All High School Math Resources
Example Questions
Example Question #2 : Solving Inequalities
Solve the inequality for x:
Subtract 4 from both sides:
Divide both sides by 2:
Example Question #51 : Basic Single Variable Algebra
Solve for .
Add 4 to both sides.
Divide both sides by –7. When dividing by a negative value, we must also change the direction of the inequality sign.
Example Question #5 : Solving Inequalities
Solve for :
Move like terms to the same sides:
Combine like terms:
Divide both sides by 3:
Example Question #11 : Solving Inequalities
Solve for :
Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that becomes , and vice versa.
Example Question #1 : Direct Proportionality
Sarah notices her map has a scale of . She measures between Beaver Falls and Chipmonk Cove. How far apart are the cities?
is the same as
So to find out the distance between the cities
Example Question #2 : Direct Proportionality
Sunshine paint is made by mixing three parts yellow paint and one part red paint. How many gallons of yellow paint should be mixed with two quarts of red paint?
(1 gallon = 4 quarts)
First set up the proportion:
x =
Then convert this to gallons:
Example Question #246 : Algebra Ii
If is directly proportional to , and when , find a formula for .
since and
Solving for , you get that
Replacing in the original equation, you get the answer of
Example Question #1 : Understanding Indirect Proportionality
If is inversely proportional to , and if when , find a formula for .
Since y is inversely porportional to x, you can use the standard equations of where k is a constant. Plugging the given x and y into this equation to solve for k, . To get the answer, substitute k into the standard equation.
Example Question #1 : Polynomials
If and , what is ?
is a composite function where is plugged into :
Example Question #1 : Intermediate Single Variable Algebra
Simplify the following expression:
.
First, multiply out the second expression so that you get .
Then, multiply your like terms, taking care to remember that when multiplying terms that have the same base, you add the exponents. Thus, you get .
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