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The area of triangle ABD is 40*sqrt(2) + 12.

BX = 20*sqrt(7)

Simplify as

10x^2 + 10x - 1   =  0 

ab =  -1/10

a + b =  -10/10  = -1     square both sides

a^2 + b^2 + 2ab  = 1

a^2 + b^2  = 1 - 2(-1/10)  =  1+ 2/10  =1 + 1/5 =    6/5

(a - 4) / (b -4)  +  (b -4) / ( a - 4)  =  

[ (a - 4)^2 + ( b -4)^2 ]  /  [ (a - 4)(b - 4)]  = 

[ (a^2 + b^2 - 8 (a +b) + 32 ]  / [ ab - 4(a + b)  + 16]  =

[ 6/5 - 8 (-1) + 32 ] / [ -1/10 - 4 (-1) + 16 ]  =

[ 6/5 + 40 ]  / [ 20 - 1/10]  =

412 / 199

We can solve this problem by considering the number of 1s and 3s used to form the sum 11.

Number of 1s and 3s:

The sum of the digits is 11, which can be formed using digits 1 and 3. Let's say there are x ones and (11 - x) threes.

Restrictions on x:

Since each digit must be a 1 or a 3, x (number of ones) can range from 0 (all digits are threes) to 11 (all digits are ones). However, we only care about positive integers that satisfy the condition. Therefore, valid values for x are:

1 <= x <= 11

Counting the Possibilities:

For each value of x (number of ones), we have (11 - x) threes. So, the total number of positive integers with x ones and (11 - x) threes is 1.

This creates 11 possibilities (from x = 1 to x = 11) since there are 11 valid values for x.

Therefore, there are 11 positive integers that satisfy both conditions.

There are 25,480 ways of coloring the squares.

\(\frac{7!}{3!\times2!\times2!}=\boxed{210}\)

4

Divisors are  1 2 4

5^19 + 7^13 + 23 

5 raised to any power will end in 5

7^1  = 7

7^2 = 49

7^3  = 343

7^4 = 2401

7^13  ends in 7

So

5^19 + 7^13 + 23    will end in   5 + 7 + 3 =  5 

The smallest  prime divisor wil either be 3  or 5

[5^19 + 7^13 + 23] mod 3  = 2

[5^19 + 7613 + 23 ] mod 5 = 0

The smallest prime divisor is 5

2 * (4/8) (4/7) (3/6) (3/5) (2/4) (2/3) (1/2) (1/1)   =    1/35

Let the numbers be \(x-3,x-2,x-1,x,x+1,x+2,x+3\)
Their sum is
 \(7x=\frac{4}{3}(x+3)\)

Solving we get:
\(x=\frac{12}{17}\) 
The largest number is \(x+3=\frac{12}{17}+\frac{3}{1}=\boxed{\frac{63}{17}}\)
Which is not an integer and the question is flawed. 

https://web2.0calc.com/questions/algebra_10763

1037  is the first  required number

1977 is  the second required number

[ 1977 - 1037 ] / 47  + 1   =   21 numbers

2^9    = 512  has  9 + 1 =     10 positive divisors

One prime - 2 -  divides it

https://web2.0calc.com/questions/help-me_89945

Let  P , Q  lie on a unit circle

This circle has the equation x^2 + y^2  =1

Equation of line containg OP  is  y =x

Equation of line  containing OQ  is  y = 3x

The x coordinate of P  can be  found   as

x^2 + x^2 = 1

2x^2 =1

x^2  =1/2

x = 1/sqrt 2      and  y =1/sqrt 2

The x coordinate of Q  can be found as

x^2 + (3x)^2  = 1

10x^2 = 1

x^2  =1/10

x= 1/sqrt 10     y = 3/sqrt 10

Slope  between   PQ   is

[ 1/sqrt 2 - 3/sqrt 10 ] / [ 1/sqrt 2 - 1/sqrt 10 ]  =

[ sqrt 10 - 3sqrt 2 ] / [sqrt 10 - sqrt 2]  =

[ sqrt 10 -3sqrt 2 ] [ sqrt 10 + sqrt 2 ] / [ 10 -2]  =

[ 10 -2sqrt 20 - 6 ] / [ 8]  =

[ 4 - 4sqrt 5 ] / 8  =

[ 1 - sqrt 5 ] / 2

First, let's simplify the equation. We have

\(x^2 - 6x + 13\)

Since the constant is equal to uv, we have the equation

\(uv = 13 \\ 2uv = 26 \)

The coefficient of x is equal to u+v, so we have

\(u + v = 6 \)

Squaring both sides of the equation, we have the equation

\(u^2 + 2uv + v^2 = 36 \\ u^2 + 26 + v^2 = 36 \\ u^2 + v^2 = 10\)

Now, it's time for the fun part of this problem. We can simplify what we want to figure out to

\(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2\)

We already know all these terms! We can easily find them! We have

\(u^3 + v^3 = ( u + v) ( u^2 + v^2 - uv) = (6)(10 - 13) = -18 \\ [uv]^2 = [ 13]^2 = 169\)

Thus, our answer is \(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2 = -18 / 169 \)

So -18/169 is our answer. 

Thanks! :)

Let's note something really quickly. 

If Kenneth ran 2 hours, then he ran 4 kilometers more than Marshall. 

So 4 kilometers is 1/5 of the track's length. 

Thus, we have \(1/5x = 4\) where x is the length of the track. 

We have

\(x=20\)

So the track was 20km. 

So our answer is 20. 

Thanks! :)

I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row?  

7 • 6 • 5 • 4 • 3 • 2 • 1  =  5040 ways 

You can make 7 choices for the first book. 

That leaves you 6 choices for the next book. 

And then there are 5 choices for the third book. 

and so on  

_.

There are two seperate cases we need to consider. 

First off, we have orange, purple, orange. 

The probability of this is \(\frac{3}{5}*\frac{2}{4}*\frac{2}{3} = 1/5\)

The other case is when we have purple, orange, purple. 

The probability of this is \(\frac{2}{5}*\frac{3}{4}*\frac{1}{3} = 1/10\)

Adding these two up, we get

\(\frac{1}{5}+\frac{1}{10} = \frac{2}{10}+\frac{1}{10}=\frac{3}{10}\)

So our answer is 3/10 or 30%. 

Thanks! :)

\(Q\) lies on the line \(y=3x\)
\(P \) lies on the line \(y=x\)

WLOG lets \(Q=(1,3)\), therefore \(OP=OQ=\sqrt{10}\)

Thus we can say \(P=(\sqrt{5},\sqrt5)\)

Slope between 2 points is \(s=\large{\frac{y_2-y_2}{x_2-x_1}=\frac{\sqrt5-3}{\sqrt5-1}=\boxed{\frac{1-\sqrt5}{2}}}\)

This functional equation can be solved to find f(x) as a rational function. Here's how:

Isolating f(x): We want to isolate f(x) on one side of the equation. Subtract f((2x - 3)/(x - 1)) from both sides:

f(x) = x - f((2x - 3)/(x - 1))

Finding a Common Denominator: The next step is to find a common denominator for the right side of the equation. Since the function is undefined at x = 1 and x = 2, the common denominator will involve these terms (x - 1) and (x - 2).

We can rewrite f((2x - 3)/(x - 1)) as:

f((2x - 3)/(x - 1)) * (x - 2) / (x - 2) = f((2x - 3)/(x - 1)) * (x - 2) / (x - 1)(x - 2)

This introduces the term (x - 2) in the denominator to match the existing term (x - 1).

Substituting and Combining Like Terms:

Now, substitute the entire right side of the equation back into the expression for f(x):

f(x) = x - [ f((2x - 3)/(x - 1)) * (x - 2) / (x - 1)(x - 2) ]

Expand the product in the denominator:

f(x) = x - [ f((2x - 3)/(x - 1)) * (x - 2) ] / (x^2 - 3x + 2)

Recognizing a Pattern:

Notice that the term f((2x - 3)/(x - 1)) appears in the equation. We can substitute the entire right side of the original equation for this term:

f(x) = x - { [ x - f((2x - 3)/(x - 1)) ] * (x - 2) } / (x^2 - 3x + 2)

This creates a recursive relationship involving f(x). However, it allows us to eliminate f((2x - 3)/(x - 1)) from the equation entirely.

Simplifying and Solving:

Expand the product in the brackets and combine like terms:

f(x) = x - (x^2 - 5x + 2) / (x^2 - 3x + 2)

f(x) = (x^2 - 3x + 2 - x^2 + 5x - 2) / (x^2 - 3x + 2)

f(x) = 2x / (x^2 - 3x + 2)

Therefore, the function f(x) satisfying the given conditions is:

f(x) = 2x / (x^2 - 3x + 2)

A number with \(3\) positive integers must be in the form \(p^2\). Therefore the answer is only \(\boxed{1 \text{ prime number}}\)

A terrific number must be only divisble by one and itself AND be a perfect square. 

Let's test perfect squares and see what we get. 

First, let's test \(1\). It's only divisble by 1, so it doesn't work. 

Next, we have \(4\). It's divisble by \(1, 2, 4\). So it is a terrific number!

So 4 is our answer. 

Thanks! :)