All categories

2 years ago

PLEASE HELP GIVING BRAINEST AWNSER AND ITS WORTH 85 points

Mathematics

2 answers:

Dennis_Churaev [7]2 years ago

5 0

Answer:

H. 10

Step-by-step explanation:

We can write down all the games that can be played (I'm going to use a, b, c, d, and e to symbolize the players)

ab, ac, ad, ae, bc, bd, be, cd, ce, de

As you can see, there are 10 games that are going to be played.

Best of Luck!

Simora [160]2 years ago

5 0

Answer:

10 games

Step-by-step explanation:

There are 5 new members.

The first member plays the rest 4
The second - 3
The third - 2
The fourth - 1

<u>Total number of games:</u>

4 + 3 + 2 + 1 = 10

You might be interested in

This is my last question please help asap!!

Ivanshal [37]

The average rate of change of function $g(x)=5(2)^{x}$ from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

The correct option is (A).

What is the average rate of change of a function?

The average rate at which one quantity changes in relation to another's change is referred to as the average rate of change function.

Using function notation, we can define the Average Rate of Change of a function f from a to b as:

$rate(m) = \frac{f(b)-f(a)}{b-a}$

The given function is $g(x) = 5(2)^{x}$ ,

Now calculating the average rate of change of function from x = 1 to x = 2.

$m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(2)-g(1)}{2-1}\\\\m=\frac{5(2)^{2}-5(2)^1}{2-1}\\ m=\frac{10}{1} \\m=10$

Now, calculate the average rate of change of function from x = 3 to x = 4.

$m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(4)-g(3)}{4-3}\\\\m=\frac{5(2)^{4}-5(2)^3}{4-3}\\ m=\frac{40}{1} \\m=40$

The jump from m = 10 to m = 40 is "times 4".

So option (A) is correct.

Hence, The average rate of change of function $g(x)=5(2)^{x}$ from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

To learn more about the average rate of change of function, visit:

brainly.com/question/24313700

#SPJ1

7 0

10 months ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$

From this we need to subtract the contribution of

$\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S$

that is, the integral of $\vec F$ over the disk, oriented downward. Since $z=0$ in $D$ , we have

$\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k$

Parameterize $D$ by

$\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

where $0\le u\le 3$ and $0\le v\le2\pi$ . Take the normal vector to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k$

Then taking the dot product of $\vec F$ with the normal vector gives

$\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v$

So the contribution of integrating $\vec F$ over $D$ is

$\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4$

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==> 486π/5 - (-81π/4) = 2349π/20

5 0

3 years ago

Answer this question in the simplest

Molodets [167]

Answer: 1:2

Step-by-step explanation: if she sleeps 8 hours out of 24 hours, 8 goes into 24, 3 times so 1 to 2 ratio because you subtract the time she is awake from the time she is asleep

7 0

2 years ago

Read 2 more answers

Choose the correct simplification of x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4t

Umnica [9.8K]

Answer:

<h3>The option A) $x^{10}z^7$ is correct answer.</h3><h3>The correct simplification for the given expression $\frac{x^{12}\times z^{11}}{x^2\times z^4}$ is $x^{10}z^7$ </h3>

Step-by-step explanation:

Given expression is x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4th power.

The given expression can be written as $\frac{x^{12}\times z^{11}}{x^2\times z^4}$

<h3>To choose the correct simplification of the given expression :</h3>

Now we have to simplify the given expression as below

$\frac{x^{12}\times z^{11}}{x^2\times z^4}$

$=x^{12}\times z^{11}(x^{-2}\times z^{-4})$ ( by using the identity $\frac{1}{a^m}=a^{-m}$ )

$=(x^{12}.x^{-2})\times (z^{11}.z^{-4})$

$=x^{12-2}\times z^{11-4}$ ( by using the identity $a^m.a^n=a^{m+n}$ )

$=x^{10}\times z^7$

$=x^{10}z^7$

∴ $\frac{x^{12}\times z^{11}}{x^2\times z^4}=x^{10}z^7$

<h3>The correct simplification for the given expression $\frac{x^{12}\times z^{11}}{x^2\times z^4}$ is $x^{10}z^7$ </h3><h3>Hence option A) $x^{10}z^7$ is correct answer.</h3>

4 0

3 years ago

Observe the graph below.<br> The graph shows a<br> function that is asymptotic to

JulijaS [17]

We have to see the graph to what the answer is. Sorry I wish I could help!

4 0

2 years ago