Algebraic Concepts - HiSET

By the two-point formula, the equation of a line through points is

,

where , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

The first formula becomes

Since , solve for . First, distribute throughout the difference at right:

Isolate by adding 10 to both sides:

Replacing, the definition of is

.

In an equation, a term is a single number or a variable. in our equation we have the following terms:

Step 1: We need to separate and arrange the terms in the jumbled expression given to us in the question..

We will separate the terms by the exponent value.

For , there is only one term:

For , there are no terms.

For , there is two terms:

For , there are two terms:

For , there are two terms:

For , there is only one term:

For , there are two terms:

For , there are three terms:

There are four constant terms:

Step 2: We will now add up the coefficients in each designation of terms. This will give us the answer.

Arrange and combine like terms (those with the same variable) as follows:

Since each term now has a different exponent for the variable, no further combining is possible. The simplified form has four terms.

It is not necessary to actually find the solutions to a quadratic equation to determine the sum of its solutions.

First, get the equation in standard form by subtracting from both sides:

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, , making the correct choice.

The Intermediate Value Theorem (IVT) states that if the graph of a function is continuous on an interval , and and differ in sign, then has a zero on . Consequently, the way to answer this question is to determine the signs of on the endpoints of the subintervals - . We can do this by substituting each value for as follows:

assumes positive values for and negative values for . By the IVT, has a zero on .

Step 1: Find the Least Common Denominator of these fraction. We will list out multiples of each denominator until we find a common number for all three fractions...

The smallest common denominator is .

Step 2: Since the denominator is , we will convert all denominators to .

Step 3: Add up all the values of x...

Step 4: The result from step is the numerator and is the denominator. We will put these together.

Final Answer:

In order to solve for the variable, , we need to isolate it on the left side of the equation. We will do this by reversing the operations done to the variable by performing the opposite of each operation on both sides of the equation.

Let's begin by rewriting the given equation.

Subtract from both sides of the equation.

Simplify.

Multiply both sides of the equation by .

Solve.

can be rewritten as the compound inequality

.

The solution set will be the union of the two individual solution sets. Find the solution set of the first inequality as follows:

Isolate by first subtracting from both sides:

Divide both sides by , reversing the direction of the inequality symbol, since you are dividing by a negative number:

.

In interval notation, this is the set .

Find the solution set of the other inequality similarly:

In interval notation, this is the set .

The union of these sets is the solution set: .

The above function can be understood as, "An employee is paid \$15 for each hour he or she works, plus a flat amount of \$10."

If an employee works 7 hours, we can find the amount that he or she is paid by plugging in 7 for "Hours" in the equation:

Adhering to order of operations, we next find the product of 15 and 7:

Finally, we find the sum of 105 and 10:

The employee is paid \$115.00 for seven hours' work.

The question asks, "Which of the following functions best models the number of paintings she will have after days?"

From this, you know that the variable represents the number of days, and that represents the number of paintings she makes as a function of days spent working.

If it takes 3 days to make a painting, each day results in paintings. Therefore, we have a linear relationship with slope .

Additionally, she begins with 6 paintings. Therefore, even when zero days are spent working on paintings, she will have 6 paintings. In other words, . This means the y-intercept is 6.

As a result, the function will be

which can be rewritten as

By the two-point formula, the equation of a line through points is

,

where , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

The first formula becomes

Since , solve for . First, distribute throughout the difference at right:

Isolate by adding 10 to both sides:

Replacing, the definition of is

.

Given a quadratic function

the vertex will always be

.

Thus, since our function is

, , and .

We plug these variables into the formula to get the vertex as

.

Hence, the vertex of

is

.

The discriminant of a quadratic polynomial

is given by

.

Thus, since our quadratic polynomial is

,

, , and .

Plugging these values into the discriminant equation, we find that the discriminant is

.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

This can be done by adding 17 to both sides:

The key to determining the nature of the solution set is to examine the discriminant

. Setting , the value of the discriminant is

This value is negative. Consequently, the solution set comprises two imaginary numbers.

A polynomial equation of the form

has one and only one (real) solution if and only if its discriminant is equal to zero - that is, if its coefficients satisfy the equation

In each of the choices, and , so it suffices to determine the value of which satisfies this equation. Substituting, we get

Solve for by first adding 400 to both sides:

Take the square root of both sides:

The choice that matches this value of is the equation

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

Collect like terms:

Now, subtract 18 from both sides:

The key to determining the nature of the solution set is to examine the discriminant . Setting , the value of the discriminant is

The discriminant is a positive number, so there are two real solutions. Since 73 is not a perfect square, the solutions are irrational.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

Collect like terms:

Add 6 to both sides:

The key to determining the nature of the solution set is to examine the discriminant . Setting , the value of the discriminant is

The discriminant is a positive number; furthermore, it is a perfect square, being equal to the square of 11. Therefore, the solution set comprises two rational solutions.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

This can be done by simply switching the first and second terms:

The key to determining the nature of the solution set is to examine the discriminant . Setting , the value of the discriminant is

The discriminant has a negative value. It follows that the solution set comprises two imaginary values.

A polynomial equation of the standard form

has one and only one (real) solution if and only if its discriminant is equal to zero - that is, if its coefficients satisfy the equation

Each of the choices can be rewritten in standard form by subtracting the term on the right from both sides. One of the choices can be rewritten as follows:

By similar reasoning, the other four choices can be written:

In each of the five standard forms, and , so it is necessary to determine the value of that produces a zero discriminant. Substituting accordingly:

Add 900 to both sides and take the square root:

Of the five standard forms,

fits this condition. This is the standard form of the equation

,

the correct choice.