View Full Version : Help with simple math

Patmian212

Nov 21, 2005, 03:01 PM

Ok guys below is a series of question I got on a paper to help study for my test,now I am horrible at math can u solve them and tell me your steps, I just really dont get it:( . Meh I can get a+ in english but im failling maths lol.

1.a. on a set of axis draw accurate graphs of y=5x-x and y=5x-x^2 (i solved this one)

b. using part a give a geometrical interpretation of the equation 5-6=5x-x^2(WHAT????)

c. solve 5-x=5x-x^2(i dunno how to solve it)

2. Solve he system of equations(i dont know the method how to solve them, is it on a graph?)

a. y= x^2

y=x+6

b. y=x^2+1

y=5

c. y=-2x+4

y=x^2-4x+5

Test is in 2 days, hope I get it by then or else im screwed :.(

adroit

Nov 21, 2005, 03:10 PM

Ok guys below is a series of question I got on a paper to help study for my test,now I am horrible at math can u solve them and tell me your steps, I just really dont get it:( . Meh I can get a+ in english but im failling maths lol.

1.a. on a set of axis draw accurate graphs of y=5x-x and y=5x-x^2 (i solved this one)

b. using part a give a geometrical interpretation of the equation 5-6=5x-x^2(WHAT????)

c. solve 5-x=5x-x^2(i dunno how to solve it)

2. Solve he system of equations(i dont know the method how to solve them, is it on a graph?)

a. y= x^2

y=x+6

b. y=x^2+1

y=5

c. y=-2x+4

y=x^2-4x+5

Test is in 2 days, hope I get it by then or else im screwed :.(

I'm too lazy to graph out the first part. But here is the rest.

1c) 5-x=5x-x^2

x^2-5x-x+5 =0 - move everything to one side

x^2-6x+5 = 0

(x-3)(x-2) = 0

x=2,3

2a) y= x^2

y=x+6

x^2=x+6 -because they both equals 'y'

x^2-x-6=0

(x-3)(x+2)

x=3,-2

you can do the rest using the same method. Try it, it really isn't that hard. PM me if you have anymore question.

good luck.

Patmian212

Nov 21, 2005, 03:13 PM

I'm too lazy to graph out the first part. But here is the rest.

1c) 5-x=5x-x^2

x^2-5x-x+5 =0 - move everything to one side

x^2-6x+5 = 0

(x-3)(x-2) = 0

x=2,3

2a) y= x^2

y=x+6

x^2=x+6 -because they both equals 'y'

x^2-x-6=0

(x-3)(x+2)

x=3,-2

you can do the rest using the same method. Try it, it really isn't that hard. PM me if you have anymore question.

good luck.

Can you explain to me how you did it? I am a real dumubass at math, maybe show working out with notes if you have the time, anyone confirm hes right?

Jaffa Cake

Nov 21, 2005, 03:15 PM

I'd love to help, but I went to art college so it's totally over my head.

Doctor Q is the man you need to ask. ;)

Sun Baked

Nov 21, 2005, 03:15 PM

Can you explain to me how you did it? I am a real dumubass at math, maybe show working out with notes if you have the time, anyone confirm hes right?Simple solutions to quadratic equations.

atszyman

Nov 21, 2005, 03:23 PM

I'm too lazy to graph out the first part. But here is the rest.

1c) 5-x=5x-x^2

x^2-5x-x+5 =0 - move everything to one side

x^2-6x+5 = 0

(x-3)(x-2) = 0

x=2,3

2a) y= x^2

y=x+6

x^2=x+6 -because they both equals 'y'

x^2-x-6=0

(x-3)(x+2)

x=3,-2

you can do the rest using the same method. Try it, it really isn't that hard. PM me if you have anymore question.

good luck.

You mis-factored part 1c it should be

(x-1)(x-5)=0

x=1,5

The basic concept is to manipulate the equations so that they contain only one variable and move the variables all to the same side of the equals sign.

From there it involves the factoring of the quadratic equations. The idea is for an equation:

x^2+b*x+c=0

You want to find two numbers that add up to b and multiply out to c. For example:

(x-5)(x+3)=x^2-2x-15

for

x^2-2x-15=0 if either of the terms in the factored equation equals 0 then this equality will hold thus the answers for x are 5 and -3.

adroit

Nov 21, 2005, 03:28 PM

You mis-factored part 1c it should be

(x-1)(x-5)=0

x=1,5

Oops!!:eek:

Yup, you're right. I'm sorta brain dead right now from writing my DC Motors Lab write up--- this thing takes forever!!! :(

freeny

Nov 21, 2005, 03:51 PM

Too many bong hits in college for me. I would just scribble in "D" for all the answers and cross your fingers.

MongoTheGeek

Nov 22, 2005, 10:16 AM

The solving of the quadratics is all well and good but it sounds like patmian might be in the dastardly clutches of new math.1c I think they want him to find the intersection of the lines on the graph since it looks like the two equations are re-represented in part 1c(although there is what appears to be a transcription error of sort in there)

Patmian212

Nov 22, 2005, 10:29 AM

Well I now understand 2 and how to draw the graph in one, im still struggling with 1b and c. I just want you all to know I appreciate the help.

MongoTheGeek

Nov 22, 2005, 12:30 PM

Well I now understand 2 and how to draw the graph in one, im still struggling with 1b and c. I just want you all to know I appreciate the help.

1b looks like circle the section of the graph between 5 and 6 on the y axis and give the x values.

1c if it truely is the two equations being equal to each other then the solution is to pick the points on the graph where the lines cross. (hence are equal.)

I had mech e profs who were big on stuff like this. Determining frequency response curves by combining templates, calculating stress vectors by assigning 1cm=1KN and then drawing it out and measuring.