The EXPAND equation of the polynomial function is f(x)= (x-2)(x-5)(x-sqrt3)(x+sqrt3) 15 POINTS Please explain as well if you can. Please

f(x) = x⁴ - 7x³ + 7x² + 21x - 30

Step-by-step explanation:

Given function is f(x) = (x - 2)(x - 5)(x - √3)(x + √3)

f(x) = (x - 2)(x - 5)[(x² - (√3)²]  {By using formula, (a - b)(a + b) = a² - b²]

     = (x - 2)(x - 5)(x² - 3)

     = (x - 2)[x(x² - 3) -5(x² - 3)] [By distributive property]

     = (x - 2)(x³ - 3x - 5x² + 15)

     = (x - 2)(x³ - 5x² - 3x + 15)

     = [x(x³ - 5x² - 3x + 15) - 2(x³ - 5x² - 3x + 15)]

     = x⁴ - 5x³- 3x² + 15x - 2x³ + 10x² + 6x - 30

     = x⁴ - 7x³ + 7x² + 21x - 30

Therefore, expanded form of the polynomial function will be,

f(x) = x⁴ - 7x³ + 7x² + 21x - 30  

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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: