CPhill

The area  of the small rectangle at the bottom right is 4 * 2  =  8 in^2

So....the rest of the shaded area must be  70 in ^2

But the dimensions the whole figure less the small rectangle at the bottom right are  10 * 8  = 80 in^2

So....the  area  of the non-shaded area in the middle must just be  80  - 70   = 10 in^2

So...since its height  = 2  ..its width must be  5 in

And its perimeter must be  2 (2 + 5 )   =  14 in

2.)The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i

The sum of the roots  =   -a / 1

So

(2 - 3i) + (2 + 3i)  = - a /1

4  =  -a

-4  = a

The product of the roots will  =  b /1

So

(2 - 3i) (2 + 3i)  =   b

4 + 9  = 13 = b

So....a + b  =   -4 + 13   =   9

4)A rectangle that is not a square is rotated counterclockwise about its center. What is the minimum positive number of degrees it must be rotated until it coincides with its original figure?

180°

1)B' is the image of B rotated 60 degrees counter-clockwise about A. Given AB = 6sqrt3 find BB'.

Let A  lie at  (0,0)

Let B lie at  (6√3 , 0)

Then  B'  will  be   (6√3 cos (60), 6√3 sin (60) )  =  (3√3, 9) 

So....BB'  is  √ [ ( 6√3 - 3√3)^2  + 9^2 ]   =  √ [ (3√3)^2  + 9^2 ]  = √ [ (√27)^2 + 81 ] =

√ [ 27 + 81 ]  =  √108  =  6√3

Thanks, tertre  and X²    !!!

Nice, tertre   !!!!!

x^2  + 16x + c

Set the discriminant to 0  and solve for  c

16^2  - 4(1)(c)  =  0

256 - 4c  = 0      add 4c to both sides

256  =  4c        divide both sides by 4

64  = c

x^2  + ax  +  25

Just like an earlier problem, set the discriminant to 0 and solve for a

So we have

a^2   -  4(1)(25)  = 0

a^2  - 100  = 0     factor this

(a - 10) (a + 10)  = 0

Setting both factors to 0 and solving for a  gives a  = 10   or a  = -10

So

a  = -10, 10

5x^2+19x+16 > 20      subtract 20  from both sides

5x^2 + 19x - 4  >  0     factor

(5x - 1) ( x + 4)  > 0

Setting both factors to 0,.....we can establish 3 possible intervals

5x - 1  = 0            x + 4  = 0

5x  = 1                 x = -4

x  =  1/5

The possible intervals that provide solutions are

(-inf, - 4)  U ( - 4, 1/5)  U ( 1/5, inf )

And either the middle interval solves the original inequality or the two outside intervals do

Testing a point in the middle interval  - I'll pick 0 -  in the original inequality we have

5(0)^2 + 19(0)   + 16  >  20      is false

So...since we are looking for the integer that make the inequality false....the solution will be all  the integers  on  ( - 4, 1/5)

And the integers in this interval  are  -3, -2, -1, 0

So.... 4 integers do not satisfy this

9x^2  + 16x + c

A perfect square binomial will have one root

So...in the quadratic formula, we will have one root when the discriminant, b^2  - 4ac, equals 0....so....

16^2  - 4* 9 * c   =  0

256   - 36c  = 0       add 36 c to both sides

256  =  36c        divide both sides by 36

64/9   =  c