All Answers 358514 Answers

thank you very much Reinout! sorry for the link!

Nope, but it wasn't that hard to go to page 801.

for 32, the center of the circle is at (p,q) = (0,0) and the radius 3

Since the general equation of a circle is given by

$$(x-p)^2+(y-q)^2 = r^2$$ where (p,q) is the center and r the radius

the equation of this circle is

$$(x-0)^2 + (y-0)^2 = 3^2 \Rightarrow x^2+ y^2 = 9$$

Similarly for 36 we find

(p,q) = (-1,1) and r = 2

Therefore

$$(x--1)^2 + (y-1)^2 = 2^2 \Rightarrow (x+1)^2 + (y-1)^2 = 4$$

Reinout 

Yes, but we can define the equation which defines all solutions to this problem.

If we add both equations we have

sf+sp = 14+21

s(f+p) = 35

This is part of the surface the equation desribes.

Reinout 

I assume you want to integrate this with respect to "x"  ??

Simplifying this , first, we have

(y + x^2*y^2) / [(x^2)(y^2)]  =

1/(x^2*y) + 1 =

(1/y)(1/x^2) +1  ........and integrating this, we have

(1/y)∫x^(-2) dx   + ∫ (1) dx =

(1/y) [-x^(-1)] + x  + C  =

x - 1/(xy) + C

@@ End of Day Wrap - Fri 13/6/14   Sydney, Australia   Time 10:10pm  

Hi Everyone,

Great answers today from CPhill, reinout-g, Rosala, zegroes, Alan, Will85237, and Bertie.

Will85237 joined a couple of days ago. Welcome to Web2.0calc forum. We hope that you enjoy yourself here.

NinjaDevo made his way onto the "Top Answerers" board today.  CONGRATULATION NINJADEVO.  I had not realised that you had answered so many questions,  (Ninja's are supposed to be stealthy) But, I have been very impressed with the quality of your answers Ninja. Thank you.

There have been a  number of questions lately which are ambiguous or simply do not make sense.  At Alan's suggestion I have posted a note question askers, requesting them to be careful about this and also giving hints to appropriate forum conduct.  This is it.

Actually we are all equally correct I think.  The question does say when the first deposit is made!

If it is made at the begining of the first month then CPhill is correct (Present value of an annuity due question)

If it is made at the end of the month then I am correct. (present value of an ordinary annuity problem)

CPhill answer is probably preferable.

Thanks Chris and Alan for making me think about it.

When you say "side," I think you mean the "opposite" side.  (It can't be the hypoteneuse because that would have to be longer than 120").

So, the tangent function would relate these. We have:

tan(θ) = 12/120     So...taking the tangent inverse to find "theta", we have

tan^-1(12/120)  ≈ 5.71°

I was only talking about when the deposit was made....at the first of the month is called an "annuity due" - if I remember correctly. Deposits made at the end of the month are known as an "ordinary" annuity. I was assuming the first situation....but  the second could be assumed also...... it's not stated in the problem....

13/6/14

Note for question askers

Ah.....I see my error here....we're picking cards one at a time....!!!

OK...I assumed "dealt"

Sorry!!

deleted deleted

It is already an algebraic expression

$$\sqrt5*x^3$$

Thanks, Bertie...I see where I made my mistake !!!!

Can you give us and example please and we will show you.

say 

$$y=3x+2$$

when x=0

y=3*0+2=2

so (0,2) is one point

When x=1

y=3*1+2=3+2=5

so (1,5) is another point.

Is this what you mean?

No it is not.  I usually just stick these things up on the forum every now and again.

There's the sticky topics things, but that's not very easy to spot if you're not looking for it.

I am  assuming that each colour is equally represented so each spin there is 1/3 chance of getting a yellow.

$$\mbox{P(7out of 10 yellow) =}10C7 * (1/3)^7 * (2/3)^3$$

$${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}^{{\mathtt{7}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{3}}} = {\mathtt{0.016\: \!257\: \!684\: \!296\: \!093\: \!1}}$$

the inverse sin, also noted as asin, arcsin or sin^-1 is the opposite of a sin

For example if we have

sin(b) = c then b = asin(c)

So if you want to type it in your calculator you can enter asin(0.8)

However sin(0.8)^-1 means something completely different than sin^-1(0.8) , but I dont think you mean the first one right?

anyway, asin(0.8) = 53.13 degrees.

there is no times in there

It is just 

sin^-10.8=Inverse sin 0.8 = arc sin 0.8 = asin0.8

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.8}}\right)} = {\mathtt{53.130\: \!102\: \!354\: \!156^{\circ}}}$$

ok so I had to put the brackets in to make it work.

I'd assumed that the lengths given were the lengths of the vertical lines. What needs clarifying is whether or not the vertical lines are equally spaced, that is, 15/6 cm apart. Even then the area will depend on the actual shape of the boundary curve. Replacing all bits of boundary curve by straight lines (as suggested by Alan) would enable you to get an approximate value.

Excellent work rosala.  

That is what I have tried to do, where do you suggest?

In sticky notes maybe?

Oh yeah, that would make sence.

Is this meaning you would need to know the angle of line A-C for example?

Or just what a strait line measurement from A-C would be?