CPhill

First bank

Interest is

25000 (1.02)^3  - 25000  = $1530. 20

Second bank

The 25000 will  be worth this much at the end  of the first year :

25000 ( 1 + .043)  = $26075

Then....this amount will be compounded for two  more years at 0.9%....so we have

26075 ( 1 + .009)^2   = $26546. 46

So...the total  amount of interest earned at the second bank  is   $26546.46 - $25000  = $1546.46

The money earns more interest at the second bank

y  = x^2 - 8x + 13

Let's put this in vertex form....complete the square on x

y  = x^2 - 8x + 16 + 13   - 16       factor the first three terms....simplify the rest

y  = (x -4)^2   - 3 

The vertex of this parabola  is    (4, -3)  .....it opens upward

Notice that the line that would be tangent to this parabola at the vertex would be the line  y  = -3 ⇒  "B"

So....this is the line that intersects the parabola in only one place

To  get a feel for this look at the following graph :  https://www.desmos.com/calculator/8gai6us8zr

Note that the other lines either intersect the graph twice....or not at all

focus   ( 4, - 3)    directrix   x = -2

Since the directrix lies to the left of the focus, this parabola  opens to the right

The y coordinate of the vertex will  be  -3

The x coordinate will be :  [x coordinate of the focus plus the directrix] / 2  =     [ 4  +  - 2] / 2   =   2/2  = 1

We have this form

4p ( x - h)  =  (y - k)^2     where  the vertex  is  (h, k)  = ( 1 , -3)

And  p  will be the distance between the vertex and the focus  =  3 units....since the parabola opens to  the right, this will be  positive

So....putting this all together, we have

4(3)(x - 1)  = ( y -  -3)^2

12 ( x - 1)  =  ( y + 3)^2

(y + 3)^2  = 12 ( x  - 1)   ⇒  "D"

center (-2,3)   point  (4, -3)

Since  (4, - 3)  is on the circle, we need to find the distance  between this point and the center....that will be the radius, r...so we have...using the distance formula

√ [ (-2-4)^2  + ( -3 - 3)^2  ]  =  √ [ (-6)^2 + (-6)^2]  = √ [36 + 36]  = √72   = r

So....the equation is  

(x  - h)^2  + (y - k)^2  = r^2      where (h, k)  is the center and  r^2  = 72

(  x -  -2)^2 + ( y - 3)^2  = 72

(x + 2)^2  + ( y - 3)^2  = 72 ⇒    "A"

4x^2 + 4y^2 - 4x + 12y - 6  = 0     add 6 to both sides

4x^2  - 4x + 4y^2 + 12y   =  6        complete the square on x and y

4(x^2 - x + 1/4)  + 4 (y^2 + 3y + 9/4)   =  6  + 1  + 9

4(x - 1/2)^2  +  4(y + 3/2)^2  =  16       divide through by 4

(x - 1/2)^2  + ( y + 3/2)^2  =   4   ⇒   "D"

C  will

(6 - 4i)  - (2 + 4i)  =

6 - 2  - 4i - 4i  =

4  - 8i

1.  The  apothem  =   side  / (2√3)

So

2  = side / (2√3)     multiply both sides by 2√3

2* 2√3   =  side  =   4√3   =   √48

So....the area is given by

(1/2) (√48)^2 * √3 / 2  =

(1/4) 48√3  =

12√3  ft^2    ≈  20.8 ft^2

2. a regular dodecagon with a side length of 5 m.

The area   is given by :

3 ( 2 + √3) * side^2  =   3 ( 2 + √3) * 5^2   =  3 (2 + √3) * 25  =

75 ( 2 + √3)  m^2   ≈  279.9 m^2

I assume that we have

  9    -       1

___         _____   =   2              get  a common denominator on the right

x+ 2        3x - 2

9(3x - 2)  - 1(x + 2)

_______________    =  2

  (x + 2) ( 3x - 2)

27x - 18 - 1x - 2

______________     =  2

   (x + 2) (3x - 2)

26x - 20  = 2 (x + 2) (3x - 2)

26x - 20  = 2(3x^2 + 6x - 2x - 4)

26x - 20  =  6x^2 + 8x - 8     rearrange as

6x^2 - 18x + 12  =  0    divide through by 6

x^2 - 3x + 2  = 0   factor

(x - 2) ( x - 1)  = 0

Setting each factor to 0  and solving for x produces  x  = 2  or x = 1

Given that $a$ and $b$ are positive integers and that $a+b=24$, what is the value of $ab$ if $2ab + 10a = 3b + 222$?

a + b  =  24      so    b = 24 - a

Sub this into the other equation

2a (24 - a)  + 10a  = 3(24 - a) + 222       simpllify

48a - 2a^2  + 10a  = 72 - 3a  + 222

-2a^2 + 58a   = 294 - 3a     rearrange as

2a^2 - 61a + 294  = 0  factor this as

(2a - 49)(a - 6)  =  0

Only the second factor will produce an integer if set to 0  ...so  a - 6  = 0  ⇒  a = 6

And b  = 24 - 6  = 18

So   ab  =  6*18   =  108

7)A circle with diameter 2 is translated 5 units. What is the perimeter of the region swept out by the circle?

I'm assuming  that the "swept out "  region  is the one between both circles

If so.....its  perimeter is  composed of the perimeter of a  circle with a radius of 1  plus  two sides of 5 units each

So we have   

2pi (1)  + 2* 10   =

[ 2pi + 10 ]  units