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is the answer right?

Very  nice  solution, Bginner  !!!!!

Please delete that.   I read the question wrong.  Sorry.

F

9

E               25   D

tan  D   =  EF  / DE

tan  25°  =  9 / DE

DE  =  9   /  tan  25°    ≈  19.3

There's nothing wrong with the theorem I showed to you.   Well, I proved it, didn't I :)
So it could be wrong only if my proof were invalid!

Now let me explain how I concluded that this problem could be solved by simply
taking the HM of the two given speeds.   

Let x be the time it takes in minutes to arrive on time.
To arrive five minutes late means the journey takes x+5 minutes.
To arrive five minutes early means the journey takes x-5 minutes.

We are given that if r1 = 45, t1 = x+5 and
if r2 = 63, t2 = x-5.`

So ta = ((x+5)+(x-5))/2 = x minutes, exactly the the AM of t1 and t2.
(It's immaterial whether you convert the times to hour or not.
Were t1 and t2 to be counted in hours, its AM would still be the on-time time.
Why?  Because the late amount and the early amount equals, so they cancel out
when you calculate the AM.  The problem poser intended it that way.)

So according to the theorem, ra should be the HM of r1 and r2.

Now back to the case of n = 3.
Were there to be a distance d, and speeds r1, r2, r3, and desired speed x
such that (t1 + t2 + t3) / 3 = x, then the AM of r1, r2, r3 would be x and
ra, the HM of the three speeds, would be the answer as well.
However, these numbers r1, r2, r3, t1, t2, t3 must have come from reality.
In a hurry I simply picked arbitrary numbers and used them in my example.
That's bad.  These numbers aren't arbitrary.  They're dependent on one another.
So that's why you get funny result.
 

ok.

Thanks for letting me know. 

Sorry, I made up those numbers and as it turned out those numbers are impossible.   They could not have occurred from a real data.  My bad.

Imagine writing down the definition of n! in full

    1 x 2 x 3 x 4 x ... x n

if it has both 14 in it then it means it's a multiple of 49 because it has one 7 (in 7) and another 7 (in 14) as prime factors.   So an an n must be at least 14.
 

thanks! but it's 60

Let's elaborate.  We define an $n$--$x$ set to be a set having $n$ as the minimum element, and $x$ as the maximum element such that $n+x=11$.   E.g., $\{4, 5, 7\}$ is a 4--7 set.  There are five possibilities according to the extreme elements.   

(i) 5--6 sets: There's only one set $\{5, 6\}$.

(ii) 4--7 sets: Besides 4 and 7 that need to belong, the numbers 5 and 6 can be chosen to be included or excluded independently.  So there are $2^2$ sets of this kind.

(iii) 3--8 sets: Besides 3 and 8 that need to belong, the numbers 4, 5, 6, and 7 can be chosen to be included or excluded independently.  So there are $2^4$ sets of this kind.

(iv) 2--9 sets: Besides 2 and 9 that need to belong, the numbers 3, 4, 5, 6, 7, and 8 can be chosen to be included or excluded independently.  So there are $2^6$ sets of this kind.

(v) 1--10 sets: Besides 1 and 10 that need to belong, the numbers 2, 3, 4, 5, 6, 7, 8, and 9 can be chosen to be included or excluded independently.  So there are $2^8$ sets of this kind.

Adding numbers from all five possibliities give the answer.
 

Solve the inequality

(x+1)(X-1) > 8

X^2 -1 >8

X^2 >9

X >3   or   x< -3       looks like 4  out of the interval of 10      40 %

no prob!

Correct, thank you so much! :D

\(\frac{2}{25}-\frac{14}{25}i\) i think

ok thanks

The answer for this problem is simple

if the sum of the digits is 2021

the smallest number would be a number, followed with all nines

2021/9=224 remandier 5, so the number is 

599999999 ect

but because we are trying to find the smallest number,

there will be 224 9s 

so, the answer is 224

lol :)

You may want to see a slick way of solving this problem.  It depends on knowing this fact.
If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.  For your problem it goes: We first note that $\frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$.  so
\begin{eqnarray*}
\dfrac{x+1}{x-1}
&=& \frac{5}{4} \\
\dfrac{(x+1)+(x-1)}{(x+1)-(x-1)}
&=& \dfrac{5+4}{5-4} \\
\dfrac{2x}{2}
&=& \dfrac{9}{1} \\
x &=& 9
\end{eqnarray*}

Let's name the train station point T, and name the point Zuleica met Wilma on her walk home M.  The amount of time saved equals the time it takes Wilma to drive from M to T and back to M again.  So driving from M to T must be equal to one half of the total time saved, and also equal to the time Zuleica had been walking.   The amount of time saved is 12 minutes (from 4:48 p.m. to 5:00 p.m.).   Therefore, Zuleica had been walking for 6 minutes.

First, solve these sub-problems:
1. How many peachy numbers start with 3? Call this number A.
2. How many peachy numbers end with 3? Call this number B.
3. How many peachy numbers start and end with 3? Call this number C.

Then, note that peachy numbers counted in 1 plus those counted in 2 cover all possiilities.   However, some peachy numbers counted in 1 or 2 include those counted in 3.   So we conclude that the answer should be A+B-C.

See link for method....

\(x=3 \text{ or } x=-3\)

Cross multiply

5x - 5/x   = 4x + 4/x

x  = 4/x + 5/x

x = 9/x

x^2 = 9

x = +- 3       I would sub in both of those values to see if true !  

nvm I messed up

Nvm I found the answer 

B = 3/4 D

Then    add 15 to D

B = 6/11(D+15)                Sub in the red equation

3/4D = 6/11D + 90/11

(33-24)/44 D = 90/11

D = 44/9 * 90/11 = 40 Dollars     (as CHris already showed !)