CPhill

Vector

TRUE>>>>!!!!!

Let the first term =  a1

5th term =     a1*r^4 =  11   ⇒    a1 =  11/r^4    (1)

11th term   = a1*r^10 = 5       (2)

Subbing (1) into (2) we have

(11/r^4) * r^10  = 5

11r^6  =  5

r^6  = 5/11       take the 6th root of both sides

r = ( 5/11)^(1/6)

a1 = 11/r^4 =  11/ [(5/11)^(1/6)]^4  =    11/ (5/11)^(4/6)

So...the 8th term is

a1*r^7  =

 ( 11/ (5/11)^(4/6) ) * [(5/11)^(1/6)]^7    =

11 [ (5/11]^(7/6)] / (5/11)^(4/6)  =

11 * (5/11)^(3/6)  =

11 * (5/11)^(1/2) =

11*√5 / √11 =

√11* √5 =

√55

OK.....!!!

3(0.4 - 0.6) * 1.5  =

3 ( -0.2) * 1.5   =

-0.6 * 1,5 =  =

-0.9

First one

Relative max is the y value where the graph is "higher" than at any other y value in the "neighborhood"

Relative min is the y value  where the graph is "lower" than at any other y value in the "neighborhood"

So....considering the "neighborhood" to be  the  interval on x  of  [1,5]

The relative min in this neighborhood is   -14

And the relative max is  -3

Does that make sense, Jenny??

Second one

First statement.....the y intercept of f(x)  = 15.....the y intercept of g(x) = 20(.1.5)^0 = 20 * 1 = 20

So this statement is true

Second statement - I cannot tell whether this is true or not ???...I don't know what the function f(x) is...

Third statement - This is true....both functons are increasing on (-inf, inf)

Last statement -  these are known as exponential functions.......they are always positive...so....this  statement is false

First one.....the function is increasing on the intervals of x where it is rising from left to right

Note these are  x =  (-inf, 0)   and x = (1,2) 

Second one

The function is positive on the intervals where it is  above the x axis

These are  x =  (-8, -6)    and  x = (-2,4)

Purplemath  explains this well :

https://www.purplemath.com/modules/ellipse2.htm

1.   This seems that, at first glance, it would lead to huge numbers...it doesn't....

Note the pattern   .......

Term

   1     13

   2     1^3 + 3^3  = 28

   3     2^3 + 8^3 =  520

   4     5^3 + 2^3 + 0^3  = 133

   5     1^3 + 3^3 + 3^3  = 55

   6     5^3 + 5^3 = 250

   7     2^3 + 5^3 + 0^3 = 133

   8     1^3 + 3^3 + 3^3 = 55

   9      5^3 + 5^3   = 250

Note

4mod 3 = 1 =  1st number in the repeating cycle   = 133

5 mod 3 = 2 =  2nd number in the repeating cycle  = 55

6 mod 3 = 0  =  3rd number in the  repeating cycle  = 250

So..for the 100th term.....

100 mod 3  = 1  =  the 1st number in the repeating pattern =  133

Let's take an example

5 / (x - 1)    +  3 / (x + 2 )   =  5

The LCM  is  (x - 1) ( x + 2)

So

5 (x - 1) ( x + 2) / (x - 1)  + 3(x - 1) ( x + 2) / (x + 2)   =  5(x - 1) ( x + 2)

5(x + 2)  + 3(x - 1)  = 5(x - 1) ( x + 2)

So...multiplying by the LCM    "cancels out" the denominators