All Answers 355921 Answers

Stephan has been busy at work writing all the possible rearrangements of the letters in his name. He is one bored boy. If he can write twelve rearrangements of his name every minute, how many hours does it take to write all the possible rearrangements of his name?

S T E P H A N   all different to there are  7! permutations

7!/12   minutes

7!/(12*60) = 7   hours

First, find the volume of the board:

V =0.3 x 2.1 x 0.1 =0.063 meters^3. Note: I took your dimentions to mean "meters" NOT "miles"

Density =Mass / Volume

Density =.07 / 0.063

Density =1.1 Kg/m^3 - density of the wooden board.

the kids

The answer to the question is 10.

I'm still trying to picture the 30 people wearing a hat and sunglasses but no bathing suit.

Step 1 The two sides we know are Opposite (300) and Adjacent (400).

Step 2 SOHCAHTOA tells us we must use Tangent.

Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.

Step 4 Find the angle from your calculator using tan-1

There are many different ways to express density, and not using or converting into the proper units will result in an incorrect value. It is useful to carefully write out whatever values are being worked with, including units, and perform dimensional analysis to ensure that the final result has units of

mass

_____

volume

. Note that density is also affected by pressure and temperature. In the case of solids and liquids, change in density is typically low. However, when regarding gases, density is largely affected by temperature and pressure. An increase in pressure decreases volume, and always increases density. Increases in temperature tend to decrease density since volume will generally increase. There are exceptions however, such as water's density increasing between 0°C and 4°C

1.1 kilogram/cubic mile

what is the volume

awesome job bro

1 kilogram is equal to 9.8066500286389 N. so 9.8066500286389x700=6864.65502

The answer is 6864.65502

dont worry..i shall be back this same year

Good Job MB see you later. I guess its bye for now becuase i have awhile before mine is over. 

For how many three-digit positive integers is the sum of the digits equal to 5

Three-digit integer: \(abc\)

\(\text{ $1\le a \le 5$ }\\ \text{ $0\le b \le 4$ }\\ \text{ $0\le c \le 4$ and $c=5-(a+b)$ } \)

\(\begin{array}{|c|c|c|c|c| } \hline & b=0 & b=1 & b=2 & b=3 & b=4 \\ \hline a=1 & c=4 & c=3 & c=2 & c=1 & c=0 \\ a=2 & c=3 & c=2 & c=1 & c=0 \\ a=3 & c=2 & c=1 & c=0 \\ a=4 & c=1 & c=0 \\ a=5 & c=0 \\ \hline \end{array} \)

15 possible integers

\(\begin{array}{ c c c c c } & 104 & 113 & 122 & 131 & 140 \\ & 203 & 212 & 221 & 230 \\ & 302 & 311 & 320 \\ & 401 & 410 \\ & 500 \\ \end{array}\)

^{I guess I submitted my answer and got it right????}

You guess you submitted your answer? Either you did or didn’t.

^{It is a bit wordy}

No kidding; who’d ever notice?

^{although GingerAle I used the fact how you used your casework and how you organized them.} 

Then you submitted my answer, not yours. Both of our answers cannot be correct.

 ^“...

^{Your casework to count the successes did work, but isn't the most efficient such approach as you noted.}

I made no comment about efficiency

^{Your second case when counting the successes was not labeled correctly.}

That was the fourth case, and it was corrected. This was an easy catch, if you had reviewed the work.

^{Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly  numbers colored:" and so on."
 That was from AoPS???}

WTF! My presentation is a concept based teaching post; it’s not something I would submit to an AoPS teacher/professor, or any teacher/professor. If you wanted to use my solution method, you should have reworded it, but not with your neurotic Sweet Polly pure piss encouragement preceding every number.

^{"My answer is spot on.  I can “prove” my answer is correct,"}

^-???

I completed the Monte Carlo simulations. As expected, it hovers near 1/3 using the “Grogg” selection method.  Randomly choosing to color or not to color numbers 1-6, gives a slight variation around a 25% probability for a factorific coloring, and the counts of colored numbers, from 0 to 6 tends toward a binomial distribution.  There are other interesting statistics based on various restrictions. 

^{And there's no need to be rude? Chill? pls?}

Yes there is. You assaulted the senses of everyone who read your post. At least, those who are not on thorazine or have a natural version of it.

 ^{I'm sorry if it is not pleasing to you}

You’re not, and that’s an understatement!

GA

Edit Corrected typo.

sin of an angle is across/hypotenuse

cos of an angle is adjacent/hypotenuse

tan of an angle is across/adjacent 

you normally get 2 values of those 3,

so you can use a calculator to find the missing value

Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.

\(\text{1. $f(x)$ is a linear function}\)

\(\begin{array}{|lrcll|} \hline f(x) =& ax +b && \quad & | \quad f(1) = 4 \\\\ f(1) =& a\cdot1 +b &=& 4 \\ & \mathbf{a +b} &\mathbf{=}& \mathbf{4 } \qquad &(1) \\ & \mathbf{a } &\mathbf{=}& \mathbf{4-b } \\ \hline \end{array} \)

\(\text{2. $f^{-1}(x)=\ ? $ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \\ y &=& ax +b \\ ax &=& y-b \quad & | \quad : a \\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\ y &=& \dfrac{x-b}{a} \\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline \end{array}\)

\(\text{3. $f(x) = 4f^{-1}(x) + 6$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& 4f^{-1}(x) + 6 \quad & | \quad x = 1 \\\\ f(1) &=& 4f^{-1}(1) + 6 \quad & | \quad f(1) = 4 \qquad f^{-1}(1) = \dfrac{1-b}{a} \\ 4 &=& 4 \left(\dfrac{1-b}{a} \right) + 6 \quad & | \quad - 6 \\ -2 &=& 4 \left(\dfrac{1-b}{a} \right) \quad & | \quad \cdot a \\ -2a &=& 4 (1-b) \\ -2a &=& 4-4b \quad & | \quad +4b \\ 4b-2a &=& 4 \quad & | \quad : 2 \\ \mathbf{2b-a} & \mathbf{=}& \mathbf{2} \qquad &(2) \\ \hline \end{array}\)

\(\text{4. $a=\ ? \qquad b=\ ?$ }\)

\(\begin{array}{|rcll|} \hline 2b-a & = & 2 \quad & | \quad a = 4-b \\ 2b-(4-b) & = & 2 \\ 2b-4+b & = & 2 \\ 3b-4 & = & 2 \quad & | \quad +4 \\ 3b & = & 6 \quad & | \quad :3 \\ \mathbf{ b} & \mathbf{=} & \mathbf{2} \\\\ a &=& 4-b \\ a &=& 4-2 \\ \mathbf{ a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

\(\text{5. $f(x)=\ ?$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \quad & | \quad a=2 \qquad b = 2 \\\\ \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \\ \hline \end{array}\)

\(\text{6. $f(2)=\ ?$ } \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \quad & | \quad x = 2 \\\\ f(2) & = & 2\cdot 2 +2 \\ \mathbf{ f(2)} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

The answer is C

There may be a correlation between eating blueberries and a lower cholesterol level.

The owner wants people to believe that dance classes are popular so that they sign up for classes

if you draw this picture you can see that you are making triangles.

distance formula or pythagorean theorem will give you the answer

\(\sqrt{}\)(10² + 14² )  or  \(\sqrt{}\)(13² + 9² ) 

\(\sqrt{}\) 296 or \(\sqrt{}\) 250 so obviously \(\sqrt{}\)250 is closer           

using the equation for graphing, we see the vertex is at (-3,-5).

thus the axis of symmetry is is vertical and this vertical line is at x = -3

distance from one x-intercept from the vertex is the same as the other x-intercept.

the distance from vertex's x coordinate 12 to 15 is 3 thus 3 in the other direction is (9,0)

I guess I submitted my answer and got it right???? It is a bit wordy 

although GingerAle I used the fact how you used your casework and how you organized them. 

And there's no need to be rude? Chill? pls? 

Uh, This work did actually belong to me, although like I said, is wordy and not the most efficient way, :/

"You correctly used constructive counting within casework to count the number of factorific colorings and total number of colorings, then used these values to calculate the desired probability.

Your casework to count the successes did work, but isn't the most efficient such approach as you noted. Plus, it is hard to check some of your cases to ensure you found all of the successful colorings and did not miss any, although your method of "combining" smaller sets could have helped you do this. Can you think of a faster and simpler approach here? You may still need some form of casework, although not necessarily based on the number of blue numbers. There are other approaches that have fewer cases and less complex ways to count the successes within each case.

Your second case when counting the successes was not labeled correctly. Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly  numbers colored:" and so on."
 That was from AoPS???

"My answer is spot on.  I can “prove” my answer is correct,"

-???

"I think I need powerful anti-nausea meds to wade through this irrelevant, wordy, neurotic swill. Jesus Christ! It’s amazing that someone far surpassed Mr. BB in pasting mathematically infused slop on this forum.  It’s interesting that someone can write reasonably tight LaTex code but doesn’t know how to implement its execution here; "

-Uh...Ok??? I'm sorry if it is not pleasing to you, but  ¯\_(ツ)_/¯ 

Well X², you discovered your own sin, repented and made reparations. You’re forgiven and absolved.