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Use calculus to determine the relative extrema of
f(x)=x-lnx
DF/DX =1-1/X = 0 AT EXTREMUM
X =1
D^F/DX^2 = 1/X^2 = 1/1=+VE
HENCE X=1 IS A MINIMUM
SLOPE OF TANGENT AT X=1 DY/DX AT X=1........THAT IS M = 0
AT X=1 ..Y = 1-LN1 =1
HENCE EQN.OF TANGENT AT X=1,Y=1 IS
Y-1=M(X-1)=0
Y = 1 IS THE EQN. OF THE TANGENT AT X=1
Find the equation of the tangent line at x=1
please help, i'm so confused!

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Find the standard equation of the circle that satisfies the conditions.
Center (-1,-3), passing through the point (3,0)
EQN. OF CIRCLE WITH CENTRE AT (H,K) AND RADIUS = R IS
(X-H)^2 + (Y-K)^2 = R^2
(X+1)^2 +(Y+3)^2 =R^2
IT PASSES THROUGH (3,0)...SO,
4^2 + 3^2 =R^2 = 25
HENCE EQN. IS
(X+1)^2 + (Y+3)^2 = 25

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22. The equation h=-16t^2+112t give the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a a height of 180 ft.
180 = -16T^2+112T
DIVIDING WITH 4
4T^2-28T+45=0
4T^2-10T-18T+45=0
2T(2T-5)-9(2T-5)=0
(2T-5)(2T-9)=0
T=2.5 OR 4.5.....
THIS MEANS THAT IT WILL BE AT 180 FEET FROM GROUND AFTER IT WAS SHOT FROM THE GROUND
1. AFTER 2.5 SECS WHILE IT IS GOING UP.
2.THEN AFTER REACHIBG THE MAXIMUM HEIGHT , IT WILL FALL BACK TO THE GROUND
AND SO AFTER 4.5 SECS REFERS TO ITS DOWN WARD FALL

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The graph of the parabola y = -2(x -1)2 + 3
HOPE YOU MEAN Y=-2(X-1)^2+3...IF SO
AXIS OF SYMMETRY IS X=1
is symmetric about a line. What is the equation of that line?
x = -1
x = 0
x = 1
x = 2
y = 0

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Find the vertex of the parabola whose equation in standard form is
y = x2 + 8x - 1
= [X^2+2(X)(4)+4^2]-4^2-1
=(X+4)^2-17
SO ITS VERTEX IS AT X=-4...AND Y=-17


(-8, -17)
(-8, -1)
(-4, -1)
(-4, -17)
(4, -17)

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I am a grade 10 students and i like trigonometry but i found this quiet tough.Plz can u solve it for me..
prove that (sec4a-1)/(sec8a-1)=tan4a/tan2a
FEW THINGS , I WANT TO KNOW..
1.IN 10 TH. GRADE ARE YOU TAUGHT MULTIPLE ANGLES.ASSUMING SO..
2.HAVE YOU TYPED THE PROBLEM CORRECTLY?IT DOES NOT APPEAR SO..GOR EXAMPLE TRY FOR A=15 DEGREES
LHS = [SEC(60)-1]/[SEC(120)-1]= 1/-3=-1/3
RHS = TAN(60)/TAN(30)= SQRT(3)/[1/SQRT(3)]=3...
SO THE PROBLEM IS NOT CORRECT....
3.ANY WAY LET US KEEP LHS IN TACT AND FIND IT OUT INTERMS OF TAN MULTIPLE ANGLES
LHS =[{1-C0S(4A)}/{COS(4A)}]/[{1-COS(8A)}/{COS(8A)}]
= 2SIN^2(2A)*COS(8A)/[2SIN^2(4A)COS(4A)]
=[{2SIN^2(2A)COS(8A)}]/[{2SIN(4A)COS(4A)}{SIN(4A)}]
=[2SIN^2(2A)COS(8A)]/[{SIN(8A)}{2SIN(2A)COS(2A)}]
= TAN(2A)/TAN(8A)....IS THE ANSWER
OK? GOT IT?.

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Solve, ln x^1/4 = (ln x)^1/2
LN(X)/4 = LN(X)/2
LN(X)=0
X=1

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Solve, ln x^1/4 = (ln x)^1/2
LN(X)/4 = LN(X)/2
LN(X)=0
X=1

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Decompose into partial fractions (check your answers)[x^2 -10x +13]/[x^2 -5x +6)(x-1)]
=[X^2-10X+13]/[(X-2)(X-3)(X-1)]
= A/(X-1) + B/(X-2) + C/(X-3)
X^2-10X+13 = A(X-2)(X-3) +B(X-1)(X-3) +C(X-1)(X-2)
PUTTING X=1,2 AND 3 IN ORDER , WE GET...
A =[1-10+13]/[(1-2)(1-3)= 4/2=2
B= [4-20+13]/(2-1)(2-3)=-3/-1=3
C=[9-30+13]/(3-1)(3-2)=-8/2 = -4
HENCE ANSWER IS
2/(X-1) +3/(X-2) - 4/(X-3)...I HOPE YOU CAN CHECK BY MULTIPLICATION.

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Find a polynomial fucntion of 5th degree given the following zeros.
-4, -2i, √6
ASSUMING THE POLYNOMIAL TO BE WITH REAL COEFFICIENTS,THE CONJUGATES OF COMPLEX AND IRRATIONAL NUMBERS WILL ALSO BE THE ROOTS. HENCE THE 5 ROOTS ARE
-4,-2I,+2I,SQRT(6),-SQRT(6)...HENCE THE POLYNOMIAL IS
P = (X+4)(X+2I)(X-2I)[X+SQRT(6)][X-SQRT(6)]=0
P=(X+4)(X^2+4)(X^2-6)=0
P=(X+4)(X^4-2X^2-24)=0
P=X^5+4X^4-2X^3-8X^2-24X-96=0

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Find the equation of the ellipse that has the four points as endpoints of the major and minor axes. (3,0), (-3,0), (0,5), (-5,0).
I tried to use the x^2/a^2 + y^2/b^2 = 1
OK...BUT MORE GENERAL EQN. IS
(X-H)^2/A^2 + (Y-K)^2/B^2=1.....CENTRE IS (H,K)
WE FIND (3,0) AND (-3,0) ARE 2 ENDS OF ONE AXIS ..ITS LENGTH IS 6
...OBVIOUSLY IT IS THE X AXIS
CHECK YOUR OTHER END POINTS DID YOU TYPE PROPERLY?IS IT (-5,0)?OR (0,-5)
ASSUMING IT TO BE (0,-5) OBVIOUSLY THE OTHER AXIS IS Y AXIS.ITS LENGTH IS 10
THEN VERTEX IS (0,0)...THAT IS H=0 AND K=0 AS YOU HAVE PUT THE EQN.
NOW COMING TO YOUR OTHER QUESTION HERE AXIS ALONG X AXIS IS SHORTER
= 6 = 2A ..OR....A =3...SO WE CALL IT MINOR AXIS
AXIS ALONG Y AXIS IS LONGER = 10 = 2B..B=5...IT IS MAJOR AXIS.
SO EQN. IS
X^2/9 + Y^2/25 =1
YOU CAN CHECK THE POINTS BY SUBSTITUTION .
but I got a lil confused as to which numbers to use for x,a,y,b

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well this is actually from algebra 2.
3 The characteristic of log 6198.....WRITE THE GIVEN NUMBER AS X*10^N
WHERE X IS A NUMBER WITH ONLY UNITS PLACE AND THEN N IS THE CHARACTERSTIC.HERE
6198 = 6.198*1000= 6.198*10^3...HENCE 3 IS THE CHARACTERSTIC


is _______
2
4
3
1

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The width ( W SAY) of a retangle is 15 feet less than its length( L SAY), and
W =L-15

the perimeter( P SAY) is 210 feet.
P= 2L + 2W
210 = 2L +2(L-15)=2L+2L-30=4L-30
4L-30=210
4L=210+30=240
L=240/4 = 60
W = 60-15 =45


Can you show me how to find the width and the length.

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The difference between two whole numbers is 8 and the difference between their reciprocals is 1/6. Find the two numbers.
LET ONE NUMBER BE X
OTHER NUMBER = X-8
THEIR RECIPROCALS ARE 1/(X-8) AND 1/X
DIFFERENCE = 1/(X-8) -1/X = 1/6.....MULTIPLY THROUGH OUT WITH 6X(X-8)
6[{X-(X-8)}=X(X-8)
6[X-X+8]=X^2-8X
X^2-8X-48=0
X^2-12X+4X-48=0
X(X-12)+4(X-12)=0
(X-12)(X+4)=0
HENCE X=12...AND THE OTHER NUMBER IS 12-8 =4
IF NEGATIVE NUMBERS ARE ALLOWED
X=-4 AND THE OTHER NUMBER IS -4-8=-12

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which one of these numbers are prime and how do i show my work?
12,31,99,1
i think 1 is prime because you only have 0,1
PRIME NUMBER IS
A NUMBER WHICH HAS NO FACTORS (THAT IS NOT DIVISIBLE LEAVING ZERO REMAINDER
-BY ANY NUMBER)EXCEPT 1 AND ITSELF.
BUT THERE IS ANEXCEPTION..AS A RULE 1 IS NOT CLASSIFIED AS A PRIME NUMBER.
THE ONLY METHOD IS TO CHECK BY DIVISION STARTING FROM 2,3,ETC...
12 IS NOT AS IT IS DIVISIBLE BY 2
31 IS PRIME AS 2,3,5,7 ARE NOT FACTORS..NEED NOT GO BEYOND SINCE 7*5=35>31
THAT IS SECOND FACTOR IS LESS THAN THE FIRST FACTOR AND TILL THIS STAGE SECOND FACTOR IS MORE THAN THE FIRST FACTOR 5*7=35>31
99 IS NOT PRIME...AS IT IS DIVISIBLE BY 3
1 IS NOT PRIME BY DEINITION

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a cylindrical tin of particular volume is to be made using as little material as possible. find the ratio of the height to the radius(the tin is closed both ends)
LET THE RATIO OF HEIGHT TO RADIUS = X =H/R.....H=RX
VOLUME OF CYLINDER = PI*(R^2)*H = V = CONSTANT.
V = PI*R^2*RX = PI*(R^3)*X
R=[V/(PI*X)]^(1/3)
A = AREA OF CYLINDER = 2PI*R*H+2PI*R^2=2PI*R[RX+R]=2PI*(R^2)[X+1]
A = 2PI*R^2[X+1]=2PI[X+1][V/(PI*X)]^(2/3)
A=[2PI*V^(2/3)/{PI}^(2/3)][(X+1)/X^(2/3)]
A = K[X^(1/3)+X^(-2/3)]
DA/DX = 0 FOR MINIMUM VALUE = K[(1/3){X^(-2/3)} -(2/3){X^(-5/3)}]=0
1/3{X^(2/3)}= 2/3{X^(5/3)}
X^(5/3)/X^(2/3)=2
X = 2

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Help Please! Find the exact an approximate solution. Find two positive real numbers that differ by one and have a product of one.
LET ONE NUMBER = X
OTHER NUMBER = X-1
PRODUCT = X(X-1)=1
X^2-X=1
X^2-X-1=0.....COMPARING WITH AX^2+BX+C=0...A=1...B=-1....C=-1




SINCE POSITIVE NUMBERS ARE ASKED FOR THEY ARE
[1+SQRT(5)]/2 AND [1+SQRT(5)]/2 -1..OR ....[SQRT(5)-1]/2

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I WAS WAITING FOR YOUR RESPONSE..ANY WAY I MADE MY OWN ASSUMPTIONS OF THE REQUIREMENTS AND WORKED OUT THE SOLUTION BELOW
----------------------------------
I need to find the basis for the null space,
NULL SPACE? WHY?ARE YOU SURE?
the range of the matrix,
RANGE? ARE YOU SURE? OR IS IT RANK?
PLEASE CONFIRM REQUIREMENTS AS THEY LOOK DOUBTFULL!
and the orthogonal basis using Gram-Schmidt.
SOLUTION:NULL SPACE , RANK ,ORTHONORMAL BASI
The matrix is:
1 -2 1 -5
2 1 7 5
1 -1 2 -2

V1= 1 -2 1 -5
V2= 2 1 7 5
V3= 1 -1 2 -2
NR2=R2-2*R1 ; NR3=R3-R1
1 -2 1 -5
0 5 5 15
0 1 1 3
NR2=R2/5
1 -2 1 -5
0 1 1 3
0 1 1 3
NR3=R3-R2
1 -2 1 -5
0 1 1 3
0 0 0 0
NR1=R1-R2
1 -3 0 -8
0 1 1 3
0 0 0 0

RANK = 2

CONSIDERING THE SET OF EQNS.
AX = 0,WE HAVE ONLY 2 INDEPENDENT EQNS.
X1-2X2+X3-5X4=0…………….1
X1-X2+2X3-2X4=0……………2
EQN.2-EQN.1 GIVES
X2+X3+3X4=0…..3
HENCE ONE SET OF SOLUTION COULD BE X2=1,X3=2,X4=-1..AND X1=-5…..THAT IS (-5,1,2,-1) = N1
AND ANOTHER SET OF SOLUTION COULD BE ..X2=2,X3=1,X4=-1 AND X1=-2..THAT IS (-5,2,1,-1)=N2
N1 AND N2 BEING 2 INDEPENDENT SOLUTIONS WILL FORM THE BASIS FOR NULL SPACE.


HENCE THE 3 VECTORS HAVE 2 INDEPENDENT VECTORS.
HENCE WE CAN CHOSE ANY 2 INDEPENDENT VECTORS TO FORM BASIS
LET US TAKE V1 AND V3 WHICH ARE INDEPENDENT TO FORM THE BASIS.
Y1=V1 = (1,-2,1,-5)
Y3 =V3+AV1…DOT WITH Y1
Y3.Y1 = V3.Y1+AV1.Y1…….SINCE Y1 AND Y3 ARE ORTHOGONAL,Y1.Y3=0
A=-(V3.Y1)/(V1.Y1) = -[1*1+(-1)*(-2)+2*1+(-5)*(-2)]/[1*1+(-2)*(-2)+1*1+(-5)*(-5)]=-15/31
Y3 = (1,-1,2,-2)-(15/31)(1,-2,1,-5) = (1/31)*(16,-1,47,13)
HENCE THE 2 ORTHOGONAL VECTORS ARE OBTAINED.
NORMALISING THEM WE GET THE ORTHONORMAL BASIS
Y1'=(1,-2,1,-5)/SQRT(31) = (1/5.568)(1,-2,1,-5)
Y3'= [16,-1,47,13]= (1/51.33)(16,-1,47,13)

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Find all real or imaginary solution to the equation.
(x-1)/(x+2)=(2x-3)/(x+4)
(X-1)(X+4)=(X+2)(2X-3)
X^2+3X-4 = 2X^2+X-6
X^2-2X-2=0
X = [2+ OR - SQRT(4+8)]/2
= 1+SQRT(3)..........OR......1-SQRT(3)

Thank you so much.

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In triangle ABC , AB=AC.
HENCE
ANGLE B = ANGLE C = X SAY
If there is a point P strictly between A and B such that AP=PC=CB,
IN TRIANGLE PCB , PC=CB....HENCE
ANGLE B = X = ANGLE BPC......................1
IN TRIANGLE PAC , AP=PC....HENCE
ANGLE A = Y SAY = ANGLE PCA
ANGLE C = X = ANGLE BCP + ANGLE PCA = ANGLE BCP + Y
ANGLE BCP = X-Y
ANGLE BPC = EXTERNAL ANGLE AT P FOR TRIANGLE APC
= SUM OF OPPOSITE INTERIOR ANGLES = ANGLE A + ANLE PCA = Y+Y = 2Y........2
FROM EQN.1 AND EQN.2.....X=2Y
SUM OF 3 ANGLES IN TRINGLE PBC =180 = ANGLE PBC + ANGLE BCP + ANGLE BPC
= X + X-Y+2Y =180
2X+Y=180
2*2Y+Y=180
5Y=180
Y=36
HENCE ANGLE A =36
then angle A = ? = 36

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The current in ther lazy river moves at a rate of 4mph. Monica'sdinghy motors 6mi upstream in the same time it take to motor 12mi downstream. What is the speed of the dinghy?
SPEED OF WATER = 4 MPH
LET SPEED OF DINGHY = S MPH
ITS SPEED UP STREAM = S-4 MPH
DISTANCE TRAVELED UPSTREAM = 6 M
TIME TAKEN = 6/(S-4) HRS..................1
ITS SPEED DOWN STREAM = S+4 MPH
DISATANCE TRAVELLED DOWN STREAM = 12 M
THEN TIME TO TRAVEL UPSTREAM = 12/(S+4) HRS...............2
TIMES ARE EQUAL
6/(S-4) = 12/(S+4)
12(S-4)=6(S+4)
12S-48=6S+24
6S=72
S=12 MPH

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Would someone help me.
Solve the following equations. Round to 4 decimals when necessary.
a. logx 729/4096=6
X^6 = 729/4096 = 3^6/4^6 = (3/4)^6
X=3/4
b. log5 325=x
5^X = 325 ..IT GIVES IRRATIOAL SOLUTION...CHEK THE PROBLEM..IS IT 625..IF O
5^X =625 = 5^4
X=4
c. log2 (x-2)+log2 (x+1)=2
LOG[(X-2)(X+1)] TO BASE 2 = 2
(X-2)(X+1)=2^2=4
X^2-X-2-4=0
X^2-3X+2X-6=0
X(X-3)+2(X-3)=0
(X-3)(X+2)=0
X=3 ....SINCE X=-2 LEADS TO LOG OF NEGATIVE NUMBER WHICH IS IMAGINARY.
d. log (2x+6)-log(x-1)=1
LOG[(2X+6)/(X-1)]=1
ASSUMING BASE TO BE 10
(2X+6)/(X-1)=10
2X+6 = 10X-10
8X = 16
X = 2


Thank you for your help!
COULD SOMEONE PLEASE HELP!!!

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use the reference angles to find the exact value of the expression:
tan -7pi/4 = -TAN [7PI/4]= - TAN [2PI-PI/4]= -TAN(-PI/4)=TAN(PI/4)=1

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To whom may it concern,
I have trouble to do this question.
From a point at the same level as the foot of a building 50 metres high, the angle of elevation of its top is found to be 52.83°. Find the distance of the point of observation from the top of the building.
Please help and thank you.
LET THE TOP OF BLDG BE 'T' AND BOTTOM BE 'B'.LET THE POINT OF OBSERVATION BE 'P'
PBT IS RIGHT ANGLED TRIANGLE WITH ANGLE PBT = 90 AND ANGLE TPB = 52.83 DEGREES AND TB=50 M.
TAN(52.83) = TB/PT = 50/PT
PT = 50/TAN(52.83)= 37.95 M

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Okay it says to solve by completeing the square.
4x^2+2x-3=0
So far I have the following:
4x^2+2x=3
x^2+2/4x=3/4
X^2+X/2-3/4=0..........MAKE PERFECT SQUARE USING X^2 AND X TERMS WITH THE FORMULA (X+A)^2 = X^2+2XA+A^2...THAT IS 2XA=X/2.....2A=1/2...A=1/4....
NOW ADD AND SUBTRACT A^2...THAT IS (1/4)^2....
=[(X)^2 + 2(X)(1/4)+(1/4)^2]-(1/4)^2-3/4=0
= [X + (1/4)]^2= 1/16 +3/4 = 13/16
X+1/4 = + OR - SQRT(13)/4
X = -1/4 + OR - SQRT(13)/4 = [-1 + OR - SQRT(13)]/4

Then I get totally lost and don't know what else to do!! HELP! PLEASE!
Thanks in advance!!!!!!

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3x^2-6x=3
Not sure if I am doing this right, but so far I have
3x^2-6x-3=0...OK
x^2-3x-1=0...DIVIDING WITH 3 WE GET X^2-2X-1=0...COMPARING WITH STD EQN.
AX^2+BX+C=0.....WE HAVE A=1,B=-2,C=-1






Am I on the right track?
Can you help me with the rest?
Thanks so much!

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Probability
I do not understand
A card is drawn from a deck of ten card numbered 1 - 10. The card is replaced in the deck and another card is drawn.
OK YOU GET PROBABILITY BY DIVIDING NUMBER OF SUCCESFULL OUT COMES
WITH TOTAL OF ALL POSSIBLE OUT COMES.
HERE THERE ARE 10 CARDS NUMBERED 1 TO 10.
HENCE THERE ARE IN TOTAL 10 POSSIBLE OUT COMES.LET US SEE SUCCESS REQUIREMENT AS ASKED FOR
P(an odd number and then an even number)
I DRAW ...
ODD NUMBER = THERE ARE 5 ODD NUMBERS = 1,3,5,7,9.
SO SUCCESSFULL OUT COMES = 5
PROBABILITY OF DRAWING ODD NUMBER = 5/10 = 1/2
II DRAW............
EVEN NUMBER...THERE ARE 5 EVEN NUMBERS = 2,4,6,8,10
SO SUCCESSFUL OUT COMES =5
PROBABILITY OF DRAWING EVEN NUMBER = 5/10=1/2
NOW PROBABILITY OF BOTH HAPPENING WHICH ARE 2 INDEPENDENT EVENTS = PRODUCT OF THEIR PROBABILITIES = (1/2)(1/2) = 1/4
GOT THE IDEA ..REST I AM GIVING ANSWERS CHECK UP

P(two numbers greater than 4)
5,6,7,8,9,10 = 6
(6/10)(6/10)=9/25
P(an odd number and then an even number)
(5/10)(5/10 =1/4
P(5 and then a 3)
5.....ONLY 1 WAY...SO P=1/10
3...ONLY 1 WAY SO...P=1/10
COMBINED PROBABILITY = (1/10)(1/10)=1/100
Please help me understand the concept....
You may edit the question. Maybe convert formulae to the same formula

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Help Please! Find the exact an approximate solution. Find two positive real numbers that differ by one and have a product of one.
LET ONE NUMBER = X
OTHER NUMBER = X-1
PRODUCT = X(X-1)=1
X^2-X=1
X^2-X-1=0.....COMPARING WITH AX^2+BX+C=0...A=1...B=-1....C=-1




SINCE POSITIVE NUMBERS ARE ASKED FOR THEY ARE
[1+SQRT(5)]/2 AND [1+SQRT(5)]/2 -1..OR ....[SQRT(5)-1]/2

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NAVIGATION: The paths of two ships are tracked on the same coordinate system. One ship is following a path described by the equation 2x+3y=6, and the other is following a path described by the equation y=2/3x-3..PLEASE PUT BRACKETS.DO YOU MEAN (2/3)X - 3 = (2X/3) -3.OR
[2/(3*X)] -3 ..OR
[2/(3-X)]...EACH WILL GIVE A DIFFERENT ANSWER.
I AM TAKING THE I ALTERNATIVE

a. Is there a possibility of a collision?
2X+3Y =6........1
Y=(2X/3)-3.....2
FROM EQN.1 DIVIDING WITH 3 WE GET
(2X/3)+ Y = 2
Y = 2 - (2X/3).......3..
SO
FROM EQNS. 2 AND 3 , WE GET
2X/3 -3 = 2 -(2X/3)
(2X/3) + (2X/3) = 2+ 3 =5
4X/3 =5
X=15/4
Y = [2*15/(4*3)]-3 = (5/2)-3=(5-6)/2 = -1/2
SINCE THE 2 LINES OF TRAVEL MEET AT A POINT THERE IS A POSSIBILITY OF COLLISION

b. What are the coordinates of the danger point?
THE COORDINATES OF COLLISION /DANGER POINT ARE [15/4 , -1/2]
c. Is a collision a certainty?
CANT SAY ..DEPENDS ON STARTING POINT ; DIRECTION OF TRAVEL ALONG THE LINE AND SPEED OF TRAVEL IN THAT DIRECTION WHICH ARE ALL NOT GIVEN.
SUPPOSE THEY ARE STARTING AT A POINT AND TRAVELLING IN THE DIRECTION AWAY FROM DANGER POINT THEN THEY DO NOT COLLIDE.
EVEN IF THEY ARE TRAVELLING TOWARDS THE DANGER POINT IF ONE IS GOING FASTER AND CROSSES DANGER POINT BEFORE THE OTHER REACHES THEY WILL NOT COLLIDE.
While the answers are in the back of the book, I do not know how to get the correct answer. Could someone help me please? Thank you, I am going to school online and I am having difficulty. Much thanks!!

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SEE MY COMMENTS BELOW...
-----------------I would like some assistance with 2 word problems please. I have tried each one about 3-4 times and keep getting stuck with non equal answers
1. The Area of a Triangle is given as A = 1/2 BH. Find the Height (H) given that the Base (B) = 14 inches and the Area (A) = 56 square inches.
56 =(1/2)14*H
H=2*56/14 = 8
2. One number is five more than twice another. The two numbers add to 29. What are those two numbers?
I come up with: 2N + 5...VERY GOOD ..BUT SAY N AND 2N+5 ARE THE 2 NUMBERS
N + 2N + 3N + 5 = 29....NO.... N+2N+5=29
3N + 5 = 29.....OK
-5 -5....NO
3N = 29 - 5 =24
N = 24/3 =8
SO THE 2 NUMBERS ARE 8 AND 2*8+5=21 WHOSE SUM IS 29
------------------------------------------------------
6N / 6 = 24 / 6
N = 8
--------

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SEE MY COMMENTS BELOW..
GOOD TO SEE YOU TRY SINCERELY.KEEP IT UP! THAT IS THE BEST WAY OF LEARNING.
--------------------------------------------------------------
Parantheses in equations:
I have tried now about 6 times working on these problems and I just can't get it all together. I get one part then mess up another, please HELP.
Problem #1
-(7D - 8) = -5(2B - 3) +1 I need to solve and do the check.
This is what I have done.
-(7D- 8) = -5(2D - 3) +1..........OK
-7D + 8 = -10D + 15 + 1..........VERY GOOD
-7D + 8 = -10D + 16..............VERY GOOD
+7D +7D
+8 = -3D + 16...........NO
TRY TO BRING UNKNOWNS TO LHS...AND KNOWNS TO RHS...
TO DO THIS FOLLOW THE RULE THAT WHEN YOU CHANGE SIDES FROM LEFT TO RIGHT OR OTHERWISE..+ WILL BECOME - AND - WILL BECOME + ON THE OTHER SIDE.
1.WE HAVE TO SEND -10D AN UNKNOWN ON RHS FROM RHS TO LHS. SO IT WILL BECOME +10D ON LHS
2.WE HAVE TO SEND +8 A KNOWN ON LHS FROM LHS TO RHS.SO IT WILL BECOME -8 ON RHS.
SO WE GET
-7D+10D = 16 - 8 =8
3D = 8
D = 8/3
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CHECK
-(7D-8) = -5(2B-3)+1
LHS = -[7*8/3 -8]= -[(7*8-8*3)/3]=-(56-24)/3=-32/3
RHS = -5(2*8/3 - 3) + 1 = -5[(2*8 - 3*3)/3]+1 = -5*7/3 +1
= (-35+3)/3=-32/3 = LHS..OK..
-16 -16
-8 = -3D
D = -8 / -3
CHECK:
-(7D - 8) = -5(2D - 3) + 1
-(7 (-8 / -3) -8) = -5 (2 (-8 / -3) -3) +1
-(56 / -3 -8) = -5 (-16 / -3) -3) +1
Please help