CPhill

Don't know about that....LOL!!!!

Thanks, Jess !!!!

2.Find the smallest real value of x such that x^2+6x+9=24.

Factor the left side

(x + 3)^2  = 24               take the negative root  [ since we want the smallest value ]

x + 3 = - √24       subtract 3 from both sides

x =  -√24 - 3  =    -2√6 - 3      and this is the smallest real value that makes the equation true

1. What is the largest value of k such that the equation 6x-x^2=k has at least one real solution?

Rearrange as    x^2 - 6x + k = 0

If this has at least one real solution, the discriminant must be ≥ 0

So

(-6)^2 -   4(1)(k) ≥ 0

36 - 4k ≥ 0

36 ≥ 4k 

9 ≥ k

So .....the max value  of k that produces at least one real solution is when k = 9

The common ratio is -1.5

To find the number of terms, we have

57875 =  8000 [ 1 - (-1.5)^n ]   / [  1 - (-1.5) ]

57875 = 3200 [ 1 - (-1.5)^n ] 

2315/128 = 1 - (-1.5)^n

(-1.5)^n =  1 - 2315/128

(-1.5)^n = -2187/128       n must be odd.....so....we can solve this

1.5^n =  2187/128     take the log of both sides

n log 1.5 =  log (2187/128)

n =    log(2187/128) / log (1.5)   =  7

So....we are looking for the 7th term which is

8000(-1.5)^(7 - 1)  =   91125

JUMBO QUESTION

Using a circle in the complex plane  centered at (0,0) with a radius of 2 , the 6 roots of 64 are

2, 1 + √3 i , -1 + √3 i , -2, -1 - √3 i , 1 - √3 i 

Here's a pic of the solutions

1. Enter the amplitude of the function f(x) .

f(x)=−2sin(3x)−1

In the form  Asin (Bx) + C

The amplitude =  l A l   =  l -2 l  = 2

2. What is the equation of the midline for the function f(x) ?

f(x)=3cos(x)−2.5

Note that the max for the function is 3(1) - 2.5   =  0.5

And the min for the function is    3(-1) - 2.5 =  -5.5

To find the midiine.....add these values and divide the sum by 2

So we have

[ . 5  + - 5.5] / 2 =     -5/2 =  -2.5

So....the equation for the midline is   y = -2.5

Nice job, tertre....!!!

The greatest possible value of xy will occur when x is most nearly equal to y

We have that

x     y            xy

17   5           85

14  10        140

11   15        165

 8    20        160

So.....xy  is largest when x =  11 and y = 15

2. In triangle ABC, \angle ABC = 90° and AD is an angle bisector. If AB=90, BC=x, and AC=2x-6, then find the area of triangle ABC. Round your answer to the nearest integer. 

Don't see where the angle bisector comes into play, here.....

We have that

BC^2 + AB^2 = AC^2

x^2 + 90^2 =  (2x - 6)^2

x^2 + 90^2 =  4x^2 - 24x + 36

3x^2 - 24x  - 8064 = 0

x^2 - 8x - 2688 = 0    this factors as

(x - 56) (x + 48) = 0

x = 56

So....the area is   (1/2)product of leg lengths  = (1/2)(AB)(BC) = (1/2)(90)(56) = 2520 units^2