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If is borrowed for years at

interest, compounded annually, andthen paid in full at the end of that period, howmuch must be paid back at that time?

Use the calculator provided and round youranswer to the nearest dollar.

After the first year, the amount due will be

;after the second year, the amount due will be

;and so on.

In general, after years the amount due will be times the original amount. Thus, theamount due will be

.

In this example, . So we have

.

Rounding to the nearest dollar , we get that the amount due will be after years.

The answer is .

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A laptop computer is purchased for .

After each year, its resale value is of itsvalue the previous year. What will be the resale

value of the computer after years?

Use the calculator provided and round youranswer to the nearest dollar.

After the first year, the computer will have a resale value of

;after the second year, the computer will have a resale value of

;

and so on.

In general, after years the resale value will be of the original value. Thus, the computer willhave a resale value of

.

In this example, . So we have

.

Rounding to the nearest dollar , we get that the computer will have a resale value of afteryears.

The answer is .

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When a ball is thrown, its height in feet after seconds is given by the equation

,

where is the initial upwards velocity in feet per second. If feet per second, find all

values of for which feet. Do not round any intermediate steps. Round your answer to

decimal places.

(If there is more than one answer, enter additional answers with the button that says "or".)

The relationship between and is a quadratic equation in . We must find the solutions of this

quadratic equation, that is, the values of for which the equation is true.

First, we substitute the numbers given for and into the equation, :

.

Next, we rewrite the equation in the standard form, :

.

Comparing this to the standard form, we see , , and . Thesolutions to this equation are given by the quadratic formula:

.

Substituting the values of , , and found above into this equation yields

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.

So, or .

Writing these two values as decimals yields or .

Rounding these values to decimal places yields seconds or seconds.

Therefore, at approximately seconds and seconds, the height of the ball is feet. The

first time ( ) occurs when the ball is on the way up, while the second time ( ) occurs whenthe ball is on the way down.

The answer is:

seconds or seconds.

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Dividing the polynomial by yields

a quotient and a remainderof . If

, find and .

The division algorithm guarantees that if the division of by yields a quotient of

and a remainderof , then

.

We can use this fact to find . Namely,

.

So .

Note that is the remainderwhen is divided by . The generalization of this result is

called the remainder theorem: if is the remainder after dividing the polynomial by ,

then . In the current problem, we have and , so by the remainder

theorem we must have , which is exactly the result obtained above.

To find , we use the division algorithm along with the given information that :

.

Thus, the answer is:

.

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Consider a triangle like

the one on the right. If ,

, and , find ,

, and . In other words, solve the

triangle. Put answers in the degree

measure to the nearest hundredth.

When three sides of an oblique triangle are given, and we are asked to find the angles, we

are said to be in the SSScase (referring to side-side-side). The procedure to solve such a triangle is

as follows.

We first solve for the largest angle by using the law of cosines. This will always be the angle oppositethe longest side.

Since is the longest side, we find the measure of by using the law of cosines:

,

and so

.

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Therefore,

.

(We will use the exact value of below rather than the approximation,

, in order to minimize rounding error.)

Second, we find the remaining sides using the law of sines. Since

,

we have

.

Hence,

.

Finally, since , we have

.

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The answer, rounded to the nearest hundredth, is , , and

.

Consider a triangle like

the one on the right. If ,

, and , find ,

, and . In other words, solve the

triangle. Round all answers to the

nearest hundredth with angles in degree

measure.

When two sides and the included angle of an oblique triangle are given, and we are asked to

find the remaining parts, we are said to be in the SAScase (referring to side-angle-side). The

procedure to solve such a triangle is as follows.

We first solve for the unknown side by using the law of cosines. Namely,

,

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so that

.

(We will use the exact value of below rather than the

approximation, , in order to minimize rounding error.)

Second, we find the measure of the angle opposite the shorter of the two given sides. Since, we solve for .Using the law of sines, we have

and hence

,

so that

.

Finally, since , we have

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.

The answer, rounded to the nearest hundredth, is , , and

.

Consider a triangle like the one shown

below. Given , , and the

measure of angle , solve the triangle

by finding , , and . If no such triangleexists, enter "No solution." If there is more thanone such triangle, use the "or" button to enteradditional solutions. Round all answers to thenearest tenth with angle measure in degrees.

When two sides and an angle not included between them are given, we are in the SSA case

(Side-Side-Angle).

In the SSA case, we always know a side and its opposite angle. In this problem, we know and its

opposite angle . We can then use the other given side to solve for its opposite angle usingthe law of sines:

.

We get the following:

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.

(We will use the exact value of below rather than the approximation,

, in order to minimize rounding error.)

Since is an angle in a triangle, we have that .

The only two possibilities for are the angles and shown on the unit circle in Figure 1:

and

.

(These values for and are approximations found using a calculator.)

Note that both and are solutions for because neither is "too large" for the triangle:

and

.

For , we have

Figure 1

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.

Using the law of sines, we have

.

For , we have

.

Using the law of sines, we have

.

Note that, in calculating , we used the values of and (given in the question) rather

than the values of and (an approximation) in order to minimize rounding error.

Here are the solutions, with values rounded to the nearest tenth.

, ,

, or

, ,

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Compute the value of the discriminant and givethe number ofrealsolutions to the quadraticequation

.

Background:

The solutions of the quadratic equation are given by the quadratic formula:

.

The expression under the square root, , is called the discriminantof the

quadratic equation. The value of the discriminant determines the number ofreal solutions to

:

y If , there are real solutions to .y If , there is real solution to .y If , there are no real solutions to (as the square

root of a negative number is not a real number).

The current problem:

For the quadratic equation , we have that , , and

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.

Thus, the value of the discriminant is

.

Since the value of the discriminant is equal to zero , the quadratic equation

has real solution .

The answer is:

Discriminant:

Number of real solutions: .

Find all the values of such that the quadraticequation

has two real solutions . Write your

answer as an equality orinequality in terms of

.

Background:

The solutions of the quadratic equation are given by the quadratic formula:

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.

The expression under the square root, , is called the discriminantof the

quadratic equation. It turns out that the value of the discriminant determines the number ofreal

solutions to :

y If , there are real solutions to .y If , there is real solution to .y If , there are no real solutions to (as the square

root of a negative number is not a real number).

The current problem:

In order for to have two real solutions , we must have the discriminant

. Plugging , , and into this inequality and

solving for , we have

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.

The answer is

.

Rewrite as a logarithmic equation.

For any numbers , , and , with and positive ( ) , we have the equivalence

if and only if .

The first is a logarithmic equation, and the second is an exponential equation.

Note that is of the form , with , , and .

We write it as a logarithmic equation as follows.

The midpoint formula states that the midpoint of the line segment joining the points and

is the point ((x1+x2)/2 , (y1+y2)/2)

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So the midpoint of the line segment joining

= and = is

.

See Figure 1.

The answer is =

.

Solve

for . Simplify your answer as much aspossible.

For any numbers , , and , with and positive ( ), we have the equivalence

if and only if .Applying this to our problem gives

if and only if .From the equation on the right, we have that

.

Thus, .

Solve

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for . Simplify your answer as much aspossible.

For any numbers , , and , with and positive ( ), we have the equivalence

if and only if .Applying this equivalence to our problem gives

if and only if .

Raising both sides of the equation on the right to the power , we get

.

The answer is .

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Find the magnitude of the vectorgiven below.Also find the measure (in degrees) of the acute

angle formed by the vector and the -axis.Do not round any intermediate computations, and

round your responses to decimal places.

Consider the right triangle with its base

along the -axis and the vector as the hypotenuse (Figure 1). The length of one of the legs of this

right triangle equals , and the length of the other leg equals .

The magnitude of a vector is the length of that vector. Therefore, we can use the Pythagorean

Theorem to find the magnitude of the given vector. Specifically, using to represent the magnitudeof the given vector, we have

.

Using a calculator, we can compute the value of (which must be nonnegative as it

represents a length):

Figure 1. Imagine a right triangle

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.

To find , we use the definition oftangent. Specifically, we have

.

Therefore, we can use arctangent to find :

.

The answer is:

magnitude of the

vector:

.

A vectorwith initial point and terminal

point is translated so that its initialpoint is at the origin. Find its new terminal point.

In translating a vector, we maintain its magnitude and direction. If we move the initial point of

a vector to the origin, we must simultaneously move its terminal point in the same way.

In other words, to move the initial point to, we must subtract from the -coordinate and

add to the -coordinate:

.

Figure 1

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We must also do this to the terminal point . Therefore, the new terminal point will

be:

.

The answer is

.

For a given arithmetic sequence, the th term,

, is equal to , and the th term,

, is equal to .

Find the value of the th term, .

A sequence is an arithmetic sequence if and only if the differences between

consecutive terms are equal to some number (called the common difference of the sequence):.Showing this another way, we have.

So, for example, we have that . More generally, for any terms and of anarithmetic sequence, we must have

.Note that if we know the value of any two terms in an arithmetic sequence, we can use this formula to

find the value of . For the current problem, we are given that and. So, we have the following.

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We can now find the value of by using this value of and the value of any other term in the

sequence. Using with , we have the following.

The answer is .

Compute . Round your answer todecimal places.

Most calculators compute logarithms only with base (denoted by ) or base (denoted by

). To compute a logarithm with a base different from or , we use the change of baseformula for logarithms.

Change of base formula for logarithms:

For any positive numbers , , and , ,

.

For base , the formula becomes

.

A proofofthisformula

In the current problem, we get the following.

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The answer is .

Consider the equation

.

Find the value of . Round your

answer to decimal places.

Using the property of logarithm of a power, we can rewrite

as

.

Solving this equation for , we have

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.

We then use the change of base formula for logarithms,

.

proofof

thisformula

The answer is

.