All Answers 358289 Answers

The smallest possible square that contains at least 12 black squares are the 5 by 5 squares; they either contain 12 or 13 squares. Therefore, we just need to find the number of 5 by 5, 6 by 6, 7 by 7, and 8 by 8 squares, since they all contain at least 12 squares.

Number of 8 by 8 squares = 1^2 = 1

Number of 7 by 7 squares = 2^2 = 4

Number of 6 by 6 squares = 3^2 = 9

Number of 5 by 5 squares = 4^2 = 16

Therefore, the total number of squares with at least 12 squares is 1+4+9+16 = 30  

Good answer, but I think you misread the problem. It said that the bigger angle should be 5 times the measure of the smaller angle. Therefore,

x+5x=180

6x=180

x=30

Let the roots be r and 2r:

(x-r)*(x-2r) = x^2 + 6x + k

x^2 - 3rx + 2r^2 =  x^2 + 6x + k

Compare coefficients

-3r = 6,  hence r = -2

k = 2r^2  Hence k = 8

Thank you,

I have deleted the old question since the answer was wrong anyway.

Thanks for the answer Alan :)

I get the following:

Factor the cubic polynomial to get \((2x+5)(2x-3)^2\)

This can be written as: \(4(2x+5)(x-\frac{3}{2})^2\)

So r = 3/2

A right circular cone and a right circular cylinder each have a radius of 4 and a height of 4. Let  A and B be the lateral surface areas of the cone and cylinder, respectively. Find A/B.

A / B = sin45^º   or    cos45º     or     √1/2     or     √50 / √100     or      0.707106781

A and B are two points on a circle with center O, and C lies outside the circle, on ray AB. Given AB = 24, BC = 30, and OA = 16, find OC.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BC = 30       AC = 54         ED = 32          EC = 32 + DC

BC * AC = DC * EC

30 * 54 = DC (DC + 32)         DC = 2(√469 - 8) = 27.31281566

OC = OD + DC        OC ≈ 43 ⁵/₁₆

https://web2.0calc.com/questions/hard-question_32

The total area of a square divided by 3

Could you please post the original address so people can answer on the original?

If you can't please say why not.

Remember DIRT - distance is rate * time

Happy new year!

Perhaps we can simplify the cases. We know that the probability the leftmost lamp is red is 1/2, so we just multiply our final answer by that.

Case 1: leftmost lamp is on (and obviously red)

5 choose 2/ 6 choose 3 = 1/2

Case 2: the second lamp is the first on and is red 

1/2 * 3/5 *2/5 = 3/25

Case 3: the third lamp is the first on and is red 

1/2 *2/5 *3/4 * 2/5 = 3/50

Case 4: the fourth lamp is the first on and is red  

1/2 * 2/5 *1/4* 1 *2/5 = 1/50

Not sure where you went wrong... though.

1,2,3,4,5,6

1 has to go first.

1,        23456

2,3, and 5 are prime so they can go in any order in relation to one another. 

4 has to go after 2

6 has to go after both 2 and 3

Each time there are 5 spots for the 5

1,2,3,4,6       5ways with the 5 included    

1,2,3,6,4       5ways with the 5 included    

1,2,4,3,6       5ways with the 5 included    

1,3,2,4,6       5ways with the 5 included    

1,3,2,6,4       5ways with the 5 included    

So I get 25 ways.

Thank you so much!!! I was having a hard time with this problem...

Ok, thanks catmg.

I misread the question.

Mine is based on the parabola going through the points  (4,3)  and  (-4,+3)

d = rt: 20 = 40t --> t = 1/2 hr = 30 mins

d = rt: 15 = 60t --> t = 1/4 hr = 15 mins + 3 + 3 = 21 mins

30-21 = 9 mins

I wrote a computer program:

For a = permutation[1,2,3,4,5,6]

good = 1;

if i > j and a[j] % a[i] = 0 then good = 0;

count = count + good

write(output,count);

output = 7.

The area is 15/2.

thank you!

Or for all the complicated equation enthusiasts. Solve this equation to get the answer: $(\sqrt{6} + \sqrt{6}-x)* x = \frac{(\sqrt{6}-x)^2}{2}$ for $(\sqrt{6} + \sqrt{6}-x)* x$