Solving a Linear Equation with Faction Coefficients

January 12, 2018 admin

Example: Solve the equation.

1 /9 (6t-9)=11/9 t -{t+9}/18

Solution:

Solve the equation by clearing fractions.
The least common multiple of the denominators is 18. Multiply the least common multiple by each term.	***
As you simplify, the fractions cancel.	***
Use the distributive property and combine like terms.
Move the variables to one side of the equation and simplify.
Get the variable by itself on one side of the equation by adding 18 on both sides and dividing by -9.

The solution to the equation is -1.

Here is a youtube video that might help as well.

4.2 Applications and Modeling of Quadratic Functions, 4.4 Build Quadratic Models from Verbal Descriptions and from Data

Application of Quadratic Function: Maximize

November 17, 2017 admin

Example:

A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by the quadratic function h(t)=-4.9t^2+17.4t+1 . How long does it take the baseball to reach its maximum height? What is the maximum height obtained by the baseball?

Solution:

h(t)=-4.9t^2+14.7t+1 is defined to be a quadratic function. The graph of a quadratic function is a parabola. This quadratic function has a leading coefficient of -4.9 and since it is negative means that the parabola is opening down.

Find the vertex to find the maximum value.

The formula for the x coordinate of the vertex
For a quadratic function a is the coefficient of the square term, b is the coefficient of the linear term, and c is the constant
Substitute the values of a and b into the formula
Simplify with a calculator

The ball reaches the maximum height 1.5 seconds after the ball was hit.

The maximum height can be found by substituting 1.5 seconds for time in the height function.

The height function
Substitute t=1.5 in the height function
Simplify using a calculator

The maximum height of the ball is 12.025 meters.

Here is a video example maximizing the position function.

3.1 Functions, 3.5 Combination of Functions; Composition of Functions

Combination of Function: Evaluation of Division

November 17, 2017 admin

Example:

Evaluate (g/f)(-3) when g(x)=x^2-5 and f(x)=sqrt{x+3}

Solution:

Evaluate the function
Rewrite the expression
Replace x with -3 in each function	$(g/f)(-3)={(-3)^2-5}/{sqrt{-3+3}}$
Simplify	$(g/f)(-3)={(-3)^2-5}/{sqrt{-3+3}}$
Division by zero is undefined	is undefined

(g/f)(-3) is undefined.

3.5 Combination of Functions; Composition of Functions, 6.1 Composite Functions

Composition of Functions

November 17, 2017 admin

Example:

Find (f circ g)(x) when

f(x)=8x-9 and g(x)={7}/{x+9}

Solution:

The definition of composition
Replace each x in f with g(x)
Simplify and write a single fraction by finding a common denominator
Simplify by using the distributive property and combine like terms.

(f circ g)(x)=-{9x+25}/{x+9}

5.3 Complex Zeros; Fundamental Theorem of Algebra

Form a Polynomial given the Degree and Zeros

November 1, 2017 admin

Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 4; Zeros -2-3i; 5 multiplicity 2

Solution:

By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.

There are three given zeros of -2-3i, 5, 5.

The remaining zero can be found using the Conjugate Pairs Theorem. f(x) is a polynomial with real coefficients. Since -2-3i is a complex zero of f(x) the conjugate pair of -2+3i is also a zero of f(x).

Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.

Since f(x) has a zero of 5, f(x) has a factor of x-5

Since f(x) has a second zero of 5, f(x) has a second factor of x-5

Since f(x) has a factor of -2-3i, f(x) has a factor of x-(-2-3i)

Since f(x) has a factor of -2+3i, f(x) has a factor of x-(-2+3i)

Form the polynomial using all of the factors. The leading coefficient will remain unknown.
Multiply the factors with complex numbers. Doing so will cancel the complex numbers from the expression Distribute the minus Multiply each term in one factor by each term in the other factor simplify combine like terms
Multiply the other pair of factors
Multiply the two trinomials by multiplying each term in the first trinomial by each term in the other trinomial and then combine like terms

The polynomial with degree 4 and zeros of -2-3i and 5 wiht multiplicity 2 is f(x)=a(x^4-6x^3-2x^2-30x+325)

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay (Half-life)

October 28, 2017 admin

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution: There is a two part process to this problem. Part 1: Use some of the information to find the decay rate of radium. Part 2: Answer the question using the rest of the given information.

Part 1: Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

$A=A _0 e^{kt}$

A _0 is the initial amount

is the decay rate

is the time

is the amount after t time has passed

Since radium has a half life of 1690 years, we know that after 1690 years there will be half of the initial amount of radium left. This allows me to establish a relationship between the initial amount and the amount after 1690. The amount after 1690 year is half of the initial amount. A=1 /2 A _0 when the t=1690 .

Substitute these values into the exponential decay formula and solve for k.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for A and t since it is known that the half-life is 1690 years and	$A=A _0 e^{kt}$ $1 /2 A _0=A _0 e^{k(1690)}$
Solve for the decay rate k:Start by dividing both sides by the coefficient to isolate the exponential factor	$1 /2 A _0=A _0 e^{k(1690)}$ ${1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0$ $1 /2= e^{k(1690)}$
Solve for the decay rate k:Take the natural log of both sides to get k out of the exponent	$1 /2= e^{k(1690)}$ $ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Use the power rule for logarithms to get k out of the exponent	$ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Simplify ln e = 1
Solve for the decay rate k:Solve for k by dividing by 1690 on both sides

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2: Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years. Symbolically that is A _0=50 when the t=630 . Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for initial amount , time (t), and the decay rate (k). , , and	$A=A _0 e^{kt}$ $A=50 e^{630{ln(1 /2)}/1690}$
Type the expression into your calculator and round to the thousandth place.	$A=50 e^{630{ln(1 /2)}/1690}$

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Here is a video example that is similar to the above example and it shows how to enter information in the calculator.

1.5 Applications of Quadratic Equations, Applications of Quadratic Equations

Application of Quadratic Equation: Translation

October 20, 2017 admin

Example: The product of some negative number and 5 less than twice that number is 273. Find the number.

Solution: Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

x (expression that comes next)

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows. Twice that number refers to 2 times the negative number that was described before. (2x-5)

x (2x-5)

“…is 273”

This translates to equals 273.

x (2x-5)=273

Solve the equation.

The translated equation
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula) I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.
Factor
Use the zero product property and set each factor equal to zero and solve.

Since the statement says that the number is negative, the number is {-21/2} .

Uncategorized

Math15Fun.com

October 13, 2017 admin

Hello all! My name is Amanda Sartor and this is my webpage. I use this to provide resources to my in person students and my online students. I don’t have resources for everything but I am always adding. At the top you will find handouts that I use during class and the solutions to the problems. On the left there are blog posts that are specific to a certain problem or problem type. I hope you find these resources helpful.

Amanda Sartor

3.2 Properties of a Function's Graph, 3.3 Properties of Functions

Using the Min and Max Feature in the Calculator

September 10, 2017 admin

8.5 Partial Fraction Decomposition

Partial Fraction Decomposition

August 12, 2017 admin

Example: Find the partial fraction decomposition for the rational expression.

$11/{x(x^2+11)}$

Solution:

The rational expression	$11/{x(x^2+11)}$
Factor the denominator completely	$11/{x(x^2+11)}$
Write the expression with each factor as a separate fraction.	$11/{x(x^2+11)}=?/x+?/{x^2+11}$
Decide what type of expression to put in the numerator x is linear so we put a constant in the numerator is a quadratic so we put a linear expression in the numerator	$11/{x(x^2+11)}=?/x+?/{x^2+11}$ $11/{x(x^2+11)}=A/x+?/{x^2+11}$ $11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}$
Now that the set up is done we need to solve for the unknowns A, B and C	$11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}$
Multiply both sides of the equation by the common denominator ${x(x^2+11)}$ and simplify by canceling out common factors	$11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}$ ${x(x^2+11)}11/{x(x^2+11)}={x(x^2+11)}(A/x+{Bx+C}/{x^2+11})$ $11={x(x^2+11)}{A/x}+{x(x^2+11)}{Bx+C}/{x^2+11}$ $11={(x^2+11)}{A}+x(Bx+C)$
Use the distributive property and collect the like terms with the x’s	$11={(x^2+11)}{A}+x(Bx+C)$
Equate coefficients to create a system of equations to solve
On the left 11 is the constant term. There is no linear term so the coefficient is zero and there is no quadratic term so the coefficient is zero. On the right the constant (no x’s) is 11A, the linear coefficient is C and the quadratic coefficient is A+B.	System of equations from equating the coefficients

Now solve the system of equations.

The first equation is 11=11A. Since it only has one variable, I can solve for A by hand.

The second equation is 0=C . It is already solved.

The third equation is 0=A+B. Since I know A=1, I can substitute and solve for B.

Using the original set up of $11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}$ , substitute the values of A, B, and C.

$11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}$ $11/{x(x^2+11)}=1/x+{-1x+0}/{x^2+11}$ $11/{x(x^2+11)}=1/x+{-x}/{x^2+11}$

The partial fraction decomposition of $11/{x(x^2+11)}$ is $1/x+{-x}/{x^2+11}$

The website of Professor Amanda Sartor

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