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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

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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)

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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

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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.

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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

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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.

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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

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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

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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.

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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

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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

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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

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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

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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.

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F=2x^3-3x^2+5x when x=-3
=2*(-3)^3-3*(-3)^2+5*(-3)=-54-27-15=-96

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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

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Please help! Find a 5th degree polynomial function with real, rational coefficients and zeros x=0, x=the square root of 3, and x=2-i.
MULTIPLY WITH ALL FACTORS OBTAINED FROM THE GIVEN ROOTS
X=0...X-0=X IS A FACTOR
X=SQRT(3)....X-SQRT(3) IS A FACTOR..NOW SINCE COEFFICIENTS ARE RATIONAL
X=-SQRT(3) SHOULD BE A ROOT....X+SQRT(3) IS A FACTOR
X=2-I........X-2+I IS A FACTOR...SINCE COEFFICIENTS ARE REAL
X=2+I SHOULD BE A ROOT ...X-2-I IS A FACTOR
SO MULTIPLYING F(X)=X{X-SQRT(3)}{X+SQRT(3)}{(X-2)+I}{(X-2)-I}=0
=X(X^2-3)({(X-2)^2-I^2}=0
X(X^2-3)(X^2-4X+5)=0...MULTIPLY TO GET THE ANSWER
I've tried and I just can't figure it out please help. It would be greatly appreciated.

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Can someone please help me with the following: What are all the possible rational roots (using synthetic division to find an actual root, then using this root to solve the equation) for x^3 + 5x^2 + 12x - 18 = 0 ?!?!?
TRY FACTORS OF CONSTANT=-18....PLUS OR MINUS 1,2,3,6,9,18 FOR X ..
HERE X=1 GIVES
1+5+12-18-0...OK...SO X=1 IS A ROOT SO X-1 IS A FACTOR.NOW DIVIDE
1|...1......5......12........-18
.....0......1.......6.........18
----------------------------------------
.....1......6.......18.........0
X^2+6X+18 IS QUOTIENT..NOW FACTORISE OR USE QUADRATIC FORMULA.
HERE B^2-4AC= 36-4*18*1=-VE...SO NO ROOTS
HENCE THERE IS ONLY 1 REAL ROOT = X=1




Can you please explain the steps involved as well as the solution (as I am fairly lost)?

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9. Factor completely. 5a2 – 125
=5(A^2-25)=5(A^2-5^2)=5(A+5)(A-5).............D IS OK
A) 5(a – 5)2 B) 5a(a – 25) C) (5a – 1)(a – 125) D) 5(a + 5)(a – 5)

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Solve by using the quadratic formula. x2 – 8x – 9 = 0
AX^2+BX+C=0 IS THE STD.EQN..A=1,B=-8,C=-9




X=9 OR -1

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Can someone please help me with this?
The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients.
since coefficients are real if 3+i is root ,then its conjugate 3-i is also a root.hence
f(x)=k(x-4)(x-3-i)(x-3+i) =0
since coefficient of x^3 is 1,k=1
f(x)=(x-4){(x-3)^2-i^2}=(x-4)(x^2-6x+9+1)=0
=x^3-6x^2+10x-4x^2+24x-40=0
=x^3-10x^2+34x-40=0
I have come up with two different answers and I am really confused:
Is one of these answers correct?
f(x)=(x+4)(x^2+6x+10)...NO..4 IS ROOT MEANS X=4...OR...X-4=0
SIMILARLY.....3+I IS A ROOT MEANS X=3+I...OR...X-3-I=0
or f(x) = (x-4)(x^2-6x+10)..THIS IS CORRECT
Thank you
V/r
Doug

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FOR THE SECOND PART YOU HAVE TO SEE WHICH LINE IS STEEPEST AND INCREASING.YOU CAN SEE THAT ALL DIVISIONS ARE FOR 1 YEAR.SO WHEREVER THERE IS MAXIMUM INCREASE THAT IS THE HIGHEST RAISE.
DURING 1990-91,91-2 AND 94-95 IT IS DECREASING..SO LEAVE THEM
92-93 IS STEEPER LINE THAN 93-94
SO 92-93 IS THE GREATEST INCREASE

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-7=-SQUARE ROOT OF (49)

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PROBLEM 7 i got wrong Factor completely. b2 – ab – 6a2
=B^2-3AB+2AB-6A^2
=B(B-3A)+2A(B-3A)
=(B-3A)(B+2A)

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How can I find the dimensions of a rectangular box with a width=W three times the height=H, the length=L is 6 inches more than the height, and the volume=V is 15 cubic inches?
W=3*H
L=H+6
V=15=LWH=(H+6)3*H*H=3H^3+18H^2
H^3+6H^2-5=0
H+1 IS A FACTOR
-1|1....6.....0.....-5
...0...-1....-5......5
--------------------------------
...1....5.....-5....0
H^2+5H-5=0
H={-5+SQRT(25-20)}/2
=-VE. CHECK THE PROBLEM.....DATA GIVEN DOES NOT LEAD TO A FEASIBLE SOLUTION

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write standard form of equation where slope of passes through points.
m=2, p=(X1,Y1)=(1,3)
EQN.IS
Y-Y1=M(X-X1)
Y-3=2(X-1)=2X-2
Y=2X+1

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USE LN(X*Y)=LN(X)+LN(Y)
LN(/Y)=LN(X)-LN(Y)
LN(X^P)=P*LN(X)
Express as a single natural log:
21nx - 51ny + 31nz
=LN(X^2)-LN(Y^5)+LN(Z^3)
=LN{(X^2)*(Z^3)/(Y^5)}

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The solution to the system of equations below is
3x+4y= -6
x-5y=17
My answer is (2,-3) is this right, thanks for checking
SUBSTITUTE AND CHECK
3*2+4*(-3)=6-12=-6...OK
2-5*(-3)=2+15=17..OK

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The system below has how many solutions.
2x-y+4z=-5..............I
x+2y-z=6...............II
x+y+z=1..................III
EQN.II-EQN.III
Y-2Z=5......................IV
2*EQN.III-EQN.I
3Y-2Z=7........................V
EQN.V-EQN.IV
2Y=2
Y=1
1-2Z=5....OR.....2Z=1-5=-4
Z=-2
X+1-2=1
X=2
IT HAS UNIQUE SOLUTION WITH X=2,Y=1,Z=-2
IF THE EQNS.ARE DEPENDENT AND CONSISTENT YOU WILL GET INFINITE SOLUTION.

My answer is infinite, but I'm not sure. thanks so much for checking.

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a telephone pole is to have a guy wire attached to its top and anchored to the ground at a point that is 34 feet less than the height of the pole=H SAY. if the wire=W SAY.. is to be 2 feet longer than the height of the pole, what is the height of the pole?
IF GUY WIRE IS ANCHORED AT G ON THE GROUND AND THE POLE IS TB...T BEING TOP AND B BEING BOTTOM OF POLE ON THE GROUND.THEN...
ANGLE GBT=90..IN THE RIGHT TRIANGLE GBT
GB=SIDE=H-34
BT=HEIGHT OF POLE=ANOTHER SIDE = H
GT=WIRE=W=H+2=HYPOTENUSE
GT^2=BT^2+GB^2
(H+2)^2=H^2+(H-34)^2
H^2+H^2-68H+34^2-H^2-4-4H=0
H^2-72H+1152=0
H^2-48H-24H+1152=0
H(H-48)-24(H-48)=0
(H-48)(H-24)=0
H=48

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The augmented matrix for the system of equations below is
3x-2y=5
4x+7y=2
My answer is
3 -2 5
4 7 2 thanks for chcking.
CORRECT