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10. The least common multiple of 210 and n is 210n, where n is an integer
greater than 1. What is the smallest possible value of n?

For  210n to be the LCM, n will be the first positive integer that does not share any factors with 210.....this is 11

LCM   210*11  =  2310

7.  The numbers 19, a, b, c, 73 form an arithmetic sequence, in that order.
Find a+b+c.

19 + 4d  =  73       subtract  19  from both sides

4d  =  54   divide both sides by 4

d  =  13.5

Calling m the first term  ....the sum  of a + b + c  is

( m + d ) + (m + 2d) + (m + 3d)  =

3m  + 6d  =

3(19) + 6(13.5)  =

57    + 81  =

138

m1    negative number  slope  may be a fraction of -1 ...depends on the scales used

m2 = 0   slope = 0      equation of the form  y= b2

m3    positive number slope  probably greater than or equal to 1

b1     y -axis intercept positive   greater than b2 or b3

b2     y - axis intercept positive  less than b1   greater than b3

b3     y-axis intercept  negative   less than b1  or b2

Sorry, ant....I would just be guessing on the rest and that usually leads to wrong answers....maybe some other people know the solutions to the rest of these   !!!!

Thanks, CPhill! Amazing! Can you answer the others, too?

1. Paula can paint a room in six hours. Paula starts painting the room at
9:00 AM. At 11:00 AM, Carlos accompanies Paula, and both of them,
working together, finish painting the room at 1:30 PM. Assuming both
painters always paint at a constant rate, how many hours can Carlos
paint the same room working by himself?

Paula paints 1/6 of the room every hour

By the time Carlos joins in, Paula has painted   2 (1/6) =  1/3 of the room

And after Carlos joins....she paints an additional 2.5 hrs

So...by  1:30 PM....she has painted  1/3  + (2.5)(1/6)  =  1/3 + (5/2)(1/6)  = 

1/3 + 5/12  =  4/12 + 5/12  = 9/12  =  3/4  of the room

So.....Carlos  must paint  the other  1/4 of the room in  2.5 hours

So...the time it would take him to paint the room by himself must be 4 times this  =

10 hrs

1)

6 - 2 = 4 hours left for Paula to finish by herself

1/4 + 1/C =2/5, where C = Carlos' time

C =20/3 = 6 2/3 - Carlos' time in painting 2/3 of the room!!!

So, to paint the whole room by himself, it would take him: 6 2/3 x 3/2 = 10 hours !!!!.

EDITED!!. I simply reversed the time they finished the room together!!. Oh, my!!

5 / 3 = 1 2/3 chickens per goat.

1 2/3 x 2 = 3 1/3 chickens per cow.

3 1/3 x 6 = 20 chickens for 6 cows!!.

Thanks to you both so much!

Solve for x:

x^3 - 2 x^2 + 3 x - 6 = 0

The left hand side factors into a product with two terms:

(x - 2) (x^2 + 3) = 0

Split into two equations:

x - 2 = 0 or x^2 + 3 = 0

Add 2 to both sides:

x = 2 or x^2 + 3 = 0

Subtract 3 from both sides:

x = 2 or x^2 = -3

Take the square root of both sides:

x = 2      or      x = i sqrt(3)      or      x = -i sqrt(3)

Solve for x:

x^6 - x^4 - x^2 + 1 = 0

The left hand side factors into a product with three terms:

(x - 1)^2 (x + 1)^2 (x^2 + 1) = 0

Split into three equations:

(x - 1)^2 = 0 or (x + 1)^2 = 0 or x^2 + 1 = 0

Take the square root of both sides:

x - 1 = 0 or (x + 1)^2 = 0 or x^2 + 1 = 0

Add 1 to both sides:

x = 1 or (x + 1)^2 = 0 or x^2 + 1 = 0

Take the square root of both sides:

x = 1 or x + 1 = 0 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x^2 + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -1 or x^2 = -1

Take the square root of both sides:

x = 1      or      x = -1      or      x = i      or      x = -i

1c = 2g

or    c = 2g

3g=5f     (f= chickens)

g=5/3f     Substitute into the first equation

c = 2 (5/3 f)       now you have 6 cows so multiply both sides by 6

6c = (6)(2)(5/3 f) = 60/3 f = 20 f   

6 cows = 20 chickens

Yep, that's what I got, too!

11.     The smallest I can find ...by trial and error    is    4  5   6

           perimeter = 15

9.  Interior angles of a regular polyhedron total   (n-2)(180)

      for an octagon    (8-2)(180) = 1080    there are 8 interior angles so each is 1080/8 = 135 degrees

     The sides of the SQUARE form isocoles triangles of sides 4, 4   and 'x' (the side of the square) with an included 135 degree angle beteen the 4 and 4 sides

   The other two angles are   (180-135)/2 = 22.5 degrees

 Use the law of sines to find 'x'

sin 135 /x  = sin22.5  / 4    results in square side length of 7.39     

 Square Area = 7.39 x 7.39 =54.63 units^2 

LOL!!!.....yep, that helps  !!!

x^4 + 7x^3  + 9x^2  - 17x - 20

The possible rational zeroes are  ±  ( 1, 2, 4, 5, 10 and 20)

We have  one sign change ..so we have 1  positive root 

To  find the number of possible negative roots look at f(-x)  =

(-x)^4 + 7(-x)^3 + 9(-x)^2 - 17(-x) -20  =

x^4 - 7x^3 + 9x^2 + 17x - 20

We have hree sign changes so we either have 3 or 1  negative roots

Let's write the above in a slightly different manner

x^4 + 7x^3 + 6x^2 + 3x^2 - 17x - 20        factor this

x^2(x^2 + 7x + 6)  +  (3x - 20)) (x + 1)  =

x^2 ( x + 1) (x + 6) + (x + 1) (3x - 20)   factor out the ( x + 1)

(x + 1)  [ x^2 (x + 6) + (3x - 20)  ]

(x + 1) [ x^3 + 6x^2 + 3x - 20 ]  the second polynomial factors as  (x + 4) (x^2 + 2x - 5)

So   we have

(x + 1) (x + 4)(x^2 + 2x - 5)

Setting the first two  factors to 0  and solving for x produces   x = - 1   and x =  4

The other two roots  come from using the quadratic formula to solve the 3rd polynomial...these are

x  = √6 - 1    (p0sitive)   and  x  =    -√ 6  -  1   (negative)

So....we have 1 positive  root and 3 negative roots

No problem, EP!

Thanx for the confirmation !  

Yes, I think both of them are correct...... 

Hello, EP!

We have 5 options for the choice of the vowel, and we must make 2 choices out of the remaining 21 letters, for a total of $\binom{21}{2} = 210$choices for the consonants. This gives a total of $5 \times 210 = \boxed{1050}$.

Here we go....my stab at it....

5 ways to choose a vowel    a e i o u

   times    21c2

5 x 21c2 = 5 x 210 = 1050 ways

Wait till CPhill comes. He's great at this.

Typo !!!!!!!!!?.

How did you get 54/3=13.5?

7)

73 -19 =54

54 / 4 = 13.5 - common difference

19, 32.5, 46, 59.5, 73.........etc.

a + b + c = 32.5 + 46 + 59.5 = 138

8)

[8! + 9! + 10!] =8! x [(1 + 9 + 90)] / 8! x n

n = 100

Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \boxed{120}.$