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We can analyze this problem by considering the divisibility rules for 2, 3, 4, 5, 6, 8, and 9:

Divisibility Rules:

A number is divisible by 2 if the last digit is even (0, 2, 4, 6, or 8).

A number is divisible by 3 if the sum of its digits is divisible by 3. (Given in the problem statement)

A number is divisible by 4 if the last two digits are divisible by 4. (Not applicable here since the number ends in 0)

A number is divisible by 5 if the last digit is a 0 or a 5.

A number is divisible by 6 if it's divisible by both 2 and 3.

A number is divisible by 8 if the last three digits are divisible by 8. (Not applicable here since the number ends in 0)

A number is divisible by 9 if the sum of its digits is divisible by 9. (This is similar to the divisibility rule for 3, but a stronger condition)

Analysis:

Ending in 0: Since the number ends in 0, it is automatically divisible by 2 and 5 (based on the divisibility rules).

Sum of Digits Divisible by 3: The problem states that the sum of the digits is divisible by 3. This means the number itself must also be divisible by 3 (based on the divisibility rule for 3).

Divisibility by Other Numbers:

Looking at the remaining numbers (4, 6, 8, and 9), we can see that:

4: Not necessarily divisible. Just because the last digit is 0 (divisible by 2), it doesn't guarantee divisibility by 4 (requires the last two digits to be divisible by 4).

6: Since the number is divisible by both 2 (ending in 0) and 3 (given in the problem statement), it is automatically divisible by 6.

8: Not necessarily divisible. Similar to 4, just because the number ends in 0 (divisible by 2), it doesn't guarantee divisibility by 8 (requires the last three digits to be divisible by 8).

9: Not necessarily divisible. The sum of digits being divisible by 3 is a weaker condition than being divisible by 9. So, a number divisible by 3 might not necessarily be divisible by 9.

Conclusion:

Therefore, a number ending in 0 with a sum of digits divisible by 3 is necessarily divisible by 2, 3, and 6 (out of the given numbers). There are 3​ such divisors.

We can solve this problem using the divisibility rule for 9. This rule states that a number is divisible by 9 if the sum of its digits is divisible by 9.

Divisibility Rule for 9: A number (like 8*10^18 + 1^18) is divisible by 9 if the sum of its digits is divisible by 9.

Simplifying the Expression:

8*10^18 can be written as 800,000,000,000,000,000 (eight hundred quadrillion).

The sum of the digits in 800,000,000,000,000,000 is 8 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 8 = 16.

Remainder When Divided by 9:

We need to find the remainder when 8*10^18 + 1^18 (which has a digit sum of 16) is divided by 9.

Since 18 is a multiple of 2, 1^18 will always be 1. So, we can focus on the divisibility of 800,000,000,000,000,000 (with a digit sum of 16).

We can find the remainder by subtracting the nearest multiple of 9 that is less than 16: 16 - 9 (nearest multiple of 9 less than 16) = 7.

Therefore, the remainder when 8*10^18 + 1^18 is divided by 9 is 7.

Given that 15 members can paint the wall in 45 minutes, we can calculate the rate at which one member paints the wall.Let's denote the rate at which one member paints the wall as R (wall per minute). If 15 members paint the wall in 45 minutes, then the combined rate of all 15 members is 15R. So, 15R = 1 wall painted in 45 minutes. Thus, the rate of one member painting the wall is R = 1/15 wall per minute.

Now, we need to find out how long it would take 20 members to paint the same wall. Let T be the time (in minutes) it would take 20 members to paint the wall. The combined rate of all 20 members is 20R, and we know that the rate of one member is R = 1/15 wall per minute.

Therefore, 20R = 20/15 = 4/3 walls per minute. We want to find out how long it would take 20 members to paint 1 wall, so we set up the equation: T * (4/3) = 1. Solving for T: T = 3/4 minutes.
 

Therefore, it would take 20 members 3/4 minutes to paint the wall.

To find the total number of students who are in the math club or the science club (or both) at Annville Junior High School, we can follow these steps:

1. There are 3 students in the math club.

2. 3 students are members of both the math and science clubs.

3. The number of students in the science club is twice the number in the math club. Given this information, we can calculate the number of students in the science club: - Number of students in the math club: 3 - Number of students in the science club = 2 number in the math club = 2 3 = 6 Next, we find the total number of students who are in the math club or the science club (or both) using the principle of inclusion-exclusion: Total students = Number in math club + Number in science club - Number in both clubs Substitute the values we found: Total students = 3 + 6 - 3 = 6

Therefore, the total number of students who are in the math club or the science club (or both) is 6.

I can answer problem 2!

This problem can be solved by analyzing the behavior of the function f(x, y) at different points and identifying repeating patterns.

Understanding the Function:

The function f(x, y) modifies the x and y coordinates based on the initial values of x and y.

If x > 4, it subtracts 4 from x and keeps y the same. (e.g., f(5, 1) = (1, 1))

If x <= 4 and y > 4, it keeps x the same and subtracts 4 from y. (e.g., f(2, 5) = (2, 1))

Otherwise (x <= 4 and y <= 4), it adds 5 to x and adds 2 to y. (e.g., f(1, 1) = (6, 3))

Analyzing the Robot's Movement:

Starting Point: The robot starts at (1, 1).

First Iteration: Applying f(1, 1) based on the third rule (x <= 4 and y <= 4), we get f(1, 1) = (6, 3).

Subsequent Iterations:

At (6, 3), x > 4 and y > 4. So, f(6, 3) = (2, -1). (This is because x is now greater than 4, and y is greater than 4 in the new position).

At (2, -1), x <= 4 and y <= 4. So, f(2, -1) = (7, 1).

At (7, 1), x > 4 and y <= 4. So, f(7, 1) = (3, 1).

Repeating Pattern:

Notice how the robot's movement falls into a repeating cycle: (1, 1) -> (6, 3) -> (2, -1) -> (7, 1) -> (3, 1) -> (8, 3) -> ...

This cycle repeats because after reaching (3, 1), the function follows the same path back to (1, 1) and continues the cycle infinitely.

Number of Unique Points:

Within this cycle, there are 4 unique points visited: (1, 1), (6, 3), (2, -1), and (7, 1).

Therefore, the robot will visit only 4 unique points regardless of how long it runs.

By my calculator, 2.5 * 0.125 * 1.6 = 0.4.

The answer is 0.4

To find the area of triangle AMN, we can use the fact that the area of a triangle is equal to half the product of its base and height. In this case, we need to find the height of triangle AMN with respect to base MN.
 

1. Let D be the point where the line through I parallel to BC intersects MN.

2. Since I is the incenter of triangle ABC, we know that ID is perpendicular to MN.

3. Therefore, triangle IDM is a right triangle with ID as its height and MD as its base.

4. By properties of parallel lines, triangles ABC and AMN are similar.

5. This means that the ratio of corresponding side lengths in these triangles is equal. So, AB/AM = BC/MN.

6. Since AB = 5 and BC = 8, we can find MN as (5/8) × MN = 5, which gives MN = 40/5 = 8.

7. Now, ID is the height of triangle IDM, and MD = MN = 8.

8. Using the Pythagorean theorem in triangle IDM, we can find ID as ID^2 + 4^2 = 5^2, which gives ID = 3.

9. The area of triangle AMN is then (1/2) × MD × ID = (1/2) × 8 × 3 = 12. Therefore, the area of triangle AMN is 12

x^2 + nx + 15 < -21 -7x + x^2       simplify as

(n + 7)x < -36

This will have no real solution if n =  -7

https://web2.0calc.com/questions/coordinates_34456

                             F                     

    A    3    E     4    B

                                     10

    D                       C

[EAD]  =  9

Therefore  

9 = (1/2) AE * BC

9 = (1/2) AE * 10

9 / 5 =  AE =  (3/7)AB

So

AB = (7/3)AE = (7/3)(9/5)  = 63/15 = 21/5 

So

[ABCD ] = AB * BC  =  (21 / 5) (10)    =  42

No problem

Just some polynomial division

                        x^2  + 4x  + 15

x^2 - 4x + 1   [  x^4  + 0x^3 + 0x^2 + 0x - 3 ] 

                         x^4 -  4x^3  + 1x^2

                       ___________________

                                   4x^3 - 1x^2 + 0x

                                   4x^3 -16x^2 + 4x

                                  ________________

                                           15x^2  - 4x - 3

                                           15x^2 - 60x +15

                                           ______________

                                                    56x - 18        (this remainder is  ignored)

The parabolis asymtote is    y =   x^2 + 4x + 15       

Here's what I found....

The prime digits wil be  2,3,5,7

Cases :

Case 1 :  all the digits are the same  =  4  ways to do this 

Case 2 :  three of the digits are the same

      4 ways to select the same three digits  

      Then, select any one of the three remaining digits  to occupy any of  four places

      4  * 3 * 4  =  48

Case 3:  two of the digts are the same

     4 ways two select the same two digits 

     Choose any two of the remaining  three primes 3C2 = 3  to occupy any two of four positions 4C2 = 6

      4 * 3 * 6  =  72

Case 4 :  all the digits are different  =  4!  = 24

Total  =   4 + 48  + 72 + 24  =   148 

Yes I'm sorry.  This is my first time using this platform.  Ain't sure how to format things.

- Chad

Do you mean                     ( 2x^4 -3)

                           y =          ____________         ????

                                          (x^2 - 4x + 1)

It should be 24 aka 4! since there are that many arrangements of four digit postivies intigers.  If you still don't understand let me know :)

- Chad

Area of quadrilateral = AD *DC =  180

Shaded area  (a trapezoid) =  (1/2) AD * ( DC + DC/3)  =  (1/2) AD *(4/3)DC = (2/3) AD * DC

So....the shaded area  =  (2/3) (180)  =  120 

https://web2.0calc.com/questions/problem-on-spheres

https://web2.0calc.com/questions/plz-help_20418