CPhill

Here's my method, QS

Forget the decimal   and write the non-repeating and repeating parts ⇒   142

(1) Subtract  the non-repeating part  from this  =   142  - 1   =   141

(2) Write this over an integer that has the same number of leading  9's as digits in the the repeating part  and the same number of 0's  as digits in the non-repeating part ⇒  990

So  we have

141  / 990    =   47 / 330

I'm waiting for the WolframAlpha website to come back up....it makes the solving of these types of problems way easier with the computations......hopefully....it will be back up soon !!!

We have  that

d  = kt^2       and we need to find  k

We know that

80  = k (4)^2

80  =  16k    divide both sides by 16

5  = k

So......we want to know this distance

[Distance it falls  in  (4 + 8) =  12 sec ]   -  [ Distance it falls in 4 sec ]  =

5(12)^2  -  5(4)^2  =

5  (12^2  - 4^2 )  =

5 (144 - 16)   =

5 (128)   =   

640 m  in the next 8 sec 

9)        Part a.   A triangle has integer length sides. If two sides of the triangle are 16 and 21, how many possible lengths are there for the third side?

Either  

Unknown Side +  16 > 21

Unknown Side >  5

or

16 + 21  > Unknown Side

37 > Unknown Side   ⇒    Unknown Side  <  37

So.....the possible integer lengths range from     6 -  36  =   36 - 6 + 1   =  31 possible lengths

If we have only one solution  ...the discriminant will = 0

So    a  =  3 ,  b  =  4    and  c  =  ???

The discriminant is  given by :

b^2  - 4ac          so.... plugging in, we have that

4^2  - 4(3)(c)  =  0

16 -  12c  = 0

16  =  12c       divide both sides by 12

16/12   = c  =    4/3

Here's the graph that confirms one solution :

 https://www.desmos.com/calculator/1ldxjsdqjn

Avg rate of change  7 to 9   =  (5760 - 1920 ) / ( 9 - 7 )  =  1920

Avg  rate of change  from  (945 - 105 )  / ( 6 - 4)  =  420

1920 / 420  ≈   4.57 times as great

All whole numbers  [ there is a possibility that nothing is purchased ]

f(x)  =  -2x^4  + 3x^2

If we just have even powers on a function [ and posssibly, a constant term ] it will always be even

Proof...if even  f(x)  =  f(-x)

f(-x)  =  - 2(-x)^4 + 3(-x)^2   =   -2x^4 + 3x^2  = f(x)

Nice job, lynx  !!!

1. What is (f−g)(x)?

f(x)=x^4−x^2+9

g(x)=x^3+3x^2+12

f - g   =   [ x^4 - x^2 + 9 ]  - [ x^3 + 3x^2 + 12 ]  =

x^4  -- x^3  - x^2 + 3x^2  + 9  -  12   =

x^4 - x^3   +  2x^2  -  3 

2. What is (f⋅g)(x)?

f(x)=x^2+2x−6

g(x)=x^4+3

f*g   =  ( x^2 + 2x + 6)  ( x^4 + 3)  =

         x^2  +  2x  + 6        multiply each term on top by 3  and then by x^4      

x                 x^4  + 3

       ____________

         3x^2  + 6x + 18

+    x^6  + 2x^5 + 6x^4

     ________________

x^6  + 2x^5 + 6x^4  + 3x^2 + 6x + 18  

3. What is (f+g)(x)?

f(x)=x^3−x

g(x)=x^3+2x^2−10   

f +  g   =   [ x^3 - x ]  +  [ x^3 + 2x^2  - 10  ]  =   2x^3 + 2x^2 - x  - 10

Do you mean this ???

ax^2 + 8x + c   =  0    ???   if so

If we have only one solution, the discriminant will = 0   .....so.....

8^2  - 4ac  = 0            and      a + c  =  10   ⇒    c  =  10 - a

So....substituting, we have

8^2 - 4a(10 - a)   = 0

8^2  - 40a + 4a^2  =  0

4a^2 - 40a +  64  =  0         divide through by 4

a^2  - 10a  + 16   =  0      factor

(a - 8) ( a - 2)  =  0

The solutions are a  = 8    or a  = 2

Note that if a  = 8, then c  = 2     or

If a  = 2  then c  = 8

But a > c  so     (a, c)   =   (  8, 2 )

Check the ggraph here that only one zero exists  at  x  = -1/2 :

https://www.desmos.com/calculator/idsuikzats