Author: Marcus Marston

Marcus Final exam review solutions

December 16, 2019 / Marcus Marston / 0 Comments

Marcus Final exam review solutions

1a)(sqrt(6)+sqrt(2))/2

1b)-(sqrt(2)+sqrt(6))/4

2a) magnitude=sqrt(10)

angle=79.1 degrees

2b) magnitude=1.479

angle=115.84

3a){x=3}

3b){x=6}

4a)domain:(0, inf)

4b)domain:(-7, inf)

4c)domain:(-5, inf)

5a)domain:(-inf,2)U(2,4)U(4,inf)

vertical:(2,0)&(4,0)

removal:none

5b)domain:(-inf,-2)U(-2,0)U(0,2)

vertical:(0,0)&(-2,0)

removal:(2,0)

6a)(3+4i)

6b)(-14+2i)

7a)(x^2-2x-2-3/(x-2))

7b)(x^2+3x-2+4/(x+3))

8a)all real roots:1/2, 2/3,…

8b)all real roots:-1/4

9a)(pi-pi/4)+2*pi*n

9b) pi/4+2*pi*n

9c) pi/6+2*pi*n

10a)magnitude=sqrt(100) , angle=53.1

10b)magnitude=111.8, angle=sqrt(29)

10c) magnitude=225, angle=4*sqrt(2)

Final exam question solutions

December 16, 2019 / Marcus Marston / 2 Comments

Jayvon Final exam Question solutions

1a) 3^5x+2=9^x+4

3^5x+2=3^2x+8

5x+2=2x+8

5x=2x+6

3x=6

x=2

1b) 5^x+3=7^x

ln5^x+3=ln7^x

(x+3)*ln5=x*ln7

x*ln5+3*ln5=x*ln7

xln5-xln7=3ln7

x(ln5-ln7)=3ln7

x=3ln5/(ln5-ln7)

2a) x^2-2x-3>=0

x^2-2x-3>=0

(x+1)=0 &. (x-3)=0

x=-1 & x=3

x>=-1 & x>=3

(-inf,-1)U(3, inf)

2b) |x-4| >= 7

(x-4)>=7 & -x+4>=7

x>=11 & x=<-3

(-inf, -3) U (12, inf)

3) y=2*sin(4x-pi)

amplitude:2

period:1/2pi

phase shift:pi/4

minima:-2

maxima:2

pi/4+2*pi*n

4a) y=5+6x

x=5+6y

f^-1(x)=(x-5)/6

4b) y=2/(8x+5)

x=2/(8y+5)

f^-1(x)=(2/x-5)/8

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

x-int:2

y-int:2/3

domain:(-inf, 3)U(3, inf)

vertical:3

horizontal:-2

6a) 3(cos315+I*sin315)/18(cos135+I*sin135)

3/18* cos(315-135)+I*sin(315-135)

1/6*cos(180)+I*sin(180)

1/6*(-1+0i)

-1/6

6b) ln* sqrt(x^3*4*sqrt(y))

ln x^3^1/2+ln y^1/4^1/2

sqrt(3)u+1/2v

7)a. a1:5

b. ratio:5

c. a_n*r^n-1=5*5^n-1

d. 5^n*5^10-1=9765625

8)2015-2003=12

80,000(1+.04)^13

=133205

3=(1+.04)^x

(ln3/ln1.04=x)

9a) sqrt((-5sqrt(3))^2+(5)^2)

sqrt(100)=10

magnitude=10

tan^-1(5/-5*sqrt(3))

tan^-1(5/-5*sqrt(3))=-30+180=

angle=(150 degrees)

10)sin(2a)=(-14*sqrt(113))/16)

cos(2a)=(-1)

tan(2a)=(sqrt(113)/-32)

final exam review questions

December 13, 2019 / Marcus Marston / 1 Comment

Final Review

1) Simplify cos and sin to get real numbers

a) cos(11*pi/12) b) sin(5*pi/12)

2)The vectors v⃗ and v⃗ below are being added. Find the approximate magnitude and directional angle of sum ⃗v = v⃗ + v⃗.

a) ||v⃗||=6, and θ =60◦, and ||v⃗||=2, andθ =180◦

b) ||v⃗||=3.7, and θ =92◦, and ||v⃗||=2.2, and θ =253◦

3) Solve for x without using a calculator.

a) ln(2x + 4) = ln(5x − 5) b) ln(x+6)=ln(x−2)+ln(3)

4) State the domain of the function f and sketch its graph.

a) f(x) = log(x) b)f(x)=log(x+7) c)f(x)=ln(x+5)−1

5) Find the domain, the vertical asymptotes and removable discontinuities of the functions

b)f(x)= x2+2 x2 −6x+8 c) f(x) = 3x+6/x^3 −4x

6)add and subtract the complex numbers

a) (5-2i)+(-2+6i) b) (-9-i)-(5-3i)

7)Divide by long division.

a) x^3−4x^2+2x+1/ x−2. b) x^3+6x^2+7x−2/x+3

8) Find all real roots

a)f(x) = 6x^4 + 25x^3 + 8x^2 − 7x − 2. b) f(x) = 4x^3 + 9x^2 + 26x + 6.

9)Find at least 5 distinct solutions of the equation

a) tan(x) = −1. b) cos(x) = √2/2 c) sin(x) = −√3/2

10) Find the magnitude and directional angle of the vector.

a) ⟨6, 8⟩. b) ⟨−2, 5⟩. c) ⟨−4, −4⟩

Test #3/ Solutions

December 11, 2019 / Marcus Marston / 0 Comments

6. (5 points) Determine the exact value of sin (pi/12)

1)First break sin(pi/12) into the subtraction of two degrees

sin(pi/12)=sin(pi/3-pi/4)

2)Then we use the angle summation formula

sin(pi/3-pi/4)=sin(pi/3)*cos(pi/3)-cos(pi/3)*sin(pi/4)

3)After find the sin and cos of each angle given

sqrt(3)/2*sqrt(2)/2-1/2*sqrt(2)/2

4) Then simplify the equation

sqrt(6)/2-sqrt(2)/2

5) simplify the equation by combining

sqrt(6)-sqrt(2)/2

Marcus Marston/Test #3 review

November 17, 2019 / Marcus Marston / 0 Comments

Webwork

Logarithmic Functions – Equations: Problem 6

solve for x: log5(x+14)-log5(x+6)=3

First combine the log5 since they both have the same base and we want to make them one log equation. Since they are subtracted then we need to combine them with division.

log5(x+14/x+6)=3

2. Afterward then we turn the equation into the exponent

5^3=(x+14/x+6)

3. Then we solve the exponent and multiply the denominator

125(x+6)=(x+14)

4. Distribute the constants

125x+750=x+14

5. Afterward then isolate the variable-x

x=-736/124

6. After you can simplify the fraction

x=-184/31

P.S: Sorry for any errors

Exam #2 solutions/ question 7

November 3, 2019 / Marcus Marston / 0 Comments

(1 point) solve -x^2-3x <= -10

2) First solve by adding +10 to both sides of the inequality

-x^2-3x+10<=0

3) Then you need to factor out -1 in the the inequality to make the leading coefficient positive.

-1(x^2+3x-10)<=0

4) After that you need to factor the inequality of what can multiply into -10 and add up to +3

-1(x+5)(x-2)<=0

5) Then you need to set them both equal to zero and place -1 in front of both of them

-1(x+5)=0 and -1(x-2)=0

6) Once you do that then factor in the -1

-x-5=0 and -x+2=0

7) Then solve for x

x=-5 and x=2

8) Make sure to switch the inequality

x>=-5 and x>=2

9) Write your answer in interval notation

Answer=(-inf, -5] U [2, inf)

Test review #2

October 20, 2019 / Marcus Marston / 0 Comments

Textbook 11.3 exercises

Exercise 11.1. Find the domain, The vertical asymptotes and removable discontinuities of the function.

Question b) (x^2+2)/(x^2+6x+8)

step 1: To find the domain we need to to set the denominator to zero

(x^2+6x+8)=0

step 2: We solve by factoring the equation to its simplest form.

2. (x+4)=0 and (x+2)=0

step 3: We solve both equations to get x by its self.

3. x=-4 and x=-2

step 4: Both -4 and -2 are the restrictions that makes the denominator zero.

4. D: {R|x not equal -4,-2}

step 5: By using the restrictions on the domain we are able to find the vertical asymptote.

5. The graph does not pass through the values of -4 and -2

step 6: To find the removable discontinuities you need to see if the equation has any factorable polynomials.

6: Since none of the polynomials can be factored then the equations has no removal discontinuities.

P.S: Sorry for any errors

Test #1 solution

September 30, 2019 / Marcus Marston / 0 Comments

(8) Let f(x)=3x^2-5 and g(x)=sqrt(x-1).

(a) Find (f of g)(x).

First plug in the g(x) equation in the x of the f(x) equation.

=4(sqrt(x-1))^2-2

2. The next step is to square the square root.

=4(x-1)-2

3. After you distribute the 4 to the x and one.

=4x-4-2

4. Then add the like terms.

=4x-6

(b)Determine the domain of (f of g)(x)

Get rid of the square root of the function and make it equal to 0

= X-1=0

2. Add 1 to the other side to isolate x

=x=1

3. After replace the equal sign with a greater than equal sign

=x is greater than or equal to 1

4. Write the answer in interval notation

Domain=[1, Inf)

Test #1 Review

September 19, 2019 / Marcus Marston / 0 Comments

Webwork: Functions – Inverse Functions: Problem 10

Computing the inverse function.

if F(x)=8e^2x-8/6e^2x+3, find the inverse function.

step 1: Switch the x and y variables.

x=8e^2y-8/6e^2y+3

step 2: Multiply the denominator to both sides.

(6e^2y+3)(x)=(8e^2y-8)

step 2: Distribute the y to both coefficients.

6xe^2y+3x=8e^2y-8

step 3: Then isolate the natural logs to one side and move the coefficients to the other.

6ye^2y-8e^2y=-8-3x

step 4: Then you factor out the e, because they have the same exponents

e^2y(6x-8)=(-8-3x)

step 5: Then you isolate e by dividing the factors to the other side.

e^2y=(-8-3x)/(6x-8)

step 6: After you have to remove e by returning it to natural log to both sides.

ln(e^2y)=ln(-8-3x)/(6x-8)

step 7: Once e is removed we divide it by 2 to isolate the y value.

y=(ln(-8-3x)/(6x-8))/2

step 8: Return the y and x- variables to the proper places.

f^-1(x)=(ln(-8-3x)/(6x-8))/2

p.s: I apologize for any errors

Marcus Marston

August 27, 2019 / Marcus Marston / 0 Comments

Hello! I look forward working with everyone this semester. I have a few hobbies that I enjoy like chess, boxing and martial arts. I am pretty good at cooking since I was doing it when I was little. One thing that I overcame was not failing my English class in 8th grade. My favorite animal is a bear because I love how it sleeps a lot. my favorite books is the Sherlock Holmes series.

Author: Marcus Marston

Marcus Final exam review solutions

Final exam question solutions

final exam review questions

Test #3/ Solutions

Marcus Marston/Test #3 review

Webwork

Logarithmic Functions – Equations: Problem 6

Exam #2 solutions/ question 7

Test review #2

Test #1 solution

Test #1 Review

Marcus Marston

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