The first few steps in deriving the quadratic formula are shown.
Which best explains why
02
422
is not added to the left
-C-ax+bx
side of the equation in the last step shown in the table?
Use the subtraction property of
equality
Factor out a
DO
-c-
-C-ax
The term
52
402
is added to the right side of the
b
2a
52
4a
Find half of the b value and square
it to determine the constant of the
perfect square trinomial
b
bi
4a
X2
a
ba
4a1
equation, so it needs to be subtracted from the left
side of the equation to balance the sides of the
equation.
The distributive property needs to be applied to
determine the value to add to the left side of the
equation to balance the sides of the equation.
b2
The term needs to be converted so it has a
492
common denominator before adding it to the left side
of the equation to balance the equation.
The square root of the term needs to be found before

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range:

$75.00

step-by-step explanation:

i = prt

i = 1,5000 * 0.025 * 2

i = 75

agree

step-by-step explanation: