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4 years ago

Look below please answer //

Mathematics

1 answer:

ExtremeBDS [4]4 years ago

5 0

[-11.2-4.7] this the awnser

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How many ways are there to fill either the pitcher or catcher field position (but not both) from among the 25 players (leaving t

Artemon [7]

Using the permutation formula, it is found that there are 600 ways to fill the pitcher and catcher positions.

The order is important, as one player is the pitcher and the other is the catcher, hence the <em>permutation formula </em>is used to solve this question.

<h3>What is the permutation formula?</h3>

The number of possible permutations of x elements from a set of n elements is given by:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

2 players are taken from a set of 25, hence the number of ways is given by:

$P_{(25,2)} = \frac{25!}{23!} = 600$ .

More can be learned about the permutation formula at brainly.com/question/25925367

#SPJ1

4 0

2 years ago

Pretty sure I’ve got the answer but anyone who uses maths watch know how to put my answer as it is not saying it’s correct

vlada-n [284]

$9(x-11)=9 \\9x-99=9 \\9x=9+99 \\9x=108 \\x=108\div9 \\x=\boxed{12} \\$

Hope this helps.

Brainliest would help me.

8 0

3 years ago

Write 120% as a fraction.<br>

ASHA 777 [7]

The answer is the first one: 1 1/5 because 120/100 = 1.2

7 0

3 years ago

The function f(x)=x2. The graph ofg(x) is f(x) translated to the left 6 units and down 5 units. What is the functionrule for g(x

Nonamiya [84]

Add 6 to x^2 to translate function to the left (x+6)^2, then minus 5 to move down: (x-6)^2-5

3 0

4 years ago

Find the area of the surface generated by revolving x=t + sqrt 2, y= (t^2)/2 + sqrt 2t+1, -sqrt 2 <= t <= sqrt about the y

EleoNora [17]

The area is given by the integral

$\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds$

where <em>C</em> is the curve and $dS$ is the line element,

$\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

We have

$x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1$

$y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2$

$\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt$

So the area is

$\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt$

Substitute $u=t^2+2\sqrt2\,t+3$ and $\mathrm du=(2t+2\sqrt 2)\,\mathrm dt$ :

$\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3$

8 0

3 years ago