Which statement is true about f(x)+2=1/6|x-3|? - id21633326112, LilFlamingo247

Ответ:

jhanes1983

07.01.2022

Mathematics

The range of f(x) is $f(x)\geq -2$ and option D is correct.

Step-by-step explanation:

The given function is

$f(x)+2=\frac{1}{6}|x-3|$

It can be written as

$f(x)=\frac{1}{6}|x-3|-2$ .... (1)

The function is in the form of

g(x)=a|x-h|+k ....(2)

Where, a is scale factor and (h,k) is vertex of the graph.

On comparing (1) and (2), we get

$a=\frac{1}{6}$

h=3

k=-2

Therefore the vertex of f(x) is (3,-2). Option A is incorrect.

The value of a is $\frac{1}{6}$ . So, the graph compressed vertically. The value of a is positive, therefore the graph of f(x) opens upward.

We know the absolute value is always greater than or equal to 0.

$|x-3|\geq 0$

$\frac{1}{6}|x-3|\geq \frac{1}{6}(0)$

$\frac{1}{6}|x-3|\geq 0$

$\frac{1}{6}|x-3|-2\geq 0-2$

$f(x)\geq -2$

Therefore the range of f(x) is $f(x)\geq -2$ and option D is correct.

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Ответ:

NeverEndingCycle

11.05.2022

Mathematics

$\begin{aligned} _{2n}c_2 = 2 ( _nc_2) + n^2 \end{aligned}$ $c(n,r) = {}_nc_r = \binom{n}{r} = \dfrac{n! }{r! (n-r)! }$ we can do $n! = n(n-1)! = n(n-1)(n-2)! = /tex]so it follows that [tex](2n)! = 2n(2n-1)! = n(2n-1)(2n-2)! = /tex]left hand side: [tex]\begin{aligned} \text{lhs} & ={} _{2n}c_2 \\ & = {} \dfrac{(2n)! }{2! (2n-2)! } \\ & = {} \dfrac{(2n)(2n-1)(2n-2)! }{2! (2n-2)! } \\ & = {} \dfrac{2n(2n-1)}{2! } \\ & = {} \dfrac{2n(2n-1)}{2} & & \text{since $2! = 2$} \\ & = {} \dfrac{n(2n-1)}{1}\\ & = {} 2n^2 - n\\ \end{aligned}$ right hand side: $\begin{aligned} \text{rhs} & = 2(_nc_2) + n^2 \\ & = 2 \left( \dfrac{n! }{2! (n-2)! } \right) + n^2 \\ & = 2 \left( \dfrac{n(n-1)(n-2)! }{2! (n-2)! } \right) + n^2 \\ & = 2 \left( \dfrac{n(n-1)}{2} \right) + n^2 \\ & = (1) \left( \dfrac{n(n-1)}{1} \right) + n^2 \\ & = n(n-1) + n^2 \\ & = n^2 - n + n^2 \\ & = 2n^2 - n \\ & = \text{lhs} \end{aligned}$

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Which statement is true about f(x)+2=1/6|x-3|?
a. the graph of f(x) has a vertex of (-3,2).
b. the graph of f(x) is horizontally compressed.
c. the graph of f(x) opens downward.
d. the graph of f(x) has range of f(x)_> -2.

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Which statement is true about f(x)+2=1/6|x-3|? a. the graph of f(x) has a vertex of (-3,2). b. the graph of f(x) is horizontally compressed. c. the graph of f(x) opens downward. d. the graph of f(x) has range of f(x)_> -2.

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Which statement is true about f(x)+2=1/6|x-3|?
a. the graph of f(x) has a vertex of (-3,2).
b. the graph of f(x) is horizontally compressed.
c. the graph of f(x) opens downward.
d. the graph of f(x) has range of f(x)_> -2.