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The above answer has it in base 10 however we need it to be in base 9. 

Thanks Chris and Ginger.

I really like questions like this one :))

The surface area of a sphere is A pi cm^2 and the volume of the sphere is B pi cm^3 where both A and B are four-digit whole numbers. Find the radius of the sphere.

The radius must be divisible by 3.

Certainly I can do that :))

NOTE: I have not looked at your answer. 

I have answered on Rosala's original thread as indicated below :)

Hi Rosala,

I have anwered on the original thread :)

1.Find the no. of permutations of letters a , b , c , d, e , f , g taken all at a time if neither  “beg” nor “cad” patterns appear in any word.

Mmm I'd put a rope around beg 

then that makes a,c,d,f (beg)  so there will be 5!= 120 permutations

If I put a rope around (cad)   that will also be 120 permutaions 

BUT some of these permutations are in both so if I put a rope around (cad) and (beg) that will be 3! =6 permutations.

so altogether the number of permutaions that include the words beg or cad is   120+120-6=  234 permutations.

Without any restrictions there are 7! = 5040 permutations

so the number of permutation without beg or cad is 5040 - 234 = 4806

2.Find the no. of ways in which letters of the word SQUARE can be arranged such that:

A) Vowels are always together.   

SQR  (UAE)   UAE can be arranged in 3!=6 ways and SQR(UAE) can be arranged in 4!=24 ways.

So altogether that is  24*6 = 144 ways

B)vowels are never together.

There are 3 vowels and 3 consonants so the choices are

CVCVCV   or   VCVCVC   

3!*3!*2 = 6*6*2 = 72 ways

C)No. of ways in which none of U , A , E are together. (in any order)

Again, UAE must be in the 1st 3rd and 5th positions   or in the 2nd 4th and 6th positions. I do not think there are any other choices so it is the same as the question above I think.  Maybe I have misunderstood the question?

72 ways

What do you think Rosala, do you question any of my answers (which is perfectly fine) or do you have any questions for me?

Yes, Lancelot uses his opposable thumbs for multiple purposes, including thumbs up or down. Nauseated, however, may give paws for dewclaw consideration, but his central tendency is toward his central digit with adjustments for any ambiguity.  

GA

A = 1296, B = 7776 ⇒ r = 18

.

4pi*18² = 1296pi

(4/3)pi*18³ = 7776pi

.

Melody: 9^(9^9) =9^(387,420,489) =387,420,489 x log(9) =369,693,099.631........etc.

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10

See Wolfram/Alpha Here:

http://www.wolframalpha.com/input/?i=9%5E9%5E9

Find how many digits in the base 10 number 9^(9^9) has when written in base nine

it will have 9^9+1 digits = 387 420 490 digits (base 10)

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simple example ( I needed this to think through my answer)

2^(2^2) base 10 written in base 2

2^4=16=10000₂

So that is 2^2+1=5 digits base 10 

OR    5₁₀ = 101₂

\(\frac{x}{x\sqrt2}\\ =\frac{1}{\sqrt2}\qquad \text{the x es cancel out, note }x\ne 0\\ =\frac{1}{\sqrt2}\times \frac{\sqrt2}{\sqrt2}\\ =\frac{\sqrt2}{\sqrt4}\\ =\frac{\sqrt2}{2} \)

Thanks Chris, it did work rather well - I thought so too   

What about this permunation?:

B-E-G-F-C-A-D

We can prove that this will always be so  no matter the current rate or Andy's rate  [ as long as his rate is greater than the current rate ]

Let Andy's normal rate be  A   and the current rate, C

Call T  the time elapsed between when his hat is knocked off and he notices that it's gone

In this time the hat has traveled  C * T

And Andy  has traveled   (A - C)T

When he turns around his rate becomes A + C

And the distance that he travels to retrieve the hat is (A + C) T

And this equals the distance he was from the bridge when he turned around plus the distance the hat has traveled from the bridge  = C *2T  ....so we have......

(A + C ) T  =  (A - C) T   + C*2T

AT + CT  =  AT - CT + 2CT

AT + CT  =  AT + CT

Note to  GA....I verified this result with Lancelot Link......he gave it a "thumbs up" ... [ at least I think that was the digit he was using....]

I like that "table" trick, Melody.....that makes the counting way easier !!!

if three straws are chosen at random from a set of 9 straws whose lengths are 1cm, 2cm 3cm, 4cm, 5cm 6cm, 7cm, 8cm, and 9cm find the probability that the three straws will form a triangle. 

There are 9C3 = 84 ways to chose 3 straws from the 9

They will from a triangle if the sum of the smallest 2 is greater than the length of the longer one.

So just count them. The trick is to dream up a good counting method.

Here is one I dreamed up. I hop I haven't made any silly errors. :)

So the probability of being able to from a triangle is    33/84

A = (3,5)   B  = (7, 11)   C  = ( 13,17)

Because of the nature of the points.....this graph will turn downward

The slope  of the line between (3,5) and ( 7,11)  is 6/4  =  3/2

And the equation of this line is

y  = (3/2) (x - 3) + 5    ⇒  (3/2)x + 1/2

And this line forms the "left " branch of the absolute graph

For the equation  of the line forming the other half of the graph...it will have a slope that is the negative of the first line and it will pass through ( 13, 17) 

So the equation of this line is just

y = (-3/2) ( x - 13) + 17    ⇒  (-3/2)x  +  73/2

The intersection of these two lines will form the "vertex" of the function....so we have....

(3/2)(x) + 1/2  =  (-3/2)x + 73/2

3x    =  72/2

3x  = 36     ⇒  x  =  12

And y  =  (3/2)(12) + 1/2  =  18.5

So   b =  12   and c =  18.5

To find "a"  we can use any of the points

So we have that

5  =  a l 3 - 12  l  + 18.5

-13.5  = a (9)

a  =  -13/5 / 9   =   -3/2  =  -1.5

Here's the graph of the lines and the absolute value graph itself :

https://www.desmos.com/calculator/1eyojys50i

3 * 5 * 6

There are a total of  90 cubes

The unpainted ones will be the ones in the middle of the prism

Let 3 be the height

There will be 2 * 6 * 5  =  60  painted cubes on the top and bottom

There will  2 * 6  +   2 * 3  =  18 more painted cubes "sandwiched" between the top and bottom surfaces

So...the total unpainted cubes  is  90  - 78   =   12

So their total surface area is just   [12 * 6 ] cm^2   =  72 cm^2

P.S.  -  I believe there is also a "formula" for this....we are cutting away 2 inches in every dimension...so  (3 - 2) (5 - 2) (6 - 2)  =  1 * 3 * 4    =   12 cubes

You could find out the price by summing up the PV of all the coupons plus the face value of the bond at maturity at the given yield of 3.5% compounded semi-annually. This is a cash flow problem and you would use the following formula:

PV = FV x [1 + R]^-N

PV =750 x [ + 0.035/2]^-1 + 750 x [1 + 0.035/2]^-2 + 875 x [1 + 0.035/2]^-3 +........and so on to + 100,000 x [1 + 0.035/2]^-14......and you should get = $94,391.03.

I have NOT included any accrued interest that may be due, because I used the term of 7 years exactly, or 14 semi-annual periods.

Call S  the radius of the smallest circle  and  M the radius of the intermediate circle

pi (12)^2  -  pi (M)^2  =  pi ( S)^2     (1)       and

pi (M)^2  - pi(S)^2   = pi (S)^2    ⇒  pi (M)^2  = 2 pi (S)^2   (2)

Sub (2)  into  (1)

pi (12)^2  - 2pi(S)^2  = pi (S)^2

12^2  - 2S^2 + S^2

12^2   =  3S^2

144  = 3S^2

48 = S^2

√48  = S

4√3  cm  = S   ≈  6.93 cm

9^(9^9) =369,693,100 base 10

=812,298,765 base 9 [I think !!!].

She used trial and error....but..with the presence of the "5"  she knew that the base had to  be  6 or greater.......so...it didn't take her long to get to 7....LOL!!!!

That is just a typo!! It should be 80 km !!! Boy, that is a Long walk!!.