CPhill

First one

a₁  = 4_n + 4

Second one

First term is 34

Each successive term is -8

So we have

a_n  =  42 - 8_n

First one :

The difference between successive terms is 9   and the first term is  21

So we must have that

a₁  = 12n + 9    ⇒   a₁  = 12 (1)  + 9   =  21

So...the last answer is correct

Second one

a_n  = 17  - 5n

The difference between successive terms is - 5   and the first term is 12

So we have

a_n  = a_n-1 - 5

a₁   = 12

Second answer

0 ≥ 54p - 144

0 > 12 - 20p

144 ≥ 54p    divide both sides by 54

144/54 ≥ p

8/3 ≥ p

p ≤ 8/3

0 > 12 - 20p

20p > 12    divide both sides by  20

p > 12/20

p > 3/5

The  answer is :

p  = (3/5, 8/3 ]

x  = 4cos t    ⇒  x/4  = cos t

y  = sin t

So...using  cos^2 t  +  sin^2 t  =  1, we have

(x/4)^2  + y^2   =  1

x^2 / 16  + y^2   =  1

This is an ellipse

When t  = 0   we have (4,0)

When t  = pi/2  we have  (0, 1)

When t  =  pi  we have (-4, 0)

When t  = 3pi/2  we have (0, -1)

When t   2pi  we have (4, 0)

So....the curve is traced out in a counter-clockwise direction

Here's the graph :  https://www.desmos.com/calculator/yscc2qxqsk

Here is a formula that will always work for these kind of problems

New Point  =  [ Old Point - Dilation Center ] * Scale Factor  + Dilation Center

Note that  (a, b) + (c - d)  = (a + c, b + d)

And that (a,b) - (c,b)  = (a - c, b - d)

Also  (a,b)*constant  =  (a*constant , b*constant )

So we have

New Point   =  [ (4,7) - (1,2) ] *10 + (1,2)  =

[(3, 5)]*10  + (1.2)  =

(30, 50)  + (1,2) =

(31, 52)

Let's find M thusy

Note that there are 5 equal sements  in PQ.....M is  2/5 of the way between these....so we have..

[  -9  + (11 - -9)(2/5)  ,  7 + (12 - 7) (2/5)  ]  =

[ -9 + (20)(2/5) , + 7 + (5)(2/5) ]=

[ -9 + 8 , 7 + 2 ]  =

(-1, 9)   =  M

The slope of PQ  is    [ 12 - 7 ] / [ 11 -  - 9 ]  =  5/20  =  1/4

So...the slope of the perpendicular line through M  =  -4/1  =  -4

So...the equation of the line L is

y = -4 (x - -1) + 9

y = -4( x + 1) + 9

y = -4x - 4 + 9

y = -4x + 5

Here's a graph :   https://www.desmos.com/calculator/axxvuzl6js

The difference between terms is 9

Consider  "C"

a₁  = 12(1)  +  9   =  21  

So.....this must be the correct choice

y  = pq^x

Note that when  x  = 0,   then y  = 5

This suggests that

5 = pq^0

5  = p (1)

5  = p

And the point  (4, 405)  is on the graph..so

405  = 5q^(4)    divide both sides by 5

81  = q^4        take the 4th root of each side and we get that

3  = q

So....to find  k   when  x  =2, we have

k  =  5*(3)^2   =  5 * 9   =  45

Thanks, guest....that's easier than my approach  !!!!

A(w)  =   -w (w  - 80)  =  -w^2  + 80w

The width that maximizes the area can  be found  by :

-80 / [2* -1)  = -80/-2  = 40

And the maximum area  is

-40 (40 - 80)  =  -40 * -40  = 1600 m^2

[ This isn't surprising....a square always maximizes the area for a given perimeter....so....really....all we need to do is to divide the perimeter by 4  and square this  result....so we would get  (160 /4)^2  =

(40)^2  =   1600 m ^2   }