anyone know the answers to these?

**jenko** · January 23rd 2008, 04:39 AM

1. A test was carried out on a electrical circuit and the following resaults obtained:

4t1+t2=6
-3t1+5t2=0

the 1s and 2s are lower case dont know how to do it sorry!

solve analytically the above simultaneous equatios and check your answer by plotting their graphs on the same set of axes

2. the area of a rectangle is 25.2m2 (squared) and the width is 3.2 metres shorter than the length. calculate the dimentions of the rectangle.

3. solve by factorising the following equations:
x(squared)-6x+9=0
6x(squared)+x-15=0

if anyone can work these out or help me i would be much app cheers!

**janvdl** · January 23rd 2008, 05:13 AM

Originally Posted by jenko

1. A test was carried out on a electrical circuit and the following resaults obtained:

4t1+t2=6
-3t1+5t2=0
by plotting their graphs on the same set of axes

Let $t_{1} = x$ and $t_{2} = y$ and then draw a graph using it as the $y = mx + c$ formula

Originally Posted by jenko

2. the area of a rectangle is 25.2m2 (squared) and the width is 3.2 metres shorter than the length. calculate the dimentions of the rectangle.

$L \times W = A$

$A = 25,2 m^2$

$L = x \ metres$

$W = (x - 3,2) \ metres$

So we know that $(x)(x - 3,2) = 25,2$

Originally Posted by jenko

3. solve by factorising the following equations:
x(squared)-6x+9=0
6x(squared)+x-15=0

$x^2 - 6x + 9 = 0$

$(x - 3)(x - 3) = 0$

$(x - 3)^2 = 0$

---

$6x^2 + x - 15 = 0$

$(3x + 5)(2x - 3) = 0$

You can solve them.

**jenko** · January 23rd 2008, 05:47 AM

are these the answers or formulas?

**janvdl** · January 23rd 2008, 05:48 AM

Originally Posted by jenko

are these the answers or formulas?

These are basically answers. You just have to finish them off.

**jenko** · January 23rd 2008, 07:18 AM

what are the final answers ive worked them out i need to check them now

**janvdl** · January 23rd 2008, 08:37 AM

Originally Posted by jenko

what are the final answers ive worked them out i need to check them now

What are your answers?

**jenko** · January 23rd 2008, 08:43 AM

i found the length and width of the square but i did not to do it the way of the question but heres what i got

Area=LxW
4.675x5.390= 25.19 is this the correct answer ? but how do i work it out as the question asked?

**janvdl** · January 23rd 2008, 08:45 AM

Originally Posted by jenko

i found the length and width of the square but i did not to do it the way of the question but heres what i got

Area=LxW
4.675x5.390= 25.19 is this the correct answer ? but how do i work it out as the question asked?

You have to expand L x W and you have to take the Area over to the left side, to get the right side equal to 0. Then solve.

**jenko** · January 23rd 2008, 08:48 AM

sorry completely confused can you give me an example using different numbers? tha i can work off? for the questions above? cheers

**janvdl** · January 23rd 2008, 08:54 AM

Originally Posted by janvdl

$L \times W = A$

$A = 25,2 m^2$

$L = x \ metres$

$W = (x - 3,2) \ metres$

So we know that $(x)(x - 3,2) = 25,2$

I am quite certain you know how to use the quadratic formula.

$x^2 - 3,2x = 25,2$

$x^2 - 3,2x - 25,2 = 0$

Now use the quadratic formula to solve for $x$ . It's that simple.

**jenko** · January 23rd 2008, 09:02 AM

no i dont this has just completly confused me ! i realy havent a clue for these ive looked at my notes and i couldnt find anything :S i need each question broken dwn induvadually so do these i cannot find any notes on these in my books from previous lessons these re only practice questions ready for my summer exam it jus the fact that im trying to get these into my head with out using notes as my exams are not open book (allowed to look at notes whilst in exam)

**janvdl** · January 23rd 2008, 09:06 AM

Originally Posted by jenko

no i dont this has just completly confused me ! i realy havent a clue for these ive looked at my notes and i couldnt find anything :S i need each question broken dwn induvadually so do these i cannot find any notes on these in my books from previous lessons these re only practice questions ready for my summer exam it jus the fact that im trying to get these into my head with out using notes as my exams are not open book (allowed to look at notes whilst in exam)

Well then it wont help to just give you the answers.

The quadratic formula is the following:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where a, b and c are co-efficients of:

$ax^2 + bx + c = 0$

**jenko** · January 23rd 2008, 09:29 AM

where shalll we start?

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