Could someone verify my answers?

[Matt.D]

Hey guys, I've been given a maths assignment that's due tomorrow and I'd just like to verify the answers before submitting. I've put the question in bold and have just put the answer where I feel confident. The rest I have shown my workings.


Q.1
The length, d m(meters), of a retangular field is 40m greater than the width.
The perimeter of the field is 400m
i) Write this info in the form of an equation for d
ii) Solve the equation and so find the area of the field

Ans i) 4d - 80 = 400

Ans ii) 4d = 480
d = 480 / 4
d = 120m
120 * 80 = 9600m2

Q.2
A) In the Formula S = ut+ 1/2 at2 make 'u' the subject
B) In the Formula T = 2Pi SQRT l/g make 'l' the subject

Ans A) u = s - 1/2 at

Ans B) T = 2Pi SQRT l/g
T / 2Pi = SQRT l /g
T2/4Pi2 = l/g
l = T2 g / 4Pi2
(sorry this looks so complicted, but I don't know how to add the right symbols)

Q.3
Write the following as simple fractions
i) x/3 + x/4
ii) 3/x + 4/x
iii) x2/x * x4/x3 - this is supposed to read xsquared over x multiplyed by x to the power of 4 divided by x to the power of 3


i) 7x/12
ii) 7/x
iii) (Not sure about this one) x2/x * x4/x3 = x6/x4


Q.4
Two resistors; r1 and r2 are placed in parallel so that their combined resistance R is given by:

1/R = 1/r1 + 1/r2

If r1 = 2x and r2 = 3x, find the formula for R in terms of x (show all your workings)

1/R = 1/r1 + 1/r2

1/R = 1/2x + 1/3x

1/R = 3/6x + 2/6x

1/R = 5/6x

1 = 5/6x *R

R = 6x/5

Q4.
Solve the following quadratic equations by either factorising or using the "b2-4ac" formula (to 3 S.F.)

i) x2 + x - 12 = 0
ii) 6x2 + x -2 = 0

Ans i) x2 + x - 12 = 0
(x -3 )(x +4) = 0
x-3 = 0 x+4 = 0
x=-4 x=-3

Ans ii) Using the formula x = -b +/- SQRTb2 - 4ac
_____________________
2a

When a = 6
b = 1
c = -2

(I can't really write the full workings because it looks too complicated without using the correct symbols ect.) The answers are:

x = -1 + 7 = 6 = 1
______ ___ __
12 12 2

x = -1 - 7 -8 -2
______ ___ __
12 12 3

Q5.
A stone this thrown into the aid and its height, h metres above the ground, is given by the equation:

h = pt - qt2

A) Where p and q are constants and t seconds is t time is has been in the air. Given that h = 40 when t = 2 and that h = 45 when t = 3, show that

p - 2q = 20

and

p-3q = 15

B) Use these equations to calculate the values of p and q Hence show that the equation for h can be expressed in the form 5t2 - 30t + h = 0

C) Use this equation to find the values of t when h = 17, gving your answers correct to two decimals places. Explain the significance of the two values of t


Ans A)
h = pt - qt2
when h = 40 and t = 2

40 = 2p - 4q

40/2 = 2p - 4q/2

p - 2q = 20

When h = 45 and t = 3

45 = 3p - 9q

45/3 = 3p - 9q/3

p - 3q = 15

Ans B)

(p- 2q = 20) - (p-3q = 15)
q = 5
p-2q=20
p-2 * 5 = 20
p - 10 = 20
p = 30

therefore h = 30t - 5t2 and 5t2 - 30t + h = 0

Ans C) (this is similar to Q4 ii using the formula
x = -b +/- SQRTb2 - 4ac
_____________________
2a

Therefore I've minimised my workings here)

5t2 - 30t + 17 = 0

t = -30 +/- 23.6
_____________
10

t = -30 - 23.6 = 0.63
__________
10

t = -30 + 23.6 = 5.37
__________
10

My own Question
I'm trying to explain the significance of the two values of t but I'm stugling to see what they are. Could someone help explain?

Well, that's it! I hope someone can give them the quick once over :)

Matt

 

[HallsofIvy]

I don't have the time to check them all!

1. Answer is correct though you skipped a few steps I would prefer to see written down.

2. S = ut+ 1/2 at²
You say u = s - 1/2 at. Did you forget to divide s by t?

2B. l = T² g / 4Pi²
Yes, that's correct. (you can get the ² by [ sup ] 2 [/ sup ] without the spaces.)

3. i and ii are correct. To do (x²/x)(x⁴/x³) I recommend cancelling first!

To answer your very last question: t₁ is when the rock passes 17 m on the way up, t₂ is when it passes that height on its way down.

 

[dextercioby]

Hey guys, I've been given a maths assignment that's due tomorrow and I'd just like to verify the answers before submitting. I've put the question in bold and have just put the answer where I feel confident. The rest I have shown my workings.


Q.1
The length, d m(meters), of a retangular field is 40m greater than the width.
The perimeter of the field is 400m
i) Write this info in the form of an equation for d
ii) Solve the equation and so find the area of the field

Ans i) 4d - 80 = 400

Ans ii) 4d = 480
d = 480 / 4
d = 120m
120 * 80 = 9600m2:

Everything is okay so far...

Q.2
A) In the Formula S = ut+ 1/2 at2 make 'u' the subject
B) In the Formula T = 2Pi SQRT l/g make 'l' the subject

Ans A) u = s - 1/2 at

Wrong,think again...It's not difficult at all...

Ans B) T = 2Pi SQRT l/g
T / 2Pi = SQRT l /g
T2/4Pi2 = l/g
l = T2 g / 4Pi2
(sorry this looks so complicted, but I don't know how to add the right symbols)

If it's
l=\frac{T^{2}g}{4\pi^{2}}

,then it's okay...

Q.3
Write the following as simple fractions
i) x/3 + x/4
ii) 3/x + 4/x
iii) x2/x * x4/x3 - this is supposed to read xsquared over x multiplyed by x to the power of 4 divided by x to the power of 3


i) 7x/12
ii) 7/x

Okay so far...

iii) (Not sure about this one) x2/x * x4/x3 = x6/x4

Me neither...It should be x^{2} and that's it...

Q.4
Two resistors; r1 and r2 are placed in parallel so that their combined resistance R is given by:

1/R = 1/r1 + 1/r2

Okay.It's good...

If r1 = 2x and r2 = 3x, find the formula for R in terms of x (show all your workings)

1/R = 1/r1 + 1/r2

1/R = 1/2x + 1/3x

1/R = 3/6x + 2/6x

1/R = 5/6x

1 = 5/6x *R

R = 6x/5

Perfect...

Q4.
Solve the following quadratic equations by either factorising or using the "b2-4ac" formula (to 3 S.F.)

i) x2 + x - 12 = 0
ii) 6x2 + x -2 = 0

Ans i) x2 + x - 12 = 0
(x -3 )(x +4) = 0
x-3 = 0 x+4 = 0
x=-4 x=-3

Correct your typo...

Ans ii) Using the formula x = -b +/- SQRTb2 - 4ac
_____________________
2a

When a = 6
b = 1
c = -2

(I can't really write the full workings because it looks too complicated without using the correct symbols ect.) The answers are:

x = -1 + 7 = 6 = 1
______ ___ __
12 12 2

x = -1 - 7 -8 -2
______ ___ __
12 12 3

Okay...Very good.

Q5.
A stone this thrown into the aid and its height, h metres above the ground, is given by the equation:

h = pt - qt2

A) Where p and q are constants and t seconds is t time is has been in the air. Given that h = 40 when t = 2 and that h = 45 when t = 3, show that

p - 2q = 20

and

p-3q = 15

B) Use these equations to calculate the values of p and q Hence show that the equation for h can be expressed in the form 5t2 - 30t + h = 0

C) Use this equation to find the values of t when h = 17, gving your answers correct to two decimals places. Explain the significance of the two values of t


Ans A)
h = pt - qt2
when h = 40 and t = 2

40 = 2p - 4q

40/2 = 2p - 4q/2

p - 2q = 20

When h = 45 and t = 3

45 = 3p - 9q

45/3 = 3p - 9q/3

p - 3q = 15

Ans B)

(p- 2q = 20) - (p-3q = 15)
q = 5
p-2q=20
p-2 * 5 = 20
p - 10 = 20
p = 30

therefore h = 30t - 5t2 and 5t2 - 30t + h = 0

Perfect...

Ans C) (this is similar to Q4 ii using the formula
x = -b +/- SQRTb2 - 4ac
_____________________
2a

Therefore I've minimised my workings here)

5t2 - 30t + 17 = 0

t = -30 +/- 23.6
_____________
10

t = -30 - 23.6 = 0.63
__________
10

t = -30 + 23.6 = 5.37
__________
10

Okay...Very good...

My own Question
I'm trying to explain the significance of the two values of t but I'm stugling to see what they are. Could someone help explain?

HINT:Plot the parabola and then draw the horizontal line y=17...

Daniel.

 

[Matt.D]

Thanks Daniel, Hallsofivy!

Regarding the question rearranging the formula s = ut 1/2 at2

would this be the correct method:

s = ut 1/2 at2

/t] s = u 1/2 at

t = s / u 1/2a

u = t / 1/2a * s

 

[dextercioby]

You forgot the signs...
h=ut+\frac{1}{2}at^{2}

Express u=u(t,a,h).

Daniel.

 

[Matt.D]

ahh, ok, if I add the signs does that make it correct (sorry, I don't understand u=u(t,a,h)

 

[dextercioby]

Of course,because "+" and "*"(multiplication) are not equivalent...

So separate "u" from that relation...

Daniel.

 

[Matt.D]

s = ut + 1/2 at2
/2] s = u + 1/2 at
t*s = u + 1/2 a
t = s / u+ 1/2 a
u = t / 1/2 a * s

 

[dextercioby]

No.
s=ut+\frac{1}{2}at^{2} \Rightarrow ut=s-\frac{1}{2}at^{2}

Can u take it from there...?Merely a division...

Daniel.

 

[Matt.D]

so u divide both sides by t to end up with

u = s - 1/2 a
________
t

 

[dextercioby]

so u divide both sides by t to end up with

u = s - 1/2 a
________
t

I sincerely hope that u meant
u=\frac{s}{t}-\frac{1}{2}at

Daniel.

 

[Matt.D]

Hi Daniel,

sorry, I got it wrong again! Thanks for all your help though.. I'm going to ask my tutor for more help in this area. Again, many thanks

Regards

Matt