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Excellent, tertre......!!!!

The shortest length should be greater than their difference. Thus, the answer is $13-9+1=4+1=\boxed{5}$ cm.

41% = .41

So.....

41% of 28244.66 =

.41 *  28244.66 =

$11580.31

Good logic PM     

Thanks for the reply! 

Height of windmill = 3cm = height of object

Height of virtual image = 90 cm = h

Lens strength = 10 dpt = focal length 10 cm

Formula: 1/f = 1/v +1/u

1/10 = 1/3 + 1/u

1/u = 1/f - 1/u

1/u = 1/10 - 1/3

1/u = -0,2333 or - 7/30

-v /u = - 7/30 / 1/3 = -0,1

so it'd be C then?

Notice that we have 3 + 4 + 5 =    12 equal parts of 120m  = 10 m each 

3 of these  =  30m

4 of these = 40m

5 of these i= 50m

A 3-4-5  triangle is a Pythagorean right triangle....so.....each side of this triangle is 10 times as much

So..We have a right triangle  with legs of 30m and 40m  and a hypotenuse of 50m

The area is     product of legs / 2  =   30 * 40 /  2    =    600 m^2

Oh and you can appologize for wasting CPhills time too! 

What do you mean 'nevermind!'

You asked a question and CPhill went to a lot of effort to answer it for you.

The next thing for you to do is to say thank you! 

Good job, PM......!!!!!

a = 9, b = 8, c = 7, and d = 1

9.87 - 0.1 = 9.77. The largest possible value is $\boxed{9.77}$.

- PM

So the answer is $\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}$.

- PM

The roots are

13 + 3√17                        13 - 3√17

_________     and          _________

     2                                       2

So

The reciprocal  sum is

      2                       2

_________  +    _________

13 + 3√17            13 - 3√17

  2 [ 13 - 3√17]  + 2 [ 13 + 2√13 ]

__________________________

        169  - 9(17)

  52              13

____  =       ___

  16               4

Note that in this equation, $x_1 + x_2 = 13$ and $x_1x_2 = 4$.

But yes, a good solution would include the use of Vieta's formula. 

In the quadratic polynomial  $P(x) = ax^2 + bx + c$, the roots $x_1$ and $x_2$ satisfy $x_1 + x_2 = -\dfrac{b}{a}$ and $x_1 x_2 = \dfrac{c}{a}$.

Using the quadratic formula, we have the two solutions $\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}$ and $\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}$. The reciprocals would be $\dfrac{1}{\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}}$  and $\dfrac{1}{\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}}$. Adding them together, we have $\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}$.

- PM

Nevermind i put the wrong question the REAL Question is 

If a, b, c, and d are replaced by four distinct digits from 1 to 9, inclusive, then what's the largest possible value of the difference a.bc - 0.d?

Hint:  Use Vieta's

Thanks Chris,

Sometimes we just get on the wrong track. 

It was a nice little question actually :)

Nice, Melody......not as hard as I was making it out to be......!!!

Good work guest!

I'll just show you how to do it formally.

Vinny wrote down all the single-digit base-$b$ numbers and added them in base $b$, getting $34_b$.What is $b$?

let the base be b then

1+2+3+.....(b-1)= 3b+4

This as an AP   

S= (b/2)(1+b) 

so

$T_1=0,\qquad T_b=b-1\\ \frac{b}{2}(0+b-1)=3b+4\\b(b-1)=6b+8\\ b^2-b-6b-8=0\\ b^2-7b-8=0\\ (b-8)(b+1)\\ b=-1\;\;or\;\;8$

But b must be a possitive integer so the base is 8

Oh yeah, I just realized that I never used it. Let me remove it. Thanks, CPhill! 

Thanks, CPhill! 

_____________________

|                                     | 

|            48 units^2        | 

|          __________         | 

|         |                   |       |

|         | 16 units^2 |       | 

|         |                   |       | 

|         |________ __|       | 

| ____________________| 

48/(48+16) = 0.75

Good job, tertre and PM!!!!

Excellent, PM.....!!!!!