College Algebra Unit 4 - Milestone 4 with answers!!
18 questions were answered correctly.
2 questions were answered incorrectly.
1
Angela is an
... [Show More] electrical engineer who is testing the voltage of a
circuit given a certain current and resistance. She uses the
following formula to calculate voltage:
The circuit she tests has a current of amps and a resistance
of ohms.
What is the voltage of the circuit?
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volts
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volts. correct
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volts
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volts
RATIONALEThe voltage of the circuit is the product of
the current and resistance. Recall that we
can
write as and as .
and can combine to . Now we
can simplify the last term, , which
contains the imaginary unit squared. Recall
that the is equivalent to , which can
be substituted in our expression.
The expression multiplies
to Next, combine like
terms.
Once we have expressed voltage in terms
of , we need to multiply these two complex
numbers by using FOIL.
Multiply the first terms , the outside
terms , the inside terms , and
the last terms . Next, evaluate
each multiplication.
is replaced by . Next, evaluate
.CONCEPT
Complex Numbers in Electrical Engineering
2
Perform the multiplication and combine like terms.
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correct
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RATIONALE
The voltage is equivalent to , or
.
times is equal to . Finally,
combine like terms and .When FOILing, multiply the first terms, outside
terms, inside terms, and last terms together.
Once the binomials have been multiplied
together, evaluate the multiplication.
The will need to be multiplied by everything
inside the parentheses.
When multiplying these two terms, we will start
by distributing into .
To multiply a set of three binomials, we can
choose any two binomials to multiply using
FOIL, and then distribute the remaining binomial
to get a final product. Here, we will use FOIL to
multiply , but you can choose any
two binomials to start.
and combined is . can
be expressed as . We still need to
distribute the third binomial, .
is equivalent to .
The next step is to combine like terms, and
.CONCEPT
Multiplying Polynomials
3
Divide the following expression.
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times equals .
This is one part of the final product. We will
distribute into as well.
times equals . This
is another part of the final product. The final step
is to add these two parts together.
When multiplying these two terms, we will start
by distributing into .
This is the final product of the three binomials,
found by adding the two parts:
and .•
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correct
RATIONALE
Start by rewriting the expression into multiple fractions
with as the denominator. Remember to use the
correct signs (addition or subtraction) between the
fractions.
Now that we have individual fractions, we can simplify
each fraction. To do this, cancel out common factors in
the numerator and denominator. Let's consider the first
set, .
simplifies to because we can factor out
from both terms. Next, consider the second set, .CONCEPT
Polynomials Divided by Monomials
4
Select the quadratic equation that has no real solution.
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correct
simplifies to because we can factor out from
both terms. Next, consider the third set, .
simplifies to because we can factor out from
both terms. The expression can be
simplified to .•
RATIONALE
We can tell if a quadratic has no real solutions by
using the quadratic formula. We can use the
discriminant, or the value underneath the square root.
Because the discriminant is underneath a square root
sign, it must not have a negative value. If it is greater
than or equal to zero, it will have real solutions. If it is
negative, it will have non-real solutions.
This is the expression for the discriminant. For each
quadratic equation, we can substitute the appropriate
values into this expression and determine if it will
have real or non-real solutions.
For this answer choice, plug the coefficients into
the expression for the discriminant, and
determine if the discriminant is negative.
In the equation , ,
, and . Now that the appropriate
values are plugged in, evaluate the discriminant.
squared is and times times is
. Next, find the difference between and
.minus is . Since this value is nonnegative, the equation has at
least one real solution.
For this answer choice, plug the coefficients into
the expression for the discriminant, and
determine if the discriminant is negative.
squared is and times times is
. Next, find the difference between and
.
minus is . Since this value is nonnegative, the equation has at
least one real solution.
In the equation , ,
, and . Now that the appropriate
values are plugged in, evaluate the discriminant.
For this answer choice, plug the coefficients into
the expression for the discriminant, and
determine if the discriminant is negative.
In the equation , ,
, and . Now that the appropriate
values are plugged in, evaluate the discriminant.CONCEPT
Quadratic Equations with No Real Solution
5
Consider the quadratic function .
squared is and times times is
. Next, find the difference between and
.
minus is . Since this value is nonnegative, the equation has at
least one real solution.
For this answer choice, plug the coefficients into
the expression for the discriminant, and
determine if the discriminant is negative.
minus is . Since this value is
negative, the equation has
no real solutions.
In the equation , ,
, and . Now that the appropriate
values are plugged in, evaluate the discriminant.
squared is and times times is
. Next, find the difference between and
.What do we know about the graph of this quadratic equation,
based on its formula?
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The vertex is and it opens upward.
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The vertex is and it opens downward. correct
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The vertex is and it opens upward.
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The vertex is and it opens downward.
RATIONALE
Compare the given equation to the
general . The values of and in
particular give us useful information about the graph.
The sign of tells us if the parabola opens upward
or downward. If is positive, the parabola opens
upward. If is negative, the parabola opens
downward.Take note of the sign in the numerator. Evaluate the
division to get the x-coordinate of the vertex.
This is the y-coordinate to the parabola's vertex.
Return to the original equation, but write in the
calculated x-coordinate, , for every instance of .
Then, evaluate the equation.
The values and can be plugged into this formula
to give us the x-coordinate of the vertex.
From the given equation, plug in for and
for . Simplify both the numerator and the
denominator.
squared is . Next, multiply this by the
coefficient, .
times is . Finally, evaluate the addition
and subtraction.
In this case, since is negative, we know the
parabola opens downward. Next, we can use the
values of and to find the x-coordinate of the
vertex.
The x-coordinate of the vertex is . To
determine the y-coordinate of the vertex, plug this xvalue into the original equation and solve for .
times is . Next, evaluate times .CONCEPT
Introduction to Parabolas
6
Consider the following rectangle.
How can we express the area of the rectangle in terms of the
variable x?
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correct
From the equation, we know that the parabola's
vertex is at and opens downward.•
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RATIONALE
When given a rectangle with unknown
measurements, we can find the area of the
rectangle by multiplying the length and the
width. We need to express the length and
width by adding the partial dimensions of
each side.
The length is , and the width is
. Next, we can find the area by using
binomial multiplication.
The area is equivalent to length times width,
or times . When
multiplying two binomials, we can use FOIL.CONCEPT
Using FOIL to Represent Area
7
Consider the quadratic inequality .
What is the solution set?
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Once all like terms have been combined,
write the expression in standard form by
decreasing degree. The area of the
rectangle is .
Multiply the first terms , the outside
terms , the inside terms , and
the last terms . Once we have
FOILed, we can evaluate each
multiplication.
The expression multiplies
to Next, combine like
terms and .•
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correct
RATIONALE
To solve a quadratic inequality, first rewrite
it as an equation set equal to zero.
Two integers that multiply to and add to
are and . Now that we have
identified and , we can substitute the
values into the factored quadratic
expression, .
Once we have a quadratic equation, we can
solve it by factoring. The expression
is a basic quadratic equation
because it is in the form and
can be factored as . Next,
identify two integers, and , whose
product is the constant term, , and
whose sum is the x-term coefficient, .If any factor of an algebraic expression
equals zero, the entire expression has a
value of zero. Next, set each factor equal to
zero and solve the simpler equations to find
the solutions to the quadratic equation.
Next, choose any value that fits within each
interval. It doesn’t matter which values we
choose as test values, but we should make
them as simple as possible. Use them as xvalues to be plugged into the inequality. For
example, use these test values: .
Setting the first factor equal to zero,
the solution is . Setting the second factor
equal to zero, the solution is .
These solutions will create three intervals
on the number line.
Substitute each test value into the original
inequality and look for
true statements.
For the test point , the inequality is a
false statement.
The solutions can be found by setting each
factor to zero. Next, solve for in each
factor.CONCEPT
Quadratic Inequalities
8
Write the following expression as a single complex number:
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correct
For the test point , the inequality is a
true statement.
For the test point , the inequality is a
false statement.
The solution to the inequality is the section
of the number line which yields true
statements. In this case, it is values
between and .RATIONALE
CONCEPT
Add and Subtract Complex Numbers
9
Write the following expression as a single complex number.
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correct
When we subtract complex numbers, we can break the
problem down into two sets of subtraction: one set for all
real numbers, and another set for the imaginary numbers.
Once the complex numbers are lined up, we can subtract
the real numbers.
minus is equal to . Next, subtract the imaginary
numbers.
The expression can be written as a
single complex number .
minus is equal to . Next, write the two parts as a
complex number in the form .•
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RATIONALE
In both the numerator and denominator, we
multiplied the first, outside, inside and last
terms together. Now, we can evaluate each
multiplication.
Once we have evaluated the multiplication, we
can combine any like terms.
When dividing complex numbers, we must
clear imaginary numbers from the
denominator. To do so, we can use the
complex conjugate of the denominator to
create a second fraction. The complex
conjugate of is .
In this second fraction, the complex
conjugate is the numerator and the
denominator. Next, multiply across the
numerators and denominators using FOIL.CONCEPT
Divide Complex Numbers
When each in the numerator and
denominator is replaced with , we can
evaluate any multiplication.
In the numerator, we can combine and
to get . In the denominator, and
cancel each other out. The next step is to
simplify the terms that contain the imaginary
unit squared. Recall that the is equivalent to
, which can be substituted in our
expression.
In the numerator, times equals and
also in the denominator, times equals
. Next, we can combine any like terms.
In the numerator, plus equals . In the
denominator, plus is equal to . Now we
can separate into two fractions.
The expression can be written as a
single complex number, .10
Write the following expression as a single complex number.
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correct
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RATIONALE
When multiplying two complex numbers, we can
use FOIL.
Multiply the first terms , the outside terms
), the inside terms , and the last
terms . Next, evaluate each multiplication.
The expression multiplies to
. Next, combine like terms.CONCEPT
Multiply Complex Numbers
11
Simplify the following expression.
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correct
and can combine to . Now we can
simplify the last term, , which contains the
imaginary unit squared. Recall that the is
equivalent to , which can be substituted in our
expression.
The expression can be written as a
single complex number, .
is replaced by . Next, evaluate .
times is equal to . Finally, combine like
terms and .•
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RATIONALE
CONCEPT
Special Products of Binomials
12
Consider the quadratic function .
Rewrite the function in vertex format.
For this expression, start by writing this as
binomial multiplication.
When FOILing, multiply the first terms, the
outside terms, the inside terms, and last terms
together. Next, evaluate the multiplication.
The expanded form of the expression
is .
Once all multiplication has been performed,
combine like terms and to simplify.
is the same as . Next,
use FOIL to simplify.•
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correct
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RATIONALE
To rewrite a quadratic equation from standard
form to vertex form
, we must recognize the perfect
square trinomial relationship by completing the
square. First, move the constant term, , to the
other side of the equation.
To move from the right side, subtract from
both sides. Next, we can factor out the xsquared coefficient, , from each term on the
right side of the equation because we want to
have just .divided by equals ; squared equals .
We will use this value to construct a perfect
square trinomial by adding it to both sides.
can be written as . Lastly,
subtract from both sides to isolate on the
left.
Factoring out from each term on the right side
results in . Next, inside the
parentheses, we need to focus on the x-term
coefficient, We need to first divide this
coefficient by 2, then square the result.
The has been added to to create a
perfect square trinomial, .
However, this is the same as adding , or
, because of the factor in front of the trinomial.
We actually needed to add to the left side
to keep it balanced. Next, simplify the left side.
The left side can simplify from to
just . On the right side, we now have a
perfect trinomial. Write this as a binomial
squared.CONCEPT
Converting a Quadratic Equation into Vertex Form
13
Write the following expression in factored form.
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correct
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RATIONALE
This is the equivalent equation written in vertex
form.CONCEPT
Basic Quadratic Factoring
14
The vertical position (in feet) of a rock seconds after it was
dropped from a cliff is given by the
Two integers that multiply to and add to
are and . Now that we have identified as
and as , we can substitute the values
into the factored quadratic expression of
.
The expression, , is a basic quadratic
equation because it is in the form and
can be factored as . We must
identify two integers, and , whose product is
the constant term, , and whose sum is the xterm coefficient, . Since our constant term is
positive, but the x-term coefficient is negative, we
know that both values will be negative since a
negative times a negative equals a positive, but a
positive plus a positive will never be negative.
The quadratic expression can be
factored as .formula . When the rock reaches the base of
the cliff, .
How many seconds will it take for the rock to hit the ground at the
base of the cliff?
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6
seconds
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8
seconds
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4 seconds correct
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3
seconds
RATIONALE
To determine how long it will take for the rock
to hit the ground, set the equation equal to
zero.
When a quadratic equation is set equal to zero,
you can use the quadratic formula to solve for
.
Use the coefficients in the general
expression to substitute
for in the quadratic formula.One equation involves subtraction, while the
other equation involves addition. Next,
evaluate the numerator in each equation.
We can start with the denominator, since it is
easily simplified by multiplying and .
Next, simplify the expression underneath the
radical in the numerator.
The square root of is . To find the
solutions, create two equations (due to the ).
minus is equivalent to .
Now that the expression underneath the
radical is simplified, take the square root of
.
In the first fraction, minus equals
. In the second fraction, plus
equals . Finally, divide each numerator by
.
In this case, , , and .
Once you have substituted values in
for , begin to solve for t.
squared equals , and times
times equals . Then, find the
difference between and .CONCEPT
Using Quadratic Equations to Represent Motion
15
Consider the quadratic function .
What is the graph of this function? C is correct
It will take 4 seconds for the rock to hit the
ground at the base of the cliff.
The two solutions to the quadratic equation are
seconds and seconds. However, in
this context, the variable represents time, and
time cannot be negative.•
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• correct
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RATIONALE
Compare the given equation to the general
. The values of and in
particular give us useful information about
the graph.CONCEPT
Take note of the sign in the numerator.
Evaluate the division to get the x-coordinate
of the vertex.
This is the parabola that opens upward with
an axis of symmetry at .
The values and can be plugged into
this formula to give us the x-coordinate of
the vertex.
In this case, since is positive, we know
the parabola opens upward. Next, we can
use the values of and to find the xcoordinate of the vertex.
From the given equation, plug in for
and for . Simplify both the numerator
and the denominator.
The sign of tells us if the parabola opens
upward or downward.
The x-coordinate of the vertex is .
This provides information about the
equation for the axis of symmetry.Graphing Parabolas
16
What is the product of and ?
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correct
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RATIONALE
When multiplying these two terms, you will multiply the
coefficients together and evaluate the multiplication of the
x-terms.
When multiplying a monomial by binomial, you will multiply
the monomial by each term in the binomial. First distribute
into each of the terms inside the parentheses. Start by
distributing into .CONCEPT
Multiplying Monomials and Binomials
17
Which of the following graphs represents the quadratic
inequality ?
Similar to above, when multiplying these two terms, you
will multiply the coefficients together and evaluate the
multiplication of the x-terms.
The final product is the sum of the two distributions.
The coefficients and multiply to . Recall that if two
terms with the same base are multiplied, simply add the
exponents, so times equals . Don't forget to
include the . Next, distribute into .
The coefficients of and multiply to . The times
equals . Don't forget to include the . Finally, add
both parts, and , together.•
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• correct
RATIONALE
Compare the given inequality to the general
. The values of and , as
well as the inequality symbol, give us useful
information about the graph.Take note of the sign in the numerator.
Evaluate the division to get the x-coordinate
of the vertex.
The inequality symbol tells us two
things: because it includes "or equal to" the
line must be solid; and because it is a
greater than symbol, the area above the
curve will be shaded.
In this case, since is positive, we know
the parabola opens upward.
The values and can be plugged into
this formula to give us the x-coordinate of
the vertex.
From the given inequality, plug in for
and for . Simplify both the numerator
and the denominator.
The x-coordinate of the vertex is . This
provides information about the equation for
the axis of symmetry.
The sign of tells us if the parabola opens
upward or downward.CONCEPT
Graphing Quadratic Inequalities
18
Write the following expression in factored form.
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correct
RATIONALE
This is the graph of . It is a
parabola that opens upward, has an axis of
symmetry at , the boundary line is
solid, and the solution region is above the
boundary line.The x-squared term, , is written in the
top-left corner. The constant term, , is
written in the bottom-right corner. Next, use
the factors and that we found above
as coefficients to x-terms to be included in
the 2x2 box.
One method to factor a quadratic expression
in the form is to use the box
method. First, multiply the -value, , and
the constant, , together.
Two factors of that also add to are
and . We will use these two values in
a 2x2 grid, along with the x-squared term
and the constant term. To start, write the xsquared term and the constant term in the
corners of a 2x2 grid.
times equals . Next, list factors
that multiply to and add them together
to see if they equal , the x-term
coefficient.CONCEPT
More Challenging Quadratic Factoring
19
In the first column, which contains and
, we can only factor from both terms. In
the second column, which contains and
, we can factor out . In the first row,
which contains and , we can factor
out from both terms. In the second row,
which contains and , we can factor
out from both terms. The values we
factored out of each column, and , and
each row, and , become the two
binomials that are multiplied together to get
the original expression.
The expression can be written
in the factored form of .
The two factors, and , are written as
the x-terms in the 2x2 box. Next, look at the
rows and columns, and factor out the
greatest common factor in each row and
write it outside the grid.Consider the quadratic equation .
Find the solutions by factoring.
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correct
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RATIONALE
Two integers that multiply to and add to are
and . Now that we have identified as and
as , we can substitute the values into the factored
quadratic expression of .
The expression, , is a basic quadratic
equation because it is in the form and can
be factored as . We must identify two
integers, and , whose product is the constant term,
, and whose sum is the x-term coefficient, .CONCEPT
Solving Quadratic Equations
20
Consider the equation .
Find the solutions by using the quadratic formula.
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and
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and
If any factor of an algebraic expression equals zero,
the entire expression has a value of zero. We can set
each factor equal to zero and solve the simpler
equations to find the solutions to the quadratic
equation.
Setting each factor to zero will allow us to find the
solutions. We can now solve for in each factor.
When we set the first factor equal to zero, the
solution is . When we set the second factor
equal to zero, the solution is . The two
solutions to the equation are
.•
and
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and correct
RATIONALE
When we have a quadratic equation, we
can find the solutions by using the quadratic
formula. However, be sure that the equation
is set equal to zero. First, subtract 5 from
both sides.
Plugging the values of , , and into the
quadratic formula produces the solutions to
the quadratic equation. Be sure to perform
the operations in the correct order. First,
simplify what is underneath the radical, or
Now that it is set equal to zero, identify the
coefficients , , and that are found in the
standard equation, .
In the equation , ,
, and . Substitute these
values for , , and , into the quadratic
formula.One equation involves subtraction, while
the other equation involves addition. Next,
evaluate the numerator in each equation.
minus is equal to . Next,
take the square root of .
The square root of is . Next,
evaluate the denominator by multiplying
and .
minus is , and plus is .
To find the solutions, divide the numerators
by the denominator, .
times equals . Also note
that in the numerator is equivalent
to . To find the solutions, create two
equations (due to the ).
The two solutions to the quadratic equation
are and
squared equals , and times
times equals . Next, find the
difference between and . [Show Less]