CPhill

This is a parabola that turns upward

We can find the x coordinate of the vertex  as  -b/2

So....we can solve this for "b"

(-b/2)^2  + b(-b/2) + 15  < 0

b^2/4  - b^2/2  + 15 < 0     multiply through by 4

b^2 - 2b^2 + 60 < 0

-b^2 + 60< 0   multiply through by - 1  and reverse the inequality sign

b^2 -60 > 0

(b + √60) (b - √60) > 0

We have three possible solution intervals

(-infinity, -√60) or ( -√60, √60)  or ( √60, infinity)

Testing z = 0 in the middle interval in the original inequality makes it untrue

So.....the number of integers between -√60  and √60  =  -7  and 7 inclusive  =  15 integers 

So....15 integer values for b produce no real solutions for z

x ( x + 7.5) > 38.5

x^2 + 7.5x > 38.5

x^2 + 7.5x - 38.5 > 0    multiply through by 2

2x^2 + 15x - 77 > 0     factor

(2x  - 7) ( x + 11) > 0    set  each factor to 0

2x - 7  = 0             x + 11 = 0

2x  = 7                   x  = - 11

x  = 7/2

We have three possible intervals that make the original inequality true

(-infinity, -11)  or  ( -11, 7/2)  or  (7/2, infinity)

Testing 0  in the middle interval makes the inequality false

So....the two intervals that make the original inequality true are

(-infinity, -11) U ( 7/2, infinity )

[ ( 3x + 2y)^2 ( 3x - 2y)^2 ]^3  =

 ( [ (3x + 2y) (3x - 2y) ]^2 ) ^3  =

[ (3x + 2y) (3x - 2y) ]^6 =

[ 9x^2 - 4y^2 ]^6 

Will have 7 terms

The pathway will consist of 4 squares each with a side of x, two rectangles with dimensions of 8 *x   and two rectangles with dimensions of 11*x ......  where x is the width of the path

4x^2  + 2(8*x)  + 2(11*x)  = 120

4x^2 + 16x + 22x  = 120

4x^2 + 38x  = 120

4x^2 + 38x - 120  = 0   factor

(x + 12)(4x - 10)= 0      set each factor to 0 ans solve for x

x + 12  = 0                 4x  - 10  = 0

x = -12  (reject)          4x  = 10

                                    x  = 10/4  = 5/2  =  2.5 ft

So.....the path width is 2.5 ft

Let  x  be the width of the border......then one dimension of the picture and frame  =  ( 10 + 2x)

And the other dimension of the picture and frame  = ( 14 + 2x)

So....the area  =

(10 + 2x) ( 14 + 2x)  = 221      simplify

140 + 28x + 20x + 4x^2  = 221

4x^2 + 48x + 140  = 221       subtract 221 from both sides

4x^2 + 48x  - 81  = 0       factor

(2x - 3) (2x+ 27)  = 0        set each factor to 0   and solve for x

(2x - 3)  = 0                         2x  + 27  = 0

2x  = 3                                  2x  = -27

x  = 3/2  = 1.5 inches            x  = -27/2  = - 13.5  reject

So.....the width is 1.5 inches

I believe that this probability  is just the ratio of the volume of a sphere with a radius of 1 inscribed in  a cube with a side of 2

So.....we have that

(4/3)pi (1)^3            (4/3) pi          4 pi            pi

__________  =     _______  =   _____  =   ____ ≈    .5236   ≈   52.36 %

    2^3                         8                 24              6

kx + y + 3z  = 10      multiply through by -2 ⇒  -2kx - 2y - 6z  =  -20      (1)

-4x + 2y + 5z  = 7   (2)

kx + z  = 3     (3)

Add (1) and (2)   and we have that

-(2k + 4)x  - z  =  -13     multiply through by -1    ⇒    (2k + 4)x  + z =  13      (4)

Note that   if      (2k + 4)  =  k,  then the system will have no solutions

So

2k + 4  = k

k  = - 4

This will be a parabola  that turns upward

To find the x coordinate of the vertex we have that    (k -3) / 2

So

[(k - 3) /2]^2  - (k -3) (k - 3)/2 - k + 6 > 0

( k^2 - 6k + 9 ] / 4   - ( k^2 - 6k+ 9)/2  - k + 6 > 0      multiply through by 4

k^2 - 6k + 9  -  2(k^2 - 6k + 9)  - 4k + 24 > 0

k^2 - 6k + 9  - 2k^2 + 12k - 18 - 4k +  24 > 0

-k^2  + 2k  + 15 > 0         mutiply through by -1 and reverse the inequality sign

k^2 - 2k - 15  < 0  factor

(k - 5) ( k + 3) < 0

We have three possible intervals that solve this

(-infinity, -3)   or  ( - 3, 5)   or  ( 5, infinity)

The outside intervals do not provide the correct solution

So.....the interval   k = (-3, 5)   is correct   

Here :  https://web2.0calc.com/questions/help-please_10141

No big deal, KLP.....we both got the correct answer.....I gave you a point, too !!!