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A school orders  textbooks, all for the same price. When the bill for the total order comes, the first and last digits are obscured. What are the missing digits?  a482.7b     

The other problem like this says that the school orders 99 textbooks.     

This problem has a blank spot where that 99 is in the other problem.     

So, I will assume the number of books in this problem should be 99.    

Consider that the price per book is multiplied by 99.   

And, multiplying by 99 incorporates multiplying by 9.    

The digits of the product of a number multiplied by 9    

will always sum to a total that is evenly divisible by 9.    

4 + 8 + 2 + 7 = 21 so the sum of the two missing     

numbers will have to add 6 (to make 27, which is     

evenly divisible by 9) or 15 (to make 36) to the 21.     

I found that 4482.72 is evenly divided by 99.   

So the missing numbers are a = 4 and b = 2.    

And the cost of each textbook was 45.28.    

_.   

I made a conversion mentally that I now feel I should have shown explicitly.  

The problem asks for 1/x + 1/y = 1/123         with x < y    

I submitted 1/369 + 2/369 = 3/369 = 1/123  

For clarity, there should have been a step before that.    

                  1/369 + 1/184.5    

Stated thus, the x and y positions are reversed.    

Pedantically, I admit I should not have done that.   

I just thought the equation looked nicer as 1, 2, 3. 

x = 184.5 and y = 369     note that x < y   

check answer       

1/x        + 1/y       =  1/123    

1/184.5 + 1/369  =?  1/123      

2/369    + 1/369  =  3/369  =  1/123       True    

_.   

Find the area of trapezoid EFGH.

The height of the trapezoid EFGH is h. 

\(A_{EFP}=\frac{\overline{EF}}{2} \cdot \frac{h}{3}=6\\ \overline{EF}=\frac{36}{h}\\ A_{GHP}=\frac{\overline{GH}}{2} \cdot \frac{2h}{3}=6\\ \overline{GH}=\frac{18}{h}\)

\(A_{EFGH}=\frac{\frac{36}{h}+\frac{18}{h}}{2}\cdot h\\ \color{blue}A_{EFGH}=27\)

The area of trapezoid EFGH is 27.

 !

You can use modular arithmetic.

15 members x 45 minutes = 675 member miunutes required 

675 member minutes / 20 members = 33 3/4 minutes will be needed.

What is the only solution to 1/x + 1/y = 1/123 where x < y?    

You could go for 3 / (3)(123) which would be    

written as 1 / 369 + 2 / 369 = 3 / 369 = 1 / 123     

but I don't guarantee that it's the only solution.   

_.   

From my response to your other post, m is either of the form \(pq^3\) or pqr, where p, q, r are distinct primes. 

By a similar approach, n is of the form \(pq^2\) where p, q are distinct primes.

Case 1: m is of the form pq^3.

Suppose n = rs^2, where r, s are distinct primes. r and s must be equal to p and q, regardless of order, since otherwise mn has more than 12 positive divisors. Then mn = p^2 q^5 or p^3 q^4. Taking the smallest values require q = 2 and p = 3. Then the minimum value of mn is min(3^2 * 2^5, 3^3 * 2^4) = 288.

Case 2: m is of the form pqr.

No matter what, mn will have more than 12 positive divisors.

Hence, the answer is 288.

Suppose the prime factorization of \(m\) is \(m = p_1^{k_1} \cdot p_2^{k_2} \cdots p_n^{k_n}\). Then the number of positive divisors is \((1 + k_1)(1 + k_2)\cdots(1 + k_n)\).

The ways of writing 8 into product of numbers are \(8 = (1 + 1) \times (1 + 3) = (1 + 1) \times (1 + 1) \times (1 + 1)\). From this and the fact above, we know that \(m\) is either of the form \(pq^3\) or \(pqr\), where p, q, r are distinct primes. The number of distinct prime factors of m is either 2 or 3. The given conditions are not sufficient to conclude definitely that the number of distinct prime factors is 2 (resp. 3).

All the possible events and their probability are as follows:

There is a probability of 1/4 + 1/4 = 1/2 that the number of runs is 1 and a probability of 1/2 that the number of runs is 2. Hence, the expected number of runs is:

(1/2)(1) + (1/2)(2) = 1.5.

Use the identity \((a + b + c)^2 = a^2 + b^2 +c^2 +2ab+2bc+2ca\).

\((n -k + 2)^2 = \textcolor{red}{1}n^2 + \textcolor{red}{1}k^2 + \textcolor{red}{4}n + \textcolor{red}{(-4)}k + \textcolor{red}{(-2)}nk + \textcolor{red}{4}\).

The options that are correct are 1, 2, 3, 4, and 5.

Analyzing the Problem

Given:

Regular hexagon ABCDEF

Area of hexagon ABCDEF = 72

P is the midpoint of EF

We need to find the area of triangle APF (the shadow)

Approach:

Find the area of triangle ADF.

Find the ratio of the area of triangle APF to triangle ADF.

Calculate the area of triangle APF.

Solution

Step 1: Find the area of triangle ADF.

Since a regular hexagon can be divided into six equilateral triangles, the area of one equilateral triangle is 72/6 = 12.

Triangle ADF is made up of two equilateral triangles.

Area of triangle ADF = 2 * 12 = 24

Step 2: Find the ratio of the area of triangle APF to triangle ADF.

Triangle APF is similar to triangle ADF (both right triangles with the same angle at A).

The ratio of their areas is the square of the ratio of their corresponding sides.

Since P is the midpoint of EF, AP = AF/2.

The ratio of the areas of triangle APF to triangle ADF is (AF/2)^2 / AF^2 = 1/4.

Step 3: Calculate the area of triangle APF.

Area of triangle APF = (1/4) * Area of triangle ADF

Area of triangle APF = (1/4) * 24 = 6

Therefore, the area of the shadow (triangle APF) is 6 square units.

To find the missing digits, we can use the fact that the total number of textbooks is 99.

We know that the total cost is a482.7b, where a and b are the missing digits. We also know that the total cost is equal to the price per textbook multiplied by the number of textbooks.

Since there are 99 textbooks, the total cost must be a multiple of 99. We can check the multiples of 99 until we find one that matches the given pattern:

99 * 1 = 99

99 * 2 = 198

99 * 3 = 297

99 * 4 = 396

99 * 5 = 495

99 * 6 = 594

99 * 7 = 693

99 * 8 = 792

99 * 9 = 891

99 * 10 = 990

From this list, we can see that the only multiple of 99 that matches the given pattern is 495. Therefore, the missing digits are a = 4 and b = 5.

Therefore, the total cost of the textbooks is $4482.75.

Because $\angle AOB = 60$, $\triangle OXA$ is a 30-60-90 triangle. This means that $AX=6$ and $OX=6 \sqrt{3}$.

Then we can use power of a point on $X$ to get the equation $6\sqrt3 \cdot XY = 36$, from which we get $XY =2 \sqrt{3}$.

There are 8 rolls = 8! = 40320 possible combinatins  IF htere were no duplicate numbers

  BUT there is   two 1's   two 2 's   two 3's  and two 4's    so you have to divide that result by  2!  2! 2! 2! 

to get   40320 / 16 = 2520 possible sequences  

Let's know that XZW will form a triangle

\(\angle ZXW = 90^\circ \\ \angle XWZ = 60^\circ\)

Thus, \(\angle XZW = 180 - 90 - 60 = 30\)

30 degrees is our answer. 

Thanks! :)

h is a function, so in order to find h(x), plug 45 into the equation :

h(45) = (1/9) * (90 - 45)

h(45) = 45/9 = 5

So the answer is B