All categories

3 years ago

If you can solve it.Thanks in advance

Mathematics

1 answer:

vivado [14]3 years ago

4 0

Answer:

$f(g(x))=\frac{1}{(x^{2}+1)^{2}} +\sqrt[3]{x^{2}+1}$

Step-by-step explanation:

we have

$f(x)=x^{2} +\frac{1}{\sqrt[3]{x}}$

$g(x)=\frac{1}{x^{2}+1}$

we know that

In the function

$f(g(x))$

The variable of the function f is now the function g(x)

substitute

$f(g(x))=(\frac{1}{x^{2}+1})^{2} +\frac{1}{\sqrt[3]{(\frac{1}{x^{2}+1})}}$

$f(g(x))=\frac{1}{(x^{2}+1)^{2}} +\sqrt[3]{x^{2}+1}$

You might be interested in

What’s 34x568 can’t find my Calculator

Brut [27]

Hi MayceeRhea15,

34x568 is 19,312 and don't worry about not being able to find your calculator, we have all been there before :)

Hope this helps

Cupkake~

4 0

2 years ago

Read 2 more answers

Write an expression of the sequence of operations described below

snow_lady [41]

Answer:

the answer is: (g-f)-10

8 0

2 years ago

A circular garden is enlarged so that the new diameter is twice the old diameter. What is the ratio of the original area to the

Amanda [17]

Answer:

Explained briefly below.

Step-by-step explanation:

<u>For old circular garden:</u>

take the radius as r.

then use the formula to find area of circle: πr² ......this is old garden area.

<u>For new enlarged garden:</u>

the radius is twice the old radius so, radius = 2 * r = 2r ......enlarged radius

now find area for this new garden: π(2r)² → 4πr²

In common fractions: (old garden)/(new garden)

: ( πr² ) / ( 4πr² )

: 1/4

7 0

2 years ago

Read 2 more answers

Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6

Alla [95]

Given:

The expression is:

$2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

$m+n=-6$

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

Let the given polynomial is:

$P(x)=2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1. By using remainder theorem, we can say that

$P(2)=P(-1)$ ...(i)

Substituting $x=-1$ in the given polynomial.

$P(-1)=2(-1)^3+m(-1)^2+n(-1)+c$

$P(-1)=-2+m-n+c$

Substituting $x=2$ in the given polynomial.

$P(2)=2(2)^3+m(2)^2+n(2)+c$

$P(2)=2(8)+m(4)+2n+c$

$P(2)=16+4m+2n+c$

Now, substitute the values of P(2) and P(-1) in (i), we get

$16+4m+2n+c=-2+m-n+c$

$16+4m+2n+c+2-m+n-c=0$

$18+3m+3n=0$

$3m+3n=-18$

Divide both sides by 3.

$\dfrac{3m+3n}{3}=\dfrac{-18}{3}$

$m+n=-6$

Hence proved.

7 0

2 years ago

How many symmetry does maple leaf have

Travka [436]

Only one Line.........

6 0

2 years ago