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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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Please show me how to factor completly 4a^4-26a^2+30
2[2a^4-13a^2+15]=2[2a^4-10a^2-3a^2+15]=2[2a^2(a^2-5)-3(a^2-5)]
=2(a^2-5)(2a^2-3)

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please factor completly x^5-7x^4+12x^3
= x^3(x^2-7x+12)=x^3{x^2-3x-4x+12}=x^3[x(x-3)-4(x-3)]
=x^3(x-3)(x-4)

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Please solve 6x^2+x=15
6x^2 +x-15=0
6x^2+10x-9x-15=0
2x(3x+5)-3(3x+5)=0
(3x+5)(2x-3)=0
3x+5=0.......that is 3x=-5......x=-5/3....or
2x-3=0......2x=3......x=3/2

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Solve the system by graphing, addition and substitution methods.
3x – y = 1 ..........1
3x – y = 2................2
eqn.1 - eqn.2
0=1-2=-1...not possible..hence no solution..they are in consistent eqns.
x...............0.......................2..............4......etc
y=3x-1.........-1.......................5..............11.....etc
y=3x-2.........-2.......................4...............10.....etc

the 2 graphs are parallel and do not intersect hence there is no solution.

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Solve the system by graphing, addition and substitution methods.
–4x + 7y = –10 ...............1
x – 5y = 9................2
x=9+5y
putting in eqn.1
-4(9+5y)+7y=-10
-36-20y+7y=-10
-13y=-10+36=26
y= -26/13 =-2
x=9+5*(-2)=9-10=-1
x=............0...................1.....................2....etc
y=(x-9)/5....-1.8................-1.6..................-1.4.. etc
Y=(4X-10)/7..-10/7................-6/7.................-2/7...ETC

the 2 graphs intersect at x=-1 and y=-2

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Express 2x^2+2y^2-12x+16y=-32.DIVIDING BY 2
x^2+y^2-6x+8y=-16
[X^2-2(X)(3)+3^2]-3^2+[Y^2+2(Y)(4)+4^2]-4^2=-16
(X-3)^2+(Y+2)^2=9=3^2
THIS IS IN THE STD.FORM OF
(X-H)^2+(Y-K)^2=R^2
HENCE CENTRE = (H,K) = (3,-2)
RADIUS = R =3
in standard form for a circle. Identify the radius and center of the circle.

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The owner of a highway construction company knows that the maximum load (L SAY)a cylindrical column of circular cross section can support varies directly as the fourth power of the diameter(D SAY) and inversely as the square of the height(H SAY). If a column 2 feet in diameter and 10 feet high supports up to 6 tons, how much of a load can a column 3 feet in diameter and 14 feet high support?
L = K*D^4*1/H^2...WHERE K IS A CONSTANT
CASE 1..
6=K*2^4/10^2=16K/100
K=6*100/16=75/2=37.5
CASE 2
L=37.5*3^4/14^2 = 37.5*81/196 = 15.5 T.

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What is the area defined by the following graphs:
|x|+|y| is greater than or equal to 2
CASE 1 ..X AND Y POSITIVE..WE GET...X+Y>=2
CASE 2...X + VE Y - VE.....WE GET ..X-Y>=2
CASE 3...X -VE..Y -VE......WE GET ..-X-Y>=2
CASE 4...X - VE..Y..+ VE...WE GET ...-X+Y>=2
and |x|+|y| is less than or equal to 5
SAME WAY WE GET HERE X+Y<=5......X-Y<=5......-X-Y<=5.......-X+Y<=5
GRAPH LOOKS LIKE THIS


it is the area of square with side of 5sqrt(2) - area os square with side of 2sqrt(2) = 5*sqrt(2)*5*sqrt(2)-2*sqrt(2)*2*sqrt(2) =50-8=42

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solve the equation ay/x=cz
AY=CXZ
Y=CXZ/A............1
CXZ=AY
Z=AY/CX...............2
CXZ=AY
X=AY/CZ...................3
SEE THE FOLLOWING EXAMPLE TO KNOW RULES ON TRANSFORMATIONS.
-----------------------------------------------------------------
I really don't understand this problem. Can you please help me?
3(1-3x)+2=4x+14........OK
3-9x+2=4x+14................GOOD
so then add or subtract from both sides?
3-9x+9x+2=13+14......HERE YOU GOT CONFUSED..THE STEPS INVOLVED ARE
1.REMOVE BRACKETS ......THIS YOU HAVE DONE.
2.TRANSFER ALL UNKNOWN TERMS TO ONE SIDE.... USUALLY L.H.S. AND KNOWN TERMS TO ONE SIDE ...USUALLY R.H.S.
3.THIS TRANSFER YOU CAN THINK OF AND DO AS FOLLOWS....
EVERY TERM HAS A + OR - SIGN INFRONT OF IT.WHEN YOU TRANSFER THAT TERM FROM ONE SIDE TO ANOTHER ,+ BECOMES - AND - BECOMES PLUS.
THIS IS BECAUSE...AS YOU WERE TRYING TO SAY 'ADD OR SUBTRACT FROM BOTH SIDES'
AN EQUATION IS UNALTERED IF WE ADD OR SUBTRACT SAME QUANTITY FROM BOTH SIDES OF THE EQN.SO IF WE WANT TO REMOVE +4X ABOVE FROM R.H.S.WE HAVE TO ADD -4X TO IT.SO TO BALANCE IT AND KEEP THE VALIDITY OF THE EQN.UNALTERED,WE ADD -4X TO L.H.S.ALSO.THEN ON R.H.S ,WE GET +4X-4X=0...THAT IS +4X DIAPPEARS AS WE WANTED.BUT -4X REMAINS ON L.H.S.....THIS IS THE FINAL EFFECT OF ADD AND SUBTRACT WHICH IS SUMMARISED FIRST ABOVE.
HENCE WE GET HERE
-9X-4X = 14-2-3
4.NOW SIMPLIFY BOTH SIDES COMBINING ALL UNKNOWNS ON L.H.S AND ALL KNOWNS ON R.H.S
WE GET HERE
-13X=9
5.NOW TRANSFER THE COEFFICIENT OF UNKNOWN ON L.H.S.TO R.H.S.
APPLYING THE SAME RULE & LOGIC MENTIONED ABOVE,HERE * WILL BECOME / AND / BECOMES *
HERE WE GET
X = -9/13....
THIS IS THE ANSWER.
6.IF YOU HAVE TIME AND WOULD LIKE TO DO....CHECK BACK THE ANSWER FOR CONFIRMATION BY SUBSTITUTION..
3{1-3*(-9/13)}+2=4*(-9/13)+ 14
3{1+27/13)+2=-36/13 +14
3*40/13 +2=(-36+14*13)/13
120/13 +2= 146/13
146/13 = 146/13.......OK

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SEE THE FOLLOWING EXAMPLE AND TRY...IF STILL IN DIFFICULTY PLEASE COME BACK.
------------------------------------------------------------------
I really don't understand this problem. Can you please help me?
3(1-3x)+2=4x+14........OK
3-9x+2=4x+14................GOOD
so then add or subtract from both sides?
3-9x+9x+2=13+14......HERE YOU GOT CONFUSED..THE STEPS INVOLVED ARE
1.REMOVE BRACKETS ......THIS YOU HAVE DONE.
2.TRANSFER ALL UNKNOWN TERMS TO ONE SIDE.... USUALLY L.H.S. AND KNOWN TERMS TO ONE SIDE ...USUALLY R.H.S.
3.THIS TRANSFER YOU CAN THINK OF AND DO AS FOLLOWS....
EVERY TERM HAS A + OR - SIGN INFRONT OF IT.WHEN YOU TRANSFER THAT TERM FROM ONE SIDE TO ANOTHER ,+ BECOMES - AND - BECOMES PLUS.
THIS IS BECAUSE...AS YOU WERE TRYING TO SAY 'ADD OR SUBTRACT FROM BOTH SIDES'
AN EQUATION IS UNALTERED IF WE ADD OR SUBTRACT SAME QUANTITY FROM BOTH SIDES OF THE EQN.SO IF WE WANT TO REMOVE +4X ABOVE FROM R.H.S.WE HAVE TO ADD -4X TO IT.SO TO BALANCE IT AND KEEP THE VALIDITY OF THE EQN.UNALTERED,WE ADD -4X TO L.H.S.ALSO.THEN ON R.H.S ,WE GET +4X-4X=0...THAT IS +4X DIAPPEARS AS WE WANTED.BUT -4X REMAINS ON L.H.S.....THIS IS THE FINAL EFFECT OF ADD AND SUBTRACT WHICH IS SUMMARISED FIRST ABOVE.
HENCE WE GET HERE
-9X-4X = 14-2-3
4.NOW SIMPLIFY BOTH SIDES COMBINING ALL UNKNOWNS ON L.H.S AND ALL KNOWNS ON R.H.S
WE GET HERE
-13X=9
5.NOW TRANSFER THE COEFFICIENT OF UNKNOWN ON L.H.S.TO R.H.S.
APPLYING THE SAME RULE & LOGIC MENTIONED ABOVE,HERE * WILL BECOME / AND / BECOMES *
HERE WE GET
X = -9/13....
THIS IS THE ANSWER.
6.IF YOU HAVE TIME AND WOULD LIKE TO DO....CHECK BACK THE ANSWER FOR CONFIRMATION BY SUBSTITUTION..
3{1-3*(-9/13)}+2=4*(-9/13)+ 14
3{1+27/13)+2=-36/13 +14
3*40/13 +2=(-36+14*13)/13
120/13 +2= 146/13
146/13 = 146/13.......OK

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Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0
[X^2+2(X)(3)+3^2]-3^2+Y^2+8=0
(X+3)^2+Y^2=1
COMPARING WITH STD EQN.
(X-H)^2+(Y-K)^2 =R^2
CENTRE = (H,K) = (-3,0)
RADIUS =R =1

Center = (_,_) Radius = _

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SEE MY COMMENTS BELOW...
I have this problem that I would like someone to check my work:
If a skydiver jumps from an airplane at a height of 8256 feet, then for the first five seconds, her height above the earth is approximated by the formula h = -16t^2 + 8256. How many seconds does it take her to reach 8000 feet?
WORK DONE :
What I have done is:
h = 16^t + 8256....HOW?TYPING ERROR..IT SHOULD BE -16T^2+8256
h = -16t^2 + 8256 = 8000...OK
h = -16t^2 + 8256 - 8000 = 0...DONT WRITE H=...IT IS NOT CORRECT.
SIMPLY WRITE
-16T^2+8256-8000=0.....GOT IT?
-16t^2 + 256 = 0...OK
-16t^2 + 256 - 256 = - 256...OK
16t^2 = -256 ......NO AGAIN TYPING ERROR? IT IS -16T^2=-256
-16t^2/16 = -256/-16 .....OK
t^2 = 16 ......OK
t = 4 .....OK....
I'm not confident about the way I have worked this out. Could someone check.
THE ANSWER IS OK.THE WORKING IS FAULTY AT SOME PLACES AS MENTIONED.
CORRECT THOSE ERRORS AND YOU WILL DO WELL.SEE THE FOLLOWING EXAMPLE ON SOLUTION OF EQNS.FOR PROPER STEPS OF WORKING.
---------------------------------------------------------------
I really don't understand this problem. Can you please help me?
3(1-3x)+2=4x+14........OK
3-9x+2=4x+14................GOOD
so then add or subtract from both sides?
3-9x+9x+2=13+14......HERE YOU GOT CONFUSED..
THE STEPS INVOLVED ARE
1.REMOVE BRACKETS ......THIS YOU HAVE DONE.
2.TRANSFER ALL UNKNOWN TERMS TO ONE SIDE.... USUALLY L.H.S. AND KNOWN TERMS TO ONE SIDE ...USUALLY R.H.S.
3.THIS TRANSFER YOU CAN THINK OF AND DO AS FOLLOWS....
EVERY TERM HAS A + OR - SIGN INFRONT OF IT.WHEN YOU TRANSFER THAT TERM FROM ONE SIDE TO ANOTHER ,+ BECOMES - AND - BECOMES PLUS.
THIS IS BECAUSE...AS YOU WERE TRYING TO SAY 'ADD OR SUBTRACT FROM BOTH SIDES'
AN EQUATION IS UNALTERED IF WE ADD OR SUBTRACT SAME QUANTITY FROM BOTH SIDES OF THE EQN.SO IF WE WANT TO REMOVE +4X ABOVE FROM R.H.S.WE HAVE TO ADD -4X TO IT.SO TO BALANCE IT AND KEEP THE VALIDITY OF THE EQN.UNALTERED,WE ADD -4X TO L.H.S.ALSO.THEN ON R.H.S ,WE GET +4X-4X=0...THAT IS +4X DIAPPEARS AS WE WANTED.BUT -4X REMAINS ON L.H.S.....THIS IS THE FINAL EFFECT OF ADD AND SUBTRACT WHICH IS SUMMARISED FIRST ABOVE.
HENCE WE GET HERE
-9X-4X = 14-2-3
4.NOW SIMPLIFY BOTH SIDES COMBINING ALL UNKNOWNS ON L.H.S AND ALL KNOWNS ON R.H.S
WE GET HERE
-13X=9
5.NOW TRANSFER THE COEFFICIENT OF UNKNOWN ON L.H.S.TO R.H.S.
APPLYING THE SAME RULE & LOGIC MENTIONED ABOVE,HERE * WILL BECOME / AND / BECOMES *
HERE WE GET
X = -9/13....
THIS IS THE ANSWER.
6.IF YOU HAVE TIME AND WOULD LIKE TO DO....CHECK BACK THE ANSWER FOR CONFIRMATION BY SUBSTITUTION..
3{1-3*(-9/13)}+2=4*(-9/13)+ 14
3{1+27/13)+2=-36/13 +14
3*40/13 +2=(-36+14*13)/13
120/13 +2= 146/13
146/13 = 146/13.......OK


You may edit the question. Maybe convert formulae

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SEE MY COMMENTS BELOW
I really don't understand this problem. Can you please help me?
3(1-3x)+2=4x+14........OK
3-9x+2=4x+14................GOOD
so then add or subtract from both sides?
3-9x+9x+2=13+14......HERE YOU GOT CONFUSED..THE STEPS INVOLVED ARE
1.REMOVE BRACKETS ......THIS YOU HAVE DONE.
2.TRANSFER ALL UNKNOWN TERMS TO ONE SIDE.... USUALLY L.H.S. AND KNOWN TERMS TO ONE SIDE ...USUALLY R.H.S.
3.THIS TRANSFER YOU CAN THINK OF AND DO AS FOLLOWS....
EVERY TERM HAS A + OR - SIGN INFRONT OF IT.WHEN YOU TRANSFER THAT TERM FROM ONE SIDE TO ANOTHER ,+ BECOMES - AND - BECOMES PLUS.
THIS IS BECAUSE...AS YOU WERE TRYING TO SAY 'ADD OR SUBTRACT FROM BOTH SIDES'
AN EQUATION IS UNALTERED IF WE ADD OR SUBTRACT SAME QUANTITY FROM BOTH SIDES OF THE EQN.SO IF WE WANT TO REMOVE +4X ABOVE FROM R.H.S.WE HAVE TO ADD -4X TO IT.SO TO BALANCE IT AND KEEP THE VALIDITY OF THE EQN.UNALTERED,WE ADD -4X TO L.H.S.ALSO.THEN ON R.H.S ,WE GET +4X-4X=0...THAT IS +4X DIAPPEARS AS WE WANTED.BUT -4X REMAINS ON L.H.S.....THIS IS THE FINAL EFFECT OF ADD AND SUBTRACT WHICH IS SUMMARISED FIRST ABOVE.
HENCE WE GET HERE
-9X-4X = 14-2-3
4.NOW SIMPLIFY BOTH SIDES COMBINING ALL UNKNOWNS ON L.H.S AND ALL KNOWNS ON R.H.S
WE GET HERE
-13X=9
5.NOW TRANSFER THE COEFFICIENT OF UNKNOWN ON L.H.S.TO R.H.S.
APPLYING THE SAME RULE & LOGIC MENTIONED ABOVE,HERE * WILL BECOME / AND / BECOMES *
HERE WE GET
X = -9/13....
THIS IS THE ANSWER.
6.IF YOU HAVE TIME AND WOULD LIKE TO DO....CHECK BACK THE ANSWER FOR CONFIRMATION BY SUBSTITUTION..
3{1-3*(-9/13)}+2=4*(-9/13)+ 14
3{1+27/13)+2=-36/13 +14
3*40/13 +2=(-36+14*13)/13
120/13 +2= 146/13
146/13 = 146/13.......OK


3-2=13x+14?
that's not right is it but then what is right and what would you even do next?

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Solve and find exact solutions.
4^(2x-1)=1/2 =2^(-1)
[2^2]^(2x-1)=2^(-1)
2^{2(2x-1)}=2^(-1)
2(2x-1)=-1
4x-2=-1
4x=1
x=1/4
book answer: 1/4

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24. The owner of a highway construction company knows that the maximum load (L SAY)a cylindrical column of circular cross section can support varies directly as the fourth power of the diameter(D SAY) and inversely as the square of the height(H SAY). If a column 2 feet in diameter and 10 feet high supports up to 6 tons, how much of a load can a column 3 feet in diameter and 14 feet high support?
L = K*D^4*1/H^2...WHERE K IS A CONSTANT
CASE 1..
6=K*2^4/10^2=16K/100
K=6*100/16=75/2=37.5
CASE 2
L=37.5*3^4/14^2 = 37.5*81/196 = 15.5 T.

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One angle of a triangle has a measure of 70 degrees. The other two angles have degree measures in a ratio of 5 to 6. What is the sum of the measures, in degrees, of the two largest angles?
I first tried 50 +60=110+70=180.
ONE ANGLE =70
HENCE SUM OF OTHER 2 ANGLES = 180-70=110...YOU ARE CORRECT!WHY DOUBT?
SINCE ANGLES ARE IN 5:6 RATIO...TAKE THEM AS 5X,6X
5X+6X=11X=110...HENCE X=110/11=10
SO THE 2 ANGLES ARE 50,60..
Then I realized that 180 is the sum of all of the angles in a triangle, so it couldn't be right.
WHY IT IS RIGHT ..SUM OF 3 ANGLES 50,60,70 IS 180
Thanks if you can help!

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SEE THE FOLLOWING EXAMPLE AND TRY FOR SLOPE AND A POINT PROBLEM
FOR THE 2 POINT PROBLEM SOLUTION IS AS FOLLOWS.
------------------------------------------------------------------
a) Write the slope-y-intercept equation form of the line which passes through the 2 points: X1,Y1=(-3,1), (0,6)=X2,Y2
FORMULA IS
Y= (Y2-Y1)(X-X1)/(X2-X1)+Y1
Y=(6-1)(X+3)/((0+3) + 1
Y=5(X+3)/3 +1
Y=(5X/3)+ 6.................M=5/3=SLOPE.........6=Y INTERCEPT
----------------------------------------------------------------
use the given conditions to write an equation in the point-slope form and the slope-intercept form...thank you
slope=-2 = M , passing through (0, -3) = X1,Y1
FORMULA IS
Y=M(X-X1)+Y1
Y=-2(X-0)+(-3)= -2X-3

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SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
-------------------------------
PLEASE HELP ME WITH THIS PROBLEM;
For f(x)=1 over x62-2x-8 find the intervals where f(x)<0.
HOPE YOU MEAN
Y = F(X)=1/ X^2-2X-8.....ASSUMING SO
Y= 1/X^2-4X+2X-8=1/[X(X-4)+2(X-4)]=1/[(X+2)(X-4)]
THE ROOTS ARE X=-2 AND X=4
FOR F(X) TO BE <0,X SHOULD LIE BETWEEN THE 2 ROOTS ..THAT IS ..BETWEEN -2 & 4 .
-2 -------------------------------------------------------
ANSWERS ARE
a) 2y^2+5y-63<0 ...(2Y-9)(Y+7)<0....Y LIES BETWEEN -7 AND 9/2...(-7,9/2)
b) 4y^2+10y-14<0 ...2(2Y+7)(Y-1)<0...Y LIES BETWEEN -7/2 AND 1 ...(-7/2,1)
c) x^2+x-30<0....(X+6)(X-5)<0....X LIES BETWEEN -6 AND 5......(-6,5)

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SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
-------------------------------
PLEASE HELP ME WITH THIS PROBLEM;
For f(x)=1 over x62-2x-8 find the intervals where f(x)<0.
HOPE YOU MEAN
Y = F(X)=1/ X^2-2X-8.....ASSUMING SO
Y= 1/X^2-4X+2X-8=1/[X(X-4)+2(X-4)]=1/[(X+2)(X-4)]
THE ROOTS ARE X=-2 AND X=4
FOR F(X) TO BE <0,X SHOULD LIE BETWEEN THE 2 ROOTS ..THAT IS ..BETWEEN -2 & 4 .
-2 -------------------------------------------------------
ANSWERS ARE
a) 2y^2+5y-63<0 ...(2Y-9)(Y+7)<0....Y LIES BETWEEN -7 AND 9/2...(-7,9/2)
b) 4y^2+10y-14<0 ...2(2Y+7)(Y-1)<0...Y LIES BETWEEN -7/2 AND 1 ...(-7/2,1)
c) x^2+x-30<0....(X+6)(X-5)<0....X LIES BETWEEN -6 AND 5......(-6,5)
You may edit the question. M

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PLEASE HELP ME WITH THIS PROBLEM;
For f(x)=1 over x62-2x-8 find the intervals where f(x)<0.
HOPE YOU MEAN
Y = F(X)= X^2-2X-8.....ASSUMING SO
Y= X^2-4X+2X-8=X(X-4)+2(X-4)=(X+2)(X-4)
THE ROOTS ARE X=-2 AND X=4
FOR F(X) TO BE <0,X SHOULD LIE BETWEEN THE 2 ROOTS ..THAT IS ..BETWEEN -2 & 4 .
-2

---

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Write an expression for the area of an isosceles triangle and simplify. The base is 1/2Y and the height is 2/5y + 4
AREA=BASE*HT/2=(Y/2)(2Y/5 +4)/2=Y(2Y+5)/(2*2)=(2Y^2+5Y)/4

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SEE THE FOLLOWING EXAMPLE AND TRY FOR SLOPE AND A POINT PROBLEM
FOR THE 2 POINT PROBLEM SOLUTION IS AS FOLLOWS.
------------------------------------------------------------------
a) Write the slope-y-intercept equation form of the line which passes through the 2 points: X1,Y1=(-3,1), (0,6)=X2,Y2
FORMULA IS
Y= (Y2-Y1)(X-X1)/(X2-X1)+Y1
Y=(6-1)(X+3)/((0+3) + 1
Y=5(X+3)/3 +1
Y=(5X/3)+ 6.................M=5/3=SLOPE.........6=Y INTERCEPT
----------------------------------------------------------------
use the given conditions to write an equation in the point-slope form and the slope-intercept form...thank you
slope=-2 = M , passing through (0, -3) = X1,Y1
FORMULA IS
Y=M(X-X1)+Y1
Y=-2(X-0)+(-3)= -2X-3

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SEE THE FOLLOWING AND TRY
use the given conditions to write an equation in the point-slope form and the slope-intercept form...thank you
slope=-2 = M , passing through (0, -3) = X1,Y1
FORMULA IS
Y=M(X-X1)+Y1
Y=-2(X-0)+(-3)= -2X-3

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SEE THE FOLLOWING AND TRY
------------------------------------------
use the given conditions to write an equation in the point-slope form and the slope-intercept form...thank you
slope=-2 = M , passing through (0, -3) = X1,Y1
FORMULA IS
Y=M(X-X1)+Y1
Y=-2(X-0)+(-3)= -2X-3

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use the given conditions to write an equation in the point-slope form and the slope-intercept form...thank you
slope=-2 = M , passing through (0, -3) = X1,Y1
FORMULA IS
Y=M(X-X1)+Y1
Y=-2(X-0)+(-3)= -2X-3

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SEE THE FOLLOWING EXAMPLE AND TRY
------------------------------------------------------------------
PLEASE HELP ASAP: 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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Given a right circular cone leaking water (original dimensions of r=6m h=8m
1. Rewrite the formula for V in terms of the radius only.
WHERE IS THE FORMULA?OK LET ME ASSUME
V=PI*R^2*H/3..IF YOU WANT ONLY INTERMS OF RADIUS WE HAVE TO BE GIVEN SOME RELATION BETWEEN HEIGHT AND RADIUS..AGAIN LET ME ASSUME SLANT HEIGHT IS GIVEN AS 10.(THIS CORRESPONDS WITH H=8 FOR R=6,SINCE R^2+H^2=6^2+8^2=36+64=10^2=100).THEN
R^2+H^2=10^2
H= SQRT (100-R^2)
HENCE
V=PI*R^2/3
2. Rewrite the formula for V in terms of height only.
SAME COMMENTS AS ABOVE APPLY..MAKING SAME ASSUMPTIONS
R^2=100-H^2
V=PI(100-H^2)*H/3

3. If the volume is decreasing at a rate of 81/32 (pie) m^3/min. how fast is the height dropping when h=3m?
V=PI(100-H^2)H/3=(100H-H^3)PI/3
DV/DT = (100-3H^2)(PI/3)DH/DT
DV/DT = 81PI/32
H=3
DH/DT = {(81PI/32)}/{(100-3*3^2)PI/3}=(81*3)/(32*73)=243/2336=0.104

Please help...i've search so many places for help. Lillian

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Help!!!!!!!!!!!!! My boyfriend (majoring in engineering) says if I can get this solved for P (Proposal) that is a hint as to the number of weeks until he proposes! I think I know the answer but can you help me?
p2 + 3p + 2
____________ = 3/2 + 3p -3
p + 1

OK THIS IS MY BIT TO YOUR HAPPINESS WITH YOUR BOY FRIEND THOUGH IT LOOKS A STRANGE TEST!!
p2 + 3p + 2
____________ = 3/2 + 3p -3 = 1.5+3P-3....MULTIPLYING BOTH SIDES WITH P+1,WE GET
p + 1
(THIS TAKES IN TO ACCOUNT THAT P+1=0...OR...P=-1 .IS NOT POSSIBLE.THAT IS HE WOULD HAVE ALREADY PROPOSED TO YOU A WEEK AGO!!!HE HAS NOT DONE SO AS PER YOU!!..SO..IT IS OK)
P^2+3P+2=(P+1)(1.5+3P-3)=1.5P+3P^2-3P+1.5+3P-3=1.5P+3P^2-1.5
1.5P+3P^2-1.5-P^2-3P-2=0
2P^2-1.5P-3.5=0............MULTIPLYING WITH 2 TO REMOVE FRACTIONS..
4P^2-3P-7=0
4P^2-7P+4P-7=0
P(4P-7)+1(4P-7)=0
(4P-7)(P+1)=0....BUT P+1 IS NOT ZERO AS WE ALREADY CONCLUDED ABOVE!
4P-7=0
4P=7
P=7/4=1.75 WEEKS ! OH GOSH! IT IS ONLY (OR TOO FAR FOR YOU?) ANOTHER 12 DAYS AWAY!!THAT IS OVER NEXT WEEK END!!DONT FORGET TO GIVE ME A TREAT THEN!!!!!!!!!!!!!!I AM WAITING!!

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I am stuck on correct answer for:
'
If x + y = 2...................1
and x - y = 6..........................2
EQN.I - EQN.2
X+Y - (X-Y) = 2-6= -4
X+Y - X + Y =-4
2Y = -4
then y = -4/2 =-2
I am not using any specific book. I am just practicing..

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Find four solutions for the equation 3x + 5y = 15.
No book. Thank you very much for your time!!
GIVE ANY VALUE FOR X AND FIND Y
FOR EX..PUT X=0...SO 3*0+5Y=15...OR..5Y=15..OR...Y=15/5=3
PUT X = 5.....SO 3*5+5Y=15...OR...5Y=0....OR....Y=0
LIKE THIS YOU CAN FIND NY NUMBER OF SOLUTIONS