All Answers 357325 Answers

The answer is $582$

Wait, but @ikleyn, check out my answer here:

Let $y=f(x)$, so $y$ can take on any value from $-3$ to $5$, inclusive. Then, $y^2$ can take on any value from 0 to 25, inclusive. )If we take any value from $-3$ to $0$ and square it, we get a value from 0 to 9. And if we take any value from 0 to 5 and square it, we get a value from 0 to 25.) Therefore, the range of $g(x)$ is $\boxed{[0,25]}$

The probability is 4/21.

"Anchor" any one of the groups  and  arrange them in two ways

Choose any one of the  three remainig groups and arrange this  group in two ways

Choose any one of the two remaing groups and arrange this  group in two ways

Arrange the remaining group in two ways

2 * C(3,1) * 2 * C(2,1)* 2 * 2  =   96 ways

7 times  =  10

6 times  =  10 * 9 * 7  =  630

10 + 630  =  640 

3210

510

420

321

60

51

42

AB / BD    =  AC / DC

6 / 3  = AC /  5

2  =   AC / 5

AC = 10

In \triangle PQR, we have \angle P = 90^\circ, PQ=3, and QR = 5. Find \tan R.   

By Pythagoras  PR² = QR² – PQ²    

                          PR  =  sqrt(25 – 9)  =  4     

                       tan R  =  opposite over adjacent  =  3 / 4      

                       tan R  =  0.7500    

If you draw / label the triangle, the answer becomes clear.    

I'd like to show you, but I don't know how to post a picture.   

_.    

The area of a cross section of a sphere is 64\% of the largest possible cross sectional area of the sphere. If the sphere has radius 1/2, what is the area of the cross section?    

A cross section of a sphere would be a circle, right?  The largest circle would be through the center, so ...     

A_{through center}  =  π r²  =  3.1416 • 0.5²  =  3.1416 • 0.25     

                       =  0.7854    

 0.7854 • 0.64  =  0.5027 unit²    

_.    

The value of a_1 is 80/7.

 If BD = 3 and CD = 5, what is AC?

I

$mx=\sqrt{8^2-(x-6)^2}$

II

$\frac{m}{2}\cdot x=\sqrt{3^2-(x-6)^2}\\ mx=2\cdot \sqrt{3^2-(x-6)^2}$

III

$\sqrt{8^2-(x-6)^2}=2\cdot \sqrt{3^2-(x-6)^2}\\ 64-x^2+12x-36=4\cdot(9-x^2+12x-36)\\ 64-x^2-12x-36=36-4x^2+48x-144\\ 3x^2-36x+136=0\\ x^2-12x+45.3333=0\\ x=6\pm \color{red}\sqrt {36-45.3333}$

$\color{blue}Under\ the\ given\ conditions\, no\ result\ is\ possible.$

 !

The probability is 7/12.

What is the probability that BD < 4?

$\color{blue}P=\dfrac{A_{DBC}}{A_{ABD}}$

$f_{AC}(x)=\sqrt{5^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-7)^2}\\ 25-x^2=9-x^2+14x-49\\ 14x=65\\ x_C=4.6429 \\ y_C=1.8558\\ C(4.6429,1.8558)\\ A(0,0)\\ B(7.0)$

$f_{AD}=\frac{1.8558}{4.6429}x\\ f_{BD}=\sqrt{4^2-(x-7)^2}\\ \frac{1.8558}{4.6429}x=4^2-(x-7)^2\\ \frac{1.8558}{4.6429}x=16-x^2+14x-49$

$x^2-13.6003x+33=0\\ x=6.8001\pm \sqrt{46.2420-33}\\ x_D=3.1661\\ y_D=1.2635\\ D(3.1661,1.2635)$

$A_{ABC}=\frac{7}{2}\cdot 1.8558\\ A_{ABC}=6.4953\\ A_{ABD}=\frac{7}{2}\cdot 1.2635\\ \color{blue}A_{ABD}=4.4224\\ A_{DBC}=A_{ABC}-A_{ABD}=6.4953-4.4224\\ \color{blue}A_{DBC}=2,0729$

$P=\dfrac{A_{DBC}}{A_{ABD}}=\frac{2.0729}{4.4224}:\frac{2.0729}{2.0729}=1:2.1334$

The probability that BD < 4 is  $\color{blue }P=1:2.1334$

 !

Let S be the set of all permutations of (1,2,3,4,5,6,7). The total number of such permutations is ∣S∣=7!

We are interested in the number of permutations that have at least one even fixed point. The even numbers in the set {1,2,3,4,5,6,7} are {2,4,6}.

Let Ai​ be the set of permutations where i is a fixed point. We want to find the number of permutations in A2​∪A4​∪A6​.

Using the Principle of Inclusion-Exclusion, we have:

∣A2​∪A4​∪A6​∣=∣A2​∣+∣A4​∣+∣A6​∣−(∣A2​∩A4​∣+∣A2​∩A6​∣+∣A4​∩A6​∣)+∣A2​∩A4​∩A6​∣

We calculate the size of each intersection:

\begin{enumerate}

\item ∣Ai​∣: If i is a fixed point, then the remaining n−1 elements can be permuted in (n−1)! ways. For n=7:

∣A2​∣=(7−1)!=6!=720

∣A4​∣=(7−1)!=6!=720

∣A6​∣=(7−1)!=6!=720

\item ∣Ai​∩Aj​∣: If i and j are fixed points, then the remaining n−2 elements can be permuted in (n−2)! ways. For n=7:

∣A2​∩A4​∣=(7−2)!=5!=120

∣A2​∩A6​∣=(7−2)!=5!=120

∣A4​∩A6​∣=(7−2)!=5!=120

\item ∣A2​∩A4​∩A6​∣: If 2,4, and 6 are fixed points, then the remaining n−3 elements can be permuted in (n−3)! ways. For n=7:

∣A2​∩A4​∩A6​∣=(7−3)!=4!=24

\end{enumerate}

Now, substitute these values into the Inclusion-Exclusion Principle formula:

∣A2​∪A4​∪A6​∣=720+720+720−(120+120+120)+24

∣A2​∪A4​∪A6​∣=2160−360+24

∣A2​∪A4​∪A6​∣=1800+24

∣A2​∪A4​∪A6​∣=1824

Thus, there are 1824 permutations of (1,2,3,4,5,6,7) that have at least one even fixed point.

Final Answer: The final answer is 1824​

What is distance from A to B along the (minor) arc of a great circle?

$s=2\\ r=2\\ b=2\cdot \frac{\alpha}{2}\cdot \frac{2\pi r}{360^{\circ}} \\sin\frac{\alpha}{2}=\frac{1}{2}\\ b=2\cdot arcsin \frac{1}{2}\cdot\frac{2\pi \cdot 2}{360^{\circ}}\\ \color{blue}b=2.0944$

$1.11111 + 0.11111 + 0.01111 + 0.00111 +0.00011 + 0.00001\color{blue} = 1.23456$

   !