All Answers 358187 Answers

\(\dfrac{5}{\dfrac{1}{6}+\dfrac{1}{x+1}} = \\ \dfrac{5}{\dfrac{(x+1)+6}{6(x+1)}} = \\ \dfrac{30(x+1)}{x+7}\)

\(N=6\times 14\times 5\times 5 \times 3 = 6300\)

-122

use Pythagorean formula

\(3\times 3\times 2 = 18\)

There's actually 36.  He's got 4 shoes total and only has to wear 2.

No one says he has to wear matching shoes.  His date might be wild and crazy and like that sort of thing!

He does however have to wear the proper shoe on the proper foot.

So we end up with 4 combos (L1,R1), (L2,R2), (L1,R2), (L2, R1)

So the real number is

\(3\times 3\times 4 = 36\)

Seeing how 36 isn't a choice I guess they are assuming he's going to wear a matching pair of shoes

and so 18 would be the correct answer.

\(8.12 \times 10^5 + 5.27 \times 10^5 = (8.12+5.27)\times 10^5 = 13.39\times 10^5~gal\)

Thanks so much for all of the help!

Thank you so much for helping me out with this one and explaining the process to me!!

\(\text{Assuming we start the sequence numbering at 1 so that }s_1 = 2\\ s_k = 2+9(k-1)\\ 227 = 2 + 9(k-1)\\ 225 = 9(k-1)\\ k -1 = 25\\ k=26\)

choice (D)

Thanks!!

Thanks for the help!

Thank you so much!

\((90^\circ - \theta) = \dfrac 1 4 (180^\circ-\theta)\\ 45^\circ = \dfrac 3 4 \theta \\ \theta = 60^\circ\)

Here's another way:

p²-1=(p-1)*(p+1). p is a prime number larger than 2, therefore p-1 and p+1 both have to be divisible by 2. Because they are 2 numbers divisible by 2 with a difference of 2 one of them is divisible by 4, meaning that one is divisible by at least 4 and one is divisible by at least 2- therefore (p-1)*(p+1) is divisible by at least 8. Because p is larger than 3, p is noy divisible by 3. The numbers p-1, p, p+1 are consecutive, meaning that at leadt one of them is divisible by 3. P is a prime larger than 3 therefore he is not divisible by 3 meaning that one of p-1, p+1 is divisible by 3 meaning that (p-1)*(p+1) is divisible by 3.

(p1)*(p+1) is divisible by 3 and by 8 so he is divisible by 3*8=24

Amen. 

we all are man

sorry about this

ok since the plane is at 1,200 ft and the goal altitude is 16.000 ft you need to subtract the goal altitude and the current altitude and that will give you 14,800 ft and since the plane is ascending at a constant rate off 925ft you divide 14,800 by 925 and you should get how many min. it takes to get to the goal altitude: 16 minutes 

Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.

\(\text{p in the form $4n+1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n+1)^2 - 1 \\ &=& 16n^2+8n+1-1 \\ &=& 8n(n+1) \quad\Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)

\(\text{p in the form $4n-1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n-1)^2 - 1 \\ &=& 16n^2-8n+1-1 \\ &=& 8n(n-1)\quad \Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)

\(\text{Euler's theorem}\\ \text{This states that if $a$ and $n$ are relatively prime $gcd(a,n) = 1$} \\ \text{then} \\ {\displaystyle a^{\varphi (n)}\equiv 1\mod n.} \)

\(\begin{array}{|rcll|} \hline a^{\varphi (3)} &\equiv& 1\pmod{3} \quad | \quad \varphi (3) = 2 \\ a^2 &\equiv& 1\pmod{3} \quad | \quad \text{if } gcd(a,3) = 1, \text{ so $a$ is a prime number and not } 3 \\ p^2 &\equiv& 1\pmod{3} \quad | \quad p\ne 3! \\ \text{so } p^2 -1 &\equiv& 0 \pmod{3} \quad\Rightarrow \quad 3 | p^2-1 \\ \hline \end{array}\)

\(\text{If $8 | p^2-1$ and $3 | p^2-1$ so also divides $24|p^2-1$}\)

I suspect this is meant to be "What two numbers when multiplied equal 4 and when added equal -9?"

If so, from x*y = 4 and x+y = -9, we find the two numbers to be:  (√(65) - 9)/2  and (-√(65) - 9)/2.

Numerically, this seems to converge (very quickly) to 9/32.  This is such a simple result I suspect there is a way of finding it analytically. However, I have so far failed to do so!!

exponent problem
solve
\(x\)  for  \(\large 3^{(2x+2)}+27^{(x+1)}=36\)

\(\text{Formula:}\\ \large{\boxed{\left(a^b\right)^c = a^{bc}=a^{cb}=\left(a^c\right)^b}}\)

\(\begin{array}{|rcll|} \hline 3^{(2x+2)}+27^{(x+1)} &=& 36 \quad | \quad 27 = 3^3 \\ 3^{2( x+1)}+\left(3^3\right)^{(x+1)} &=& 36 \quad | \quad \left(3^3\right)^{(x+1)} = 3^{3(x+1)}\\ 3^{2( x+1)}+ 3^{3(x+1)} &=& 36 \\ \left(3^{(x+1)}\right)^{2} + \left(3^{(x+1)}\right)^{3} &=& 36 \\ && \text{we substitute:} ~ \boxed{u = 3^{(x+1)}} \\\\ u^2+u^3&=& 36 \\ u^3+u^2-36 &=& 0 \\\\ u_1 &=& 3 \\ u_2 &=& -2-2i\sqrt{2} \\ u_3 &=& -2+2i\sqrt{2} \\\\ u &=& 3^{(x+1)} \quad | \quad u = 3 \\ 3 &=& 3^{(x+1)} \\ 3^1 &=& 3^{(x+1)} \\ 1 &=& x+1 \\ \mathbf{ x } & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

very amusing

how do you expand, then express in a single trig ratio and/or constant?
\(4 \sin(x) \cos(x) \Big(\sec^3(x) \cos(x) + \csc(x) \sec(x)\Big)\)

\(\text{Formula:}\\ \text{$\csc(x) = \dfrac{1}{\sin(x)} $} \\ \text{$\sec(x) = \dfrac{1}{\cos(x)} $} \\\)

\(\begin{array}{|rcll|} \hline && 4 \cdot\sin(x) \cos(x) \Big(\sec^3(x) \cos(x) + \csc(x) \sec(x)\Big) \\\\ &=& 4 \cdot\sin(x) \cos(x) \left( \dfrac{\cos(x)}{\cos^3(x)} + \dfrac{1}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\sin(x) \cos(x) \left( \dfrac{1}{\cos^2(x)} + \dfrac{1}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\left( \dfrac{\sin(x) \cos(x)}{\cos^2(x)} + \dfrac{\sin(x) \cos(x)}{\cos(x)\sin(x)} \right) \\\\ &=& 4 \cdot\left( \dfrac{\sin(x)}{\cos(x)} + 1 \right) \quad | \quad \dfrac{\sin(x)}{\cos(x)}=\tan(x) \\\\ &\mathbf{=}& \mathbf{ 4 \cdot \Big( \tan(x) + 1 \Big )} \\ \hline \end{array}\)