Anyone take A Level Math first year? Or did?

[LiesThatBind]

So i need some help in A Level maths, i got a couple of questions which i can't do, and i got a mock test tomorrow, and if we totally screw it up i don't think we will be entered for final exam.
So does anyone take a level math (UK)? The questions are about Simaltanoues Equations by subsitution, Sequences, and 2 other questions which i dont know what they're about.

Help.
Rep+ for anyone that does.

 

[Thomas]

I am no authority on math, but from what I remember, a substitution is something you can do when you have two different equations, let's call them a and b for reference, in which there are to variables, e.g. x and y.

Isolate y in equation b (so that you have y = ...) and substitute the y in equation a with the '...'. You can now find the value of x by solving equation a. You can also find the value of y since you know the value of x.

I hope this is at least slightly helpful. If not, maybe someone more savvy than me can chime in (JBroll?).

 

[Nerina]

I took UK A level Math and Statistics a few years ago, I still have the books from it, I never throw math books away, what exactly do you have a question about?

 

[LiesThatBind]

I am no authority on math, but from what I remember, a substitution is something you can do when you have two different equations, let's call them a and b for reference, in which there are to variables, e.g. x and y.

Isolate y in equation b (so that you have y = ...) and substitute the y in equation a with the '...'. You can now find the value of x by solving equation a. You can also find the value of y since you know the value of x.

I hope this is at least slightly helpful. If not, maybe someone more savvy than me can chime in (JBroll?).

Yeah i can remember that, but its just the final bit :S i get an equation that i cant figure out;

9y^2 + 21y - 18 = 0. And to do this without a calculator. I think i have done wrong :S

 

[Emiliano]

try posting what you need, maybe i can give you some help





lol, my post got older, i tought it was the first one! shame on me!

[action=Emiliano]flex at himself and explode[/action]

 

[Drew]

Ok, my algebra is like SERIOUSLY rusty...

But there's no reason you can't start by dividing both sides of the equation by 3, right? Giving you...

3y^2 + 7y - 6

And shouldn't that just simplify to something like (3y - 2)*(y + 3) ?

I mean, I haven't done this stuff since high school so that's, um, like 8 or 9 years, but...

 

[LiesThatBind]

Ok, here it is.

x - 3y = 6

3xy + x = 24



and the other question; In the year 2001, a car dealer sold 400 new cars, a model for future sales assumes that sales will increase by x cars per year for the next 10 years, so that (400 + x) cars are sold in 2002, (400 +2x) cars are sold in 2003, and so on:

Using this model with x = 30, calculate

a. The number of cars sold in the year 2010.

b. To total number of cars sold over the 10 years from 2001 to 2010

The dealer wants to sell at least 6000 cars over the 10 year period. Using the same model:

c. Find the least value of x required to achieve this target

Ok, my algebra is like SERIOUSLY rusty...

But there's no reason you can't start by dividing both sides of the equation by 3, right? Giving you...

3y^2 + 7y - 6

And shouldn't that just simplify to something like (3y - 2)*(y + 3) ?

I mean, I haven't done this stuff since high school so that's, um, like 8 or 9 years, but...

OMG i totally forgot about doing that! Rep+ Equation is done now

 

[Emiliano]

9y^2 + 21y - 18 = 0



apply solving formula

x = (-b +- sqrt(b^2-4ac ))/2a ( i know it's not easy to read it in ascii )

you obtain

(-21 +- sqrt( 1089 ) )/ 18 ==> x1 = -54/18 = -3
x2 = 12/18 = 2/3


first solved, let me see the second

 

[LiesThatBind]

yah i got the first one now, but thanks for trying. Drew got there before yah.

But second one..
In the year 2001, a car dealer sold 400 new cars, a model for future sales assumes that sales will increase by x cars per year for the next 10 years, so that (400 + x) cars are sold in 2002, (400 +2x) cars are sold in 2003, and so on:

Using this model with x = 30, calculate

a. The number of cars sold in the year 2010.

b. To total number of cars sold over the 10 years from 2001 to 2010

The dealer wants to sell at least 6000 cars over the 10 year period. Using the same model:

c. Find the least value of x required to achieve this target

 

[Emiliano]

you have to use summation to solve the second problem ( the operation with the capital sigma before the number)

the first question is easy, you simply have to remember

400 car sold plus N times X car where

X is the number that the problem gives you, in this case is 30
N is the number of the year minus one
( if you look => in 2001 n=0, 2002 n=1, 2003 n=2, etc ) so in 2010 n=9

so for the first question you have
400+(9*30)= 670 cars


working on the second question....

 

[LiesThatBind]

omg your a saviuor. Rep +

 

[Emiliano]

we were saying

for the second question you have to use summation because you have to take into consideration each year sale, so

400 + (400 +30 ) + ( 400 + 2*30 ) etc

the formula is

SUM ( 400 +n30 ) with n that goes from 0 to 9

is that clear?

it's hard to translate from italian!


and i really do not know how to solve the third part

 

[jaxadam]

Your basic formula from the assumptions are as follows:

2001 -> 400

2002 -> 400 + x

2003 -> 400 + 2x ... and so on, so that generically, year "y" becomes:

y -> 400 + (y - 2001)x

So for part a:

In 2010 (or y = 2010), the number of cars "n" can be denoted as

n = 400 + (y - 2001)x

In this case, y = 2010 and x = 30, therefore you have

n = 400 + (2010-2001)(30)

or

n = 670

Now for part b:

Follow this same logic for years 2001 through 2010.

2001 = 400 cars
2002 = 430 cars
2003 = 460 cars, etc...

which eventually gives you 5350.

Part C:

Solve for x where n = 6000 over the period from 2001 to 2010.

n = 6000.

6000 = 400 + (400 + x) + (400 + 2x) + (400 + 3x) ... etc...

which will eventually give you

6000 = 4000 + 45x

where

x = 44.4

 

[Sebastian]

hmm.
year 01 - 400
year 02 - 400 + 30
year 03 - 400 + 60
year 04 - 400 + 120
year 05 - 400 + 240
year 06 - 400 + 480
year 07 - 400 + 960
year 08 - 400 + 1920
year 09 - 400 + 3840
year 10 - 400 + 7680

a - 8080 cars sold in 2010

b-Cars sold 01-10 = 19,330

c- x to 6000 sold needed = 8x

I can be wrong

EDIT: Shit.. it s wrong.. or It can be wrong... ehhhh ... maths.,...
EDIT 2: Well.. it can be good also

I dont know if with every year its only one more x - or its two times x of the year before

 

[LiesThatBind]

I got the answers to the questions, so i can check, but i just needed to know how to get it, a is 670, b is 5350 and c is 45

Adam your c is wrong but others are correct. Rep+

Sebastian thanks for trying but your a bit out =[

 

[jaxadam]

I got the answers to the questions, so i can check, but i just needed to know how to get it, a is 670, b is 5350 and c is 45

Adam your c is wrong but others are correct. Rep+

Sebastian thanks for trying but your a bit out =[

Well, 44.4 is pretty close to 45.

Since you can't sell 0.4 of a car, this would be a case of a whole round up to 45.

 

[LiesThatBind]

Wait, how did you get the 44.4 from the 6000 = 4000 + 45x?

Oh never mind got it, i thought simple and i got it

And shall i ask the other 2 questions?

 

[jaxadam]

Wait, how did you get the 44.4 from the 6000 = 4000 + 45x?

And shall i ask the other 2 questions?

For part c:

x = (6000-4000)/45

so

x = 44.4, so rounded up it becomes 45.

Yes, ask the other two questions...

 

[JBroll]

I'm up for trying this stuff, all sorts of ways to solve this sort of thing.

I've been tutoring for about as long as I've been in school, and I'm currently tutoring at a school that caught on fire the first day I was there (and I wasn't even involved...), so you've got another one who doesn't mind doing this.

Jeff

 

[LiesThatBind]

9a. Given that
x^2 + 4x + c = (x + a)^2 + b

where a, b and c are constants:
i. find the value of a
ii. find b in terms of c

Given also that the equation x^2 + 4x + c = 0 has unequal real roots:
iii. find the range of possible values of c.

b. Find the set of values of x for which:

i. 3x < 20 - x
ii. X^2 + 4x - 21 > 0
iii. both 3x < 20 - x and x ^2 + 4x -21 > 0


10a. Show that (3x - 4)^2 over x^ 2 may be written as P + Q over x + R over x^2 where p, Q and R are constants to be found.


b. The curve C has equation y = (3x - 4)^2 over x^2 , x =/= 0. Find the gradient of the tangent to C at the point on C where x = -2


c. Find the equation of the normal to C at the point on C where x = -2, giving your answer in the form ax + by + c = 0, where a, b and c are integers.


These 2 really got me stuck. Thanks for help man. And thanks Jeff