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\( 9/(1 - \sqrt[3]{3} - \sqrt[3]{3} + \sqrt[3]{9}) \)

9  / ( 1  - 2(3)^(1/3) + 9^(1/3))  

9 / (1  - 2(3)^(1/3) + (3^2)^(1/3) )

9  / ( 1 - 2(3)^(1/3) + 3^(2/3) ]

9 / ( 1 - 3^(1/3) )^2

Let  (1 - 3)  =  [ 1 - 3^(1/3) ] [ 1 + 3^(1/3) + 3^(2/3) ] 

-2  = [ 1 - 3^(1/3) ] [ 1 + 3^(1/3) + 3^(2/3) ] 

1 / [ 1 -3^(1/3) ]   =   [ 1 + 3^(1/3) + 3^(2/3) ] / -2            square both sides

1 / [ 1 - 3^(1/3)]^2  =  [7 + 5*3^(1/3)  + 3*3^(2/3)] / 4      multiply both sides by 9

9 / [ 1 - 3^(1/3) ] ^2   =   (9/4) [ 7 + 5*3^(1/3) + 3*3^(2/3) ]

Simplify

y = 5x^2 -10x + c

y = 6x^2 - x 

Set the equations equal

6x^2  - x =  5x^2  -10x + c

x^2 + 9x  - c  =  0

One solution

9^2  + 4c   = 0

81  =  -4c

c = -81/4

Two solutions

9^2 + 4c >  0

81 > -4c

c < -81/4

No solutions

9^2  + 4c  < 0

81  < -4c

c > -81/4

a^3/b^3  + b^3/a^3  =  ( a^6 + b^6) / (ab)^3

x^2 + 5x +  46   =   0 

ab = 46

a + b = -5    square both sides

a^2 + 2ab + b^2 =  25

a^2 + 2(46) + b^2  =  25

a^2 + b^2  = -67

a^3 + b^3  =  (a + b) (a^2 + b^2  - ab)  =  (-5)(-67 - 46)  =  565

(a^3 + b^3)^2  = a^6 + 2a^3b^3 + b^3  =

a^6 + 2 (ab)^3 + b^6  =  565^2

a^6 + 2(46)^3 + b^6  = 565^2

a^6  + b^6  =  565^2  - 2(46)^3

a^3/b^3  + b^3/a^3  =  ( a^6 + b^6) / (ab)^3 =  [ 565^2 -2(46)^3 ] / [ 46^3 ] = 

124553 / 97336

a^2/b + b^2/a =  (a^3 + b^3)  (ab)

x^2  - 5x + 3 = 0

ab = 3

a + b = 5    square both sides

a^2 + 2ab + b^2 = 25

a^2 + 2(3) + b^2  = 25

a^2 + b^2  = 19

a^3 + b^3  =  (a + b)(a^2  - ab + b^2)  = (5) ( 19 - 3) =  80

So

a^2/b  + b^2/a =  (a^3 + b^3)  / ( ab)  =  80  / 3

Suppose that on the trip from the cafe to the boba shop a total of U metres is uphill, ( either in a single stretch or made up in bits), and similarly L metres is on the level, and D metres is downhill, then the time taken for the trip will be U/100 + L/120 + D/150.

On the return trip uphill becomes downhill and downhill becomes uphill so the time taken will be U/150 + L/120 + D/100.

Total time there and back will be U(1/100 + 1/150) + L(1/120 +1/120) + D(1/150 + 1/100) = (U + L + D)/60.

The total distance covered is 2(U + L + D) metres, so dividing distance by time, the average speed will be 120 m/hr.

First one

3x^2 - 18x + 13 =  0

ab  = 13/3

a + b =  18/3 =  6

1/a  + 1/b =   (a + b)  / (ab)  =  6 / (13/3)  =  18 / 13

Second one

4x^2 + x + 11  = 0

uv = 11/4

u + v = - 1/4     square both sides

u^2 + 2(uv) + v^2  = 1/16

u^2 + 2(11/4) + v^2  = 1/16

u^2 + v^2   = 1/16 - 22/4

u^2 + v^2  = -87 / 16

u / v  +  v /u   = 

[ u^2 + v^2  ]  / [ uv ]  =    (-87/16) /( 11/4) =   (-87/16)(4/11)  = -87/44

Because \(P_1P_2P_3\dots {P}_{10}\)is a regular polygon, all the side lengths are equal.

Therefore we only need to calculate \({(\overline{P_1P_2)}}^{2}*10\).

Suppose the center of the circle is O, then \(\angle P_1OP_2 = {36}^{\circ}\).

Apply law of cosines:

\({(\overline{P_1P_2)}}^{2}={1}^{2}+{1}^{2}-2*\frac{\sqrt{5}+1}{4}=2-\frac{\sqrt{5}+1}{2}\).

\({(\overline{P_1P_2)}}^{2}*10=20-10(\frac{\sqrt{5}+1}{2})=20-5\sqrt{5}-5=15-5\sqrt{5}\).

So our answer is \(15-5\sqrt{5}\).

Amt you have at first  =  A

Amt  your brother has at first  = B

After you give him $5

A - 5 =  B + 5

A  = B + 10      (1)

If he gives you $10

A + 10 =  2(B - 10)

(B + 10) + 10  = 2B - 20

B + 20  = 2B - 20

B = 40

He has  $40

You have $50

                            A

                                       8

        B         2       K        3           C

AK =  sqrt (AC^2 - CK^2 )  =  sqrt (8^2 - 3^2 )  = sqrt (55)

AB =  sqrt (BK^2 + AK^2 )

AB = sqrt ( 2^2  + 55)

AB = sqrt [ 59 ] ≈ 7.68

                                 A

                            10         15

                  B      1     D       3                  C

AD^2  =    AB^2  - (BC/4)^2

AD =   AC^2  - [(3/4)BC]^2 

So

AB^2  - (BC/4)^2  =  AC^2  - [(3/4)BC]^2

10^2  - BC^2 / 16  = 15^2  - (9/16)BC^2

15^2  - 10^2  =  (9/16 - 1/16)BC^2

125  = (1/2)BC^2

250  = BC^2

AD^2  =  10^2  - (BC/4)^2

AD^2  = 100 -  (1/16)BC^2

AD^2   =100  -  250/16

AD^2   =    675 / 8

AD  = sqrt (675 / 8)   = (15/4)sqrt 6  ≈ 9.19

A

B               C

AC =  sqrt [ (4sqrt 2)^2 + 5^2  ] =  sqrt  [ (sqrt (32))^2 + 5^2 ] =  sqrt (57)

cos (ACB)  =  5/sqrt (57)

Law of Cosines

BM^2   = (AC / 2)^2  + BC^2  - 2 ( AC/2 * BC) (5/ sqrt 57)

BM^2  = AC^2 /  4  + BC^2  - (AC*BC) ( 5/sqrt 57)

BM^2  =   57 / 4  + 25 - (sqrt (57) * 5) (5  /sqrt 57)

BM^2  = 57/4 + 25 - 25

BM  =  sqrt (57)  /  2 =  AC / 2

\(6x^2 + \frac{3}{5}y^2 - 2x + 8y \)

xy = 2/5  →  y = 2 / (5x)

6x^2  + (3/5)(2 / (5x)^2 - 2x + 8(2 / (5x))

6x^2  + (12/125)x^(-2) - 2x + (16/5)x^(-1)

The graph shows that the minimum ≈ -26.525  when  x ≈ -.06

\(f(x) = \frac{(2x + 3)(x - 7)}{\sqrt{x}} \)

The denominator  cannot  =  0 

In the radical, x  must be >  0

Domain  = (0 , infinity)

1-9  = 9 digits

10 - 99 = 180 digits

100 - 199 = 300 digits

200 - 250  = 151 digits

9 + 180 + 300 + 151 =     640 digits

No. of 2's

1-100  =  10 + 10  = 20

101 - 199 = 10 + 10  = 20

200 - 250  = 5 + 10 + 51  = 66

20 + 20 + 66  =  106

Prob of 2  =   106  / 640 =  53 /  320    (if I didn't mis-count !!! )

5n / (n + 2)  =  6

Can you solve this ???

(n - 3) + (n - 2) + (n -1) + n + (n + 1) + (n + 2) + (n + 3)  = 7(n + 3)

7n = 7(n + 3)

7n = 7n + 21

No solution

Triangle COB is similar to Triangle DOB  by HL

tan (angle  COB) =   1/sqrt(3) = sqrt (3) / 3   = tan angle DOB

tan angle DBC = tan angle ABC

tan (angle ABC)  = tan (angle COB + angle DOB)  = tan (2 angle COB) = AC / BC

tan (angle COB + angle DOB )  =   [ tan COB + tan DOB ] / [ 1 - tan COB*tan DOB] =

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

sqrt 3  =  AC / BC

sqrt 3  = AC / sqrt 3

sqrt 3 * sqrt 3  = AC  =  3

6  3  9

1  8  7

1  1  2

No other possibilities.

The diameter is 4. 

Any perpendicular lines, assuming that they connect to both ends of a cricle, will meet at the center, which is X in this case. Now that we know X is the center of the circle and that 2 is from D to X, we can conclude that the radius is 2. Therefore, the diameter is 4.

25 min =   5/12 hr

Call your  normal time at 40 mph = T

So    rate * time =  distance

Equating  distances

40T = 55 (T - 5/12)

40T =  55T  - 275/12

275 /12  =  (55 - 40)T

275 / 12  = 15T

T = 275/(15*12)   = 55/36 hr

Miles driven =   40 (55/36)  ≈  61.1 miles  

Call the sq ft of the wall =  S

S / [ 45 * 60 ] = S  / [ 12 * x]

45*60  =  12 * x

x =  45 * 60  /  12   =    45 * 5  =   225 min

100 /[ 40 * 50 ] =  480 / [ 40 ( x ] 

100 / 50  = 480 / x

100x = 480 * 50

100 / 50  * x   = 480

2x  = 480

x =  480  / 2  =   240 min

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