All categories

10 months ago

RCX)=2VX SCx) = x(RSX4=

Mathematics

1 answer:

aev [14]10 months ago

3 0

Given the two functions:

$\begin{gathered} R(x)=2\sqrt[]{x} \\ S(x)=\sqrt[]{x} \end{gathered}$

We need to find (RoS)(4). THis is the functional composition. We take S(x) and put it into R(x) and then put "4" into that composed function. Shown below is the process:

$(RoS)(x)=2\sqrt[]{\sqrt[]{x}}$

When we plug in "4", into "x", we have:

$\begin{gathered} (RoS)(x)=2\sqrt[]{\sqrt[]{x}} \\ (RoS)(4)=2\sqrt[]{\sqrt[]{4}} \\ =2\sqrt[]{2} \end{gathered}$

You might be interested in

$300, 5%, 2 years explain

dsp73

<h2>ANSWER:</h2><h2 /><h2>16,000</h2>

<h2>EXPLANATION:</h2>

300+5%+2YEARS=16,000

HOPE IT HELPS YOU

CAN YOU BRAINLEST ME

6 0

2 years ago

Read 2 more answers

Algebra problem below.

umka21 [38]

$A= \frac{1}{2}bh= \frac{1}{2}*18*11=9*11=99 \ ft^2$

3 0

3 years ago

two pages that are back to back in a text have 201as the sum of their page numbers what are the page numbers.

REY [17]

Answer:

Pages 100 and 101

Step-by-step explanation:

Round 201 to 200 (to make it easier).

Divide 200 by 2, because it is 2 pages.

You are left with 100.

100 is one of the pages, but because you rounded down 1 number, the other number must be 101, because the pages are back to back.

I hope this made sense, this is just how I figured it out. :)

6 0

3 years ago

Eric drew a scale drawing of the high school. The parking lot, which is 54 meters wide in real l

DaniilM [7]

Answer:

9 meters

Step-by-step explanation:

54/6 = 9

8 0

2 years ago

How do you find absolute extrema for a function? f(x)= (8+x)/(8-x); Interval of [4,6]

Mkey [24]

$\bf f(x)=\cfrac{8+x}{8-x}\implies \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{1(8-x)-(8+x)(-1)}{(8-x)^2}}\implies \cfrac{dy}{dx}=\cfrac{16}{(8-x)^2}$

now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.

so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.

therefore, we settle for the endpoints, 4 and 6,

f(4) =3 and f(6) = 7

doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.

3 0

3 years ago