You can put this solution on YOUR website!
Mrs. Johnson bought ten cans of vegetables - 2 cans of corn, 1 of beans, 5 of tomatoes, and 2 of beets. If she reaches into her cupboard and pulls out one without looking, what is the chance of it being beans?
TOTAL NUMBER OF CANS = 2+1+5+2=10
NUMBER OF BEAN CANS = 1
PROBABILITY OF DRAWING BEAN CAN = 1/10

You can put this solution on YOUR website!
SEE FOLLOWING EXAMPLES WHICH ARE SIMILAR AND TRY. IF YOU STILL HAVE PROBLEM COME BACK
----------------------------------------------------------------
A motorboat starts out and travels 9 miles an hour. In 3 hours another motorboat rraveling 18 miles an hour starts out to overtake the first one. In how many hours will thei sceond boat overtake the first.
1 solutions
Answer 13785 by venugopalramana(735) About Me on 2006-01-27 00:19:56 (Show Source):
A motorboat starts out and travels 9 miles an hour. In 3 hours another motorboat rraveling 18 miles an hour starts out to overtake the first one. In how many hours will thei sceond boat overtake the first.
RELATIVE SPEED WHILE TRAVELLING IN SAME DIRECTIONS = S2-S1=18-9=9 MPH.
INITIAL DISTANCE OF SEPERATION = DISTANCE TRAVELLED BY FIRST BOAT IN 3 HOURS
= 3*9=27 MILES.
FINAL DISTANCE OF SEPERATION=0
HENCE DISTANCE TO BE COVERED WHEN BOTH BOATS ARE TRAVELLING IN SAME DIRECTION =27-0=27
TIME TAKEN =DISTANCE / RELATIVE SPEED = 27 / 9 = 3 HRS.
Travel_Word_Problems/25659: C and D start from two points 480 miles apart and travel toward eash other. They meet in 8 hours. If C travels 6 miles per hour faster than D, find their rates.
1 solutions
Answer 13784 by venugopalramana(735) About Me on 2006-01-27 00:06:59 (Show Source):
SEE THE FOLLOWING PROBLEM WHICH IS SIMILAR
--------------------------------------------------------------------------
Two men, A and B, start toward each other at the same time from points 510 miles apart. If they travel 40 and 4 milea an hour respectively, in how many hours will they meet.
RELATIVE SPEED WHILE TRAVELLING IN OPPOSITE DIRECTION TO EACH OTHER = S1+S2=40+4=44 MPH.
INTIAL DISTANCE OF SEPERATION =510 MILES.
FINAL DISTANCE OF SEPERATION =0
HENCE DISTANCE TO BE COVERED TOGETHER = 510 -0 =510 MILES
TIME NEEDED =DISTANCE / RELATIVE SPEED = 510/44 HRS
----------------------------------------------------------------------------------
IN YOU CASE ,WE GET..WITH X AND X+6 SPEEDS AS GIVEN,
8= 480/(X+X+6)
8(2X+6)=480
2X+6=480/8=60
2X=60-6=54
54/2=27.
HENCE THEIR SPEEDS ARE 27 AND 33 MPH.
Travel_Word_Problems/25660: Two men, A and B, start toward each other at t

You can put this solution on YOUR website!
A worker can cover a parking lot with asphalt in ten hours. With the help of an assistant, they can do the job in six hours. How long would it take the assistant, working alone, to cover the parking lot with asphalt?
----------------------------------------------------------------------------------------
A worker can cover a parking lot with asphalt in ten hours.
HENCE IN 1 HOUR HE CAN DO 1/10 JOB.
With the help of an assistant, they can do the job in six hours.
HENCE IN ONE HOUR BOTH (HE+ASSISTANT)CAN DO 1/6 JOB
How long would it take the assistant, working alone, to cover the parking lot with asphalt
HENCE IN 1 HOUR ASSISTANT ALONE CAN DO 1/6 - 1/10 JOB =(5-3)/30=2/30=1/15 JOB
HENCE ASSISTANT ALONE CAN DO 1 JOB IN 1/(1/15) HOURS = 15 HOURS.

You can put this solution on YOUR website!
A worker can cover a parking lot with asphalt in ten hours.
HENCE IN 1 HOUR HE CAN DO 1/10 JOB.
With the help of an assistant, they can do the job in six hours.
HENCE IN ONE HOUR BOTH (HE+ASSISTANT)CAN DO 1/6 JOB
How long would it take the assistant, working alone, to cover the parking lot with asphalt
HENCE IN 1 HOUR ASSISTANT ALONE CAN DO 1/6 - 1/10 JOB =(5-3)/30=2/30=1/15 JOB
HENCE ASSISTANT ALONE CAN DO 1 JOB IN 1/(1/15) HOURS = 15 HOURS.

You can put this solution on YOUR website!
A motorboat starts out and travels 9 miles an hour. In 3 hours another motorboat rraveling 18 miles an hour starts out to overtake the first one. In how many hours will thei sceond boat overtake the first.
RELATIVE SPEED WHILE TRAVELLING IN SAME DIRECTIONS = S2-S1=18-9=9 MPH.
INITIAL DISTANCE OF SEPERATION = DISTANCE TRAVELLED BY FIRST BOAT IN 3 HOURS
= 3*9=27 MILES.
FINAL DISTANCE OF SEPERATION=0
HENCE DISTANCE TO BE COVERED WHEN BOTH BOATS ARE TRAVELLING IN SAME DIRECTION =27-0=27
TIME TAKEN =DISTANCE / RELATIVE SPEED = 27 / 9 = 3 HRS.

You can put this solution on YOUR website!
SEE THE FOLLOWING PROBLEM WHICH IS SIMILAR
--------------------------------------------------------------------------
Two men, A and B, start toward each other at the same time from points 510 miles apart. If they travel 40 and 4 milea an hour respectively, in how many hours will they meet.
RELATIVE SPEED WHILE TRAVELLING IN OPPOSITE DIRECTION TO EACH OTHER = S1+S2=40+4=44 MPH.
INTIAL DISTANCE OF SEPERATION =510 MILES.
FINAL DISTANCE OF SEPERATION =0
HENCE DISTANCE TO BE COVERED TOGETHER = 510 -0 =510 MILES
TIME NEEDED =DISTANCE / RELATIVE SPEED = 510/44 HRS
----------------------------------------------------------------------------------
IN YOU CASE ,WE GET..WITH X AND X+6 SPEEDS AS GIVEN,
8= 480/(X+X+6)
8(2X+6)=480
2X+6=480/8=60
2X=60-6=54
54/2=27.
HENCE THEIR SPEEDS ARE 27 AND 33 MPH.

You can put this solution on YOUR website!
Two men, A and B, start toward each other at the same time from points 510 miles apart. If they travel 40 and 4 milea an hour respectively, in how many hours will they meet.
RELATIVE SPEED WHILE TRAVELLING IN OPPOSITE DIRECTION TO EACH OTHER = S1+S2=40+4=44 MPH.
INTIAL DISTANCE OF SEPERATION =510 MILES.
FINAL DISTANCE OF SEPERATION =0
HENCE DISTANCE TO BE COVERED TOGETHER = 510 -0 =510 MILES
TIME NEEDED =DISTANCE / RELATIVE SPEED = 510/44 HRS

You can put this solution on YOUR website!
)prove that cosA + sinA tanA = secA
LHS=COS(A)+SIN A * SIN(A) / COS(A)= {COS(A) * COS(A) + SIN(A)*SIN(A)} / COS(A)
=1/COS(A)=SEC(A) ..SINCE {COS (A)}^2 +{(SIN (A)}^2 = 1
2)prove that cos(180-y) = -cos y
GO BACK TO THE DEFINITION OF TRIGNOMETRIC RATIOS.
WE START WITH THE ORIGIN AND THE 2 AXES X'OX AND Y'OY
WE START WITH A LINE SEGMENT OP ON X AXIS CALLED RADIUS VECTOR.BY CONVENTION THIS IS ALWAYS POSITIVE.WE LET IT TRAVEL IN COUNTERCLOCKWISE DIRECTION (CCD IN SHORT)WHICH IS CONSIDERED THE POSITIVE DIRECTION (PD IN SHORT ), MAKING AN ANGLE ..SAY...A... WITH OX IN THE CCD OR PD.DEPENDING ON THE ANGLE TRAVELLED IT MAY END UP IN ANY OF THE 4 QUADRANTS....YOX (I QUADRANT...0 TO 90 DEGREES),YOX'(II QUADRANT...90 TO 180 DEGREES),Y'OX'(III QUADRANT....180 TO 270 DEGREES),Y'OX (IV QUADRANT...270 TO 360 DEGREES).
THEN WE DROP A PERPENDICULAR PQ FROM P TO X'OX.WE CONSIDER THE RIGHT ANGLED TRIANGLE OPQ TO DEFINE THE TRIGNOMETRIC RATIOS AS FOLLOWS, WITH ANGLE AT O THAT IS ANGLE POQ BEING A
SIN A = QP/OP
COS A =OQ /OP
TAN A =QP/OQ
AS GIVEN IN THE BEGINING OP IS ALWAYS POSITIVE .SO SIGN OF A RATIO DEPENDS ONLY ON SIGN OF OQ AND QP.
NOW IN YOUR PROBLEM
COS (180-Y)...OP STARTED ON X'OX AND TRAVELLED 180 DEGREES IN CCD ENDING UP ON OX'.THEN IT TRAVELLED BACK BY Y DEGREES IN THE NEGATIVE OR CLOCK WISE DIRECTION TO END UP IN II QUADRANT.
SO IN OUR CASE IN THE RIGHT ANGLED TRIANGLE OPQ ANGLE POQ = Y....OQ IS NEGATIVE SINCE AT THIS POSITION OQ IS TOWARDS OX' THE NEGATIVE X AXIS.
HENCE COS (180-Y)= OQ / OP = - COS Y.
3)use the sum and difference identities to find the exact value of tan 165 degrees
TAN 165 = TAN (180-15)= - TAN 15 = - TAN (45-30)
= - {(TAN 45 - TAN 30 )/ ( 1 + TAN 45 * TAN 30 ) }
= - { (1-1/SQ.RT.3)/ ( 1+1*1/SQ.RT.3)}
= - { (SQ.RT.3-1)/(SQ.RT.3 +1)}= - { ( SQRT.3 - 1 )^2/((SQRT.3+1)*(SQRT.3-1))}
= - (3+1-2*SQRT.3)/(3-1)= - (2-SQRT.3)=SQRT.3-2

You can put this solution on YOUR website!
Ok, this is one that has baffled not only myself but my parents and boyfriend as well. This is a problem that no one seems to be able to figure out. Do you think you could help me?
GOOD PROBLEM YOUNG GIRL!NOW THAT YOU SAY THAT YOUR BOY FRIEND IS ALSO FOXED BY THE PROBLEM ,YOU CAN USE IT TO GOOD EFFECT TO PUT ACROSS A PROPOSAL TO HIM ...
YOU PROMISS HIM A 100,000..POUNDS EVERY DAY FOR 30 DAYS.LET HIM GIVE YOU IN RETURN JUST A PENNY ON THE FIRST DAY,2 PENNIES ON THE SECOND DAY ,4 PENNIES ON THE THIRD DAY,......ETC FOR 30 DAYS ..AS IN YOUR PROBLEM...BASED ON YOUR FEED BACK , I PRESUME HE WILL LAP ON THE PROPOSAL AND I CAN ASSURE YOU THAT YOU WONT REGRET THE PROPOSAL!!!!
OK NOW LET US GET BACK TO YOUR PROBLEM AS WELL AS MY SUGGESTION ,WHICH IS JUST A PART OF IT .I AM GIVING BELOW THE MONEY TO BE PAID ON THIS BASIS FOR 30 AND 64 DAYS ,ROUNDED OFF TO POUNDS OR MILLION POUNDS AT THE LATER DAYS.
DAY.....MONEY TO BE PAID........MONEY TO BE PAID BY YOUR FRIEND TO YOU IN
............TO YOUR FRIEND..... CENTS...POUNDS...MILLION POUNDS
................POUNDS
1 100000 1 0 0
2 100000 2 0 0
3 100000 4 0 0
4 100000 8 0 0
5 100000 16 0 0
6 100000 32 0 0
7 100000 64 1 0
8 100000 128 1 0
9 100000 256 3 0
10 100000 512 5 0
11 100000 1024 10 0
12 100000 2048 20 0
13 100000 4096 41 0
14 100000 8192 82 0
15 100000 16384 164 0
16 100000 32768 328 0
17 100000 65536 655 0
18 100000 131072 1311 0
19 100000 262144 2621 0
20 100000 524288 5243 0
21 100000 1048576 10486 0
22 100000 2097152 20972 0
23 100000 4194304 41943 0
24 100000 8388608 83886 0
25 100000 16777216 167772 0
26 100000 33554432 335544 0
27 100000 67108864 671089 1
28 100000 134217728 1342177 1
29 100000 268435456 2684355 3
30 100000 536870912 5368709 5
31 100000 1073741824 10737418 11
32 100000 2147483648 21474836 21
33 100000 4294967296 42949673 43
34 100000 8589934592 85899346 86
35 100000 17179869184 171798692 172
36 100000 34359738368 343597384 344
37 100000 68719476736 687194767 687
38 100000 1.37439E+11 1374389535 1374
39 100000 2.74878E+11 2748779069 2749
40 100000 5.49756E+11 5497558139 5498
41 100000 1.09951E+12 10995116278 10995
42 100000 2.19902E+12 21990232556 21990
43 100000 4.39805E+12 43980465111 43980
44 100000 8.79609E+12 87960930222 87961
45 100000 1.75922E+13 175921860444 175922
46 100000 3.51844E+13 351843720888 351844
47 100000 7.03687E+13 703687441777 703687
48 100000 1.40737E+14 1407374883553 1407375
49 100000 2.81475E+14 2814749767107 2814750
50 100000 5.6295E+14 5629499534213 5629500
51 100000 1.1259E+15 11258999068426 11258999
52 100000 2.2518E+15 22517998136853 22517998
53 100000 4.5036E+15 45035996273705 45035996
54 100000 9.0072E+15 90071992547410 90071993
55 100000 1.80144E+16 180143985094820 180143985
56 100000 3.60288E+16 360287970189640 360287970
57 100000 7.20576E+16 720575940379279 720575940
58 100000 1.44115E+17 1441151880758560 1441151881
59 100000 2.8823E+17 2882303761517120 2882303762
60 100000 5.76461E+17 5764607523034230 5764607523
61 100000 1.15292E+18 11529215046068500 11529215046
62 100000 2.30584E+18 23058430092136900 23058430092
63 100000 4.61169E+18 46116860184273900 46116860184
64 100000 9.22337E+18 92233720368547800 92233720369

30 DAY..3000000.........1073741823......10737418.....................11
TOTAL

64 DAY..6400000.........1.84467E+19.....184467440737095000......184467440737
TOTAL

YOUR GAIN IN 30 DAYS = 8 MILLION POUNDS..GOT IT BUT ONLY BE CAREFULL THAT YOUR FRIEND WONT RUN OUT OF YOU ON THE 25 TH. DAY.!!!
YOUR GAIN IN 30 DAYS = 184467440731 MILLION POUNDS
NOW COMING TO THE MATHS PART OF THIS ,THIS SEQUENCE WHERE EACH NUMBER BEARS A CONSTANT RATIO TO ITS PREDECESSOR IS CALLED GEOMETRIC PROGRESSION..HERE YOU FIND EACH NUMBER IS OBTAINED FROM THE PREVIOUS ONE BY MULTIPLYING WITH 2 , CALLED COMMON RATIO.
THE LAST NUMBER ON NTH. DAY AND SUM OF SUCH SERIES OF NUMBERS UPTO THE N TH.DAY IS GIVEN BY THE FOLLOWING FORMULAE...
N TH. NUMBER = FIRST NUMBER *(C0MM0N RATIO )^(N-1)=A*(R)^(N-1)=(2)^(N-1) IN THIS CASE.
SUM UP TO N TH.NUMBER =A*{((R)^N - 1))/(R-1)}={(2)^N-1}IN THIS CASE

You can put this solution on YOUR website!
YOU ARE CORRECT. HAVE CONFIDENCE..BELIEVE IN YOUR SELF..THE PROBLEM HAS NO REAL NUMBER SOLUTION
ITS DISCRIMINANT =(-2)^2-4*3*5=4-60=-56... IS NEGATIVE.SO IT HAS NO REAL SOLUTION.
ON COMPLETING THE SQUARE YOU GET NEGATIVE VALUE FOR A PERFECT SQUARE WHICH IS NOT POSSIBLE IN REAL NUMBERS

You can put this solution on YOUR website!
i need help with the equation 10^(4log10^3). it asks me to evaluate it, and the anwer is 81. but when i did it i set it = to x, and got 10^x=log10^12.then i got 10^10X=log10^12, in which x=12.CORRECT...SEE BELOW...BUT I THINK YOU MIGHT HAVE TYPED THE PROBLEM WRONGLY
IF IT IS 10^(4LOG(3))=X,YOU WILL GET X=81...YOU MIGHT HAVE SEEN BASE 10 ..LOG 3 TO BASE 10...ANY WAY CHECK THE PROBLEM
LET 10^(4log10^3)=X....TAKING LOGS
4*{LOG((10)^3)}LOG(10)=LOG(X)
4*3*LOG(10)=LOG(X)
LOG((10)^12)=LOG(X)
X=10^12

You can put this solution on YOUR website!
ln ((square root x) divided by (y square root z)).
PUT BRACKETS PROPERLY
LN{SQ.RT(X)/(Y*SQ.RT(Z))}IT MEANS SQ.RT.X IS DIVIDED BY BOTH Y AND SQ.RT.Z.
SO WE GET -O.5 LN Z AND NOT +0.5 LN Z
WHERE AS IF IT HAD BEEN
LN{SQ.RT(X)/(Y)*(SQ.RT(Z)},IT COULD BE TAKEN AS
LN{(SQ.RT(X))*(SQ.RT(Z))/(Y),WHEN YOUR ANSWER IS OK. SO PROPER USE OF BRACKETS IS VERY IMPORTANT.HOPE YOU UNDERSTOOD.
i got .5lnx-lny+.5lnz, but the last part is supposed to be -.5lnz. what did i do wrong? thanks

You can put this solution on YOUR website!
condense 6lnx-1-ln3.=LN (X)^6 -LN (E)- LN 3..
i got ln(x^6/3)
YOU MISSED TO PUT 1 = LN E..YOU CAN COMBINE BY ADDITION OR SUBTRACTION ONLY IF ALL TERMS ARE LOGS .YOU CANNOT COMBINE LOGS WITH ORDINARY NUMBERS BY ADDITION OR SUBTRACTION.
but the answer says ln(x^6/3e).
NOW IF YOU PUT 1 =LN E ,THEN YOU WILL GET LN(X^6/3E)
where did the e come from?
HOPE IT IS CLEAR.

You can put this solution on YOUR website!
There is a figure drawn with Trapezoid ABCD with Median MN. Then there are diagonals AC and DB with the Intersecting points, E and F, respectively. The problem is: If DC=3x, AB=2xsquared, and EF=7, find the value of X.
THOUGH , I DONOT HAVE YOUR BOOK ,YOU SHOULD BE HAVING IT.SO FOLLOW THE DIAGRAM FROM THE BOOK.I SHALL FOLLOW THE SAME NOMENCLATURE GIVEN BY YOU.O.K?
ABCD IS THE TRAPEZIUM.LET AB BE THE BOTTOM SIDE AND DC THE TOP SIDE.M IS THE MID POINT OF AD,AND N IS THE MID POINT OF BC.MN IS THE MEDIAN.JOIN AC AND BD THE DIAGONALS.LET AC CUT MN AT E AND BD CUT MN AT F.WE ARE GIVEN
EF=7
AB=2X^2
DC=3X
NOW MEDIAN IN A TRAPEZIUM IS PARALLEL TO THE 2 PARALLEL SIDES AB AND DC AND IS HALF THE SUM OF THE 2 PARALLEL SIDES .THAT IS
MN = (AB+DC)/2=(2X^2+3X)/2
AGAIN ,WE HAVE 3 PARALLEL LINES AB,DC AND MN.ONE TRANSVERSAL AD CUTS THEM IN EQUAL PARTS.(AM=MD).HENCE ANY OTHER TRANSVERSAL ALSO CUTS THEM IN EQUAL PARTS.
HENCE FOR TRANSVERSAL DFB WE HAVE
DF=FB
AND FOR TRANSVERSAL CEA WE HAVE
CE=EA.
NOW IN TRIANGLE DAB , MF IS THE LINE JOINING MIDPOINTS OF DA AND DB.HENCE
MF=AB/2=2X^2/2=X^2....
SIMILARLY IN TRIANGLE CAB , EN IS THE LINE JOINING MIDPOINTS OF CA AND AB.HENCE
EN=AB/2=2X^2/2=X^2.......NOW
MN=(2X^2+3X)/2=MF+FN=MF+EN-EF= X^2+X^2-7 =2X^2-7....OR
2X^2+3X=2(2X^2-7)=4X^2-14
4X^2-14-2X^2-3X=0
2X^2-3X-14=0
2X^2-7X+4X-14=0
X(2X-7)+2(2X-7)=0
(2X-7)(X+2)=0
2X-7=0
X=7/2=3.5

You can put this solution on YOUR website!
ABSOLUTE VALUE OF A COMPLEX NUMBER Z = A+iB IS GIVEN BY
|Z|=SQUARE ROOT OF (A^2+B^2)...HENCE FOR Z = 5-4i.WE GET
|Z|=SQUARE ROOT OF (5^2+(-4)^2)= SQUARE ROOT OF (25+16)= SQUARE ROOT OF (41)

You can put this solution on YOUR website!
When using gaussian elimination, can an inconsistant solution have a general solution? i worked out the matrix and came to a spot where 1=-3 and 1=10.4 (inconsistant 1 can't equal 2 different solutions); would there still be a way to formulate a general solution such as x=(2-4x,2-3x,x,0) = (2,2,0,0) + x(-4,-3,1,0)
I hope this is enough information to clearly explain my question. thanks
I THINK YOU ARE CONFUSED BETWEEN ...INCONSISTENT,AND DEPENDENT EQUATIONS.
AS THE NAME IMPLIES AN INCONSISTENT SET OF EQUATIONS HAS NO SOLUTION AS THEY ARE INCONSISTENT.FOR EXAMPLE
X+Y=2
2X+2Y=3....HENCE THERE IS NO QUESTION OF GETTING A GENERAL SOLUTION FOR THAT BY ANY METHOD GAUSSIAN ELIMINATION OR ANY..
CONSISTENT AND DEPENDENT EQNS.HAVE INFINITE SOLUTIONS GIVEN BY A GENERAL FORMULA AS YOU HAVE MENTIONED..FOR EX.
X+Y=1
2X+2Y=2....THE GERAL FORMULA IS Y=1-X...OR (X,1-X)IS A SOLUTION SET..LIKE (1,0),(2,-1) ETC....
FINALLY CONSISTENT AND INDEPENDENT EQNS.HAVE UNIQUE SOLUTION..FOR EXAMPLE
X+Y=2
X-Y=0...HAVE ONE UNIQUE SOLUTION X=1 AND Y=1
NOW YOU CAN USE GAUSSIAN ELIMINATION TO THE 3 EXAMPLES ABOVE AND SEE WHAT YOU GET.IF YOU STILL AVE DIFFICULTY COME BACK

You can put this solution on YOUR website!
the arithmetic series of numbers 1,3,5,7,9,......find the difference between any 2 terms.
DIFFERENCE =D=3-1=2
using the formula for nth term
TN=A+(N-1)*D...WHERE A=FIRST TERM OF SERIES =1
what is the 101 term
T101=1+(101-1)*2=1+200=201
what is the sum of the first 20 terms.
SN=(N/2){2*A+(N-1)*D}
S20=(20/2){2*1+(20-1)*2}=10*(2+38)=10*40=400
------------------------------------------------------------------------
use the geometric series 1,2,4,8,16.find the ratio between 2 terms.
COMMON RATIO =R = 2/1=2
with the formula for nth of the geo series what is the 24th term.
TN=A*R^(N-1) WHERE A=FIRST TERM OF SERIES =1
T24=1*2^(24-1)=2^23

You can put this solution on YOUR website!
An open-top box is to be connected from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and then folding up the flaps. Let x denote the length of each side of the square to be cut out. Find the function V that represents the volume of the box in terms of x.
This is what I got:
6x+8x+x=14x²
x=14
LENGTH OF BOARD=6 FT....WIDTH OF BOARD =8 FT.
IF X LONG SQUARE PIECES ARE CUT AT 4 CORNERS WE SHALL HAVE THE BOX DIMENSIONS AS
HEIGHT =X.....LENGTH = 6-2X.....WIDTH =8-2X
VOLUME = V = L*W*H =X(6-2X)(8-2X)=X(48-28X+4X^2)=48X-28X^2+4X^3

You can put this solution on YOUR website!
CONSIDER THE FOLLOWING EQUALITY
(N+1)!-N!=(N+1)*N!-N!=N!*(N+1-1)=N*N!...THAT IS
(N+1)!-N!=N*N!....PUT IN THIS EQUALITY N=1,2,3,....ETC...TILL N AND ADD THEM UP
N=1....WE GET 2!-1!=1*1!
N=2...........3!-2!=2*2!
N=3...........4!-3!=3*3!
........................
.........................
N=N.......(N+1)!-N!=N*N!
-----------------------------ADDING ALLTHE ABOVE,WE GET
..........(N+1)!-1!=1*1!+2*2!+3*3!+...........N*N!............PROVED

You can put this solution on YOUR website!
In triangle ABC, AC=7, BC=12, and angle B=35 degrees. solve the triangle
AC/SIN B = BC/SIN A
7/SIN 35 = 12/SIN A
SIN A = 12*SIN 35/7=0.982838969
A = 79.4 DEG.....OR...... 100.6 DEG.
C=180-35-79.4=65.6 DEG.OR....180-35-100.6=44.4
AC/SIN B = AB/SIN C
7/SIN 35 = AB /SIN 65.6....OR.....7/SIN 35 = AB /SIN 44.4
AB = 7*SIN 65.6/SIN 35....OR..........7*SIN 44.4/SIN 35
AB = 11.12..........OR............8.54

You can put this solution on YOUR website!
In triangle ABC, b=11, c=18, and angle B=40 degrees. solve the triangle.
b/SIN B =c/SIN C..HENCE
SIN C = c*SIN B / b
SIN C = 18*SIN 40 /11= 1.051390552
THE PROBLEM IS NOT CORRECT .PLEASE CHECK UP THE VALUES OF b AND c

You can put this solution on YOUR website!
Write the equation of a line parallel to y=-3/4x-2, which has a y-intercept of 5.
EQN.OF LINE PARALLEL WILL BE OF THE FORM
Y=-3/4 *X+K.....WHERE K IS TO BE FOUND.FOR THIS WE ARE GIVEN THAY Y INTERCEPT IS 5 .THAT IS WHEN X=0,Y=5.HENCE
5=-3/4 *0+K....OR...K=5...HENCE THE EQ2N.IS
Y=(-3/4)X+5

You can put this solution on YOUR website!
THIS IS A STANDARD FORMULA AND IS EQUAL TO COS (2X)...THERE IS NOTHING MORE TO IT UNLESS YOU ARE TO PROVE THIS FORMULA.PLEASE GIVE ME YOUR COURSE OF STUDY AND YOUR BACK GROUND KNOWLEDGE SO THAT I CAN GIVE FURTHER ADVICE.
cos^2(x) - sin^2 (x)=?=COS(2X)
Thanks

You can put this solution on YOUR website!
ARRANGE THE SCORES IN INCREASING ORDER...WE GET
79,83,86,92,93
NOW
MEAN =(79+83+86+92+93)/5=433/5=86.6
MEDIAN=MIDLE SCORE=86
MODE=SCORE WITH MAXIMUM FREQUENCY...HERE ALL HAVE SAME FREQUENCY ...ONE ONLY ...IN CASE 2 OR MORE SCORES HAVE SAME FREQUENCY WE TAKE THEIR AVERAGE AS THE MODE...HERE MODE COINCIDES WITH MEAN AS ALL SCORES ARE HAVING SAME FREQUENCY.
MEAN IS CONSIDERED THE BEST REPRESENTATION FOR CENTRAL TENDENCY.

You can put this solution on YOUR website!
I know that I need three equations to solve the system, but it seems like I don't have enough information to solve. I can do x+y+z=40 for one equation. But I'm not sure how to get the other two equations. Thanks for the help.
The college bookstore sells 3 types of calculators. They have a total of 40 calculators in stock. The total amount of revenue from these calculators will be $3590. The most expensive type costs $75 more than the cheapest. Find the cost of each type.
THERE ARE NOT 3 BUT 4 UNKNOWNS HERE INCLUDING THE RATES
LET X =NUMBER OF CHEAPEST CALCULATORS SOLD AT RATE OF A PER CALCULATOR.
LET Y =NUMBER OF COSTLIEST CALCULATORS SOLD AT RATE OF A+75 PER CALCULATOR.
LET 40-X-Y =NUMBER OF MIDCOST CALCULATORS SOLD AT RATE OF B PER CALCULATOR.
SO THE EQUATIONS IS
AX+(A+75)Y+B(40-X-Y)=3590
HENCE THEWRE ARE 4 UNKNOWNS A,X,Y,B WITH 1 EQN.BUT THERE IS ONE CONDITION THAT X AND Y SHOULD BE INTEGERS AS THEY ARE NUMBER OF CALCULATORS SOLD.
WE CAN ONLY TRY BY GUESS OR USE ADVANCED NUMBER THEORY TECHNIQUES TO REDUCE NUMBER OF TRIALS TO ANSWER.
I DO NOT KNOW YOUR COURSE OF STUDY OR YOUR BACK GROUND KNOWLEDGE TO DWELL ON THESE THINGS.PLEASE INFORM .OR...
GET BACK TO YOUR TEACHER WITH THIS REPORT.

You can put this solution on YOUR website!
The perimeter of a rhombus is 40cm, and one diagonal is 12cm long. How long is the other diagonal?
RHOMBUS PERIMETER =4*SIDE=40
SIDE=40/4=10
DIAGONALS BISECT EACH OTHER AT RIGHT ANGLES IN RHOMBUS.HENCE
SIDE^2 = (D1/2)^2+(D2/2)^2
10*10 = (12/2)^2+(D2/2)^2
(D2/2)^2=100-36=64
D2/2=8
D2=16

You can put this solution on YOUR website!
LET THE NUMBERS BE 2N,2N+2,2N+4
THEIR SUM IS
2N+2N+2+2N+4=6N+6=-72
6N=-72-6=-78
N=-78/6=-13
HENCE THE NUMBERS ARE -26,-24,-22

You can put this solution on YOUR website!
WE HAVE 1/2*2/3*3/4*......*99/100
2,3,4,...TILL 99..IN NUMERATOR AND DENOMINATOR CANCEL EACH OTHER .SO WE GET
ANSWER AS 1/100

You can put this solution on YOUR website!
WE HAVE 1/2*2/3*3/4*......*99/100
2,3,4,...TILL 99..IN NUMERATOR AND DENOMINATOR CANCEL EACH OTHER .SO WE GET
ANSWER AS 1/100

You can put this solution on YOUR website!
Are the lines y= 3x + 2 and x= 3y + 2 parallel?
NO THE COEFFICIENTS OF X AND Y IN THE 2 EQUATIONS IN STANDARD FORM MUST BE IN SAME RATIO FOR THE LINES TO BE PARALLEL.HERE WE HAVE
3X-Y+2=0....AND
-X+3Y+2=0
THE RATIOS ARE FOR X...... -3/1 AND FOR Y....... -1/3 WHICH ARE NOT EQUAL.HENCE THEY ARE NOT PARALLEL

You can put this solution on YOUR website!
A window in a bulinging is 18m above the ground. From this window, the angle of elevation to the top of the building across the street is 40° and the angle of depression to the bottom of the building acress the street is 42°. How tall is the building across the street? Include a labelled diagram with you solution
I AM GIVING A GRAPH BELOW FOR YOUR GUIDANCE.YOU CAN COMPLETE THE DRAWING AND LABEL THE DIAGRAM ON THE BASIS GIVEN BELOW.
LET O BE THE ORIGIN;LET THE Y AXIS BE THE BUILDING , THE WINDOW AT 18M.ABOVE GROUND BE ON THE Y AXIS AT W ; AND X AXIS THE GROUND.
HENCE OW = 18 M.
THE LINE OF ELEVATION OF TOP OF THE BUILDIND ACROSS THE STREET FROM WINDOW ON Y AXIS IS SHOWN BY WT;AND THE LINE OF DEPRESSION OF THE BUILDING FROM THIS WINDOW IS SHOWN BY WB.JOIN BT WHICH WILL BE PERPENDICULAR TO THE GROUND OR X AXIS.DRAW A HORIZONTAL W H FROM WIDOW W TO BUILDING LINE BT.

WE HAVE 2 RIGHT ANGLE TRIANGLES WTH AND WHB.IN TRIANGLE WTH
ANGLE TWH = 40 DEGREES THE ANGLE OF ELEVATION
HENCE TAN 40 = TH/WH OR TH =WH * TAN 40.............I
IN TRIANGLE WHB
ANGLE HWB = 42 DEGREES THE ANGLE OF DEPRESSION.
HENCE TAN 42 = HB/WH OR WH=HB/TAN 42
BUT HB = OW =18 M ..SO..WH=18 /TAN 42.................II
SUBSTITUTING IN EQN.I WE GET
TH = 18*TAN 40 / TAN 42
SO HEIGHT OF BUILDING = TH +HB =18*TAN 40 / TAN 42 +18
=18*{(TAN 40 / TAN 42 ) + 1 }..YOU CAN SUBSTITUTE THE VALUES FOR TAN 40 AN TAN 42 AND GET