mathematics

Discussion

First study Algebra tab, then the next tab and so on....Please use scientific calculatorAny question please email me with your concern

a scientific calculator

versionupdated 3/7/2012tuned 1.96

My E-Mail is [email protected] the text is colored maroon it is a formula

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Discussion

ALGEBRA

A. Algebrathe part of Mathematics which investigates the relations and properties of numbers or other mathematical structures by means of general symbols; a system of this based on given axioms.

a*a*a*a 3*3*3*3*3 =

y*y*y mn= m*n

xy= x*y (x multiply by y) x*x*x*x*x*x*x*x

(x)(y)= x*y (x multiply by y) 9.72(m)(n)= m*n (m multiply by n)

(m)n= m*nx(y)= x*y (x multiply by y)

m(n)= m*nLaws of Exponent

iff n>m

1 iff n=m

1

SAMPLE=base exponent

3 4answer= 81

Properties of Radicals

where:

m= index

√(a) a= radicand

√ = radical symbol

1

m√[n√(a)]

SAMPLE=base exponent index27 1 3

answer= 3Binomial Expansion

a4= 35=

y3=

x8=

35/52=

an * am= am+n

an / am= an-m

an / am=

(a*b)n= an * b n

(a/b)n= an / b n

a0=

a-n= 1/an

(an)m= (am)n= a mn

Computerized Formula A.1(do not edit red text, input value at blue text)34

an/m= m√(an)

a1/m= mth root of a =m√(a)a1/2=

n√(an)

n√(a*b) =n√(a) – n√(b)

n√(a/b) =n√(a) / n√(b)

=m*n√(a)Computerized Formula A.2(do not edit red text, input value at blue text)

271/3

(a+b)n= an +(n/1)a(n-1)/2 b1+ (n/1)*((n-1)/2)a(n-2) b2

+{[n(n-1)(n-2)]/[1(2)(3)]}(a(n-3)-b3)

+{[n(n-1)(n-2)........(n-(w-1)]/[1(2)(3)......(q)]}(a(n-w)-bw)

rth term of (a+b)n

ALGEBRA

where: w=r-1

Example:

Factorial (!)5!= 5*4*3*2*10!= 11!= 17!= 5040

where:w=r-1

SAMPLE= 4 !base

4answer= 24

SAMPLE PROBLEMS

( 2 x - 1 y ) to the power of 12exponent= 12

term=r= 6w= 5

answer

792 2 x7

y5

-

-101376 x7

y5

the 6th term

r= (n/2)+1r= (12/2)+1

r=using formula A.4:

=

={[r(n-1)(n-2)........(n-(w-1)]/[1(2)(3)......(w)]}(a(n-w)-bw)

(a+b)2= a2+2ab+b2

(a+b)3= a3+3a2b+3ab2+b3

(a+b)4= a4+4a3b+6a2b2+4ab3+b4

(a+b)5= a5+5a4b+10a3b2+10a2b3+5ab4+b5

rth term of (a+b)n

={n!/[(n-w)!(w!)]}(a(n-w)-bw)

Computerized Formula A.3(do not edit red text, input value at blue text)

1. Find the 6th term of the expansion of (2x-y)12


2. Find the middle term in the expansion of (x+y)12.

7th term

924x6y6

ALGEBRA

1/x=

( 1 x2

1 x-1

)10

+n= 10w= 4

Input= x 8Required

Output=

210 1 x6

x2

-

-210 x8

210 x8

A.i Exponential and Logarithmic EquationsExponential Equation=balancing something multiplied by a constant factor in successive equal periods of timeLogarithmic Equation=balancing fixed number or base raised to a power in order to produce any given numberLogarithmic Equations=balancing the simplifications of computation by replacing multiplication and division of numbers by addition and subtraction of their correspondent exponents

Exponential Form Logarithmic Form

a. log=logarithm

b. natural logwhere e= 2.718000

Example:

2 100

3 8

x

48=

log 82Input Base= 2

Input Number= 8Output= 3

Example: =

1

a. log m = x

m=

3. Find the term containing x8 in the expansion of (x2+(1/x))10.

x-1


bN*M --------------------------------------> logb M=N

log10 M = log M common log

loge M = ln M ln=natural logarithm

log10 100= 102=

log2 8= 23=

Log8 48=

8x

Computerized Formula B.1(do not edit red text, input value at blue text)

lne= logee=

10x

ALGEBRA

b. ln 8= x

8=

ln 8Input Number= 8

Output= 2.079442

Properties of Logarithm

1=

2=

3=

4= ex.

5=

6= 1

7= 0

SAMPLE PROBLEMS1. Solve for y in the expression:

y= ln

y=y= x ln e – (x-1) ln e

y= x – (x-1)

or another solution:

y=y= lne=1

3

8

8+1

9x= (+3)x= -3

log ( x2

11

) = 32 -

+ value = 3

ex


logbMN= logbM + logbN

logb(M/N)= logbM - logbN

logbMP= plogbM

logbM= (logaM)/(logab) =(logm)/(logb) log848=(log48)/(log8)

logbM= 1/(logmb)

logbb=

logb1=

ex

ex-1

ln ex – ln e(x-1)

but ln e=logee=1

e(x-1)= ex/e1

ln ex/(ex/e)

2. Solve for x in the expression log2 (x2-1)=3

Log2(x2-1)=

(x2-1)= 23

x2-1=

x2=

x2=


ALGEBRA

- value = -3

2

4= =a= √(4) = 3a= 2

log 4 = 2a

log 8 = xa

a = 3

3. Solve for x in the equation=

ln(x+2)(x-1)=

ln(x+2)+ln(x-1)=

ln(x+2)=ln10= ln10

(x+2)=

0(x-2)(x+1)= 0

x= -2x= (+2)

A.ii ProgressionArithmetic Progression (A.P.)

an AP is a sequence of numbers formed by adding a constant number called the common difference to the immediately preceding term.

Example: 2,4,6,8,10.............

Finding the nth term in an AP

+ d

+ d

+ d......

nth term

.Sum of nth term in AP

3. If loga4 = 2, Find loga8

loga4 = loga8= log 8 / log a

a2 log 8 / log 2


ln(x2+x-2)= lnx2+ln(x-1)

lnx2+ln(x-1)

lnx2+ln(x-1)

lnx2

x2

(x2-x-2)=

1st term a1

2nd Term a2 = a1

3rd term a3 = a2 =a1+2d

4th term a4 = a3 =a1+3d

an = a1 + (n-1)d

an = am + (n-m)d

Sn=(n/2) (a1+an) or Sn=(n/2) (2a1+n-1)d

ALGEBRA

SAMPLE PROBLEMS1. The sum of an A.P. Is 196. If the first term is 52 and the last term is 4, determine the number of arithmetic means between 52 & 4?

52 ---extremes

4 ---extremes

196

Figure:

52,___,___,___,___,___,4196= (n/2) (52+4)

n= 7

Input 52

Input 4

Input 196

Output n= 7 Arithmetic MeansInput n= 7

Input 4

Input 196

Output 52 First term in APInput n= 7

Input 196

Input 52

Output 4 Last term in APInput n= 10

Input 10.5

Input -3

Output 37.5 Sum of AP

2. Find the quotient of the sum of all odd integers between 100 & 1000 when it is divided by 9.Given: 100,101,103,105,107,109,111.....999,1000

difference between odd integers is 2

d= 2999= 101 +[(n-1)2]

898/2= n-1n= 450

(n/2) (formula A.1)

450/2 [101+999]

247500

divided by 9R= 27500

a1=

an=

Sn=

Sn=(n/2)(a1+an)

Computerized Formula C.1 (do not edit red text, input value at blue text)

Sn= (n/2)(a1-an)

a1=

an=

Sn=

an=

Sn=

a1=

Sn=

a1=

an=

an=

a1=

Sn=

an=a1+[(n-1)d]

Sn= [a1+a2]

Sn=

Sn=

ALGEBRA

Input 101

Input 999

Input d= 2Output n= 450 number of progression

Input 101

Input 999

Input n= 450Output d= 2 common differenceInput d= 2

Input 999

Input n= 450

Output 101 first term in AP

Input 101

Input d= 2Input n= 450

Output 999 nth term in AP

Given: ___,___,___,___,3,___,___,___,9,___first term tenth term

9= 3 + 4dd ### or 1.5

3=

3=

-3

9+(1)d

10

using computerized formula A.1

37.5000

Input 3

Input 9

Computerized Formula C.2(do not edit red text, input value at blue text)

an=a1+[(n-1)d]

a1=

an=

a1=

an=

an=

a1=

a1=

an=

3. The 5th term of an AP is 3 and the 9th term is 9. Find the sum of the first 10 terms of this AP.

an=am+[(n-m)d]

a9= a5+(9-5)d

a5= a1+(n-1)d

a1+(5 -1)1.5

a1+(4)1.5

a1=

a10= a9+(10 -9)d

a10=

a10=

Sn=


an=am+[(n-m)d]

am=

an=

ALGEBRA

Input d= 1.5Input m= 5Output n= 9 nth value

Input 3

Input 9

Input n= 9Input m= 5Output d= 1.5 common differenceInput d= 1.5

Input 9

Input n= 9Input m= 5

Output 3 mth term in AP

Input 3

Input d= 1.5Input n= 9

Input 9

Output m= 5 first term in AP

Geometric Progression (A.P.)a GP is a sequence of numbers formed by multiplying a constant number, called the common ratio, by the immediately preceding term.

Example: 2,4,8,16,32.....

Finding the nth term in a GP

..

..

..

nth term

or

Sum of nth term in a GP

Fractional Progression (F.P.)- 1.0 < 1.0

Sum of FP

where: -1.0 < r < 1.0

SAMPLE PROBLEMS for GP and FP

am=

an=

an=

am=

am=

an=

1st term a1

2nd Term a2 = a1r

3rd term a3 = a2r =a1r2

4th term a4 = a3r =a2r2 =a1r2

an = a n t n-1 =a3r1 =a2r2 =a1r3

an = a 1 r n-1

an = a m r n-m

Sn=[a1(1-r n)]/(1-r)

S= [a1/(1-r)]

ALGEBRA

112=

4r= 2

x+2=

112 x+2=

5628 x+2 28 28

= 3136 = 2

x+2 = 56

14336

inputa 28

1input

a x + 2note= you can substitute any value to x+2 depending on the problem

2 subtract the given number by 2 you can find the value x+2input

a 1123

inputn= 10

output r= 2

a 14336output 10

2. The numbers x, 2x+7,10x-7 form a GP. Find the value of the sum of the first 7 terms.2x+7

=10x-7

x 2x+7=

=

= 0by Quadratic Formula:

x = [-b(+/-) √(b2-4(a)(c))]/2(a)

x = 7 7,21,63..... therefore r=7

= 7651

2 x + 7=

10 x - 7x 2 x + 7

a= 6b= -35 x

1. The number 28, x+2,112..... form a GP. What is the 10th term?

an= a1r n-1

a3= a1r3-1

28r3-1

r2=

(x+2)2

a10= a1r 10-1

a10= 28(2)9

a10=Computerized Formula C.4(do not edit red text, input value at blue text)

an= a1r n-1

(2x+7)2 10x2-7x

4x2+28x+49 10x2-7x

6x2-35x-49

Sn=[a1(1-r n)]/(1-r)

Sn


x2

ALGEBRA

c= -49

x= 7 using sign +

by substituting= 7 , 21 , 63 …..therefore r= 3

7651

7

r= 3n= 7 term

3. The first term of a GP is 6 and the last term is 486. If there are 3 terms, determine the sum of the series.

6

486

n= 3

486=

81=9= r

Sum of Series in GP

548

factor r

6

486

n= 3r= 9

r= 9n= 3

6

548

4. Find the value of “x” in the GP (1/3),(2/x),(4/27). Also compute for the sum of the series.2 4x 27

------------- = ------------- from: Computerized formula C.9


Sn=[a1(1-r n)]/(1-r)

Sn=

a1=

a1=

an=

a3= a1r 3-1

a1 r 2

r 2

Sn= {[a1(1-rn)]/(1-r)}

Sn= [6(1-93)]/(1-9)

Sn=


a1=

an=


Sn= {[a1(1-rn)]/(1-r)}

a1=

Sn=

ALGEBRA

1 2 13 x

( 2 ) 2=

4x 81

2=

2x 9x = 9

r= (2/9)/(1/3)

r= 0.66666666667

a1=1 = 0.333 n= 3

a2=2 = 0.22x

a3=4 = 0.15

27x= 9r= 0.66666666667

Sn= 1

ProbabilityP(E)= f favorable outcomes

T total possible outcomesSAMPLE PROBLEM1. Roll a pair of dice one time. What is the probability that the sum of two numbers is 10?

T= 6(6)= 36sum= 10 1 2 3 4 5

12345 10

P 0.08 6 10P 12/30/99

Err:509

2. What is the probability that you can win swertres with 3 straight combination numbers out of 3 numbers?T= 0 to 1000T= 1000 possible outcomes

f= 1 favorable outcomesP(E)= 1/(1000-1)P(E)= (1)/(999)P(E)= 1 of 999 results.

3. What is the probability that you can win swertres with 3 rumble combination numbers out of 3 numbers?T= 0 to 1000

Sn=

r =a w /(1/n)

where: aw is the midterm


ALGEBRA

T= 1000 possible outcomesf= 9f= 1000-9f= 991 favorable outcomes

P(E)= (1)/(991)P(E)= 1 of 991 results.

MiscellaneousSAMPLE PROBLEM1. The sum of scores of Team 1 and Team 2 is equal to 75. If the score of Team 1 is twice than Team 2. Find their respective scores.

let x = Team 1let y = Team 2

x+y= 75 ------------------equation 1the of team 1 is twice than team 2

x= 2y ------------------equation 2from equation 1 substitute equation 2x+y= 7575= x+y75= 2y+y75= 3y

y= 25from equation 2x= 2yx= 2*(25)x= 50therefor:

Team 1= 50Team 2= 25

2. Eight years ago the sum of the ages of Jun & Jess is equal to 26. Five years from now Jess age will be equal to twice Jun’s age less than 35. How old is Jess now.let x = Jess agelet y = Jun age

Past Present Futurex-8 x x+5y-8 y (2(y+5))-3526

equation 1(x-8)+(y-8)= 26

equation 2(2(y+5))-35= x+5

(2y+10)-35-5= x(2y+10)-40= x(2y+10)-40= x

2y+10-40= x2y-30= x

from equation 1 substitute equation 2(x-8)+(y-8)= 26

((2y-30)-8)+(y-8)= 26(2y-30-8)+y-8)= 26

(2y-38+y-8)= 26

ALGEBRA

(2y-46+y)= 263y-46= 26

3y= 72y= 24 Jun’s age

from equation 2(2(y+5))-35= x+5

(2(24+5))-35= x+5(2(29))-35= x+5

(58)-35= x+558-35= x+5

23= x+523-5= x

x= 18 Jess’s age

3. Jose can paint the house in 40 days, George can do the same task in 50 days. If Jose and George work together, How long would it take them to finish the job?

Work Rate= 1/C (formula)let c= no. of days needed by Jose and George to complete the task

1/40= Jose’s rate1/50= George’s rate

1/x= combined rate1/x= (1/40)+(1/50)1/x= 0.025 + 0.021/x= 0.045

1/0.045= xx= 22.2 days

4. A certain paint job could be finished in 150 days if 50 men were working full time. In the actual implementation, 60 men started working but after 20 days 20 more men were added, after 80 days from the very start, 50 men quit the job. Determine the total number of the days for the completion of the job.Work Completion Rate= w*e

where:w= completion dayse= number of workers

20 days (80-20) days (x-80) days--------------------------------------------------------------

80 days60 men (60+20) men (80-50) men

let x= number of completion days(50*150)= ((20*60)+(60*80)+((x-80)*30)

(7500)= 6000+(30x-2400)7500= 3600+3x3900= 3x

x= 1300 days

5. The boat travels to Pagsanghan at 2/3 the time than going to Gandara (current is flowing to Gandara) If the velocity of the boat in still water is 40kph, determine the velocity of the river current.Distance= Speed * Time

let:t= time

ALGEBRA

v= velocity (speed)d= t*v (formula)

upstream

downstream

: river current (velocity)

: boat velocity

d= t*v

d=equation 1

equation 2

equating 1 and 2

(2/3)(d/d)

(2/3)(1)

120-80

40

(40/5)

8 river current

6. A kilo of onion and a kilo of garlic is worth 75 pesos, if you bought an onion which is thrice than garlic in quantity and paid 225 for it, determine the price of onion per kilo.let:

x= oniony= garlic

x+y= 65 a kilo of onion and a kilo of garlic worth 753x+y= 65 onion which is thrice than garlic in quantity

x= 65 – y Equation 1y= 65-3x Equation 2

Equating 1 and 2y= 65-3(35-y)

vc

vb

vc

vb

vc

vb

vn : vb + vc

vn : vb - vc

tn*vn

tn= d/vn

tn1= d/(vb+vc)

tn= d/vn

tn2= d/(vb-vc)

tn1=(2/3)tn2

d/(vb+vc)= (2/3)(d/(vb-vc))

d/(40+vc)= (2/3)(d/(40-vc))

(40-vc)/(40+vc)=

(40-vc)/(40+vc)=

3*(40-vc)= 2*(40+vc)

120-3vc= 80+2vc

2vc+3vc=

5vc=

vc=

vc=

ALGEBRA

y= 65-105+3yy= -105+3y

3y-y= 1052y= 105y= 52.50 garlic per kilox= 65 – yx= 65 – 52.50x= 12.50 onion per kilo

7. A chemical engineer mixed 40ml of 35% HCL(hydrochloride) solution with 20ml of 50% HCL solution. What is the percentage of HCL in the hew solution?40ml + 20ml= 60ml

35(40)+50(20)= 60x1400+1000=. 60x 1400

2400= 60x2400/60= x

40= x

8. A student wants to form a 32ml mixture from two solutions to contain 30% and solution A contains 42% acid and solution B contains 18% acid. How many ml(milliliter) of each solution must be used?Soln A Soln B Final Soln

42 18 30x y 32

x+y= 32 Equation 142x+18y= 30(32)42x+18y= 960 Equation 2

substituting eq1 to eq2:42(32-y)+18y= 960

1344-42y+18y= 9601344-24y= 960

-24y= 960-1344-24y= -384

reversing the equation.384= 24y

384/24= y16= y

solution B is 16mlfrom equation 1

x+y= 32x+16= 32

x= 32-16x= 16

solution A is 16ml

Ax+By=Cz (Formula)Soln A Soln B Final Soln

A B Cx y z

9. From an observation, the value of C varies directly with x and the square of y but inversely with z. When x=2, y=1 and z=4; c=100. Find the value of C when x=3,y=2 and z=5

ALGEBRA

C=

100=k= 200

C=C= 400

10. From table below Team 1 to Team 3 has this score (triple tie), who is the winner thru quotient system?

Team1 81Team2 78

Team1 83Team3 84

Team2 82Team3 78

thru quotient systemTeam1= 1.01234568Team2= 1.00628931 .Team3= 0.98181818

thru point systemTeam1= 164.00Team2= 160.00Team3= 162.00

highest qoutient is the winner on triple tie thru qoutient systemhighest point is the winner on triple tie thru point system

C ἄ (xy2)/z(kxy2)/z

K(2*12)/4

(200(3)(2)2)/5


ALGEBRA

the part of Mathematics which investigates the relations and properties of numbers or other mathematical structures by means of general symbols; a system of this based on given axioms.243 where= a,y,x,m,n is a variable

a= base

division 4= exponent

ALGEBRA

Exponential Equation=balancing something multiplied by a constant factor in successive equal periods of timeLogarithmic Equation=balancing fixed number or base raised to a power in order to produce any given numberLogarithmic Equations=balancing the simplifications of computation by replacing multiplication and division of numbers by addition and subtraction of their correspondent exponents

Logarithmic Form

2.718000=natural logarithm

ALGEBRA

log848=(log48)/(log8)

ALGEBRA

an AP is a sequence of numbers formed by adding a constant number called the common difference to the immediately preceding term.

ALGEBRA

1. The sum of an A.P. Is 196. If the first term is 52 and the last term is 4, determine the number of arithmetic means between 52 & 4?

2. Find the quotient of the sum of all odd integers between 100 & 1000 when it is divided by 9.

ALGEBRA

term is 9. Find the sum of the first 10 terms of this AP.

ALGEBRA

a GP is a sequence of numbers formed by multiplying a constant number, called the common ratio, by the immediately preceding term.

ALGEBRA

you can substitute any value to x+2 depending on the problemsubtract the given number by 2 you can find the value x+2

ALGEBRA

3. The first term of a GP is 6 and the last term is 486. If there are 3 terms, determine the sum of the series.

4. Find the value of “x” in the GP (1/3),(2/x),(4/27). Also compute for the sum of the series.

ALGEBRA

1. Roll a pair of dice one time. What is the probability that the sum of two numbers is 10?

6

10 6+4= 105+5= 104+6= 10

f= 3

2. What is the probability that you can win swertres with 3 straight combination numbers out of 3 numbers?

3. What is the probability that you can win swertres with 3 rumble combination numbers out of 3 numbers?

ALGEBRA

1. The sum of scores of Team 1 and Team 2 is equal to 75. If the score of Team 1 is twice than Team 2. Find their respective scores.

2. Eight years ago the sum of the ages of Jun & Jess is equal to 26. Five years from now Jess age will be equal to twice Jun’s age less than 35. How old is Jess now.

ALGEBRA

3. Jose can paint the house in 40 days, George can do the same task in 50 days. If Jose and George work together, How long would it take them to finish the job?

no. of days needed by Jose and George to complete the task

4. A certain paint job could be finished in 150 days if 50 men were working full time. In the actual implementation, 60 men started working but after 20 days 20 more men were added, after 80 days from the very start, 50 men quit the job. Determine the total number of the days for the completion of the job.

5. The boat travels to Pagsanghan at 2/3 the time than going to Gandara (current is flowing to Gandara) If the velocity of the boat in still water is 40kph, determine the velocity of the river current.

ALGEBRA

6. A kilo of onion and a kilo of garlic is worth 75 pesos, if you bought an onion which is thrice than garlic in quantity and paid 225 for it, determine the price of onion per kilo.

ALGEBRA

7. A chemical engineer mixed 40ml of 35% HCL(hydrochloride) solution with 20ml of 50% HCL solution. What is the percentage of HCL in the hew solution?

8. A student wants to form a 32ml mixture from two solutions to contain 30% and solution A contains 42% acid and solution B contains 18% acid. How many ml(milliliter) of each solution must be used?

9. From an observation, the value of C varies directly with x and the square of y but inversely with z. When x=2, y=1 and z=4; c=100. Find the value of C when x=3,y=2 and z=5

ALGEBRA

10. From table below Team 1 to Team 3 has this score (triple tie), who is the winner thru quotient system?

ALGEBRA

4. A certain paint job could be finished in 150 days if 50 men were working full time. In the actual implementation, 60 men started working but after 20 days 20 more men were added, after 80 days from the very start, 50 men quit the job. Determine the total number of the days for the completion of the job.

ALGEBRA

8. A student wants to form a 32ml mixture from two solutions to contain 30% and solution A contains 42% acid and solution B contains 18% acid. How many ml(milliliter) of each solution must be used?

ANALYTIC GEOMETRY

B. Analytic Geometry

Cartesian Coordinate System

where: ∞ Err:508

P1,P2..... ###

Distance between 2(two) points

(formula)Example:

Plot x=6,y=-7 and x=-4,y=+3, connect two points and compute for distance.between two pointslet (6,-7) be point 1 and (-4,3) point 2

D2= (x2-x1)2+(y1-y2)2

ANALYTIC GEOMETRY

100 + 100

D=D= 14.142135624 units

6.00000000000000000000

-7.00000000000000000000

-4.00000000000000000000

3.00000000000000000000

D2= (x2-x1)2+(y1-y2)2

D2= (-4-(+6))2+(-7-(3))2

D2= (-10)2+(-10)2

D2=

2√100


x1=

y1=

x2=

y2=

ANALYTIC GEOMETRY

D= 14.14213562373100000000 units

Midpoint Formula

x=

y=

6.00000000000000000000

-7.00000000000000000000

-4.00000000000000000000

3.00000000000000000000x= 1.00000000000000000000y= -2.00000000000000000000

Straight Line General EquationAx+By+C=0

where: A,B & C are constants

Standard FormsSlope-Intercept Form

y= mx+b

Point-Slope Form

Two Point-Slope Form

Intercept Form(x/a)+(y/b)= 1

Area of a Triangle

1

A= ½ 1

1

A= ½

Example: Find the area of a triangle formed by this three points; pt.1(5,2),pt.2(-2,4) and pt.3(1,-1)

5 2 1 5 2

(x1+x2)/2

(y.1+y2)/2


x1=

y1=

x2=

y2=

y-y1= m(x-x1)

y-y1= [(y2-y1)/x2-x1)](x-x1)

Distance from a Point P(x0,y0) to a line Ax0+by0+C=0

x1 y1 x1 y1

x2 y2 x2 y2

x3 y3 x3 y3

x1 x2 x3 x1

y1 y2 y3 y1

ANALYTIC GEOMETRY

A= ½ -2 4 1 -2 41 -1 1 1 -1

5 2 1 5 2A= ½ -2 4 1 -2 4

1 -1 1 1 -1

A= ½ ((20+2+2)-(4-5-4))A= ½ ((24-(-5))A= ½ 29A= 29/2 sq unitsA= 14.5 sq units

or

A= ½ 5 -2 1 52 4 -1 2

A= ½ 5 -2 1 52 4 -1 2

A= ½ ((20+2+2)-(-4+4-5))A= ½ ((24-(-5))A= ½ 29A= 29/2 sq unitsA= 14.5 sq units

5.00000000000000000000

2.00000000000000000000

-2.00000000000000000000

4.00000000000000000000

1.00000000000000000000

-1.00000000000000000000A = 14.50000000000000000000 sq. units


x1=

y1=

x2=

y2=

x3=

y3=

ANALYTIC GEOMETRY

Plot x=6,y=-7 and x=-4,y=+3, connect two points and compute for distance.between two points

ANALYTIC GEOMETRY

ANALYTIC GEOMETRY

12

Find the area of a triangle formed by this three points; pt.1(5,2),pt.2(-2,4) and pt.3(1,-1)

ECONOMICS

C. Economicsthe part of Mathematics which deals with the financial considerations attaching to a particular activitybased on Gregorian calendar one (1) year is equal to 12 months, 1 year is equal to 365 and ¼ daysFebruary 29 occurs every 4 years

Simple Interest:I= S(i)N where: I= total interest earned or paid

F= S+ I S= principal amount lent or loanedi= interest rate per interest period

N= number of interest periodsF= total amount to be received or paid at the end of N time

Compound Interest:F= NOTE: FIGURES ARE LOCATED AT CELL DA1

SAMPLE PROBLEMS1. Draw a cash flow diagram for P 10,500 being loaned out at an interest rate of 15% per annum over a period of 6 years. How much simple interest would be repaid as a lump sum amount at the end of the sixth year?What will be the interest rate if paid lump sum at the end of sixth months?

see figure 1Solution:

1I= S(i)NI= 10,500 (0.15) 6I= 9450

F= 10500+9450F= 19950 sixth year

2i= 15/12 :12 since one year in a gregorian calendar is 12 monthsi= 1.25I= S(i)NI= 10,500 (0.01) 6I= 630

F= 10500+630F= 11130 sixth month

1input S= 10,500.00input N= 6.00input i= 15.00 0.15

output I= 9,450.00 total interest earned or paidoutput F= 19,950.00 total amount to be received or owed at the end of N years

2input S= 10,500.00input F= 19,950.00input i= 15.00 0.15

output I= 9,450.00 total interest earned or paidoutput N= 6.00 number of interest periods

3input N= 6.00input I= 9,450.00input i= 15.00 0.15

S (1+i)N

Computerized Formula A.1(do not edit red text input value at blue text)

ECONOMICS

output S= 10,500.00 principal amount lent or loanedoutput F= 19,950.00 total amount to be received or owed at the end of N years

4input I= 9,450.00input N= 6.00input S= 10,500.00

output i= 0.15i= 15.00 Percent interest rate per interest period

2. How much interest is payable each year on a loan of P2,000 if the interest rate is 10% per year when half of the loan principal will be repaid as a lump sum at the end of 3 years and the other half will be repaid in one lump sum amount at the end of six years? How much interest will be paid over the 6-year period?see figure 2

2000*(0.1)

200

(2000-1000)*0.1

100

I=I= (3*(200+100))I= 900

1

3.00

3.00N= 6.00i= 10.00

S= 2000.00Z= 1000.00

200.00

100.00

I= 900.002

3.00

3.00N= 6.00I= 900.00

100.00

200.00

S= 2000.00Z= 1000.00i= 10.00

3

3.00

3.00

i= 10.00I= 900.00

Ia=

Ia=

Ib=

Ib=

(3(Ia))+(3(Ib))


Ia=

Ib=

Ian=

Ibn=

Ia=

Ib=

Ibn=

Ian=

Ia=

Ib=

ECONOMICS

100.00

200.00

S= 2000.00N= 6.00Z= 2000.304n= 6.00Z= 1000.00i= 10.00I= 900.00

100.00

200.00

S= 2000.00

3.00

100.00

3. A future amount “F” is equivalent to P1,500.00 now when 6 years separates the amount and the annual compounded interest is 12%. What is the value of “F”?

F=

F=F= 2960.73

see figure 3

1N= 6.00i= 12.00

S= 1500.00F= 2,961.002N= 6.00i= 12.00

F= 2,961S= 1,500.003N= 6.00S= 1,500.00F= 2,961.00i= 12.00

4i= 12.00

S= 1,500F= 2,961N= 6.00

4. You have used your credit card to purchase mobile phone battery worth 340 pesos. Unable to make payments for 7 months, you then write a letter of apology to pay your bill in full. The credit card company’s nominal interest rate is 18% compounded monthly. For what amount should you write the check?i= 18/12

Ibn=

Ian=

Ibn=

Ian=

Ia=

Ib=

S(1+i)N

1500(1+(12/100))6


ECONOMICS

i= 1.50%i= 0.015

F=

F=F= 377.35 pesos

TEST YOUR SELF1. You have just learned that ABC corporation has an investment opportunity that costs 500 pesos and 1017 months later pays a lump sum amount of 1,000,000.00 pesos. The cash flow diagram looks like this:

see figure 4

What interest rate would be earned on this investment? Caculate your answer..

2. Suppose that you have 500 pesos cash today and can invest it at 0.75% compound interest each year. How many years will it take you to become a millionaire?

Business Interest:P=

A=

SAMPLE PROBLEMS1. It is estimated that a certain business like Mlhuillier can save 60,000 pesos per year on pawning and fund transfering. The business has a lot contract of 6 years. If the business must earn a 20% annual return, how much could be justified for the construction of such establishment? Draw a cash flow diagram.

see figure 5

P=P= 199531

1A= 60000.00N= 6.00i= 20.00

P= 199530.61

2. A proposed development plan for a water district to avoid difficulties will require an immediate expenditures of 5,000,000 pesos to rehabilitate the water district facilities. What annual savings must be realized to recover this expenditure in 4 years with annual return of 10%see figure 6

A=P= 1577354.0185

1P= 1577354.02

N= 4.00i= 10.00

A= 497609.14

3. A certain jewelry cost 7,000 pesos now. If it will be appraised at 10,000 pesos after 5 months period. What will be the interest per month?

P(1+i)N

340(1+0.015)7

A ((((1+i)N) -1))/(i(1+i)N)))

P(((i(1+i)N))/(((1+i)N)-1)

60000((((1+0.2)6)-1))/(0.2(1+0.2)6)


5000000(((0.1(1+0.1)4))/(((1+0.1)4)-1)


ECONOMICS

F=

10000=

i/12

from computerized formula A.3-3answer i=7.3941

(7.3941/12)

0.616175

What will be the monthly payment that the pawner will pay?

(0.616175*12)i= 7.3941

from computerized formula A.5A= 2085.01343

TEST YOUR SELF1. A certain land cost 28,000 pesos now. If it will be appraised at 60,000 pesos after 10 years period. What will be the interest rate per year?What will be the annual amount that the said land is appreciating?

Methods used in computing depreciation.Depreciation =the action or process of lowering in value; fall in the exchange value of currency.

NOTE: SEE CHART AT CELL DA31Straight Line Method

kdt

where:N= depreciable life of the asset in yearsB= cost basis

book value at end of year k

estimated salvage value in year N

cumulative depreciation through year k

SAMPLE PROBLEMS1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a straight line.

((500,000.00-200,000.00)/5) from table: monetary value at third year is 380,000.00 pesos

((300,000.00)/5)

60,000.00

S(1+i)N

(7000*((1+i)5))

im=

im=

im=

im=

dk= ((B-ESN)/N)

dk*= For 1 ≤ k ≤ N

Bvk= B-dk*

dk= annual depreciation deduction in year k (1 ≤ k ≤ N)Bvk=

ESN=

dk*=

dk= ((B-ESN)/N)

dk=

dk=

dk=

Computerized Straight Line Depreciation Table (do not edit red text input value at blue text)

ECONOMICS

input= 200,000.00input= B= 500,000.00input= N= 5.00

k

(years) (Php)1 500,000.002 440,000.003 380,000.004 320,000.005 260,000.006 200,000.007 140,000.008 80,000.009 20,000.00

10 -40,000.00

Table Chart: SEE CHART 1

Declining Balance Method/Fixed or Constant Percentage Method/Peakzone Formula-in the declining balance method, sometimes called the constant percentage methodor the Peakzone formula, it is assumed that the annual costof depreciation is a fixed percentage of the BV at the beginning of the year. The ratio of the depreciation in any one year to the BV at he beginningof the year is constant throughout the life of the asset and is designated by r(0<R<1) but not including 0 and 1.

B(R)

where:N= depreciable life of the asset in yearsB= cost basis

book value at end of year k

estimated salvage value in year N

cumulative depreciation through year k

cumulative depreciation on the first year

book value at end of year NR=

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a declining line.

ESN=

BVSTART

dk= B(1-R)(k-1)(R)

d1=

dk*= B(1-(1-R)k)

BvN= B(1-R)N =ESN

Bvk= B(1-R)k

dk= annual depreciation deduction in year k (1 ≤ k ≤ N)Bvk=

ESN=

dk*=

d1*=

BvN=

BvN= (B(1-R)N)=ESN

ECONOMICS

1-R=

1-R= from table: monetary value at third year is 346,572.42 pesos1-R= 0.832553207401873

R= 0.167446792598127


R= 0.16744679259812700000

k

(years) (Php)1 500,000.002 416,276.603 346,572.424 288,539.985 240,224.896 200,000.007 166,510.648 138,628.979 115,415.99

10 96,089.95

Table Chart:SEE CHART 2

Double Rate Declining Balance MethodThis method is a declining balance method with R= 2/N

R= 2/N

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a double declining line.from table: monetary value at third year is 180,000.00 pesos


R= 0.40000000000000000000

k

(years) (Php)

(ESN/B)(1/N)

(200,000.00/500,000.00)(1/5)

Computerized Declining Balance method Depreciation Table (do not edit red text input value at blue text)

ESN=

BVSTART

Computerized Double Declining Balance method Depreciation Table (do not edit red text input value at blue text)

ESN=

BVSTART

ECONOMICS

1 500,000.002 300,000.003 180,000.004 108,000.005 64,800.006 38,880.007 23,328.008 13,996.809 8,398.08

10 5,038.85

Table Chart: SEE CHART 3

ECONOMICS

the part of Mathematics which deals with the financial considerations attaching to a particular activitybased on Gregorian calendar one (1) year is equal to 12 months, 1 year is equal to 365 and ¼ days

1. Draw a cash flow diagram for P 10,500 being loaned out at an interest rate of 15% per annum over a period of 6 years. How much simple interest would be repaid as a lump sum amount at the end of the sixth year?What will be the interest rate if paid lump sum at the end of sixth months?

:12 since one year in a gregorian calendar is 12 months

total amount to be received or owed at the end of N years

ECONOMICS

total amount to be received or owed at the end of N years

2. How much interest is payable each year on a loan of P2,000 if the interest rate is 10% per year when half of the loan principal will be repaid as a lump sum at the end of 3 years and the other half will be repaid in one lump sum amount at the end of six years? How much interest will be paid over the 6-year period?

time to pay part of the principal

remaining time for interest to take effectnumber of lending/loaning periodPercent interest rate per interest periodprincipal amount lent/loanamount paid within lending/loaning period

interest per month before grace period

interest per month after grace periodtotal interest earned or paid


remaining time for interest to take effectnumber of lending/loaning periodtotal interest earned or paid

interest per month after grace period

interest per month before grace periodprincipal amount lent/loanamount paid within lending/loaning periodPercent interest rate per interest period


grace periodPercent interest rate per interest periodtotal interest earned or paid

ECONOMICS


interest per month before grace periodprincipal amount lent/loannumber of lending/loaning periodamount paid within lending/loaning period

number of lending/loaning periodamount paid within lending/loaning periodPercent interest rate per interest periodtotal interest earned or paid


interest per month before grace periodprincipal amount lent/loan


grace period

3. A future amount “F” is equivalent to P1,500.00 now when 6 years separates the amount and the annual compounded interest is 12%. What is the value of “F”?

number of periodPercent interest rate per interest periodamount at the start of a periodamount at the end of the period

number of periodPercent interest rate per interest periodamount at the end of the periodamount at the start of a period

number of periodamount at the start of a periodamount at the end of the periodPercent interest rate per interest period

Percent interest rate per interest periodamount at the start of a periodamount at the end of the periodnumber of period

4. You have used your credit card to purchase mobile phone battery worth 340 pesos. Unable to make payments for 7 months, you then write a letter of apology to pay your bill in full. The credit card company’s nominal interest rate is 18% compounded monthly. For what amount should you write the check?

ECONOMICS

1. You have just learned that ABC corporation has an investment opportunity that costs 500 pesos and 1017 months later pays a lump sum amount of 1,000,000.00 pesos. The cash flow diagram looks like this:

2. Suppose that you have 500 pesos cash today and can invest it at 0.75% compound interest each year. How many years will it take you to become a millionaire?

1. It is estimated that a certain business like Mlhuillier can save 60,000 pesos per year on pawning and fund transfering. The business has a lot contract of 6 years. If the business must earn a 20% annual return, how much could be justified for the construction of such establishment? Draw a cash flow diagram.

appreciation valuenumber of periodannual returninvested amount

2. A proposed development plan for a water district to avoid difficulties will require an immediate expenditures of 5,000,000 pesos to rehabilitate the water district facilities. What annual savings must be realized to recover this expenditure in 4 years with annual return of 10%

invested amountnumber of periodannual returnappreciation value

3. A certain jewelry cost 7,000 pesos now. If it will be appraised at 10,000 pesos after 5 months period. What will be the interest per month?

ECONOMICS

1. A certain land cost 28,000 pesos now. If it will be appraised at 60,000 pesos after 10 years period. What will be the interest rate per year?What will be the annual amount that the said land is appreciating?

Depreciation =the action or process of lowering in value; fall in the exchange value of currency.

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a straight line.

monetary value at third year is 380,000.00 pesos

text input value at blue text)

ECONOMICS

200,000.00500,000.005.00

(Php) (Php) (Php)500,000.00 60,000.00 60,000.00440,000.00 60,000.00 120,000.00380,000.00 60,000.00 180,000.00320,000.00 60,000.00 240,000.00260,000.00 60,000.00 300,000.00200,000.00 60,000.00 360,000.00140,000.00 60,000.00 420,000.0080,000.00 60,000.00 480,000.0020,000.00 60,000.00 540,000.00-40,000.00 60,000.00 600,000.00

-in the declining balance method, sometimes called the constant percentage methodor the Peakzone formula, it is assumed that the annual costof depreciation is a fixed percentage of the BV at the beginning of the year. The ratio of the depreciation in any one year to the BV at he beginningof the year is constant throughout the life of the asset and is designated by r(0<R<1) but not including 0 and 1.


dk dk*

ECONOMICS

monetary value at third year is 346,572.42 pesos

200,000.00500,000.005.000.16744679259812700000

(Php) (Php) (Php)500,000.00 83,723.40 83,723.40416,276.60 69,704.18 153,427.58346,572.42 58,032.44 211,460.02288,539.98 48,315.09 259,775.11240,224.89 40,224.89 300,000.00200,000.00 33,489.36 333,489.36166,510.64 27,881.67 361,371.03138,628.97 23,212.98 384,584.01115,415.99 19,326.04 403,910.0596,089.95 16,089.95 420,000.00

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a double declining line.monetary value at third year is 180,000.00 pesos

200,000.00500,000.005.000.40000000000000000000

(Php) (Php) (Php)

red text input value at blue text)

dk dk*

Computerized Double Declining Balance method Depreciation Table (do not edit red text input value at blue text)

dk dk*

ECONOMICS

500,000.00 200,000.00 200,000.00300,000.00 120,000.00 320,000.00180,000.00 72,000.00 392,000.00108,000.00 43,200.00 435,200.0064,800.00 25,920.00 461,120.0038,880.00 15,552.00 476,672.0023,328.00 9,331.20 486,003.2013,996.80 5,598.72 491,601.928,398.08 3,359.23 494,961.155,038.85 2,015.54 496,976.69

ECONOMICS

1. Draw a cash flow diagram for P 10,500 being loaned out at an interest rate of 15% per annum over a period of 6 years. How much simple interest would be repaid as a lump sum amount at the end of the sixth year?What will be the interest rate if paid lump sum at the end of sixth months?

ECONOMICS


ECONOMICS

4. You have used your credit card to purchase mobile phone battery worth 340 pesos. Unable to make payments for 7 months, you then write a letter of apology to pay your bill in full. The credit card company’s nominal interest rate is 18% compounded monthly. For what amount should you write the check?

ECONOMICS

1. You have just learned that ABC corporation has an investment opportunity that costs 500 pesos and 1017 months later pays a lump sum amount of 1,000,000.00 pesos. The cash flow diagram looks like this:

1. It is estimated that a certain business like Mlhuillier can save 60,000 pesos per year on pawning and fund transfering. The business has a lot contract of 6 years. If the business must earn a 20% annual return, how much could be justified for the construction of such establishment? Draw a cash flow diagram.

2. A proposed development plan for a water district to avoid difficulties will require an immediate expenditures of 5,000,000 pesos to rehabilitate the water district facilities. What annual savings must be realized to recover this expenditure in 4 years with annual return of 10%

ECONOMICS

1. A certain land cost 28,000 pesos now. If it will be appraised at 60,000 pesos after 10 years period. What will be the interest rate per year?What will be the annual amount that the said land is appreciating?

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a straight line.

ECONOMICS

(Php) (Php)60,000.00 440,000.00

120,000.00 380,000.00180,000.00 320,000.00240,000.00 260,000.00300,000.00 200,000.00360,000.00 140,000.00420,000.00 80,000.00480,000.00 20,000.00540,000.00 -40,000.00600,000.00 -100,000.00


dk* BVk

ECONOMICS

(Php) (Php)83,723.40 416,276.60

153,427.58 346,572.42211,460.02 288,539.98259,775.11 240,224.89300,000.00 200,000.00333,489.36 166,510.64361,371.03 138,628.97384,584.01 115,415.99403,910.05 96,089.95420,000.00 80,000.00

1. A certain submersible motor pump costs 500,000.00 pesos, its warranty is 5 years to run smoothly, if its estimated salvage value is 200,000.00 tabulate and find its monetary value at the third year. Graph and check if it is a double declining line.

(Php) (Php)

dk* BVk

dk* BVk

ECONOMICS

200,000.00 300,000.00320,000.00 180,000.00392,000.00 108,000.00435,200.00 240,224.89461,120.00 200,000.00476,672.00 166,510.64486,003.20 138,628.97491,601.92 115,415.99494,961.15 96,089.95496,976.69 80,000.00

ECONOMICS


ECONOMICS

(Php)440,000.00380,000.00320,000.00260,000.00200,000.00140,000.0080,000.0020,000.00-40,000.00

-100,000.00

ECONOMICS

(Php)416,276.60346,572.42288,539.98240,224.89200,000.00166,510.64138,628.97115,415.9996,089.9580,000.00

(Php)

ECONOMICS

300,000.00180,000.00108,000.00240,224.89200,000.00166,510.64138,628.97115,415.9996,089.9580,000.00

ECONOMICS

FIGURES ------------->

FIGURE 1

CHARTS---------------->

1 2 3 4 5 6 7 8 9 10

-200,000.00

-100,000.00

0.00

100,000.00

200,000.00

300,000.00

400,000.00

500,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

CHART 1

1 2 3 4 5 6 7 8 9 10

-200,000.00

-100,000.00

0.00

100,000.00

200,000.00

300,000.00

400,000.00

500,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

FIGURE 2

1 2 3 4 5 6 7 8 9 10

-200,000.00

-100,000.00

0.00

100,000.00

200,000.00

300,000.00

400,000.00

500,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

400,000.00

450,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

CHART 2

1 2 3 4 5 6 7 8 9 10

-200,000.00

-100,000.00

0.00

100,000.00

200,000.00

300,000.00

400,000.00

500,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

400,000.00

450,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

FIGURE 3 FIGURE 4

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

400,000.00

450,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

CHART 3

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

400,000.00

450,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

FIGURE 5

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

FIGURE 6

1 2 3 4 5 6 7 8 9 100.00

50,000.00

100,000.00

150,000.00

200,000.00

250,000.00

300,000.00

350,000.00

1 2 3 4 5 6 7 8 9 10

Years

Book Value

ECONOMICS

1 2 3 4 5 6 7 8 9 100.00

50,000.00

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ECONOMICS

PLANE GEOMETRY

D. Plane GeometryCircle

a graph of all points equidistant from a fixed points called the center. The fixed distance is called the radius R

Area= ¶= 3.1415926535897900000 unitsCircumference= 2 ¶ R

Diameter= 2R distance between two points in a circle that passes the center point this is a straight line

Polygonstriangle= a polygon with 3 sidessquare= a polygon with 4 equal sides

rectangle= a polygon with 2 equal sidesquadrilateral= a polygon with 4 unequal sides

trapezoid= a polygon with 4 unequal sides but 2 sides are parallelpentagon= a polygon with 5 sideshexagon= a polygon with 6 sides

heptagon= a polygon with 7 sidesoctagon= a polygon with 8 sides

nonagon= a polygon with 9 sidesdecagon= a polygon with 10 sides

undecagon= a polygon with 11 sides

¶ R2

PLANE GEOMETRY

dodecagon= a polygon with 12 sides

Ellipsea graph of all points the sum of whose distances from two fixed points (foci) is constant.

Square:Area:

A= where: s= length of sides (formula)Perimeter

P= 4s where: s= length of sides (formula)

Rectangle:Area:

A= lh where: l= length of 2 even sides (formula)h= length of 2 even sides

PerimeterP= 2(l+h) where: l= length of 2 even sides (formula)

h= length of 2 even sides

Quadrilateral:Area:

A=

w= (a+b+c+d)/2

ø= (Ὤ+ɸ)/2 where: w= sum of 4 sides divided by two (2)ø= sum of opposite internal angle over two (2)

Perimeter a+b+c+d= length of sides of a quadrilateralP= a+b+c+d

Trapezoid:Area:

A= ((a+b)h)/2a= length of one of the short parallel sideb= length of one of the long parallel sideh= distance between two parrallel sides

Cyclic Quadrilateral:-a quadrilateral inside a circleArea:

A=where:

w= (a+b+c+d)/2 a+b+c+d= length of sides of a quadrilateralw= sum of 4 sides divided by two (2)

Ptolemy’s Theorem:d1d2= ac+bd

Radius of Circle:

s2

2√{[(w-a)(w-b)(w-c)(w-d)]-abcd2ø}

2√[(w-a)(w-b)(w-c)(w-d)]

PLANE GEOMETRY

R=

example:1. What is the radius of the circle that circumscribed a quadrilateral with this sides?

a= 2.00000 Unit/sb= 2.00000 Unit/sc= 2.00000 Unit/sd= 2.00000 Unit/sS= 4.00000 Unit/s

4.00000 Square unit/sR= 2.59808 Unit/s

Segment of a Circle:

A=

Sector of a Circle:

A=

Rɸ

(2√[(ab-cd)(ac-bd)(ad-cb)])/4AqQ


AQ=

(1/2)R2(ɸ-sinɸ)

R2(ɸ/2)

LP=

PLANE GEOMETRY

SAMPLE PROBLEMS1. A lot is surveyed and yielded the table below. What is the area of the lot?

Lot 84 CAD 444-D has this corrected data:Line Bearing Distance Degrees

1 to 2 S 5 º 8 ‘ 0 “ E 16.98 S 5.1333 E

2 to 3 N 87 º 41 ‘ 0 “ W 15.36 N 87.6833 W

3 to 4 N 1 º 59 ‘ 0 “ W 3.47 N 1.9833 W

4 to 5 N 11 º 34 ‘ 0 “ W 13.76 N 11.5667 W

5 to 1 N 87 º 59 ‘ 49 “ E 16.72 N 87.9969 E

just add A1 A2 A3 A4 A5 A6 and A7area is 269 sq.m.

PLANE GEOMETRY

a graph of all points equidistant from a fixed points called the center. The fixed distance is called the radius R

Diameter

distance between two points in a circle that passes the center point this is a straight line

PLANE GEOMETRY

sum of opposite internal angle over two (2)

PLANE GEOMETRY

TRIGONOMETRY

E. Trigonometrythe part of Mathematics which deals with the sides and angles of a triangle as expressed by the trigonometric functions

Angle in degrees

degrees=A= any number

Trianglea polygon with three (3) sidesRight Triangle

a polygon with three (3) sides and a corner with angle of 90 degrees, sum of all angles is 180 degrees

sign for degrees is º

One complete circle is equal to 360 degrees (360º)Half circle is 180º, full circle is 360º, ¼ of circle is 90º

TRIGONOMETRY

Phytagorean Theorem:

(formula) where:a= side 1b= side 2c= hypotenuse

Example:1. If side 1 is 2 and side 2 is 3, what is the right triangle hypotenuse?

4 + 9=

13=

cc= 3.605551275

Six (6) Trigonometric Function:sinø cscøcosø secøtanø cotø

TriviaRemember SOH, CAH, TOA

SOH= sinø=opposite over hypotenuseCAH= cosø=adjacent over hypotenuseTOA= tanø=opposite over adjacent

Example:1. What is the angle between a triangle that has a hypotenuse of 5 and an adjacent side of the angle is 3?

take CAH

a2 +b2= c2

a2 +b2= c2

22 +32= c2

c2

c2

2√(13)=

TRIGONOMETRY

cosø= adjacent over hypotenusecosø 12/30/99

cosø= 0.6

ø=ø= 53.130102354156

Scalene Trianglea polygon with three (3) sides, sum of all angles is 180 degrees

Ice breakerAt what time after 12 noon will the hour and the minute hands of the clock form an angle of 120 degrees for the first time?

Remember when the minute hand has moved x minute spaces, the hour hand has moved x/60 minute spaces.

m/120= 15/90m= (15(120))/90

m= x-(x/60)(x/60)= x-m

x= 60x-60m60m= 60x-x60m= 59x

60/59m= x

Cos-1 (0.6)

TRIGONOMETRY

x= 1.017mm= 20x= 20.34

therefore time is:12:20.34

12:20:20 PM

Trivia2.5 kilos on a 10 kilo weighing scale forms a 90 degrees angle

and 180 degrees weighs 5 kilos, 270 degrees weighs 9 kilos

Remember:Sine Law:

(a/sinɸ) = =

Cosine Law:

A= s= (a+b+c)/2

where:abc= sides of the triangle

s= perimeter over two

Triangle with Inscribed Circle

A= rswhere:

s= perimeter over two

Triangle Inside a Circle

A= abc/4R

where:R= radius

abc= sides of the triangle

SAMPLE PROBLEMS

(b/sinβ) (c/sinὨ)

a2= b2+c2-(2bc)cosɸ

b2= a2+c2-(2ac)cosβ

c2= a2+b2-(2ab)cosὨ

2√(s(s-a)(s-b)(s-c))

1. The speed of an airplane is 300 nautical miles per hour in a direction N60ºE. The wind velocity at that instant is 5 nautical miles per hour coming from the west. Compute the actual speed of the plane and its direction relative to the ground.

TRIGONOMETRY

cosine law

90025-((3000)*(-0.642787609)

90025+1928.36282991953.362829

R= 303.24 mph actual speed of the plane

sin ɸ= sine law

300 303.24

sin ɸ =(sin130º)300

303.24

sin ɸ = 0.757859559ɸ =

ɸ =

direction of the plane = 90-49.276direction of the plane = N40.72E

2.Points A & B, which are 100m apart, are of the same elevation as the foot of the building. The angles of elevation of the building top from points A & B are 21 and 32 degrees respectively. How far is the building from point B. Assume that the points A & B are of the same line.

R2= 52+3002.-(2(5)(300)cos130º)

R2= 25+90000-((3000)cos130º)

R2=

R2=

R2=

sin130º

Sin-1 (0.757859559)

49.276º

TRIGONOMETRY

180 – 148 – 21 = 11degrees since total interior angle of a triangle is 180 degrees

sine law:

S=

100

sin21 sin11

S = 187.82m

Cos 32 = x/(187.82)

x = 159.28m distance of the building from point B

3.A bus travels from point M Northward for 30 min, then eastward for 1 hour, then shifted N-30degrees-W. If the speed is constant at 40kph, how far directly from M in km will the bus after 2 hrs?

TRIGONOMETRY

2hours= 30min north+1hour east then 30 min NW

H= 40-(20(sin30))H= 30km

V= 20+(20(cos30)V= 37.32km

2292.7824D= 47.88km

E.1 Spherical TrigonometrySpherical Triangle

A= Area

E= Spherical Excess

Remember:Sine Law:

sin(a)/asin(A) = (sin(b)/sin(B) =

Cosine Law:

D2= (302+37.322)

D2=

(¶R2E)/180º

A + B + C -180º

sin(c)/sin(C)

TRIGONOMETRY

Cos a= cosbcosc+sinbsinccosACos b= cosacosc+sinasinccosBCos c= cosacosb+sinasinbcosC

Cos A= -cosBcosC+sinBsinCcosa

where:abc= sides of the triangle

TRIGONOMETRY

the part of Mathematics which deals with the sides and angles of a triangle as expressed by the trigonometric functions

a polygon with three (3) sides and a corner with angle of 90 degrees, sum of all angles is 180 degrees

TRIGONOMETRY

1. What is the angle between a triangle that has a hypotenuse of 5 and an adjacent side of the angle is 3?

TRIGONOMETRY

At what time after 12 noon will the hour and the minute hands of the clock form an angle of 120 degrees for the first time?Remember when the minute hand has moved x minute spaces, the hour hand has moved x/60 minute spaces.

TRIGONOMETRY

1. The speed of an airplane is 300 nautical miles per hour in a direction N60ºE. The wind velocity at that instant is 5 nautical miles per hour coming from the west. Compute the actual speed of the plane and its direction relative to the ground.

TRIGONOMETRY


TRIGONOMETRY

since total interior angle of a triangle is 180 degrees


TRIGONOMETRY

TRIGONOMETRY

E. The wind velocity at that instant is 5 nautical miles per hour coming from the west. Compute the actual speed of the plane and its direction relative to the ground.

TRIGONOMETRY


SOLID GEOMETRY

F. Solid Geometry

Sphere

Volume of a Sphere:

V =4 where:3 ¶= (pi) circular constant equals to 3.1415926535897932384626433832795

r= radiusSurface Area:

SA = 4

Prisms

¶(r2)

¶(r2)

SOLID GEOMETRY

Volume= A*hLSA= p*h

where:LSA= least surface area

p= perimeter

Cones and Pyramids

Volume= (1/3)(A*h)LSA= (1/2)(p*L)

where:L= slant heightp= Base perimeter

Frustum of a Cone and Pyramid

SOLID GEOMETRY

Volume=

LSA=

where:L= slant heightp= perimeter

Spherical Segment (1 base)

Volume= ¶(h/3)[((3RA)-h)]SA= 2¶R

where:R= radius of the main sphere

Spherical Segment (2 bases)

Volume=SA= 2¶RH

where:R= radius of the main spherer= radius of the segment in the sphere

(h/3)[(A1+A2 + ((A1A2)(1/2))]

(1/2)(P1+P2)L

¶(h/6)[(3r12+3r2

2+h)]

SOLID GEOMETRY

like a billiard ball, basketball, tennis ball any ball

radius (r)

(pi) circular constant equals to 3.1415926535897932384626433832795

SOLID GEOMETRY

DIFFERENTIAL CALCULUS

G. DIFFERENTIAL CALCULUS

Newtons Method

from f(x) = 0.13x3 + 0 x2 + 10053.12x - 4423000.00

f(x)=

A= 0.13333333333333300000

B= 0.00000000000000000000C= 10053.12000000000000000000 xD= -4423000.00000000000000000000

f’(x)=

A= 0.39999999999999900000B= 0.00000000000000000000 xC= 10053.12000000000000000000

244.97500000000000000000

244.97560044492300000000


x3

x2

x2

x1= trial and error to equalize x1 and x2

x2=

DIFFERENTIAL CALCULUS

trial and error to equalize x1 and x2

mathematics

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