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x^2 + 6x + y^2 - 14y = -c         complete the square on x,y

x^2 + 6x + 9 + y^2 - 14y + 49  = -c + 9 + 49

To have a radius of 4

-c + 9 + 49   must  = 16

So

-c  =  16 - 9 - 49

-c = -24

c =  24

The line that cuts line L at a 45^o angle also makes a 45^o angle with K in the little triangle. 

In that little triangle, one angle is 90^o.  The sum of all angles of a triangle is 180^o.  

The angle in the triangle opposite x equals 180^o – (90^o + 45^o) = 45^o   therefore x = 45^o  

Let p be a prime number between 40 and 60. What is the probability that p+2 is also a prime number? Express your answer as a common fraction.  

Primes between                            

40 and 60                p + 2             Is p+2 prime  

       41                       43                     Yes  

       43                       45                     No  

       47                       49                     No  

       53                       55                     No  

       59                       61                     Yes  

Five primes p, two of p+2 are also prime        The fraction is 2/5.  

The length of a rectangle is increased by 25% , but the width of the rectangle is decreased by 25%.  By what percent was the rectangle's area decreased?  

Let's do it with an example. 

Start with a rectangle that's 8 x 4.  The area of this rectangle is 32.  

8 + 25% = 10 

4 – 25% =   3                                  The area of this rectangle is 30. 

The area was decreased by 2. 

To get the percent of decrease, divide this by the size of the original rectangle. 

2/32 = 0.0625   so the percent of decrease is    6.25%   

This was a special case, using specific numbers. 

Try it with a few different size original rectangles, and see if the answer is consistently the same.  

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A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, two coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?  

Let's try it with 22 coins. 

22/6 = 3 with 4 left over 

22/5 = 4 with 2 left over 

Okay, so that part is good. 

Now 22/7 = 3 with 1 left over.  

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I think that 28 does NOT satisfy the requirement that "If the coins are equally divided among five people, two coins are left over"  

A bag of grapes is to be distributed evenly to 6 kids in a class, and the grapes that are left over will be thrown out. If each student receives the greatest possible number of grapes, what is the greatest possible number of grapes that could be thrown out? 

Well, it's 5.  If there were more left over than that, each kid could get another grape(s) and it starts over. 

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c = 7t

s = 2t

c + t + s = 140

7t + t + 2t = 140

t = 14

c = 98

c - t = 98 - 14 = 84

84 stamps Terry collected than Clay

20! + 25! ==1.551121247623299416064e+25 = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19 * 73 * 87337

There are 10 DISTINCT primes

1116==4! / 3!==4 permutations
1123==4! / 2! ==12 permutations

The probability is: [4 + 12] / 6^4 ==16 / 1296==1 / 81

Solution by HELPMEEEEEEEEEEEEE:

The smallest number that satisfies both conditions is 28, so when divided among 7 people, there are 0 coins left.

To find the smallest number you'll need to know something called modular arithmetic which is something I cannot explain so you may have to look that up yourself.

Total number of animals: \(110 \)

Number of rabbits: \(90 \)

Number of chickens: \(110-90=20\)

Fraction representation of chickens over the total number of animals: \(\frac{20}{110}\)

Percentage :\(\frac{20}{110}\times100=\frac{(20\times100)}{110}=2000/110=18.18 \)%

I understood what you did but when i put the answer in it said it was incorrect so I'm not sure if its right?

The area is decreased by 4 percent.

I think its 108

The answer is [3,24].

I don't think this is right?

Since f(x) = (ax + b)/(cx + d), the equation f(f(x)) = x becomes

(a*(ax + b)/(cx + d) + b)/(c*(ax + b)/(cx + d) + d) = x.

Cross-multiplying and simplifying, we end up with

a^2*x + ab - acx^2 - ax + bd - b - cdx^2 + cx^2 - d^2 x + dx = 0.

This factors as (a + d - 1)(-cx^2 + ax - dx + b) = 0.

We can't have -cx^2 + ax - dx + b constantly 0, so a + d - 1 = 0.  Therefore, a + d = 1.

The number of cookies  before Mrs. Johnson ate any =  (3)*(4/3)  = 4

The number of cookies  before the class ate any =  4 / (1/6)  = 24  = what Mrs. Johnson brought

Proof

24  (5/6)  =  20   what the class ate

4 left

4( 3/4)  = 3 what Mrs. Johnson ate

Let z_1 = a + bi and z_2 = c + di.  Then z_2/z_1 = pure imaginary gives you d/a = -c/b.  Let d/a = k and c/b = -k.

Then d = ka and c = -bk.  A little bit of algebra then gives you

\(\left| \frac{2z_1 + 7z_2}{2z_1 - 7z_2} \right| = \frac{2}{7}\)

There are 15,425 possible passcodes.

The number of possible combinations is 8.

The largest angle is 120 degrees.

Only three factors of 4  =   1, 2 , 4

P(factor of 4)  =  3  / 24    =    1 / 8