Equations - ISEE Upper Level Mathematics Achievement

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

Now factor the quadratic expression into two binomial factors , replacing the question marks with two integers whose product is and whose sum is . These numbers are , so:

or

The solution set is .

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

Rewrite this quadratic equation in standard form:

Factor the expression on the left. We want two integers whose sum is and whose product is . These numbers are , so the equation becomes

.

Set each factor equal to 0 separately, then solve:

Rewrite this as a compound statement and solve each separately:

FOIL each of the two expressions, then solve:

Solve the resulting linear equation:

Simplify both sides, then solve:

Simplify both sides, then solve:

This is an identically true statement, so the original equation has the set of all real numbers as its solution set.

With this problem, you'll need to test each of the possible answers to see if they are equivalent to .

The value of the fraction, if you try and divide it out, has a repeating three after the decimal - which is shown as . Therefore you can eliminate this answer choice, since it is the same value.

You can also eliminate because when you reduce that fraction, by dividing the numerator and denominator by 15, you will end up with .

You can also eliminate . If you multiply the numerator and denominator by a special value of 1 to get rid of the decimal points - in this case multiplying each by 10 - you will get a fraction of . This can more easily be reduced to the fraction in question of .

This leaves us with the correct answer - the value that is not equal to the fraction in question: . While this value is extremely close to the repeating decimal, it is not exactly the same because the fraction does terminate - it has a final ending point. Therefore, it is not the exact same value as .

The first step here is to recognize you need to simplify within the radical before moving outside of the radical (avoid a common mistake of getting the square root of 169 and 144 before subtracting). Thus, you subtract 144 from 169, which gives you a difference of 25:

= =

Then you can take the square root of 25, which gives you the answer of 5.

First, you need to idenitfy what an integer is: any whole number. That can be any positive number, negative number, or zero. A non-integer is a number that cannot be written as a whole number, such as a fraction or decimal number.

You can simplify each expression to determine which one does not simplify into an integer. Three expressions result in integers, while one does not:

Each of those three expressions does simplify into an integer, so they are not the correct answers. The following results in a fraction that cannot be simplified into a whole number, so it is not an integer, and thus, the correct answer:

The first step is to combine like terms. In this case, we need to subtract from both sides.

Next we add to both sides:

Divide both sides by :

Finally, take the square root of both sides:

The first step is to rewrite the problem by combining the like terms.

Next we add to both sides.

Then we add to both sides to isolate the term with the variable.

Finally, we divide both sides by .

To find of one minute in terms of seconds, you need to first convert one minute to its value in seconds, which is 60 seconds.

Now, you just need to find the value for of 60, which you determine by multiplying the two values:

You can simplify the above fraction by division, which will you give you your correct answer in seconds:

Your answer is 4 seconds.

When adding fractions, generally you determine the least common multiple of the two denominators given to modify the fractions and successfully add. However, this problem gives you a specific denominator to reach: 24.

You will need to change both and to fractions that have a denominator of 24. This will require you to modify their numerators to keep the fractions the same. You will then add the two fractions, and the numerator in the sum is your answer.

First, change to a fraction of equal value with a denominator of 24. To go from 3 to 24 in the denominator, you must mutliply by 8, so you perform the same with the numerator to keep the fraction at the same value. (You are multiplying by a fraction that is equal to 1, so you do not change the value of the fraction.):

Next, you'll do the same thing with . To go from 4 to 24 in the denominator, you must multiply by 6, so you perform the same operation with the numerator, as described above:

Now that you have two fractions with common denominators, you can add them together:

Adding 16 and 6 together gives you the value of , which is 22:

You can think of the total number of students in the classroom as an unknown variable, .

You can set up an equation that includes and the information given. Normally in a percent problem, you would start with the total amount of something and multiply it by the percent to get the part of that total amount. You can set this up the same way, but the total amount is the missing information:

Total Amount x Percent = Part of Total Amount

You use to numerically represent 15 percent.

Now you can algebraically solve for , which will give you the total number of students in the class. You do this by dividing each side by .

Therefore there are 60 total students.