Pre-Calculus - Math

Card 0 of 228

Question

$f(x)=\frac{x^{2}-x-6}{x+2}$

$\lim_{x\rightarrow -2}f(x) =$

Answer

A limit describes what -value a function approaches as approaches a certain value (in this case, ). The easiest way to find what -value a function approaches is to substitute the -value into the equation.

$f(-2)=\frac{(-2)^{2}-(-2)-6}{-2+2}=\frac{4+2-6}{0}=\frac{0}{0}$

Substituting for gives us an undefined value (which is NOT the same thing as 0). This means the function is not defined at that point. However, just because a function is undefined at a point doesn't mean it doesn't have a limit. The limit is simply whichever value the function is getting close to.

One method of finding the limit is to try and simplify the equation as much as possible:

$f(x)=\frac{(x-3)(x+2)}{x+2}$

As you can see, there are common factors between the numerator and the denominator that can be canceled out. (Remember, when you cancel out a factor from a rational equation, it means that the function has a hole -- an undefined point -- where that factor equals zero.)

After canceling out the common factors, we're left with:

Even though the domain of the original function is restricted ( cannot equal ), we can still substitute into this simplified equation to find the limit at

Compare your answer with the correct one above

Question

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$

What is the fifteenth term in the sequence?

Answer

The sequence can be described by the equation $\small 2+3(n-1)$ , where is the term in the sequence.

For the 15th term, .

$\small 2+3(n-1)=2+3(15-1)$

Compare your answer with the correct one above

Question

What is the sixth term when $\left (\frac{1}{2}x-2 \right )^{10}$ is expanded?

Answer

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of , where n is an integer. The rth term of this expansion is given by the following formula:

$\binom{n}{r-1}a^{n-r+1}b^{r-1}$ ,

where $\binom{n}{r-1}$ is a combination. In general, if x and y are nonnegative integers such that x > y, then the combination of x and y is defined as follows: $\binom{x}{y}=_xC_y=\frac{x!}{(y)!(x-y)!}$ .

We are asked to find the sixth term of $\left (\frac{1}{2}x-2 \right )^{10}$ , which means that in this case r = 6 and n = 10. Also, we will let $a=\frac{1}{2}x$ and . We can now apply the Binomial Theorem to determine the sixth term, which is as follows:

$\binom{10}{6-1}\left (\frac{1}{2}x \right )^{10-6+1}\cdot(-2) ^{6-1}$

$=\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$

Next, let's find the value of $\binom{10}{5}$ . According to the definition of a combination,

$\binom{10}{5}=\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!}{5!5!}$

$=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=252$ .

Remember that, if n is a positive integer, then $n!=n\cdot(n-1)\cdot(n-2)\cdot\cdot\cdot3\cdot2\cdot1$ . This is called a factorial.

Let's go back to simplifying $\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$ .

$\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}=252\cdot\left (\frac{1}{2}x \right )^{5}\cdot-32$

$=252\cdot\left (\frac{1}{2} \right )^5\cdot x^5\cdot (-32) =252\cdot \frac{1}{32}\cdot-3 2\cdot x^5$

The answer is .

Compare your answer with the correct one above

Question

What are the first three terms in the series?

$\sum_{n=1}^{7}(4n-5)$

Answer

To find the first three terms, replace with , , and .

The first three terms are , , and .

Compare your answer with the correct one above

Question

Find the first three terms in the series.

$\sum_{n=3}^{10}(8-5n)$

Answer

To find the first three terms, replace with , , and .

The first three terms are , , and .

Compare your answer with the correct one above

Question

Indicate the first three terms of the following series:

$\sum_{x=1}^{7}(4x-5)$

Answer

In the arithmetic series, the first terms can be found by plugging , , and into the equation.

Compare your answer with the correct one above

Question

Indicate the first three terms of the following series:

$\sum_{c=3}^{10}(8-5c)$

Answer

In the arithmetic series, the first terms can be found by plugging in , , and for .

Compare your answer with the correct one above

Question

Indicate the first three terms of the following series:

$\sum_{m=-2}^{9}(2)^m$

Answer

The first terms can be found by substituting , , and for into the sum formula.

$(2)^{-2}=\frac{1}{4}$

$(2)^{-1}=\frac{1}{2}$

$(2)^{0}=1$

Compare your answer with the correct one above

Question

Indicate the first three terms of the following series.

$\sum_{b=2}^{11} \frac{1}{2}(4)^{b-2}$

Answer

The first terms can be found by substituting , , and in for .

$\frac{1}{2}(4)^{2-2}=\frac{1}{2}$

$\frac{1}{2}(4)^{3-2}=2$

$\frac{1}{2}(4)^{4-2}=8$

Compare your answer with the correct one above

Question

Evaluate: $1 - \frac{2}{3} + \frac{4}{9} - \frac{8}{27} + ...$

Answer

This sum can be determined using the formula for the sum of an infinite geometric series, with initial term $a_{0} = 1$ and common ratio $r = -\frac{2}{3}$ :

$S = \frac{a_{0}} {1-r} = \frac{1} {1- \left ( -\frac{2}{3}\right ) } = \frac{1} {1+ \frac{2}{3} }= \frac{1} { \frac{5}{3} } =\frac{3}{5}$

Compare your answer with the correct one above

Question

The fourth term in an arithmetic sequence is -20, and the eighth term is -10. What is the hundredth term in the sequence?

Answer

An arithmetic sequence is one in which there is a common difference between consecutive terms. For example, the sequence {2, 5, 8, 11} is an arithmetic sequence, because each term can be found by adding three to the term before it.

Let $a_{n}$ denote the nth term of the sequence. Then the following formula can be used for arithmetic sequences in general:

, where d is the common difference between two consecutive terms.

We are given the 4th and 8th terms in the sequence, so we can write the following equations:

.

We now have a system of two equations with two unknowns:

Let us solve this system by subtracting the equation from the equation . The result of this subtraction is

.

This means that d = 2.5.

Using the equation , we can find the first term of the sequence.

Ultimately, we are asked to find the hundredth term of the sequence.

$a_{100}=a_1+(100-1)d=-27.5+99(2.5)=220$

The answer is 220.

Compare your answer with the correct one above

Question

Find the sum, if possible:

$3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...$

Answer

The formula for the summation of an infinite geometric series is

$S= \frac{a_1}{1-r}$ ,

where is the first term in the series and is the rate of change between succesive terms. The key here is finding the rate, or pattern, between the terms. Because this is a geometric sequence, the rate is the constant by which each new term is multiplied.

Plugging in our values, we get:

$S = \frac{3}{1-\frac{1}{3}}$

$S=\frac{3}{\frac{2}{3}}$

$S = \frac{9}{2}$

Compare your answer with the correct one above

Question

Find the sum, if possible:

$8-6+\frac{9}{2}-\frac{27}{8}+...$

Answer

The formula for the summation of an infinite geometric series is

$S= \frac{a_1}{1-r}$ ,

where is the first term in the series and is the rate of change between succesive terms in a series

Because the terms switch sign, we know that the rate must be negative.

Plugging in our values, we get:

$S = \frac{8}{1-(\frac{-3}{4})}$

$S = \frac{8}{\frac{7}{4}}$

$S = \frac{32}{7}$

Compare your answer with the correct one above

Question

Find the sum, if possible:

Answer

The formula for the summation of an infinite geometric series is

$S= \frac{a_1}{1-r}$ ,

where is the first term in the series and is the rate of change between succesive terms in a series.

In order for an infinite geometric series to have a sum, needs to be greater than and less than , i.e. .

Since , there is no solution.

Compare your answer with the correct one above

Question

Factor the polynomial if the expression is equal to zero when $x=-6, -4,\ or\ 2$ .

Answer

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

$\small (x^2+2x-8)(x+6)=x^3+6x^2+2x^2+12x-8x-48=x^3+8x^2+4x-48$

We are able to get back to the original expression, meaning that the answer is .

Compare your answer with the correct one above

Question

A polyomial with leading term $x^ {3}$ has 5 and 7 as roots; 7 is a double root. What is this polynomial?

Answer

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is $(x-5)(x-7)^{2}$ . To put this in expanded form:

$(x-7)^{2} = x^{2} - 2 \cdotx\cdot 7 + 7^{2} = x^{2} - 14x + 49$

$(x-5)(x-7)^{2} =(x-5) \left ( x^{2} - 14x + 49 \right )$

$=x \left ( x^{2} - 14x + 49 \right ) -5\left( x^{2} - 14x + 4 9 \right )$

$= x^{3} - 14x^{2} + 49x - \left( 5x^{2} - 70x + 245 \right )$

$= x^{3} - 14x^{2} + 49x - 5x^{2} + 70x -245$

$= x^{3} - 19x^{2} + 119x -245$

Compare your answer with the correct one above

Question

A polyomial with leading term $x^{3}$ has 6 as a triple root. What is this polynomial?

Answer

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is $(x-6)^{3}$ , which we can expland using the cube of a binomial pattern.

$(x-6)^{3} =x ^{3} - 3\cdot x^{2} \cdot 6 + 3\cdot x\cdot 6 ^{2} -6^{3}$

$x ^{3} - 18 x^{2}+ 108x -216$

Compare your answer with the correct one above

Question

What are the solutions to $y = x^{2} + x -6$ ?

Answer

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

$0 = x^{2} + x - 6 = (x+3)(x-2)$

Then, we are looking for the values where each of these factors are equal to zero.

implies

and implies

Thus, these are our solutions.

Compare your answer with the correct one above

Question

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

Answer

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Compare your answer with the correct one above

Question

Simplify the following polynomial:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

Answer

To simplify the polynomial, begin by combining like terms:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

$(\frac{-a^{2}b^{4}}{3c^{4}})^{-2}$

Compare your answer with the correct one above

Tap the card to reveal the answer