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The answer is "C"

I don't understand the question......

2² + 5²   =

4 + 25  =

29

I)     5/ √2   multiply top/bottom by √2  = 5√2 / [√2 * √2]  = [5√2] / 2  

II)  10/ ∛4   multiply top/bottom by  4^2/3  = 10 * 4^2/3 / 4  = [ 5 * 4^2/3 ] / 2  

III)  [ 5 + √2] / [ 5 - √2]     multiply top/bottom by [5 + √2]

     ([5 + √2] [ 5 + √2]) / [ 25 - 2]   =  [25 + 10√2 + 2] / 23  = [ 27 + 10√2] / 23

IV)  √3 / arctan (√3)  =   √3 / [pi/ 3] =   [3 √3] / pi

V)   3 / [ 6^1/5 / 6^1/7 ]   = [3 * 6 ^1/7 ] / 6^1/5      multilply top/bottom by 6^4/5 =

     [ 3 * 6^1/7 * 6^4/5] / [6^1/5 * 6^4/5]  = [ 3 * 6^33/35]  / [ 6]  =     6^33/35 / 2

NO!!

15.0   = 15

150  = 150

No.

Hello Guest!

What percent is 1000 from 2 200 000?

x : 1 000 = 100% : 2 200 000

x = 1 000 * 100% / 2 200 000

x = 0,04545454%

Greeting asinus :- )

  !

New Area : Old Area  =  9 : 1

Hello DamienDenninson!

An jet flies between two points on the ground that are 500 km apart with an airspeed of 200 m/s. The destination is directly north of the point of origin of the flight.. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination?

A.5.9° east of north

B.2.9° east of north

C.2.9° west of north

D.5.9° west of north

E.85° west of north

sin α = ( t * 10 m/s) / (t * 200 m/s)

sin α = 10 / 200 = 0.05

α = arcsin 0.05

Answer B is correct.

The heading of the aircraft is

α = 2.866° east of north

Greeting asinus :- )

  !

An jet flies between two points on the ground that are 500 km apart with an airspeed of 200 m/s. The destination is directly north of the point of origin of the flight.. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination?

A.5.9° east of north

B.2.9° east of north

C.2.9° west of north

D.5.9° west of north

E.85° west of north

I've just noticed this question has been asked and answered earlier.

Please don't post the same question twice!!

1).       w + y + z  = 6   

2).       2w + x + z = 8

3).      3w + x + y =10

4).       x + y + z = 9

5).          3). - 4).     3w - z = 1     

6).          1) + 2).     3w + x+ y + 2z = 14   

From 4) we have x+ y = 9 - z  so, using this in 6). we have:    3w + 9 - z + 2z = 14  or:

7)         3w + z = 5

8).         7). - 6).        2z = 4    hence  z = 2

Put this into 7) to get  3w + 2 = 5  or  3w = 3   or w = 1

Put these back into 1) to get  1 + y + 2 = 6   so  y = 3

and into 2) to get  2 + x + 2 = 8  or  x = 4

.

7P3    is defined as the total number of ordered sets that can be constucted by choosing any 3 elements from a set of 7 elements.....

The number of possible sets in this case  = 210

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I don't see any figure.......

x^2+7x=-8  add 7 to both sides

x^2 + 7x + 8  = 0       this will not factor....using the quadratic formula......the solutions are :

x  =   [ - 7 + √17]/ 2    and  x  = [ - 7 - √17]/ 2

Hre's a couple of practice websites:

http://regentsprep.org/Regents/math/algtrig/ATE3/quadpractice.htm

http://regentsprep.org/Regents/math/ALGEBRA/AE5/PFacEq.htm

I am going to rewrite the equations because I prefer them in alphabetical order:

1)     w +      y + z  =  6

2)   2w + x +      z  =  8

3)   3w + x + y       =  10

4)            x + y + z  =  9

Combine equations 1) and 2) to remove the w-term:

     w +      y + z  =  6     --->   multiply by 2    --->    2w     2y + 2z  =  12

   2w + x +      z  =  8     --->   multiply by -1   --->   -2w - x -      z  =    -8

Add down the columns                                  --->         -x + 2y + z  =   4      (equation 5)

Combine equations 1) and 3) to remove the w-term:

     w +       y + z  =    6    --->   multiply by 3   --->     3w +   3y +  3z  =    18   

   3w + x + y        =  10    --->  multiply by -1   --->   -3w - x - y           =   -10

Add down the columns                                   --->          -x + 2y + 3z =      8

Combine these two new equations:

    -x + 2y +   z  =   4                                   --->   -x + 2y +  z  =   4

    -x + 2y + 3z  =   8   --->   multiply by -1  --->    x -  2y - 3z  =  -8

Add down the columns:                             --->                -2z  =  -4     --->   z  =  2

Combine equations 5) and 4) to remove the x-term:

                                             -x + 2y + z  =  4

                                              x +   y + z  =  9

Add down the columns:              3y + 2z  =  13

Since z = 2:                                3y + 2(2)  =  13     --->     3y + 4  =  13     --->     3y  =  9     --->     y  =  3

Using equation 4):            x + y + z  =  9

Since z = 2 and y = 3:      x + 3 + 2  =  9     --->     x + 5  =  9     --->      x  =  4

Using equation 1):             w + y + z  =  6     --->      w + 3 + 2  =  6     --->     w + 5  =  6     --->   w  =  1

yes what do you need help with? that is the quadratic formula?.

is it multiple choice?