All categories

2 years ago

Please help me struggling

Mathematics

1 answer:

Liula [17]2 years ago

5 0

Answer:

Step-by-step explanation:

find the composition of g(x)=x−1 and h(x)= $\sqrt{x}$

hence we will get,

(g∘h)(x)=g(h(x))= g(x−1)= f(√x)= -1+√x = √x -1

Now to find f(g(h(x))),

f(x)=x4+6 and g(h(x))=√x -1

hence , putt g(h(x)) in f(x) ,

f(g(h(x)))= $(\sqrt{x}-1)^{4}$ +6 -

Plzzzz Brain-list it or subscribe to my channel " ZK SOFT&GAMING"

You might be interested in

Use the distance formula, show that the points (4,0), (2,1), and (-1,-5) form the vertices of a right triangle

mario62 [17]

Step-by-step explanation:

The distance formula between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points.

$A(4,\ 0),\ B(2,\ 1),\ C(-1,\ -5)\\\\AB=\sqrt{(2-4)^2+(1-0)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4+1}=\sqrt5\\\\AC=\sqrt{(-1-4)^2+(-5-0)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}\\\\BC=\sqrt{(-1-2)^2+(-5-1)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45}$

If a ≤ b < c are the sides of the right triangle, then

a² + b² = c²

$\sqrt5$

used $(\sqrt{a})^2=a$ for a ≥ 0.

$AB^2+BC^2=AC^2$ therefore ΔABC is a right triangle.

6 0

3 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

Maria, a biologist, is observing the growth pattern of a virus. She starts with 100 of the virus that grows at a rate of 10% per

Leni [432]

<h3>Answer: 985</h3>

=============================================

Work Shown:

A = amount after t hours

P = initial amount = 100

r = growth rate in decimal form = 0.10

t = number of hours = 24

--------

A = P*(1+r)^t

A = 100(1+0.10)^24

A = 100*(1.10)^24

A = 100*9.84973267580762

A = 984.973267580762

A = 985

5 0

3 years ago

Read 2 more answers

If varies inversely as . It is known that = 10 when = 3, find the value of when = 5.

Leto [7]

Answer:

y = 6

Step-by-step explanation:

If x varies inversely proportional as y.

$x=\dfrac{k}{y}$ , k is constant of proportionality

k = xy

When x = 3 and y = 10

k = 3×10

k = 30

Put x = 5,

$y=\dfrac{k}{x}\\\\y=\dfrac{30}{5}\\\\y=6$

So, the value of y is 6 when x is 5.

8 0

2 years ago

Drag and drop an answer to each box to correctly complete the explanation for deriving the formula for the volume of a sphere

denis-greek [22]

The answers that would fill in the blanks are

2r
a circle
an annulus
1/3πr³
4/3πr³

<h3>What is the Cavalier's principle?</h3>

This principle states that if two solids are of equal altitude then the sections that the planes would make would have to be parallel and also be at the same distances from their bases which are equal such that the volumes of the solids would be equal.

Now we have to fill in the blanks with the solution.

For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius r and height<u> 2r</u> minus the volume of two cones, each with a radius and height of r. A cross section of the sphere is a <u>circle</u> base of cylinder, is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is an <u>annulus_ </u>.The volume of the cylinder with radius r and height 2r is 2πr³, and the volume of each cone with radius r and height r is 1/3πr³. So the volume of the cylinder minus the two cones is 4/3πr³. Therefore, the volume of the sphere is by Cavalieri's principle

Read more on Cavalieri's principle here

brainly.com/question/22431955

#SPJ1

5 0

1 year ago