Answer is C
16. The price for which supplies equals
the demand is 96.236 (see graph)
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
-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 -
B) 2i, 7, -7
19. The asymptotes are y = 2 , x = -2, x= 2
C) x = 2, x = -2, y = 2
five pi divided by six
360 degrees ………….2pi radians
x = 150°
sin B = 21/75 = 7/25
tan B=21/72 = 7/24
C) sin B = seven divided by twenty five ;
B = seven divided by twenty four
22. X = 17/cos(58°) = 32.080
Coterminal angles 202°
202° +360° = 562°
202° -360° = -158°
D) 562°; -158°
24. period of the function.
y = 5 cos (x/2) (see graph3)
y = csc-1(-1) = -pi/2
B) negative pi divided by two
arcos( cosine of pi divided by two)
arcos(cos(pi/2)) = arcos(0) = pi/2
D) pi divided by two
sin45 = 0.7071
cos45 = 0.7071
B) θ = 45°
c**2 = a**2 + b**2
sqrt(17**2 +7**2) = 18.384 ft
C) 18.4 ft
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
4 sin2 x - 4 sin x + 1 = 0
D) pi divided by six , five pi divided by six
f(x) = 3x**2 - 1 (1 point)
f(-x) = 3(-x)**2 -1 = 3x**2 -1 = f(x)
f(x) IS 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
is Not a factor
k = 2; f(x) = 2x3 + 3x2 - 4x + 4; Lower bound?
Using synthetic division, the results are (see image 4)
And since they
are all positive values, k= 2 is an upper bound.
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 -
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
cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x
If we use the following identity
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)
If the demand and supply are equal, then we equate the two functions in p and solve for p.
We can rearrange to obtain,
The real roots of this polynomial equation are:
Since price can not be negative, we discard the negative value ,
The correct answer for question 16 is C.
We were given the solution to this polynomial as
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
is a solution to
then the complex conjugate
is also a root.
is a solution then,
is also a solution.
Therefore we have
We expand to obtain,
We now expand to obtain,
We simplify further to obtain,
The correct answer for question 17 is A.
is a zero of the polynomial,
then the complex conjugate
is also a zero,
This means that ,
are factors of the polynomial.
The product of these two factors,
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,
Applying our knowledge from difference of two squares, we obtain,
Hence all the zeroes of these polynomial can be found by setting
The correct answer for question 18 is B
We were asked to find the horizontal and vertical asymptote of
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,
For vertical asymptote, equate the denominator to zero and solve for x.
None of the options is correct, so the correct answer for question 19 is A.
We are converting,
to degrees .
To convert from radians to degrees, multiply by,
We simplify to obtain,
The correct answer is B.
Recall the mnemonics, SOH CAH TOA
The sine ratio is given by,
From the diagram,
The correct answer is C.
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,
We can simply switch positions to make x the subject.
Hence the correct answer is A.
Coterminal angles have the same terminal sides.
To find coterminal angles, we keep adding or subtracting 360 degrees.
is coterminal with
The correct answer is D.
See the attached file for continuation.
Coefficient of x, in this case, is '2'
coefficient of x³y²=80
using pascal triangle to expand we get
the one in bracket is the answer so the coefficient of x³y²=80
If it's x²y³ then we know it's the second term of the expansion, that known we can use the combination
C(5, 2) = 5!/(2!.3!) = 10
Then if we had something like
(a + b)^5 our second term would be 10a²b³ but as we can see it's "a²"
And in our case we have 2x as a
So we must do 2² too
2² = 4
10 . 4 = 40
Then our second term of the expansion would be
The solution of this system is x=9/4, y=5/2, and z=-13/8
1. Writing the equations in matrix form
The system of linear equations given can be written in matrix form as
This is the matrix form of the simultaneous equations.
2. Solving the simultaneous equations
we can multiply both sides by the inverse of A
We know that , the identity matrix, so
All we need to do is calculate the inverse of the matrix of coefficients, and finally perform matrix multiplication.
3. Calculate the inverse of the matrix of coefficients
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 2 from row 2
Make zeros in column 2 except the entry at row 2, column 2. Add row 2 multiplied by 3 to row 3
Multiply row 2 by −1
Make zeros in column 3 except the entry at row 3, column 3. Divide row 3 by −8
Subtract row 3 multiplied by 2 from row 1
Subtract row 3 multiplied by 4 from row 2
As can be seen, we have obtained the identity matrix to the left. So, we are done.
4. Find the solution
A coefficient is a number that comes before a variable and is used to multiply the variable.
A variable can never be a coefficient, so we already know that the answer cannot be B or D.
Now we are left with A and C. 4 is the exponent, not the coefficient, so the answer can't be A.
So, we are left with C. 2 multiplies x and comes before x. Therefore, our answer is C.
I hope I helped!
Let me know if you need anything else!