All Answers 352749 Answers

thank you math god

Any product of a non-zero real and an irrational is irrational   ( I believe)

c.) focus (–2, 1) and directrix y = 13

turns downward

vertex =   ( - 2 ,  (13 + 1) / 2 )   =  ( -2, 14/2)   = ( -2, 7)

p =   6

4p ( y - 7) = - (x - - 2 )^2

4(6) ( y - 7) = - ( x + 2)^2

24 ( y - 7) = - ( x + 2)^2

y - 7=  - (1/24) ( x + 2)^2

y = - (1/24)( x + 2)^2 + 7

Look at the graph:

at 0.25   to  2.25 (2 seconds)

      the ball travels up 3 cm  then down 3cm then down 3cm more then back up 3 cm = 12 cm

From 0.25 s to 1 sec is 0.75 sec .....the ball travels up 3cm and begins to travel down.....a total of more than 3 cm

Amplitude is height from the midline ....or total height / 2  =   3 cm

From highest point at 0.75 sec to lowest point at 1.75 sec   IS   1 second....

b.) focus (3, 5) and directrix y = 7

Again....the directrix is above the focus....so this turns downward

The vertex is hafway between the focus and directrix = (3, 6)

p =   distance between focus and vertex = 1

So we have

4p (y - k) =  -(x - h)^2             (h, k) = (3, 6)   and p = 1

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

4 ( y - 6) = - ( x - 3)^2

y - 6 =  - ( 1/4)(x - 3)^2

y =   - (1/4) ( x - 3)^2 + 6

See:

https://web2.0calc.com/questions/amount-of-real-solutions

a.) focus (0, –2) and directrix y = 2

The directrix is above the focus so this parabola opens downward

The vertex is   halfway between the focus and the directrix....so it must be (0, 0)

"p" is the distance between the vertex and the focus [ or the vertex and the directrix ] =  2

So.....we have this form

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

So we have

4(2) ( y - 0) = - ( x - 0)^2          simplify

8y = -x^2         divide both sides by 8

y = -(1/8)x^2

Umm

The asyptotes are true for both graphs so they do not effect the identity.

The holes are only present for the right hand side so they are the ones that should be stated as non-permissable for the identity.

The graph is vertically stretched by a factor of   [ 2 / (1/2)  ]  = [ 2 * (2/1) ]   =  4

And the graph is shifted up by     8 - 2  =   6 units

We have that

2200 =  500 +  cost per foot * feet of fencing      ......subtract 500 from both sides

1700 =   cost per foot * feet of fencing

1700 = cost per foot * 200

1700 /200 =  cost per foot = 8.5   =  $8.50

f(x) =  2cos (x) - 4

We have the form

f(x) =   Acos [ Bx ] + C

Note that

B = 1    =   2pi / period

This implies that the period must be 2pi

The frequency  =    1 /period =    1 / [ 2 pi ]

f(x) = sin x

g(x)  =    4sin (x - 3)

See the graphs here : https://www.desmos.com/calculator/js4vt924c0

C- the graph is shifted 4 units right

See the graphs here :   https://www.desmos.com/calculator/xaomug9is8

What's the question ???

f(x)=x^2-2x How would you find f(f(f(f(f(f(-1))))))  ?

We (seemingly) have six evaluations to make

(1)  f ( - 1) =  (-1)^2 -2(-1)  = 1 + 2  = 3

(2)   f ( f  (-1) ) =  f (3) =    (3)^2 - 2(3) =  9 - 6 =  3

(3)  f (  f ( f  (-1) ) ) =  f(3)  = 3

Note that this pattern continues....so

f(f(f(f(f(f(-1))))))  =  3

B- the graph shifts 5 units left and 3 units up 

1.

 ax-b=c, for x         add b to both sides

ax = b + c               divide both sides by 

x = [ b + c ] / a

2.

15x+1=y, for x          subtract 1 from both sides

15x = y - 1                divide both sides by 15

x = [ y - 1 ] / 15

3.

(x+f)+2=j, for x       subtract 2 from both sides

x + f  = j - 2             subtract f  from both sides

 x = j - 2 - f

Let the ordering of the integers from lowest to highest be a, b, c, d and e

And we know that  [a + b + c + d + e ] / 5    =  6

a + b + c + d + e  = 30

And the median is 6, so....c = 6

And   ...    e - a =  6  →  e = 6 + a

So....

a + b + c +  d + e =  30

a + b + 6 + d +( 6 + a)  = 30

2a + b + d =   18

b + d =  18 - 2a

[ b + d ] / 2  =   9 - a

"a" must be odd.....so...."e" must be odd

And b, d must be of the same parity....both even or both odd

And since the mode is 6, either b or d  (or both ) are  6

Let a = 1 and b = 6... then d would have to be 10  

But e = 7....which means that d > e......not possible

And if d = 6, then b = 10....but b > c   ....not possible

So a cannot be 1

Next....let a = 3

[ b + d  ] / 2 = 9 - 3

b + d = 12        

So....if either b, d  = 6....then the other must also = 6

So....

3 + 6 + 6 +  6 + 9 = 30        is true

The mean, mode, range and median = 6

So....the numbers are just as EP found  !!!!

\(\text{A little toying with this and lookup at OEIS show that this sum is equal to}\\ \dfrac 3 4\sum \limits_{k=1}^\infty~\dbinom{k+1}{2}^{-1} = \dfrac 3 2\)

\(\text{a little toying with it shows that }\\ f^{\circ k}(x) = c,~\forall k >0\\ \text{here }f^{\circ k} \text{ means applying the function }f, ~k \text{ times, i.e.}\\ f^{\circ 2}(x) = f(f(x))\\ \text{I leave it to you to determine }c \text{ and convince yourself that above is true}\)

EP has the correct answer with 3 6 6 6 9

x^2 - 3x + 4  = -3x-1      add 3x to both sides

x^2+4 = -1                     add 1 to both sides

x^2 + 5 = 0

x^2 = -5

x =+- sqrt( -5)           No REAL solutions to the system...

Thank you for checking