CPhill

The point is    (2, -2)

r  = √ [ (2)^2 + (-2)^2 ]  = √8  = 2√2

θ  =  arctan ( -2/2)  = arctan (-1)  =   -45°   = -pi/4   =  (2pi - pi/4)  =  (8pi/4 - pi/4)  = 7pi/4

So

2√2 ( cos  7pi/4  +  i sin 7pi/4 )

8(cos 240°   +   i sin 240°  )   =

8 ( -√3/2  - 1/2 i )  =

 - 4√3  -  4 i

( 7 cos pi/2, 7sin pi/2 )  =

(  7 * 0 ,  7 * 1 )  =

(0 , 7)

r  = √[2^2 + 2^2]  =  √8  =  2√2

θ  =  arctan (2/2)   =  arctan (1)  =  45°

So

2√2 ( cos 45°  + i sin 45° )

THX, EP!!!!

4x^2 - 2y^2 = 4  ⇒  2x^2 -y^2 = 2   (1)

x^2 + y^2  = 10   (2)

Add (1) and (2)   to get

3x^2 =  12         divide both sides by 3

x^2  = 4           take both roots

x = ±√4

So

x = 2        and     x  = -2

And when x is either of these values, we can find y  as

(2)^2 + y^2  = 10

4 + y^2  = 10

y^2  =  6

y  = ±√6

So.....we have these 4 solutions

(2, √6)  (2, -√6)  ( -2, √6)   (-2, - √6 )

s  =  [ u^2 + v^2]  / [ 2uv]  =  [ u^2 + v^2] [ 2uv]^(-1)

ds/ du     [ Here...we treat  v as a constant]....so  we have [  using the product and chain rules ]

[ 2u + v^2] [ 2uv]^(-1)   - (2v)  [ u^2 + v^2] [ 2uv]^(-2)  

Similarly

ds /dv  [ we treat u as a constant ]

[ u^2 + 2v] [ 2uv]^(-1) - (2u) [ u^2 + v^2] [ 2uv]^(-2)

P ( z < 1.33)  = 0.9082

If P (not A)  = 3/4....then P(A) = 1/4

If independent

P(A and B ) = P(A) * P(B)

3/40  =  1/4 * 3/10

3/40  = 3/40    Independent....good job !!!

Second one

P(A and B)  = P(A) * P(B)

9/100  =  1/ 5  *  9 /20

x^2 + 6x + 18 > 2 - (1/2)x

2x^2 + 12x  + 36 >  4 - x

2x^2 + 13x + 32 > 0

This is a parabola that opens upward.....

The discriminant is

13^2 - 4(2)(32)  =  -87

Since the discriminant is negative.....we have no real roots....which means that this parabola lies above the x axis for all x's

Since we worked down to a logical conclusion....then it must be that our assumption was true....that is...

x^2 + 6x + 18 > 2 - (1/2)x    for all  x's     !!!

[Your method is completely valid, YEEEEEET....!!!!]