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What figure? 

How many integers are in the sequence?

Hello Guest!

\(\dfrac{a+z}{2}=2043\\ z=4086-a\\ 2042-a+1=2(z-2044+1)\\ 2042-a+1=2(4086-a-2044+1)\\ 2042-a+1=8172-2a-4088+2\\ \color{red}a=2043\\ \color{red}z=2043 \)

There are the same number of integers before and after the middle number.

The task cannot be solved.

The "finite sequence" consists of only one integer.

  !

Ah an inverse question.

set f(x) as y for simplicity.

y = 2/(x+1), the goal for finding the inverse of the function is to isolate x, so the first step: yx + y = 2. 

next, move y to the right hand side to reach yx = 2 - y. 

finally, divide y over to the right hand side to reach x = (2-y)/y

Now as the core concept of the inverse, we switch the positions of x and y by the time we reach this step.

y = (2-x)/x, and plug in 1/5 to get y = (2-1/5)/(1/5) and that is 9/5 * 5 = 9 

3 + 3 + 2 + 4==12 - total number of crayons

Probability of picking a purple crayon ==4 / 12 ==1 / 3

Probability of picking a red crayon == 2 / 12 ==1 / 6

Since the first crayon was replaced, therefore:

The probability of picking a purple crayon followed by a red crayon is:

1/3 x  1/6 = 1/18

Expand the right hand side:

\((-16x^4+x^2+2x+1)=adx^4+aex^3+afx^2+bdx^3+bex^2+bfx+cdx^2+cex+cf\)

\(\hspace{4.4cm}=adx^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf\)

So:

\(ad=-16 \\ ae+bd=0 \\ af+be+cd=1 \\ bf+ce=2 \\ cf=1 \\\)

The possible combinations of a and d: \((a,d)=(4,-4),(16,-1),(8,-2)\)

\(c=1,f=1\) is a must by the last equation.

So our system becomes:

\(ad=-16 \\ ae+bd=0 \\ a+be+d = 1 \\ b+e = 2 \\\)

It is good idea to try one of the combinations of the above. 

Suppose (a,d) = (4,-4)
Then the first equation is satisfied. The second equation is 4e-4b = 0, which only works if e-b=0 or e=b, but from the last equation, b+e = 2 , hence 2e=2 meaning e=1 and thus b=1.

so we solved the system. 

Therefore: \(-16x^4+x^2+2x+1=(4x^2+x+1)(-4x^2+x+1)\)

I hope this helps :)! 

M==Total money
K==cost of 1 key chain.

1/2*9K  +  3K ==1/4M
7.5K == 1/4M
K==M / 30 - price of 1 key chain

2/9 * 3/4M ==1/6M - what he spent on buying additional key chains.

1/6M / [M / 30] ==1/6 x 30==5 additional key chains he putchased

3 + 5 ==8 total number of key chains he purchased.

Hi sherwyo!

Notice, x=0 is a solution, but it is not what we want, so now suppose \(x \neq 0\), and divide both sides by x:

\(\dfrac{1}{x+1}+\dfrac{1}{x+2}=k \\ \iff \dfrac{2x+3}{(x+1)(x+2)} = k \hspace{0.5cm} \text{Common denominator} \\ \text{Next, multiply both sides by (x+1)(x+2):} \\ \iff 2x+3 = k(x+1)(x+2) \hspace{0.5cm} \text{Expand the R.H.S:} \\ \iff 2x+3=k(x^2+3x+2) \space \space \space \text{Simplify} \\ \iff 2x+3=kx^2+3kx+2k \\ \iff kx^2+(3kx-2x)+(2k-3)=0 \\ \iff kx^2+(3k-2)x+(2k-3)=0 \\ \)

Now this is a quadratic equation. For this quadratic equation to have two complex roots, the disciminant: \(\Delta =b^2-4ac <0\) (Must be less than zero.)

So:

\((3k-2)^2-4(k)(2k-3) < 0 \\ \iff (9k^2-12k+4)-8k^2+12 <0 \\ \iff k^2-12k+16 <0 \\ \text{The critical points (when the above inequality is = 0) }\space \space \space k_{1,2}=6 \pm 2\sqrt{5}\)

So the desired solution of the inequality:    \(6-2\sqrt{5}

1, 1, 2, 2, 3, 3 ==6! / [2!2!2!] ==720 / 8 ==90 - permutations

[(1, 1, 2, 2, 3, 3), (1, 1, 2, 3, 2, 3), (1, 1, 2, 3, 3, 2), (1, 1, 3, 2, 2, 3), (1, 1, 3, 2, 3, 2), (1, 1, 3, 3, 2, 2), (1, 2, 1, 2, 3, 3), (1, 2, 1, 3, 2, 3), (1, 2, 1, 3, 3, 2), (1, 2, 2, 1, 3, 3), (1, 2, 2, 3, 1, 3), (1, 2, 2, 3, 3, 1), (1, 2, 3, 1, 2, 3), (1, 2, 3, 1, 3, 2), (1, 2, 3, 2, 1, 3), (1, 2, 3, 2, 3, 1), (1, 2, 3, 3, 1, 2), (1, 2, 3, 3, 2, 1), (1, 3, 1, 2, 2, 3), (1, 3, 1, 2, 3, 2), (1, 3, 1, 3, 2, 2), (1, 3, 2, 1, 2, 3), (1, 3, 2, 1, 3, 2), (1, 3, 2, 2, 1, 3), (1, 3, 2, 2, 3, 1), (1, 3, 2, 3, 1, 2), (1, 3, 2, 3, 2, 1), (1, 3, 3, 1, 2, 2), (1, 3, 3, 2, 1, 2), (1, 3, 3, 2, 2, 1), (2, 1, 1, 2, 3, 3), (2, 1, 1, 3, 2, 3), (2, 1, 1, 3, 3, 2), (2, 1, 2, 1, 3, 3), (2, 1, 2, 3, 1, 3), (2, 1, 2, 3, 3, 1), (2, 1, 3, 1, 2, 3), (2, 1, 3, 1, 3, 2), (2, 1, 3, 2, 1, 3), (2, 1, 3, 2, 3, 1), (2, 1, 3, 3, 1, 2), (2, 1, 3, 3, 2, 1), (2, 2, 1, 1, 3, 3), (2, 2, 1, 3, 1, 3), (2, 2, 1, 3, 3, 1), (2, 2, 3, 1, 1, 3), (2, 2, 3, 1, 3, 1), (2, 2, 3, 3, 1, 1), (2, 3, 1, 1, 2, 3), (2, 3, 1, 1, 3, 2), (2, 3, 1, 2, 1, 3), (2, 3, 1, 2, 3, 1), (2, 3, 1, 3, 1, 2), (2, 3, 1, 3, 2, 1), (2, 3, 2, 1, 1, 3), (2, 3, 2, 1, 3, 1), (2, 3, 2, 3, 1, 1), (2, 3, 3, 1, 1, 2), (2, 3, 3, 1, 2, 1), (2, 3, 3, 2, 1, 1), (3, 1, 1, 2, 2, 3), (3, 1, 1, 2, 3, 2), (3, 1, 1, 3, 2, 2), (3, 1, 2, 1, 2, 3), (3, 1, 2, 1, 3, 2), (3, 1, 2, 2, 1, 3), (3, 1, 2, 2, 3, 1), (3, 1, 2, 3, 1, 2), (3, 1, 2, 3, 2, 1), (3, 1, 3, 1, 2, 2), (3, 1, 3, 2, 1, 2), (3, 1, 3, 2, 2, 1), (3, 2, 1, 1, 2, 3), (3, 2, 1, 1, 3, 2), (3, 2, 1, 2, 1, 3), (3, 2, 1, 2, 3, 1), (3, 2, 1, 3, 1, 2), (3, 2, 1, 3, 2, 1), (3, 2, 2, 1, 1, 3), (3, 2, 2, 1, 3, 1), (3, 2, 2, 3, 1, 1), (3, 2, 3, 1, 1, 2), (3, 2, 3, 1, 2, 1), (3, 2, 3, 2, 1, 1), (3, 3, 1, 1, 2, 2), (3, 3, 1, 2, 1, 2), (3, 3, 1, 2, 2, 1), (3, 3, 2, 1, 1, 2), (3, 3, 2, 1, 2, 1), (3, 3, 2, 2, 1, 1)] >Total distinct permutations = 90

Sorry, that's wrong. Thanks for trying though.

Thank you!

Thank you!

\(400,000 = 4 \times 10^5\)

\((4 \times 10^5)(4 \times 10^5) = (4 \times 4)(10^5 \times 10^5)=16 \times 10^{10}\)

\(16 \times 10^{10} = 1.6 \times 10^{11}\)

\(n = \color{brown}\boxed{11}\)

How many numbers between 10 and 100 end in zero??! If you can answer that correctly, you should be able to answer your question.

He wants to split it evenly between his 7 friends.  

That should be "... split it evenly among his 7 friends." 

When there are two, use between. 

When more than two, use among. 

You learn more than math here.    

Que pasa, amigo.  

By the Angle Bisector Theorem, we have \({AY \over AB} = {YC \over CB}\). Plugging in the given info, we have \({6 \over 8} = {3 \over CB}\), meaning \(CB = 4\)

Now, apply the Angle Bisector Theorem again: \({AZ \over AC} = {ZB \over CB}\). Subbing in what we know, we have \({AB \over 9} = {ZB \over 4} \ \ \ (i)\).

For simplicity, let \(AB = x \) and \(ZB = y \). We also know that \(x + y = 8 \ \ \ (ii)\).

Solving this system, we find \(x = AB = {72 \over 13}\) and \(y = ZB = \color{brown}\boxed{32 \over 13}\)

From the second sentence, we know \(o + d = 18\)

From the first sentence, we know \(o = 5d\).

Now, just solve or o and d and see how many games Oliver gave to Darius so they have the same number of games. 

Note that because \(\triangle STU\) is isosceles, \(SN\) is also a median. 

Now, draw the median \(TK\). This intersects at \(X\), and by the properties of medians, we know that \(SX = {2 \over 3}\times SN\).

\(\triangle SNU \) is a right triangle with \(SU = 22\) and \(NU = 8 \div 2 = 4\), meaning \(SN = \sqrt{22^2-4^2} = \sqrt{468} = 6 \sqrt{13}\)

So, \(SX = {2 \over 3} \times 6 \sqrt{13} = \color{brown}\boxed{4 \sqrt {13}}\)

Draw altitude \(CF\). This will intersect at the orthocenter, which is H. This means that \(\triangle AFC \) is a right triangle. 

If we know what \(\angle A\) was, we can figure out what \(\angle HCA \) is, because \(\angle FCA = \angle HCA\) and \(\angle F\) is a right angle. 

Now, note that \(\triangle ADB \) is a right triangle, and since we know \(\angle BAH = 28\) and \(\angle ADB = 90\), \(\angle ABD = 62\).

Also, we know that \(\angle ABH = 24\), meaning \(\angle HBD = \angle ABD - \angle ABH = 62 - 24 = 38\).

Now, \(\triangle BEC\) is a right triangle, so we know that \(\angle BCA = 42\).

This means that \(\angle A = 180 - 62 - 42 = 76\), so \(\angle HCA = \angle FCA = 180 - 90 - 76 = \color{brown}\boxed{14}\)

well

\(f(6)=-\frac{1}{6}\\~\\ f(f(6))=f(\frac{-1}{6})=6\\~\\ etc. \)

You can take if from there.

plug in 2x for every x in f(x) = x^2-3x+4

That leaves us with f(2x) = (2x)^2 -3(2x) + 4 = 4x^2 - 6x + 4

Like so:

Can you take it from here?

1/5, 3/25, 7/125, 3/125, 31/3125, 63/15625, 127/78125, 51/78125, 511/1953125, 1023/9765625, ...etc.

The "closed form" of the sequence is:

a(n) = 5^(-n) (2^n - 1) (for any term)

∑[ 5^(-n) (2^n - 1), n,  1,  ∞] ==It converges to 5 / 12