You can put this solution on YOUR website!
HELP i have two triangles ans the order from the bottom left corner is A the point is B and the right corner C inbetween the B AND C line there is the number 18 and between A AND C there is an x the other one is a triangle same order but D E AND F between E AND F There is a 12 and between D AND F there is a 6 . . WHAT DO THEY WANT?? the directions say triangle ABC~ triangle DEF and solve for x
THIS MEANS BC=18......AC=X.........EF=12.......DF=6
TRIANGLE ABC IS SIMILAR TO TRIANGLE DEF...
THEN AB/DE=BC/EF=CA/FD..HENCE..
18/12=X/6
12X=6*18
X=18*6/12=9

You can put this solution on YOUR website!
complete the square, and find the roots of the quadratic equation.
x2 + 16x = 0
X^2+2*X*8+8^2-8^2=0
(X+8)^2=8^2
X+8=+8...........OT..-8
X=0........OR........-16

You can put this solution on YOUR website!
apply the quadratic formula to find the roots of the given function, and then graph the function.
f(x) = x2 - 4=0




X=2..OR...-2
X.....0.......1......2.......3........4.....ETC
Y.....-4......-3.....0.......5........12

You can put this solution on YOUR website!
use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
x2 + 6x - 7 = 0
DEL=6^2-4*1*(-7)=36+28=64..2 SOLUTIONS,REAL
z2 + z + 1 = 0
DEL=1-4=-3..-VE........IMAGINARY SOLUTION ..2
(3)1/2y2 - 4y - 7(3)1/2 = 0...????PLEASE TYPE PROPERLY
2x2 - 10x + 25 = 0
DEL=100-4*2*25=-VE......IMAGINARY..2
2x2 - 6x + 5 = 0
DEL=36-40=-VE...IMAGINARY..2

You can put this solution on YOUR website!
The product of two consecutive positive even integers is 360. Find the integers.
LET ANY EVEN INTEGER=2N...NEXT EVEN INTEGER =2N+2
2N(2N+2)=360
N(N+1)=90
N^2+N-90=0
(N+10)(N-9)=0
N=9
THE 2 EVEN INTEGERS ARE 2*9=18 AND 20

You can put this solution on YOUR website!
Write the equation of the line that passes through point (6,4) =(X1,Y1)with a slope of 2 =M
Y-Y1=M(X-X1)
Y-4=2(X-6)
Y=2X-8

You can put this solution on YOUR website!
Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
let his speed be =x mph
speed at 10 mph faster=x+10 mph
time to go 200 miles
1.at normal speed =200/x
at faster speed =200/(x+10)
difference = 200[(1/x)-1/(x+10)]=200(x+10-x)/x(x+10)=1
2000=x^2+10x
x^2+10x-2000=0
(x+50)(x-40)=0
x=40 mph
x=

You can put this solution on YOUR website!
The Hudson River flows at a rate of 3 miles per hour. A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?
let the boat speed in still water =b
speed of water be =3
speed up river =b-3..time to travel 60 miles=60/(b-3)
speed down river =b+3..time to travel down river=60/(b+3)
total time =60[{1/(b+3)}+{1/(b-3)}]=9
60*(b+3+b-3)=9(b+3)(b-3)
40b=3(b^2-9)=3b^2-27
3b^2-40b-27=0
b=13.978 mph

You can put this solution on YOUR website!
In ratios, how would you represent the following:
4 minutes to 1 hour
in general as a rule ratios are dimensionless.so if n.r is in minutes d.r should also be in minutes..hence=4 mts/1 hr=4mts/60 mts=1/15
and
125 hits to 500 times at bat
=125/500=1/4..ratios are usually simplified to get smallest possible integers in n.r.and d.r.

You can put this solution on YOUR website!
A solar collector is 2.5m long by 2.0m wide. It is held in place by a frame of uniform width around its outside edge. If the exposed collector area is 2.5m^2 , what is the width of the frame, to the nearest tenth of a centimeter?
let the width of frame = x
length of frame=x+2.5+x=2.5+2x
width of frame =x+2+x=2x+2
area of frame & reflector assembly=(2x+2.5)(2x+2)=4x^2+9x+5
area of collector=2.5*2=5
exposed area =4x^2+9x+5-5=2.5
4x^2+9x-2.5=0
8x^2+18x-5=0
x={-18+sqrt(18^2+4*5*8)}/16=1/4 metre=100/4 cm = 25 cm.

You can put this solution on YOUR website!
I need help, I can't seem to grasp this concept. The question is.. identify the center, vertices, foci, and equations of the asymptotes of the hyperbola: 25y^2 - 9x^2 - 100y - 54x - 206 = 0. If you can help it would be greatly appreciated.
----------------------------------------------------------------------------------
FIRST COMPLETE SQUARES
25(Y^2-4y)-9(X^2+6X)-206=0
25{Y^2-2*Y*2+2^2-2^2}-9{X^2+2*X*3+3^2-3^2}-206=0
25(Y-2)^2-100-9(X+3)^2+81-206=0
-9(X+3)^2+25(Y-2)^2=-81+100+206=225
-[{(X+3)^2}/(225/9)]+[{(Y-2)^2}/(225/25)]=1
[(Y-2)^2/5^2]-[(X+3)^2/3^2]=1
NOW
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF..
Quadratic-relations-and-conic-sections/28184: What is the vertices,
foci, and slope of the asymptotes for the hyperbola whose equation is,
y^2/16 - x^2/25 =25?
1 solutions
Answer 15970 by venugopalramana(1088) About Me on 2006-03-04 03:12:07
(Show Source):
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
THE ANSWERS FOR YOUR CASE...H=0...K=0..A=4...B=5....EQN IS OF THE TYPE
(Y-K)^2/B^2-(X-H)^2/A^2=1....
SO VERTICES ARE...(H,(K-B)) AND (H,(K+B)) ...(0,-5) AND (0,5)
FOCI ARE {H,(K-BE)} AND {H,(K+BE)}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/B^2}=SQRT((16+25)/25)=SQRT(41/25)
SO FOCI ARE =(0,-5SQRT(41/25) AND (0,5SQRT(41/25)...
OR....(0,-SQRT(41)) AND (0,SQRT(41)
ASYMPTOTES ARE GIVEN BY
Y^2/16-X^2/25=K
25Y^2-16X^2-400K=0
(5Y+4X+A)(5Y-4X+B)=0
SLOPES OF ASYMPTOTES ARE
-4/5 AND 4/5
THE GRAPHS LOOK LIKE THIS

You can put this solution on YOUR website!
Use the discriminant to determine if the equation can be solved by factoring. If the equation can be solved by factoring, then factor it.
9k^2+6k+1=0
discr.=(6)^2-4*9*1=36-36=0....perfect square...equal roots ....equal to ...-6/2*9=-6/18=-1/3
hence.....9k^2+6k+1=(3k+1)^2

You can put this solution on YOUR website!
Find the imaginary solutions of the equation.
5x^2+3x+6=0
discriminant=3^2-4*5*6=9-120=-111
x=[{-3+i(sq.rt(111)}/10]
or
x=[{-3-i(sq.rt(111)}/10]

You can put this solution on YOUR website!
Hi i have a question concerning similar triangles:
A six foot man is standing 10 feet away from a 20 foot lamppost. What is the lenght of his shadow?
i tried to work it out and thought maybe i should use a proportion
20/10 X 6/a ----= 3 ft
LET AB BE THE LAMP POST WITH A ON GROUND AND B AT THE TOP.
LET CD BE THE MAN WITH C...FEET ON GROUND AND D ..HEAD
LET THE SHADOW BE CE.
WE HAVE 2 SIMILAR TRIANGLES.ABE AND CDE.
AB/CD=AE/CE=(AC+CE)/CE=(10+CE)/CE=1+10/CE
WE ARE GIVEN AB=20...CD=6.......AC=10
20/6=1+10/CE
10/CE=20/6 -1 =14/6=7/3
CE=3*10/7=30/7=SHADOW=4.285 FEET

thats what i got is that the correct answer? if not can you show me what is ? thanks!

You can put this solution on YOUR website!
Show that each equattion is true by simplifying the following:
USE A^-B^2=(A+B)(A-B)...RATIONALISE THE IRRATIONAL N.R....OR...D.R BY MULTIPLYING WITH CONJUGATE FACTOR.
a.) 1/(sqrt(x)+sqrt(y)) = (sqrt(x)-sqrt(y))/(x-y)
MULTIPLY N.R AND D.R WITH {SQRT(X)-SQRT(Y)}
{SQRT(X)-SQRT(Y)}/[{SQRT(X)}^2-{SQRT(Y)}^2]=(sqrt(x)-sqrt(y))/(x-y)
b.) (sqrt(x+h)- sqrt(x))/h = 1/(sqrt(x+h)+sqrt(x))
SAME WAY AS ABOVE ..MULTIPLY N.R AND D.R WITH (sqrt(x+h)+ sqrt(x))

You can put this solution on YOUR website!
Find the x & y intercepts of:
a.) (x^2)/4 -(y^2)/9 = 1
PUT Y=0 AND FIND X TO GET X INTERCEPTS
X^2/=1....X=+2..AND...-2...ARE THE 2 X INTERCEPTS.
PUT X=0 AND FIND Y TO GET X INTERCEPTS
Y^2/9=-1
Y^2=-9
Y IS IMAGINARY...HENCE THERE ARE NO Y INTERCEPTS.
b.) x^2+y^2 = 9
PUT Y=0
X=+3...AND -3 ARE THE X INTERCEPTS
PUT X=0
Y=+3...AND...-3 ARE THE Y INTERCEPTS.

You can put this solution on YOUR website!
The length of a rectangle is 1 cm longer than its width.
LET WIDTH =W
LENGTH=W+1
If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
DIAGONAL^2=L^2+W^2
4^2=(W+1)^2+W^2
W^2+2W+1+W^2=16
2W^2+2W-15=0



=2.28
W=2.28 AND L=3.28
Thank You!

You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
----------------------
Y INTERCEPT IS DETERMINED BY THE POINTS WHERE THE CURVE INTERSECTS X AXIS.HENCE THIS IS
OBTAINED BY EQUATING THE EQN. TO ZERO AND FINDING VALUES OF X SATISFYING THE EQN.
What is the y-intercept of the function, y=f(x)=2x^2+4x-6?=0
x^2+2x-3=0
x^2+3x-x-3=0
x(x+3)-1(x+3)=0
(x+3)(x-1)=0
x=-3 and x=1 ARE THE 2 Y INTERCEPTS.
--------------------
i need help with this quadratic using the formula.. i did the work and my answer came out to be irrational is this true?
a=3x^2+
b=8x+
c=2
3x^2+8x+2=0





yes you are correct.the roots are irrational

You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
----------------------
Y INTERCEPT IS DETERMINED BY THE POINTS WHERE THE CURVE INTERSECTS X AXIS.HENCE THIS IS
OBTAINED BY EQUATING THE EQN. TO ZERO AND FINDING VALUES OF X SATISFYING THE EQN.
What is the y-intercept of the function, y=f(x)=2x^2+4x-6?=0
x^2+2x-3=0
x^2+3x-x-3=0
x(x+3)-1(x+3)=0
(x+3)(x-1)=0
x=-3 and x=1 ARE THE 2 Y INTERCEPTS.
--------------------
i need help with this quadratic using the formula.. i did the work and my answer came out to be irrational is this true?
a=3x^2+
b=8x+
c=2
3x^2+8x+2=0





yes you are correct.the roots are irrational

You can put this solution on YOUR website!
Find the equations for the horizontal and vertical asypmtotes of each function:
a.) y=1/(x-1).........D.R=0 WHEN X-1=0..AND... X=1 IS THE ASYMPTOTE
b.) y=x/(sqrt(x^2-9))...D.R=0 WHEN X^2=9...OR...X=3..AND..X=-3 ARE THE 2 ASYMPTOTES.
FURTHER AS X TENDS TO ZERO Y TENDS TO ZERO.....Y=0 IS ANOTHER ASYMPTOTE
c.) y=x^3/(x^3-1)...D.R=0 WHEN X^3-1=0...OR...X=1..IS THE ASYMPTOTE
FURTHER AS X TENDS TO ZERO Y TENDS TO ZERO....Y=0 IS ANOTHER ASYMPTOTE.

You can put this solution on YOUR website!
Evaluate the factorial expression
100!/98! 2!
100!=[100*99*{98*97*96*.........*3*2*1}]
98!=98*97*96*.........*3*2*1
HENCE 100!/98!=100*99....AS 98,97,96,......3,2,1 CANCEL EACH OTHER
IN FACT WE CAN WRITE 100!=100*99*98!
SO WE GET 100!/98!2! = 100*99/2*1=50*99=4950

You can put this solution on YOUR website!
Y INTERCEPT IS DETERMINED BY THE POINTS WHERE THE CURVE INTERSECTS X AXIS.HENCE THIS IS
OBTAINED BY EQUATING THE EQN. TO ZERO AND FINDING VALUES OF X SATISFYING THE EQN.
What is the y-intercept of the function, y=f(x)=2x^2+4x-6?=0
x^2+2x-3=0
x^2+3x-x-3=0
x(x+3)-1(x+3)=0
(x+3)(x-1)=0
x=-3 and x=1 ARE THE 2 Y INTERCEPTS.

You can put this solution on YOUR website!
find the asymptotes: f (x)=2
---
x squared -9
i hope that equation made sense
it should be right on top of eachother but i cant seem to do that.
thanks for your help
y=2/(x^2-9)=2/(x+3)(x-3)
hence y tends to infinity as x+3 tends to zero or x tends to -3
and also as x-3 tends to zero...that is x tends to 3.
hence the asymptotes are
x=-3 and x=3

You can put this solution on YOUR website!
a motorboat travels 60 kms upstream in 10 hrs if it returns in 8 hrs. . Find th rate of the current and the boat rates are constant during the trip.
LET SPEED OF BOAT IN STILL WATER=B
SPEED OF WATER=W
UP STREAM SPEED =B-W
T=10 HRS....D=60 KM
10(B-W)=60
B-W=6................I
DOWN STREAM SPEED = B+W
T=8 HRS.....D=60 KM
8(B+W)=60
B+W=60/8=7.5.................II
EQN.I+EQN.II
2B=13.5
B=6.75 KMPH
W=7.5-6.75=0.75 KMPH

You can put this solution on YOUR website!
For what values of x is the following expression undefined:
(3x - 2)(5x - 4)
________________
(9x - 6)(7x - 1)
IT IS UNDEFINED WHEN DENOMINATOR IS ZERO
9X-6=0...OR...X=6/9=2/3...AND
7X-1=0....OR...X=1/7

You can put this solution on YOUR website!
1. From a box containing 6 red bulbs, 8 yellow bulbs and (??????HOW MANY )blue bulbs, two bulbs is drawn. Dteremine the probability that the first is a red and the second is yellow.

2. Five girls and 3 boys are to be seated in a row. If seats are taken randomly, waht is the probability that the girls will be seated together?
TOTAL NUMBER OF POSSIBILITIES TO SEAT 8 PERSONS=8!
FOR THE 3 GIRLS TO BE SEATED TOGETHER,CONSIDER 3 GIRLS TOGETHER AS ONE UNIT..THEN TOTAL =5+1=6
THEY CAN BE SEATED IN 6! WAYS
HENCE P=6!/8!=1/(8*7)=1/56

3. A box of 15 batteries contains( ?????HOW MANY ??)defectives, what is the probability that out of the 3 baterries selected without replacement, two is not defective?



4. A box contains 5 white and 3 green marbles. A second box contains 4 white and 6 green marbles. Suppose that a marble is drawn from the first box and placed unseen in the second box. What is the probability that a marble now drawn from the second box is green?
CASE 1....LET A GREEN MARBLE BE DRAWN FROM BOX 1
ITS PROBABILITY = 3/8
NOW WHEN IT IS PUT IN THE SECOND BOX THERE ARE 4 WHITE AND 7 GREEN MARBLES
P OF DRAWING GREEN MARBLE = 7/11
SO COMBINED PROBABILITY = (3/8)(7/11)=21/88
CASE 2......LET A WHITE MARBLE BE PICKED UP FROM BOX 1
ITS PROBABILITY = 5/8
NOW WHEN IT IS PUT IN THE SECOND BOX THERE ARE 5 WHITE AND 6 GREEN MARBLES
P OF DRAWING GREEN MARBLE = 6/11
SO COMBINED PROBABILITY = (5/8)(6/11)=30/88
HENCE P OF EITHER HAPPENING = 21/88 + 30/88 = 51/88
You may edit the question. Maybe convert formulae

You can put this solution on YOUR website!
travelling at the full speed against the current of a river a motorboat moves at the rate of 12 kilometers per hour relative to the land.Travelling at a half speed with the current,it moves at 8 kilometers per hour. Find the maximum speed of the boat in still water.
FULL SPEED OF BOAT IN STILL WATER =B
SPEED OF CURRENT OF WATER =W
SPEED OF BOAT AT FULL SPEED AGAINST CURRENT =B-W=12..............I
HALF SPEED OF BOAT IN STILL WATER =B/2=0.5B
SPEED OF BOAT AT HALF SPEED WITH THE CURRENT = 0.5B+W=8..............II
EQN.I+EQN.II
1.5B=20
B=20/1.5=13.333

You can put this solution on YOUR website!
Given f(x)=3x+1 and g(x)=1/2, what is the domain of f,g & fog
DOMAIN OF F(X) IS ALL REAL NUMBERS
DOMAIN OF G(X) IS ALL REAL NUMBERS
FOG=3*1/2+1=2.5
ITS DOMAIN IS ALL REAL NUMBERS

You can put this solution on YOUR website!
The sum of the digits of a 2 digit no. is 9, when the digits are reversed the new no. is 27 more than the original no. Find the no.
LET XY BE THE NUMBER...SUM OF DIGITS =X+Y=9.....................I
ITS VALUE =10X+Y
WHEN DIGITS ARE REVERSED WE GET YX...ITS VALUE = 10Y+X
DIFFERENCE BETWEEN THE 2 NUMBERS =27=10Y+X-10X-Y=9Y-9X=9(Y-X)
Y-X=3.............II
EQN.I+EQN.II....
2Y=9+3=12
Y=12/2=6
X=9-Y=9-6=3
THE NUMBER IS 36

You can put this solution on YOUR website!
SUBSTITUTION METHOD 3X - 3Y = 2 EQUATION 1
X + 2 = 3Y EQUATION 2
X=3Y-2
PUT IN EQN.I
3(3Y-2)-3Y=2
9Y-6-3Y=2
6Y=6+2=8
Y=8/6=4/3
X=3*(4/3)-2=4-2=2

You can put this solution on YOUR website!
i need help with this quadratic using the formula.. i did the work and my answer came out to be irrational is this true?
a=3x^2+
b=8x+
c=2
3x^2+8x+2=0





yes you are correct.the roots are irrational