All Answers 358093 Answers

I got 8.

First, you add the fractions. 1/4 + 3/4 = 1 whole.

Then, you add the whole numbers. 4 + 3 = 7.

Finally, you add the fractions and whole numbers. 

7 + 1 = 8.

That's how I got it! 

NanaLlama? Is that you?

13*13=169

so just divide the two values by 13.

Here are 4 square pies but I do not know how to find the diameter       Sorry!

@@ End of Day Wrap       Wed 26/11/14       Sydney,   Australia     Time 10:30 pm          ♬

Hi all,

As always there were lots of great questions and answers today.  Our fabulous answerers were CPhill, Tiddle, alijamali, happy7, Anonymouss4338, Heureka, Alan, SelenaPubblT, MrBalloon and SevenUp.  Thank you I really appreciate everything that each of you contribute.

Now for some interest posts:

1) It is really nice when people voice there appreciations.  Thank you anons.  

Thurs 27/11/14

1) Another limit to memorise      Thanks Alan

I want the answer for 16 

find using first principles the derivative of cosx

$$\boxed{
e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)}
}$$

$$e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} \quad | \quad \frac {d}{dx}$$

$$\left(
e^{i\phi}
\right)'
= ie^{i\phi}
=
\left( \cos{(\phi)}+i\sin{(\phi)} \right) ' = (\cos{(\phi)})'+i(\sin{(\phi)})'$$

$$ie^{i\phi}
= (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\left( \cos{(\phi)}+i\sin{(\phi)}
\right) = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\cos{(\phi)}+\underbrace{i^2}_{i^2=-1}\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\cos{(\phi)}-\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
\textcolor[rgb]{1,0,0}{ -\sin{(\phi)} } + i\textcolor[rgb]{0,0,1}{\cos{(\phi)} }= \textcolor[rgb]{1,0,0}{ (\cos{(\phi)})'}+i\textcolor[rgb]{0,0,1} {(\sin{(\phi)})'}\\ \\
\textcolor[rgb]{1,0,0}{\boxed{ (\cos{(\phi)})'= -\sin{(\phi)} }} \\
\textcolor[rgb]{0,0,1}{\boxed{ (\sin{(\phi)})'= \cos{(\phi)} }} \\$$

10)     Okay Chris I have accepted your challenge.

I had to really think!

Now we know that    $$\frac{dx}{dt}=5$$

$$\\sin\theta=\frac{x+y}{15}=\frac{y}{6}\\\\
\begin{array}{rll}
\frac{x+y}{15}&=&\frac{y}{6}\\\\
6(x+y)&=&15y\\
6x+6y&=&15y\\
6x&=&9y\\
y&=&\frac{6x}{9}=\frac{2x}{3}\\
\frac{dy}{dx}&=&\frac{2}{3}\\
\end{array}$$

Now we need to find      $$\frac{dy}{dt}$$

$$\\\frac{dy}{dt}=\frac{dy}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}=\frac{2}{3}\times 5\\\\
\frac{dy}{dt}=\frac{10}{3}\\\\$$

So the tip of the shadow is moving away at the constant speed of  10/3   feet per second

I hope that is correct.  (I think that it is)     

Payment formula for ordinary annuities.

P= Payment per period (n).

PV= Present Valve. 

r =Rate per period.

P = r(PV)/(1-(1+r)^-n)

0.07*(160000)/(1-(1+0.07)^-11)=£21337.10

                                                 £21337

 _7UP_

Here's (11)

Let the side of the square cut out = x

So, if you can imagine it....the area of the bottom of the box will be (20 -2x)(20-2x)....and the height of the box = x

So....the volume will be

V = x*(20 - 2x)^2    ..... let's just expand this out

V = x(4 x^2 - 80x + 400)

V = 4x^3 - 80x^2 + 400x

V' = 12x^2 - 160x + 400    divide through by 4 

V' = 3x^2 - 40x + 100   let's see if this will factor (and set it to 0)

(3x -10)(x - 10) = 0

Reject the second factor...it would make a side length at the bottom = 0 !!

So we have

3x - 10 = 0

3x = 10

x = 10/3  ....  this is the height

And the length of each side of the bottom = (20 - 2(10/3)) = (20 - 20/3) = (60/3 - 20/3) =40/3

And the volume is just...h x area of the base = (10/3)(40/3)^2  = about 592.6 cm^3

Omg.. you guys are amazing.. seriously.

@Melody....Yes, I have......I might need you to do (10)....I never can quite remember what to do here.....I remember that it's a similar triangle type problem, and I know where to find a similar problem, but I think you can do it off the top of your head......!!!!

I believe I can handle (11)........

WOW Chris - you have been a busy wolf !     

Just do in on a calculator

You should show how you get your answer Tiddle.  :)

It is the other way around Selena  :)

cost/bag = money/number of bags    

Good answer Selena  

Here's (9)

We can use the idea of a right triangle for this one.  We have

r^2 = x^2 + y^2       and differentiating this with respect to time (t), we have...

2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)

We are looking for dr/dt

dx/dt  = 500mi/h      dy/dt = 0   (the plane is not rising or falling)......so we can ignore the last term above..... and r = 2  and x = 1 at the time of interest .......so we have

2(2mi)(dr/dt) = 2(1mile)(500 m/hr)

4mi(dr/dt) = 1000m²/hr        divide both sides by (4mi)

(dr/dt) = (1000m²/hr) / (4mi)   =  250 m/hr

I had never thought about that before.  Thanks Chris :))

step1:  Subtract 12.4x from both sides.

step2:  Add 17.2 to both sides

then think about what to do next

$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{f(x+h)-(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cos(x+h)-cos(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx*cosh-sinx*sinh-cos(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)-sinx*sinh}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)}{h}-\frac{sinx*sinh}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;cosx*\frac{(cosh-1)}{h}-sinx*\frac{sinh}{h}\\\\
=cosx*0-sinx*1\\\\
=0-sin(x)\\\\
=-sin(x)$$

If you need these proved that also be done

$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosh-1}{h}=0\\\\
\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{sinh}{h}=1\\\\$$

Here's 8

Let x be the width parallel to the river...so the amount remaing for two equal sides is  b - x ....... so...the length of one side is just(b - x) /2

So the area is

A = x ( (b - x) / 2 )   = (1/2)bx - (1/2)x^2

A ' = (1/2)b - x      set to 0   and we have

x = (1/2)b

And length of the side is  (b - (1/2)b) / 2  = (1/2)b / 2  = (1 / 4) b

And the max area is just  (1/2)b * (1 /4)b =  (1/8)b^2