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120/12=10 ft.
42*12=504 inches.

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4+X=2X OR 4=2X-X OR 4=X

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SUE HAS 4 STAMPS MARY HAS 4TIMES AS MANY OR 4*4=16 STAMPS JOHN HAS 4 TIMES AS MANY STAMPS AS MARY 4*16=54 IS THE NUMBER OF JOHN'S STAMPS.

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L*W=74 & L=4W+1 OR (4W+1)W=74 OR 4W^2+W-74=0 OR USING THE QUADRATIC FORMULA WE GET W=4.178 & -4.428 SO THE WIDTH IS 4.178 & THE LENGTH IS 4*4.178+1 OR 17.712
PROOF 4.178*17.712=74 OR 74=74

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THE EQUATION FOR THE SLOPE INTERCEPT IS Y=mX+b OR IN THIS CASE WE HAVE
2X+3Y+32=0 OR 3Y=-2X-32 OR Y=-2X/3-32/3 THUS THE SLOPE (m)IS -2/3 &NTHE Y INTERCEPT (b) IS -32/3.

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Y=mX+B OR IN THIS CASE WE HAVE Y=-X/2+b & -3=-2/2+b OR -3=-1+b OR b=-2 THUS
WE HAVE Y=-X/2-2
YOU WERE ON THE RIGHT TRACK.

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YOU HAVE (X1,Y1) & (X2,Y2) OR (-6,-5) & (4,1) SO ALL WE NEED TO DO IS FIND
Y2-Y1 & X2-X1 SO WE HAVE FOR THE Y VALUES (1-[-5]) OR 1+5=6 FOR THE Y VALUE
FOR THE X VALUE WE HAVE (4-[-6]) OR 4+6=10
NOW WE DIVIDE THE Y VALUE BY THE X VALUE THUS 6/10 OR .6 FOR THE SLOPE OF A LINE PASSING THROUGH THE POINTS (-6,-5) & (4,1)

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7 3/10-2 3/4 OR 7.3-2.75=4.55 MILES TO BE PAVED. DECIMAL SOLUTION
FRACTION SOLUTION 7 3/10=73/10 & 2 3/4=11/4 SO 73/10-11/4 REQUIRES FINDING A COMMON DENOMINATOR THAT 4 & 10 is evenly divisable by. finding the factors or 4 & 10 helps find the lowest commmon denominator thus 2*2=4 & 2*5=10 both denominators have a common factor of 2 therefotre we only need one of the 2s but we need the other 2 & the 5 thus 2*2*5=20 which is the lowest common denominator. you could have just multiplied 4*10=40 but this would not be the lowest common denominator but a common denominator that works.
now change each fraction into a fraction with a denominator of 20 by multiplying the 73/10 by 2/2 we get 146/20 & multiplying 11/4 by 5/5 we get 55/20. note that 2/2 & 5/5 is just another way of expressing the number 1 thus any number multiplied by 1 is the same number just in a different form.
now we can subtract 55/20 from 146/20 thus (146-55)/20 or 91/20 or 4 11/20 & 11/20=.55 so we still have an answer of 4.55

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Y=X^2-2X-15 OR Y=(X-5)(X+3) OR X=5 & X=-3

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3^5/3^2 OR 3*3*3*3*3/3*3 OR 3*3*3 OR 3^3 OR 27 ALSO 3^(5-2)=3^3

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2X^2-4X-11=0 OR USING THE QUADRATIC EQUATION WE GET X=3.5495 & X=-1.5495

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3X-5(3X-7)=2(X+9)+45 OR 3X-15X+35=2X+18+45 OR -12X+35=2X+63 OR -14X=28 OR
14X=-28 OR X=-28/14 OR X=-2

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3X+5Y=54 & 5Y=3X-2 THUS 3X=54-5Y THUS 5Y=(54-5Y)-2 OR 5Y=54-5Y-2 OR 10Y=52 OR Y=52/10 OR Y=5.2 THEN 5*5.2=3X-2 OR 26=3X-2 OR 3X=28 OR X=28/3

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THE DIAGONAL OF A RECTANGLE FORMS A RIGHT TRIANGLE THUS WE HAVE A WAY TO FIND THE DIAGONAL OR THE HYPOTENUSE (LONG SIDE) OF THE TRIANGLE.
IT'S CALLED THE PYTHAGORIAN THEROM A^2+B^2=C^2. WE HAVE THE A & B VALUES
(30 & 40) NOW WE SOLVE FOR C THE LONG SIDE. 30^2+40^2=C^2 OR 900+1600=C^2 OR
C^2=2500 OR C=SQRT2500 OR C=50 INCHES FOR THE DIAGONAL.

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07.80
78.00
00.78
----
86.58
SECRET IS TO LINE UP ALL THE DECIMALS, FILL IN THE MISSING ZEROS AND JUST ADD.

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2L+2W=300 OR 2(L+W)=300 OR (L+W)=300/2 OR L+W=150 THUS TO MAXIMIZE THE AREA L SHOULD = W SO L&W ARE 75 FEET.
THE MAXIMUM AREA WITH THE MINIMUM PERIMETER IS A SQUARE (ALL SIDES ARE EQUAL).

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N+8

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FIRST GET SOME GRAPH PAPER. NOW DRAW A HORIZONTAL LINE ACROSS THE MIDDLE OF THE PAGE. THIS IS THE X AXIS. NOW DRAW A VERTICLE LINE DOWN THE MIDDLE OF THE PAGE. THIS IS THE Y AXIS. WHERE THESE TWO LINES INTERSECT (CROSS) IS THE (X,Y) POINTS (0,0). NOW MARK EACH SQUARE TO THE RIGHT OF THE (0,0) INTERSECTION ON THE X AXIS 1,2,3,4,5,6,7,8,9 & 10. NOW MARK THE LEFT SIDE OF THIS X LINE STARTING AT (0,0) -1,-2,-3,-4,-5,-6,-7,-8,-9 & -10. NOW MARK THE SQUARES ABOVE THE INTERSECTION OF 0,0 ON THE Y AXIS 1,2,3,4,5,6,7,8,9,10,11,12,13 & 14. NOW MARK ALL THE SQUARES BELOW THE (0,0) INTERSECTION ON THE Y AXIS -1,-2,-3,-4,
-5,-6,-7,-8,-9, & -10.
YOU NOW HAVE A GRAPH FOR PLOTTING THE LINES REPRESENTED BY THE EQUATIONS YOUR TRYING TO GRAPH IN THE FORM Y=mX+b WHERE m=THE SLOPE & b=THE Y INTERCEPT.
2X+Y=13 MUST NOW BE TRANFORMED TO THE STANDARD LINE FORMULA THUS Y=-2X+13.
THIS EQUATION HAS A SLOPE (m) OF -2 & a y intercept of 13 (0,13). NOW TO FIND THE X INTERCEPT TO HAVE TWO POINTS TO DRAW A THE LINE THROUGH WE SET Y=0 THUS
0=-2X+13 OR -2X=-13 OR 2X=13 OR X=13/2 OR X=6.5 (6.5,0).
TO PLOT THE POINT (0,13) START AT THE INTRSECTION (0,0) COUNT UP 13 SQUARES AND MARK THIS POINT(0,13). NOW STARTING AGAIN AT THE INTERSECTION (0,0) AND COUNT 6 AND 1/2 SQUARES TO THE RIGHT OF THIS POINT AND MARK THIS POINT (6.5,0). NOW DRAW A LINE THROUGH THESE 2 POINTS. YOU NOW HAVE A LINE REPRESENTED BY THE EQUATION 2X+Y=13 WITH A SLOPE OF -2 AND A Y INTERCEPT OF 13.
NOW YOU TRY THE NEXT EQUATION USING THE SAME GRAPH.
5X-2Y=1 OR -2Y=-5X+1 OR 2Y=5X-1 OR Y=5/2X-1/2 WHICH HAS A SLOPE OF 5/2 OR 2.5 AND A Y INTERCEPT OF -1/2 OR .5 (0,-.5) NOW SET Y=0 WE GET 5X-2*0=1 OR 5X=1 OR X=1/5 OR .2 (.2,0)

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FILL THE 5 QUART PITCHER THEN FILL THE 3 QUART PITCHER. YOU NOW HAVE 2 QUARTS LEFT IN THE 5 QUART PITCHER. EMPTY THE 3 QUART PITCHER & POUR THE 2 QUARTS LEFT IN THE 5 QUART PIRCHER INTO THE 3 QUART PITCHER. NOW REFILL THE 5 QUART PITCHER. NOW FILL THE 3 QUART PITCHER WITH 1 QUART FROM THE 5 QUART PITCHER. YOU NOW HAVE EXACTLY 4 QUARTS IN THE 5 QUART PITCHER.

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8-9-T

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12 CLASSES @ 10.00=120.00 PLUS 1 MAT FOR 19.95 FOR A TOTAL OF 120.00+19.95=139.95 THUS WE HAVE 12 CLASSES + 1 MAT

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X^2-24=2X OR X^2+2X-24=0 OR (X-6)(X+4)=0 OR X=6 & X=-4
2X^2+X=3 OR 2X^2+X-3=0 OR (2X+3)(X-1)=0 OR 2X+3=0 OR 2X=-3 OR X=-3/2 & X-1=0 OR X=1

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X+Y=2 & X*Y=-63 OR X=2-Y THEN (2-Y)Y=-63 OR 2Y-Y^2=-63 OR Y^2-2Y-63 OR
(Y-9)(Y+7)=0 THEN Y=9 & Y=-7 THEN X=-7 & X=9
X+Y=-2 & X*Y=-80 OR X=-2-Y THEN (-2-Y)Y=-80 OR -2Y-Y^2=-80 OR Y^2+2Y-80=0 OR
(Y+10)(Y-8)=0 OR Y=-10 & Y=8 THEN X=8 & X=-10
X+Y=4 & X*Y=-60 OR X=4-Y THEN (4-Y)Y=-60 OR 4Y-Y^2=-60 OR Y^2-4Y-60=0 OR
(Y-10)(Y+6)=0 OR Y=10 & Y=-6 THEN X=-6 & X=10
X+Y=11 & X*Y=28 OR X=11-Y THEN (11-Y)Y=28 OR 11Y-Y^2=28 OR Y^2-11+28=0 OR
(Y-7)(Y-4)=0 OR Y=7 & Y=4 THEN X=4 & X=7

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14 8 FOOT TIES CAN MAKE A RECTANGLE OF L+W=14/2 OR L+W=7 OR MAX VALUES OF X&Y ARE 3 & 4 THUS A 3 BY 4 TIE RECTANGLE IS THE MAX SIZE OR (3*8)(4*8)=24*32=768 SQFT.
PROOF DRAW RECTANGLE USING 8 FOOT LINES FOR EACH TIE AND YOU'LL FIND A 3 BY 4
RECTANGLE IS THE LARGEST AREA.

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(X+5)+4 OR X+5+4 OR X+9

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X*Y=AREA OF ORIGINAL RECTANGLE SO IF WE DOUBLE BOTH DIMENTIONS WE HAVE
2X*2Y= NEW AREA OR 4X*Y THUS THE ORIGINAL AREA IS INCREASED BY A FACTOR OF 4

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D+Q=40 & 10D+25Q=490 THEN D=40-Q THUS 10(40-Q)+25Q=490 OR 400-10Q+25Q=490 OR
15Q=90 OR Q=90/15 OR Q=6 THEN D+6=40 OR D=40-6 OR D=34
PROOF 10*34+25*6=340+150=490 THUS THERE ARE 6 QUARTERS & 34 DIMES IN HIS JAR.

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PICTURE A GRAPH WITH AN X AXIS OF 1000 METERS AND A Y AXIS OF AT LEAST 3 METERS.
THE FIRST POINT ON THE LINE FOR THE AQUEDUCT IS (0,0) AND THE SECOND POINT IS (1000,3) THUS YOU HAVE A SLOPE OF (0-3)/(0-1000) OR -3/-1000 OR .003 FOR THE SLOPE OF THE AQUADUCT.

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X/4=-1 OR X=-4

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40/5=8 FEET PER SECOND OR 8*3600=28,800 FEET PER HOUR OR 28,800/5,280=5.4545 MPH.

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3X+2Y=9 NOW MULTIPLY THE SECOND EQUATION BY 3 TO GET
-3X+9Y=24 NOW ADD THESE TWO EQUATIONS
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+11Y=33 OR Y=33/11 OR Y=3 THEN 3X+2*3=9 OR 3X+6=9 OR 3X=3 OR X=1
PROOF -1*1+3*3=8 OR -1+9=8 OR 8=8