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Example Questions
Example Question #1 : Simplifying Polynomials
Evaluate the following:
To subtract these two trinomials, you first need to flip the sign on every term in the second trinomial, since it is being subtrated:
Next you can combine like terms. You have two terms with , two terms with , and two terms with no variable:
Example Question #11 : Trinomials
Find the difference:
Find the difference:
Distribute the negative to the second trinomial:
Combine like terms:
Example Question #1 : How To Subtract Trinomials
Subtract:
When subtracting trinomials, first distribute the negative sign to the expression being subtracted, and then remove the parentheses:
Next, identify and group the like terms in order to combine them: .
Example Question #1 : How To Divide Trinomials
Find the quotient:
Find the quotient:
Step one: Factor the numerator
Step two: Simplify
Example Question #1 : How To Use The Direct Variation Formula
Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?
Every minute Phillip completes another square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes square feet per minute, then we can multiply by the total minutes to find the final answer.
Example Question #2 : How To Use The Direct Variation Formula
The value of varies directly with the square of and the cube of . If when and , then what is the value of when and ?
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = kx2z3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8k
24 = 8k
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = kx2z3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
Example Question #2 : How To Use The Direct Variation Formula
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.
In this case, the equation will be .
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
Example Question #63 : Variables
varies directly as and inversely as ; and . Assuming that no other variables affect , which statement is true of concerning its relationship to ?
varies directly as the fifth power of .
varies inversely as the fifth power of .
None of the other statements are correct.
varies inversely as .
varies directly as .
varies inversely as .
varies directly as and inversely as , meaning that for some constant of variation ,
.
Setting and , the formula becomes
Setting as the new constant of variation, the variation equation becomes
,
so varies inversely as .
Example Question #1 : How To Use The Direct Variation Formula
varies directly as both and the cube of . Which statement is true of concerning its relationship to ?
varies inversely as the cube of .
varies inversely as the cube root of .
None of the other statements is true.
varies directly as the cube root of .
varies directly as the cube of .
varies inversely as the cube root of .
varies directly as both and the cube of , meaning that for some constant of variation ,
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
takes the role of the new constant of variation here, and we now have
so varies inversely as the cube root of .
Example Question #1 : How To Use The Inverse Variation Formula
The square of varies inversely with the cube of . If when , then what is the value of when ?
When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.
x2y3 = k
We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.
82(83) = k
k = 82(83)
Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.
k = 82(83) = 82+3 = 85.
Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.
x2y3 = 85
(1)2y3 = 85
y3 = 85
In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.
y3 = (23)5
We can now apply the property of exponents that states that (ab)c = abc.
y3 = 23•5 = 215
In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.
(y3)(1/3) = (215)(1/3)
Once again, let's apply the property (ab)c = abc.
y(3 • 1/3) = 2(15 • 1/3)
y = 25 = 32
The answer is 32.
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