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I need help with this problem i have in Algebra 1.Here is the instructions: The measure of two angles of a triangle are given. Find the measure of the third angle. Problem:x, y Please help me.
2 ANGLES ARE X AND Y ..LET THIRD ANGLE BE CALLED A
THEN WE HAVE IN A TRIANGLE =SUM OF 3 ANGLES =180.SO
X+Y+A=180
A = 180-X-Y
SINCE WE KNOW X AND Y WE CAN FIND THE THIRD ANGLE A FROM THIS RELATION/EQUATION..
SUPPOSE X= 50 AND Y =70 DEG.THEN
A = 180-50-7-=180-120=60
THIRD ANGLE =60

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The rectangle ABCD is inscribed in the circle with center(0,0) and radius 6. The x-coordinate of the vertex A is expressed as x. (Shows a diagram with an x and y axis. A circle, with a square inside it where each of the 4 corners of the square touches the circle. And a point A at the top right corner labeled as point A (x,y). Where A is the vertex.)
OK ..BUT THE DIAGRAM IS NEEDED AS THE DATA GIVEN BY YOU IS NOT FULLY CLEAR.IS THE RECTANGLE AND CIRCLE HAVE SAME CENTRE IS A VERY IMPORTANT POINT.ANY WAY ASSUMING THAT CENTRE IS SAME AND THE AXES ARE PARALLEL TO SIDES,I AM GIVING THE ANSWER.PLEASE CONFIRM THE ASSUMPTIONJS OR GIVE THE DRAWING.
SO AS PER OUR ASSUMPTION
AB || DC || Y AXIS
BC || AD || X AXIS
CENTRE OF RECTANGLE IS O(0,0) AS IS THE CENTRE OF CIRCLE.
HENCE DIAGONAL = AC = DIAMETER = 2*6=12 AND OA = RADIUS = 6
SINCE A IS (X,Y)
OA^2 = 6^2 = X^2+Y^2
Y=SQRT(36-X^2)
SIDES OF RECTANGLE ARE
AB = 2Y=2SQRT(36-X^2)
BC = 2X
AREA OF RECTANGLE = AB*BC = 2X*2SQRT(36-X^2)=4XSQRT(36-X^2)
B=4.....C=36 ...IN YOUR EQN.



Then express the area S of the rectangle ABCD as a function of x in the form: A(x) = b*x*sqr(c-x^2)
Where b and c are coefficients.
I have tried area=base*height which gives the formula area= x^2+Y^2=12^2. I just am not sure where to go from here or if I am even on the right track.

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Sorry, asked the wrong distance question before. I am trying to find distance for (-3, -2)=(X1,Y1)..and (1,4)=(X2,Y2). I get 10.
HOW DID YOU GET 10? FORMULA IS D = SQRT[(X2-X1)^2 + (Y2-Y1)^2]
D = SQRT[(1+3)^2 + (4+2)^2]=SQRT(16+36)=SQRT(52)
Thanks for the help

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Amanda has 400 feet of lumber to frame a
rectangular patio (the perimeter of a rectangle is 2
times length plus 2 times width). She wants to
maximize the area of her patio (area of a rectangle is
length times width). What should the dimensions of the
patio be, and show how the maximum area of the patio
is calculated from the algebraic equation.
Answer:
IF L AND B ARE DIMENSIONS WE HAVE
PERIMETER=2(L+B)=400.....OR.....L+B=200..OR......B=200-L.................I
AREA=A
=LB=L(200-L)=200L-L^2=-{L^2-200L}=-{(L^2)-2(L)(100)+100^2-100^2}
A=10000-(L-100)^2
(L-100)^2 BEING PERFECT SQUARE,ITS MINIMUM VALUE IS
ZERO.
HENCE AREA IS MAXIMUM WHEN L-100 IS ZERO,OR WHEN L=100
AND THEN THE MAXIMUM AREA WOULD BE
A-MAX.=10000-0=10000
DIMENSIONS ARE 100*100

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f(x) = x2 - kx + 9, where k is a constant
find the set of values of k for which the equation f(x) = 0 has no real solutions
I know that b2 - 4ac<0 has no real solutions
CORRECT SO FIND B^2-4AC...AND......MAKE IT < 0
B^2-4AC = K^2-4*1*9=K^2-36<0
K^2<36
|K|< 6..THAT IS K LIES BETWEEN -6 AND +6.....(-6,6)...

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4. If h(x) = x^3 - x, what is the relationship between h(-3) and -h(-3)?
h(x) is an odd function ...since
H(A)= A^3-A....AND
H(-A) = (-A)^3-(-A)= -A^3+A= -(A^3-A)= -H(A)
HENCE
H(3) = 3^3 - 3 =27-3 =24
H(-3) = (-3)^3-(-3)= -27+3 = -24
SO....H(3) = -H(-3)

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I am sorry but that is all he gave us. The question is come up with you rown example of a prime quadratic polynomial in one variable, then compute its discrimination to prove that it is prime. Choose coefficients so that your discriminant is positive. I have no clue on this at all.
DO YOU MEAN THE ROOTS SHALL BE PRIME NUMBERS?..OR THE QUADRATIC SHOULD GENERATE PRIMES?ASSUMING THE FORMER..TAKE 2 PRIME NUMBERS SAY 2 AND 3 AS ROOTS AND FRAME THE EQN.
THE EQN IS (X-2)(X-3)=0
X^2-5X+6=0
DISCRIMINANT = 5^2-4*1*6=25-24=1
HENCE THE ROOTS ARE
[5+ OR - SQRT(1)]/2=3 OR 2
I HOPE HIS IS WHAT HE WANTS.IF HE WANTS THE LATER ONE IT IS QUITE COMPLEX FOR YOUR COURSE WORK I FEEL

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Can someone help me solve this problem?
I need to write the equation of the line L satisfying the givengeometric conditions:
L has y-intercept (0, -3) and is parallel to the line with equation y= 2/3x + 1
EQN.OF PARALLEL LINE IS
Y=2X/3 +K.....IT PASSES THROUGH.(0,-3)...SO
-3=2*0/3+K=K=-3
HENCE EQN. OF LINE IS
Y=2X/3 - 3
Thanks, Ashley

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Can someone help me on this problem?
I need to write the equation of the line L satisfying the given geometric conditions
L has y-intercept (0, 2) and is perpendicular to the line with equation 2x - 3y = 6
PERPENDICULAR EN. IS
2Y+3X=K.....IT GOES THROUGH (0,2).SO
2*2+3*0=K=4
HENCE EQN. OF PERPENDICULAR IS
2Y+3X=4
Thanks, Ashley

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see the following example and try
---------------------------------------------------------------------
Function y = x2 - 4x - 5, perform the
following tasks:
a) Put the function in the form y = a(x - h)2 + k.
b) What is the line of symmetry?
c) Graph the function using the equation in part
a. Explain why it is not necessary to plot points to
graph when using y = a (x – h)2 + k.
Show graph here.
Explanation of graphing.
d) In your own words, describe how this graph compares
to the graph of y = x2?
1 solutions
Answer 13849 by venugopalramana(1619) About Me on
2006-01-28 11:20:53 (Show Source):
Y=X^2-4X-5={(X-2)^2-4-5}=(X-2)^2-9
COMPARING WITH THE GIVEN EQN .
y=a(x-h)^2+k,WE INFER THAT
A=1,H=2 AND K=-9....THE LINE OF SYMMETRY IS X-2=0 AS
YOU WILL GET SAME VALUE OF Y WHETHER X-2=+4 SAY OR
-4...NAMELY,Y=7.
hence using the property of symmetry , vertex coordinates ,we can plot the graph without plotting all points
COMPARISON WITH Y=X^2 IS SHOWN BELOW

YOU CAN SEE THAT LINE OF SYMMETRY IS X=0 HERE.
ALSO THE MINIMUM VALUE OR VERTEX AT 0,0 IN CASE OF
Y=X^2,WHERE AS IT WAS AT (2,-9)
FOR THE GIVEN EQUATION---------------------------------------

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the product of two consecutive positive integers is 288. Find the integers.
LET THE 2 NUMBERS BE N,N+1
N(N+1)=288
N^2+N-288=0
NOT POSSIBLE .
THE PROBLEM SHOULD BE PRODUCT OF 2 CONSECUTIVE 'EVEN'INTEGERS IS 288..THEN
ANY INTEGER IS N....ANY EVEN INTEGER IS 2N...NEXT EVEN INTEGER IS 2N+2
2N(2N+2)=288
4N(N+1)=288
N^2+N-72=0
N^2+9N-8N-72=0
N(N+9)-8(N+9)=0
(N-8)(N+9)=0
N=8
SO THE 2 EVEN INTEGERS ARE 2*8=16 AND 16+2=18

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ok this is the right problem now, please help to solve 3x-6y=9
6Y=3X-9
Y=(3X-9)/6......................1
x-2y=3
2Y=X-3
Y=(X-3)/2
X..................0.............3.................6..........ETC
Y=(3X-9)/6........-9/6...........0.................9/6....ETC
Y=(X-3)/2..........-3/2..........0.................3/2.....ETC...
SEE GRAPH BELOW

please solve by graphing, thank you in advance nightowl!
BOTH THE GRAPHS COINCIDE.HENCE THE EQNS.ARE DEPENDENT.HENCE THERE ARE INFINTE SOLUTIONS.ANY POINT ON THE GRAPH IS A SOLUTION.VIZ..(3,0),(0,-1.5),(6,1.5)...ETC......IN GENEARAL..{X,(X-3)/2}FOR ANY VALUE OF X IS A SOLUTION.

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A grain-storage warehouse has a total of 30 bins. Some hold 20 tons of grain each ( T SAY), and the rest hold 15 tons each( F SAY ). How many of each type of bin are there if the capacity of the warehouse is 510 bins?
T+F = 30..................1
CAPACITY OF TWENTY TON BINS = 20T
CAPACITY OF FIFTEEN TON BINS = 15F
TOTAL = 20T+15F = 510..OR.....4T+3F=102....................2
3*EQN1-EQN.2 GIVES
3T+3F-4T-3F=3*30-102=90-102=-12
-T=-12
T=12
F=30-12=18

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SEE THE FOLLOWING EXAMPLE AND TRY
-----------------------------------------
With a head wind, a plane traveled 1000 miles in 4 hours. With the same wind as a tail wind, the return trip took 3 hours 20 minutes. Find the plane's air speed( P SAY ) and the wind speed ( W SAY)
LET DISTANCE TRAVELLED = 1000 MILES
RELATIVE SPEED WHILE GOING HEAD WIND = P-W MPH
TIME NEEDED = T = D/R.S = 1000/(P-W) = 4
P-W=250.................1
R.S. WHILE GOING TAIL WIND = P+W
TIME NEEDED = 1000/(P+W)=3 HRS 20 MTS = 3+20/60 = 10/3 HRS
P+W =300.......................2
EQN.1+EQN.2
2P=550
P=275 MPH
W = 300-275 = 25 MPH

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With a head wind, a plane traveled 1000 miles in 4 hours. With the same wind as a tail wind, the return trip took 3 hours 20 minutes. Find the plane's air speed( P SAY ) and the wind speed ( W SAY)
LET DISTANCE TRAVELLED = 1000 MILES
RELATIVE SPEED WHILE GOING HEAD WIND = P-W MPH
TIME NEEDED = T = D/R.S = 1000/(P-W) = 4
P-W=250.................1
R.S. WHILE GOING TAIL WIND = P+W
TIME NEEDED = 1000/(P+W)=3 HRS 20 MTS = 3+20/60 = 10/3 HRS
P+W =300.......................2
EQN.1+EQN.2
2P=550
P=275 MPH
W = 300-275 = 25 MPH

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5a^4+7a^3+a^2 please factor
=a^2{5a^2+7a+1}...there are no more rational factors.

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what quadrants will this line intersect
y=-4x+1
its intercepts are
at x=0..........y=1
at y=0..........x=1/4
hence it will intersect I quadrant.see graph below
x........0.............1..............-1...........etc
y=-4x+1..1.............-3..............5.........etc

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Prove: A parallelogram is a rhombus if and only if its diagonal bisect the oppostie angles.
1..ABCD IS A PARALLELOGRAM(P.G.) AND AC IS ANGLE BISECTOR OF ANGLES A AND C.... AND BD IS ANGLE BISECTORS OF ANGLES B AND D
T.S.T.ABCD IS A RHOMBUS
ANGLE DAB = ANGLE DCB .OPPOSITE ANGLES IN A P.G.ARE EQUAL
AC BISECTS ANGLES DAB AND DCB...GIVEN
HENCE IN TRIANGLE ABC
ANGLE CAB = ANGLE DAB/2 = ANGLE DCB/2=ANGLE ACB
HENCE ABC IS ISOCELLES TRIANGLE
HENCE AB=BC .SIDES OPPOSITE EQUAL ANGLES
BUT AB=CD AND BC=AD....OPPOSITE SIDES OF P.G. ARE EQUAL
HENCE AB=BC=CD=DA
HENCE ABCD IS RHOMBUS
2.ABCD IS A RHOMBUS ..T.P.T DIAGONALS BISECT OPPOSITE ANGLES.
WRITE THE PROOF IN REVERSE MANNER TO THAT GIVEN ABOVE
IF IN DIFFICULTY PLEASE COME BACK

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Please help w/ the following:
M(6,-5) is the midpoint of RS. If S has coordinates (14,0), find the coordinates of R.
Thank you.
FORMULA FOR COORDINATES OF MID POINT (X,Y) BETWEEN (X1,Y1) AND (X2,Y2) IS
X=(X1+X2)/2.....Y=(Y1+Y2)/2.SO
6=(14+X2)/2..........-5=(0+Y2)/2
12=14+X2............-10=Y2
X2=-2............Y2=-10
R IS (-2,-10)

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1) A ball is dropped from a height of 18 ft. If on it rebound, it rises to a height 2/3 the distance from which it fell, how far ( up and and down ) will the ball have traveled when it hits the ground for the sixth times? (Give the approximately answer)
HT. DURING I DROP = H1 = 18'
HT DURING II DROP = H2 = (2/3)*H1=2*18/3=12
HT.DURING III DROP = H3 = (2/3)*H2=[(2/3)^2]H1
.........................................
..........................................
HT DURING N TH. DROP = HN = [(2/3)^(N-1)]H1
IT HIT GROUND 6 TIMES..HENCE WE WANT HT FOR THE 7 TH. DROP.
THAT IS FOR N=7 WE HAVE H7 = 18=1.58'


2) A pendulum is released and swings through an arc measuring 15 inche. Each swing thererater it travels a distance 0.85 times the length of the previous pass. Approximately how far will the pendulum swing on the eighth pass?
PROCEEDING IN THE SAME MANNER AS ABOVE,WE GET
SN=15

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15. Two cars with average speeds of 40 and 50 km/h respectively moved toward each other at 4:00 AM. If the square of the distance traveled by the second car is 250 times the distance traveled by the first car, at what time will they meet? how far has each traveled by then?
LET THEIR TIME OF TRAVEL FROM START TO TILL THEY MEET = T HRS
DISTANCE TRAVELLED BY I CAR = 40T
DISTANCE TRAVELLED BY SECOND CAR = 50T
SQUARE OF DISTANCE TRAVELLED BY II CAR = 50T*50T=2500T^2
250 TIMES DISTANCE TRAVELLED BY I CAR = 250*40T=10000T
HENCE
10000T = 2500T^2
4T=T^2
T^2-4T=0
T(T-4)=0
T=4 HRS.....THAT IS THEY MEET AT 4:00 AM+4 = 8:00 AM
BY THIS TIME
I CAR TAVELLED = 4*40 = 160 KM
II CAR TRAVELLED = 4*50 = 200 KM.
HENCE THEIR INTIAL DISTANCE OF SEPERATION = 160+200 = 360 KM.

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Please give me an example proof where you use the reflexive property(a=a) correctly.
EXAMPLE
N IS A SET OF ALL NATURAL NUMBERS
THEN THE RELATION 'X IS A DIVISOR Y' IS REFLEXIVE SINCE ANY AND EVERY NATURAL NUMBER IS A DIVISOR OF IT-SELF X IS A DIVISOR OF X OR 5 IS A DIVISOR OF 5 ETC....

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SEE THE FOLLOWING AND TRY
--------------------------------------------------------------
Ziggy's famous yogurt blends regular
yogurt that is 3% fat with its no fat yogurt to obtain
low fat yogurt that is 1% fat. How many pounds of
regular and how many pounds of non-fat yogurt should
be mixed to obtain 60 pounds of lowfat yogurt.
PLEASE HELP ASAP. thank you
THESE ARE MATERIAL BALANCE PROBLEMS.THE PRINCIPLE IS
TO APPLY
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS..
THIS PRINCIPLE CAN BE APPLIED TO TOTAL MIXTURE AS A
WHOLE AS WELL AS INDIVIDUAL COMPONENTS OF THE
MIXTURE.LET US SEE THE APPLICATION USING YOUR PROBLEM.
HERE THE MIXTURE COMPRISES 2 INPUTS-REGULAR YOGURT
(RY) & NO FAT YOGURT (NFY)
AND ONE OUT PUT-LOW FAT YOGURT (LFY).THE COMPONENT OF
IMPORTANCE IN THE MIXTURE IS FAT CONTENT.SO WE TAKE 2
BALANCES HERE ..ONE FOR THE TOTAL MIXTURE AND ANOTHER
FOR COMPONENT OF FAT IN THE MIXTURE.
I..TOTAL BALANCE...
INPUTS
1.QTY.OF.RY=X POUNDS
2.QTY OF NFY=Y POUNDS
OUT PUT
1.QTY.OF LFY=60 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
X+Y=60.............................I
II..COMPONENT BALANCE..HERE IT IS FAT .
INPUTS
1.QTY.OF FAT IN RY=X*3/100=3X/100 POUNDS
2.QTY OF FAT IN NFY=Y*0/100=0 POUNDS
OUT PUT
1.QTY.OF FAT IN LFY=60*1/100=60/100 POUNDS
SO APPLYING
TOTAL OF ALL INPUTS =TOTAL OF ALL OUTPUTS.....WE GET
3X/100 + 0=60/100.............................II
3X=60
X=20 POUNDS. OF REGULAR YOGURT
Y=60-20=40 POUNDS OF NO FAT YOGURT.

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write the slope-y-intercept equation form of the line which has slope 2/3 =M SAY and passes through the point:P = (X1,Y1)SAY (-3,3)
FORMULA IS
Y= MX -MX1+Y1
Y= 2X/3 -(2/3)(-3)+3=2X/3 +5
write the point-slope equation form of the line which passes through 2 points: P= (X1,Y1) SAY=(-9,0), Q = (X2,Y2) SAY= (-5,3)
FORMULA IS
Y = (Y2-Y1)(X-X1)/(X2-X1) +Y1=(3-0)(X+9)/(-5-0) + 0
Y= -3X/5 -27/5

write the general equation form of the line which passes through the 2 points:(1,2), (6,7)
SAME AS ABOVE

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The length =L SAY of a rectangle is 4 cm more than the width =W SAY. The area is 45 cm ^ 2. Find the length and width.
L=W+4
AREA = LW = (W+4)W =45
W^2+4W-45=0
W^2+9W-5W-45=0
W(2+9)-5(W+9)=0
(W-5)(W+9)=0
W-5=0...OR...W=5
HENCE L=4+5=9

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Solve the system by graphing, addition and substitution methods.
4x – 4y = –8 .....1
y = 2 + x ................2
PUTTING 2 IN EQN.1
4X-4(2+X)=-8
4X-8-4X=-8
0=0
HENCE THE EQNS. ARE DEPENDENT/ CONSISTENT.SO THERE WILL BE INFINITE SOLUTIONS..SAY
X=0... Y=2...OR
X=1..Y=3..OR
X=2..Y=4..ETC..IN FACT
X,X+2.WITH ANY VALUE FOR X IS A SOLUTION SET.
X....................0.................1......................2...ETC
Y=2+X................2..................3......................4...ETC
Y=(4X+8)/4...........2..................3.....................4....ETC

THE 2 GRAPHS ARE IDENTICAL OR COINCIDENT.SO INFINITE SOLUTIONS ,WITH ANY POINT ON THE GRAPH BEING A SOLUTION.

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I am stuck on correct answer for:
if a triangle had these degrees of measurement what would be the degree measure for B?
A = left side of trangle is degree measure of 30
C = top of triangle is degree measure of 100
B = ?
What is correct resolve to this?
SUM OF 3 ANGLES IN A TRIANGLE = 180
HERE SUM OF 2 ANGLES = 30+100=130
HENCE THE THIRD ANGLE = B = 180-130 = 50

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I am stuck on correct answer for:
on a right triangle, what is length of AC?
A = top of triangle
B = bottom left corner
C= bottom right corner
Between line between B and C is the length of 6
Between line between A and B is the length of 10
THERE IS STILL ONE THING AMISS...YOU HAVE NOT SAID WHICH ANGLE IS RIGHT ANGLE!!ASSUMING IT TO BE ANGLE C,WE HAVE BY PYTHOGARUS THEOREM
HYPOTENUSE ^2= AB^2=SUM OF SQUARES OF OTHER 2 SIDES FORMING THE RIGHT ANGLE = AC^2+BC^2
10^2=AC^2+6^2
AC^2=100-36=64
AC=8

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Travis made 44 and 72 on the first two tests in algebra and has one test remaining. The average on the three tests must be at least 60 for Travis to pass the course. Let s represent his score on the last test.
Write an inequality to describe each situation.
total score in 3 tests = 44+72+s = 116+s
average = (116+s)/3...this should be atleast 60..so
(116+s)/3 >=60
116+s>=180
s>=180-116=64
s>=64

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Travis made 44 and 72 on the first two tests in algebra and has one test remaining. The average on the three tests must be at least 60 for Travis to pass the course. Let s represent his score on the last test.
Write an inequality to describe each situation.
total score in 3 tests = 44+72+s = 116+s
average = (116+s)/3...this should be atleast 60..so
(116+s)/3 >=60
116+s>=180
s>=180-116=64
s>=64