NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math. Let f(x) = 1/x-3 and g(x) = √x + 5.What is the domain of ( f o g) (x)?

A: [-5,3]U[3,&]

B: [-5,-4]U[-4,&)

C:[-5,4]U[4,&]

D: [-5,-3]U [-3,&]

The & equals
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Given

\begin{gathered}f(x)=\dfrac{1}{x-3}g(x)=\sqrt{x+5}\end{gathered}

f(x)=

x−3

1

g(x)=

x+5

Here, \begin{gathered}\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)\end{gathered}

x+5

is always greater than equal to 0

⇒x+5≥0

⇒x≥−5…(i)

To get f\left(g(x)\right)f(g(x)) , replace xx in f(x)f(x) by g(x)\ \text{i.e. by}\ \sqrt{x+5}g(x) i.e. by

x+5

\begin{gathered}\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}text{Denominator must not be equal to 0}therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)\end{gathered}

⇒f(g(x))=

x+5

−3

1

Denominator must not be equal to 0

∴

x+5

−3



=0

⇒

x+5



=3

⇒x+5



=9

⇒x



=4…(ii)

Using (i)(i) and (ii)(ii) it can be concluded that the domain of f\left(g(x)\right)f(g(x)) is all real numbers except 0.

Therefore, its domain is given by

x\in [-5,4)\cup (4,\infty)x∈[−5,4)∪(4,∞)

Option (c) is correct.

Step-by-step explanation:

that's so hard to answerhope it's help

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)