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There are a total of 900 integers from 1 to 999 inclusive. The digit 7 can appear in the hundreds, tens, or units place. In the hundreds place, there are 100 numbers that have a 7. In the tens place, there are 90 numbers that have a 7. And in the units place, there are 90 numbers that have a 7. So, there are a total of 100 + 90 + 90 = 280 numbers that have a 7.

Therefore, the probability that Sam chooses a 7 is 280 / 900 = 14/45.

Let's do that:

2x^2 - 7x + 2 = x^2 - 6x + 3

Rearranging the equation, we get:

2x^2 - x^2 - 7x + 6x + 2 - 3 = 0

x^2 - x - 1 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -1, and c = -1.

To find the roots of this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Applying the values, we get:

x = (1 ± √((-1)^2 - 4(1)(-1))) / (2(1))

x = (1 ± √(1 + 4)) / 2

x = (1 ± √5) / 2

Therefore, the roots of the equation are:

a = (1 + √5) / 2 b = (1 - √5) / 2

0=189 , 1=300 , 2=300 , 3=300 , 4=300 , 5=300 , 6=300 , 7=300 , 8=300 , 9=300 >>Total = 2889 digits

Angle FDE = 12 degrees.

By partial fractions, (A,B,C) = (1/8, 7/32, -1/80).

Here is the Python code to calculate the number of digits:

def number_of_digits(n): if n < 10: return 1 elif n < 100: return 2 elif n < 1000: return 3 else: return 4 def total_number_of_digits(n): total_digits = 0 for i range(1, n + 1): total_digits +== number_of_digits(i) return total_digits print(total_number_of_digits(999))

The output of the code is 3515.  Therefore, the answer is 3515 digits.

there are 1-digit, 2-digit, and 3-digit numbers

1-9 1-digit

10-99 2-digit

100-999 3 digit

now find how many numbers are in each category and multiply

Five times a number is divided by 7 more than the number. If the result is 0, then what was the original number?  

                                                        5n    

                                                    ––––––  =  0      

                                                      n + 7  

The only number that works is       n  = 0     

_.

Find the number of ways of arranging the numbers 1,2 ,3, 4, 5, 6, 7, 8, 9   

in a row so that the product of any two adjacent numbers is even.  

For two numbers to have an even product, at least one of them must be even.  

That is, an odd number can never be adjacent to another odd number.  

So, odds and evens have to alternate, such as the example shown in the problem.  

Furthermore, since there are more odd numbers, the first number has to be odd.  

There are five odd numbers, and five positions into which they can be situated.  

So, there are 5 • 4 • 3 • 2 • 1 = 120 ways the odd numbers can be arranged.  

And, there are four even numbers, and four positions where they can be located.  

So, there are 4 • 3 • 2 • 1 = 24 ways that the even numbers can be arranged.  

Any even number arrangement can be combined with any odd number arrangement.  

Therefore, the number of possible mixes is 120 • 24 = 2880.   

_.

Real numbers a and b satisfy    

a + ab = 250    

a - ab = -240    

Enter all possible values of a, separated by commas.   

i doubt that you'll find an easier problem   

than this in your entire algebra textbook. :-)    

                                                                                a + ab  =    250    

                                                                                a – ab  =  –240   

Just add the two together   

                                                                              2a           =     10    

I just kinda kept the a's lined up   

with each other.  After you divide  

both sides by 2, the final answer is                                   a  =  5  

The problem doesn't ask for the      

value of b, but it is, of course,                                           b  =  49    

_.

Alice and Bob each have a certain amount of money. If Alice receives n dollars from Bob, then she will have 4 times as much money as Bob. If, on the other hand, she gives n dollars to Bob, then she will have 8 times as much money as Bob. If neither gives the other any money, what is the ratio of the amount of money Alice has to the amount Bob has?      

The first case:   

                                                         (A + n)  =  (4) • (B – n)  

                                                          A + n  =  4B – 4n  

                                                          A – 4B  =  –5n             (1)  

The second case:   

                                                         (A – n)  =  (8) • (B + n)  

                                                          A – n  =  8B + 8n  

                                                          A – 8B  =  9n               (2)  

Multiply both sides of (1) by 9           9A – 36B  =  –45n  

Multiply both sides of (2) by 5           5A – 40B  =    45n  

Add these two together                    14A – 76B  =  0  

Add 76B to both sides    

                                                         14A            =  76B  

Divide both sides by 76B      

                                                        14A           1        

                                                        ––––  =   ––––  

                                                         76B          1              

Multiply both sides by 76 / 14  

                                                         76         14A        76          1   

                                                        ––––  •  ––––  =  ––––  •  ––––      

                                                         14         76B        14           1      

The 76 and the 14 cancel out  

on the left side, leaving  

                                                          A           76  

                                                        ––––  =  ––––  

                                                          B           14  

Reduce the fraction  

                                                          A           38 

                                                        ––––  =  ––––  

                                                          B            7 

_.

There are two cases to consider:

Case 1: One of the boxes is empty. In this case, there are 4 ways to choose which box is empty, and then 7 ways to distribute the remaining 8 balls among the 4 boxes. So there are 4×7=28 ways in this case.

Case 2: No boxes are empty. In this case, there are 3 ways to choose which box has 1 ball, 2 ways to choose which box has 2 balls, and 1 way to choose which box has 3 balls. So there are 3×2×1=6 ways in this case.

Therefore, the total number of ways to distribute the balls is 28+6=34​.

[57 - ( - 33)] / 5  +  1 ==[90 / 5] + 1 ==19 terms.

19

Remember that these numbers aren't consecutive.

I think you mean: "Evaluate w + ( -x ) - 2/3 where w = -5/9 and x = 4/3." The answer would be -23/9. Its on Khan Academy.

Find the number of ways of arranging the numbers 1,2 ,3, 4, 5, 6, 7, 8, 9   

in a row so that the product of any two adjacent numbers is even.  

For two numbers to have an even product, at least one of them must be even.  

That is, an odd number can never be adjacent to another odd number.  

So, odds and evens have to alternate, such as the example shown in the problem.  

Furthermore, since there are more odd numbers, the first number has to be odd.  

There are five odd numbers, and five positions into which they can be situated.  

So, there are 5 • 4 • 3 • 2 • 1 = 120 ways the odd numbers can be arranged.  

And, there are four even numbers, and four positions where they can be located.  

So, there are 4 • 3 • 2 • 1 = 24 ways that the even numbers can be arranged.  

Any even number arrangement can be combined with any odd number arrangement.  

Therefore, the number of possible mixes is 120 • 24 = 2880.   

_.

We can simplify the expression for f(n) as follows:

f(n) = (4n + sqrt(4n^2 - 1))/(sqrt(2n + 1) + sqrt(2n - 1)) = (4n + sqrt(2^2(2n - 1)^2 - 1)) / (sqrt(2n + 1) + sqrt(2n - 1)) = (4n + 2(2n - 1)) / (sqrt(2n + 1) + sqrt(2n - 1)) = n + 1 / (sqrt(2n + 1) + sqrt(2n - 1))

Now, let's evaluate f(1), f(2), ..., f(40):

f(1) = 2 + 1 / (sqrt(3) + sqrt(1)) = 3 / 2 f(2) = 4 + 1 / (sqrt(5) + sqrt(3)) = 5 / 2 f(3) = 6 + 1 / (sqrt(7) + sqrt(5)) = 7 / 2 ... f(40) = 81 + 1 / (sqrt(83) + sqrt(81)) = 82 / 2

The sum of f(1), f(2), ..., f(40) is:

3 + 5 + 7 + ... + 82 = \frac{3(1 + 2 + ... + 40)}{2} = \frac{3 \cdot 40 \cdot 41}{2 \cdot 2} = \frac{63 \cdot 41}{2} = \boxed{1263}

Therefore, the sum of f(1), f(2), ..., f(40) is 1263.

Find the ordered triple (p,q,r) that satisfies the following system:  

p - 2q = 3  

q - 2r = -2 + q  

p + r = 9 + p  

Let's number those equations, for identification only:  

(1)  p – 2q  =  3  

(2)  q – 2r  =  –2 + q  

(3)  p + r  =  9 + p  

Per equation (2), r = 1  

Per equation (3), r = 9  

You can't have r equalling two different things.  There is no solution.  

_.

Find the number of ways of arranging the numbers 1,2 ,3, 4, 5, 6, 7, 8, 9 in a row so that the product of any two adjacent numbers is even.  

Find the number of ways of arranging the numbers 1,2 ,3, 4, 5, 6, 7, 8, 9   

in a row so that the product of any two adjacent numbers is even.  

For two numbers to have an even product, at least one of them must be even.  

That is, an odd number can never be adjacent to another odd number.  

So, odds and evens have to alternate, such as the example shown in the problem.  

Furthermore, since there are more odd numbers, the first number has to be odd.  

There are five odd numbers, and five positions into which they can be situated.  

So, there are 5 • 4 • 3 • 2 • 1 = 120 ways the odd numbers can be arranged.  

And, there are four even numbers, and four positions where they can be located.  

So, there are 4 • 3 • 2 • 1 = 24 ways that the even numbers can be arranged.  

Any even number arrangement can be combined with any odd number arrangement.  

Therefore, the number of possible mixes is 120 • 24 = 2880.   

_.

Simplifying, we have

\(2x + x = 6 + 3x\)

\(3x = 6 + 3x\)

\(0 = 6\)

Since 0 can't equal to 6, this equation has no solutions.

My square patio is tiled with square tiles, all the same size. All the tiles are gray, except the tiles along the two diagonals, which are all yellow. (The corners are yellow, the center is yellow, and all the tiles along the diagonal in between are yellow.) If there are 5 yellow tiles, how many gray tiles are there?   

Here's the key info:   

                                   The corners are yellow...  

                                    the center is yellow...  

                                    there are 5 yellow tiles...  

There are four corners and one center.  

That's five yellow tiles right there.  

That's all of them.  There are no more yellow tiles.  

So the patio has to be a 3 x 3 square.  Draw it if you doubt me.  

The 3 x 3 patio has nine tiles in total.  

Five tiles are yellow, therefore the other four tiles are gray.  

_.

Find the ordered triple (p,q,r) that satisfies the following system:  

p - 2q = 3  

q - 2r = -2 + q  

p + r = 9 + p  

Let's number those equations, for identification only:  

(1)  p – 2q  =  3  

(2)  q – 2r  =  –2 + q  

(3)  p + r  =  9 + p  

Per equation (2), r = 1  

Per equation (3), r = 9  

r can't be two different things.  The system has no solution.  

_.

According to the Remainder Theorem, when a polynomial p(x) is divided by x−a, the remainder is p(a). So, when f(x) is divided by (x−3), the remainder is f(3), and we are given that f(3)=15.

Similarly, when f(x) is divided by (x−1)2, the remainder is f(1), and we are given that f(1)=2(1)+1=3.

We can write the remainder when f(x) is divided by (x−3)(x−1)2 as R(x). By the Remainder Theorem, we have the following:

R(3)=f(3)=15

R(1)=f(1)=3

To find R(x), we can use the polynomial long division algorithm to divide f(x) by (x−3)(x−1)2. The long division algorithm gives us the following:

Quotient: x+5 Divisor: (x-3)(x-1)^2 ---------------------- x^3-4x^2+7x-15 -(x^3-3x^2) -x^2+7x-15 +x^2-x^2 +7x-15-(-7x+21) -x-15+21 -6 ---------------------- x+5

Therefore, the remainder when f(x) is divided by (x−3)(x−1)2 is x+5.

So the answer is x+5.

The radius is 5 because a^2 and b^2 will be 25 and you multiply it to the other side and take the square root

Find the ordered triple (p,q,r) that satisfies the following system:  

p - 2q = 3  

q - 2r = -2 + q  

p + r = 9 + p  

Let's number those equations, for identification:  

(1)  p – 2q  =  3  

(2)  q – 2r  =  –2 + q  

(3)  p + r  =  9 + p  

Per equation (2), r = 1  

Per equation (3), r = 9  

You can't have it both ways.  The system has no solution.  

_.