Algebra II - Math

By definition, .

An arithmetic sequence is one in which the common difference is added to one term to get the next term. Thus, if the first term is 3, we add 5 to get the second term, and continue in this manner.

Thus, the first four terms are:

A geometric sequence is one in which the next term is found by mutlplying the previous term by a particular constant. Thus, we look for an implicit definition which involves multiplication of the previous term. The only possibility is:

We are asked to solve the equation .

Often, when solving equations involving rational expressions, it helps to elminate fractions by multiplying both sides of the equation by the denominators of each term intervolved.

In the context of this problem, we can first multiply both sides of the equation by x+2 to eliminate the denominator of the first term.

Be sure to distribute the x+2 to each term on the left side of the equation.

Next, multiply both sides of the equation by x to elminate the term with an x in the denominator.

Then, multiply both sides of the equation by 2.

In order to solve this polynomial equation, we can use the Rational Root Theorem. According to this theorem, if there is a rational root to a polynomial equation, then that root must be in the form p/q, where p is a factor of the constant, and q is the factor of the coefficient of the highest term.

In the context of this problem, p will be a factor of 4 (which is the constant), and q will be a factor of 2 (which is the coefficient of the highest term).

The factors of 4 are . The factors of 2 are .

This means that if there is a rational root for the equation, then the numerator must be either , and the denominator must be one of . In other words, the possible rational roots are , because these are all the possible ratios we can make with a numerator of and a denominator of one of .

At this point, we have narrowed down the rational roots of the equation to the following eight choices: .

Notice, that if we were to let , then the polynomial would evaluate to 4. Also, if we were to let , then the polynomial would evaluate to -3. Because the value of the polynomial switches from a postive value of 4 when x = 0 to a negative value of -3 when x = -1, there must be a root somewhere between and .

Let's see what happens when , because -1/2 is between 0 and -1.

We will put -1/2 into the polynomial to see if it evaluates to zero.

Thus is indeed a root to the equation.

However, the question asks us to find the largest real root; this means that there could be other roots larger than . We will need to continue solving the equation to see what other roots are possible.

We can make use of the Factor Theorem of polynomials, which states that if is a root of a polynomial equation, where a is constant, then x-a must be a factor of the polynomial.

This means that for our polynomial, is a factor. We can divide the polynomial by using long division.

This means that

In order to find the remaining roots of the equation, we now need to solve the quadratic equation .

First, divide both sides of the equation by two.

Perhaps the most straightforward way to solve this is to use the quadratic formula:

When we evaluate this, we will be forced to take the square root of . Because this will result in an imaginary number, has no real solutions.

Thus, returning to our original polynomial equation , it turns out that -1/2 is indeed the only root of the equation.

The question asks us to find the largest real value of x that solves the equation. Because -1/2 is the only real value that solves the equation, the answer must be -1/2.

Begin by multiplying the numerator and denominator by the complement of the denominator:

Combine like terms:

Simplify. Remember that is equivalent to

To find the distance on a number line:

We are looking for a number that is not real.

, , and are irrational numbers, but they are still real.

Then, is equivalent to by the rules of complex numbers. Thus, it is also real.

That leaves us with: which in fact is imaginary (since no real number multiplied by itself yields a negative number) and simplifies to .

How many cards in the deck are either a spade or a 3?

There are thirteen spades, including a 3 of spades.

There are four 3's, including a 3 of spades.

Since we are counting the same card (3 of spades) twice, there are actually

distinct cards that fit the criteria of being either a spade or a 3.

Since any of the 52 cards is equally likely to be drawn, the probability that it is a spade or a 3, is

We know that and .

Because 90 falls is approximately halfway between 81 and 100, the square root of 90 is approximately halfway between 9 and 10, or 9.5.

Real numbers can be found anywhere on a continuous number line ranging from negative infinity to positive infinity; therefore, all of the numbers are real numbers.

To place in order, first we must find a common denominator and convert all fractions to that denominator.

have a common denominator of .

have a common denominator of .

have a common denominator of .

Therefore we can use common denominators to make all of the fractions look similar. Then the ordering becomes trivial.

For percent problems there are verbal cues:

"IS" means equals and "OF" means multiplication.

Then the equation to solve becomes:

Find the least common denominator (LCD) and convert each fraction to the LCD and then add. Simplify as necessary.

The result is an improper fraction because the numerator is larger than the denominator and can be simplified and converted to a mix numeral.

Every matrix has a dimension, which is represented as the number of rows and columns. For example, a matrix with three rows and two columns is said to have dimension 3 x 2.

The matrix

has two rows and three columns, so its dimension is 2 x 3. (Remember that rows go from left to right, while columns run up and down.)

Matrix multiplication is defined only if the number of columns in the first matrix is equal to the number of rows on the second matrix. The easiest way to determine this is to write the dimension of each matrix. For example, let's say that one matrix has dimension a x b, and the second matrix has dimension c x d. We can only multiply the first matrix by the second matrix if the values of b and c are equal. It doesn't matter what the values of a and d are, as long as b (the number of columns in the first matrix) matches c (the number of rows in the second matrix).

Let's go back to the problem and analyze the choice .

The dimension of the first matrix is 2 x 3, because it has two rows and three columns. The second matrix has dimension 2 x 2, because it has two rows and two columns.

We can't multiply these matrices because the number of columns in the first matrix (3) is not equal to the number of rows in the second matrix (2). Thus, this product is not defined.

The answer is .

Convert the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator to get

Dividing by a fraction is the same as multiplying by its reciprocal so the problem becomes

Chenge the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator to get

.

To find the -intercepts, we must set .

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

Now we must set up our two scenarios:

and

and

and

Divide both sides by 3.

Consider both the negative and positive values for the absolute value term.

Subtract 2 from both sides to solve both scenarios for .

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

and

This gives us:

and

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as cannot equal a negative number.

Our final solution is then