Single-Variable Algebra - GED Math

Let be the number of quarters. Then there are dimes.

An equation can be set up and solved for for the amount of money in dollars:

To answer this question, we will solve for y. We get

To answer the question, we will solve for x. So, we get

To solve for k, we want k to stand alone. So, we get

To answer the question, we will solve for y. So, we get

To answer the question, we will solve for x, So, we get

To solve for b, we want b to stand alone. So, we get

To answer this, we will solve for y. So, we get

Every second, you know that Marge and William will become a total of or . Now, you can use the simple work formula for distance:

or, for our data,

(Remember kilometers is meters.)

Thus, solving for you get:

This is in seconds, though. You need minutes. To convert, you need to divide by :

Timmy makes ten dollars per hour for eight hours.

For five days:

Timmy also will pay for the week to get to work and back.

Subtract his expense from his earnings for the week.

Timmy will have by the end of the week.

where

The numbers and fit those criteria. Therefore,

You can double check the answer using the FOIL method

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that is a factor of polynomial if and only if . We substitute 1, 2, 4, and 9 for in the polynomial to identify the factor.

:

:

:

:

Only makes the polynomial equal to 0, so among the choices, only is a factor.

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that is a factor of polynomial if and only if . We substitute and for in the polynomial to identify the factor.

:

:

:

:

Only makes the polynomial equal to 0, so of the four choices, only is a factor of the polynomial.

Factor all the way to its prime factorization.

can be factored as the difference of two perfect square terms as follows:

is a factor, and, as the sum of squares, it is a prime. is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

Therefore, all of the given polynomials are factors of , but is the correct choice, as it is not a prime factor.

is the sum of two cubes:

As such, it can be factored using the pattern

where ;

The first factor,as the sum of squares, is a prime.

We try to factor the second by noting that it is "quadratic-style" based on . and can be written as

;

we seek to factor it as

We want a pair of integers whose product is 1 and whose sum is . These integers do not exist, so is a prime.

is the prime factorization and the correct response is .

This can be most easily solved by first substituting for , and, subsequently, for :

This becomes quadratic in the new variable, and can be factored as

,

filling out the blanks with two numbers whose sum is and whose product is . Through some trial and error, the numbers can be seen to be .

Therefore, after factoring and substituting back,

The first factor, the sum of squares, is prime. The second factors as the difference of squares, so the final factorization is

.

Of the choices given, is correct.

can be seen to fit the pattern

:

where

can be factored as , so

.

does not fit into any factorization pattern, so it is prime, and the above is the complete factorization of the polynomial. Therefore, is the correct choice.

The area of a triangle is:

Use the FOIL Method to simplify.