All Answers 358514 Answers

2 hrs 40 min =  2 + 2/3  hrs =  7/3 hrs

3 hrs 30min = 3 + 1/2 hrs  =  7/2 hrs

Distance =   Rate * Time

Let  the speed of the plane in still air =  S

When there is no wind it is traveling  (S)  mph    and against the wind it is traveling (S - 50) mph

So.....equating the same distance

S * ( 7/3)  =  (S - 50) ( 7/2)

(7/3)S  = (7/2)S -  175

175 = ( 7/2 - 7/3)S

175  =  (7/6)S

S = 175 ( 6/7)  =    150 mph in still air

So....the distance between cities  is   (150) (7/3)  =  350  miles

\( \dfrac{1}{\sqrt2+sqrt3}+\dfrac{1}{\sqrt2-sqrt3} \)

1 [ sqrt2 - sqrt 3]   +   1 [sqrt2 + sqrt 3]

_________________________________  =

  (sqrt 2 + sqrt 3) ( sqrt 2 - sqrt 3)

2sqrt 2

______  =

2 - 3

-2 sqrt 2

\( \frac{4}{1 - \sqrt[3]{3}} + \sqrt[3]{9} \)

Note that    1 - 3    can be factored as a difference of  cubes  =   ( 1 - ∛3) ( 1 + ∛3 + ∛9)

Multiplting the denominator of the first fraction by  ( 1 + ∛3 + ∛9)  we have

[ 4  ( 1 + ∛3 + ∛9) ] /  [ 1 - 3 ]   +  ∛9   =

[ -2 (  ( 1 + ∛3 +∛9) ] + ∛9  =

-2 - 2 ∛3 - 3∛9

2ab -12a + 3a + 3b - 8b - ab       simplify

ab -9a - 5b

a ( b - 9)  - 5 ( b - 9)   =

ab  - 9a  - 5b  + 45

mn = 3m + 6n -2m + 4n

mn = m + 10n

mn - m  = 10n

m ( n - 1)   = 10n

m =  10n  / [ n - 1]

m =  10  [ n]/[n -1]

m        n

20       2

15       3

12       6

11      11

 0        0

-1        5

-4        8

-9        9

 (1-3i)(2-i)(2+i-7+5i)(1+3i-2+4i)

(1 -3i) (2-i) (-5 + 6i) (-1 + 7i)

[ 2 - 6i -i +3i^2 ] [ 5 - 6i  -35i + 42i^2]

[ -1 - 7i ] [ -37 - 41i]  =

[ 37 + 259i + 41ii + 287i^2]  =

[ 37 + 300i - 287 ]  =

[ -250 + 300i]

[15 - 16i]  / [ 2 -3i + 5 -7i]  =

[15 -16i] / [ 7 -10i]    =   

[ 15 -16i] [ 7 + 10i]

_______________   =

[ 7 -10i] [ 7 +10i]

[ 105 - 112i + 150i - 160i^2 ]

_______________________  =

49 - 100i^2

[ 105 + 160 + 38i ]

------------------------  =

49 + 100

[ 265 + 38i ]

__________

     149

One root of the quadratic equation $x^2 + bx -28 = 4x + 14$ is $-7$. What is the value of $b$?

b=3

a11 = a10 + a9

1 = 4  + a9

a9 = -3

a10 = a9 + a8

4 = -3  + a8

a8 = 7

a9  = a8 + a7

-3 = 7 + a7

a7 =  -10

a8  = a7 + a6

7 = -10 + a6

a6 = 17

By inspection  ( a,b,c,d,e) =  (1,0,0,0,0)

A first letter  =  15 ways

E first letter = 11 ways

I first letter = 7 ways

O first letter = 1 way

So  34 total ways

4 + 1/4 =  17/4

4 + 4/17  =  72/17

4 + 17/72  = 305/72

1 + 72 / 305  =  377/305

1/ [ 377/305] =   305 / 377

Here's how to find all possible values of a:

Add the two equations: Adding both equations eliminates the term ab: 2a = 490

Solve for a: Divide both sides by 2: a = 245

Therefore, the only possible value of a is 245.

Note: We don't need to worry about finding possible values of b, as the problem only asks for a. However, if you're interested, you can substitute a = 245 back into either of the original equations to find that b = 5.

So, the answer is simply: 245

\(\frac{1}{\sqrt{36} + \sqrt{39}} + \frac{1}{\sqrt{42} +\sqrt{45}} + \frac{1}{\sqrt{45} + \sqrt{49}} \)

[ sqrt 36 - sqrt 39] / [ 36 -39 ]    +  [ sqrt 42  -sqrt 45 ] / [42 - 45]  +  [ sqrt 45 - sqrt 49] / [45 - 49 ]  =

[   6 - sqrt 39 ] / -3     +  [ sqrt 42 - sqrt 45 ] / -3   +  [ sqrt 45 - 7 ] / - 4  =

[ sqrt 39 - 6 ] / 3  + [ sqrt 45 - sqrt 42 ] / 3   + [7  - sqrt 45 ] / 4

[ 4 sqrt 39 - 24 + 4sqrt 45 - 4sqrt 42 + 21 - 3sqrt 45 ]  /   12

[ -3  + sqrt 45  + 4sqrt 39  - 4sqrt 42 ] / 12  =

[  -3 + 3sqrt 5  + 4sqrt 39 -  4sqrt 42 ]  / 12

Identifiable sequences  = 

7!  / 2!  =     2520

Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 + 5x - 19.  Find \frac{u}{v} + \frac{v}{u}. 

                                                 3x² + 5x + 7  =  2x² + 5x – 19  

Combine like terms                     x²  =  –26  

therefore                                     x  =  + sqrt(–26)

                                                   Call u  =  + sqrt(–26)  

                                                   Call v  =  – sqrt(–26) 

 u       + sqrt(–26)                               v       – sqrt(–26)  

––  =  ––––––––––  =  –1                 ––  =  ––––––––––  =  –1  

 v        – sqrt(–26)                              u        + sqrt(–26)  

therefore                                       u/v + v/u  =  –1 + (–1)  

                                                      u/v + v/u  =  –2   

_.

Since the quadratic's roots are +5 and –5 then the factors are (x – 5)(x + 5).  

That makes the product x² – 25.  Therefore the blanks are filled by 0 and –25.  

_.

                         20 A

                                      E                    

                      144 H        

        B                  D               C

ABH forms a triangle

Angle AHB = 144

Angle BAH = 20

So angle ABH =  180 - 144 - 20  =  16° 

Triangle HQN similar to Triangle FQG

HN / FG =  2 / 7

Therefore  HQ / FQ  = 2 / 7

Therefore  HQ / (HQ + FQ) =  HQ / HF  =  2 / ( 2 + 7)  = 2/9  ⇒ HQ = (2/9) HF

Triangle PFM similar to Triangle PHG

FM / HG =  3 / 8

Therefore PF / PH = 3 / 8

Therfore PF / (PF + PH)  = PF / HF  = 3 /( 3 + 8) = 3 / 11 ⇒  PF = (3/11) HF 

Getting a common denominator  between  9  and 11

HQ = 22/99FH

PF  = 27/99 FH

Therefore

FH - HQ - PF  =  PQ

FH - (22/99)FH - (27/99)FH  = PQ

(50/99)FH  = PQ

Therefore

PQ / FH  =  50 / 99

See this

AE = EB  =  2.5

Height of EAD = Height of EBF =  4

Area of each triangle =  (1/2) ( 4) (2.5)  =  5

Triangle FDC  has a base and  height twice that of EAD.....so....its area = 2^2 * 5   = 20

[ FDC ]  - [ EBF]   =  20  -  5 =   15   = area of EBCD

[ EAD ] +  [ EBCD ]  =  5 + 15  =    20    =   [ ABCD  ]

Simplify as

21mn - 2m - 8n + 8

Add  -19m

So we have

21mn - 21m - 8n + 8  =

21m ( n - 1)  -  8 (n - 1)  =

(21m - 8) ( n - 1)

                      X                          Y

                      90                     120

      60

      W                             Z

XZW  will form a triangle

Angle ZXW  = 90

Angle XWZ = 60

So XZW =  180 - 90 - 60  =   30

Simplify  as

(7 + sqrt 5 + 2sqrt 3)^2 =    (7 + sqrt (5) + sqrt (12))^2

[ (7 + sqrt 5) + sqrt (12) ]  [ (7 + sqrt (5) + sqrt (12) ] =

(7 + sqrt 5)^2 + 2 [ (7 + sqrt (5)) * sqrt (12) ]  +  (sqrt (12))^2  =

49 + 14 sqrt 5 + 5  + 2*7 sqrt (12) + 2sqrt (60) + 12  =

66 + 14sqrt (5) + 14 sqrt (12) + 4sqrt (15)  =

66 + 14sqrt (5) + 28sqrt (3) + 4sqrt(15)

Simplify  as  ( 3y - 11)^4

Expansion

C(4,0)*(3y)^4 =   1* 81 * y^4  =   81 y^4

Check your question.....if there are only 12 students in the school, 15 can't take only Spanish....