All Answers 358090 Answers

Mon 1/12/14

1) Graphing equations by layering. I made that term up but it does describe the process. There is probably  proper word. 

This question actually laid around unanswered for aa couple of days but it is a good question. 

@@ End of Day Wrap       Sun 30/11/14       Sydney,   Australia        Time 11:30 pm          ♬

Hi all,

Lots of great answers today from TakahiroMaeda, saseflower, SelenaPubblT, geno3141, BrittanyJ, CPhill, Alan, afroninja5959 and NathanDeAlien.  Thank you all.   

Now for the interest posts:

1) Beal's conjecture     Thanks TakahiroMaeda.

18/2=9m/hour with the current        P+C=9   (1)

18/4.5=4m/h against the current      P-C=4    (2)

(1)+(2)

2P+C-C=13

2P=13

P=6.5m/h

sub into (1)

6.5+C=9

C=2.5m/h

so Pratap can row 6.5m/h and the current is flowing at 2.5m/h

The part of the question 'stillwater and current' can be interpretted to ways:

1) The current is pushing him in the right direction getting him a time of 2 hours.

 OR

2) The current is pushing against him getting him a time of 4 and a half hours.

I would guess that the question is refering to the view 2. So all you have to do is 18m divided by 2hrs and 18m divided by 4 and a half hrs. Getting 9m/h for still water and 4m/h for fighting against the curent.

(4a+3)(4a -3)=(4a)²-(3)²=16a²-9

If you did  this the long way you would get

$$\\(4a+3)(4a -3)\\
=4a(4a-3)+3(4a-3)\\
=(4a)^2-12a+12a-3^2\\
=(4a)^2-3^2\\
=16a^2-9$$

NOW when you have the same thing in both brackets except one has a minus in the middle and the other has a plus in the middle they always work out this way.

They are called a difference of two squares because when they are expanded and simplified you will end up with

the first one squared - the second on squared

so

$$\\(4a+3)(4a -3)\\
=(4a)^2-3^2\\
=16a^2-9$$

Does that help lessen your confusion?  

No Cue :)

$${\frac{{\mathtt{34}}}{{\mathtt{3}}}} = {\mathtt{11.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$    the threes keep going for ever

this can be written as     $$11.\bar3$$  In some countries they use a dot instead of a bar.  :)

No Cue

7954 rounded to the nearest hundred is  8000

This is because you have to use the 5 for rounding.  Since it is big 5 and bigger are 'big' the 9 must be rounded up 

BUT

you cannot round up 9 so you have to round up the 79 that is why it becomes 8000    

No that is not correct Cue 

I like to do these in boxes.  Unfortunately I cannot put a frame around my boxes so you will have to imagine that they are there.   

A more normal way of doing this is to start by saying

S=140% of A

S=1.4A

sub in S=70

70=1.4A

A=50

She ate 50 apples.  I bet she got a big belly ache from that!

You would do 4a+3 times 4a-3 which gets you $${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}$$ for the fist part!

Then for the secound part $${{\mathtt{4}}}^{{\mathtt{2}}} = {\mathtt{16}}$$, $${{\mathtt{a}}}^{{\mathtt{2}}}$$ is just $${{\mathtt{a}}}^{{\mathtt{2}}}$$ and $${{\mathtt{3}}}^{{\mathtt{2}}} = {\mathtt{9}}$$ so you get the same eqaution as above!

And finally for the last part there is no working out (simplifying) as it's the answer!

All that the equation is doing is showing you how to go from (4a+3)(4a -3) to a more simplified answer.

Ok thanks!

You would do 4 times 5 and 4 times x wich gets you 20 and 4x so your answer is 20+4x

Hi Nathan,

i is the basic unit of the imaginary number system.

Nothing multiplied by itself can be negative.  That is impossible in the REAL world.

So    $${\sqrt{-{\mathtt{1}}}}$$    does not exist in the real world.

BUT there is a whole other system of numbers called imaginary OR complex numbers and in that system

$${\sqrt{-{\mathtt{1}}}}$$     is assigned the letter  $${i}$$

Hi Scrutinizer,

Long time, no see :)

I'm not a computing person but there are 8 bits in a Byte

So if I wanted to change 9.98 Mb/s into MB/s I would divide by 8

I think mine is right Chris.  

You are right I did describe ABD staying in the horizontal plane - not BDC as described - but this makes no difference to the volume.  I did it because it made the calculations easier.

I have decided to stick by my answer.  For now anyway.  (I always leave my options open LOL )

Maybe Gino or Alan could arbitrate here ?

H-16=J

H+J=276

H+(H-16)=276

2H-16=276

2H=260

H=130

J=130-16=114

$$\\sin15^0=\frac{h}{325}\\\\
h=325sin15^0$$

I'm going to try this too and I am going to use Chris's pic too.

Find the equation of line CD

C(20,0)   D(13,4)

$$\\\frac{4-0}{13-20}=\frac{y-0}{x-20}\\\\
\frac{-4}{7}=\frac{y}{x-20}\\\\
-4x+80=7y\\\\
4x+7y-80=0\\\\$$

Now find the perpendicular distance of this line to the point (8,0)

$$\\d=\frac{|4*8+7*0-80|}{\sqrt{16+49}}\\\\
d=\frac{48}{\sqrt{65}}\\\\
d=\frac{48\sqrt{65}}{65}\\\\
d\approx 5.95\;cm$$

Interesting - 3 mathematicians and 3 different answers.   

Alan has answered this here

Did you want to do that on the site calc?

Look here

See if this helps:

(1)  10° + 32° + θ + Φ = 180°

(2) sin(32°)/CD = sin(θ)/12

(3) sin(Φ)/CD = sin(10°)/5

Three equations in three unknowns should enable you to find the angles.

Then use the sin rule again to find the lengths of the other sides, AC and BC.

Then use Herons formula to find the area:

Area = √[s(s - AB)(s - AC)(s - BC)]  where s = (AB + BC + AC)/2

. 

3A+4C=159

2A+3C=112

This is the same as the question that I just answered

Try working through these units. :)

3A+4C=159   (1a)→ x 2  →  6A+8C=318   (1B)

2A+3C=112  (2a) → x 3  →   6A+9C=336    (2B)

(2B)-(1B)   

C=18

Sub C=18 into (2a)

You should be able to finish it.  But if you have problems add a post to this thread.  :)

$$\int\;x^{-\sqrt2} \;dx \\\\
= \dfrac{x^{-\sqrt2+1}}{-\sqrt2+1}\;+c\\\\\\
= \dfrac{x^{1-\sqrt2}}{1-\sqrt2}\;+c$$

I have been mucking around with this one this afternoon and no solution has jumped out at me.

A part of the problem is I am not sure how the question should be interpreted but even so the things I try just lead to added confusion.

I got a new idea - Maybe that first x is not an x at all but a times sign - then it is easy!

maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]

$$y=asin(bx)+1.2$$

360/b will give the wavelength (in degrees)  

360/b=4*9

360/36=b

b=10

$$y=asin(10x)+1.2$$

Sub in (12,3.25)

$$\\3.25=asin120+1.2\\
2.05=asin60\\
2.05=a*\frac{\sqrt3}{2}\\
\mathbf{a=\frac{4.1}{\sqrt3}}\\$$

$$y=\frac{4.1}{\sqrt3}\;sin(10x)+1.2$$

Check by subbing in (9,5.5)

$$\\RHS=\frac{4.1}{\sqrt3}\;sin(90)+1.2\\
RHS=\frac{4.1}{\sqrt3}+1.2\\$$

$${\frac{{\mathtt{4.1}}}{{\sqrt{{\mathtt{3}}}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.2}} = {\mathtt{3.567\: \!136\: \!103\: \!677\: \!465\: \!6}}$$

Well now that did not work - bummer!

It does not work because it cannot work - those conditions cannot be met on the same graph.

Maybe my interpretation is wrong again.!

Her is the graph I have.  

(It passes through (12,3.25) and it has a maximum at x=9 BUT the maximum value is not 5.5 )