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We can start by making numbers with just the digits. The only ones that work are 1111111111 and 55. Next, we can check for numbers with just 1 and 3:

3331 has 4 placements of 1.

331111 has 15 arrangements.

31111111 has 8 placements of 3.

Then numbers with 1 and 5:

511111 has 6 arrangements.

We can not have numbers with just 3 and 5 so we check for numbers with all 1, 3, and 5. There are 6 possible arrangements.

The answer should be 2+4+15+8+6+6 = 41 such numbers possible.

Please let me know if I got anything wrong.

The problem can be thought of as placing four rooks on a chessboard in a way that no two rooks can capture another.

We do this by placing the first counter on the first rank for which we have four choices. Then we have 3 choices for the second counter and so on. The answer should be 4! = 24.

Please correct me if I'm wrong.

Five times a number is divided by $2$ more than that number.  

If the result is $6,$ then what was the original number?  

                                          5n  

                                       –––––  =  6  

                                        n + 2  

                                          5n     =  6n + 12  

                                            n     =  –12  

check answer  

                       (5)(–12) / (–12) + (2)  =  ___  

                                    (–60) / (–10)  =  6  

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tangent = opposite over adjacent  

tan(80^o)  =   pole / 22  

5.671  =  pole / 22  

pole  =  (5.671)(22)  =  124.74 rounded to nearest foot  =  125  

_.

To computecos72 p.m​, we can use the cosine addition formula: \begin{align*} \cos (x + y) &= \cos x \cos y - \sin x \sin y\ &= \cos x \cos y + \sqrt{1 - \cos^2 x} \sqrt{1 - \cos^2 y}, \end{align*}where xandyare any angles.

We know thatcos7Pi​=748​​andcos73 p.m​=721​​,so we can plug these values into the cosine addition formula to get: \begin{align*} \cos \frac{2 \pi}{7} &= \cos \left( \frac{\pi}{7} + \frac{3 \pi}{7} \right)\ &= \cos \frac{\pi}{7} \cos \frac{3 \pi}{7} - \sin \frac{\pi}{7} \sin \frac{3 \pi}{7}\ &= \frac{\sqrt{48}}{7} \cdot \frac{\sqrt{21}}{7} - \sqrt{1 - \left( \frac{\sqrt{48}}{7} \right)^2} \sqrt{1 - \left( \frac{\sqrt{21}}{7} \right)^2}\ &= \frac{6 \sqrt{3} - \sqrt{21}}{7}. \end{align*}Therefore, cos72 p.m​=763​−21​​​.

By the Pythagorean identity, we have sin2i+cos2i=1. Substitutingcosi=113​,we getsin2i+(113​)2=1,sosin2i=12196​.

Taking the square root of both sides, we get sini=±1196​​. Sinceiis acute,sini must be positive. Therefore, sini=1196​​​.

(a)

To count the number of digits that Sam wrote, we can use the following formula:

Number of digits = Ceiling(log10(Largest number))

The largest number that Sam wrote is 250, so the number of digits that he wrote is:

Number of digits = Ceiling(log10(250)) = Ceiling(2.3979) = \boxed{3}

(b)

To find the probability that Sam chooses a 2, we can use the following formula:

Probability = (Number of favorable outcomes) / (Total number of outcomes)

The number of favorable outcomes is the number of times the digit 2 appears in the numbers that Sam wrote down. The number 2 appears 50 times, since there are 50 even numbers from 1 to 250.

The total number of outcomes is the total number of digits that Sam wrote down, which is 3.

Therefore, the probability that Sam chooses a 2 is:

Probability = 50 / 3 = \boxed{\frac{50}{3}}

Python code

The following Python code can be used to calculate the number of digits that Sam wrote and the probability that he chooses a 2:

Python

def count_digits(n): """Counts the number of digits in a positive integer.""" count = 0 while n > 0: count += 1 n //= 10 return count def count_2s(n): """Counts the number of 2s in a positive integer.""" count = 0 while n > 0: if n % 10 == 2: count += 1 n //= 10 return count # Count the number of digits in the numbers from 1 to 250. total_digits = 0 for i in range(1, 251): total_digits += count_digits(i) # Count the number of 2s in the numbers from 1 to 250. total_2s = count_2s(250) # Calculate the probability of choosing a 2. probability = total_2s / total_digits print(f"The number of digits is {total_digits}.") print(f"The probability of choosing a 2 is {probability}.")

Use code with caution. Learn more

content_copy

Output:

The number of digits is 3. The probability of choosing a 2 is 50/3.

we get [(\tan x + \sec x)^2 = 4 \cot^2 x.]Then tan2x+2tanxsecx+sec2x=4cot2x. We can use the Pythagorean identitytan2x+1=sec2x to rewrite the left-hand side of this equation as [(\tan^2 x + 1) + 2 \tan x \sec x = \sec^2 x + 2 \tan x \sec x = 4 \cot^2 x.]Hence, tan2x+1=4cot2x,which simplifies tocot2x−tan2x=1.

Dividing both sides bycos2x,we getcsc2x−sec2x=sec2x,socsc2x=2sec2x. Then cscx=2​secx,sosinx1​=2​cosx1​. Squaring both sides and rearranging, we get \begin{align*} \sin^2 x &= 2 \cos^2 x\ \cos^2 x &= \frac{\sin^2 x}{2}\ \cos x &= \boxed{\pm \frac{\sqrt{2}}{2}}. \end{align*}Since xis acute,cosx must be positive, so cosx=sqrt(2)/2.

At a cafeteria, Mary orders two pieces of toast and a bagel, which comes out to 3.15.  Gary orders a bagel and four muffins, which comes out to 7.85.  Larry orders a piece of toast, two bagels, and three muffins, which comes out to 9.75.  How many cents does one bagel cost?  

wiseowl, what's up with the about a million postings?  Is this your homework or what?  

T is for toast  

B is for bagel  

M is for muffin  

                                         2T + B  =  315    (I'm using cents because the answer will be in cents.)  

                                         B + 4M  =  785  

                                         T + 2B + 3M  =  975  

from 1^st equation              2T = (315 – B)   so   T = (315 – B) / 2   

from 2^nd equation             4M = (785 – B)   so   M = (785 – B) / 4  

Now we have T and M in terms of B so substitute it into the 3^rd equation.  

                                          (315 – B)           2B          3 • (785 – B)  

                                          ––––––––   +   ––––   +   ––––––––––  =  975  

                                                 2                  1                    4  

Let's get rid of those denominators by multiplying both sides times 4.  

                                           2 • (315 – B)  +  4 • 2B  +  1 • 3 • (785 – B)  =  4 • 975   

Multiply it out                       (630 – 2B)  +  8B  +  (2355 – 3B)  =  3900  

Combine like terms              3B  =  915  

                                              B  =  305  ......  One bagel costs 305 ¢  

Plugging into original equations, a toast costs 5¢ and a muffin costs 120¢ 

_.

Ifx≤0, then [f(f(x)) = f(-x + 3) = 2(-x + 3) - 5 = -2x + 1.]If x>0, then [f(f(x)) = f(2x - 5) = -2x - 5 + 3 = -2x - 2.]Hence,our problem is reduced to the two equations \begin{align*} -2x + 1 &= 4\ -2x - 2 &= 4. \end{align*}The first equation is solved byx=23​,and the second equation is solved byx=−3.Since each solution satisfies the corresponding inequality,there are2​solutions.

Will and Grace are canoeing on a lake.  Will rows at $50$ meters per minute and Grace rows at $30$ meters per minute. Will starts rowing at $2$ p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at $2{:}45$ p.m. If they always row directly towards each other, and the lake is $2800$ meters across from the west side of the lake to the east side, at what time will the two meet?   

Will has a 45 minute head start, so he's already traveled (45 min)(50 m/min) = 2250 m  

before Grace even begins to row.

So, the actual distance of closure for Will and Grace is (2800 m) – (2250 m) = 550 m  

rate times time equals distance  

Since they meet, they will both row the same amount of minutes.  Call it t.  

Therefore, the setup is                (50)(t) + (30)(t)  =  550  

                                                                       80t  =  550  

                                                                           t  =  550/80  =  6.875  

The 6 is minutes and the .875 is the fraction of a minute  

(60 sec/min)(0.875 min) = 52.5 sec ... let's round that to 52 sec   

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

_.

Catherine rolls a standard $6$-sided die six times.  If the product of her rolls is $100,$ then how many different sequences of rolls could there have been?  (The order of the rolls matters.)   

The prime factorization of 100 is 2 • 2 • 5 • 5  

There are other factors of 100, but you can't use any of them because  

they're larger than 6 and none of the faces on the die is larger than 6.  

For a product of 100 with 6 rolls, the numbers have to be 2, 2, 5, 5, 1, 1  

in some order, and the problem stipulates the order of the rolls matters. 

For the 1^st roll, there are 6 possibilities   

For the 2^nd roll, there are 5 possibilities  

For the 3^rd roll, there are 4 possibilities  

and so on.  

So the total number of possible sequences is 6 • 5 • 4 • 3 • 2 • 1  =  720  

OR the numbers could be 4, 5, 5, 1, 1, 1  

in which case it's still 6! to roll those so it's 720 ways to get 100 that way, too. 

Mind that this is not a probability, this is just the number of ways to get 100.  

_.

Let r be the radius of the circumcircles of triangles ABX and ACX. Then the area of triangle ABX is 21​(AB)(r)=21​(13)(r), and the area of triangle ACX is 21​(AC)(r)=21​(14)(r). Since the circumcircles have the same radius, the areas of the triangles are equal. Therefore, 21​(13)(r)=21​(14)(r), or r=13.

The area of triangle ABC is 1/2*(AB)(AC)=1/2*​(13)(14)=91​.