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x^3 - x^2 -21x + 45  factors as   (x-3)^2 (x + 5)

Multiply through by this factorization  and  we  get

1  =  A(x-3)^2 + B (x-3) ( x + 5)  + C(x + 5)

Let x = -5

Then  A(-8)^2  = 1  →  A = 1/64

\(f(x) \text{ being invertible means that if }f(x)=f(y) \text{ then }x=y\\ \text{so we are looking for points where }x^2 = x^4\\ \)

There are 3 such points. 

We have \(x=0, 1, -1\)

This means they intersect at 3 points. 

I'm not sure if this is correct, though. 

Thanks! :)

f(a)  =  (13 + a) / (3a + 7)

There is  no max value

At  a = 0   f(a)  =  13/7

As a →  inf,    f(a) →  1/3

See the graph here : https://www.desmos.com/calculator/reaplfkcny

Hello!

So, first, let's review. We know that 

\(f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7}\) . I believe you understand this part!

Now, for the 1/3 in the second part where it starts getting confusing, let's first note that \(\frac{(ax+b)}{(x+b)} = \frac{a}{1} = a\). Since the polynomial is equivalent, we can cancel them out. That's what we are doing here. Canceling out the 3a + 7, we get the remaining 1/3. 

Now, for the \(\frac{32}{9a+21}\). Let's note that \(\frac{a}{b} = \frac{a}{b} \cdot \frac{c}{c}\). We want to get rid of the 3 in the numerator, so we multiplythe numerator and denominator by 3. 

We get \(\frac{32/3}{3a+7} = \frac{32/3}{3a+7} \cdot \frac{3}{3} = \frac{32}{3(3a+7)} = \frac{32}{9a+21}\)

So there, that's how we got from the second step to the third step. 

Good luck from here!

I hope I answered your question!

Thanks! :)

The first function will  be <  0  on [-4,  -3]

x^2 + 5x + 6  = ( x + 3) ( x + 2)

At  x = -3 ....   x^2 + 5x + 6  =  (-3 + 3) (-3 + 2)  =  0

At x = -4........x^2 + 5x + 6  =   (-4 + 3) (-4 + 2)  = 2

So  x = -4  is  correct

See the graph here : https://www.desmos.com/calculator/nf9utyhcye

https://web2.0calc.com/questions/quadratic_39762

PT = PU  )tangents form a point outside a circle are equal_

PO = PO

TO = UO  =   radii of the same  circle

So, by SSS  triangle TOP   is congruent to triangle UOP

So angleTOP  = angle UOP  = 45

So angle TOU = 90

This is a  central angle intercepting minor arc TU  ...so minor arc TU  = 90

We get two seperate equations from the problem. 

We have

\( a^3 + 3ab^2 = 679 \\ 3a^3 - ab^2 = 615\)

Let's multiply the second equation by 3 so that we can get 3a^3 in both equations

\(3a^3+9a^2=2037\)

\(3a^3-ab^2=615\)

Now, subtract the second equation from the first equation. We get

\(10ab^2 = 1422 \\ ab^2=142.2\)

Now, sub this value back into the first equation to get that

\(a^3+3(142.2)=679 \\ a^3+426.6=679 \\ 252.4=a^3 \\ a=\sqrt[3]{252.4} \\ a \approx 6.19\)

Now we have a, we can find b. 

\(b^2 = (142.2)/(6.19) \\ b^2 \approx 22.97 \\ b \approx \sqrt{22.97} \\ b \approx 4.79\)

So, we have

 \(a - b \approx 6.19 - 4.79 \\ a - b \approx 1.40\)

So 1.4 is our answer

Thanks! :)

https://web2.0calc.com/questions/triangle_4602

https://web2.0calc.com/questions/geometry-question_42617

Here's what I think......

https://web2.0calc.com/questions/coordinates_87801#r1

\( \frac{a}{3} + 1 = \frac{a + 3}{a} + 1 \)

a/3 = (a + 3) / a       cross-multiply

a^2  = 3(a + 3)

a^2  = 3a + 9

a^2 - 3a  = 9

a^2 -3a + 9/4 = 9 + 9/4

(a - 3/2)^2  = 45 / 4      take both roots

a - 3/2  =  + / -  3sqrt 5 / 2

a = [ 3 +/- 3sqrt 5 ] / 2

\(xy + x = \frac{3x + 2y + z + y + 2z}{3} \)

3x (y + 1) =  3x + 2y + z + y + 2z

3x (y+ 1) - 3x = 3y + 3z

3x (y + 1 - 1)  = 3 (y + z)

x (y )  = (y + z)

         ( y + z)

x =     _________

               y 

Let  ABCDEFGH  be the octagon  and let it be inscribed in a circle

We can  find   the radius of the circle as follows

(2sqrt 2)^2  = 2r^2 - 2r^2* (1/sqrt2)

8 = r^2 (2 - 2/sqrt 2)

8 = r^2 (2 -sqrt (2))

8 / (2 -sqrt 2) = r^2

8 (2 + sqrt 2) / (4 -2)  = r^2

4 (2 + sqrt 2) = r^2

r = 2sqrt [ 2 +sqrt 2] =  sqrt [ 8 + sqrt 32]

Let A=   (0 , sqrt [8  + sqrt 32[ )

Let E = ( 0 . -sqrt [8 +sqrt 32] )

Let C = (sqrt [8 + sqrt 32] , 0]

Let D =  ( sqrt [8 + sqrt32] /sqrt 2 , -sqrt[8-sqrt32] /sqrt 2)  = (sqrt [4 + sqrt 8] , -sqrt [4 +sqrt 8] )

There are only four diagonals of differing lengths

The easiest one  to  calculate   is AE  = 2r =   2sqrt [ 8 +  sqrt32]   = 2sqrt [8 + 4sqrt 2 ]

AD =  sqrt [  ( sqrt [4 + sqrt 8[)^2 +  ( sqrt [8 +sqrt 32) ]  + sqrt [4 +sqrt 8] ^2 ] = sqrt [ 24 + 24sqrt 2]

AB = sqrt (8) 

AC =  sqrt [  ( sqrt [8 + sqrt 32 ] )^2 +  ( sqrt [8 +sqrt 32 ] )^2 ]= sqrt [ 16 + 8sqrt 2 ]

Sum of squares of differing length diagonals

AE^2 + AD^2 + AB^2 + AC^2  = 

4 [8 + 4sqrt 2] + [24 + 24sqrt 2]  + 8  + 16 + 8sqrt 2 = 80 + 48sqrt 2  

First, we have \(16*5*1/10 = 8\)

\( \log_{1/2} (8 ) =-3\)

so -3 is our answer

Thanks! :)

\( \log_{8a} 4$ if $a = 1/4\)

log _8(1/4) 4  =

log ₂  4 =               {use the change-of-base theorem }

log 4  / log 2   =

log 2^2 / log 2  =

2log 2  / log 2  =

2

Very nice.....I would have to rely on a graph to solve this.....your approach is WAY more elegant !!!

Thanks, NotThat Smart !!!!

\(\log_2 (\log_3 x) = 2 \log_4 (x) \)

To elaborate a little, we can use the change of base theorem to write

log ( logx / log 3 )            2 log (x)

______________    =    ________                { log 4  = log 2^2  = 2log 2 }

  log 2                             log 4

log ( log x / log 3 )           2log (x)

______________  =     _______

      log 2                        2log 2

log ( logx / log 3 ) =      log (x)

This implies that

logx / log 3  = x

log x = xlog 3

log x  = log 3^x

x = 3^x

Note here that the graphs of both functions never intersect

https://www.desmos.com/calculator/82kerdz2ip

So, like NotThatSmart, I don't find any solutions, either !!!

First, let's note that

\(2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11 \cdot {67} \\ 2^3 + 3^{11} = 177155 = 16105 \cdot 11\)

Now, from this, we can tell that

\(m = 10a + 3 \\ n = 5b + 1\)

where a and b are different numbers. 

This means m has \( 3, 13, 23, ..., 593 = 60 \) values

And n has \(1, 6, 11, ..., 596 = 120 \) values. 

This means that there are \(60 * 120 = 7,200 \) cases that work.

In total, there are \(600 * 600 = 360,000\) different cases. 

Now, we just have \(7200/360000 = 72/3600 = 2/100 = 1/50\)

This is a 2% chance. 

Thanks! :)

First, let's apply some log rules to simplify our equation. We get

\(\log _3\left(x\right)=x\) from this. 

We know \(\mathrm{If}\:f\left(x\right)=g\left(x\right),\:\mathrm{then}\:a^{f\left(x\right)}=a^{g\left(x\right)}\)

From this, we have \(3^{\log _3\left(x\right)}=3^x\)

Which is the same as \(x=3^x\)

There are NO SOLUTIONS to this equation. 

Thanks! :)

We can probably write some equations with the information we have. 

We have:

\(a_n = a_{n - 1} + a_{n - 2}\)  Subtracting both sides by a_n-1, we have

\(a_{n-2}=a_n-a_{n-1}\)

Now, let's plug in numbers we are given in the question. Plugging in 11 gets us \(a_9=4-1=3\)

Plugging in 10 gets us \(a_8=1-3=-2\)

Ok, let's see what happens when we plug in 9. 

We get \(a_7 = a_9 - a_8\). We already know these two values!

We have \(a_7=3-(-2)=5\)

Now, we plug in 8. We get \(a_6=-2-5=-7\)

So -7 is our final answer.

Thanks! :)

First, let's find every possible interval with this equation. 

We have \(c<-5,\:\:-5\le \:c<0,\:\:0\le \:c<4,\:\:c\ge \:4\)

Now, we solve the equation for each inequality. 

Case 1 - \(c<-5\)

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

This root is greater than -5, so it is invalid. 

Case 2 - \(-5 \leq c < 0\)

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This works!

Case 3 - \(0 \leq c < 4\)

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This works!

Case 4 - \(c>4\)

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

This is invalid. 

Now, we have \(c = -\frac{13}6, \frac{13}{6}\)

Thanks! :)

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}}.\)

Workng from the bottom-up

4 + 1/2  =  9/2

4 + 1/ (9/2)  =   4 + 2/9  =  (38/9)

1 + 1/(38/9)  =    1 + 9/38  = 47/38

1 / (47/38)  =   

38 / 47