All Answers 358514 Answers

Simplify as

5x^4  + 5x^3  - 13x^2  + 19x -  47  = 0

No rational roots are possible

WolframAlpha shows the real  roots to be     x ≈ -2.8195   and x ≈ 1.5644

The non-real roots are  ≈ .1276 +/- 1.4542i

If 1 -sqrt 2  is a root, so is 1 + sqrt 2

So

(x - (1 -sqrt 2)) ( x - (1 + sqrt 2)) (x - 2) ( x -7)

Taking this  by parts

(x - (1 - sqrt 2)) ( x -(1 + sqrt 2) = 

x^2  - (1-sqrt 2)x - (1 + sqrt 2)x  +  ( 1-sqrt 2)(1 + sqrt 2)   =

x^2  -2x  + 1 - 2  =

x^2 -2x - 1

And ( x -2) (x -7) =  x^2  - 9x + 14

So

x^2  - 2x   -1

x^2 -9x + 14     

___________    =

x^4 -9x^3 + 14x^2

     -2x^3 +  18x^2  - 28x

                -   x^2    + 9x  - 14   =

x^4 -11x^3 + 31x^2 - 19x - 14   

f(sqrt 5)  =  0

So

(sqrt 5)^3  + b(sqrt 5) + c =  0

5sqrt 5  + bsqrt (5)  + c   = 0

One  possibility

b = -5     c = 0

b + c  =  -5

The solution is x = -3/11.

The largest possible value of x + y is 11 + 5*sqrt(3).

First, reduce/simplify the quadratic to:

-x^2 -25x +4      This is a dome shaped parabola 

put it into vertex form to find the highest point

-1 ( x^2 + 25x) =  y -4 

-1 ( x+12.5)^2   = y-4 - 156.25

-1 ( x+12.5)^2  = y + 160.25          shows vertex is   -12.5, 160.25

So the y-values this parabola can have ( the range)   is     ( - inf, 160.25]             

The answer is f(x) = -24x^5 + 40x^4 + 66x^3.

No  AI....I'm the real  thing

4.

y  = -4sin x

This is  just the   graph  of   y =  4sin (x)   reflected across the x axis

The amplitude = 4

The midline =  0

High point  =  4

Low point = -4

Here's the graph

What the heck?!?! You're really smart 🥹🥹 TYSM

also, just to make sure these aren't answers from ai? ☹️

1.  We have the form

y  = A sin (1/3 * x)  - 1

A  = the amplitude  =  l  -5 l  =   5

2. Midline

y = (High point + Low point) / 2

Low point  = -5sin (pi/2)  - 1=   -5 * ( 1 )  -1  =  -6

High point  = -5 (sin (3pi/2)  - 1 =  -5 * (-1) - 1  =  4

Midline =  y =  ( 4  + - 6)  / 2   =  -2  / 2  =  -1

3. Frequency =  1/ period  

Period  =  2pi  / B  =  2pi / (1/3)  =  6pi

So the frequency =  1/ (6pi)

3. At a dock on the east coast, low tide occurs at 3 p.m. with a water depth of 5ft. The depth at high tide is 83 ft. High tide occurs every 6 hours. Explain how to find the sinusoidal function that models the depth in terms of time, X.

The function will be

y =  A cos (Bx + C)  +  D

High tide will occur at  noon  and  6 PM

Low tide occurs at 3PM and 9PM

The period =    6

So

2pi / B = 6

B = (2/6)pi  =  pi/3

Amplitude = A  = (high point -low point)/ 2 = (83 - 5) / 2  =  39

D =   (high point + low point) / 2 =    (83 + 5)  / 2 =  44 

Let noon be  x  =  0

The  hardest thing to figure is "C"

When x=  0, y= 83   we can  solve for "C"  thusly

y = 39sin (pi/3 *x + C) + D

83   = 39 sin (pi/3 * 0  + C)   + 44

39  = 39 sin (C)

39/39 = sin (C)

1  = sin (C)

arcsin (1)   = C

We are asking....where is the sin   = 1 ???     answer...at pi/2

arcsin (1)  = pi/2  =  C

So...our  function  is

y = 39 sin (pi/3 * x  + pi/2) + 44

Here's the graph

Sorry to bother 😭 but, can you also do question 3. 

2.

We have

y= A cos (Bx)  + C

y = 3cos (4x)  +  5

A= the amplitude  = 3    (vertical stretching)

B = 4

The   period = 2pi/ B =  (2pi) / 4  =  pi/2  =  ( horizontal compression from  normal period of 2pi)

5  just shifts  the normal  graph up 5 units

Here's  the normal graph  and our transformed graph 

1.

y = sin ( (pi/3 * x)

Just a normal  sine graph with a different  period from normal  

This has the form

y= sin  (Bx)     where  B = pi/3

Period  = 2pi / B  =     2pi / (pi/3)  =  2pi * 3/pi  =  6

Here's the  graph

Okay! The questions should be posted! ❤️

3x^2 + y^2 + 9x - 5y - 20 = 8x^2 + 6x + 47

Simplify as

-5x^2 + 3x  + y^2 - 5y   =  67

y^2 - 5y  - 5x^2  + 3x  = 67

Hyperbola  

y^2  term positive

x^2 term  negative

Simplify as

2x^2 + 4x  + 4y^2  + 20y   =  -k    complete the square on  x and  y

2(x^2 + 2x  + 1)  + 4(y^2 + 5y + 25/4)  =  -k  + 2  + 25

2 ( x + 1)^2   +  4( y + 5/2)^2  =  27  - k

(x + 1)^2 / (1/2)  + (y + 5/2)^2  / (1/4)  = 27  - k     divide both sides by (27 -k)

( x + 1)^2 / [ (1/2)(27 -k)]  +  (y+ 5/2)^2 / [(1/4) (27 -k) ]  =  1

a^2  =  (27 - k) / 2       b^2  = (27 - k) / 4

a =  sqrt [(27 -k) /2 ]    b = sqrt [ (27 -k) /4)]

Area =  pi * a * b   =  6  pi ......so

pi * a * b   = 6 pi

  a *b   = 6

sqrt  [ (27 -k) / 2   *  (27 -k) /4 ]  =  6

(27 - k) / sqrt 8  =  6

(27 - k) / (2sqrt 2) = 6

27 -k  =  6 * 2sqrt 2

27 - k  = 12sqrt 2

k  =  27 - 12sqrt 2

Using polynomial division

                          x^2 - 2x + 8

x^2  + 2x  - 5  [  x^4  + 0x^3  - x^2  + 3x  -7 ]

                          x^4   +2x^3  - 5x^2

                         _______________________

                                   -2x^3  + 4x^2  + 3x

                                  -2x^3   - 4x^2  + 10x

                                 ___________________

                                               8x^2   - 7x  - 7

                                               8x^2   +16x -40

                                               _____________

                                                        -23x  + 33

Q = x^2 -2x + 8

R = 33 - 23x

I'll try  !!!

Let A  =(-3,3)  B =  (3,3)   C  =(3,-3)  D = (-3,-3)

Let  the  center of  the ellipse = (0,0) ....let  one of the axis lie along x....such that (a,0)    = P 

The equation of the ellipse is x^2/a^2 + y^2/b^2  =1

Constructing circles at  (3,3)  and (3,-3)  both with radii = 8   we have the equations

(x - 3)^2  + ( y -3)*2   =8^2

(x -3)^2 + (y +3)^2  = 8^2

Let y = 0  and we can  find a  as

(x-3)^2 + 9 = 8^2

(x -3)^2 = 55

x- 3  = sqrt (55)

x = sqrt(55) + 3  = a  ≈  10.42

To find b^2  we can use this equation

x^2 / 10.42^2  +  y^2 / b^2   =1

Since (x,y)  =  (3,3) is on the ellipse  then

3^2 /10.42^2   +  3^2 / b^2  = 1

9 / 10.42^2  + 9/b^2  =1

9 / b^2  =  1 - 9/10.42^2

9/b^2 = .917

b^2  = 9/9.17

b = sqrt (9/.917)  ≈  3.13

The  approximate  area of  the ellipse =  pi * a * b   = pi * 10.42 * 3.13 = 102.46