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tan A =  2 sin (A/2)cos (A/2)  = 2 * 1/sqrt 17 * 4/sqrt 17 =  8/17  =  24 / 51

tan B = 2 sin (B/2) cos (B/2) = 2 * 1/sqrt 10 * 3/sqrt 10 =  6/10 = 3/5  = 24/ 40

AB = ( 51 + 40) =  91

AC =  sqrt [ 24^2 + 51^2] = 3sqrt (353)

sin B =  24/ [8 sqrt (34)] = 3 /sqrt 34

Law of Sines

sin C  / AB =  sin B / AC

sin C /  91 = [3/sqrt 34] / [ 3sqrt (353)]

sin C  = 91 / sqrt [ 34* 353] =  91 / sqrt [12002]

cos C  =  sqrt [ 12002 - 91^2  ] / sqrt [ 12002[  =  61/ sqrt [12002]

tan C =  91/61

arctan (91/61)  = C =  56.16°   or   180 - 56.16  = 123.84°

tan (C/2)  =  tan (123.84 / 2) ≈  1.8744

Call the original price = P

He pays   ( P -.20P) = .80P for the first pair of shoes  and .50P  for the second pair

So

.80P + .50P  =  156

1.3P =156

P = 156 / 1.3  =  120

The first pair cost  120 (.80)  =  $96

And the second piar cost  120 (.50) = $60

4/5 - 1/2 - (1/4)(4/5)  = 

4/5 - 1/2 - 4/20

4/5 - 1/2 - 1/5

3/5 - 1/2

6/10 - 5/10

1/10 kg was left

Call the total number of  coins = C

Originally, (2/7) of the coins were 50 cent  coins and (5/7)were dollar coins

When his mother gives him 10 one dollar coins he has  10 + (5/7)C dollar coins

When his mother gives him 5 dollars of 50 cent coins he  has 10 of them in addition to the (2/7)C worth of  50 cent coins

So

In the end, he had twice the number of  one dollar coins as  the number of  fifty cent coins 

2 [ (2/7C + 10 ] =  [ 5/7C + 10]     simplify

(4/7)C + 20  = (5/7)C + 10  

20 -10  =  (5/7)C - (4/7)C

10 = (1/7)C

70 = C = the number he had at first

tan (theta)  =  sqrt [2^2 - 1^2] / 1  =  sqrt 3

cos α =  -2 / sqrt [ 2^2 + 1^2]   =   -2 / sqrt 5

cos A = 2/5     →  tan A =  sqrt 21 / 2 →  sqrt (63) / sqrt (12)

cos B = 1/2    →   tan B =  sqrt 3 / 1  →   sqrt (63) / sqrt (21)

AC =  sqrt (63 +12)  = sqrt (75)

Law of Sines

sin C / AB =  sin B / AC

sin C / ( sqrt (12) +sqrt (21))  =  (sqrt (3) / 2) / sqrt (75)

sin C / ( sqrt (12) + sqrt (21))  =  1 / [ 2 sqrt 25 ]

sin C / (sqrt (12) + sqrt (21)) = 1/10

sin C =  [ sqrt (12) + sqrt (21) ] /10

cos C  =   sqrt [ 10^2 - ( sqrt 12 + sqrt 21)^2] /10   =

sqrt [ 100 - ( 12 + 21 + 2sqrt (12)sqrt (21) ] /10   =

sqrt [ 67 - 12sqrt 7] /10

AB  = sqrt [ (4K)^2 - (3K)^2 ] =  K sqrt  7

sin C =   K sqrt 7  / 4K  =  sqrt (7) / 4 =  0.66

By the Pythagorean Theorem, AB = sqrt(16,000,000 – 9,000,000)    

AB  =  sqrt(7,000,000)    

sin C  =  sqrt(7,000,000) / 4,000  ≈  2,645.75 / 4,000  ≈  0.66    

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What is the sine of an acute angle whose cosine is \frac{5}{7}?    

cosine is adjacent over hypotenuse    

given that the adjacent side is 5 and the hypotenuse is 7    

by Pythagoras, the opposide side is sqrt(7² – 5²) = sqrt(24)    

sine is opposite over hypotenuse    

so the sine is sqrt(24) / 7  =  2 • sqrt(6) / 7  or  ≈ 0.6999    

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The minimum distance is sqrt(2) - 1.  The maximum distance is 1 + sqrt(2).

The set of all possible values of (x + y + z)/(|x| + |y| + |z|) is (0,inf).

1. Understand the Relationship Between Pens and Rulers:

"The cost of 2 pens was the same as that of 3 rulers"

This means the cost of 1 pen is equal to the cost of 2 rulers.

2. Let's use variables:

Let 'x' be the cost of 1 ruler.

Then, the cost of 1 pen is 2x.

3. Calculate the total cost of pens and rulers:

Cost of 3 pens: 3 * 2x = 6x

Cost of 8 rulers: 8x

Total cost of pens and rulers: 6x + 8x = 14x

4. Determine the fraction of money spent:

Suresh spent 2/5 of his money on the pens and rulers.

Therefore, 14x represents 2/5 of his total money.

5. Find the total amount of money Suresh had:

Let 'y' be the total amount of money Suresh had initially.

(2/5)y = 14x

y = (14x * 5) / 2

y = 35x

6. Find the money left:

Money left = Total money - Money spent

Money left = 35x - 14x

Money left = 21x

7. Find the maximum number of rulers Suresh can buy with the remaining money:

Number of rulers = Money left / Cost per ruler

Number of rulers = 21x / x

Number of rulers = 21

Therefore, the greatest number of rulers Suresh could buy with the money he had left is 21.

a) Fraction of Salary Spent on Television

Let Ali's initial salary be 'x'

Amount spent on the table: (1/3)x

Remaining salary after buying the table: x - (1/3)x = (2/3)x

Amount spent on the television: (5/8) * (2/3)x = (5/12)x

Therefore, the fraction of Ali's salary spent on the television is 5/12.

b) Fraction of Salary Spent on Plates

Remaining money after buying the table and television: $350

Total cost of plates and bowls: $316

Let 'p' be the number of plates and 'b' be the number of bowls.

p + b = 20

10p + 25b = 316

Solve the system of equations:

From the first equation, p = 20 - b

Substitute 'p' in the second equation: 10(20 - b) + 25b = 316

200 - 10b + 25b = 316

15b = 116

b = 7.73 (Since the number of bowls must be a whole number, we can round down to 7)

p = 20 - 7 = 13

Cost of plates: 13 plates * $10/plate = $130

Total salary: $350 (remaining) + $316 (spent on plates and bowls) = $666

Fraction of salary spent on plates: $130 / $666 ≈ 0.1955

Therefore, the fraction of Ali's salary spent on plates is approximately 0.1955 or about 19.55%.

Find the number of bases n \ge 2 such that 100_n + 1_n is prime.    

This is another way of saying    

How many numbers have the property that (n² + 1) is prime?    

They've been asking that question for ages.    

No one has ever been able to answer it, yet.    

I think mathematicians say that it is infinite.     

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Any line of the form y = ax  will go through the origin and bisect the area of the square

So, b = 0

The coordinates of B are (5/7, -15/7).

Slope of diagonal AC =  [ 6-3] / [ 3 -7]  = -3/4

Equation of line through AC is

y =(-3/4)(x -3) + 6

y = (-3/4)x + 9/4 + 6

y = (-3/4)x + 33/4       (1)

Slope of diagonal  BD = [ 6-1] / [ 9-1] = 5/8

Equation of line through BD is

y = (5/8)(x -1) + 1

y = (5/8)x + 3/8         (2)

Set (1)  = (2)   to find the x intersection pt

(-3/4)x + 33/4  = (5/8)x + 3/8

33/4 - 3/8 = (5/8 + 3/4)x

63/8 = (11/8)x

x = 63/11  

y = (5/8) ( 63/11 ) + 3.8

y = 87/22

Intersection pt =  ( 63/11, 87/22 )

The distance AB is 50/3.

3x - 4y  = 5

x axis intersection , let y = 0

3x = 5

x = 5/3

y axis intersection, let x =0

-4y = 5

y = -5/4

AB  = sqrt  [ (5/3)^2  + (-5/4)^2 ] =   sqrt [ 25/9 + 25/ 16 ]  = [ sqrt [ (9 + 16)(25) / (16 * 9) ] = 

sqrt [ 25^2 / 144]  =   25 / 12

The area of the region is 304*pi.

Originally we have that :

(2/5)X =Y      (1)

After he sells 44 marvles from bag X, we have this situation :

(5/7) (X -44) = Y      (2)  

Sub (2)  into (1)

(5/7)(X -44)  =(2/5)X

(5/7)X  - 220/7 = (2/5)X

[ 5/7 - 2/5 ] X  =  220/7

[ 11/35]X = 220/7

X = 220/7 * 35/11  =100 marbles originally in X

Check  

Otiginally

X =100

Y = (2/5) 100 =  40

After he sells 44 marbles from X  we have that

X = (5/7)(100 - 44)  = (5/7) (56) = 40  =  the number in Y = (2/5)*100

We cut a regular octagon ABCDEFGH out of a piece of cardboard.    

If $AB = 1$, then what is the area of the octagon?    

There's a formula for the area of a regular octagon in terms of its side "s" 

A  =  2s² • (1 + sqrt(2))    

A  =  (2)(1²)(1 + 1.414)    

A  =  (2)(2.414)    

A  =  4.828     same as Chris ... I knew it would be.    

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