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# Recent problems solved by 'venugopalramana'

venugopalramana answered: 3288 problems
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 Angles/55992: This question is from textbook Algebra 1 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.1 solutions Answer 38043 by venugopalramana(3286)   on 2006-10-13 11:15:32 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/56077: 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.) 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. 1 solutions Answer 38041 by venugopalramana(3286)   on 2006-10-13 11:04:32 (Show Source): You can put this solution on YOUR website!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.
 Length-and-distance/56030: Sorry, asked the wrong distance question before. I am trying to find distance for (-3, -2)and (1,4). I get 10. Thanks for the help.1 solutions Answer 38040 by venugopalramana(3286)   on 2006-10-13 10:49:44 (Show Source): You can put this solution on YOUR website!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
 Graphs/56047: 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. Use the vertex form to find the maximum area. Thank you so much 1 solutions Answer 38026 by venugopalramana(3286)   on 2006-10-13 09:21:53 (Show Source): You can put this solution on YOUR website!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
 Functions/56052: 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 solutions1 solutions Answer 38007 by venugopalramana(3286)   on 2006-10-13 07:13:10 (Show Source): You can put this solution on YOUR website!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)...
 Linear_Equations_And_Systems_Word_Problems/56039: This question is from textbook Applied College Algebra 4. If h(x) = x^3 - x, what is the relationship between h(-3) and -h(-3)?1 solutions Answer 38006 by venugopalramana(3286)   on 2006-10-13 06:24:53 (Show Source): You can put this solution on YOUR website!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)
 Polynomials-and-rational-expressions/55911: This question is from textbook Elementary and Intermediate Algebra 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.1 solutions Answer 37931 by venugopalramana(3286)   on 2006-10-12 12:22:07 (Show Source): You can put this solution on YOUR website!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
 Geometric_formulas/55914: This question is from textbook Beginning Algebra 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 Thanks, Ashley1 solutions Answer 37929 by venugopalramana(3286)   on 2006-10-12 12:15:05 (Show Source): You can put this solution on YOUR website!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
 Geometric_formulas/55915: This question is from textbook Beginning Algebra 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 Thanks, Ashley1 solutions Answer 37928 by venugopalramana(3286)   on 2006-10-12 12:12:44 (Show Source): You can put this solution on YOUR website!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
 Graphs/55873: For the function y = x2 - 6x + 8, perform the following tasks: a) Put the function in the form y = a(x - h)2 + k. Answer: Show work in this space. b) What is the equation for the line of symmetry for the graph of this function? Answer: 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? Answer: Thank you tine 1 solutions Answer 37927 by venugopalramana(3286)   on 2006-10-12 12:09:53 (Show Source): You can put this solution on YOUR website!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---------------------------------------
 real-numbers/55896: the product of two consecutive positive integers is 288. Find the integers.1 solutions Answer 37900 by venugopalramana(3286)   on 2006-10-12 08:57:53 (Show Source): You can put this solution on YOUR website!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
 Graphs/55844: ok this is the right problem now, please help to solve 3x-6y=9 x-2y=3 please solve by graphing, thank you in advance nightowl!1 solutions Answer 37899 by venugopalramana(3286)   on 2006-10-12 08:49:32 (Show Source): You can put this solution on YOUR website!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.
 Miscellaneous_Word_Problems/55633: A grain-storage warehouse has a total of 30 bins. Some hold 20 tons of grain each, and the rest hold 15 tons each. How many of each type of bin are there if the capacity of the warehouse is 510 bins?1 solutions Answer 37754 by venugopalramana(3286)   on 2006-10-11 06:40:36 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/55634: with a tail wind, a helicopter traveled 300 miles in 1 hour 20 minutes. The return trip against the same wind took 20 minutes longer. Find the plane's air speed and the speed of the helicopter.1 solutions Answer 37753 by venugopalramana(3286)   on 2006-10-11 06:36:01 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/55635: 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 and the wind speed.1 solutions Answer 37752 by venugopalramana(3286)   on 2006-10-11 06:34:59 (Show Source): You can put this solution on YOUR website!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
 Numeric_Fractions/55644: 5a^4+7a^3+a^2 please factor1 solutions Answer 37751 by venugopalramana(3286)   on 2006-10-11 06:26:44 (Show Source): You can put this solution on YOUR website!5a^4+7a^3+a^2 please factor =a^2{5a^2+7a+1}...there are no more rational factors.
 Linear-systems/55625: what quadrants will this line intersect y=-4x+11 solutions Answer 37750 by venugopalramana(3286)   on 2006-10-11 06:23:47 (Show Source): You can put this solution on YOUR website!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
 Proofs/55418: Prove: A parallelogram is a rhombus if and only if its diagonal bisect the oppostie angles.1 solutions Answer 37641 by venugopalramana(3286)   on 2006-10-10 12:35:02 (Show Source): You can put this solution on YOUR website!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
 Linear-equations/55449: 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.1 solutions Answer 37639 by venugopalramana(3286)   on 2006-10-10 12:23:01 (Show Source): You can put this solution on YOUR website!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)
 Sequences-and-series/55465: Good evening, I have the problems about the GP and Ap. Please help me in..... 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) 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? 3) A readioactive dye is injected into a system in a medical test. After one hour, 60% of the dye remains. At the end of two hours 60% of the remaining dye remains, and so on. If one unit of dye is injected, approximately what percent will remain after 12 hours?1 solutions Answer 37634 by venugopalramana(3286)   on 2006-10-10 11:57:03 (Show Source): You can put this solution on YOUR website!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
 Rational-functions/55443: Hi there. I submitted some problems last Friday, but havent heard anything. If someone has time and is willing to lend soem brain power, I would really appreciate the help. 1. Consider the function f(x)=2.0x^3 + 4.5x^2. What are the coordinate pairs of the relative maximum and the relative minimum of f(x) in the interval {xI-3 2.In what interval or intervals is the function in problem #1 increasing or decreasing? Use set notation as in question # to indicate the answer. 3. What is the greatest integer less than -10.5? 4. Given that f(x)=x^2+6 and g(x)=x-4, find the composite function f(g(x)). Thank you for your time. Angela1 solutions Answer 37633 by venugopalramana(3286)   on 2006-10-10 11:41:22 (Show Source): You can put this solution on YOUR website!Hi there. I submitted some problems last Friday, but havent heard anything. If someone has time and is willing to lend soem brain power, I would really appreciate the help. 1. Consider the function f(x)=2.0x^3 + 4.5x^2. What are the coordinate pairs of the relative maximum and the relative minimum of f(x) in the interval {xI-3 F(X)=X^2(2X+4.5).....THE ZEROS ARE X=0...AND 2X+4.5=0...OR..X=-4.5/2=-2.25 IN INCREASING ORDER THEY ARE -2.25 AND 0... HENCE F(X) IS + VE AND INCREASING FOR X>0 F(X)=0 AT X=0 F(X) IS -VE IN THE INTERVAL (-2.25,0) F(X)=0 AT X=-2.25 F(X) IS POSITIVE FOR X<-2.25 F'(X)= 6X^2+9X =0......3X(2X+3)=0......2 EXTREMA ARE AT X=0 AND X=-3/2 SEE GRAPH BELOW..F(X) HAS A LOCAL MAXIMUM FOR X IN (-2,0)... MAXIMUM POINT IS (-3/2 ,27/8) AND A LOCAL MINIMUM FOR X IN (-1,1)...MINIMUM POINT IS (0,0) 2.In what interval or intervals is the function in problem #1 increasing or decreasing? Use set notation as in question # to indicate the answer. ALREADY INDICATED ABOVE 3. What is the greatest integer less than -10.5? PUT ON NUMBER LINE -11........-10.5.........-10 HENCE -11 IS THE GREATEST INTEGER LESS THAN -10.5 4. Given that f(x)=x^2+6 and g(x)=x-4, find the composite function f(g(x)). PUT G(X) =Y =X-4 F(G(X)) = F(Y)= Y^2+6 = (X-4)^2+6=X^2+16-8X+6= X^2-8X+22 Thank you for your time. Angela
 Travel_Word_Problems/55445: Dearest tutors.,, i hope you can dtill remember last time when i sent you my problem, i told you that my 'teacher' would made me squat on the floor if i cant get the right answer?.. i hope you still remember that one... well, as for now, she might do that again.. by the way thanks for solving my previous preoblem., pls help me with this other one... 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?1 solutions Answer 37616 by venugopalramana(3286)   on 2006-10-10 08:02:42 (Show Source): You can put this solution on YOUR website!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.
 Geometry_proofs/55262: Please give me an example proof where you use the reflexive property(a=a) correctly.1 solutions Answer 37512 by venugopalramana(3286)   on 2006-10-09 12:07:45 (Show Source): You can put this solution on YOUR website!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....
 Mixture_Word_Problems/55259: What quantity of 60% acid solution must be mixed with 30% solution to produce 300mL of a 50% solution? 1 solutions Answer 37510 by venugopalramana(3286)   on 2006-10-09 12:02:25 (Show Source): You can put this solution on YOUR website!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.
 Linear-systems/55257: write the slope-y-intercept equation form of the line which has slope 2/3 and passes through the point: (-3,3) write the point-slope equation form of the line which passes through 2 points: (-9,0), (-5,3) write the general equation form of the line which passes through the 2 points:(1,2), (6,7) 1 solutions Answer 37509 by venugopalramana(3286)   on 2006-10-09 11:59:25 (Show Source): You can put this solution on YOUR website!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
 Quadratic_Equations/55246: The length of a rectangle is 4 cm more than the width. The area is 45 cm ^ 2. Find the length and width.1 solutions Answer 37504 by venugopalramana(3286)   on 2006-10-09 11:27:25 (Show Source): You can put this solution on YOUR website!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
 Graphs/55214: Solve the system by graphing, addition and substitution methods. 4x – 4y = –8 y = 2 + x 1 solutions Answer 37486 by venugopalramana(3286)   on 2006-10-09 06:03:10 (Show Source): You can put this solution on YOUR website!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.
 Coordinate-system/55220: 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?1 solutions Answer 37485 by venugopalramana(3286)   on 2006-10-09 05:54:16 (Show Source): You can put this solution on YOUR website!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
 Coordinate-system/55221: 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 What is correct resolve to this?1 solutions Answer 37484 by venugopalramana(3286)   on 2006-10-09 05:52:06 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/55226: Tilak made 44 and 72 on the first two tests in algebra, and has one test remaining. For Tilak to pass the course, the average on the three tests must be at least 60. For what range of scores on his last test will Tilak pass the course? Solve the problem by using an inequality.1 solutions Answer 37483 by venugopalramana(3286)   on 2006-10-09 05:45:26 (Show Source): You can put this solution on YOUR website!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
 Miscellaneous_Word_Problems/55225: 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.1 solutions Answer 37482 by venugopalramana(3286)   on 2006-10-09 05:42:00 (Show Source): You can put this solution on YOUR website!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