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write an equation for the line with slope 1/3 and y-intercept 4.
write an equation for the line with slope 3 going through the point (1,-2)
write an equation for the line going through the points (3,2) and (-2,4) 1 solutions
Answer 90746 by checkley71(8403) on 2008-01-30 15:20:14 (Show Source):
You can put this solution on YOUR website!Y=mX+b IS THE LINE EQUATION.
Y=X/3+4 ANSWER TO THE FIRST PROBLEM.
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-2=3*1+b
-2=3+b
b=-2-3
b=-5 OR
Y=3X-5 ANSWER.
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SLOPE=(4-2)/(-2-3)=2/-5=-2/5
2=3*-2/5+b
2=-6/5+b
b=2+6/5
b=(10+6)/5
b=16/5
Y=-2X/5+16/5 ANSWER.
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Problems-with-consecutive-odd-even-integers/123685: Find three consecutive even integers such that the sum of the twice the first, six times the second and three times the third is 133. 1 solutions
Answer 90738 by checkley71(8403) on 2008-01-30 15:01:25 (Show Source):
You can put this solution on YOUR website!x, x+2, x+4
2x+6(x+2)+3(x+4)=133
2x+6x+12+3x+12=133
11x=133-24
11x=109
x=109/11
x=9.9090909 not an even number.
---------------------------------------------------------------
ARE YOU SURE YOU HAVE THE PROPER WORDING IN THIS PROBLEM??????
THREE EVEN NUMBERS WHEN ADDED ADD OR MULTIPLIED MUST BE AN EVEN NUMBER 133 IS NOT AN EVEN NUMBER.
IF THESE ARE NOT EVEN NINTERGERS THEN:
--------------------------------------------------------------
x, x+1, x+2 are the integers.
2x+6(x+1)+3(x+2)=133
2x+6x+6+3x+6=133
11x=133-12
11x=121
x=121
11x=11 answer for the first integer.
11=1=12
11+2=13
proof:
2*11+6*12+3*13=133
22+72+39=133
133=133
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Miscellaneous_Word_Problems/123683: i have a total of 47,000 products shipped out over the past 3 years. out of that 47,000 i have had a total of 282 defects. what would be the average ratio of number of defects per 1,000 products shipped over the past 3 years? 1 solutions
Answer 90735 by checkley71(8403) on 2008-01-30 14:54:20 (Show Source):
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Quadratic_Equations/123616: Local building codes require that a rectangular factory building must be surrounded by a parking lot of uniform width. The total area occupied by the factory and the parking lot must be double the factory area. If the factory is to be 80m wide and 120m long. How wide must the parking lot be? HELP:) 1 solutions
Answer 90673 by checkley71(8403) on 2008-01-30 09:38:13 (Show Source):
You can put this solution on YOUR website!80*120=9,600 m^2 is the area of the factory.
therefore the total area of the factory & the parking lot must be: 2*9,600=19,200.
the formula for this problem is:
(80+2x)(120+2x)=19,200
9,600+240x+160x+4x^2=19,200
4x^2+400x+9,600-19,200=0
4x^2+400x-9,600=0
4(x^2+100-2,400)=0
4(x-20)(x+120)=0
x-20=0
x=20 answer for the width of the parking lot.
proof
(80+40)(120+40)=19,200
120*160=19,200
19,200=19,200
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Linear-equations/123620: I am suppose to draw the graph of each line using its y-intercept and its slope..I have gone over the chapter and am just confused on how exactly i find that using this equation. I am so ready to tear my hair out...can someone help?
thank you 1 solutions
Answer 90671 by checkley71(8403) on 2008-01-30 09:29:23 (Show Source):
You can put this solution on YOUR website!THE STANDARD LINE FORMULA IS:
Y=mX+b WHERE m=THE SLOPE & b=Y INTERCEPT.
NOW CONVERT YOUR EQUATION TO THIS FORM THUS:
4X+3Y=9
3Y=-4X+9
Y=-4X/3+9/3
Y=-4X/3+3 ANSWER
THIS REPRESENTS A LINE WITH A SLOPE OF -4/3 & THE Y INTERCEPT IS 3.
 (graph 300x300 pixels, x from -5 to 5, y from -5 to 5, y = -4x/3 +3).
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Evaluation_Word_Problems/123423:
PROBLEM SOLVING: hOW TO SoLve
Suppose your class recieves a $1084 for seeling 205 packages of greeting cards and gift wrap. If tyhe girft wrap is 4 dollars a package and the greeting cards are 10 a package, find the number of each type of package sold. 1 solutions
Answer 90594 by checkley71(8403) on 2008-01-29 17:26:15 (Show Source):
You can put this solution on YOUR website!C+W=205 OR C=205-W
4C+10W=1084
4(205-W)+10W=1084
820-4W+10W=1084
6W=1084-820
6W=264
W=264/6
W=44 NUMBER OF WRAP ITEMS.
C+44=205
C=205-44
C=161 NUMBER OF CARDS SOLD.
PROOF
4*161+10*44=1084
644+440=1084
1084=1084
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Evaluation_Word_Problems/123424: Probleam solving: How To SoLve:
Last week, Rebas Reptile store recived 1975 dollars for selling 35 reptiles. Iguanas sell for $75 and corn snakes sell for $25 how many iguanas and how many corn snake did Reba sell 1 solutions
Answer 90589 by checkley71(8403) on 2008-01-29 17:12:49 (Show Source):
You can put this solution on YOUR website!C+I=35 OR C=35-I
75C+25I=1975
75(35-I)+25I=1975
2625-75I+25I=1975
-50I=1975-2625
-50I=-650
I=-650/-50
I=13 NUMBER OF IGUANAS.
35-13=22 NUMBER OF CORN SNAKES.
PROOF.
75*22+25*13=1975
1650+325=1975
1975=1975
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Evaluation_Word_Problems/123414: PROBLEM SOLVING: HOW TO SOLV???
A weekend at the Beach Bay Hotel in Florida incledues 2 nights and 4 meals at the cost of $194. A week cost $650 for 7 nights and 10 meals. Fiind the cost for one night then the cost of one meal. 1 solutions
Answer 90587 by checkley71(8403) on 2008-01-29 17:04:49 (Show Source):
You can put this solution on YOUR website!2N+4M=194 NOW MULTIPLY THIS EQUATION BY -2.5 THEN ADD THEM.
7N+10M=650
-5N-10M=-485
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2N=650-485
2N=165
N=165/2
N=$82.50 IS THE NIGHT RATE.
2*82.5+4M=194
165+4M=194
4M=194-165
4M=29
M=29/4
M=$7.25 IS THE MEAL RATE.
PROOF.
7*82.50+10*7.25=650
577.5+72.5=650
650=850
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Mixture_Word_Problems/123412:
A seedman has seeds worth $0.50 per pound mixed with seeds worth $0.80 per pound. If he wishes to make a mixture of 30 lbs to sell at $0.75 per pound, how many pounds of each should he use?
5 lbs of $0.50
25 lbs of $0.80 1 solutions
Answer 90585 by checkley71(8403) on 2008-01-29 16:52:28 (Show Source):
You can put this solution on YOUR website!.8X+.5(30-X)=.75*30
.8X+15-.5X=22.5
.3X=22.5-15
.3X=7.5
X=7.5/.3
X=25 POUNDS OF $.80 MIX IS USED.
30-25=5 POUNDS OF $.50 MIX IS USED.
PROOF.
.8*25+.5*5=22.5
20+2.5=22.5
22.5=22.5
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Mixture_Word_Problems/123407: An alloy of tin is 16% tin. How much tin must be added to make 820 lbs of alloy that is 18% tin? Answer to the nearest tenth. 1 solutions
Answer 90583 by checkley71(8403) on 2008-01-29 16:42:57 (Show Source):
You can put this solution on YOUR website!.16X+(820-X)=.18*820
.16X+820-X=147.6
-.84X=147.6-820
-.84=-672.4
X=-672.4/-.84
X=800.5 ANSWER FOR THE AMOUNT OF 18% TIN.
820-800=20 LBS. OF PURE TIN IS USED.
PROOF.
.16*800.5+19.5=.18*820
128.08+19.5=147.6
147.6~147.6
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Mixture_Word_Problems/123383: This problem has me perplexed. Help me please!
How many pounds of a chicken feed that sells for $5 per pound must be mixed with
400 pounds of a feed that sells for $8 per pound to make a chicken feed that sells for $7.50? 1 solutions
Answer 90574 by checkley71(8403) on 2008-01-29 16:02:25 (Show Source):
You can put this solution on YOUR website!5X+400*8=7.5(400+X)
5X+3200=3000+7.5X
5X-7.5X=3000-3200
-2.5X=-200
X=-200/-2.5
X=80 POUNDS OF CHICKEN IS NEEDED.
PROOF.
5*80+3200=7.5(480)
400+3200=3600
3600=3600
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Quadratic_Equations/123394: I am having a very hard time trying to figure this out. Can you please help me?
I am sorry, I forgot to put in there what it is I need to do with the problem.
Find the axis of symmetry.
y = x squared - 5x + 3 1 solutions
Answer 90572 by checkley71(8403) on 2008-01-29 15:57:32 (Show Source):
You can put this solution on YOUR website!Y=X^2-5X+3
 (graph 300x200 pixels, x from -6 to 5, y from -10 to 10, y = x^2-5x +3).
IT LOOKS LIKE X=2.5 IS THE AXIS OF SYMMETRY.
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Linear-systems/123287: I have a problem (8th grade) that I can't get started on. "a group of friends
eat at a restaurant and agreed to split the $6.00 bill equally. Two men left without paying and the remaining others were assessed .25 more. How many men in the party originally?"
It seems like a linear equation that has 2 variables. How much did each spend and how many men. I can't put it into equation form. 1 solutions
Answer 90470 by checkley71(8403) on 2008-01-28 21:49:03 (Show Source):
You can put this solution on YOUR website!6/X=[6/(X-2)]-.25
6/X=[6+.25X+.5]/(X-2)
6(X-2)=X(6-.25X+.5)
6X-12=6X-.25X^2+.5X
-.25X^2+.5X+12=0
.25X^2-.5X-12=0
(.25X-2)(X+6)=0
.25X-2=0
.25X=2
X=2/.25
X=8 ANSWER FOR THE NUMBER OF MEN BEFORE 2 LEFT.
PROOF
6/8=6/(8-2)-.25
.75=6/6-.25
.75=1-.25
.75=.75
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Quadratic_Equations/123260: A 4 by 6 inch photograph is to be matted such that the area of the mat is double the area of the photo. Assuming that the mat is the same width all around the picture, what is the width of the mat ? 1 solutions
Answer 90450 by checkley71(8403) on 2008-01-28 21:02:56 (Show Source):
You can put this solution on YOUR website!THE BEST WAY TO SOLVE THESE TYPE OF PROBLEMS IS TO DRAW A SKETCH ( NOT NECESSARY TO SCALE), ASIGN VALUES FOR THE LENGTH & WIDTH, THEN ADD THE MAT AREA & ASSIGN THE WIDTH=X. YOU'LL GET 4*6=24 FOR THE ORIGINAL PHOTO &
(4+2X)(6+2X)=2+24 OR TWICE THE ORIGINAL AREA.
DOES THIS HELP VISUALIZE THE PROBLEM AND PUT IT IN PERSPECTIVE???????
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2*4*6=(4+2X)(6+2X)
48=24+12X+8X+4X^2
4X^2+20X+24-48=0
4X^2+20X-24=0
4(X^2+5X-6)=0
4(X+6)(X-1)=0
X=1 ANSWER FOR THE WIDTH OF THE MAT.
PROOF
(4+2)(6+2)=48
6*8=48
48=48
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Travel_Word_Problems/123279: a family was driving 270 miles from miami to tampa.after 5 hours of driving they saw a sign that said tampa 45 miles.what was the average number of miles per hour they drove for the first five hours? 1 solutions
Answer 90445 by checkley71(8403) on 2008-01-28 20:57:28 (Show Source):
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Quadratic_Equations/123264: A rectangular lawn has dimensions 30 metres by 40 metres. A strip of uniform width is to be mowed around the perimeter.How wide should the strip be so that exactly half of the lawn gets mowed ? 1 solutions
Answer 90440 by checkley71(8403) on 2008-01-28 20:47:45 (Show Source):
You can put this solution on YOUR website!30*40=2(30-2X)(40-2X)
1200=2(1200-80X-60X+4X^2)
1200=2(1200-140X+4X^2)
1200=2400-280X+8X^2
8X^2-280X+2400-1200=0
8X^2-280X+1200
8(X^2-35X+150)
8(X-5)(X-30)=0
X-5=0
X=5 ANSWER FOR THE WIDTH OF THE STRIP MOWED TO REDUCE THE LAWN TO 50%.
PROOF
(30-2*5)(40-2*5)=1200/2
20*30=600
600=600
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