Need help for mathematics. - 7 questions! :D they are easy! (1 Viewer)

Joshrerts · Aug 20, 2011

So I have some question I just cannot do!

so I was wondering if any of you great people put their would be able to help me!!!! There are 7 questions. So I was wondering if you could post e answers here, or if you could pm me, or even call me - 0425201713! I need these by Monday so please help!

Thanks so much:

Here they are ( they are pretty easy)

1. The equation of a parabola is given by (x^2) -4x - 2y + 8 =0boredofstudies.org

Find the equation of the normal at (0,4)boredofstudies.org

2. A woman walks 120 meters on a bearing of 312 then turns and walks for a further 96 meters on a bearing of 056boredofstudies.org

How far is she from her starting point,

And find the bearing of the woman from her starting point.boredofstudies.org

3.
Y=-(x-1)^2boredofstudies.org
Y= 2x-10boredofstudies.org
Intersect in forth quadrant, find the point of intersection.boredofstudies.org
Find x intercepts and hence find the area under the curveboredofstudies.org

4. The series 1 + 5 + 7 + 11 + ... + 199 is formed by omitting from the first 200 positive interfere all those which are multiples of 2 or 3boredofstudies.org

Find the sum.boredofstudies.org

5. Two straight roads run from point O. One heads west while the other heads south.boredofstudies.org
Person A is 24km west of point O whilst person A is 18km south of O.boredofstudies.org
Person A walks at 2km/h towards o, where as person B walks 4km/h away from O.boredofstudies.org

Show that after t hours, the area A km^2 of triangle AOB is given by:boredofstudies.org
A=216+30t-4t^2boredofstudies.org

Calculate rate of change in the area after 1 hour

When does this triangles are start to decrease?boredofstudies.org

6. Peter deposits $5000 at the end of the ear into a superannuation fund with an interest rate of 10% p.a. Compounded half yearly. He withdraws his investment after making the 30th payment. boredofstudies.org
What is his total investment?boredofstudies.org

(b) the area of the sector of a circle, radius r units, subtending an angle of (theta) at the center is 100(pie)cm^2boredofstudies.org

Show that the perimeter of the sector can be given byboredofstudies.org
P= 2r + 200/rboredofstudies.org

Find the radius of the circle that makes the perimeter of the sector a minimum.boredofstudies.org

7. In a given state there are 1000 post codes ranging from 2000 to 2999 inclusive:
How many post codes end in 0.boredofstudies.org

When addressing a letter to a friend who lives in this state, Sophie forgot the Post-code, however she remembered that it ended with 0 and that at least two consecutive digits in the number are the same.

What is the probability the post code is 2880?boredofstudies.org

Hermes1 · Aug 20, 2011

why do you have that boredofstudies.org thing next to each question. it is really off putting.

Joshrerts · Aug 20, 2011

To be honest I just copied and pasted them off my iPad I had no idea how they got there :$!!!! Sorry

Hermes1 · Aug 20, 2011

1. y = 0.5x + 4

Joshrerts · Aug 20, 2011

Awesome thank you so much!!! I had no idea how to do it, what was the process?

Hermes1 · Aug 20, 2011

Joshrerts said:
Awesome thank you so much!!! I had no idea how to do it, what was the process?

you make y the subject.
then you find dy/dx by differentiating
then you sub into dy/dx, x = 0
this will give you gradient of tangent at point
then divide -1 by the gradient of tangent to find gradient of normal.
then simply use point gradient form to find equation of normal.

nightweaver066 · Aug 20, 2011

1. $x^2 - 4x - 2y + 8 = 0$
$x^2 - 4x + 8 = 2y$
$y = \frac{x^2}{2} - 2x + 4$
$\frac{dy}{dx}=x - 2$
$\textup{At (0,4), } \frac{dy}{dx}=-2$
$\therefore \textup{Gradient of normal = } \frac{1}{2}$
$\textup{Equation of normal: } y-4 = \frac{1}{2}(x-0)$
$2y-8 = x$
$x-2y+8=0$

2. We end up with a triangle with 76 degrees and 96m and 120m on the two adjacent sides.
Now using cos rule.
$d = \sqrt{96^2 + 120^2 - 2(96)(120)cos76}$
$d \approx 132.32m \uptext{(2d.p)}$

3. Unsure where you want the area..

4. First we find the sum of 1 to 200.
$S(200) = \frac{200}{2}(1 + 200))$
$= 100(201)$
$= 20100$

Sum of all multiples of 2 between 1 to 200, using the arithmetic series 2 + 4 + 6... 200.
2 goes in to 200 one hundred times so,
$S(100) = \frac{100}{2}(2+200)$
$= 10100$

Sum of all multiples of 3 between 1 to 200, using the arithmetic series 3 + 6 + 9 ... 198
3 goes in to 200 sixty six times so,
$S(66) = \frac{66}{2}(3+198)$
$= 6633$

Now, sum of the first 200 - sum of multiples of 2 - sum of multiples of 3 = 3367

5. Since person A is walking towards O, this length is $24-2t$
Since person B is walking away from O, this length is $18 + 4t$

Area of this triangle = $\frac{1}{2}(18 + 4t)(24-2t)$
$=\frac{432-60t-8t^2}{2}$
$=216 + 30t - 4t^2$

$\frac{dA}{dt}=30 - 8t$
$\textup{At one hour, t = 1, } \frac{dA}{dt} = 22\frac{km^2}{hr}$

$\textup{When it decreases, }\frac{dA}{dt}<0$
$30 - 8t < 0$
$8t > 30$
$t > \frac{15}{4}hr$
$\therefore \textup{Triangle starts to decrease after 3hrs 45 mins}$

6. Someone else do this..

b) Area of Sector = $\frac{r^2\theta}{2} = 100\pi cm^2$
$\theta = \frac{200\pi}{r^2} cm^2$

Perimeter = $2r + r\theta$
$=2r + r(\frac{200\pi}{r^2})$
$=2r + \frac{200\pi}{r}$

Unsure why it's different? :S

Working using my equation..

$\frac{dP}{dr}=2 - \frac{200\pi}{r^2}$
$\textup{For minimum or maximum P, }\frac{dP}{dr} = 0$
$2 - \frac{200\pi}{r^2} = 0$
$2r^2 - 200\pi = 0$
$r^2 = 100\pi$
$r = 10\sqrt{\pi}\textup{,} r>0$
$\frac{d^2P}{dr^2} = \frac{400\pi}{r^3}$
$\textup{Subbing} r = 10\sqrt{\pi}\textup{ in to second derivative}$

$\frac{d^P}{dr^2} = \frac{400\pi}{10\sqrt{\pi}} > 0$
$\therefore r = 10\sqrt{\pi}\textup{ gives minimum perimeter}$

7. Someone else do this..

Going to get started on some other work now. Someone check my working out as i may have made silly mistakes because i have a tendency to do so.

Joshrerts · Aug 20, 2011

Whoa thanks dude!!!!

Aykay · Aug 20, 2011

7 part 1.
I can do the second part but there has to be a better way so I wont show it.

You can think of this as an arithmetic series, with difference of 10.

$a_{1} = 2000$

$a_{n} = 2990$

$d = 10$

$a_{n} = a_{1}(n-1)d$

$2990 = 2000 + (n-1)10$

$1000 = 10n$

$n = 100$

Had a typo earlier, sorry.

SpiralFlex · Aug 20, 2011

Finishing off.

2. Consider the curve: $y=-(x-1)^2$

It has an $x$ intercept at $(1,0)$

Consider the line: $y=2x-10$

It has an $x$ intercept at $(5,0)$

For intersection points,

$-(x-1)^2=2x-10$

$-x^2+2x-1=2x-10$

$x=3$ . Since $x>0$

When $x=3, y= -4$

I drew you a pretty diagram,

Now let's integrate!

$A=\int_{1}^{3}-(x-1)^2 \: {\mathrm{d} x}$

$=-[\frac{(x-1)^3}{3}]$ with limits of 1 to 3.

$=\frac{8}{3} \textup{units}^2$

$B=\frac{1}{2}*4*2$

$=4 \textup{units}^2$

Area= $A+B$

$=6\frac{2}{3}\textup{units}^2$

6. Now we will need to form a GP. So how do we do that? Let's discuss. Firstly you should notice that the question says $10 \: \% \: \textup{p.a}$ . We can change this to $5 \: \% \: \textup{p.h.y}$ . The 30th payment is the 30th year.

Now,

There are $60$ half years in your $30$ years.

After the first year,

$5000*1.05^6^0$

After the second year,

$5000*1.05^5^8$

And so on.

This will continue until the last year,

$5000*1.05^2$

$\textup{Total}= 5000(1.05^2+1.05^4...1.05^6^0)$

$a = 1.05^2, r= 1.05^2, n = 30.$

Our GP formula,

$Sn=\frac{a(1-r^n)}{1-r}$

$=5000[\frac{(1-(1.05^2)^3^0)}{(1-1.05^2)}]$

$= \$950 795.241$

Seems a bit big. Did I make an error? Or is Peter that rich?

7. a. We can see that the first number must be 2 and the last must be 0.

So,

$2 [] [] 0$

But, the only numbers that can go in there are $0-9$ . That's $10$ possible numbers for each box.

Hence,

$10*10=100$

b. Now, poor Sophie.

For at least $2$ consecutive digits,

A possible consecutive form is,

$2[]00$

We can have the obvious $2000, 2100, 2200.$ etc...

This gets you $10$ figures.

Another possible consecutive form is,

$22[]0$

Now we can have $2210, 2220.$ etc...

This gives you another $9$ figures.

Finally, you can have,

$2[][]0$

$2110, 2230, 2440.$ etc...

This gives you a further $8$ figures.

So all together $27.$

$2880$ is a number so the probability is $\frac{1}{27}$

This paper looks familiar to me. Is it South Strathfield?

EDIT: I am now sure it's the Strathfield South paper. The OP typed question 6b incorrectly. The area of the sector should indeed be $100\: \textup{cm}^2.$

Need help for mathematics. - 7 questions! :D they are easy! (1 Viewer)

Joshrerts

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