expressed in sine and cosine of angle in product form as follows.
$\implies$ $\sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$
#### Usage
The sine of double angle identity is mainly used in two different cases in mathematics.
##### Expansion
It is used to expand the sine of double angle functions in sine and cosine functions.
$\implies$ $\sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$
##### Simplified form
It is used to simplify the product of sine and cosine functions as sine of double angle function.
$\implies$ $2\sin{\theta}\cos{\theta} \,=\, \sin{2\theta}$
#### Other forms
The angle in sine of double angle formula can be denoted by any symbol. So, the sine of double angle identity can be expressed in terms of any variable. It is also usually expressed in three other popular forms.
$(1). \,\,\,\,\,\,$ $\sin{2x} \,=\, 2\sin{x}\cos{x}$
$(2). \,\,\,\,\,\,$ $\sin{2A} \,=\, 2\sin{A}\cos{A}$
$(3). \,\,\,\,\,\,$ $\sin{2\alpha} \,=\, 2\sin{\alpha}\cos{\alpha}$
#### Proof
Learn how to derive the rule of sin double angle formula in trigonometry by geometric method.
##### Another form
The sin of double angle formula can also be expanded in terms of tan of angle mathematically.
$\sin{(2\theta)} \,=\, \dfrac{2\tan{\theta}}{1+\tan^2{\theta}}$
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