Find the equation of the line that best fits the following set of points

a.      Determine the value of the correlation coefficient

In the start, find the mean value of X

∑_(i=1)^n▒(x_i-X ̅ )(y_i-Y ̅ ) =13

∑_(i=1)^n▒(x_i-X ̅ )^2 =4+1+1+4

∑_(i=1)^n▒(x_i-X ̅ )^2 =10

now slope will become 

m=13/10

m=1.3

Now work for the y-intercept 

b=Y ̅-mX ̅

putting the values 

b=4.25-1.3×4

b=-9.5

Now the equation of line will become like this 

y=mx+b

y=1.3x-0.95

It is showing that value of coefficient is -0.95

Predict the value of  y when x=4

According to the given data the equation of line is 

y=1.3x-0.95 

y=1.3×4-0.95

y=4.25

Question 2

Consider the parabola that is the graph of quadratic function

 f(x)=-2x^2+12x-10=-2(x-3)^2-2(x-1)(x-5). 

Identify the:

X-intercepts of y=f(x)

f(x)=-2x^2+12x-10=-2(x^2+9-6x)-2(x-1)(x-5).

0=-2x^2+12x-10=〖2x〗^2+18-12x- 2(x^2-6x+5)

0=-2x^2+12x-10=〖2x〗^2+18-12x- 2x^2+12x-10

0=-2x^2+12x-10=9

-2x^2+12x-18=0

x=3

Y-intercepts of  y=f(x)

Put x=0

y=-2x^2+12x-10=-2(x-3)^2-2(x-1)(x-5). 

y=-10=18-10

y=-10-8

y=-18

Vertexs, or vertices of y=f(x)

The vertices of x-intercept are 

(3,0)

The vertices of y-intercept are 

0.-18

Question 3

Consider f(x)=〖2x〗^3+x^2-15x-18

Determine the set of possible rational zero of f(x)

According to the given question it can be noted that all coefficients are integers so rational theorem can be applied

The least coefficient is about -18  now according to this factors will become 

±1,±2,±3,±6,±9,±18

this shows all possible values of p

Now according to this fact, the leading coefficient of highest degree is about 2

Now its factors will be 

±1,±2 

according to this possible values of p/q will become 

p/q=±1/1±1/2,±2/1,±2/2  ±3/1,±3/2  ±6/1.±6/2,±9/1,±9/2±18/1,±18/2

P/q=1±1/2,±2,±1 ±3,±3/2  ±6,±3,±9,±9/2±18,±9

Now remove repeating numbers it will become 

p/q= ±1,1/2,±2,3,±3/2  ±6,9,±9/2±18,

Is X=2 a zero of f(x)?

Yes according to the given polynomials, it can be noted that x=2 is a zero of f(x)

It can be prove easily through synthetic division

▭2  ▭(■(x^3&x^2&■(x^1&x^0 ))) 

▭(■(■(2&1&-15)&-18))

After simplifying it we will get 

〖2x〗^2+5x-5  Is the quotient

-28 is the reminder 

Is x=-2 A zero of f(x)

No , it can be seen from the polynomials. Moreover if synthetic division is performed on it, it will get 2x^2-3x-9 as quotient and 0 is remainder. This is why it is no considered a zero of f(x) because it is actual zero.

Determine a complete factorization of f(x)

〖2x〗^3+x^2-15x-18

(x+2)(2x+3)(x-3)

Determine the set of actual rational zero of f(x)

According to given f(x) the actual rational zero can be found with synthetic division. This shows that if its remainder is zero then it is considered as actual rational zero. 

The actual zero are

actual zero=-2.3.-3/2

Question 4

Give an example of a polynomial, f(x) of lowers degree that has zero of 

x=-2,5/3  and 4+i 

You may leave your answer in factored form 

According to this fact, the lowest polynomial is given by a, b and c

The function will become like this 

f(x)=(x-a)(x-b)(x-c)

a=-2

b=5/3

c=4+i

f(x)=(x+2)(x-5/3)(x-4-i)

f(x)=(3x-5)(x+2)/3(x-4-i)

f(x)=((3x-5)(x+2)  (x-4-i))/3

Give an example of a different polynomial, g(x), of lowest degree that has zero of 

x=-2,5/3  and 4+i 

g(x)=(x-a)(x-b)(x-c)

a=-2

b=5/3

c=4+i

g(x)=(x+2)(x-5/3)(x-4-i)

g(x)=(3x-5)(x+2)/3(x-4-i)

g(x)=((3x-5)(x+2)  (x-4-i))/3

Give an example of a different polynomial g(x), of highest degree that has zero of 

x=-2,5/3  and 4+i 

g(x)=(x+a)(x+b)(x+c)

a=-2

b=5/3

c=4+i

g(x)=(x-2)(x+5/3)(x+4+i)

g(x)=(3x-5)(x-2)/3(x+4+i)

g(x)=((3x-5)(x+2)  (x-4-i))/3

Give an example of a polynomial, p(x) of lowest degree that has zero of 

x=-2,5/3  and 4+i 

And has rational coefficients

p(x)=((3x-5)(x+2)  (x-4-i))/3

It can be noted that the number of zero are 3 of the lowest degree

The rational coefficients are -2, 5/3 and 4+i

Question 5

Find all solutions for 

X=x^2+10x+13=0

for that case, use the quadratic formula it will become like this 

x=(-b±√(b^2-4ac))/2a

Putting the values 

x=(-10±√(100-4(1)(13)))/(2(1))

x=(-10±√(100-52))/2

x=(-10±√48)/2

x=(-10±4√3)/2

x=-5±2√3

x=-5+2√3,-5-2√3

Question 6

Consider 

f(x)=(9-3x+2x^2)/(3+2x-x^2 )=(2x+3)(x-3)/(-1)(x-3)(x+1) =(2x+3)/(-1(x+1) )=g(x)

Determine the domain of f(x). Give your answers using interval notation

According to the given function 

f(x)=(2x+3)/(-1(x+1) )

It can be noted that the domain of the function will be 

-x-1>0 

-x>1

x>-1

Its domain will become 

x ϵ (-1,∞)

Determine the equation of any vertical asymptotes of f(x)

For the vertical asymptotes take the f(x)

f(x)=(2x+3)/(-1(x+1) )

Put denominator =0

-x-1=0

-x=1

x=-1

This shows the equation of vertical asymptotes will become 

-x-1=0

Determine the coordinates of any holes for f(x)

For that case, put limit on the given function 

f(x)=(2x+3)/(-1(x+1) )

The limit will be approaches to -1

lim┬(x=-1)⁡〖(2x+3)/(-1(x+1) )〗

=⁡〖(2(-1)+3)/(-1(-1+1) )〗=-∞

The limit will be approaches to 1

lim┬(x=1)⁡〖(2x+3)/(-1(x+1) )〗

=(2(1)+3)/(-1(1+1) )=-5/2

Coordinates of any holes will be 

-5/2,-∞

Determine the equation of any horizontal asymptotes of f(x)

For the equation of horizontal asymptotes its function will be 

f(x)=(2x+3)/(-1(x+1) )

lim┬(x-∞)⁡〖x(2+3/x)/x(-1-1/x) 〗

lim┬(x-∞)⁡〖((2+3/x))/((-1-1/x) )〗

=((2+0))/((-1-0) )

this shows that the horizontal asymptote will become 

y=-2

Question 7

Mary invests $ 20000 in an account that earns 2.75% annual interest compounded monthly. How long will it take care for mary’s investment to double in value?

Given data 

Invested amount = 20000 dollars

Annual interest rate =2.75

To find

Time for double investment 

Solution 

Apply the interest formula it will become 

A=P(1+r/n)^nt

A=2P

r=2.75/100

r=0.0275

n=1

2P=P(1+0.0275/1)^1t

2=(1+0.0275/1)^1t

Apply log on both sides

log⁡2=log⁡〖(1+0.0275/1)^1t 〗

log⁡〖2=1t  log〗 (1+0.0275/1)

t=(log2 )/log⁡(1+0.0275/1) 

t=0.693/0.01179

t=58.77

It shows that it will take 59 years almost to reach this value

Question 8

Solve analytically

〖log〗_5  (2x+1)+〖log〗_5 (x+2)=1

As we know that the log rule is 

log_( c)⁡〖 ( a )〗   +   log_( c)⁡〖 ( b )〗   =    log_( c)⁡〖 ( ab )〗  

Using the multiple rules that provides 

〖log〗_5  ( 2 x + 1)  (x + 2)  = 1

Using the logarithmic conversion rules as if log_( c)⁡〖 ( a )〗    =    b then it can be converted to the equation form as a =  〖 c〗^( b) and equation according to the rule gives

( 2 x + 1)  (x + 2)   =   5^( 1)

( 2 x + 1)  (x + 2)    =   5 

2x^( 2)   +   5x  +  2   =   5

2x^( 2)   +   5x    =   5  -   2

2x^( 2)   +   5x    =   3

2x^( 2)   +   5x  -  3   =   0

Now using quadratic equation formula, it gives 

x  =    (-  b ±  √(b^( 2)   -   4ac ))/( 2a)

a  =  2   ,   b  =5 ,    c  = -3 

Inserting the values back in the equation gives 

x  =    (-  5 ±  √(〖 (5)〗^( 2)   -   4 (2)(-3) ))/( 2 (2))

x_( 1)   =    (-  5 +  √( 25   -    (8)(-3) ))/( 4)

x_( 1)   =    (-  5 +  √( 49 ))/( 4)   

 =     (-  5 +   7)/( 4)     =   ( 2)/( 4)  

 =   ( 1)/( 2)

x_( 2)   =    (-  5  -   √( 25   -    (8)(-3) ))/( 4)

x_( 2)   =    (-  5  -   √( 49 ))/( 4)    

=     (-  5  -    7)/( 4)     =   (- 12)/( 4)

  =   -3

There are two possible solutions as ½ and -3 and the reliable solution is x = 1/2.

Question 9

Determine the centre and radius of the circle whose equation is 

x^2+y^2+5x-y-1=0

Writing the equation in the standard form as (x -  a)^( 2)  +  ( y  - b)^( 2)   =  r^2 where radius of the circle is r and it is centered at a and b two coordinates. Therefore, rewriting the equation in the standard form. 

x^( 2)  + y^( 2)  + 5x - y - 1  = 0

x^( 2)  + y^( 2)  + 5x - y  = 1

Organizing the variables according to x and y variables. Converting the equation in the standard form gives 

(x^( 2)  + 5x)  + (y^( 2)  - y )  =  1

Converting the x equation to the square form 

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

(x^( 2)  + 5x +  25/( 4))     =  (x  +  ( 5)/( 2))^( 2)    

(y^( 2)  - y  +  1/( 4))   =  (y -  ( 1)/( 2))^( 2)   

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

(x  -(- ( 5)/( 2)))^( 2)    +   (y -  ( 1)/( 2))^( 2)   =   ( √(( 15)/( 2)))^( 2)  

From the circle properties equation, it gives 

radius of the circule    =   r   =    √(( 15)/( 2))

center ofcircule  =  c =    ( - ( 5)/( 2),   ( 1)/( 2)  )

Question 10

A guy wire 400 feet long is attached to the top of a tower when pulled Taut it makes a 55 degree angle with the ground.  How tall is the tower? How far away from the base of the tower does the wire hit the ground.

length of wir =   hypotenus  =  400 ft 

angle with base =   θ  =   55

height of the tower=   h  =  ?  

Solving the problem by using trigonometric relation,

sin  θ  =    ( height of the tower )/( length of wire)

To find the height of the tower, we need to modify the equation for the suitable solution 

 lenght of wire ×   sin⁡〖  θ〗      =  height of tower 

400  ×    sin⁡〖 55〗   =  height of tower  =   h 

h  =   328 ft

to find the horizontal distance the equation becomes 

cos⁡〖 θ  〗  =       (horizontal distance )/( length of wire )

Rearranging the equation gives 

cos⁡〖 θ 〗  ×  length of wire  =  horizontal distance 

horizontal distance  =  a  =  400 〖 cos〗⁡〖  55〗  

a =    229.43 

Question 11

Let A be the angle in standard position whose terminal side contains the point Q(4, -3) 

Determine the value of the six basic trigonometric function evaluated at θ

In case of the linear line the equation becomes ax +  by  =   c and the equation is valid for the terminal side of the angle in this form. The slope of the equation becomes 

tan⁡〖θ  〗   = -   ( a)/( b)

The slope changes with the values of the equation as 

tan⁡〖θ  〗   = -   ( a)/( b)   =   -( 4)/( - 3 )    =    ( 4)/( 3)

r  =  √( x^( 2)   +    y^( 2) )    =   √( 4^( 2)   +    (  - 3 )^( 2) )     

r  =   √( 16  +   9)  

r   =  √( 25)    =  5  

The statement defines that relation between angle and process, 

sin⁡〖 θ〗   =    ( y)/( r)

sin⁡〖 θ〗   =    ( - 3)/( 5)

cos⁡〖 θ〗   =    ( x)/( r)

cos⁡〖 θ〗   =    ( 4)/( 5)

〖tan 〗⁡〖 θ〗   =    ( y)/( x)

    〖tan 〗⁡〖 θ〗  =  -  ( 3)/( 4)  

csc⁡〖 θ〗   =   ( 1 )/sin⁡〖 θ  〗       

csc⁡〖 θ〗  =    ( r)/( y)   

csc⁡〖 θ〗  =    ( 5)/( -3)

sec⁡〖 θ  〗  =     ( 1)/〖 cos〗⁡〖 θ〗     =   1/(( x)/( r))

sec⁡〖 θ  〗  =    ( r)/( x)     

sec⁡〖 θ  〗  =   ( 5)/( 4)  

cot⁡〖 θ〗   =   ( 1)/tan⁡〖  θ〗 

   cot⁡〖 θ〗   =  1/(( y)/( x))   

cot⁡〖 θ〗   =     ( x)/( y)

cot⁡〖 θ〗   =   ( 4)/( - 3)

Determine α such that it is coterminal with θ and \ 0  <  α <  360    

Coterminal angle for  0  <   α  <  360

coterminal angle  =  β =  α  +   θ  

coterminal angle  =  β =  360 +   θ    

    〖tan 〗⁡〖 θ〗  =  -  ( 3)/( 4)

     ⁡〖 θ〗  =  tan^( - 1)⁡   -  ( 3)/( 4)

θ   =   36.86

coterminal angle  =  β =  360   -   36.86   = 323.14 

Question 12

Determine the amplitude, period and phase shift of y =f(x)  =  2 sin ( x - ( π)/( 3)). Sketch the graph, including the full cycles, of y =f(x). Label your axes sufficiently to indicate the scale and location of the graph.  

y =f(x)  =  2 sin⁡( x - ( π)/( 3))

Amplitude 

The standard form of equation is f( x )   =   A  sin⁡( B x -  C)   +  D and the standard equation shows |A| as amplitude. Now comparing the equation y = f( x )   =  2 〖 sin〗⁡( x  -  (  π)/(  3)) with the standard equation. It gives that amplitude of the function is 2.

 | A |   =   2 

Period 

The standard equation can be used to compare the equation and extract information. The standard equation of 

a  sin⁡〖 (bx  +   c)〗   +   d  =   (periodicity of sin⁡〖 ( x )〗)/( | b |)

Now comparing our given equation, 

y = f( x )   =  2 〖 sin〗⁡( x  -  (  π)/(  3))

The period of sin⁡〖 ( x )〗 is 2 π and it becomes 

periodicity  =    ( |  2  π |)/(| 1 |)

periodicity  =    2  π

Phase shift 

To find the phase shift the standard relation becomes 

f( x )   =  A g  (Bx   -  c )  +  D

While in the equation g(x) can be described as one of the basic trig. function and C/B can be described as phase shift and D becomes the vertical shift. In the present given function comparing it with the other systems. 

g  (Bx   -  c )   =  2 〖 sin〗⁡( x  -  (  π)/(  3))

Comparing both sides of the equation gives 

B  =   1

C  =   ( π )/( 3)  

D  =   0 

The phase shift of the equation becomes 

( C)/( B)   =    (( π)/( 3))/( 1)  

phase shift  =   ( π)/( 3s)  

vertical shift  D =   0 

Question 13

Find all solution in [0,4π] cos⁡θ=2/5

cos⁡θ  =  ( 2)/( 5)

As we know that 

sin^( 2)⁡〖 θ〗   +   cos^( 2)⁡〖 θ〗   =   1

sin⁡〖 θ〗   =    √(1 -  cos^( 2)⁡〖 θ〗   )

  sin⁡〖 θ〗   =    √(1 -   (( 2)/( 5))^( 2)  )

sin⁡〖 θ〗   =    √(1 -   (( 4)/( 25))^   )

sin⁡〖 θ〗   =    √(((25  -  4))/( 25)   )

sin⁡〖 θ〗   =    √(( 21)/( 25)   )   =  ( √( 21  ))/( 5)

tan⁡〖 θ〗   =   〖sin 〗⁡θ/cos⁡〖 θ 〗    =    (( √( 21  ))/( 5))/(( 2)/( 5))  

tan⁡〖 θ〗   =   ( √( 21  ))/( 2)

csc⁡〖 θ〗   =   ( 1 )/sin⁡〖 θ  〗       

csc⁡〖 θ〗  =    1/( ( √( 21  ))/( 5))

   csc⁡〖 θ〗  =    ( 5)/( √( 21  )  )

sec⁡〖 θ  〗  =     ( 1)/〖 cos〗⁡〖 θ〗     

sec⁡〖 θ  〗  =    1/(( 5)/( 2))     

sec⁡〖 θ  〗  =   ( 2)/( 5)

cot⁡〖 θ〗   =   ( 1)/tan⁡〖  θ〗 

   cot⁡〖 θ〗   =  ( 1)/(( √( 21  ))/( 2))   

cot⁡〖 θ〗   =     ( 2)/( √( 21  )  )

Question 14

Verify the following identity

sin⁡〖θ (tan⁡〖θ+cot⁡θ 〗 )=sec⁡θ 〗

solving left hand side of the equation 

sin⁡〖θ (tan⁡〖θ + cot⁡θ 〗 )  = sec⁡θ 〗     .    .   .    .   .   .   .   .   .   .   ( 1 )

tan⁡〖 θ〗   =  ( 〖sin 〗⁡θ)/〖 cos〗⁡〖 θ 〗   .    .   .    .   .   .   .   .   .   .   ( 2 )

cot⁡〖 θ〗   =   〖 cos 〗⁡θ/( sin⁡〖 θ 〗 )  .    .   .    .   .   .   .   .   .   .   ( 3 )

Inserting equation 2 and 3 in equation 1 gives

sin⁡〖θ (( 〖sin 〗⁡θ)/〖 cos〗⁡〖 θ 〗 ⁡〖 +  〖 cos 〗⁡θ/( sin⁡〖 θ 〗 )〗 )=〖 sec〗⁡θ 〗

Solving the equation 

sin⁡〖θ ((〖  sin〗⁡〖 θ〗 (sin⁡〖 θ〗 )     +  〖 cos〗⁡〖 θ 〗 (〖 cos〗⁡〖 θ 〗 )   )/(sin⁡〖 θ〗  cos⁡〖 θ〗 ))=〖 sec〗⁡θ 〗

sin⁡〖θ ((sin^( 2)⁡〖 θ〗   +  cos^( 2)⁡〖 θ〗    )/(sin⁡〖 θ〗  cos⁡〖 θ〗 ))= sec⁡θ 〗

As we know that 

sin^( 2)⁡〖 θ〗   +   cos^( 2)⁡〖 θ〗   =   1

sin⁡〖θ (( 1   )/(sin⁡〖 θ〗  cos⁡〖 θ〗 ))= sec⁡θ 〗

(( 1   )/( cos⁡〖 θ〗 ))    =   sec⁡θ    

As we know that sec⁡θ    =   (( 1   )/( cos⁡〖 θ〗 ))      then equation becomes 

sec⁡θ    =   sec⁡θ

Hence provided that left hand side of the equation is equal to right hand side of the equation.