Negative five teen divided by three

16. The price for which supplies equals the demand is 96.236 (see graph)

C) $96.24

17.

 Roots   2, -4, and 1 + 3i

The minimum degree polynomial would be

(x-2)*(x+4)*(x-(1+3i))(x-(1-3i)) = f(x)

f(x) = x4 - 2x2 + 36x - 80

A) f(x) = x4 - 2x2 + 36x - 80

18.
-2i is a zero of f(x) = x4 - 45x2 - 196 

Another zero must be 2i (conjugate) 

f(x) = (x2 +4)(x+7)(x-7) = x4 - 45x2 - 196

2i,7,-7

B) 2i, 7, -7

19. The asymptotes are y = 2 , x = -2, x= 2

(See graph2)

C) x = 2, x = -2, y = 2

20.
           five pi divided by six
360 degrees ………….2pi radians

x……………………..5pi/6 radians

x = 150°

B) 150°

21.

         sin B = 21/75 = 7/25

 

    tan B=21/72 = 7/24

C) sin B = seven divided by twenty five ;

tan B = seven divided by twenty four

22.  X = 17/cos(58°) = 32.080

A) 32.08

23.
    Coterminal angles 202°

202° +360° = 562°

202° -360° = -158°

D) 562°; -158°

24. period of the function.
y = 5 cos (x/2) (see graph3)

A) 4π

25.

y = csc-1(-1) = -pi/2

B) negative pi divided by two

26.
arcos( cosine of pi divided by two)
arcos(cos(pi/2)) = arcos(0) = pi/2

D) pi divided by two

27.

sin45 = 0.7071

cos45 = 0.7071
B) θ = 45°

28.

Pythagoras theorem

c**2 = a**2 + b**2

c = sqrt(17**2 +7**2) = 18.384 ft

C) 18.4 ft

29.
tan (7*pi/8)

 

tan(x/2)  = ±sqrt((1-cosx)/( 1+cosx))
tan(7*pi/8)  = -sqrt((1-cos7*pi/4)/( 1+cos7*pi/4)) = -0.414 = 1-sqrt(2)

B) 1 - square root of two

30. See graph4

 

[0, 2π).
4 sin2 x - 4 sin x + 1 = 0

D) pi divided by six , five pi divided by six

31.

f(x) = 3x**2 - 1 (1 point)
f(-x) = 3(-x)**2 -1 = 3x**2 -1 = f(x)

 

f(x) IS EVEN

B) Even

32. The first polynomial is a factor of the second if the division does not have any remainder.

This can be checked by substituting x = 5 in the second polynomial and verifying if it is a root

 

 3(5)**2 + 5(5) + 50 =  150

 

It is Not a factor

B) No

33.
k = 2; f(x) = 2x3 + 3x2 - 4x + 4; Lower bound?

Using synthetic division, the results are (see image 4)

 

2,7,10,24

 

And since they are all positive values, k= 2 is an upper bound.

B) No

34. f(g(x)) = x , g(f(x)) = x.

 

f(x) = (x -7)/(x+3)

g(x) = (-3x - 7)/(x-1)

 

f(g(x)) = [(-3x - 7)/(x-1) -7] / [[(-3x - 7)/(x-1) +3]
f(g(x)) = [-10x/(x-1)] / [-10/(x-1)] = x

 

g(f(x)) = [(-3(x -7)/(x+3) - 7)]/[((x -7)/(x+3)-1)]

g(f(x)) = [-10x/(x+3)] / [-10/(x+3)] = x

35. Verify
cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

 

 

If we use the following identity

cos(2a)=1−2.sin2(x)

>> 

cos(4x) = 1-2*sin2(2x)

cos (2x) = 1-2*sin2(x)

If we add them ………..> cos (2x)+ cos (4x) = 2 -2*sin2(x) -2*sin2(2x)

[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]

[tex]QUESTION 16[/tex]

If the demand and supply are equal, then we equate the two functions in p and solve for p.

That is

[tex]S(p)=D(p)[/tex]

[tex]400 - 4p + + 0.00002 {p}^{4} = 2800 - 0.0012 {p}^{3}[/tex]

We can rearrange to obtain,

[tex]0.00002 {x}^{4} + 0.0012 {x}^{3} - 4x - 2400 = 0[/tex]

[tex]2 {p}^{2} + 1200 {p}^{3} - 400000p - 240000000 = 0[/tex]

The real roots of this polynomial equation are:

[tex]p = - 118.26 \: p = 96.24[/tex]

Since price can not be negative, we discard the negative value ,

[tex]p = 96.24[/tex]

The correct answer for question 16 is C.

[tex]QUESTION 17[/tex]

We were given the solution to this polynomial as

[tex]x=2,x=-4, x=1+3i[/tex]

We need to recognize the presence of the complex root and treat it nicely.

There is one property about complex roots of polynomial equations called the complex conjugate property. According to this property, if

[tex]a + bi[/tex]

is a solution to

[tex]p(x)[/tex]

then the complex conjugate

[tex]a - bi[/tex]

is also a root.

Since

[tex]x = 1 + 3i[/tex]

is a solution then,

[tex]x = 1 - 3i[/tex]

is also a solution.

Therefore we have

[tex]f(x) = (x - 2)(x + 4)(x - (1+3i))(x - (1 - 3i)[/tex]

[tex]f(x) = ( {x}^{2} + 4x - 2x - 8)( {x}^{2} - (1 - 3i)x - (1 + 3i)x + (1 + 3i)(1 - 3i))[/tex]

We expand to obtain,

[tex]( {x}^{2} + 4x - 2x - 8)( {x}^{2} - x + 3xi - x - 3xi + 1 + 9)[/tex]

Note that:

[tex]{i}^{2} = - 1[/tex]

[tex]f(x) = ( {x}^{2} + 2x - 8)( {x}^{2} - 2x + 10)[/tex]

We now expand to obtain,

[tex]f(x) = {x}^{4} - 2 {x}^{3} + 10 {x}^{2} + 2 {x}^{3} - 4 {x }^{2} + 20x - 8 {x}^{2} + 16x - 80[/tex]

We simplify further to obtain,

[tex]f(x) = {x}^{4} + 2 {x}^{2} + 36x - 80[/tex]

The correct answer for question 17 is A.

[tex]QUESTION 18[/tex]

If

[tex]-2i[/tex]

is a zero of the polynomial,

[tex]f(x)=x^4-45x^2-196[/tex]

then the complex conjugate

[tex]2i[/tex]

is also a zero,

This means that ,

[tex](x+2i), (x-2i)[/tex]

are factors of the polynomial.

The product of these two factors,

[tex](x+2i), (x-2i) =x^2-(2i)^2=x^2+4[/tex]

is also a factor , so we use it to divide and get the remaining factors.

see diagram for long division.

The above polynomial can therefore factored completely as,

[tex]f(x)=(x^2+4)(x^2-49)[/tex]

Applying our knowledge from difference of two squares, we obtain,

[tex]f(x)=(x+2i)(x-2i))(x-7)(x+7)[/tex]

Hence all the zeroes of these polynomial can be found by setting

[tex](x+2i)(x-2i))(x-7)(x+7)=0[/tex]

This implies,

[tex]x=-2i,x=2i,x=7,x=-7[/tex]

The correct answer for question 18 is B

[tex]QUESTION 19[/tex]

We were asked to find the horizontal and vertical asymptote of

[tex]f(x)=\frac{2x^2+1}{x^2-1}[/tex]

To find the horizontal asymptote, divide the term with the highest degree in the numerator by the term with the highest degree in the denominators. That is the horizontal asymptote is given by,

[tex]y=\frac{2x^2}{x^2}=2[/tex]

For vertical asymptote, equate the denominator to zero and solve for x.

[tex]x^2-1=[/tex]

[tex]\Rightarrow x=-1, x=1[/tex]

None of the options is correct, so the correct answer for question 19 is A.

[tex]QUESTION 20[/tex]

We are converting,

[tex]\frac{5\pi}{6}[/tex]

to degrees .

To convert from radians to degrees, multiply by,

[tex]\frac{180\degree}{\pi}[/tex]

That is,

[tex]\frac{5\pi}{6}=\frac{5\pi}{6} \times \frac{180\degree}{\pi}[/tex]

We simplify to obtain,

[tex]\frac{5\pi}{6}=\frac{5}{1} \times \frac{180\degree}{1}=5\times30\degree =150\degree[/tex]

The correct answer is B.

[tex]QUESTION 21[/tex]

Recall the mnemonics, SOH CAH TOA

The sine ratio is given by,

[tex]\sin(B)=\frac{21}{75}=\frac{7}{25}[/tex]

From the diagram,

[tex]\tan(B)=\frac{21}{72}=\frac{7}{24}[/tex]

The correct answer is C.

[tex]QUESTION 22[/tex]

From the above diagram, We can determine the value of x using the sine or cosine ratio, depending on where the 17 is placed.

Using the cosine ratio, we obtain,

[tex]\cos(58\degree)=\frac{17}{x}[/tex]

We can simply switch positions to make x the subject.

[tex]x=\frac{17}{\cos(58\degree)}[/tex]

[tex]x=\frac{17}{0.5299}[/tex]

[tex]x=32.08[/tex]

Hence the correct answer is A.

[tex]QUESTION 23[/tex]

Coterminal angles have the same terminal sides.

To find coterminal angles, we keep adding or subtracting 360 degrees.

See diagram.

[tex]x=202\degree[/tex]

is coterminal with

[tex]202\degree +360\degree =562\degree[/tex]

or

[tex]202\degree -360\degree =-158\degree[/tex]

The correct answer is D.

See the attached file for continuation.

[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]
[tex]16. photon lighting company determines that the supply and demand functions for its most popular lam[/tex]

-5

Step-by-step explanation:

-15/3=-5