domain: find the domain of the function. in the space provided, type the answer without any spaces between the letters, numbers, or symbols.
let x and y be directly related. if x is 18 when y is 6, find y when x is 21.
i'm going to assume that "directly related" means a linear relationship. otherwise, there is no way to write an equation based on the information provided.assuming a linear function, use the form y = mx + b. with the information given, m and b cannot be determined. however, the question is simply the domain. for a linear equation, the domain is simply "all real numbers".in the space provided, type the answer without any spaces between the letters, numbers, or symbols.i have no idea how that relates to the stated question.......
?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a for this, just use an if function. see https://macmost.com/learn-how-to-use-the-if-function-in-numbers.html so something like this in a2: if(isblank(b2),c2,b2) the condition is to check if b2 is blank. if it is, get the value of c2. if it isn’t blank, then get the value of b2. f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> Let $y=f(x)$ be a function. Suppose that $a$ and $\mathbf{L}$ are numbers such that:


 


* whenever $\boldsymbol{x}$ is close to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ is close to $\mathbf{L} ;$
* as $\boldsymbol{x}$ gets closer and closer to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ gets closer and closer to $\mathbf{L}$; and
* suppose that $f(x)$ can be made as close as we want to $\mathbf{L}$ by making $x$ close to $a$ but not equal to $a$.


Then we say that the limit of $f(x)$ as $x$ approaches a is $L$ and we write $\lim \_{x \rightarrow a} f(x)=\mathbf{L}$ given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region