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# Recent problems solved by 'venugopalramana'

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 Functions/47565: For the equation x-2sqrt.x=0 , perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. b) Graph the functions y = x and x-2sqrt.x=0 on the same graph (by plotting points if necessary). Show the points of intersection of these two graphs. Graph: Points of intersection: 1 solutions Answer 31406 by venugopalramana(3286)   on 2006-07-28 23:32:52 (Show Source): You can put this solution on YOUR website!For the equation x-2sqrt.x=0 , perform the following: a) Solve for all values of x that satisfies the equation. Answer: Show work in this space. x-2sqrt(x)=0 sqrt(x){sqrt(x)-2}=0 sqrt(x)=0.....or....sqrt(x)=2 hence x=0 or .....x=4 b) Graph the functions y = x and x-2sqrt.x=0...I DONT THINK THIS IS CORRECT.IT IS PROBABLY Y-2SQRT(X)=0... on the same graph (by plotting points if necessary). Show the points of intersection of these two graphs. Graph: x...................0..............1............4...............9..........etc... y=x.................0..............1............4...............9..........etc y=2sqrt(x)..........0..............2............4...............6..........etc Points of intersection: x=0,x=4............y=0...........y=4
 Square-cubic-other-roots/47524: simplify by combining like terms sqrt 63 - 2 sqrt 28 + 5 sqrt 71 solutions Answer 31372 by venugopalramana(3286)   on 2006-07-28 12:14:07 (Show Source): You can put this solution on YOUR website!simplify by combining like terms sqrt 63 - 2 sqrt 28 + 5 sqrt 7=sqrt(9*7)-2sqrt(4*7)+5sqrt(7) =sqrt(9)sqrt(7)-2sqrt(4)sqrt(7)+5sqrt(7)=3sqrt(7)-2*2sqrt(7)+5sqrt(7) =sqrt(7){3-4+5}=4sqrt(7)
 Probability-and-statistics/47507: Find the odds against obtaining two (2) heads in a single toss of two coins. Is it 1/21 solutions Answer 31358 by venugopalramana(3286)   on 2006-07-28 09:37:14 (Show Source): You can put this solution on YOUR website!Find the odds against obtaining two (2) heads in a single toss of two coins. Is it 1/2...NO... THERE ARE 2 POSSIBILITIES IN EACH TOSS ...PROBABILITY OF HEAD IN ONE TOSS=1/2 SAME IS CASE IN SECOND TOSS =1/2 HENCE PROBABILITY OF GETTING 2 HEADS IN 2 TOSSES =(1/2)*(1/2)=1/4 PROBABILITY AGAINST 2 HEADS =1-1/4=3/4 ODDS AGAINST 2 HEADS ARE 3/4:1/4=3:1
 Mixture_Word_Problems/47508: Hydrochloric acid and distilled water were mixed to produce 210mL of solution. If 120mL more H2O was used than HCI, how many mL of each were mixed? tnx!.. pls answer as soon as possible!.. tnx!1 solutions Answer 31355 by venugopalramana(3286)   on 2006-07-28 09:31:28 (Show Source): You can put this solution on YOUR website!Hydrochloric acid and distilled water were mixed to produce 210mL of solution. If 120mL more H2O was used than HCI, how many mL of each were mixed? tnx!.. pls answer as soon as possible!.. tnx! let hcl used =x ml. 120 more =x+120 ml. total mix =x+x+120=2x+120=210 2x=210-120=90 x=90/2=45 ml. hcl= 45 ml water = 45+120=165 ml.
 Probability-and-statistics/47461: 40 students enrolled in three areas of study. 23 students in music 25 students in math 27 students in comp. science 7 took mathematics and music only 5 took music and computer science only 6 took math and comp. sci. only 7 took all three subjects What is the probability a student took exactly one of the three courses?1 solutions Answer 31333 by venugopalramana(3286)   on 2006-07-27 12:01:07 (Show Source): You can put this solution on YOUR website! 40 students enrolled in three areas of study. LET US USE A ^ B TO SHOW A INTERSECTION B 23 students in music(S..SAY) 25 students in math(M..SAY) 27 students in comp. science(C...SAY) ------------------------------------------------------------------------------ 7 took mathematics and music only=M&S ONLY 5 took music and computer science only=S&C ONLY 6 took math and comp. sci. only=M&C ONLY 7 took all three subjects=M&C&S=7 =================================================== HENCE THOSE WHO TOOK MATHS ONLY = M-M&S ONLY-M&C ONLY - M&S&C=25-7-6-7=5 THOSE WHO TOOK MUSIC ONLY = S-S&M ONLY-S&C ONLY-S&C&M=23-7-5-7=4 THOSE WHO TOOK COMPUTRERS ONLY =C-C&M ONLY-C&S ONLY-C&S&M=27-6-5-7=9 HENCE PROBABILITY OF HAVING THOSE WHO TOOK ONLY ONE SUBJECT=(5+4+9)/40=18/40=9/20 What is the probability a student took exactly one of the three courses? You may edit the question. Maybe convert formulae to the same formula notatio
 Money_Word_Problems/47455: A candy store sells malted milk balls for \$2.50 per pound, and chocolate-covered raisins for \$3.75 per pound. How many pounds of each should be used to make a 10 pound mixture that sells for \$3.00 per pound?1 solutions Answer 31331 by venugopalramana(3286)   on 2006-07-27 11:26:56 (Show Source): You can put this solution on YOUR website!A candy store sells malted milk balls for \$2.50 per pound, and chocolate-covered raisins for \$3.75 per pound. How many pounds of each should be used to make a 10 pound mixture that sells for \$3.00 per pound? LET MILK BALLS USED BE M LB. RAISINS USED =10-M LB. COST OF MILK BALLS = M*2.5 COST OF RAISINS = (10-M)*3.75=37.5-3.75M TOTAL COST =2.5M+37.5-3.75M=10*3=30 3.75M-2.5M=37.5-30=7.5 1.25M=7.5 M=7.5/1.25=6 MILK BALLS USED =6 LB. RAISINS USED =10-6=4 LB.
 Equations/46980: John has 24 coins in his pocket-nickels, dimes and quarters-amounting to \$3.80. If he has twice as many nickels as quarters, how many dimes does he have?1 solutions Answer 31157 by venugopalramana(3286)   on 2006-07-23 12:41:04 (Show Source): You can put this solution on YOUR website!John has 24 coins in his pocket-nickels=N SAY, dimes=D SAY and quarters=Q SAY-amounting to \$3.80. If he has twice as many nickels as quarters, how many dimes does he have? N+D+Q=24............I N+5D+25Q=380...........II EQN.II-EQN.I... 4D+24Q=380-24=356.DIVIDING WITH 4... D+6Q=89......................III THERE ARE 2 UNKNOWNS AND ONLY ONE EQN. BUT THERE IS A PHYSICL FACT THAT N,D,Q MUST BE POSITIVE INTEGERS..HENCE TRY IN EQN.III...D=1,2,3 ETC,TO GET INTEGRAL ANSWERS D=1..1+6Q=89...........6Q=88...BUT THIS WILL MAKE Q FRACTION.HENCE NOT POSSIBLE SIMILARLY.....D=2,3,4 ARE NOT POSSIBLE D=5.....5+6Q=89....6Q=84....Q=84/6=14...SO D=5,Q=14 AND N=24-14-5=5 FURTHER TRIALS FOR D=11....WILL LEAD TO 11+6Q=89....6Q=78...Q=13...THEN N=24-13-11=0..BUT WE ARE GIVEN THAT THERE ARE SOME NICKELS..HENCE ONLY SOLUTION IS N=5=5 CENTS D=5=5*5=25 CENTS Q=14=14*25=350 CENTS ------------------------ TOTAL=24 COINS WITH VALUE OF 380 CENTS=\$3.80
 Equations/46982: Solve the equation for the indicated variable. (Leave ± in your answer, when appropriate.) 1 solutions Answer 31156 by venugopalramana(3286)   on 2006-07-23 12:28:19 (Show Source): You can put this solution on YOUR website!Solve the equation for the indicated variable. (Leave ± in your answer, when appropriate.) rm=t^2-mt for t T^2-MT-RM=0.......COMPRING WITH STD.EQN....AX^2+BX+C=0....A=1...B=-M....C=-RM
 Sequences-and-series/46984: The sequence is defined recursively. Write the first four terms. I really need some help with this problem please. 1 solutions Answer 31155 by venugopalramana(3286)   on 2006-07-23 12:22:46 (Show Source): You can put this solution on YOUR website!The sequence is defined recursively. Write the first four terms. a[1]=2 a[2]=5 and a[n]=a[n-2] - 3a[n-1] for n >= 3 PUT N=3 A3=A1-3A2=2-3*5=2-15=-13 PUT N=4 A4=A2-3A3=5-3*(-13)=5+39=44 I really need some help with this problem please.
 Systems-of-equations/46992: This question is from textbook CLEP Official Study Guide 1. What is the determinant of the matrix [-1 2][1 3]. The [1 3] is directly under the [-1 2] in the problem. I have no idea where to begin to solve this problem. 2. The below table gives some of the values of a 5th degree polynominal p(x). Based on the values shown, what it the minimum number of real roots fot the equation p(x)=0 x 0 1 2 3 4 5 6 7 p(x) -30 22 110 150 34 -130 222 2,350 I have no idea how to begin to solve this problem. 1 solutions Answer 31154 by venugopalramana(3286)   on 2006-07-23 12:19:23 (Show Source): You can put this solution on YOUR website!1. What is the determinant of the matrix [-1 2][1 3]. The [1 3] is directly under the [-1 2] in the problem. I have no idea where to begin to solve this problem. |-1,2|=(-1)(3)-(2)(1)=-3-2=-5 | 1,3| IN GENERAL |A,B| |C,D|=A*D-B*C 2. The below table gives some of the values of a 5th degree polynominal p(x). Based on the values shown, what it the minimum number of real roots fot the equation p(x)=0 x 0 1 2 3 4 5 6 7 p(x) -30 22 110 150 34 -130 222 2,350 I have no idea how to begin to solve this problem. REAL ROOTS ARE THOSE VALUES OF X AT WHICH P(X) BECOMES ZERO. WE FIND THAT AT X=0.....P(X)=-30..AND AT X=1...P(X)=+22.....SO P(X) WOULD BE ZERO BETWEN X=0 AND X=1..THIS IS ONE REAL ROOT SIMILARLY...AT X=4.....P(X)=34 AT X=5....P(X)=-130.....SO P(X) WOULD BE ZERO BETWEEN X=4 AND X=5...THI9S IS ANOTHER REAL ROOT SIMILARLY...AT X=5.....P(X)=-130 AT X=6............P(X)=222..HENCE P(X)WOULD BE ZERO BETWEEN X=5 AND X=6...THIS IS ANOTHER REAL ROOT. HENCE THERE ARE 3 REAL ROOTS IN ALL
 Distributive-associative-commutative-properties/47009: geometry problems: I'm uncertain as how to set these up. the length of a rectangle is two feet more than twice the width. the area is 144ft^2. find the length and the width of the rectangle.1 solutions Answer 31152 by venugopalramana(3286)   on 2006-07-23 12:09:53 (Show Source): You can put this solution on YOUR website!the length=L SAY of a rectangle is two feet more than twice the width=W SAY. the area is 144ft^2. find the length and the width of the rectangle. TWICE WIDTH =2W 2 FEET MORE =2W+2=2(W+1) L=2W+2 AREA=A=L*W=2(W+1)W=144 W(W+1)=144/2=72 W^2+W-72=0 W^2+9W-8W-72=0 W(W+9)-8(W+9)=0 (W-8)(W+9)=0 W=8 L=2(8+1)=18
 Exponential-and-logarithmic-functions/47010: use properties of logarithms to condense the logarithmic expression log3 54-log3 21 solutions Answer 31151 by venugopalramana(3286)   on 2006-07-23 12:05:37 (Show Source): You can put this solution on YOUR website!use properties of logarithms to condense the logarithmic expression X=log3 54-log3 2 SINCE BOTH ARE TO BASE 3 WE CAN OMIT THE BASE LOG(54)-LOG(2)=LOG(54/2)=LOG(27) TO BASE 3 LET IT BE =X SO X^3=27=3^3 X=3
 Rational-functions/47011: use the given root to find the solution set of the polynomial equation x4-45x2-196=0 : 2i1 solutions Answer 31150 by venugopalramana(3286)   on 2006-07-23 12:03:07 (Show Source): You can put this solution on YOUR website!use the given root to find the solution set of the polynomial equation F=x4-45x2-196=0 : 2i.SINCE THE EQN.HAS REAL COEFFICIENTS ,IF 2I IS A ROOT ,ITS CONJUGATE -2I IS ALSO A ROOT.SO LET G=(X+2I)(X-2I)=X^2-4I^2=X^2+4 IS A FACTOR.DIVIDING F WITH G X^2+4)X^4-45X^2-196(X^2-49 ......X^4+4X^2 ------------------------- ..........-49X^2-196 ..........-49X^2-196 ------------------------------ ..............0................ SO QUOTIENT IS X^2-49=(X)^2-7^2=(X+7)(X-7) HENCE F=(X+2i)(X-2I)(X+7)(X-7) SOLUTION SET IS 2I,-2I,7,-7
 Distributive-associative-commutative-properties/47013: physics: an object is released from a plane at an altitude of 1600 ft. the initial velocity is 0ft/s. how many seconds will the object hit the ground? d=vt+16t^2 is the formula but I'm just not seeing where this is supposed to all come together. my guess is that it will fall very darn fast.1 solutions Answer 31149 by venugopalramana(3286)   on 2006-07-23 11:55:59 (Show Source): You can put this solution on YOUR website!an object is released from a plane at an altitude of 1600 ft. the initial velocity is 0ft/s. how many seconds will the object hit the ground? d=vt+16t^2 is the formula but I'm just not seeing where this is supposed to all come together. my guess is that it will fall very darn fast. D=1600...V=0 1600=0*T+16*T^2 T^2=1600/16=100 T=10 SEC.
 Exponents-negative-and-fractional/47015: 2x^3-3x^2+5x when x=-31 solutions Answer 31148 by venugopalramana(3286)   on 2006-07-23 11:54:07 (Show Source): You can put this solution on YOUR website!F=2x^3-3x^2+5x when x=-3 =2*(-3)^3-3*(-3)^2+5*(-3)=-54-27-15=-96
 expressions/47016: 3ab^2+5a^2b-1 when a=2 and b=-21 solutions Answer 31147 by venugopalramana(3286)   on 2006-07-23 11:51:51 (Show Source): You can put this solution on YOUR website!F=3ab^2+5a^2b-1 when a=2 and b=-2 JUST SUBSTITUTE THE GIVEN VALUES FOR A AND B AND COMPUTE THE ANSWER AS SHOWN BELOW F=3*2*(-2)^2+5(2^2)*(-2)-1=6*4+5*4*(-2)-1=24-40-1=-17