CPhill

You're welcome  !!!!

(x^4y^7)^2/3  =

 ( [ x^4 y^7] ^2 ) ^1/3   =

( x^8 y^14 ) ^1/3  =

∛  [  x^8  y^14 ]  =

To determine  the powers that come out and those that stay in, do this

8/3  =   2  + 2/3          2 powers of x come out, 2 stay in

14/3  =  4  + 2/3         4 powers of x come out, 2 stay  in

x^2 y ^4  ∛  [x^2  y ^2 ]   

       .

x^2  +  bx +  16

If it has one (repeated) root  then  the discriminant  is  = 0

b^2   -  4(1)(16)   =  0

b^2   -  64  =  0            add 64 to both sides

b^2  =   64                    take both roots

b  =  ±√64

b  = ±8

If it has two  roots, then the discriminant  is >  0

b^2  -  64  >  0

b^2  > 64

So.....   b < -8    or  b >  8

In interval notation, we have     (-inf, -8)   U  (8, inf)

The correct  equation in the first part should be

A  =   100(1/2)^t/5.27   

To  see that this is correct....let  t  =  5.27  and we have that

A  = 100(1/2) ^5.27/5.27   =  100  (1/2)¹  =  100 (1/2)   =  50%  left  after 5.27 years, the half-life  

Your   second answer is correct

I can't tell exactly what the third one is....sorry....!!!

sin (x)  =   1  - 2cos^2 (x)

Note  that   cos^2(x)   =  1 - sin^2(x)

So  we have

sin x  =   1 - 2  [  1  - sin^2 (x) ]

sin x  =  1  - 2  +  2sin^2(x)     

sin (x)  =  - 1  +  2sin^2(x)         rearrange as

2sin^2(x)  -  sin (x)  - 1  =  0        factor as

(2sin (x)  + 1) (sin (x) - 1)  =  0

Se each factor to 0 and solve

2sin x  + 1  =  0      subtract  1 from  both sides

2sin x  = -1         divide both sides by 2

sinx  =  -1/2    and this occurs  at    7 pi / 6    + n*2pi   and  11pi / 6  + n*2pi     

[ where n is an integer]

And..for the other solution we have that

sin x  - 1   = 0   add 1 to both sides

sin x  = 1     and this occurs at   pi/2 + n*2pi 

Possible outcomes

P          N          Q

H        H           H

H        H           T

H        T           T 

T         T          T

T         T          H  

H        T           H

T        H           T

T        H           H

Note there is only one outcme with no heads....so...the probability of at least one head is :

7/8

Here's the "formula"  to compute each new point

New Point  =  [ Given Point  - Dilation Center ] * Scale Factor  + Dilation Center

Note that   

(a, b )  - (c, d)  =  ( a - c, b - d)

(a, b)   + ( c, d)  =  ( a + c, b + d)

(a, b) * N  =   ( N*a, N*b )

So....the new point associated with ( 3, - 3)  is

[  (3, - 3)  - (2, -7) ] * 3  +  (2, -7)  =

[  ( 1, 4 ) ]  *3   +  (2, - 7)  =

 (3, 12 )  +  (2, - 7)  =

(5, 5 )  

You should be able to compute the other three

Total possibilities for 1st row

We have  10  choices for each   square  in the row

So...   10⁵  =   100,000 possibilities

%  of  Ni  =  33.1 / 41.9   ≈   79%

%  of O   =   100 -  79  ≈  21%