All categories

3 years ago

If f(x) = 3(x + 4), find f(2). 729 243 81 18

Mathematics

2 answers:

bonufazy [111]3 years ago

7 0

The answer is 18.

Just replace the x with 2

Serga [27]3 years ago

4 0

Answer:

f(2) = 18

Option 4 is correct

Step-by-step explanation:

Given: f(x)=3(x+4)

To find: f(2)

The given function is linear function with variable x to find the value of function at x=2

So, put x=2 into the function and simplify it

f(x) = 3(x+4)

f(2) = 3(2+4)

f(2) = 3(6) [ multiply 3 and 6 = 18]

f(2) = 16

Hence, The value of f(2) is 18 of the given function.

You might be interested in

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

Question above please help due in a few minutes

nexus9112 [7]

Answer:

Step-by-step explanation:

here x $\geq$ 3

the circle is filled so it includes the value

7 0

3 years ago

Why is the number 6 scared of the number 7?

Lady_Fox [76]

This is a well known joke:

Why six afraid of seven? Because seven ate nine!

It is supposed to be a "word game" where instead of writing eight, the authour wrote ate, which sound the same but means something different.

Hope it helped,

Happy homework/ study/ exam!

8 0

3 years ago

Read 2 more answers

Complete the pattern 9.5 950 95,000

Sati [7]

<span>9,500,000 is the next number in the pattern. Each time you multiply the number on the left by 100 to get the number on the right.</span>

4 0

3 years ago

PLEASE HELP ME SOLVE THIS Find all x-values for which f ( x ) = 2 for the function below.

alisha [4.7K]

Answer:

Step-by-step explanation:

First let us get the equation of the coordinates

y-y0 = m(x-x0)

Using the coordinates ( - 3, 2 ), ( - 1, 0 )

m = 0-2/-1-(-3)

m = -2/2

m = -1

Substitute m = -1 and (-1, 0) into the formula

y - 0 = -1(x+1)

y = -x-1

f(x) = -x-1

Since f(x) = 2

2 = -x-1

-x = 2+1

-x = 3

x = -3

Hence the value of x is -3

3 0

3 years ago